College algebra

College Algebra The qualif1 of the materials used in the manufacture of this book is governed by cQnUnuedpostwar shorta...

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College Algebra

The qualif1 of the materials used in the manufacture of this book is governed by cQnUnuedpostwar shortages.

College Alge bra BY

A. ADRIAN ALBERT Professor of Mathematics The University of Chicago

First Edition Second Impression

London

New York McGRAW-HILL

BOOK

1946

COMPANY,

INC.

COLLEGE ALGEBRA COPYRIGHT,

1946, BY TEE

MCGRAW-HILL BOOK COMPANY, INC. PRINTED IN TEE UNITED STATES OF AMERICA

All rights reseTlJea. This book, or parts thereof, may not be reproduced in any form 'Without permis8ion of the publishers.

TEE IUPLE PRESS COMPANY, YOBlt, PA••

PREFACE The mathematical material presented in standard texts on college algebra is a fundamental part of mathematics. The concepts involved are basic for an understanding of all more advanced mathematical topics, and the techniques are used in virtually all subsequent courses in algebra and analysis. Nevertheless, college algebra has been a most abused subje,ct. The time allotted to it is frequently inadequate for a genuinely good treatment, and indeed the entire course is sometimes omitted. This is due partly to a desire to bring students to a study of the calculus as early as possible. It is also due partly to the presentation of college algebra, in all texts thus far published, as a collection of seemingly unrelated topics. The desire to teach the calculus as early as possible tends to defeat its own ends. The building of a course in the calculus on what must be a weak foundation cannot result in a good student understanding of the subject. There is also no reason why the :ql8.terial of college algebra cannot be cohesively organized. ' About fifteen y~ars ago the author began to study the possibility of reorganizing the material of college algebra so as to present it as a sound and unified whole. The result of that study is the present text. He began with a formulation of what he felt should be standard minimum requirements for all college texts. These requirements are the following: 1. The material should be presented as a unified and compact body of mathematical theory. 2. The definitions and theorems should be stated accurately. Proofs of results should be given whenever it is reasonable to expect that the better students will be able to grasp them and provided that their inclusion will add to t4e understanding of the results. Whenever a proof is omitted for any reason it should be made clear that the v

-VI

proof is omitted. Proofs of special cases of theorems should not be presented as proofs of the theorems. 3. The important techniques and concepts of the text should be emphasized by the presentation of an adequat~· number of illustrative examples and exercises. Exercises should be given in sufficient numbers so that there are enough both for additional classroom illustration and for student home assignment. They should be solvable by" the use of the techniques of the text. 4. Where such material exists, the text should be rich in additional material on the same subject for the better student. It seems wasteful to use the trick problem as a device to test the student's native ability as a mathematician. His time can be employed far more profitably iu the learning of more advanced topics which are a part of the text subject and which are given in the text itself. The reader may judge for himself how well the present text lives up to these standards. It would seem that they should be met and that few college texts of today meet them. ' College algebra has a basic unity. It should consist of, a study of the number systems of elementary mathematics, polynomials and allied functions, algebraic identities, equations, and systems of equations. The unity of the present text is achieved by fitting the standard topics of college algebra into this pattern. Thus the text begins with three, chapters on number systems. The first chapter on Natural Numbers introduces the counting concept and so the fundamental idea of a sequence which appears so frequently in mathematics. As permutations and combinations involve nothing but b ~d b > a. then c > a. This remark is verified by observing that c > b means that c == b + g where g is a positive inte~r; b > a means that b == a + h where h is a positive integer. Then c == b + g == (a + h) + g == a + (h +" g). Since h + g is positive, c > a. It is customary to indicate that c > b and b > a by writing

c· > b > a, (read HC greater than b 'greater than a"), and i1r is then important to know that this means also ·that c >. a. A relation haying the 'properties described in the paragraph above is called an order relation and a ~t, for which an order relation holds for all pairs of its elemeJ;l.ts, is called.

SEC.

6]

THE ORDERING OF NATURAL NUMBERS

9

an ordered 8et. All but the final one of the number systems we shall study will be ordered. This last unordered number system will be called the 8et of all complex number8. A notation indicating the property that a is not greater than c is a:l> c (read "a not greater than c"). This' property occurs in the case where a and' c are natural numbers (as well as in the other cases of ordered number systems which we shall study) only when either c = a or c > a. We then write c~a

(read "c greater than or equal to a"). In the case of natural numbers the condition c ~ a is equivalent to the property that c = a +'b where b is also a natural number. The property is identical with a~c

(read "a less than or equal to c"). we have c ~ a and so write

c

~

b

~

Also if c

~

band b

~

a

a.

When we apply the notations c ~ 0 and c > 0 to natural numbers c, they mean only that c is a natural number in the former case and that c is a positive integer in the latter case. The relation c ~ a is called an inequality. When we have such a relation and show that it can be replaced by c > a, we say that we have 8trengthened the inequality. Thus c > a is called a 8trong inequality, c ~ a a weak inequality. The inequalities c ~ b ~ a imply the equality c = a if and only if neither of the inequalities c ~ b, b ~ a is a strong inequality. , The principle of mathematical induction can be stated in two other useful forms which should be learned by the student of the FULL COURSE. They are as follows: PRINCIPLE 2. Let K be a 8et of positive integer8 and 1 be in K. Suppo8e that.whenever an integer n i8 such that all integer~

10

'N ATUR:A:L

N-UM'BERIil

[CHAP. '1

l'688 than n are in K, then n i8 in K. 'i'hen K i8 the 8et of all positive integer8. PRINCIPLE 3. Every 8et of p08itive integer8 contain8 a lea8t integer. Pririciple 3 states that L"l every set K cf positive ip.teger-s there is an integer c such that if k is ail integer of K then k ~ c. We shall not try to show how to obtain each of the three forms of the principle of mathematical induction from anyone of them. 7. Subtraction (FULL COURSE). Whenever c ~ a, there' is a natural number b such that c = a + b. We call this . number the difference of c and a, write (4)

b = c - a,

(read" b equals c minus a"), and say that b is the result of subtracting a from c. If a = c, the difference (5)

c- a

=

O.

Otherwise b = c - a is a positive integer. If c = b + g and b = a + h then c = (g + h) + 'a and c - a =:= g + h. When h > 0, we have g + h > .g, b - a = h, c - a > b - a. In any case c - a ~ b, - a and we have a result which we shall state as follows: Theorem 3. Let a, b, c be natural number.s 8uch tha~ c ~ b ~ a. Then c - a ~ b - a ~ 0 and c - a > b - a ,if

> b.

. It is most important for us to keep the fact in mind that if a > c the difference c - a, as a natural number" dO~8 not exi8t. In Chap. II we shall so extend our number system that in the new system all differences b - a will exist. Remember that c - a'means a number b such that c = a + b. 8. Division. Let a and b be natural numbers. Then we say that b divides a if there exists a natural number c such that

'c

(6)

a :;= bc.

SEc. 9]

POWERS

11

When this oocurs, we call b either a factor or a divisor of a and say that a has the factorization a = bc. Then c is also a factor of a. If b = 0, the only natural number a such that a- = Oc is a = O. Hence 0 does not divide any number but itself. On the other hand if a = 0, then a = bO for every b. Hence every natural number divides zero. It follows that the symbol a

'b' for what we call the quotient c of a = bc by b, has no meaning when b = 0 and a ¢' O. Moreover, when a = b = 0, this quotient represents all natural numbers. In the solution of a problem by division we want our answer both to exist and to be unique, and we do not satisfy one or the other of these requirements when b = O. This is what is meant by the statement that division by zero is impossible. The quotient c is unique when b ¢' O. For the cancellation law for multiplication states that if b ¢' 0 and a = bc = bd then c = d. However c may not exist in the set of natural numbers. - For example, take a ~ 1 and b = 2. This will be our motivation in the extension we shall make in Chap. III of our set of integers to a number system including all quotients and called the set of all rational numbers or all fractions. 9. Powers. The formal distinction between the products ah and ba is in the order of the two factors a and b. The distinction between a(bc) and (ab)c is in the grouping of the factors. The commutative and associative laws may be combined and extended to products with any number of factors and the combined extension -reads as follows: Theorem 4. T,he value of any product is independent both of order and of grouping. We see similarly that the value of any sum is independent both of order and of grouping. If sums and products are both involved, the value depends upon the reading of the expres-

12

[CHAP. 1

sion as a sum. of products (the products might involve sums!) and the use of the distributive law.' . In a product of n factors all equal to a the independence of grouping implies that there is a common value for all such products. We call this value the nth power of a and designate it by an (read "a to the nth"). Here n is any positive integer, and we call n the exponent of the power an. For example, we are stating that the common value of (00) (00) == a[a(oo)] == a[(oo)a] == [a(oo)] a is a4• Every such power of 0 is O. We shall not need to consider powers of zero further, since if even one factor of a product is zero the product is zero. Henceforth let the products of this section be products of positive integers. We extend our definition of powers by defining (not by proving) (7)

aO

== 1

(a

;;&!O

0).

Then we have defined an for every natural number n and all positive integers a. A product be of b· == and e == am is a product of m + n factors all equal to a. By Theorem 4 we have the first of our laws of exponents given simply as

or

(8)

anam == an+tA.

We ~y translate this formula into words by saying that the exponent oj a product oj pOW6r8 oj the 8ame number is the wm oj the e:z;ponenta oj the Jactor8. A product (fA of m factors all equal to an is a product of n + n"+ • • • + n (read "n plus n plus and so on plus n") factors a. There are m terms in this sum and hence its value is nm. Then (9)

(an)m == a-.

Thus, to Jorm the mth power oj the nth power oj a we multipl1/ the exponenta m and n.

BlDC.

10]

-

SUMMARY

13

We notice finally that a"'b'" is a product with n factors a and the same number of factors b. The product (ab)'" has exactly the same factors as a"'b'" but with a different order and grouping. By Theorem 4 we have the third la}V of exponents (10)

a"'b'" == (ab)"'.

