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National Council of Teachers of Mathematics Principles and Standards for School Mathematics Principles for School Mathematics • EQUITY. Excellence in mathematics education requires equity — high expectations and strong support for all students. • CURRICULUM. A curriculum is more than a collection of activities: it must be coherent, focused on important mathematics, and well articulated across the grades. • TEACHING. Effective mathematics teaching requires understanding what students know and need to learn and then challenging and supporting them to learn it well.
• LEARNING. Students must learn mathematics with understanding, actively building new knowledge from experience and prior knowledge. • ASSESSMENT. Assessment should support the learning of important mathematics and furnish useful information to both teachers and students. • TECHNOLOGY. Technology is essential in teaching and learning mathematics; it influences the mathematics that is taught and enhances students’ learning.
Standards for School Mathematics NUMBER AND OPERATIONS
DATA ANALYSIS AND PROBABILITY
Instructional programs from prekindergarten through grade 12 should enable all students to —
Instructional programs from prekindergarten through grade 12 should enable all students to — • formulate questions that can be addressed with data and collect, organize, and display relevant data to answer them; • select and use appropriate statistical methods to analyze data; • develop and evaluate inferences and predictions that are based on data; • understand and apply basic concepts of probability.
• understand numbers, ways of representing numbers, relationships among numbers, and number systems; • understand meanings of operations and how they relate to one another; • compute fluently and make reasonable estimates. ALGEBRA Instructional programs from prekindergarten through grade 12 should enable all students to — • understand patterns, relations, and functions; • represent and analyze mathematical situations and structures using algebraic symbols; • use mathematical models to represent and understand quantitative relationships; • analyze change in various contexts. GEOMETRY Instructional programs from prekindergarten through grade 12 should enable all students to — • analyze characteristics and properties of two and threedimensional geometric shapes and develop mathematical arguments about geometric relationships; • specify locations and describe spatial relationships using coordinate geometry and other representational systems; • apply transformations and use symmetry to analyze mathematical situations; • use visualization, spatial reasoning, and geometric modeling to solve problems. MEASUREMENT Instructional programs from prekindergarten through grade 12 should enable all students to — • understand measurable attributes of objects and the units, systems, and processes of measurement; • apply appropriate techniques, tools, and formulas to determine measurements.
PROBLEM SOLVING Instructional programs from prekindergarten through grade 12 should enable all students to — • build new mathematical knowledge through problem solving; • solve problems that arise in mathematics and in other contexts; • apply and adapt a variety of appropriate strategies to solve problems; • monitor and reflect on the process of mathematical problem solving. REASONING AND PROOF Instructional programs from prekindergarten through grade 12 should enable all students to — • recognize reasoning and proof as fundamental aspects of mathematics; • make and investigate mathematical conjectures; • develop and evaluate mathematical arguments and proofs; • select and use various types of reasoning and methods of proof. COMMUNICATION Instructional programs from prekindergarten through grade 12 should enable all students to — • organize and consolidate their mathematical thinking through communication; • communicate their mathematical thinking coherently and clearly to peers, teachers, and others; • analyze and evaluate the mathematical thinking and strategies of others; • use the language of mathematics to express mathematical ideas precisely.
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CONNECTIONS
REPRESENTATION
Instructional programs from prekindergarten through grade 12 should enable all students to — • recognize and use connections among mathematical ideas; • understand how mathematical ideas interconnect and build on one another to produce a coherent whole; • recognize and apply mathematics in contexts outside of mathematics.
Instructional programs from prekindergarten through grade 12 should enable all students to — • create and use representations to organize, record, and communicate mathematical ideas; • select, apply, and translate among mathematical representations to solve problems; • use representations to model and interpret physical, social, and mathematical phenomena.
Curriculum Focal Points for Prekindergarten through Grade 8 Mathematics PREKINDERGARTEN Number and Operations: Developing an understanding of whole numbers, including concepts of correspondence, counting, cardinality, and comparison. Geometry: Identifying shapes and describing spatial relationships. Measurement: Identifying measurable attributes and comparing objects by using these attributes. KINDERGARTEN Number and Operations: Representing, comparing and ordering whole numbers, and joining and separating sets. Geometry: Describing shapes and space. Measurement: Ordering objects by measurable attributes. GRADE 1 Number and Operations and Algebra: Developing understandings of addition and subtraction and strategies for basic addition facts and related subtraction facts. Number and Operations: Developing an understanding of whole number relationships, including grouping in tens and ones. Geometry: Composing and decomposing geometric shapes. GRADE 2 Number and Operations: Developing an understanding of the baseten numeration system and placevalue concepts. Number and Operations and Algebra: Developing quick recall of addition facts and related subtraction facts and fluency with multidigit addition and subtraction. Measurement: Developing an understanding of linear measurement and facility in measuring lengths. GRADE 3 Number and Operations and Algebra: Developing understandings of multiplication and division and strategies for basic multiplication facts and related division facts. Number and Operations: Developing an understanding of fractions and fraction equivalence. Geometry: Describing and analyzing properties of twodimensional shapes. GRADE 4 Number and Operations and Algebra: Developing quick recall of multiplication facts and related division facts and fluency with whole number multiplication.
Number and Operations: Developing an understanding of decimals, including the connections between fractions and decimals. Measurement: Developing an understanding of area and determining the areas of twodimensional shapes. GRADE 5 Number and Operations and Algebra: Developing an understanding of and fluency with division of whole numbers. Number and Operations: Developing an understanding of and fluency with addition and subtraction of fractions and decimals. Geometry and Measurement and Algebra: Describing threedimensional shapes and analyzing their properties, including volume and surface area. GRADE 6 Number and Operations: Developing an understanding of and fluency with multiplication and division of fractions and decimals. Number and Operations: Connecting ratio and rate to multiplication and division. Algebra: Writing, interpreting, and using mathematical expressions and equations. GRADE 7 Number and Operations and Algebra and Geometry: Developing an understanding of and applying proportionality, including similarity. Measurement and Geometry and Algebra: Developing an understanding of and using formulas to determine surface areas and volumes of threedimensional shapes. Number and Operations and Algebra: Developing an understanding of operations on all rational numbers and solving linear equations. GRADE 8 Algebra: Analyzing and representing linear functions and solving linear equations and systems of linear equations. Geometry and Measurement: Analyzing two and threedimensional space and figures by using distance and angle. Data Analysis and Number and Operations and Algebra: Analyzing and summarizing data sets.
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The first person to invent a car that runs on water… … may be sitting right in your classroom! Every one of your students has the potential to make a difference. And realizing that potential starts right here, in your course. When students succeed in your course—when they stay ontask and make the breakthrough that turns confusion into confidence—they are empowered to realize the possibilities for greatness that lie within each of them. We know your goal is to create an environment where students reach their full potential and experience the exhilaration of academic success that will last them a lifetime. WileyPLUS can help you reach that goal.
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athematics M F o r E l e m e n t a r y Te a c h e r s
E I G H T H EDITION
A CONTEMPORARY APPROACH
Gary L. Musser Oregon State University
William F. Burger Blake E. Peterson Brigham Young Univeristy
John Wiley & Sons, Inc.
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To: Irene, my supportive wife of over 45 years; Greg, my son, for his continuing progress in life; Maranda, my granddaughter, for her enthusiasm and appreciation of her love; my parents, who have both passed away, but are always on my mind and in my heart; and Mary Burger, Bill Burger’s wonderful daughter. G.L.M. Shauna, my eternal companion and best friend, for making me smile along this wonderful journey called life; Quinn, Joelle, Taren, and Riley, my four children, for choosing the right; Mark, Kent, and Miles, my brothers, for their examples and support. B.E.P. PUBLISHER ACQUISITIONS EDITOR ASSISTANT EDITOR SENIOR PRODUCTION EDITOR MARKETING MANAGER CREATIVE DIRECTOR SENIOR DESIGNER PRODUCTION MANAGEMENT SERVICES SENIOR PHOTO EDITOR EDITORIAL ASSISTANT MEDIA EDITOR COVER & TEXT DESIGN COVER IMAGE BY BICENTENNIAL LOGO DESIGN
Laurie Rosatone Jessica Jacobs Michael Shroff Valerie A. Vargas Jaclyn Elkins Harry Nolan Kevin Murphy mb editorial services Lisa Gee Jeffrey Benson Stefanie Liebman Michael Jung Miao Jin, Junho Kim, and Xianfeng David Gu Richard J. Pacifico
This book was set in 10/12 Times New Roman by GGS Book Services and printed and bound by RRD–Jefferson City. The cover was printed by RRD–Jefferson City. This book is printed on acidfree paper. ˆ Copyright © 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate percopy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 070305774, (201) 7486011, fax (201) 7486008, website http://www.wiley.com/go/permissions. To order books or for customer service please, call 1800CALL WILEY (2255945). ISBN13 9780470105832 Printed in the United States of America 1 09 8 7 6 5 4 3 2 1
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About the Authors GARY L. MUSSER is Professor Emeritus from Oregon State University. He earned both his B.S. in Mathematics Education in 1961 and his M.S. in Mathematics in 1963 at the University of Michigan and his Ph.D. in Mathematics (Radical Theory) in 1970 at the University of Miami in Florida. He taught at the junior and senior high, junior college, college, and university levels for more than 30 years. He served his last 24 years teaching prospective teachers in the Department of Mathematics at Oregon State University. While at OSU, Dr. Musser developed the mathematics component of the elementary teacher program. Soon after Professor William F. Burger joined the OSU Department of Mathematics in a similar capacity, the two of them began to write the first edition of this book. Professor Burger passed away during the preparation of the second edition, and Professor Blake E. Peterson was hired at OSU as his replacement. Professor Peterson joined Professor Musser as a coauthor beginning with the fifth edition. Professor Musser has published 40 papers in many journals, including the Pacific Journal of Mathematics, Canadian Journal of Mathematics, The Mathematics Association of America Monthly, the NCTM’s The Mathematics Teacher, the NCTM’s The Arithmetic Teacher, School Science and Mathematics, The Oregon Mathematics Teacher, and The Computing Teacher. In addition, he is a coauthor of two other college mathematics books: College Geometry—A ProblemSolving Approach with Applications (2008) and A Mathematical View of Our World (2007). He also coauthored the K–8 series Mathematics in Action. He has given more than 65 invited lectures/workshops at a variety of conferences, including NCTM and MAA conferences, and was awarded 15 federal, state, and local grants to improve the teaching of mathematics. While Professor Musser was at OSU, he was awarded the university’s prestigious College of Science Carter Award for Teaching. He is currently living in sunny Las Vegas, where he continues to write, ponder the mysteries of the stock market, and entertain both his wife and his faithful yellow lab, Zoey. BLAKE E. PETERSON is currently a Professor in the Department of Mathematics Education at Brigham Young University. He was born and raised in Logan, Utah, where he graduated from Logan High School. Before completing his B.A. in secondary mathematics education at Utah State University, he spent two years in Japan as a missionary for The Church of Jesus Christ of Latter Day Saints. After graduation, he took his new wife, Shauna, to southern California, where he taught and coached at Chino High School for two years. In 1988, he began graduate school at Washington State University, where he later completed a M.S. and Ph.D. in pure mathematics. After completing his Ph.D., Dr. Peterson was hired as a mathematics educator in the Department of Mathematics at Oregon State University in Corvallis, Oregon, where he taught for three years. It was at OSU that he met Gary Musser. He has since moved his wife and four children to Provo, Utah, to assume his position at Brigham Young University where he is currently a full professor. As a professor, his first love is teaching, for which he has received a College Teaching Award in the College of Science. Dr. Peterson has published papers in Rocky Mountain Mathematics Journal, The American Mathematical Monthly, The Mathematical Gazette, Mathematics Magazine, The New England Mathematics Journal, and The Journal of Mathematics Teacher Education as well as NCTM’s Mathematics Teacher, and Mathematics Teaching in the Middle School. After studying mathematics student teachers at a Japanese junior high school, he implemented some elements he observed into the student teaching structure at BYU. In addition to teaching, research, and writing, Dr. Peterson has done consulting for the College Board, founded the Utah Association of Mathematics Teacher Educators, is on the editorial panel for the Mathematics Teacher, and is the associate chair of the department of mathematics education at BYU. Aside from his academic interests, Dr. Peterson enjoys spending time with his family, fulfilling his church responsibilities, playing basketball, mountain biking, water skiing, and working in the yard.
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About the Cover The checkered figure on the cover, which is called Costa’s minimal surface, was discovered in 1982 by Celso Costa. It is studied in the field of mathematics called differential geometry. There are many different fields of mathematics and in each field there are different tools used to solve problems. Particularly difficult problems, however, may require reaching from one field of mathematics into another to find the tools to solve it. One such problem was posed in 1904 by Henri Poincaré and is in a branch of mathematics called topology. The problem, stated as a conjecture, is that the threesphere is the only compact threemanifold which has the property that each simple closed curve can be contracted. While this conjecture is likely not understandable to one not well versed in topology, the story surrounding its eventual proof is quite interesting. This conjecture, stated in three dimensions, had several proof attempts in the early 1900s that were initially thought to be true only to be proven false later. In 1960, Stephen Smale proved an equivalent conjecture for dimensions 5 and higher. For this he received the Fields medal for it in 1966. The Fields medal is the equivalent of the Nobel prize for mathematics and to receive it, the recipient must be 40 years of age or younger. The medal is awarded every 4 years to between two and four mathematicians. In 1983, Michael Freedman proved the equivalent conjecture for the 4th dimension and he also received a Fields Medal in 1986. In 2003, Grigory “Grisha” Perelman of St. Petersburg, Russia, claimed to have proved the Poincaré conjecture as stated for threedimensions when he posted three short papers on the internet. These postings were followed by a series of lectures in the United States discussing the papers. Typically, a proof like this would be carefully written and submitted to a prestigious journal for peer review. The brevity of these papers left the rest of the mathematics community wondering if the proof was correct. As other
vi
mathematicians have filled in the gaps, their resulting papers (3 in total) were about 1000 pages long of dense mathematics. One of the creative aspects of Perelman’s proof is the tools that he used. He reached beyond the field of topology into the field of differential geometry and used a tool called a Ricci flow. In August of 2006, Dr. Perelman was awarded the Fields medal along with three other mathematicians. However, he did not attend the awards ceremony in Spain and declined to accept the medal along with its $13,400 stipend. In the 70year history of the Fields medal, there have only been 48 recipients of the award and none have refused it before Perelman. His colleagues indicate that he is only interested in knowledge and not in awards or money. Such an attitude is even more remarkable when you consider the $1,000,000 award that is also available for proving the Poincaré conjecture. In 2000, the Clay Mathematics Institute in Cambridge, Massachusetts, identified seven historic, unsolved mathematics problems that they would offer a $1 million prize for the proof of each. The Poincaré conjecture is one of those 7 historic unsolved problems. Each proof requires a verification period before the prize would be awarded. As of this writing, it is unknown if Dr. Perelman would accept the $1 million prize. There is one other interesting twist to this story. Because of the brevity of Perelman’s proofs, other mathematicians filled in some details. In particular, two Chinese mathematicians, Professors Cao and Zhu, wrote a paper on this subject entitled “A Complete Proof of the Poincaré and Geometrization Conjectures—Application of the HamiltonPerelman Theory of Ricci Flow”. This paper was 327 pages long! But what about the $1,000,000? If you are interested, search the internet periodically to see if Perelman accepts all or part of the $1,000,000. The image on the cover was created by Miao Jin, Junho Kim and Xianfeng David Gu.
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Brief Contents
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Introduction to Problem Solving 1 Sets, Whole Numbers, and Numeration 43 Whole Numbers: Operations and Properties 107 Whole Number Computation: Mental, Electronic, and Written Number Theory 203 Fractions 237 Decimals, Ratio, Proportion, and Percent 285 Integers 341 Rational Numbers, Real Numbers, and Algebra 379 Statistics 439 Probability 513 Geometric Shapes 581 Measurement 665 Geometry Using Triangle Congruence and Similarity 739 Geometry Using Coordinates 807 Geometry Using Transformations 849 Epilogue: An Eclectic Approach to Geometry 909 Topic 1 Elementary Logic 912 Topic 2 Clock Arithmetic: A Mathematical System 923 Answers to Exercise/Problem SetsPart A, Chapter Tests, A1 and Topics Index I1 Contents of Book Companion Web Site Resources for Technology Problems Technology Tutorials Webmodules Additional Resources
155
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Preface
1
Introduction to Problem Solving 1.1 1.2
2
171
Primes, Composites, and Tests for Divisibility 205 Counting Factors, Greatest Common Factor, and Least Common Multiple
237
The Sets of Fractions 239 Fractions: Addition and Subtraction 255 Fractions: Multiplication and Division 266
Decimals 287 Operations with Decimals 297 Ratios and Proportion 310 Percent 320
Integers 8.1 8.2
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107
203
Decimals, Ratio, Proportion, and Percent 7.1 7.2 7.3 7.4
8
Mental Math, Estimation, and Calculators 157 Written Algorithms for WholeNumber Operations Algorithms in Other Bases 192
Fractions 6.1 6.2 6.3
7
45 59
Addition and Subtraction 109 Multiplication and Division 123 Ordering and Exponents 140
Number Theory 5.1 5.2
6
Sets as a Basis for Whole Numbers Whole Numbers and Numeration The Hindu–Arabic System 70 Relations and Functions 82
43
Whole Number Computation—Mental, Electronic, 155 and Written 4.1 4.2 4.3
5
3
Whole Numbers: Operations and Properties 3.1 3.2 3.3
4
The Problem Solving Process and Strategies Three Additional Strategies 20
1
Sets, Whole Numbers, and Numeration 2.1 2.2 2.3 2.4
3
xi
341
Addition and Subtraction 343 Multiplication, Division, and Order
357
285
219
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Contents
9
Rational Numbers, Real Numbers, and Algebra 9.1 9.2 9.3
10
11
581
Recognizing Geometric Shapes 583 Analyzing Shapes 600 Properties of Geometric Shapes: Lines and Angles Regular Polygons and Tessellations 628 Describing ThreeDimensional Shapes 640
615
665
Measurement with Nonstandard and Standard Units Length and Area 686 Surface Area 707 Volume 717
667
Congruence of Triangles 741 Similarity of Triangles 752 Basic Euclidean Constructions 765 Additional Euclidean Constructions 777 Geometric Problem Solving Using Triangle Congruence and Similarity
Geometry Using Coordinates 15.1 15.2 15.3
16
560
Geometry Using Triangle Congruence 739 and Similarity 14.1 14.2 14.3 14.4 14.5
15
513
Probability and Simple Experiments 515 Probability and Complex Experiments 532 Additional Counting Techniques 549 Simulation, Expected Value, Odds, and Conditional Probability
Measurement 13.1 13.2 13.3 13.4
14
Organizing and Picturing Information 441 Misleading Graphs and Statistics 464 Analyzing Data 484
Geometric Shapes 12.1 12.2 12.3 12.4 12.5
13
439
Probability 11.1 11.2 11.3 11.4
12
The Rational Numbers 381 The Real Numbers 399 Functions and Their Graphs 417
Statistics 10.1 10.2 10.3
379
807
Distance and Slope in the Coordinate Plane 809 Equations and Coordinates 822 Geometric Problem Solving Using Coordinates 834
Geometry Using Transformations 16.1 16.2 16.3
849
Transformations 851 Congruence and Similarity Using Transformations Geometric Problem Solving Using Transformations
875 893
790
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Epilogue: An Eclectic Approach to Geometry Topic 1. Elementary Logic
909
912
Topic 2. Clock Arithmetic: A Mathematical System Answers to Exercise/Problem Sets—Part A, Chapter Tests, and Topics A1 Photograph Credits Index
P1
I1
Contents of Book Companion Web Site Resources for Technology Problems • eManipulatives • Spreadsheets • Geometer’s Sketchpad
Technology Tutorials • • • •
Spreadsheets Geometer’s Sketchpad Programming in Logo Graphing Calculators
Webmodules • • • •
Algebraic Reasoning Using Children’s Literature Introduction to Graph Theory Guide to Problem Solving
Additional Resources • Research Articles • Web Links
923
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Preface elcome to the study of the foundations of elementary school mathematics. We hope you will find your studies enlightening, useful, and fun. We salute you for choosing teaching as a profession and hope that your experiences with this book will help prepare you to be the best possible teacher of mathematics that you can be. We have presented this elementary mathematics material from a variety of perspectives so that you will be better equipped to address the broad range of learning styles that you will encounter in your future students. This book also encourages prospective teachers to gain the ability to do the mathematics of elementary school and to understand the underlying concepts so they will be able to assist their students, in turn, to gain a deep understanding of mathematics. We have also sought to present this material in a manner consistent with the recommendations in (1) The Mathematical Education of Teachers prepared by the Conference Board of the Mathematical Sciences; and (2) the National Council of Teachers of Mathematics’ Principles and Standards for School Mathematics, and Curriculum Focal Points. In addition, we have received valuable advice from many of our colleagues around the United States through questionnaires, reviews, focus groups, and personal communications. We have taken great care to respect this advice and to ensure that the content of the book has mathematical integrity and is accessible and helpful to the variety of students who will use it. As always, we look forward to hearing from you about your experiences with our text.
W
GARY L. MUSSER,
[email protected] BLAKE E. PETERSON,
[email protected] Unique Content Features Number Systems The order in which we present the number systems in this book is unique and most relevant to elementary school teachers. The topics are covered to parallel their evolution historically and their development in the elementary/middle school curriculum. Fractions and integers are treated separately as an extension of the whole numbers. Then rational numbers can be treated at a brisk pace as extensions of both fractions (by adjoining their opposites) and integers (by adjoining their appropriate quotients) since students have a mastery of the concepts of reciprocals from fractions (and quotients) and opposites from integers from preceding chapters. Longtime users of this book have commented to us that this whole numbersfractionsintegersrationalsreals approach is clearly superior to the seemingly more efficient sequence of whole numbersintegersrationalsreals that is more appropriate to use when teaching high school mathematics.
Approach to Geometry Geometry is organized from the point of view of the fivelevel van Hiele model of a child’s development in geometry. After studying shapes and measurement, geometry is approached more formally through Euclidean congruence and similarity, coordinates, and transformations. The Epilogue provides an eclectic approach by solving geometry problems using a variety of techniques.
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Preface
Additional Topics
• •
Topic 1, “Elementary Logic,” may be used anywhere in a course. Topic 2, “Clock Arithmetic: A Mathematical System,” uses the concepts of opposite and reciprocal and hence may be most instructive after Chapter 6, “Fractions,” and Chapter 8, “Integers,” have been completed. This section also contains an introduction to modular arithmetic.
Underlying Themes Problem Solving An extensive collection of problemsolving strategies is developed throughout the book; these strategies can be applied to a generous supply of problems in the exercise/problem sets. The depth of problemsolving coverage can be varied by the number of strategies selected throughout the book and by the problems assigned.
Deductive Reasoning The use of deduction is promoted throughout the book. The approach is gradual, with later chapters having more multistep problems. In particular, the last sections of Chapters 14, 15, and 16 and the Epilogue offer a rich source of interesting theorems and problems in geometry. Technology Various forms of technology are an integral part of society and can enrich the mathematical understanding of students when used appropriately. Thus, calculators and their capabilities (long division with remainders, fraction calculations, and more) are introduced throughout the book within the body of the text. In addition, the book companion Web site has eManipulatives, spreadsheets, and sketches from Geometer’s Sketchpad®. The eManipulatives are electronic versions of the manipulatives commonly used in the elementary classroom, such as the geoboard, base ten blocks, black and red chips, and pattern blocks. The spreadsheets contain dynamic representations of functions, statistics, and probability simulations. The sketches in Geometer’s Sketchpad® are dynamic representations of geometric relationships that allow exploration. Exercises and problems that involve eManipulatives, spreadsheets, and Geometer’s Sketchpad® sketches have been integrated into the problem sets throughout the text.
Course Options We recognize that the structure of the mathematics for elementary teachers course will vary depending upon the college or university. Thus, we have organized this text so that it may be adapted to accommodate these differences. Basic course: Chapters 1–7 Basic course with logic: Topic 1, Chapters 1–7 Basic course with informal geometry: Chapters 1–7, 12. Basic course with introduction to geometry and measurement: Chapters 1–7, 12, 13
Summary of Changes to the Eighth Edition
• • •
Exercise sets have been revised and enriched, where necessary, to assure that they are closely aligned with and provide complete coverage of the section material. In addition, the exercises are in matched pairs between Part A and Part B. New problems have been added and Problems for Writing/Discussion at the end of the sections in the Seventh Edition have been appended to the end of the problem sets. All Spotlights in Technology that were in Seventh Edition, other than those involving calculators, have been converted to exercises or problems.
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Preface
• • • • • • • •
xiii
Sections 10.2 and 10.3 have been interchanged. NCTM’s “Curriculum Focal Points” are listed at the beginning of the book and cited in each chapter introduction. A set of problems based on the NCTM Standards and Focal Points has been added to the end of each section. Several changes have been made in the body of the text throughout the book based on recommendations of our reviewers. New Mathematical Morsels have been added where appropriate. The Table of Contents now includes a listing of resources on the Web site. Topic 3, “Introduction to Graph Theory,” has been moved to our Web site. Complete reference lists for both Reflections from Research and Children’s Literature are located in the Web site.
Pedagogy The general organization of the book was motivated by the following mathematics learning cube:
Applications Problem solving Understanding
Goal of Mathematics Instruction
Skill Knowledge START
Concrete Pictorial Abstract Measurement Number systems Geometry
The three dimensions of the cube—cognitive levels, representational levels, and mathematical content—are integrated throughout the textual material as well as in the problem sets and chapter tests. Problem sets are organized into exercises (to support knowledge, skill, and understanding) and problems (to support problems solving and applications). We have developed new pedagogical features to implement and reinforce the goals discussed above and to address the many challenges in the course.
Summary of Pedagogical Changes to the Eighth Edition
• • •
Student Page Snapshots have been updated. Reflections from Research have been edited and updated. Children’s Literature references have been edited and updated. Also, there is additional material offered on the Web site on this topic.
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Preface
Key Features ProblemSolving Strategies are integrated throughout the book. Six strategies are introduced in Chapter 1. The last strategy in the strategy box at the top of the second page of each chapter after Chapter One contains a new strategy.
ProblemSolving Strategies 1. Guess and Test
STRATEGY
3. Use a Variable 4. Look for a Pattern 5. Make a List 6. Solve a Simpler Problem 7. Draw a Diagram 8. Use Direct Reasoning
One’s point of view or interpretation of a problem can often change a seemingly difficult problem into one that is easily solvable. One way to solve the next problem is by drawing a picture or, perhaps, by actually finding some representative blocks to try various combinations. On the other hand, another approach is to see whether the problem can be restated in an equivalent form, say, using numbers. Then if the equivalent problem can be solved, the solution can be interpreted to yield an answer to the original problem.
9. Use Indirect Reasoning 10. Use Properties of Numbers 11. Solve an Equivalent Problem
11
Solve an Equivalent Problem
2. Draw a Picture
Mathematical Structure reveals the mathematical ideas of the book. Main Definitions, Theorems, and Properties in each section are highlighted in boxes for quick review.
DEFINITION Addition of Fractions with Common Denominators Let
a c and be any fractions. Then b b a c a + c + = . b b b
INITIAL PROBLEM A child has a set of 10 cubical blocks. The lengths of the edges are 1 cm, 2 cm, 3 cm, . . . , 10 cm. Using all the cubes, can the child build two towers of the same height by stacking one cube upon another? Why or why not?
THEOREM CLUES The Solve an Equivalent Problem strategy may be appropriate when
• • • • •
You can find an equivalent problem that is easier to solve.
Addition of Fractions with Unlike Denominators a c Let and be any fractions. Then b d
A problem is related to another problem you have solved previously. A problem can be represented in a more familiar setting. A geometric problem can be represented algebraically, or vice versa.
a c ad + bc + = . b d bd
Physical problems can easily be represented with numbers or symbols.
A solution of this Initial Problem is on page 281.
PROPERTY Commutative Property for Fraction Addition a c Let and be any fractions. Then b b a c c a + = + . b b b b
Starting Points are located at the beginning of each
S TA RTI N G P OI N T
Following recess, the 1000 students of Wilson School lined up for the following activity: The first student opened all of the 1000 lockers in the school. The second student closed all lockers with even numbers. The third student “changed” all lockers that were numbered with multiples of 3 by closing those that were open and opening those that were closed. The fourth student changed each locker whose number was a multiple of 4, and so on. After all 1000 students had entered completed the activity, which lockers were open? Why?
section. These Starting Points can be used in a variety of ways. First, they can be used by an instructor at the beginning of class to have students engage in some novel thinking and/or discussion about forthcoming material. Second, they can be used in small groups where students discuss the query presented. Third, they can be used as an advanced organizer homework piece where a class begins with a discussion of what individual students have discovered.
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Technology Problems appear in the Exercise/Problem sets through the book. 8. Using the Chapter 6 eManipulative activity Dividing Fractions on our Web site, construct representations of the following division problems. Sketch each representation. a. 74 , 12 b. 34 , 23 c. 214 , 58
These problems rely on and are enriched by the use of technology. The technology used includes activities from the eManipulatives (virtual manipulatives), spreadsheets, Geometer’s Sketchpad®, and the TI34 II calculator. Most of these technological resources can be accessed through the accompanying book companion Web site.
8. The Chapter 11 dynamic spreadsheet Roll the Dice on our Web site simulates the rolling of 2 dice and computing the sum. Use this spreadsheet to simulate rolling a pair of dice 100 times and find the experimental probability for the events in parts a–e. a. The sum is even. b. The sum is not 10. c. The sum is a prime. d. The sum is less than 9. e. The sum is not less than 9. f. Repeat parts a–e with 500 rolls.
17. The Chapter 12 Geometer’s Sketchpad® activity Name That Quadrilateral on our Web site displays seven different quadrilaterals in the shape of a square. However, each quadrilateral is constructed with different properties. Some have right angles, some have congruent sides, and some have parallel sides. By dragging each of the points on each of the quadrilaterals, you can determine the most general name of each quadrilateral. Name all seven of the quadrilaterals.
Student Page Snapshots have been updated. Each chapter has a page from an elementary school textbook relevant to the material being studied.
S T U D E N T PA G E S N A P S H O T
22. The fraction 12 18 is simplified on a fraction calculator and the result is 23. Explain how this result can be used to find the GCF(12, 18). Use this method to find the following. a. GCF(72, 168) b. GCF(234, 442)
Exercise/Problem Sets are separated into Part A (all answers are provided in the back of the book and all solutions are provided in our supplement Hints and Solutions for Part A Problems) and Part B (answers are only provided in the Instructors Resource Manual). In addition, exercises and problems are distinguished so that students can learn how they differ. Problems for Writing/Discussion have been integrated into the problem sets throughout the book and are designated by a writing icon. They are also included as part of the chapter review.
From Harcourt Mathematics, Level 5, p. 500. Copyright 2004 by Harcourt.
NCTM Standards and Curriculum Focal Points In previous editions the NCTM Standards that have been listed at the beginning of the book and then highlighted in margin notes throughout the book. The eighth edition also lists the Curriculum Focal Points from NCTM at the beginning of the book. At the beginning of each chapter, the Curriculum Focal Points that are relevant to that particular chapter are listed again.
Key Concepts from NCTM Curriculum Focal Points
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G RAD E 1 : Developing an understanding of whole number relationships, including grouping in tens and ones.
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G RAD E 2 : Developing quick recall of addition facts and related subtraction facts
and fluency with multidigit addition and subtraction.
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Problems Relating to the NCTM Standards and Curriculum Focal Points 1. The Focal Points for Grade 3 state “Developing an understanding of and fluency with addition and subtraction of fractions and decimals.” Based on the discussions in this section, explain at least one main concept essential to understanding addition and subtraction of fractions. 2. The NCTM Standards state “All students use visual models, benchmarks, and equivalent forms to add and subtract commonly used fractions and decimals.” Explain what is
meant by “visual models” when adding and subtracting fractions. 3. The NCTM Standards state “All students should develop and use strategies to estimate computations involving fractions and decimals in situations relevant to students’ experience.” List and explain some examples of strategies to estimate fraction computations.
Problems from the NCTM Standards and Curriculum Focal Points To further help students understand and be aware of these documents from the National Council of Teachers of Mathematics, new problems have been added at the end of every section. These problems ask students to connect the mathematics being learned from the book with the K–8 mathematics outlined by NCTM.
Reflections from Research Extensive research has been done in the mathematics education community that focuses on the teaching and learning of elementary mathematics. Many important quotations from research are given in the margins to support the content nearby.
FOCUS ON
Famous Unsolved Problems
umber theory provides a rich source of intriguing problems. Interestingly, many problems in number theory are easily understood, but still have never been solved. Most of these problems are statements or conjectures that have never been proven right or wrong. The most famous “unsolved” problem, known as Fermat’s Last Theorem, is named after Pierre de Fermat who is pictured below. It states “There are no nonzero whole numbn cn, for n a whole number bers a, b, c, where an greater than two.”
N
Fermat left a note in the margin of a book saying that he did not have room to write up a proof of what is now called Fermat’s Last Theorem. However, it remained an unsolved problem for over 350 years because mathematicians were unable to prove it. In 1993, Andrew Wiles, an English mathematician on the Princeton faculty, presented a “proof ” at a conference at Cambridge University. However, there was a hole in his proof. Happily, Wiles and Richard Taylor produced a valid proof in 1995, which followed from work done by Serre, Mazur, and Ribet beginning in 1985.
Reflection from Research Given the proper experiences, children as young as eight and nine years of age can learn to comfortably use letters to represent unknown values and can operate on representations involving letters and numbers while fully realizing that they did not know the values of the unknowns (Carraher, Schliemann, Brizuela, & Earnest, 2006).
Historical vignettes open each chapter and introduce ideas and concepts central to each chapter.
The following list contains several such problems that are still unsolved. If you can solve any of them, you will surely become famous, at least among mathematicians. 1. Goldbach’s conjecture. Every even number greater than 4 can be expressed as the sum of two odd primes. For example, 6 3 3, 8 3 5, 10 5 5, 12 5 7, and so on. It is interesting to note that if Goldbach’s conjecture is true, then every odd number greater than 7 can be written as the sum of three odd primes. 2. Twin prime conjecture. There is an infinite number of pairs of primes whose difference is two. For example, (3, 5), (5, 7), and (11, 13) are such prime pairs. Notice that 3, 5, and 7 are three prime numbers where 5 3 2 and 7 5 2. It can easily be shown that this is the only such triple of primes. 3. Odd perfect number conjecture. There is no odd perfect number; that is, there is no odd number that is the sum of its proper factors. For example, 6 1 2 3; hence 6 is a perfect number. It has been shown that the even perfect numbers are all of the form 2 p 1 (2p 1), where 2 p 1 is a prime. 4. Ulam’s conjecture. If a nonzero whole number is even, divide it by 2. If a nonzero whole number is odd, multiply it by 3 and add 1. If this process is applied repeatedly to each answer, eventually you will arrive at 1. For example, the number 7 yields this sequence of numbers: 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Interestingly, there is a whole number less than 30 that requires at least 100 steps before it arrives at 1. It can be seen that 2n requires n steps to arrive at 1. Hence one can find numbers with as many steps (finitely many) as one wishes.
Mathematical Morsels end every section with an interesting historical tidbit. One of our students referred to these as a reward for completing the section.
MATHEMATIC AL MORSEL The University of Oregon football team has developed quite a wardrobe. Most football teams have two different uniforms: one for home games and one for away games. The University of Oregon team will have as many as 384 different uniform combinations from which to choose. Rather than the usual light and dark jerseys, they have 4 different colored jerseys: white, yellow, green, and black. Beyond that, however, they have 4 different colored pants, 4 different colored pairs of socks, 2 different colored pairs of shoes and 2 different colored helmets with a 3rd one on the way. If all color combinations are allowed, the fundamental counting principle would suggest that they have 4 ⫻ 4 ⫻ 4 ⫻ 2 ⫻ 3 ⫽ 384 possible uniform combinations. Whether a uniform consisting of a green helmet, black jersey, yellow pants, white socks and black shoes would look stylish is debatable.
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People in Mathematics John Von Neumann (1903–1957) John von Neumann was one of the most remarkable mathematicians of the twentieth century. His logical power was legendary. It is said that during and after World War II the U.S. government reached many scientific decisions simply by asking von Neumann for his opinion. Paul Halmos, his onetime assistant, said, “The most spectacular thing about Johnny was not his power as a mathematician, which was great, but his rapidity; he was very, very fast. And like the modern computer, which doesn’t memorize logarithms, but computes them, Johnny didn’t bother to memorize things. He computed them.” Appropriately, von Neu
Julia Bowman Robinson (1919–1985) Julia Bowman Robinson spent her early years in Arizona, near Phoenix. She said that one of her earliest memories was of arranging pebbles in the shadow of a giant saguaro—”I’ve always had a basic liking for the natural numbers.” In 1948, Robinson earned her doctorate in mathematics at Berkeley; she went on to contribute to the solution of “Hilbert’s tenth problem.” In 1975 she became the first woman mathematician elected to the prestigious National Academy of Sciences. Robinson also served as president of the American Mathematical Society, the main professional organization for research mathematicians. “Rather than
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People in Mathematics, a feature near the end of each chapter, highlights many of the giants in mathematics throughout history.
A Chapter Review is located at the end of each chapter. A Chapter Test is found at the end of each chapter. An Epilogue, following Chapter 16, provides a rich eclectic approach to geometry. Logic and Clock Arithmetic are developed in topic sections near the end of the book.
Supplements for Students Student Activity Manual This activity manual is designed to enhance student learning as well as to model effective classroom practices. Since many instructors are working with students to create a personalized journal, this edition of the manual is shrinkwrapped and threehole punched for easy customization. This supplement is an extensive revision of the Student Resource Handbook that was authored by Karen Swenson and Marcia Swanson for the first six editions of this book. ISBN 9780470105849
FEATURES INCLUDE • HandsOn Activities: Activities that help develop initial understandings at the concrete level. • Exercises: Additional practice for building skills in concepts. • Connections to the Classroom: Classroomlike questions to provoke original thought. • Mental Math: Short activities to help develop mental math skills. • Directions in Education: Specially written articles that provide insights into major issues of the day, including the Standards of the National Council of Teachers of Mathematics. • Solutions: Solutions to all items in the handbook to enhance selfstudy. • TwoDimensional Manipulatives: Cutouts are provided on cardstock —Prepared by Lyn Riverstone of Oregon State University
The ETA Cuisenaire® Physical Manipulative Kit A generous assortment of manipulatives (including blocks, tiles, geoboards, and so forth) has been created to accompany the text as well as the Student Activity Manual. It is available to be packaged with the text. Please contact your local Wiley representative for ordering information. ISBN 9780470135525
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State Correlation Guidebooks In an attempt to help preservice teachers prepare for state licensing exams and to inform their future teaching, Wiley has updated seven completely unique statespecific correlation guidebooks. These 35page pamphlets provide a detailed correlation between the textbook and key supplements with state standards for the following states: CA, FL, IL, MI, NY, TX, VA. Each guidebook may be packaged with the text. Please contact your local Wiley representative for further information. —Prepared by Chris Awalt of the Princeton Review [CA: 9780470231722; FL: 9780470231739; IL: 9780470231715; MI: 9780470231746; NY: 9780470231753; TX: 9780470231760; VA: 9780470231777]
Student Hints and Solutions Manual for Part A Problems This manual contains hints and solutions to all of the Part A problems. It can be used to help students develop problemsolving proficiency in a selfstudy mode. The features include:
• • •
Hints: Gives students a start on all Part A problems in the text. Additional Hints: A second hint is provided for more challenging problems. Complete Solutions to Part A Problems: Carefully writtenout solutions are provided to model one correct solution. —Developed by Lynn Trimpe, Vikki Maurer, and Roger Maurer of LinnBenton Community College. ISBN 9780470105856
Companion Web site http://www.wiley.com/college/musser The companion Web site provides a wealth of resources for students.
Resources for Technology Problems These problems are integrated into the problem sets throughout the book and are denoted by a mouse icon.
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eManipulatives mirror physical manipulatives as well as provide dynamic representations of other mathematical situations. The goal of using the eManipulatives is to engage learners in a way that will lead to a more indepth understanding of the concepts and to give them experience thinking about the mathematics that underlies the manipulatives. —Prepared by Lawrence O. Cannon, E. Robert Heal, and Joel Duffin of Utah State University, Richard Wellman of Westminster College, and Ethalinda K. S. Cannon of A415software.com. This project is supported by the National Science Foundation. ISBN 9780470135518
• •
The Geometer’s Sketchpad® activities allow students to use the dynamic capabilities of this software to investigate geometric properties and relationships. They are accessible through a Web browser so having the software is not necessary. The Spreadsheet activities utilize the iterative properties of spreadsheets and the userfriendly interface to investigate problems ranging from graphs of functions to standard deviation to simulations of rolling dice.
Tutorials
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The Geometer’s Sketchpad® tutorial is written for those students who have access to the software and who are interested in investigating problems of their own choosing. The tutorial gives basic instruction on how to use the software and includes some sample problems that
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will help the students gain a better understanding of the software and the geometry that could be learned by using it. —Prepared by Armando MartinezCruz, California State University, Fullerton.
•
The Spreadsheet Tutorial is written for students who are interested in learning how to use spreadsheets to investigate mathematical problems. The tutorial describes some of the functions of the software and provides exercises for students to investigate mathematics using the software. —Prepared by Keith Leatham, Brigham Young University.
Webmodules
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The Algebraic Reasoning Webmodule helps students understand the critical transition from arithmetic to algebra. It also highlights situations when algebra is, or can be, used. Marginal notes are placed in the text at the appropriate locations to direct students to the webmodule. —Prepared by Keith Leatham, Brigham Young University.
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The Children’s Literature Webmodule provides references to many mathematically related examples of children’s books for each chapter. These references are noted in the margins near the mathematics that corresponds to the content of the book. The webmodule also contains ideas about using children’s literature in the classroom. —Prepared by Joan Cohen Jones, Eastern Michigan University.
•
The Introduction to Graph Theory Webmodule has been moved from the Topics to the companion Web Site to save space in the book and yet allow professors the flexibility to download it from the Web if they choose to use it.
The companion Web site also includes:
• • • •
Links to NCTM Standards A Logo and TI83 graphing calculator tutorial Four cumulative tests covering material up to the end of Chapters 4, 9, 12, and 16. Research Article References: A complete list of references for the research articles that are mentioned in the Reflections from Research margin notes throughout the book.
Guide to Problem Solving This valuable resource, available as a webmodule on the companion Web site, contains more than 200 creative problems keyed to the problem solving strategies in the textbook and includes:
• • • •
Opening Problem: an introductory problem to motivate the need for a strategy. Solution/Discussion/Clues: A workedout solution of the opening problem together with a discussion of the strategy and some clues on when to select this strategy. Practice Problems: A second problem that uses the same strategy together with a workedout solution and two practice problems. Mixed Strategy Practice: Four practice problems that can be solved using one or more of the strategies introduced to that point.
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Additional Practice Problems and Additional Mixed Strategy Problems: Sections that provide more practice for particular strategies as well as many problems for which students need to identify appropriate strategies. —Prepared by Don Miller, who retired as a professor of mathematics at St. Cloud State University.
The Geometer’s Sketchpad© Developed by Key Curriculum Press, this dynamic geometry construction and exploration tool allows users to create and manipulate precise figures while preserving geometric relationships. This software is only available when packaged with the text. Please contact your local Wiley representative for further details. WileyPLUS WileyPLUS is a powerful online tool that will help you study more effectively, get immediate feedback when you practice on your own, complete assignments and get help with problem solving, and keep track of how you’re doing—all at one easytouse Web site.
Resources for the Instructor Companion Web Site The companion Web site is available to text adopters and provides a wealth of resources including:
• •
PowerPoint Slides of more than 190 images that include figures from the text and several generic masters for dot paper, grids, and other formats. Instructors also have access to all student Web site features. See above for more details.
Instructor Resource Manual This manual contains chapterbychapter discussions of the text material, student “expectations” (objectives) for each chapter, answers for all Part B exercises and problems, and answers for all of the evennumbered problems in the Guide to ProblemSolving. —Prepared by Lyn Riverstone, Oregon State University ISBN 9780470233023
NEW! Computerized/Print Test Bank The Computerized/Printed Test Bank includes a collection of over 1,100 open response, multiplechoice, true/false, and freeresponse questions, nearly 80% of which are algorithmic. —Prepared by Mark McKibben, Goucher College Computerized Test Bank ISBN 9780470292969 Printed Test Bank ISBN 9780470292952
WileyPLUS WileyPLUS is a powerful online tool that provides instructors with an integrated suite of resources, including an online version of the text, in one easytouse Web site. Organized around the essential activities you perform in class, WileyPLUS allows you to create class presentations, assign homework and quizzes for automatic grading, and track student progress. Please visit http://edugen.wiley.com or contact your local Wiley representative for a demonstration and further details.
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Acknowledgments During the development of Mathematics for Elementary Teachers, Eighth Edition, we benefited from comments, suggestions, and evaluations from many of our colleagues. We would like to acknowledge the contributions made by the following people: Reviewers for the Eighth Edition Seth Armstrong, Southern Utah University Elayne Bowman, University of Oklahoma Anne Brown, Indiana University, South Bend David C. Buck, Elizabethtown Alison Carter, Montgomery College Janet Cater, California State University, Bakersfield Darwyn Cook, Alfred University Christopher Danielson, Minnesota State University Mankato Linda DeGuire, California State University, Long Beach Cristina Domokos, California State University, Sacramento Scott Fallstrom, University of Oregon Teresa Floyd, Mississippi College Rohitha Goonatilake, Texas A&M International University Margaret Gruenwald, University of Southern Indiana Joan Cohen Jones, Eastern Michigan University Joe Kemble, Lamar University Margaret Kinzel, Boise State University J. Lyn Miller, Slippery Rock University Girija NairHart, Ohio State University, Newark Sandra Nite, Texas A&M University Sally Robinson, University of Arkansas, Little Rock Nancy Schoolcraft, Indiana University, Bloomington Karen E. Spike, University of North Carolina, Wilmington Brian Travers, Salem State Mary Wiest, Minnesota State University, Mankato Mark A. Zuiker, Minnesota State University, Mankato Student Activity Manual Reviewers Kathleen Almy, Rock Valley College Margaret Gruenwald, University of Southern Indiana Kate Riley, California Polytechnic State University Robyn Sibley, Montgomery County Public Schools State Standards Reviewers Joanne C. Basta, Niagara University Joyce Bishop, Eastern Illinois University Tom Fox, University of Houston, Clear Lake Joan C. Jones, Eastern Michigan University Kate Riley, California Polytechnic State University Janine Scott, Sam Houston State University Murray Siegel, Sam Houston State University Rebecca Wong, West Valley College
In addition, we would like to acknowledge the contributions made by colleagues from earlier editions. Reviewers Paul Ache, Kutztown University Scott Barnett, Henry Ford Community College Chuck Beals, Hartnell College Peter Braunfeld, University of Illinois Tom Briske, Georgia State University Anne Brown, Indiana University, South Bend Christine Browning, Western Michigan University Tommy Bryan, Baylor University Lucille Bullock, University of Texas Thomas Butts, University of Texas, Dallas Dana S. Craig, University of Central Oklahoma Ann Dinkheller, Xavier University John Dossey, Illinois State University Carol Dyas, University of Texas, San Antonio Donna Erwin, Salt Lake Community College Sheryl Ettlich, Southern Oregon State College Ruhama Even, Michigan State University Iris B. Fetta, Clemson University Majorie Fitting, San Jose State University Susan Friel, Math/Science Education Network, University of North Carolina Gerald Gannon, California State University, Fullerton Joyce Rodgers Griffin, Auburn University Jerrold W. Grossman, Oakland University Virginia Ellen Hanks, Western Kentucky University John G. Harvey, University of Wisconsin, Madison Patricia L. Hayes, Utah State University, Uintah Basin Branch Campus Alan Hoffer, University of California, Irvine Barnabas Hughes, California State University, Northridge Joan Cohen Jones, Eastern Michigan University Marilyn L. Keir, University of Utah Joe Kennedy, Miami University Dottie King, Indiana State University Richard Kinson, University of South Alabama Margaret Kinzel, Boise State University John Koker, University of Wisconsin David E. Koslakiewicz, University of Wisconsin, Milwaukee Raimundo M. Kovac, Rhode Island College Josephine Lane, Eastern Kentucky University Louise Lataille, Springfield College Roberts S. Matulis, Millersville University Mercedes McGowen, Harper College Flora Alice Metz, Jackson State Community College J. Lyn Miller, Slippery Rock University Barbara Moses, Bowling Green State University Maura Murray, University of Massachusetts Kathy Nickell, College of DuPage Dennis Parker, The University of the Pacific William Regonini, California State University, Fresno
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James Riley, Western Michigan University Kate Riley, California Polytechnic State University Eric Rowley, Utah State University Peggy Sacher, University of Delaware Janine Scott, Sam Houston State University Lawrence Small, L.A. Pierce College Joe K. Smith, Northern Kentucky University J. Phillip Smith, Southern Connecticut State University Judy Sowder, San Diego State University Larry Sowder, San Diego State University Karen Spike, University of Northern Carolina, Wilmington Debra S. Stokes, East Carolina University Jo Temple, Texas Tech University Lynn Trimpe, Linn–Benton Community College Jeannine G. Vigerust, New Mexico State University Bruce Vogeli, Columbia University Kenneth C. Washinger, Shippensburg University Brad Whitaker, Point Loma Nazarene University John Wilkins, California State University, Dominguez Hills Questionnaire Respondents Mary Alter, University of Maryland Dr. J. Altinger, Youngstown State University Jamie Whitehead Ashby, Texarkana College Dr. Donald Balka, Saint Mary’s College Jim Ballard, Montana State University Jane Baldwin, Capital University Susan Baniak, Otterbein College James Barnard, Western Oregon State College Chuck Beals, Hartnell College Judy Bergman, University of Houston, Clearlake James Bierden, Rhode Island College Neil K. Bishop, The University of Southern Mississippi Gulf Coast Jonathan Bodrero, Snow College Dianne Bolen, Northeast Mississippi Community College Peter Braunfeld, University of Illinois Harold Brockman, Capital University Judith Brower, North Idaho College Anne E. Brown, Indiana University, South Bend Harmon Brown, Harding University Christine Browning, Western Michigan University Joyce W. Bryant, St. Martin’s College R. Elaine Carbone, Clarion University Randall Charles, San Jose State University Deann Christianson, University of the Pacific Lynn Cleary, University of Maryland Judith Colburn, Lindenwood College Sister Marie Condon, Xavier University Lynda Cones, Rend Lake College Sister Judith Costello, Regis College H. Coulson, California State University Dana S. Craig, University of Central Oklahoma Greg Crow, John Carroll University Henry A. Culbreth, Southern Arkansas University, El Dorado
Carl Cuneo, Essex Community College Cynthia Davis, Truckee Meadows Community College Gregory Davis, University of Wisconsin, Green Bay Jennifer Davis, Ulster County Community College Dennis De Jong, Dordt College Mary De Young, Hop College Louise Deaton, Johnson Community College Shobha Deshmukh, College of Saint Benedict/St. John’s University Sheila Doran, Xavier University Randall L. Drum, Texas A&M University P. R. Dwarka, Howard University Doris Edwards, Northern State College Roger Engle, Clarion University Kathy Ernie, University of Wisconsin Ron Falkenstein, Mott Community College Ann Farrell, Wright State University Francis Fennell, Western Maryland College Joseph Ferrar, Ohio State University Chris Ferris, University of Akron Fay Fester, The Pennsylvania State University Marie Franzosa, Oregon State University Margaret Friar, Grand Valley State College Cathey Funk, Valencia Community College Dr. Amy Gaskins, Northwest Missouri State University Judy Gibbs, West Virginia University Daniel Green, Olivet Nazarene University Anna Mae Greiner, Eisenhower Middle School Julie Guelich, Normandale Community College Ginny Hamilton, Shawnee State University Virginia Hanks, Western Kentucky University Dave Hansmire, College of the Mainland Brother Joseph Harris, C.S.C., St. Edward’s University John Harvey, University of Wisconsin Kathy E. Hays, Anne Arundel Community College Patricia Henry, Weber State College Dr. Noal Herbertson, California State University Ina Lee Herer, TriState University Linda Hill, Idaho State University Scott H. Hochwald, University of North Florida Susan S. Hollar, Kalamazoo Valley Community College Holly M. Hoover, Montana State University, Billings WeiShen Hsia, University of Alabama Sandra Hsieh, Pasadena City College Jo Johnson, Southwestern College Patricia Johnson, Ohio State University Pat Jones, Methodist College Judy Kasabian, El Camino College Vincent Kayes, Mt. St. Mary College Julie Keener, Central Oregon Community College Joe Kennedy, Miami University Susan Key, Meridien Community College Mary Kilbridge, Augustana College Mike Kilgallen, Lincoln Christian College Judith Koenig, California State University, Dominguez Hills Josephine Lane, Eastern Kentucky University
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Don Larsen, Buena Vista College Louise Lataille, Westfield State College Vernon Leitch, St. Cloud State University Steven C. Leth, University of Northern Colorado Lawrence Levy, University of Wisconsin Robert Lewis, LinnBenton Community College Lois Linnan, Clarion University Jack Lombard, Harold Washington College Betty Long, Appalachian State University Ann Louis, College of the Canyons C. A. Lubinski, Illinois State University Pamela Lundin, Lakeland College Charles R. Luttrell, Frederick Community College Carl Maneri, Wright State University Nancy Maushak, William Penn College Edith Maxwell, West Georgia College Jeffery T. McLean, University of St. Thomas George F. Mead, McNeese State University Wilbur Mellema, San Jose City College Diane Miller, Middle Tennessee State University Clarence E. Miller, Jr. Johns Hopkins University Ken Monks, University of Scranton Bill Moody, University of Delaware Kent Morris, Cameron University Lisa Morrison, Western Michigan University Barbara Moses, Bowling Green State University Fran Moss, Nicholls State University Mike Mourer, Johnston Community College Katherine Muhs, St. Norbert College Gale Nash, Western State College of Colorado T. Neelor, California State University Jerry Neft, University of Dayton Gary Nelson, Central Community College, Columbus Campus James A. Nickel, University of Texas, Permian Basin Kathy Nickell, College of DuPage Susan Novelli, Kellogg Community College Jon O’Dell, Richland Community College Jane Odell, Richland College Bill W. Oldham, Harding University Jim Paige, Wayne State College Wing Park, College of Lake County Susan Patterson, Erskine College (retired) Shahla Peterman, University of Missouri Gary D. Peterson, Pacific Lutheran University Debra Pharo, Northwestern Michigan College Tammy PowellKopilak, Dutchess Community College Christy Preis, Arkansas State University, Mountain Home Robert Preller, Illinois Central College Dr. William Price, Niagara University Kim Prichard, University of North Carolina Stephen Prothero, Williamette University Janice Rech, University of Nebraska Tom Richard, Bemidji State University Jan Rizzuti, Central Washington University Anne D. Roberts, University of Utah David Roland, University of Mary Hardin–Baylor
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Frances Rosamond, National University Richard Ross, Southeast Community College Albert Roy, Bristol Community College Bill Rudolph, Iowa State University Bernadette Russell, Plymouth State College Lee K. Sanders, Miami University, Hamilton Ann Savonen, Monroe County Community College Rebecca Seaberg, Bethel College Karen Sharp, Mott Community College Marie Sheckels, Mary Washington College Melissa Shepard Loe, University of St. Thomas Joseph Shields, St. Mary’s College, MN Lawrence Shirley, Towson State University Keith Shuert, Oakland Community College B. Signer, St. John’s University Rick Simon, Idaho State University James Smart, San Jose State University Ron Smit, University of Portland Gayle Smith, Lane Community College Larry Sowder, San Diego State University Raymond E. Spaulding, Radford University William Speer, University of Nevada, Las Vegas Sister Carol Speigel, BVM, Clarke College Karen E. Spike, University of North Carolina, Wilmington Ruth Ann Stefanussen, University of Utah Carol Steiner, Kent State University Debbie Stokes, East Carolina University Ruthi Sturdevant, Lincoln University, MO Viji Sundar, California State University, Stanislaus Ann Sweeney, College of St. Catherine, MN Karen Swenson, George Fox College Carla Tayeh, Eastern Michigan University Janet Thomas, Garrett Community College S. Thomas, University of Oregon Mary Beth Ulrich, Pikeville College Martha Van Cleave, Linfield College Dr. Howard Wachtel, Bowie State University Dr. Mary WagnerKrankel, St. Mary’s University Barbara Walters, Ashland Community College Bill Weber, Eastern Arizona College Joyce Wellington, Southeastern Community College Paula White, Marshall University Heide G. Wiegel, University of Georgia Jane Wilburne, West Chester University Jerry Wilkerson, Missouri Western State College Jack D. Wilkinson, University of Northern Iowa Carole Williams, Seminole Community College Delbert Williams, University of Mary Hardin–Baylor Chris Wise, University of Southwestern Louisiana John L. Wisthoff, Anne Arundel Community College (retired) Lohra Wolden, Southern Utah University Mary Wolfe, University of Rio Grande Vernon E. Wolff, Moorhead State University Maria Zack, Point Loma Nazarene College Stanley 1. Zehm, Heritage College Makia Zimmer, Bethany College
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Focus Group Participants
Content Connections Survey Respondents
Mara Alagic, Wichita State University Robin L. Ayers, Western Kentucky University Elaine Carbone, Clarion University of Pennsylvania Janis Cimperman, St. Cloud State University Richard DeCesare, Southern Connecticut State University Maria Diamantis, Southern Connecticut State University Jerrold W. Grossman, Oakland University Richard H. Hudson, University of South Carolina, Columbia Carol Kahle, Shippensburg University Jane Keiser, Miami University Catherine Carroll Kiaie, Cardinal Stritch University Cynthia Y. Naples, St. Edward’s University Armando M. MartinezCruz, California State University, Fullerton David L. Pagni, Fullerton University Melanie Parker, Clarion University of Pennsylvania Carol PhillipsBey, Cleveland State University
Marc Campbell, Daytona Beach Community College Porter Coggins, University of Wisconsin–Stevens Point Don Collins, Western Kentucky University Allan Danuff, Central Florida Community College Birdeena Dapples, Rocky Mountain College Nancy Drickey, Linfield College Thea Dunn, University of Wisconsin–River Falls Mark Freitag, East Stroudsberg University Paula Gregg, University of South Carolina Aiken Brian Karasek, Arizona Western College Chris Kolaczewski, Ferris University of Akron R. Michael Krach, Towson University Randa Lee Kress, Idaho State University Marshall Lassak, Eastern Illinois University Katherine Muhs, St. Norbert College Bethany Noblitt, Northern Kentucky University
We would like to acknowledge the following people for their assistance in the preparation of the first seven editions of this book: Ron Bagwell, Jerry Becker, Julie Borden, Sue Borden, Tommy Bryan, Juli Dixon, Christie Gilliland, Dale Green, Kathleen Seagraves Higdon, Hester Lewellen, Roger Maurer, David Metz, Naomi Munton, Tilda Runner, Karen Swenson, Donna Templeton, Lynn Trimpe, Rosemary Troxel, Virginia Usnick, and Kris Warloe. We thank Robyn Silbey for her expert review of several of the features in our seventh edition and Becky Gwilliam for her research contributions to Chapter 10 and the Reflections from Research. We also thank Lyn Riverstone and Vikki Maurer for their careful checking of the accuracy of the answers. We also want to acknowledge Marcia Swanson and Karen Swenson for their creation of and contribution to our Student Resource Handbook during the first seven editions with a special thanks to Lyn Riverstone for her expert revision of the new Student Activity Manual for the seventh edition. Thanks are also due to Don Miller for his Guide to Problem Solving, to Lyn Trimpe, Roger Maurer, and Vikki Maurer, for their longtime authorship of our Student Hints and Solutions Manual, to Keith Leathem for the Spreadsheet Tutorial and Algebraic Reasoning Web Module, Armando MartinezCruz for The Geometer’s Sketchpad® Tutorial, to Joan Cohen Jones for the Children’s Literature Webmodule, and to Lawrence O. Cannon, E. Robert Heal, Joel Duffin, Richard Wellman, and Ethalinda K. S. Cannon for the eManipulatives activities. We are very grateful to our publisher, Laurie Rosatone, and our acquisitions editor, Jessica Jacobs, for their commitment and super teamwork, to our senior production editor, Valerie A. Vargas, for attending to the details we missed, to Martha Beyerlein, our fullservice representative and copyeditor, for lighting the path as we went from manuscript to the final book, and to Melody Englund for creating the index. Other Wiley staff who helped bring this book and its print and media supplements to fruition are: Christopher Ruel, Executive Marketing Manager; Stefanie Liebman, media editor; Ann Berlin, Vice President, Production and Manufacturing; Dorothy Sinclair, Production Services Manager; Kevin Murphy, Senior Designer; Lisa Gee, Photo Researcher; Michael Shroff, Assistant Editor; Jeffrey Benson, Editorial Assistant; and Matt Winslow, Production Assistant. They have been uniformly wonderful to work with— John Wiley would have been proud of them. Finally, we welcome comments from colleagues and students. Please feel free to send suggestions to Gary at
[email protected] and Blake at
[email protected] Please include both of us in any communications. G.L.M. B.E.P.
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FOCUS ON
CHAPTER
1
George Pólya—The Father of Modern Problem Solving
eorge Pólya was born in Hungary in 1887. He received his Ph.D. at the University of Budapest. In 1940 he came to Brown University and then joined the faculty at Stanford University in 1942.
G
In his studies, he became interested in the process of discovery, which led to his famous fourstep process for solving problems: 1. Understand the problem. 2. Devise a plan. 3. Carry out the plan. 4. Look back. Pólya wrote over 250 mathematical papers and three books that promote problem solving. His most famous
book, How to Solve It, which has been translated into 15 languages, introduced his fourstep approach together with heuristics, or strategies, which are helpful in solving problems. Other important works by Pólya are Mathematical Discovery, Volumes 1 and 2, and Mathematics and Plausible Reasoning, Volumes 1 and 2. He died in 1985, leaving mathematics with the important legacy of teaching problem solving. His “Ten Commandments for Teachers” are as follows: 1. Be interested in your subject. 2. Know your subject. 3. Try to read the faces of your students; try to see their expectations and difficulties; put yourself in their place. 4. Realize that the best way to learn anything is to discover it by yourself. 5. Give your students not only information, but also knowhow, mental attitudes, the habit of methodical work. 6. Let them learn guessing. 7. Let them learn proving. 8. Look out for such features of the problem at hand as may be useful in solving the problems to come—try to disclose the general pattern that lies behind the present concrete situation. 9. Do not give away your whole secret at once—let the students guess before you tell it—let them find out by themselves as much as is feasible. 10. Suggest; do not force information down their throats.
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ProblemSolving Strategies 1. Guess and Test 2. Draw a Picture 3. Use a Variable 4. Look for a Pattern 5. Make a List
Because problem solving is the main goal of mathematics, this chapter introduces the six strategies listed in the ProblemSolving Strategies box that are helpful in solving problems. Then, at the beginning of each chapter, an initial problem is posed that can be solved by using the strategy introduced in that chapter. As you move through this book, the ProblemSolving Strategies boxes at the beginning of each chapter expand, as should your ability to solve problems.
INITIAL PROBLEM
6. Solve a Simpler Problem Place the whole numbers 1 through 9 in the circles in the accompanying triangle so that the sum of the numbers on each side is 17.
A solution to this Initial Problem is on page 38.
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INTRODUCTION nce, at an informal meeting, a social scientist asked of a mathematics professor, “What’s the main goal of teaching mathematics?” The reply was, “Problem solving.” In return, the mathematician asked, “What is the main goal of teaching the social sciences?” Once more the answer was “Problem solving.” All successful engineers, scientists, social scientists, lawyers, accountants, doctors, business managers, and so on have to be good problem solvers. Although the problems that people encounter may be very diverse, there are common elements and an underlying structure that can help to facilitate problem solving. Because of the universal importance of problem solving, the main professional group in mathematics education, the National Council of Teachers of Mathematics (NCTM), recommended in its 1980 An Agenda for Action that “problem solving be the focus of school mathematics in the 1980s.” The National Council of Teachers of Mathematics’ 1989 Curriculum and Evaluation Standards for School Mathematics called for increased attention to the teaching of problem solving in K–8 mathematics. Areas of emphasis include word problems, applications, patterns and relationships, openended problems, and problem situations represented verbally, numerically, graphically, geometrically, or symbolically. The NCTM’s 2000 Principles and Standards for School Mathematics identified problem solving as one of the processes by which all mathematics should be taught. This chapter introduces a problemsolving process together with six strategies that will aid you in solving problems.
O
Key Concepts from NCTM Curriculum Focal Points
• • • • • •
1.1 S TART I NG POINT
K I N D E RGAR TEN : Choose, combine, and apply effective strategies for answering quantitative questions. GRA D E 1: Develop an understanding of the meanings of addition and subtraction and strategies to solve such arithmetic problems. Solve problems involving the relative sizes of whole numbers. GRA D E 3: Apply increasingly sophisticated strategies . . . to solve multiplication and division problems. GRADE 4 AND 5: Select appropriate units, strategies, and tools for solving problems. GRA D E 6: Solve a wide variety of problems involving ratios and rates. GRADE 7: Use ratio and proportionality to solve a wide variety of percent problems.
THE PROBLEMSOLVING PROCESS AND STRATEGIES Use any strategy you know to solve the problem below. As you solve the problem below, pay close attention to the thought processes and steps that you use. Write down these strategies and compare them to a classmate’s. Are there any similarities in your approaches to solving the problem below? Problem: Lin’s garden has an area of 78 square yards. The length of the garden is 5 less than three times its width. What are the dimensions of Lin’s garden?
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Children’s Literature www.wiley.com/college/musser See “The Math Curse” by Jon Sciezke.
Pólya’s Four Steps In this book we often distinguish between “exercises” and “problems.” Unfortunately, the distinction cannot be made precise. To solve an exercise, one applies a routine procedure to arrive at an answer. To solve a problem, one has to pause, reflect, and perhaps take some original step never taken before to arrive at a solution. This need for some sort of creative step on the solver’s part, however minor, is what distinguishes a problem from an exercise. To a young child, finding 3 ⫹ 2 might be a problem, whereas it is a fact for you. For a child in the early grades, the question “How do you divide 96 pencils equally among 16 children?” might pose a problem, but for you it suggests the exercise “find 96 ⫼ 16.” These two examples illustrate how the distinction between an exercise and a problem can vary, since it depends on the state of mind of the person who is to solve it. Doing exercises is a very valuable aid in learning mathematics. Exercises help you to learn concepts, properties, procedures, and so on, which you can then apply when solving problems. This chapter provides an introduction to the process of problem solving. The techniques that you learn in this chapter should help you to become a better problem solver and should show you how to help others develop their problemsolving skills. A famous mathematician, George Pólya, devoted much of his teaching to helping students become better problem solvers. His major contribution is what has become known as Pólya’s fourstep process for solving problems.
Reflection from Research
Step 1
Many children believe that the answer to a word problem can always be found by adding, subtracting, multiplying, or dividing two numbers. Little thought is given to understanding the context of the problem (Verschaffel, De Corte, & Vierstraete, 1999).
• • • • • • •
Do you understand all the words? Can you restate the problem in your own words? Do you know what is given? Do you know what the goal is? Is there enough information? Is there extraneous information? Is this problem similar to another problem you have solved?
Step 2 Developing Algebraic Reasoning www.wiley.com/college/musser See “Mathematizing.”
Understand the Problem
Devise a Plan
Can one of the following strategies (heuristics) be used? (A strategy is defined as an artful means to an end.) 1. Guess and test. 2. Draw a picture. 3. Use a variable. 4. Look for a pattern. 5. Make a list. 6. Solve a simpler problem. 7. Draw a diagram. 8. Use direct reasoning. 9. Use indirect reasoning. 10. Use properties of numbers. 11. Solve an equivalent problem.
12. Work backward. 13. Use cases. 14. Solve an equation. 15. Look for a formula. 16. Do a simulation. 17. Use a model. 18. Use dimensional analysis. 19. Identify subgoals. 20. Use coordinates. 21. Use symmetry.
The first six strategies are discussed in this chapter; the others are introduced in subsequent chapters.
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S T U D E N T PA G E S N A P S H O T
From Mathematics, Grade 2 Pupil Edition, p. 233. Copyright 2005 by Scott ForesmanAddison Wesley.
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Step 3
• • •
Implement the strategy or strategies that you have chosen until the problem is solved or until a new course of action is suggested. Give yourself a reasonable amount of time in which to solve the problem. If you are not successful, seek hints from others or put the problem aside for a while. (You may have a flash of insight when you least expect it!) Do not be afraid of starting over. Often, a fresh start and a new strategy will lead to success.
Step 4
• • •
Carry Out the Plan
Look Back
Is your solution correct? Does your answer satisfy the statement of the problem? Can you see an easier solution? Can you see how you can extend your solution to a more general case?
Usually, a problem is stated in words, either orally or written. Then, to solve the problem, one translates the words into an equivalent problem using mathematical symbols, solves this equivalent problem, and then interprets the answer. This process is summarized in Figure 1.1.
Original problem
Translate
Solve
Check Answer to original problem
Mathematical version of the problem
Interpret
Solution to the mathematical version
Figure 1.1
Reflection from Research Researchers suggest that teachers think aloud when solving problems for the first time in front of the class. In so doing, teachers will be modeling successful problemsolving behaviors for their students (Schoenfeld, 1985).
NCTM Standard Instructional programs should enable all students to apply and adapt a variety of appropriate strategies to solve problems.
Learning to utilize Pólya’s four steps and the diagram in Figure 1.1 are first steps in becoming a good problem solver. In particular, the “Devise a Plan” step is very important. In this chapter and throughout the book, you will learn the strategies listed under the “Devise a Plan” step, which in turn help you decide how to proceed to solve problems. However, selecting an appropriate strategy is critical! As we worked with students who were successful problem solvers, we asked them to share “clues” that they observed in statements of problems that helped them select appropriate strategies. Their clues are listed after each corresponding strategy. Thus, in addition to learning how to use the various strategies herein, these clues can help you decide when to select an appropriate strategy or combination of strategies. Problem solving is as much an art as it is a science. Therefore, you will find that with experience you will develop a feeling for when to use one strategy over another by recognizing certain clues, perhaps subconsciously. Also, you will find that some problems may be solved in several ways using different strategies. In summary, this initial material on problem solving is a foundation for your success in problem solving. Review this material on Pólya’s four steps as well as the strategies and clues as you continue to develop your expertise in solving problems.
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7
ProblemSolving Strategies The remainder of this chapter is devoted to introducing several problemsolving strategies.
Strategy 1
Guess and Test
Problem Place the digits 1, 2, 3, 4, 5, 6 in the circles in Figure 1.2 so that the sum of the three numbers on each side of the triangle is 12. We will solve the problem in three ways to illustrate three different approaches to the Guess and Test strategy. As its name suggests, to use the Guess and Test strategy, you guess at a solution and test whether you are correct. If you are incorrect, you refine your guess and test again. This process is repeated until you obtain a solution.
Figure 1.2
Step 1
Understand the Problem
Each number must be used exactly one time when arranging the numbers in the triangle. The sum of the three numbers on each side must be 12.
First Approach: Random Guess and Test Step 2
Devise a Plan
Tear off six pieces of paper and mark the numbers 1 through 6 on them and then try combinations until one works.
Step 3
Carry Out the Plan
Arrange the pieces of paper in the shape of an equilateral triangle and check sums. Keep rearranging until three sums of 12 are found.
Second Approach: Systematic Guess and Test Step 2
Devise a Plan
Rather than randomly moving the numbers around, begin by placing the smallest numbers—namely, 1, 2, 3—in the corners. If that does not work, try increasing the numbers to 1, 2, 4, and so on.
Step 3
1 11
11
Carry Out the Plan
With 1, 2, 3 in the corners, the side sums are too small; similarly with 1, 2, 4. Try 1, 2, 5 and 1, 2, 6. The side sums are still too small. Next try 2, 3, 4, then 2, 3, 5, and so on, until a solution is found. One also could begin with 4, 5, 6 in the corners, then try 3, 4, 5, and so on.
Third Approach: Inferential Guess and Test Step 2
Figure 1.3
Devise a Plan
Start by assuming that 1 must be in a corner and explore the consequences. 2 10
Figure 1.4
Step 3 10
Carry Out the Plan
If 1 is placed in a corner, we must find two pairs out of the remaining five numbers whose sum is 11 (Figure 1.3). However, out of 2, 3, 4, 5, and 6, only 6 ⫹ 5 ⫽ 11. Thus, we conclude that 1 cannot be in a corner. If 2 is in a corner, there must be two pairs left that add to 10 (Figure 1.4). But only 6 ⫹ 4 ⫽ 10. Therefore, 2
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3 9
9
Step 4
Figure 1.5 4 3 5
2 1
cannot be in a corner. Finally, suppose that 3 is in a corner. Then we must satisfy Figure 1.5. However, only 5 ⫹ 4 ⫽ 9 of the remaining numbers. Thus, if there is a solution, 4, 5, and 6 will have to be in the corners (Figure 1.6). By placing 1 between 5 and 6, 2 between 4 and 6, and 3 between 4 and 5, we have a solution.
6
Look Back
Notice how we have solved this problem in three different ways using Guess and Test. Random Guess and Test is often used to get started, but it is easy to lose track of the various trials. Systematic Guess and Test is better because you develop a scheme to ensure that you have tested all possibilities. Generally, Inferential Guess and Test is superior to both of the previous methods because it usually saves time and provides more information regarding possible solutions.
Figure 1.6 NCTM Standard Instructional programs should enable all students to monitor and reflect on the process of mathematical problem solving.
Additional Problems Where the Strategy “Guess and Test” Is Useful 1. In the following cryptarithm—that is, a collection of words where the letters represent numbers—sun and fun represent two threedigit numbers, and swim is their fourdigit sum. Using all of the digits 0, 1, 2, 3, 6, 7, and 9 in place of the letters where no letter represents two different digits, determine the value of each letter. sun + fun swim
Step 1
Understand the Problem
Each of the letters in sun, fun, and swim must be replaced with the numbers 0, 1, 2, 3, 6, 7, and 9, so that a correct sum results after each letter is replaced with its associated digit. When the letter n is replaced by one of the digits, then n ⫹ n must be m or 10 ⫹ m, where the 1 in the 10 is carried to the tens column. Since 1 ⫹ 1 ⫽ 2, 3 ⫹ 3 ⫽ 6, and 6 ⫹ 6 ⫽ 12, there are three possibilities for n, namely, 1, 3, or 6. Now we can try various combinations in an attempt to obtain the correct sum.
Step 2
Devise a Plan
Use Inferential Guess and Test. There are three choices for n. Observe that sun and fun are threedigit numbers and that swim is a fourdigit number. Thus we have to carry when we add s and f. Therefore, the value for s in swim is 1. This limits the choices of n to 3 or 6.
Step 3
Carry Ot the Plan
Since s ⫽ 1 and s ⫹ f leads to a twodigit number, f must be 9. Thus there are two possibilities: (a)
1u3 + 9u3 1wi6
(b)
1u6 + 9u6 1wi2
In (a), if u ⫽ 0, 2, or 7, there is no value possible for i among the remaining digits. In (b), if u ⫽ 3, then u ⫹ u plus the carry from 6 ⫹ 6 yields i ⫽ 7. This leaves w ⫽ 0 for a solution.
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9
Look Back
The reasoning used here shows that there is one and only one solution to this problem. When solving problems of this type, one could randomly substitute digits until a solution is found. However, Inferential Guess and Test simplifies the solution process by looking for unique aspects of the problem. Here the natural places to start are n ⫹ n, u ⫹ u, and the fact that s ⫹ f yields a twodigit number. 2. Use four 4s and some of the symbols ⫹, ⫻, ⫺, ⫼, ( ) to give expressions for the whole numbers from 0 through 9: for example, 5 ⫽ (4 ⫻ 4 ⫹ 4) ⫼ 4. 3. For each shape in Figure 1.7, make one straight cut so that each of the two pieces of the shape can be rearranged to form a square. (NOTE: Answers for these problems are given after the Solution of the Initial Problem near the end of this chapter.) Figure 1.7
CLUES The Guess and Test strategy may be appropriate when
• • • • • •
There is a limited number of possible answers to test. You want to gain a better understanding of the problem. You have a good idea of what the answer is. You can systematically try possible answers. Your choices have been narrowed down by the use of other strategies. There is no other obvious strategy to try.
Review the preceding three problems to see how these clues may have helped you select the Guess and Test strategy to solve these problems.
Strategy 2 Reflection from Research Training children in the process of using pictures to solve problems results in more improved problemsolving performance than training students in any other strategy (Yancey, Thompson, & Yancey, 1989).
Draw a Picture
Often problems involve physical situations. In these situations, drawing a picture can help you better understand the problem so that you can formulate a plan to solve the problem. As you proceed to solve the following “pizza” problem, see whether you can visualize the solution without looking at any pictures first. Then work through the given solution using pictures to see how helpful they can be.
Problem Can you cut a pizza into 11 pieces with four straight cuts?
Step 1
Understand the Problem
Do the pieces have to be the same size and shape?
Step 2
Devise a Plan
An obvious beginning would be to draw a picture showing how a pizza is usually cut and to count the pieces. If we do not get 11, we have to try something else (Figure 1.8). Unfortunately, we get only eight pieces this way. NCTM Standard All students should describe, extend, and make generalizations about geometric and numeric patterns.
Figure 1.8
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Step 3
Carry Out the Plan
See Figure 1.9
1 Figure 1.9
Step 4
2
4
7
1
1
Look Back
Were you concerned about cutting equal pieces when you started? That is normal. In the context of cutting a pizza, the focus is usually on trying to cut equal pieces rather than the number of pieces. Suppose that circular cuts were allowed. Does it matter whether the pizza is circular or is square? How many pieces can you get with five straight cuts? n straight cuts?
Additional Problems Where the Strategy “Draw a Picture” Is Useful Not a tetromino
1. A tetromino is a shape made up of four squares where the squares must be joined along an entire side (Figure 1.10). How many different tetromino shapes are possible?
A tetromino
Step 1
Figure 1.10
The solution of this problem is easier if we make a set of pictures of all possible arrangements of four squares of the same size.
Step 2
Understand the Problem
Devise a Plan
Let’s start with the longest and narrowest configuration and work toward the most compact.
Step 3
Carry Out the Plan
Four in a row Three in a row with one on top of (or below) the end square. (Note: The upper square can be at either end — these two are considered to be equivalent.)
Three in a row, with one on top of (or below) the center square.
Two in a row, with two above.
Two in a row, with one above and one below the two.
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Look Back
Many similar problems can be posed using fewer or more squares. The problems become much more complex as the number of squares increases. Also, new problems can be posed using patterns of equilateral triangles. 2. If you have a chain saw with a bar 18 inches long, determine whether a 16foot log, 8 inches in diameter, can be cut into 4foot pieces by making only two cuts. 3. It takes 64 cubes to fill a cubical box that has no top. How many cubes are not touching a side or the bottom?
CLUES The Draw a Picture strategy may be appropriate when
• • • •
A physical situation is involved. Geometric figures or measurements are involved. You want to gain a better understanding of the problem. A visual representation of the problem is possible.
Review the preceding three problems to see how these clues may have helped you select the Draw a Picture strategy to solve these problems. NCTM Standard All students should represent the idea of a variable as an unknown quantity using a letter or a symbol.
Strategy 3
Use a Variable
Observe how letters were used in place of numbers in the previous “sun ⫹ fun ⫽ swim” cryptarithm. Letters used in place of numbers are called variables or unknowns. The Use a Variable strategy, which is one of the most useful problemsolving strategies, is used extensively in algebra and in mathematics that involves algebra.
Developing Algebraic Reasoning
Problem
www.wiley.com/college/musser See “Variable.”
What is the greatest number that evenly divides the sum of any three consecutive whole numbers? By trying several examples, you might guess that 3 is the greatest such number. However, it is necessary to use a variable to account for all possible instances of three consecutive numbers.
Reflection from Research Given the proper experiences, children as young as eight and nine years of age can learn to comfortably use letters to represent unknown values and can operate on representations involving letters and numbers while fully realizing that they did not know the values of the unknowns (Carraher, Schliemann, Brizuela, & Earnest, 2006).
Step 1
Understand the Problem
The whole numbers are 0, 1, 2, 3, . . . , so that consecutive whole numbers differ by 1. Thus an example of three consecutive whole numbers is the triple 3, 4, and 5. The sum of three consecutive whole numbers has a factor of 3 if 3 multiplied by another whole number produces the given sum. In the example of 3, 4, and 5, the sum is 12 and 3 ⫻ 4 equals 12. Thus 3 ⫹ 4 ⫹ 5 has a factor of 3.
Step 2
Devise a Plan
Since we can use a variable, say x, to represent any whole number, we can represent every triple of consecutive whole numbers as follows: x, x ⫹ 1, x ⫹ 2. Now we can proceed to see whether the sum has a factor of 3.
Step 3
Carry Out the Plan
The sum of x, x ⫹ 1, and x ⫹ 2 is x + (x + 1) + (x + 2) = 3x + 3 = 3(x + 1).
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Thus x ⫹ (x ⫹ 1) ⫹ (x ⫹ 2) is three times x ⫹ 1. Therefore, we have shown that the sum of any three consecutive whole numbers has a factor of 3. The case of x ⫽ 0 shows that 3 is the greatest such number.
Step 4
Look Back
Is it also true that the sum of any five consecutive whole numbers has a factor of 5? Or, more generally, will the sum of any n consecutive whole numbers have a factor of n? Can you think of any other generalizations?
Additional Problems Where the Strategy “Use a Variable” Is Useful 1. Find the sum of the first 10, 100, and 500 counting numbers.
Step 1
Understand the Problem
Since counting numbers are the numbers 1, 2, 3, 4, . . . , the sum of the first 10 counting numbers would be 1 ⫹ 2 ⫹ 3 ⫹ . . . ⫹ 8 ⫹ 9 ⫹ 10. Similarly, the sum of the first 100 counting numbers would be 1 ⫹ 2 ⫹ 3 ⫹ . . . ⫹ 98 ⫹ 99 ⫹ 100 and the sum of the first 500 counting numbers would be 1 ⫹ 2 ⫹ 3 ⫹ . . . 498 ⫹ 499 ⫹ 500.
Step 2
Devise a Plan
Rather than solve three different problems, the “Use a Variable” strategy can be used to find a general method for computing the sum in all three situations. Thus, the sum of the first n counting numbers would be expressed as 1 ⫹ 2 ⫹ 3 ⫹ . . . ⫹ (n ⫺ 2) ⫹ (n ⫺ 1) ⫹ n. The sum of these numbers can be found by noticing that the first number 1 added to the last number n is n ⫹ 1, which is the same as (n ⫺ 1) ⫹ 2 and (n ⫺ 2) ⫹ 3. Adding up all such pairs can be done by adding up all of the numbers twice.
Step 3
Carry Out the Plan
Reflection from Research “Students’ problemsolving performance was highly correlated with their problemposing performance.” Compared to less successful problem solvers, good problem solvers generated problems that were more mathematical, and their problems were more mathematically complex (Silver & Cai, 1996).
1 + 2 + 3 + . . . + (n  2) + (n  1) + n + n + (n  1) + (n  2) + . . . + 3 + 2 + 1 (n + 1) + (n + 1) + (n + 1) + . . . + (n + 1) + (n + 1) + (n + 1) = n # (n + 1) Since each number was added twice, the desired sum is obtained by dividing n 䡠 (n ⫹ 1) by 2 which yields 1 + 2 + 3 + . . . + (n  2) + (n  1) + n =
n # (n + 1) 2
The numbers 10, 100, and 500 can now replace the variable n to find our desired solutions: 1 + 2 + 3 + . . . + 8 + 9 + 10 =
10 # (10 + 1) = 55 2
1 + 2 + 3 + . . . + 98 + 99 + 100 =
100 # (101) = 5050 2
1 + 2 + 3 + . . . 498 + 499 + 500 =
500 # 501 = 125,250 2
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Look Back
Since the method for solving this problem is quite unique could it be used to solve other similar looking problems like: i. 3 ⫹ 6 ⫹ 9 ⫹ . . . ⫹ (3n ⫺ 6) ⫹ (3n ⫺ 3) ⫹ 3n ii. 21 ⫹ 25 ⫹ 29 ⫹ . . . ⫹ 113 ⫹ 117 ⫹ 121 2. Show that the sum of any five consecutive odd whole numbers has a factor of 5. 3. The measure of the largest angle of a triangle is nine times the measure of the smallest angle. The measure of the third angle is equal to the difference of the largest and the smallest. What are the measures of the angles? (Recall that the sum of the measures of the angles in a triangle is 180⬚.)
CLUES The Use a Variable strategy may be appropriate when
• • • • • • • •
A phrase similar to “for any number” is present or implied. A problem suggests an equation. A proof or a general solution is required. A problem contains phrases such as “consecutive,” “even,” or “odd” whole numbers. There is a large number of cases. There is an unknown quantity related to known quantities. There is an infinite number of numbers involved. You are trying to develop a general formula.
Review the preceding three problems to see how these clues may have helped you select the Use a Variable strategy to solve these problems.
Using Algebra to Solve Problems
NCTM Standard All students should develop an initial conceptual understanding of different uses of variables.
To effectively employ the Use a Variable strategy, students need to have a clear understanding of what a variable is and how to write and simplify equations containing variables. This subsection addresses these issues in an elementary introduction to algebra. There will be an expanded treatment of solving equations and inequalities in Chapter 9 after the real number system has been developed. A common way to introduce the use of variables is to find a general formula for a pattern of numbers such as 3, 6, 9, . . . 3n. One of the challenges for students is to see the role that each number plays in the expression. For example, the pattern 5, 8, 11, . . . is similar to the previous pattern, but it is more difficult to see that each term is two greater than a multiple of 3 and, thus, can be expressed in general as 3n ⫹ 2. Sometimes it is easier for students to use a variable to generalize a geometric pattern such as the one shown in the following example. This type of example may be used to introduce seventhgrade students to the concept of a variable. Following are four typical student solutions.
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Example 1.1
Describe at least four different ways to count the dots in Figure 1.11.
S O LU T I O N The obvious method of solution is to count the dots—there are 16. Another student’s method is illustrated in Figure 1.12. Figure 1.11
4 * 3 + 4 f ¡ 4 * (5  2) + 4
Figure 1.12
The student counts the number of interior dots on each side, 3, and multiplies by the number of sides, 4, and then adds the dots in the corners, 4. This method generates the expression 4 ⫻ 3 ⫹ 4 ⫽ 16. A second way to write this expression is 4 ⫻ (5 ⫺ 2) ⫹ 4 ⫽ 16 since the 3 interior dots can be determined by subtracting the two corners from the 5 dots on a side. Both of these methods are shown in Figure 1.12. A third method is to count all of the dots on a side, 5, and multiply by the number of sides. Four must then be subtracted because each corner has been counted twice, once for each side it belongs to. This method is illustrated in Figure 1.13 and generates the expression shown.
4 * 5  4 = 166 ¡
Figure 1.13 Reflection from Research Sixthgrade students with no formal instruction in algebra are “generally able to solve problems involving specific cases and showed remarkable ability to generalize the problem situations and to write equations using variables. However they rarely used their equations to solve related problems” (Swafford & Langrall, 2000).
In the two previous methods, either corner dots are not counted (so they must be added on) or they are counted twice (so they must be subtracted to avoid double counting). The following fourth method assigns each corner to only one side (Figure 1.14).
4 * 4 = 16 s ¡ 4 * (5  1) = 16
Figure 1.14
Thus, we encircle 4 dots on each side and multiply by the number of sides. This yields the expression 4 ⫻ 4 ⫽ 16. Because the 4 dots on each side come from the 5 total dots on a side minus 1 corner, this expression could also be written as 4 ⫻ (5 ⫺ 1) ⫽ 16 (see Figure 1.14). ■
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There are many different methods for counting the dots in the previous example and each method has a geometric interpretation as well as a corresponding arithmetic expression. Could these methods be generalized to 50, 100, 1000 or even n dots on a side? The next example discusses how these generalizations can be viewed as well as displays the generalized solutions of seventhgrade students. = n dots
Suppose the square arrangement of dots in Example 1.1 had n dots on each side. Write an algebraic expression that would describe the total number of dots in such a figure (Figure 1.15).
Example 1.2
Figure 1.15
S O LU T I O N It is easier to write a general expression for those in Example 1.1 when you understand the origins of the numbers in each expression. In all such cases, there will be 4 corners and 4 sides, so the values that represent corners and sides will stay fixed at 4. On the other hand, in Figure 1.15, any value that was determined based on the number of dots on the side will have to reflect the value of n. Thus, the expressions that represent the total number of dots on a square figure with n dots on a side are generalized as shown next.
4 * 3 + 4 f ¡ 4(n  2) + 4 4 * (5  2) + 4
Connection to Algebra In algebra, the letter “x” is most commonly used for a variable. However, any letter (even Greek letters, for example) can be used as a variable.
Developing Algebraic Reasoning www.wiley.com/college/musser See “Variable” and “Equations.”
Connection to Algebra Variable is a central concept in algebra. Students may struggle with the idea that the letter x represents many numbers in y ⫽ 3x ⫹ 4 but only one or two numbers in other situations. For example, the solution of the equation x2 ⫽ 9 consists of the numbers 3 and ⫺3 since each of these numbers squared is 9. This means, that the equation is true whenever x ⫽ 3 or x ⫽ ⫺3.
4 * 5  4
¡ 4n  4
4 * 4 f 4 * (5  1)
¡ 4(n  1)
■
Since each expression on the right represents the total number of dots in Figure 1.15, they are all equal to each other. Using properties of numbers and equations, each equation can be rewritten as the same expression. Learning to simplify expressions and equations with variables is one of the most important processes in mathematics. Traditionally, this topic has represented a substantial portion of an entire course in introductory algebra. An equation is a sentence involving numbers, or symbols representing numbers where the verb is equals (⫽). There are various types of equations: 3 + 4 = 7 3 + 4 = 9 2x + 5x = 7x x + 4 = 9
True equation False equation Identity equation Conditional equation
A true or false equation needs no explanation, but an identity equation is always true no matter what numerical value is used for x. A conditional equation is an equation that is only true for certain values of x. For example, the equation x ⫹ 4 ⫽ 9 is true when x ⫽ 5, but false when x is any other value. In this chapter, we will restrict the variables to only whole numbers. For a conditional equation, a value of the variable that makes the equation true is called the solution. To solve an equation means to find all of the solutions. The following example shows three different ways to solve equations of the form ax ⫹ b ⫽ c.
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Example 1.3
Suppose the square arrangement of dots in Example 1.2 had 84 total dots (Figure 1.16). How many dots are there
on each side? Reflection from Research Even 6yearolds can solve algebraic equations when they are rewritten as a story problem, logic puzzle, or some other problem with meaning (Femiano, 2003).