In words, the nth power oj a product i8 the product oj the nth power8 oj the Jactor8. This law has been derived here for all natural numbers n. EXBRClSES

1. The following products may be expressed as products of powers of 2, 3, 0, 7 by factoring and the use of the laws of exponents. Give the e",pressions. (a) 2" 5' • 4 • (14)' . (35) (b) (10)4. 8' • 91 (e) (28 • 31 • 4' . 5' . 71)0 (d) (2" 3' . 5 • 71)1 • [2 . 9 • (70)]8 (6) 3· 5 • 7 • 9 • (49) . (25) • 8 • 6 . (24) • (21) • (28) • (105) I. Use formula (9) and find a positive integer z such that (a) 2'3851 = Zl (b) 2'3851& = zI (e) 16(2'3')' = z4 (d) 2'3'(zl)' '= 3(Z8)' (e) 8' 38(y',8)' = z', y' = 71, .' = 25

10. Summary. In the sections just completed we defined the concept of natural numbers and told how the operations of addition, multiplication, subtraction, and division on natural numbers are performed. We also presented the fundamental properties of our number system consisting of all natural numbers as nine laws of addition and multiplication, and derived three consequent laws of exponents. We showed that this number system was ordered and gave the principle of mathematical induction and two consequences. Finally we defined the terms difference and Jactor (or diviaor) and observed that subtraction and division of natural numbers were not always possible. No problem 'Solving techniques were given in these sections except for the bit of drill of the exercise just

completed on the laws of exponents and the use of the' distributive . law. We shall give some drill later on the removal of parentheses, brackets, and braces by the use of our fundamental laws. ORAL REVIEW EXERCISES

1. State the principle of mathematical induction and its two sub.. stitutes. 2. Let K be a set of even integers. From what two hypotheses 'Would it follow, by the principle of mathematical induction, that K is the set of all even integers? S. From what two hypotheses would it follow that a set of odd integers is the set of all odd integers? 4. State the four laws of addition. 6. StQ.te the four laws of multiplication. 6. Give all the possible groupings in a product abed.

11. Finite sequences. The remainder of this chapter will be devoted to topics .involving nothing but counting, and so requiring the use of no numbers but natural numbers. The concepts presented in the present section on counting are of major importance in mathematics. The procedure used in counting the objects of a set of five objects may be described in terms of a J;lotation that has many uses in our subject. We begin by selecting a first object and represent it by the symbol al (read" a sub, one" or simply "a one"). Then we might think of al as a name for the first object. We name a second object a2 and then label a third as, a fourth a" and the remaining oneaG. We shall then have exh~usted the set, that is, used up all its elements (i.e., objects) and so shall have completed the count. We call this counting process that of setting up a correspondence between the five objects and the integers 1, 2, 3, 4,5. The correspondence is one to one, that is, each object has been made to correspond to precisely one positive integer, each of the five positive integers to only o~' object. The notation just used may be employed when we have a set of n o!>jects. Here n is any unspecified positive integer.

Bl!IC.

11J

FINITE SEQUENOES

15

When we do this, the set is labeled so that the names of its objects are (11) (read It aI, aI, to a. "). Here al is the first object, as is the second object, an is the last or nth object. The three dots represent objects which are n~mbered but which cannot be enumerated (i.e., written out) because 11. is umpecified. A line of symbols like formula (11) is usually called a finite sequence (or simply a sequence) and the symbols aI, a.~ and so on, its terms. We could use the letter b instead of a and so em,ploy the notation bl , bl ; • • • , btl or even bl ,' bs, • • • ,b... Of course other letters can also be used. When we wish to refer to an arbitrary term of a sequence (11), we shall speak of the ith term. This is the term as where i is a symbol representing anyone of the integers 1, 2, .•. ,11.. It is also called the general term of the sequence. The letter i need not be used and we might speak of the jth term ai or the kth term all. It is important for us to realize that, althou,gh as is anything, we are using i to represent a counting SUbscript, that is, a positive integer. Symbols that have the connotation of integers are usually limited to i, j, k, m, 11., p, q, r, s, or t. Other letters may be used but x, y, s are rarely employed for this purpose. We shall sometimes refer to the terms of a sequence aI, as, • . • , an as the as. For example; we might wish to' indicate that the terms ale all integers. Then we shall say that the as are aU integers. Later on we shall say that the as 'have a certain property from a certain place on. ThEm we shall mean that there is a positive integer k such that every as' has the prescribed property for all integers i ~ k. , Our sequence notation has the term as in it. However we must agree that if we specify later that 11. = 1 the sequence collapses to its first term al = an. In the theory of polynomials w:hich we treat l8.ter on it will 'be 'desirable to consider sequences that begin with the

16

NATURAL NUMBERS

[CHAP. I

syinbol ao rather than al. Then the notation for such a sequence will be (12)

ao, at, •.. ,

term:

an.

Here ao is the first al is the second term, CIt is the (i + 1)st term, an is. the (n + 1)st term. The sequence has n 1 terms. There is also no reason why we could not write

+

(13)

to represent a sequence in which n ~ 5 and the number of terms is not n but n - 4. For example, we might wish to speak of th~ sequence consisting of all those terms in formula (11) which appear after Th~ notation (14)

CIt

a,. "

(i = 1, . . . , n),

(read "CIt for i equals one to n") may be used instead of ai, a2, • • • , an, and the notation CIt

instead of ao, ai, ..• (15)

,an.

(i = 0, 1, . . . , n),

We could then use the notation CIt (i = 5, ••• ,n)

for the sequence as, as, • • • ,an. Sequence notations are used in many" connections. One example is that of the notation which could be employed for the digits of an n-digit positive integer a. We write a "= (ai, a2, ••• , an). Then al is the extreme left-hand digit, a2 is the next digit, and so on. In this case every CIt is an integer taken from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and al is not zero. In the case abpve a is a symbol for the whole sequence of n objects, and we have indicated this by putting parentheses around the sequence. We sometimes call such an object an n-tuple. If n = 2, it is a pair of objects; it is a triple of objects when n = 3.

smc. 12]

INFINITE SEQUENCES

17

Exercises designed to test the student's underStanding of the sequence notation are given at the close of the next section. 12. Infinite sequences. It may be desirable to consider sequences that do not terminate. Such sequences are called infinite 86quence8. One such sequence is the sequence of all natural numbers given in. formula (1). Infinite sequences represent sets of objects which may not be counted in the sense of a terminating counting process, but which are countable in that they may be put into one-to-one correspondence with the set of all positive integers. Such sets are called denumerable and their objects form infinite sequences (16) al, all .•• (read "al, as, and so on"). We shall usually designate the general term of an infinite sequence by tin (rather than ~). As before we could use hlJ hI ••• , or ao, al . . . , or (17) tin (11, = 1, 2 •.. ) (read "11, equals one, two, and so on"), or finally (18) tin (11, = 0, 1 . . . ). Sequence notations are used so frequently in algebra that it is of great importance that students be perfectly at home in their use. The following exercises have been constructed to give the student· drill work which should result in the desired facility. ORAL EXERCISES

1. Wha.t is n a.il.d wha.t are al, az • • . , tit. for the following finite sequences: (a) 1, 3, 5, 7, 9, 11 (b) 1, 2, 2, 3, 3, 4, 5, 6 (c) '3, 3, 3, 3, 3, 3, 3 (d) 1, 2, 1, 2, 1, 2, 1, 2, 1, 2 I. Give a formula for as for the following infinite sequences: (a) 1, 2, 3, 4, 5 •.. (the sequence of positive integers) (b) 2, 4, 6, 8, 10 .•. (the sequence of positive even integers) (c) 0,2, 4, 6 . . . (the sequence of even natural numbers) (d) 1, 6, 11, 16' •.. (the sequence whose members differ by 5).

18

N ATlJRAL NTiTM:6~'Ri;S

(c.Hap. I

EXERCISES

1. Enumerate the values of a1, a2, aa, a4, ai; in the sequences whose' general terms are given by the following formulas: (a) a,. = 2n (b) a .. = 3n {c) a,. = 3n (d) a.. n(n (e) an (2n

= =

+2 1 + 1) + 1)(3n -

+

(f) a" = £1,,_1 2n - 1 for n > 1, a1 = 1 (g) a" = na.. _1 for' n > 1, a1 = 2 (h) a,.

1)

(i) a"

= =

a2 =

2an _1 a,,_1

2

(3) a,. = a1

for n

> 1, a1 =

2

+ a"_2 for n > 2, a1 =: 1,

+ a2 fo:r n > 2, a1 =

a2

=1

2. Let bo, b1 • • • be a sequence such that bo = a1, b" = a,,+l for each of the sequences of Exercise 1. Write out the corresponding formul9.!! ~or b... 3. Write out the fifth, sixth, seventh, eighth, and ninth terms of a sequence b1, b2 • • • formed from each of the sequences of Exercise 1 by the use of the formulas b1 = a1, b.. = 2a" - a..-1 for every n > 1. 4. The first six terms of a sequence are given below. FInd a formula for a,. in terms of n which will fit these terms. (a) (b) (c) (d)

1,4, 7, 10, 13, 16 5, 9,13, 17, 21, 25 2' 1, 3 . 2, 4 . 3, 5 . 4, 6 . 5 0, :3 . 1, 4 . 2, 5 . 3, 6 . ~, 7'5

(e) 2' 1, 4· 3, 6· 5, 8· 7, 10· 9

12' 11 (f) 6, 6, 6, 6, 6, 6 (g) 1, 3, 5, 7, 9, 11 (h) 3, 5, 7, 9, 11, 13

5. Use the formulas obtained in each part of Exercise 4 and find tlie values of a10 and the formula for aH1. 6. The terms of a finite sequence of six terms are 1, 2, 5, 3, 4, 2. Define a new sequence b1, b2, • • • , ba by b1 = a1, b.. = a" rb.._1 for n > 1. Find the terms of the sequence of b" if

+

(a) r = 1 (b) r = 2

(c) r

(d) r

=4 = 10

(e) r = 0 (f) r = 5

13. Counting of pairs. The sequence notation may be used in counting the number of pair8 of objects if the first ,object in the pair may be anyone of n objects and the second anyone of m objects. Designat,e the objects of the first set by aI, a2, • . . ,an and $.ose of the second set b~ bl, b2 , • • • ,bm • Then the pairs are Ca., bi ) where i h~'S any value from 1 to nand j any value from 1 to m. There are then 11,1, 8et8 of pairs and each consists of the n pairs

SJDC.

13J

.