— 84 total dots Figure 1.16 SOLUTION From Example 1.2, a square with n dots on a side has 4n ⫺ 4 total dots.
Thus, we have the equation 4n ⫺ 4 ⫽ 84. Three elementary methods that can be used to solve equations such as 4n ⫺ 4 ⫽ 84 are Guess and Test, Cover Up, and Work Backward.
Guess and Test As the name of this method suggests, one guesses values for the variable in the equation 4n ⫺ 4 ⫽ 84 and substitutes to see if a true equation results. Try n ⫽ 10: 4(10) ⫺ 4 ⫽ 36 ⫽ 84 Try n ⫽ 25: 4(25) ⫺ 4 ⫽ 96 ⫽ 84 Try n ⫽ 22: 4(22) ⫺ 4 ⫽ 84. Therefore, 22 is the solution of the equation.
Cover Up In this method, we cover up the term with the variable: ×4 22 ÷4 Figure 1.17
n  4 = 84. To make a true equation, the n must be 88. Thus 4n = 88. Since 4 # 22 = 88, we have n = 22.
–4 84
88 +4
Work Backward The left side of the equation shows that n is multiplied by 4 and then 4 is subtracted to obtain 84. Thus, working backward, if we add 4 to 84 and divide by 4, we reach the value of n. Here 84 ⫹ 4 ⫽ 88 and 88⫼4 ⫽ 22 so n ⫽ 22 (Figure 1.17). ■
MATHE MATH E M AT ICA I CA L MO RSE R SE L There is a story about Sir Isaac Newton, coinventor of the calculus, who, as a youngster, was sent out to cut a hole in the barn door for the cats to go in and out. With great pride he admitted to cutting two holes, a larger one for the cat and a smaller one for the kittens.
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EXERCISE / PROBLEM SET A
Section 1.1
1. a. If the diagonals of a square are drawn in, how many triangles of all sizes are formed? b. Describe how Pólya’s four steps were used to solve part a. 2. Scott and Greg were asked to add two whole numbers. Instead, Scott subtracted the two numbers and got 10, and Greg multiplied them and got 651. What was the correct sum? 3. The distance around a standard tennis court is 228 feet. If the length of the court is 6 feet more than twice the width, find the dimensions of the tennis court.
11. Find a set of consecutive counting numbers whose sum is each of the following. Each set may consist of 2, 3, 4, 5, or 6 consecutive integers. Use the spreadsheet activity Consecutive Integer Sum on our Web site to assist you. a. 84 b. 312 c. 154 12. Place the digits 1 through 9 so that you can count from 1 to 9 by following the arrows in the diagram.
4. A multiple of 11 I be, not odd, but even, you see. My digits, a pair, when multiplied there, make a cube and a square out of me. Who am I? 5. Show how 9 can be expressed as the sum of two consecutive numbers. Then decide whether every odd number can be expressed as the sum of two consecutive counting numbers. Explain your reasoning. 6. Using the symbols ⫹, ⫺, ⫻, and ⫼, fill in the following three blanks to make a true equation. (A symbol may be used more than once.) 6
6
6
14. Using the numbers 9, 8, 7, 6, 5, and 4 once each, find the following: a. The largest possible sum:
6 = 13
7. In the accompanying figure (called an arithmogon), the number that appears in a square is the sum of the numbers in the circles on each side of it. Determine what numbers belong in the circles.
41
13. Using a 5minute and an 8minute hourglass timer, how can you measure 1 minute?
49
36
8. Place 10 stools along four walls of a room so that each of the four walls has the same number of stools.
+ b. The smallest possible (positive) difference:
– 15. Using the numbers 1 through 8, place them in the following eight squares so that no two consecutive numbers are in touching squares (touching includes entire sides or simply one point).
9. Susan has 10 pockets and 44 dollar bills. She wants to arrange the money so that there are a different number of dollars in each pocket. Can she do it? Explain. 10. Arrange the numbers 2, 3, . . . , 10 in the accompanying triangle so that each side sums to 21. 16. Solve this cryptarithm, where each letter represents a digit and no digit represents two different letters: USSR + USA PEACE 17. On a balance scale, two spools and one thimble balance eight buttons. Also, one spool balances one thimble and one button. How many buttons will balance one spool?
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18. Place the numbers 1 through 8 in the circles on the vertices of the accompanying cube so that the difference of any two connecting circles is greater than 1.
through 14 in the 7 circles below so that the sum of the three numbers in each circle is 21.
19. Think of a number. Add 10. Multiply by 4. Add 200. Divide by 4. Subtract your original number. Your result should be 60. Why? Show why this would work for any number. 20. The digits 1 through 9 can be used in decreasing order, with ⫹ and ⫺ signs, to produce 100 as shown: 98 ⫺ 76 ⫹ 54 ⫹ 3 ⫹ 21 ⫽ 100. Find two other such combinations that will produce 100.
24. The hexagon below has a total of 126 dots and an equal number of dots on each side. How many dots are on each side?
21. The Indian mathematician Ramanujan observed that the taxi number 1729 was very interesting because it was the smallest counting number that could be expressed as the sum of cubes in two different ways. Find a, b, c, and d such that a3 ⫹ b3 ⫽ 1729 and c 3 ⫹ d 3 ⫽ 1729. 22. Using the Chapter 1 eManipulative activity Number Puzzles, Exercise 2 on our Web site, arrange the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 in the following circles so the sum of the numbers along each line of four is 23.
25. Explain advantages and disadvantages of the Guess and Test method for solving an equation such as 3x ⫺ 8 ⫽ 37. 26. Some students feel “Guess and Test” is a waste of time; they just want to get an answer. Think of some reasons, other than those mentioned in the text, why “Guess and Test” is a good strategy to use.
23. Using the Chapter 1 eManipulative activity Circle 21 on our Web site, find an arrangement of the numbers 1
Section 1.1
27. Research has shown that some better math students tend not to draw pictures in their work. Yet future teachers are encouraged to draw pictures when solving problems. Is there a conflict here?
EXERCISE / PROBLEM SET B
1. Find the largest eightdigit number made up of the digits 1, 1, 2, 2, 3, 3, 4, and 4 such that the 1s are separated by one digit, the 2s by two digits, the 3s by three digits, and the 4s by four digits. 2. Think of a number. Multiply by 5. Add 8. Multiply by 4. Add 9. Multiply by 5. Subtract 105. Divide by 100.
Subtract 1. How does your result compare with your original number? Explain. 3. Carol bought some items at a variety store. All the items were the same price, and she bought as many items as the price of each item in cents. (For example, if the items cost 10 cents, she would have bought 10 of them.) Her bill was $2.25. How many items did Carol buy?
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4. You can make one square with four toothpicks. Show how you can make two squares with seven toothpicks (breaking toothpicks is not allowed), three squares with 10 toothpicks, and five squares with 12 toothpicks. 5. A textbook is opened and the product of the page numbers of the two facing pages is 6162. What are the numbers of the pages?
The ProblemSolving Process and Strategies
19
at the regular price, a third pair (of lower or equal value) will be free. Neither friend wants three pairs of shoes, but Pat would like to buy a $56 and a $39 pair while Chris is interested in a $45 pair. If they buy the shoes together to take advantage of the sale, what is the fairest share for each to pay? 14. Find digits A, B, C, and D that solve the following cryptarithm. ABCD * 4 DCBA
6. Place numbers 1 through 19 into the 19 circles below so that any three numbers in a line through the center will give the same sum.
15. If possible, find an odd number that can be expressed as the sum of four consecutive counting numbers. If impossible, explain why. 16. Five friends were sitting on one side of a table. Gary sat next to Bill. Mike sat next to Tom. Howard sat in the third seat from Bill. Gary sat in the third seat from Mike. Who sat on the other side of Tom? 17. In the following square array on the left, the corner numbers were given and the boldface numbers were found by adding the adjacent corner numbers. Following the same rules, find the corner numbers for the other square array. 7. Using three of the symbols ⫹, ⫺, ⫻, and ⫼ once each, fill in the following three blanks to make a true equation. (Parentheses are allowed.) 6
6
6
6 = 66
8. A water main for a street is being laid using a particular kind of pipe that comes in either 18foot sections or 20foot sections. The designer has determined that the water main would require 14 fewer sections of 20foot pipe than if 18foot sections were used. Find the total length of the water main. 9. Mike said that when he opened his book, the product of the page numbers of the two facing pages was 7007. Without performing any calculations, prove that he was wrong. 10. The Smiths were about to start on an 18,000mile automobile trip. They had their tires checked and found that each was good for only 12,000 miles. What is the smallest number of spares that they will need to take along with them to make the trip without having to buy a new tire? 11. What is the maximum number of pieces of pizza that can result from 4 straight cuts? 12. Given: Six arrows arranged as follows:
6 19 13 8 14 2 3 1
— 10 15 — 16
— 11 —
18. Together, a baseball and a football weigh 1.25 pounds, the baseball and a soccer ball weigh 1.35 pounds, and the football and the soccer ball weigh 1.9 pounds. How much does each of the balls weigh? 19. Pick any two consecutive numbers. Add them. Then add 9 to the sum. Divide by 2. Subtract the smaller of the original numbers from the answer. What did you get? Repeat this process with two other consecutive numbers. Make a conjecture (educated guess) about the answer, and prove it. 20. An additive magic square has the same sum in each row, column, and diagonal. Find the error in this magic square and correct it. 47 24 29 17 97 35
56 67 52 49 6 19
34 44 3 89 3 46
22 26 99 4 11 87
83 13 18 53 74 8
7 75 48 37 28 54
c c c T T T Goal: By inverting two adjacent arrows at a time, rearrange to the following: c T c T c T Can you find a minimum number of moves? 13. Two friends are shopping together when they encounter a special “3 for 2” shoe sale. If they purchase two pairs of shoes
21. Two points are placed on the same side of a square. A segment is drawn from each of these points to each of the 2 vertices (corners) on the opposite side of the square. How many triangles of all sizes are formed? 22. Using the triangle in Problem 10 in Part A, determine whether you can make similar triangles using the digits 1, 2, . . . , 9, where the side sums are 18, 19, 20, 21, and 22.
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23. Using the Chapter 1 eManipulative activity, Number Puzzles, Exercise 4 on our Web site, arrange the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 in the circles below so the sum of the numbers along each line of four is 20.
24. Using the Chapter 1 eManipulative activity Circle 99 on our Web site, find an arrangement of the numbers provided in the 7 circles below so that the sum of the three numbers in each circle is 99.
25. An arrangement of dots forms the perimeter of an equilateral triangle. There are 87 evenly spaced dots on each side including the dots at the vertices. How many dots are there altogether? y 26. The equation + 12 = 23 can be solved by subtracting 12 5 y from both sides of the equation to yield + 12  12 = 5 y 23  12. Similarly, the resulting equation = 11 can be 5 solved by multiplying both sides of the equation by 5 to obtain y ⫽ 55. Explain how this process is related to the Work Backward method described in Example 1.3. 27. When college students hear the phrase “Use a variable,” they usually think of algebra, which makes them think of using the letter x to represent the unknown. But first graders are often given problems like n + 3 = 5
Is this the same as x ⫹ 3 ⫽ 5? Do you think first graders can do simple algebra?
Problems Related to the NCTM Standards and Curriculum Focal Points 1. The Focal Points for Grades 4 and 5 state “Select appropriate units, strategies, and tools for solving problems.” Find a problem or example in this section that illustrates this statement and explain your reasoning. 2. The NCTM Standards state “All students should describe, extend, and make generalizations about geometric and
1.2 S TART I NG POINT
NCTM Standard All students should represent, analyze, and generalize a variety of patterns with tables, graphs, words, and, when possible, symbolic rules.
numeric patterns.” Find a problem in this problem set that illustrates this statement and explain your reasoning. 3. The NCTM Standards state “All students should develop an initial conceptual understanding of different uses of variables.” Find a problem or example in this section that illustrates this statement and explain your reasoning.
THREE ADDITIONAL STRATEGIES Solve the problem below using Pólya’s four steps and any other strategy. Describe how you used the four steps, focusing on any new insights that you gained as a result of looking back. How many rectangles of all shapes and sizes are in the figure at the right?
Strategy 4
Look for a Pattern
When using the Look for a Pattern strategy, one usually lists several specific instances of a problem and then looks to see whether a pattern emerges that suggests a solution to the entire problem. For example, consider the sums produced by adding consecutive odd numbers starting with 1: 1, 1 ⫹ 3 ⫽ 4 (⫽ 2 ⫻ 2), 1 ⫹ 3 ⫹ 5 ⫽ 9 (⫽ 3 ⫻ 3),
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Developing Algebraic Reasoning www.wiley.com/college/musser See “Pattern Generalization.”
Three Additional Strategies
21
1 ⫹ 3 ⫹ 5 ⫹ 7 ⫽ 16 (⫽ 4 ⫻ 4), 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫽ 25 (⫽ 5 ⫻ 5), and so on. Based on the pattern generated by these five examples, one might expect that such a sum will always be a perfect square. The justification of this pattern is suggested by the following figure.
1
1 + 3
1 + 3 + 5
1 + 3 + 5 + 7
Each consecutive odd number of dots can be added to the previous square arrangement to form another square. Thus, the sum of the first n odd numbers is n2. Generalizing patterns, however, must be done with caution because with a sequence of only 3 or 4 numbers, a case could be made for more than one pattern. For example, consider the sequence 1, 2, 4, . . . . What are the next 4 numbers in the sequence? It can be seen that 1 is doubled to get 2 and 2 is doubled to get 4. Following that pattern, the next four numbers would be 8, 16, 32, 64. If, however, it is noted that the difference between the first and second term is 1 and the difference between the second and third term is 2, then a case could be made that the difference is increasing by one. Thus, the next four terms would be 7, 11, 16, 22. Another case could be made for the differences alternating between 1 and 2. In that case, the next four terms would be 5, 7, 8, 10. Thus, from the initial three numbers of 1, 2, 4, at least three different patterns are possible:
A
1, 2, 4, 8, 16, 32, 64, . . . 1, 2, 4, 7, 11, 16, 22, . . . 1, 2, 4, 5, 7, 8, 10, . . . B
Doubling Di f erence increasing by 1 Difference alternating between 1 and 2
Problem
Figure 1.18
How many different downward paths are there from A to B in the grid in Figure 1.18? A path must travel on the lines.
A
Step 1
Understand the Problem
What do we mean by different and downward? Figure 1.19 illustrates two paths. Notice that each such path will be 6 units long. Different means that they are not exactly the same; that is, some part or parts are different.
Step 2
B Figure 1.19
1 1
1 2
1
Let’s look at each point of intersection in the grid and see how many different ways we can get to each point. Then perhaps we will notice a pattern (Figure 1.20). For example, there is only one way to reach each of the points on the two outside edges; there are two ways to reach the middle point in the row of points labeled 1, 2, 1; and so on. Observe that the point labeled 2 in Figure 1.20 can be found by adding the two 1s above it.
Step 3 Figure 1.20
Devise a Plan
Carry Out the Plan
To see how many paths there are to any point, observe that you need only add the number of paths required to arrive at the point or points immediately above. To reach
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1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 Figure 1.21
a point beneath the pair 1 and 2, the paths to 1 and 2 are extended downward, resulting in 1 ⫹ 2 ⫽ 3 paths to that point. The resulting number pattern is shown in Figure 1.21. Notice, for example, that 4 ⫹ 6 ⫽ 10 and 20 ⫹ 15 ⫽ 35. (This pattern is part of what is called Pascal’s triangle. It is used again in Chapter 11.) The surrounded portion of this pattern applies to the given problem; thus the answer to the problem is 20.
Step 4
Look Back
Can you see how to solve a similar problem involving a larger square array, say a 4 ⫻ 4 grid? How about a 10 ⫻ 10 grid? How about a rectangular grid? Children’s Literature www.wiley.com/college/musser See “Math for All Seasons” by Greg Tang.
NCTM Standard All students should analyze how both repeating and growing patterns are generated.
Reflection from Research In classrooms where problem solving is valued, where instruction reflects the spirit of the Standards (NCTM, 1989), and where teachers have knowledge of children’s mathematical thinking, children perceive engaging in mathematics as a problemsolving endeavor in which communicating mathematical thinking is important (Franke & Carey, 1997).
A pattern of numbers arranged in a particular order is called a number sequence, and the individual numbers in the sequence are called terms of the sequence. The counting numbers, 1, 2, 3, 4, . . . , give rise to many sequences. (An ellipsis, the three periods after the 4, means “and so on.”) Several sequences of counting numbers follow. SEQUENCE
NAME
2, 4, 6, 8, . . . 1, 3, 5, 7, . . . 1, 4, 9, 16, . . . 1, 3, 32, 33, . . . 1, 1, 2, 3, 5, 8, . . .
The even (counting) numbers The odd (counting) numbers The square (counting) numbers The powers of three The Fibonacci sequence (after the two 1s, each term is the sum of the two preceding terms)
Inductive reasoning is used to draw conclusions or make predictions about a large collection of objects or numbers, based on a small representative subcollection. For example, inductive reasoning can be used to find the ones digit of the 400th term of the sequence 8, 12, 16, 20, 24, . . . . By continuing this sequence for a few more terms, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, . . . , one can observe that the ones digit of every fifth term starting with the term 24 is a four. Thus, the ones digit of the 400th term must be a four.
Additional Problems Where the Strategy “Look for a Pattern” Is Useful 1. Find the ones digit in 399.
Step 1
Understand the Problem
The number 399 is the product of 99 threes. Using the exponent key on one type of scientific calculator yields the result 1.71792506547 . This shows the first digit, but not the ones (last) digit, since the 47 indicates that there are 47 places to the right of the decimal. (See the discussion on scientific notation in Chapter 4 for further explanation.) Therefore, we will need to use another method.
Step 2
Devise a Plan 1
Consider 3 , 32, 33, 34, 35, 36, 37, 38. Perhaps the ones digits of these numbers form a pattern that can be used to predict the ones digit of 399.
Step 3
Carry Out the Plan
3 ⫽ 3, 3 ⫽ 9, 33 ⫽ 27, 34 ⫽ 81, 35 ⫽ 243, 36 ⫽ 729, 37 ⫽ 2187, 38 ⫽ 6561. The ones digits form the sequence 3, 9, 7, 1, 3, 9, 7, 1. Whenever the exponent of the 3 has a factor of 4, the ones digit is a 1. Since 100 has a factor of 4, 3100 must have a 1
2
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ones digit of 1. Therefore, the ones digit of 399 must be 7, since 399 precedes 3100 and 7 precedes 1 in the sequence 3, 9, 7, 1.
Step 4
Look Back
Ones digits of other numbers involving exponents might be found in a similar fashion. Check this for several of the numbers from 4 to 9. 2. Which whole numbers, from 1 to 50, have an odd number of factors? For example, 15 has 1, 3, 5, and 15 as factors, and hence has an even number of factors: four. 3. In the next diagram, the left “H”shaped array is called the 32H and the right array is the 58H. 0 10 20 30 40 50 60 70 80 90
1 11 21 31 41 51 61 71 81 91
2 12 22 32 42 52 62 72 82 92
3 13 23 33 43 53 63 73 83 93
4 14 24 34 44 54 64 74 84 94
5 15 25 35 45 55 65 75 85 95
6 16 26 36 46 56 66 76 86 96
7 17 27 37 47 57 67 77 87 97
8 18 28 38 48 58 68 78 88 98
9 19 29 39 49 59 69 79 89 99
a. Find the sums of the numbers in the 32H. Do the same for the 58H and the 74H. What do you observe? b. Find an H whose sum is 497. c. Can you predict the sum in any H if you know the middle number? Explain.
CLUES The Look for a Pattern strategy may be appropriate when
• • • • •
A list of data is given. A sequence of numbers is involved. Listing special cases helps you deal with complex problems. You are asked to make a prediction or generalization. Information can be expressed and viewed in an organized manner, such as in a table.
Review the preceding three problems to see how these clues may have helped you select the Look for a Pattern strategy to solve these problems. Reflection from Research Problemsolving abililty develops with age, but the relative difficulty inherent in each problem is grade independent (Christou & Philippou, 1998).
Strategy 5
Make a List
The Make a List strategy is often combined with the Look for a Pattern strategy to suggest a solution to a problem. For example, here is a list of all the squares of the numbers 1 to 20 with their ones digits in boldface. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400 The pattern in this list can be used to see that the ones digits of squares must be one of 0, 1, 4, 5, 6, or 9. This list suggests that a perfect square can never end in a 2, 3, 7, or 8.
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NCTM Standard Instructional programs should enable all students to build new mathematical knowledge through problem solving.
Problem The number 10 can be expressed as the sum of four odd numbers in three ways: (i) 10 ⫽ 7 ⫹ 1 ⫹ 1 ⫹ 1, (ii) 10 ⫽ 5 ⫹ 3 ⫹ 1 ⫹ 1, and (iii) 10 ⫽ 3 ⫹ 3 ⫹ 3 ⫹ 1. In how many ways can 20 be expressed as the sum of eight odd numbers?
Step 1
Understand the Problem
Recall that the odd numbers are the numbers 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, . . . . Using the fact that 10 can be expressed as the sum of four odd numbers, we can form various combinations of those sums to obtain eight odd numbers whose sum is 20. But does this account for all possibilities?
Step 2
Devise a Plan
Instead, let’s make a list starting with the largest possible odd number in the sum and work our way down to the smallest.
Step 3
Carry Out the Plan 20 20 20 20 20 20 20 20 20 20 20
= = = = = = = = = = =
13 + 1 + 1 + 1 + 1 + 1 + 1 + 1 11 + 3 + 1 + 1 + 1 + 1 + 1 + 1 9 + 5 + 1 + 1 + 1 + 1 + 1 + 1 9 + 3 + 3 + 1 + 1 + 1 + 1 + 1 7 + 7 + 1 + 1 + 1 + 1 + 1 + 1 7 + 5 + 3 + 1 + 1 + 1 + 1 + 1 7 + 3 + 3 + 3 + 1 + 1 + 1 + 1 5 + 5 + 5 + 1 + 1 + 1 + 1 + 1 5 + 5 + 3 + 3 + 1 + 1 + 1 + 1 5 + 3 + 3 + 3 + 3 + 1 + 1 + 1 3 + 3 + 3 + 3 + 3 + 3 + 1 + 1
Reflection from Research
Step 4
Correct answers are not a safe indicator of good thinking. Teachers must examine more than answers and must demand from students more than answers (Sowder, ThreadgillSowder, Moyer, & Moyer, 1983).
Could you have used the three sums to 10 to help find these 11 sums to 20? Can you think of similar problems to solve? For example, an easier one would be to express 8 as the sum of four odd numbers, and a more difficult one would be to express 40 as the sum of 16 odd numbers. We could also consider sums of even numbers, expressing 20 as the sum of six even numbers.
Look Back
Additional Problems Where the Strategy “Make a List” Is Useful 1. In a dart game, three darts are thrown. All hit the target (Figure 1.22). What scores are possible? 0 1 4 16
Step 1
Step 2 Figure 1.22
Understand the Problem
Assume that all three darts hit the board. Since there are four different numbers on the board, namely, 0, 1, 4, and 16, three of these numbers, with repetitions allowed, must be hit.
Devise a Plan
We should make a systematic list by beginning with the smallest (or largest) possible sum. In this way we will be more likely to find all sums.
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Step 3 0 1 1 4
+ + + +
Step 4
Three Additional Strategies
25
Carry Out the Plan 0 1 1 4
+ + + +
0 1 4 4
= = = =
0 3, 6, 12,
0 + 0 + 1 = 1, 0 + 0 + 4 = 4, 0 + 4 + 4 = 8, ...,
0 + 1 + 0 + 1 + 1 + 4 + 16 + 16
1 4 4 +
= 2, = 5, = 9, 16 = 48
Look Back
Several similar problems could be posed by changing the numbers on the dartboard, the number of rings, or the number of darts. Also, using geometric probability, one could ask how to design and label such a game to make it a fair skill game. That is, what points should be assigned to the various regions to reward one fairly for hitting that region? Figure 1.23
2. How many squares, of all sizes, are there on an 8 ⫻ 8 checkerboard? (See Figure 1.23; the sides of the squares are on the lines.) 3. It takes 1230 numerical characters to number the pages of a book. How many pages does the book contain?
CLUES The Make a List strategy may be appropriate when
• • • • •
Information can easily be organized and presented. Data can easily be generated. Listing the results obtained by using Guess and Test. Asked “in how many ways” something can be done. Trying to learn about a collection of numbers generated by a rule or formula.
Review the preceding three problems to see how these clues may have helped you select the Make a List strategy to solve these problems. The problemsolving strategy illustrated next could have been employed in conjunction with the Make a List strategy in the preceding problem.
Strategy 6 Children’s Literature www.wiley.com/college/musser See “How Much Is a Million?” by David M. Schwartz.
Solve a Simpler Problem
Like the Make a List strategy, the Solve a Simpler Problem strategy is frequently used in conjunction with the Look for a Pattern strategy. The Solve a Simpler Problem strategy involves reducing the size of the problem at hand and making it more manageable to solve. The simpler problem is then generalized to the original problem.
Problem In a group of nine coins, eight weigh the same and the ninth is heavier. Assume that the coins are identical in appearance. Using a pan balance, what is the smallest number of balancings needed to identify the heavy coin?
Step 1
Understand the Problem
Coins may be placed on both pans. If one side of the balance is lower than the other, that side contains the heavier coin. If a coin is placed in each pan and the pans balance, the heavier coin is in the remaining seven. We could continue in this way, but if we missed the heavier coin each time we tried two more coins, the last
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coin would be the heavy one. This would require four balancings. Can we find the heavier coin in fewer balancings?
Step 2
Devise a Plan
To find a more efficient method, let’s examine the cases of three coins and five coins before moving to the case of nine coins.
Step 3 Figure 1.24
Figure 1.25
Figure 1.26
Carry Out the Plan
Three coins: Put one coin on each pan (Figure 1.24). If the pans balance, the third coin is the heavier one. If they don’t, the one in the lower pan is the heavier one. Thus, it only takes one balancing to find the heavier coin. Five coins: Put two coins on each pan (Figure 1.25). If the pans balance, the fifth coin is the heavier one. If they don’t, the heavier one is in the lower pan. Remove the two coins in the higher pan and put one of the two coins in the lower pan on the other pan. In this case, the lower pan will have the heavier coin. Thus, it takes at most two balancings to find the heavier coin. Nine coins: At this point, patterns should have been identified that will make this solution easier. In the threecoin problem, it was seen that a heavy coin can be found in a group of three as easily as it can in a group of two. From the fivecoin problem, we know that by balancing groups of coins together, we could quickly reduce the number of coins that needed to be examined. These ideas are combined in the ninecoin problem by breaking the nine coins into three groups of three and balancing two groups against each other (Figure 1.26). In this first balancing, the group with the heavy coin is identified. Once the heavy coin has been narrowed to three choices, then the threecoin balancing described above can be used. The minimum number of balancings needed to locate the heavy coin out of a set of nine coins is two.
Step 4
Look Back
In solving this problem by using simpler problems, no numerical patterns emerged. However, patterns in the balancing process that could be repeated with a larger number of coins did emerge.
Additional Problems Where the Strategy “Solve a Simpler Problem” Is Useful 1. Find the sum
1 1 1 1 + 2 + 3 + Á + 10 . 2 2 2 2
Step 1
Understand the Problem
This problem can be solved directly by getting a common denominator, here 210, and finding the sum of the numerators.
Step 2
Devise a Plan
Instead of doing a direct calculation, let’s combine some previous strategies. Namely, make a list of the first few sums and look for a pattern.
Step 3
Carry Out the Plan 1 1 1 3 1 1 1 7 1 1 1 1 15 , + = , + + = , + + + = 2 2 4 4 2 4 8 8 2 4 8 16 16
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The pattern of sums, 210  1 1023 or . 1024 210,
Step 4
Three Additional Strategies
27
1 3 7 15 , , , , suggests that the sum of the 10 fractions is 2 4 8 16
Look Back
This method of combining the strategy of Solve a Simpler Problem with Make a List and Look for a Pattern is very useful. For example, what is the sum 1 1 1 + 2 + Á + 100 ? Because of the large denominators, you wouldn’t want to 2 2 2 add these fractions directly.
Connection to Algebra Using a variable, the sum to the right can be expressed more generally as follows: 1 2n  1 1 1 . + 2 + Á + n = 2 2 2n 2
2. Following the arrows in Figure 1.27, how many paths are there from A to B? B
A Figure 1.27
3. There are 20 people at a party. If each person shakes hands with each other person, how many handshakes will there be?
CLUES The Solve a Simpler Problem strategy may be appropriate when
• • • • •
The problem involves complicated computations. The problem involves very large or very small numbers. A direct solution is too complex. You want to gain a better understanding of the problem. The problem involves a large array or diagram.
Review the preceding three problems to see how these clues may have helped you select the Solve a Simpler Problem strategy to solve these problems.
Combining Strategies to Solve Problems Reflection from Research The development of a disposition toward realistic mathematical modeling and interpreting of word problems should permeate the entire curriculum from the outset (Verschaffel & DeCorte,1997).
As shown in the previous fourstep solution, it is often useful to employ several strategies to solve a problem. For example, in Section 1.1, a pizza problem similar to the following was posed: What is the maximum number of pieces you can cut a pizza into using four straight cuts? This question can be extended to the more general question: What is the maximum number of pieces you can cut a pizza into using n straight cuts? To answer this, consider the sequence in Figure 1.9: 1, 2, 4, 7, 11. To identify patterns in a sequence, observing how successive terms are related can be helpful. In this case, the second term of 2 can be obtained from the first term of 1 by either adding 1 or multiplying by 2. The third term, 4, can be obtained from the second term, 2, by adding 2 or multiplying by 2. Although multiplying by 2 appears to be a pattern, it fails as we
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move from the third term to the fourth term. The fourth term can be found by adding 3 to the third term. Thus, the sequence appears to be the following: Sequence
1
Differences
4
2 2
1
7 3
11 4
Extending the difference sequence, we obtain the following Term Sequence Differences
1 1
2 2 1
3 4 2
4 7 3
5 11 4
6 16 5
7 22 6
8 ... n 29 7
Starting with 1 in the sequence, dropping down to the difference line, then back up to the number in the sequence line, we find the following: 1st term: 1 ⫽ 1 2nd Term: 2 ⫽ 1 ⫹ 1 3rd Term: 4 ⫽ 1 ⫹ (1 ⫹ 2) 4th Term: 7 ⫽ 1 ⫹ (1 ⫹ 2 ⫹ 3) 5th Term: 11 ⫽ 1 ⫹ (1 ⫹ 2 ⫹ 3 ⫹ 4), and so forth (n  1)n . Thus, the nth 2 (n  1)n . Notice that term in the sequence is 1 + 31 + 2 + . . . + (n  1)4 = 1 + 2 7#8 = 1 + 28 = 29. Hence, to as a check, the eighth term in the sequence is 1 + 2 solve the original problem, we used Draw a Picture, Look for a Pattern, and Use a Variable. It may be that a pattern does not become obvious after one set of differences. Consider the following problem where several differences are required to expose the pattern. Recall that earlier we saw that 1 + 2 + 3 + . . . + n =
Problem If 10 points are placed on a circle and each pair of points is connected with a segment, what is the maximum number of regions created by these segments?
Step 1
Understand the Problem
This problem can be better understood by drawing a picture. Since drawing 10 points and all of the joining segments may be overwhelming, looking at a simpler problem of circles with 1, 2, or 3 points on them may help in further understanding the problem. The first three cases are in Figure 1.28.
Figure 1.28
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Step 2
Three Additional Strategies
29
Devise a Plan
So far, the number of regions are 1, 2, and 4 respectively. Here, again, this could be the start of the pattern, 1, 2, 4, 8, 16, . . . . Let’s draw three more pictures to see if this is the case. Then the pattern can be generalized to the case of 10 points.
Step 3
Carry Out the Plan
The next three cases are shown in Figure 1.29. 6 5 1
22
2
3
4
7 8
24 25 23 30 21
27 31 28 29
20
9
26
19
16 15
10
11 12 13 14
17 18 Figure 1.29
Making a list of the number of points on the circle and the corresponding number of regions will help us see the pattern. Points Regions
1 1
2 2
3 4
4 8
5 16
6 31
While the pattern for the first five cases makes it appear as if the number of regions are just doubling with each additional point, the 31 regions with 6 points ruins this pattern. Consider the differences between the numbers in the pattern and look for a pattern in the differences. Points Regions 1st Difference 2nd Difference 3rd Difference 4th Difference
1 1
2 2 1
3 4 2
1
4 8 4
2 1
8 4
2 1
5 16
6 31 15
7 3
1
Because the first, second, and third difference did not indicate a clear pattern, the fourth difference was computed and revealed a pattern of all ones. This observation is used to extend the pattern by adding four 1s to the two 1s in the fourth difference to make a sequence of six 1s. Then we work up until the Regions sequence has ten numbers as shown next.
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1 1
Points Regions 1st Difference
2 2 1
2nd Difference 3rd Difference
3 4 2
1
4 8 4
2 1
4th Difference
8 4
2 1
5 16 15 7 3
1
6 31 26 11 4
1
7 57 42 16 5
1
8 9 1 0 99 163 256 64 22 6
1
93 29
7 1
By finding successive differences, it can be seen that the solution to our problem is 256 regions.
Step 4
Look Back
By using a combination of Draw a Picture, Solve a Simpler Problem, Make a List, and Look for a Pattern, the solution was found. It can also be seen that it is important when looking for such patterns to realize that we may have to look at many terms and many differences to be able to find the pattern.
Recapitulation When presenting the problems in this chapter, we took great care in organizing the solutions using Pólya’s fourstep approach. However, it is not necessary to label and display each of the four steps every time you work a problem. On the other hand, it is good to get into the habit of recalling the four steps as you plan and as you work through a problem. In this chapter we have introduced several useful problemsolving strategies. In each of the following chapters, a new problemsolving strategy is introduced. These strategies will be especially helpful when you are making a plan. As you are planning to solve a problem, think of the strategies as a collection of tools. Then an important part of solving a problem can be viewed as selecting an appropriate tool or strategy. We end this chapter with a list of suggestions that students who have successfully completed a course on problem solving felt were helpful tips. Reread this list periodically as you progress through the book.
Suggestions from Successful Problem Solvers
• • • • • • • • •
Accept the challenge of solving a problem. Rewrite the problem in your own words. Take time to explore, reflect, think. . . . Talk to yourself. Ask yourself lots of questions. If appropriate, try the problem using simple numbers. Many problems require an incubation period. If you get frustrated, do not hesitate to take a break—your subconscious may take over. But do return to try again. Look at the problem in a variety of ways. Run through your list of strategies to see whether one (or more) can help you get a start. Many problems can be solved in a variety of ways—you only need to find one solution to be successful.
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Reflection from Research The unrealistic expectations of teachers, namely lack of time and support, can cause young students to struggle with problem solving (Buschman, 2002).
• • • • • • • • • • •
Three Additional Strategies
31
Do not be afraid to change your approach, strategy, and so on. Organization can be helpful in problem solving. Use the Pólya fourstep approach with a variety of strategies. Experience in problem solving is very valuable. Work lots of problems; your confidence will grow. If you are not making much progress, do not hesitate to go back to make sure that you really understand the problem. This review process may happen two or three times in a problem since understanding usually grows as you work toward a solution. There is nothing like a breakthrough, a small aha!, as you solve a problem. Always, always look back. Try to see precisely what the key step was in your solution. Make up and solve problems of your own. Write up your solutions neatly and clearly enough so that you will be able to understand your solution if you reread it in 10 years. Develop good problemsolving helper skills when assisting others in solving problems. Do not give out solutions; instead, provide meaningful hints. By helping and giving hints to others, you will find that you will develop many new insights. Enjoy yourself! Solving a problem is a positive experience.
MATH EM MATHE M AT ICA L MO R SEL RSE L Sophie Germain was born in Paris in 1776, the daughter of a silk merchant. At the age of 13, she found a book on the history of mathematics in her father’s library. She became enthralled with the study of mathematics. Even though her parents disapproved of this pursuit, nothing daunted her—she studied at night wrapped in a blanket, because her parents had taken her clothing away from her to keep her from getting up. They also took away her heat and light. This only hardened her resolve until her father finally gave in and she, at last, was allowed to study to become a mathematician.
Section 1.2
EXERCISE / PROBLEM SET A
Use any of the six problemsolving strategies introduced thus far to solve the following. 1. a. Complete this table and describe the pattern in the ‘Answer’ column. SUM 1 1⫹3 1⫹3⫹5 1⫹3⫹5⫹7 1⫹3⫹5⫹7⫹9
ANSWER 1 4
b. How many odd whole numbers would have to be added to get a sum of 81? Check your guess by adding them. c. How many odd whole numbers would have to be added to get a sum of 169? Check your guess by adding them. d. How many odd whole numbers would have to be added to get a sum of 529? (You do not need to check.) 2. Find the missing term in each pattern. a. 256, 128, 64, _____, 16, 8 1 1 1 b. 1, , , _____, 3 9 81 c. 7, 9, 12, 16, _____ d. 127,863; 12,789; _____; 135; 18
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3. Sketch a figure that is next in each sequence. a. A A B B B
NUMBER 1 2 3 4 5 6
C C C
NUMBER OF DOTS (TRIANGULAR NUMBERS) 1 3
C b. Make a sketch to represent the seventh triangular number. c. How many dots will be in the tenth triangular number? d. Is there a triangular number that has 91 dots in its shape? If so, which one? e. Is there a triangular number that has 150 dots in its shape? If so, which one? f. Write a formula for the number of dots in the nth triangular number. g. When the famous mathematician Carl Friedrich Gauss was in fourth grade, his teacher challenged him to add the first one hundred counting numbers. Find this sum.
b.
4. Consider the following differences. Use your calculator to verify that the statements are true. 2
2
6  5 = 11 562  452 = 1111 5562  4452 = 111,111 a. Predict the next line in the sequence of differences. Use your calculator to check your answer. b. What do you think the eighth line will be? 5. Look for a pattern in the first two number grids. Then use the pattern you observed to fill in the missing numbers of the third grid. 21 7
3
72 36 2
3
1
3
8
4
2
7
7
1
9
9
1
60 6 5 2
6. The triangular numbers are the whole numbers that are represented by certain triangular arrays of dots. The first five triangular numbers are shown.
1
2
3
4
5
a. Complete the following table and describe the pattern in the ‘Number of Dots’ column.
1 + 2 + 3 + . . . + 100 7. In a group of 12 coins identical in appearance, all weigh the same except one that is heavier. What is the minimum number of weighings required to determine the counterfeit coin? Use the Chapter 1 eManipulative activity Counterfeit Coin on our Web site for eight or nine coins to better understand the problem. 8. If 20 points are placed on a circle and every pair of points is joined with a segment, what is the total number of segments drawn? 9. Find reasonable sixth, seventh, and eighth terms of the following sequences: a. 1, 4, 9, 17, 29, _____, _____, _____ b. 3, 7, 13, 21, 31, _____, _____, _____ 10. As mentioned in this section, the square numbers are the counting numbers 1, 4, 9, 16, 25, 36, . . . . Each square number can be represented by a square array of dots as shown in the following figure, where the second square number has four dots, and so on. The first four square numbers are shown.
a. Find two triangular numbers (refer to Problem 6) whose sum equals the third square number. b. Find two triangular numbers whose sum equals the fifth square number.
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c. What two triangular numbers have a sum that equals the 10th square number? the 20th square number? the nth square number? d. Find a triangular number that is also a square number. e. Find five pairs of square numbers whose difference is a triangular number. 11. Would you rather work for a month (30 days) and get paid 1 million dollars or be paid 1 cent the first day, 2 cents the second day, 4 cents the third day, 8 cents the fourth day, and so on? Explain.
1
1 1
1
1
1 number of triangles
1
1 1
1
2
1 1
1 1 1
1 3
4
1 5
6
1 10
perimeter
33
16. Write out 16 terms of the Fibonacci sequence and observe the following pattern: 1 + 2 = 3 1 + 2 + 5 = 8 1 + 2 + 5 + 13 = 21 Use the pattern you observed to predict the sum 1 + 2 + 5 + 13 + . . . + 610 without actually computing the sum. Then use your calculator to check your result. 17. Pascal’s triangle is where each entry other than a 1 is obtained by adding the two entries in the row above it.
12. Find the perimeters and then complete the table. 1
Three Additional Strategies
n 40
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 • • • • • • • a. Find the sums of the numbers on the diagonals in Pascal’s triangle as are indicated in the following figure. 1
13. The integers greater than 1 are arranged as shown.
1
2 3 4 5 9 8 7 6 10 11 12 13 17 16 15 14
#
#
1 2
14. How many cubes are in the 100th collection of cubes in this sequence?
15. The Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, . . . , where each successive number beginning with 2 is the sum of the preceding two; for example, 13 ⫽ 5 ⫹ 8, 21 ⫽ 8 ⫹ 13, and so on. Observe the following pattern. 12 + 12 = 1 * 2 12 + 12 + 22 = 2 * 3 12 + 12 + 22 + 32 = 3 * 5 Write out six more terms of the Fibonacci sequence and use the sequence to predict what 12 ⫹ 12 ⫹ 22 ⫹ 32 ⫹ . . . ⫹ 1442 is without actually computing the sum. Then use your calculator to check your result.
3
1 4 1 1
5 6
1
3
1
#
a. In which column will 100 fall? b. In which column will 1000 fall? c. How about 1999? d. How about 99,997?
1
6
4
1
10
5
10 15
1
20
15
1 6
1
• • • • • • • • • b. Predict the sums along the next three diagonals in Pascal’s triangle without actually adding the entries. Check your answers by adding entries on your calculator. 18. Answer the following questions about Pascal’s triangle (see Problem 17). a. In the triangle shown here, one number, namely 3, and the six numbers immediately surrounding it are encircled. Find the sum of the encircled seven numbers. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 • • • • • • • • • b. Extend Pascal’s triangle by adding a few rows. Then draw several more circles anywhere in the triangle like the one shown in part (a). Explain how the sums obtained by adding the seven numbers inside the circle are related to one of the numbers outside the circle.
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19. Consider the following sequence of shapes. The sequence starts with one square. Then at each step squares are attached around the outside of the figure, one square per exposed edge in the figure.
Step 1
Step 2
Step 3
Step 4
20. In a dart game, only 4 points or 9 points can be scored on each dart. What is the largest score that it is not possible to obtain? (Assume that you have an unlimited number of darts.) 21. If the following four figures are referred to as stars, the first one is a threepointed star and the second one is a sixpointed star. (Note: If this pattern of constructing a new equilateral triangle on each side of the existing equilateral triangle is continued indefinitely, the resulting figure is called the Koch curve or Koch snowflake.)
a. How many points are there in the third star? b. How many points are there in the fourth star? a. Draw the next two figures in the sequence. b. Make a table listing the number of unit squares in the figure at each step. Look for a pattern in the number of unit squares. (Hint: Consider the number of squares attached at each step.) c. Based on the pattern you observed, predict the number of squares in the figure at step 7. Draw the figure to check your answer. d. How many squares would there be in the 10th figure? in the 20th figure? in the 50th figure?
Section 1.2
22. Using the Chapter 1 eManipulative activity Color Patterns on our Web site, describe the color patterns for the first three computer exercises. 23. Looking for a pattern can be frustrating if the pattern is not immediately obvious. Create your own sequence of numbers that follows a pattern but that has the capacity to stump some of your fellow students. Then write an explanation of how they might have been able to discover your pattern.
EXERCISE / PROBLEM SET B
1. Find the missing term in each pattern. a. 10, 17, _____, 37, 50, 65 3 7 9 b. 1, , _____, , 2 8 16 c. 243, 324, 405, _____, 567 d. 234; _____; 23,481; 234,819; 2,348,200 2. Sketch a figure that is next in each sequence. a.
b.
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3. The rectangular numbers are whole numbers that are represented by certain rectangular arrays of dots. The first five rectangular numbers are shown.
1 • • • • 4 • •
• • • •
• • • •
• • • •
• • • •
• • 5 • • •
• • • • •
• • • • •
• • • • •
• • • • •
• • • • •
a. Complete the following table and describe the pattern in the ‘Number of Dots’ column.
NUMBER
NUMBER OF DOTS (RECTANGULAR NUMBERS)
1 2 3 4 5 6
2 6
b. Make a sketch to represent the seventh rectangular number. c. How many dots will be in the tenth rectangular number? d. Is there a rectangular number that has 380 dots in its shape? If so, which one? e. Write a formula for the number of dots in the nth rectangular number. f. What is the connection between triangular numbers (see Problem 6 in Part A) and rectangular numbers? 4. The pentagonal numbers are whole numbers that are represented by pentagonal shapes. The first four pentagonal numbers are shown.
1
2
35
a. Complete the following table and describe the pattern in the ‘Number of Dots’ column.
NUMBER
• • • • 3 • • • • • • • •
• • • 2 • • •
Three Additional Strategies
NUMBER OF DOTS (PENTAGONAL NUMBERS)
1 2 3 4 5
1 5
b. Make a sketch to represent the fifth pentagonal number. c. How many dots will be in the ninth pentagonal number? d. Is there a pentagonal number that has 200 dots in its shape? If so, which one? e. Write a formula for the number of dots in the nth pentagonal number. 5. Consider the following process. 1. Choose a whole number. 2. Add the squares of the digits of the number to get a new number. Repeat step 2 several times. a. Apply the procedure described to the numbers 12, 13, 19, 21, and 127. b. What pattern do you observe as you repeat the steps over and over? c. Check your answer for part (b) with a number of your choice. 6. How many triangles are in the picture?
7. What is the smallest number that can be expressed as the sum of two squares in two different ways? (You may use one square twice.) 8. How many cubes are in the 10th collection of cubes in this sequence?
3
4
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9. The 2 ⫻ 2 array of numbers c
4 5
5 d has a sum of 4 ⫻ 5, 6
6 7 8 and the 3 ⫻ 3 array C 7 8 9 S has a sum of 9 ⫻ 8. 8 9 10 a. What will be the sum of the similar 4 ⫻ 4 array starting with 7? b. What will be the sum of a similar 100 ⫻ 100 array starting with 100? 10. The Fibonacci sequence was defined to be the sequence 1, 1, 2, 3, 5, 8, 13, 21, . . . , where each successive number is the sum of the preceding two. Observe the following pattern. 1 1 1 1
+ + + +
1 1 1 1
= + + +
3 2 2 2
= + +
14. While only 19 years old, Carl Friedrich Gauss proved in 1796 that every positive integer is the sum of at the most three triangular numbers (see Problem 6 in Part A). a. Express each of the numbers 25 to 35 as a sum of no more than three triangular numbers. b. Express the numbers 74, 81, and 90 as sums of no more than three triangular numbers. 15. Answer the following for Pascal’s triangle. a. In the following triangle, six numbers surrounding a central number, 4, are circled. Compare the products of alternate numbers moving around the circle; that is, compare 3 䡠 1 䡠 10 and 6 䡠 1 䡠 5. 1
1 5  1 3 = 8  1 3 + 5 = 13  1
1 1
1
1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
Write out six more terms of the Fibonacci sequence, and use the sequence to predict the answer to
2
1 + 1 + 2 + 3 + 5 + . . . + 144 without actually computing the sum. Then use your calculator to check your result. 11. Write out 16 terms of the Fibonacci sequence. a. Notice that the fourth term in the sequence (called F4) is odd: F4 ⫽ 3. The sixth term in the sequence (called F6) is even: F6 ⫽ 8. Look for a pattern in the terms of the sequence, and describe which terms are even and which are odd. b. Which of the following terms of the Fibonacci sequence are even and which are odd: F38, F51, F150, F200, F300? c. Look for a pattern in the terms of the sequence and describe which terms are divisible by 3. d. Which of the following terms of the Fibonacci sequence are multiples of 3: F48, F75, F196, F379, F1000? 12. Write out 16 terms of the Fibonacci sequence and observe the following pattern.
b. Extend Pascal’s triangle by adding a few rows. Then draw several more circles like the one shown in part (a) anywhere in the triangle. Find the products as described in part (a). What patterns do you see in the products? 16. A certain type of gutter comes in 6foot, 8foot, and 10foot sections. How many different lengths can be formed using three sections of gutter? 17. Consider the sequence of shapes shown in the following figure. The sequence starts with one triangle. Then at each step, triangles are attached to the outside of the preceding figure, one triangle per exposed edge.
Step 1
Step 2
Step 3
Step 4
1 + 3 = 5  1 1 + 3 + 8 = 13  1 1 + 3 + 8 + 21 = 34  1 Use the pattern you observed to predict the answer to 1 + 3 + 8 + 21 + . . . + 377 without actually computing the sum. Then use your calculator to check your result. 13. Investigate the “Tower of Hanoi” problem on the Chapter 1 eManipulative activity Tower of Hanoi on our Web site to answer the following questions: a. Determine the fewest number of moves required when you start with two, three, and four disks. b. Describe the general process to move the disks in the fewest number of moves. c. What is the minimum number of moves that it should take to move six disks?
a. Draw the next two figures in the sequence. b. Make a table listing the number of triangles in the figure at each step. Look for a pattern in the number of triangles. (Hint:Consider the number of triangles added at each step.) c. Based on the pattern you observed, predict the number of triangles in the figure at step 7. Draw the figure to check your answer. d. How many triangles would there be in the 10th figure? in the 20th figure? in the 50th figure?
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Section 1.2
18. How many equilateral triangles of all sizes are there in the 3 ⫻ 3 ⫻ 3 equilateral triangle shown next?
19. Refer to the following figures to answer the questions. (NOTE: If this pattern is continued indefinitely, the resulting figure is called the Sierpinski triangle or the Sierpinski gasket.)
a. How many black triangles are there in the fourth figure? b. How many white triangles are there in the fourth figure? c. If the pattern is continued, how many black triangles are there in the nth figure? d. If the pattern is continued, how many white triangles are there in the nth figure? 20. If the pattern illustrated next is continued, a. find the total number of 1 by 1 squares in the thirtieth figure.
Three Additional Strategies
b. find the perimeter of the twentyfifth figure. c. find the total number of toothpicks used to construct the twentieth figure.
21. Find reasonable sixth, seventh, and eighth terms of the following sequences: a. 1, 3, 4, 7, 11, _____, _____, _____ b. 0, 1, 4, 12, 29, _____, _____, _____ 22. Many board games involve throwing two dice and summing of the numbers that come up to determine how many squares to move. Make a list of all the different sums that can appear. Then write down how many ways each different sum can be formed. For example, 11 can be formed in two ways: from a 5 on the first die and a 6 on the second OR a 6 on the first die and a 5 on the second. Which sum has the greatest number of combinations? What conclusion could you draw from that? 23. There is an old riddle about a frog at the bottom of a 20foot well. If he climbs up 3 feet each day and slips back 2 feet each night, how many days will it take him to reach the 20foot mark and climb out of the well? The answer isn’t 20. Try doing the problem with a well that is only 5 feet deep, and keep track of all the frog’s moves. What strategy are you using?
Problems Related to the NCTM Standards and Curriculum Focal Points 1. The Focal Points for Grade 3 state “Apply increasingly sophisticated strategies . . . to solve multiplication and division problems.” Find a problem in this problem set that illustrates this statement and explain your reasoning.
2. The NCTM Standards state “Instructional programs should enable all students to build new mathematical knowledge through problem solving.” Explain some new mathematical knowledge that you have learned by solving problems in this section.
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3. The NCTM Standards state “All students should represent, analyze, and generalize a variety of patterns with tables, graphs, and, when possible, symbolic rules.” Explain how tables, graphs, and symbolic rules could be used to represent and analyze the pattern in Problem 11 of Problem Set A.
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Introduction to Problem Solving
EN D OF C H APTER M AT E RIAL 6 ⫽ ((4 ⫹ 4) ⫼ 4) ⫹ 4 7 ⫽ 4 ⫹ 4 ⫺ (4 ⫼ 4) 8 ⫽ ((4 ⫻ 4) ⫼ 4) ⫹ 4 9 ⫽ 4 ⫹ 4 ⫹ (4 ⫼ 4) There are many other possible answers.
Solution of Initial Problem Place the whole numbers 1 through 9 in the circles in the accompanying triangle so that the sum of the numbers on each side is 17. 3.
Draw a Picture
Strategy: Guess and Test Having solved a simpler problem in this chapter, you might easily be able to conclude that 1, 2, and 3 must be in the corners. Then the remaining six numbers, 4, 5, 6, 7, 8, and 9, must produce three pairs of numbers whose sums are 12, 13, and 14. The only two possible solutions are as shown. 1 6
1 4
8 2
9 9
5
7
6
5 3
2
8
Solution for Additional Problems Guess and Test 1. s ⫽ 1, u ⫽ 3, n ⫽ 6, f ⫽ 9, w ⫽ 0, i ⫽ 7, m ⫽ 2 2. 0 ⫽ (4 ⫺ 4) ⫹ (4 ⫺ 4) 1 ⫽ (4 ⫹ 4) ⫼ (4 ⫹ 4) 2 ⫽ (4 ⫼ 4) ⫹ (4 ⫼ 4) 3 ⫽ (4 ⫹ 4 ⫹ 4) ⫼ 4 4 ⫽ 4 ⫹ 4 ⫻ (4 ⫺ 4) 5 ⫽ (4 ⫻ 4 ⫹ 4) ⫼ 4
Use a Variable 1. 55, 5050, 125,250 2. (2m ⫹ 1) ⫹ (2m ⫹ 3) ⫹ (2m ⫹ 5) ⫹ (2m ⫹ 7) ⫹ (2m ⫹ 9) ⫽ 10m ⫹ 25 ⫽ 5(2m ⫹ 5) 3. 10⬚, 80⬚, 90⬚
Look for a Pattern 7
4
1. 5 2. Yes; make one cut, then lay the logs side by side for the second cut. 3. 12
3
1. 7 2. Square numbers 3. a. 224; 406; 518 b. 71 c. The sum is seven times the middle number.
Make a List 1. 48, 36, 33, 32, 24, 21, 20, 18, 17, 16, 12, 9, 8, 6, 5, 4, 3, 2, 1, 0 2. 204 3. 446
Solve a Simpler Problem 1023 1024 2. 377 3. 190 1.
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Chapter Review
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People in Mathematics Carl Friedrich Gauss (1777–1885) Carl Friedrich Gauss, according to the historian E. T. Bell, “lives everywhere in mathematics.” His contributions to geometry, number theory, and analysis were deep and wideranging. Yet he also made crucial contributions in applied mathematics. When the tiny planet Ceres was discovered in 1800, Gauss developed a technique for calculating its orbit, based on meager observations of its direction from Earth at several known times. Gauss contributed to the modern theory of electricity and magnetism, and with the physicist W. E. Weber constructed one of the first practical electric telegraphs. In 1807 he became director of the astronomical observatory at Gottingen, where he served until his death. At age 18, Gauss devised a method for constructing a 17sided regular polygon, using only a compass and straightedge. Remarkably, he then derived a general rule that predicted which regular polygons are likewise constructible.
Sophie Germain (1776–1831) Sophie Germain, as a teenager in Paris, discovered mathematics by reading books from her father’s library. At age 18, Germain wished to attend the prestigious Ecole Polytechnique in Paris, but women were not admitted. So she studied from classroom notes supplied by sympathetic male colleagues, and she began submitting written work using the pen name Antoine LeBlanc. This work won her high praise, and eventually she was able to reveal her true identity. Germain is noted for her theory of the vibration patterns of elastic plates and for her proof of Fermat’s last theorem in some special cases. Of Sophie Germain, Carl Gauss wrote, “When a woman, because of her sex, encounters infinitely more obstacles than men . . . yet overcomes these fetters and penetrates that which is most hidden, she doubtless has the most noble courage, extraordinary talent, and superior genius.”
CHAPTER REVIEW Review the following terms and problems to determine which require learning or relearning—page numbers are provided for easy reference.
SE CTI ON 1.1 The ProblemSolving Process and Strategies VOCABULARY/NOTATION Exercise 4 Problem 4 Pólya’s fourstep process Strategy 4 Random Guess and Test
4 7
Systematic Guess and Test Inferential Guess and Test Cryptarithm 8 Tetromino 10 Variable or unknown 11
PROBLEMS For each of the following, (i) determine a reasonable strategy to use to solve the problem, (ii) state a clue that suggested the strategy, and (iii) write out a solution using Pólya’s fourstep process. 1. Fill in the circles using the numbers 1 through 9 once each where the sum along each of the five rows totals 17.
7 7
Equation 15 Solution of an equation Solve an equation 15
15
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2. In the following arithmagon, the number that appears in a square is the product of the numbers in the circles on each side of it. Determine what numbers belong in the circles.
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3. The floor of a square room is covered with square tiles. Walking diagonally across the room from corner to corner, Susan counted a total of 33 tiles on the two diagonals. What is the total number of tiles covering the floor of the room?
60 135
SE CTI O N 1.2 Three Additional Strategies VOCABULARY/NOTATION Pascal’s triangle 22 Sequence 22 Terms 22 Counting numbers 22
Ellipsis 22 Even numbers 22 Odd numbers 22 Square numbers 22
Powers 22 Fibonacci sequence Inductive reasoning
22 22
PROBLEMS For each of the following, (i) determine a reasonable strategy to use to solve the problem, (ii) state a clue that suggested the strategy, and (iii) write out a solution using Pólya’s fourstep process.
2. a. How many cubes of all sizes are in a 2 ⫻ 2 ⫻ 2 cube composed of eight 1 ⫻ 1 ⫻ 1 cubes? b. How many cubes of all sizes are in an 8 ⫻ 8 ⫻ 8 cube composed of 512 1 ⫻ 1 ⫻ 1 cubes?
1. Consider the following products. Use your calculator to verify that the statements are true.
3. a. What is the smallest number of wholenumber gram weights needed to weigh any wholenumber amount from 1 to 12 grams on a scale allowing the weights to be placed on either or both sides? b. How about from 1 to 37 grams? c. What is the most you can weigh using six weights in this way?
1 * (1) = 12 121 * (1 + 2 + 1) = 222 12321 * (1 + 2 + 3 + 2 + 1) = 3332 Predict the next line in the sequence of products. Use your calculator to check your answer.
P R O B L EM S FOR W R IT I N G/ DI S C U SS I O N 1. Describe reasoning that can be used to place the numbers 3, 4, 5, 6, 7, 8, 9, 10, 11 in the diagram to form a 3by3 magic square.
big bang. Unfortunately, the villain had left the hero only two jugs, one that would hold exactly 3 gallons and one that would hold exactly 5 gallons. There was a fountain nearby with lots of water in it. Explain how to get exactly 4 gallons into the 5gallon container. (No, there is no measuring cup.) 3. Show three different methods for solving the following problem: Find three consecutive counting numbers whose sum is 78.
(NOTE: There may be more than one answer.) 2. In the movie Die Hard with a Vengeance, the hero was told that in order to stop an explosion from taking place he would have to place a plastic jug with exactly 4 gallons of water in it on a scale. The wrong weight would set off the
4. Show why the following problem has no solution: Find three consecutive odd whole numbers whose sum is 102. (Hint: There is more than one way to demonstrate the answer.) 5. The following problem can be solved in more than one way. Find at least one way to solve it without algebra: Mary Kay
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Chapter Test
wanted to buy some makeup. She spent $28 of her paycheck on foundations, 23 of the rest for eye shadows, and 12 of what was left after that for a lipstick. She had $12 left over. How much was her paycheck? (For the purposes of this problem, we’ll ignore sales tax.) 6. Create a new problem similar to the preceding problem. Show a solution as well. 7. Try to discover a pattern in problems involving consecutive whole numbers. Focus on problems that involve finding three numbers that are either consecutive, consecutive even, or consecutive odd and that add up to some specific number. If you find the pattern, you will understand how teachers go about making up such problems. 8. Try to extend your patternfinding skills to consecutive wholenumber problems involving the sum of four consecutive (or consecutive odd or consecutive even) whole numbers. What pattern did you find? 9. Explain why the following problems are unsolvable. Then change each problem in such a way that it would be solvable. a. The sum of two numbers is 87. What are the numbers?
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b. The perimeter of a rectangular garden is 58 feet. The sum of the length and width is 29 feet. Find the length and width. 10. Mr. Nelson manages a shoe store in the mall. One day a man came into the store right after Mr. Nelson opened up and before he had a lot of change in the cash register. This man wanted to buy a pair of athletic shoes that cost $80, and he gave Mr. Nelson a $100 bill. Mr. Nelson did not have change for $100, so he ran next door and exchanged the $100 bill for five $20 bills from the cashier at The Gap. He then gave his customer the $20 in change, and the man left with the shoes. Later, the cashier at The Gap heard there were some counterfeit bills being passed in the mall. She went over to the shoe store, gave Mr. Nelson the $100 bill, and said she wanted five $20s. By now Mr. Nelson had change, so he took back the $100 bill and gave her five $20s. Shortly after that the police arrived and checked the $100 bill. Sure enough, it was counterfeit and the police confiscated it. Can you figure out how much money Mr. Nelson is out altogether? Be prepared to explain your reasoning!
CHAPTER TEST KNOWLEDGE
UNDERSTANDING
1. List the four steps of Pólya’s problemsolving process.
6. Explain the difference between an exercise and a problem.
2. List the six problemsolving strategies you have learned in this chapter.
7. List at least two characteristics of a problem that would suggest using the Guess and Test strategy.
SKILLS 3. Identify the unneeded information in the following problem. Birgit took her $5 allowance to the bookstore to buy some backtoschool supplies. The pencils cost $.10, the erasers cost $.05 each, and the clips cost 2 for $.01. If she bought 100 items altogether at a total cost of $1, how many of each item did she buy? 4. Rewrite the following problem in your own words. If you add the square of Ruben’s age to the age of Angelita, the sum is 62; but if you add the square of Angelita’s age to the age of Ruben, the sum is 176. Can you say what the ages of Ruben and Angelita are? 5. Given the following problem and its numerical answer, write the solution in a complete sentence. Amanda leaves with a basket of hardboiled eggs to sell. At her first stop she sold half her eggs plus half an egg. At her second stop she sold half her eggs plus half an egg. The same thing occurs at her third, fourth, and fifth stops. When she finishes, she has no eggs in her basket. How many eggs did she start with? Answer: 31
8. List at least two characteristics of a problem that would suggest using the Use a Variable strategy.
PROBLEM SOLVING / APPLICATION For each of the following problems, read the problem carefully and solve it. Identify the strategy you used. 9. Can you rearrange the 16 numbers in this 4 ⫻ 4 array so that each row, each column, and each of the two diagonals total 10? How about a 2 ⫻ 2 array containing two 1s and two 2s? How about the corresponding 3 ⫻ 3 array? 1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
10. In three years, Chad will be three times my present age. I will then be half as old as he. How old am I now? 11. There are six baseball teams in a tournament. The teams are lettered A through F. Each team plays each of the other teams twice. How many games are played altogether?
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12. A fish is 30 inches long. The head is as long as the tail. If the head was twice as long and the tail was its present length, the body would be 18 inches long. How long is each portion of the fish? 13. The Orchard brothers always plant their apple trees in square arrays, like those illustrated. This year they planted 31 more apple trees in their square orchard than last year. If the orchard is still square, how many apple trees are there in the orchard this year?
1
4
9
14. Arrange 10 people so that there are five rows each containing 4 persons. 15. A milk crate holds 24 bottles and is shaped like the one shown below. The crate has four rows and six columns. Is it possible to put 18 bottles of milk in the crate so that each row and each column of the crate has an even number of bottles in it? If so, how? (Hint: One row has 6 bottles in it and the other three rows have 4 bottles in them.)
Milk
lk
Mi
16. Otis has 12 coins in his pocket worth $1.10. If he only has nickels, dimes, and quarters, what are all of the possible coin combinations? 17. Show why 3 always divides evenly into the sum of any three consecutive whole numbers. 18. If 14 toothpicks are arranged to form a triangle so none of the toothpicks are broken or bent and all 14 toothpicks are used, how many differentshaped triangles can be formed? 19. Together a baseball and a football weigh 1.25 pounds, the baseball and a soccer ball weigh 1.35 pounds, and the football and the soccer ball weigh 1.6 pounds. How much does each of the balls weigh? Explain your reasoning. 20. In the figure below, there are 7 chairs arranged around 5 tables. How many chairs could be placed around a similar arrangement of 31 triangular tables?
21. Carlos’ father pays Carlos his allowance each week using patterns. He pays a different amount each day according to some pattern. Carlos must identify the pattern in order to receive his allowance. Help Carlos complete the pattern for the missing days in each week below. a. 5¢, 9¢, 16¢, 26¢, _____, _____, _____ b. 1¢, 6¢, 15¢, 30¢, 53¢, _____, _____ c. 4¢, 8¢, 16¢, 28¢, _____, _____, _____
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CHAPTER
Sets, Whole Numbers, and Numeration
2
The Mayan Numeration System
FOCUS ON
he Maya people lived mainly in southeastern Mexico, including the Yucatan Peninsula, and in much of northwestern Central America, including Guatemala and parts of Honduras and El Salvador. Earliest archaeological evidence of the Maya civilization dates to 9000 B.C.E. , with the principal epochs of the Maya cultural development occurring between 2000 B.C.E. and C.E. 1700. Knowledge of arithmetic, calendrical, and astronomical matters was more highly developed by the ancient Maya than by any other New World peoples. Their numeration system was simple, yet sophisticated. Their system utilized three basic numerals: a dot, ●, to represent 1; a horizontal bar, —, to represent 5; and a conch shell, , to represent 0. They used these three symbols, in combination, to represent the numbers 0 through 19.
T
0
1
2
3
4
5
6
7
8
9
10
11
12
13
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For numbers greater than 19, they initially used a base 20 system. That is, they grouped in twenties and displayed their numerals vertically. Three Mayan numerals are shown together with their values in our system and the place values initially used by the Mayans.
The sun, and hence the solar calendar, was very important to the Maya. They calculated that a year consisted of 365.2420 days. (Present calculations measure our year as 365.2422 days long.) Since 360 had convenient factors and was close to 365 days in their year and 400 in their numeration system, they made a place value system where the column values from right to left were 1, 20, 20 䡠 18 (⫽ 360), 202 䡠 18 (⫽ 7200), 203 䡠 18 (⫽ 144,000), and so on. Interestingly, the Maya could record all the days of their history simply by using the place values through 144,000. The Maya were also able to use larger numbers. One Mayan hieroglyphic text recorded a number equivalent to 1,841,641,600. Finally, the Maya, famous for their hieroglyphic writing, also used the 20 ideograms pictured here, called head variants, to represent the numbers 0–19.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
The Mayan numeration system is studied in this chapter along with other ancient numeration systems.
8000 (=203) 400 (=202) 20 1 20
806
10871
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ProblemSolving Strategies
7
1. Guess and Test
STRATEGY
2. Draw a Picture
Draw a Diagram
3. Use a Variable 4. Look for a Pattern 5. Make a List 6. Solve a Simpler Problem 7. Draw a Diagram
Often there are problems where, although it is not necessary to draw an actual picture to represent the problem situation, a diagram that represents the essence of the problem is useful. For example, if we wish to determine the number of times two heads can turn up when we toss two coins, we could literally draw pictures of all possible arrangements of two coins turning up heads or tails. However, in practice, a simple tree diagram is used like the one shown next. Outcomes when two coins are tossed H
HH
T H
HT TH
T
TT
H
T
Coin 1
Coin 2
This diagram shows that there is one way to obtain two heads out of four possible outcomes. Another type of diagram is helpful in solving the next problem.
INITIAL PROBLEM A survey was taken of 150 college freshmen. Forty of them were majoring in mathematics, 30 of them were majoring in English, 20 were majoring in science, 7 had a double major of mathematics and English, and none had a double (or triple) major with science. How many students had majors other than mathematics, English, or science?
CLUES The Draw a Diagram strategy may be appropriate when
• • •
The problem involves sets, ratios, or probabilities. An actual picture can be drawn, but a diagram is more efficient. Relationships among quantities are represented. A solution of this Initial Problem is on page 99.
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Section 2.1
Sets As a Basis for Whole Numbers
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INTRODUCTION uch of elementary school mathematics is devoted to the study of numbers. Children first learn to count using the natural numbers or counting numbers 1, 2, 3, . . . (the ellipsis, or three periods, means “and so on”). This chapter develops the ideas that lead to the concepts central to the system of whole numbers 0, 1, 2, 3, . . . (the counting numbers together with zero) and the symbols that are used to represent them. First, the notion of a onetoone correspondence between two sets is shown to be the idea central to the formation of the concept of number. Then operations on sets are discussed. These operations form the foundation of addition, subtraction, multiplication, and division of whole numbers. Finally, the Hindu–Arabic numeration system, our system of symbols that represent numbers, is presented after its various attributes are introduced by considering other ancient numeration systems.
M
Key Concepts from NCTM Curriculum Focal Points
•
P RE K I N D ER GAR TEN : Developing an understanding of whole numbers, includ
•
K I N D E RGAR TEN : Representing, comparing, and ordering whole numbers and
ing concepts of correspondence, counting, cardinality, and comparison.
• • •
2.1 S TART I NG POINT
joining and separating sets. Ordering objects by measurable attributes. GRA D E 1: Developing an understanding of whole number relationships, including grouping in tens and ones. GRA D E 2: Developing an understanding of the baseten numeration system and placevalue concepts. GRADE 6: Writing, interpreting, and using mathematical expressions and equations.
SETS AS A BASIS FOR WHOLE NUMBERS After forming a group of students, use a diagram like the one at the right to place the names of each member of your group in the appropriate region. All members of the group will fit somewhere in the rectangle. Discuss the attributes of a person whose name is in the shaded region. Discuss the attributes of a person whose name is not in any of the circles.
Students in Group Brown Eyes
Brown Hair
Curly Hair
Sets A collection of objects is called a set and the objects are called elements or members of the set. Sets can be defined in three common ways: (1) a verbal description, (2) a listing of the members separated by commas, with braces (“{” and “}”) used to enclose the list of elements, and (3) setbuilder notation. For example, the verbal description “the set of all states in the United States that border the Pacific Ocean” can be represented in the other two ways as follows:
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Sets, Whole Numbers, and Numeration
1. Listing: {Alaska, California, Hawaii, Oregon, Washington}. 2. Setbuilder: {x 兩 x is a U.S. state that borders the Pacific Ocean}. (This setbuilder notation is read: “The set of all x such that x is a U.S. state that borders the Pacific Ocean.”) Sets are usually denoted by capital letters such as A, B, C, and so on. The symbols ““ and “” are used to indicate that an object is or is not an element of a set, respectively. For example, if S represents the set of all U.S. states bordering the Pacific, then Alaska S and Michigan S. The set without elements is called the empty set (or null set) and is denoted by { } or the symbol . The set of all U.S. states bordering Antarctica is the empty set. Two sets A and B are equal, written A B, if and only if they have precisely the same elements. Thus {x 兩 x is a state that borders Lake Michigan} {Illinois, Indiana, Michigan, Wisconsin}. Notice that two sets, A and B, are equal if every element of A is in B, and vice versa. If A does not equal B, we write A ⫽ B. There are two inherent rules regarding sets: (1) The same element is not listed more than once within a set, and (2) the order of the elements in a set is immaterial. Thus, by rule 1, the set {a, a, b} would be written as {a, b} and by rule 2. {a, b} {b, a}, {x, y, z} {y, z, x}, and so on. The concept of a 11 correspondence, read “onetoone correspondence,” is needed to formalize the meaning of a whole number.
DEFINITION OnetoOne Correspondence A 11 correspondence between two sets A and B is a pairing of the elements of A with the elements of B so that each element of A corresponds to exactly one element of B, and vice versa. If there is a 11 correspondence between sets A and B, we write A ⬃ B and say that A and B are equivalent or matching sets. A x y z A x y z Figure 2.1
B a b c B a b c
Figure 2.1 shows two 11 correspondences between two sets, A and B. There are four other possible 11 correspondences between A and B. Notice that equal sets are always equivalent, since each element can be matched with itself, but that equivalent sets are not necessarily equal. For example, {1, 2} ⬃ {a, b}, but {1, 2} ⫽ {a, b}. The two sets A {a, b} and B {a, b, c} are not equivalent. However, they do satisfy the relationship defined next.
DEFINITION Subset of a Set: A B Set A is said to be a subset of B, written A B, if and only if every element of A is also an element of B. The set consisting of New Hampshire is a subset of the set of all New England states and {a, b, c} {a, b, c, d, e, f}. Since every element in a set A is in A, A A for all sets A. Also, {a, b, c} {a, b, d} because c is in the set {a, b, c} but not in the set {a, b, d}. Using similar reasoning, you can argue that A for any set A since it is impossible to find an element in that is not in A.
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Section 2.1
Sets As a Basis for Whole Numbers
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If A 債 B and B has an element that is not in A, we write A 傺 B and say that A is a proper subset of B. Thus {a, b} 傺 {a, b, c}, since {a, b} 債 {a, b, c} and c is in the second set but not in the first. Circles or other closed curves are used in Venn diagrams (named after the English logician John Venn) to illustrate relationships between sets. These circles are usually pictured within a rectangle, U, where the rectangle represents the universal set or universe, the set comprised of all elements being considered in a particular discussion. Figure 2.2 displays sets A and B inside a universal set U. U B x A
Figure 2.2
Set A is comprised of everything inside circle A, and set B is comprised of everything inside circle B, including set A. Hence A is a proper subset of B since x 僆 B, but x 僆 A. The idea of proper subset will be used later to help establish the meaning of the concept “less than” for whole numbers.
Finite and Infinite Sets
Developing Algebraic Reasoning www.wiley.com/college/musser See “Variable.”
There are two broad categories of sets: finite and infinite. Informally, a set is finite if it is empty or can have its elements listed (where the list eventually ends), and a set is infinite if it goes on without end. A little more formally, a set is finite if (1) it is empty or (2) it can be put into a 11 correspondence with a set of the form {1, 2, 3, . . . , n}, where n is a counting number. On the other hand, a set is infinite if it is not finite.
Example 2.1
Determine whether the following sets are finite or infinite.
a. {a, b, c} b. {1, 2, 3, . . . } c. {2, 4, 6, . . . , 2 0 } Reflection from Research While many students will agree that two infinite sets such as the set of counting numbers and the set of even numbers are equivalent, the same students will argue that the set of counting numbers is larger due to its inclusion of the odd numbers (Wheeler, 1987).
S O LU T I O N
a. {a, b, c} is finite since it can be matched with the set {1, 2, 3}. b. {1, 2, 3, . . .} is an infinite set. c. {2, 4, 6, . . . , 20} is a finite set since it can be matched with the set {1, 2, 3, . . . , 10}. (Here, the ellipsis means to continue the pattern until the last element is reached.) ■ NOTE: The small solid square ( ■ ) is used to mark the end of an example or mathematical argument.
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An interesting property of every infinite set is that it can be matched with a proper subset of itself. For example, consider the following 11 correspondence: A ⫽ {1, 2, 3, 4, . . . , n, . . .}
dddd
d
B ⫽ {2, 4, 6, 8, . . . , 2n, . . .}. Connection to Algebra Due to the nature of infinite sets, a variable is commonly used to illustrate matching elements in two sets when showing 11 correspondences.
Note that B A and that each element in A is paired with exactly one element in B, and vice versa. Notice that matching n with 2n indicates that we never “run out” of elements from B to match with the elements from set A. Thus an alternative definition is that a set is infinite if it is equivalent to a proper subset of itself. In this case, a set is finite if it is not infinite.
Operations on Sets Two sets A and B that have no elements in common are called disjoint sets. The sets {a, b, c} and {d, e, f } are disjoint (Figure 2.3), whereas {x, y} and {y, z} are not disjoint, since y is an element in both sets. U A
B b
a
e
d
c
f
Figure 2.3
There are many ways to construct a new set from two or more sets. The following operations on sets will be very useful in clarifying our understanding of whole numbers and their operations.
DEFINITION Union of Sets: A B The union of two sets A and B, written A B, is the set that consists of all elements belonging either to A or to B (or to both). Informally, A B is formed by putting all the elements of A and B together. The next example illustrates this definition.
Example 2.2
Find the union of the given pairs of sets.
a. {a, b} {c, d, e} b. {1, 2, 3, 4, 5} c. {m, n, q} {m, n, p} S O LU T I O N
a. {a, b} {c, d, e} {a, b, c, d, e} b. {1, 2, 3, 4, 5} {1, 2, 3, 4, 5} c. {m, n, q} {m, n, p} {m, n, p, q}
■
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Notice that although m is a member of both sets in Example 2.2(c), it is listed only once in the union of the two sets. The union of sets A and B is displayed in a Venn diagram by shading the portion of the diagram that represents A B (Figure 2.4). U A
B
Figure 2.4 Shaded region is A B.
The notion of set union is the basis for the addition of whole numbers, but only when disjoint sets are used. Notice how the sets in Example 2.2(a) can be used to show that 2 3 5. Another useful set operation is the intersection of sets.
DEFINITION Intersection of Sets: A B The intersection of sets A and B, written A B, is the set of all elements common to sets A and B. Thus A B is the set of elements shared by A and B. Example 2.3 illustrates this definition. Find the intersection of the given pairs of sets.
Example 2.3 a. {a, b, c} {b, d, f } b. {a, b, c} {a, b, c} c. {a, b} {c, d} S O LU T I O N
a. {a, b, c} {b, d, f } {b} since b is the only element in both sets. b. {a, b, c} {a, b, c} {a, b, c} since a, b, c are in both sets. c. {a, b} {c, d} since there are no elements common to the given two sets. ■ Figure 2.5 displays A B. Observe that two sets are disjoint if and only if their intersection is the empty set. Figure 2.3 shows a Venn diagram of two sets whose intersection is the empty set. U A
B
Figure 2.5 Shaded region is A B.
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In many situations, instead of considering elements of a set A, it is more productive to consider all elements in the universal set other than those in A. This set is defined next.
DEFINITION Complement of a Set: A The complement of a set A, written A, is the set of all elements in the universe, U, that are not in A. The set A is shaded in Figure 2.6. U A
A Figure 2.6 Shaded region is A.
Example 2.4
Find the following sets.
a. A where U {a, b, c, d} and A {a} b. B where U {1, 2, 3, . . .} and B {2, 4, 6, . . . } c. A B and A B where U {1, 2, 3, 4, 5}, A {1, 2, 3}, and B {3, 4} S O LU T I O N
a. A {b, c, d} b. B {1, 3, 5, . . . } c. A B {4, 5} {1, 2, 5} {1, 2, 4, 5} A B = {3} {1, 2, 4, 5}
■
The next set operation forms the basis for subtraction.
DEFINITION Difference of Sets: A B The set difference (or relative complement) of set B from set A, written A B, is the set of all elements in A that are not in B. In setbuilder notation, A B {x 兩 x A and x B}. Also, as can be seen in Figure 2.7, A B can be viewed as A B. Example 2.5 provides some examples of the difference of one set from another. U A
B
Figure 2.7 Shaded region is A B
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Find the difference of the given pairs of sets.
a. {a, b, c} {b, d} b. {a, b, c} {e} c. {a, b, c, d} {b, c, d} S O LU T I O N
a. {a, b, c} {b, d} {a, c} b. {a, b, c} {e} {a, b, c} c. {a, b, c, d} {b, c, d} {a}
■
In Example 2.5(c), the second set is a subset of the first. These sets can be used to show that 4 3 1. Another way of combining two sets to form a third set is called the Cartesian product. The Cartesian product, named after the French mathematician René Descartes, forms the basis of wholenumber multiplication and is also useful in probability and geometry. To define the Cartesian product, we need to have the concept of ordered pair. An ordered pair, written (a, b), is a pair of elements where one of the elements is designated as first (a in this case) and the other is second (b here). The notion of an ordered pair differs from that of simply a set of two elements because of the preference in order. For example, {1, 2} {2, 1} as sets, because they have the same elements. But (1, 2) ⫽ (2, 1) as ordered pairs, since the order of the elements is different. Two ordered pairs (a, b) and (c, d) are equal if and only if a c and b d.
DEFINITION Cartesian Product of Sets: A B The Cartesian product of set A with set B, written A B and read “A cross B,” is the set of all ordered pairs (a, b), where a A and b B. In setbuilder notation, A B {(a, b) 兩 a A and b B}.
Example 2.6
Find the Cartesian product of the given pairs of sets.
a. {x, y, z} {m, n} b. {7} {a, b, c} S O LU T I O N
a. {x, y, z} {m, n} {(x, m), (x, n), (y, m), (y, n), (z, m), (z, n)} b. {7} {a, b, c} {(7, a), (7, b), (7, c)}
■
Notice that when finding a Cartesian product, all possible pairs are formed where the first element comes from the first set and the second element comes from the second set. Also observe that in part (a) of Example 2.6, there are three elements in the first set, two in the second, and six in their Cartesian product, and that 3 2 6. Similarly, in part (b), these sets can be used to find the wholenumber product 1 3 3. All of the operations on sets that have been introduced in this subsection result in a new set. For example, A B is an operation on two sets that results in the set of all elements that are common to set A and B. Similarly, C D is an operation on set C that results in the set of elements C that are not also in set D. On the other hand,
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expressions such as A B or x C are not operations; they are statements that are either true or false. Table 2.1 lists all such set statements and operations. TABLE 2.1 STATEMENTS
OPERATIONS
AB AB A⬃B xB AB AXB xA A⫽B
AB A B A AB AB
Venn diagrams are often used to solve problems, as shown next. Thirty elementary teachers were asked which high school courses they appreciated: algebra or geometry. Seventeen appreciated algebra and 15 appreciated geometry; of these, 5 said that they appreciated both. How many appreciated neither?
Example 2.7
S O LU T I O N Since there are two courses, we draw a Venn diagram with two overlapping circles labeled A for algebra and G for geometry [Figure 2.8(a)]. Since 5 teachers appreciated both algebra and geometry, we place a 5 in the intersection of A and G [Figure 2.8(b)]. Seventeen must be in the A circle; thus 12 must be in the remaining part of A [Figure 2.8(c)]. Similarly, 10 must be in the G circle outside the intersection [Figure 2.8(d)]. Thus we have accounted for 12 5 10 27 teachers. This leaves 3 of the 30 teachers who appreciated neither algebra nor geometry. ■ U A
G
5
(a) Figure 2.8 (a)(d)
(b)
U
U
U G
A
G
A 12
5
(c)
MATHE MATH E M AT ICA I CA L MO RSE R SE L There are several theories concerning the rationale behind the shapes of the 10 digits in our Hindu–Arabic numeration system. One is that the number represented by each digit is given by the number of angles in the original digit. Count the “angles” in each digit below. (Here an “angle” is less than 180⬚.) Of course, zero is round, so it has no angles.
G
A 12
5
(d)
10
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EXERCISE/PROBLEM SET A
EXERCISES 1. Indicate the following sets by the listing method. a. Whole numbers between 5 and 9 b. Even counting numbers less than 15 c. Even counting numbers less than 151 2. Which of the following sets are equal to {4, 5, 6}? a. {5, 6} b. {5, 4, 6} c. Whole numbers greater than 3 d. Whole numbers less than 7 e. Whole numbers greater than 3 or less than 7 f. Whole numbers greater than 3 and less than 8 g. {e, f, g} h. {4, 5, 6, 5} 3. True or false? a. 7 {6, 7, 8, 9} b. 5 {2, 3, 4, 6} c. {1, 2, 3} {1, 2, 3} d. {1, 2, 5} {1, 2, 5} e. {2} {1, 2} 4. Find four 11 correspondences between A and B other than the two given in Figure 2.1. 5. Determine which of the following sets are equivalent to the set {x, y, z}. a. {w, x, y, z} b. {1, 5, 17} c. {w, x, z} d. {a, b, a} e. {䡺, 䉭, O} 6. Which of the following pairs of sets are equal? a. {a, b} and “First two letters of the alphabet” b. {7, 8, 9, 10} and “Whole numbers greater than 7” c. {7, a, b, 8} and {a, b, c, 7, 8} d. {x, y, x, z} and {z, x, y, z}
12. Given A {2, 4, 6, 8, 10, . . .}, and B {4, 8, 12, 16, 20, . . .}, answer the following questions. a. Find the set A B. b. Find the set A B. c. Is B a subset of A? Explain. d. Is A equivalent to B? Explain. e. Is A equal to B? Explain. f. Is B a proper subset of A? Explain. 13. True or false? a. For all sets X and Y, either X Y or Y X. b. If A is an infinite set and B A, then B also is an infinite set. c. For all finite sets A and B, if A B , then the number of elements in A B equals the number of elements in A plus the number of elements in B. 14. The regions A B and A B can each be formed by shading the set A in one direction and the set B in another direction. The region with shading in both directions is A B and the entire shaded region is A B. Use the Chapter 2 eManipulative activity Venn Diagrams on our Web site to do the following: a. Find the region (A B) C. b. Find the region A (B C). c. Based on the results of parts a and b, what conclusions can you draw about the placement of parentheses? 15. Draw a Venn diagram like the following one for each part. Then shade each Venn diagram to represent each of the sets indicated. U T
S
7. List all the subsets of {a, b, c}. 8. List the proper subsets of {嘷, 䉭}. 9. Let A {v, x, z}, B {w, x, y}, and C {v, w, x, y, z}. In the following, choose all of the possible symbols (, , , , , ⬃, or ) that would make a true statement. a. z _____ B b. B _____ C c. _____ A d. A _____ B e. v _____ C f. B _____ B
a. S
b. S T
16. A Venn diagram can be used to illustrate more than two sets. Shade the regions that represent each of the following sets. The Chapter 2 eManipulative activity Venn Diagrams on our Web site may help in solving this problem. U
10. Determine which of the following sets are finite. For those sets that are finite, how many elements are in the set? a. {ears on a typical elephant} b. {1, 2, 3, . . . , 9 9 } c. {0, 1, 2, 3, . . . , 200} d. Set of points belonging to a line segment e. Set of points belonging to a circle 11. Since the set A {3, 6, 9, 12, 15, . . .} is infinite, there is a proper subset of A that can be matched to A. Find two such proper subsets and describe how each is matched to A.
c. (S T) (T S )
B
A
C a. A (B C)
b. A (B C)
c. A (B C)
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17. Represent the following shaded regions using the symbols A, B, C, , , and . The Chapter 2 eManipulative activity Venn Diagrams on our Web site can be used in solving this problem. a.
U
b.
B
A
U B
A
C
C c.
U B
A
C 18. Draw Venn diagrams that represent sets A and B as described as follows: a. A B b. A B 19. In the drawing, C is the interior of the circle, T is the interior of the triangle, and R is the interior of the rectangle. Copy the drawing on a sheet of paper, then shade in each of the following regions.
a. C T d. (C R) T
b. C R e. (C R) T
c. (C T) R f. C (R T)
20. Let W {women who have won Nobel Prizes}, A {Americans who have won Nobel Prizes}, and C {winners of the Nobel Prize in chemistry}. Describe the elements of the following sets. a. W A b. W A c. A C 21. Let A {a, b, c}, B {b, c}, and C {e}. Find each of the following. a. A B b. B C c. A B 22. Find each of the following differences. a. {嘷, 䉭, /, 䡺} {䉭, 䡺} b. {0, 1, 2, . . . } {12, 13, 14, . . . } c. {people} {married people} d. {a, b, c, d} { }
23. Let A {2, 4, 6, 8, 10, . . .} and B {4, 8, 12, 16, . . . }. a. Find A B. b. Find B A c. Based on your observations in part a and b, describe a general case for which A B . 24. Given A {0, 1, 2, 3, 4, 5}, B {0, 2, 4, 6, 8, 10}, and C {0, 4, 8}, find each of the following. a. A B b. B C c. A B d. B C e. B C f. (A B) C g. A 25. Let M the set of the months of the year, J {January, June, July}, S {June, July, August}, W {December, January, February}. List the members of each of the following: a. J S b. J W c. S W d. J (S W) e. M (S W) f. J S 26. Let A {3, 6, 9, 12, 15, 18, 21, 24, . . .} and B {6, 12, 18, 24, . . . }. a. Is B A? b. Find A B. c. Find A B. d. In general, when B A, what is true about A B? about A B? 27. Verify that A B = A B in two different ways as follows: a. Let U {1, 2, 3, 4, 5, 6}, A {2, 3, 5}, and B {1, 4}. List the elements of the sets A B and A B. Do the two sets have the same members? b. Draw and shade a Venn diagram for each of the sets A B and A B. Do the two Venn diagrams look the same? NOTE: The equation A B = A B is one of two laws called DeMorgan’s laws. 28. Find the following Cartesian products. a. {a} {b, c} b. {5} {a, b, c} c. {a, b} {1, 2, 3} d. {2, 3} {1, 4} e. {a, b, c} {5} f. {1, 2, 3} {a, b} 29. Determine how many ordered pairs will be in the following sets. a. {1, 2, 3, 4} {a, b} b. {m, n, o} {1, 2, 3, 4} 30. If A has two members, how many members does B have when A B has the following number of members? If an answer is not possible, explain why. a. 4 b. 8 c. 9 d. 50 e. 0 f. 23
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31. The Cartesian product, A B, is given in each of the following parts. Find A and B. a. {(a, 2), (a, 4), (a, 6)} b. {(a, b), (b, b), (b, a), (a, a)}
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55
32. True or false? a. {(4, 5), (6, 7)} {(6, 7), (4, 5)} b. {(a, b), (c, d)} {(b, a), (c, d)} c. {(4, 5), (7, 6)} {(7, 6), (5, 4)} d. {(5, 6), (7, 4)} {4, 5, 6} {5, 7, 9}
PROBLEMS 33. a. If X has five elements and Y has three elements, what is the greatest number of elements possible in X Y ? in X Y ? b. If X has x elements and Y has y elements with x greater than or equal to y, what is the greatest number of elements possible in X Y ? in X Y ? 34. How many different 11 correspondences are possible between A {1, 2, 3, 4} and B {a, b, c, d}? 35. How many subsets does a set with the following number of members have? a. 0 b. 1 c. 2 d. 3 e. 5 f. n 36. If it is possible, give examples of the following. If it is not possible, explain why. a. Two sets that are not equal but are equivalent b. Two sets that are not equivalent but are equal 37. a. When does D E D? b. When does D E D? c. When does D E D E? 38. Carmen has 8 skirts and 7 blouses. Show how the concept of Cartesian product can be used to determine how many different outfits she has. 39. How many matches are there if 32 participants enter a singleelimination tennis tournament (one loss eliminates a participant)? 40. Can you show a 11 correspondence between the points on base AB of the given triangle and the points on the two sides AC and CB? Explain how you can or why you cannot. C
42. A poll of 100 registered voters designed to find out how voters kept up with current events revealed the following facts. 65 watched the news on television. 39 read the newspaper. 39 listened to radio news. 20 watched TV news and read the newspaper. 27 watched TV news and listened to radio news. 9 read the newspaper and listened to radio news. 6 watched TV news, read the newspaper, and listened to radio news. a. How many of the 100 people surveyed kept up with current events by some means other than the three sources listed? b. How many of the 100 people surveyed read the paper but did not watch TV news? c. How many of the 100 people surveyed used only one of the three sources listed to keep up with current events? 43. At a convention of 375 butchers (B), bakers (A), and candlestick makers (C), there were 50 who were both B and A but not C 70 who were B but neither A nor C 60 who were A but neither B nor C 40 who were both A and C but not B 50 who were both B and C but not A 80 who were C but neither A nor B How many at the convention were A, B, and C?
A
B
41. Can you show a 11 correspondence between the points on ២? Explain how you can chord AB and the points on arc ACB or why you cannot. A
C B
44. A student says that A B means you start with all the elements of A and you take away all the elements of B. So A B must mean you take all the elements of A and multiply them times all the elements of B. Do you agree with the student? How would you explain your reasoning?
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EXERCISE / PROBLEM SET B
EXERCISES 1. Indicate the following sets by the listing method. a. Whole numbers greater than 8 b. Odd whole numbers less than 100 c. Whole numbers less than 0 2. Represent the following sets using setbuilder notation: a. {Alabama, Alaska, . . . Maine, Maryland, . . . Wisconsin, Wyoming} b. {1, 6, 11, 16, 21, . . . } c. {a, b, c, d, . . . x, y, z} d. {1, 3, 5, 7, 9} 3. True or false? a. 23 {1, 2, 3} b. 1 {0, 1, 2} c. {4, 3} {2, 3, 4} d. { } e. {1, 2} h {2} 4. Show three different 11 correspondences between {1, 2, 3, 4} and {x, y, z, w}. 5. Write a set that is equivalent to, but not equal to the set {a, b, c, d, e, f }. 6. Which of the following sets are equal to {4, 5, 6}? a. {5, 6} b. {3, 4, 5} c. Whole numbers greater than 3 d. Whole numbers less than 7 e. Whole numbers greater than 3 or less than 7 f. Whole numbers greater than 3 and less than 8 g. {e, f, g} h. {4, 5, 6, 5}
12. Let A {0, 10, 20, 30, . . .} and B {5, 15, 25, 35, . . . }. Decide which of the following are true and which are false. Explain your answers. a. A equals B. b. B is equivalent to A. c. A B ⫽ . d. B is equivalent to a proper subset of A. e. There is a proper subset of B that is equivalent to a proper subset of A. f. A B is all multiples of 5. 13. True or false? a. The empty set is a subset of every set. b. The set {105, 110, 115, 120, . . .} is an infinite set. c. If A is an infinite set and B A, then B is a finite set. 14. Use Venn diagrams to determine which, if any, of the following statements are true for all sets A, B, and C: The Chapter 2 eManipulative Venn Diagrams on our Web site may help in determining which statements are true. a. A (B C ) (A B) C b. A (B C ) (A B) C c. A (B C ) (A B) C d. A (B C ) (A B) C 15. Draw a Venn diagram like the following for each part. Then shade each Venn diagram to represent each of the sets indicated. U T
S
7. List all subsets of {嘷, 䉭, 䡺}. Which are proper subsets? 8. How many proper subsets does R {r, s, t, u, v} have? 9. Let A {1, 2, 3, 4, 5}, B {3, 4, 5}, and C {4, 5,6}. In the following, choose all of the possible symbols (, , , , h , ⬃, or ) that would make a true statement. a. 2 _____ A b. B _____ A c. C _____ B d. 6 _____ C e. A _____ A f. B C _____ A 10. Show that the following sets are finite by giving the set of the form {1, 2, 3, . . . , n} that matches the given set. a. {121, 122, 123, . . . , 139} b. {1, 3, 5, . . . , 2 7 } 11. Show that the following sets are infinite by matching each with a proper subset of itself. a. {2, 4, 6, . . . , n, . . . } b. {50, 51, 52, 53, . . . , n, . . . }
a. T S
b. S T
c. (S T) (T S)
16. A Venn diagram can be used to illustrate more than two sets. Shade the regions that represent each of the following sets. The Chapter 2 eManipulative activity Venn Diagrams on our Web site may help in solving this problem. U B
A
C a. (A B) C b. A (B C)
c. (A B) (B C)
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17. Represent the following shaded regions using the symbols A, B, C, , , and . The Chapter 2 eManipulative activity Venn Diagrams on our Web site can be used in solving this problem. a.
b.
U B
A
U B
A
C
C
c.
U
57
21. If A is the set of all sophomores in a school and B is the set of students who belong to the orchestra, describe the following sets in words. a. A B b. A B c. A B d. B A 22. Find each of the following differences. a. {h, i, j, k} {k} b. {3, 10, 13} { } c. {twowheeled vehicles} {twowheeled vehicles that are not bicycles} d. {0, 2, 4, 6, . . . , 20} {12, 14, 16, 18, 20} 23. In each of the following cases, find B A. a. A B b. A B c. B A 24. Let R {a, b, c}, S {c, d, e, f}, T {x, y, z}. List the elements of the following sets. a. R S b. R S c. R T d. R T e. S T f. S T g. T
B
A
Sets As a Basis for Whole Numbers
18. Draw Venn diagrams that represent sets A and B as described below. a. A B ⫽ b. A B A
25. Let A {50, 55, 60, 65, 70, 75, 80} B {50, 60, 70, 80} C {60, 70, 80} D {55, 65} List the members of each set. a. A (B C) b. (A B) C c. (A C) (C D) d. (A C) (C D) e. (B C ) A f. (A D) (B C)
19. Make 6 copies of the diagram and shade regions in parts a–f.
26. a. If x X Y, is x X Y ? Justify your answer. b. If x X Y, is x X Y ? Justify your answer.
C
A
B
C
D a. (A B) C b. (C D) (A B) c. (A B) (C D) d. (B C ) D e. (A D) (C B) f. (A B) (C A) 20. Let A {a, b, c, d, e}, B {c, d, e, f, g}, and C {a, e, f, h}. List the members of each set. a. A B b. A B c. (A B) C d. A (B C )
27. Verify that A B = A B in two different ways as follows: a. Let U {2, 4, 6, 8, 10, 12, 14, 16}, A {2, 4, 8, 16}, and B {4, 8, 12, 16}. List the elements of the sets A B and A B. Do the two sets have the same members? b. Draw and shade a Venn diagram for each of the sets A B = A B. Do the two Venn diagrams look the same? NOTE: The equation A B = A B is one of the two laws called DeMorgan’s laws. 28. Find the following Cartesian products. a. {a, b, c} {1} b. {1, 2} {p, q, r} c. {p, q, r} {1, 2} d. {a} {1} 29. Determine how many ordered pairs will be in A B under the following conditions. a. A has one member and B has four members. b. A has two members and B has four members. c. A has three members and B has seven members.
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30. Find sets A and B so that A B has the following number of members. a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 0 31. The Cartesian product, X Y, is given in each of the following parts. Find X and Y. a. {(b, c), (c, c)} b. {(2, 1), (2, 2), (2, 3), (5, 1), (5, 2), (5, 3)}
32. True or false? a. {(3, 4), (5, 6)} {(3, 4), (6, 5)} b. {(a, c), (d, b)} {a, c, d, b} c. {(c, d), (a, b)} {(c, a), (d, b)} d. {(4, 5), (6, 7)} {4, 5, 6} {5, 7, 9}
PROBLEMS 33. How many 11 correspondences are there between the following pairs of sets? a. Two 2member sets b. Two 4member sets c. Two 6member sets d. Two sets each having m members 34. Find sets (when possible) satisfying each of the following conditions. a. Number of elements in A plus number of elements in B is greater than number of elements in A B. b. Number of elements in I plus number of elements in J is less than number of elements in I J. c. Number of elements in E plus number of elements in F equals number of elements in E F. d. Number of elements in G plus number of elements in K equals number of elements in G K. 35. Your house can be painted in a choice of 7 exterior colors and 15 interior colors. Assuming that you choose only 1 color for the exterior and 1 color for the interior, how many different ways of painting your house are there? 36. Show a 11 correspondence between the points on the given circle and the triangle in which it is inscribed. Explain your procedure.
37. Show a 11 correspondence between the points on the given triangle and the circle that circumscribes it. Explain your procedure.
38. A schoolroom has 13 desks and 13 chairs. You want to arrange the desks and chairs so that each desk has a chair with it. How many such arrangements are there? 39. A university professor asked his class of 42 students when they had studied for his class the previous weekend. Their responses were as follows: 9 had studied on Friday. 18 had studied on Saturday. 30 had studied on Sunday. 3 had studied on both Friday and Saturday. 10 had studied on both Saturday and Sunday. 6 had studied on both Friday and Sunday. 2 had studied on Friday, Saturday, and Sunday. Assuming that all 42 students responded and answered honestly, answer the following questions. a. How many students studied on Sunday but not on either Friday or Saturday? b. How many students did all of their studying on one day? c. How many students did not study at all for this class last weekend? 40. At an automotive repair shop, 50 cars were inspected. Suppose that 23 cars needed new brakes and 34 cars needed new exhaust systems. a. What is the least number of cars that could have needed both? b. What is the greatest number of cars that could have needed both? c. What is the greatest number of cars that could have needed neither? 41. If 70% of all students take science, 75% take social science, 80% take mathematics, and 85% take English, at least what percent take all four? 42. If two sets are equal, does that mean they are equivalent? If two sets are equivalent, does that mean they are equal? Explain. 43. If A is a proper subset of B, and A has 23 elements, how many elements does B have? Explain.
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Problems Relating to the NCTM Standards and Curriculum Focal Points 1. The Focal Points for Prekindergarten state “Developing an understanding of whole numbers, including concepts of correspondence, counting, cardinality, and comparions.” How does the idea of 11 correspondence introduced in this chapter relate to this focal point for young children? 2. The NCTM Standards state “students should count with understanding and recognize ‘how many’ in sets of objects.”
2.2 S TAR T I NG POINT
Children’s Literature www.wiley.com/college/musser See “Moja Means One” by Muriel Feelings.
NCTM Standard All students should develop understanding of the relative position and magnitude of whole numbers and of ordinal and cardinal numbers and their connections.
Reflection from Reserch There is a clear separation of development in young children concerning the cardinal and ordinal aspects of numbers. Despite the fact that they could utilize the same counting skills, the understanding of ordinality by young children lags well behind the understanding of cardinality (Bruce & Threlfall, 2004).
Discuss how the concept of “how many” is highlighted as students count objects in equivalent sets. 3. The NCTM Standards state “Instructional programs should enable students to select, apply, and translate among mathematical representations to solve problems.” Explain how Venn diagrams can be used as a representation to solve problems.
WHOLE NUMBERS AND NUMERATION Today our numeration system has the symbol “2” to represent the number of eyes a person has. The symbols “1” and “0” combine to represent the number of toes a person has, “10.” The Roman numeration system used the symbol “X” to represent the number of toes and “C” to represent the number of years in a century. Using only the three symbols at the right, devise your own numeration system and show how you can use your system to represent all of the quantities 0, 1, 2, 3, 4, . . . , 100.
Numbers and Numerals As mentioned earlier, the study of the set of whole numbers, W {0, 1, 2, 3, 4, . . .}, is the foundation of elementary school mathematics. But what precisely do we mean by the whole number 3? A number is an idea, or an abstraction, that represents a quantity. The symbols that we see, write, or touch when representing numbers are called numerals. There are three common uses of numbers. The most common use of whole numbers is to describe how many elements are in a finite set. When used in this manner, the number is referred to as a cardinal number. A second use is concerned with order. For example, you may be second in line, or your team may be fourth in the standings. Numbers used in this way are called ordinal numbers. Finally, identification numbers are used to name such things as telephone numbers, bank account numbers, and social security numbers. In this case, the numbers are used in a numeral sense in that only the symbols, rather than their values, are important. Before discussing our system of numeration or symbolization, the concept of cardinal number will be considered. What is the number 3? What do you think of when you see the sets in Figure 2.9?
α
a
β
b c Figure 2.9
γ
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First, there are no common elements. One set is made up of letters, one of shapes, one of Greek letters, and so on. Second, each set can be matched with every other set. Now imagine all the infinitely many sets that can be matched with these sets. Even though the sets will be made up of various elements, they will all share the common attribute that they are equivalent to the set {a, b, c}. The common idea that is associated with all of these equivalent sets is the number 3. That is, the number 3 is the attribute common to all sets that match the set {a, b, c}. Similarly, the whole number 2 is the common idea associated with all sets equivalent to the set {a, b}. All other nonzero whole numbers can be conceptualized in a similar manner. Zero is the idea, or number, one imagines when asked: “How many elements are in the empty set?” Although the preceding discussion regarding the concept of a whole number may seem routine for you, there are many pitfalls for children who are learning the concept of numerousness for the first time. Chronologically, children first learn how to say the counting chant “one, two, three, . . .”. However, saying the chant and understanding the concept of number are not the same thing. Next, children must learn how to match the counting chant words they are saying with the objects they are counting. For example, to count the objects in the set {䉭, 嘷, 䡺}, a child must correctly assign the words “one, two, three” to the objects in a 11 fashion. Actually, children first learning to count objects fail this task in two ways: (1) They fail to assign a word to each object, and hence their count is too small; or (2) they count one or more objects at least twice and end up with a number that is too large. To reach the final stage in understanding the concept of number, children must be able to observe several equivalent sets, as in Figure 2.9, and realize that, when they count each set, they arrive at the same word. Thus this word is used to name the attribute common to all such sets. The symbol n(A) is used to represent the number of elements in a finite set A. More precisely, (1) n(A) m if A ⬃ {1, 2, . . . , m}, where m is a counting number, and (2) n( ) 0. Thus n ({a, b, c}) 3 since {a, b, c} ⬃ {1, 2, 3}, and n ({a, b, c, . . . , z}) 26 since {a, b, c, . . . , z} ⬃ {1, 2, 3, . . . , 26}, and so on, for other finite sets.
Ordering Whole Numbers NCTM Standard All students should count with understanding and recognize “how many” in sets of objects.
Children may get their first introduction to ordering whole numbers through the counting chant “one, two, three, . . . .” For example, “two” is less than “five,” since “two” comes before “five” in the counting chant. A more meaningful way of comparing two whole numbers is to use 11 correspondences. We can say that 2 is less than 5, since any set with two elements matches a proper subset of any set with five elements (Figure 2.10).
20 O (b)
P
TAB (P ) [Figure 16.25(a)] RO,a (P ) [Figure 16.25(b)] Ml (P ) [Figure 16.25(c)] Ml (TAB (P )) [Figure 16.25(d)]
P
A
Ml (P)
B
P l (c)
l
Ml (TAB (P)) (d)
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For example, TAB (P) is read “T sub AB of P” or “the image of point P under the translation TAB” [Figure 16.25(a)]. Similarly, RO,a(P) is read “R sub O, a of P” and denotes the image of point P under the rotation RO,a [Figure 16.25(b)]. The reflection image of point P in line l is denoted by Ml (P) and read “M sub l of P” [Figure 16.25(c)]. Finally, the glide reflection image of point P, Ml (TAB(P)), is read “M sub l of T sub AB of P” [Figure 16.25(d)]. Note that in the notation for the glide reflection, the transformation TAB is applied first. Next, we will investigate properties of the transformations listed in Table 16.1. In particular, we first will determine whether they preserve distance; that is, whether the distance between two points is the same as the distance between their images. Suppose that we have a translation TAB and that we consider the effect of TAB on a line segment PQ (Figure 16.26). Let P⬘ ⫽ TAB(P) and Q⬘ ⫽ TAB(Q). According to the definition of a translation, PP ¿ ‘ AB and PP⬘ ⫽ AB. Similarly, QQ ¿ ‘ AB and QQ⬘ ⫽ AB. Combining our results, we see that PP ¿ ‘ QQ ¿ and PP⬘ ⫽ QQ⬘. B
A P´
Q´
Q
P Figure 16.26
Thus PP⬘Q⬘Q is a parallelogram, since opposite sides, PP ¿ and QQ ¿ , are parallel and congruent. Consequently, PQ ⫽ P⬘Q⬘. We have verified that translations preserve distance; that is, that the distance P⬘Q⬘ is equal to the distance PQ. (Also, since PQ ‘ P ¿ Q ¿ , we can deduce that a translation maps a line to a line parallel to the original line.) Next, we show that rotations also preserve distance. Suppose that RO,a is a rotation with center O and angle with measure a. Consider the effect of RO,a on a line segment PQ (Figure 16.27). We wish to show that PQ ⫽ P⬘Q⬘. To do this, we will establish O
b P
Connection to Algebra By using variables for this situation, we have shown that the results hold for any values that we substitute in for the variables. Thus, the geometric relationship always holds.
c
d
Q´
a a
P´
Q Figure 16.27
that 䉭POQ 艑 䉭P⬘OQ⬘ using the SAS congruence property. Since OP ⫽ OP⬘ and OQ ⫽ OQ⬘ by the definition of rotation RO,a, all that remains is to show that ∠POQ 艑 ∠P⬘OQ⬘. Let b ⫽ m(∠POQ), c ⫽ m(∠P⬘OQ⬘), and d ⫽ m(∠QOP⬘) (Figure 16.27).
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Here we are assuming that P rotates counterclockwise to P⬘ and Q to Q⬘. (If it does not, a similar argument can be developed.) Then we have b + d = a = d + c
so that
b + d = d + c
Hence b ⫽ c, so that m(∠POQ) ⫽ m(∠P⬘OQ⬘). Thus 䉭POQ 艑 䉭P⬘OQ⬘, by the SAS congruence property. Consequently, we have that PQ ⬵ P ¿ Q ¿ , since they are corresponding sides in the congruent triangles. This shows that rotations preserve distance. Unlike translations, a rotation, in general, does not map a line to a line parallel to the original line. In fact, only rotations through multiples of 180⬚ do. Using triangle congruence, we can show that reflections also preserve distance. There are four cases to consider: (1) P and Q are on l; (2) only one of P or Q is on l; (3) P and Q are on the same side of l; and (4) P and Q are on opposite sides of l.
Q l S
CASE 1: P and Q are fixed by Ml; that is, Ml (P) ⫽ P⬘ ⫽ P, and Ml (Q) ⫽ Q⬘ ⫽ Q, so that P⬘Q⬘ ⫽ PQ.
CASE 2: Suppose that P is on line l and Q is not on l (Figure 16.28). Let Q⬘ ⫽
P
Ml (Q) and let S be the intersection of segment QQ ¿ and line l. Then QS ⫽ Q⬘S and ∠QSP is a right angle, by the definition of Ml (Q). Hence 䉭QSP 艑 䉭Q⬘SP by the SAS congruence condition. Consequently, QP ⫽ Q⬘P⬘ as desired. Cases 3 and 4 are somewhat more complicated and are left for Part A Problem 24 and Part B Problem 24, respectively, in the Problem Set.
Q´ Figure 16.28
In the case of a glide reflection, since both the translation and the reflection comprising the glide reflection preserve distance, the glide reflection must also preserve distance. In the first section, a transformation that preserved size and shape was called an isometry. More formally, an isometry is a transformation that preserves distance (iso ⫽ “equal,” metry ⫽ “measure”). We have just shown that translations, rotations, reflections, and glide reflections are isometries. It can also be shown that isometries map lines to lines; that is, the image of a line l under an isometry is another line l⬘. Figure 16.29 illustrates this result for a translation, rotation, and reflection. O A
B
r
l´ = R0,a(l)
l
l
a
l
l´ = TAB(l) l´ = Mr(l) (a)
(b)
(c)
Figure 16.29 P
Q
P´
Q´
R
R´ Figure 16.30
We can show that every isometry also preserves angle measure; that is, an isometry maps an angle to an angle that is congruent to the original angle (Figure 16.30). Consider ∠PQR and its image, ∠P⬘Q⬘R⬘, under an isometry. Form 䉭PQR and its image 䉭P⬘Q⬘R⬘ (Figure 16.30). We know that PQ ⫽ P⬘Q⬘, QR ⫽ Q⬘R⬘, and PR ⫽ P⬘R⬘. Thus 䉭PQR 艑 䉭P⬘Q⬘R⬘ by the SSS congruence property, so that ∠PQR 艑 ∠P⬘Q⬘R⬘. Because isometries preserve distance, we know that isometries map triangles to congruent triangles. Also, using the corresponding angles property, it can be shown
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that isometries preserve parallelism. That is, isometries map parallel lines to parallel lines. A verification of this is left for Part B Problem 20 in the Problem Set. We can summarize properties of isometries as follows.
THEOREM Properties of Isometries 1. Isometries map lines to lines, segments to segments, rays to rays, angles to angles, and polygons to polygons. 2. Isometries preserve angle measure. 3. Isometries map triangles to congruent triangles. 4. Isometries preserve parallelism.
We can analyze congruent triangles by means of isometries. Suppose, for example, that 䉭ABC and 䉭A⬘B⬘C⬘ are congruent, they have the same orientation, and the three pairs of corresponding sides are parallel, as in Figure 16.31. Since the corresponding sides are parallel and congruent, TAA⬘ maps 䉭ABC to 䉭A⬘B⬘C⬘. A
A´
B
C
B´
C´ Figure 16.31
If 䉭ABC and 䉭A⬘B⬘C⬘ are congruent, they have the same orientation, and if AB ‘ A ¿ B ¿ , there is a rotation that maps 䉭ABC to 䉭A⬘B⬘C⬘ (Figure 16.32).
C´
A
B´ C
B O
Figure 16.32
A´
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In Figure 16.33, 䉭ABC 艑 䉭A⬘B⬘C⬘. Find a rotation that maps 䉭ABC to 䉭A⬘B⬘C⬘.
Example 16.8
S O LU T I O N To find the center, O, of the rotation, use the fact that O is equidistant from A and A⬘. Hence, O is on the perpendicular bisector, l, of AA ¿ (Figure 16.34). But O is also equidistant from B and B⬘, so O is on m, the perpendicular bisector of BB ¿ . Hence {O } ⫽ l 艚 m. The angle of the rotation is ∠AOA⬘ (Figure 16.35). m B´ A
A
B
l O B´
O B´ A
B C
C
Figure 16.33
A´
Figure 16.34
■
Figure 16.35
Thus, if two congruent triangles have the same orientation, and a translation will not map one to the other, there is a rotation that will. Next, suppose that 䉭ABC and 䉭A⬘B⬘C⬘ are congruent and have opposite orientation. Further, assume that there is a line l that is the perpendicular bisector of AA ¿ , BB ¿ , and CC ¿ (Figure 16.36). Then, by the definition of a reflection, Ml maps 䉭ABC to 䉭A⬘B⬘C⬘. If there is no reflection mapping 䉭ABC to 䉭A⬘B⬘C⬘, then a glide reflection maps 䉭ABC to 䉭A⬘B⬘C⬘. The next example shows how to find the glide axis of a glide reflection.
A´ B´ P Q A
B
C´
A´
A´
R
C
C´
C´
B
C´
C
In Figure 16.37, 䉭ABC 艑 䉭A⬘B⬘C⬘. Find a glide reflection that maps 䉭ABC to 䉭A⬘B⬘C⬘.
Example 16.9
l Figure 16.36
S O LU T I O N Since 䉭ABC and 䉭A⬘B⬘C⬘ have opposite orientation, we know that
either a reflection or glide reflection is needed. Since 䉭A⬘B⬘C⬘ does not appear to be a reflection image of 䉭ABC, we will find a glide reflection. Find the midpoints P, Q, and R of segments AA ¿ , BB ¿ , and CC ¿ , respectively (Figure 16.38). A*
B*
A´
B´
A´
P A'
A
B'
B´
P A
Q
Q
C* A
C B
B Figure 16.37
C
C
R C´
B
l
C' Figure 16.38
R C´ l
Figure 16.39
Í ! Then P, Q, and R are collinear and the line PQ is the glide axis of the glide reflection (Figure 16.39). (This result follows from Hjelmslev’s Ítheorem, which is too technical ! for us to prove here.) We reflect 䉭ABC across line PQ to obtain 䉭A*B*C*. Then 䉭A*B*C* is mapped to 䉭A⬘B⬘C⬘ by TA*A⬘. Thus we see the effect of the glide reflection, MPQ followed by TA*A⬘, which maps 䉭ABC to 䉭A⬘B⬘C⬘.
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Thus, if two congruent triangles 䉭ABC and 䉭A⬘B⬘C⬘ have opposite orientation, there exists either a reflection or a glide reflection that maps one to the other. We can summarize our results about triangle congruence and isometries as follows.
THEOREM Triangle Congruence and Isometries 䉭ABC 艑 䉭A⬘B⬘C⬘ if and only if there is an isometry that maps 䉭ABC to 䉭A⬘B⬘C⬘. If the triangles have the same orientation, the isometry is either a translation or a rotation. If the triangles have opposite orientation, the isometry is either a reflection or a glide reflection. NCTM Standard All students should examine the congruence, similarity, and line or rotational symmetry of objects using transformations.
Based on the foregoing discussion, it can be shown that there are only four types of isometries in the plane: translations, rotations, reflections, and glide reflections. Isometries allow us to define congruence between general types of shapes (i.e., collections of points) in the plane.
DEFINITION Developing Algebraic Reasoning www.wiley.com/college/musser See “Equality.”
Congruent Shapes Two shapes, and ⬘, in the plane are congruent if and only if there is an isometry that maps onto ⬘. Figure 16.40 shows several pairs of congruent shapes. Can you identify the type of isometry, in each case, that maps shape to shape ⬘? In parts (a), (b), (c), and (d), respectively, of Figure 16.40, a translation, rotation, reflection, and glide reflection will map shape to shape ⬘.
A
B´
B
A´ (a)
(b)
D´ C´
C (c)
D (d)
Figure 16.40
We can determine when two polygons are congruent by means of properties of their sides and angles. Suppose that there is a onetoone correspondence between
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polygons ᏼ and ᏼ⬘ such that all corresponding sides are congruent, as are all corresponding angles. We can show that ᏼ is congruent to ᏼ⬘. To demonstrate this, consider the polygons in Figure 16.41(a). B´
C D
D
A´
B
D´
D´
A E
E´
C´
A´
B
A E
B´
C
C´
(a)
E´ (b)
Figure 16.41
Suppose that the vertices correspond as indicated and that all pairs of corresponding sides and angles are congruent. Choose three vertices in ᏼ—say A, B, and C—and their corresponding vertices A⬘, B⬘, and C⬘ in ᏼ⬘ [Figure 16.41(b)]. By the side–angle–side congruence condition, 䉭ABC 艑 䉭A⬘B⬘C⬘. Thus there is an isometry T that maps 䉭ABC to 䉭A⬘B⬘C⬘. But then T preserves ∠BCD and distance CD. Since ᏼ and ᏼ⬘ have the same orientation, T must map segment CD to segment C ¿ D ¿ . Similarly, T maps segment DE to segment D ¿ E ¿ and segment EA to segment E ¿ A ¿ . From this we see that T maps polygon ᏼ to polygon ᏼ⬘ so that the polygons are congruent.
THEOREM Congruent Polygons Suppose that there is a onetoone correspondence between polygons ᏼ and ᏼ⬘ such that all pairs of corresponding sides and angles are congruent. Then ᏼ 艑 ᏼ⬘.
Similarity and Similitudes
P´ P
O Figure 16.42
Q
Q´
Next we study properties of size transformations and similitudes. Isometries preserve distance, whereas size transformations, hence similitudes, preserve ratios of distances. To see this, consider PQ and its image P ¿ Q ¿ under the size transformation SO,k, where k ⬎ 1 and where O, P, and Q are not collinear (Figure 16.42). By definition, OQ ¿ OQ ¿ OP ¿ OP ¿ = k, and OQ⬘ ⫽ k 䡠 OQ, so = k. Thus = . OP⬘ ⫽ k 䡠 OP, so OP OQ OP OQ OQ ¿ OP ¿ = , we can conclude that 䉭POQ ⬃ 䉭P⬘OQ⬘ Since ∠POQ ⫽ ∠P⬘OQ⬘ and OP OQ P¿Q¿ = k. In the case when O, P, and Q by SAS similarity. By corresponding parts, PQ are not collinear, the size transformation SO,k takes every segment PQ to a segment k times as long. This is also true in the case when O, P, and Q are collinear. Thus, in general, size transformations preserve ratios of distances. Since size transformations preserve ratios of distances, the image of any triangle is similar to the original triangle by SSS similarity. Thus size transformations also preserve angle measure. A proof of this result and others in the following theorem are left for the problem set.
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THEOREM Properties of Size Transformations 1. The size transformation SO,k maps a line segment PQ to a parallel line segment P ¿ Q ¿ . In general, size transformations map lines to parallel lines and rays to parallel rays. 2. Size transformations preserve ratios of distances. 3. Size transformations preserve angle measure. 4. Size transformations preserve parallelism. 5. Size transformations preserve orientation. Recall that a similitude is a combination of a size transformation and an isometry. Thus the following properties of similitudes can be obtained by reasoning from the properties of size transformations and isometries. The verification of these properties is left for the problem set.
THEOREM Properties of Similitudes 1. Similitudes map lines to lines, segments to segments, rays to rays, angles to angles, and polygons to polygons. 2. Similitudes preserve ratios of distances. 3. Similitudes preserve angle measure. 4. Similitudes map triangles to similar triangles. 5. Similitudes preserve parallelism. The next theorem and following example display the connection between the general notion of similarity and similar triangles.
THEOREM Triangle Similarity and Similitudes 䉭ABC ⬃ 䉭A⬘B⬘C⬘ if and only if there is a similitude that maps 䉭ABC to 䉭A⬘B⬘C⬘. Suppose 䉭ABC and 䉭A⬘B⬘C⬘ in Figure 16.43 are similar under the correspondence A ↔ A⬘, B ↔ B⬘, and A´ C ↔ C⬘. Suppose also that AC ⫽ 4, BC ⫽ 9, AB ⫽ 12, and A⬘C⬘ ⫽ 7.
Example 16.10
7 C´
A 4 C
a. Find B⬘C⬘ and A⬘B⬘. b. Find a similitude that maps 䉭ABC to 䉭A⬘B⬘C⬘.
12 9
B
S O LU T I O N
B¿C¿ A¿C¿ = , or BC AC B¿C¿ 7 63 3 A¿C¿ A¿B¿ = . Solving for B⬘C⬘, we find B ¿ C ¿ = = 15 . Similarly, = , 9 4 4 4 AC AB A¿B¿ 7 so that = , from which we find A⬘B⬘ ⫽ 21. 12 4
a. Since corresponding sides are proportional, we must have
B´ Figure 16.43
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A´ C´
A
7
12
4
B
C
9 B*
C*
B´ Figure 16.44 A´
Congruence and Similarity Using Transformations
883
b. Consider the size transformation SA,7/4. Figure 16.44 shows the effect of SA,7/4 on 䉭ABC. Note that B* and C* are the images of B and C, respectively, under this size transformation. Calculate that AC* ⫽ 7, AB* ⫽ 21, and B*C* = 1534. Hence 䉭AB*C* 艑 䉭A⬘B⬘C⬘ by the SSS congruence property. (In fact, any size transformation with scale factor 74 will map 䉭ABC to a triangle congruent to 䉭A⬘B⬘C⬘. For convenience, we chose point A as the center of the size transformation. However, we could have chosen any point as the center.) Observe that 䉭AB*C* and 䉭A⬘B⬘C⬘ have the same orientation. However, 䉭A⬘B⬘C⬘ is not a translation image of 䉭AB⬘C⬘. Therefore, there is a rotation that maps 䉭AB*C* to 䉭A⬘B⬘C⬘. Thus SA,7/4 followed by a rotation is a similitude that maps 䉭ABC to 䉭A⬘B⬘C⬘ (Figure 16.45). ■ We can define similarity of shapes in the plane via similitudes.
C´ A
DEFINITION Similar Shapes
B
C C*
B*
O
B´ Figure 16.45
Two shapes, and ⬘, in the plane are similar if and only if there is a similitude that maps to ⬘. Using the triangle similarity and similitudes theorem, we can show that two polygons are similar if there is a onetoone correspondence between them such that all pairs of corresponding sides are proportional and all pairs of corresponding angles are congruent (Figure 16.46).
Children’s Literature www.wiley.com/college/musser See “The Quiltmaker’s Gift” by Jeff Brumbeau.
ᏼ ᏼ⬘ Figure 16.46
THEOREM Similar Polygons Suppose that there is a onetoone correspondence between polygons ᏼ and ᏼ⬘ such that all pairs of corresponding sides are proportional and all pairs of corresponding angles are congruent. Then ᏼ ⬃ ᏼ⬘. From the definition of similar shapes, we see that all congruent shapes are similar, since we can use the size transformation with scale factor 1 to serve as the similitude in mapping one congruent shape to the other. Hence all transformations that we have studied in this chapter—namely, isometries (translations, rotations, reflections, glide reflections), size transformations, and combinations of these transformations—are similitudes. Thus all the transformations that we have studied preserve shapes of figures.
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Finally, since there are only four types of isometries in the plane, it follows that every similitude is a combination of a size transformation and one of the four types of isometries: a translation, rotation, reflection, or glide reflection. The approach to congruence and similarity of geometric shapes via isometries and similitudes is called transformation geometry. Transformation geometry provides additional problemsolving techniques in geometry that effectively complement approaches using triangle congruence and similarity and approaches using coordinates. We explore the use of transformations to solve geometry problems in the next section.
MATHE MATH E M AT ICA I CA L MO RSE R SE L When Memphis, Tennessee, officials wanted an eyecatching new sports arena, they decided to take a page from the book on their sister city, Memphis, Egypt. A $52 million stainlesssteel clad pyramid, twothirds the size of the world’s largest pyramid, the Great Pyramid of Cheops, was the result. The square pyramid is 321 feet, or 32 stories, high and its base covers 300,000 square feet, or about six football fields. In addition to housing sporting events and circuses, the pyramid houses a museum of American music.
Section 16.2
EXERCISE / PROBLEM SET A
EXERCISES 1. a. What are the coordinates of points A and B?
2. Consider P(p, q), TOP, X(x, y), and Y(x ⫹ p, y ⫹ q).
X (x, y) A Y (x + p, y + q) P (0, 1)
B
O (1, 0)
b. What are the coordinates of A⬘ and B⬘, where A⬘ ⫽ TOP(A) and B⬘ ⫽ TOP(B)? c. Use the distance formula to verify that this translation has preserved distances (i.e., show that AB ⫽ A⬘B⬘).
O
P(p,q)
1 1 a. Are the directed line segments OP and XY parallel? 1 1 b. Do OP and XY have the same length? Explain. c. Is Y ⫽ TOP(X)?
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a. Verify that the midpoint of XY lies on line l whose equation is y ⫽ x. b. Verify that XY ⬜ l. c. Is Y ⫽ Ml(X)? Why or why not?
3. Consider X(x, y) and Y(⫺y, x). Y( – y, x)
90° X (x, y)
8. a. What are the coordinates of points A and B? b. What are the coordinates of A⬘ and B⬘ where A⬘ ⫽ Mx (TPQ(A)) and B⬘ ⫽ Mx(TPQ(B))? c. Use the distance formula to verify that this glide reflection has preserved the distance AB (i.e., show that AB ⫽ A⬘ B⬘).
O
B
a. Verify that m(∠XOY) ⫽ 90⬚. b. Verify that OX ⫽ OY. c. Is Y ⫽ RO,90⬚(X)? Why or why not? 4. Given that RO,90⬚ (x, y) ⫽ (⫺y, x), find the coordinates of A⬘ and B⬘, where A⬘ ⫽ RO,90⬚(A), B⬘ ⫽ RO,90⬚(B), and A and B have coordinates (5, 2) and (3, ⫺4), respectively. Verify that RO,90⬚ preserves the distance AB.
A
P
5. a. What are the coordinates of points A and B?
Q
y 9. Determine the type of isometry that maps the shape on the left onto the shape on the right. a.
A
(0, 1) (1, 0)
B
b. What are the coordinates of A⬘ and B⬘ for A⬘ ⫽ My(A), B⬘ ⫽ My(B), where My is the reflection in the yaxis? c. Use the distance formula to verify that this reflection has preserved the distance AB (i.e., show that AB ⫽ A⬘B⬘).
b.
6. Let A be a point in the plane with coordinates (a, b). a. What are the coordinates of A⬘ ⫽ Mx(A), where Mx is the reflection in the xaxis? b. If A and B have coordinates (a, b) and (c, d), respectively, verify that Mx preserves distances. 7. Consider X(x, y) and Y(y, x). Y(y, x)
y=x c.
X(x, y)
l
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10. Find a translation, rotation, reflection, or glide reflection that maps ABCD onto A⬘B⬘C⬘D⬘ in each case. a. A B´ B A´
12. a. Find the image of ∠ABC under SP,2. b. How do the measures of ∠ABC and ∠A⬘B⬘C⬘ compare?
A D
C
D´
C´ P
b. A
B
B´
A´
D
C
C´
D´
c. A
B
D´
C´
B C
13. Is there a size transformation that maps 䉭RST to 䉭R⬘S⬘T⬘? If so, find its center and scale factor. If not, explain why not.
T
R D
C
A´
B´ S
d. A
B
C´
D´
D
C
B´
A´
11. a. Find the image of 䉭ABC under the transformation SP,2.
T´
R´ S´
14. For each part, find a similitude that maps 䉭ABC to 䉭A⬘B⬘C⬘. Describe each similitude as completely as you can. a. C´ B 3
2
4.5
A A
C
4
P
6 B´
B
C 3 A´
b. Find the lengths of AB and A ¿ B ¿ . What is the ratio A⬘B⬘/AB? c. Find the lengths of BC and B ¿ C ¿ . What is the ratio B⬘C⬘/BC? d. Find the lengths of AC and A ¿ C ¿ . What is the ratio A⬘C⬘/AC? e. In addition to their lengths, what other relationship exists between a segment and its image?
b.
B 6 2 B´
6 2 2 A
6
C
2
C´ 2 A´
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PROBLEMS 15. Let A, X, and B be points on line l with X between A and B and let A⬘, X⬘, and B⬘ be the images of A, X, and B, respectively, under a translation TPQ. P
18. Suppose there is a reflection that maps A to A⬘, B to B⬘, and C to C⬘. Verify that reflections preserve collinearity; that is, if A, B, and C are collinear, show that A⬘, B⬘, and C⬘ are collinear.
Q
A
B
B
B´
C
X
19. Verify that reflections preserve parallelism; that is, if p 储 q, show that p⬘ 储 q⬘. (Hint: Draw a reflection line and the images p⬘, q⬘, and m⬘.)
X´
m A
A´
q l
1
p 2
m
a. Show that BB⬘X⬘X and BB⬘A⬘A are both parallelograms. b. Combine conclusions from part (a) and the uniqueness of line m through B⬘ such that l 储 m to verify that A⬘, X⬘, and B⬘ are collinear. This verifies that the translation image of a line is a set of collinear points. [NOTE: To show that m ⫽ TPQ(l) goes beyond the scope of this book.]
20. Given is 䉭ABC. An isometry is applied yielding images A⬘, B⬘, and C⬘ of points A, B, and C, respectively. Verify that isometries map triangles to congruent triangles (i.e., show that 䉭ABC 艑 䉭A⬘B⬘C⬘). B
16. Suppose there is a translation that maps P to P⬘, Q to Q⬘, and R to R⬘. Verify that translations preserve angle measure; that is, show that ∠PQR 艑 ∠P⬘Q⬘R⬘. (Hint: Consider 䉭PQR and 䉭P⬘Q⬘R⬘.)
C B´ A
P P´
C´
R R´
Q
Q´ 17. Given are parallel lines p and q. Suppose that a rotation maps p to p⬘ and q to q⬘. Verify that p⬘ 储 q⬘ (i.e., that rotations preserve parallelism).
A´
21. Let P and Q be any two points. a. How many translations are possible that map P to Q? Describe them. b. How many rotations are possible that map P to Q? Describe them. 22. a. Find the center of the magnification that maps P to P⬘ and Q to Q⬘.
m 1 2
•Q⬘ p
4
m´
3
q p´ q´
P⬘•
•Q P•
b. Is the scale factor less than 1 or greater than 1?
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c. Repeat parts (a) and (b) for the following points.
24. Suppose that P and Q are on the same side of line l, P⬘ ⫽ Ml (P), and Q⬘ ⫽ Ml (Q). Show that P⬘Q⬘ ⫽ PQ.
•P
Q
•P⬘ P
l
•Q⬘ •Q 23. The size transformation SP,k maps Q to Q⬘. If P is on line l, describe how you would construct SP,k(R).
Q´ Q R
l
Section 16.2
Q´
P´
25. Use the Chapter 16 eManipulative activity Composition of Transformations on our Web site to create a reflection about two parallel lines. Describe how the original object and the image appear to be related. Is there a single transformation that would map the original object to its image? If yes, what is it? 26. Monica asks you this question about isometries and parallel lines: “If a figure has parallel sides, will they still be parallel after the transformation? And what about the corresponding parts of a figure and its image? Will they be parallel?” Discuss.
EXERCISE / PROBLEM SET B
EXERCISES 1. a. What are the coordinates of points A and B? b. Point P has coordinates (4, 5). Let A⬘ ⫽ TOP(A) and B⬘ ⫽ TOP(B). What are the coordinates of A⬘ and B⬘? c. Verify that TOP has preserved the length of AB.
3. Consider 䉭ABC pictured on the graph. The rotation RO,⫺90⬚ takes a point with coordinates (x, y) to a point with coordinates (y, ⫺x).
B B A A C
O
O 2. Given that TOP(x, y) ⫽ (x ⫹ p, y ⫹ q) and that points A and B have coordinates (a, b) and (c, d), respectively, answer the following. a. Find AB. b. What are the coordinates of A⬘ and B⬘ where A⬘ ⫽ TOP(A) and B⬘ ⫽ TOP(B)? c. Find A⬘B⬘. d. Does this general translation preserve distances?
a. Find the coordinates of A⬘, B⬘, and C⬘ where A⬘ ⫽ RO,⫺90⬚(A), B⬘ ⫽ RO,⫺90⬚(B), and C⬘ ⫽ RO,⫺90⬚(C). b. Is 䉭ABC 艑 䉭A⬘B⬘C⬘? Verify your answer. c. Is ∠ABC 艑 ∠A⬘B⬘C⬘? Verify. 4. Given that RO,180⬚(x, y) ⫽ (⫺x, ⫺y), find the coordinates of A⬘ and B⬘ where A⬘ ⫽ RO,180⬚(A) and B⬘ ⫽ RO,180⬚(B) and A and B have coordinates (⫺2, 3) and (⫺3, ⫺1), respectively. Verify that RO,180⬚ preserves the distance AB.
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5. a. What are the coordinates of points A and B? b. What are the coordinates of A⬘ and B⬘ where A⬘ ⫽ Mx(A) and B⬘ ⫽ Mx(B)? c. Use the distance formula to verify that this reflection has preserved the distance AB (i.e., show that AB ⫽ A⬘B⬘).
Congruence and Similarity Using Transformations
889
c. Verify that Mx (TOP) preserves the distance AB.
B
y A A O
x
x
9. Determine the type of isometry that maps the shape on the left onto the shape on the right. a.
B
6. Let A be a point in the plane with coordinates (a, b). a. What are the coordinates of A⬘ ⫽ My(A)? b. If points A and B have coordinates (a, b) and (c, d), respectively, verify that My preserves the distance AB.
b.
7. Consider 䉭ABC pictured on the following graph. The reflection Ml takes a point with coordinates (x, y) to a point with coordinates (y, x).
B A
y=x
C
l a. Find the coordinates of A⬘, B⬘, and C⬘, where A⬘ ⫽ Ml (A), B⬘ ⫽ Ml (B), and C⬘ ⫽ Ml (C). b. Is 䉭ABC 艑 䉭A⬘B⬘C⬘? Verify. c. Is 䉭ABC 艑 䉭A⬘B⬘C⬘? Verify. 8. a. What are the coordinates of points A and B? b. For P (p, 0), what are the coordinates of A⬘ and B⬘, where A ⫽ Mx (TOP(A)) and B⬘ ⫽ Mx(TOP(B))?
c.
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10. Find a translation, rotation, reflection, or glide reflection that maps 䉭ABC to 䉭A⬘B⬘C⬘ in each case. a. C B'
C'
B
b.
13. a. Describe an isometry followed by a size transformation that maps circle A1 to circle A2. b. Are all circles similar? Explain why or why not.
A
A' C'
C
C2
B'
A1
B A
C1 A' A2
11. a. Find the image of AB under the transformation SP,3.
14. In each part, find a similitude that maps 䉭ABC to 䉭A⬘B⬘C⬘. Describe each similitude as completely as possible. a. 5 B´ A´ A A P
B
2 B
5.2 12 4.8
b. Verify that PA⬘ ⫽ 3PA. c. Verify that PB⬘ ⫽ 3PB. d. Find the length of AB and A ¿ B ¿ . How do these lengths compare? 12. a. Find the center P and a scale factor k such that the transformation SP,k maps the small figure to the large figure. b. If SP,h maps the large figure onto the small figure, find the scale factor h. c. How are k and h related?
13
C
C´
b.
4
A´
A 10
20 3
6
B
B´
16 3
C
8
C´
c.
A´ A
2 29 29
4
2 B
5
C
C´
10
B´
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PROBLEMS 15. Suppose there exists a translation that maps line p to p⬘ and line q to q⬘. Verify that translations preserve parallelism; that is, if p 储 q, show that p⬘ 储 q⬘. p q
19. Given are two points A and B. A right triangle is drawn that has AB as its hypotenuse and point C at the vertex of the right angle. Line l is the perpendicular bisector of AC. A
p´ q´
C
16. Let RO,a be a rotation that maps A to A⬘, B to B⬘, and C to C⬘. Verify that rotations preserve collinearity; that is, if points A, B, and C are also collinear, show that points A⬘, B⬘, and C⬘ are also collinear. (Hint: Show A⬘B⬘ ⫹ B⬘C⬘ ⫽ A⬘C⬘.) C´
A´
B´
C
B
l
B
a. Verify that l ‘ CB. b. Verify that Ml followed by TCB maps A to B. 1 c. Find another directed line segment NO and line m such that Mm followed by TNO maps A to B. (Hint: Draw another right triangle with AB as its hypotenuse.) d. At what point do line l and line m intersect? Will this be true for other glide axes?
A 17. Let RO,a be a rotation that maps A to A⬘, B to B⬘, and C to C⬘. Verify that rotations preserve angle measure; that is, show that ∠BAC 艑 ∠B⬘A⬘C⬘. (Hint: Consider 䉭ABC and 䉭A⬘B⬘C⬘.)
B
C
20. Given are parallel lines p and q. Line p⬘ is the image of line p under a certain isometry, and line q⬘ is the image of line q under the same isometry. Verify that the isometry preserves parallelism (i.e., show that p⬘ 储 q⬘). (Hint: Use corresponding angles.)
1
B´ A´
A
2
C´ 18. Let line l be chosen such that B⬘ ⫽ Ml (B). Let P be a point on l and Q be the point where BB ¿ intersects l. Show that ∠BPQ 艑 ∠B⬘PQ. B
p´
p
q´
3
q
4
21. Let P and Q be any two points. a. How many reflections map P to Q ? Describe them. b. How many glide reflections map P to Q? Describe them. 22. Construct the image of quadrilateral ABCD under SP,k, where k ⫽ PA⬘/PA. Describe your procedure.
l A´ A
Q
P
B' P
B
D
C
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23. Given a size transformation that maps P to P⬘ and Q to Q⬘, describe how to construct the image of R under the same size transformation.
3. Let a ⫽ m(∠AXP) and b ⫽ m(∠A⬘XB). Use the fact that A, X, and B are collinear to show that m(∠AXP) ⫹ m (∠PXA⬘) ⫹ m(∠A⬘XB) ⫽ a ⫹ a ⫹ b ⫽ 180⬚. Determine the value for m (∠A⬘X⬘B⬘) and justify your answer. Can we conclude that A⬘, X⬘, and B⬘ are collinear?
R P´ P Q
4. Since X is on line l, we can conclude that AX ⫽ A⬘X⬘. Similarly, BX ⫽ B⬘X⬘. What is the reason for these conclusions?
Q´
24. Given that A and B are on opposite sides of line l, Ml (A) ⫽ A⬘, and Ml (B) ⫽ B⬘, follow the steps to show that AB ⫽ A⬘B⬘.
l A
2. Next, we wish to show that A⬘, X⬘, and B⬘ are collinear. Give reasons why the following angles are congruent (one reason for each congruent symbol): ∠BXQ 艑 ∠AXP 艑 ∠A⬘X⬘P 艑 ∠B⬘X⬘Q.
P
5. Use what has been shown to verify AB ⫽ A⬘B⬘ and justify. 25. Use the Chapter 16 eManipulative activity Composition of Transformations on our Web site to create a reflection about two intersecting lines. Describe how the original object and the image appear to be related. Is there a single transformation that would map the original object to its image? If yes, what is it?
A´ 26. Thomas wants to know: “In a size transformation, would the corresponding line segments of a figure and its image be parallel?” How would you answer? What would you say if Thomas asked this question about a similitude?
X
B´ Q B 1. Let P be the intersection of l and AA ¿ and Q be the intersection of l and BB ¿ . Let X be the point where AB intersects line l. Let X⬘ ⫽ Ml (X). Where is X⬘?
27. Kristie wants to know: “If there are two congruent shapes anywhere on my paper, will I always be able to find an isometry that maps one onto the other, even if one is the flip image of the other, like two mittens where the righthand one is horizontal and the lefthand one is vertical?” Discuss.
Problems Relating to the NCTM Standards and Curriculum Focal Points 1. The Focal Points for Kindergarten state “Describing shapes and space.” Explain how “slides,” “flips,” and “turns” can be used to describe the relationship between two congruent shapes. 2. The Focal Points for Grade 7 state “Students solve problems about similar objects (including figures) by using scale factors that relate corresponding lengths of the objects or by
using the fact that relationships of lengths within an object are preserved in similar objects.” 2. The NCTM Standards state “All students should examine the congruence, similarity, and line or rotational symmetry of objects using transformations.” Discuss how the similarity of two polygons can be examined using transformations.
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16.3 S TAR T I NG POINT
Geometric Problem Solving Using Transformations
893
GEOMETRIC PROBLEM SOLVING USING TRANSFORMATIONS You and your friends are shooting pool on the pool table as shown when you are confronted with three collinear balls. One of the rules of the game is that you cannot hit the eightball (black ball) first. Also, you are penalized if you don’t hit another ball with the cue ball (white ball). Describe a path that the cue ball (the white one) can take so that it will miss the eightball (the black one) but hit the other ball. Find another such path.
Using Transformations to Solve Problems The use of transformations provides an alternative approach to geometry and gives us additional problemsolving techniques. Our first example is a transformational proof of a familiar property of isosceles triangles. A
Use isometries to show that the base angles of an isosceles triangle are congruent.
Example 16.11
Let 䉭ABC be isosceles with AB ⬵ AC (Figure 16.47). Reflect 䉭ABC Í ! B´ SOLUTION
B
C
Figure 16.47
in AC forming 䉭AB⬘C. As a result, we know that ∠ABC 艑 ∠AB⬘C and ∠BAC 艑 ∠B⬘AC, since reflections preserve angle measure. Next, consider the rotation with center A and directed angle ⭿B⬘AC. Since ∠B⬘AC 艑 ∠BAC, and since AB ⬵ AC ⬵ AB ¿ , we know that this rotation maps 䉭AB⬘C to 䉭ACB. Hence ∠AB⬘C 艑 ∠ACB. Combining this with our preceding observation that ∠AB⬘C 艑 ∠ABC, we have that ∠ABC 艑 ∠ACB. ■
A particular type of rotation, called a halfturn, is especially useful in verifying properties of polygons. A halfturn, HO, is a rotation through 180⬚ with any point, O, as the center. Figure 16.48, shows a halfturn image of 䉭ABC, with point O as the center. In Figure 16.48, AB is rotated by HO to A ¿ B ¿ , where it appears that A ¿ B ¿ ‘ AB. Next we verify that the halfturn image of a line is indeed parallel to the original line.
Reflection from Research Students’ development of understanding of concepts in transformation geometry is consistent with the van Hiele theory (Soon, 1989).
C´
A
B´
O B Figure 16.48
C
A´
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A
In Figure 16.49, let l be a line and let O be the center of a halfturn. Notice that in this case O is not on l. Since HO is an isometry, it maps l to a line l⬘. Let A and B be points on l. Then A⬘ and B⬘, the images of A and B under the halfturn Í ! HO, are on l⬘. Also, and most important, A, O, and A⬘ are collinear, so that line AA ¿ is a transversal for lines l and l⬘ containing point O. Since HO is an isometry, ∠BAO 艑 ∠B⬘A⬘O. Thus, by the alternate interior angles theorem, l 储 l⬘. In summary, we have shown that a halfturn, HO, maps a line l to a line l⬘ such that l⬘ 储 l. Figure 16.50 shows the two possible cases; namely, O is on l [Figure 16.50(a)] or O is not on l [Figure 16.50(b)]. If O is on l, the result follows immediately, since every line is parallel to itself.
l
O
B´
A´ Figure 16.49
l´
l´ l
O l
O
(a)
(b)
Figure 16.50
Example 16.12 illustrates the use of halfturns to verify a property of quadrilaterals. ProblemSolving Strategy
Example 16.12
Draw a Picture
Show that if the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.
B S O LU T I O N Suppose that ABCD is a quadrilateral with diagonals AC and BD inter
A
secting at point O, the midpoint of each diagonal (Figure 16.51). Consider the halfturn HO. Then HO(A) ⫽ C, HO(B) ⫽ D, HO(C) ⫽ A, and HO(D) ⫽ B (verify). Thus HO maps side AB of quadrilateral ABCD to side CD so that AB ‘ CD. Also, HO maps AD to CB so that AD ‘ CB. Hence quadrilateral ABCD is a parallelogram, since both pairs of its opposite sides are parallel. (NOTE: Since HO is an isometry, this also shows that opposite sides of a parallelogram are congruent.) ■
O
D
C
Figure 16.51
Transformations can also be used to solve certain applied problems. Consider the following pool table problem. Ml(A) = A' 2 P 1 3
l
A B Figure 16.52
Cue ball A is to hit cushion l, then strike object ball B (Figure 16.52). Assuming that there is no “spin” on the cue ball, show how to find the desired point P on cushion l at which to aim the cue ball.
Example 16.13
S O LU T I O N Reflect the cue ball, A, in line l. That is, find Ml (A) ⫽ A⬘ (Figure 16.52). Let P be the point at which the line A ¿ B intersects cushion l. Then ∠1 艑 ∠2, since an angle is congruent to its reflection image, and ∠2 艑 ∠3, because they are vertical angles. Since the angle of incidence is congruent to the angle of reflection, we have that the cue ball should be aimed at point P. ■
Examples of pool shot paths caroming, or “banking,” off several cushions appear in the Exercise/Problem Set. Our next example concerns a minimal distance problem involving a translation.
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Towns A and B are on opposite sides of a river (Figure 16.53). The towns are to be connected with a bridge, CD, perpendicular to the river, so that the distance AC ⫹ CD ⫹ DB is as small as possible. Where should the bridge be located?
A
Example 16.14
B Figure 16.53
ProblemSolving Strategy Look for a Pattern
S O LU T I O N No matter where the bridge is located, distance CD, the width of the river, will be a constant in the sum AC ⫹ CD ⫹ DB [Figure 16.54(a)]. Hence we wish to minimize the sum AC ⫹ DB. Let S be a translation in a direction from B toward the river and perpendicular to it, for a distance equal to the width of the river, d. Let B⬘ ⫽ S(B). Then the segment AB ¿ is the shortest path from A to B⬘. Let point C be the intersection of AB ¿ and line m, one side of the river [Figure 16.54(b)].
A
A C
d
C d
D
D
B (a)
m B´ B
(b)
Figure 16.54
Let point D be the point opposite C on the other side of the river, where CD ⬜ m. That is, CD ‘ B ¿ B and CD ⫽ B⬘B. Hence quadrilateral BB⬘CD is a parallelogram, since the opposite sides CD and B ¿ B are congruent and parallel. Thus the sum AC ⫹ CD ⫹ DB is as small as possible. ■
MATHE MATH E M AT ICA L MO RSE R SE L Have you ever seen a rabbit or a bear in the clouds? How about a hexagon? In March of 2007, NASA’s Cassini spacecraft transmitted a picture of a hexagonal cloud formation circling the north pole of Saturn. Two decades earlier NASA’s Voyager 1 and 2 transmitted a similar image indicating this regular hexagon cloud is a longterm phenomena. Such an oddly shaped cloud has not been seen near any other planet. This hexagon is about 15,000 miles across, which is about twice the diameter of the earth. The thermal imaging that was used to capture the image also determined that the hexagon goes 60 miles down into the clouds. The hexagon appears to be moving with the rotation and axis of the planet. Since the rotation rate of Saturn is still unknown, this cloud may shed light on the rotation of Saturn.
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EXERCISE / PROBLEM SET A
PROBLEMS 1. On the lattice shown. ABCD is a parallelogram.
4. Figure ABCD is a kite. Point E is the intersection of the diagonals. Which of the following transformations can be used to show that 䉭ABC 艑 䉭ADC using the transformation definition of congruence? Explain. A
C
B Q A
D
P
E
B
D
a. Find the image of point P under the following sequence of halfturns: HA, then HB, then HC, then HD. That is, find HD(HC(HB(HA(P)))). b. Find HD(HC(HB(HA(Q)))). c. Write a conjecture based on your observations. 2. Find a combination of isometries that map ABCD to A⬘B⬘C⬘D⬘, or explain why this is impossible.
a. HE
b. TBE
c. RE,90⬚
C
A
D´
F
B´ A
d. MAC
5. ABCD is a rectangle. Points E, F, G, H are the midpoints of the sides. Find all the reflections and rotations that map ABCD onto itself.
C´ B
C
D
B
P
E
G
A´ 3. a. Show how the combination of RB,90⬚ followed by SO,1/2 will map 䉭ABC onto 䉭A⬙B⬙C⬙. b. Is 䉭ABC ⬃ 䉭A⬙B⬙C⬙? Explain.
H
D
C
6. ABCDEF is a regular hexagon with center O. A
B
C
F A
B
C
O
B´´ O E
C´´
A´´
D
a. List three reflections and three rotations that map the hexagon onto itself. b. How many isometries are there that map the hexagon onto itself?
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Geometric Problem Solving Using Transformations
7. ABCD is an isosceles trapezoid. Are there two isometries that map ABCD onto itself? Explain. A
B
897
c. Why is the red area in figure (3) the same size as in figures (4) and (5)? (Think parallelograms again.) d. How does the combination of parts a, b, and c prove the Pythagorean theorem? 9. ABCD is a parallelogram. Show that the diagonal AC divides ABCD into two congruent triangles using the transformation definition of congruence. (Hint: Use a halfturn.)
C
D
B
A
8. The following images illustrate one of the proofs of the Pythagorean theorem that utilizes transformation. Answer the following questions about how the red area in each image is the same size. D
C
10. Use the results of Problem 9 to show that the following statements are true. a. Opposite angles of a parallelogram are congruent. b. Opposite sides of a parallelogram are congruent. 11. Suppose that point B is equidistant from points A and C. Show that B is on the perpendicular bisector of AC. (Hint: Let P be a point on AC so that BP is the bisector of ∠ABC. Then use MBP.) (1)
(2) B
A
(3)
(4)
C
12. Let ABCD be a kite with AB ⫽ AD and BC ⫽ DC. Explain how Problem 11 shows that the following statements are true. A
B
D
(5) a. Why is the red area in figure (1) the same size as in figure (2)? (Hint: Think about the area of a parallelogram.) b. Why is the red area in figure (2) the same size as in figure (3)?
C
a. The diagonals of a kite are perpendicular. b. A kite has reflection symmetry.
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13. a. Suppose that r and s are lines such that r 储 s. Show that the combination of Mr followed by Ms is equivalent to a translation. [Hint: Let A be any point that is x units from r. Let A⬘ ⫽ Mr (A) be y units from s. Consider the distance from A to A⬘.] b. How are the distance and direction of the translation related to lines r and s?
15. ABCD is a square with side length a, while EFGH is a square with side length b. Describe a similarity transformation that will map ABCD to EFGH. B
a
C G b
A´´= Ms (A´) F s
A
H
D
E
y A´ r x A
16. Jaime drew this picture of one hole at a miniature golf course. He says he can hit the ball from point A and have it follow the path he showed and end up in the hole at point B. How accurate is his thinking on this problem? Discuss.
14. On the billiard table, ball A is to carom off two of the rails (sides) and strike ball B. a. Draw the path for a successful shot. b. Using a reflection and congruent triangles, give an argument justifying your drawing in part (a). B
B A
A
EXERCISE / PROBLEM SET B
Section 16.3
PROBLEMS 1. Triangle ABC is equilateral. Point G is the circumcenter (also the incenter). Which of the following transformations will map 䉭ABC onto itself? a. RG,120⬚ b. RG,60⬚ c. MAF d. SC,1
2. For equilateral triangle ABC in Exercise 1, list three different rotations and three different reflections that map 䉭ABC onto itself.
A
A
3. ABCDE is a regular pentagon with center O. Points F, G, H, I, J are the midpoints of the sides. List all the reflections and all the rotations that map the pentagon onto itself.
G
F
B
E E
D
O
G H
J
B
F
C
D
I
C
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4. ABCD is a parallelogram. List all the isometries that map ABCD onto itself.
A
Geometric Problem Solving Using Transformations
7. Show how the combination of Mr, followed by SO,1/2 will map 䉭ABC onto 䉭A⬘B⬘C⬘.
C
B
O
r P
C´
B´ A
D
899
B
C A´
5. a. Using the triangle ABC, find the image of point P under the following sequence of halfturns: HC(HB(HA(HC(HB(HA(P)))))) b. Apply the same sequence to point Q. c. Write a conjecture based on your observations.
8. Points P, Q, and R are the midpoints of the sides of 䉭ABC.
B B Q
P Q A
C
A
P
C
R
a. Show SA,2 (䉭APR) ⫽ 䉭ABC, SB,2 (䉭PBQ) ⫽ 䉭ABC, and SC,2 (䉭QCR) ⫽ 䉭BCA. b. How does part (a) show that PBQR is a parallelogram? 9. Suppose that ABCD is a parallelogram. Show that the diagonals AC and BD bisect each other. 6. Find a combination of isometries that will map 䉭ABC to 䉭A⬘B⬘C⬘, or explain why this is impossible. B
C
B´
C A´ A C´ B
D
A [Hint: Let P be the midpoint of AC and show that Hp(B) ⫽ D.]
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10. Suppose that point B is on the perpendicular bisector l of AC. Use Ml to show that AB ⫽ BC (i.e., that B is equidistant from A and C).
13. Suppose that ABCD is a rhombus. Use reflections MAC and MBD to show that ABCD is a parallelogram.
B
A
A
C
D
B
l 11. Suppose that lines r and s intersect at point P. C A´´
s 14. Euclid’s proof of the Pythagorean theorem involves the following diagram. In the proof, Euclid states that 䉭ABD 艑 䉭FBC and that 䉭ACE 艑 䉭KCB. Find two rotations that demonstrate these congruences.
y° P
A´ x° A
r G
a. Show that Mr followed by Ms is a rotation. b. How are the center and angle of the rotation related to r and s? [Hint: Let A⬘ ! ⫽ Mr(A) and suppose that the angle formed by PA and r measures x. Let ! A⬙ ⫽ Ms(A⬘) and suppose that the angle formed by PA and s measures y.] 12. On the billiard table, ball A is to carom off three rails, then strike ball B.
H F
A K C
B
B A
a. Draw the path for a successful shot. b. Using three reflections, give an argument justifying your drawing in part (a).
D
L
E
15. A Reuleaux triangle is a curved threesided shape (see the figure). Each curved side is a part of a circle whose radius is the length of the side of the equilateral triangle. Find the
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End of Chapter Material
area of a Reuleaux triangle if the length of the side of the triangle is 1.
901
16. Annie has reflected an isosceles triangle over its nonequal side. She says the triangle together with its image seems to have the shape of a rhombus. Is she correct? Explain. 17. Lee says a parallelogram has a line of symmetry in between two of the parallel sides. What properties of reflection could you use to explain that this is not a line of symmetry?
S S
Problems Relating to the NCTM Standards and Curriculum Focal Points 1. The Focal Points for Grade 4 state “By using transformations to design and analyze simple tilings and tessellations, students deepen their understanding of twodimensional space.” Describe an example from this section where transformations are used to analyze and prove properties of quadrilaterals.
2. The NCTM Standards state “All students should describe a motion or series of motions that will show that two shapes are congruent.” Find an example from this section where a series of motions is used to show two shapes are congruent.
E N D OF C H APTER M AT E RIAL sum AP ⫹ PB⬘ is as small as possible. By the triangle inequality, we must have A, P, and B⬘ collinear.
Solution of Initial Problem Houses A and B are to be connected to a television cable line l, at a transformer point P. Where should P be located so that the sum AP ⫹ PB is as small as possible?
B
A B
P
a R
A
P
b
x
d – x
Q
l
l B´
Strategy: Use Symmetry Reflect point B across line l to point B⬘. Then 䉭BPB⬘ is isosceles, and line l is a line of reflection symmetry for 䉭BPB⬘. Hence BP ⫽ B⬘P, so the problem is equivalent to locating P so that the
Let point Q be the intersection of BB ¿ and l. Let R be the point on l such that AR ⬜ l. Since points A, P, and B⬘ are collinear and 䉭BPB⬘ is isosceles, we have ∠APR 艑 ∠B⬘PQ 艑 ∠BPQ. But then, 䉭APR ⬃ 䉭BPQ by the AA similarity property. Thus correspon
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ding sides are proportional in these triangles. Let RP ⫽ x and RQ ⫽ d, so that PQ ⫽ d ⫺ x. Then, by similar triangles,
Additional Problems Where the Strategy “Use Symmetry” Is Useful 1. The 4 ⫻ 4 quilt of squares shown here has reflective symmetry.
a x = , d  x b so that bx = ad  ax ax + bx = ad x(a + b) = ad ad x = , a + b Hence, locate point P along RQ so that RP =
ad . a + b
Explain how to find all such quilts having reflective symmetry that are made up of 12 light squares and 4 dark squares. 2. Find all threedigit numbers that can serve as house numbers when the numbers are installed upsidedown or rightsideup. Assume that the 1 and 8 are read the same both ways. 3. The 4th number in the 41st row of Pascal’s triangle is 9880. What is the 38th number of the 41st row?
People in Mathematics Edward G. Begle (1914–1978) Edward G. Begle could be called the father of the “New Math.” Influenced by the success of Sputnik in 1957, the U.S. government, through its National Science Foundation, embarked on a systematic revision of the K–12 curriculum in the late 1950s. Although Begle was well on his way to becoming a firstrate research mathematician, he was asked to lead this effort, called the School Mathematics Study Group (SMSG), due to his unique talent for administration. He moved to Stanford, where he coordinated the SMSG for many years. In addition to this project, he initiated the National Longitudinal Study of Mathematical Abilities (NLSMA), which set a standard for subsequent such studies. Begle, whose respect for mathematics education grew as he worked with SMSG, stated that “Mathematics education is more complicated than you expected even though you expected it to be more complicated than you expected.”
Jaime Escalante (1930– ) Jaime Escalante began teaching at Garfield High in East Los Angeles as a seasoned mathematics teacher, having taught for 11 years in Bolivia. But he was not prepared for this rundown innercity school, where gang violence abounded and neither parents nor students were interested in education. After the discouraging first day on the job, he vowed, “First I’m going to teach them responsibility, and I’m going to teach them respect, and then I’m going to quit.” He succeeded but stayed on to build, year after year, a team of mathematics students with the Advanced Placement calculus exam as their Olympic competition. He built their selfconfidence and trained, coached, and prodded them as a coach might train athletes. Students would go to the Advanced Placement exam wearing their school jackets, yelling “Defense, Defense!” The movie Stand and Deliver was about Escalante’s successes as a teacher at Garfield.
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Chapter Review
903
CHAPTER REVIEW Review the following terms and exercises to determine which require learning or relearning—page numbers are provided for easy reference.
SE CTI ON 16.1 Transformations VOCABULARY/NOTATION Transformation, T 852 Image of a point, P⬘ 852 Isometry 852 Rigid motion 852 1 Directed line segment, AB 853 Equivalent directed line segments 853 Translation, TAB 853 Directed angle, ∠ABC 854 Measure of a directed angle 854 Initial side 855
Terminal side 855 Rotation with center O and angle with measure a, RO, a 855 Í ! Reflection in line l, Ml (or MAB if l = AB) 856 Glide reflection determined by directed 1 line segment AB and glide axis l (or TAB followed by MAB) 859 Clockwise orientation 860 Counterclockwise orientation 860 Translation symmetry 860
Rotation symmetry 860 Reflection symmetry 860 Glide reflection symmetry 861 Eschertype patterns 861 Size transformation SO, k with center O and scale factor k 863 Magnification 863 Dilation 863 Dilitation 863 Similitude 864
EXERCISES 1 1. Draw directed line segment AB and P anywhere. Show how to find the image of P determined by TAB.
6. Sketch a pattern that has a rotation symmetry of less than 180⬚.
2. Draw a directed angle of measure a and vertex O and P anywhere. Show how to find the image of P determined by RO, a.
7. Sketch a pattern that has reflection symmetry.
3. Draw a line l, Q on l, and P anywhere. Show how to find the images of Q and P determined by Ml. 1 4. Draw a line l, a directed line segment AB parallel to l, and P anywhere. Show how to find the image of P determined by 1 the glide reflection l and AB.
9. Draw 䉭ABC and point O anywhere. Show how to find the image of 䉭ABC determined by SO,2.
8. Sketch a pattern that has glide reflection symmetry. What must be true about such patterns?
10. Using the idea of a similitude, show that any two equilateral triangles are similar.
5. Sketch a pattern that has translation symmetry. What must be true about such patterns?
SECTION 16.2 Congruence and Similarity Using Transformations VOCABULARY/NOTATION Image of a point P under a translation, TAB(P) 875 Image of a point P under a rotation, RO,a(P) 875 Image of a point P under a reflection, Ml (P) 875
Image of a point P under a glide reflection: TAB followed by Ml, Ml (TAB(P)) 875 Isometry 877 Congruent shapes 艑 ⬘ 880 Congruent polygons, ᏼ 艑 ᏼ⬘ 881
Similar shapes, ⬃ ⬘ 883 Similar polygons, ᏼ ⬃ ᏼ⬘ 883 Transformation geometry 884
EXERCISES 1. Name the four types of isometries. 2. Given that an isometry maps line segments to line segments and preserves angle measure, show that the isometry maps any triangle to a congruent triangle.
3. Given that an isometry maps lines to lines and preserves angle measure, show that the isometry maps parallel lines to parallel lines.
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4. Describe how isometries are used to prove that two triangles are congruent. 5. Given that size transformations map line segments to line segments and preserve angle measure, show that size transformations map any triangle to a similar triangle.
6. Explain why size transformations preserve parallelism. 7. Describe how similarity is used to prove that two triangles are similar.
SE CTI O N 16.3 Geometric Problem Solving Using Transformations VOCABULARY/NOTATION Halfturn with center O, HO
893
EXERCISE 1. Given a triangle and any point O, find the halfturn image of the triangle determined by the point O.
P R O B L EM S FOR W R IT I N G/ DI S C U SS I O N 1. If you saw a figure, and its image after a translation, explain how you would go about reconstructing the direction and distance of the slide.
p
2. If you saw a figure, and its image after a reflection, explain how you would go about reconstructing the flip line. 3. If you saw a figure, and its image after a rotation, explain how you would go about reconstructing its center and the degree and direction of the turn.
40° F
q
4. If you saw a figure, and its image after a glide reflection, explain how you would go about reconstructing its glide axis.
7. Mackie says any slide can be replaced by one flip if a figure is symmetric, two flips if it isn’t. Is he correct? Explain. What are the restrictions on the flip lines?
5. If you saw a figure, and its image after a size transformation, explain how you would go about reconstructing the center of the transformation and the scale factor.
8. Penny says any turn can be replaced by two flips. If you asked her to elaborate, what specifics would you expect her to include?
6. Chungsim has taken an Lshaped figure and reflected it over a line, p. She then took the image and reflected that over a second line, q. The two lines meet in a 40⬚ angle at point F. Chungsim asks you, “Did I have to do two motions? Could I have done it with just one flip?” How would you respond? Be specific.
9. Emir says a glide reflection could not be replaced by flips. Do you agree? Explain. 10. Chris says he can move any figure in the plane onto its congruent image using only flips, no matter how difficult the task may seem. What is the most challenging problem you can think of to give him? Include your solution, of course.
CHAPTER TEST KNOWLEDGE 1. True or false? a. A translation maps a line l to a line parallel to l. b. If the direction of a turn is clockwise, the measure of the directed angle associated with the turn is positive. c. A reflection reverses orientation. d. A regular tessellation of square tiles has translation, rotation, reflection, and glide reflection symmetry.
e. An isometry preserves distance and angle measure. f. An isometry preserves orientation. g. A size transformation preserves angle measure and ratios of length. h. A similarity transformation is a size transformation followed by an isometry. 2. List the properties that are preserved by isometries but not size transformations.
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Chapter Test
3. Which of the transformations leaves exactly one point fixed when performed on the entire plane?
b. Rotate 90⬚ about point O.
4. Which of the following transformations map the segment AB to A ¿ B ¿ in such a way that AB ‘ A ¿ B ¿ ? (There may be more than one correct answer.) a. Translation b. Rotation c. Reflection d. Glide reflection e. Size transformation
O
5. Which of the following transformations map the segment AB to A ¿ B ¿ in such a way that AA ¿ ‘ BB ¿ ? (There may be more than one correct answer.) a. Translation b. Rotation c. Reflection d. Glide reflection e. Size transformation
1 c. Translate parallel to the directed line segment MN.
6. Each of the following notations describes a specific transformation. Identify the general type of transformation corresponding to each notational description.
M
a. SM,3(Q) b. RN,b(Q) c. Mn(Q) d. Mn(Txy(Q)) e. Txy(Q)
N
SKILL 7. For each of the following, trace the figure onto a piece of paper and perform the indicated transformation on the quadrilateral. a. Reflect about line m. 8. Find the following points on the square lattice.
m
a. TAB(P) b. RC,90⬚(P) c. MOC(P) d. TPB(MOC(P))
A
P
B C
O E
D
905
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9. Trace the following grid and triangle on a piece of paper and find SO,3(䉭ABC).
UNDERSTANDING 12. Explain how to prove that the square ABCD is similar to the square EFGH D
H
C
E
G
C A
F
B
B 13. Explain why performing two glide reflections, one after the other, yields either a translation or a rotation.
A O 10. Determine which of the following types of symmetry apply to the tessellation shown here: translation, rotation, reflection, glide reflection (assume that the tessellation fills the plane).
14. Given that l, m, and n are perpendicular bisectors of sides AB, BC, and AC, respectively, find the image of B after applying successively Ml followed by Mn, followed by Mm. l
C
m
n
B
A
11. Describe the following isometries as they relate to the triangles shown.
15. In each of the following cases 䉭XYZ 艑 䉭X⬘Y⬘Z⬘. Identify the type of isometry that maps 䉭XYZ to 䉭X⬘Y⬘Z⬘ as either rotation, translation, reflection, glide reflection, or none. Y a.
F X
4
3 C
A
B Z 2
1
Z′
D a. The reflection that maps 3 to 2 b. The rotation that maps 3 to 4 c. The translation that maps 3 to 1
Y′ X′
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Chapter Test
b.
907
19. Given isosceles Í ! 䉭ABC, where M is the midpoint of AB, prove that CM is a symmetry line for 䉭ABC.
Y X Y′
C
X′ Z
Z′ c.
Y
A
B
M
X 20. In the following figure, 䉭ABC 艑 䉭A⬘B⬘C⬘. Completely describe a transformation that maps 䉭ABC to 䉭A⬘B⬘C⬘ (i.e., if the transformation is a reflection, then construct the line of reflection; if the transformation is a translation, then construct the directed line segment, etc.).
Z Y′
Z′ X′
B′
A′
16. Trace the following figure onto a piece of paper. Sketch the approximate location of the image of 䉭XYZ under the rotation RO,60⬚.
C′ B
Y
C
X
A Z O 17. Let M ⫽ (⫺1, 5) and N ⫽ (3, ⫺1). Give the coordinates of the images of the points A ⫽ (1, 1), B ⫽ (⫺2, ⫺4), and C ⫽ (x, y) under the transformation TMN.
21. In the following figure, is 䉭ABC the image of 䉭XYZ under a size transformation? Justify your conclusion. Z
PROBLEM SOLVING/APPLICATION Y
18. Find AB, AC, and CE for the figure shown. A
B
C 3 2
C
D X 4 E
A
B
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Epilogue
AN ECLECTIC APPROACH TO GEOMETRY Introduction NCTM Standard
Three approaches to geometry were presented in Chapters 14–16:
Instructional programs should enable all students to select and use various types of reasoning and methods of proof.
1. The traditional Euclidean approach using congruence and similarity 2. The coordinate approach 3. The transformation approach A fourth common approach is the vector approach. The value of multiple approaches to problem solving is that a theorem may be proved using any of several methods, one of which may lead to an easy proof. Facility using these various approaches gives one “mathematical power” when making proofs in geometry. This section gives a proof of the midsegment theorem using each of the three approaches. The problem set contains problems where you may choose the approach that you feel will lead to an easy solution.
The Midsegment Theorem In Section 14.5 it was shown that the midsegment of a triangle is parallel and equal to onehalf the length of the third side. Following are three different proofs, the first using congruence, the second using coordinates, and the third using transformations. Let PQ be a midsegment of ⌬ABC as shown in Figure E.1(a). Extend PQ through Q to R so that PQ ⫽ QR , and draw CR [Figure E.1(b)]. C O N G RU EN C E PR OOF
B
B Q
P A
(a)
B Q
P C
A
(b)
R C
Q
P A
(c)
R C
Figure E.1
By SAS, ⌬PBQ 艑 ⌬RCQ, since ∠PQB and ∠RQC are vertical angles. Consequently, ∠PBQ 艑 ∠RCQ by corresponding parts. Viewing these parts as a pair of congruent alternate interior angles, we have AP ‘ CR. Again, by corresponding parts, PB ⬵ CR. Also, since AP ⬵ PB, we have AP ⬵ CR [Figure E.1(c)]. Thus, by Example 14.13, APRC is a parallelogram, since AP ‘ CR and A ⬵ CR. It follows that PQ ‘ AC and PQ = 12AC. ■
909
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Epilogue
Let PQ be a midsegment of ⌬ABC with coordinates as shown in Figure E.2. By the midpoint formula, the coordinates of P are (a2, b2) and of Q are (a +2 c, b2). Since P and Q have the same ycoordinate, namely b2, PQ is horizontal; hence it is parallel to the xaxis and thus to AC. Also, PQ = a +2 c  a2 = 2c and AC ⫽ c. Thus PQ = 12AC. C O O RD I N ATE PR OOF
B(a,b) Q
P A
C(c,0)
■
Figure E.2
T RA N S FOR MATION PR OOF Let PQ be a midsegment of ⌬ABC as shown in Figure E.3. Consider the size transformation SB,2. Under SB,2, the image of P is A and the image of Q is C. Thus, according to the properties of size transformations, PQ ‘ AC and PQ = 12AC, since the scale factor is 2. B P
Q
A Figure E.3
C
■
If the coordinate and transformation proofs seem to be much easier, it is because much work went into developing concepts before we applied them in these proofs. For example, we applied the midpoint formula in the coordinate proof and properties of size transformations in the transformation proof. The following Problem Set will provide practice in proving geometric relationships using the three approaches. Your choice of approach is a personal one. You may find one approach to be preferable (easier?) to a friend’s. You will find that comparing the various proofs and discussing the merits of the various approaches is worthwhile.
E
EXERCISE / PROBLEM SET A
1. Prove: Consecutive angles of a parallelogram ABCD are supplementary. (Use congruence geometry.)
F(c,d) N E(a,b)
2. The sides of DEFG have midpoints M, N, O, and P, as shown. Verify that MNOP is a parallelogram.
O M
D(0,0)
P
G(e,0)
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3. Suppose that ⌬ABC is isosceles with AB ⫽ AC. Let P be the point on BC so that AP bisects ∠BAC. Use the transformation MAP to show that ∠ABC 艑 ∠ACB. 4. In quadrilateral ABCD, both pairs of opposite sides are congruent. Prove that ABCD is a parallelogram. (Hint: Draw diagonal BD.) (Do two proofs, one congruence and one coordinate.)
911
Prove Problems 6–8 using any approach. 6. A rectangle is sometimes defined as a parallelogram with at least one right angle. If parallelogram PQRS has a right angle at P, verify that PQRS has four right angles. 7. Prove: The diagonals of a square are perpendicular. 8. In quadrilateral PQRS, both pairs of opposite angles are congruent. Prove that PQRS is a parallelogram.
5. Prove: In an isosceles triangle, the medians to the congruent sides are congruent. (Do two proofs, one congruence and one coordinate.)
E
EXERCISE / PROBLEM SET B
1. If a pair of opposite sides of ABCD are parallel and congruent, then it is a parallelogram. (Hint: Let AB ‘ DC, AB ⬵ DC and draw BD.) (Use congruence geometry.) 2. Given is trapezoid ABCD with AB 储 CD. M is the midpoint of AD and N is the midpoint of BC. Show that MN = 12(AB + DC) and MN ‘ AB. (Use coordinates.)
3. Suppose that ABCD is a parallelogram and P is the intersection of the diagonals. a. Show that HP (A) ⫽ C and HP (B) ⫽ D. b. How does part (a) show that a parallelogram has rotation symmetry? 4. In ABCD, the diagonals bisect each other at E. Prove that ABCD is a parallelogram. (Use coordinates.) 5. Verify that the diagonals of a rhombus are perpendicular. (Do two proofs, one congruence and one coordinate.) Prove Problems 6–8 using any approach.
D(a,b) M
A(0,0)
C(c,b)
6. Given parallelogram LMNO with perpendicular diagonals LN and MO intersecting at P, prove that LMNO is a rhombus. N
7. Show that the diagonals of an isosceles trapezoid are congruent. B(e,0)
8. Quadrilateral HIJK is a rectangle where HJ and IK intersect at L and are perpendicular. Prove that HIJK is a square.
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ELEMENTARY LOGIC
y
x
xy
x2
x
y2
xy
y
Figure T1.1
Logic allows us to determine the validity of arguments, in and out of mathematics. The validity of an argument depends on its logical form, not on the particular meaning of the terms it contains. For example, the argument, “All X’s are Y’s; all Y’s are Z’s; therefore all X’s are Z’s” is valid no matter what X, Y, and Z are. In this topic section we will study how logic can be used to represent arguments symbolically and to analyze arguments using tables and diagrams.
Statements Often, ideas in mathematics can be made clearer through the use of variables and diagrams. For example, the equation 2m ⫹ 2n ⫽ 2(m ⫹ n), where the variables m and n are whole numbers, can be used to show that the sum of any two arbitrary even numbers, 2m and 2n here, is the even number 2(m ⫹ n). Figure T1.1 shows that (x ⫹ y)2 ⫽ x2 ⫹ 2xy ⫹ y2, where each term in the expanded product is the area of the rectangular region so designated. In a similar fashion, symbols and diagrams can be used to clarify logic. Statements are the building blocks on which logic is built. A statement is a declarative sentence that is true or false but not both. Examples of statements include the following: 1. Alaska is geographically the largest state of the United States. (True) 2. Texas is the largest state of the United States in population. (False) 3. 2 ⫹ 3 ⫽ 5. (True) 4. 3 ⬍ 0. (False) The following are not statements. 1. Oregon is the best state. (Subjective) 2. Help! (An exclamation) 3. Where were you? (A question) 4. The rain in Spain. (Not a sentence) 5. This sentence is false. (Neither true nor false!) Statements are usually represented symbolically by lowercase letters (e.g., p, q, r, and s). New statements can be created from existing statements in several ways. For example, if p represents the statement “The sun is shining,” then the negation of p, written ⬃p and read “not p,” is the statement “The sun is not shining.” When a statement
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is true, its negation is false and when a statement is false, its negation is true; that is, a statement and its negation have opposite truth values. This relationship between a statement and its negation is summarized using a truth table: p
⬃p
T F
F T
This table shows that when the statement p is T, then ⬃ p is F and when p is F, ⬃ p is T.
Logical Connectives Two or more statements can be joined, or connected, to form compound statements. The four commonly used logical connectives “and,” “or,” “if–then,” and “if and only if ” are studied next.
AND If p is the statement “It is raining” and q is the statement “The sun is shining,” then the conjunction of p and q is the statement “It is raining and the sun is shining” or, symbolically, “p ¿ q.” The conjunction of two statements p and q is true exactly when both p and q are true. This relationship is displayed in the next truth table. p
q
p¿q
T T F F
T F T F
T F F F
Notice that the two statements p and q each have two possible truth values, T and F. Hence there are four possible combinations of T and F to consider.
OR The disjunction of statements p and q is the statement “p or q,” symbolically,
“p ¡ q.” In practice, there are two common uses of “or”: the exclusive “or” and the inclusive “or.” The statement “I will go or I will not go” is an example of the use of the exclusive “or,” since either “I will go” is true or “I will not go” is true, but both cannot be true at the same time. The inclusive “or” (called “and/or” in everyday language) allows for the situation in which both parts are true. For example, the statement “It will rain or the sun will shine” uses the inclusive “or”; it is true if (1) it rains, (2) the sun shines, or (3) it rains and the sun shines. That is, the inclusive “or” in p ¡ q allows for both p and q to be true. In mathematics, we agree to use the inclusive “or,” whose truth values are summarized in the next truth table. p
q
p¡q
T T F F
T F T F
T T T F
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Example T1.1
Determine whether the following statements are true or false, where p represents “Rain is wet” and q represents
“Black is white.” a. ⬃p b. p ¿ q d. p ¿ (⬃q) e. ⬃( p ¿ q)
c. (⬃p) ¡ q f. ⬃[ p ¡ (⬃ q)]
S O LU T I O N
a. p is T, so ⬃p is F. b. p is T and q is F, so p ¿ q is F. c. ⬃p is F and q is F, so (⬃p) ¡ q is F. d. p is T and ⬃q is T, so p ¿ (⬃q) is T. e. p is T and q is F, so p ¿ q is F and ⬃( p ¿ q) is T. f. p is T and ⬃q is T, so p ¡ (⬃ q) is T and ⬃[p ¡ (⬃q)] is F. ■
IFTHEN One of the most important compound statements is the implication. The statement “If p, then q, “ denoted by “p : q,” is called an implication or conditional statement; p is called the hypothesis, and q is called the conclusion. To determine the truth table for p : q, consider the following conditional promise given to a math class: “If you average at least 90% on all tests, then you will earn an A.” Let p represent “Your average is at least 90% on all tests” and q represent “You earn an A.” Then there are four possibilities: AVERAGE AT LEAST 90% Yes Yes No No
EARN AN A
PROMISE KEPT
Yes No Yes No
Yes No Yes Yes
Notice that the only way the promise can be broken is in line 2. In lines 3 and 4, the promise is not broken, since an average of at least 90% was not attained. (In these cases, a student may still earn an A—it does not affect the promise either way.) This example suggests the following truth table for the conditional. p
q
p:q
T T F F
T F T F
T F T T
One can observe that the truth values for p ¿ q and q ¿ p are always the same. Also, the truth tables for p ¡ q and q ¡ p are identical. However, it is not the case that the truth tables of p : q and q : p are identical. Consider this example: Let p be “You live in New York City” and q be “You live in New York State.” Then p : q is true, whereas q : p is not true, since you may live in Albany, for example. The conditional q : p is called the converse of p : q. As the example shows, a conditional may be true, whereas its converse may be false. On the other hand, a conditional and its converse may both be true. Two other variants of a conditional occur in mathematics, the contrapositive and the inverse. Given conditional: p : q The converse of p : q is q : p. The inverse of p : q is (⬃p) : (⬃q). The contrapositive of p : q is (⬃q) : (⬃p).
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The following truth table displays the various truth values for these four conditionals.
p
q
⬃p
⬃q
CONDITIONAL p:q
CONTRAPOSITIVE ⬃q : ⬃p
CONVERSE q:p
INVERSE ⬃p : ⬃q
T T F F
T F T F
F F T T
F T F T
T F T T
T F T T
T T F T
T T F T
Notice that the columns of truth values under the conditional p : q and its contrapositive are the same. When this is the case, we say that the two statements are logically equivalent. In general, two statements are logically equivalent when they have the same truth tables. Similarly, the converse of p : q and the inverse of p : q have the same truth table; hence, they, too, are logically equivalent. In mathematics, replacing a conditional with a logically equivalent conditional often facilitates the solution of a problem.
Example T1.2
Prove that if x2 is odd, then x is odd.
S O LU T I O N Rather than trying to prove that the given conditional is true, consider its logically equivalent contrapositive: If x is not odd (i.e., x is even), then x2 is not odd (i.e., x2 is even). Even numbers are of the form 2m, where m is a whole number. Thus the square of 2m, (2m)2 ⫽ 4m2 ⫽ 2(2m2), is also an even number since it is of the form 2n, where n ⫽ 2m2. Thus if x is even, then x2 is even. Therefore, the contrapositive of this conditional, our original problem, is also true. ■
IF AND ONLY IF The connective “p if and only if q, “ called a biconditional and written p 4 q, is the conjunction of p : q and its converse q : p. That is, p 4 q is logically equivalent to ( p : q) ¿ (q : p). The truth table of p 4 q follows. p
q
p:q
q:p
(p : q) ¿ (q : p)
p4q
T T F F
T F T F
T F T T
T T F T
T F F T
T F F T
Notice that the biconditional p 4 q is true when p and q have the same truth values and false otherwise. Often in mathematics the words necessary and sufficient are used to describe conditionals and biconditionals. For example, the statement “Water is necessary for the formation of ice” means “If there is ice, then there is water.” Similarly, the statement “A rectangle with two adjacent sides the same length is a sufficient condition to determine a square” means “If a rectangle has two adjacent sides the same length, then it is a square.” Symbolically we have the following: p : q means q is necessary for p p : q means p is sufficient for q p 4 q means p is necessary and sufficient for q
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Arguments Deductive or direct reasoning is a process of reaching a conclusion from one (or more) statements, called the hypothesis (or hypotheses). This somewhat informal definition can be rephrased using the language and symbolism in the preceding section. An argument is a set of statements in which one of the statements is called the conclusion and the rest comprise the hypothesis. A valid argument is an argument in which the conclusion must be true whenever the hypothesis is true. In the case of a valid argument, we say that the conclusion follows from the hypothesis. For example, consider the following argument: “If it is snowing, then it is cold. It is snowing. Therefore, it is cold.” In this argument, when the two statements in the hypothesis— namely “If it is snowing, then it is cold” and “It is snowing”—are both true, then one can conclude that “It is cold.” That is, this argument is valid since the conclusion follows from the hypothesis. An argument is said to be an invalid argument if its conclusion can be false when its hypothesis is true. An example of an invalid argument is the following: “If it is raining, then the streets are wet. The streets are wet. Therefore, it is raining.” For convenience, we will represent this argument symbolically as [( p : q) ¿ q ] : p. This is an invalid argument, since the streets could be wet from a variety of causes (e.g., a street cleaner, an open fire hydrant, etc.) without having had any rain. In this example, p : q is true and q may be true, while p is false. The next truth table also shows that this argument is invalid, since it is possible to have the hypothesis [( p : q) ¿ q ] true with the conclusion p false. p
q
p:q
(p : q) ¿ q
T T F F
T F T F
T F T T
T F T F
The argument with hypothesis [( p : q) ¿ ⬃p] and conclusion ⬃q is another example of a common invalid argument form, since when p is F and q is T, [( p : q) ¿ ⬃p] is T and ⬃q is F. Three important valid argument forms, used repeatedly in logic, are discussed next.