COUNTING OF PAIRS

19

(al, bi), (a2, bi), •.• , (an, bi)' Hence there are n~ pairs altogether. In a special case we may enumerate the pairs (a, b). Let us do so in the case where a may be anyone of the integers 1, 2, 3, 4, 5 and b anyone of the integel'B 6, 7, 8. Then the pairs occur in three sets, each 'of which consists of five pairs. The first set consists of the pairs (1, 6), (2, 6), (3,6), (4,6), (5,6), the second set of (1,7), (2,7), (3,7), (4, 7), (5, 7), and the last of (1, 8), (2, 8), (3, 8), (4,8), (5, 8). There are 15 == 5 . 3 pairs all told, and we have liste!i them. The result above is used in ordinary experience in counting pairs of events. We state it then in the following form: FUNDAMENTAL COUNTING PRINCIPLE. Let it be p08Bible Jor an event to take place in anyone oj n way8 'and, aJter it take8 place, Jdr a.8econd event to take place in anyone of m waY8. Then the number oj waY8 in which the sequence oj two events can take place is nm. The result may be extended to triples of events, quadruples of events, etc. In every case the number of ways in which the sequence of events can occur is the product of the numbers of ways in which each entry can occur. It cannot be overemphasized that the way we do any .job of counting depends upon what we are counting. If we have two mutually exclusive sets of events and we are merely interested in counting these events, we add the numbers of events in each set. But, if we are counting pair8 of events, we multiply the numbers. In the special case above there are altogether 8 == 5· + 3 integers 1, 2; 3, 4, 5, 6, 7, 8. But there are 15 == 5 . 3 pairs. Some of the examples given below will combine both types of co.unting. mustrative ExamPles I. A man living in Chicago can go to Minneapolis by plane, train, or bus, but can return only by train or bus. He can also go to Milwaukee by train, boa.t, bus, automobile, or plane, and can return by train, automobile, or plane. How many round trips would it be possible for him to take?

20

NATURAL NUMBERS

[CHAP. I

Solution

We are counting the total number of round trips. By the fundamental counting principle there are 6 = 3 . 2 round trips possible to Minneapolis and back and there are 15 = 5 . 3 round trips to Milwaukee and back. • The total number of round trips is 21 ... 6 + 15. II. Ho:w many different trips would it be possible for the man in the example above to take? Here we are agreeing that a trip from Minneapolis to Chicago by train is not the same as one from Chicago to Minneapolis by train. Solution

In this problem we are counting trips, not round trips. = 13 trips.

There are

3+2+5+3

EXERCISES

1. The first' digit of an ,n-digit natural number a may not be zero unless n = 1 and a = O. Use this to find the number of (a) Four-digit numbers (b) Four-digit even numbers (i.e., final digit 0,2,4,6,8) (0) Even numbers with three, two, or one digit 2. A license plate is to have on it two letters followed by a four-digit number, the first digit being not zero. How many plates with different labels can be made? Am. 6,084,000. 8. How many four-letter code words can be made if the first two letters are any consonants and the . last two any vowels? (Consider a, e, i, 0, v, and 1/ as vowels.) " How many five-letter code words can be made if the first, and last letters are any one of the letters b, 0, d, p, q, r, 8, t, the second and fourth letters any vowels, and the middle letter anyone of the letters g, h, j, Te, l, tn. Am. 13,824. 6. A collection ~f pictures contains five nineteenth-century portraits, eight'landscapes, and six modern paintings. It is desired to make an exhibit consisting of one painting of each kind. How many exhibits differing in at least one painting can be selected? 6. A student is taking English, chemistry, and mathematics and can get anyone of the grades A, B, C, D, F in each subject. How many different sets of three grades could he get? Am. 125. 7. A signal device consists of four rows of lights each consisting of a red light, an amber light, and a green light. A si.gna.l consists of. a light in the top row, or a light in each of the top two rows, or a light in each of the top three rows, or one in all four rows. How many different signals can be made ~th this, system?

SEC.

21

FACTORIALS

14J

8. A company comma.nder has under his comma.nd five platoons each of nine men. He details one man out of each platoon for special A.na: 9&. duty. How many different groups can he select? 9. A woman ha.s 15 different handkerchiefs, 8 different hats, and 4 different pairs of gloves. She wishes to select one of each for a given day. How ~y different selections could she make? 10. How many two-digit numbers can be made from the digits 1, 2, 3, 4,7,8 if the two digits must be different? A.na.30. 11. Two men cross a river separately. The first man can use any one of five different boats and the second man anyone of those remaining. How many different choices of boats can the two men make?

14. Factorials. It is useful to have a symbol to indicate the product of the first n positive integers. We therefore define (19) nl = n(n - l)(n - 2) • • . 3' 2 . 1 and read the symbol nl as "nJactoriaZ." Then we have the values . 1!=1, 21=2, 81=6, 41=24, 51=120, and so forth. Some authors use the symbol ~ rather then' nl. It may also be desirable to introduce the convention that zero factorial be defined to have the value 1. Since nl is the product of the first n integers, it should be' clear that if r is a positive integer less than n the product rl of ~he first r integers is a factor of nl. Indeed we have (20)

nl == n(n - 1) .•. (r

+ 1) • rl.

Then the number n(n - 1) ... (r + 1) is the product of the n - r Zarge8t displayed factors of n I. The difference n - r is also less than n and we may replace r by n - r in the formula above to obtain (21) nl = n(n - 1) . . . (n - r

+ 1) . (n -

r)!.

We shall use the symbolP",r (readP, n, r) for the product of the factors to the left of (n - r) 1in formula. (21) and hence define the integer (22)

n!

P",r == (n _ r)l = n(n - 1) ••• (n - r

+ 1).

22

NATURAL NUMBERS

It is the product of the r integers n, n - 1, n - 2, . . . , 'n - (r - 1), and we emphasize that the last 9f these factors

isn-r+1. We have now seen that n! is divisible both by rt and by' (n ....,. r)! for any r < n. Indeed n! is divisible by thejr product, a result we shall derive in the next section. Let us introduce the, notation

c

(23)

_

n,r ~

nl r!(n - r)1

for the integral quotient. If we replace r by n - r in Cn,r we obtain (fn,n-r' becomes (n - r)! and (n - r) I becomes 8! where 8

=

Then r

r

n - (n - r) = r.

Thus, (24)

for all values of n > 1 and r (25)

P n,r

Cn,r

=-, r. =

< n, The formula

n(n - 1) . . . (n - r r.I

+ 1)

is an immediate consequence- of our definitioI!l.S. The numerator is the product of the r integers obtained by beginning with n and diminishing by 1 a total of r - 1 ti~es. It should be compiled by the simple process of , enumerating these r factors, Similarly the denominator is the product of r factors obtained by beginning with rand diminishing by one r - 1 times. For example,

2 3

C11,6 =

11 . 10 . 9 . g . 7 . ~ ~

•g. 2

. is"

=

11 . 7 . 6 = 462 = C11, D·

How then should we compute the 'quotient C80,78 of two products each with 78 factors? We simply use formula (24) to write C 80,78

=

C80,2

80' 79

= 2~

=

40' 79 = 3,160.

SEC.

14]

23,

FACTORIALS

I. Find 11. if C"' 3

nlustrative Examples . = 84.

Remarks: The problem requires that 11. be 11.(11. - 1)(11. - 2) _ 84 '

6.

-,

a natural number such that

11.(11. - 1)(11. - 2) = 504.

Then 11. 3 > 11.(11. - 1)(11. - 2) .> (11. - 2)3 and 11. is that natural number for which 504 lies between (11. - 2)3 and 11.3• Note that a problem such as this need not have a solution. For example, the problem of finding an 11. such that C... 3 = 85 has no solution. Solution

We list the cubes

33

=

43 = 64,

27,

53 = 125, 68 = 216, 88 = 512, 93 = 729 78 = 343,

The only pos81ole solutions of this problem are 11. = 8, 9 and we verify that .

CU• 3 = II. Find 11. if P ...9

9·8'7

~ =

3·4'7

=

84.

= 42P... 7•

Solution The integer P ... 9 has nine factors, the product of the first seven of which is P ... 7• Then P •• 9 = P ... 7(n - 7)(11. - 8). Here the final factor is 11. - 'I' + 1 where 'I' = 9. Then P ... 7(n - 7)(11. - 8) = 42P... 7 , 7)(11. - 8) = 42. The pro<Juct of two consecutive integers equal to 42 is 7' 6 and 11. - 7 = 7,11. = 14. III. Fmd 11. and 'I' if P ... r = 15,120. C... r = 126.

en -

Solution

P ... r = (rl)C... r so that '1'1

15,120 = ---r26 = 120 = 2 . 60 = 2 . 3 • 20 = 2 . 3 • 4 . 5,

= 5. Then P ... 5 = 11.(11. - 1)(11. - 2)(11. - 3)(11. - 4) = 15,120. A table of values of P ... & for 11. ~ 5 yields P s.& = 120; P a.& = 720; P 7•S = 2,520; P a.& = 6,720; P 9• S = 15,120. Hence 11. = 9, 'I' = 5. 'I'

ORAL EXERCISES

Express the ·first of the following numbers in terms of the second (using products or quotients): ·

24

NATURAL NUMBERS

L 81,61 2. 61,81 3. 101, 91 4. 101,81 6. 91, 61

6. 61, 71 7. P •. G, P •. 4

8. P ...., P".•- l 9. nl, P ..."-l

,10. p".s, 0.... & EXERCISES

1. Simplify the following by writing the result as an integral multiple of nl for n as large as possible: (a) (b) (0) (d)

81 - 4' 3' 91 8' 51 12' 81 -

(6) 3' 71 - 2 . 61 (f) 4· 71 - 6 . 51 (g)'3161 71 (h) (41)2 2 . 61

61 12' 81 61 7·61

+ +

Am. Am. Am. Am.

19·61 2,7' 61 13.' 61 84 '41

2. Compute (a) P&'8 (b) o&'& (0) P S.7 P •. 4 (d) 2P10.& P S•8 (6) PU.8 33P7•8 (f) 50s. 8 0S.4

Am.10·

Am. 1.

(g) 70s. 7 0 7•8 (h) 50s.6 3C10•4 (.1.) 91P12 .& P 10.&014.& (~ 80U.aPI.8 :J P 14.•

Am. 1.

Am. 10.

Am.6.

3. Find n if (a) (b) (0) (d) (6)

= 110 = 28 = 60 p".• = 1,716

P ... 2 0"'2 P ....

Am. 8. Am. 13.

0 ... 8 = 364

Am. 12. (f) 0";8 = 220 (g) P ".8 = 42P... 8 Am. 6. (h) P" .• = 120,,_1.& (i) 50".& = 80,,-1.4 W 180..- 1.4 = 15(J".• Am. 9.

4. Find r if (a) P19 .• (b) P17 .•

= 5,814

= 272

Am. 2.

(0) 017 .• = 12,376 (d) 40018 •• = 51016 ••

Am. 2.

6. Find'n and r if (a) P .... = 840, 0 .... = 35 (b) P .... = 11,880, a.... = 495 (0) rl = 6,P.... = 210 (d) nl. = 5,040, 0" .• = 35

Am. n = 7, r = 3. Am. n = 7, r = 3.