Modus Ponens: [(p : q) ¿ p] : q In words modus ponens, which is also
called the law of detachment, says that whenever a conditional statement and its hypothesis are true, the conclusion is also true. That is, the conclusion can be “detached” from the conditional. An example of the use of this law follows. If a number ends in zero, then it is a multiple of 10. Forty is a number that ends in zero. Therefore, 40 is a multiple of 10.
(NOTE: Strictly speaking, the sentence “a number ends in zero” is an “open” sentence, since no particular number is specified; hence the sentence is neither true nor false as given. The sentence “Forty is a number that ends in zero” is a true statement. Since the use of open sentences is prevalent throughout mathematics, we will permit such “open” sentences in conditional statements without pursuing an indepth study of such sentences.)
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The following truth table verifies that the law of detachment is a valid argument form.
Logical people
Mathematicians • Pólya
Figure T1.2
Multiple of 2 Multiple of 4
Multiple of 8
Figure T1.3
p
q
p:q
(p : q) ¿ p
T T F F
T F T F
T F T T
T F F F
Notice that in line 1 in the preceding truth table, when the hypothesis ( p : q) ¿ p is true, the conclusion, q, is also true. This law of detachment is used in everyday language and thought. Diagrams can also be used to determine the validity of arguments. Consider the following argument. All mathematicians are logical. Pólya is a mathematician. Therefore, Pólya is logical. This argument can be pictured using an Euler diagram (Figure T1.2). The “mathematician” circle within the “logical people” circle represents the statement “All mathematicians are logical.” The point labeled “Pólya” in the “mathematician” circle represents “Pólya is a mathematician.” Since the “Pólya” point is within the “logical people” circle, we conclude that “Pólya is logical.” The second common valid argument form follows.
Hypothetical Syllogism: [(p : q) ¿ (q : r)] : (p : r) The following argument is an application of this law:
If a number is a multiple of 8, then it is a multiple of 4. If a number is a multiple of 4, then it is a multiple of 2. Therefore, if a number is a multiple of 8, it is a multiple of 2. Hypothetical syllogism, also called the chain rule, can be verified using an Euler diagram (Figure T1.3). The circle within the “multiples of 4” circle represents the “multiples of 8” circle. Then the “multiples of 4” circle is within the “multiples of 2” circle. Thus, from the diagram, it must follow that “If a number is a multiple of 8, then it is a multiple of 2,” since all the multiples of 8 are within the “multiples of 2” circle. The following truth table also proves the validity of hypothetical syllogism. p
q
r
p:q
q:r
p:r
(p : q) ¿ (q : r)
T T T T F F F F
T T F F T T F F
T F T F T F T F
T T F F T T T T
T F T T T F T T
T F T F T T T T
T F F F T F T T
Observe that in rows 1, 5, 7, and 8 the hypothesis ( p : q) ¿ (q : r) is true. In both of these cases, the conclusion, p : r, is also true; thus the argument is valid. The final valid argument we study here is used often in mathematical reasoning.
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Modus Tollens: [(p : q) ¿ ⬃q] : ⬃p Consider the following argument: Numbers that end in 1, 3, 7, 9
Numbers that are powers of 3
Figure T1.4
If a number is a power of 3, then it ends in a 9, 7, 1, or 3. The number 3124 does not end in a 9, 7, 1, or 3. Therefore, 3124 is not a power of 3. This argument is an application of modus tollens. Figure T1.4 illustrates this argument. All points outside the larger circle represent numbers not ending in 1, 3, 7, or 9. Clearly, any point outside the larger circle must be outside the smaller circle. Thus, since 3124 is outside the “powers of 3” circle, it is not a power of 3. In words, modus tollens says that whenever a conditional is true and its conclusion is false, the hypothesis is also false. The next truth table provides a verification of the validity of this argument form. p
q
p:q
(p : q) ¿ ⬃q
⬃p
T T F F
T F T F
T F T T
F F F T
F F T T
Notice that row 4 is the only instance when the hypothesis ( p : q) ¿ ⬃q is true. In this case, the conclusion of [( p : q) ¿ ⬃q] : ⬃p, namely ⬃p, is also true. Hence the argument is valid. Notice how the validity of modus tollens also can be shown using modus ponens with a contrapositive: 3( p : q) ¿ ' q4 4 3( ' q : ' p) ¿ ' q4 and 3( ' q : ' p) ¿ ' q4 : ' p.
Mathematicians
All three of these valid argument forms are used repeatedly when reasoning, especially in mathematics. The first two should seem quite natural, since we are schooled in them informally from the time we are young children. For example, a parent might say to a child: “If you are a good child, then you will receive presents.” Needless to say, every little child who wants presents learns to be good. This, of course, is an application of the law of detachment. Similarly, consider the following statements: If you are a good child, then you will get a new bicycle. If you get a new bicycle, then you will have fun.
Logicians
Figure T1.5
P
M L Figure T1.6
The conclusion children arrive at is “If I am good, then I will have fun,” an application of the law of syllogism. The three argument forms we have been studying can also be applied to statements that are modified by “quantifiers,” that is, words such as all, some, every, or their equivalents. Here, again, Euler diagrams can be used to determine the validity or invalidity of various arguments. Consider the following argument: All logicians are mathematicians. Some philosophers are not mathematicians. Therefore, some philosophers are not logicians. The first line of this argument is represented by the Euler diagram in Figure T1.5. However, since the second line guarantees that there are “some” philosophers outside the “mathematician” circle, a dot is used to represent at least one philosopher who is not a mathematician (Figure T1.6). Observe that, due to the dot, there is always a
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philosopher who is not a logician; hence the argument is valid. (Note that the word some means “at least one.”) Next consider the argument: All rock stars have green hair. No presidents of banks are rock stars. Therefore, no presidents of banks have green hair.
G R
P
Figure T1.7
T1
An Euler diagram that represents this argument is shown in Figure T1.7, where G represents all people with green hair, R represents all rock stars, and P represents all bank presidents. Note that Figure T1.7 allows for presidents of banks to have green hair, since the circles G and P may have an element in common. Thus the argument, as stated, is invalid since the hypothesis can be true while the conclusion is false. The validity or invalidity of the arguments given in the Exercise/Problem Set can be determined in this way using Euler diagrams. Be sure to consider all possible relationships among the sets before drawing any conclusions.
EXERCISE / PROBLEM SET A
EXERCISES 1. Determine which of the following are statements. a. What’s your name? b. The rain in Spain falls mainly in the plain. c. Happy New Year! d. Five is an odd number. 2. Write the following in symbolic form using p, q, r, ⬃, ¿, ¡, :, 4, where p, q, and r represent the following statements: p: The sun is shining. q: It is raining. r: The grass is green. a. If it is raining, then the sun is not shining. b. It is raining and the grass is green. c. The grass is green if and only if it is raining and the sun is shining. d. Either the sun is shining or it is raining.
3. If p is T, q is F, and r is T, find the truth values for the following: a. p ¿ ⬃ q b. ⬃( p ¡ q) c. (⬃p) : r d. (⬃p ¿ r) 4 q e. (⬃q ¿ p) ¡ r f. p ¡ (q 4 r) g. (r ¿ ⬃p) ¡ (r ¿ ⬃q) h. ( p ¿ q) : (q ¡ ⬃r) 4. Write the converse, inverse, and contrapositive for each of the following statements. a. If I teach third grade, then I am an elementary school teacher. b. If a number has a factor of 4, then it has a factor of 2. 5. Construct one truth table that contains truth values for all of the following statements and determine which are logically equivalent. a. (⬃ p) ¡ (⬃ q) b. (⬃ p) ¡ q c. (⬃p) ¿ (⬃ q) d. p : q e. ⬃( p ¿ q) f. ⬃( p ¡ q)
PROBLEMS 6. Determine the validity of the following arguments. a. All professors are handsome. Some professors are tall. Therefore, some handsome people are tall. b. If I can’t go to the movie, then I’ll go to the park. I can go to the movie. Therefore, I will not go to the park.
c. If you score at least 90%, then you’ll earn an A. If you earn an A, then your parents will be proud. You have proud parents. Therefore, you scored at least 90%. d. Some arps are bomps. All bomps are cirts. Therefore, some arps are cirts.
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e. All equilateral triangles are equiangular. All equiangular triangles are isosceles. Therefore, all isosceles triangles are equilateral. f. If you work hard, then you will succeed. You do not work hard. Therefore, you will not succeed. g. Some girls are teachers. All teachers are college graduates. Therefore, all girls are college graduates. h. If it doesn’t rain, then the street won’t be wet. The street is wet. Therefore, it rained. 7. Determine a valid conclusion that follows from each of the following statements and explain your reasoning. a. If you study hard, then you will be popular. You will study hard. b. If Scott is quick, then he is a basketball star. Scott is not a basketball star. c. All friends are respectful. All respectful people are trustworthy. d. Every square is a rectangle. Some parallelograms are rhombuses. Every rectangle is a parallelogram. 8. Which of the laws (modus ponens, hypothetical syllogism, or modus tollens) is being used in each of the following arguments?
T1
a. If Joe is a professor, then he is learned. If you are learned, then you went to college. Joe is a professor, so he went to college. b. All women are smart. Helen of Troy was a woman. So Helen of Troy was smart. c. If you have children, then you are an adult. Bob is not an adult, so he has no children. d. If today is Tuesday, then tomorrow is Wednesday. Tomorrow is Saturday, so today is not Tuesday. e. If I am broke, I will ride the bus. When I ride the bus, I am always late. I’m broke, so I am going to be late. 9. Decide the truth value of each of the following. a. Alexander Hamilton was once president of the United States. b. The world is flat. c. If dogs are cats, then the sky is blue. d. If Tuesday follows Monday, then the sun is hot. e. If Christmas day is December 25, then Texas is the largest state in the United States. 10. Draw an Euler diagram to represent the following argument and decide whether it is valid. All timid creatures (T) are bunnies (B). All timid creatures are furry (F). Some cows (C) are furry. Therefore, all cows are timid creatures.
EXERCISE / PROBLEM SET B
EXERCISES
4. Prove that the conditional p : q is logically equivalent to ⬃p ¡ q.
1. Let r, s, and t be the following statements:
5. State the hypothesis (or hypotheses) and conclusion for each of the following arguments. a. All football players are introverts. Tony is a football player, so Tony is an introvert. b. Bob is taller than Jim, and Jim is taller than Sue. So Bob is taller than Sue. c. penguins are elegant swimmers. No elegant swimmers fly, so penguins don’t fly.
r: Roses are red. s: The sky is blue. t: Turtles are green. Translate the following statements into English. a. r ¿ s b. r ¿ (s ¡ t) c. s : (r ¿ t) d. (⬃t ¿ t) : ⬃r
2. Fill in the headings of the following table using p, q, ¿, ¡, ⬃, and :. p
q
T T F F
T F T F
T F T T
F T T T
T F F F
T T F T
3. Suppose that p : q is known to be false. Give the truth values for the following: a. p ¡ q b. p ¿ q c. q : p d. ⬃q : p
6. Use a truth table to determine which of the following are always true. a. ( p : q) : (q : p) b. ⬃p : p c. [p ¿ ( p : q)] : q d. ( p ¡ q) : ( p ¿ q) e. ( p ¿ q) : p 7. Using each pair of statements, determine whether (i) p is necessary for q; (ii) p is sufficient for q; (iii) p is necessary and sufficient for q. a. p: Bob has some water. q: Bob has some ice, composed of water. b. p: It is snowing. q: It is cold. c. p: It is December. q: 31 days from today it is January.
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PROBLEMS 8. If possible, determine the truth value of each statement. Assume that a and b are true, p and q are false, and x and y have unknown truth values. If a value can’t be determined, write “unknown.” a. p : (a ¡ b) b. b : ( p ¡ a) c. x : p d. a ¡ p e. b ¿ q f. b : x g. a ¿ (b ¡ x) h. (y ¡ x) : a i. ( y ¿ b) : p j. (a ¡ x) : (b ¿ q) k. x : a l. x ¡ p m. ⬃x : x n. x ¡ (⬃x) o. ⬃[y ¿ (⬃y)]
10. Use the following Euler diagram to determine which of the following statements are true. (Assume that there is at least one person in every region within the circles.) a. All women are mathematicians. b. Euclid was a woman. c. All mathematicians are men. d. All professors are humans. e. Some professors are mathematicians. f. Euclid was a mathematician and human.
Mathematicians
9. Rewrite each argument in symbolic form, then check the validity of the argument. a. If today is Wednesday (w), then yesterday was Tuesday (t). Yesterday was Tuesday, so today is Wednesday. b. The plane is late (l) if it snows (s). It is not snowing. Therefore, the plane is not late. c. If I do not study (s), then I will eat (e). I will not eat if I am worried (w). Hence, if I am worried, I will study. d. Meg is married (m) and Sarah is single (s). If Bob has a job ( j), then Meg is married. Hence Bob has a job.
Humans Euclid Professors
Women
TOPIC REVIEW VOCABULARY/NOTATION Statement, p 912 Negation, ⬃p 912 Truth table 913 Compound statements 913 Logical connectives 913 Conjunction (and), p ¿ q 913 Disjunction (or), p ¡ q 913 Exclusive “or” 913 Inclusive “or” 913 Implication/conditional (if . . . then), p : q 914
Hypothesis 914 Conclusion 914 Converse 914 Inverse 914 Contrapositive 914 Logically equivalent statements 915 Biconditional (if and only if), p 4 q 915 Is necessary for 915 Is sufficient for 915 Is necessary and sufficient for 915
Deductive/direct reasoning 916 Argument 916 Valid argument 916 Invalid argument 916 Modus ponens (law of detachment) 916 Euler diagram 917 Hypothetical syllogism (chain rule) 917 Modus tollens 918
ELEMENTARY LOGIC TEST KNOWLEDGE 1. True or false? a. The disjunction of p and q is true whenever p is true and q is false. b. If p : q is true, then ⬃p : ⬃q is true. c. In the implication q : p, the hypothesis is p.
d. [( p : q) ¿ (q : r)] : ( p : r) is the modus tollens. e. “I am older than 20 or younger than 30” is an example of an exclusive “or.” f. A statement is a sentence that is true or false, but not both. g. The converse of p : q is ⬃p : ⬃q. h. p : q means p is necessary for q.
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Elementary Logic
SKILL
UNDERSTANDING
2. Find the converse, inverse, and contrapositive of each.
6. Using an Euler diagram, display an invalid argument. Explain.
a. p : ⬃q
b. ⬃p : q
c. ⬃q : ⬃p
7. Is it ever the case that the conjunction, disjunction, and implication of two statements are all true at the same time? all false? If so, what are the truth values of each statement? If not, explain why not.
3. Decide the truth value of each statement. a. 4 ⫹ 7 ⫽ 11 and 1 ⫹ 5 ⫽ 6. b. 2 ⫹ 5 ⫽ 7 4 4 ⫹ 2 ⫽ 8. c. 3 䡠 5 ⫽ 12 or 2 䡠 6 ⫽ 11. d. If 2 ⫹ 3 ⫽ 5, then 1 ⫹ 2 ⫽ 4. e. If 3 ⫹ 4 ⫽ 6, then 8 䡠 4 ⫽ 31. f. If 7 is even, then 8 is even.
PROBLEM SOLVING/APPLICATION
4. Use Euler diagrams to check the validity of each argument. a. Some men are teachers. Sam Jones is a teacher. Therefore, Sam Jones is a man. b. Gold is heavy. Nothing but gold will satisfy Amy. Hence nothing that is light will satisfy Amy. c. No cats are dogs. All poodles are dogs. So some poodles are not cats. d. Some cows eat hay. All horses eat hay. Only cows and horses eat hay. Frank eats hay, so Frank is a horse. e. All chimpanzees are monkeys. All monkeys are animals. Some animals have two legs. So some chimpanzees have two legs.
8. In a certain land, every resident either always lies or always tells the truth. You happen to run into two residents, Bob and Sam. Bob says, “If I am a truth teller, then Sam is a truth teller.” Is Bob a truth teller? What about Sam? 9. The binary connectiveris defined by the following truth table:
p
q
prq
T T F F
T F T F
F T T T
5. Complete the following truth table. p
q
p¿q
T
F
p¡q
T F
p:q
⬃p ⬃q
⬃q 4 p ⬃q : ⬃p
T F T
T F
Compose a statement using only the connectiverthat is logically equivalent to a. ⬃p b. p ¿ q c. p ¡ q
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CLOCK ARITHMETIC: A MATHEMATICAL SYSTEM
The mathematical systems that we studied in Chapters 1 through 8 consisted of the infinite sets of whole numbers, fractions, and integers, together with their usual operations and properties. However, there are also mathematical systems involving finite sets.
Clock Arithmetic The hours of a 12clock are represented by the finite set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. The problem “If it is 7 o’clock, what time will it be in 8 hours?” can be represented as the addition problem 7 䊝 8 (we use a circle around the “plus” sign here to distinguish this clock addition from the usual addition). Since 8 hours after 7 o’clock is 3 o’clock, we write 7 䊝 8 3. Notice that 7 䊝 8 can also be found simply by adding 7 and 8, then subtracting 12, the clock number, from the sum, 15. Instead of continuing to study 12clock arithmetic, we will simplify our discussion about clock arithmetic by considering the 5clock next (Figure T2.1). In the 5clock, the sum of two numbers is found by adding the two numbers as whole numbers, except that when this sum is greater than 5, 5 is subtracted. Thus, in the 5clock, 1 䊝 2 3, 3 䊝 4 2 (i.e., 3 4 5), and 3 䊝 3 1. Since 1 䊝 5 1, 2 䊝 5 2, 3 䊝 5 3, 4 䊝 5 4, and 5 䊝 5 5, the clock number 5 acts like the additive identity. For this reason, it is common to replace the clock number with a zero. Henceforth, 0 will be used to designate the clock number. Addition in the 5clock is summarized in the table in Figure T2.2. 5 4
1
3 Figure T2.1
2
0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3
Figure T2.2
It can be shown that 5clock addition is a commutative, associative, and closed binary operation. Also, 0 is the additive identity, since a 䊝 0 a for all numbers a in the 5clock. Finally, every 5clock number has an opposite or additive inverse: 1 䊝 4 0 (the identity), so 4 and 1 are opposites of each other; 2 䊝 3 0, so 2 and 3 are opposites of each other; and 0 䊝 0 0, so 0 is its own opposite. Subtraction in the 5clock can be defined in three equivalent ways. First, similar to the takeaway approach for whole numbers, a number can be subtracted by counting backward. For example, 2 䊞 4 3 on the 5clock, since counting backward 4 from 2 yields 1, 0, 4, 3. One can also use the missingaddend approach, namely 2 䊞 4 x if and only if 2 4 䊝 x. Since 4 䊝 3 2, it follows that x 3. Finally, 2 䊞 4 can be found by the addingtheopposite method; that is, 2 䊞 4 2 䊝 1 3, since 1 is the opposite of 4. 923
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Example T2.1 a. 6 䊝 8 (12clock) d. 8 䊞 2 (12clock)
Calculate in the indicated clock arithmetic. b. 4 䊝 4 (5clock) e. 1 䊞 4 (5clock)
c. 7 䊝 4 (9clock) f. 2 䊞 5 (7clock)
S O LU T I O N
a. In the 12clock, 6 䊝 8 6 8 12 2. b. In the 5clock, 4 䊝 4 4 4 5 3. c. In the 9clock, 7 䊝 4 7 4 9 2. d. In the 12clock, 8 䊞 2 6 since 8 2 䊝 6. e. In the 5clock, 1 䊞 4 1 1 2 by adding the opposite. f. In the 7clock, 2 䊞 5 2 2 4. 0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1
2 0 2 4 0 2 4
3 0 3 0 3 0 3
4 0 4 2 0 4 2
Figure T2.3
0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 1 2 3 4 5
Figure T2.4
5 0 5 4 3 2 1
■
Multiplication in clock arithmetic is viewed as repeated addition. In the 5clock, 3 䊟 4 4 䊝 4 䊝 4 2. The 5clock multiplication table is shown in Figure T2.3. As with addition, there is a shortcut for finding products. For example, to find 3 䊟 4 in the 5clock, first multiply 3 and 4 as whole numbers. This result, 12, exceeds 5, the number of the clock. In the 5clock, imagine counting 12 starting with 1, namely, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2. Here you must go around the circle twice (2 5 10) plus two more clock numbers. Thus 3 䊟 4 2. Also, notice that 2 is the remainder when 12 is divided by 5. In general, to multiply in any clock, first take the wholenumber product of the two clock numbers. If this product exceeds the clock number, divide by the clock number—the remainder will be the clock product. Thus 7 䊟 9 in the 12clock is 3, since 63 leaves a remainder of 3 when divided by 12. As with clock addition, clock multiplication is a commutative and associative closed binary operation. Also, 1 䊟 n n 䊟 1 n, for all n, so 1 is the multiplicative identity. Since 1 䊟 1 1, 2 䊟 3 1, and 4 䊟 4 1, every nonzero element of the 5clock has a reciprocal or multiplicative inverse. Notice that 0 䊟 n 0 for all n, since 0 is the additive identity (zero); this is consistent with all of our previous number systems, namely zero times any clock number is zero. Division in the 5clock can be viewed using either of the following two equivalent approaches: (1) missing factor or (2) multiplying by the reciprocal of the divisor. For example, using (1), 2 3 n if and only if 2 3 䊟 n. Since 3 䊟 4 2, it follows that n 4. Alternatively, using (2), 2 3 2 䊟 2 4, since 2 is the reciprocal of 3 in the 5clock. Although every nonzero number in the 5clock has a reciprocal, this property does not hold in every clock. For example, consider the multiplication table for the 6clock (Figure T2.4). Notice that the number 1 does not appear in the “2” row. This means that there is no number n in the 6clock such that 2 䊟 n 1. Also consider 2 in the 12clock and the various multiples of 2. Observe that they are always even; hence 1 is never a multiple of 2. Thus 2 has no reciprocal in the 12clock. This lack of reciprocals applies to every nclock, where n is a composite number. For example, in the 9clock, the number 3 (as well as 6) does not have a reciprocal. Thus, in composite number clocks, some divisions are impossible.
Example T2.2 a. 5 䊟 7 (12clock) c. 6 䊟 5 (8clock) e. 2 5 (7clock)
Calculate in the indicated clock arithmetic (if possible). b. 4 䊟 2 (5clock) d. 1 3 (5clock) f. 2 6 (12clock)
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925
S O LU T I O N
a. 5 䊟 7 11 in the 12clock, since 5 7 35 and 35 divided by 12 has a remainder of 11. b. 4 䊟 2 3 in the 5clock, since 4 2 8 and 8 divided by 5 has a remainder of 3. c. 6 䊟 5 6 in the 8clock, since 30 divided by 8 has a remainder of 6. d. 1 3 1 䊟 2 2 in the 5clock, since 2 is the reciprocal of 3. e. 2 5 2 䊟 3 6 in the 7clock, since 3 is the reciprocal of 5. f. 2 6 in the 12clock is not possible, because 6 has no reciprocal. ■ Other aspects of various clock arithmetics that can be studied, such as ordering, fractions, and equations, are covered in the Exercise/Problem Set.
Congruence Modulo m 5 1
1 4
2
1
3
Figure T2.5
2
2
Clock arithmetics are examples of finite mathematical systems. Interestingly, some of the ideas found in clock arithmetics can be extended to the (infinite) set of integers. In clock arithmetic, the clock number is the additive identity (or the zero). Thus a natural association of the integers with the 5clock, say, can be obtained by wrapping the integer number line around the 5clock, where 0 corresponds to 5 on the 5clock, 1 with 1 on the clock, 1 with 4 on the clock, and so on (Figure T2.5). In this way, there are infinitely many integers associated with each clock number. For example, in the 5clock in Figure T2.5, the set of integers associated with 1 is { .. . , 14, 9, 4, 1, 6, 11, . . .}. It is interesting to note that the difference of any two of the integers in this set is a multiple of 5. In general, this fact is expressed symbolically as follows:
DEFINITION Congruence Mod m Let a, b, and m be integers, m 2. Then a ⬅ b mod m if and only if m 兩 (a b). In this definition, we need to use an extended definition of divides to the system of integers. We say that a 兩 b, for integers a (⫽ 0) and b, if there is an integer x such that ax b. The expression a ⬅ b mod m is read a is congruent to b mod m. The term “mod m” is an abbreviation for “modulo m.” Using the definition, determine which are true. Justify your conclusion. b. 5 ⬅ 11 mod 6 c. 5 ⬅ 14 mod 6 d. 7 ⬅ 22 mod 5
Example T2.3 a. 13 ⬅ 7 mod 2 S O LU T I O N
a. 13 ⬅ 7 mod 2 is true, since 13 7 6 and 2 兩 6. b. 5 ⬅ 11 mod 6 is true, since 5 11 6 and 6 兩 6. c. 5 ⬅ 14 mod 6 is false, since 5 14 19 and 6 ⱈ 19. d. 7 ⬅ 22 mod 5 is true, since 7 (22) 15 and 5 兩 15.
■
If the “mod m” is omitted from the congruence relation a ⬅ b mod m, the resulting expression, a ⬅ b, looks much like the equation a b. In fact, congruences and equations have many similarities, as can be seen in the following seven results. (For simplicity, we will omit the “mod m” henceforth unless a particular m needs to be specified. As before, m 2.)
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Clock Arithmetic: A Mathematical System 1. a ⬅ a for all clock numbers a. This is true for any m, since a a 0 and m 兩 0. 2. If a ⬅ b, then b ⬅ a. This is true since if m 兩 (a b), then m 兩 (a b) or m 兩 (b a). 3. If a ⬅ b and b ⬅ c, then a ⬅ c. The justification of this is left for the Problem Set, Part A, 10. 4. If a ⬅ b, then a c ⬅ b c. If m 兩 (a b), then m 兩 (a b c c), or m 兩 [(a c) (b c)]; that is, a c ⬅ b c. 5. If a ⬅ b, then ac ⬅ bc. The justification of this is left for the Problem Set, Part B, 12. 6. If a ⬅ b and c ⬅ d, then ac ⬅ bd. Results 5 and 3 can be used to justify this as follows: If a ⬅ b, then ac ⬅ bc by result 5. Also, if c ⬅ d, then bc ⬅ bd by result 5. Since ac ⬅ bc and bc ⬅ bd, we have ac ⬅ bd by result 3. 7. If a ⬅ b and n is a whole number, then an ⬅ bn. This can be justified by using result 6 repeatedly. For example, since a ⬅ b and a ⬅ b (using a ⬅ b and c ⬅ d in result 6), we have aa ⬅ bb or a2 ⬅ b2. Continuing, we obtain a3 ⬅ b3, a4 ⬅ b4, and so on. Congruence mod m can be used to solve a variety of problems. We close this section with one such problem.
Example T2.4
What are the last two digits of 330?
S O LU T I O N The number 330 is a large number, and its standard form will not fit on
calculator displays. However, suppose that we could find a smaller number, say n, that did fit on a calculator display so that n and 330 have the same last two digits. If n and 330 have the same last two digits, then 330 n has zeros in its last two digits, and vice versa. Thus we have that 100 兩 (330 n), or 330 ⬅ n mod 100. We now proceed to find such an n. Since 330 can be written as (36)5, let’s first consider 36 729. Because the last two digits of 729 are 29, we can write 36 ⬅ 29 mod 100. Then, from result 7, (36)5 ⬅ 295 mod 100. Since 295 20,511,149 and 20,511,149 ⬅ 49 mod 100, by result 3 we can conclude that (36)5 ⬅ 49 mod 100. Thus, 330 ends in 49. ■
T2
EXERCISE / PROBLEM SET A
EXERCISES 1. Calculate in the clock arithmetics indicated. a. 8 䊝 11 (12clock) b. 1 䊞 5 (7clock) c. 3 䊟 4 (6clock) d. 3 2 (5clock) e. 7 䊝 6 (10clock) f. 5 䊞 7 (9clock) g. 4 䊟 7 (11clock) h. 2 9 (13clock) 2. Find the opposite and reciprocal (if it exists) for each of the following. a. 3 (7clock) b. 5 (12clock) c. 7 (8clock) d. 4 (8clock)
3. In clock arithmetics, an means a a . . . a (n factors of a). Calculate in the clocks indicated. a. 73 (8clock) b. 45 (5clock) c. 26 (7clock) d. 94 (12clock) 4. Determine whether these congruences are true or are false. a. 14 ⬅ 3 mod 3 b. 3 ⬅ 7 mod 4 c. 43 ⬅ 13 mod 14 d. 7 ⬅ 13 mod 2 e. 23 ⬅ 19 mod 7 f. 11 ⬅ 7 mod 8 5. Explain how to use the 5clock addition table to find 1 䊞 4 in the 5clock.
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6. Using the 5clock table, explain why 5clock addition is commutative. 7. In the 5clock, 13 is defined to be 1 3, which equals 1 䊟 2 2. Using this definition of a clock fraction, calculate
Clock Arithmetic: A Mathematical System
927
1 3
1 1 1 䊝 2. Then add 3 and 2 as you would fractions, except do it in 5clock arithmetic. Are your answers the same in both cases? Try adding, subtracting, multiplying, and dividing 34 and 23 in 5clock in this way.
PROBLEMS 8. Suppose that “less than” is defined in the 5clock as follows: a ⬍ b if and only if a 䊝 c b for some nonzero number c. Then 1 ⬍ 3 since 1 䊝 2 3. However, 3 ⬍ 1 also since 3 䊝 3 1. Thus, although this definition is consistent with our usual definition, it produces a result very different from what happens in the system of whole numbers. For each of the following, find an example that is inconsistent with what one would expect to find for whole numbers. a. If a ⬍ b, then a 䊝 c ⬍ b 䊝 c. b. If a ⬍ b and c ⫽ 0, then a 䊟 c ⬍ b 䊟 c.
T2
9. Find all possible replacements for x to make the following true. a. 3 䊟 x 2 in the 7clock b. 2 䊟 x 0 in the 12clock c. 5 䊟 x 0 in the 10clock d. 4 䊟 x 5 in the 8clock 10. Prove: If a c ⬅ b c, then a ⬅ b. 11. Prove: If a ⬅ b and b ⬅ c, then a ⬅ c. 12. Find the last two digits of 348 and 349. 13. Find the last three digits of 4101.
EXERCISE / PROBLEM SET B
EXERCISES 1. Calculate in the clock arithmetics indicated. a. 3 䊟 (4 䊝 5) and (3 䊟 4) 䊝 (3 䊟 5) in the 7clock b. 2 䊟 (3 䊝 6) and (2 䊟 3) 䊝 (2 䊟 6) in the 12clock c. 5 䊟 (7 䊞 3) and (5 䊟 7) 䊞 (5 䊟 3) in the 9clock d. 4 䊟 (3 䊞 5) and (4 䊟 3) 䊞 (4 䊟 5) in the 6clock e. What do parts (a) to (d) suggest? 2. Calculate as indicated [i.e., in 24 䊟 34, calculate 24, then 34, then multiply your results, and in (2 䊟 3)4, calculate 2 䊟 3, then find the fourth power of your product]. a. 32 䊟 52 and (3 䊟 5)2 in the 7clock b. 23 䊟 33 and (2 䊟 3)3 in the 6clock c. 54 䊟 64 and (5 䊟 6)4 in the 10clock d. What do parts (a) to (c) suggest? 3. Make a 7clock multiplication table and use it to find the reciprocals of 1, 2, 3, 4, 5, and 6.
4. Find the following in the 6clock. a. 2 b. 5 c. (2) 䊟 (5) d. 2 䊟 5 e. 3 f. 4 g. (3) 䊟 (4) h. 3 䊟 4 What general result similar to one in the integers is suggested by parts (c), (d), (g), and (h)? 5. In each part, describe all integers n , where 20 n 20, which make these congruences true. a. n ⬅ 3 mod 5 b. 4 ⬅ n mod 7 c. 12 ⬅ 4 mod n d. 7 ⬅ 7 mod n 6. Show, by using an example in the 12clock, that the product of two nonzero numbers may be zero. 7. List all of the numbers that do not have reciprocals in the clock given. a. 8clock b. 10clock c. 12clock d. Based on your findings, predict the numbers in the 36clock that don’t have reciprocals. Check your prediction.
PROBLEMS 8. Find reciprocals of the following: a. 7 in the 8clock b. 4 in the 5clock c. 11 in the 12clock d. 9 in the 10clock e. What general idea is suggested by parts (a) to (d)? 9. State a definition of “square root” for clock arithmetic that is consistent with our usual definition. Then find all square roots of the following if they exist.
a. 4 in the 5clock b. 1 in the 8clock c. 3 in the 6clock d. 7 in the 12clock e. What do you notice that is different or similar about square roots in clock arithmetics? 10. The system of rational numbers was divided into the three disjoint sets: (i) negatives, (ii) zero, and (iii) positives. The set of positives was closed under both addition and
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multiplication. Show that it is impossible to find two disjoint nonempty sets to serve as positives and negatives in the 5clock. (Hint: Let 1 be positive and another number be negative, say 1 ⬅ 4. Then show that if the set of positive 5clock numbers is closed under addition, this situation is impossible. Observe that this holds in all clock arithmetics.) 11. Explain why multiplication is closed in any clock. 12. Prove: If a ⬅ b, then ac ⬅ bc.
14. Suppose that you want to know the remainder when 3100 is divided by 7. If r is the remainder and q is the quotient, then 3100 7q r or 3100 r 7q. Thus 7 兩 (3100 r) or 3100 ⬅ r mod 7. (Recall that for the remainder r, we have 0 r ⬍ 7.) Now 35 243 ⬅ 5 mod 7. So (35)2 ⬅ 52 mod 7 or 310 ⬅ 25 ⬅ 4 mod 7. Thus 3100 (310)10 ⬅ 410 mod 7. But 410 165 ⬅ 25 ⬅ 4 mod 7. Thus the remainder is 4. Find the remainder when 7101 is divided by 8. (Hint: 7101 7100 䡠 7.)
13. Prove or disprove: If ac ⬅ bc mod 6 and c ⫽ 0 mod 6, then a ⬅ b mod 6.
TOPIC REVIEW VOCABULARY/NOTATION Clock arithmetic 923 Addition, a 䊝 b 923 Subtraction, a 䊞 b 923
Multiplication, a 䊟 b 924 Division, a b 924
Congruence modulo, m, a ⬅ b, mod m 925
CLOCK ARITHMETIC TEST KNOWLEDGE 1. True or false? a. The 12clock is comprised of the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. b. Addition in the 5clock is associative. c. The number 1 is the additive identity in the 7clock. d. The number 4 is its own multiplicative inverse in the 5clock. e. Every clock has an additive inverse for each of its elements. f. Every number is congruent to itself mod m. g. If a ⬅ b mod 7, then either a b 7 or b a 7. h. If a ⬅ b and c ⬅ b, then a ⬅ c.
SKILL 2. Calculate. a. 5 䊝 9 in the 11clock b. 8 䊟 8 in the 9clock c. 3 䊞 7 in the 10clock d. 4 9 in the 13clock e. 24 in the 5clock f. (2 䊞 5)3 䊟 6 in the 7clock 3. Show how to do the following calculations easily mentally by applying the commutative, associative, identity, inverse, or distributive properties. a. 3 䊝 (9 䊝 7) in the 10clock b. (8 䊟 3) 䊟 4 in the 11clock
c. (5 䊟 4) 䊝 (5 䊟 11) in the 15clock d. (6 䊟 3) 䊝 (3 䊟 4) 䊝 (3 䊟 3) in the 13clock 4. Find the opposite and the reciprocal (if they exist) of the following numbers in the indicated clocks. Explain. a. 4 in the 7clock c. 0 in the 5clock
b. 4 in the 8clock d. 5 in the 12clock
5. In each part, describe the set of all numbers n that make the congruence true. a. n ⬅ 4 mod 9 where 15 n 15 b. 15 ⬅ 3 mod n where 1 ⬍ n ⬍ 20 c. 8 ⬅ n mod 7
UNDERSTANDING 6. Using a clock as a model, explain why 4 does not have a reciprocal in the 12clock. 7. Explain (a) why 0 cannot be used for m and (b) why 1 is not used for m in the definition of a ⬅ b mod m. 8. Explain why a ⬅ a mod m.
PROBLEM SOLVING/APPLICATION 9. If January 1 of a nonleap year falls on a Monday, show how congruence mod 7 can be used to determine the day of the week for January 1 of the next year.
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EXERCISE/PROBLEM SETS—PART A, CHAPTER REVIEWS, CHAPTER TESTS, AND TOPICS SECTION Section 1.1A 1. a. 8 b. 1. Understand: Triangles have 3 sides and are formed by the sides and parts of the diagonals. 2. Devise a plan: Sketch various triangles. 3. Carry out the plan: Make sure that all possible triangles are sketched. 4. Look back: Check by sketching the triangles on another piece of paper. Compare with your original sketches. Do you have the same triangles in both cases?
c. 2: None 3: None 4: 37, 38, 39, 40 5: None 6: None 12. Row 1: 9, 3, 4; row 2: 8, 2, 5; row 3: 7, 6, 1 13. Run three consecutive 5minute timers concurrently with two consecutive 8minute timers. There will be 1 minute left on the 8minute timer. 14. a. 1839
b. 47
2. 52
15. For example, row 1: 3, 5; row 2: 7, 1, 8, 2; row 3: 4, 6
3. 36 ft 78 ft
16. U 9, S 3, R 8, A 2, P 1, E 0, C 7
4. 88
17. 3
5. 9 4 5. If n is odd, then both n 1 and n 1 are even and n  1 n + 1 n + . 2 2 6. 6 6 6 6 13
18. Top pair: 8, 5; second pair from the top: 4, 1; third pair from the top: 6, 3; bottom pair: 2, 7. There are other correct answers. 19.
4(n + 10) + 200
20. For example, 98 7 6 5 4 3 2 1 100, and 9 8 76 5 4 3 21 100.
27 41
21. 1 and 12, 9 and 10
49
22.
14
36
n (n 10) 50 n n 60 n 60
4
7.
7
22
2
3
1
5
6
4
8
8.
9
23.
4 14
6
3
13
1 9. No; 0 1 2 3 4 5 6 7 8 9 45, which is too large.
12
10. For example, beginning at corner: 8, 2, 6, 7, 3, 4, 9, 5, 1 11. a. 2: None 3: 27, 28, 29 4: None 5: None 6: None b. 2: 106, 107 3: 70, 71, 72 4: None 5: None 6: 33, 34, 35, 36, 37, 38
5 8 11
9 2
7
10
24. 22 25. Advantages: reemphasizes that a solution is the value of x that makes an equation true. Disadvantages: slow, inefficient, tedious.
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26. By working with a number rather than a variable, students can get a feeling for what operations are needed. Also, if they guess a number that yields something close to the desired answer, it can serve as an estimate to check their final answer. 27. No. Many students likely profit from drawing pictures to solve problems. However, students need not be forced to draw pictures.
Section 1.2A 1. a. 9, 16, 25 The Answer column is comprised of perfect squares. b. 9 c. 13 d. 23 [1 3 5 䡠 䡠 䡠 (2n 1) n2] 1 2. a. 32 b. c. 21 d. 1287 27 3. a.
D
D
D
D
D
19 # 20 190 2 b. 43, 57, 73
8. 19 18 䡠 䡠 䡠 2 1 9. a. 46, 69, 99
10. a. 3 6 9 b. 10 15 25 c. 45 55 100, 190 210 400, (n 1)st nth d. 36 is the 8th triangular number and is the 6th square number. e. Some possible answers: 4 1 3, 49 4 45, 64 49 15, 64 36 28, 64 9 55, 169 64 105. 11. $10,737,418.23 if paid the second way 12. For n triangles, perimeter n 2 NUMBER OF TRIANGLES
1
2
3
4
5
6
10
38
n
PERIMETER
3
4
5
6
7
8
12
40
n2
13. a. In the 4, 6, 12, 14, . . . column c. In the 3, 7, 11, 15, . . . column
b. In the 2, 8, 10, 16, . . . column d. In the 5, 13, . . . column
14. 1003 1,000,000 b.
15. 34, 55, 89, 144, 233, 377; 144 233 33,552 16. 987 17. a. Sums are 1, 1, 2, 3, 5, 8, 13. Each sum is a Fibonacci number. b. Next three sums: 21, 34, 55
4. a. 55562 44452 11,111,111 b. 55,555,5562 44,444,4452 1,111,111,111,111,111.
18. a. Sum is 20. b. Sum of numbers inside the circle is always twice the number directly below the center of the circle.
5.
19. a. Step 5
60 6 10 30 6
5
2
2
6. a.
1
TRIANGLE NUMBER
NUMBER OF DOTS IN SHAPE
1
1
2
3
3
6
4
10
5
15
6
21
Step 6
The numbers in the ‘Number of Dots’ column satisfy the formula n(n + 1) where n is the number in the first column. 2 b.
b.
STEP 1
c. 55 dots d. Yes, the 13th number. e. No, the 16th number has 136 and the 17th number has 153. 100(101) n(n + 1) f. g. 5050 2 2 7. 3 weighings
c. 85
NUMBER OF SQUARES 1
2
5
3
13
4
25
5
41
6
61
d. 10th, 181; 20th, 761; 50th, 4901
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20. 23
11. 30
21. a. 18
b. 66
12. Head and tail are 4 inches long and the body is 22 inches long.
22. a. Lt Blue, Lt Blue, Blue, Lt Blue b. Green, Blue, Green, Yellow c. Red, Blue, Blue, Purple
13. 256 14.
3 5 7 23. There are many answers possible. For example 1, , , , . . . . 2 4 8 The numerators are odd numbers—they increase by twos. The denominators of the fractions are powers of 2: 20 1, 21 2, 22 4, etc.
1. Try Guess and Test
4 6
7
3
16. 2 nickels & 10 dimes; 5 nickels, 6 dimes, & 1 quarter; 8 nickels, 2 dimes, & 2 quarters.
1 9
2 8
15. X X X X X X
X X X X X X X X X X X X
CHAPTER REVIEW Section 1.1
17. Let x, x 1, and x 2 be any three consecutive numbers. Their sum is (x) (x 1) (x 2) 3x 3 3(x 1).
5
18. 4 different triangles. 2, 6, 6; 3, 5, 6; 4, 5, 5; 4, 4, 6
2. Use a Variable or Guess and Test
19. Baseball is 0.5 pounds, football is 0.75 pounds, soccer ball is 0.85 pounds.
4 60
36 9
135
15
3. Draw a Picture—289 tiles
Section 1.2 1. Look for a Pattern 1234321 (1 2 3 4 3 2 1) 44442 2. Solve a Simpler Problem a. 9 b. 1296 3. Make a List a. 3: 1, 3, 9 b. 4: 1, 3, 9, 27 c. 364 grams
Chapter 1 Test 1. Understand the problem; devise a plan; carry out the plan; look back 2. Guess and Test; Use a Variable; Look for a Pattern; Make a List; Solve a Simpler Problem; Draw a Picture 3. $5 allowance. 4. Answers may vary. 5. Amanda had 31 hardboiled eggs to sell. 6. Exercises are routine applications of known procedures, whereas problems require the solver to take an original mental step. 7. Any of the 6 clues under Guess and Test. 8. Any of the 8 clues under Use a Variable. 9. 4 4: 1, 2, 3, 4; 3, 4, 1, 2; 4, 3, 2, 1; 2, 1, 4, 3. The 2 2 is impossible. 3 3: 1, 3, 2; 3, 2, 1; 2, 1, 3. 10. 6
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20. By using a simpler problem and looking for a pattern. 1 table ⇒ 3 chairs 2 tables ⇒ 4 chairs 3 tables ⇒ 5 chairs 31 tables ⇒ 31 2 chairs 21. a. 39¢, 55¢, 74¢
b. 86¢, 131¢
c. 44¢, 64¢, 88¢
Section 2.1A 1. a. {6, 7, 8} 2. a. No d. No g. No 3. a. T d. F 4. x
y z
b. {2, 4, 6, 8, 10, 12, 14} b. Yes e. No h. Yes
b. T e. F
a b c
c. {2, 4, 6, . . . , 150}
c. No f. No c. T
x y z
a b c
x y z
a b c
a b c
x y z
5. b, c, and e 6. a. Yes
b. No
c. No
d. Yes
7. ⭋, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} 8. ⭋, {}, {嘷} 9. a. 僆 10. a. 2
b. 傺, 債 b. 99
c. 傺, 債 c. 201
d. 菹, ⬃ d. infinite set
e. 僆
f. 債, , ⬃ e. infinite set
11. {6, 9, 12, 15, 18, . . .} where 6 4 3, 9 4 6, 12 4 9, a 4 a 3. {9, 12, 15, 18, 21, . . .} where 9 4 3, 12 4 6, 15 4 9, a 4 a 6. There are other correct answers 12. a. {2, 4, 6, 8, 10, . . .} A b. {4, 8, 12, 16, 20, . . .} B c. Yes. Since all of the numbers in set B are even, they are also in set A. d. Yes. x 僆 A maps to 2x 僆 B so 2 4 4, 4 4 8, 6 4 12, etc. e. No. 2 僆 A but 2 僆 B. f. Yes. B 債 A and 2 僆 A but 2 僆 B. 13. a. F
b. F
c. T
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14. a.
A
19. a.
b.
c.
d.
e.
f.
B
C b.
A
B
C c. Parentheses placement is important. 15. a.
c.
b.
U S
T
S
T
U T
S
20. a. Women or Americans who have won Nobel Prizes b. Nobel Prize winners who are American women c. American winners of the Nobel Prize in chemistry
U
21. a. {b, c} 16. a.
A
B
c. {a}
23. a. {2, 6, 10, 14, . . .} b. ⭋ c. A B ⭋ if A 債 B
C b.
b. {b, c, e}
22. a. {嘷, /} b. {0, 1, 2, . . . , 11} c. All single people d. {a, b, c, d}
A
24. a. {0, 1, 2, 3, 4, 5, 6, 8, 10} c. {0, 2, 4} e. {2, 6, 10} g. ⭋
B
b. {0, 2, 4, 6, 8, 10} d. {0, 4, 8} f. {1, 2, 3, 5, 6, 10}
25. a. {January, June, July, August} b. {January} c. ⭋ d. {January, June, July} e. {March, April, May, September, October, November} f. {January}
C c.
A
B
26. a. Yes b. {3, 6, 9, 12, 15, 18, 21, 24, . . .} c. {6, 12, 18, 24, . . .} d. A 艛 B A, A 艚 B B 27. a. A 艛 B A 傽 B {6}. Yes, the sets are the same. b. A 艛 B and A 傽 B are both represented by
C
U A
17. a. (B 艚 C) A b. C (A 艛 B) c. [A 艛 (B 艚 C)] (A 艚 B 艚 C). Note: There are many other correct answers. 18. a.
b.
B A
A
B
B
Yes, the diagrams are the same. 28. a. {(a, b), (a, c)} b. {(5, a), (5, b), (5, c)} c. {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)} d. {(2, 1), (2, 4), (3, 1), (3, 4)} e. {(a, 5), (b, 5), (c, 5)} f. {(1, a), (2, a), (3, a), (1, b), (2, b), (3, b)}
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29. a. 8
b. 12
30. a. 2 d. 25
b. 4 e. 0
b. F
b.
c.
d.
c. Not possible f. Not possible
31. a. A {a}, B {2, 4, 6} b. A B {a, b} 32. a. T
9. a.
A5
c. F
d. F
33. a. Three elements; eight elements b. y elements; x y elements
10. a.
b.
c.
d.
34. 24 35. a. 1 b. 2 c. 4 d. 8 f. 2 䡠 2 䡠 2 䡠 䡠 䡠 2 (2 appears n times) 36. a. Possible 37. a. When D 債 E
e. 32
b. Not possible b. When E 債 D
c. When E D
38. The Cartesian product of the set of skirts with the set of blouses will determine how many outfits can be formed—56 in this case. 39. 31 matches 40. Yes. Use lines perpendicular to the base. 41. Yes. Use lines perpendicular to the chord. 42. a. 7
b. 19
c. 49
43. Twentyfive were butchers, bakers, and candlestick makers. 44. Analogies between set operations and arithmetic operations are not direct. One could say that A B means you take all the elements of B away from A. However, A B is a set of ordered pairs such that the first elements come from A and the second elements come from B. So all the elements of A are “paired” with all the elements of B, but not multiplied.
Section 2.2A
11. a. 12 e. 976 i. 3010
b. 4270 f. 3245 j. 14
c. 3614 g. 42 k. 52
12. a.
d. 1991 h. 404 l. 333
b. CXXXI
c.
13. a. Egyptian b. Mayan c. No. For example, to represent 10 requires two symbols in the Mayan system, but only one in either the Egyptian or Babylonian systems. 14. 1967 15. IV and VI, IX and XI, and so on; the Egyptian system was not positional, so there should not be a problem with reversals.
1. 3147819804, identification: 13905, identification; 6, cardinal; 7814, identification; 28, ordinal; $10, cardinal; 20, ordinal
16. a. (i) 30, (ii) 24, (iii) 47, (iv) 57 b. Add the digits of the addends (or substrahend and minuend).
2. a. Attribute common to all sets that match or are equivalent to the set {a, b, c, d, e, f, g} b. Attribute common to all sets that match or are equivalent to {a} c. Impossible d. How many elements are in the empty set?
17. MCMXCIX. It was introduced in the fall of 1998.
3. Put in onetoone correspondence with the set {1, 2, 3, 4, 5, 6}. 4. 8; 5 5. A set containing 3 elements, such as {a, b, c}, can be matched with a proper subset of a set with 7 elements, such as {t, u, v, w, x, y, z}. 6. a. ⬍ b. ⬎ c. ⬎ All solutions can be found by plotting the numbers on a number line. 7. a.
18. 18 pages 19. 1993 (1 2 3 4 䡠 䡠 䡠 1994) 20. Do a threecoin version first. For five coins, start by comparing two coins. If they balance, use a threecoin test on the remaining coins. If they do not balance, add in one of the good coins and use a threecoin test. 21. a. (i) 42, (ii) 625, (iii) 3533, (iv) 89,801 b. (i) , (ii) , (iii) ⬘ , (iv) ⬘⬘ c. No. 22. A numeral zero is necessary in our system. For example, if one writes 11 (with a space between the two 1s, do we mean eleven or 101? For a similar reason, we need to use the symbol zero in the Mayan system.
SECTION 2.3A c.
1. a. 7(10) 0(1) b. 3(100) 0(10) 0(1) c. 9(100) 8(10) 4(1) d. 6(107) 6(103) 6(10)
d.
2. a. 1207
b.
8. a. LXXVI b. XLIX c. CXCII d. MDCCXLI
3. a. 100 4. a. 12 d. septillion g. 30
b. 500,300 b. 1
c. 8,070,605
d. 2,000,033,040
c. 1000 b. 4 e. 24 h. decillion
c. quintillion f. 9
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5. a. Two billion b. Eightyseven trillion c. Fiftytwo trillion six hundred seventytwo billion four hundred five million one hundred twentythree thousand one hundred thirtynine
c.
6. a. 7,603,059 b. 206,000,453,000 7. Any three of the following five attributes: digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; grouping by tens; place value; additive; multiplicative 8. 3223four 9. a.
82
81
1
12. a. 222five b. 333five; 32143five
b.
13. 23 14. Improper digit symbols for the given bases—can’t have an 8 in base eight or a 4 in base three
c.
15. a. T
b. F
c. F
16. a. In base five, 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44, 100 b. In base two, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000 c. In base three, 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000 d. 255, 300, 301, 302 (in base six) e. 310four 17. a. 15seven 1(7) 5(1) b. 123seven 1(72) 2(7) 3(1) c. 5046seven 5(73) 0(72) 4(7) 6(1)
10. 2, 1, 2, 0 11. a.
10
18. a. 333four b. 1000four, 1001four, 1002four, 1003four, 1010four
1
19. 2400 6666seven 2402 10001seven 10001seven has more digits because it is greater than 74.
b.
20. a. 613eight
b. 23230four
21. a. 194 c. 723 e. 1451 22. a. 531
52
51
1
b. 328 d. 129 f. 20,590 b. 7211
23. a. 177sixteen b. B7Dsixteen c. 2530sixteen d. 5ED2sixteen 24. a. 202six; 62twelve b. 332six; T8twelve c. 550six; 156twelve d. 15142six; 14E2twelve
c. 110110two
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25. a. five
b. nine
26. a. Seven c. Twelve
c. eleven
10. a. (0, 0), (2, 10), (4, 116). Range: {0, 10, 116} b. (1, 3), (2, 4), (9, 11). Range: {3, 4, 11}
b. Fortyseven d. x 7, x 3y 5
9 64 9 64 c. (1, 2), ¢ 2, ≤ , ¢ 3, ≤ . Range: b 2, , r 4 27 4 27
27. a. It must be 0, 2, 4, 6, or 8. b. It must be 0 or 2. c. It must be a 0. d. It may be any digit in base 5.
11. a. 100
29. 57 30. a. If final answer is abcdef, then the first two digits (ab) give the month of birth, the second two (cd ) give the date of birth, and the last two (ef ) give the year of birth. b. If birthday is ab/cd/ef, then the result will be 100,000a 10,000b 1,000c 100d 10e f. For example, begin by calculating 4(10a b) 13. 31. Studying other bases provides insight into grouping. 32. 159. There are 60 seconds in a minute, so one less than 60 is 59.
1. a. {(a, a), (a, b), (b, c), (c, b)} b. {(1, x), (2, y), (3, y), (4, z)} c. {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (2, 3), (3, 2), (2, 5), (5, 2), (3, 5), (5, 3)} d. {(2, 2), (4, 4), (6, 6), (8, 8), (10, 10), (12, 12), (4, 2), (6, 2), (8, 2), (10, 2), (12, 2), (8, 4), (12, 4), (12, 6)} b.
1 3 5 7
Aaron, 23 Sam, 17 Judy, 29 Amy, 14 Sarah, 30
b. S
5. a. Reflexive only c. Equivalence relation
d. 2
c. T b. None d. None
6. a. Transitive only b. Equivalence relation. Partition: Each set in the partition will have numbers with the same number of factors. c. Equivalence relation. Partition: Each set in the partition will have numbers with the same tens digits. 7. a. Not a function, since b is paired with two different numbers b. Function c. Function d. Not a function, since 3 is paired with two different numbers 8. a. Function b. Function c. Not a function, since some college graduates have more than one degree d. Function 9. a. Yes b. Yes c. No, since the number 1 is paired with two different numbers. d. Yes e. No, since the letter b is paired with both b and c (and the letter d is paired with both e and f ).
x
f (x)
0
0
1
0
4
60
x
f (x)
1
1
4
2
9
3
c. f (x) 2x for x 僆 {1, 2, 10}, {(1, 2), (2, 4), (10, 20)}
20 25 27 11
3. a. “was president number” b. “is the capital of ” 4. a. R, T
c. 3
1: 1 b. f (x) 2x for x 僆 {1, 4, 9}, 4 : 2. 9: 3
Section 2.4A
0 2 4
b. 81
0: 0 12. a. {(0, 0), (1, 0), (4, 60)}, 1 : 0 4 : 60
28. 1024 pages
2. a.
A7
x
f (x) 1
2
2
4
10
20
5 : 55 d. f (x) 11x for x 僆 {5, 6, 7}, {(5, 55), (6, 66), (7, 77)}, 6 : 66. 7 : 77 13. a. 29, 18, 17
b. 7, 10, 4
7 6 5 c. , , 9 9 6
d. 5, 3, 4
14. a. $240, $89.75, $2240.50, $100 b. $4321.30, $181.34 c. $22,780 15. a. 0.44
b. 0.56
c. 0.67
d. 4.61
16. Fraction equality is an equivalence relation. The equivalence class 5 , . . .6. containing 12 is 512 , 24 , 36 , 48 , 10 17. a. 32; 212; 122; 40 b. 0; 100; 40; 40 c. Yes; 40 18. 3677 19. a. C(x) 85 35x b. C(18) 715. The total amount spent by a member after 18 months is $715. c. After 27 months 20. a. r
1 2
b. 2400, 1200, 600, 300, 150, 75
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24. a.
n
T (n)
1
4
n NUMBER OF YEARS
ANNUAL INTEREST EARNED
2
12
0
0
100
3
20
1
5
105
4
28
2
5.25
110.25
5
36
3
5.51
115.76
6
44
4
5.79
121.55
7
52
5
6.08
127.63
8
60
6
6.38
134.01
7
6.70
140.71
8
7.04
147.75
9
7.39
155.13
10
7.76
162.89
b. Arithmetic sequence with a 4 and d 8 c. T(n) 4 (n 1)8 or T (n) 8n 4 d. T(20) 156; T(150) 1196 e. Domain: {1, 2, 3, 4, . . .} Range: {4, 12, 20, 28, . . .}
VALUE OF ACCOUNT
b. Geometric sequence with a 100 and r 1.05. A(n) 100(1.05)n c. $12.89 25. h(1) 48, h(2) 64, h(3) 48; 4 seconds
22. a.
n
T (n)
1
3
26. Convert the number to base two. Since the largest possible telephone number, 9999999, is between 223 8388608 and 224 16777216, the base two numeral will have at most 24 digits. Ask, in order, whether each digit is 1. This process takes 24 questions. Then convert back to base ten.
2
9
3
18
4
30
5
45
27. a. Arithmetic, 5, 1002 c. Arithmetic, 10, 1994
6
63
28. 228
7
81
29. Answers will vary.
8
108
b. Geometric, 2, 14 2199 d. Neither
30. No. Consecutive terms vary by a common multiple, not a common difference, in a geometric sequence.
b. Neither c. T(n)
3n(n + 1)
PROBLEMS WHERE THE STRATEGY “DRAW A DIAGRAM” IS USEFUL
2 d. T(15) 360. T(100) 15,150 e. Domain: {1, 2, 3, 4, . . .} Range: {3, 9, 18, 30, . . .}
1. Draw a tree diagram. There are 3 䡠 2 䡠 2 12 combinations. 2. Draw a diagram tracing out the taxi’s path: 5 blocks north and 2 blocks east. 3. Draw a diagram showing the four cities: 15,000 arrive at Canton.
23. a.
n NUMBER OF YEARS
ANNUAL INTEREST EARNED
VALUE OF ACCOUNT
0
0
100
1. Verbal description, listing, setbuilder notation
1
5
105
2
5
110
2. a. T g. F
3
5
115
3. {1, 2, 3, 4, 5, 6, 7}
4
5
120
5
5
125
6
5
130
7
5
135
8
5
140
1. No—it should be house numeral.
9
5
145
10
5
150
2. a. How much money is in your bank account? b. Which place did you finish in the relay race? c. What is your telephone number?
b. Arithmetic sequence with a 100 and d 5. A(n) 100 5n
CHAPTER REVIEW Section 2.1 b. T h. F
c. T i. F
d. T j. F
e. T k. T
f. T l. F
4. An infinite set can be matched with a proper subset of itself. 5. 7
Section 2.2
3. a. T 4. a. 111
b. T
c. F
b. 114
d. T c. 168
e. T
f. T
g. F
h. F
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5. a.
b. XXXVII
c.
6. IV ⫽ VI shows that the Roman system is positional. However, this system does not have place value. Every place value system is positional.
Section 2.3 1. a. Digits tell how many of each place value is required. b. Grouping by ten establishes the place values. c. Place values allow for large numbers with few numerals. d. Digits are multiplied by the place values and then all resulting values are added. 2. The names of 11 and 12 are unique; the names of 13–19 read the ones digits first, then say “teen.” The numerals 21–29 are read from left to right, where twenty means 2 tens and the second digit is the number of ones. 3. a. F
b. T
c. F
d. T
A9
Chapter 2 Test 1. a. F g. T
b. T h. F
c. T i. F
d. T j. F
e. F k. F
f. F l. T
2. 19 3. The intersection is an empty set. 4. Arrow Diagrams, Tables, Machines, Ordered Pairs, Graphs, Formulas, Geometric Transformations (any 6 is sufficient) 5. a. {a, b, c, d, e} b. { } or ⭋ c. {b, c} d. {(a, e), (a, f ), (a, g), (b, e), (b, f ), (b, g), (c, e), (c, f ), (c, g)} e. {d} f. {e} 6. a. 32
b. 944
c. 381
d. 106
e. 21
f. 142
7. a. 7 ⫻ 100 ⫹ 5 ⫻ 10 ⫹ 9 b. 7 ⫻ 1000 ⫹ 2 c. 1 ⫻ 26 ⫹ 1 ⫻ 23 ⫹ 1 8.
4. It gives you insight into base 10.
Section 2.4 1. a. Yes. b. Neither symmetric nor transitive c. Not transitive 2. a. {(1, 10), (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1)}, symmetric b. {(12, 8), (11, 7), (10, 6), (9, 5), (8, 4), (7, 3), (6, 2), (5, 1)} c. {(1, 12), (12, 1), (2, 6), (6, 2), (3, 4), (4, 3)}, symmetric d. {(12, 6), (10, 5), (8, 4), (6, 3), (4, 2), (2, 1)}
9.
3. a. (i) 1, 7, 13, 19, 25; (ii) 2, 8, 32, 128, 512 b. (i) 6; (ii) 4 4. a. T
b. F
c. F
d. F
5. a. Dawn : Jones, Jose : Ortiz, Amad : Rasheed b. FIRST NAME
SURNAME
Dawn
Jones
Jose
Ortiz
Amad
Rasheed
c. (Dawn, Jones), (Jose, Ortiz), (Amad, Rasheed)
10. 4034five 11. [A ⫺ (B 艛 C)] 艛 (B 艚 C) There are other correct answers. 12. a. {(a, c), (b, c), (c, a), (c, d), (d, a), (d, e), (e, b)} b. {(a, a), (a, b), (b, a), (c, e), (d, d), (e, c), (e, e)}, symmetric c. {(1, 2), (2, 1), (⫺2, 3), (⫺1, ⫺1), (⫺3, ⫺2), (⫺2, ⫺3), (3, ⫺2)}, symmetric
6.
13. IV ⫽ VI; thus position is important, but there is no place value as in 31 ⫽ 13, where the first 3 means “3 tens” and the second three means “3 ones.”
Range ⫽ {3, 4, 5, 7} 7. For example, the area of a circle with radius r is r2, the circumference of a circle with radius r is 2r, and the volume of a cube having side length s is s3.
14. a. True for all A and B b. True for all A and B c. True whenever A ⫽ B d. True whenever A ⫽ B or where A or B is empty 15. (b, a), (b, c), (d, a), (d, c) 16. 2, 6, 10, 14, . . . is an arithmetic sequence, and 2, 6, 18, 54, . . . is geometric. 17. Zero, based on groups of 20, 18 䡠 20, etc.
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Section 3.1A
18.
A 4 0 3 5
U
2
10
B
1. a.
4
3
7 6 11 14 12 20 16 9 8 19 18 17 C
13 15
19. No. There are 26 letters and 40 numbers. 20. Yes; No, 3 has no image; No, two arrows from 2.
4⫹3⫽7
21.
33 32 31 30
22. a. {(1, 1), (2, 2), (3, 3), (4, 4)} b. None c. {(1, 1), (2, 2), (3, 3), (4, 4), (1, 4), (4, 1), (3, 2), (2, 3)} d. Same as part (c)
b.
0
1
2
3
4
5
6
7
8
23. 97 24. a 6, b 8
2. Only a. and b. are true. c. is false because D 艚 E ⫽ ⭋.
25. 3 b. C(n) 4(n 1)2 3(2n 1)
26. a. $241
27. 187, 3 (n 1) 䡠 4 28. a.
1 2 3 4 5 6 b.
2 4 8 16 32 64
n
2n
1
2
2
4
3
8
4
16
5
32
6
64
3. a. Closed b. Closed c. Not closed, 1 2 3 d. Not closed, 1 2 3 e. Closed 4. a. Closure b. Commutativity c. Associativity d. Identity e. Commutativity f. Associativity and commutativity 5.
2 >
0
2
4
6
2 > 8
10
They differ in the order of the bars, but are the same because the total length in both cases is 9. 6. Associative property and commutative property for wholenumber addition 8. a.
70 60 50 40 30 20 10
>
7
7. a. 67, 107
c.
>
7
b. 51, 81
c. 20, 70
0
1
2
3
0
0
1
2
3
4
1
1
2
3
4
10
4
2
2
3
4
10
11
3
3
4
10
11
12
4
4
10
11
12
13
1 2 3 4 5 6 d. f(n) 2n 29. f (x) 0.75(220 x)
b. (i) 4five ? 13five, ? 4five (iii) 4five ? 12five, ? 3five
(ii) 3five ? 11five, ? 3five (iv) 2five ? 10five, ? 3five
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9. a. 4five 3five 12five, 12five 4five 3five, 12five 3five 4five b. 1five 4five 10five, 10five 1five 4five, 10five 4five 1five c. 2five 4five 11five, 4five 2five 11five, 11five 2five 4five 10. 11 3 8 and 11 8 3. Measurement TakeAway or Missing Addend 11. a. 7 2 5, set model, takeaway b. 7 3 4, measurement model, missing addend c. 6 4 2, set model, comparison approach
A11
19. a. (i) 363, (ii) 4884, (iii) 55 b. 69, 78, 79, or 89 20. 4, 8, 12, 16, 20 21. One arrangement has these sides: 9, 3, 4, 7; 7, 2, 6, 8; 8, 1, 5, 9. 22. Karen’s analogy is good. {0, 1) is not closed under addition. An operation usually “links” two different numbers that are “inside the room.” It is also allowed to link a number to itself, for example, 1 1 2, which is “outside the room.”
12. a. No; that is, 3 5 is not a whole number. b. No; that is, 3 5 ⫽ 5 3. c. No; that is, (6 3) 1 ⫽ 6 (3 1). d. No, 5 0 5 but 0 5 ⫽ 5.
23. 3 7 4. To find the answer 4 we had to go “outside the room,” as the student in the previous problem said. Even if only one answer takes us “outside the room,” that is enough to prove that the closure property does not hold for the whole numbers under subtraction by showing one counterexample.
13. a. Measurement, Takeaway model; no comparison since $120 has been removed from the original amount of $200:
Section 3.2A
200  120 = x
1. a. 2 4 2. a.
b. Set, missing addend, comparison, since two different sets of tomato plants are being compared:
a b c
24  18 = x c. Measurement, missingaddend model; no comparison since we need to know what additional amount will make savings equal $1795
b. 4 2
c. 3 7 b.
d e (a, d) (a, e) (b, d) (b, e) (c, d) (c, e)
3 2
c.
1240 + x = 1795 14. a. Kofi has 3 dollars and needs 8 dollars to go to the movie. How much more money does he need? b. Tabitha and Salina both had 8 yards of fabric and Tabitha used 3 yards to make a skirt. How much does she have left? c. How many?
8
d.
2
2
2
3 0
2
4
6
2 3 d.
3. a. Rectangular array approach, since there are rows and columns of tiles.
3 8 8
n = 15 # 12
012345678
b. Cartesian product approach, since we are looking at a set of 3 pairs of shorts, each of which is paired with one of a set of eight tee shirts.
15. The rest of the counting numbers 16. 123 45 67 89 100 17.
25
11
12
22
14
20
19
17
18
16
15
21
13
23
24
10
18. Sums of numbers on all six sides are the same as are sums on the “halfdiagonals,” lines from center to edge.
x = 3#8 c. Repeated addition, since the number of pencils could be found by adding 3 3 䡠 䡠 䡠 3 where the sum has 36 terms. p = 36 # 3 4. a. 36
b. 68
5. a. No. 2 4 8 c. No. 3 3 9 e. Yes f. Yes h. Yes i. Yes
c. 651
d. 858
b. Yes d. Yes g. Yes j. Yes
6. a. Commutative b. Commutative c. Distributive property of multiplication over addition d. Identity
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7. a. 4 䡠 60 4 䡠 37 b. 21 䡠 6 35 䡠 6 c. 37 䡠 60 37 䡠 22 d. (5 2)x e. (5 3)(a 1)
25. 31 33 35 37 39 41 216; 43 45 47 49 51 53 55 343; 57 59 61 63 65 67 69 71 512
8. a. 45(11) 45(10 1) c. 23(21) 23(20 1)
b. 39(102) 39(100 2) d. 97(101) 97(100 1)
9. a. (372 2) 12 4488 There are other possible ways. b. 374 (11 1) 4488 There are other possible ways. 10. a. 5(23 4) 23(5 4) 23 20 460, There are other possible ways. b. 12 25 3(4 25) 300, There are other possible ways.
26. 9, 5, 4, 6, 3, 2, 1, 7, 8 2(n + 10) + 100 27. n 60 2 28. 1st place: Pounce, Michelle 2nd place: Hippy, Kevin 2nd place: Hoppy, Jason 3rd place: Bounce, Wendy 29. 3 cups of tea, 2 cakes, and 7 people 30. 329
0
1
2
3
4
0
0
0
0
0
0
31. Put three on each pan. If they balance, the lighter one of the other two will be found in the next weighing. If three of the coins are lighter, then weigh one of these three against another of these three. If they balance, the third coin is the lighter one. If they do not balance, choose the lighter one.
1
0
1
2
3
4
32. Yes, providing c ⫽ 0.
2
0
2
4
11
13
3
0
3
11
14
22
4
0
4
13
22
31
11. a. (In base five)
33. No. “4 divided by 12” leads to a fraction and is written 4 12 whereas 4 divided into 12 could be written 12 4.
Section 3.3A 1. a. 19
b. (i) 3five 2five 11five, 11five 3five 2five, 11five 2five 3five (ii) 3five 4five 22five, 4five 3five 22five, 22five 4five 3five (iii) 2five 4five 13five, 4five 2five 13five, 13five 4five 2five 12. a. Measurement—How many 3 pints can be measured out? b. Partitive—How many tarts can be partitioned to each of 4 members. c. Measurement—How many groups of 3 straws each can be formed? 13. a. Fifteen students are to be divided into 3 teams. How many on each team? Answers may vary. b. Fifteen students are put into teams of 3 each. How many teams? Answers may vary. 14. a. 48 8 6
b. 51 3 䡠 x
15. a. 0
c. 12
b. 2
d. 8
c. x 5 䡠 13 e. 32
f. 13
16. If the remainder were larger than the divisor, then the remainder column would be taller than the rest of the rectangle and the extra portion above the top could be taken off to create a new remainder column. In other words, the divisor could divide into the dividend more times until the remainder is smaller than the divisor. 17. a. 3 2 ⫽ whole number b. 4 2 ⫽ 2 4 c. (12 3) 2 ⫽ 12 (3 2) d. 5 1 5, but 1 5 ⫽ 5 e. 12 (4 2) ⫽ 12 4 12 2 18. 1704 19. $3.84 20. Approximately $2.48. 22. Push 1; multiplicative identity 23. Yes, because multiplication is repeated addition 24. a. Row 1: 4, 3, 8; row 2: 9, 5, 1; row 3: 2, 7, 6 b. Row 1: 24, 23, 28; row 2: 29, 25, 21; row 3: 22, 27, 26
3. Yes 4. a. Yes
b. No. 2 ⫽ 3 and 3 ⫽ 2, but 2 2.
5. a. 34 d. x2y4
b. 24 䡠 32 e. a2b2 or (ab)2
c. 63 䡠 72 f. 5363 or (5 䡠 6)3
6. 34 ⬎ 43 ⬎ 25 ⬎ 52 7. a. 3 䡠 x 䡠 x 䡠 y 䡠 y 䡠 y 䡠 y 䡠 y 䡠 z b. 7 䡠 5 䡠 5 䡠 5 c. 7 䡠 5 䡠 7 䡠 5 䡠 7 䡠 5 8. a. 57
b. 310
c. 107
d. 28
e. 54
f. 67
9. 32 䡠 34 is an abbreviated form of (3 䡠 3)(3 䡠 3 䡠 3 䡠 3), so there is a total of 6 factors of 3, which can be written as 36. The student’s answer means a product of six 9s. 10. 52䡠7, (52)7, (57)2, 577, 57 䡠 57. There are other possible answers. 11. Always true when n 1. If n ⬎ 1, true only when a 0 or b 0. 12. a. x 6 b. x 5 c. x can be any whole number. 13. a. 1,679,616
b. 50,625
c. 1875
14. a. 15 3 䡠 5 0 b. 2 䡠 25 50 c. 9 䡠 4 2 䡠 8 36 16 20 6 + 2(27  16)2 + 16 6 + 2 # 121 + 16 d. 4 6(27  16) 66 15. This is the accepted convention in mathematics and is related to the fact that multiplication is repeated addition. 16.
(a m ) n ⫽ a m ⭈ a m ⭈ ⭈ ⭈ a m
}
21. $90,000
b. 16
2. 2 ⬍ 10, 8 ⬍ 10, 10 ⬎ 2, 10 ⬎ 8
n factors n addends
}
bansw01.qxd
⫽ a m⫹ m⫹ ⭈ ⭈ ⭈ ⫹ m ⫽ a mn
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17. a. 610 (2 䡠 3)10 210 䡠 310 ⬍ 310 䡠 310 320 b. 99 (32)9 318 ⬍ 320 c. 1210 (4 䡠 3)10 410 䡠 310 ⬎ 310 䡠 310 320 18. a. 200¢ or $2.00 b. 6400¢ or $64.00 c. Price 25 䡠 2n cents 19. When a n(A) and b n(B), a ⬍ b means A can be matched to a proper subset of B. Also, b ⬍ c when c n(C) means the B can be matched to a proper subset of C. In that matching, the proper subset of B that matches A is matched to a proper subset of a proper subset of C. Thus, since A can be matched to a proper subset of C, a ⬍ c. 20. a. 17 18 19 䡠 䡠 䡠 25 43 53; 26 27 28 䡠 䡠 䡠 36 53 63 b. 93 103 82 83 䡠 䡠 䡠 100 c. 123 133 145 䡠 䡠 䡠 169 d. n3 (n 1)3 (n2 1) (n2 2) 䡠 䡠 䡠 (n 1)2 21. 4(24) 64 22. a. The only onedigit squares are 1, 4, and 9. The only combination of these that is a perfect square is 49. b. 169, 361 c. 1600, 1936, 2500, 3600, 4900, 6400, 8100, 9025 d. 1225, 1444, 4225, 4900 e. 1681 f. 1444, 4900, 9409
A13
b.
0
1
2
3
2. a. Associative
4
5
6
b. Identity
7
8
9
c. Commutative
d. Closure
3. a. 5 6 5 (5 1) (5 5) 1 10 1 11; associativity for addition, doubles, adding ten b. 7 9 (6 1) 9 6 (1 9) 6 10 16; associativity, combinations to ten, adding ten 4. a. 7 3 n({a, b, c, d, e, f, g} {e, f, g}) n({a, b, c, d}) 4. b. 7 3 n if and only if 7 3 n. Thus, n 4. 5. To find 7 2, find 7 in the “2” column. The answer is in the row containing 7, namely 5. 6. None
Section 3.2 1. a. 15 altogether:
b. 15 altogether:
0
5
10
15
23. The second one 24. a. If a ⬍ b, then a n b for some nonzero n. Then (a c) n b c, or a c ⬍ b c. b. If a ⬍ b, then a c ⬍ b c for all c, where c a, c b. Proof: If a ⬍ b, then a n b for some nonzero n. Hence (a c) n b c, or a c ⬍ b c. 25. There is no shortcut for this problem. Although some people might prefer to write it as 213 32, the answer to 73 35 is still 343 243 83,349. If the bases match, the exponents may be added. If the exponents match, the bases may be multiplied. But if neither matches, you have to perform the calculations as shown. 26. These two numbers are the only whole numbers for which, using the same number twice, the sum and product are equal. That can be determined by considering all instances where x x x x, or 2x x2. The only numbers that make this sentence true are 0 and 2. Multiplication is sometimes defined as repeated addition, but it is clear these operations are not the same if you consider 3 4 and 3 4, for example.
c.
d.
5 3
5 3
2. a. Identity
b. Commutative
c. Associative
d. Closure
3. a. 7 27 7 13 7(27 13) 7(40) 280 b. 8 17 8 7 8(17 7) 8(10) 80 4. a. 6 7 6(6 1) 6 6 6 1 36 6 42; distributivity b. 9 7 (10 1)7 10 7 7 70 7 63; distributivity 5. a. 17 3 5 remainder 2:
b. 17 3 5 remainder 2:
PROBLEMS WHERE THE STRATEGY “USE DIRECT REASONING” IS USEFUL 1. Since the first and last digits are the same, their sum is even. Since the sum of the three digits is odd, the middle digit must be odd. 2. Michael, Clyde, Jose, Andre, Ralph 3. The following triples represent the amount in the 8, 3, and 5liter jugs, respectively, after various pourings: (8, 0, 0), (5, 3, 0), (5, 0, 3), (2, 3, 3), (2, 1, 5), (7, 1, 0), (7, 0, 1), (4, 3, 1), (4, 0, 4).
CHAPTER REVIEW Section 3.1 1. a. 5 4 n({a, b, c, d, e}) n({f, g, h, i}) n({a, b, c, d, e} 艛 {f, g, h, i}) 9
0
3
6
9
12
1517
6. To find 63 7, look for 63 in the “7” column. The answer is the row containing 63, namely 9. 7. a. 7 0 n if and only if 7 0 n. Since 0 n 0 for all n, there is no answer for 7 0. b. 0 7 n if and only if 0 7 n. Therefore, n 0. c. 0 0 n if and only if 0 0 n. Since any number will work for n, there is no unique answer for 0 0. 8. 39 7:39 7 5 4 where 5 is the quotient and 4 is the remainder 9. None
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10. This diagram shows how subtraction, multiplication, and division are all connected to addition.
Repeated addition
+
15. {2, 3, 4, 5, . . .} 16. One number is even and one number is odd.
x
Missing addend
17. a.
Missing factor –
14. (2 䡠 3)2 36 but 2 䡠 32 2 䡠 9 18
3
3
3
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
÷
Repeated subtraction
Three added together 4 times is 12. b. Four rows of three 3 make 12 squares.
Section 3.3 1. a ⬍ b (b ⬎ a) if there is nonzero whole number n such that a n b. 2. a. a c ⬍ b c
3
4
b. a c ⬍ b c
3. 5 5 5 5 5 4
4. a. 734 712 c. 5(73) 54
b. (3 7)5 215 d. 41213 425
18.
5. Use a pattern: 53 125, 52 25, 51 5, 50 1, since we divided by 5 each time.
(2 + 3) + 1 = 6
Chapter 3 Test 1. a. F f. F
b. T g. F
c. T h. T
2.
d. F i. F
e. F j. T
same
ADD
SUBTRACT
MULTIPLY
DIVIDE
Closure
T
F
T
F
Commutative
T
F
T
F
Associative
T
F
T
F
Identity
T
F
T
F
2 + (3 + 1) = 6
19. a. 3. a. Commutative for multiplication (CM) b. IM c. AA d. AM e. D 4. a. 30 10 9 2 c. (5 䡠 2)73 d. 10 33 2 33
? How many more need to
f. CA
be added to the set of 5 to make it a set of 8?
5
b. 40 80 7 7
8 b.
5. 64 R1 19
6. a. 3
b. 5
24
15
c. 7
2
d. 2
15
e. 14
f. 36
7. a. 13(97 3); distributivity b. (194 6) 86; associativity and commutativity c. 23(7 3); commutativity and distributivity d. (25 䡠 8)123; commutativity and associativity 8. a. partitive
b. measurement
10. A 11. a. (73)4 (7 䡠 7 䡠 7)4 (7 䡠 7 䡠 7)(7 䡠 7 䡠 7)(7 䡠 7 䡠 7)(7 䡠 7 䡠 7) 712 b. (73)4 73 䡠 73 䡠 73 䡠 73 73333 712 12. 3 0 n if and only if n 䡠 0 3. But n 䡠 0 0
} } m
}
ab ⭈ab ⭈ ⭈ ⭈ ab ⫽ (a ⭈b)m m
m
5
?
0 1 2 3 4 5 6 7 8 9 10 20.
3
4
c. measurement
9. a. Since there are two sets, comparison can be used with either missingaddend, 137 x 163, or takeaway, 163 137 26. b. missingaddend, 973 x 1500 c. takeaway, $5 $1.43 $3.57
13. am ⭈bm ⫽ a ⭈a ⭈ ⭈ ⭈ a ⭈b ⭈b ⭈ ⭈ ⭈ b ⫽
8
12
4
= 3
21. 64 82 43 22. 2 䡠 2 2 2 and 0 䡠 0 0 0
Section 4.1A 1. a. 105
b. 4700
c. 1300
2. a. 43 17 46 20 26 b. 62 39 63 40 23 c. 132 96 136 100 36 d. 250 167 283 200 83
d. 120
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3. a. 579 c. 215
17. a. 45
b. 903 d. 333
4. a. 198 387 200 385 585 b. 84 5 42 10 420 c. 99 53 5300 53 5247 4 d. 4125 25 4125 100 165 5. a. 290,000,000,000 b. 14,700,000,000 c. 91,000,000,000 d. 84 1014 e. 140 1015 f. 102 1015
b. 37
c. 47
d. 36
18. a. 17 817 100 1,388,900 b. 10 98 673 659,540 c. 50 4 674 899 2 100 674 899 119,837,200 d. 8 125 783 79 1000 783 79 61,857,000 19. a. 12, 7
b. 31, 16
c. 6, 111
d. 119, 828
20. 10 100 10,100; 588 2353 5,882,353 2
2
2
2
21. Yes, yes, no 22. a. Yes
b. Yes
c. Yes
23. Yes, yes, yes
6. a. 32 20 52, 52 9 61, 61 50 111, 111 6 117 b. 54 20 74, 74 8 82, 82 60 142, 142 7 149 c. 19 60 79, 79 6 85, 85 40 125, 125 9 134 d. 62 80 142, 142 4 146, 146 20 166, 166 7 173, 173 80 253, 253 1 254 7. a. 84 14 42 7 6 b. 234 26 117 13 9 c. 120 15 240 30 8 d. 168 14 84 7 12
24. a. 30
b. 83,232
c. 4210
d. 32
25. 7782 26. True 27. a. 1357 90 b. 6666 66 c. 78 93 456 d. 123 45 67 28. One possible method to find a range for 742 281 is 700 200 500 and 700 300 400. The answer is between 400 and 500.
8. Underestimate so that fewer than the designated amount of pollutants will be discharged.
29. 177,777,768,888,889
9. a. (i) 4000 to 6000, (ii) 4000, (iii) 4900, (iv) about 5000 b. (i) 1000 to 5000, (ii) 1000, (iii) 2400, (iv) about 2700 c. (i) 7000 to 11,000, (ii) 7000, (iii) 8100, (iv) about 8400
31. a. 4225, 5625, 9025 b. (10a 5)2 100a2 100a 25 100a(a 1) 25
10. a. 600 to 1200 b. 20,000 to 60,000 c. 3200 to 4000
b. 700 e. 3000
c. 1130
13. a. 4 350 1400 b. 603 c. 5004 d. 5 800 4000 14. a. 30 20 600, 31 20 620 30 23 690, 30 25 750 b. 35 40 1400, 40 40 1600 30 40 1200, 30 45 1350 c. 50 27 1350, 50 25 1250 50 30 1500, 45 30 1350 d. 75 10 750, 70 12 840 75 12 900, 80 10 800 15. There are many acceptable estimates. One reasonable one is listed for each part. a. 42 and 56 b. 12 and 20 c. 1 and 4 16. a. 4
b. 13
30. 5643, 6237
32. Yes 33. The first factor probably ends in 9 rather than 8.
11. a. 63 97 艐 63 100 6300 b. 51 212 艐 50 200 10,000 c. 3112 62 艐 3000 60 50 d. 103 87 艐 100 87 8700 e. 62 58 艐 60 60 3600 f. 4254 68 艐 4200 70 60 12. a. 370 d. 460
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34. a. 54 46 502 42 2484 b. 81 79 802 1 6399 c. 122 118 1202 22 14,396 d. 1210 1190 12002 102 1,439,900 35. True, Express the product as (898,000 423) (112,000 303). Use distributivity twice, then add. 36. (439 6852) 1000 268 6852 3,009,864,336 37. a. 76 (54 97) b. (4 13)2 c. 13 (592 47) d. (79 43) 2 172 38. a. 57 53 3021 b. (10a b)(10a 10 b) 100a2 100a 10b b2 100a(a 1) b(10 b). c. Problem 32 is the special case when b 5. 39. (i) Identify the digit in the place to which you are rounding. (ii) If the digit in the place to its right is a 5, 6, 7, 8, or 9, add one to the digit to which you are rounding. Otherwise, leave the digit as it is. (iii) Put zeros in all the places to the right of the digit to which you are rounding. 40. 3 䡠 5 䡠 7 䡠 9 䡠 11 䡠 13 䡠 15 䡠 17 34,459,425 41. Pour as much as possible from two of the 3liter pails into the 5liter pail; one liter will be left in the 3liter pail. 42. If you round the first number down to 3000 and the second number up to 300, you can see that an estimate of the answer would be 900,000 whereas the student’s answer is close to 90,000. The error was that a repeated “9” was omitted.
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43. This will always work if the student remembers to affix the zero at the end. The reason it works is because multiplying by 5 is the same as multiplying by 10 and then dividing by 2. This student is just dividing by 2 first, then multiplying by 10.
3. Expanded form; commutative and associative properties of addition; distributive property of multiplication over addition; singledigit addition facts; place value. 4. a. 986 5. a. 598 +396 14 180 800 994
Section 4.2A 1. a.
b. 909 b.
322 799 +572 13 180 1500 1693
6. a. 1229
b. 13,434
7. a. 751
b. 1332
8. a. Simple; requires more writing b. Simple; requires more space; requires drawing lattice
+
9. The sum is 5074 both ways! This works because the numerals 1, 8 stay as 1 and 8 while 6 and 9 trade places 10. Lefthand sum; compare sum of each column, from left to right. 11. a. ABC 12. a.
b.
⫺
+
b.
= 433
2. a.
10
1
4
7
15 ⫹ 32 b.
15
= 47 + 32
b. CBA
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Answers
b.
13. a.
⫺
33 *27 21 210 60 +600 891
c. The 21 is the 21 unit squares in the lower right. The 210 is the 21 longs at the bottom. The 60 is the 6 longs in the upper right. The 600 is the 6 flats.
= 20 21.
10
A17
34
1
57
100
100
100
40
100
100
100
40
80
80
80
32
⫺37 2
0 28
b.
600 ⫹ 240 ⫹ 80 ⫹ 32 ⫽ 952
34 ⫺ 29 ⫽ 5 22. a.
10
1
10
1
15
29 10
5
450
300
150
30
90
60
30
6
34
14 8, 12, 13, 12 15. (b), (a), (c) 16. a. 477
540 b. 776
c. 1818
17. 62, 63, 64, 65, 70, 80, 100; $38 Other answers are possible. 18. a. 358
b. 47,365
19. a. 135
b. 17,476
20. a.
36
b.
33
62 60
2
1860
1800
60
30
310
300
10
5
35
2170 23. Expanded form; distributivity; expanded form; associativity for ; place value; place value; addition
27
24. a. 1426
b. 765
25. a. 3525 b. 153,244 c. 684,288 26. a. 2380 27. a. 56
891
28. a. 1550
b. 2356 b. 42 b. 2030
c. 60 c. 7592
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29. a. 13 冄 899 b. 23 冄 5697 650 50(13)  4600 200(23) 249 1097 10(13) 30(23) 130  690 119 407 7(13) 10(23) 91  230 28 177 2(13) 7(23) 26  161 2 16 2 69(13) 899 16 247(23) 5697 30. Subtract 6 seven times to reach 0. 31. a. 6: 24 4 4 4 4 4 4 0 b. 8: 56 7 7 7 7 7 7 7 7 0 c. Subtract the divisor from the dividend until the difference is 0. The total number of subtractions is the quotient. 32. a. (i) q: 15 r: 74
(ii) q: 499 r: 70
(iii) q: 3336 r: 223
33.
34. Bringing down the 3 is modeled by exchanging the 7 leftover flats for 70 longs, making a total of 73 longs. 35. Larry is not carrying properly; Curly carries the wrong digit; Moe forgets to carry. 36. One answer is 359 127 486. 37. a. For example, 863 742 1605 b. For example, 347 268 615 38. 1 2 34 56 7 100; also, 1 23 4 5 67 100 39. a. 990 077 000 033 011 b. (i) 990 007 000 003 111; (ii) 000 770 000 330 011 (iii) 000 700 000 300 111 40. a. Equal b. Differ by 2 c. Yes d. Difference of products is 10 times the vertical 10splace difference. 41. a.
Break 2 flats and one long down
x x x x
x x x x
x x x x
x x x x
x x x x
b. 1 2 3 4 12 (4 * 5) c. 12 (50 * 51) 1275; 12 (75 * 76) 2850 42. 888 777 444 2190; 888 666 555 2109 43. Bob: 184; Jennifer: 120; Suzie: 206; Tom: 2081 44. 5314 79 28 736 2109 3527 45 419806 45. All eventually arrive at 6174, then these digits are repeated.
Put in groups of 5
46. Many answers are possible. Most future teachers enjoy seeing different methods from the ones they are familiar with. Yet sometimes the same teachers feel these other methods should only be used as enrichment for gifted students. Ironically enough, this means that the gifted students get to use the concrete methods in addition to the abstract (standard algorithms) while the students who are struggling, and who might benefit the most from concrete methods, are forced to work only with the abstract methods. 47. The distributive property is used here. 345 * 27 = = = =
(300 + 40 + 5)27 300 # 27 + 40 # 27 + 5.27 300(20 + 7) + 40(20 + 7) + 5(20 + 7) 300 # 20 + 300 # 7 + 40 # 20 + 40 # 7 + 5 # 20 + 5 # 7
Section 4.3A 1. a.
1
5
2
R2
13seven
5seven
0 1 2 3 4 5 6 10 11 12 13 14 15 16 20 21 22 23 24 25 26 30 31 32 33 BASE SEVEN
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b.
14. a. 3four b. 5six c. 4eight
4seven 21seven 0 1 2 3 4 5 6 10 11 12 13 14 15 16 20 21 22 23 24 25 26 30 31 32 33 BASE SEVEN
c.
3seven 3seven
3seven
3seven
3seven
15. Thought One
3seven
0 1 2 3 4 5 6 10 11 12 13 14 15 16 20 21 22 23 24 25 26 30 31 32 33 BASE SEVEN
d.
6seven
6seven
6seven
0 1 2 3 4 5 6 10 11 12 13 14 15 16 20 21 22 23 24 25 26 30 31 32 33 BASE SEVEN
One group of 4 flats.
2. a. 3four b. 100four c. 331four d. 12013four 3. a. 114six b. 654seven c. 10012four 4. a. 55six c. 1130six
b. 123six d. 1010six
5. a. 62seven
b. 102four
Thought Two
6. a. 1201five b. 1575eight 7. a. 13four b. 31four c. 103four 8. a. 4six b. 456seven c. 2322four 9. a. 17eight
b. 101two
Exchange one flat for 7 longs.
c. 13four
10. 10201three 2122three 10201three 100three 10000three 1three 1002three; the sums, in columns, of a number and its complement must be all twos. 11. a.
3four 23four
⫽ 201four Thought Three
b.
12five
23five
12. a. 122four b. 234five c. 132four 13. a. 11six b. 54seven c. 132four
Two groups of 4 longs.
331five
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Answers 3. Assume that both x and y are odd. Then x 2m 1 and y 2n 1 for whole numbers m and n. Thus x2 4m2 4m 1 and y 2 4n2 4n 1, or x2 y 2 4(m2 n2 m n) 2, which has a factor of 2 but not 4. Therefore, it cannot be a square, since it only has one factor of 2.
Thought Four
Exchange two longs for 14 ones.
CHAPTER REVIEW Section 4.1
Thought Five Remainder Four groups of 4 with 2 left over yields a quotient of 124seven remainder 2seven.
1. a. 97 78 97 (3 75) (97 3) 75 100 75 175; associativity b. 267 3 (270 3) 3 (270 3) (3 3) 90 1 89; right distributivitity of division over subtraction c. (16 7) 25 25 (16 7) (25 16) 7 400 7 2800; commutativity, associativity, compatible numbers d. 16 9 6 9 (16 6) 9 10 9 90; distributivity e. 92 15 92(10 5) 920 460 1380; distributivity f. 17 99 17(100 1) 1700 17 1683; distributivity g. 720 5 1440 10 144; compensaton h. 81 39 82 40 42; compensation 2. a. 400 ⬍ 157 371 ⬍ 600 b. 720,000 c. 1400 d. 25 56 5600 4 1400 3. a. 47,900
b. 4750
c. 570
4. a. Not necessary b. 7 (5 2) 3 c. 15 48 (3 4) 5. a. 11
b. 6
c. 10
d. 8
e. 10
f. 19
Section 4.2 1. 982 in all parts 2. 172 in all parts 3. 3096 in all parts 4. 9 R 10 in all parts 16. Six
Section 4.3
17. Steve has $2, Tricia has $23, Bill has $40, Jane has $50.
1. 1111six in all parts
18. 39,037,066,084
2. 136seven in all parts
19. 125
3. 1332four in all parts
20. The only digits in base four are 0, 1, 2, 3. In any number base, the number of digits possible is equal to the number used as the base. Since 0 is always the first digit, the last digit that can be used for any number base is one less than the number. For example, in base ten (our number base) there are ten different digits and the greatest is 9, which is one less than ten.
4. 2five R 31five in all parts
21. To subtract 4 from 1, you need to regroup from the fives place. That makes the 1 a 6, and when you subtract 4, the answer is 2. To subtract the 3 from the 1 (which was previously a 2) in the fives place, you need to regroup from the twentyfives place. That makes the 1 a 6, and when you subtract 3 the answer is 3. In the twentyfives place you subtract 2 from 3 (which was previously a 4) and the answer is 1. So the final answer is 132five.
PROBLEMS WHERE THE STRATEGY “USE INDIRECT REASONING” IS USEFUL
Chapter 4 Test 1. a. F
2. Suppose that n is odd. Then n is odd; therefore, n cannot be odd.
c. F
2. a. One possibility is 376 +594 10 160 800 970 3. a.
1. Assume that x can be even. Then x2 2x 1 is odd. Therefore, x cannot be even. 4
b. F
0 1
d. F b. One possibility is 56 *73 18 150 420 3500 4088 b.
1
5
6
8
4
9
3
0
1 1
7 0
9 0
1
5 6
1
3
7 2
9 2 6
7 3 5
6 1 4
8 2 2
3 7
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Answers 4. a. 54 93 16 47 54 16 93 47 70 140 210 b. 9225 2000 7225 c. 3497 1362 2135 52 d. 25 52 100 4 52 100 4 1300
Thought Two
5. 234 R 8 6. a. (i) 2500, (ii) 2500 to 2900, (iii) 2660, (iv) 2600 b. (i) 350,000, (ii) 350,000 to 480,000, (iii) 420,000, (iv) 420,000 7. 32 21 (30 2) (20 1) (30 2)20 (30 2)1 30 䡠 20 2 䡠 20 30 䡠 1 2 䡠 1 600 40 30 2 672 8. Since we are finding 321 20, not simply 321 2
Exchange one flat for 10 longs.
9. Commutativity and associativity 10.
2
10
1
10
10
0
2
1
0
10
10
10
6
4
1
374 267
11. Answers may vary 168 R 37 8 60 100 43 冄 7261 4300 2961 2580 381 344 37
Thought Three
Three groups of 4 longs.
12. Thought One
One group of 4 flats.
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14.
Thought Four
2
9
9
1
0
9
2
1
0
9
9
9
3
2
6nine
278nine 37nine
Exchange one long for 10 ones.
15. a.
Exchange 10 units for 1 long.
Remainder Thought Five Exchange 10 longs for 1 flat. Four groups of 4 units.
Yields 403 b.
Exchange
Quotient 134, Remainder: 2 13.
Note enough units to take away three units
Can't take away 8 units from 3 so exchange 1 long for 10 units. Now take away 8 units. Exchange
Exchange a long for four units.
Take away one long
Take away 6.
Take away three units Leaves 13four
Can't take away 6 longs from 4 so exchange 1flat for 10 longs. Now take away 6 longs to leave 185.
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c.
A23
Thought Three
Thought One
Five groups of 5 longs. One group of 5 flats.
Exchange 3 longs for 30 units. Bring down the 9.
Thought Two Thought Four
Seven groups of 5 units with 4 remaining yields a quotient of 157 and remainder of 4. Exchange 2 flats for 20 longs. Bring down the 8.
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16. Intermediate
Section 5.1A
Standard
492 * 37 14 630 2800 60 2700 12000 18204
1. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
2 61
2. a.
492 * 37 3444 14760 18204
b.
36 2
18 2
2
c.
3 3
2
51 3
10
10
3
1000 2
17
500 2
250 5
50 5
18. The subtractfromthe base algorithm appears to be more natural for young students. The only disadvantage is its lack of use because of the tradition of the standard algorithm. 19.
9 3
d.
102
17. The advantage of the standard algorithm is that it is short. The disadvantage is that because of its brevity, it loses some meaning. The advantage of the lattice is that all of the multiplication is done first and then all of the addition, which eliminates some confusion. The disadvantage is its length.
27
9 3
In both cases the ones (7) in the number 37 is multiplied by each of the ones, tens, and hundreds of 492. Similarly the tens (3) of 37 is multiplied by each of the ones, tens, and hundreds of 492.
54
10 2
5
3. a. 23 33 b. 22 3 5 72 c. 3 52 11 d. 2 32 7 112 13
3
4. a. T, ● ● ●
10
●●● ●●●
b. F, There is no whole number x such that 12x 6. c. T, ● ● ● ● ● ● ● ●●●●●●● ●●●●●●●
d. F, There is no whole number x such that 6x 3. e. T, ● ● ● ●
7
●●●● ●●●● ●●●●
f. F, There is no whole number x such that 0x 5. h. T, ● ● ● g. T, ●
200
140
30
21
200 140 30 + 21 391
●
●●●
●
●●●
●
●●●
●
●●●
●
●●●
●
●●●
●
●●●
●
●●●
●
●●●
●
●●● ●●● ●●● ●●● ●●●
20. H 2, E 5, S 6
●●●
21. a 7, b 5, c 9; 62,015
5. 1, 3, and 7
22. A 5, B 6; A 4, B 7; A 3, B 8; A 2, B 9. The roles of A and B can be reversed. In all cases C 1, D 2.
6. a. 8 兩 152, since 8 19 152 b. x 15,394 by long division c. Yes. 15,394 8 123,152
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7. a. 4 only
b. 3, 4
32. 22 3 5 60
8. a. and c. 9. a. Yes
33. 3. Proof: n (n 1) (n 2) 3n 3 3(n 1) b. No
c. Yes
10. a. T, 3 is a factor of 9 b. T, 3 and 11 are different prime factors of 33 11. a. 4 ⱈ 5 and 4 ⱈ 3, but 4 兩 (5 3) 12. a. F; 3 ⱈ 80 c. T; 4 兩 00 13. a. 2 兩 12
A25
b. T
b. F; 3 ⱈ 10,000 d. T; 4 兩 32,304 and 3 兩 32,304 b. 3 兩 123
c. 2 兩 1234
d. 5 兩 12,345
14. 2, 3, 5, 7, 11, 13, 17, 19. No others need to be checked. 15. a. 22 32 b. 1, 2, 3, 4, 6, 9, 12, 18, 36 c. 4 22, 6 2 3, 9 32, 12 22 3, 18 2 32, 36 22 32 d. The divisors of 36 have the same prime factors as 36, and they appear at most as many times as they appear in the prime factorization of 36. e. It has at most two 13s and five 29s and has no other prime factors. 16. For 5, 5 兩 10(a 䡠 10 b) or 5 兩 (a 䡠 102 b 䡠 10); therefore, if 5 兩 c, then 5 兩 (a 䡠 102 b 䡠 10 c). Similar for 10.
34. Use a variable; the numbers a, b, a b, a 2b, 2a 3b, 3a 5b, 5a 8b, 8a 13b, 13a 21b , 21a 34b have a sum of 55a 88b, which is 11(5a 8b), or 11 times the seventh number. 35. a. Use distributivity. b. 1001! 2, 1001! 3, . . . , 1001! 1001 36. $3.52 $2.91 $.61 61¢ is not a multiple of 3. 37. 504. Since the number is a multiple of 7, 8, and 9, the only threedigit multiple is 7 䡠 8 䡠 9. 38. 61 39. abcabc abc(1001) abc(7 䡠 11 䡠 13), 7 and 11 40. a. Apply the test for divisibility by 11 to any fourdigit palindrome. b. A similar proof applies to every palindrome with an even number of digits. 41. 151 and 251 42. Conjecture: If 7 兩 abcd, then 7 兩 bcd,00a. Proof: Using expanded form, we know: bcd,00a + abcd = (100,000b + 10,000c + 1000d + a) + (1000a + 100b + 10c + d) = 100,100b + 10,010c + 1001d + 1001a = 7(14,300b + 1430c + 143d + 143a)
17. (a), (b), (d), and (e) 18. a. Yes b. No c. Composite numbers greater than 4 19. 333,333,331 has a factor of 17. 20. p(0) 17, p(1) 19, p(2) 23, p(3) 29: p(16) 162 16 17 16(17) 17 is not prime. 21. a. They are all primes. b. The diagonal is made up of the numbers from the formula n2 n 41. 22. The numbers with an even number of ones have 11 as a factor. Also, numbers that have a multipleofthree number of ones (e.g., 111) have 3 as a factor. That leaves the numbers with 5, 7, 11, 13, and 17 ones to factor. 23. Only 7( 5 2). For the rest, since one of the two primes would have to be even, 2 is the only candidate, but the other summand would then be a multiple of 5. 24. a. 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, and 97. b. 5 1 4, 13 9 4, 17 1 16, 29 4 25, 37 1 36, 41 16 25, 53 4 49, 61 25 36, 73 9 64, 89 25 64, 97 16 81 25. There are no other pairs, since every even number besides 2 is composite. 26. 3 and 5, 17 and 19, 41 and 43, 59 and 61, 71 and 73, 101 and 103, 107 and 109, 137 and 139, 149 and 151, 179 and 181, 191 and 193, 197 and 199
Thus, 7 兩 (bcd,00a abcd). Since 7 兩 abcd, we know 7 兩 bcd,00a. 43. 289 172 is the first nonprime. 44. 11 兩 [a (1001) b(99) c(11) a b c d] if and only if 11 兩 (a b c d). Therefore, we only need to check the a b c d part. 45. 11 101,010,101 1,111,111,111 13 8,547,008,547 111,111,111,111 17 65,359,477,124,183 1,111,111,111,111,111 46. a. n 10
b. n 15
c. n 10
47. n 16; p(17) 323 17 䡠 19, composite, p(18) 359, prime 48. When we say 4 兩 12, we are making the statement “4 divides evenly into 12” or “4 is a factor of 12.” When we say 12/4, we are indicating an operation that has an answer of 3. Similarly, we can say “24/7” and get the answer 33冫7, but we would not say “7 兩 24,” because 7 is not a factor of 24. 49. For this rule to work, the two numbers that divide 36 must be “relatively prime,” that is, have no factors in common greater than 1. The numbers 2 and 9 meet this criterion, so 18 兩 36, but 4 and 6 have 2 as a common factor, which explains why we cannot conclude that 24 兩 36.
Section 5.2A 1. a. 6
b. 12
c. 60
28. Yes
2. a. 2 䡠 2 䡠 3 䡠 3 b. 1, 2, 3, 4 2 䡠 2, 6 2 䡠 3, 9 3 䡠 3, 12 2 䡠 2 䡠 3, 18 2 䡠 3 䡠 3, 36 2 䡠 2 䡠 3 䡠 3 c. Every prime factor of a divisor of 36 is a prime factor of 36. d. It contains only factors of 74 or 172.
29. 34,227 and 36,070
3. a. 2
b. 6
c. 6
30. a. 6, Each has a factor of 2 and 3. b. 3, The only common factor is 3.
4. a. 6
b. 121
c. 3
d. 2
5. a. 6
b. 13
c. 8
d. 37
31. 2520
6. a. 6
b. 1
c. 5
d. 14
27. a. Many correct answers are possible. b. Let n be an odd whole number greater than 6. Take prime p (not 2) less then n. n p is an even number that is a sum of primes a and b. Then n a b p.
e. 13
f. 29
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7. a. 8. a. 24 b. 20 c. 63 d. 30 9. a. 2  21 24 63 70 b. 2  20 36 2  21 12 63 35 2  10 18  2 21 6 63 35 3 5 9 3  21 3 63 35 3 5 3 3 7 5 7 7 7 1
1 7
7
5 5 7 1 11  1
1
1
1
1 21 35 1 7 35 1
23 䡠 32 䡠 5 䡠 7 10. a. 360
e. 40
f. 72 c. 2  15 35 42 80 2  15 35 21 40 2  15 35 21 20
42 33 21 33 21 33
2  15 35 21 10 3  15 35 21 5 5  5 35 7 5 7 1 7 7 1
7 11
1
7 11
1
7 11
1
1 11
1
1
1
1
22 䡠 32 䡠 5 䡠 7 䡠 11
b. 770
1
1
24 䡠 3 䡠 5 䡠 7
c. 135
1
18. a. 2, 3, 5, 7, 11, 13; primes b. 4, 9, 25, 49, 121, 169; primes squared c. 6, 10, 14, 15, 21, 22; the product of two primes, or 8, 27, 125 or any prime to the third power d. 24, 34, 54, 74, 114, 134; a prime to the fourth power 19. 6, 28, 496, 8128 20. 1, 5, 7, 11, 13, 17, 19, 23 21. 16 candy bars 22. Chickens, $2; ducks, $4; and geese, $5 23. None, since each has only one factor of 5 24. 773 25. a. 11, 101, 1111
b. 1111
26. 41, 7, 11, 73, 67, 17, 13
11. a.
27. 31 (i)
Factors of a
28. 343 7 䡠 49. Let a b 7. Then 7 兩 (10a 10b). But 7 兩 91, so 7 兩 91a. Then 7 兩 (10a 10b 91a) or 7 兩 (100a 10b a).
Factors of b 3
7
29. GCF(54, 27) 27, LCM(54, 27) LCM(54, 18) LCM(18, 27) 54
2
3
5
30. a. Fill 8, pour 8 into 12, fill 8, pour 8 into 12, leaves 4 ounces in the 8 ounce container. b. Fill 11, pour 11 into 7, empty 7, pour 4 into 7, fill 11, pour 11 into 7, leaves 8 ounces in 11 ounce container, empty 7, pour 8 into 7, leave 1 ounce in 11 ounce container.
(ii) GCF (63, 90) 3 䡠 3 LCM (63, 90) 2 䡠 3 䡠 3 䡠 5 䡠 7 b.
31. The Euclidean algorithm for finding the greatest common factor of two numbers is the second idea. The third is the proof about perfect numbers which is described in Part A Problem 19. That is, the expression 2n 1(2n 1) will yield a perfect number whenever 2n 1 is prime.
(i)
Factors of a
Factors of b
2 2
2
32. When you are trying to find the Least Common Multiple, you are looking among all the numbers that are multiples of your original numbers. So that makes them all “big” to start with. You are only looking for the smallest of the “big” numbers. When you are looking for the Greatest Common Factor, you are looking among all the numbers that divide into your original numbers. So you are looking for the biggest among the “small” numbers.
5
2
3
(ii) GCF (48, 40) 2 䡠 2 䡠 2 LCM (48, 40) 2 䡠 2 䡠 2 䡠 2 䡠 3 䡠 5 c. (i)
Factors of a
PROBLEMS WHERE THE STRATEGY “USE PROPERTIES OF NUMBERS” IS USEFUL
Factors of b
2
1. Mary is 71 unless each generation married and had children very young.
7
2 2 2
2. Four folding machines and three stamp machines, since the LCM of 45 and 60 is 180
7
3. Let p and q be any two primes. Then p6q12 will have 7 䡠 13 91 factors.
(ii) GCF (16, 49) 1 LCM (16, 49) 2 䡠 2 䡠 2 䡠 2 䡠 7 䡠 7 12. LCM, (12, 18) 13. a. All except 6, 12, 18, 20, 24 14. a. amicable
b. 12, 18, 20, 24
b. not amicable
c. 6
c. amicable
CHAPTER REVIEW Section 5.1 1. 17 13 11 7 2. 90, 91, 92, 93, 94, 95, 96, 98, 99, 100
15. All 16. a. a 2 3
3
17. a. 1 d. 21 31 6 g. 26 64
b. a 2 7 11 2
3
b. 2 2 e. 24 16 h. 23 3 24 1
2
c. 2 4 f. 22 3 12 2
3. a. F g. T
b. F h. T
c. T
d. F
e. T
f. T
4. All are factors. 5. Check to see whether the last two digits are 00, 25, 50, or 75.
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Section 5.2
A27
Section 6.1A
1. 24
1. a. 31
2. 36
2. a. (i)
(ii)
b. (i)
(ii)
c. (i)
(ii)
d. (i)
(ii)
b. 73
d. 64 or 23
7 c.10
3. 27 4. 432 5. Multiply each prime by 2. 6. 81 135 GCF(81, 135) LCM(81, 135)
Chapter 5 Test 1. a. F g. T
b. T h. F
c. T i. F
d. F j. T
e. T
f. T
2. a. Three can be divided into 6 evenly. b. Six divided by 3 is 2. c. The statement 3 divides 6 is true. (Answers may vary) 3. a. 23 䡠 5 䡠 3
b. 24 䡠 52 䡠 33
4. a. 2, 4, 8, 11 5. a. 24
b. 2, 3, 5, 6, 10
b. 16
6. a. 23 䡠 3; 24 䡠 32 䡠 5 d. 41; 128,207 7.
c. 32 䡠 7 䡠 13 c. 2, 3, 4, 5, 6, 8, 9, 10
1
1
3. a.
b.
c. 27 b. 7; 2 䡠 3 䡠 5 䡠 72 e. 1; 6300
c. 23 䡠 34 䡠 53; 27 䡠 35 䡠 57 c.
6 R 123 1025 冄 6273 8 R 41 123 冄 1025 3R0 41 冄 123 GCF(1025, 6273) 41
4. No. The regions are not the same size. 15 5. a. 45 or 31 1 6. a. 10
26 b. 45
b. 41
c. 41 45 c. 31
d. 13
7.
8. LCM(18, 24) 72
=
9. All the crossedout numbers greater than 1 are composite. 10. No, because if two numbers are equal, they must have the same prime factorization. 11. x (x 1) (x 2) (x 3) 4x 6 2(2x 3)
The computer allows you to change the number of dividing lines of the area. Do this until the new dividing lines match up with the old ones so that either each original piece is cut into more pieces or the same number of original pieces are combined into newer equalsized larger pieces.
12. a. 4 兩 36 and 6 兩 36 but 24 ⱈ 36. b. If 2 兩 m and 9 兩 m then 18 兩 m. 13. LCM(a, b) (a 䡠 b)/GCD(a, b) 270/3 90. 14. ● ●
●●●
8. (a), (b), and (d)
●●
●●●
9 a. Not equal
●●
●●
10.
●●
4兩8
3ⱈ8
15. n m because they have the same prime factorization; Fundamental Theorem of Arithmetic
a. 43
b.
11. a. 515 32
b. Not equal 7 8
c.
9 13
c. Equal
d.
d. Equal
11 5
b. 2185 216
11 13 6 12 12. a. (i) 17 (ii) 17 6 16 6 15 17 6 17 b. Increasing numerators, decreasing denominators
17. If two other prime numbers differ by 3, one is odd and one is even. The even one must have a factor of 2.
a + c 13. Based on a few examples, it appears as if is always between b + d c a and . b d
18. a. 23 䡠 5
14. a. 17 23 6
16. 23 䡠 32 䡠 5 䡠 7 2520
b. 23 䡠 33 䡠 5
51 68 68 , 91 811 6 214 897 , 3408
19. 24, 25, 26, 27, 28; or 32, 33, 34, 35, 36
597 c. 2511
20. 1, 4, 9, 16; They are perfect squares.
58 51 , 100 a. 113
21. n 11
15.
b.
43 93 567 , 1254 3 96 93 2811 6 87 , 2898
50 6 b. 687
d. 30 27 113 , 100
14 12 , 100 c. 113
22. a 15, b 180; and a 45, b 60
16. a. The fractions are decreasing. b. The fraction may be more than 1.
23. 36
17. 2003
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Answers
18. a.
b.
b.
2 5
0 c.
d. 2.
e.
f.
5 9 5 9 5 9 5 9
7 12 7 + 12 7 + 12 7 + 12 +
1 3
+
2 5
1
11 15
20 21 + = 36 36 40 42 = + = 72 72 60 63 = + 108 108 80 84 = + 144 144
41 36 82 72
=
123 108 164 = 144 =
Other correct answers are possible. 3. a. 19. Yes, since
26 # 11 Á 1 26 288 Á 86 = = . 588 Á 83 53 # 11 Á 1 53
+
=
=
5 6
+
=
=
5 8
20. 81 21. a. False. Explanations will vary. b. False. Explanations will vary. c. True. Explanations will vary. d. True. Explanations will vary.
b.
22. Both 12 2 and 18 3 are the GCF(12, 18) a. 24 b. 26
c.
+
=
+
23. Only (c) is correct. Explanations will vary. 4. a. 3/4 b. 3/4 c. 16/21 d. 167/144 e. 460/663 f. 277/242 g. 617/1000 h. 9/10 i. 15,059/100,000
24. 15 25. 562,389 26. 41 27. 2200 28. 15 26. a. 50, 500, etc. b. 25, 250, etc. c. 4, 40, 400, etc. d. 75, 750, etc.
5. a.
30. There are infinitely many such fractions. Two are
15 28
and
31. When the numerator is larger than the denominator, you need more than one circle. She might draw three halfcircles, then fit them together to see how many wholes she has—in this case, one whole and one half. The denominator tells you what size pieces you need; the numerator tells you how many of those pieces are gathered.
0
b.
32. When you are trying to work backward from a cross multiplication to a fraction equation, you have to remember that the numbers multiplied together were diagonally across from each other in the a d a c original. In this case, ab cd is equivalent to = or = . c b d b
Section 6.2A
4 10
+
1 12
0
11 15
1
4 5 12 12
1 4
=
1 3
1
7 10
c.
1. a.
2 5
3 10
31 56 .
0
5 12
2 3
1
= 15
12
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6. a. 4/11 d. 23/54
b. 13/63 e. 154/663
7. a. 23 6
b. 23 8
b. 256
1
 7 =
d.
826
=
 4 =
413
1
b. (i) 13 to 15, (ii) 1412
12. a. (i) 13 to 15, (ii) 14 c. (i) 15 to 18, (ii) 16 312
1 5⫻ : 3
b. 948  3 = 612
457
2. a.
1312
b.
912
512
+
5 2 3
3
4
5
5 2 3
3
4
5
b.
= 15
c. 7 + 5 + 212 = 1412
5 6
14. 15 9 15. a. 29/20 or 120
16. 17.
a. 73 5 12
+
2 7
=
b. 13/45
5 7
b. 31 +
1 6
=
2 5
1 2
18. Let t years of lifetime, t 72 years
3 8
1 3
7 19. 12
c.
20. 13 30 21.
A29
1 1 1 1 1 3 3 3 3 3
c. 1015
11. a. 837  3 = 537
13. a. 10 +
1 c. 3812 , 614
1 ⫻ 5: 3
1 b. 35
3 9. a. 59 56 = 156
10. a. 119
1.
d. 64 9
8 1 b. 1321 ; 221
Answers
Section 6.3A
c. 1/20 f. 11/1000
c. 26 5
5 8. a. 311 12 , 112
c. 1157
Page A29
1 36
22. When borrowing 1, he does not think of it as 55 . Have him use blocks (i.e., base five pieces could be used with long 1). 23. a. 1
b. 1
c. Yes; only perfect numbers
1 + 3 + 5 + 7 + 9 1 + 3 + 5 + 7 + 9 + 11 24. Yes. , . 11 + 13 + 15 + 17 + 19 13 + 15 + 17 + 19 + 21 + 23 In general, the numerator is 1 3 … (2n 1) n2 and the denominator is [1 3 … (2m 1)] [1 3 … (2n 1)], where m 2n. This difference is m2 n2 4n2 n2 3n2. Thus the fraction is always n2/3n2 = 13 . 25. 1 26.
2100
2 3
2100  1
1
a. 63
7 10
=
2100
b. 18
3 1 e. No, 10 {2 =
4 12
=
1 3
Z
3. a. 53 *
18 d. 56
c. 56 9 26
=
6 20
4.
3
{6
3 4
9 20 b. 31 b. 43
5. a. ⬍
27. The sum is 1.
b. 47 *
=
21 a. 11
c. 7
2 3
9 121
1 6
=
4 42
=
2 21
1 d.108
c. Order is reversed
29. 28 matches
6. a. Associativity and commutativity for fraction multiplication b. Distributivity for multiplication over addition c. Associativity for fraction multiplication
30. 1299 0s are necessary. 300 9s are necessary.
7. a. 4 ÷ 1 = 12.
28. a. 51 =
1 6
+
1 30
b. 71 =
1 8
+
1 56
1 c. 17 =
1 18
+
1 306
31. It depends on the problem. Many students find the idea of like denominators difficult. So for them, multiplying would be easier than adding if the adding involved two different denominators. For example, 3/4 4/10 would be a harder problem for most students 3 4 than * . 4 10 32. You might start by using examples that agree with the student’s conjecture. For example, if you have 3/10 and 4/10, then 4/10 is larger. If you have 3/7 and 3/8, then 3/7 is larger. So the student’s rules work if the denominators are equal and the numerators are different, and vice versa. But what if you have 4/5 and 5/6? Now the student’s two rules are in conflict. At this point the student needs to be able to express the two fractions with common denominators (or numerators) or reach for a calculator and compare decimals.
3
b. 2
1 2
c. 3 ÷
÷
1 4
3 = 4
= 10.
4.
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Answers
8. a.
28. 60 employees 29. 8:00 30. a. 212 cups b. 85 cup 1 c. 25 12 or 212 cups
31. Gale, 9 games; Ruth, 5 games; Sandy, 10 games 32. a. $54,150
b. After 8 years
33. 1
b.
34. a.
11 16 * 5
b. Yes
11 9 9 c. 411 8 , 2 = 216 ; 1016 , 2 = 532
1 35. Sam: 18 12 = 12 , addition (getting common denominator); 2 Sandy: 20 9 = 29 , division (using reciprocal).
36. 60 apples 37. 21 years old 38. Even though the remaining part of a group is twofifths of the whole, it is only two out of three parts needed to make another threefifths. Thus, we have twothirds of a group of threefifths.
c.
9. a. 5
b. 34
33 c. 39 or 11 13
10. a. 54
b. 1
c. 13 7
11. a. 21/10
13.
b. 112
3 a. 14 40 f. 189
b. g.
c. h.
15. a. 1159 d. 16.
=
361 63
98 a. 117
b.
17. a. 21 *
3 7
d. 15 26
6 e. 49
5 18
17 j. 24
i.
PROBLEMS WHERE THE STRATEGY “SOLVE AN EQUIVALENT PROBLEM” IS USEFUL 1. Solve by finding how many numbers are in {7, 14, 21, . . . , 392} or in {1, 2, 3, . . . , 56}.
21 b. 341 32 = 1032
c. 9
3 e. 15 4 = 34
2. Rewrite 230 as 810 and 320 as 910. Since 8 ⬍ 9, we have 810 ⬍ 910.
f. 6
3. First find eight such fractions between 0 and 1, namely 19 , 29 , 39 , . . . , 89 .
4 9
1 2 3 8 Then divide each of these fractions by 3: 27 , 27 , 27 , . . . , 27 .
= 9
d. 3 * 54 +
5 9
b. 35 *
c. 1 *
c. 12,200
=
3 8
CHAPTER REVIEW Section 6.1 1. Because 4 ⬎ 2 2.
b. 53 ⫽ 125
b. 3000
21. a. No
= 15
3 8
b. 56 ⫼ 8 ⫽ 7 d. 31 ⫻ 6 ⫽ 186
19. a. 12 ⫻ 12 ⫽ 144 20. a. 1100
3 7
* 54 = 162 + 30 = 192
18. a. 30 ⫻ 5 ⫽ 150 c. 72 ⫼ 9 ⫽ 8
d. 31,200
b. No
15 22. a. 31 231 c. 102
d. 3/4
2 5 49 121
b. 14/5 or 245
14. a. 14/15 250 63
c. 10/13 c. 123
1 5 49 50
40. Draw 8 small rectangles composed of 4 squares each. Then count how many groups of 3 squares there are—namely, 10 with 2 left over. 3 2 Thus, 8 , equals 10 . 4 3
d. 21
b. 91/18
12. a. 117 28
5 39. First draw a rectangle shading of it one way (say horizontally) 7 3 and shade of the shaded region the other way (vertically). There 4 should be 15 squares shaded twice out of 28 squares. Thus the 15 product is . 28
3. The numerator of the improper fraction is greater than the denominator. A mixed number is the sum of a whole number and a proper fraction.
b. 56 67 or
77 34
23. a. 127/144
b. 61/96
24. 32 loads
4. a.
25. Approximately 14,483,000 barrels per day 26. 432 27. a. 1245 gallons, or nearly 13 gallons b. About 512 gallons more
8 24 6 19 56
b. Equal
5.
3 8 12 15 3 24 = , , = = 56 7 19 28 35 7
6.
2 + 5 5 2 6 6 5 5 + 12 12
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Answers
Section 6.2
6. a. 43
1. a.
b. 37
16 c. 92
31 7. a. 36
8. 7 12
11 64 b. 75 c. 43 d. 49 5 2 3 3 5 3 2 a. 2 (4 5 ) = (2 5 ) 4 = 4 3 144 b. 47 35 + 45 35 = (47 + 45 )35 = 48 35 5 = 175 13 5 4 13 4 5 5 c. (17 + 11 ) + 17 = (17 + 17 ) + 11 = 111 51 17 d. 38 57  49 38 = 38 (57  49 ) = 38 17 63 = 504 = 168 4 2 5 a. 355 , 97 L 36 , 9 = 4 b. 38 * 1423 L 4 * 15 = 60 4 1 3 3 c. 39 + 135 + 13 = (3 + 13) + (49 + 15 + 13 ) L 16 + 1 = 17
# # # # #
9.
1 8
2. a. 114 45
b. Identity
c. Associative
d. Closure
4. None 5. a. 528 +
1 8
# #
#
#
#
Answers may vary.
b. 16 75
3. a. Commutative
6.
+ 378 = 914 . Compensation, associativity
b. 3118  5 = 2618 . Equal additions c. 135 . Commutativity, associativity a. 12 to 14 b. 17 + 2412 = 4112
c. 23
6 10. The fraction 12 represents 6 of 12 equivalent parts (or 6 eggs), whereas 12 represents 12 of 24 equivalent parts (or 12 halves of 24 eggs). NOTE: This works best when the eggs are hardboiled.
11.
a(b  d) a b d a b  d ab  ab = ¢  ≤ = ¢ ≤ = c c c c c c2 c2 ad a#b a#d ab . = 2  2 = c c c c c c
1. 8 15
4 5
a c bc ad 6 if and only if 6 if and only if ad ⬍ bc b d bd bd
12. NOTE: For simplicity we will express our fractions using a common denominator.
Section 6.3
13.
2 3
=
10 15
3 5
=
9 15
1 4
+
2 3
3 12
+
8 12
2 3 4 2. a. 15
b. 23
3. a. Inverse d. Closure 4.
b. Associative e. Associative
c. Identity
a c a e a c e ¢ + ≤ = * + * ; f b d b b d f 4 14 4 4 4 3 3 14 * + * = ¢ + ≤ * = 17 9 17 19 9 9 17 17
5. 12 25 , 12 25
,
1 5 1 5
= =
12 25 12 25
, *
5 25 25 5
= =
14.
12 5 12 5
6. None 7. a. (23 , 9) * 5 = 6 * 5 = 30. Commutativity and associativity b. (25 * 2) + (25 * 25 ) = 50 + 10 = 60. Distributivity b. 312 * 4 = 14
8. a. 35 to 48
Chapter 6 Test 1. a. T e. T
b. F f. F
c. F g. T
d. T h. F
2. Number: relative amount represented Numeral: representing a parttowhole relationship 3. Closure for division of nonzero elements or multiplicative inverse of nonzero elements 4. a. 32 5.
A31
a. 38 11
17 b. 18
b.
511 16
c. 52 c.
37 7
41 d. 189 2 d. 1111
15.
3 4
3 5
=
9 20
=
11 12
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Answers
16. a. Problem should include 2 groups of size 34 . How much all together? b. Problem should include 2 wholes being broken into groups of size 31 . How may groups? c. Problem should include 25 of a whole being broken into 3 groups. How big is each group? 17. a.
13. a. 14 * 44 = 11 c. 35 * e. 65 *
2 5 4 5
b. 34 * 80 = 60
= 14
d. 15 * 65 = 13
= 52
f. 380 *
1 20
14. a. 6,750,000 c. 296 followed by 26 zeros
= 19 b. 0.00019514 d. 29,600
15. a. 16 to 19; 5 6 7 18 b. 420 to 560; 75 6 450 c. 10 d. 40 b. 14 * 88 = 22
16. a. 48 3 16 c. 125 *
b.
1 5
= 25
d. 56,000 *
e. 15,000 750 20 17. a. 97.3 e. 0.0183
f.
b. 350 f. 0.5
3 5
1 4
= 14,000
* 500 = 300
c. 350 g. 0.50
d. 0.018
18. One possible answer: n n + 1 if and only if n(n 2) ⬍ (n 1)2. 6 n + 1 n + 2 However, since n2 2n ⬍ n2 2n 1, the latter inequality is always true when n 0. 19. 90 20. $240,000
18.
21. 25 = 22. 4 ,
56 57 58 59 140 6 140 , 140 , 140 2 3 = 6
6
60 140
=
15.72 41.92 36.68 19. At least 90 cents per hour
b. 0.063
c. 306.042
2. a. (i) 4(1/10) 5(1/100); (ii) 45/100 b. (i) 3 1(1/10) 8(1/100) 3(1/1000); (ii) 3183/1000 c. 2(10) 4 2(1/10) 5(1/10,000); (ii) 242,005/10,000 3. 746,000
52.4 31.44 10.48
3 7
Section 7.1A 1. a. 75.603
26.2 20.96 47.16
b. 0.746
c. 746,000,000
4. a. Thirteen thousandths b. Sixtyeight thousand four hundred eightyfive and five hundred thirtytwo thousandths c. Eightytwo ten thousandths d. Eight hundred fiftynine and eighty thousand five hundred nine millionths
20. If everyone knows what he means, then there is no problem. However, it is correct to use “and” only to signal a decimal point. 21.
6 783 29 600 783 290 1673 + + = + + = = 1.673. 10 1000 100 1000 1000 1000 1000 Because the least common multiple of all the denominators (in this case, 1000) is always present; one only has to multiply the numerator and denominator of the other fractions by an appropriate power of 10. With the other problem, finding the LCM or LCD is more complicated.
Section 7.2A 1. a. (i) 47.771, (ii) 485.84
5. There should be no “and.” The word “and” is reserved for indicating the location of the decimal point.
2. a. (i) 0.17782, (ii) 4.7
6. (b), (c), (d), and (e)
4. None
7. a. R b. T, 3 places c. R d. T, 4 places Explanation: Highest power of 2 and/or 5
5. a. 0.72
10. a. 54 6
19 34 7 9 8 6 10
18 b. 25 6
b.
38 52 43 539 40 6 500
b. 3.41
6. a. 5.9 10 d. 1.0 106
b. 4.326 103 e. 6.402 107
8. a. 3.658 10 c. 35 6
5 8
11. Lucas and Amy were arrested. 12. a. 18.47 10 8.47; equal additions b. 1.3 70 91; commutativity and distributivity c. 7 5.8 12.8; commutativity and associativity d. 17 2 34; associativity and commutativity e. 0.05124; powers of ten f. 39.07, left to right g. 72 4 76; distributivity h. 15,000; powers of ten
6
7 9
d. 6908.3
c. 9.7 104 f. 7.1 1010
b. 3.5 109m b. 5.893 109
6
27 25
c. 5844.237 c. 36.9
7. a. 4.16 1016m
6
b. Same as in part (a)
b. 37.6
1
8. a. 0.085, 0.58, 0.85 b. 780.9999, 781.345, 781.354 c. 4.09, 4.099, 4.9, 4.99 9. a. 59 6
3. a. 2562.274
b. Same as in part (a)
9. a. 5.2 10
b. 8.1 103
2
10. a. 1.0066 10
15
c. 4.1 102
b. 1.28 10
14
11. a. 1.286 109, 3.5 107
b. About 36.7 times greater
12. a. Approximately 4.5 10 hours b. Approximately 514 years c. Approximately 4.6 104 km/hr. 6
13. a. 0.7
b. 0.4712
14. a. 0.317417417417 16 15. a. 99
43 b. 111
c. 0.18 b. 0.317474747474 c. 359 495
c. 0.317444444444
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Answers
16. Approximately 1600 lightyears 17. Nonterminating; denominator is not divisible by only 2 or 5 after simplification. 18. a. 81 19. 20. 21. 22.
1 b. 16 3 1 a. 9 = 3 3 1 a. 99 = 33 3 1 a. 999 = 333 10 a. (i) 23 99 , (ii) 999 ,
c. 1/58 b. 59 5 b. 99 5 b. 999 (iii) 769 999 ,
c. c. c.
(iv) 99 = 1, (v)
d. 289 d. 37 99 19 d. 999 57 19 99 = 33 ,
13. 1612 days
e. 6 e. 64 99
f. 597 99
e. 827 999
f.3217 999
(vi) 1827 9999 =
203 1111
23. a. 1/99999 = 0.00001 b. x/99999 where 1 x 99998, where digits in x are not all the same 24. a. 2
b. 6
c. 0
26. 0.94376 艐 364 365 363 365 and so on 28. $31,250 29. $8.91 31. $82.64 32. 20.32 cm by 25.4 cm 33. Approximately $100 34. 2.4liter 4cylinder is 0.6 liter per cylinder. 3.5liter V6 is 0.583 liter per cylinder. 4.9liter V8 is 0.6125 liter per cylinder. 6.8 liter V10 is 0.68 liter per cylinder.
17. 4.8 pounds 18. 24,530 miles 19. About 1613 feet 21. a. 30 teachers b. $942.86 c. $1650
24. 120 miles away 25. 1, 2, 4, 8, 16, 32, 2816 cents
35. 14 moves 468 50 50 36. 50 * equals one half. This is correct = * 468, and 100 100 100 due to the commutative and associative properties of multiplication. 500 8.52 can be changed to 0.5 times 8520 which equals 4260.
Section 7.3A
5 b. 11
1. a. 56
c. Cannot be determined
d. 15
2. Each is an ordered pair of numbers. For example, a. Measures efficiency of an engine b. Measures pay per time c. Currency conversion rate d. Currency conversion rate b.
2 5
c. or 5
b. Yes b. 18
c. 8
d. 30
7. a. 0.64
b. 8.4
c. 0.93
d. 1.71
e. 2.17
36¢ 18 oz 18 oz 21 oz 42¢ 21 oz 8. = , = , = 42¢ 21 oz 36¢ 42¢ 36¢ 18 oz 9. a. 242 484 968 19216 b. 13.501 272 816 c. 30012 1004 2008 d. 2015 43 1612 e. 328 164 4812
27. 81 dimes 28. $17.50 29. 119¢ (50, 25, 10, 10, 10, 10, 1, 1, 1, 1); 219¢ if a silver dollar is used. 30. Yes; the first player always wins by going to a number with ones digit one. 31. Start both timers at the same time. Start cooking the object when the 7minute timer runs out—there will be 4 minutes left on the 11minute timer. When the 11minute timer runs out, turn it over to complete the 15 minutes. 33. There are other ratios equal to 54 that might also represent the number of students in the class, for example, 108 or 1512. Those equivalent ratios would represent 10 boys and 15 boys respectively. If this were a college class with 72 students, there would be 40 “boys” and 32 “girls.”
5 1
6. a. 20
26. Eric: .333 Morgan: .333
32. 87 * 11 = 1247 seconds
5. 3.8, yes
10. 62.5 mph
16. About 3023 years
23. a. Approximately 416,666,667 b. Approximately 6,944,444 c. Approximately 115,741 d. About 12 noon e. About 22.5 seconds before midnight 1 f. Less than 10 of a second before midnight (0.0864 second before midnight)
30. 32
4. a. No
15. 1024 ounces
22. a. 2.48 AU b. 93,000,000 miles or 9.3 107 miles c. 2.31 108 miles
27. $7.93
3.
14. 15 peaches
20. About 29⬙
d. 7
25. Disregarding the first two digits to the right of the decimal point in 1 the decimal expansion of 71 , they are the same.
a. 41
11. a. 17 cents for 15 ounces b. 29 ounces for 13 cents c. 73 ounces for 96 cents 12. 40 ounces
d. 1/217 7 9 7 99 7 999
A33
f. 0.11
34. If you scale down the 180 miles in 4 hours to 45 miles in 1 hour, then scale up, that would mean 450 miles in 10 hours. But that would be the total amount; Melvina shouldn’t add that to the 180 miles she had initially.
Section 7.4A 1. 1/2, 0.5; 7/20, 35%; 0.25, 25%; 0.125, 12.5%; 1/80, 1.25%; 5/4, 1.25; 3/4, 75% 2. a. (i) 5, (ii) 6.87, (iii) 0.458, (iv) 3290 b. (i) 12, (ii) 0.93, (iii) 600, (iv) 3.128 c. (i) 13.5, (ii) 7560, (iii) 1.08, (iv) 0.0099
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Answers
3. a. 252
b. 144
c. 231
d. 195
e. 40
f. 80
4. a. 56
b. 76
c. 68
d. 37.5
e. 150
f. 13313
5. a. 32
b. 37
c. 183
d. 70
e. 122
f. 270
6. a. 40% 70 28 c. 125% * 60 = 54 * 60 = 75 e. 20% 70 14 g. 1% 60 0.6 7. a. $4.70
b. $2.60
e. 5
13 27
(i)
b. 60% 30 18 d. 50% 200 100 f. 10% 300 30 h. 400% 180 4 180 720
c. $13.50
d. $6.50
1
12 3
8. a. 56% (i) How many squares are equal to 42?
9. a. 33.6 f. 123.5
75 x 42 = (ii) 75 100
(iii) 42 x 䡠 75
12
225 = (ii) 100
(iii) 2.25 # x = 12
x b. 2.7 g. 33.3%
c. 82.4 h. 213.3
10. a. 56.7 d. 350
b. 115.5 e. 2850
c. 375.6 f. 15,680
11. a. 129 d. 62.5%
b. 30.87 e. $5.55
c. 187.5 f. $14
12. a. $20.90
b. 163.88 (i)
1 3
b. $7.76
1 3 d. 48.7% i. 0.5
c. $494.92
e. 55.5% j. 0.1
d. $249.91
13. 8.5% 14. 1946 students 15. a. $5120.76
b. About $53.05 more
16. (100)(1.0004839)15 100.73, about 73 cents 17. 5.7%; 94.3% 18. a. About $5.72 1010
964 17 x (ii) = 100 964 c. 423
b. 55.3%
19. a. 72.5 quadrillion BTU b. Nuclear, 10.6%; crude oil, 17.2%; natural gas, 30.1%; renewables 9.9%; coal, 32.1%. Percentages don’t add up to 100% due to rounding.
(iii) 0.17 䡠 964 = x
9 37
20. The discount is 13% or the sales price should be $97.75. 21. $5000
(i)
22. a. 250
b. 480
c. 15
23. $4875 24. a. About 16.3% c. 3.64 1014 square meters
25. No; 3 grams is 4% of 75 grams, but 7 grams is 15% of 46.6 grams, giving different U.S. RDA of protein.
156.6 (ii)
b. About 34.8% d. 0.44% or about 11 25 of 1%
156.67 37 = x 100
26. 31.68 inches
(iii) 0.37 䡠 x = 156.6
27. The result is the same
d. approximately 71%
28. 32% 29. 119 to 136
(i)
How many squares are equal to
30. If the competition had x outputs, then x 0.4x 6, or 1.4x 6. There is no whole number for x, therefore, the competition couldn’t have had x outputs. 31. 10% 5% 15% off, whereas 10% off, then 5% off is equivalent to finding 90% 95% 85.5%, or 14.5% off. Conclusion: Add the percents.
1
12 3 3 8 4 x (ii) = 1 100 12 3
834 ?
32. An increase of about 4.9% (iii) 8
3 1 = x # 12 4 3
33. The player who is faced with 3 petals loses. Reasoning backward, so is the one faced with 6, since whatever she takes, the opponent can force her to 3. The same for 9. Thus the first player will lose. The key to this game is to leave the opponent on a multiple 3.
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Answers 34. Let P be the price. Option (i) is P ⫻ 80% ⫻ 106%, whereas (ii) is P ⫻ 106% ⫻ 80%. By commutativity, they are equal. 35. $14,751.06 36. $59,000
Section 7.2 1. a. 21.009 b. 36.489 2. a. 0.384615 b. 0.396 3. a. 3674 b. 23,891 999 990
c. 153.55 c. 21
d. 36.9
Section 7.3
37. $9052.13
1. A ratio is an ordered pair, and a proportion is a statement saying that two ratios are equal.
4.7 L 2.72% 38. 172.8 39. $50,000. If you make a table, you can see that you will keep more of your money up to $50,000, namely 50% of $50,000, or $25,000. However, at $51,000 you keep only 49% of $51,000, or $24,990, and it goes down from there until you keep $0 at $100,000! Observe that there is symmetry around $50,000. 40. $9.25
7 2. a. No, since 13 Z 35 .
b. Yes, since 12 ⫻ 25 ⫽ 15 ⫻ 20.
3. ab = dc if and only if (i) ad ⫽ bc or (ii) ba and dc and are equivalent to the same fraction. 4. a. 58 24 6 b.
3.45 7
47 16 , so 58¢ for 24 oz is the better buy. 7 5.11 11 , so $5.11 for 11 pounds is the better
buy.
5. 413 cups.
41. 35 or 64 years old.
Section 7.4
42. 34, 36, 44, 54, 76, 146 43. Pilot should fly when there is no wind. 44. The student is correct. If the sale was for 60% off, then the amount you are paying for the shirt is 40% off the original price. So the original price is 27.88 ⫼ 0.40. But the amount saved is 60% of the 27.88 original price. That is, 0.60 ⫻ = $41.82 is the amount saved. 0.40 45. The two percents are being taken on two different bases. The first 20% is based on $17,888, the cost of the GM. 20% of $17,888 equals $3,577.60, which means the car was listed at $21,465.60. The second 20% is based on that list price. 20% of $21,465.60 is $4.293.12. This is more than $3,577.50. If the dealer takes that off the list price, he will be selling the car for $17,172.48. This is below cost—a good deal for you!
PROBLEMS WHERE THE STRATEGY “WORK BACKWARD” IS USEFUL 1. Working backward, we have 87 ⫺ 59 ⫹ 18 ⫽ 46. So they must have gone 46 floors the first time.
56 100 ( = 48 100 ( =
1. a. 56% = 0.56 = b. 0.48 = 48% =
14 25 ) 12 25 )
c. 18 = 0.125 = 12.5% 1 4
2. a. 48 * c. 34
b. 13 * 72 = 24
= 12
d. 15 * 55 = 11
* 72 = 54
3. a. 25% ⫻ 80 ⫽ 20 c. 3313 % * 60 = 20 4. a. $16,000
b. 50% ⫻ 200 ⫽ 100 d. 6623 % * 300 = 200
b. 59%
Chapter 7 Test 1. a. F
b. T
c. T
d. F
2. a. 3 # 101 + 2 # 1 + 1 # ¢ b. 3 # ¢
1 104
≤ + 4# ¢
1 105
1 101
e. F
≤ + 9¢
≤ + 2# ¢
f. T
1 102
1 106
g. T
≤ + 8¢
1 103
≤
≤
2. If she ended up with 473 cards, she brought 473 ⫺ 9 ⫹ 2 ⫺ 4 ⫺ 2 ⫹ 7 ⫺ 5 ⫹ 3 ⫽ 465.
3. Hundred
3. Working backward, {[(13 䡠 4 ⫹ 6)3 ⫼ 2] ⫺ 9} ⫼ 6 ⫽ 13 is the original number.
4. a. The ratio of red to green is 9:14, or the ratio of green to red is 14 9. b. 923 or 1423 5. a. 17.519
CHAPTER REVIEW Section 7.1
b. 6.339
c. 83.293
d. 500
400 1000 and 0.1 ⬍ 0.4; therefore, 0.103 ⬍ 0.4 997 1000 10,000 6 10,000 and 0.09 ⬍ 0.10; therefore, 0.0997
103 6. a. 1000 6
1. 3 ⫻ 10 ⫹ 7 ⫹ 1 ⫻ (1/10) ⫹ 4 ⫻ (1/100) ⫹ 9 ⫻ (1/1000)
b.
2. Two and three thousand seven hundred ninetyeight ten thousandths
7. a. 0.285714
3. (a) and (c)
8. a. Terminating b. Nonterminating c. Terminating
4. a. Shade 24 small squares and 3 strips of ten squares. Thus, 0.24 ⬍ 0.3. b. 0.24 is to the left of 0.3. 24 30 3 6 100 ( = 10 ) c. 100 d. Since 2 ⬍ 3, 0.24 ⬍ 0.3. 5. a. 24.6. Use commutativity and associativity to find 0.25 ⫻ 8 ⫽ 2. b. 9.6. Use commutativity and distributivity to find 2.4(1.3 ⫹ 2.7). c. 18.72. Use compensation to find 15.72 ⫹ 3.00. d. 7.53. Use equal additions to find 27.53 ⫺ 20.00. 6. a. Between 48 and 108 c. 8.5 ⫺ 2.4 ⫽ 6.1
A35
b. 14 d. 400 ⫼ 50 ⫽ 8
4 9. a. 11
10. a. 0.52,
11 b. 30 52 100
b. 0.625
c. 0.14583
⬍ 0.1
d. 0.4
909 c. 2500
b. 125%, 125 100
c. 0.68, 68%
11. 18 12. a. 53 ⫻ 0.48 艐 52 ⫻ 0.5 ⫽ 52 ⫼ 2 ⫽ 26 b. 1469.2 , 26.57 = 14.692 , 0.2657 L 16 , 0.25 = 16 , 14 = 16 * 4 = 64 c. 33 , 0.76 L 33 , 0.75 = 33 *
4 3
d. 442.78 ⫻ 18.7 艐 450 ⫻ 20 ⫽ 9000
= 44
h. T
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Answers
13. 3%, 27 , 0.3, 13
7. a. ⫺14 ⫹ 6 ⫽ [⫺8 ⫹ (⫺6)] ⫹ 6 ⫽ ⫺8 ⫹ (⫺6 ⫹ 6) ⫽ ⫺8 ⫹ 0 ⫽ ⫺8 b. 17 ⫹ (⫺3) ⫽ (14 ⫹ 3) ⫹ (⫺3) ⫽ 14 ⫹ (3 ⫹ (⫺3)) ⫽ 14 ⫹ 0 ⫽ 14
14. 123,456,789 has prime factors other than 2 or 5. 15. 37 , 100 ⫻ 58 is 37% of 58. 16. a. Bernard got 80% of the questions correct on his math test. If he got 48 correct, how many questions were on the test? b. Of his 140 times at bat for the season, Jose got a hit 35 times. What percent of the time did he get a hit? (Answers may vary.)
8. a. Commutative property of addition b. Additive inverse property b. ⫺17
9. a. 635 10. a.
b. BBBB BBBBB
BBB BBBB
17. If we convert 1.3 and 0.2 to fractions before adding, the denominator of the sum is 10 and thus the sum has one digit to the right of the
RRRR
13 2 15 + = = 1.5 ≤ . When multiplying, the decimal. ¢ 10 10 10
11. a. ⫺4
b. 12
denominators are multiplied, giving a denominator of 102 (two digits to
12. a. 26
b. ⫺5
c. 370
13. a. ⫺5
b. 12
c. ⫺78
the right of the decimal) in the product ¢
13 2 26 * = 2 = 0.26 ≤ . 10 10 10
19. $2520 20. 522
b. F
16. a. 5
21. 9.6 inches
c. F
d. 1 d. ⫺128 d. 1
e. 9
f. 8
d. T
24. If the competition has 4 new styles, then 6 new styles is 50% more. If the competition has 5 new styles, then 6 new styles is 20% more. In other words, 6 styles is 40% more than 4.28 styles and 0.28 of a style makes no sense. 25. $870
c. Neither d.
b. BRRBBBBB
f. 2
19. a. Yes; ⫺3 ⫺7 is an integer b. No; 3 ⫺ 2 ⫽ 2 ⫺ 3 c. No; 5 ⫺ (4 ⫺ 1) ⫽ (5 ⫺ 4) ⫺ 1 d. No; 5 ⫺ 0 ⫽ 0 ⫺ 5 20. If a ⫺ b ⫽ c, then a ⫹ (⫺b) ⫽ c. Then a ⫹ (⫺b) ⫹ b ⫽ c ⫹ b, or a ⫽ b ⫹ c.
23. a. (i) 9 ⫺ 4, (ii) 4 ⫺ 9, (iii) 4 ⫺ 4, (iv) 9 ⫺ [(4 ⫺ 4) ⫺ 4], (v) (9 ⫺ 4) ⫺ 4, (vi) [(9 ⫺ 4) ⫺ 4] ⫺ 4 b. All integers c. Any integer that is multiple of 4 d. If GCF(a, b ) ⫽ 1, then all integers; otherwise, just multiples of GCF
0 b. RRRRR
0
0
24. Second: 2, ⫺24; third: 4, ⫺2, ⫺22; bottom: ⫺7
c. BR
25. As long as we have integers, this algorithm is correct. Justification: 72 ⫺ 38 ⫽ (70 ⫹ 2) ⫺ (30 ⫹ 8) ⫽ (70 ⫺ 30) ⫹ (2 ⫺ 8) ⫽ 40 ⫹ (⫺6) ⫽ 34.
0 b. 4
e. ⫺2
22. First row: 7, ⫺14, 1; second row: ⫺8, ⫺2, 4; third row: ⫺5, 10, ⫺11
0 3. a. BBB
d. ⫺2
21. a. (i) When a and b have the same sign or when one or both are 0, (ii) when a and b are nonzero and have opposite signs, (iii) never, (iv) all integers will work. b. Only condition (iv)
Section 8.1A 2. a. RRRRB c.
c. 2
18. $64
23. 28
1. All are integers a. Positive b. Negative
b. 17
17. Top: 5; second: ⫺1, 6; third: ⫺2, 1
22. 5.37501, 5.37502, 5.37503 (Answers may vary.)
5. a. I
c. 1
15. a. Five minus two b. Negative six or opposite of six (both equivalent) c. Negative three or opposite of three
18. 7
4. a. ⫺3
14. a. T
=9
RRRRR
=_4
c. 0
d. 168
e. ⫺56
b. { . . . , ⫺4, ⫺3, ⫺2, ⫺1, 1, 2, 3, 4, . . .}
26. f. 1235 c. ⭋
14
6. a.
18
0 1 2 3
4 5
or BBBBBRRR = BB; 2.
4 10 7
b.
1
_5 _4 _3 _2 _1 0 or RRR RR = _ 5.
9
15 8
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Answers 22. a. ⬍
27.
–3 7
–1
–4
6
24. a. 10 and 8 25. a. ?– b. (i) ? ?
–2
1
0
2 –5
4
Section 8.2A 1. a. 2 2 2 2 8 or 4 2 8 b. (3) (3) (3) 9 or 3 (3) 9 c. (1) (1) (1) (1) (1) (1) 6 or 6 (1) 6
c. 15
d. 39
36. Assume ab 0 and b ⫽ 0. Then ab 0 䡠 b. Since ac bc and c ⫽ 0 implies that a b, we can cancel the b’s in ab 0 䡠 b. Hence a 0. Similarly, if we assume a ⫽ 0.
b. 86 c. 15 b. 1938 e. 156
c. 97, 920 f. 1489
10. Yes to all parts b. 27
11. a. 16
c. 16
d. 25
e. 243
f. 64
12. Positive: (c), (d); negative: (e) 1 13. a. 100
1 b. 64
1 d. 125
1 c. 64
1 # 6 1 1 1 b. 426 44 c. 4 # 2 = 6 = 56 4 = 44 42 5 5 5 1/32 1 1 15. a. 5 = 7 b. 325 = 37 = 7 c. 610, 610 3 3 3 16. a. 33 27 b. 6 c. 38 6561 14. a.
17. a. 0.000037
b. 0.0000000245
18. a. 4 104
b. 1.6 106
19. a. 7.22 1025 37
d. 4 10
20. a. 3 is left of 2.
d. Yes d. Yes
c. 4.95 1010
b. 8.28 1026
c. 2.5 1010
e. 8 10
f. 2.05 107
14
b. 6 is left of 2.
21. a. 5, 2, 0, 2, 5 c. 11, 8, 5, 3, 2
29. First row: 2, 9, 12; second row: 36, 6, 1; third row: 3, 4, 18 30. x ⬍ y means y x p for some p ⬎ 0, y 2 (x p)2 x2 2xp p2. Since x ⬎ 0 and p ⬎ 0, 2xp p2 ⬎ 0. Therefore, x2 ⬍ y 2.
35. 30 cents
6. Adding the opposite approach to subtraction; distributivity of multiplication over addition; (a)b (ab); adding opposite approach; distributivity of multiplication over subtraction. b. 5
28. Put the amounts on a number line, where positive numbers represent assets and negative numbers represent liabilities. Clearly, 10 ⬍ 5.
34. True. Every whole number can be expressed in the form 3n, 3n 1, or 3n 2. If these three forms are squared, the squares will be of the form 3m or 3m 1.
5. Distributivity of multiplication over addition; additive inverse; multiplication by 0
9. a. 2592 d. 47
27. This is correct, by a(1) a.
33. 100 sheep, 0 cows, and 0 rabbits or 1 sheep, 19 cows, and 80 rabbits
RRRRRRRRRRRR
8. a. 6
(iii)
32. a. 1.11 102 b. About 2.33 108 seconds, or 7.4 years
4. a. RR RR RR 6 b. BBBBBBBBBBBB = 12
7. a. 3
(ii)
c. No
31. 1.99 1023 grams per atom of carbon
2. a. (i) 6 (1) 6, 6 (2) 12, 6 (3) 18; (ii) 9 (1) 9, 9 (2) 18, 9 (3) 27 b. Positive times negative equals negative. b. 32
b. 8
26. a. (i) When x is negative, (ii) when x is nonnegative (zero or positive), (iii) never, (iv) all integers b. Only (iv)
3
28. If you have 3 black chips in the circle and you need to subtract 8, there aren’t enough black chips there. You are always allowed to add a “neutral” set of chips (zero) to the circle, that is, a pair consisting of one black and one red. By adding 5 black and 5 red you can subtract 8 black chips, leaving 5 red chips, or 5.
3. a. 30
b. ⬎
23. a. 43,200, 240, 180, 12, 5, 3
–6 5
A37
c. 12 is the left of 3.
b. 8, 6, 5, 3, 12 d. 108, 72, 36, 23, 45
37. The student’s assumption that xyz is negative is a problem. The answer may be positive, negative, or zero depending on x, y, and z. The negative sign in front of the xyz should be read “opposite” to try to eliminate confusion. 38. It is true that the Zero Divisors Property says that when ab 0 then a 0 or b 0. In mathematics that means any of three situations: (1) a 0 and b ⫽ 0, (2) a ⫽ 0 and b 0, or (3) a and b both equal 0. However, when we say 3 0 or x 0, we can use the fact that we know 3 ⫽ 0 to conclude that we have situation (2), so x must be 0.
PROBLEMS WHERE THE STRATEGY “USE CASES” IS USEFUL 1. Case 1: odd even odd even. Case 2: even odd even odd. Since only Case 1 has an even sum, two of the numbers must be odd. 2. m2 n2 is positive when m2 ⬎ n2. Case 1: m ⬎ 0, n ⬎ 0. Here m must be greater than n. Case 2: m ⬎ 0, n ⬍ 0. Here m ⬎ n. Case 3: m ⬍ 0, n ⬎ 0. Here m ⬎ n. Case 4: m ⬍ 0, n ⬍ 0. Here m ⬍ n. 3. Case 1: If n 5m, then n2 25m2 and hence is a multiple of 5. Case 2: If n 5m 1, then n2 25m2 10m 1, which is one more than a multiple of 5. Case 3: If n 5m 2, then n2 25m2 20m 4, which is 4 more than (hence one less than) a multiple of 5. Case 4: If n 5m 3, then n2 25m2 30m 9, which is one less than a multiple of 5. Case 5: If n 5m 4, then n2 25m2 40m 16, which is one more than a multiple of 5.
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Answers
CHAPTER REVIEW Section 8.1 1. a. Use black chips for positive integers and red chips for negative integers. b. Arrows representing positive integers point to the right, and arrows representing negative integers point to the left.
7. a. 8.2 1012
b. 6 106
8. n 4 9. a. Associativity b. Associativity and commutativity c. Distributivity d. Commutativity and distributivity
2. a. BBBBBBBRRRR BBB b.
01234567
10. a. (i) BBBBBBBBBBBBBRRRRR = 13 (ii) 8 (5) 8 5 13 (iii) 8 (5) c if and only if 8 c (5); c 13.
3. a. Identity b. Inverse c. Commutativity d. Associativity e. Closure
b. (i) RR : BBBBBRRRRRRR = 5 (ii) (2) (7) 2 7 5 (iii) (2) (7) c if and only if 2 c (7); c 5.
4. a. 3 (2) 5
BBBBB RR
6. a. 3(4 2) 3(2) 6, 3(4) 3(2) 12 6 6 b. 3[5 (2)] 3(7) 21, (3)(5) (3)(2) 15 6 21
b. 3 (2) 3 2 5 c. 3 (2) n if and only if 3 (2) n; therefore, n 5. 5. (a) only
11. a. Negative b. Negative c. Positive d. Positive 12. a.
–3 8
Section 8.2 1. a. (2) (2) (2) (2) (2) 10 b. (5)2 10, (5)1 5, (5)0 0, (5)(1) 5, (5)(2) 10 2. a. Commutativity d. Identity
b. Associativity e. Cancellation
–2 –1 0 1 2 3 4 5 6 7 8 9
c. Closure
3. Let a 3 and b 4.
=
4. n 0; zero divisors 5. a b c if and only if a bc. 6. None 7. a. Positive—even number of negative numbers b. Negative—odd number of negative numbers c. 0—zero is a factor
b.
8. 73 7 7 7, 72 7 7, 71 7, 70 1, 71 = 17 , etc. 9. a. 7.9 105 d. 23,900,000
b. 0.0003
10. a. 21 is to the left of 17
–4 –3 –2 –1 0 1 2 3 4
c. 4.58127 102 b. 21 4 17
11. a. ⬎: Property of less than and multiplication by a negative b. ⬍: Transitivity c. ⬍: Property of less than and multiplication by a positive d. ⬍: Property of less than and addition
2. an
b. F f. F
=
c.
Chapter 8 Test 1. a. T e. F
4 –2
–5 3
c. F g. F
d. T h. T
–4 –3 –2 –1 0 1 2 3 4
1 = n a
3. (b) and (c) 4. Takeaway, missingaddend, addtheopposite 5. a. 6 e. 30
b. 42 f. 3
c. 48 g. 52
d. 8 h. 12
=
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Answers
13. (i) takeaway
takeaway positive add 3 zero pairs three
b.
6.
b. b.
8. a. 23
b. 2
10. 11. 12. (ii) missingaddend What needs to be added to a set of 3 to get 2?
13. 14. 15.
a. 14 b. 1835 27 a. 1711 b. 143 4 a. 3 b. 61 24 a. 143 b. 1123 245  455 a. 3456 b. 1032 703 a. 119 6 113 b. c. 109 6 65 d. 152 231 a. 201 6 356
16. a. I, N, Q 14. a. 3 4 12 248 144 040 1 4 4 2 4 8
17.
b. 120, 60
9 _12 _6
0
c. 900, 3600
18.
_256
12
d. 1, 2
19. a b b a is the same as a b (a b). The only number that is equal to its opposite is 0, so a b 0 which means a b. 20. a. 20
b. 18
Section 9.1A
2. 3.
c. 6
2 3 where 2, 3 are integers b. 631 where 31, 6 are integers c. 10 1 where 10, 1 are integers 1 a. 9 b. 34 c. 41 d. 81 3 3 3 3 1 ,  1 , 1 ,  1
1. a.
4. (a) and (b)
 31 36 1 4 2 7 7 5
c. c.
d. d. d.
2 5 6 910  500  761 345 6 532
1 3 9 8
6
b.
c. W, F, I, N, Q
d. I, Q
e. W, F, I, Q
4 3
19. a. x 6
3 2
b. x 6
20. a. x 7
5 4
b. x 7 
21. a. 90,687 22. a.
 43 88  59 97
6 6
1 12
b. x 6
3 4 3 2
b. 77
 37 76 ,  68 113 ,
 80  20 164 = 41  127 210
 6  11 7 , 13 ,
and
 13 15
a c 25. a. , are rational numbers so b and d are not zero (definition of b d a c ac rational numbers), # = (definition of multiplication), b d bd ac and bd are integers (closure of integer multiplication), bd ⫽ 0 ac (zero divisors property); therefore, is a rational number. bd Similar types of arguments hold for parts (b) to (e).
2 _64
_4
d.
5 18 31 21 9 20
24. a/b an/bn if and only if a(bn) b(an). The last equation is true due to associativity and commutativity of integer multiplication.
_9
_16 512
c. 75
23. There are many correct answers. a. For example, 32 , 75 , and 53
6
128
4 5
18. a. x 6
b.
3
32
8
d.
c.
b. Q
b. For example,
_3
c. 52
17. a. Associative—addition b. Commutative—multiplication c. Distributive—multiplication over addition d. Property of less than and addition
b. 2 4 8 2 3 6 2 2 4 2 1 2 2 0 0 2 1 2 2 2 4 2 3 6 2 4 8
15. No; let a 2, b 3, c 4, then a (b 䡠 c) 24, but a 䡠 b a 䡠 c 48. 16. a. 30, 30
3 5 135  53
7. a. 2 9.
leaves –5
5 7 a. 258
5. a.
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26. a. a/b c/d e/f b. If a/b c/d e/f, then a/b (c/d) e/f. Add c/d to both sides. Then a/b c/d e/f. Also, if a/b c/d e/f, add c/d to both sides. Then a/b (c/d) e/f or a/b c/d e/f. c. If a/b c/d e/f, then a/b c/d e/f. Adding c/d to both sides will yield a/b (c/d) e/f. Hence, a/b c/d a/b (c/d). a (cf + de) a c a cf + de e = 27. ¢ + ≤ = ¢ ≤ = b d f b df bdf acf + ade
acf
ade ac ae + = + = bdf bdf bdf bd bf a e a#c + # , using addition and multiplication of rational numbers b d b f and distributivity of integers =
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Answers
28. If a/b ⬍ c/d, then a/b p/q c/d for some positive p/q. Therefore, a/b p/q e/f c/d e/f , or a/b e/f p/q c/d e/f for positive p/q. Thus a/b e/f ⬍ c/d e/f.
7. a. 19
a a a a 29. ¢  ≤ + B  ¢  ≤ R = ¢  ≤ + . Therefore, by additive b b b b
9. a. 2 3, 6 b. 2 5, 10 c. 3 4, 12 d. 3 5, 15 e. 2a * 2b = 2a * b
a a cancellation,  ¢  ≤ = . b b 30. Start both timers. When the 5minute timer expires, start it again. When the 8minute timer expires, start it again; the 5minute timer will have 2 minutes left on it. When the 5minute timer expires, start measuring, since the 8minute timer will have 6 minutes left. 31. Both are correct since 1 is the greatest factor of the numerators and denominators. Maria’s answer may be more useful within a mathematical computation. However, Karl’s answer, the mixed number, is more useful in everyday commerce. 32. Maria and Billy have the same answer, but in a different form. Karl’s answer equals 3/4, hence it is not equal to theirs.
b. 27
8. a. Distributive of multiplication over addition b. Yes; 8 c. No, the numbers under the radical are not the same.
4 10. a. , 2 2
6 b. , 3 2
c.
16 ,8 8
d.
21 ,3 7
e.
2a 2b
12. There are many correct answers. One is 0.37414243 . . . 13. For example, 210, 211, 212, 3.060060006 . . . 14. a. 2.65
b. 3.95
c. 0.19
15. For example, if r1 3.5, then r3 3.6055516 and s3 3.6055509. Therefore 213 L 3.60555. 17. a. 0.3 ⬍ 0.5477225 b. 0.5 ⬍ 0.7071067 Square root is larger than number.
1. a. Irrational b. Rational c. Irrational d. Rational e. Irrational f. Irrational g. Rational h. Rational
18. a. 5
c. 243
b. Not real
d. 81
e. 8
c. 2
20. a. 25 b. 3.804 (rounded) c. 7,547,104.282 (rounded) d. 9,018,968.484 (rounded) 21. a.
3. a. 25
22 7
b. 2 3 37
22. a.
b. 28 = 2 22
X X
c. 220 = 2 25 4. a. 22
b. 2
19. a. 3
2. No; if it did, would be a rational number. This is an approximation to .
b. 23
c.
1
1 1 1
1 1 1 1
1 1 1
1 1 1 1
1 1 1
1 2x + 3 = 7
1 1 √3 1
√2
√4
X X
√5 √6
1
5. a. 234
1
1 1 1
√7
2x + 3 – 3 = 7 – 3 2x = 4 X
1 1
X
b. 220 = 2 25 c. 6 6. a. 423 b. 327 c. 922
a Ab
11. 0.56, 0.565565556 . . . , 0.565566555666 . . . , 0.56, 0.566, 0.56656665 . . . , 0.566
16. The numbers decrease in size, 1.
Section 9.2A
=
2x = __ 4 ___ 2 2 x=2
1 1
1 f. 125
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b.
X X X
1 1 1 1
30. a. 1 + 23 = a/b so 23 = (a  b)/b, which is a rational number, and this is a contradiction because 23 is an irrational number. b. Argue as in part (a); assume that the number is rational and solve for 23.
1 1 1 1 1 1 1 1
X
31. a. Apply 29(b).
1 1 1 1
b. Apply 30(b).
c. Apply 30(b).
32. 2a + 2b Z 2a + b except when a 0 or b 0. There is no consistent analogy between multiplication and addition. 33. 2a # 2b = 2ab is true for all a and b, where a 0 and b 0.
3x + 4 = x + 8
X X X
A41
34. {(3n, 4n, 5n) 兩 is a nonzero whole number} is an infinite set of Pythagorean triples.
1 1 1 1 1 1 1 1
X
35. For example, if u 2, v 1, then a 4, b 3, c 5. If u 3, v 2, then a 12, b 5, c 13. u 8, v 3, (18, 55, 73) u 8, v 5, (39, 80, 89) u 8, v 7, (15, 112, 113) 36. 3, 4, 5; 1, 2, 3; 2, 3, 4; 1, 0, 1 37. Yes; 1 or 1
3x – x + 4 = x – x + 8 2x + 4 = 8
38. Cut from the longer wire a piece that is 13 the sum of the lengths of the original pieces. 1 1 1 1 1 1 1 1
1 1 1 1
X X
39. Mr. Milne 40. 20 and 64 41. It is true that 5 and 5 are both square roots of 25. However, the symbol “ 2 ” represents only the positive square root by definition. Thus, 225 = 5.
Section 9.3A 1.
2x + 4 – 4 = 8 – 4 2x = 4
4
f. (0,3) X
1 1
X
1 1
b. (–3,2)
a. (3,2)
2
d. (3,0) 2x = __ 4 ___ 2 2 x=2
–5
5 –2
e. (–3,–2) 23. a. 8 24. a. 9
b. 5 b. 5
11 25. a. x 7 3
c.
5 9
c. 5 b. x 4
–4
d. 322 d.
1 2
e.
9 c. x 7 5
c. (3,–2)
7 6
f. 2 d. x …
a. 1st quadrant c. 4th quadrant e. 3rd quadrant
22 27
26. Let 23 = a/b. Then 3 a2/b2 or a2 3b2. Count prime factors. 27. When you get to the step a2 9 䡠 b2, this can be written as a2 32 䡠 b2. Thus both sides have an even number of prime factors and no contradiction arises. 28. Assume not. Then (a/b)3 2 for some rational a/b. Count prime factors. 29. a. By closure of realnumber multiplication, 5 23 is a real number, and thus a rational or an irrational number. Assume that it is rational, say m = 5 23. Since m/5 is rational and 23 is irrational, we have a contradiction. Therefore, 5 23 must be irrational. b. Argue as in part (a); replace 5 with any nonzero rational and 23 with any irrational.
2. a. III and IV 3.
b. 2nd quadrant d. xaxis f. yaxis b. IV
c. III
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b.
4. a. (i) 3, (ii) 4, (iii) 0 b. Domain {x 兩 3 x 6} Range {y 兩 4 y 3} c. 0, 2, 5 5. a. d(4) 2.4 miles, d(5.5) 艐 2.81 miles b. Approximately 2.16 miles c. Domain is the set of all nonnegative real numbers. Range is the set of all nonnegative real numbers up to the farthest number of miles one can see. 6. a.
x
f (x )
2
1
3
1
1
2
0
3
1
5
2
7
–3
–2
–1
2
1 9
1
1 3
0
1
1
3
2
9
3 2 1 –3
3
–2
x
m (x )
2
50
1
45
0
40
1
35
2
30
x
s(x)
2
10
1
3
0
2
3 2 –3
–2
–1
1
1
–1
2
6
–2
x
g (x )
2
18.9
1
11.7
0
4.5
1
2.7
2
9.9
4 3 2
30
1
20
–30
c.
–20
–10
20
1
–4 –3 –2 –1
30
2
3
4
2
3
4
–1 –2
–10
–3
–20
–4
–30
–5
(ii)
20
4 3 2
10
1
5 10 15 20 –20
–10
1
–4 –3 –2 –1
–5
–1
–10
–2 –3
–20
–4
(iii) 7. a.
x
h (x )
2
0
3
1
0.5
2
4 3
4
0
0
1
1.5
2
4
1
–3
10
1 –4
–3
–2
–1
2 1
1
2
3
3
–1
8. a. (i)
10
2
–1
–2
–3
b.
1
–3
1 1
r (x )
–2
c.
–1
x 2
–3 –2 –1
1 –1 –2
–1
–3
–2
–4
–3
–5
2
3
5
6
1
2
3
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Answers
(iv)
(ii)
2 1 –3 –2 –1
1
2
3
4
5
6
A43
10 9
7
8
–1
7
–2
6
–3
5
–4
4 3 2
b. The larger the coefficient of x, the greater the slope or steeper the slant. c. It changes the slant from lower left to upper right to a slant from upper left to lower right.
1 –4 –3 –2 –1
4
2
3
4
4
5
–1
5
2
–4 –3 –2 –1
3
6
3 1
2
–2
(iii)
(i)
1
4 1
2
3
3 2
–1
1
–2 –3
1
–4 –3 –2 –1
–1
–4
–2
–5 –6
(iv)
9 8 7
(ii)
3
6
2
5
1 –14–13–12–11–10 –9 –8 –7 –6 –5 –4 –3 –2 –1
4 1
2
3
–1
2
–2
1
–3
–2 –1
–4
1
2
3
–1 –2
9. a. The line gets closer to being vertical. b. The line is horizontal. c. The line slants from upper left to lower right.
(v)
7 6 5 4
10. a. (i)
9
3
8
2
7
1
6
–4 –3 –2 –1
5
1 –1 –2
4 3 2 1 –4 –3 –2 –1
1 –1 –2
2
3
4
b. (ii) shifts the graph of (i) two units up, (iii) shifts the graph of (i) two units down, (iv) shifts the graph of (i) two units to the right, (v) shifts the graph of (i) two units to the left. c. The graph of f (x) x2 4 should be the same as the graph in (i) except it is shifted up 4 units. The graph of f (x) (x 3)2 should be the same as the graph in (i) except it is shifted 3 units to the right.
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Answers
11. a. Changing b has the effect of moving the parabola to the left or right. b. Changing c has the effect of moving the parabola up or down.
c.
Cost in cents
bansw01.qxd
12. a. (i)
175 150 125 100 75 50 25 1 2 3 4 5 6 Ounces Graph for first class mail up to 6 oz.
(ii)
d. 20 39 cents $7.80 for 20 pieces or 20 * 15. a. (i) 2, (ii) 7, (iii) 5, (iv) 0 b. 3
2 –3 –2 –1 1 (iii)
–11 –2 –3
2
3
16. a.
5 4 3 2 1
(iv)
1
2
3
4
b. b. When the base is greater than 1, the larger the base, the steeper the rise of its graph from left to right, especially in the first quadrant. When the base is between 0 and 1, the closer to zero, the steeper the fall of its graph from left to right, especially in the second quadrant. c. x x
f(x) = 10
3 2 1 1
–1
2
3
f(x) = (0.95)
c.
5 4 3 2 1 13. a. The right part comes closer to the yaxis and the left part gets closer to the xaxis b. It is a horizontal line. c. The graph is decreasing from left to right instead of increasing. 14. a. (i) P(0.5) $0.39 (ii) P(5.5) $1.59 (iii) P(11.9) $3.03 (iv) P(12.1) $3.27 b. Domain: 0 oz. ⬍ w ⬍ 13 oz Range {39¢, 63¢, 87¢, $1.11, $1.35, $1.59, $1.83, $2.07, $2.31, $2.55, $2.79, $3.03, $3.27}
1
2
3
4
5
d. 18
15 12 9 6 3 1
2
3
4
5
6
3 4
= 15 oz is $3.75.
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17. a. Exponential
b. Quadratic
c. Cubic
18. a. Domain {x 兩 2 x 3} Range { y 兩 1 y 3} b. Not a function c. Not a function d. Domain is the set of all real numbers. Range is the set of all positive real numbers. 19. a. The length of the shadow varies as time passes. Exponential. b. L(5) 150, L(8) 1100, L(2.5) 50 c. 4.5, 6 d. It is too dark to cast a shadow. 20. a. 135 115 95 75 55 35 15 5 b. Approximately 0.58 second and 3.8 seconds c. In approximately 5.1 seconds d. Approximately 131 feet 21. a.
A45
a are nonzero integers for rationals but are whole b numbers for fractions.
2. The a and b in
3. The restriction that denominators are positive must be stated when dealing with rational numbers. 4. The number  34 is the additive inverse of 43 , and three over four”; however, they are equal. 5. a. T d. T g. T j. T
b. T e. T h. F k. T
3 4
is read “negative
c. F f. T i. F l. T
6. a. Commutativity for addition b. Associativity for multiplication c. Multiplicative identity d. Distributivity e. Associativity for addition f. Multiplicative inverse g. Closure for addition h. Additive inverse i. Additive identity j. Closure for multiplication k. Commutativity for multiplication l. Additive cancellation 5 11 , is to the  35  33 77 6 77 is false. 3 5 2 7 = 11 + 77
7. a. No, since b. c.
left of
3 7 .
d. 33 ⬍ 35 is false.
b. 6.6 billion c. Approximately 29.8 years d. Approximately 50 years
8. a. 32 6 75 ; transitivity b. ⬍; property of less than and multiplication by a positive c. 58 ; property of less than and addition d. ⬍; property of less than and multiplication by a negative e. Between; density property
22. (a) 23. 130 drops 160 off on the top floor and returns. 210 takes the elevator to the top while 130 stays behind. 160 returns and comes up to the top with 130. 24. As b gets larger in a positive direction, the graph near the yaxis looks more like a parabola opening up. Similarly b is negative but as 兩 b 兩 gets larger, the graph near the yaxis looks more like a parabola opening down. 25. She is correct. Whenever the ratio of the change in two x values to the change of the corresponding y values is the same, the graph will be a line (or set of points along a line in the case of a sequence).
PROBLEMS WHERE THE STRATEGY “SOLVE AN EQUATION” IS USEFUL 1. $27,000 2. 840 3. 4
CHAPTER REVIEW Section 9.1 1. Every fraction and every integer is a rational number and the operations on fractions and integers are the same as the corresponding operations on rational numbers.
Section 9.2 1. Every rational number is a real number, and operations on rational numbers as real numbers are the same as rationalnumber operations. a 2. Rational numbers can be expressed in the form , where a and b are b integers, b ⫽ 0; irrational numbers cannot. Also rational numbers have repeating decimal representations, whereas irrational numbers do not. 3. None 4. Completeness. Real numbers fill the entire number line, whereas the rationalnumber line has “holes” where the irrationals are. 5. a. T d. F g. T
b. T e. T h. F
c. T f. F
6. (i) aman amn (ii) ambm (ab)m (iii) (am)n amn (iv) am an amn 7. x =  13 3 in all four cases. 8. a. 73 b. x 6
29 22
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Answers
Section 9.3
3.
1. a. Cubic
100 90 80 70 60 50 40 30 20 10 0
1
2
4
3
b. Exponential
3000
Chapter 9 Test
2500
1. a. F
c. F
d. F
e. T
f. F
g. T
h. F
2. a. i, ii, iii
1500
3. a.
1000
4. a. Commutativity and associativity b. Commutativity, distributivity, and identity for multiplication
500 0
1
2
c. Quadratic
1
–2 –4 –6 d. Linear
4
b. i, ii, iii
 23 21
b.
6 55
c. i, iii
c.
12 r 7
b.
6. a. 729
b. 128
1 c. 243
7. a.
177 98
b.
c.
7 8. 14.1%, , 1.41411411 . . . , 1.41, 22, 1.4142 5
2
3
4
c. 7 25 d. 1 ( 3)(  1) 1 3 = = 1 7( 1) 7
b. 6 23
9. a. 16
#
11. No. For example, 12.
1
1 =
57
(1/57)
1 2
6
1 3 ;
however, 3 ⬎ 2.
= 57
13.
1
2
–1 –1 –2 –3 –4
2x + 3 = 9
3 2x + 3 –3 = 9 – 3 2 1 0
1
2
d. i, iii
3
4
e. i, ii, iii
1 28
5. a. b x ƒ x 7
3 3 10. = 7 7
4 3 2 1 –2
3
14 12 10 8 6 4 2
–3 –2 –1
2.
b. T
2000
2x = 6 x=3
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Answers
i. For increment 5, 73 to 78 and 78 to 83 both have 6. For increment 8, 74 to 82 has 11. For increment 5, the 12 in 73–83 is close to the 11 in 74 to 82 for increment 8.
a a , where is a rational number. Then 8b2 a2. But b b this is impossible, since 8b2 23b2 has an odd number of prime factors, whereas a2 has an even number.
16. Suppose 28 =
17. 105 18. Only when x 0 or when a b 3. a.
21. Any a and b where both a and b are not zero. 22. F(C) 1.8C 32
Section 10.1A
4 3 2 1
e.
70
50 59
1
60 69
4
70 79
12
80 89
8
90 99
5
80
90
05
18.
12555
19.
24
20.
23699
21.
148
22.
001
3 2 1 0
20
30
40
50
60
7 6 5 4 3 2 1
1. a. 58, 63, 65, 67, 69, 70, 72, 72, 72, 74, 74, 76, 76, 76, 76, 78, 78, 80, 80, 80, 82, 85, 85, 86, 88, 92, 92, 93, 95, 98 b. 58, 98 c. 76 d.
60
134
17.
100
4. a.
CLASS 2 876 9775 7755432 522200
5059
20. 0.45455455545555 . . . , 0.45616116111 . . . , 0.46363663666 . . . (Answers may vary.)
89
16.
4049
19. t 3
15.
3039
2.
2029
14. Since 217 is irrational, it has a nonrepeating, nonterminating decimal representation. But 4.12310562 is repeating, so it is rational. Thus, the two numbers cannot be equal. 15. a. Exponential b. Quadratic c. Cubic
CLASS 1 5 6 7 8 9
789 02 25669 0016667 333
f.
14 12 10 8 6 4 2
b. Class 2
80–89 90–99
70–79
50–59 60–69
5. a. Portland b. 4 months; 0 month c. December (6.0 inches); July (0.5 inch) d. August (4.0 inches); January (2.7 inches) e. New York City (40.3 inches) (Portland’s total 37.6 inches) 6. a.
Buick BMW
g. 70–79 h. 6
4 2
Honda Civic Geo 11 9 7 5 3 1
A47
Neon Land Rover 10 20 30 40 50 Fuel consumption
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b. Land Rover; Geo c. Geo; Land Rover d. Buick, $8266.67 BMW, $7971.43 Honda Civic, $6377.14 Geo, $4852.17 Neon, $5873.68 Land Rover, $13,950 e. A histogram could not be used because the categories on the horizontal axis are not numbers that can be broken into different intervals.
Population in millions
7. a.
30 27 24 21 18 15 12 9 6 3 0
1990 2000
9. a. Taxes b. 20.3% c. 64⬚ d. Natural resources e. Social assistance, transportation, health and rehabilitation, and natural resources f. 23⬚, 26⬚ 10. Grants to local governments, $1,269,000,000; salaries and fringe benefits, $1,161,000,000; grants to organizations and individuals, $873,000,000; operating, $621,000,000; other, $576,000,000 11. a. $15,000,000 b. $5,000,000 c. $85,000,000 12. a.
Students 1920 1930 1940
ires s A iro e eno Bu e Jan d Rio tta lcu Ca ay mb Bo ork wY Ne l ou Se aulo y oP Sa o Cit ma ha xic Me /Yoko yo Tok
1950 1960 1970 1980
World's Largest Urban Areas (Source: World Almanac)
1990 2000
c. New York
8 7 6 5 4 3 2 1 0
Each stick figure represents 5,000,000 students b. 1910s, 1920s, 1950s, 1960s, 1980s, 1990s c. Bar or line graph 13. a.
Freshman GPA
Expenditure (%)
b. Sao Paulo 8. a.
1950 1960 1970 1980 1990 2000 Year b.
8 7 6
4 3.5 3 2.5 2
2
5 4
b. No outliers
3
c.
2 1 0 1950 1960 1970 1980 1990 2000 Year
c. The first; the second
Freshman GPA
Expenditure (%)
= 5,000,000
1910
3 High school GPA
4
3 High school GPA
4
4 3.5 3 2.5 2
2
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Answers
19. a. Double bar graph or pictograph for comparing two sets of data.
Speed gain
250 200
b.
150
Projected enrollment (in thousands)
14. a.
100 50 2
4 6 Weeks in program
8
b. No outliers c.
Speed gain
250
40 35 30 25 20 15 10 5 0
Public Private
Elementary Secondary College
200 Type of school
150 100 50 2
4 6 Weeks in program
8
20. a. Circle graph—compare parts of a whole b.
Average score
15. a. and b.
100 95 90 85 80 75 70
Social insurance receipts 39%
1
3 5 Practice time
Individual income taxes 43%
7 Corporate income taxes 10.1%
c. They should look similar.
Excise taxes 3.7% Estate and gift taxes 4.2%
21. a. Double line graph of bar graph to show trends. b.
0
10 Test results
20
c. They should look similar. 17. a. 84
b. ⬃89
18. a. The number of bars increases because the range is still the same but divided into smaller intervals. b. Because of the gaps in the data such as from 7 to 12 and from 14 to 22, when the cell width gets small, the cells in those intervals will have no values in them.
$25000 Tuition and fees
Weekly output
16. a. and b.
60 50 40 30 20 10 0
A49
$20000
Private
$15000 $10000 $5000
Public
95 96 97 98 99 00 01 02 03 04 Year c. Private—The graph is generally steeper.
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Answers
25. a–b.
22. a. Multiple bar graph to allow comparison. b.
10 Miles
20
Absences 26.
Not High school high school graduate graduate
Attended college
College graduate
23. a. Bar graph or line graph to show a trend. b.
250,000 Number of cell phone subscribers (x1000)
0
c. 6
Women’s earnings
100 90 80 70 60 50 40 30 20 10 0
200,000
1000 900 800 700 600 500 400 300 200 100 0
0
200 400 600 800 100012001400 Men’s earnings
150,000 The corresponding weekly salary for a woman is about $620.
100,000 27. There are 30 students in the class, so each student is represented by 360 ⫽ 12 degrees. Thus, her use of 10⬚ for each student was off by 30 2⬚ explaining why she had extra space.
50,000 0
97 98 99 00 01 02 03 04 05 06 Year
Percentage
24. a. Multiple bar graph or line graph to show a trend. b.
100 90 80 70 60 50 40 30 20 10 0
= Federal sources = State sources = Local sources
1920193019401950196019701980 1990 2000 Year
c. Federal funds increased steadily until sometime during the 1980s, then decreased. State funds increased steadily. Local funds decreased steadily until the 1980s and provide less than half of school funds.
28. A line graph is more appropriate when there is a clear connection between the data measured, for example, the cumulative total of rainfall in inches throughout the year. The color choices made by students are not related so a circle graph is more appropriate.
Section 10.2A 1. a.
Time in Minutes: Seconds
Percent viewing or reading
Television viewing Newspaper reading Accessed internet
8 7 6 5 4 3 2 1 0
4:06 4:00 3:54 3:48 3:42 3:36 3:30 50 55 60 65 70 75 80 85 90 95 00 Year
b. It makes the downward trend more apparent.
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Answers
2. a.
800 700 600 500 400 300 200 100 0
b.
800 700 600 500 400 300 200 100
6.
Cardiovascular Death Rate per 100,000
Red Meat 56%
1960
1980 1990 Year
Seafood 8%
2000
7. a.