BEc.15]

PERMUTATIONS AND COMBINATIONS

25

6. Show that 0 .... + 0 ....- 1 .. 0,,+1 .•• 'I. Compute the values of the expression 21" + 1 for n = 1, 2, 3, 4. (These integers are connected with the number of sides in a regular polygon constructible with ruler and compass.) 8. If the legs of a right triangle have lengths a and b and if c is the length of its hypotenuse, the theorem of Pythagoras states that as + bl = cl • It is known that a.ll cases where a, b, c are positive relatively prime integers are given by the formulas

+

a ... 2mn, b = m' - nl, c ... ml nl, where m and n are positive integers;" one of them is even, and the other odd, m > n. Compute at least four difierent sets of values of a, b, c.

15. Permutations and combinations. The theory of permutations and combinations is concerned with methods of counting the number of ways in which a subset of objects of a given finite set may be selected, or selected and arranged, in a sequence. A combination of n objects r at a time is a 8election of r of the n objects. A permutation of n objects r at a time is an arranged 8election. It is important for the student to be alert to the fact that combination meam 8election (i.e., choice), permutation meam arranged 8election. The number of combinations of n objects n at a time is evidently precisely one. Then we may refer to a permutation of n objects n at a time as simply a permutation of this single choice of n objects. We shall show shortly that the number of permutations of n objects is nl. Let us now consider the simple example of three objects a, b, c. Their permutations are (a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), (c, b, a). Thus there are 6 = 31 permutations. If we count permutations two at a time, we observe first that we may select a, b or a, c or b, c for our pairs. The corresponding permutations are (a, b), (b, a), (a, c), (c, a), (b, c), (p, b), and there are 3 . 2 = Pa,s such permutations. We proceed to count the number of permutatiom of n object8 r at a time. As first object in the sequence of r objects we may select anyone of the n objects". When it has been selected and used, only n - 1 objects are left to be chosen as a second object. By the fundamental counting principle there are n(n - 1) ways to fill the first two

26

NATURAL NUMBERS

[CHAP. I

places in our sequence. There remain n - 2 objects and so the third term may be selected ip n - 2 = n - (3 - 1) ways. There are then n(n --'- l)(n - 2) ways to select tlie first three terms of our sequence. Mter r such steps we arrive at a final factor of n - (r - 1) = n ~ r + 1 and SQ the number of permutations of n obfects r at a time is the integer P n,r of formula (22). In the special case r = n we have Pn,n =;: n!. Note that nand r are integers and that

o a. The only remaining case is that where a « 0 and b ~ O. In this case b + a is already defined, and we define a + b to be b + a. As in the case of natural numbers we say that an integer c is greater than a if c == a + b where b is a positive integer. Then c will be to the right of a in the line (1). Principle 3 in the FULL-COURSE Sec. 6 of Chap. I may now be generalized. , Principle 4. EVtII"Jj 8et oj integer8 all oj which are greater . than 80me fixed integer a contai'f1.8 a least integer. ' Principle 6. . Every set of integer8 all oj which, are les8 than 80me,fixed integer a contains a greate8t integer. We shall not try to show how these two principles are de~ved from that of Sec. 6 of Chap.!. S. The law of subtraction. It can be verified that all the nine laws of addition and multiplication are properties' of the domain. of all integers. An additional property is given as the X. LAW OF SUBTRACTION. 1J a and c are any number8, there is a unique number b which we call the cWference oj c and a and write b==c-a 8'Uch that c == a + b. The definition of addition implies that if n is any natural number the sum n + (-n) 0= O. If a < 0 and we write a == -n, we have a + n == O. We may call the number g such that a + g == 0 the mgative of a and have - ( -'7 n) ....; n. Then it may be verified that in all cases (2) -(-a) == a, c - a == c + (-a). We shall also state without verification the properties a(-b) == (-a)b == -Cab), (':"'a)(-b) == ab, -"(a + b) == (-I)(a + b) == -a - b == (-a) + '( -b), -(a - b) == b - a, -(-a - b) == a + b, . -(-a + b) == a - b, (-a)- == a-, (3) , (-a)ln+l= -(a--H),

for all integers a and b and ·positive integers n.'

37

ABSOLUTE VALUES

SEC. 4]

The processes of addition, subtraction, and multiplication will be referred to henceforth as the integral operati0n8. All the symbols used in this text and combined by integral operations will be ~sumed to o~ey the 10 laws we have set down. They will also obey the laws for nonnegative integral exponents of Sec. 9 of Chap. I as well as formulas (2) and (3). . ORAL BDRCISES

1. Compute (-1)&, (-1)8, (-2)8, (-2)'. a. Simplify the following expressions: (a) -2(z - 211) (b) 5z - [-2(71 - 2z)] (c) (-z + 1)(z 1) (d) 1 - (-2)(5 - 3) (e) 1 - 1 1 - 1 + 1 (f) 1 - (1 + 1) - (1

+ + (-z) (z)( -2)

+

3 - 1 1) - 1

+

BDRCISES

.

1. Use the method of the Oral ltxercises to simplify the following , expressIOns : (a) -( -[7(4 - 6 (b) 2[z (c) [3z -

+ 1)] -

[(5

+ 2)(-3 + 1)

+ (3 + 2)(-3 - 1)(-2) + (-3)(-4 + 8)]} (3z - 71)) + 3[ -2(z + 371) + 4(z - 71) + (-2)(2z - By)) (z + 271)2 -. (271 + z) + (-2)( -3y)][z(y + 2z)

+ zy(1 - 2) - zl) (d) (z + 71)1 - (z - 71)1 - (2z)(2y) . (e) -2[Z(ZI - 3z) + (z + 1)(z - 1)] 3z[(z 1)(z - 2) + 1] (f) (z 71)( -1 z)[z Zl - (zy + 71)] - ZI(ZI - 711) , a. Write the following expressions as the difference of the first term. minus a parenthesis enclosing the remaining terms:

+

+

+

+

(a) Zl - 718 - 2z + 371 (b) -a + 2b + 3c - d (c) 2z - 2(71 + ,) + Zl.

+

+

(d) 4z1 - Z - 71 2zy (e) z 2(71' - ,) (f) 2z - (-71 -,) - Zl

+

4. Absolute values. The ab80lute value of any natural number a is defined to be a itself. H a < 0 we define its absolute value to be the positive integer -a. Then the absolute value of a, designated by

lal

38

INTEGERS

[CHAP. IX

is a positive integer unless a = 0, and in this case We shall st,ate without proof the properties

101 = O.

la + bl ~ lal + Ibl· We also define the int~ger c = lal - Ibl and have (5) lei ~ la - bl. (4)

labl = lal Ibl,

OlUL EXERCISES

1. Find 1a:1, if (a) a: = (-1)(-1)(-2)(-3) (b) a: = (-1)(-1)(-3)(-4)

2. Compute

=3

- (4

= (-3)8

+ 6)

la :+ bl~ lal + Ibl in the following cases:

= 7, b = (b) a = 7, b == (c) a = -7, b (d) a = -7, b (a) a

(c) a: (d) a:

(e) a = b = 7 (f) a = b = -7 (g) a = 7 = -b (h) a = -7 = ~b

3

-3 =3 = -3

5. Divisibility. .As in the case of natural numbers we say that an integer b divides an integer a if a = bq where q is Btn integer. Then if b ~ 0 t~e quotient , q =

a

b

is a unique integer.' We !,!hall call b afaetor or divisor of a. If a = bq then a == (-b)( -q). Hence, if. b is a factor of a, so is -'- b. It follows that with every factorization 'of a as It product of integers there are associated other factorizations in which we insert an even number of minus signs. 'Also each factorization a = bq of a results in a factorization -a = b( -q) of -a. It follows that in order to give a complete theory of the factorizations of all integers it is sufficient to give such a theory of the factorizations of posi-' tifJe integers as products of positive factors. Eve:vy positive integer a has the trivial factorization q = a . 1 and we shall call a and 1 the trivial factors of a. An integer p > 1 will be called a prime if it has no nontrivial factors. An integer a > 1 will be called composite if it is not a prime. We now prove the following:

SlIIC.

6]

39

THE DIVISION ALGORITHM

)

Lemma 1. EVfJry nontrivial Jactor b > 0 oj a poBiti1J6 integer a iB less than a. . For if a = bqwhere b ~ a,.b ~ I we have q.~ l~q = 1+ d where d > o. Thena = b(l + rI) = b + bdwhere btl > 0, a> b. AB an immediate corollary of Lemma I we have the following: Lemma~. Let b > a ~ O. Then b divides a onZy if a = o. . Lemma S. The only divisors oJ.1 are 1 and -1. 6. The division algorithm (FULL COURSE). The division process of elementary arithmetic may be extended to hold for all integers. We state the result as the following: \ DIVISION ALGORITHM FOR INTEGERS. Let a and b be any two integers BUch that b ~ o. Th;en there are unit[U6 integers q and r Jor which 0 :ii r < Ibl BUch that (6)

a = qb + r.

To prove this result we consider the set of all of those integral multiples of h = Ibl which are greater then a. By Principle 4 of Sec. 2 there is a least such multiple. Designate it by (t + l)h. . This defines t and, since th < (t + l)h, we have th :ii a < (t + l)h. Then 0 :ii a - th < h. Put r = a - th and JJ.ave a =:= th + r = qb + r. Here q = t if b = Ibl, q = -tif b = -Ibl. To prove q and r unique we suppose that we have a second set of such integers, a = ub + v. Choose the notation so that v ~ r. Then h > v ~ r, h - r > v - r ~ 0, h ~ h - r > v - r ~ o. By Lemma 2 it is possible for h to divide v - r only when v - r = 0, v = r. Since ub + v = a = qb + r we have v - r = qb - ub = (q - u)b, Ibl divides v .:... r, v = r. Also (q - u)b = 0 for b ~ 0 and q = u. I

nlustrative E:J6ample

Compute q and r if a - -249, b = 17.

40

INTEGERS

[CRAP, II

Solution

+ 11. Hence (-14)(17) - 11 + 17 - 17 = (-15)(17) + 6, q = -15, r

By the usual division process 249 = (14)(17) -249

'.

= (-14)(17)

- 11 = .

"" 6.

EXERCISES C~mpute

q and r for the following integers:

(a) a = -180, b = 9

(b) a

= -169, b = 11

(c) a = -982, b = 21 (d) a = 279, b = -11 (e) a = 247, b =- -27

(f) a = 11,692, b = -245 (g) a = -6,942, b = -111 (h) a = -6,721, b = -100 (~') a = -4,127, b = -121 W a = -17;115, b = -489

7. The Euclidean greatest-common-divisor process If a and b are integers and c is an integer dividing a' and b, we say that.c is a common divisor of a and b. If a .;.. b = 0, any integer is a common divisor of a and b. Otherwise there is a largest positive common divisor d of a and b which we shall call their greatest common divisor (abbreviated g.c.d.). It is sometimes referred to as their highest common factor (abbreviated h.c.f.). We first prove the following: Lemma 4. Let b ~ 0 divide a. Then the g.c.d. of a and b is Ibl. , For the largest divisor of b is Ibl and Ibl divides a. We next prove the following: . Lemma 6. Let b ¢ 0 and a = qb + r. Then the com(FULL COURSE).

mon factors of a and b coincide with the common factors of b and r; a and b have the same g.c.d. as band r. For, if c divides a and b, we may write a = if, b = cg and have a - qb = if - qcg = c(J - qg) = r, f - qg is an integer, c divides r. Conversely, if c divides band r, we write b = cg, r = ch, and have a = qcg 4- ch = (qg h)c,

+

c divides a. Lemmas 4 and 5 yield a process for finding the g.c.d. of any two integers a and b not both zero. We suppose that we have labeled our integers so that b is the one not zero. Then Lemma 4 gives their g.c.d. if b divides a. Otherwise a = qb + r where 0 < r < Ibl, and if r divides b it is the g.c.d. of a and b. The process of determining

SIlO.

7]

41

THE EUCLIDEA.N PROCESS

whether or not r divides h is the division process which gives h == qlr + rl for 0 ~ rl < r. H r is not the g.e.d. of a and h, we have 0 < Tl and the g.e.d. of a and h is the g.e.d. of rand Tl. We have thereby described a division process for determining the g.c.d. with a sequence of diminishing positive divisors. Mter at most Ihl steps this· process terminates and the last nonzero remainder is our g.e.d. We shall use the notation {a, hI for the g.c.d. of a and h in the remainder of this chapter. nlustrative Examples I. Fb;I.d th~ g.c.d. of 8,381 and 1,015.

Solution We arrange the computations as follows:

29 -

8 1 3 8 232 261 1015 8381 232 232 ~ 8120 -0- -W 232 261

Am. {8381, 1015} - 29. II. Show that 14233, 884} - 17. Solution By division we find that 4,233 = 249 . 17 and that 884 = 52 • 17. Then 14233,884} - 17 if and only if 1249, 52} = 1. This latter fact follows from the computation

1 2:~' : 1~1~ ~ 24: 2683341208

0 1 2 - 3 81141

This computation may be abbreviated when we reach the remainder 11 whose only factors are 1 and 11. BXERCISBS

1. Find (a) 1191, 78} (b) {l98, 62} (c) {253,29} (d) 184, 276} (6) 192, 876} . (f) 1147,637} (0) 111682, 1072} (It) 14M,2288} (,') 12904, 312}

W 1528, 2343} Am. 2.

Am. 33.

(k) 1648, 2997}

(l) {894,3278}

Am. 298.

Am. 12. (m) 11442, 6489} (~)

Am. 49. Am. 16.

(0) (p) (q) (r)

{l432, 8469} 11134, 8019} 11608,17523} 1768, 9552} 1266664, 877769}

Am. 1. Am. 3. Am. 11,111.

42

[CHAP.

INTEGER,S

'II

2. Use the method of Inustrative Example II to show that (a) {21924,2144) = 4

(d) {5103, 7614) = 81 (e) {4526,9344) = 146 (f) {7992, 17640} = 72

(b) {179,21) = 1 (c) {4078, 814) = 2

8. Linear combinations ar'e integers, the sum

(FULL COUR,SE).

If a, b, m, n,

ma+nb

is called a linear combination of a and b. It is also, of course, a linear combination of m andn. We may prove Lemma 6. Let f and g be linear combinations of a and b. Then any linear combination of f and g is a linear combination of a and b. For we have assumed that f = ma nb, g = sa tb.

+

+

Then

+ kg = h(ma + nb) + k(sa + tb) = hma + hnb + ksa + ktb = (hm + ks)a + (hn + kt)b.

hf

We now prove Theorem 1. The g.c.d. oj a and b is the least positive linear combination

(7) of a and b.

d=ma+nb Every common divisor of a and b divides d.

For we have seen that if b divides a the number Ibl = ± 1 . b + 0 . a = d is such a linear combination. Otherwise a = qb + r, r = 1 '. a + (-q)b is a linear combination of a and b. If r ~ d, we form b = qlr + rl and rl is a linear combination of band r. :Sut rl is a linear combination of band r and hence of a and b by Lemma 6. Thus the process of Euclid gives a' sequence. of remainders each a linear combination of a and b. The last such nonzero remainder is d. Hence d = ma + nb. If w = sa + tb we see that a = aId, b = bi.d, w = (sal + tbl)d is divisible by d . . But then d is the least such positive linear c'ombination.

DC.

8]

LINEAR COMBINATIONS

43

Two integers a and' b are called relatitJely prime if their only positive common divisor is 1. Then we shall say that "a is prime to b" and, of course, that "b is prime to a." The g.c.d. of a ~d b is 1 and, by Theorem 1, there are integers m and n such that ,

(8)

ma+nb==1.

COD,versely, if ma + nb == 1, every common divisor of a and b divides 1, and a and b are relatively prime. , Theorem 2. Two integer8 a and b are relatively prime when and only when there are integer8 m and n 8Uch that ma+nb==1.' . If d is the g.c.d. of two integers. a and b, we may write d == ma + nb. But a == gd, b == hd, and d == mgd

+ nhd ==

(mg

+ nh)d,

mg +.M == 1. We thus have,the following: Theorem S. Let d be the g.c.d. of a and b 80 that a == dg, b == dh~ Then g and h are relatively prime. To find the integers m and n of formula (8) we need only find an integer m such that ma - 1 is divisible by b. rhe quotient will be -no It is only necessary to try the integers 1,2, . . . ,b - 1. When a and ~ are nQt relatively prime and we' wish to find m and n such that ma + nb == d, we use the process above for mg + nh == 1. Theorem 2 may be used to derive the following important lemma: Lemma 7. Let b divide ac and let b be prime to a. Then b divides c. For ma + nb == 1. Multiply by c to obtain mac + nbc == c. Since b divi~es ac, we have ac == bq, c == mbq

+ nbc ==

Hence b is a factor of c,

(mq

+ nc)b.

INTEGERS

[CHAP. II

~ The g.c.d. of any integer a and a prime P divides P and is .either 1 or Ipl. Hence P divides a or is prime to a. By Lemma 7, if P divides ac and does not divide a, it does divide c. If P divides abc and does not divide a it must divide bc and hence either b or c. This result may be extended to any number of factors and we state it as Theorem 4. .A. prime P divide8 a product abo if and only if it divide8 one oj the fact(1f'8 a (1f' b. 9. The fundamental theorem of arithmetic (FULL COURSE) .. The most important theorem on. positive integers may be stated as follows: . Fundamental theorem of arithmetic. Every integer a > 1 can be ~pr688ed aB· a product PiP2 • • • p, of potJitive prime facf(1f'8 Pi. ThiB factorisation i8 unique apart from the qrder oj the facf(1f'8. If a is itself It prime, our notation means that t.= 1 and we stop with the fll:st factor, a = Pl. In any factorization of a we can order our prime factors Pi so that Pi ~ P2 ~ • • • ~ p,.

Suppose then that we have a second factorization such that a = ~l1g2 ••• g. and we have ordered the prime factors so that gl :! g2 • • • ~ g•• Then the fundamental theorem states that t = 8 and that Pi = gi for every value of i from 1 to t. To prove this result we let Pi be the least positive nontrivial factor of a. Every divisor of Pi divides a and, if nontrivial, is less than Pl. Since Pi is the least ~uch divisor, it must be a prime. Then either a = Pi and we have the desired factorization or a = pial where a > al > 1. We factor out the least nontrivial factor P2 of al and have a = PiP2 or ~ = P1Plia2. where al = P2a2, al > a2 >.1. This process must terminate after a finite number of steps and yields the' desired factorization. We now assume two factorizations as above. By Theorem 4 the prime Pi divides gl • • ; g. only if it divides

PC. 10]

THE GREATEST COMMON DIVISOR

45

one of the q's. But then PI is equal to this q. Let the notation be chosen so that PI == ql' Then al == PI . • . p, == q:a • • • q. By the same argument we obtain PI == ql. After a finite number ot such steps we end with the integer 1 on one side of our equality and would end with a product of primes on the other side unless t == 8. This proves our theorem. 10. The greatest common divisor and least common multiple of several integers (JIULL COURSE). The greatest common divisor of n integers ai, aI, • • • ,a,..is defined to be the largest poSitive integer dividing all of them. It is possible to give an inductive proof of the following result: Theorem 5. Let ai, aI, • • • ,tlra be a sequence of n positive integers and designate the g.e.d. of ai, aI, ••• ,alb by dlb. Then dil.j-I = {dlb, ail.j-I} (k == 2 . . . n - 1), .

every common divisor of ai, aI, • • ., tlra divides d,.. The proof is made by observing that the common divisors of dlb and ail.j-I are common divisors of ai, as, • • • ,alb, tIlb.t-l, the common divisors of ai, • • • ,alb, ail.j-I divide dlb and tIlb.t-l. Tl;len d,; and alb+1 have the sam~ g.c.d. We shall not go into further details about the rigorous formulation of the inductive proof. . The least eommon muZtiple (abbreviated l.c.m.) of ai, aI, • • • ,tlra is the least positive integer divisible by every a. We shall prove Theorem 6. The l.e.m. of a and b is the quotient ab (9)

d

of their product by their greatest common divisor. For a == gd, b = hd where g and h are relatively prime. If m is the l.c.m. of a and b, we may write m == sa = tb, sgd == thd, sg == tho Hence h is a factor of sg and is prime to g; h divides s. Since.m == sa, we see that ha = ghd is a factor of m. But ghd is divisible by a == gd and b == hd, m

46' "

must be equal to ghd,. This product is the quotient of' formula (9). As in the case of the g.c.d. we have a result for the l.c.m. which we shall state without proof. Theorem 7. Denote the l.c.m. of aI, 'a~, . . . ak by mk. Then mk+l is equal to the l.c.m. of ak,J-I and mk. fllustrative Examples

I. Find the g.c.d. of 8,381, 1,015, 87. Solution .By a previous illustrative example {8381, 1015} = 29. divides 87, the answer is 29. II. Find the l.c.m. of 528, 792, 132.

Since 29,

Solution By our process we find that \.528, 792) = 264 and that 528 = 264· 2, 792 = 264 . 3. Hence the l.c.m. of 528 and 792 is 264 . 3 . 2 = 1,584. Also 132 is a divisor of 264 and our answer is 1,584. EXERCISES

1. Find the g.c.d. of each of the following sets of integers: (a) 102, 174, 112

(b) 378, 462, 399 • (0-) 1,296, 1,584, 936, 1,558 ~.