Cardiovascular Death Rate per 100,000
Pampers 30%
Percent increase in cost
b. Different vertical scale
c. i.
Luvs 20%
Huggies 31%
52 weeks ending December 11, 1993 Luvs Private label 17% 22%
d. ii.
Changing Health Care Costs 8.0% 7.7% 7.9% 7.0% 6.9% 6.0% 5.4% 5.7% 5.0% 4.9% 4.0% 4.5% 3.0% 2.0% 1.0% 0.0% 96 97 98 99 00 01 02 Year
5.
52 weeks ending June 13, 1992 Private label 19%
c. Answers will vary 4.
2004 Poultry 36%
1950 1960 1970 1980 1990 2000 2010 Year 3. a. Yes
A51
New Car Sales
7.3%
Pampers 27%
6.9%
Huggies 34%
b. New ones represent relative amounts more accurately. 8.
1991 $1.2 billion
1992 $1.7 billion
8% 03
c. No
14%
04
92%
86%
Sales (x1000)
9000 8500
1993 $2.4 billion
8000 18% 7500 7000 6500
1994 1996 1998 2000 2002 2004 Year
82%
It gives the impression that every graph represents the same amount of money.
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9.
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Answers
18.
Cracked Nest Eggs Distributed Reserved
Federal Tax Burden per Capita
3
Taxes
2 1 1991
1992 Year
1993
10. The height of “Dad’s” sack should be close to three times as tall as the “other” sack and it is not even twice as tall. The height of “The Kid’s” sack should be close to twice as tall (200%) and it is only about 30% taller. Finally, the height of “Mom’s” sack should be more than 3 times (300%) as tall as “The Kid’s” sack and it is only about 50% taller. The graph could be more mathematically correct if the sack heights were all proportional to the percent that they represented while keeping the width and depth of all of the sacks constant. 11. a. (iii) and (iv). b. In (i), the volume represented on the right is actually 8 times as large. 12. Cropped vertical axis, horizontal instead of vertical bars, reverse the order of the categories. 13. Population ⫽ set of lightbulbs manufactured. Sample ⫽ package of 8 chosen. 14. Population ⫽ set of fulltime students enrolled at the university. Sample ⫽ set of 100 students chosen to be interviewed. 15. a. 22 or about 1.4 in. Because the graphs are twodimensional, their revenues vary as the square of their radii and 12⬊ 222 ⫽ 1⬊2 ⫽ 5,000,000⬊10,000,000. b. 2 3 2 or about 1.3 in. Because the graphs are threedimensional, their revenues vary as the cube of their radii and 13⬊ 2 3 23 ⫽ 1⬊2 ⫽ 5,000,000⬊10,000,000. 16.
Help Wanted Ads 130,000 120,000 110,000 100,000 90,000 80,000 17.
Taxes
Indices of Products for Which a U.S. Farmer would have received $100 in 1992. $140 $120 $100 $80 $60 $40 $20 $0
Crops
Livestock
96 97 98 99 00 01 02 03 04 05 Year
20. Population ⫽ the set of fish in the lake. Sample ⫽ the 500 fish that are caught and are examined for tags. Bias results from the fact that some of the tagged fish may be caught or die before the sample is taken and the fish might not redistribute throughout the lake. 21. Population ⫽ set of all doctors. Sample ⫽ the set of 20 doctors chosen. Bias results from the fact that they will commission studies until they get the result they want. 22. Cropping is not always intended to deceive. There are many cases when a small change is significant, for example, in the stock market index or the prime interest rate. In order to emphasize the significance of these small changes graphically, cropping the vertical axis may be appropriate.
1992
1993 Year
1994
Federal Tax Burden per Capita 7400 7200 7000 6800 6600 6400 6200 6000
1999 2000 2001 2002 2003 2004 Year
19.
Index in dollars
0
10000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0
Section 10.3A 1. a. 9.83; 9.5; 9 c. 0.483; 1.9; no mode
b. 14.16; 13.5; no mode d. 4.2; 0; 0
2. a. Median: 32 + 27 Mode: 3 ⫹ 27 Mean: 56 + 27 b. Median: 4 Mode: 4 Mean: 11 3p c. Median: 5.37 Mode: 6.37 Mean: 37 6 ⫹ .37 艐 6.54
1999 2000 2001 2002 2003 2004 Year
3. No student is average overall. On the math test, Doug is closest to the mean; on the reading test, Rob is closest.
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4. 584 students 5. Yes but the remaining 6 students must all have perfect scores. 6. a. {1}, Answers may vary
b. {1, 7, 7}, Answers may vary
7.
120
130
120 124
140
150
133
147
A53
17. A: test score 95 B; 90 test score ⱕ 95 C: 80 test score ⱕ 90 D: 75 test score ⱕ 80 F: test score ⱕ 75
160
18. a. 87th percentile b. 87%
160
19. Approximately 38.3 20. 23.94 or 24
8. a.
50
60
70
57
80
67
90
79.5 86
100 93
21. Set 1: 1, 3, 3, 5 SD ⫽ 22 Set 2: 1, 5, 5, 9 SD ⫽ 2 22 Answers may vary 22. 293 23. 76.81 to two places 24. 1456
66
77
25. Test 1: her zscore (0.65) is slightly higher than on test 2 (0.63).
84.5 90 95
26. Mode, since this represents the most frequently sold size
b. Class 2; all five statistics are higher than their counterparts for Class 1.
27. a. The distribution with the smaller variance
9. a. The upper quartile: 86 b. The median: 84.5 c. The lower quartile: 67 10. a. 0; 0 b. 6.6; 2.58 (to two places) c. 371.61; 19.28 (to two places) 11. a. 2; 22 b. 18; 3 22 c. 50; 522 d. 72; 622. If the variance is v and the standard deviation is s, and if all data are multiplied by r, the new variance is r 2v and the new standard deviation is 2r 2 䡠 s. e. Their standard deviations are all the same and all three sets of data are arithmetic sequences with a difference of 5.
28. a. 1.14 (to two places) b. 1.99 (to two places) c. 97.5%
12. 18, 18.5, 19, 2.6, 1.61 13. ⫺0.55, ⫺1.76, 0.36, 1.71, ⫺0.32, ⫺0.10, 0.66 14.
–20 –10
0
10
20
30
40
29. Spike looks at the middle number without first ordering the numbers from smallest to largest.
50 East West
A trend toward greater growth west of the Mississippi 15. a.
7 6 5 4
b. The distribution with the larger mean
30. A box and whisker plot for each grade would provide better comparisons. Although the mean and median for each grade may be in the forties, the box and whisker plots will display the progression in heights from grade to grade.
PROBLEMS WHERE THE STRATEGY “LOOK FOR A FORMULA” IS USEFUL 14 34 55 60 04 11 12 12 21 21 34 36 48 63 63 69 73 83 86 88 52 55 62 64 65 70 72 75 75 87 93 93
1. The first allowance yields 7 ⫹ 21 ⫹ 35 ⫹ 䡠 䡠 䡠 ⫹ 7(2 䡠 30 ⫺ 1), which equals 7(1 ⫹ 3 ⫹ 5 ⫹ 䡠 䡠 䡠 ⫹ 59). Since 1 ⫹ 3 ⫹ 5 ⫹ 䡠 䡠 䡠 ⫹ (2n ⫺ 1) ⫽ n2 and 59 is 2 䡠 30 ⫺ 1, the total is 7 䡠 302 ⫽ $63. The other way, the total is 30 䡠 2 ⫽ $60. He should choose the first way. 2. Let x be its original height. Its height after several days would be given 3 4 5 by x ¢ ≤ ¢ ≤ ¢ ≤ Á . One can see that the product of these fractions 2 3 4
400 450 500 550 600 650 700 750 800 b. Aaron’s, Bond’s, and Ruth’s totals are mild outliers. 16. a. 84%, zscore ⫽ 1 b. 97.5%, zscore ⫽ 2
3 4 5 n + 1 n + 1 ≤⫽ leads to this formula: x ¢ ≤ ¢ ≤ ¢ ≤ Á ¢ . 2 3 4 n 2 n + 1 100 when n ⫹ 1 200, or when n 199, 2 it would take 199 days.
Therefore, since
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3. Pairing the first ray on the right with the remaining rays would produce 99 angles. Pairing the second ray on the right with the remaining ones would produce 98 angles. Continuing in this way, we obtain 99 ⫹ 98 ⫹ 97 ⫹ 䡠 䡠 䡠 ⫹ 1 such angles. But this sum is (100 䡠 99)/2 or 4950. Thus 4950 different angles are formed.
4.
CHAPTER REVIEW Section 10.1 1.
CLASS 1
Mean Teacher Salary
Salary
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CLASS 2 33 97 1
986442 210 64440
1 2 3 4 5 6 7 8 9
7 59 9 00009 12
48 47 46 45 44 43 42 41 40
Elementary Secondary
2001
2002
2003
2004
Year 5.
Others Milk 10% 10%
059 25779
Meat 10%
Fruit 30%
Vegetable 40% 2.
7 6.
Sporting Good Sales vs. Rain
5 4
Sales (in thousands)
Number of scores
6
3 2
0
0–9 10–19 20–29 30–39 40–49 50–59 60–69 70–79 80–89 90–100
1
400 350 300 250 200 150 100 50 0 0
10 20 Rain (in inches)
The point (18, 340) is an outlier. Fifteen inches of rain should correspond to $245,000 in sales. For sales of $260,000, the rain should be around 13.5 inches.
Section 10.2 1.
3.
600
Elementary Secondary
Average Weekly Manufacturing Earnings in Oregon, 1993—1994
500 Dollars
Salary
Mean Teacher Salary 48 47 46 45 44 43 42 41 40
400 300 200 100
2001 2002 2003 2004 Year
30
0
AMJ J A S O N D J F M A Month
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2. a.
6. Local dentists may be more likely to use a local product than dentists across the country.
Percent of GDP
Health Care Spending 14 13 12 11 10 9 8 7 6 5
A55
Section 10.3 1. Mode ⫽ 14, median ⫽ 9, mean ⫽ 8.4 2. 1 3
9
14
3. Range ⫽ 13, variance ⫽ 23.42, standard deviation ⫽ 4.84 4. 2⬊⫺1.27; 5⬊⫺0.68; 14⬊1.11
1960
1970
1980 Year
1990
2000
b.
5. It represents a data point’s number of standard deviations away from the mean in which above the mean is positive and below the mean is negative. 6. 68%
Health Care Spending
7. 95%
Percent of GDP
30
8. 2 is the 10th or 11th percentile. 5 is the 25th percentile. 14 is the 86th or 87th percentile.
25 20 15
Chapter 10 Test
10 5 0
1960
1970
1980 Year
1990
2000
3.
Dad 11%
Other 4%
The Kids 21%
1. a. F d. F g. F j. F
b. F e. T h. T k. F
c. F f. T i. F
2. Measures of central tendency: mean, median, mode Measures of dispersion: variance, standard deviation 3. Bar and line graphs are good for comparisons and trends and circle graphs are not. Circle graphs are good for relative amounts, not for trends. 4. 108⬚ 5. Mean is 6; median is 6; mode is 3; range is 7. 6. 11, 14, 20, 23
Mom 64%
7. 5.55
Who makes the kids' lunch?
8. The lineman is in the 98th percentile and the receiver is in the 8th percentile. 9.
Dad 11%
Other 4%
50 The Kids 21%
60 54
70
80
90
70.5
80 85
100 97
10.
200
Brand of golf ball
Magna
0
Topflite
100 Titleist
5. Population: all voters in town. Sample: adult passersby near high school. Bias: Since most inline skaters are high schoolaged students, people near the high school are more likely to have a polarized opinion about the issue.
300
Maxfli
4. Because the size of the people in the “marry” category is larger, it gives the section a much more dominant appearance over the small person representing the “live alone” category. Answers may vary.
400
Dunlop
Who makes the kids' lunch?
500
Ultra
Mom 64%
Combined Distance (in yards)
600
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11. a.
19. A line graph is good to display this data because it shows the trend over time.
7
High School Graduates Enrolled in College
6 5 4 Percent
3 2
71–80
61–70
51–60
41–50
0
31–40
1
74 72 70 68 66 64 62 60 58 56 54
Male Female
95
96
97
98
99
00
01
02
03
04
Year
b.
20.
Federal Tax Burden per Capita
7 6 Year
5 4 3 2 1 73–80
65–72
57–64
49–56
41–48
33–40
$0 25–32
0
2004 2003 2002 2001 2000 1999
$6369 $6229 $6632 $7441 $7404 $6796 $2000 $4000 $6000 $8000 $10000 Taxes
21. 15 22. 23
12. Population: families with schoolaged children in school system Sample: families of 200 students with home addresses 13. 0, 6, 6, 7, 8, 9. There are many other possibilities. 14. 3, 3 (Actually, any single nonzero number is a correct answer.) 15. The professor may look at both of them to determine how to assign grades depending on the distribution. The histogram in part (a) would indicate 4 A grades and 7 B grades while the histogram in part (b) would indicate 1 A grade and 4 B grades. The histogram in part (b) also gives a better sense of how dispersed the scores are. Answers may vary. 16. To indicate an increase (or decrease) in the data being measured, sometimes the size of the picture in the pictograph is incorrectly increased instead of the number of pictures being increased. Answers may vary. 17. The data set {4, 5, 6} has a mean of 5 and a standard 2 deviation of , while the data set {0, 5, 10} has a mean A3 2 of 5 and a standard deviation of . A3 Answers may vary. 18. Line graphs can be deceptive by cropping either the vertical or horizontal axis. They are also distorted by making the graph excessively narrow or wide. Answers may vary.
23. Science, since its zscore is the highest 24. Customers are more likely to prefer the lemonlime so that they will be on television and to please the people making the commercial. 25. Because the bars are on an angle, the bar for the Oakland Athletics is longer than the bar for the Atlanta Braves even though the bars represent the same number. It is unclear where the end of each bar is. If measured to the end of the ball, the bar for the Los Angeles Dodgers represents 0.125 inches per playoff. With this scale, the bars for the top teams should be about a quarter of an inch longer than they are. Answers may vary.
Section 11.1A 1. (c) 2. a. {H, T} b. {A, B, C, D, E, F} c. {1, 2, 3, 4} d. {red, yellow, blue} 3. a. {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT} b. {HHHH, HHHT, HHTH, HTHH, HHTT, HTHT, HTTH, HTTT} c. {HHHT, HHTH, HTHH, THHH} d. Same as part (a) e. {HHTH, HHTT, THTH, THTT}
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4. a. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} b. {2, 4, 6, 8, 10, 12} c. {1, 2, 3, 4, 5, 6, 7} d. {6, 12} e. ⭋ 5. a. P
b. I
c. I
6. {(H, 1), (H, 2), (H, 3), (H, 4), (T, 1), (T, 2), (T, 3), (T, 4)} 12 7. a. 60 =
28 b. 60 =
1 5
8. a. Answers will be b. Answers will be c. Answers will be d. Answers will be e. Answers will be
7 c. 31 15 60 near 12 but will vary. near 11 12 but will vary. 5 near 12 but will vary. 13 near 18 but will vary. 5 near 18 but will vary.
26. Jennifer may believe that the “law of averages” is about to catch up with Melissa; since the probability of heads or tails is 50–50, getting 5 tails in a row means it is “time” to get a head. Karen may be thinking that if Melissa got 5 tails in a row, the coin may not be a fair coin, which would make the probability of getting a tail much greater than the probability of getting a head. The fact is, if the coin is a fair coin, the odds of getting a head or a tail on the next toss is still 1/2.
Section 11.2A 1. a.
b.
H
T
R G
c.
f. The answers for 500 tosses will be similar to those found with 100 tosses but may vary.
9
12 13
1
2. a.
7 8 3 8
c. Answers will be near but will vary. d. Answers will be near but will vary.
T
e. The answers for 500 tosses will be similar to those found with 100 tosses but may vary. 10. a. 61 11.
8 a. 13
b. 14 b.
9 c. 13
d. 12 13
T H TT TH
12. a. Point up is generally more likely 42 7 18 3 b. 60 = 10 ; 60 = 10 c. 70; 30 13. a. 51
b. 52
1 14. a. 19
1 b. 38
15. 16. 17. 18.
a. 83 a. 13 a. 34 7 a. 16
b. b. b. b.
1 2 (or 2 3 4 13 12 16 =
c. 35
d. 52
9 c. 19
d. 21 38
c. 32
d. 31
c. 11 26
4 d. 13
T H HT HH
b.
G
Y
3 6)
B R Y GB GR GY
3 4
G
20. a. Probability that the student is a sophomore or is taking English b. Probability that the student is a sophomore taking English c. Probability that the student is not a sophomore 21. a. The sum should be 100% b. There are 16 face cards. Thus, the probability of drawing a face 4 card is 16 52 ⫽ 13 . c. A probability is between 0 and 1, inclusive. Thus, one can’t have a probability of 1.5. 22. a. 4, 5, 6, 5, 4, 3, 2, 1 b. 20
c.
5 1 1 11 b. 12 , 3 , 4 , 12 20 1 ⫽ 100 5
G GG
B
B GB
3.
G BG H
H T H
1 24. 16
25. Actually there are four possibilities in the sample space, BB, GG, BG, or GB, depending on the order of arrival of the children. So the probability of 2 boys is 1/4, the probability of 2 girls is 1/4, but the probability of having one child of each sex is 1/2.
B R Y YB YR YY
c.
10 19. a. Getting a blue on the first spin or a yellow on one spin; 16 ⫽ 85 1 b. Getting a yellow on both spins: 16 9 c. Not getting a yellow on either spin: 16
23. a. 100
H
c. 1
4 13
B W
d.
2 6
9. a. Answers will be near 18 but will vary. b. Answers will be near 12 but will vary.
A57
T T
B BB H T H T H T H T
HHH HHT HTH HTT THH THT TTH TTT
2
3
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4. a–d.
W RW
R
8. a.
b.
1 – 2
1 – 4
R
1 – 4
B RB 1 – 2
R WR
W
1 – 4
G
1 – 4
R W B G
B WB c.
R BR B
1 – 5
W BW
R
1 – 5
1 – 5
1 – 5
1 – 5
W
Y
B
G
e. 6 outcomes 5. 9. a.
1
P G B
2
P G B H
3 T H
2 __ 5
R
_3_ 5
W
b.
10. a. A: 13 , B: 31 , C: 13 b. A: 13 , B: 61 , C: 12 c. A: 14 , B: 41 , C: 12
11. a.
A
4
A
T
B C
H
A
5 B
T
B C
H
A
6
C
T
B C
6. a.
C
T
S CS P CP S TS P TP S BS
B
P b. 8 7. 300
c. 4 䡠 2 ⫽ 8, yes
b.
A 1 3
1 6
A B C
B
P BP S PS
1 3 1 6 1 2
A B C
1 2
P PP
1 3 1 6 1 2
C
1 3 1 6 1 2
A B C
_1_ 3
B
2 __ 3
G
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c.
1 3 1 6 1 2
A 1 3
1 6
1 3 1 6 1 2
B
C
d. 19 +
1 36
+
1 4
=
14 36
1 9
B
1 18
C
1 6 1 18
A
1 3 1 6 1 2
1 2
A
B
1 36
C
1 12 1 6
A B
1 12
C
1 4
c.
W 1 2
1 3
W
1 6
R
1 6
B
1 6 1 6
W 1 4
R
2 3 1 3
W
1 12 1 6
B
1 12
B 1 4
B
7 18
=
1 3 1 3
d. 61 +
1 6
=
2 3 1 3
A59
1 3
13. a. Number of ways of getting 0 heads (1), 1 head (4), 2 heads (6), 3 heads (4), or 4 heads (1) 4 b. 16 = 14 12. a.
c. 15 16 d. Against
W W
14. a. 10 ⫻ 10 ⫽ 100 b. 9 ⫻ 10 ⫻ 10 ⫽ 900 c. 10 ⫻ 9 ⫻ 8 ⫻ 7 ⫽ 5040 d. 9 ⫻ 10 ⫻ 10 ⫻ 10 ⫻ 10 ⫽ 90,000
R B W
W
R
B
15. a. 8 䡠 7 䡠 6 ⫽ 336 b. 3 䡠 2 䡠 1 ⫽ 6 6 1 c. 336 = 56
W
16. a.
R
B
W
G BBG
B W B
B BBB
B
W
B BGB
R
G B
G BGG B GBB G GBG
b.
W 1 2
1 3 1 3
1 3
R
B GGB G
R
2 3 1 3
1 1 3 1 1 , 4 , 8 , 4 , 16 17. a. 1, 4, 6, 4, 1; 16 b. 3 hits and 1 miss
18. a. Each branch has probability of 12 b. 21 *
B 1 4
2 3 1 3
19.
R
1 2
=
1 8
1 4
(top branch), 21 *
1 1 1 2 * 2 ⫽ 8 (LPP) 1 1 1 c. (i) + = 4 (PLL and LPL), (ii) 8 + 8 + 14 ⫽ 21 (LPP, PLP, and PP), (iii) 81 + 18 + 18 + 18 ⫽ 21 (LPL, LPP, PLL, and PLP) 4 a. 25 b. L ⫽ 53 , P ⫽ 52 c. 25 18 12 81 d. 125 e. 25 f. 125 1 8
W B
G GGG
b. 1; 3; 3; 1 c. They are the 1, 3, 3, 1, row
B W
1 4
G
W
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2 8 for each branch; 3 27
2 3 1 8 b. Both paths have probability ¢ ≤ ¢ ≤ ⫽ ; 3 3 81 8 8 8 8 + + ⫽ 81 81 81 27 c. BBAAA, BABAA, BAABA, ABBAA, ABABA, AABBA; 2 3
3
1 3
2
¢ ≤ ¢ ≤ ⫽ d.
8 16 ; 243 81
T T
F
T FFT F FFF
b. 8 22. a.
1 3
E
E
1 5 2 5
8. 7! 9. a. 12C5 ⫽ 792
TFT TFF FTT FTF
5! ⫽1 5!0! 5! ⫽5 5C1 ⫽ 4!1! 5! ⫽ 10 5C2 ⫽ 3!2! 5! ⫽ 10 5C3 ⫽ 2!3! 5! ⫽5 5C4 ⫽ 1!4! 5! ⫽1 5C5 ⫽ 0!5! 6! b. 6C3 ⫽ ⫽ 20 3!3! 5! ⫽ 10 5C2 ⫽ 3!2! 5C3 ⫽ 10 20 ⫽ 10 ⫹ 10
11. a. 5C0 ⫽
2 5
N
K E
1 3
1 3 2 5 1 5
N
N
K
2 5
K E
2 5 2 5
N
1 5
K
b. {(E, E), (E, N ), (E, K), (N, E ), (N, N ), (N, K ), (K, E), (K, N ), (K, K)} c. {(E, E), (N, N ), (K, K )} 1 d. 51 e. 15 23. 4 socks
12. a. 13.
25. 37 55
14. a. b. 170
27. Herman’s Dad will have 16 different possible outfits, if varying just one item of clothing qualifies as a different outfit. This can be illustrated with a tree diagram. 28. Because the events of drawing a king and a heart are not mutually exclusive, the probabilities cannot simply be added as Freda suggests. Since only one action is taking place, Mattie’s suggestion of multiplying the probabilities would be incorrect. The correct answer would be 1/52 because there is only one king of hearts in a deck of 52 cards.
Section 11.3A 1. a. 10 䡠 9 ⫽ 90 b. 9 䡠 8 䡠 7 䡠 6 䡠 5 䡠 4 ⫽ 60,480 c. 102 䡠 101 䡠 100 ⫽ 1,030,200 2. a. m ⫽ 9, n ⫽ 3 b. m ⫽ 19, n ⫽ 5
10C9 10
b.
2
10C7 10
2
c.
10C5 10
2
d.
10C3 10
2
e.
10C1 10
2
4
7 24. ⫽ 1 ⫺ P(3 females) 8 26. a. 78
b. 52C5 ⫽ 2,598,960
10. 8C5 ⫽ 56
d. 18
c. 1
b. 11,232,000
7. a. 64,000 b. 59,280 c. 60,840 d. The order of the numbers is important
T TTT F TTF
T F
b. 12C9
6. a. 15,600,000
T F T F
F
4. a. m ⫽ 13, n ⫽ 12 b. m ⫽ 10, n ⫽ 7 or m ⫽ 10, n ⫽ 3 5. a. 12P2
8 8 16 64 17 + + ⫽ ; 27 27 81 81 81
21. a.
12 # 11 # 10 # 9 ⫽ 495 4#3#2#1 # 6 5 b. ⫽ 15 2 15 # 14 # 13 c. ⫽ 455 3#2#1
3. a.
266 10C4
7 585 238 17C4 b. ⫽ 1755 27C4 27C4
⫽
n! n! + (r  1)!(n  r + 1)! r!(n  4)! n!(n  r + 1) n!r = + r!(n  r + 1)! r!(n  r + 1)! (n + 1)! = = n + 1Cr r!(n + 1  r)!
15. a. nCr  1 + nCr =
b. Each “inside” entry is equal to the sum of the nearest two entries above it. 16. a. 1680 6 17. a. 3 26 18. 4! ⫽ 24
3 b. 56 1 b. 3 26
19. a. m ⫽ n ⫽ 5
b. m ⫽ 10, n ⫽ 4
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20. a. 376,992
b. 1,947,792
21. a. 10!
b. 10!
c. 10C8 䡠 8!
d. 10C5 䡠 5!
22. a. 10!
b. 9!
c. 5 䡠 9!
d. 25 䡠 8!
23. a. 5!
b. 4!
c. 2 䡠 4!
d. 2 䡠 3 䡠 3!
e. 10Cr 䡠 r!
20! 20! ⫽ 15! 5! 5! 15! n! n! b. ⫽ (n  r)! r! r!(n  r)! c. 99,884,400
25. a. 9!
b. 3! 䡠 6!
27. a.
48 52C5
b.
b.
d. 8 c.
c.
3 # 17C4 b. 艐 0.46 20C5
28. a. 20C5
6 b. 15 ⫽ 52
19. a. 61
b. 13
c. 0
a. 72 a. 31 a. 53
b. 57
c. 74
22.
364 1365
24 52C5
8 18. a. 15
21.
c. 2
26. a. 15C4 ⫽ 1365
17. For example, the sum is a. even. b. 7. c. 5 or 6.
20.
24. a.
e. 96
78 1365
13C5
d.
52C5
45 52C5
#
c.
+ 3C2 # 17C3 ⫽ 0.59 20C5
3C1 17C4
29. It appears that the error is due to a misunderstanding of factorials or in the simplification of them. By having the student explain their understanding of factorials, the misunderstanding could be revealed and dealt with. It would also be important to have the student go through the process of simplifying the expression to assure that there are not any further misunderstandings.
Section 11.4A 1. a. Answers will vary. b. Answers will vary, but theoretically it should take about 22 attempts. 2. Average should be near 22. 3. Answers will vary; theoretical value is 0.63.
A61
b. b.
1 4 11 12
c. 21
d. 83
c. 51 c.
23. a. 15/105 ⫽ 1/7 b. 60/105 ⫽ 4/7 c. 61/105 e. 29/105 g. 24/45 ⫽ 8/15 i. 32/53
31 60
d. 12
e. 1
f. 0
d. 1
e. 31 55
f. 31 36
d. 53/105 f. 7/15 h. 32/61
24. a. Disagree to be correct. b. p4 c. q4 d. p4 ⫹ q4 1 1 1 e. For 1⬊1—12 , 12 , 16 , 16 , 8 ⫽ 0.13;
2 1 16 1 17
for 2⬊1—3 , 3 , 81 , 81 , 81 ⫽ 0.21; 3 1 81
1
41
3 2 81
16
97
for 3⬊1—4 , 4 , 256 , 256 , 128 ⫽ 0.32; for 3⬊2—5 , 5 , 625 , 625 , 625 ⫽ 0.15. f. The prospect for series longer than 4 games is greater when teams are evenly matched. 25. a. p4q; 4p4q b. 4pq4 c. 4p4q ⫹ 4pq4 26. a. 10p4q2 b. 10p2q4 c. 10p4q2 ⫹ 10p2q4
22 4. a. 100 = 11 50 b. Answers will vary.
27. a. 20p4q3 b. 20p3q4 c. 20p4q3 ⫹ 20p3q4
5. a. On a standard die, let a boy be an even number and a girl be an odd number. Toss a die until you get one of each and count the number of tosses. b. 3, Answers will vary c. Let a boy be 0 and a girl be 1. “Draw” until you get one of each. Count the number of draws.
28. a. For X ⫽ 4, p4, q4, p4 ⫹ q4; for X ⫽ 5, 4p4q, 4pq4, 4p4q ⫹ 4pq4; for X ⫽ 6, 10p4q2, 10p2q4, 10p4q2 ⫹ 10p2q4; for X ⫽ 7, 20p4q3, 20p3q4, 20p4q3 ⫹ 20p3q4 b. 0.125, 0.25, 0.3125, 0.3125 c. 5.8 games
6. In any twodigit number, let an even digit be H and an odd digit be T. Then 15 ⫽ TT. Using the first column plus ten in the second column we have HH—6 HT—4 TH—6 TT—9 8. 47,350 people b. $1.50
c. $3.50
10. $36.39 11. They each equal 1⬊1, so are equivalent. 12. a. 16
b. 1⬊5
c. 5⬊1
13. a. 1⬊51
b. 40⬊12
14. a. 7⬊1
b. 3⬊5
15. a. 3⬊2, 2⬊3 9 16. a. 10
b. 72
30. 28 segments 31. If Julio has hit 3 times out of 7 at bats, his batting average is 3/7 or .429. But the odds that he would not get a hit are figured as a ratio of P(no hit)⬊P(hit). So the odds would be 4⬊3, not 7⬊3.
7. 750 9. a. $2.50
29. a. Select numbers 1–5. Have the computer draw until all 5 numbers appear. Record the number of draws needed. This is one trial. Repeat at least 100 times and average the number of rolls needed in all the trials. Your average should be around 11. Theoretical expected value ⫽ 11.42. b. Answers will vary.
b. 1⬊3, 3⬊1 c. 12 17
PROBLEMS WHERE THE STRATEGY “DO A SIMULATION” IS USEFUL 1. Sketch a hexagon, flip coins, and play the game several times to determine the experimental probability of winning.
c. 5⬊1, 1⬊5
2. Toss a die twice. If a 1 or a 2 turns up on the die, the man received his coat. Toss it again. If a 1 or 2 turns up, the woman received her coat. Repeat several times. Another simulation could be done using pieces of paper labeled 1 through 6.
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3. Using your textbook, perform several trials of this situation to determine the experimental probability.
CHAPTER REVIEW Section 11.1 1. a. S ⫽ {(H, A), (H, B), (H, C), (T, A), (T, B), (T, C)} b. E ⫽ {(H, A), (H, B)} c. P(E) ⫽ 62 ⫽ 13 d. E ⫽ {(H, C), (T, A), (T, B), (T, C)} e. P(E) ⫽ 64 ⫽ 23 2. a. E ⫽ toss a sum less than 13 on a pair of standard dice b. E ⫽ toss a sum of 13 on a pair of standard dice. 3. Theoretical probability is the probability that should occur under perfect conditions. Experimental probability is the probability that occurs when an experiment is performed. 4. For example, E and E as listed in Exercise 1 are mutually exclusive, since E 艚 E ⫽ ⭋.
Section 11.2 R G
G R
2. There are 3 equally likely outcomes for each of the two spins. Thus there are 3 ⫻ 3 ⫽ 9 outcomes.
1 1
3 4
5
1 2
1
1 3 4 10
1 5
1
b. Toss 5 coins. c. The number of ways to obtain 3 heads and 2 tails when tossing 5 coins 4. a.
2 3 2 3
1 3
R
G
R
6. a. 5 䡠 8 ⫽ 40 b. 5 䡠 8 䡠 2 ⫽ 80
6. 11 30 7 7. $112 , or about $1.58
1
6 10
5. a. 9C2 䡠 5C4 ⫽ 36 䡠 5 ⫽ 180 b. Any team of 6 players can be seated in 6P6 ⫽ 6! ⫽ 720 ways in a line. # 32 8 8C1 4C3 c. # ⫽ 180 ⫽ 45 C 9 2 5C4
8. a. Testing to see how many times one must shoot an arrow to hit an apple on your professor’s head b. Use the digits 1, 2, 3, 4, 5, 6 to represent the faces of the dice, disregarding 0, 7, 8, 9.
1 1
4 9
Chapter 11 Test 1. a. F g. F
b. T h. F
2 3
1 3
G
2 9
R
2 9
1
G 9
c. F i. F
d. F j. F
e. T k. T
f. F
2. The event is a subset of the sample space. 3. When drawing from a container, an object is drawn and not replaced before drawing a second object. 4. 144
1 3
d. 495
5. When a special condition is imposed on the sample space
G
1
c. 120
4. 43⬊57
R
3. a.
b. 5005
4. 10P10 ⫽ 10! ⫽ 3,628,800
1. If a⬊b are the odds in favor of an event, then b⬊a are the odds against the event. a 2. If P(E) ⫽ , then the odds in favor of E are a⬊(b ⫺ a). b 3. 12⬊24 or 1⬊2
R
G
c. F d. F e. T b. 9C6 ⫹ 11C7 ⫽ 414
Section 11.4
R R
1. a. F b. T 2. a. 12C7 ⫽ 792 252 10C5 c. ⫽ 792 12C7
7. a. 8 for n ⫽ 3, 16 for n ⫽ 4, 32 for n ⫽ 5 b. 64, 1024, 220 ⫽ 1,048,576, 2n c. It is the sum of the entries in row n.
R R
Section 11.3
3. a. 5040
f. P(E) ⫹ P(E) ⫽ 1
1.
b. The probabilities along successive branches are multiplied to obtain the 49 , 29 , 29 , and 19 . c. To find, for example, the probability of spinning RG or GR, you find 29 + 29 ⫽ 49 .
5. 47 56 5 6. 12 5 7. 11
8. a. 120 b. 2520 c. 495 9. a. 19,656,000 1 b. 650
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10. a.
1 8 3 8 4 8
R 2 9
15. In a bag place 4 pieces of paper numbered 1–4. Draw from the bag 7 times (with replacement), keeping track of which numbers are drawn. At the end of the 7 draws, if all 4 numbers have been drawn, record a “yes” for the trial. Repeat this process at least 20 times, recording “yes” or “no” at the end of each trial. After all of the trials, compute the number of “yeses” divided by the number of trials. This will be the experimental probability of getting all 4 prizes in 7 tries.
2
R 72 6
W 72 8
B 72
3 9
2 8 2 8 4 8
W
6
R 72 6
W 72
16. The odds of getting the numbers 1, 2, 3, or 4. Answers may vary. 17. Merle is correct because in order to have a onethird probability in this situation, each outcome would have to be equally likely, and that is not the case.
12
2 8 3 8 3 8
4 9
B
B 72 8
R 72
b.
2 9
c.
30 72
d.
11. a.
B 1 6
2 6
G
1 6 2 6 3 6 1 6 2 6 3 6
1 36
G
2 36
R
3 36 2 36
3 6
R
1 b. 36
11 12
B
6 36
R
9 36
1 4
6
7
8
9
10
11
12
4
5
6
5
4
3
2
1
12
11
10
9
8
7
6
5
4
3
2
26
⫽
1 262
1. a. Level 0 b. Level 0 c. Level 1 2. a. 2, 7 b. 2, 7 c. 2, 7, 5 d. 2, 7, 5, 8, 3, 6 e. Level 2—relationships
1
2 3
B 3
1 3 2 3
W6 B 1
1
3. a. {2, 4, 5} and {1, 3, 6, 7, 8}. Shapes 2, 4, and 5 all have a right angle and the rest of the shapes do not. Answers may vary. b. {2, 6, 7, 8} and {1, 3, 4, 5}. Shapes 2, 6, 7, and 8 all have two congruent sides and the rest of the shapes do not. Answers may vary. c. Level 0 d. Level 1 4. 12 triangles 5. 18 6.
6
V
1 4
5
3
Section 12.1A
13. Both can be used to find the total number of outcomes in an event.
R
4
263 23. $.50
12. 1 ⫽ P(A 艛 A) ⫽ P(A) ⫹ P(A); therefore, P(A) ⫽ 1 ⫺ P(A)
1 2
3 2
14 ⫺ n
22.
c. 11 36
14.
2 1
Note the symmetry in this table.
6 36 3 36
G
SUMS n WAYS
21. 14
4 G 36
R 1 6 2 6 3 6
537 1024
20. e.
B
B
19.
12 72
52 72
b. 11 24
18. a. 24
12
W 72 B
N 1
W 12
III
1 3
II
B 1 6
IV
G
1 3 2 3
A63
1
W 12
I
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7. a.
10. a.
b. Several possibilities exist for each part. (i) (ii)
(iii)
b.
(iv)
11. a. b, d c. a, e 8. a.
b.
b. a, c, e, f d. d, f
12. a. FHQO b. ADPK, and so on c. KQHI, and so on d. ABL, FGH, OPQ and so on. e. MNO, MOS, COM f. MNOS g. COSM, CONM h. CLK, and so on i. CAL, ROP j. DEFO, MORK, KQHJ, and so on k. BCKL 13. a. (ii), draw a diagonal
b. (i), rotate 12 turn
14. Arrow D. Use a ruler. 15. Same length in both cases. 16. a. 3
b. 2
17. Both types of lines 18. All but (c) 19. a. 12
b. 4
20. a.
9.
c. 1
d. 1
b.
1
2
21. There is nothing in the mathematical definition of a rectangle that says its length and width have to be different. Since a square satisfies the definition of a rectangle, it is a rectangle. For emphasis we say every square is a rectangle, but not all rectangles are squares.
Section 12.2A 1. a. Reflection over vertical line through the center of the square b. Rotations of 1/4, 1/2, 3/4 turns around the center of the square
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2. a. 5 lines b. 6 lines c. 7 lines d. 8 lines e. n lines f. Vertex, midpoint g. Vertex, vertex, midpoint, midpoint 3. H, I, N, O, S, X, Z 4. a. 2
b. Both have 180⬚ rotation symmetry
5. Yes. If the paper is folded along one line, the ends of the other line match.
18. Gail is thinking that if she slides or rotates one half of the parallelogram on top of the other, she has met the criterion for reflection. She needs to think of the symmetry line as a reflection line. It will help if she thinks of it as a fold line, not a cut line.
Section 12.3A
8. No
1. a. {D, B, E}, {A, E, F}, {D, C, F}, {A, B, C} Í ! Í ! Í ! b. DE , AE , and CE , Í ! Í ! Í ! CE , CA , and CD ! ! ! ! 2. LP, MP, NP, OP, ! ! ! ! ML, NL, OL, PL,
9.
3. a. 10
6. No. If the paper is folded so the ends of one line match, the ends of the other do not. 7. a. S, N
b. S
c. None
d. C, S, N
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4. a. Right
b. 4
c. 6
b. Acute
5. a. m(∠C) ⫽ 62⬚, AC ⫽ 6.3 cm b. m(∠C) ⫽ 110⬚, AC ⫽ 3.8 cm 6. a. ∠AIH and ∠AJG, ∠BIC and ∠BKD ∠GKF and ∠HIF. are three of many. b. ∠HIK and ∠JKI, ∠KJI and ∠CIJ, ∠GKI and ∠CIK ∠DJI and ∠HIJ
10.
7. a. 30⬚ 8. a. 20
b. 20⬚ b. 6
9. a.
c. 60⬚ c. 10
d. Not possible d. 2
e. 8
b.
c. Not possible 11. a. (i) b. (iii), equilateral triangles inside isosceles c. (ii), intersection is “isosceles right triangles” d. (i), equilateral triangles have 60⬚ angles 12. Flip the tracing over so that point A of the tracing is matched with point D and point B of the tracing is matched with point C. Then diagonal AC of the tracing will coincide with diagonal DB. 13. Rotate 14 , 12 , 34 of a full turn around the center. 14. Rotate 13 , 23 of a full turn around the center. 15. Fold one diagonal along itself so that opposite vertices coincide. Then the other diagonal lies along the fold line. 16. a. Fold on diagonal AC. Then ∠ADC coincides with ∠ABC. ∠A is not necessarily congruent to ∠C. b. Both pairs of opposite angles are congruent, since a rhombus is a kite in two ways. 17. It sounds as though each group has drawn 4 lines of symmetry. In fact, all 8 lines are lines of symmetry for the regular octagon.
Kites
Triangles
10. m(∠AFB) ⫽ 60⬚, m(∠CFD) ⫽ 35⬚ 11. m(∠1) ⫽ 54⬚, m(∠2) ⫽ 126⬚ 12. m(∠1) ⫽ 80⬚ m(∠2) ⫽ 100⬚ m(∠3) ⫽ 135⬚ m(∠4) ⫽ 125⬚ m(∠5) ⫽ 55⬚ m(∠6) ⫽ 125⬚ m(∠7) ⫽ 55⬚ m(∠8) ⫽ 45⬚ m(∠9) ⫽ 135⬚ m(∠10) ⫽ 45⬚
m(∠11) ⫽ 80⬚ m(∠12) ⫽ 100⬚ m(∠13) ⫽ 125⬚ m(∠14) ⫽ 55⬚ m(∠15) ⫽ 100⬚ m(∠16) ⫽ 80⬚ m(∠17) ⫽ 100⬚ m(∠18) ⫽ 80⬚ m(∠19) ⫽ 125⬚ m(∠20) ⫽ 55⬚
13. m(∠1) ⫽ m(∠2), given; m(∠2) ⫽ m(∠3), vertical angles have the same measure; m(∠1) ⫽ m(∠3); l ‘ m, corresponding angles property 14. a. m(∠1) ⫽ m(∠6), given; m(∠1) ⫽ m(∠3), vertical angles have the same measure; m(∠3) ⫽ m(∠6); l ‘ m, corresponding angles property b. l ‘ m, given; m(∠1) ⫽ m(∠4), corresponding angles property; m(∠4) ⫽ m(∠6), vertical angles have the same measure; m(∠1) ⫽ m(∠6) c. Two lines are parallel if and only if at least one pair of alternate exterior angles formed have the same measure. Í ! Í ! 15. Since ∠DAB is a right angle, DA ⊥ AB and ∠1 is a rightÍ angle ! Í also. ! However, since both are right angles, ∠1 艑 ∠ADC and AB ‘ DC by the corresponding angles property. Similarly, show AD ‘ BC.
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16. a. (i)
(ii)
(iii)
(iv)
13.
a. Hexagons
b. Squares
c. Triangles
2 2a 2 14. All ratios equal ¢ = ≤ . They are proportional. 3 3a 3
b. 8 points 17. All measure 90⬚. Any angle drawn from endpoints of a diameter of a circle and with its vertex on the circle will measure 90⬚.
15. a. Yes; point C will have all triangles if pattern is continued. b. No. Point C will have a different arrangement than A and B. c. (5, 5, 10)
18. 3 lines : 7 regions, 4 lines : 11 regions, 5 lines : 16 regions, n(n + 1) 10 lines : 56 regions, n lines : ⫹ 1 regions 2
16. a. See Table 12.4 b. 4 ways c.
19. All measure 90⬚. Yes. 20. Different geometry books can choose to define certain words differently because of the particular aims they have in mind. The definitions cited in the problem are common, as are the definitions in this text. Neither is “right” or “wrong”; they simply indicate a choice on the part of the author(s). 21. At first glance these symbols seem to represent 6 different rays. ! ! However, they are not all distinct. Here, RS and RT are two names for the same ray. So there are only 4 different rays.
(3, 12, 12)
Section 12.4A 1. a. 540⬚
b. 1080⬚
2. a. 105⬚
b. 108⬚
3. a. 150⬚, 30⬚, 30⬚ c. 144⬚, 36⬚, 36⬚
c. 110⬚, 60⬚, 80⬚, 110⬚ b. 157.5⬚, 22.5⬚, 22.5⬚ d. 162⬚, 18⬚, 18⬚
4. 21 5. a. 9
b. 20
c. 180
6. a. 12
b. 5
c. 72
7. a. 40
b. 8
c. 36
8. a. 90⬚
b. 4⬚
c. 30⬚
9. a. 108⬚
b. 170⬚
c. 178⬚
10. a.
(4, 6, 12)
b.
(4, 8, 8)
These patterns can be continued. 11. a. Yes c. Have same measure
b. Have equal measure d. Equals 180⬚, supplementary
12. a. Square vertex figure, triangular vertex figure
(6, 6, 6) 17. The triangle surrounded by the shaded regions is a right triangle. The two small squares include a total of four triangles—the same area covered by the larger square. Thus, if two short sides of the triangle are a and b, and the hypotenuse is c, we have a2 ⫹ b2 ⫽ c2.
b. All
18. a ⫽ 70⬚, b ⫽ 130⬚, c ⫽ 120⬚, d ⫽ 20⬚, e ⫽ 20⬚, f ⫽ 80⬚, g ⫽ 60⬚, h ⫽ 100⬚
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19. 3, 0, 0; 4, 1, 2; 5, 2, 5; 6, 3, 9; 7, 4, 14; 8, 5, 20; n, n ⫺ 3;
n(n  3) 2
20. 190 handshakes 21. 180⬚
4. a. (i) ABC, DEF, (ii) ACFD, BCFE, ABED, (iii) DEF, ABED, (iv) right triangular prism b. (i) MPTQ, NOSR, (ii) MNOP, OSTP, MNRQ, RSTQ, (iii) MNRQ, NOSR, RSTQ, (iv) right trapezoidal prism 5. a. Right square pyramid b. Oblique triangular pyramid c. Oblique octagonal pyramid 6. a. Right square pyramid b. Right hexagonal prism c. Regular octahedron d. TA B
22. Each small tile is a square with vertex angles of 90⬚. Each large tile is a nonregular octagon. Each pair of octagon vertex angles meeting a vertex of the square must add up to 360⬚ ⫺ 90⬚ ⫽ 270⬚. So each angle measures 135⬚. Thus all the angles in this octagon measure 135⬚. The measures of all vertex angles in a regular octagon are equal so they must also be 135⬚.
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TA B
TA B TA B
23. a. 6 points b. 6 points c. 8 points d. If n ⱕ p, then 2n points; otherwise 2p points 24. a. 180⬚ b. The sum of the pentagon’s interior angles is 540⬚. The sum of their vertical angles is thus 540⬚. The sum of the base angles of the triangles on the pentagon is 5 䡠 360⬚ ⫺ (540⬚ ⫹ 540⬚) ⫽ 720⬚. The sum of the angles in question is 5 䡠 180⬚ ⫺ 720⬚ ⫽ 180⬚.
TAB
TAB TA B
TAB
B TA
7. a.
b. TAB B TA
B TA
TAB
TAB B TA
TA B
c. 108⬚
TAB
b. No
TAB
8. a. Yes; 5
b. Yes; 4
c. No
d. Yes; 1
e. No
f. Yes; 2
TAB
c. 5ON, PQ6 5QR, OL6 5SR, LM6 5PS, NM6 other answers are possible. d. The planes LMNO and ONPS with edge ON. Other answers are possible. e. The planes LMNO and MNSR with edge NM. Other answers are possible. 3. a. No because it has a hole. b. Yes. 6 total faces—3 triangles, 2 quadrilaterals, 1 pentagon. c. No. Faces aren’t polygons.
TAB
TAB
TA B
B TA
5PQ, NM6 5OL, SR6 5NM, SR6 5PQ, SR6 5ON, LM6 5LM, QR6 5QR, PS6 5PS, ON6 5ON, QR6 5PS, LM6
2. a. Yes; F, G, H, I, J
TA B
Section 12.5A 1. a. LMNO and PQRS b. 5PQ, OL6 5OL, NM6
TA B
B TA
B TA
25. Donna is correct that any triangle can tessellate the plane, but it is also true that any quadrilateral can tessellate the plane, even the concave quadrilateral shown. The easiest way to explain this is by rotating the original figure 180⬚ about each side. This process can be continued in every direction. The fact that the four angles of a quadrilateral add up to 360⬚ means that if you have all four angles represented in each vertex of the tessellation, there will be no gaps.
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9. a.
PROBLEMS WHERE THE STRATEGY “USE A MODEL” IS USEFUL
b.
1. Trace, cut out, and fold these shapes as a check. Only (b) forms a closed box. 2. Try this arrangement with pennies. Six can be placed around one. 10. a. Triangle: 5, 6, 11, 9; quadrilateral: 6, 8, 14, 12; pentagon: 7, 10, 17, 15; hexagon: 8, 12, 20, 18; ngon: n ⫹ 2, 2n, 3n ⫹ 2, 3n b. Yes 11. a. (i) 7, 10, 17, 15
(ii) 12, 17, 29, 27
b. Yes
12. Pentagonal Prism: 4–4–5 Truncated tetrahedron: 6–6–3 Truncated cube: 8–8–3 Truncated octahedron: 6–6–4 On each polyhedron all vertex arrangements are the same so they are all semiregular. 13. a. Rectangle
b. Circle
14. a. Rectangle
b. Kite
15. a. Circles
b. Infinitely many.
16. a. A cube in a corner or a cube outside on the outside corner of a cube b. One of the corners 17. 14 18. (c) 19. a.
Front
Top
Right Side
Front
Right Side
c.
Top
Front
Right Side
20. a. (ii) b. (i)
3 1 1 1
3 2 2 1
21. a. (iii) 22. a. 3
3 3 2 1
3 3 3 3
(ii)
3 2 4 2 3 2 4 2
b. (v) b. 6
c. 9
23. 3 24. a. Order 3
b. 4
25. a. Order 2
b. 6
CHAPTER REVIEW Section 12.1 1. a. Recognition: A person can recognize a geometric shape but does not know any of its attributes. b. Analysis: A person can analyze a geometric shape for its various attributes. c. Relationships: A person can see relationships among geometric shapes. For example, a square is a rectangle, a rectangle is a parallelogram, etc. d. Deduction: A person can deduce relationships. For example, a person can prove that a quadrilateral with four right angles must have opposite sides parallel. Hence, a rectangle is a parallelogram. 2. a. Pencil d. Intersecting roads g. Window j. Kite
b. Corner e. Railroad tracks h. Stair railing k. Silhouette of a water glass
c. Yield sign f. Tiles i. Diamond
Section 12.2
b.
Top
3. Try this with several tennis balls. The maximum number that can be stacked into a pyramid is 25 ⫹ 16 ⫹ 9 ⫹ 4 ⫹ 1 ⫽ 55.
26. A parallelogram 27. Rene is making a good attempt at analogy. In fact, planes in space do correspond in many ways with lines in a plane: either they intersect or they are parallel. But lines in space have one other possible relationship. They may be skew. That is, they do not intersect, but they’re not parallel either. They are not contained in the same plane. 28. The “polygonal sides” should be referred to as “polygonal faces,” which means the polygons together with their interiors.
1. a. Perpendicular bisector of the base b. Perpendicular bisector of each side c. Perpendicular bisector of each side d. Perpendicular bisector of each side and angle bisector of each angle e. Angle bisector f. None g. None h. Perpendicular bisector of the bases 2. a. None b. 120⬚ and 240⬚ around its center c. 180⬚ around its center d. 90⬚, 180⬚, and 270⬚ around its center e. 180⬚ around its center f. None g. None h. None (n  1)360° 360° 720° i. , ,..., n n n 3. a. If one line is the fold line, all perpendicular lines must fold onto themselves. b. If one line folds onto itself, so must any parallel line (other than the fold line). 4. A geometric shape is convex if any line segment is in the interior of the shape whenever its endpoints are in the interior of the shape. 5. a.
a
b.
c.
b
b d
d
c
c
a
a
d
b
c h
e g
6. (a) and (b) are both infinite.
f
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Section 12.3
2. a. 1, 5 or 3, 7 or 2, 6 or 4, 8.
1. a. A dot b. An arrow c. A stiff piece of wire d. A taut sheet e. A pencil f. Blades of an open pair of scissors
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b. 3, 6, or 4, 5
3. The set of all points a fixed distance from a given point. 4. a. Right circular cylinder b. Oblique pentagonal prism 5. a. ∠ABC is obtuse
2.
b
B
c
a
A
b. AB ‘ CD, AC ‘ BD, AC ⬵ BD
a ⫹ b ⫽ 180⬚ ⫽ c ⫹ b; therefore a ⫽ c. 3.
C
A
c
B
a b C
(i) : (ii): Given: c ⫽ b. Show: a ⫽ b. First, a ⫽ c, since they are vertical angles. From a ⫽ c and c ⫽ b, we have a ⫽ b. (ii) : (i): Given: a ⫽ b. Show: b ⫽ c. First, a ⫽ c, since they are vertical angles. From a ⫽ c and a ⫽ b, we have b ⫽ c. 4.
6. Fold so that point A folds onto point C. Observe that BD lies along the fold line. Thus AC ⊥ BD. 7. 90⬚ and 135⬚ 8. 144⬚ 9. 9; 12 (not including a 0⬚ or 360⬚ rotation) 10. a ⫽ 13⬚, b ⫽ 66⬚, c ⫽ 101⬚, d ⫽ 79⬚, e ⫽ 53⬚, f ⫽ 48⬚
b
a
11. F ⫽ 7, V ⫽ 7, E ⫽ 12; Euler’s formula: 7 ⫹ 7 ⫽ 12 ⫹ 2
c
a ⫹ b ⫹ c ⫽ 180⬚. Thus the sum of the angle measures in a triangle is 180⬚.
Section 12.4 1.
D
15. a. (ii)
2. Equal 360° n
4. Regular 3gon, 4gon, and 6gon, since the measure of their vertex angles divided evenly into 360⬚.
Section 12.5 1. a. Two floors in a building b. An intersecting wall and floor c. Two intersecting walls and their ceiling d. The angle at which two walls intersect e. Telephone pole and telephone wire attached to a crossbar f. A cube
b. Ice cream cone
c. Ball
Chapter 12 Test b. T i. F
c. T j. F
d. T k. F
e. T l. F
f. T m. T
b. (i)
16. By drawing diagonals from one vertex of an n sided polygon, (n ⫺ 2) triangles are created. The angles of all of the triangles make up all of the interior angles of the original polygon. Since the sum of the angles of a triangle is 180, the sum of the interior angles of the ngon is (n ⫺ 2)180. Since all of the angles in the ngon are (n  2)180 . congruent, the measure of each angle is n 17. (b) 19. Since there are 7 triangles, the sum of all the angle measures is 7 䡠 180⬚ ⫽ 1260⬚. If the measures of the central angles are subtracted, the result, 1260⬚ ⫺ 360⬚ ⫽ 900⬚, yields the sum of the measures of all the vertex angles.
3. F ⫹ V ⫽ E ⫹ 2. For a cube, F ⫽ 6, V ⫽ 8, and E ⫽ 12. Thus, 6 ⫹ 8 ⫽ 12 ⫹ 2.
1. a. F h. T
c. F
18. 64 vertices, 34 faces
2. The five polyhedron all of whose faces are congruent regular polygons.
4. a. Can
b. A
13. m(∠3) ⫹ m(∠4) ⫽ 180⬚. Proof: 180⬚ ⫽ m(∠2) ⫹ m(∠4) ⫽ m(∠3) ⫹ m(∠4) since m(∠1) ⫽ m(∠2) and m(∠1) ⫽ m(∠3). 14. By 13, any two consecutive interior angles are supplementary. Thus if one angle measures 90⬚, they all must.
360° n
3. 180⬚ ⫺
12. a. N; 180⬚ rotation
g. T n. T
20. The measures of the vertex angles of regular 5gons and regular 7gons are 108⬚ and 12847 °, respectively, and there is no combination of them that will total 360⬚. Thus it is impossible to tessellate the plane using only regular 5gons and regular 7gons. 21. Four squares meet at a point with angles that add up to 4 䡠 90⬚ ⫽ 360⬚. Two octagons and a square meet at a point with angles that add up to 2 䡠 135⬚ ⫹ 90⬚ ⫽ 360⬚. However, the angles of two octagons are only 270⬚, while the angles of 3 octagons meeting at a point add up to 405⬚, neither of which is exactly 360⬚. 22. 540⬚
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Section 13.1A 1. a. 512 , 414 , 1
b. 4, 234 , 23 c. Not necessarily since the sizes of the paper clips and sticks of gum may vary. 2. a. Height, width (with ruler); weight (scale); how much weight it will hold (by experiment) b. Diameter, height (with ruler); volume (pouring water into it); weight (scale) c. Height, length (with ruler); surface area (cover with sheets of paper); weight (scale) d. Height, length, width (with ruler); volume (pour water into it); surface area (cover it); weight (scale) 3. a. 63,360 inches d. 46,656 in3 g. 16 cups 4. a. 28 mm
b. 148 cm
5. a. 900 mL
b. 15 mL
6. a. 23 kg
b. 10 g
7. a. 22 ⬚C
b. 5 ⬚C
d. 103
e. 10⫺2
f. 10⫺12
9
c. 4
d. $108.80
31. About 19 years 32. About 263% 33. Many students believe this because they don’t understand that when we go from one dimension to two dimensions, the original ratio needs to be squared. Here is a picture of one square yard. Its dimensions are 3 ft by 3 ft, but it contains 9 ft2.
3 ft
3 ft
b. 103; kilometer d. 10⫺3; millimeter
9. a. 10 ; gigameter c. 10⫺12; picometer b. 75 f. 0.0093 j. 0.764
12. a. 1000 dm3
b. $18.75/hr
30. 3 km/hour
c. 305 mg c. 10⫺6
b. 6.1 f. 950
29. a. $300
c. 473 mL
b. 10⫺3
11. a. 100 e. 382,000
28. a. 1, 4, 6 are measured directly. 2 is 4 to 6, 3 is 1 to 4, 5 is 1 to 6, 7 is 1 to 8, and 8 is 4 measured twice. b. 4. There are several ways. One is 1, 4, 6, 7. c. 4. There are several ways. One is 1, 4, 6, 8.
c. 27 cm
8. a. 106
10. a. 100 e. 76 i. 0.125
c. 4840 yd2 f. 4 cups
b. 1760 yards e. 8000 ounces h. 16 tablespoons
27. a. 6,272,640 cubic inches, 3630 cubic feet b. 225,060 pounds c. 27,116 gallons
c. 3100 g. 2.3
34. If there are 14 acres and each acre is 43,560 ft2, Monique needs to multiply 14 by 43,560 to find the total number of ft2 (609,840). Then she would take the square root of that number, which equals 780.9 ft. So the plot of land is about 781 ft by 781 ft.
d. 3060 h. 0.035
Section 13.2A c. 0.000564 g. 6.54
b. 3 places right
d. 8.21 h. 0.00961
1. a. 4.34 units
b. 1.91 units
c. 6.65 units
c. 3 places left
13. a. 9.5
b. 7000
c. 0.94
3. a. 9.8
14. a. 500
b. 5.3
c. 4600
15. a. 34
b. 18, 18
c. mL
4. a. 24 units b. m ⫽ 3 ⫹ 8 ⫽ 11 units, n ⫽ 3 ⫹ 16 ⫽ 19 units c. 3 d. r ⫽ p ⫹ 14 PQ ⫽ 9 units,
16. a. 177⬚C
b. 16⬚C
17. a. 57.6 oz 18. a. 240 pence 19. a. 15.24 cm
c. 68⬚F
b. 4840 ft/min
d. ⫺18⬚C
c. 616 in./sec
e. 23⬚F d. $0.40/min
b. 480 halfpennies
21. 7.5 gallons
s ⫽ p ⫹ 42 PQ ⫽ 15 units, 5. a. 17.7
c. 7.8 g/cm3
24. The length of a foot is based on a prototype and is not exactly reproducible. Converting linear measure, for example, uses ratios: 12 inches:1 foot; 3 feet:1 yard; 1760 yards:1 mile; so measures are not easily convertible. Volumes of in3, quarts, and pounds are not related directly. 25. a. Approximately 5,874,600,000,000 miles b. Approximately 446,470,000,000,000 miles c. Approximately 43 minutes 1 26. 733 ft. 3
b. ⫺8.5
6. a. 15.82 units, 12.936 square units b. 151.6 units, 1436.41 square units 7. a. 5.5 square units c. 12 square units 8. a. 29 units
22. 2.7 ⫻ 1013 cells b. 0.695 g/cm3
b. 2.95
t ⫽ p ⫹ 34 PQ ⫽ 21 units
b. 91.44 m
20. a. The student has used an inappropriate ratio by using 1 ft/12 in. b. Treat the units as fractions. The unwanted units should cancel, leaving just desired units.
23. a. 8.94 kg/dm3
d. 4.22 units
2. a. 18 units b. 9 units c. 3 d. PM ⫽ 3 ⫺ (⫺6) ⫽ 9 units, QM ⫽ 12 ⫺ 3 ⫽ 9 units
b. 9 square units d. c, 3 square units
b. 456 units
c. 1414 units
10 ft
9 in.
9. No
8 ft
6 in. Shaded area would be carpeted. Unshaded would not.
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10. a. 30 cm2
b. 9.9 m2
32. a. Obtuse e. Acute
11. 49.1 units, 111.9 square units
b. Right f. Obtuse
c. Acute
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d. Right
33. 46.7 square units
12.
34. a. 7.5⬚; when lines are parallel, alternate interior angles have the same measure. 500 x b. = , x ⫽ 24,000 miles 7.5 360 c. He was off by 901.55 miles. 35. 2.23 cm Figures may vary. a. 10 square units b. 14.93 units, 15.40 units, 16.17 units. Answers may vary. c. Triangles with the same area can have different perimeters. 13. 40.2 units, 68.8 square units 14. a. 213 units, 4 units, A1 ⫽ 12 square units; A2 ⫽ 8 square units b. Statement is false since parallelograms with the same length sides had different areas. 15. a. b 䡠 h
b.
1 2 (b
+ b) 䡠 h ⫽
1 2 (2b)
䡠h⫽b䡠h
c. Yes
37. 32 ⫺ 64 square units 38. a. /4 or 78.5% b. /4 or 78.5% c. They both irrigate the same amount. d. It doesn’t pay to use more sprinklers with a smaller radius. Check this in the n ⫻ n case. 39. They are the same. 40. Same 41. Equal
b. 2 215 square units
16. a. 823 square units
36. Area of region ⫽ square units
17. The areas vary for a fixed perimeter. The largest area occurs when the rectangle is a square. The area is 25 square centimeters.
42. a. 1534 acres b. $325,982.58 per acre
18. a. 32 square units
43.
b. 6 square units
c. 1⬊4
19. a. 9.6 艐 30.16 units; (4.8) 艐 72.38 square units b. 3 2 艐 29.61 units; (3/2)2 艐 69.76 square units
a
2
20. a. 210 units 21. a. 15 units 22. a. Yes
c
b. 213 units
b
a
b
b. 104 units b. No.
c. Yes
c
23. 25 feet
Because the areas of the two figures are covered with the same square and 4 triangles, they are the same. The area of the left figure is c 2 and the area of the figure on the right is a2 ⫹ b2 where a and b are the lengths of the legs of the triangle and c is the length of the hypotenuse of the triangle. Thus, a2 ⫹ b2 ⫽ c 2.
24. 24 m by 48 m 25. Yes 26. Area of large square ⫽ sum of areas of smaller pieces: c 2 = 4(12 ab) + (b  a)2 c 2 = 2ab + b2  2ab + a2 c 2 = a2 + b2 27. a. 2l 2 + w 2
b. 2l 2 + w 2 + h 2
c. 25600 艐 74.8 cm
28. (b) and (c).
44. Many possible answers. One idea would be to drop an altitude from one corner of the parallelogram and then slide the triangle formed to the right side of the parallelogram in order to form a rectangle. The area of this rectangle would be 20 times h. Since the length of h must be less than 16, the area is less than 320 in2.
29. a. Subdivide. Areas are indicated. Total area ⫽ 13 square units. 1 2
h 3 2
2
1 2
16 in.
45°
1
2
20 in.
1 2
45. Dana is correct about the squares, but not about rectangles. In the examples below, the perimeter of the “skinny” rectangle is 26 whereas the perimeter of the square rectangle is 24. Yet the area of the square is 36 cm2 while the area of the skinny rectangle is only 12 cm2.
1 12
12 b. (i)
14 2
⫹ 0 ⫺ 1 ⫽ 6 square units (ii)
35 2
⫹ 68 ⫺ 1 ⫽
8412
6
square units
30. 8 cm by 5 cm 31. No. Try all cases.
6
1
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Section 13.3A 1. a. 800 square units
b. 216 square units
2. a. SA ⫽ 3223 ⫹ 240 square units b. SA ⫽ 360 ⫹ 24 210 square units 3. a. B ⫽
75 2 23
12. a. 545 in3
5. a. 1141 cm
2
b. 130.2 square units
b. 276 cm
6. a. 360 square units
b. 101 in3
13. a. Circumference (2r) is greater than height (6r). b. 3313 % (onethird) of the can
square units
b. A ⫽ 7523 ⫹ 300 square units 4. a. 150 square units
11. a. Volume ⫽ 14 cubic units; surface area ⫽ 46 square units b. Volume ⫽ 7 cubic units; surface area ⫽ 30 square units c. Volume ⫽ 21 cubic units; surface area ⫽ 54 square units
2
b. 896 square units
14. a. V ⫽ 160 23 cubic units c. V ⫽ 600 cubic units
b. V ⫽ 576 cubic units
15. About 9.42 (exactly 3) ft 3
7. S ⫽ 21623 ⫹ 144 219 square units
16. 1.47 ⫻ 108 yd3
8. a. 300 square units
17. a. Approximately 32,100,000 ft3 b. Approximately 463,000 ft2
b. 6 square units
9. 84 ⫻ 264 in2 10. a. 452 square units c. 1810 square units
b. 66 square units d. 141 square units
11. 144 cm2
18. a. V 艐 3.88 yd3 b. V 艐 0.5 m3 c. The computation in (b) is easier. 19. 85,536 m3 20. a. h ⫽ 40 feet b. Approximately 848,000 gallons c. The sphere has the smaller surface area.
12. Sphere 13. 5 liters
21. 36.3 kg
14. 10,044 m2 15. a. A 6 ⫻ 6 ⫻ 1 prism of cubes b. A 9 ⫻ 2 ⫻ 2 prism of cubes c. A 3 ⫻ 4 ⫻ 3 prism of cubes. Surface area is 66 square units. d. A 36 ⫻ 1 ⫻ 1 prism of cubes. Surface area is 146 square units. 16. The new box will require 4 times as much cardboard as the old box. 17. a. 6380 km
b. 5.12 ⫻ 10 km 8
2
c. 26.5%
18. Surface area is 14 of the original. 18 19. SA ⫽ 36 ⫹ cm2 p 20. 150 square units 21. If Jalen could visualize what an aluminum can looked like before it was shaped into a cylinder, this might help. For example, show him a soup can. When you cut off the paper label perpendicular to the base and lay it flat, you will see that it is a rectangle. 22. One way to get something close is to peel an orange with one continuous peel. Then trace the peel onto a piece of paper. Also, you could cut the orange into slices, and make patterns to match the skin on each slice.
Section 13.4A 1. a. 1500 cubic units
b. 144 cubic units
2. Fruity O’s 3. 8172 ft3 4. a. 250 cubic units
b. 172.019 cubic units
3
5. a. 2958 cm b. 342 cm3
6. a. 400 cubic units
b. 1568 cubic units
7. a. 800 cubic units b. 100 3 2651 cubic units c. Part b is larger 8. a. 240 cubic units
b. 1.5 cubic units
b. No
23. a. 8 times
b. 4 times
24. a. 4 times
b. 2 times
25. 12cm pipes 26. a. 8 m3
3 16 ⫽ 2 2 3 2m b. 2
27. 78% 10 艐 7.07 ft 22 29. cylinder height ⫽ 43 units, cone slant height ⫽ 27 units 28.
30. a. 2.625 board feet d. 15 board feet 31. a.
b. 9.375 board feet
b. 42 square units. Make a 1 ⫻ 1 ⫻ 10 stack. c. 30 square units. A 2 ⫻ 2 ⫻ 2 stack with 2 on the top d. 110 and 54; 258 and 96 e. A 1 ⫻ 1 ⫻ n stack for n cubes f. The most cubical arrangement g. Maximal surface area is coolest; minimal surface is warmest. 32. To find volume, Nedra will need to use h, the height of the pyramid, which is perpendicular to the base of the pyramid. To find surface area, she would need to use 1, the slant height, which is perpendicular to the base of the face. By considering two right triangles, one can see that e 1 h because in any right triangle the hypotenuse is always the longest side.
ADDITIONAL PROBLEMS WHERE THE STRATEGY “USE DIMENSIONAL ANALYSIS” IS USEFUL 2. $14.40
b. 7238 cubic units
c. 24 board feet
Many arrangements are possible.
1. 6.3 m
9. a. Doubles b. Quadruples 10. a. 905 cubic units
22. a. 90 in3
3. Light is 1,079,270 times faster than sound.
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CHAPTER REVIEW Section 13.1
4. 100 cubic units
1. (i) Select an attribute to be measured. (ii) Select a unit. (iii) Determine the number of units in the object to be measured. 2. Units are based on some convenient unit rather than a scientifically based unit. 3. a. Inch, foot, yard b. Square inch, square foot, acre c. Cubic inch, cubic foot, cubic yard d. Teaspoon, cup, pint e. Ounce, pound, ton
Chapter 13 Test 1. a. F e. F
b. F f. T
c. T g. F
3. a. squares
d. F h. T
b. cubes
5280 feet # 12 inches # 2.54 cm # 1 km 4. 1 mile ⫽ ⫽ 1.609 km 1 mile 1 foot 1 inch 100,000 cm
5. a. Meter, kilometer, centimeter b. Square meter, square kilometer, square centimeter c. Cubic meter, cubic kilometer, cubic centimeter d. Liter, kiloliter, milliliter e. Gram, kilogram, milligram
5. 7,000,000,000 dm3; from hm3 to dm3 is a move right of three steps on the metric converter, and each step involves moving the decimal point three places. 6. Area is 1/ or approximately 0.318 cubic unit. 7. Volume ⫽ (7.2)(3.4)(5.9) cm3 ⫽ 144.432 cm3 (13)(12) 8. Volume ⫽ ⫽ 52 cm3 3
6. Kilo, centi, milli, hecto, deci7. 129.2⬚F 8. 112.65 kph
9. 27 times bigger
Section 13.2 b. 50
5. 288 cubic units
2. 1 gram
4. (i) Portability, (ii) convertibility, (ii) interrelatedness
1. a. 24
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c. 14
d. 18.8
e. 24
f. 10
b. 25 square units d. 12.25 square units f. 54 square units
2. a. 10.865 square units c. 48 square units e. 40.5 square units
3. a. The sum of the areas of the squares on the sides of a right triangle equals the area of the square on the hypotenuse. b. If the sides of a triangle have lengths a and b and the hypotenuse has length c, then a2 ⫹ b2 ⫽ c 2. 4. The length of the third side is less than 7 ⫹ 9 ⫽ 16.
10. 6 ⫹ 3 25 ⫹ 213 units 11. a. 36 e. 0.000543
b. 8 f. 0.015
c. 216 g. 225
d. 5.43 h. 3780
12. Cutting off one piece with a cut perpendicular between two bases and reassembling gives us a rectangle. The two sides of the rectangle correspond to the base and height of the parallelogram and thus the formula follows. 13. It aids in comparing and converting measurements of volume, capacity, and mass; for example, 1 cm3 of volume equals 1 mL and, if water, weighs 1 g. 14. The convertibility of the metric system makes it easier to learn because the prefixes have the same meaning for all measurements, and converting between measurements involves factors of 10 and thus just movement of the decimal point.
Section 13.3 1.
15. iii i ii
4
6 5
16. a. 10 kg
b. 6 m
c. 500 mL
17. Place an identical copy of the triangle next to the original as shown. The new figure has a side AC at the top that is congruent to AC at the bottom. In the figure AB on the right is congruent to AB on the left. Thus the new figure is a parallelogram which has an area of bh. Since the original triangle is exactly half of the parallelogram, the area of the triangle is 12 bh.
SA ⫽ 148 square units 2. 60 square units 3. 96 square units 4. 90 square units 5. 144 square units
B
Section 13.4
A´
C´
1.
A 4
C
B´
6 5
V ⫽ 120 cubic units 2. 63 cubic units 3. 48 cubic units
18. The height of the cone is 3 times larger. 19. a. Area ⫽ 23s2/2 square units; 䉭ABC is an equilateral triangle with sides of length s22 and height 26s/2. b. Use 䉭ADC as a base. Its area is s2/2 and the height BD from this base is s, so the volume equals s3/6 cubic units.
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20. The circumference flown is 7933 miles; 7933p miles #
5280 feet # 1 second # 1 hour L 33 hours. 1 mile 1100 feet 3600 seconds
21. a. Surface area ⫽ 300 square units; volume ⫽ 300 cubic units. b. Surface area ⫽ 936 ⫹ 1352/ 艐 1366 square units; volume ⫽ 12,168/ 艐 3873 cubic units 22. 4 ⫺ ft2. 3 23. 365 81 艐 4.5 yd 35 24. a. miles 2p
b. Between 29,311 and 29,512 feet
Section 14.1A 1. a. KJL
b. KLJ
c. LJK
d. RTS
2. a. 䉭ABC 艑 䉭EFD or equivalent statement b. 䉭AXD 艑 䉭CXB or equivalent statement 3. RS ⬵ XY, ST ⬵ YZ 4. a. WX; ∠X; XY b. BC ⬵ XY, ∠C, CA ⬵ YW; CA ⬵ YW, ∠A 艑 ∠W, AB ⬵ WX 5. ST ⬵ YZ, ∠T 艑 ∠Z or RS ⬵ XY, ∠R 艑 ∠X 6. a. ∠P, PQ, ∠Q
16. a. 䉭BCD 艑 䉭XYZ, by SAS congruence property b. ∠ABC 艑 ∠WXY (given), ∠2 艑 ∠6 [corresponding parts of congruent triangles in part (a)], thus m(∠ABC) ⫺ m(∠2) ⫽ m(∠WXY) ⫺ m(∠6) or m(∠1) ⫽ m(∠5), so ∠1 艑 ∠5. c. AB ⬵ WX (given), ∠1 艑 ∠5 [part (b)], BD ⬵ XZ [corresponding parts of congruent triangles in part (a)], so 䉭ABD 艑 䉭WXZ by SAS congruence property; ∠A 艑 ∠W, AD ⬵ WZ. d. ∠3 艑 ∠7 [corresponding parts of congruent triangles in part (c)], ∠4 艑 ∠8 [corresponding parts of congruent triangles in part (a)], so m(∠3) ⫹ m(∠4) ⫽ m(∠7) ⫹ m(∠8) and m(∠ADC) ⫽ m(∠WZY), so ∠ADC 艑 ∠WZY. e. Yes, and the quadrilaterals are thus congruent. 17. m(∠D) ⫹ m(∠E) ⫽ 180⬚ (supplementary), m(∠D) ⫽ m(∠E) (congruent), m(∠D) ⫹ m(∠D) ⫽ 180⬚ (substitution), 2m(∠D) ⫽ 180⬚, so m(∠D) ⫽ 90⬚ and m(∠E) ⫽ 90⬚. 18. a. Show 䉭ABD 艑 䉭CDB as in Example 14.3. Then AB ⬵ CD and BC ⬵ DA by corresponding parts. b. ∠A 艑 ∠C follow because they deal with corresponding parts. Show 䉭ACD 艑 䉭CAB similarly, show ∠B 艑 ∠D. 19. a. The two triangles are congruent because of the SAS property. b. ∠A 艑 ∠D, AB ⬵ DE, BC ⬵ EF but 䉭ABC is not congruent to 䉭DEF
C
b. ∠M 艑 ∠Q, MN ⬵ QR, ∠N 艑 ∠R or ∠N 艑 ∠R, NL ⬵ RP, ∠L 艑 ∠P
E
7. ∠R 艑 ∠Y, ∠S 艑 ∠Z, ∠T 艑 ∠X, RT ⬵ YX, and ST ⬵ ZX.
B
8. Use SSS. 9. a. Yes c. Yes
b. No, congruent angles are not included angles.
F
10. a. Yes; m(∠C) ⫽ 70⬚, m(∠A) ⫽ 40⬚, m(∠E) ⫽ m(∠F) ⫽ 70⬚, so 䉭ABC 艑 䉭DEF by SAS or ASA. b. No; m(∠T) ⫽ 50⬚, VU ⫽ 4, m(∠V) ⫽ 50⬚, but there is no correspondence of the sides. c. Yes, m(∠K) ⫽ 60⬚, XZ ⫽ 7, so 䉭JKL 艑 䉭YZX by ASA. 11. a. SAS
b. ASA
C Z 4 cm 70° A 5 cm
B
A
c. ASA
12. a. ∠ A 艑 ∠D, AB ⬵ DE, BC ⬵ EF b. No c. No; this example is a counterexample. 13.
D
4 cm 40° X 5 cm
20. The parallelogram can be split into two congruent triangles by a diagonal; congruent by SSS. Therefore, the rectangle, rhombus, and square can also be split into two congruent triangles by a diagonal, since they are each parallelograms. A trapezoid that is not a parallelogram cannot be divided into two congruent triangles.
Section 14.2A Y
BC > YZ 14. a. AB ⬵ CB (they measured it off), ∠TAB and ∠DCB are both right angles (directly across and directly away from), ∠TBA 艑 ∠DBC (vertical angles). b. Yes; ASA c. DC ⬵ TA, so they can measure DC and that equals the width of the river. 15. a. For example, have 3 pairs of angles congruent. b. Not possible. Consider possible cases: (1) 3 sides, 1 angle, apply SSS; (2) 2 sides, 2 angles (third angle would also have to be), apply ASA; (3) 1 side, 3 angles, apply ASA. c. Not possible, 艑 by either SSS or ASA. d. Not possible, 艑 by definition.
1. a. 䉭ABC ⬃ 䉭DEC, by AA similarity b. 䉭FGH ⬃ 䉭IJK, by SSS similarity c. 䉭XYZ ⬃ 䉭VWZ, by SAS similarity d. Not similar, because 23 Z 35 . 2. a. EF ⫽ 10, DF ⫽ 14 b. RS ⫽ 6, RT ⫽ 6 210, LN ⫽ 210 3. a. EI ⫽ 10, HI ⫽ 3 23, FI ⫽ 5 23 b. TV ⫽ 16 3 , WX ⫽ 20 4. a. F; triangle may have sides with different lengths b. T; by AA c. F; see part (a) d. T; by AA 5. 9 6. a. AA
b. 25 m
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7. a. 16.8 m b. 6 m c. 9.6 m
22. a. 3 units b. 4 units c. 513 units
8.
1 1 1 + 4 ¢ 2 ≤ + Á + 4n  2 ¢ n  1 ≤ R units , n ⬎ 1 3 3 3 19. a. 12 square units b. 1313 square units c. 1325 27 square units d. 3 B 1 +
d. 3 B 3 + 1 + 4 ¢
1 32
1 ≤ + Á + 4n  2 ¢ 2n  4 ≤ R square units , n ⬎ 1. 3
24. The ratios of the sides of the rectangles are 4⬊6 and 5⬊8. These ratios are not equal, thus a true enlargement could not be done. 9. a. AB ‘ DE and AD ‘ BE from given information, and therefore ABED is a parallelogram by definition. Similarly, BC ‘ EF, BE ‘ CF, and BCFE is a parallelogram b. Opposite sides of a parallelogram are congruent. c. DE/EF d. Yes 10. a. ∠P 艑 ∠P 艑 ∠P, ∠PAD 艑 ∠PBE 艑 ∠PCF (corresponding angles property) so that 䉭PAD ⬃ 䉭PBE ⬃ 䉭PCF by AA similarity property b. a/(a ⫹ b) ⫽ x/(x ⫹ y), so a(x ⫹ y) ⫽ x(a ⫹ b), ax ⫹ ay ⫽ xa ⫹ xb, or ay ⫽ bx c. a/(a ⫹ b ⫹ c) ⫽ x/(x ⫹ y ⫹ z), so a(x ⫹ y ⫹ z) ⫽ x(a ⫹ b ⫹ c), ax ⫹ ay ⫹ az ⫽ ax ⫹ bx ⫹ cx, or ay ⫹ az ⫽ bx ⫹ cx d. Combining parts (b) and (c) yields bx ⫹ az ⫽ bx ⫹ cx or az ⫽ cx. e. bx/cx ⫽ ay/az or b/c ⫽ y/z f. Yes
Section 14.3A 1. Follow the steps of the 1. Copy a Line Segment construction. 2. Follow the steps of the 2. Copy an Angle construction. 3. Follow the steps of the 3. Construct a Perpendicular Bisector construction. 4. Follow the steps of the 4. Bisect an Angle construction. 5. Follow the steps of the 5. Construct a Perpendicular Line through a Point on a Line construction. 6. Follow the steps of the 6. Construct a Perpendicular Line through a Point not on a Line construction. 7. Follow the steps of the 7. Construct a Line Parallel to a given Line through a Point not on the Line construction. 8.
b C
c
B
11. a. EF/FG ⫽ KJ/JG b. BC/CD ⫽ JG/GI c. DB/BA ⫽ HF/FE
a
12. 22 meters 13. a. 䉭ABC ⬃ 䉭EDC, by AA similarity. We assume that both the tree and the person are perpendicular to the ground, and the angle of incidence, ∠ECD, is congruent to the angle of reflection, ∠ACB. b. 27 meters 14. a. 150 feet b. 46 feet 15. a. 9.6 inches b. Image would be half as long (i.e., the image of his thumb would be 4.8 inches long). 16. 䉭ABC ⬃ 䉭DEF, so 䉭ABC is also a right isosceles triangle. Therefore, the height BC is the same length as the leg AC. 17. BD ⫽ 40. 18. The area of the larger triangle will be four times the area of the smaller triangle. 19. 䉭ADC ⬃ 䉭CDB; therefore,
a x ⫽ , or a ⫽ x2 x 1
20. AB ⫽ 2208 ⫽ 4 213. a c 21. The three triangles are similar by AA similarity. Therefore, ⫽ , x a b c or (i) a2 ⫽ cx and ⫽ , or (ii) b2 ⫽ c2 ⫺ cx. c  x b Substituting a2 for cx in (ii) we have b2 ⫽ c2 ⫺ a2, or a2 ⫹ b2 ⫽ c2.
d A D a. Follow the steps of the 7. Construct a Line Parallel to a given Line through a Point not on the Line construction. b. Follow the steps of the 7. Construct a Line Parallel to a given Line through a point not on the Line construction. c. Follow the steps of the 6. Construct a Perpendicular Line through a point not on a Line construction. d. Follow the steps of the 6. Construct a Perpendicular Line through a point not on a Line through a Point not on a Line construction. 9. a. Construct a pair of perpendicular lines using any of the perpendicular line constructions. b. Bisect the angle constructed in part a. c. Copy the angles constructed in parts a and b so that they are adjacent to each other because the sum of their angle measures is 135⬚. d. Bisect the angle in part b and copy it to be adjacent to the angle in part b. 10. c. k ‘ l since corresponding angles are congruent (right angles). 11. They all intersect in a single point. 12. They all intersect in a single point.
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13. a.
b.
5
13 5
3
12 13 units
4 5 units
3. a. Follow the steps of the Inscribed Circle of a Triangle construction using the angle bisector tool and perpendicular tool on The Geometer’s Sketchpad® to find the center and radius respectively. b. Inside 4.
c.
17 8 15 15 units 14. a. Construction b. ∠A 艑 ∠P, ∠B 艑 ∠Q, ∠C 艑 ∠R c. 䉭ABC ⬃ 䉭PQR, since angles are congruent and all sides are proportional d. SSS similarity 15. a. Ratios are the same: PQ/AB ⫽ PR/ AC⫽ QR/BC. b. 䉭ABC ⬃ 䉭PQR, since angles are congruent and sides are proportional. c. AA similarity 16. They do in an equilateral triangle. In every isosceles triangle the median to the base and the angle bisector of the vertex angle coincide. 17. An isosceles triangle that is not equilateral a. inside
b. vertex of the right angle
c. outside
18. They do in an equilateral triangle. In every isosceles triangle the perpendicular bisector of the base and median to the base coincide.
5. Follow the steps of the Equilateral Triangle construction
19. An isosceles triangle that is not equilateral
6. Yes, regular hexagon.
20. a. PA ⬵ PB, QA ⬵ QB (by construction), PQ ⬵ PQ. b. PA ⬵ PB, ∠APR 艑 ∠BPR [corresponding angles of congruent triangles in part (a)], PR ⬵ PR. c. ∠PRA and ∠PRB are supplementary by definition since A, R, and B are collinear, ∠PRA 艑 ∠PRB since they are corresponding angles of congruent triangles in part (b). Since these angles are supplementary and congruent, they are right angles. d. AR ⬵ BR since they are corresponding sides of congruent triangles in part (b). Hence, by definition, PQ bisects AB. 21. Bisect ∠A and call the intersection of the angle bisector with BC point D. Then ∠B 艑 ∠C (given), ∠BAD 艑 ∠CAD (angle bisector), and AD ⬵ AD. Therefore, 䉭BAD 艑 䉭CAD and corresponding sides AB and AC are congruent. By definition 䉭ABC is isosceles.
7. b. Regular pentagon. 8. a. Construct equilateral triangle, bisect one angle, and bisect one of the bisected angles. b. Add 15⬚ and 60⬚ angles. c. Add 15⬚ angle to a right angle. 9. (a), (b), (d), (h), and (i). 10. a. Yes. r ⫽ 0, s ⫽ 0, u ⫽ 1, v ⫽ 1 b. No c. Yes. r ⫽ 1, s ⫽ 1, u ⫽ 2, v ⫽ 1 11. The resulting segments AT, TS, and SB are congruent, thus dividing AB into three congruent segments. Below is the start of the construction for dividing AB into 6 congruent pieces.
22. If Tammy makes her first two arcs from one endpoint with one compass size and the second two arcs from the other endpoint with a different compass opening, she will have constructed an (invisible) kite. The original segment and the constructed line are diagonals of the kite; they are perpendicular, but they do not bisect each other. So Tammy is correct that the constructed line is perpendicular to the segment, but it is not a bisector. 23. Look at three examples: an acute triangle, an obtuse triangle, and a right triangle. In an acute triangle, all of the altitudes are inside the triangle, for an obtuse triangle, two of the altitudes are outside of the triangle, and in a right triangle, one altitude is inside the triangle and the other two are the sides.
S
B
A
Q R 12.
3 2
3
Section 14.4A 1. Follow the steps of the Circumscribed Circle of a Triangle construction. 2. Follow the steps of the Inscribed Circle of a Triangle construction.
d. Yes
3
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13. The hypotenuse is twice as long as the shortest leg. 14. The ratio length⬊width should be close to the golden 1 + 25 , which is about 1.618. 2 15. c. It lies inside the triangle. ratio ⫽
16. c. It is outside the triangle. 17. 䉭C1C2C3 is always an equilateral triangle (Napoleon’s theorem). 18. Case 1: a 1.
1
B
AB =
1
a
A
a B 1
AB =
1
a
A a 1 Case 3: If a ⫽ 1, then
2. AB ⬵ CB (rhombus), ∠ABE 艑 ∠CBE (䉭ABD 艑 䉭CBD, so vertex angle bisected by diagonal), BE ⬵ BE (common side), 䉭ABE 艑 䉭CBE (SAS), so ∠AEB 艑 ∠CEB (corresponding parts). Thus AC is perpendicular to BD, since ∠AEB and ∠CEB are congruent and supplementary. 3. Since STUV is a rhombus, it has 4 congruent sides. We need to show that it has 4 right angles. Since a rhombus is a parallelogram, opposite angles ∠S and ∠U are congruent (right angles) and adjacent angles are supplementary; ∠V and ∠T are thus right angles. 4. Since AB ‘ CD, we have ∠EBA 艑 ∠EDC and ∠EAB 艑 ∠ECD. Also, AB ⬵ CD. Therefore, 䉭EAB 艑 䉭ECD (ASA congruence property). By corresponding parts, AE ⬵ CE and BE ⬵ DE. 5. Using the SSS congruence property, construct 䉭ABC, where AB ⫽ a, BC ⫽ b, and AC ⫽ d. Using AC as one side, similarly construct 䉭CDA, where CD ⫽ a and DA ⫽ b. ABCD will be the desired parallelogram.
1
Case 2: a 1.
A77
1 ⫽ 1. a
19. a. 22 b. 䉭AEC and 䉭EBC are both isosceles triangles sharing base angle ∠ECB and are therefore similar by AA similarity property. Corresponding sides are thus proportional, so AE/EB ⫽ EC/BC or a/x ⫽ x/b. 20. Lay the piece diagonally across the ruled paper so that the edge of the piece acts as a transversal. If the piece crosses six parallel lines, portions of the strip between them will be congruent.
6. SV ⬵ TU (all sides are congruent), ST ⬵ TS, and TV ⬵ SU (diagonals congruent), so 䉭STV 艑 䉭TSU (SSS congruence property). Thus supplementary angles ∠TSV and ∠STU (adjacent angles of parallelogram are supplementary) are also congruent (corresponding parts of congruent triangles). Therefore, ∠TSV and ∠STU are both right angles. Similarly, ∠SVU and ∠TUV are right angles. Thus, STUV is a square. 7. Construct an equilateral 䉭ABC using side length a. Using AC as one side, construct equilateral 䉭CDA. ABCD is the desired rhombus. 8. Construct AB, the length of the longer base. Construct a segment whose length is a ⫺ b and bisect this segment. Mark off an arc centered at A with length (a ⫺ b)/2, intersecting AB at R, and an arc centered at B with the same length, intersecting AB at S. At R and S construct segments perpendicular to AB. With compass set at length c, mark off an arc centered at B, intersecting perpendicular from S at point C. Similarly, draw an arc from A, intersecting the perpendicular from R at point D. ABCD is the desired isosceles trapezoid. 9. AD/AC ⫽ AC/AB (AC is geometric mean) and ∠A 艑 ∠A, so 䉭ADC ⬃ 䉭ACB (SAS similarity property). Therefore, ∠ADC 艑 ∠ACB and since ∠ADC is a right angle, so is ∠ACB. Thus 䉭ABC is a right triangle. 10. a. M1M2M3M4 will be a rectangle when the diagonals PR and QS are perpendicular. b. M1M2M3M4 will be a rhombus when the diagonals PR and QS are congruent. c. M1M2M3M4 will be a square when the diagonals PR and QS are both perpendicular and congruent.
21. Two are sufficient, assuming they are correct. Some teachers may recommend constructing the third one as a check.
Section 14.5A 1. a. AC ⬵ AB, ∠CAD 艑 ∠BAD b. ∠C 艑 ∠B, opposite congruent sides c. ASA d. Corresponding parts of congruent triangles are congruent. e. ∠ADC and ∠ADB are right angles, since they are congruent and supplementary. f. AD ⬜ BC and AD bisects BC (divides it into 2 congruent pieces).