(d) 968, 2,057, 1,276, 348 (e) 460, 644, 1,012, 598 (f) 4,329, 8,991, 2,553, 2,109

Find the l.c.m. of each of the following sets of integers:

(a) 2, 6, 8, 10

(b) 3, 81, 15 (c) 87, 58, 174

(d) 108, 90, 135, 315 (e) 81, 54, 135, 225

11. The factors of an integer. If p is a positive prime factor of an integer a 'F 0 and e is the largest positive integer such that pB divides a, we shall say that a is exactly divisible by pB. Then in the factorization of a as plus or minus a product of positive primes, exactly e,of the prime factors will be equal to p. We may group the prime factors of an integer a so as to write ±a as a product of powers of distinct prime~. Then the fundamental theorem of arithmetic states that. these primes and their exponents are unique.

SEC.

11]

THE FACTORS OF AN INTEGER

47

The process of factoring an integer is illustrated in the example below. A systematic method of listing divisors is also given and will be useful in Chap. VI. lll~strative

°

Examples

I. Sh,ow that if n > 1 and b ;oE are integers and p is a prime there is no integer a such that aa = ph-.

SoZution Suppose that there is such an integer a. Then pba ¢

°and a cannot

be zero. Let a be exactly divisible by p" and b by pl. Then a- will be exactly divisible by p"-, b- by pI_, and pb- by pia+!. Since pb- = aa we have en ... In + 1, en - In = (6 - f)n - 1. This is impossible since n > 1 cannot be a factor of 1. . . II. Factor the integer 80,784.

SoZution We list the positive primes less than 50 for use in the exercises below. Thel are 2,3,5,7, 11, 13, 17, 19,23,29,31,37,41,43,47. Testing 2, 4, 8, 16, 3, 9, 27 in turn we find that 80,784 = 4(20,196)

= 2'(5,049} = 2' . 9(561} = 2' . 3 1 • 187

=0

2' • 31 • 11 '17.

III. Let a = plqlr where p, q, and r are distinct primes., List aU positive divisors of a. Solution The divisors of a are 1, p, pi, pI, q, pq, pig, plq, q', pq', plq', p8q', r, pr, plr, par, qr, pqr, plqr, p'qr, qlr, pqlr, p'qlr, plqlr. There are (3 + 1)(2 + 1)(1 + I) = 24 divisors. IV~ List the positive divisors of 1,800.

Solution 1,800 = 2 3 ' 51. The 36 divisors are 1, 2, 4, 8, 3, 6, 12, 24, 9, 18, 36, 72, 5, 10, 20, 40, 15, 30, 60, 120, 45, 90, 180, 360, 25, 50, 100, 200, 75, 150, 300, 600, 225, 450, 900, 1,800. Listing. them in increasing order we have 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 25, 30, 36, 40, 45, 50, 50, 72, 75, 90, 100, 120, 150, 180, 200, 225, 300, 360, 450, 600, 900,1,800. 1

INTEGEBS

[CJUP. n

BDRCIDS i~ :Faotor the following integers:

Ca) 504 (b) 2,310 (0) 1,078 (d) 32,175

(,,") 8,281 65,875 (Tc) 1,120,581 (0 15,750

(e) 5,712 (f) 31,603 (g) 1,271

W

(h) 22,607

t. List all positive integral divisors of the following numbers: (a) 504. (b) 5,712 (0) 21 • 31 • 11

(d) 28 • 31 • 71 (6) 1,084 (f) 512

(g) 1,944 (h) 936

(,1 588

W 420 (Tc) 486 (0 384

CHAPTER

m

RATIONAL, REAL, AND COMPLEX NUMBERS 1. The rea1line. The intuitive idea of ·a .positive real number is that of a magnitude, i.e., a size or a measurement of length. Basically this idea is that of regarding positive real numbers as corresponding, in a one-to-one fashion, to the points on a ray (a half line) such as in Fig. 1. The point marked 0 is called the origin of our system of measurement, i.e., the point from which we measure. The point marked 1

o

1

Q

. FIG. 1.

is our unit point and the distance from 0 to 1 is our unit of

length. The ray is unending on the right and each of its Points determines a length from the origin to the point. The number a of units of length is then the positive real number a corresponding to the point, and we label the point with the symbol a.

o FIG. 2.

1

•

It should be observed that a change in the unit point changes the correspondence between points and positive real numbers. For example, if the unit is an inch and we change it to a foot, the point marked 24 (inches) becomes 2 (feet). However, the real numbers are fixed. OnIy the corresponde~ce between point and number is variable. H we extend our ray to be a line (Fig. 2), we have wh~t is normally the intuitive idea of the set of all real numbers. Then we have both magnitude and direction. Points to the left of the origin now correspond to negative real numbers. 49

50

REAt AND COMPLEX NUMBERS

[CBAP.. m

They are stiU magnitudes but are directed in a direction opposite to. that of positive real numbers. Real nUmbers may now be added graphicaJ.ly. lor suppose that a and b are any two real numbers. Then we add a to b by taking the line segment from 0 to b and laying it off on the line segment from 0 to a so that the point on the

. o

•.

•

11+6

FIG. 3.

former segment which was at 0 is now at a. Then the point which was 'at b will be at a + 'b. For example, when a and b are positive, we have the diagram shown in Fig. 3. In case a is positive and b is negatiye, we have the diagram as in Fig. 4. Similar~y we may form a - b, by laying off the segment from 0 to b on the segment lJ:om 0 to a so that the point which was at b is at a. Then the point foI"ID.6rly at 0

.

b

o

6

G

FIG. 4.

will be at a-b. In case both a and b are positive, the diagram is as in Fig. 5. The result is then the same as that of adding a to -b. Multiplica~on of' real numbers cannot be described by this graphic method, however, and we shaJ.l pass to the , 0'

.

b

t

b G-b FIG.

o.

G

more abstract and rigorous algebraic definition of the real number system. 2. Rational numbers. If b is any positive integer, the intuitive idea. of the rational number (also caJ.led fraction or rational fraction) 1

b

51

RATIONAL NUMBERS

nc.2]

is that of the length of the line segment obtained. by dividing the segment of unit length (i.e., length 1) into b equal parts. Then if a is a natural number, the r~tional number alb is the length of a of these parts. It follows that if a is zero alb is zero. Siniilarly if a > 0 the number -alb is the length of a of these parts but directed to the left. Examples are given in Fig. 6. Then the lengths t and f are the same, - t is the negative of t in the sense that their sum is zero. The intuitive idea above may be regarded. as the inspiration of the general description we shall now present. ,

,

-I

-~

,

~

,

,

-"

o FIG.

,

, 1

6.

A rational number a is a pair of integers a and b where we write the pair as a

a='b

(re8.d. "a equals a over b"). Here a is any integer and is called. the numerator of a; b is any nonzero integer and is called the denominator of a. Our first definition of equality (i.e., of being the same pair) is now extended. and we define (1)

if and only if (2)

ad = be.

If b is a factor of a so that a = bq, the rational number alb

Is identified with the integer q. Thus our set of rational numbers is conceived. of as a number system containing the set of all integers. In particular each integer a may be replaced at any time by the rational number

a

1" Finally our definition of equality is really equivalent to the property that

52

REAL AND COMPLEX' NUMBERS

[CHAP. m

(3)

for every nonzero integer c. . In order to obtain the intuitively desirable property (4)

we define the

BUm

of any two rational numbers by ~

( 5)

b

+ d~ -_ ad bd + bc.

We also define products by (6)

We have now completed the definition of the number system of all rational numbers. It may be verified that all the laws fo:r integral operations hold and that a c ad-bc

(7)

b-a=

bd

Here the rational num1;>er

o

O=b has the property a where

+0 =a

for every a, a .

+ (-a)

=0

-(-i)=i·

(8)

Observe that the definition of addition in formula (5) and that of multiplication in formula. (6) have meaning only because a product bd of denominators b ~ 0 and d ~ 0 is not zero and so may be used as a denominator. Notice also that formula (4) may be obtained from formulas (5) and (3) by writing a +c ab"+ cb (a' + c)b a+c

b

b=

bb

=

bb

= -b-·

SEC.

53

THE DIVISION LAW

3]

Conversely we may obtain formula (5) from formula (3) and (4). For

All rational numbers may be written in one and only one of the forms

a -a 0

(9)

b' T' b ==

0,

a

where and b are positive integers. The numbers of the first kind are called the positive rational numbers (or positive fractions), the numbers of the second kind the negative rational numbers, and the single last number is the rational number zero. The notion of absolute value which we .gave for integers in Chap. II carries over and we define lal == a if a is a positive or zero rational number, lal == I-PI == Pif a is a negative rational number. ORAL EXERCISES

1. Express the following fractions in one of the forms of formula (9): (a)

1 --=3

(b) -

~ := :

(e)!

:= :

(d) 1-~ 3+2

2. What is the absolute value of each of the following fractions?

4-2

(a) 3 - 1

(b) :

:= ~

(e)!

:= :

5-5

(d) 3 - 11

8+1 +2

(e) 1 - 3

8. The division law. The rational number system has one property iIi addition to those which we have already listed in Chaps. I and II as properties of integers. We state this property as a law.

XI. LAW OF DIVISION. If a andp are any two numbers such that P is not zero, there is a unique number 'Y in the number system such that a == fJ'Y. We shall call 'Y the quotient of a by P and write a

-• . 'Y .== P

54

REAL AND COMPLEX NUMBERS

[CRAP. m

Then every rational number is the quotient of two rational numbers which happen also to be integers. Thus the rational number system may be thought of as resulting from the system of integers by adjoining to it all quotients of integers. The verification of our new law is a consequence of the . formula (10) when c ~ O.

In particular b

1

(11)

alb =

a'

for all rational numbers a and fJ ~ 0 and all integers a ~ 0, b ~ O. It is sometimes convenient to use the special case~,

a

(12)

cld

all

= c/~ =

ad

ad

Ie = c'

for 8.l1. integers a, c ~ 0, d ~ O. Formulas (10) to (12) remain correct when we replace these integers by fractions. ORAL EXERCISES

Express the following as quotients alb where a a.nd b are integers: (a)

~

(d)

~

(J)3t

(~)

l

. t

(g) 1 -

1 (0) -1-

t (e) -5t

t

1" . 1 1---

1-+

3+t ,. Fractions .in least terms. Every ·rational number except zero may be expressed as a quotient alb where b Is a positive integer and a js positive or negative according as alb is positive or negative. We say that alb is in lea8t termB if the g.c.d. of a and b ~ 1. If alb is not in least terms

SEC.-4]

55

FRACTIONS IN LEAST TERMS

and d > 1 is the g.c.d. of a and b, then a the g.c.d. of g and h is 1. Then

a

b=

= dg, b =

dh where

dg g dh = h'

and we have expressed the fraction alb as the fraction glh which is in least terms. The expression of a rational number in least terms is unique. mustrative Examl'es I. Express 11,604/15,472 in least terms. Solution We find the g.c.d. by the computation

3

3868

1

I 11604 I 15472 11604 I 11604 3868

and see that 15,472 ... 4(3,868).