11. a. AA similarity property b. AA similarity property c. a2 ⫹ b2 ⫽ x2 ⫹ xy ⫹ xy ⫹ y 2 ⫽ x2 ⫹ 2xy ⫹ y 2 ⫽ (x ⫹ y)2 ⫽ c2. 12. a. By SSS, 䉭ABC 艑 䉭ADC . Hence ∠BAC 艑 ∠DAC, so 䉭AEB 艑 䉭AED by SAS. Thus, ∠BEA and ∠DEA are congruent supplementary angles, so each is a right angle. b. Area ⫽ (AE ⫻ BD)/2 ⫹ (CE ⫻ BD)/2 ⫽ (AC ⫻ BD)/2. 13. 䉭BAD 艑 䉭BAC by SAS. Therefore, BD ⫽ BC ⫽ BC. Since 䉭ACD and 䉭BDC are both isosceles, their base angles are congruent. By addition, ∠ACB 艑 ∠ADB. Thus 䉭BAD 艑 䉭BAC by SAS. Therefore, 䉭BAC 艑 䉭BAC, since they are both congruent to 䉭BAD.
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b. If two angles of one triangle are congruent, respectively, to two angles of another triangle, the triangles are similar. c. If three sides of one triangle are proportional, respectively, to three sides of another triangle, the triangles are similar.
14. 䉭ABC ⬃ 䉭ABD. Thus ∠ABC 艑 ∠ABD . Also m∠ABD, m∠ABC, so m∠ABC m∠ABC. But ∠ABC 艑 ∠ABC was given. Thus we have a contradiction. A similar contradiction is reached if m∠ABD m∠ABC. 15. a. 䉭ABD 艑 䉭FBC and 䉭ACE 艑 䉭KCB by SAS. b. The triangles have congruent bases and the sum of their heights is BE. The sum of their areas is 1/2 the area of the square. c. This verification uses the same method as in part (b). d. Combine parts (a), (b), and (c). e. The sum of the area of the squares on the legs of a right 䉭ABC is equal to the area of the square on the hypotenuse. This is the geometric statement of the Pythogorean theorem.
3. a. Not necessarily similar.
B
E
16. The area of the midquad is 12 the area of the original quadrilateral.
A
17. Students whose Van Hiele level is less than 3 often do not understand the need for proof. The teacher’s job at this point would be to try to help Willard raise his Van Hiele level.
ADDITIONAL PROBLEMS WHERE THE STRATEGY “IDENTIFY SUBGOALS” IS USEFUL
C
D
F
b. SSS c. AA 4.
1. Factoring 360 into its prime factorization, we have 360 ⫽ 23 䡠 32 䡠 5. To find the smallest square having 23 䡠 32 䡠 5 as a factor, we need to find the smallest number with the same prime factors but with even exponents. Thus 24 䡠 32 䡠 52 ⫽ 3600 is the smallest such number. 2. Subgoal: Find the lengths of the sides. Volume is 192 cm3. 3. If 32% of n is 128, then 16% of n is 64, or 8% of n is 32, or 1% of n is 4. Therefore, n ⫽ 400.
Shadow and stick
Direct Sighting
Section 14.3
CHAPTER REVIEW Section 14.1
1. See Figure 14.22.
1. 䉭ABC 艑 䉭DEF if and only if AB ⬵ DE, BC ⬵ EF, AC ⬵ DF, ∠A 艑 ∠D, ∠B 艑 ∠E, ∠C 艑 ∠F.
2. See Figure 14.24.
2. a. If two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of another triangle, the triangles are congruent. b. If two angles and the included side of one triangle are congruent, respectively, to two angles and the included side of another triangle, the triangles are congruent. c. If three sides of one triangle are congruent, respectively, to three sides of another triangle, the triangles are congruent.
4. See Figure 14.30.
3. a. SSS b. Not necessarily congruent
B
Reflection in a mirror
3. See Figure 14.27. 5. See Figure 14.32. 6. See Figure 14.34. 7. See Figure 14.37.
Section 14.4 1. Construct the perpendicular bisectors of two of the sides. Use the distance from their intersection to any vertex as a radius to draw the circle. 2. Construct the angle bisectors of two angles. Then construct a perpendicular from their intersection to any of the sides. Use the distance from the intersection of the angle bisectors to the selected side as the radius of the circle.
E
3. See Figure 14.45.
A
C D
F
c. ASA d. SAS
Section 14.2 1. 䉭ABC ⬃ 䉭DEF if and only if ∠A 艑 ∠D, ∠B 艑 ∠E, ∠C 艑 ∠F, AB BC AC ⫽ ⫽ . DE EF DF 2. a. If two sides of one triangle are proportional, respectively, to two sides of another triangle and their included angles are congruent, the triangles are similar.
4. a. Construct a pair of perpendicular lines. Mark off a pair of congruent segments having endpoints at the point of intersection. Then construct two lines perpendicular at the other ends of the segments. A square should be apparent. b. Draw a circle. Using the radius, mark off six points around the circle, putting the compass point at the next pencil point. Connect the six resulting points consecutively on the circle to form a hexagon. c. Construct a square, its circumscribed circle, and the perpendicular bisectors of two adjacent sides of the square extended to intersect the circle. Connect the eight consecutive points on the circle to form an octagon.
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9. Construct a right triangle with sides of length 1 and 2. The hypotenuse will have length 25. Construct a square with sides of length 25.
5.
10. From 䉭ABC 艑 䉭CBA, we have ∠A 艑 ∠C. Sides opposite congruent angles are congruent and thus 䉭ABC is isosceles. 11. From 䉭ABC 艑 䉭BCA, we have ∠A 艑 ∠B and ∠B 艑 ∠C. Thus ∠A 艑 ∠B 艑 ∠C and 䉭ABC is equilateral.
AB 3 AB 2
12.
A
B 75°
Section 14.5
75° 1.5 in.
1. See Example 14.15.
1.5 in.
13. Measure the height of a person (hp) and the length of the person’s shadow (sp) and the length of the shadow of the tree (st). Set up the hp ht proportion ⫽ and solve for ht, which is the height of the tree. sp st
2. See the discussion preceding Figure 14.59.
Chapter 14 Test 1. a. T b. F c. T d. T e. T f. T g. F h. F
14. It means that if two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of another triangle, then all sides and angles of one triangle are congruent to the corresponding parts of the other triangle. 15. Since PA ⬵ PB and AC ⬵ BC,Í PACB ! Í is! a kite that has perpendicular diagonals. Thus, PC ⊥ AB. 16. a. Two objects are similar if all pairs of corresponding angles are congruent and all pairs of corresponding sides are proportional (i.e., the objects are the same shape). b. (i) F (ii) F (iii) T
2. SAS, ASA, SSS 3. (b) and (d) 4. Construct an equilateral triangle and circumscribe a circle about it. Then bisect the central angles twice. 5. The circumcenter and incenter for an equilateral triangle are the same point. 6. a. Yes. The triangle on the left is isosceles so the base angles are 75⬚ and the triangle on the right has a third angle of 75⬚. Thus, the triangles are congruent by ASA. b. No. In the triangle on the right, the congruent angle isn’t between the congruent sides. BE 9 AE 15 7. Since ⫽ ⫽ ⫽ and ∠AEB 艑 ∠DEC, the triangles are ED 3 EC 5 similar by SAS. 8.
17. a. Let P and Q be points on AB such that DP ⬜ AB and CQ ⬜ AB. Since parallel lines are everywhere equidistant, DP ⬵ CQ. Also ∠APD 艑 ∠BQC, since they are both right angles and ∠A 艑 ∠B (given). Thus ∠ADP 艑 ∠BCQ. Hence 䉭APD 艑 䉭BQC by the ASA congruence property and AD ⬵ BC. b. As in part (a), draw DP and CQ. Then show that AP ⬵ BQ by the Pythagorean theorem. Thus 䉭APD 艑 䉭BQC by the SSS congruence property and ∠A 艑 ∠B. 18. a. Because the prime factorization of 9 includes two 3s b. 40⬚ c. 20⬚ d. It is impossible. 19. a. 4 b. 147 8 20. 22 feet tall
P
Section 15.1A 1. a. 213 b. 25 2. a. 22 + 2 22 ⫽ 3 22; yes b. 234 + 2117 2277; no 3. a. ¢
3 , 2≤ 2
b. (⫺1, 2) c. ¢
l
5 9 , ≤ 2 2
d. (3, ⫺6)
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4. a. 21
b. 43
18. a. ⫺1.062
1 , yes 5. a.  2
19. a. 20 m
b. 1, yes
7. Yes 8. a. Slopes of AB and PQ equal 1; parallel b. Slopes equal 23 ; parallel 12 c. Slope of AB is 25 51 , and slope of PQ is 25 . Thus, they are not parallel.
21. a. Since PO ‘ RT (both horizontal), ∠QPO 艑 ∠SRT. Also ∠O 艑 ∠T (both right angles). Thus 䉭PQO ⬃ 䉭RST by AA similarity. b. OP and TR are corresponding sides, as are OQ and TS; since the triangles are similar, corresponding sides are proportional. Interchange means of proportion. c. y1/x1 is the slope of PQ and y2/x2 is the slope of RS. By part (b) they are equal. y2  y1 2 x2  x1 2 ≤ + ¢ ≤ ⫹ 22. PM ⫹ MQ ⫽ ¢ 2 2 B y2  y1 2 x2  x1 2 ≤ + ¢ ≤ 2 2 B
9. a. Top and bottom are both horizontal, 4  1 4  1 ⫽ ⫺3, ⫽ ⫺3; parallelogram 1  2 4  5 b. Top, bottom are both horizontal, sides are vertical; parallelogram 2 b. 3
11. a. Yes
c. No slope
b. Yes
c. No
d. No
13. a. AD and BC have slope 52 , AB and CD have slope  25 , so AB ⬜ AD, AB ⬜ BC, BC ⬜ CD, and AD ⬜ CD.
b. AC ⫽ 2(5  13) + 37  (  2)4 ⫽ 2145; BD ⫽ 2(15  3)2 + (3  2)2 ⫽ 2145; so AC ⫽ BD. c. Both pairs are congruent. 1 d. No, the slope of AC ⫽  98 , the slope of BD ⫽ 12 . e. Diagonals are congruent and bisect each other but are not necessarily perpendicular. 2
⫽2
(x2  x1)2 + ( y2  y1)2
B
4
⫽ 2(x2  x1)2 + (y2  y1)2 PQ ⫽ 2(x2  x1)2 + (y2  y1)2; and since PM ⫹ MQ ⫽ PQ, P, M, and Q are collinear. Also, 2 2 y1 + y2 x1 + x2 PM 2 ⫽ ¢  x1 ≤ + ¢  y1 ≤ 2 2 ⫽¢
2
14. A and B with the following pairs: a. (1, 5), (3, 5); (2, 5), (4, 5); (3, 5), (5, 5); (1, 3), (3, 3); (2, 3), (4, 3); (3, 3), (5, 3); (1, 2), (3, 2); (2, 2), (4, 2); (3, 2), (5, 2); (1, 1), (3, 1); (2, 1), (4, 1); (3, 1), (5, 1) b. (2, 5), (4, 5); (2, 3), (4, 3); (2, 2), (4, 2); (2, 1), (4, 1) c. (2, 2), (4, 2) d. (2, 2), (4, 2) 15. a. Isosceles right
¢
d. 0
12. a. Yes, a rectangle. b. Not a rectangle.
x1 + x2 2

2
+ ¢
y1 + y2 2

2y1 2
≤
2
MQ 2 ⫽ ¢ x2 ⫽¢
y1 + y2 2 x1 + x2 2 ≤ + ¢ y2 ≤ 2 2
y2  y1 2 x2  x1 2 ≤ + ¢ ≤ ⫽ PM2 2 2
Hence PM ⫽ MQ. 23. a.
z
b.
z
b. Obtuse isosceles
17. Sides of ABCD are 13 the length of the corresponding sides of ABCD.
y
y
x
x
c.
z
C D
2x1 2
y2  y1 2 x2  x1 2 ⫽¢ ≤ + ¢ ≤ 2 2
16. a. AB ⫽ 240, BC ⫽ 2160, AC ⫽ 2200. Since AC 2 ⫽ AB2 ⫹ BC 2, 䉭ABC is a right triangle. b. DE ⫽ 252, EF ⫽ 232, DF ⫽ 268. Since DF 2 DE 2 ⫹ EF 2, 䉭DEF is not a right triangle.
D´
d. 3.18
c. 4770 cm
20. 6.667%
6. a. Steep slope up to the right. b. Gradual slope down to the right. c. Steep slope down to the right. d. About a 45⬚ slope up to the right.
10. a. 1
c. ⫺0.104
b. 2.814 b. 25 m
y
B C´ B´ A´ A
x 24. No. Janiece is using the incorrect fact 2x2 + y 2 ⫽ x ⫹ y. This is true only when x 2 ⫹ y 2 ⫽ (x ⫹ y)2, which is true only when one of x or y is zero.
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25. In this case, it doesn’t matter due to the squaring. But it is good practice to subtract the numbers in the same order since in some formulas, such as the slope formula, it is critical to follow this practice.
14. a. (2, 5)
8
Section 15.2A
b. 9 ⫽  23 (  6) ⫹ 5 ⫽ 4 ⫹ 5 ⫽ 9
1. a. 5 ⫽ 7(1) ⫺ 2 ⫽ 5
6
2. There are three points in each answer, however, there are infinitely many correct possibilities in each case. a. (3, 0), (0, ⫺2), (6, 2) b. (0, 0), (4, ⫺1), (2, ⫺1/2) 3. a. 3/2, ⫺5
4 2
b. 2/7, ⫺8/7
4. a. y ⫽ 3x ⫹ 6; 3; 6 5. a. y ⫽ 3x ⫹ 7
b. y ⫽
3 3 4 x; 4 ;
0 c. y ⫽ 23 x ⫹ 5
b. y ⫽ 2x ⫺ 3
⫺5
5
10
5
10
6. Any point of form (x, 3) for any x b. No solution
7. a.
6 4 2 ⫺5 b. They all pass through (0, 3). c. A family of lines passing through (0, d). 8. a. x ⫽ 1
b. x ⫽ ⫺5
9. a. ( y ⫹ 3) ⫽ ⫺2(x ⫺ 2) 10. a. y ⫽ 2x ⫺ 11 11. a. y ⫽ ( 32 )x + 9
c. y ⫽ 6
⫺4 d. y ⫽ ⫺3
3 2
c. (⫺3, 1)
b. y ⫽ 2x ⫺ 5
1 6 (x
⫺6
b. ( y ⫹ 4) ⫽ 32 (x + 1)
b. y ⫽  32 x 
⫺2
1 6 (x
1 6x
 6), ( y ⫺ 2) ⫽  0), and y ⫽ + 2 12. a. y ⫺ 3 ⫽ b. y ⫺ 8 ⫽ ⫺2(x ⫹ 4) or y ⫹ 6 ⫽ ⫺2(x ⫺ 3) and y ⫽ ⫺2x 13. a.
6
b.
4 2 ⫺10
c.
d.
y
⫺2
5
⫺4 y
⫺6 x
e.
⫺5
d. Infinitely many solutions of the form (x, 23 x ⫹ 3).
4
f.
2 ⫺10 x
⫺5
⫺2 ⫺4
5
A81
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15. a. x ⫽ 58 , y ⫺  15 16. a. 6x ⫹ 2y ⫽ 7; no 17. a. (⫺3, ⫺16)
Section 15.3A
b. x ⫽  13 , y ⫽  20 3 c. x ⫽
b. Yes
y ⫽ ⫺1
c. (0, ⫺4)
b. (⫺4, 0)
18. a. No solution c. Unique solution e. No solution
3 2,
d.
(157 44 ,
 15 44 )
b. Infinitely many d. Unique solution f. Infinitely many
2  0 ⫽ ⫺1; 5  3 1 + 3 slope of TU ⫽ ⫽ ⫺1; RS ‘ TU; 4 + 2 0 + 1 1 slope of ST ⫽ ⫽ ; 3 + 4 7 2 + 3 1 slope of RU ⫽ ⫽ ; ST ‘ RU. Since both pairs of opposite 5 + 2 7 sides are parallel, RSTU is a parallelogram.
6. Slope of RS ⫽
22. (x ⫹ 2)2 ⫹ ( y ⫺ 5)2 ⫽ 10 23. y ⫽ 5x, y ⫽  34 x, y ⫽ 52 x ⫹ 23 5 24. Point D is (6, 3); equation of median is y ⫽  43 x ⫹ 11 25. y ⫽ 2x ⫺ 1 26. a. $335, $425, $537.50, $650, $4.50n ⫹ 200 b. y ⫽ 4.5x ⫹ 200 c. 4.5; cost per person d. 200; the fixed costs
b. II
b. G(a, b)
5. a. X ⫽ (0, 0), Y ⫽ (8, 0), Z ⫽ (4, 5) b. X ⫽ (⫺4, 0), Y ⫽ (4, 0), Z ⫽ (0, 5)
21. a. (x ⫹ 2)2 ⫹ ( y ⫺ 3)2 ⫽ 4 b. (x ⫺ 3)2 ⫹ ( y ⫹ 4)2 ⫽ 5 c. (x ⫹ 3)2 ⫹ ( y ⫹ 5)2 ⫽ 34 d. (x ⫺ 2)2 ⫹ ( y ⫹ 2)2 ⫽ 17
28. a. I
b. (⫺1, 3)
4. a. Q(0, 0), R(6, 0), S(0, 4) b. Q(0, 0), R(a, 0), S(0, b)
b. (⫺1, 3); r ⫽ 7
27. a. ae ⫽ bd and ce ⫽ bf c. ae bd
2. a. (⫺2, 4)
3. a. C(a, a), D(0, a)
19. a. 32 ⫹ 42 ⫽ 9 ⫹ 16 ⫽ 25 b. (⫺3)2 ⫹ 52 ⫽ 9 ⫹ 25 ⫽ 34 20. a. (3, 2); r ⫽ 5
1. a. (3, 5), (⫺3, 5), or (3, ⫺5) b. (1, ⫺2), (⫺5, ⫺2), or (9, 8)
7. Slope of AB ⫽  32 and slope of CD ⫽  32 , so AB ‘ CD. Slope of BC ⫽ 23 and slope of AD ⫽ 23 , so BC ‘ AD. Since (  32 )(23 ) ⫽ ⫺1, AB ⬜ BC. Thus ABCD is a parallelogram (opposite sides parallel) with a right angle, and thus a rectangle.
b. ae ⫽ bd and ce bf
c. III
d. III
e. IV
f. yaxis
8. a. (0, 4) b. (4, 5) c. Both equal 1/4, MN ‘ AB d. MN ⫽ 217, AB ⫽ 268 ⫽ 2 217, MN ⫽ (1/2)AB
4
f
9. AC ⫽ 2a2 + b2; BD ⫽ 2(0  a)2 + (b  0)3 ⫽ 2a2 + b2
a
10. a. y ⫽ 72 x
b 2
⫺5
c d ⫺2
5 e
⫺4
29. AP: x 2 ⫹ y 2 ⫽ 9 ① BP: (a ⫺ x)2 ⫹ y 2 ⫽ 16 ② CP: (a ⫺ x)2 ⫹ (b ⫺ y)2 ⫽ 25 ③ DP: Want to find x 2 ⫹ (b ⫺ y)2 ⫽ DP 2 ③ ⫺ ② (b ⫺ y)2 ⫺ y 2 ⫽ 9 ⫹① x2 ⫹ y2 ⫽ 9 (b ⫺ y)2 ⫹ x2 ⫽ 18, so DP ⫽ 218 30. a. 100 people b. 125 people c. 225 people 31. a. 2 b. (0, 1) and ( 45 , 35 ) 32. He can use either point, so he should choose the point that looks like it would make his calculations easier.
b. y ⫽  23 x + 48 5 c. y ⫽ 2x ⫺ 24 d. (14, 4); yes e. The medians are concurrent (meet at a single point). f. centroid 11. Yes 12. a. If AB ⫽ BC, then 2a2 + b2 ⫽ 2(a  c)2 + b2 or a2 ⫽ (a ⫺ c)2, a2 ⫽ a2 ⫺ 2ac ⫹ c2, c2 ⫽ 2ac, and then c ⫽ 2a. b. The midpoint of AC has coordinates (a, 0), and thus the median from B is vertical and thereby perpendicular to horizontal AC 13. a. y ⫽ 21 x b. x ⫽ 8 c. y ⫽  43 x + d. (8, 4), yes e. orthocenter
44 3
14. The midpoint of AC is ¢
¢
a + b c , ≤ . The midpoint of BD is 2 2
b + a c , ≤ . The diagonals meet at the midpoint of each, thus 2 2
bisecting each other. 15. a. The slope of QR is is  ¢
b , so the slope of l (perpendicular to QR) a  c
a  c c  a . ≤ or b b
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b. Since the point has an xcoordinate of a, it is on line m. c  a)a c  a c  a x yields Substituting into y ⫽ ⫽¢ ≤ a. b b b which is satisfied, so the point lies on line l also. b a . c. The slope of PQ is , so the slope of n (perpendicular to PQ) is a b 16. Impossible. The length of the horizontal side is 2a. The height is b, a whole number. Also, 2a2 + b2 ⫽ 2a, so a2 ⫹ b2 ⫽ 4a2, b2 ⫽ 3a2, b ⫽ 23, which is irrational. a 17. 7 and 11 years old 18. 5 tricycles and 2 bicycles 19. Mike, $7000; Joan, $4000 20. 75 dimes and 35 quarters 21. a. 35 paths b. 18 paths c. 18 35
A83
Section 15.3 1. (4, 53 )
2. (0, 0) 3. (6, 212 )
Chapter 15 Test 1. a. T b. F c. T d. F e. F f. T g. T h. F 2. Slopeintercept: y ⫽ mx ⫹ b Pointslope: ( y ⫺ y0) ⫽ m(x ⫺ x0) 3. Length is 241; midpoint is (3, 4.5); slope is 54 .
22. Her point (b, c) could be renamed (a, c), since it is directly above (a, 0). That would reduce the number of variables, which is one of the criteria of good placement. Jolene may place her triangle as she did, although it generally simplifies computations if the right angle is placed at the origin.
4. a. x ⫽ ⫺1 b. y ⫽ 7 c. y ⫽ 3x ⫹ 10 d. 2y ⫺ 3 x ⫽ 17
23. Herbert has the equation of the line containing the median to the hypotenuse, not the perpendicular bisector of the hypotenuse. He needs to find the slope of the hypotenuse, then take its negative reciprocal to get the slope of a line perpendicular to it. Using this slope and the midpoint, he can then find the equation of the perpendicular bisector.
6. a. 0
ADDITIONAL PROBLEMS WHERE THE STRATEGY “USE COORDINATES” IS USEFUL
5. (x ⫹ 3)2 ⫹ ( y ⫺ 4)2 ⫽ 52, or x2 ⫹ 6x ⫹ y2 ⫺ 8y ⫽ 0 b. 1 c. Infinitely many d. 1 7. y ⫽ 34 x ⫺ 2 8. ( y  412 ) ⫽  45 (x  3) 9. 0, 1, or 2, since a circle and a line may meet in 0, 1, or 2 points
1. One of the diagonals of the kite is on the yaxis and the other is parallel to the xaxis. Thus they are perpendicular.
10. The first pair of lines is parallel, the last pair is parallel, but no two of the lines are perpendicular.
2. On a coordinate map, starting at (0, 0), where north is up, he goes to (0, 4), then to (⫺3, 4), then to (⫺3, 2). Thus his distance from base
11. (⫺3, ⫺1), (⫺3, 9), (⫺8, 4), (2, 4), any values of x and y that satisfy the equation.
camp is 232 + 22 ⫽ 213 km.
12. As seen in the following figure
3. Place the figure on a coordinate system with A at the origin, B ⫽ (0, 2m), and C ⫽ (2m, 0). Then E ⫽ (m, ⫺m) and D ⫽ (⫺m, m). The slope of BC ⫽ ⫺1, as does the slope of DE. Thus they are parallel.
(x2, y2)
CHAPTER REVIEW Section 15.1 1. 240 ⫽ 2210 2. (4, 1) 3. Yes
(x1, y1)
4. P, Q, and M are collinear. 5. Yes
a right triangle can be constructed between the pair of points and the legs of the triangle have lengths 兩x2 ⫺ x1兩 and 兩y2 ⫺ y1兩 . Using the Pythagorean theorem, we can find the length of the hypotenuse as 2(x2  x1)2 + ( y2  y1)2, which is the distance formula.
6. No
Section 15.2 1. y ⫽  52 x ⫹ 3 2. y ⫽ 2 ⫽
( 32 )(x
 3)
13. All sides have length 210. The slope of LM is  13 and the slope of MN is 3, so ∠LMN is a right angle. Thus, LMNO is a square.
3. There are one, none, or infinitely many solutions.
14. (7, 3)
4. (x ⫹ 2)2 ⫹ ( y ⫺ 5)2 ⫽ 25
15. Any point on the line y ⫽ 2x [e.g., (0, 0), (1, 2), (2, 4)]
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16. a.
5. a. ⫺80⬚ 6. a.
b. 120⬚ b.
A
B (3, 6)
A 75°
C (6, 5) A (2, 3)
–90°
O O
A´ c. A
d. A
O
–130°
A´
180°
A´
O 7.
B´ C´
b. Isosceles triangle: AB ⫽ 210, BC ⫽ 210, AC ⫽ 220 c. Right triangle: AB 2 ⫹ BC 2 ⫽ AC 2 17. ABCD is a square, since it is a parallelogram whose diagonals are perpendicular and congruent (verify). 18. a. The slopes of DE and AC are both zero. a + 2b 2b a 1 b. DE ⫽ ⫽ . Therefore, DE ⫽ AC. 3 3 3 3 1 1 1 In 䉭ABC, if DB ⫽ AB and EB ⫽ CB, then DE ⫽ AC and 3 3 3 DE ‘ AC. 19. a. N ⫽ (a ⫹ c, b) a + c b , ≤ and the midpoint of MO is b. The midpoint of LN is ¢ 2 2 a + c b , ≤ . Since the midpoints coincide, the diagonals bisect 2 2 each other.
B O A´
C
A 8. a.
b.
A
B A´
B´
B A´
9. a.
A´
Section 16.1A
B´
b.
O D´
2.
A
O
¢
1. a.
C´
A P
Q
B
P´
Q´
b.
A´
3. a. Any directed line segment that goes right 4, up 2 b.
O B´
D´
C´
c. Down 4, right 5 4. Construct a line through P parallel to AB. Then with a compass, mark off PP ¿ having the length of AB. NOTE: P should be to the right and above B.
A´
10. (⫺1, 4) 11. a. (⫺3, 2) d. (2, ⫺4)
b. (⫺3, ⫺1) e. (4, 2)
c. (4, ⫺1) f. (⫺y, x)
O B´
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12. a. (⫺3, 1) c. (4, ⫺2)
b. (6, 3) d. (⫺x, ⫺y)
13. a. (⫺y, x)
b. (⫺x, ⫺y)
14. a.
A85
18. Construct a line through A perpendicular to l (to point P on l ), extend beyond l, and mark off A such that PA ⫽ AP. 19. a. b.
A
A´
(
A
A´ (
45° ( (
B
(
O
C
A (
O
A´ (
c.
C´
A B´
(
O
15. P
b. A ⫽ (3, ⫺1), B ⫽ (5, ⫺3), C ⫽ (8, ⫺2) c. (a ⫹ 5, ⫺b) Í ! 20. Construct lines, through A and B, parallel to XY and mark off the translation. Then construct perpendiculars to l through A and B to find the reflection of the translated image.
A´
l
21. a. Same; A to B to C is counterclockwise in each case. b. Same c. Opposite 22. a.
P´
C 16. a.
A´
b.
D´
A D
A´
l c.
l
B
A´
C´ C
R B´
A
l A
S
P
Q
l b.
A
A´
S C´
R
A
C
B B´
17. a.
D´ l
A´ D
P
Q
B
A
C
A´
C´
c.
S R C
A B
l
B´
C´ D
P D´
b. A ⫽ (1, ⫺2), B ⫽ (3, ⫺5), C ⫽ (6, ⫺1)
c. (a, ⫺b)
Q
B´ A´
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Answers
d.
S
Page A86
b. B
R A
C B´
C B
C´ P
l
D´
A´
C´ A
D
P D´
Q B´
D
27. a. TAA
A´ A 23. a. Translation, rotation, reflection, glide reflection b. Translation, rotation, reflection, glide reflection c. Translation, rotation, reflection, glide reflection d. Translation, rotation, reflection, glide reflection 24. a.
A´
b.
b. R0,⫺90 is one possibility.
A c.
d.
A´ c. Ml
25. a.
A O
P
l
P´
A´
b.
d. TXY followed by Ml is one possibility.
O P
X A P´
c. O
Y
P
A´
l
28. a. Not possible b. RA,⫺90
P´
c. Ml
B
26. a.
B´ B
A
A
A´
A´
l
P
A
B´
C
A´
B
C´
d. Not possible
B´
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29. a. Reflection
A87
3. a. OX ⬜ OY since (slope of OX)(slope of OY) ⫽
¢ ≤#¢ y
B
x
A
x ≤ ⫽ ⫺1. y
b. OX ⫽ 2x2 + y2, OY ⫽ 2(  y)2 + x2 ⫽ 2y2 + x2 c. Yes, by definition.
A´ C B´ C´
4. A⬘ ⫽ (⫺2, 5), B⬘ ⫽ (4, 3); AB ⫽ 222 + 62 ⫽ 2 210 and A⬘B⬘ ⫽ 2( 6)2 + 22 ⫽ 2 210, so AB ⫽ A⬘B⬘. 5. a. A ⫽ (1, 2), B ⫽ (2, 0) b. A⬘ ⫽ (⫺1, 2), B⬘ ⫽ (⫺2, 0) c. AB ⫽ 2(1  2)2 + (2  0)2 ⫽ 25 d. A⬘B⬘ ⫽ 2(  1 + 2)2 + (2  0)2 ⫽ 25
b. Rotation
6. a. (a, ⫺b) b. AB ⫽ 2(a  c)2 + (b  d)2;
B
A⬘ ⫽ (a, ⫺b), B⬘ ⫽ (c, ⫺d), and
A
A⬘B⬘ ⫽ 2(a  c)2 + ( b + d)2
C´
C
B´
⫽ 2(a  c)2 + (b  d)2 x + y y + x , ≤ . Since its x and 2 2
7. a. The midpoint of XY ⫽ ¢
A´
ycoordinates are equal, it lies on line l. x  y b. The slope of XY ⫽ ⫽ ⫺1, and the slope of line l is 1. y  x
c. Translation
Since (⫺1)(1) ⫽ ⫺1, XY ⊥ l. c. Yes, line l is the perpendicular bisector of XY.
B A B´
C
8. a. A ⫽ (2, 1), B ⫽ (⫺2, 3) b. A⬘ ⫽ (4, ⫺1), B⬘ (0, ⫺3)
A´
c. AB ⫽ 2(2 + 2)2 + (1  3)2 ⫽ 2 25; A⬘B⬘ ⫽ 2(4  0)2 + (  1 + 3)2 ⫽ 2 25
C´
9. a. Rotation
b. Translation
10. a. Translation
b. Reflection
30. The farther the center is from the object, the greater the image is from the original.
P
31. Moving the center of a size transformation does not effect the size of the image. It only effects the location of the image.
C
B
C´
B´ b. 25, 2 25; 2 c. 3, 6; 2 e. They are parallel. 12. a.
A
b. A⬘ ⫽ (⫺1, 4); B⬘ ⫽ (0, 2)
c. AB ⫽ 2(1  2)2 + (3  1)2 ⫽ 25; A⬘B⬘ ⫽ 2( 1  0)2 + (4  2)2 ⫽ 25 q 2. a. Yes, both have slope . p b. Yes, both have length 2p2 + q2. c. Yes, by definition.
d. 2 22, 422; 2
A´
Section 16.2A 1. a. A ⫽ (1, 3); B ⫽ (2, 1)
c. Glide reflection
A´
11. a.
32. If Marjorie rotates a figure 180⬚ clockwise, it will end up in the same location as if she rotated it 180⬚ counterclockwise, assuming she is using the same turn center. This is because 180⬚ ⫹ 180⬚ ⫽ 360⬚. A turn of 90⬚ clockwise would be the same as a turn of 20⬚ counterclockwise.
c. Glide reflection
P
B
B´ C
b. They are equal.
C´
d. Rotation
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Í ! Í ! Í ! 13. No; the lines SS, RRS and TT ¿ do not meet at a common point. (If it were a size transformation, they would meet at the center.) Also RS/RS ⫽ 2 but RT/RT ⫽ 35 2, so the ratios of sides are not equal. 14. a. A size transformation of scale factor 32 followed by a rotation. b. A size transformation of scale factor 13 followed by a reflection or glide reflection. 1 1 1 1 15. a. Since PQ , BB , XX , and AA are equivalent directed line segments by the definition of TPQ, PQ ‘ BB ¿ ‘ XX ¿ ‘ AA ¿ and PQ 艑 BB ¿ 艑 XX ¿ 艑 AA ¿ . In each case, one pair of opposite sides is congruent and Í parallel, ! Í ! so ÍBBXX ! Í and ! BBAA are both parallelograms. b. B ¿ X ¿ ‘ BA and B ¿ A ¿ ‘ BA, since they contain opposite sides of parallelograms.Í However, B there can only be one ! Í ! since Í through ! line parallel to BA, B ¿ X ¿ and B ¿ A ¿ must be the same line. Therefore, A, X, and B are collinear. 16. Since translations preserve distances, PQ ⫽ PQ, QR ⫽ QR, and RP ⫽ RP. Thus 䉭PQR 艑 䉭PQR. Therefore, ∠PQR 艑 ∠PQR, since they are corresponding parts of congruent triangles.
Section 16.3A 1. a. HD(HC (HB(HA(P)))) ⫽ P b. HD(HC (HB(HA(Q)))) ⫽ Q c. This combination of four halfturns around the vertices of a parallelogram maps each point to itself. 2. Impossible. Even though corresponding sides have the same length, corresponding diagonals do not. Yet isometries preserve distance. 3. a.
C
A
19. q ‘ p implies that ∠1 艑 ∠2. Since reflections preserve angle measures, ∠1 艑 ∠1 and ∠2 艑 ∠2, where ∠1 and ∠2 are the images, respectively, of ∠1 and ∠2. Thus ∠2 艑 ∠1 and p ‘ q by the corresponding angles property. 20. Since isometries map segments to segments and preserve distances, AB 艑 A ¿ B ¿ , BC 艑 B ¿ C ¿ , and CA 艑 C ¿ A ¿ . Thus, by the SSS congruence property, 䉭ABC 艑 䉭ABC. 21. a. One, just TPQ b. Infinitely many; the center C may be any point on the perpendicular bisector of PQ, and kPCQ is the rotation angle. Í ! Í ! 22. a. The point where PP ¿ and QQ ¿ intersect b. GreaterÍ than ! 1 Í ! c. Where PP ¿ and QQ ¿ intersect, less than 1 23. Construct line through Q parallel to QR. The intersection of that line and l is R. 24. Let point R be the intersection of line l and segment PP ¿ and let S be the intersection of l and QQ ¿ . Then 䉭QRS 艑 䉭QRS by SAS congruence. Hence RQ ⫽ RQ and ∠QRS 艑 ∠QRS. Thus 䉭PRQ 艑 䉭PRQ. We can see this by subtracting m(∠QRS) from 90⬚, which is the measure of ∠PRS), and subtracting m(∠QRS from 90⬚, which is the measure of ∠PRS. Thus 䉭PRQ 艑 䉭PRQ by the SAS congruence property. Consequently, PQ 艑 P ¿ Q ¿ , since these sides correspond in 䉭PRQ and 䉭PRQ. Thus PQ ⫽ PQ, as desired. 25. Yes, a translation 26. Because isometries preserve angle measure, parts of a figure that are parallel before transformation will still be parallel after. As for the corresponding segments of a figure and its image after transformation, if the transformation is a translation, the corresponding segments will be parallel. In the case of a rotation, reflection, or glide reflection, there is no guarantee that they will be parallel.
B´ O
17. p ‘ q implies that ∠1 艑 ∠2. Since rotations preserve angle measures, ∠1 艑 ∠3 and ∠2 艑 ∠4. Thus ∠3 艑 ∠4 and p ‘ q by the corresponding angles property. 18. Since reflections preserve distances, AB ⫽ AB, BC ⫽ BC, and AC ⫽ AC . Since A, B, and C are collinear, AB ⫹ BC ⫽ AC. Therefore, AB ⫹ BC ⫽ AB ⫹ BC ⫽ AC ⫽ AC and A, B, and C are collinear.
B
C´
A´
b. Yes; RB,90⬚ followed of SO,1/2 is a similitude. 4. (d), since A : A, B : D, C : C, D : B 5. MEG, MFH, HP, RP,360⬚ 6. a. For example, MAD, MBE, MCF, RO,60⬚, RO,120⬚, and RO,180⬚, RO,240⬚, RO,300⬚, RO,360⬚ b. 11 (6 reflections and 6 rotations) 7. Yes. Let E and F be the midpoints of sides AB and CD. Then MEF and RA,360⬚ will work. 8. a. The sum of the areas of the two red rectangles equals the areas of the two red parallelograms. b. The red figure has been translated which preserves areas. c. The areas of the two red parallelograms in (3) are equal to the areas of the two respective red parallelograms in (4) and then the ones in (5) d. The area of the red square region in (1) equals the sum of the areas of the two red square regions in (5). 9. Let M be the midpoint Íof AC. Hence H ! Í M!(A)Í ⫽! C, HM(C) ⫽ A. We Í ! know that HM(B) is on CD, since HB(AB) ‘ AB and C is on HM(AB). Í ! Í ! Í ! Similarly, HM(B) is on AD, so that HM(B) is on CD, 艚 AD. That is, HM(B) ⫽ D. Hence (䉭ABC) ⫽ 䉭CDA so that 䉭ABC 艑 䉭CDA. 10. a. HM(∠ADC) ⫽ ∠CBA, so that ∠ADC 艑 ∠CBA Similarly, ∠BAD 艑 ∠DCB. b. HM(AB) ⫽ (DC), so AB 艑 DC, and so on. 11. Let l ⫽ BP. Then Ml (A) is on BC, but since AB ⫽ BC, we must have Ml (A) ⫽ C. Thus Ml (䉭ABP) ⫽ 䉭CBP, so 䉭ABP 艑 䉭CBP. Hence ∠BPA 艑 ∠BPC, but since these angles are supplementary, each is a right angle. Since AP ⫽ CP, BP is the perpendicular bisector of AC. 12. a. From Problem 11, points A and C are on the perpendicular bisector of BD, so AC ⊥ BD. b. Since AC is the perpendicular bisector of BD, MAC (B) ⫽ D, MAC (D) ⫽ B, MAC (A) ⫽ A, and MAC (C) ⫽ C. Hence MAC (ABCD) ⫽ ADCB, so the kite has reflection symmetry.
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13. a. AA ⫽ 2x and AA ⫽ 2y, so that AA ⫽ 2(x ⫹ y). Also, A, A, and A are collinear since AA ¿ ⊥ r, r ‘ s and A ¿ A– ⊥ s. Hence, Mr , followed by Ms is equivalent to the translation TAA. Since A was arbitrary, Mr followed by Ms is TAA since orientation is preserved. b. The direction of the translation is perpendicular to r and s, from r toward s, and the distance is twice the distance between r and s. 14. a. (Any two intersecting rails can be used.) Let B ⫽Í M!l (B), and B ⫽ Mm(B). Let P ⫽ AB Í !– 艚 m. Shoot ball A toward point P. b. Let Q ⫽ PB ¿ 艚 l . Then use an argument as in Example 16.13.
2.
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P′
P P′
3.
Q′
l
l
Q P 4.
P
B´
B A
P′ Q
A
m
P
R
B
5. • • •
B´´ 15. Consider SA,bla. This size transformation will map ABCD to a square ABCD congruent to EFGH. Then from Section 16.2, we know that there is an isometry J that maps ABCD to EFGH. Hence the combination of SA,bla and J is the desired similarity transformation. 16. Jaime’s path, as drawn, is not possible since when the ball hits a wooden board at the edge of the green, it must bounce away at an angle that equals the angle at which it approached the board. Jaime will need a more complicated path, which can be found by using reflections.
••••••••
They must be infinite in one direction. 6.
7.
ADDITIONAL PROBLEMS WHERE THE STRATEGY “USE SYMMETRY” IS USEFUL 1. Place any two squares on a 4 ⫻ 4 square grid. Reflect those two squares across a diagonal and across a vertical line through the center as shown. If two more squares are produced in either of these cases, that pattern will have reflective symmetry with four dark squares.
8.
• • •
••••••••
They must be infinite in one direction. 9.
2. There are three digits that read the same upside down: 0, 1, and 8. Since house numbers do not begin with zero, there are 2 䡠 3 䡠 3 ⫽ 18 different such house numbers. 3. Pascal’s triangle has reflective symmetry along a vertical line through its middle. In the 41st row, the 1st and 41st numbers are equal, as are the 2nd and 40th, the 3rd and 39th, and 4th and 38th. Thus the 38th number is 9880.
CHAPTER REVIEW Section 16.1 1. A
A′
C′ B A
C
O
B P
B′
P′
10. Apply a size transformation to get the triangles congruent, then use an isometry to map one to the other.
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Section 16.2
b.
1. Translations, rotations, reflections, and glide reflections 2. The image of a triangle will be a triangle with sides and angles congruent to the original triangle, hence congruent to it.
O
3. The corresponding alternate interior angles formed by the transversals will be congruent, hence the image of parallel lines will be parallel. 4. One triangle is mapped to the other using a combination of a translation, rotation, or reflection.
c.
5. A triangle will map to a triangle with corresponding angles congruent. Thus, the triangles are similar by AA.
M N
6. Similitudes preserve angle measure. Congruent alternate interior angles formed by a translation will map to congruent alternate interior angles. 7. Using a size transformation, the triangles can be made to be congruent. Then an isometry can be used to map one triangle to the other. The combination of the size transformation and the isometry is a similarity that takes one triangle to the other. 8. a. D
Section 16.3 1.
b. E
c. E
d. D
9.
B
A
C
C′
A C B B′
A O 10. Translation, reflection, and glide reflection
Chapter 16 Test 1. a. T g. T
b. F h. T
c. T
11. a. MAB d. T
e. T
2. Distance
b. RF,⫺90⬚
c. TCD
12. Let P be the intersection of EG and FH. Then RP,45⬚ (SP,22 (EFGH))⫽ ABCD; thus ABCD is similar to EFGH. 13. The result of two glide reflections is an isometry that preserves orientation. Therefore, the isometry must be a translation or a rotation.
3. Rotation 4. (a) and (e)
14. B
5. (a) and (c)
15. a. Reflection b. Translation c. Rotation
6. a. Size transformation b. Rotation c. Reflection d. Glide reflection e. Translation 7. a.
f. F
16. Y´
Y X
m X´
Z´ O Z
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5. Consider the following isosceles triangle with the appropriate coordinates.
17. A ⫽ (5, ⫺5) B ⫽ (2, ⫺10) C ⫽ (x ⫹ 4, y ⫺ 6) 18. AB ⫽ 83 ; AC ⫽ 10 3 ; CE ⫽ 5 19. 䉭AMC 艑 䉭BMC by SSS. Therefore, ∠AMC 艑 ∠BMC, or m(∠AMC) 艑 m(∠BMC) ⫽ 90⬚. Since MCM(A) ⫽ B and MCM(C) ⫽ C, we have MCM (䉭ABC) ⫽ Í ! 䉭BAC. This means that CM is a symmetry line for 䉭ABC. 20.
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C (a, b)
(
a b M 2,2
)
(
3a b N 2 ,2
)
B´
A´
A (0, 0) C´ B
B (2a, 0)
Congruence proof: AM ⫽ BN, ∠A 艑 ∠B, and AB ⫽ BA. Thus 䉭ABN 艑 䉭BAM by SAS. By corresponding parts, AN ⫽ BM. Coordinate proof:
C
M
AN ⫽ BM ⫽
B
¢ 2a 
B
¢a 
a 2 3a 2 b 2 b 2 ≤ + ¢ ≤ = ¢ ≤ + ¢ ≤ ; 2 2 2 2 B b 2 b 2 a 2 3a 2 ¢ ≤ + ¢ ≤ ; ≤ + ¢ ≤ = 2 2 2 B 2
The transformation is a rotation with the center labeled M and ∠CMC is the angle of rotation. Í ! Í ! Í ! 21. No, because the lines XA, ZC, and BY are not concurrent.
6. Because PQ ‘ SR and PS ‘ QR and interior angles on the same side of the transversal are supplementary, ∠Q and ∠S must also be right angles since they are supplementary to ∠P. Further, since opposite angles of a parallelogram are congruent, ∠R must also be a right angle.
Section EA
7. Slope of AC ⫽
A
1. AB ‘ DC by definition of a parallelogram; ∠A and ∠D are supplementary, since they are interior angles on the same side of a transversal. Since AD ‘ BC, ∠A and ∠B are supplementary. Similarly, ∠C is supplementary to ∠B and ∠D. d 2. Slope of MN ⫽ slope of OP ⫽ , so MN ‘ OP. slope of MN ⫽ slope c b of ON ⫽ , so MP ‘ ON. Therefore MNOP is a parallelogram. a  e Í ! ! 3. Let l ⫽ AP. Then Ml (B) is on AC, since ∠BAP 艑 ∠CAP. Because AB ⫽ AC, however, we must have Ml(B) ⫽ C. Thus Ml (C) ⫽ B, so that Ml (䉭ABP) ⫽ 䉭ACP Hence ∠ABP 艑 ∠ACP. 4. Congruence proof: AB ⫽ CD, AD ⫽ CB, and BD ⫽ DB. Therefore 䉭ABD 艑 䉭CDB by SSS. By corresponding parts, ∠ADB 艑 ∠CBD, so AD ‘ CB. Since AD and CB are parallel and congruent, ABCD is a parallelogram. Coordinate proof: If A, B, and D have coordinates (0, 0), (a, b), and (c, 0), respectively, then C must have coordinates (a ⫹ c, b), since both pairs of opposite sides are congruent.
B (a, b)
C (a + c, b)
a  0 ⫽ ⫺1; 0  a since 1(⫺1) ⫽ ⫺1, AC ⬜ BD. slope of BD ⫽
D (0, a)
C (a, a)
A (0, 0)
B (a, 0)
8. m(∠P) ⫽ m(∠R) and m(∠Q) ⫽ m(∠S) (opposite angles are congruent); m(∠P) ⫹ m(∠Q) ⫹ m(∠R) ⫹ m(∠S) ⫽ 360⬚ (vertex angles of quadrilateral total 360⬚); m(∠P) ⫹ m(∠Q) ⫹ m(∠P) ⫹ m(∠Q) ⫽ 360⬚, so m(∠P) ⫹ m(∠Q) ⫽ 180⬚; thus ∠P and ∠Q are supplementary and PS ‘ QR (interior angles on same side of transversal are supplementary). Similarly, m(∠P) ⫹ m(∠S) ⫹ m(∠P) ⫹ m(∠S) ⫽ 360⬚, m(∠P) ⫹ m(∠S) ⫽ 180⬚, P and S are supplementary, and PQ ‘ SR. Since both pairs of opposite sides are parallel, PQRS is a parallelogram.
Section T1A 1. (b) and (d)
A (0, 0)
D (c, 0)
Using slopes, we have AB ‘ DC and AD ‘ BC.
a  0 ⫽ 1; a  0
2. a. q : ⬃p b. q ¿ r c. r 4 (q ¿ q) d. p ¡ q
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3. a. T d. T g. T
b. F e. T h. T
c. T f. T
4. a. Invalid d. Invalid
5.
p
q
⬃p
⬃q
(⬃p) (⬃q)
T T F F
T F T F
F F T T
F T F T
F T T T
(⬃p) q
(⬃p) (⬃q)
T F T T
F F F T
p : q ⬃(p q) T F T T
F T T T
(⬃p) ¡ (⬃q) is logically equivalent to ⬃(p ¿ q). (⬃p) ¡ q is logically equivalent to p : q (⬃p) ¿ (⬃q) is logically equivalent to ⬃(p ¡ q). b. Invalid f. Invalid
c. Invalid g. Invalid
d. Valid h. Valid
7. a. You will be popular. b. Scott is not quick. c. All friends are trustworthy. d. Every square is a parallelogram.
b. F e. F
q
p q
p q
p:q
⬃p
⬃q
⬃q 4 p
⬃q : ⬃p
T F F T
T T F F
T F F F
T T F T
T T T F
F T T F
F F T T
F T F T
T T T F
6. For example: Some B’s are A’s. All C’s are B’s. Therefore, all C’s are A’s.
B
⬃(p q)
A
F F F T
C
7. All are true when both the hypothesis and conclusion are true. The only time an implication is false, the disjunction is true. Therefore, they are not all false at the same time. 8. Both Bob and Sam are truth tellers. 9. a. p r p
1. a. 7 c. 0 e. 3 g. 6
b. (p r q) r (p r q)
c. (p r p) r (q r q)
b. 3 d. 4 f. 7 h. 6
2. a. 4, 5 b. 7, 5 c. 1, 7 d. 4, does not exist
c. T
10. Invalid
B
p
Section T2A
8. a. Hypothetical syllogism b. Modus ponens c. Modus tollens d. Modus tollens e. Hypothetical syllogism 9. a. F d. T
c. Valid
5.
4. a. If I am an elementary school teacher, then I teach third grade. If I do not teach third grade, then I am not an elementary school teacher. If I am not an elementary school teacher, then I do not teach third grade. b. If a number has a factor of 2, then it has a factor of 4. If a number does not have a factor of 4, then it does not have a factor of 2. If a number does not have a factor of 2, then it does not have a factor of 4.
6. a. Valid e. Invalid
b. Valid e. Invalid
T
F
3. a. 7 c. 1
b. 4 d. 9
4. a. F d. T
b. F e. T
c. T f. F
5. 1 ⫺ 4 ⫽ n if and only if 1 ⫽ 4 ⫹ n. Look for a “1” in the “4” row; the number heading the column that the “1” is in represents n, namely 2. 6. The table is symmetric across the upper left/lower right diagonal.
C
7. 13 䊝 12 ⫽ 2 䊝 3 ⫽ 5, 31 䊝 12 ⫽ 21 䊝 31 ⫽ 5 (NOTE: Normally 13 ⫽ 62 , but 6 ⫽ 1 in the 5clock.) Yes. For example, 34 䊝 23 ⫽ 2 䊝 4 ⫽ 1.
ELEMENTARY LOGIC TEST 1. a. T c. F e. F g. F
b. F d. F f. T h. F
b. F e. T
8. a. If 1 2, then 1 ⫹ 3 2 ⫹ 3. But 4 v 0 in the whole numbers. b. If 2 3 and 2 0, then 2 䡠 2 2 䡠 3. But 4 v 1 in the whole numbers. 9. a. {3}
2. a. ⬃q : p, ⬃p : q, q : ⬃p b. q : ⬃p, p : ⬃q, ⬃q : p c. ⬃p : ⬃q, q : p, p : q 3. a. T d. F
17 17 , and 12 On the other hand, 34 䊝 23 ⫽ 12 ⫽ 22 ⫽ 1 in 5clock arithmetic.
c. F f. T
b. {0, 6}
c. {0, 2, 4, 6, 8}
d. { }
10. If a ⫹ c ⬅ b ⫹ c, then a ⫹ c ⫹ (⫺c) ⬅ b ⫹ c ⫹ (⫺c), or a ⬅ b. 11. If a ⬅ b and b ⬅ c, then m 兩 (a ⫺ b) and m 兩 (b ⫺ c). Therefore, m 兩 [(a ⫺ b) ⫹ (b ⫺ c)] or m 兩 (a ⫺ c). Thus a ⬅ c. 12. 61, 83 13. 504
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CLOCK ARITHMETIC TEST 1. a. F c. F e. F g. F
b. T d. T f. T h. T
2. a. 3 c. 6 e. 1
b. 1 d. 12 f. 6
3. a. 3 䊝 (9 䊝 7) ⫽ 10 䊝 9 ⫽ 9 b. (8 䊟 3) 䊟 4 ⫽ 8 䊟 1 ⫽ 8 c. (5 䊟 4) 䊝 (5 䊟 11) ⫽ 5 䊟 0 ⫽ 0 d. (6 䊟 3) 䊝 (3 䊟 4) 䊝 (3 䊟 3) ⫽ 3 䊟 0 ⫽ 0 4. a. 3, 2 b. 4; the reciprocal doesn’t exist since all multiples of 4 are either 4 or 8 in the 8clock. c. 0; the reciprocal doesn’t exist since all multiples of 0 are 0 in the 5clock. d. 7, 5
5. a. {⫺14, ⫺5, 4, 13} b. {2, 3, 4, 6, 12} c. (7k ⫹ 1; k is any integer} 6. All multiples of 4 land on 4, 8, or 12, never on 1. 7. a. 0 cannot be a divisor. b. 1 divides everything, thus all numbers would be congruent. 8. Because m divides a ⫺ a. 9. There are 365 days in a nonleap year and 365 ⫽ 1 mod 7. Thus January 1 will fall one day later, or on a Tuesday.
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Chapter 1 Page 1: ©AP/Wide World Photos. Page 39 (left and right): The Granger Collection, New York. Chapter 2 Page 100 (left): Courtesy of the Bryn Mawr College Library. Page 100 (right): Courtesy of Professor G. L. Alexanderson. Chapter 3 Page 107: JL Charmet/Science Photo Library/Photo Researchers, Inc. Page 149 (left): Bettmann Archive, New York. Page 149 (right): Courtesy of George M. Bergman, University of California, Berkeley. Chapter 4 Page 155 (right, bottom): ThinkStock LLC/Index Stock. Page 155 (left, top): Photo Researchers. Page 155 (left, upper center) and Page 155 (left, lower center): National Museum of American History ©2002 Smithsonian Institution. Page 155 (left, bottom) and Page 155 (right, top): NMPFT/Science & Society Picture Library. Page 155 (right, center): Corbis Images. Page 163: Courtesy of Texas Instruments Incorporated. Page 198 (left): ACME/UPI/The Bettmann Archive, New York. Page 198 (right): Courtesy of Dartmouth College Library, Hanover, New Hampshire. Chapter 5 Page 203: The Granger Collection, New York. Page 232 (left): Reproduced by permission of the Syndics of Cambridge University Library. Page 232 (right): Courtesy of Constance Reid.
Chapter 8 Page 341: The Needham Research Institute, Cambridge, England. Page 374 (left): Courtesy of Sylvia Wiegand. Page 374 (right): Courtesy Kadon Enterprises, Inc. Chapter 9 Page 379: Newberry Library, Chicago/Superstock. Page 433 (left): Courtesy of Paul Cohen. Page 433 (right): Courtesy of Leon Harkleroad. Chapter 10 Page 507 (left): UPI/CorbisBettmann. Page 507 (right): Laura Wulf/Harvard University News Office. Chapter 11 Page 513 (left): The Granger Collection, New York. Page 513 (top right and top left): PhotoDisc, Inc./Getty Images. Page 513 (bottom left and bottom right): Bettmann/ Corbis. Page 575 (left): Photo by Joe Munroe/Ohio Historical Society/Courtesy Caltech Archives. Page 575 (right): UPI/CorbisBettmann. Chapter 12 Page 581 (left and right): Courtesy of Dr. Pierre van Hiele. Page 658 (left): Courtesy of Cathleen Synge Morawetz. Photo by James Hamilton. Page 658 (right): The Center for American History, The University of Texas at Austin. Chapter 13 Page 665: The Granger Collection, New York. Page 733 (left): CorbisBettmann. Page 733 (right): Culver Pictures,Inc.
Chapter 6 Page 237: The Art Archive/British Museum/The Kobal Collection, Ltd. Page 281 (left): Courtesy of Evelyn Granville. Page 281 (right): Courtesy of Professor G. L. Alexanderson.
Chapter 14 Page 739: The Granger Collection, New York. Page 800 (left): CorbisBettmann. Page 800 (right): Courtesy of International Business Machines Corporation. Unauthorized use not permitted.
Chapter 7 Page 285: Eberhard Otto. Page 336 (left): George M. Bergman, University of California, Berkeley. Page 336 (right): The MittagLeffler Institute, Djursholm, Sweden.
Chapter 15 Page 807: Photo Researchers. Page 845 (left): Courtesy of George Bergman. Page 845 (right): Courtesy of H.S.M. Coxeter, University of Toronto, Canada. P1
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Chapter 16 Page 849 (left): Photo by Ralph A. Raimi. Page 849 (right, top): M.C. Escher’s “Circle Limit III” ©2004 The M.C. Escher CompanyBaarnHolland. All rights reserved. Page 849 (right, center): M.C. Escher’s “Shells and Starfish” ©2004 The M.C. Escher CompanyBaarnHolland. All rights reserved. Page 849 (right, bottom):
M.C. Escher’s “Design Drawing for Intarsia Wood Panel with Fish” ©2004 The M.C. Escher CompanyBaarnHolland. All rights reserved. Page 902 (left): News and Publications Service, Stanford University, Stanford, California. Page 902 (right): FASE Corporate (Foundation for Advancements in Science & Education).
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Index
A AA (angleangle) similarity property, 754–55, 783 AAS (angleangleside) triangle congruence, 746 abacus, 71, 107, 155, 172–75 absolute value, 354 absolute vs. relative in circle graphs, 472 abstractions of geometric shapes, 588–90 abstract representation, 406, 408 abstract thinking, 585, 588–90 abundant numbers, 228 acre, 669, 670–71 acute angles, 617, 621 acute triangle, 620, 621, 630 addends, 110. See also missingaddend subtraction adding the complement, subtraction by, 185 adding the opposite integers, 349–50 rational numbers, 387 addition, 109–15, 255–59 algorithms, 172–73, 184, 185, 192, 297 associative property clock arithmetic, 923, 924 decimals, 291 fractions, 244, 258, 260 integers, 347 rational numbers, 386–87 real numbers, 402 whole numbers, 112–15, 158–59, 172–73 in base five, 192 as binary operation, 110 cancellation property, 123, 347, 387 with chip abacus, 72, 172–73 clock arithmetic, 923, 924 closure property, 111, 257, 347, 386 common denominators, 255–57 commutative property clock arithmetic, 923, 924 decimals, 291 fractions, 244, 258, 260 integers, 347 rational numbers, 386–87 real numbers, 402 whole numbers, 111–15, 158–59, 172–73
compensation, 158–59, 261, 291 decimals, 291, 297 denominators common, 255–57 unlike, 256–57 exponent properties, 404 facts, 111–15, 174, 176 for base five subtraction, 192–93 fractions, 255–59, 260–61 identity property fractions, 258 integers, 347 rational numbers, 384, 386–87 real numbers, 402 integer, 345–48, 352, 365–67 inverse property integers, 347, 352 rational numbers, 384–87 lattice method algorithm, 173, 192 lefttoright method, 159 mental math, 157–59 multidigit, 114–15 negative integer, 348 nonstandard algorithms, 183, 184 numerators, 249 obtaining ranges for estimation, 160 order of operations, 144–45, 163–64 with placevalue pieces, 172–73 probability property, 536–38 rational numbers, 384–87, 392 real numbers, 402 repeated addition for multiplication, 123–24, 135, 266, 357, 924 sum of consecutive whole numbers, 11–12 sum of counting numbers, 12–13, 20–21 unlike denominators, 256–57 whole numbers, 109–15, 158–59, 172–73 additive magic squares, 19, 217, 295, 296 additive numeration systems, 62–66, 72 adjacent angles, 617, 621 Adventures of a Mathematician (Ulam), 575
aesthetics, golden ratio, 285 Agnesi, Maria, 733 Ahmes Papyrus, 237 a is congruent to b mod multiplication, 925–26 algebra, 13–16, 405–11 balancing method, 405–6, 408 coefficients, 408 concept of variables, 15 derivation of term, 830 geometric problems, 807 graphing integers on the coordinate plane, 815 Guess and Test strategy with, 16 and onetoone correspondence, 48 pan balance, 407 solving equations, 405–6, 408, 826–29 solving inequalities, 409–11 transposing, 409 Use a Variable strategy with, 13–16, 27–28 variables (See variables) algebraic logic, 163–64 algorithms, 107, 171–83, 297–301 addition, 172–73, 184, 185 base five, 192 lattice method, 173, 192 base five, 192–95 CaleyPurser, 368 cashier’s, 184–85 decimal, 297–301 derivation of term, 830 division, 133, 178–83 base five, 194–95 duplication, 186 equaladditions, 185 Euclidean, 223, 225 German lowstress, 188 intermediate, 173, 176–77, 180, 182, 192, 194 lattice method, 173, 177, 192, 194 multiplication, 176–77 base five, 194 RSA, 368 Russian peasant, 185 scratch addition, 184 Sieve of Eratosthenes, 206, 234 standard addition, 172–73, 192 division, 182–83, 302–3
multiplication, 176–77, 194 subtraction, 174–75, 192–93 subtractfromthebase, 175–76, 193 subtraction, 174–75, 298 base five, 192–93 wholenumber operations, 171–83 alKhowarizimi, 107, 830 alternate exterior angles, 624 alternate interior angles, 619–20, 743, 744, 771, 791–92 altitude of a triangle, 773, 779, 837–38, 839 amicable whole numbers, 228 amoebas, exponential growth of, 87 amounts, relative vs. absolute, 472 analysis—level 1, 584–85 analytic geometry, 816 analyzing data, 484–500 box and whisker plot, 489–92, 503 dispersion, 492–96 distribution, 496–500 measures of central tendency, 484–88 Oregon rain, 501 organizing, 442–44 (See also graphs) standard deviation, 493–96, 498–500 variance, 493–94 and connective, 913 angle(s), 588 AA similarity property, 754–55, 783 acute, 617, 621 adjacent, 617, 621 alternate exterior, 624 alternate interior, 619–20, 743, 744, 771, 791–92 ASA congruence, 744, 746, 791, 794 base, 602, 604 bisectors, 746, 769–70, 779 central, 607–8, 628–30 complementary, 618, 620, 621 congruent, 588, 590, 606, 741–42, 742–46, 791 consecutive, 627 copying, 767 corresponding, 618–19 creating shapes with, 584, 587
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angle(s) (continued) dihedral, 641–44 directed, 854–55 Euclidean constructions, 769–70 exterior, 607–8, 617, 630 included, 743–44, 753–54 interior, 608, 619–20, 627, 769 interior of the, 617 measurement of, 617–18 dihedral, 641–42 in polygons, 628–31 model and abstraction, 589 nonadjacent, 617, 621 notation, 616–17 obtuse, 617, 620, 621 parallel lines and, 618–22 perpendicular bisector, 769–70 polar coordinates, 832 properties, 617, 619–20 reflex, 617, 621 right, 589, 617 SAS congruence property, 743–44, 746, 778, 790–92, 793 straight, 617, 620, 621, 631 sum in a triangle, 620, 629–30 supplementary, 618, 621 types of, 616–18 vertex of the, 616–17 vertical, 618, 619, 621 See also vertex (interior) angles angleangleside (AAS) triangle congruence, 746 angleangle (AA) similarity property, 754–55, 783 angle bisector, 746, 769–70, 779 anglesideangle (ASA) congruence property, 744, 746, 791, 794 angle sum in a triangle, 620, 629–30 apex, 643–44, 646, 647 Appel, Kenneth, 713 applied problems in transformation, 894–95 Arabic numeration. See HinduArabic numeration system arc(s) of a circle, 766, 767, 770 intersecting, 766–70 passim, 780 of radius, 767, 769, 780 Archimedean method for deriving volume of a sphere, 665, 712 Archimedes, 411, 665, 712 are (metric area of a square), 675 area, 670–71, 675–76, 689–93 abstract, 689–93 circles, 693 English system, 670–71 formulas (See inside back cover) height, 692, 710, 718, 720–26
Hero’s formula, 697 lateral surface, 707 metric system, 675–76 parallelogram, 692 rectangle, 689–91 trapezoid, 692 triangle, 691, 692 argument, 916 arguments, logical, 916–19 Aristotle, 379 Arithmetic, Fundamental Theorem of, 206, 219, 399–400 Arithmetica (Diophantus), 341 arithmetic average. See mean arithmetic logic, 163 arithmetic sequences, 86–87 arithmogon, 17 array 8by8 square, 118 calendar, 416 coordinates and, 818 of points, 818, 820 rectangular division, 207, 208, 210 multiplication, 124–25, 267 numbers, 87 Sieve of Eratosthenes, 206, 234 square lattice, 588, 615, 695, 700 triangular lattice, 588 arrow diagrams, 83 functions as, 89 in sequences, 86 Ars Magna (Cardano), 341 ASA (anglesideangle) congruence property, 744, 746, 791, 794 associative property, 112, 126 of addition clock arithmetic, 923, 924 decimals, 291 fractions, 244, 258, 260 integers, 347 rational numbers, 386–87 real numbers, 402 whole numbers, 112–15, 158–59, 172–73 in mental math, 158–59, 291 of multiplication clock arithmetic, 924–25 decimals, 291 fractions, 269–70, 275 integers, 360–61 rational numbers, 389 real numbers, 402 whole numbers, 126, 129, 158–59 astronomical unit (AU), 316 asymmetrical distribution, 497–98 Aubel’s theorem, 788 avoirdupois ounces and pounds, 672 axiomatics – level 4, 585 axis manipulation and scaling, 465–67
axis of rotational symmetry, 652 axis of symmetry, 600
B Babbage, Charles, 155 Babylonian numeration system, 64–65, 66, 795 backtoback stem and leaf plot, 443 balancing method of solving equations, 405–6, 408 bar graphs, 444–46 doublebar, 446 histograms compared to, 445 misleading distortion of compressing yaxis, 444–45, 447, 465–67 pictorial embellishments, 474 threedimensional effects, 470–72 multiplebar, 445 with pictorial embellishments, 450–51 SAT scores, 446 when to use, 454 barrel, 671 Barry, Rick, 353 base, 71–73, 77–78 exponent of the, 142 subtractfromthebase algorithm, 175–76, 193 base (geometric) angles of an isosceles triangle, 602 angles of a trapezoid, 614 of a cone, 646 of a cylinder, 645 of a parallelogram, 692 of a prism, 650 of a pyramid, 647, 654 threedimensional figures, 707–12 of a triangle, 691 in volume, 718, 720–26 base design, 651–52 base two, 78, 110, 923, 924 base three, 77 base five, 75–77 converting from base ten, 76–77 converting to, 75–76 multiplication facts, 194 operations, 192–95 subtractfromthebase algorithm, 193 subtraction facts, 193 base ten blocks/pieces, 71, 287 addition, 172–73 Dienes blocks, 71 long division, 178–79, 181 multiplication, 177 subtraction, 175–76
converting to, 76–77 converting to base five, 75–76 grouping by tens, 71–73 long division algorithm, 178–79, 181 powers of 10, 62, 159, 289, 291–92, 301 subtractfromthebase algorithm, 175–76 base twelve, 78 Begle, Edward G., 902 bellshaped curves, 497–99 Bernoulli, 411 Bertrand’s conjecture, 217 betrothed numbers, 228 Bhaskara, 276, 699 bias and samples, 475–77 biased surveys, sources of, 475–77 biconditional statement, 915 Bidder, George Parker, 168, 195 billion, 74, 79 binary numeration system, 77, 78 binary operations, 78, 110 clock arithmetic, 923, 924 biorhythm, theory of, 231 birthday probability problem, 513 bisector of an angle, 746, 769–70, 779. See also perpendicular bisector black and red chips model, 382 black and red chips model for integers, 344, 345, 348, 349–50, 352 Blackwell, David, 336 blocks, in base ten pieces, 71 board foot, 729 borrowing subtraction, 174 box and whisker plot, 489–92 interquartile range, 489–92 outliers, 489–92, 503 Brahmagupta, 237, 341 Braille numerals, 69 Buffon, Georges, 567 bundlesofsticks model, 71, 75 buoyancy, principle of, 665
C calculator activities comparing two fractions, 294 crossmultiplication, 248 crossmultiplication of fraction inequality, 248 decimals, dividing, 302, 303 exponents, 404 fraction equality, 245 fractions to percents, 322 greatest common factor, 222, 223 integer computation, 352 long division, 302–3 notation, standard to scientific, 364–65 percents, 322, 326–27
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proportions, 326–27 repeatedaddition approach, 135 simplifying fractions, 257 sums and differences of rational numbers, 388 calculators, 163–66 algebraic logic, 163–64 arithmetic logic, 163 constant functions, 164–65 DEL key, 163 distributivity and, 164, 166 exponent keys, 165 factorial key, 550 graphing, 420 history of, 155 improper fractions, 268 logic, types of, 163–64 long division with, 163, 164–65 memory functions, 165 M⫹ key, 165 M– key, 165 MR key, 165 ON key, 163 order of operations, 163–64 parentheses, 163–64 RCL key, 165 Reverse Polish notation, 163 scientific notation, 165, 301–2, 364–65, 369 STO key, 165 SUM key, 165 TI34 II, 163 calendar array, 416 Caley, Arthur, 368 CaleyPurser algorithm, 368 cancellation property of addition integers, 347 rational numbers, 387 whole numbers, 123 of multiplication integers, 362 whole numbers, 137 zero divisor, 362 Canterbury Puzzles, The (Dudeney), 215 capacity, 668–69, 671 Cardano, 341 card deck probabilities, 517, 518–19, 555, 565 cardinal number, 59 carrying, in addition, 172, 173 Cartesian coordinate system, 417–19, 807, 819 Cartesian product of sets, 51–52, 125, 526–27 cartographers, 419 cashier’s algorithm, 184–85 Cavalieri’s principle, 724–26 Celsius, degrees, 88, 678–79 center of a size transformation, 863 center of a sphere, 646 centimeter, 667, 675–76 central angles, 607–8, 628–30
central tendency, measures of, 484–88 mean, 486–88 median, 485–86, 488 mode, 485, 486, 488 centroid of a triangle, 779, 787, 835–37 certain events, 519, 523 cevian line segment, 335 chain rule, 917 chant, counting, 60, 129, 140 chaos theory, 756 characteristic, 301 charts. See bar graphs; graphs; misleading graphs and statistics chess problem, 590 children abstract thinking, 585, 588–90 holistic thinking, 581, 583–84, 586 “Child’s Thought and Geometry, The” (van Hiele), 581 Chinese abacus, 155 Chinese numeration system, 69 chip abacus, 72, 172–75 Chudnovsky, David V., 411 Chudnovsky, Gregory V., 411 circle(s), 608–9, 777–79, 829 analyzing shape, 585–86, 608 arc, 766, 767, 770 area, 693 circumference, 689 circumscribed, 777–79, 838–39 diameter, 608 equations of, 829, 839 Euclidean construction, 777–79 hexafoils, 706 inscribed, 777–78, 779 noncollinear, 839 radius, 608 spheres (See spheres) symmetry, 608 tangent line to, 777 circle graphs (pie charts), 448–49, 451 exploding, 472 misleading, 472–73 multiplecircle, 448–49 relative vs. absolute, 472 when to use, 454 Circle Limit III (Escher), 849 circumcenter of a triangle, 777 circumference, 689. See also inside back cover circumscribed circles, 777–79, 838–39 Clinton, William Jefferson, 475 clock arithmetic, 923–25 clockwise orientation, 854–55, 860 closure property of addition fractions, 257 integers, 347
rational numbers, 386 real numbers, 402 whole numbers, 111 of multiplication fractions, 269 integers, 360–61 rational numbers, 389 real numbers, 402 whole numbers, 125 cluster estimation, 167 clusters and gaps, 443 codomain, 88–89 coefficients, 408 solving simultaneous equations and, 827 Cohen, Paul, 433 coin tossing probabilities combinations, 554 conditional, 566–67 experiments, 516, 518–20 tree diagrams, 540–41 Cole, Frank Nelson, 213 collinearity, reflections and, 879 collinearity, slope and, 813–14, 836 collinearity test for distance, 810 collinear medians of a triangle, 836 collinear points, 615, 863, 879, 881, 893, 894 combinations, 552–54 and Pascal’s triangle, 553–54 of r objects chosen from n objects, 553 commission, 333 common denominators, 247–48 addition, 255–57 division, 270–72, 273, 391 equality of rational numbers, 383, 387–88, 391 subtraction, 259–60, 387–88 common difference of a sequence, 86 commonpositivedenominator approach to ordering rational numbers, 392 common ratio of a sequence, 86–87 commutative property, 111, 125–26 of addition clock arithmetic, 923, 924 decimals, 291 fractions, 244, 258, 260 integers, 347 rational numbers, 386–87 real numbers, 402 whole numbers, 111–15, 158–59, 172–73 calculators and, 164 of multiplication clock arithmetic, 924–25 decimals, 291 fractions, 266, 269 integers, 360–61 in mental math, 158–59, 291 rational numbers, 389
I3
real numbers, 402 whole numbers, 125–26, 127, 128, 158–59, 266 comparison approach to subtraction, 117–18 compass, 608, 781, 865 compass constructions, 772, 780, 782 compass properties, 766, 767 compatible fractions, 260, 275, 322–23 compatible numbers, 158 decimals, 291, 293 rounding to, 162 whole numbers, 158, 162 compensation, 158–59 addition, 261, 291 additive, 159 decimals, 291, 292, 293 division, 167 multiplication, 159, 292, 293 subtraction, 159, 261 complement odds, 563 of a set, 50 subtraction by adding the, 185 complementary angles, 618, 620, 621 complementary triangles, 620 complex experiments, probability and, 532–42 complex fractions, 274 composite numbers, 205–7, 206, 214, 217 compound inequality, 145 compound interest, 421, 431 compound statements, 913 computational devices, 155 computational estimation, 160–63 cluster, 167 decimals, 291–93 fraction equivalents, 291, 322 fractions, 250–51, 275 frontend, 160–61, 261 percents as fraction equivalents, 322–23 range, 160, 261, 275, 292 rounding, 161–63, 182, 261, 292–93 of volume, 719 whole number operations, 160–63 See also mental math computations, 155, 157 with calculators, 163–66 estimation, 157–63 See also calculator activities; computational estimation; mental math computers, 155, 225, 305 concave (nonconvex) shapes, 607, 617, 631 conclusion, 914 concrete representation, 406, 408
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concurrent lines, 615 concurrent medians of a triangle, 836 conditional equations, 15, 405 conditional probability, 565–67 conditional statements, 914 cones, 646, 647, 711–12, 716, 722–23 congruence angles, 588, 590, 606, 741–42, 742–46 line segments, 588–90, 741 medians of isosceles triangle, 835 midsegment theory, 909 parallelograms, 604 parallelograms and rhombuses, 744, 791 polygons, 881, 883–84 shapes, 880–81 transformations, 875–81 triangles (See triangle congruence) congruence modulo m, 925–26 conjectures Bertrand’s, 217 Goldbach’s, 203 Hamann’s, 217 odd perfect number, 203 twin prime, 203 Ulam’s, 203 conjunction of statements, 913 connectives, logical, 913–15 consecutive angles, 627 consecutive sides, 608 consecutive vertices, 608 constant function with calculator, 164–65 constructions, 781. See also Euclidean constructions continuous set of numbers, 445 continuum hypotheses, 433 contrapositive, 914 converse, 914 converse of the Pythagorean theorem, 792–93 conversions between bases, 75–78 Celsius to Fahrenheit, 88, 678–79 measurements, 680–81 convertibility of measurement systems, 672–73 convex shapes, 607–8, 643–44 coordinate geometry, 809–16, 822–29, 834–39 Cartesian coordinate system, 417–19, 807, 819 collinearity test, 810 coordinates on a line, 616 coordinate system, 418–19 distance in coordinate plane, 809–12 Earth’s surface, 821
equations of circles, 829, 839 of lines, 822–26, 838–39 simultaneous, 826–29, 830–31 mapping history, 815 midpoint formula, 811–12 midsegment proof, 910 orthocenter, 837–38 percent grade, 818, 821 polar coordinates, 832 problem solving, 834–39 slope, 812–16, 826, 836 slopeintercept equation of a line, 823–25 See also planes coordinate plane distance in, 809–12 graphing integers on, 815 correlation relationship, 453 correspondence angles, 618–19 congruent triangles, 742 onetoone, 46, 60 points, 645, 647 similar triangles, 752–56 counterclockwise orientation, 854–55, 860, 877 counting factors, 219–20 counting numbers, 22, 45, 381 composite, 205–7, 214 as distinct primes with respective exponents, 220 doubling function, 88–89 factors of, 219–20 number systems diagram, 381, 382, 392, 401 prime, 205, 212–13 relatively prime, 227 sequences of, 20–21 set of, 45 Smith number, 231 sum of, 12–13, 20–21 counting techniques, 14–15, 549–55 by 1 and 2, 113 chant, 60, 129, 140 combinations, 552–54 factors, 219–20 fundamental counting property, 534, 536, 540, 542, 549–52 permutations, 549–53, 555 and probability, 555 and tree diagrams, 532–34, 540–41 Cover Up strategy, 16 Coxeter, H. S. M., 845 cropping graphs, 468–70 crossmultiplication, 244–45 of fraction inequality, 247–48, 249 of rational number inequality, 393 of ratios, 312–13
crossproducts, 244 cryptarithm, 8–9 cryptographic codes, 134 cubed numbers, 88 cubes (geometric), 640–42, 671, 718 cubic centimeter, 675–76 decimeter, 675–77 foot, 671 inch, 671 meter, 675–76 unit, 87, 717 yard, 671 cubic function, graphs of, 422–23 cup (unit of measure), 671 curved shapes in threedimensional space, 645–47 curves, bellshaped, 497–99 cylinders, 645–47, 709, 720–21, 723, 724
D data, 484–500 analyzing (See analyzing data) cluster, 443 comparison with multiple bar graphs, 445 discrete values, 445, 446 gap, 443 grouping in intervals, 443–44 histogram vs. bar graph, 445 inaccurate, 475–77 intervals, 445 organizing, 442–44 (See also graphs; statistics) picturing information, 449–54 utilizing, 441 debugging, 305 decay, 422 decimal point, 288, 297–300, 301 decimals, 71–73, 287–93, 297–305 additive properties, 291 algorithms, 297–301 addition, 297 division, 300–301 multiplication, 298, 300 subtraction, 298 associative property, 291 with calculators, 300, 302–3 commutative property, 291 compatible numbers, 291, 293 compensation, 291, 292, 293 concept of, 287–89 converting to and from fractions, 289, 291–93, 298, 300–301 distributive property, 291 dividing by powers of 10, 291–92, 301 Draw a Picture strategy, 288 equaladditions, 291, 301 estimation, 291–93 expanded form, 288, 297, 300
fraction approach to addition, 297 fraction method for ordering, 290, 300 fractions as, 287–89 frontend estimation, 292 hundreds square, 288, 289–90 Look for a Pattern strategy, 303–4 mental math, 291–93 metric system and, 673, 676, 677–78 monetary system, 249 multiplying by powers of 10, 291–92, 301 nondecimal systems vs., 75–78 nonrepeating, 400–402 nonterminating, 304–5 notation, 294 number lines, 288, 289–90 number system diagram, 400 operations with, 297–301 ordering, 289–90, 291 percents and, 321–22 placevalue for decimal algorithm, 297 for ordering, 289–90, 291 properties, 291 repeating (See repeating decimals) rounding, 292–93 squeeze method, 412 terminating, 289, 305 decimeter, 667, 673–74, 675–76, 676–77 deduction—level 3, 585 deductive arguments, 916–19. See also direct reasoning deductive reasoning, 916 deficient numbers, 228 definitions of geometric shapes, 621, 647 degrees Celsius, 88, 678–79 degrees Fahrenheit, 88, 672, 678–79 dekameter, 667, 673–74 Deming, W. Edwards, 441 Deming Award, 441 DeMorgan’s laws, 54, 57 denominator(s), 240 common (See common denominators) Egyptians, Mayans, and, 237 of equivalent fractions, 244 of improper fractions, 246 least common, 256–57 and numerators approach to division, 272–73 in rational numbers, 382–83 reducing fractions and, 244–45 unlike denominators addition, 256–57 division, 272–73 subtraction, 260, 388
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density property, 403 density property of fractions, 249 denying the consequent (modus tollens), 917–18 Descartes, René, 51, 807, 816 description of geometric shapes, 589 diagonals, 605, 606, 629 of a kite, 746, 768 of a quadrilateral, 894 of a rectangle, 743, 744, 790–91 of a rhombus, 768–69, 770, 791–92 slopes for analyzing, 816 of a square, 795 diagrams arrow, 83, 89 functions as, 89 in sequences, 86 graphs (See graphs) number systems, 381, 382, 392, 400, 401 tree (See tree diagrams) Venn (See Venn diagrams) See also Draw a Diagram strategy diameter, 608, 646 dice throwing, 516, 518–19, 520–21, 534, 564 Dienes blocks, 71 difference common, of a sequence, 86 minuend minus subtrahend, 116 order of operations, 144–45, 163–64 set, 110 of sets, 50–51 successive, 27–30 digits, 52, 67, 71, 74–75 digits, random, 561, 568 dihedral angles, 641–44 dilations/dilitations. See transformations dimensional analysis, 680–81 Diophantus, 341 directed angles, 854–55 directed line segments, 853 direct reasoning, 108, 148, 916 Direct Reasoning strategy, 108, 148 discounts, finding, 326–27 discrete values, 445, 446 disjoint sets, 48, 109–10, 345 disjunction of statements, 913–14 dispersion, measures of standard deviation, 493–96 variance, 493–94 distance(s) circumcenter of an angle, 770 coordinate distance formula, 810 in coordinate plane, 809–12 equidistant from endpoints, 770, 778
from perpendicular bisector, 778 from sides of a triangle, 779 equidistant from endpoints, 616 isometries and, 853, 876–77, 895 on a number line, 616, 686–87 parsec, 685 from point P to point Q, PQ, 686–87, 766, 770–71, 777–79, 809–10, 876–77 ratio of, 881–84 unit distance, 687 distribution(s), 489–500, 496–500 bellshaped, 497–99 mean, 497–500 (See also mean) medians, 497–99 (See also medians) mode, 496–99 (See also mode) normal, 498–500 outliers, 489–92, 500 relative frequency, 496 standard deviation, 498–500 symmetrical and asymmetrical, 496–99 zscores, 495–96, 498–500, 499–500 distributive property, 127 decimals, 291 division over addition, 158 multiplication over addition fractions, 270 integers, 360–61 rational numbers, 390 real numbers, 402 whole numbers, 135, 176 multiplication over subtraction fractions, 270 rational numbers, 390 whole numbers, 128 distributivity, calculators and, 164, 166 divide and average for square roots, 412 dividends, 132, 182 and “division makes smaller” misconception, 297 divides, 207 divine proportion, 285 divisibility. See tests for divisibility division, 123, 129–34, 270–75 algorithm, 133 base five algorithm, 194–95 compensation, 167 decimals, 292, 300–301 denominators (See common denominators; denominators) denominators and numerators approach, 272–73 dividend, 132 divides, 207 divisors, 132, 182, 207, 270–73, 300–301, 362 does not divide, 207
estimation for fractions, 275 fractions, 270–75, 300–301 invert and multiply, 390, 409 integers, 357, 362–63 invert and multiply, 272–74 long (See long division) measurement, 129–30, 270–73 mental math, 158, 275, 292 missing factor approach, 132, 182–83, 924–25 missing factors (See quotients) nonstandard algorithms, 186 numerators and denominators approach, 272–73 order of operations, 144–45, 163–64 partitive, 129–30 by powers of 2, 78 by powers of 10, 291–92, 301 quotients (See quotients) rational numbers, 390–92 real numbers, 402–3 remainders, 133, 164–65, 182–83 repeatedsubtraction approach, 133–34 standard algorithm, 182–83 unlike denominators, 272–73 whole numbers, 123, 129–34, 274 zero, property of, 132 See also tests for divisibility “division makes smaller” misconception, 297 divisor(s), 132, 207 decimals into whole numbers, 300–301 division of fractions, 270–73 rounding up, 182 zero divisors property of integers, 362 Do a Simulation strategy, 514, 574 dodecahedron, 644–45 does not divide, 207 domain, 88–89 dot plot, 442 doublebar graphs, 446 doubles, 114 doubling function, 88–89 dram, 672 Draw a Diagram strategy factor trees, 206 fractions, 240, 241–42, 243 measurement division, 129–30 outcome probabilities, 533, 537, 539, 562 partitive division, 129–30 percent problems, 324 sample space, 524, 566 sets, ratios, and probabilities, 44, 99 sets, ratios and probabilities, 99–100
I5
whole numbers less than and addition, 141 transitive property of “less than,” 140–41 Draw a Picture strategy, 9–11 addition, 112–13, 114 algebraic equations, 406, 408 angle sum of a pentagon, 629 centroid of a triangle, 836 circumcenter of a triangle, 838–39 combining with other strategies, 27–30 congruence properties, 746, 790–91 congruent medians of a triangle, 835 converse of the Pythagorean theorem, 792–93 cubic function, 422 decimals, 288 division, 133 fourfact families, 116–17 fraction strips, 242, 247, 255, 257 indirect measurement, 755 initial problem, 2, 38 orthocenter of a triangle, 837 pizzacutting, 9–10 repeatedsubtraction, 133–34 tetromino shapes, 10–11 See also number lines drawing with/without replacement, 534–35, 536–39 dual of a tessellation, 635 Dudeney, Henry, 215, 628 duplication algorithm, 186 dynamic spreadsheet activities on the Web site Base Converter, 79, 81 Circle Graph Budget, 462 Coin Toss, 527 Cubic, 428 Euclidean, 231 Function Machines and Tables, 95 Roll the Dice, 527 Scaffold Division, 186 Standard Deviation, 503, 505
E earned run average, 319 earth’s surface, 821 edges of a cube, 642 edges of a polyhedron, 643, 644 Egyptian numeration system, 62–63, 66, 237, 411 Einstein, 198 elementary logic. See logic/logical Elements, The (Euclid), 739, 784 elements in a finite set, 59, 89, 110 elements of a set, 45–46, 59–60 elimination method of solving simultaneous equations, 831
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Elliott wave theory, 425 ellipse, 834 ellipsis, 22 eManipulatives on the Web site Balance Beam Algebra, 413, 415 Base Blocks—Addition, 183, 188 Base Blocks—Subtraction, 184, 189 Chip Abacus, 79, 81 Chips Minus, 353, 356 Chips Plus, 353, 355 Circle 0, 355 Circle 21, 18 Circle 99, 20 Coin Toss—Heads in a Row, 532, 574 Color Patterns, 34 Comparing Fractions, 250, 252, 253 Composition of Transformations, 888, 892 Congruence, 749 Coordinate Geoboard, 817, 818, 820, 872–73 Counterfeit Coin, 32 Dividing Fractions, 276, 278, 279 Equivalent Fractions, 250, 253 Factor Tree, 228, 230 Fill ‘n Pour, 229 Function Grapher, 426, 429, 430 Function Machine, 95 Geoboard, 595, 623, 626, 634, 637, 697, 698, 700, 703, 865, 866, 868, 870, 871 Geoboard—Triangular Lattice, 594, 598 Histogram, 454, 457, 459 Let’s Make a Deal, 548 Multibase Blocks, 79, 80–81, 195, 196, 197 Multiplying Fractions, 276, 278 Number Bars, 119 Number Puzzles, 18, 20, 120, 123 Parts of a Whole, 254 Pattern Blocks, 254 Percent Gauge, 329, 330, 332, 333 Perpendicular Lines, 818 Pythagorean Theorem, 701 Rectangle Multiplication, 191 Rectangular Division, 136 Scatterplot, 456, 457, 461 Sieve of Eratosthenes, 214, 215, 217 Simulation, 568, 571, 572, 574 Slicing Solids, 655, 656 Tessellation’s, 635, 638 Tower of Hanoi, 36 Venn Diagrams, 53–54, 56, 57 Visualizing Fractions, 254
embedded graphs, 473–75 empty set, 46, 112 endpoints of a line segment, 601, 608, 616 English system of measurement, 669–72 area, 670–71 avoirdupois ounces and pounds, 672 length, 669–70 metric system compared, 674, 678, 680–81 notation, 669–72 ratios, 669–70 temperature, 672, 678–79 volume, 671 weight, 672 English system units of measurement acre, 669, 670–71 barrel, 671 board foot, 729 cup (unit of measure), 671 degrees Fahrenheit, 672, 678–79 dram, 672 foot, 669, 670–71 furlong, 669 gallon, 671 grain, 672 inch, 669, 670–71 mile, 669, 670–71 ounce, 671, 672 pint, 671 quart, 671 rod, 669 tablespoon, 671 teaspoon, 671 volume, 671 weight, 672 yard, 669, 670–71 ENIAC, 155 equaladditions algorithm, 185 compensation in subtraction method, 159 decimals, 291, 301 subtraction, 261 equality equal groups and equal shares, 131 equal sets, 46 of fractions, 244, 322–23 of rational numbers, 383, 387–88, 391 of ratios, 311 equallylikely events additive property of probability, 536–38 children as male or female, 560–61 conditional probability, 566–67 drawing with/without replacement, 535
marble drawing, 521 multiplicative property of probability, 536–38 odds and, 563–65 Pascal’s triangle, 542 probability property for, 517–18 sample space of, 524 unequally likely events vs., 520–21, 525, 565, 566 Venn diagram of conditional probability, 566–67 equal sets, 46 equals sign, 742 equal to, mode and median, 486 equation(s), 15 abstract representation, 406, 408 Cartesian coordinate system, 417–19, 807, 819 of circles, 829, 838–39 concrete representation, 406, 408 conditional, 15, 405 extremes and means, 311 false, 15 identity, 15 linear, 830–31 of lines, 822–26, 838–39 percents, 325–28 pointslope of a line, 825–26 simultaneous, 826–29, 830–31, 831 slopeintercept of a line, 823–25 solving (See solving equations) true, 15 equiangular, 590 equiangular polygons, 628–30 equidistant from endpoints, 770, 778 from perpendicular bisector, 778 from sides of a triangle, 779 equidistant from endpoints, 616 equilateral, 588 polygons, 589, 628–30, 632–33 squares as, 590 triangle, 588, 589, 601, 602, 632 equilateral triangles Euclidean construction, 780, 782 Koch curve, 756 equivalence relation, 85 equivalent directed line segments, 853 fractions, 242, 244–45, 256, 289, 291, 322–23 logically equivalent, 915 parts, 240 ratios, 311 sets, 46 Eratosthenes, 700–701
Eratosthenes, Sieve of, 206, 234 Erdos, Paul, 281, 713 Escalante, Jaime, 902 Escher, M. C., 849 Eschertype tessellations, 849, 861–62 estimation. See computational estimation; frontend estimation Euclid, 739, 784 Euclidean algorithm, 223, 225 Euclidean constructions, 765–71, 777–80, 782 bisect an angle, 769–70 circumscribed circles, 777–79 compass properties, 766, 767 copy a line segment, 766–67 copy an angle, 767 equilateral triangle, 780, 782 inscribed circles, 777–78, 779 line parallel to a given line through a point not on the line, 771 parallel lines, 771 perpendicular bisector, 768 perpendicular line through a point on the line, 770 perpendicular line to a given line through a point not on the line, 770–71 regular polygons, 780, 782 segment lengths, 783 straightedge properties, 766–67 Euclid of Alexandria, 739 Euclid’s perfect number, 228 Eudoxus, 379 Euler, Leonhard, 213, 411, 733 Euler diagram, 917 Euler’s formula polyhedra, 644, 733 even counting numbers, 22 n th root and, 403–4 number of factors, 363 number of prime factors, 399–400 numbers, 88 odd and, making groups of two, 209 rounding to nearest, 162 event(s), 516 pairwise mutually exclusive simpler, 536 probability of, P(E), 517–24, 529 See also equallylikely events; outcomes exchanging addition, 172–73 subtraction, 174–75 exclusive “or,” 913 exercises, problems vs., 4 expanded form/notation, 72 decimals, 288, 297, 300
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exponents, 142–43 place value and, 72, 76, 172–73 whole numbers, 72, 76, 172–73 expected outcome, 560 expected value, 562–63 experiment(s), 516 Cartesian product of sets, 526–27 complex, 532–42 simple, 515–25 computing probabilities in, 517–24 experimental probability, 519–20, 532–42 simulation, 560–62 exploding circle graphs, 472 exponent(s), 142–44 additive property, 404 base systems and, 77–78 with calculators, 165, 404 division, 143–44 integers and, 357 multiplication, 142–43 negative integer, 363–65 and ones digit, 22–24 powers of 10, 62, 159, 289, 291–92, 301 powers of three, 22, 88 prime factorization method greatest common factor, 221–23 least common multiple, 224, 225 product of primes with, 219–20 rational, 403–5 realnumber, 405 scientific notation of, 363–65 unit fraction, 404 variable as, 421 whole number, 77–78, 142–44 zero as, 144 exponential decay, 422 exponential functions, graphs of, 420–22 exponential growth, 87, 421, 422 compound interest and, 431 exponential notation, 142–43, 225, 363–65 exterior angles, 607–8, 617 measures, 630 exterior of the angle, 617 extreme outliers, 503 extremes, 311
F faces of a dihedral angle, 641–42 faces of a polyhedron, 643, 647 factorial, 214 n factorial, 550–51 factorization. See prime factorization factors, 124 counting, 219–20 of integers, 352
missing (See quotients) number of, 219–20 prime factors vs., 219–20 primes and composites, 205–7 proper, 203, 228 special, 159 test for, 213 (See also tests for divisibility) factor trees, 206 facts addition, 111–15, 174, 176 for base five subtraction, 192–93 fourfact families, 116–17 multiplication, 128–29, 176 base five, 194 subtraction, 116–17 base five, 193 Fahrenheit, degrees, 88, 672, 678–79 Fahrenheit, Gabriel, 672 false equations, 15 Fermat, Pierre de, 203, 213, 513 Fermat primes, 782 Fermat’s Last Theorem, 203 Fermi problems, 201 Feuerbach, Karl, 839 Feuerbach circle, 839 Fibonacci sequence counting numbers and, 22 golden ratio and, 285 patterns and, 33, 36 Pythagorean triples and, 705 sum of ten consecutive, 218 finger multiplication, 186 finite sets, 47–48, 59, 60, 89, 110 Fish (Escher), 849 5con triangle pairs, 746 Flannery, Sarah, 368 flats, in base ten pieces, 71 foot, 669, 670–71 formulas coordinate distance, 810 coordinate midpoint, 811–12 Euler’s formula polyhedra, 644, 733 functions as, 90–91 Hero’s formula, 697 population predictions, 428 as problem solving strategy, 440, 506–7 See also inside back cover fourcolor problem, 713 fourfact families, 116–17 fourstep process. See Pólya’s fourstep process Fractal Geometry of Nature, The (Mandelbrot), 755 fractals and selfsimilarity, 755–56 fraction(s), 239–49, 255–61, 266–75, 381 addition, 255–57 additive properties associative, 244, 258, 260
closure, 257 commutative, 244, 258, 260 identity, 258 associative property of addition, 260 commutative property of addition, 260 comparing two, 294, 393 complex, 274 concept of, 239–46 crossmultiplication, 244–45, 247–48, 249 decimals, 287–89 denominator (See denominators) density property, 249 dimensional analysis, 680 division, 270–75, 300–301 with common denominators, 270–72, 273 complex fractions, 274 invert and multiply, 272–74, 390, 409 numerators and denominators approach, 272–73 unlike denominators, 272–73 Draw a Diagram strategy, 240, 241–42, 243 Egyptian numeration, 237 equivalent, 242, 244–45, 256, 289, 291, 322–23 equivalent parts, 240 estimation, 250–51, 275 greater than, 246, 247–48 history of, 237 improper, 246, 258–59 calculator, 268 infinite number of, 245, 248, 383, 385 least common multiple, 223–26 less than, 247–48 lowest terms, 244, 245, 253 mental math, 260–61, 275 missingaddend subtraction, 259–60 mixed numbers, 246, 257, 258–59 multiplying, 267–68, 275 monetary system and, 249 multiplication, 266–70, 275 decimals and, 298, 300 multiplicative properties associative, 269–70, 275 closure, 269 commutative, 266, 269 distributive, over addition, 270 identity, 269–70 inverse, 269–70, 274 number lines, 247, 248, 255, 266
I7
as numbers, 240–41, 242–46 number systems diagram, 381, 382, 392, 401 as numerals, 240, 241–42, 245 numerator, 237, 240 numerousness, 241 ordering, 247–49 parttowhole, 241–42 percents and, 321, 322 prime factors, 245 as rational numbers, 382 reciprocals, 269, 382, 389, 390–91 “reducing,” 244 relative amount, 240–42, 244 repeating decimal representation, 304–5 repeating decimals and, 304–5 simplified form, 242, 244, 245, 253 with calculator, 257 LCD, 256 Solve an Equivalent Problem strategy, 238, 281 subtraction, 259–61 terminating decimals, 289 “to break,” 237 unitary, 263 See also decimals; rational numbers; ratios fraction method for ordering decimals, 290, 300 fractions compatible, 322–23 fraction strips, 242, 247, 255, 257 Franklin, Benjamin, 118 frequency of a number, 442 histograms, 443–44 stem and leaf plots, 442–43 frontend estimation, 160 with adjustment, 160–61, 292 decimals, 292 one and two column, 160 one/two column, 160 range, 160 frustum of a right circular cone, 716 fuel gauge model, 324 function(s), 82, 85–91, 417–25 arithmetic sequence, 86–87 as arrow diagrams, 89 calculator memory, 165 Cartesian coordinate system, 417–19 codomain, 88–89 common difference of a sequence, 86 common ratio of a sequence, 86–87 cubic, 422–23 defined, 85 domain, 88–89 doubling, 88–89 exponential, 421