Am.... II. Express the following fraction as a quotient alb in least terms: Or!

= 27 _

123

6-

1 5-

1

1 4- 3---t

Solution

We simplify as follows: 1 6-1 5 12 220-218 3 - '2 = -2- = 2' t = 5' 4 - '5 ... - 5 - == 1f' 1 5 5 90 - 5 85 1 18 ¥ = I8' 5 - 18 = ---"IS = 18' it - 85' 18 510 - 18 492 123 85 85 6 - 85 = 85 == "85' W = 123· 492 ... 4' 85 108 - 85 23 27 - 4 4 - TAm. III. Let 110 ... 15, m = 9. List all positive fractions alb where a divides 110 and b divides m.

56

REAL AND COMPLEX lfUMBER8

(OlW', IiJ

Solution

The factors of 15 are 1, 3, 5, 15 and those -of 9 are 1, 3, 9, Hence our answers are 1, 3, 5, 15, t, t, t, .., EXERCISES 1. Write the following fractions in least terms:

(J)

() 812 a 1044 , (b) () c (d) () 6

6

1,038 -2,249 -2,730 -1274 , 1,027 - 832 2,268 812

Ana. - 13'

888 744 62 57 (0) 31 108 (1&) 147 _ 101 21 105 (10') 2,175 _ 872 1,450 1,308

+

91

Ana, 36' 5

Ana'6'

81

Ana, 29'

I. Simplify the following and write in least terms: 9

Ana. - 59' 5

Ana. 28'

1

(0) 1 - 1 _

1

(1&)

Ana. -1.

1

-

1

1 1-1-+

(2-~):(3-~) 2 a 2-

3-

57

THE LAWS OF EXPONENTS

SIlO. 5)

(!_.!X~~l~ (!+-~!+D--- ---+~ 11 1

!.....J! 1

+

~ .1 2

(.' 3 7 ~V \)11 G:V\l~;

.,

_7_ 1

1--

1

.Am. 5.

3. List all positive fractions alb such that a divides n and b divides m in the following cases: (a) (b) (c) (d)

n = 9, m = 15

n = n = n = (6) n =

24, m = 8 16, m = 12 180, m = 30

-48, m

= 20

(f) n = 90, m = -42 (g) n - 504, m = 60 (/l) n = -512, m = 48 (or.") n = -1080, m = -63 W n = 240, m = 143

6. The laws of exponents. The laws of exponents which we have already given for integers hold also for rational numbers. They may be repeated as ..

= a"+tn, (an)m = a"m, an{J" = (afJ)n. Here a and P are any rational numbers, nand m are any (13)

and"

rational numbers, aD is defined to be 1 for any a ~ o. A new phenomenon now arises since every rational number a = alb has an inverse P = bla such that afJ = 1. We define the symbol. 1 a l = -ab = a-, (14) for every a = alb p!i 0 and every natural number n. It follows from our multiplication definition in formula (6) that (15)

or'" = (an)-l,

and that (16)

58


[CHAP.

m

It may now be shown that formulas (13) hold for all integers nand m. We also have the additional laws 1 _an = an-m -- a - --. (17) am It will be convenient to observe that anp p p a.-p (18) -=-, -=-, 'Y a.-'Y an'Y 'Y that is, afactor of the numerator (denominator) may be carried into the denominator (numerator) by changing the sign of its exponent. We shall not carry out the verifications of these properties. EXERCISES

1. Express the following rational -numbers as products of powers (with integer exponents) of the integers 2, 3, 5: (28 • 32 • 5)4 (a) (2 2 • 3 8 • 54)8 Ana. 26 • 3-1 • 5-8• (b)

8' (75) (2 . 4 . 3 . 45)8 (28)4 • 3-8 • 56

-(c) 2- 4 • 86 (3- 10 • 52)4

Ana. 2 . 382 • 514•

2. 8. 4)8

(2 3 5 (d) 2'34.57 (e) 54 (37 • 2 6)8(3-14 • 4-2 • 5-1) (J) (24 • 58 • 3-&)1(2-8 . 5-4 . 36)-3

Ana. 211 . 3 7 • 58.

2. Express the following numbers as quotients 2»7'" /3r 5' for integers n, m, r, t: 24.7-9 Ana. 3-9 . 5-2'

24 • 716 Ana. 38. 5-4

59

DECIMALS

SJIC.6J

6. Decimals. A finite decimal is a finite sequence of digits with a decimal point, such as 1,472.697, or the negative of such a sequence. The decimal point has the property that the integer obtained by deleting it divided by an appropriate power of 10 is the given decimal. For example, -1,472.697 = -1,~~~,697.

Since finite decimals are rational numbers, their sums, products, differences, and q'Llotients are also rational numbers. Their sums, products, and differences are also finite decimals obtainable by the procedures of elementary arithmetic. The quotient of two finite decimals is expressible as the quotient of two integers and, when reduced to least terms, is expressible as a finite decimal if and only if all prime factors of the denominator are 2 or 5. If not, the division process of elementary arithmetic yields an infinite decimal. For example t = 0.333 . . • • An infinite decimal is an infinite sequence of digits with a decimal point and a sign. Infinite decimals may be described in terms of a concept called approximation. We define the first approximation of any decimal a to be the integer that appears to the left of the decimal point in a. It is an r digit integer which may be zero, positive, or negative. Designate this approximation by ao. Define ak to be the finite decimal of r k digits formed by the first r + k digits of a. When a is also a finite decimal such as

+

643.29654,

will

the adjoined digits from some point on be zero and the approximations from some point on will be a. Indeed they are

ao

= 643, a.

al = 643.2,

= 643.296,

a4

a2

= 643.29,

= 643.2965,

G.

= 643.29654,

a=a,=a'···.

60


[cJ:UP. m

Note that our approximation a1 is 643.2 and not 643.3 even though as = 643.29. We are not rounding off digits but taking them as they actually occur. .Any decimal a may now be conceived of as being an infinite sequence

of the finite decimals ak which are its approximations. Moreover this sequence has the property that the first r + k digits of all approximations ak, aJ:+1, aAl+s • • • are exactly the same and all have the same sign. For exain.ple if a = -462.3578965942 . . • the first eight digits of all approximations ai, aa .•. are -462.35789. Let us agree finally that any infinite decimal a, in which all the digits from a certain place on are 9's, will be identified with the finite decimal obtained by deleting all these 9's and adding 1 to the digit to the left of the first in the unending sequence of 9's. For example, . -199.98999 . . . = -199.99;

0.999 . . . = 1.

.7. The real number system. We shall now define a real number to be any positive, zero, or negative infinite decimal. Let us agree that two decimals are equal real numbers if and only if their digits are either the same or one is obtained from the other by the proceSs above of deletion of 9's. The definition of real numbers which we have now given is complete and the set of all real numbers will become what is called the r6(J,l number 8Y8tem when we tell how real numbers are added, subtracted, multiplied, and divided. We shall make these definitions in terms of the following very important concept. Consider an infinite sequence act,

a1 •••

whose members (i.e., terms) are real numbers. Suppose ~hat.for every poBititJe integer k there is' a place in the sequence

SEC.

7]

THE REAL NUMBER SYSTEM:

61

(the place depends upon k) beyond which all the members have the same sign and the same first k digits: Then we shall call this real number sequence a confJergent 8equenc6. Every convergent sequence defines an infinite decimal and hence a real number whose digits are obtainable from the digits of the terms of the convergent sequence sufficiently far out. This decimal is called the limit of the sequence. The simplest example of a convergent sequence is of course the approximation sequence of a real number, and the real number is then the limit. More complicated illustrations are furnished below. Our operations on real numbers are now easily defined. Let ao, a1 ••• be the approximation sequence for a and bo, b1 • • • be the approximation sequence for a real number {3. Then it will be true that the sequence Co, C1 • • • of sums c, = ~. + b, is a convergent sequence and we define a + {3 to be its limit. Similarly a - {3 is the limit of the sequence of differences ~ - b" a(3 is the limit of the sequence of products a,.:fJ,. Finally if {3 is not zero we may delete a finite number of its terms and obtain a convergent sequence with the same limit {3 but in which all terms are not zero. For example the sequence for t is 0, 0.3, 0.33, 0.333, etc. and we would replace it by 0.3, 0.33, 0.333, 0.3333, etc. If the members of the new convergent sequence for {3 are designated by {30, {31 • • • ,the number {3, is a nonzero finite decimal and the sequence of rational numbers ~/b,' is a convergent sequence whose limit is the quotient a/{3. Let us observe that the sequences for a and {3 which we have used are sequences of rational numbers and that we have defined a + {3, a - {3, a(3 and a/{3 in terms of the limit concept and the corresponding operations for rational numbers. It is not at all obvious that the sum, difference, product, and quotient sequences defined above are convergent but we shall not try to give the proofs here. We shall also wish to state without proof that the real number system obeys all the eleven laws we have put down as well as the exponent laws. Moreover the real number

62


[CHAP.

m

system is ordered by the property that a is greater than b (b is less than a) if a - b is a positive real number. mutT«tive EumPles I. Let a "" 3.165 • . • , fJ - 2.153 . . •• Compute as much of a fJ-as the information given p8l'Qlits.

+

Solution We observe that ao at

la,

+ bo = 3 + 2 = 5,

a1

+ b1 = 3.1 + 2.1

- 5.2,

+ bs .. 3.16 + 2.15 "" 5.31, + b. "" 3.165 + 2.153 = 5.318.

Hence from the information given we are almost certain that

= 5.31 •..• Indeed, if a .. 3.1659 • • • and fJ = 2.153999 . . . ,we have a +fJ

a. + b,

= 5.3198

and so 5.318 is false. In the most extreme case where a = 3.165999 • • • , fJ = 2.153999 • . • ,

+

we have a = 3.166, fJ = 2.154 and so a fJ = 5.320 but otherwise 5.31 • • • is correct. Remark: It is true, in general, that the use of the (k l)st approximations of a and fJ will yield the true value of the kth approximation Ci of a fJ or a - fJ. However this is not true for products. Indeed suppose that a "" 300, fJ = t = 0.333 • • • so that afJ = 100. Then ao "" a1'" • • • "" 300, aobo "" 0, a1b1 = 90, asb. = 99, a,ba = 99.9, a.b, = 99.99, and we see that aobo and a1b1 are not close to the correct value of Co = ~ "" 100. n. Define a real number by the formula ao = 0.4,

+

+

"'+1 III ac

+ (0.8)(0.5 -

a,).