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compass, 608 geometric shape concepts, 583–90, coordinates (See coordinate 600–609, 615–22 geometry) adding zing to, 845 diagonals, 605, 606, 629 analytic geometry, 816 diameter, 608, 646 congruent, 880–81 (See also Escher’s art, 849, 860–61 congruence) Euclidean constructions, definitions of shapes, 621, 647 765–71, 777–80, 782 describing shapes, 588–90 Euclid’s organization of, 739 models of shapes, 589 fractals and selfsimilarity, van Hieles’ stages of learning, 755–56 581 great circle of a sphere, 651, See also base (geometric) 712 geometric shape(s) height (See height) angles (See angles) length (See length) circles (See circles) mean proportional, 786 concave, 607, 617, 631 median, interpretation of, 497–99 cones, 646, 647, 711–12, 716, Mira®, 602, 608 723 models and abstractions, 589 congruent (See congruence; orthocenter, 779, 784, 837–38 triangle congruence) parallel lines and transversals, convex, 607–8, 643–44 618–19 cubes, 640–42, 718 patterns, 13–15 (See also cylinders, 645–47, 709, patterns; tessellations) 720–21, 723, 724 perpendicular bisector (See dodecahedron, 644–45 perpendicular bisector) ellipse, 834 polygonal regions, 630, 643–45 frustum, 716 probability of concept of hexafoils, 706 G measure, 529 hexagons, 607, 632, 633, 781 gallon, 671 problem solving hexahedron, 644 Gallup poll, 477 with algebra, 807 hyperbola, 834 gaps and clusters, 443 with coordinates, 834–39 icosahedron, 644–45 Gardner, Martin, 374 with transformations, kites (See kites) Garfield, James, 705 893–95 ngons (See polygons) Gauss, Carl Friedrich, 36, 39, 665 with triangle congruence octahedron, 644–45 Gauss’s theorem for constructible and similarity, 790–95 parabola, 834 regular ngons, 782–83 protractor, 617–18 parallelograms (See GCF. See greatest common factor Pythagorean theorem (See parallelograms) Geometer’s Sketchpad on the Web Pythagorean theorem) pentagon, 605, 607–8 site sequence, 86–87 pentominoes, 342, 373, 613 Centroid, 787 shape (See geometric shapes) planes (See planes) Circumcenter, 787 skew lines, 642 polygons (See polygons) Midquad, 797 slant height, 710, 711–12 polyhedra (See polyhedra) Name That Quadrilateral, 614 surface area, 707–12 prisms (See prisms) Orthocenter, 784 symmetry, 600–604 properties Parallelogram Area, 704 tangent, 777 angles, 617, 619–20 Rectangle Area, 698 teaching, 845 convex attribute, 607 Same Base, Same Height, Same terminology, 589, 606, 621, 647 diagonals of a kite, 605 Area, 703 timeline for, 807 isosceles triangles and Size Transformation, 869 transformation (See trapezoids, 601–2 Slope, 817 transformation geometry) parallelograms, 604 Tree Height, 758 vector geometry, 909 points and lines, 616 Triangle Inequality, 707 vertex arrangement, 632–33, pyramids (See pyramids) geometric concepts 638 quadrilaterals (See analytic geometry, 816 vertices, 588, 602–3, 605, quadrilaterals) analyzing shapes, 600–609 643–44 recognizing, 583–90, 585–87 circles, 608–9 volume (See volume) rhombicuboctahedron, 645 parallel line segments, See also coordinate geometry; rhombus (See rhombus) 605–7 number lines; transformation shape identification, 585–87 perpendicular line geometry simple closed curves, 607, segments, 605–7 geometric fallacy, 285 645–46, 657 regular polygons, 607–8 geometric mean, 786 sphere, 646, 648, 651, 712 symmetry, 600–604 geometric representation of real square (See square) angle bisector, 746, 769–70, 779 numbers, 401 table of, 589 function(s) (continued) exponential growth, 87 as formulas, 90–91 geometric sequences, 86–87 as geometric transformations, 91 from graphs, 424–25 graphs of, 90, 417–24 greatest integer, 427 initial term, 86 inputs into, 90, 93 linear, 419 as machines, 90 notation, 87–88 as ordered pairs, 90 quadratic, 420 range of, 88–89, 91 representations of, 88–91 step, 423–24 stock market and, 425 as tables, 89 fundamental counting property, 534, 536, 540, 542, 549–52 Fundamental Theorem of Arithmetic, 206, 219, 399–400 furlong, 669
tessellations (See tessellations) tetrahedron, 644–45 tetromino, 10–11, 611 threedimensional (See threedimensional shapes) trapezoid (See trapezoids) triangles (See triangles) Germain, Sophie, 31, 39 German lowstress algorithm, 188 Gleason, Andrew, 507 glide reflection, 858–60, 875–77 collinearity and, 879 congruent shapes, 880–81 distance and, 877 glide axis of, 859, 879–80 image, 859 isometries, 858–60 notation, 875, 876 symmetry, 861 transformations, 858–60 Goldbach’s conjecture, 203, 215 golden ratio, 285 golden rectangle, 285, 786 grain (unit of measure), 672 gram, 678 Granville, Evelyn Boyd, 281 graphing calculator, 420 graphs, 417–25, 444–54, 464–77 backtoback stem and leaf plot, 443 bar (See bar graphs) bellshaped curves, 497–99 box and whisker plot, 489–92, 503 Cartesian coordinate system, 417–19, 807, 819 circle, 448–49, 451, 454, 472–73 clusters and gaps, 443 comparison of, 454 coordinate system, 418–19 cropping, 468–70 cubic functions, 422–23 distortion (See misleading graphs and statistics) as distributions, 489–500 dot plot, 442 ellipse, 834 exponential functions, 420–22 functions, 90 of functions, 417–24 functions from, 424–25 gaps and clusters, 443 histograms, 443–44, 454 hyperbola, 834 line (See line graphs) linear functions, 419–20 line plot, 442 misleading (See misleading graphs and statistics) multipleline, 447–48 outliers, 453, 489–92, 503 parabola, 834 pictographs, 449–51, 454, 471–72
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pictorial embellishments, 450–51 pie charts, 448–49, 451, 454, 472–73 plot, 442 quadratic functions, 420 regression lines, 453 scatterplot, 452–54 solving simultaneous equations, 830–31 stem and leaf plot, 442–43, 490–91, 503 step functions, 423–24 vertexedge (See vertexedge graphs) vertical line test, 424–25 See also diagrams great circle of a sphere, 651, 712 greater than addition inequalities, 410 fractions, 246, 247–48 general set formulation of, 62 inequalities, 405, 410 integers, 365–67 mode and medians, 486 multiplication by a negative inequalities, 410 multiplication by a positive inequalities, 410 rational numbers, 392 real numbers, 402–3 transitive property of less than and, 141 whole numbers, 62 greater than or equal to, 62, 140 fractions, 246, 247–48 properties for solving inequalities, 410 greatest common factor (GCF), 220–23 calculator method, 222, 223 Euclidean algorithm, 223, 225 least common multiple and, 225–26 prime factorization method, 221–23 set intersection method, 220–21 Venn diagrams, finding with, 228, 234 greatest integer function, 427 Great Mental Calculators, The (Smith), 166 Great Pyramid of Giza, 261 Gregory, 411 grid approach to percent problem solving, 324 grouping data in intervals, 443–44 equal groups and equal shares, 131 by fives, 62, 75–76 nondecimal systems, 75–78 for ratios, 311
subtractive pairs, 63 in tally numeration system, 62 by tens, 62–63, 71–73, 74 Guess and Test strategy, 5, 9, 38 for algebraic equations, 16 cryptarithm, 8–9 Inferential, 6, 8 Random, 6, 8 Systematic, 6, 8 gumball probabilities, 525, 539–40
H Haken, Wolfgang, 713 halfturn rotation, 893–94 Halmos, Paul, 149 halving and doubling algorithm, 169, 185 Hamann, Arthur, 217 Hamann’s conjecture, 217 hamburger condiments probability, 534 Harding, Warren G., 513 Hardy, George, 232 harmonic triangles, 265 hectare, 675–76 hectometer, 667, 673–74 height cone, 711–12 cylinder, 709 parallelogram, 692 prism, 718, 720 slant, of a pyramid, 710 trapezoid, 692 triangle, 691 Hero’s formula, 697 hexafoils, 706 hexagonal cloud, 895 hexagons, 607, 632, 633, 781 hexahedron, 644 Hiele, Pierre van, 581, 583 HieleGeldorf, Dieke van, 581, 583 Hilbert, David, 100, 433 HinduArabic numeration system, 71–75, 71–78 base, 71–73, 77–78 chip abacus, 72, 172–75 concepts, 74–75 digits, 71 digit shapes, 52 expanded notation/form, 72, 76 fractions, 237 history of, 107 nondecimal systems, 75–78 numeral word names, 74–75 place value, 72, 73 See also base ten Hippasus, 379 histograms, 443–44, 445 holistic measurement, 667 holistic thinking, 581, 583–84, 585–86, 586 Honea, David, 726 Hoppe, Oscar, 215
Hopper, Grace Brewster Murray, 198, 305 horizontal axis. See xaxis horsepower, 685 How to Solve It (Pólya), 1 hundreds square, 288, 289–90 Hypatia, 800 hyperbola, 834 hypotenuse of a right triangle, 401, 693–95 hypothesis, 914, 916 hypothetical syllogism, 917
I9
Inferential Guess and Test strategy, 7–9 infinite geometric series, 265 infinite number of primes, 226, 402 infinite sets, 47–48 informal measurement. See nonstandard units of measurement information, 442–54 organizing, 442–44 picturing, 449–54 See also data; graphs initial side of a directed angle, 855 I initial term, 86 icosahedron, 644–45 inscribed circles, 777–78, 779 identification numbers, 59 inscribed hexagon, constructing, Identify Subgoals strategy, 740, 781 799 Instituzioni Analitiche (Agnesi), 733 identity element (one), 126 insurance company’s expected identity element (zero), 112–13 value, 562–63 identity equations, 15 integer(s), 343–52, 357–67, 381 identity property absolute value, 354 of addition addition, 345–48 fractions, 258 negative integers, 348 integers, 347, 384 additive properties rational numbers, 384, associative, 347 386–87 cancellation, 347 real numbers, 402 closure, 347 whole numbers, 112–13 commutative, 347 of multiplication identity, 347, 384 fractions, 269–70 inverse, 347, 352, 384–85 integers, 360–61 black and red chips model, 344, rational numbers, 389 345, 348, 349–50, 352, 382 real numbers, 402 calculators, 352 whole numbers, 126–27 distributivity of multiplication zero, 112–13 over addition, 360–61 if and only if connective, 915 division, 357, 362–63 ifthen connective, 914–15 exponents and, 357 image factor of, 352 halfturn, 893 graphing on coordinate plane, of points, 852, 861–62, 875 815 reflection and, 852, 859 greatest integer function, 427 translations and, 853–55 measurement model, 345, 346 image of P, 852, 875 multiplication, 357–62 implication of statements, 914 look for a pattern, 358–59 impossible events, 519, 521, 523 zero divisors of, 362 improper fractions, 245, 258–59 multiplicative properties calculator and, 268 associative, 360–61 incenter of a triangle, 777–78, 779 cancellation, 362 inch, 669, 670–71 closure, 360–61 included angles, 743–44, 753–54 commutative, 360–61 included side of a triangle, 744 distributive, over addition, inclusive “or,” 913 360–61 index, 404 distributive, over indirect measurement, 754–55 subtraction, 360–61 indirect reasoning, 156, 198 identity, 360–61 problem solving with, 198, 226, negative (See negative integers) 399–400 negative exponents, 363–65 inductive reasoning, 22 number lines, 344, 365–67 inequalities number systems diagram, 381, compound, 145 382, 392, 401 defined, 405 opposites of, 344, 352 solving, 409–11 ordering, less than properties, triangle, 695 365–67
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integer(s) (continued) positive, 343 as rational numbers, 382 scientific notation, 363–65 set model, 344, 345 sets of, 343 subtraction, 348–52 adding the opposite, 349–50, 352 missing addend, 350 pattern, 348 takeaway, 349, 350, 352 Use a Variable strategy, 350 zero as, 343, 362 See also rational numbers interior angles, 608, 619–20, 627, 769. See also vertex angles interior of the angle, 617 intermediate algorithms addition, 173, 192 long division, 180, 182 multiplication, 177, 194 interquartile range (IQR), 489–92 interrelatedness of measurement systems, 672, 678, 679 intersecting arcs, 766–70 passim, 780 intersecting lines, 618, 621, 826–29, 830–31 intersecting planes, 641, 651 intersection of sets, 49, 110 probability and, 522, 524–25 intervals, data, 443–44, 445 invalid argument, 916 inverse, 914 inverse property of addition integers, 347, 350, 352, 384–85 rational numbers, 384–87 real numbers, 402 of multiplication fractions, 269–70, 274 rational numbers, 389 real numbers, 402 invert and multiply, 272–74, 390, 409 Ionian numeration system, 68 irrational numbers defined, 379 infinite number of, 402 number systems diagram, 401 pi, 402, 411, 567, 689, 723–24 set of, 400–402 is an element of a set, 46 is divisible by, 207 is necessary and sufficient for, 915 is necessary for, 915 is not an element of a set, 46 isometries, 852–60, 864, 875–81 congruence polygons, 881, 883–84 shapes, 880–81 transformations, 875–81
distance and, 876–77 glide reflection (See glide reflection) properties of, 878 reflection (See reflection) rotation (See rotation) similitudes, 864 size transformation prior to, 864 translation (See translation) isosceles trapezoids, 589 reflection symmetry, 601–2 isosceles triangle, 589, 644 acute, 630–31 congruent medians and, 835 reflection symmetry, 601–2, 604 isosceles triangles Pons Asinorum Theorem, 739 is sufficient for, 915
J Japan, Deming and, 441 Jefferson, Thomas, 454, 696 Johnson, C., 789 joule, 685
K Kemeny, John, 198 kilogram, 678 kiloliter, 676–77 kilometer, 667 kites attributes of, 589, 606 creating shape of, 590 diagonal of a, 746, 768 model and abstraction, 589 onetoone correspondence, 852 perimeter, 688–89 perpendicular diagonals, 605, 606 reflection symmetry, 852 SSS congruence property, 746 tessellations with, 631 Koch curve, 756 Koch curve (or snowflake), 34 Kovalevskaya, Sonya, 336
L lateral faces, 643, 647. See also quadrilaterals lateral surface area, 707 of a cone, 711–12 of a cylinder, 709 of a prism, 708 of a pyramid, 710 latitude, 821 lattice, square, 588, 615, 695, 700 lattice, triangular, 588 lattice method algorithm addition, 173, 192 multiplication, 177, 194
law of detachment (modus ponens), 916–17 least common denominator (LCD), 256–57 least common multiple (LCM), 223–26 buildup method, 224–26 greatest common factor and, 225–26 least common denominator, 256–57 number lines, 224 prime factorization method, 224, 225 set intersection method, 224, 225 Venn diagrams, finding with, 228, 234 LeBlanc, Antoine (Sophie Germain), 31, 39 lefttoright methods, 159 legs of a right triangle, 693–95 Lehmer, D. H., 217 Leibniz, Gottfried, 155 Leibniz, Gottfried Wilhelm, 665 length, 669–70, 673–74, 686–87 abstract, 686–87 circumference, 689 distance (See distance) English system, 669–70 metric system, 673–74 perimeter, 688–89 length of a line segment, 588, 616 less than, 140–42, 365–67, 392 addition inequalities, 409–11 integers, 365 rational numbers, 393 real numbers, 403 whole numbers, 141 fractions, 247–48 inequalities, 405 mode and median, 486 multiplication rational numbers, 393 whole numbers, 141–42 multiplication by a negative inequalities, 409–11 integers, 366–67 rational numbers, 393 real numbers, 402–3 multiplication by a positive inequalities, 409–11 integers, 366–67 rational numbers, 393 real numbers, 402–3 transitive property inequalities, 409–11 integers, 366 rational numbers, 393 real numbers, 403 whole numbers, 140–42 less than or equal to, 140–42 properties for solving inequalities, 410
lightyear, 683 Lilavat (Bhaskara), 276 line(s), 822–27 angle bisector, 746, 769–70, 779 concurrent, 615 equations of, 822–27, 837–38, 838–39 graphs (See line graphs) intersection of, 618, 621, 826–29, 830–31 number (See number lines) parallel (See parallel lines) perpendicular, 587, 605–7, 618, 621, 642, 770–71, 814, 816 in a plane, 615–16 pointslope equation, 825–26 properties of points and, 616 reflection in, 856, 858 regression, 453 skew, 642 slopeintercept equation, 823–25 slope of, 812–13, 826, 837–38 in threedimensional space, 642 See also parallel lines; perpendicular bisector linear equations, 830–31 linear functions, graphs of, 419–20 line graphs, 447–48, 468–70 and cropping, 468–70 distorting, 471 multipleline, 447–48 with pictorial embellishments, 451 plot and, 442 scatterplot and, 453 threedimensional effects, 471 when to use, 454 line of symmetry, 600 line plot, 442 line segment(s), 588 AA similarity property, 783 bisector, 746, 768, 770–71, 778 cevian, 335 congruent, 588–90, 741 copying, 766–67 creating shapes with, 588–90 definition, 588 diagonal, 605, 606 directed, 853 endpoints, 601, 608, 616 equivalent directed, 853 Euclidean constructions, 766–67, 783 length of, 686–87 length of a, 588, 616 midpoint, 616, 811–12 midsegment, 704, 794 model and abstraction, 589 parallel, 606, 615–16, 642 perpendicular, 589 perpendicular bisector, 746, 768, 770–71, 778 slope of, 813
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Mandelbrot, Benoit, 755, 800 mantissa, 301 mapping. See transformation geometry mapping history on the coordinate plane, 815 maps, geographical, 419, 594 Marathe’s triangle, 123 marbles probabilities, 521–24, 532–34, 535–38 matching sets, 46 mathematical art, 849 mathematical games, 373 Mathematical Sciences Research Institute (MSRI), 845 mathematics of chance. See probability Mayan numeration system, 43, 65–66, 70 Maya people, 43 mean (arithmetic average), 486–88 central tendency, 486–88 distribution, 497–500 median and, 485 standard deviation, 493–96, 498–500 variance, 493 zscore, 495–96 mean proportional, 786 means, extremes and, 311 measurement, 667–81, 686–95, 707–12, 717–26 of angles, 617–18 dihedral, 641–42 directed, 854 in polygons, 628–31 angle sum in a triangle, 619–20 area (See area) capacity vs. volume, 671 central tendency (See measures of central tendency) circumference, 689 dispersion, 492–96 (See also measures of dispersion) distance (See distance) division, 129–30, 270–73 English (See English system of measurement) holistic, 667 indirect, 754–55 informal (See nonstandard units of measurement) length, 669–70, 673–74, 686–87 M liquid measures of capacity, machines, functions as, 90 671 magic squares, additive, 19, 217, metric (See metric system) 295, 296 models magnifications. See transformations addition, 110–11 Make a List strategy, 23–25 fractions, 255 combining with other strategies, integers, 345, 346 26–30 takeaway, 115–16
liquid measures of capacity, 671 Literary Digest poll, 477 liters, 676–77 logical connectives, 913 logically equivalent, 915 logic/logical algebraic, 163–64 arguments, 916–19 arithmetic, 163 connectives, 913–15 statements, 912–13 long division, 178–83 base five, 77 base five algorithm, 194–95 base ten algorithm, 178–79, 181 calculator activities, 302, 303 with calculators, 163, 164–65 decimals, 300–301 intermediate algorithm, 180, 182 missing factor approach, 178–80, 182, 195 missing factors approach, 301, 303 remainders, 178–80, 182 repeating decimals, 300–304 scaffold method, 180, 195 standard algorithm, 182–83, 302–3 thinking strategy, 180, 182 longitude, 821 longs, in base ten pieces, 71 Look for a Formula strategy, 440, 506–7 Look for a Pattern strategy, 20–23, 27–28 combining with other strategies, 27–30, 440 counting factors, 219–20 decimals, 303–4 distance involving translations, 895 downward paths in grid, 21–22 find ones digits in 399 problem, 22–23 integer multiplication, 358–59 negative exponents, 363–64 repeating decimals, 303–4 Loomis, Elisha, 784 lower quartile, 489–92 lowest terms fractions (simplified), 242, 244, 245, 253, 256, 257 rational numbers, 383
number lines (See number lines) perimeter, 687–89 process, 667 with a protractor, 617–18 with square units, 689 standard deviation, 493–96, 498–500 surface area, 707–12 system of units, ideal features of, 672 temperature, 672, 678–79 variance, 493–94 volume, 671, 676–77, 717–26 measurement process, 667 measures of central tendency, 484–88 mean, 486–88 median, 485–86, 488 mode, 485, 486, 488 measures of dispersion standard deviation, 493–96 variance, 493–94 median(s), 485–86 box and whisker plot, 489–92 central tendency, 485–86 centroid, 779, 787 distribution, 497–99 finding, 488 geometric interpretation of, 497–99 mean and, 485 of a triangle, 773, 779 collinear, 836 concurrent, 836 congruent, 835 ratio of, 837 members of a set, 45–46 memory function of calculators, 165 mental math, 157–59, 163 addition, 158–59 of fractions, 260–61 compatible numbers, 158 compensation, 158–59, 167 decimals, 291–93 developing child’s ability, 163 division, 158 of fractions, 275 fraction equivalents, 322–23 fractions, 260–61, 275 halving and doubling, 169 lefttoright method, 159 multiplication, 158–59 of fractions, 275 order of operations, 158–59 percents, 322–28 percents as fraction equivalents, 322–23 powers of 10 multiplication, 159 properties, 158 scaling up/down, 314
I11
subtraction, 158–59 of fractions, 260–61 whole numbers, 158–59 See also computational estimation; thinking strategies Mere, Chevalier de, 513 meridians, 821 Mersenne number 267–1, 213, 217 meter, 667, 673–74, 675–76 meter prototype, 673 metric system, 662–79, 672–79 area, 675–76 converter diagram, 673–74 convertibility of, 672–73, 676–78 decimals, 673, 676, 677–78 English system compared to, 674, 678, 680–81 ideal features of, 672 interrelatedness of, 672, 678, 679 length, 673–74 mass, 678 notation, 673, 676, 677, 678 powers of 10, 62, 159, 289, 291–92, 301 prefixes, 673–74 temperature, 678–79 volume, 676–77 world acceptance of, 696 metric system units of measurement are (metric area of a square), 675 centimeter, 667, 675–76 decimeter, 667, 673–74, 675–76, 675–77, 676–77 degrees Celsius, 678–79 dekameter, 667, 673–74 dram, 672 gram, 678 hectare, 675–76 hectometer, 667, 673–74 kilogram, 678 kiloliter, 676–77 liters, 676–77 meter, 667, 673–74, 675–76 metric ton, 678 milliliter, 676–77 millimeter, 667, 673–74 temperature, 678–79 metric ton, 678 midpoint formula, 811–12 midpoint of a line, 616 midquad theorem, 794–95 midsegment, 704, 794, 909–10 mild outliers, 503 mile, 669, 670–71 milliliter, 676–77 millimeter, 667, 673–74 million, 74, 104 minuend, 116 Mira®, 602, 608
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Index
misleading graphs and statistics, 439, 464–77 bar graph distortion compressing yaxis, 444–45, 447, 465–67 pictorial embellishments, 474 threedimensional effects, 470–72 circle graph distortion, 472–73 line graph distortion, 468–70, 471 pictorial embellishments, 473–75 samples and bias, 475–77 scaling and axis manipulation, 465–67 threedimensional effects, 470–72 missingaddend subtraction fractions, 259–60 integers, 350 whole numbers, 116–17, 128, 192–93 missing factor approach division, 132, 182–83, 924 missing factors, 132, 300, 326. See also quotients mixed numbers, 246, 257, 258–59 Killie’s Way solution, 394 multiplying, 267–68, 275 rational numbers, 382 mode central tendency, 485, 486 distribution, 496–99 finding, 488 models black and red chips, 382 blocks/pieces (See under base ten: blocks/pieces) bundles of sticks, 71, 75 chip abacus, 172–75 fractions as parttowhole, 241–42 fraction strips, 242, 247, 255, 257 fuel gauge, 324 geometric shapes, 589, 616 for integers, 344, 345, 348, 349–50, 352 measurement, 110–11, 115–16, 255, 345, 346 number lines (See number lines) place value (See place value) region, 255 set (See set models) solving problems with, 582, 657 See also number lines modus ponens (law of detachment), 916–17 modus tollens (denying the consequent), 917–18
Moebius, A. F., 622 Moore, R. L., 658 Morawetz, Cathleen Synge, 658 MSRI (Mathematical Sciences Research Institute), 845 multibase pieces, 71, 72, 76 multidigit addition, 114–15 multiple, 207 common, 224, 225 least common, 223–26 multiplebar graphs, 445 multiplecircle graphs, 448–49 multipleline graphs, 447–48 multiplication, 123–29, 159, 266–70 by 0, 1, 2, 5, or 9, 128–29 associative property decimals, 291 fractions, 269–70, 275 integers, 360–61 rational numbers, 389 real numbers, 402 whole numbers, 126, 129, 158–59 base five, 194 cancellation property, 137, 362 Cartesian product approach, 51, 125 Cartesian product of sets, 526–27 closure property, 125, 269, 360–61, 389 coefficients, 408 commutative property decimals, 291 fractions, 266, 269 integers, 360–61 rational numbers, 389 real numbers, 402 whole numbers, 125–26, 127, 128, 158–59, 266 decimals, 291–92, 298–300 distributive property, over addition, 135 fractions, 270 integers, 360–61 rational numbers, 390 real numbers, 402 whole numbers, 127, 129, 176 distributive property, over subtraction, 128 duplication algorithm, 186 estimation for fractions, 275 of exponents, 142–43 facts, 128–29 base five, 194 fingers, using, 186 fraction of a fraction, 267 fractions, 266–70, 275 German lowstress algorithm, 188 halving and doubling algorithm, 169, 185
identity property, 126, 269–70, 360–61, 389 integers, 357–62, 366–67 inverse property, 269–70, 274, 389 lattice method algorithm, 177, 194 less than (See under less than) mental math, 158–59, 275 mixed numbers, 267–68, 275 multidigit by multidigit numbers, 176–77 by a negative, less than properties, 366–67, 393, 409–11 negative integers, 357–62 nonstandard algorithms, 185, 186, 188 obtaining ranges for estimation, 160 order of operations, 144–45, 163–64 powers of 10, 159 by powers of 10, 159, 291–92, 301 probability property, 536–38 products (See products) properties (See under properties: multiplication) rational numbers, 388–90 real numbers, 402 rectangular array, 124–25, 267 repeated addition approach, 123–24 with calculator, 135 clock arithmetic, 924–25 with fractions, 266 negative integers, 357 properties, 123–24 Russian peasant algorithm, 185 singledigit by multipledigit numbers, 129 by special factors, 159 standard algorithms, 176–77, 194 thinking strategies, 128–29 tree diagram approach, 125 whole numbers, 123–29 algorithms, 169, 176–77, 185, 186, 188, 194 as product of primes with their respective exponents, 219–20 set models, 123–24 zero, property of, 128 zero divisor property, 362 “multiplication makes bigger” misconception, 297 multiplicative compensation, 159 multiplicative numeration system, 72 multiplicative numeration systems, 63–66
multiplicative properties. See under properties musical scale, 416 mutually exclusive events, 524–25, 536
N naming numbers, 74–75 Napier, John, 155, 294 Napier’s “decimal point” notation, 294 NASA, 895 National Council of Teachers of Mathematics (NCTM), 3 necessary, 915 Needle Problem, Buffon’s, 567 negation of a statement, 912–13 negative integers, 343–45, 352 addition, 346, 348 additive inverse vs., 352 division, 362–63 exponents, 363–65 multiplication, 357–62 number lines, 344, 345, 346, 358 opposite vs., 352 negative numbers, 341, 343 rational, 385 real, 400–401 See also negative integers Newton, Sir Isaac, 16, 665 n factorial, 550–51 n gons. See polygons Noether, Emmy, 100, 575 nonadjacent angles, 617, 621 nonadjacent vertices, 605 nondecimal numeration systems, 75–78 nonillion, 79 nonintersecting lines, 642 nonnegative (principal) square root, 402 nonrepeating decimals, 400–402 nonstandard algorithms addition, 183, 184 division, 186 multiplication, 185, 186, 188 subtraction, 184–85 nonstandard units of measurement, 667–69, 681, 685 board foot, 729 horsepower, 685 joule, 685 lightyear, 683 parsec, 685 Smoots, 681 nonterminating decimals, 305, 400 nonzero numbers, 382, 924 normal distribution, 497–99 notation absolute value, 354 angles, 616–17 conversions with calculators, 364–65
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decimals, 294, 300 directed line segments, 853 elements in a finite set, 60 English system of measurement, 669–72 equals sign, 742 expanded (See expanded form/notation) expected value, 562–63 exponents, 142–44, 225, 363–65 factorial, 214 fractions, 241 function, 87–88 intersection of sets, 522, 524–25 is congruent to, 742 isometries, 875–76 metric system, 673, 676, 677, 678 negative number, 341, 352 negative signs and subtraction, 351 pi, 402 points and lines, 601, 615 scientific, 165, 301–2, 363–65, 369 set, 241, 405 setbuilder, 45–47 subtraction, 350, 352 sum of a ⫹ b, 111 union of sets, 522, 524–25 See also inside back cover not equally likely events, 520–21, 525, 565, 566 nth percentile, 492 nth root, defined, 403–4 null set, 46 number(s), 59–61 abundant, 228 amicable, 228 base, 71 betrothed, 228 cardinal, 59 compatible, 158, 291, 293 counting (See counting numbers) decimal (See decimals) deficient, 228 diagrams of relationship between, 381, 382, 401 rational numbers, fractions, and integers, 392 real numbers, 400 even, 22, 88 fraction (See fractions) frequency of, 442–44 identification, 59 integer (See integers) natural, 45 negative, 341, 343, 348, 357–62, 363–65, 400–401 numerals vs., 59 odd, 22, 203, 363, 400–402
ordering (See ordering) ordinal, 59 perfect, 228 positive, 343, 346, 357–62, 363, 385, 400–401 powers of, 22–24 prime (See prime numbers) properties as problem solving strategies, 204, 231–32 rational (See rational numbers) real (See real numbers) rectangular, 35, 87 relatively prime, 227 Smith, 231 square, 22 systems (See numeration systems) theory (See number theory) whole (See whole numbers) zero (See zero) number line(s) base five, 192–94 decimals, 288, 289–90 distance on, 616, 686–87 fractions, 247, 248, 255, 266 integers, 344, 358, 365–67 least common multiple, 224 metric converter diagram, 673–74 plotting statistics on, 489–90 rational numbers, 381, 385, 392 real numbers, 401, 403 transitive property of less than, 140–41 whole number, 110–11, 140–41, 248 ordering with, 61 number of a set, 59–60 number sense, developing, 163 number sequence, 20–22. See also sequences number theory composite numbers, 205–7 counting factors, 219–20 Euclidean algorithm, 223, 225 greatest common factor, 220–23, 225–26 prime factorization method, 221–23 set intersection method, 220–21 least common multiple, 223–26 buildup method, 224–26 prime factorization method, 224, 225 set intersection method, 224, 225 prime numbers, 213, 226 infinite number of, 226, 402 tests for divisibility, 207–13 numerals, 59 Braille, 69 fractions as, 240, 241–42, 245
numbers and, 59 word names for, 74–75 numeration systems, 61–66, 381 additive, 62–66, 72 Babylonian, 64–65, 66 base, 71–73 binary, 77, 78 Braille, 69 Chinese, 69 comparison of, 66 converting between bases, 75–78 decimal (See decimals) digit, 71 Egyptian, 62–63, 66 expanded notation (See expanded form/notation) grouping, 62, 75–78 HinduArabic, 52 Ionian, 68 Mayan, 43, 65–66, 70 multiplicative, 63–66, 72 nondecimal, 75–78 pictographic, 62–63 placeholder, 64, 66 positional, 43, 64–66 (See also place value) Roman, 63–64, 66 subtractive, 63–64, 66 tally, 62, 66 with zero, 65–66 See also HinduArabic numeration system numerators, 237, 240 comparing, 247–48 and denominators, 249, 271–73 in rational numbers, 382, 388–89, 391 numerousness, 241
O objects no particular order of (combinations), 552–54 ordered arrangement of (permutation), 549–52 oblique circular cone, 646, 647 oblique cylinder, 645, 647 oblique prism, 643, 647 oblique pyramid, 644, 647 obtuse angle, 617, 620, 621 obtuse triangle, 620, 621, 630–31 octahedron, 644–45 octillion, 79 odd counting numbers, 22 even and, making groups of two, 209 nth Root and, 403–4 number of factors, 363 number of prime factors, 400–402 perfect number conjecture, 203 odds, 563–65
I13
O’Neal, Shaquille, 353 onecolumn frontend estimation, 160 ones digit, 22–24 onestage tree diagrams, 542–43 onetoone correspondence, 46, 60, 852, 880–81, 883 operations in base five, 192–95 binary, 78, 110, 923, 924 compatibility with respect to, 158 order of, 144–45, 158–59, 163–64 real numbers, 402, 403 on sets, 48–52 whole number, four basic, 134 See also addition; division; multiplication; subtraction opinion polls, 475–77 opposite of the opposite for rational numbers, 387 opposites of integers, 344, 352 of rational numbers, 387 or connective, 913–14 ordered arrangement of objects (permutation), 549–52 ordered pairs, 51 fractions as, 247 functions as, 90 graph coordinates, 417–18 relations and, 82–83 ordering counting chant, 60, 129 decimals, 289–90, 291 fractions, 247–49 greater than (See greater than) inequalities, 409–11 integers, 365–67 less than (See less than) onetoone correspondence, 46, 60 PEMDAS, 145 rational numbers, 392–93 real numbers, 403 whole numbers, 60–61, 140–42 See also number lines order of operations, 144–45, 158–59, 163–64 ordinal numbers, 59 Oregon rain data, 501 organizing information, 442–44. See also graphs origin, 417 orthocenter, 779, 784, 837–38 ounce, 671, 672 outcome(s), 516 card decks, 517, 518–19, 555, 565 certain, 519 certain or impossible, 523 coin tossing (See coin tossing probabilities)
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model and abstraction, 589 perimeter, 687–89 rectangles as, 619 rhombuses as, 606 symmetry, 601, 604 tessellations of a plane with, 860–61 tessellations with, 631 translations, distance, and, 876 trapezoids as, 698 parallel planes, 645, 647 parallels of latitude, 821 parentheses on calculators, 163–64 parsec, 685 Parthenon (Athens), 285 partition of a set, 85 partitive division, 129–30 parttopart ratios, 311 parttowhole model, 241–42 parttowhole ratios, 311 Pascal, Blaise, 155, 513 Pascal’s triangle combinations, 553–54 and different downward paths, 22 Fibonacci sequence and, 285 predicting sums of diagonals, 33 probability and, 541–42 patterns counting dots using, 14–15 Eschertype, 861–62 and Fibonacci sequence, 33, 36 integer subtraction, 348 P See also Look for a Pattern P(E), 517–24, 529 strategy; tessellations pairwise mutually exclusive PEMDAS (mnemonic for simpler events, 536 ordering), 145 palindromes, 120, 308 pentagon, 604, 607–8, 629 pan balance, 25–26, 108, 148, 407 pentagonal numbers, 35 parabola, 834 pentagonal prism, 645 parallel lines, 615–16, 618–22 pentagonal pyramid, 644 angles associated with, 618–22 pentominoes, 342, 373, 613 Euclidean construction, 771 percent grade, 818, 821 perpendicular lines and, 587, percentile, 492 605–7 percents, 320–28 slopes of, 814, 816 with calculators, 322, 326–27 in threedimensional space, 642 converting, 320–22 and transversals, 618–19 decimals and, 321–22 parallel line segments, 606, Draw a Diagram strategy, 324 615–16, 642 fractions and, 321, 322–23 directed, 853 percent grade, 818, 821 parallelogram as ratios, 324–25 base, 692 solving problems, 322–23 parallelograms equation approach, area, 692 325–28 attributes of, 606 grid approach, 324 congruent, 604, 744, 791 proportion approach, creating shape of, 589, 590 324–25, 326–27 defined, 744, 791 perfect numbers, 203, 228 Eschertype patterns, 861–62 perfect square, 146 height, 692 perimeter(s), 687–89. See also lateral faces of polyhedra, 643, inside back cover 647 period of a decimal, 303 outcome(s) (continued) dice throwing, 516, 518–19, 520–21, 534, 564 drawing gumballs, 525, 539–40 drawing marbles, 521–24, 532–34, 535–38 equallylikely (See equallylikely events) expected, 560 experiments with two outcomes, 540–41 fundamental counting property, 534 impossible, 519, 521 mutually exclusive, 524–25, 536 odds, 563–65 pairwise mutually exclusive simpler, 536 permutations, 549–53, 555 Pick Six wager, 566 possible, 516 probability of, 517–24, 529 spinning spinners, 516–17, 518–19, 525, 538–39 unequally likely, 520–21, 525, 565, 566 See also events; probability outliers, 453, 489–92 box and whisker plot, 489–92, 503 explaining occurrence of, 489–90 extreme and mild, 503
permutations, 549–53, 555 counting techniques, 549–53, 555 n factorial, 550–51 of r objects chosen from n objects, 551 perpendicular bisector, 768 angles, 769–70 construction of circumscribed circles, 777–79 Euclidean construction, 768 of line segments, 746, 768, 770–71, 778 in reflection transformations, 856, 879 rhombus properties and, 768–71 passim sides of the triangle, 778, 779 through a point not on a line, 770–71 through a point on a line, 770 triangle, 856, 879 perpendicular diagonals, 605, 606 perpendicular lines, 618, 621 parallel lines and, 587, 605–7 slopes of, 814, 816 in threedimensional space, 642 through a point on the line, 770 not on the line, 770–71 perpendicular line segments, 589 Peter, Rozsa, 433 pi area of a circle, 693 Buffon’s Needle Problem, 567 circumference of a circle, 689 mnemonic, 411 notation, 402 perimeter, 689 volume of a sphere, 723–24 Pick Six wager, 566 Pick’s theorem, 700 pictogram numeration systems, 62–63 pictographs, 449–51, 454, 471–72 pictorial embellishments of graphs, 450–51 deceptive, 473–75 picturing information, 449–54. See also graphs pie charts, 448–49, 451, 454, 472–73 pint, 671 pizza order probability, 534 placeholder, 64, 65 place value, 72, 73 addition algorithm, 172–73 decimals and, 289–90, 291 expanded notation, 72, 76, 172–73 HinduArabic system, 72, 73 multidigit numbers, 115 multiplication algorithm, 177 nondecimal systems, 75–78
patterns, 73 positional numeration systems, 43, 64–66 subtraction algorithm, 174–75 plane(s), 615–16, 640–42 coordinate distance in, 809–12 graphing integers on, 815 parallel, 645, 647 points in, 615–16 circle, 608 points on, 807 similar shapes via similitudes, 883 slope (See slope) of symmetry, 652 tessellations and, 630–31, 632, 860–61 threedimensional, 640–47 Platonic solids, 644 point(s) arrays of, 818, 820 collinear, 615, 863, 879, 881, 893, 894 corresponding, 645, 647 distance equidistant from endpoints, 770, 778 equidistant from perpendicular bisector, 778 equidistant from sides of a triangle, 779 from point P to point Q, PQ, 766, 770–71, 777–79 distance from point P to point Q, PQ, 686–87, 809–10, 876–77 endpoints of a line, 601, 608, 616 equidistant from endpoints, 616 image of, 852, 861–62, 875 image of P, 852, 875 midpoint formula, 811–12 midpoint of a line, 616 noncollinear, 839 perpendicular line through, 770–71 in a plane, 608, 615–16 on a plane, 807 properties of lines and, 616 pointslope equation of a line, 825–26 polar coordinates, 832 Polk, James K., 513 Pólya, George, 1, 145 Pólya’s fourstep problemsolving process, 1, 4–6, 30 utilizing with strategies, 7–13, 21–28 polygon(s), 419 angles of, 607–8 arbitrary, 630
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concave and convex, 607 congruent, 880–81, 883–84 equilateral, 589, 628–30, 632–33 equilateral triangles, 780, 782 Fermat prime, 782 as fundamental units of area, 670 onetoone correspondence, 880–81, 883 perimeter of, 687–89 regular, 607–8, 628–30, 632–33, 780, 782–83 angle measures, 628–30 as equiangular and equilateral, 628 Euclidean constructions, 780, 782 Gauss’s theorem for constructible ngons, 782–83 tessellations with, 632–33 similar, 883 size transformations, 863 tessellations, 632–33 polygonal regions, 630, 643–45. See also polyhedra polyhedra, 642–45 cubes, 640–42 dodecahedron, 644–45 Euler’s formula, 644, 733 hexahedron, 644 icosahedron, 644–45 octahedron, 644–45 prism, 643, 645, 650 pyramid, 643–44, 645, 654 regular, 644 rhombicuboctahedron, 645 semiregular, 645 tetrahedron, 644–45 threedimensional aspect, 642–45 vertices, 643–44 Pons Asinorum Theorem, 739 pool shot paths, 893, 894 population, in statistics, 475–77 population predictions, 428 portability of measurement systems, 672, 679 positional numeration systems, 43, 64–66. See also place value position function, 420 positive numbers integers, 343, 346, 357–62, 363 rational, 385 real, 400–401 possible events, 516. See also outcomes pound (weight), 672 power of. See exponents powers of three, 22, 88 powers of 10 decimals, 289, 291–92, 301 dividing by, 291–92, 301
Egyptian numeration system, 62–63 HinduArabic numeration system, 71–73 multiplying by, 159, 291–92, 301 Precious Mirror of the Four Elements (Chu), 341 prime factorization of composite number, 206, 214 counting factors and, 219–20 divisibility tests for, 212–13 fundamental theorem of arithmetic and, 206 greatest common factor (GCF), 221–23 least common multiple (LCM), 224, 225 prime factors counting factors vs., 219 fractions, 245 odd number of, 399–400 rational numbers, 383 test for, 213 prime meridian, 821 prime numbers, 205–13 2, nature of, 399–400 composites and, 205–7 factorization (See prime factorization) factor test, 213 Fermat prime, 782 finding, 212–13 infinite number of, 226 largest known, 227 Mersenne number, 213, 217 relatively, 227 Sieve of Eratosthenes, 206, 234 square root, 213 with their respective exponents, 219–20 triples, 217, 413, 705 twins, 215 primitive Pythagorean triple, 413 principal square root, 402 prisms, 643, 645, 650, 718 Cavalieri’s principle, 724 height, 718, 720 oblique, 643 right, 708, 718, 724 right rectangular volume, 718 right regular, 709 right square, 643 right triangular, 720 surface area, 708 volume, 718 probability, 515–25, 532–42, 549–55, 560–67 additive probability, 536–38 additive property, 536–38 coin tossing (See coin tossing probabilities) complex experiments, 532–42 conditional probability, 565–67
counting techniques, 549–55 dice throwing, 516, 518–19, 520–21, 534, 564 drawing with/without replacement, 534–35, 536–39 events (See events) expected value, 562–63 experimental, 519–20 simulation, 560–62 experiments with two outcomes, 540–41 fundamental counting property, 534, 536, 540, 542, 549–52 intersection of sets, 522 multiplicative probability, 536–38 multiplicative property, 536–38 odds, 563–65 outcomes (See outcomes) Pascal’s triangle, 541–42, 553–54 properties additive probability, 536–38 fundamental counting property, 534, 536, 540, 542, 549–52 multiplicative probability, 536–38 as ratios, 525 relative frequency, 496, 516, 517, 566 simple experiments, 515–25 computing probabilities in, 517–24 simulation, 560–62 theoretical, 519–20, 560, 574 tree diagrams and, 532–42 unequally likely outcomes, 520–21, 525, 565, 566 Venn diagrams and, 566 probability of an event, P(E), 517–24, 529 problem of the points, 513 problems, exercises vs., 4 problem solvers, suggestions from, 6, 30–31 problem solving algebraic, 13–16 geometric with algebra, 807 with coordinates, 834–39 with transformations, 893–95 with triangle congruence and similarity, 790–95 importance of, 3 Pólya (See Pólya’s fourstep process) strategies (See problemsolving strategies) suggestions for, 30–31 See also algorithms; properties; solving equations
I15
problemsolving process, Pólya’s, 1, 4, 6, 30 utilizing with strategies, 7–13, 21–28 problemsolving strategies, 4 combining, 1–2, 26–30 Cover Up, 16 Do a Simulation, 514, 574 Draw a Diagram (See Draw a Diagram strategy) Draw a Picture (See Draw a Picture strategy) Guess and Test, 7–9 Identify Subgoals, 740, 799 Look for a Formula, 440, 506–7 Look for a Pattern (See Look for a Pattern strategy) Solve an Equation, 380, 432, 838, 839 Solve an Equivalent Problem, 238, 281 Solve a Simpler Problem, 6 Use a Model, 582, 657, 695, 707 Use a Variable, 11–13, 350, 695 Use a Variable with algebra, 13–16, 27–28 Use Cases, 342, 373 Use Coordinates, 808, 845 Use Dimensional Analysis, 666, 733 Use Direct Reasoning, 108, 148 Use Indirect Reasoning, 156, 198, 226, 399–400 Use Properties of Numbers, 204, 231–32 Use Symmetry, 850, 901–2 Work Backward, 286, 335 See also Look for a Pattern strategy products, 124 counting factors and, 219–20 crossmultiplication of fractions, 244–45, 247–48, 249 from decimal to fraction conversions, 292 exponents of the, 142–43 and “multiplication makes bigger” misconception, 297 order of operations, 144–45, 163–64 in probability, 536–38 See also multiplication proper factors, 203, 228 proper subsets, 47, 60–61 properties AA similarity, 754–55, 783 additive cancellation, 123, 347, 387 additive probability, 536–38 aesthetics property of golden ratio, 285 angle sum in a triangle, 620, 629–30
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properties (continued) ASA congruence, 744, 746, 791, 794 associative (See associative property) closure (See closure property) commutative (See commutative property) compass, 766, 767 crossmultiplication, of ratios, 312–13 directed angles, 854 distance on a line, 687 distributive (See distributive property) of exponents, 142–44, 405 of fractions additive, 244, 257, 258, 260 density, 249 multiplicative, 266, 269–70, 274, 275 fundamental counting property, 534, 536, 540, 542, 549–52 geometric shapes angles, 617, 619–20 convex attribute, 607 diagonals of a kite, 605 isosceles triangles and trapezoids, 601–2 parallelograms, 604 points and lines, 616 greater than, 402–3, 410 identity (See identity property) of integers, 347, 360–61 inverse additive, 347, 350, 352, 384–87 multiplicative, 389 of isometries, 878 less than, 366–67, 393, 402–3, 409–11 for mental math, 158 midquad, 794–95 multiplication property of zero, 128 multiplicative cancellation, 137, 362 multiplicative probability, 536–38 multiplicative rectangular array, 124–25 of points and lines, 616 of probability, 524, 534, 536–38, 540, 542, 549–52 as problem solving strategy, 204, 231–32 of rational exponents, 405 of rational numbers, 384–87, 389, 393 of real numbers, 402, 403 reflexive, 83–84 SAS congruence for quadrilaterals, 790–92, 793
for triangles, 743–44, 746, 778 SAS similarity, 754, 793–94, 881 similarity of triangles, 752–56 of similitudes, 882 of size transformations, 882 SSS congruence angles, 767 converse of Pythagorean theorem, 792–93 kites, 746 parallelograms, 791 triangles, 745–46 SSS similarity, 754, 793–94, 881 straightedge, 766–67 symmetric, 84 transitive (See transitive property) trapezoids (See trapezoids) triangle similarity of, 752–56 SSS congruence, 745–46 of triangles, 600–601, 619–20, 834–39 whole number (See under whole numbers) of zero, 112–13, 128, 132, 362 zero divisor of integers, 362 proportion, 312–15 with calculators, 326–27 percent problemsolving, 324–25 scaling up/scaling down, 314 See also ratios protractor, 617–18 public opinion polls, 475–77 Purser, Michael, 368 pyramids, 654, 709–10, 721–22 apex, 643–44, 654 Cavalieri’s principle, 726 slant height, 710 square, 721–22 surface area, 710 volume, 721–22 Pythagoras, 315, 379 Pythagorean theorem, 379, 693–95 application to nontriangular shapes, 757 area of the square interpretation, 757, 797 Bhaskara’s proof, 699 converse of, 792–93 Euclid’s proof, 739, 784, 797, 900–901 Garfield’s proof, 705 and irrational numbers, 379 length of PQ, 809 length of the hypotenuse, 401 origin of, 795 proof, 586–87, 694 transformational proof, 897 Pythagorean triples, 398, 413, 705
Q quadrants, 418 quadratic functions, 420 quadrilateral(s), 588 arbitrary, 631 attributes of, 606 congruence, 749, 790, 793 creating shape of, 589, 590 diagonals of, 894 halfturn rotations, 894 midquad, 794–95 parallelograms (See parallelograms) perimeter, 687–89 rectangles (See rectangles) rhombus (See rhombus) SASAS congruence property, 749 SAS congruence property, 790–92, 793 size transformation of, 863 square (See square) tessellations from, 631 translation and, 853, 895 quadrillion, 74, 79 quality control, 441 quart, 671 quartile, upper/lower, 489–92 quotients (missing factors) calculators and, 163–65 decimals, long division, and, 301, 303 from decimal to fraction conversions, 292 defined, 132 denominators and numerators approach to division, 271–73 division and, 182–83 and “division makes smaller” misconception, 297 long division and, 178–80, 182, 195 ratio and, 310 simplified form, 391
R r objects chosen from n objects, 551, 553 radical, 404 radicand, 404 radius arc of radius, 767, 769, 780 of a circle, 608 of a sphere, 646 Ramanujan, Srinivasa, 18, 231, 232 random digits, 561, 568 Random Guess and Test strategy, 7, 8 randomnumber table, 561, 568 range estimation, 160, 261, 275, 292 functions, 88–90, 91 interquartile, 489–92
rates, ratio, 310 ratio(s), 310–11 avoirdupois measures of weight, 672 crossmultiplication property, 312–13 dimensional analysis, 680 of distances, 881–84 English system of measurement, 669–70 equality of, 311 extremes and means, 311 golden ratio, 285 liquid measures of capacity, 671 medians of a triangle, 837 metric convertibility of, 673, 674, 676 odds and, 563–65 parttopart comparison, 311 parttowhole comparison, 311 percents as, 324–25 probabilities as, 525 rates, 310 scaling up/down, 314 wholetopart comparison, 311 See also proportions rational exponents, 403–5 properties, 405 rational number(s), 382–93 addition, 384–87 additive properties, 384, 386–87 identity, 383, 387–88 inverse, 384–87 comparing two, 393 crossmultiplication of inequality, 393 division, 390–92 invert and multiply, 390, 409 equality of, 383, 387–88, 391 as exponents, 403–5 fractions as, 382 infinite number of fractions, 383, 385 integers (nonzero) as, 382 multiplication, 388–90 multiplicative properties, 389–90 nonzero integers as, 382 number lines, 381, 385, 392 products with fraction calculator, 389 set models, 382 set of, 382 simplifying, 383, 388, 389, 390, 391 subtraction, 387–88 sums and differences with fraction calculator, 388 ratio of distances, 881–84 ray, 616, 855 reading numbers, 74–75
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real number(s), 399–405 additive properties, 402 diagram of decimals and, 400 exponents, 405 geometric representation, 379, 401 multiplicative properties, 402 n th root, 403–4 number line, 401, 403 number systems diagram, 401 ordering properties, 403 roots of, with calculators, 404 set of, 400–402 reasoning direct, 108, 148 problem solving with, 108, 148 indirect, 156, 198 problem solving with, 198, 226, 399–400 inductive, 22 reciprocals, 269, 382, 389, 390–91 fractions, 269 rational numbers, 389 recognition—level 0, 583–84 rectangle area, 689–91 attributes of, 606 creating shape of, 589, 590 diagonals of a, 743, 744, 790–91 golden, 285, 786 horizontal vs. vertical, 586 model and abstraction, 589 as parallelograms, 619 perimeter, 687–89 squared, 355 symmetry, 601, 604 rectangular array description of divides, 208, 210 fraction multiplication, 267 multiplication, 124–25 rectangular numbers, 35, 87, 88 rectangular regions, 641 “reducing” fractions, 244 Rees, Mina, 507 reflection, 856–60 distance and, 877 image, 852, 859 notation, 875–76 symmetry, 600, 602, 604, 606, 652, 852, 860–61 See also glide reflection reflex angles, 617, 621 reflexive property, 83–84 region model for fraction addition, 255 regression lines, 453 regrouping in addition, 172, 183 in subtraction, 174, 184 regular inscribed hexagon, 781 regular pentagon, 604, 607–8 regular polyhedra, 644
Reid, Constance Bowman, 232 relations, 82–91 arrow diagrams, 83 defined, 72–73 equivalence, 85 functions, 85–91 number systems diagrams, 381, 382, 392, 401 partition, 85 reflexive, 83–84 symmetric, 84 transitive, 84–85 relationships – level 2, 585 relative amount, 240–42, 244, 311 relative complement of sets, 50–51 relative frequency, 496, 516, 517, 566. See also probability relatively prime numbers, 227 relative vs. absolute in circle graphs, 472 remainders, 133, 164–65, 178–80, 182–83 repeated addition approach to multiplication, 123–24 with calculator, 135 clock arithmetic, 924 with fractions, 266 negative integers, 357 properties, 123–24 repeated subtraction, division by, 133–34 repeating decimals, 300–304 calculator, 302–3 fraction representation, 304–5 long division algorithm, 302–3 nonterminating, 304–5 real numbers, 399 repetend, 303, 304–5, 400 repetend, 303, 304–5, 400 replacement, probability and, 534–35, 536–39 Reuleaux triangle, 900–901 reverse Polish notation, 163 Rhind Papyrus, 237 rhombicuboctahedron, 645 rhombus, 768, 769 attributes of, 606 children’s recognition of, 585 coordinate geometry, 834 creating shape of, 589, 590 diagonals of a, 768–69, 770, 791–92 model and abstraction, 589 as parallelograms, 606, 744, 791 perimeter, 687–89 perpendicular bisector construction and, 768–71 passim sides of a, 768 slopes for analyzing diagonals, 816 symmetry, 601, 604 Rick, Killie, 394
Riese, Adam, 294 right angles, 589, 617 right circular cone, 646, 711, 716, 724 right circular cylinder, 645–46, 709, 724 right distributivity of division over addition, 158 right prism, 708, 718, 724 right rectangular prism volume, 718 right regular prism, 709 right regular pyramid, 710, 711, 724 right square pyramid, 710 right triangle, 401, 589, 620, 621, 693–95. See also Pythagorean theorem right triangular prism, 720 rigid motion. See isometries rise over the run, 812 road signs, 609 Robinson, Julia Bowman, 149, 232 rod, 669 rolling dice probabilities, 516, 518–19, 520–21, 534, 564 Roman numeration system, 63–64, 66 rotation, 853–55, 857 axis of rotational symmetry, 652 distance and, 876 Eschertype drawings, 862 halfturn, 893–94 image, 893 notation, 875–76 symmetry, 602–4, 606, 608, 860–61 of triangles and quadrilaterals, 631 rotation symmetry, 860–61 rounding, 161–63 to compatible numbers, 162 computational estimation, 292–93, 323 decimals, 292–93 division, 182 down, 161–62, 182 a five up, 161–62 to nearest even, 162 percents, 327 truncate, 161–62 up, 161–62, 182 whole numbers, 161–63 RSA algorithm, 368 Rudin, Mary Ellen, 658 Russian peasant algorithm, 185
I17
Cartesian product of sets, 526–27 conditional probability, 565–67 counting techniques in place of, 549–55 equallylikely outcomes, 524–25 large, dealing with, 524, 549, 551 not equallylikely probability, 525 properties of probability, 524 See also tree diagrams SAS. See sideangleside SAT scores, 446 scaffold method, long division, 180, 195 scale factor, 863, 883 scalene triangle, 589, 630 scaling and axis manipulation, 465–67 scaling up/scaling down, 314 scatterplot, 452–54 School Mathematics Study Group, 902 scientific notation calculators and, 165, 301–2, 364–65, 369 characteristic, 301 mantissa, 301 negative integer exponents, 363–65 standard notation vs., 364 scratch addition, 184 segment construction, 588, 783. See also line segments selfsimilarity, 425 fractals and, 755–56 Koch curve, 756 semiregular polyhedra, 645 semiregular tessellations, 633, 635, 638 sequences, 22 arithmetic, 86–87 common difference, 86–87 common ratio of, 86–87 Fibonacci (See Fibonacci sequence) geometric, 86–87 identifying (See Look for a Pattern strategy) initial term, 86 number, 20–22 set(s), 45–52 as basis for whole numbers, 45–52 binary operations using, 110 Cartesian product, 51–52, 125 S Cartesian product of, 526–27 sample, defined, 475 combinations and, 552–54 samples and bias, 475–77 complement of, 50 sample space, 516–21 continuous set of numbers, 445 additive property of probability, counting numbers, 45 536 DeMorgan’s laws, 54, 56, 57
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set(s) (continued) difference, 110 difference of, 50–51 disjoint, 48, 109–10, 345 elements/members of, 45–46 empty, 46, 112 equal, 46 equivalent, 46 finite, 47–48, 59, 60, 89, 110 fractions concept, 246 ordering, 247–49 functions (See functions) infinite, 47–48 inherent rules regarding, 46 integers, 343–45 intersection method, 220–21, 224 intersection of, 49, 110 probability and, 522, 524–25 matching, 46 null, 46 number of a, 59–60 operations on, 48–52 ordered pair, 51, 82–83 partition of a, 85 probability and, 521–25 proper subset, 47, 60–61 rational numbers, 382 real numbers, 400–402 relations (See relations) relative complement, 50–51 sample space, 516–21, 549–51 setbuilder notation, 45–47 solution, 405 subset of, 46 theory, 79–80 union of, 48–49, 110, 522, 524–25 universal, 47 Venn diagrams of, 47–52 setbuilder notation, 45–47 set intersection method, 220–21, 224 greatest common factor, 220–21 least common multiple, 224, 225 set models integers, 344, 345 rational numbers, 382 takeaway, 115–16 whole numbers addition, 109–10 division, 130 multiplication, 123–24 subtraction, 115–16 set notation, 241, 405 sextillion, 79 shape identification, 585–87. See also geometric shape concepts Shells and Starfish (Escher), 849
ShiingShen Chern, 845 ShiKu Chu, 341 sideangleside (SAS) congruence property, 743–44 for kites, 746 for quadrilaterals, 790–92, 793 for rotation transformation, 876–77 for triangles, 743–44, 746, 778 similarity property, 753, 754, 793–94, 881 sidesideside (SSS) congruence property, 745–46, 767, 791, 792–93 similarity property, 754, 793–94, 881 sides of a parallelogram, 790, 792 sides of a rhombus, 768 sides of the triangle, 588 congruent, 745 copying an angle, 767 included, 744 midpoint of, 793–95 perpendicular bisector, 778, 779 Sierpinski triangle (or gasket), 37 Sieve of Eratosthenes, 206, 234 similarity, 881–84 AA similarity property, 754–55, 783 fractals and selfsimilarity, 755–56 geometric problemsolving with, 793–95 SAS similarity property, 754, 793–94, 881 selfsimilarity, 425, 755–56 of shapes, 883 SSS similarity property, 754, 793–94, 881 of triangles, 752–56 and similitudes, 882, 883 similar shapes, 883 similitudes, 862–64, 882 properties, 882 similar shapes in the plane, 883 size, 862–63 size transformations, 862–64 center, 863 scale factor, 863, 883 simple closed curves, 607, 645–46, 657 simple experiments, 515–25 computing probabilities in, 517–24 simplified method for integer subtraction, 350 simplifying fractions, 242, 244, 245, 253, 256, 257 rational numbers, 383, 388, 389, 390, 391 simulation, 514, 560–62, 574
simultaneous equations, 827 solutions, 828 elimination method, 831 graphical method, 830–31 substitution method, 831 simultaneous equations, solutions of, 826–29 size transformations, 862–64, 881–83 scale factor, 863, 883 skew lines, 642 slant height, 710 slide, 852–53, 859–60 slope, 812–16 collinearity, 813–14, 836 in coordinate plane, 812–16 lines, 826, 837–38 of lines, 812–13 parallel lines, 814, 816 percent grade, 818, 821 perpendicular lines, 814, 816 pointslope equation of a line, 825–26 Smith, Steven B., 166 Smith number, 231 Smoot, Oliver, 681 solution of an equation, 15 solution set, 405 solutions of simultaneous equations, 828 Solve an Equation strategy, 380, 432, 838, 839 Solve an Equivalent Problem strategy, 238, 281 Solve a Simpler Problem strategy, 6, 25–30, 38 solving equations, 15 algebraically, 13–16, 405–6, 408 Balance Beam Algebra, 405–6, 408 balancing method, 405–6, 408 Cover Up method, 16 inequalities, 409–11 percents, 325–28 simultaneous, 826–29 solutions, 831 strategies (See problemsolving strategies) transposing, 409 Work Backward method, 16 See also problem solving; problemsolving strategies special factors, 159 sphere(s), 646, 648, 723–24 Archimedean method for deriving volume, 665 Cavalieri’s principle, 725 diameter, 646 great circle of the, 651 surface area, 712 volume, 723–24 spinning spinners probabilities, 516–17, 518–19, 525, 538–39
spreadsheet activities Base Converter, 79, 81 Consecutive Integer Sum, 17 Function Machines and Tables, 97 spreadsheet activities on the Web site Consecutive Integer Sum, 17 Standard Deviation, 502 square (geometric) 8by8, 118 acre, 670–71 additive magic, 19, 217, 295, 296 area English system, 689–91 metric system, 675–76 area of a, 757, 797 attributes of, 606 as base of pyramids, 654 centimeter, 675–76 creating shape of, 589, 590 decimeter, 675–76 dekameter, 675–76 diagonal of a, 795 as equilateral and equiangular, 590 foot, 670–71 hectometer, 675–76 inch, 670–71 kilometer, 675–76 meter, 675–76 mile, 670–71 millimeter, 675–76 model and abstraction, 589 oblique prism, 643 perimeter, 687–89 right square prism, 643 squared square, 357 symmetry, 601, 604 tessellations of, 632 units, 689 yard, 670–71 square (of a number) of 2, 399–400 counting numbers, 22 perfect, 146 squared rectangle, 355 squared square, 357 square lattice, 588, 615, 695, 700 square pyramid, 721–22 square root with calculators, 213, 402 defined, 402 divide and average method, 412 irrational, 412 primes and, 213 principal, 402 squeeze method for decimals, 412 SSS. See sidesideside standard algorithms addition, 172–73, 192 division, 182–83 long division, 195, 302–3
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multiplication, 176–77, 194 subtraction, 174–75, 192–93 standard deviation, 493–96, 498–500 bellshaped curves, 497–99 dispersion, 493–96 distribution, 498–500 mean, 493–96, 498–500 unbiased, 506 zscores, 495–96, 498–500 standard meter prototypes, 673 standard units of measurement. See English system of measurement statements, logical, 912–13 statistics, 439, 441–54, 464–77, 484–500 analyzing data (See analyzing data) central tendency, 484–88 data (See data) distribution, 489–500, 496–500 graphs (See graphs) interquartile range, 489–92 lower quartile, 489–92 mean (See mean) median, 485–86, 488, 489–92, 497–99 misleading (See misleading graphs and statistics) mode, 485, 486, 488, 496–99 normal distribution, 497–99 plotting on a number line, 489–90 samples and bias, 475–77 standard deviation (See standard deviation) trends, 439, 447–49, 454 upper quartile, 489–92 visual comparison (See graphs) zscore, 495–96 stem and leaf plot, 442–43, 490–91 backtoback, 443 outliers, 503 step functions, graphs of, 423–24 Stevin, Simon, 294 Stiller, Lewis, 590 stock market, 425 straight angles, 617, 620, 621, 631 straightedge constructions, 772, 780, 781, 782, 865 straightedge properties, 766–67 strategy, defined, 4. See also problemsolving strategies subset combinations, 552–54 events, 516 of populations, 475 proper, 47, 60–61 subset of a set, 46 substitution method of solving simultaneous equations, 831 subtractfromthebase algorithm, 175–76, 193
subtraction, 115–18, 259–61, 348–52 by adding the complement, 185 adding the opposite approach, 349–50, 352, 387 algorithms, 174–75, 184–85, 192–93 base five algorithm, 192–93 cashier’s algorithm, 184–85 clock arithmetic, 923–24 comparison approach, 118 compensation, 159, 261 decimal algorithm, 298 difference (See difference) eliminating extra information, 5 equaladditions method, 159, 185, 261 estimation for fractions, 260–61 fractions, 259–61 integers, 348–52 lefttoright method, 159 mental computation, 158–59 missingaddend approach, 192–93 fractions, 259–60 integers, 350 whole numbers, 116–17, 128 nonstandard algorithms, 184–85 notation, 351, 352 order of operations, 144–45, 163–64 pattern, 348 rational numbers, 387–88 real numbers, 402–3 regrouping, 174, 184 simplified method, 350 standard algorithm, 174–75 takeaway approach clock arithmetic, 923–24 fractions, 259–60 integers, 349, 350, 352, 358–59 whole numbers, 115–16 unlike denominators, 260, 388 whole number set models, 115–16 subtractive principle of Roman numeration system, 63–64, 66 subtrahend, 116 successive differences, 27–30 sufficient, 915 sum consecutive whole numbers, 11–12 first n counting numbers, 12–13, 20–21 infinite geometric series vs., 265 order of operations, 144–45, 163–64 of a plus b, 110 of probabilities, 563
summands, 110 supplementary angles, 618, 621 surface area, 707–12 cones, 711–12 cylinders, 709 formulas (See inside back cover) prisms, 708 pyramids, 710 spheres, 712 survey of college freshmen Draw a Diagram strategy, 44, 99 symbols, list of. See inside back cover symmetrical distribution, 496–97, 498–500 symmetry, 600–604, 860–61 analyzing shapes, 600–604 of circles, 608 in Escher’s art, 849, 861–62 glide reflection, 861 line of, 600 Mira®, 602, 608 plane of, 652 property, 84 reflection, 600, 602, 604, 606, 652, 852, 860, 861, 862 rotation, 602–4, 606, 608, 652, 860, 862 translation, 860–61, 862 triangle congruence and, 588, 601–2 Use Symmetry strategy, 850, 901–2 symmetry transformations, 860–61 Systematic Guess and Test strategy, 7, 8 system of equations. See simultaneous equations system of units, ideal features of, 673
T table(s) addition, 113–15 for base five operations, 192–93, 194 bias survey sources, 477 box and whiskers data, 492 box volumes, 423 decimals and fractions, 291 earthquakes of the 1960s, 452 educational level vs. income, 453 English System units, 669, 670, 671, 672 exponential functions, 420 exponential growth, 87 functions as, 89 function values of t, 420 geometric shape definitions, 621, 647 descriptions, 589
I19
images of points under transformations, 875 linear functions, 419 measure of a vertex angle in regular ngon, 629, 632 metric prefixes, 674 metric system units, 673, 676, 677, 678 number system summary, 66 operations on sets, 52 outcome frequency, 520, 521, 524 percents and fractions, 322 Platonic solids, 644 quadrilateral attributes, 606 randomnumber, 561, 568 sequences, arithmetic and geometric, 86 stem and leaf data, 442–43, 490–91 summary of grafts and charts, 454 surface area, 724 truth table, 913 unit cubes, 87 zscores, 495, 500 tablespoon, 671 takeaway approach, 115–16 clock arithmetic, 923–24 fractions, 259–60 integers, 349, 350, 352, 358–59 set models, 115–16 whole number subtraction, 115–16 tally numeration system, 62, 66 tangent line to a circle, 777 tangram puzzles, 586–87 Taylor, Richard, 203 Tchebyshev, 217 teachers and teaching children’s “reversals stage ,” 67 of geometry, 845 Killie’s [teacher’s] Way, 394 “Ten Commandments for Teachers” (Pólya), 1 thinking aloud, 6 teaspoon, 671 temperature English system, 672 metric system, 678–79 “Ten Commandments for Teachers” (Pólya), 1 term, 86 terminal side of a directed angle, 855 terminating decimals, 289, 305, 399, 400 terms of a sequence, 22 tessellations, 630–33 of arbitrary quadrilaterals, 631 of arbitrary triangles, 630 dual of, 635 Eschertype patterns, 849, 861–62
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tessellations (continued) plane and, 630–31, 632 plane with parallelograms, 860–61 plane with squares, 670 with regular polygons, 632–33 rotation transformation of a triangle, 862 semiregular, 633, 635, 638 tests for divisibility, 207–13 by 2, 207–8, 210 by 3, 210–11 by 4, 210 by 5, 207–8, 210 by 6, 211–12 by 8, 210 by 9, 210–11 by 10, 207–8, 210 by 11, 211 prime factorization, 212–13 tetrahedron, 644–45 tetromino shapes, 10–11, 611 Theon of Alexandria, 800 theorem, defined, 142 theoretical probability, 519–20, 574 experimental probability vs., 560 thermal imaging, 895 thinking strategies addition facts, 113–15 facts for base five, 192–93, 194 long division, 180, 182 multiplication facts, 128–29 See also mental math thousand, 74 threedimensional effects, 470–72 threedimensional shapes, 640–47 curved, 645–47 dihedral angles, 641–44 planes, 640–47 polyhedra, 642–45 skew lines, 642 surface area, 707–12 volume, 671, 676–77, 717–26 threedimensional space Cartesian coordinates in, 819 Todd, Olga Taussky, 575 ton, 672 transformation geometry, 851–64, 875–84, 893–95 applied problems, 894–95 clockwise/counterclockwise orientation, 854–55, 860, 877 congruence isometries, 875–81 polygons, 881, 883–84 shapes, 880–81 Eschertype patterns, 849, 861–62 functions as, 91 geometric problemsolving using, 893–95 glide axis, 859, 879–80 halfturn, 893–94
image of P, 852 isometries (See isometries) midsegment proof, 910 notation, 875–76 similarity, 881–84 similitudes, 862–64, 881–84, 882 properties, 882 similar shapes in the plane, 883 size, 862–63 size transformations, 862–64, 863, 883 slides, 852–53, 859–60 symmetry, 860–61 in Escher’s art, 849, 861–62 isometries and, 852, 860, 861, 862 Use Symmetry strategy, 850, 901–2 transformations, 852–64, 875–84, 893–95 isometries, 852–60 congruence and, 875–81 making Eschertype patterns, 861–62 notation, 875–76 similitudes, 862–64 similarity and, 881–84 size transformations, 881–83 solving problems with, 893–95 symmetry, 860–61 transitive property less than inequalities, 409–11 integers, 366 rational numbers, 393 real numbers, 403 whole numbers, 140–41 relations, 84–85 translation, 852–53, 857 distance and, 853, 876, 895 notation, 875–76 quadrilaterals, 853, 895 symmetry, 860–61, 862 transposing, 409 transversals, 618–19, 627 trapezoid area, 692 base angles of, 614 creating shape of, 589, 590 height, 692 isosceles, 589, 601–2 midsegment, 704 model and abstraction, 589 parallelograms as, 698 symmetry, lack of, 601, 604 tessellations with, 631 tree diagrams, 532–42 as approach to multiplication, 125 counting techniques, 532–34 expected value, 562–63
one and twostage, 542–43 probability, 532–42 trends, statistics and, 439, 447–49, 454 triangle(s), 588, 752–56, 778–82 AA similarity property, 754–55 acute, 620, 621, 630 altitude, 773, 779, 837–38, 839 angle sum property, 620 angle sums in ngons and, 629–30 arbitrary, 630 area, 691 base, 691 centroid, 779, 787 centroid of a, 835–37 cevian line segments, 335 circumcenter, 777 circumscribed circle and, 777–79, 838–39 compass properties, 766, 767 complementary, 620 congruence (See triangle congruence) coordinate system, 834–39 creating shape of, 589, 590 equilateral, 588, 589, 601, 602, 632, 780, 782 5con pairs, 746 fractals and selfsimilarity, 755–56 geometric problem solving, 793–95, 834–39, 893–95 harmonic, 265 height, 691, 692 Hero’s formula, 697 hypotenuse, 401, 693–95 incenter, 777–78, 779 inequality theorem, 695 inscribed circles, 777–78, 779 isosceles, 589, 644, 739 acute, 630–31 reflection symmetry, 601–2, 604 legs, 693–95 Marathe’s, 123 median, 773, 779 medians of a collinear, 836 congruent, 835 ratio of, 837 multidigit midsegment, 794 obtuse, 620, 621, 630–31 orthocenter, 779, 784, 837–38 Pascal’s, 285, 541–42, 553–54 perpendicular bisector, 856, 879 perpendicular bisector of sides, 778, 779 properties of, 600–601, 619–20, 834–39 Reuleaux, 900–901 right, 401, 589, 620, 621, 693–95
right triangular prism, 643, 720 rotation transformation, 862 scalene, 589, 630 sides (See sides of the triangle) Sierpinski, 37 similarity, 752–56, 793–95 similarity and similitudes, 882, 883 square lattice and, 588, 695, 698 tessellations with, 630, 632 triangle congruence, 741–46, 790–93 AAside property, 746 analysis with isometries, 878–80 ASA property, 744, 746, 791, 794 correspondence, 742 geometric problem solving, 790–93 reflections, distance, and, 877 reflections and, 875, 877, 880 SAS property, 743–44, 746, 778 SSS property, 745–46 symmetry, 588, 601–2 triangular lattice, 588 triangular numbers, 32–33 trillion, 74, 79, 104 triples, prime, 217, 413, 705 troy ounce, 672 true equations, 15 truncate, 161–62 truncated cube, 645 truth table, 913 twin prime conjecture, 203 twin primes, 215 twocolumn frontend estimation, 160 twostage tree diagram, 542–43
U Ulam, Stanislaw, 575 Ulam’s conjecture, 203 unbiased standard deviation, 506 unequally likely events, 520–21, 525, 565, 566 union of sets, 48–49, 110 probability and, 522, 524–25 unitary fractions, 263 unit cubes, 87, 717 unit distance, 687 unit fraction exponents, 404 unit fractions, 237 units, in base ten pieces, 71 universal set, 47 universe, 47 unknowns. See variables unlike denominators addition, 256–57 division, 272–73 subtraction, 260, 388 upper quartile, 489–92
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Use a Model, 582, 657, 695, 707 Use a Variable strategy, 11–13, 350, 695 Use a Variable with algebra, 13–16, 27–28 Use Cases, 342, 373 Use Coordinates, 808, 845 Use Dimensional Analysis, 666, 733 Use Direct Reasoning, 108, 148 Use Indirect Reasoning, 156, 198, 226, 399–400 Use Properties of Numbers, 204, 231–32 Use Symmetry strategy, 850, 901–2
V valid argument, 916 value absolute, 354 discrete, 445, 446 expected, 562–63 place (See place value) van Hiele Theory, 581, 583–85 level 0—recognition, 583–84 level 1—analysis, 584–85 level 2—relationships, 585 level 3—deduction, 585 level 4—axiomatics, 585 variable(s), 15 coefficients, 408 concept of variables in algebra, 15 exponent as, 421 greatest integer function, 427 problem solving with, 11–13, 350, 695 problem solving with algebra, 13–16, 27–28 Use a Variable strategy, 11–13, 350, 695 Use a Variable with algebra, 13–16, 27–28 variance, 493–94 vector geometry, 909 Venn diagrams, 47–52 conditional probability, 566 DeMorgan’s laws, 54, 57 difference of sets, 50–51 freshmen survey problem, 99 GCF and LCM, finding, 228, 234 intersection of sets, 49 problemsolving with, 52 union of sets, 49 vertex (interior) angles, 607–8 adjacent, 617 alternate, 619–20 angle sum in a triangle, 620 measure of, 617 in regular polygons, 632–33 on the same side of the transversal, 627 See also vertices
vertex arrangement, 632–33, 638 vertex figures, 635 vertex of the angle, 616–17 vertical angles, 618, 619, 621 vertical axis. See yaxis vertical line test, 424–25 vertices, 588 congruence and, 602–3 nonadjacent, 605 of a polyhedron, 643–44 Vieta, Francois, 294 volume, 676–77, 717–26 abstract, 717–26 capacity vs., 671 Cavalieri’s principle, 724–26 cone, 723 cylinder, 720–21 English system, 671 estimation, 719 formulas (See inside back cover) metric system, 676–77 prism, 717–18, 720 pyramid, 721–22 sphere, 723–24 von Neumann, John, 145, 149 voter opinion polls, 475–77
W Wallis, 411 Web site activities prime number competition updates, 227 spreadsheets Consecutive Integer Sum, 17 Standard Deviation, 502 See also dynamic spreadsheet activities on the Web site; eManipulatives on the Web site; Geometer’s Sketchpad on the Web site Weierstrass, Karl, 336 weight English system, 672 metric system, 678 Wells, H. G., 439 whole number(s), 45–52, 59–61, 125–27, 134, 381 addition, 109–15, 158–59, 172–73 algorithms, 172–73, 184, 185, 192 commutative property, 257 additive properties associative, 112–15, 158–59, 172–73 cancellation, 123 closure, 111 commutative, 111–15, 158–59, 172–73 identity, 112–13 algorithms, 171–83 addition, 172–73, 184, 185, 192
division, 178–83, 194–95, 302–3 multiplication, 169, 176–77, 185, 186, 188, 194 subtraction, 174–75, 192–93 amicable, 228 cardinal, 59 computation, 157–66 with calculators, 163–66 estimation, 160–63 decimal point and, 287–88 decimals and, 287–88 division, 123, 129–34, 274 decimals, 300–301 fractions, 274 division algorithms, 178–83, 194–95 elements in a set, 59–60 estimation, 160–63 expanded notation/form, 72, 76, 172–73 exponents, 77–78, 142–44 factors, 205–7 a divides b, 207 four basic operations, 134 greater than, 61 greatest common factor, 220–23 identification, 59 intermediate algorithms addition, 173, 192 long division, 180, 182 multiplication, 177, 194 least common multiple (LCM), 223–26 less than, 61 measurement division, 129–30 mental math, 158–59, 163 missingaddend subtraction, 192–93 multiplication, 123–29 algorithms, 169, 176–77, 185, 186, 188, 194 as product of primes with their respective exponents, 219–20 properties associative, 126, 129, 158–59 cancellation, 137 closure, 125 commutative, 125–26, 127, 128, 158–59, 266 distributive, over addition, 127, 129, 135, 176 distributive, over subtraction, 128 identity, 126 rectangular array, 124–25 multiplying decimals as, 299 naming, 74–75 number lines, 61, 110–11, 141–42, 248
I21
number systems diagram, 381, 382, 392, 401 numerals and, 59 ordering, 60–61, 140–42 ordinal, 59 pentagonal, 35 primes with their respective exponents, 219–20 reading, 74–75 rectangular, 35 rounding, 161–63 set models addition, 109–10 division, 130 multiplication, 123–24 subtraction, 115–16 sets as basis for, 45–52 square roots, 213 subtraction, 115–18 algorithms, 174–75, 192–93 missingaddend approach, 116–17, 128 takeaway approach, 115–16 triangular, 32–33 in Use a Variable strategy, 11–13 writing, 74–75 zero, property of, 128 wholetopart ratios, 311 Who Wants to Be a Millionaire?, 726 Wiles, Andrew, 203 Work Backward strategy, 16, 286, 335 World Cup Soccer Championships, 633
X xaxis, 418 of bar graphs, 444–45 of histograms, 445 polar coordinates, 832 xcoordinate, 417–18 equations of lines, 822–23, 826 simultaneous equations, 826–29, 831
Y yard, 669, 670–71 yaxis, 418 compressing, 444–45, 447, 465–67 histograms vs. bar graphs, 445 manipulation of, 444–45, 447, 465–67 ycoordinate, 417–18 equations of lines, 822–23 midpoint formula, 811 orthocenter of a triangle, 837–38 simultaneous equations, 826–29 slopeintercept equation, 823–25
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yintercept, 823 pointslope equation of a line, 825–26 slopeintercept equation of a line, 823–25 Young, Grace Chisholm, 373 Young, William, 373
Z zero as additive identity, 112–13 division property of, 132 divisor property of integers, 362 as exponent, 144
factorial, 550 as identity property, 112–13 as integer, 343, 352 in Mayan numeration system, 65–66 multiplication property of, 128 repetend not, 400
zero pair, 344 zscores, 503 dispersion and, 495–96, 499–500 mean, 495–96
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List of Symbols Symbol {. . .} {x. . .} 僆 僆 {} or ⭋ ⫽ ⫽ ⬃ 債 債 傺 U 傼 傽 A – (a, b) ⫻ n(. . .) ⬍ ⬎ ⱕ ⱖ 3five (35) am f (a) ⬇ 
⁄ n! GCF LCM .abcd a⬊b % ⫺3 ⫺a a a⫺n √ n √
Meaning set braces set builder notation is an element of is not an element of empty set equal to is not equal to is equivalent to (sets) is a subset of is not a subset of is a proper subset of universal set union of sets intersection of sets complement of a set difference of sets ordered pair Cartesian product of sets number of elements in a set is less than is greater than less than or equal to greater than or equal to three base five exponent (m) image of a under the function f is approximately divides does not divide n factorial greatest common factor least common multiple repeating decimal ratio percent negative number opposite of a number absolute value negative integer exponent square root pi (3.14159 . . .) nth root
Page 45 46 46 46 46 46 46 46 46 46 47 47 48 49 50 50 51 51 60 61 61 61 61 75 77, 142 88 162 207 207 214 220 223 303 310 321 343 344 354 364 402 402 403
Symbol 1/n
Meaning
Page
a am/n x P(E) nPr nCr P(AB)
nth root of a mth power of the nth root of a mean probability of event E number of permutations number of combinations probability of A given B
404 404 486 517 551 553 566
AB ⬔ABC 䉭ABC 4 AB AB ml : AB m(⬔ABC ) l⬜ m 4 艑 ⬃ 1 AB TAB ⭿ABC RO,a
line segment AB angle ABC triangle ABC
601 601 601
line AB the length of segment AB m is parallel to l
616 616 616
ray AB measure of angle ABC l is perpendicular to m correspondence is congruent to is similar to directed line segment from A to B 1 translation determined by AB directed angle ABC rotation around O with directed angle of measure a Ml reflection in line l MAB reflection in line containing AB SO,k size transformation with center O and scale factor k TAB(P) image of P under TAB RO,a(P) image of P under RO,a Ml (P) image of P under Ml Ml (TAB(P)) image of P under TAB followed by Ml HO halfturn with center O ⬃ negation (logic) conjunction (and) disjunction (or) : implication (ifthen) 4 biconditional (if and only if) 䊝, 䊞 , 䊟 , 䊞 ⫼ clock arithmetic operations a ⬅ b mod m a is congruent to b modulo m
616 617 618 742 742 752 853 853 854 855 856 861 863 875 875 875 875 893 912 913 913 914 915 923–24 925
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Geometry Formulas for Perimeter, Circumference, Area, Volume, and Surface Area Rectangle P ⫽ 2a ⫹ 2b A ⫽ ab
b Cube V ⫽ s3 S ⫽ 6s2
a
Square P ⫽ 4s A ⫽ s2
s
s
s B Triangle P⫽a⫹b⫹c A ⫽ 12 bh
c
Right Prism V ⫽ Ah S ⫽ 2A ⫹ Ph
a
h
h A A
b
Parallelogram P ⫽ 2a ⫹ 2b A ⫽ bh
P
C
h
a
l
Right Regular Pyramid V ⫽ 13 Ah S ⫽ A ⫹ 12 Pl
h
P
A b r b
Trapezoid P⫽a⫹b⫹c⫹d A ⫽ 12 (a ⫹ b)h
c
Right Circular Cylinder V ⫽ Ah ⫽ (r 2)h S ⫽ 2A ⫹ Ch ⫽ 2(r 2) ⫹ (2r)h
d
h a
h A C
Right Circular Cone V ⫽ 13 Ah
Regular ngon P ⫽ ns A ⫽ 12 rP
r
⫽ (r )h S ⫽ A ⫹ 12 Cl ⫽ r 2 ⫹ r √h2 ⫹ r 2 1 3
s
Circle C ⫽ 2r A ⫽ r 2
r
Right Rectangular Prism V ⫽ abc S ⫽ 2(ab ⫹ ac ⫹ bc)
Sphere V ⫽ 43 r 3 S ⫽ 4r 2
c b a
l
2
r
A
C
r
h