Compute the first five approximations of a and estimate its value. Solution a1 = 0.4 (0.8)(0.1) = 0.48, as = 0.48 (0.8)(0.02) = 0.496, a. = 0.496 0.8(0.004) = 0.4992, a. = 0.4992 (0.8)(0.0008) = 0.49984, at = 0.49984 0.8(0.00016) = 0.499968.

+ + + + +

Am. a = 0.5.

Ro.8]

POSITIVE REAL POWERS

63

Remark: If r " 0 is any finite decimal between 0 and 0.5 and we define ao = r, a'+1 = ex. + 2r(0.5 - a,), the sequence is convergent and has 0.5 as limit. It converges slowly when r is near zero, that is, we need to take many terms to get near the limit t. As we have seen above, it converges rapidly when r is close to 0.5. EXERCISES

•

1. Compute the first eight approximations to the number defined by ao'" 0.3, ",+1 = a, + 0.6(0.5 - a,). z. Compute the first eleven approximations to the number defined byao ... 0.1, ",+1 = a, + 0.~(0.5 - a.).

S. Positive real powers of positive real numbers. Every Positive real number a determines a unique positive real number b ==

Va

(read" b equals the nth root of a") for every positive integer n. Then b is defined to be that positive real number whose nth power 1i'is a. We shall give a formula for b in Sec. 13. We shall prove that there is only one such number for every 11. and a in Chap. VII. Note that it is customary to write vb instead of ~b and that ~ == b. The symbol Vb is called a radical. Computations with radicals are not so simple as those with exponents and we prefer to use the fractional exponent form given by (19)

(read the "nth root of a equals a to the one over nth power"). We also define the fractional power (2O)

amlf' == (a1!"}m

(read am!" as "a to the m over nth power") for all rational nl1mbers min, and see that also am!" == (am}1!". (21)

For [(alJf'}m]" == (a1lf')- == [(alJf'}"]m == am and so (a1lf'}m is.the nth root (amp'" of am. We may now define the positive real power a" for every positive real number a and every real number n. Let no,

64


[CHAP. III

nl, n2 • • . be an approximation sequence for n so that every Then every ~ is a rational number and defines a positive real number bi = an, The sequence bo, bl • • • can be shown to be a convergent sequence with a positive limit b. We define an to be this limit . . - The student will have observed by now that, as we have pointed out where they occur, our exposition lacks a number of verifications. The material left out is suitable only for much more advanced courses and it would not be worth while to present it here. However, we do feel that what we have given should do much to give the student a clear picture of the meaning of the basic concept of a positive real power of a real number. We shall now state the following important extension of our laws of exponents without proof: Theorem 1. The positive real powers of positive real numni is a finite decimal.

bers satisfy the laws of exponents

(an)m

= anm ;

anbn

= (ab)n,

for all positive real numbers a and b and all real numbers n andm.

The laws of exponents for the cases where the exponents are fractions lin and where n ¢ 0 is an integer may be expressed as certain formulas for operation with radicals. These are (23)

_" '!'7:.

-V'ab = yaW;

VVa

=

_"I-~r: _ vava -

,,~ r:

va;

"Van+m •.

It is important to observe that

bVa = ',ybna. So, for example, ~ = ~ = 2~ since 8 = 28. Also we may wish to insert a number within a radical and so write' .

SEC.

8)

POSITIVE REAL POWERS

65

2 ~ == -¢'2i ~ == {I2 8a/2 == -¢Ta Of course, in all of these formulas the numbers a and b are positive and we cannot overemphasize that the radical8

are all positive. If a is negative and n is an odd integer, we may write a'== -d where d is positive.

Then if n is odd,

a l '" == (-d)l'" == -d l '" is a unique negative real number. However a l is not a real number since (al )2 is positive whenever al is real and must be negative if a is negative. The meaning of afl for a negative and n real is obviously an exceedingly complicated notion and we shall not discuss it further in this text. We shall discuss the meaning of a l ' " for integers n and negative real numbers a in Chap. VIII. nlustTative Emmples I. Express the number a ... -¢'2 V5 {13 as an integer raised to fractional power. Solution

-¢'2 = 21, V5 = 51, {13 = 31. The least common denominator of the exponents is 12 and we write a = 2ft . 5& . 3& = (2" 51 . 31)0. II. Simplify a =

V8 + 3 V2 -

4 vTs

+ 5 V'4.

Solution a

V8 = Vi V2 = 2 .y2, vTs = 3 V2, -¢'4 = (21)1 = 21 ... V2, = 2V2+3V2-12V2+5V2 = (2+3-12+5)V2= -2V2.

•

EXERCISES

1. Express the following numbers -(a) ..v27 {164 Am. (72)1. (b) V3-¢'2 (c) V2 V3 {/4' 27 Am. (3,888)t. (d) ~ Vi5 -¢'8

-¢'36

(6) V2

Am. (~62)1.

88

powers of a single integer-:

66


l. Write the following in the form

[CB.A.P. m

Va where n is a positive integer:

(a) 2V2 (b) 6 Vi (c) 10~ (d) 6 {1 (e)

~~

S. Simplify the following expressions:

+ v'24

(a) 2 Vi2 - 4 V27 4 (b) -6~+ lO~ + {IIOO (c) (~+ {12)(~ - ~) -?"i2 (d) (VS - VW)(Vi VW) (e) (~-~)~+~

+

+

-

{liS

~

{'6 {II6O -. (J) v'2 + -¢'3 - ~

a

9. Logarithms (FULL COURSE). Let be any fixed real number greater than unity. Then it can be shown that for every positive real nlimber c there exists an exponent n such that

a" = c. We shall give a formula for computing n in Sec. 12. The exponent n is determined uniquely by c (and the fixed a) and is called the logarithm oj c to the baBe a. We shall desip,ate it by log. c and so have the definitive property The most commonly used base for computational purposes is the base a = 10. Thus a table of common logarithms is a table of numbers c and the corresponding exponents n = 10glo c such that 10" = c. For example 101 = 100 so that 10glo 100 = 2. Also the logarithm of 2 to the base 10 is approximately 0.30103, that is, 100.80108 is approximately equal to 2. Since logarithms are exponents, the laws of exponents may be translated into laws for logarithms. Let e arid d

SEC.

9]

LOGARITHMS ,

67

be any positive . real numbers and write c = an, d = am where the n = log" c, m = log" d. Then cd = anam = an+tn. Thus log" cd = n + m and we have the formula log" cd

(24)

=

log" c + log" d.

In words, we have the foUowing property: PROPERTY 1. The logarithm oj a product is the sum oj the logarithms oj its Jactors. We similarly have c/d = an/am = a-. From the meanings of nand m we have (25)

log"

c

d

=

loga c - log" d.

This result may be stated as follows: PROPERTY II. The logarithm oj a quotient is the logarithm oj the numerator minus the logarithm· oj the denominator. If g = r!" and loga c = n then g = (an)m = anm so that log" r!" = mn; i.e., log" r!"

(26)

=

m log" c.

This result may be stated as follows: PROPERTY III. The logarithm oj a power- r!" is the product oj the exponent m by the logarithm oj c. Property III holds for all real exponents and may be· used in particular for fractional exponents. For example, it states that loglo v'2 = loglo 2! = t 10glO 2 = 0.15051 approximately. . . Let: us observe now that every positive real number c may be expressed· as a product c. = 10th, where h is a real number between 1 and 10. Indeed h may be obtained from a decimal expression for c by moving the decimal point to the right of the first (beginning at the left) nonzero digit of c. For example, 145.632

= 102 (1.45632),

0.00435 = lO-a(4.35).

68


Then loglo

1()& = t

m

and loglo c = t

(27)

[CRAP.

+ 10glO h.

It would then be sufficient to construct a table of approximate values of the l~garithms to the base 10 of all k-place decimals between 1 and 10 in order to have a table of approximate values of th,e logarit~ .of all numbers. Such tables are used in trigonometry. The integer t is called the characteristic of loglo c. It may be any positive or negative' integer. The real number loglo h is a positive number between 0 and 1 .called the manti8sa of loglo c. For example, loglo 0.2 = -1 + 0.30103, loglo 2,000,000

loglo 200 = 2.30103,

= 6.30103.

Computation with logarithms is best treated in a course in trigo:Q,ometry and we shall not discuss it further here. In that branch of mathematics called the calculus a base other than 10 is more convenient. The base used is the real number 6

= 2.7182818285

. . . •

We shall give a definition of e in Sec. 12. Logarithms to the base 6 are called N apierian or natural logarithms and are sometimes designated by the convenient notation J.D. c (read simply as "log natural of c"), while common logarithms are usually designated by log a. This latter notation is used in some calculus texts instead of In c. nlustrative Examples

I. What is the characteristic of log 341.621

Solution The characteristic of loglo a is 1 less, algebraically, than the number of digits to the left of the decimal point in a. Hence our answer is 2. II. Give the five-place logarithms of 2, 0.2, 0.02, 0.00002.

Solution The numb'er of digits to the left of the decimal point in 2 is 1. Hence the cl,1aracteristic is O. The digits to the left of the decimal point in

SEC.

69

LOGARITHMS

9]

0.2, 0.02, 0.00002 are, respectively, 0, -1, -4, and their characteristics are thus -1, :""2, -5. Indeed 0.2 = 10-1 (2), 0.02 = 10-'(2), 0.00002 = 10-1(2). Hence log 2 = 0.30103, log 0.2 = 0.30103 - 1, log 0.02 = 0.30103 - 2, log 0.00002 = 0.30103 - 5. ORAL EXERCISES 1. Give the characteristics of the common logarithms of the following numbers: (a) (b) (c) (d)

34,156.2 32,134,569 0.00123 0.142

(e) (I) (g) (h)

1.423 0.0213 0.0000003 56.0000012

(I) (g)

IOg10

2. Give the following logarithms: (a) log, 8

(b) log2 V8 (c) log2~ (d) log, {14 1 ,(e) log2

.y2

1000 IOg10 (0.001) (h) IOg10 (0.01) ViO . (i) loglo (0.001) ~

W log. ev'2

3; Find :!lif (a) log,:lJ = 3 (b) loga:lJ = 2 (c) log,:lJ = -2

(d) logs:lJ = t (e) log,7:lJ = t (f) log,,:lJ = -l

4. Find:lJ if (a) (b) (c) (d)

log., log., log., log.,

100 = 2 e' = 2 e' = 3 100 ,.;, 3

(e) log., 64 = 3 (f) log., 64 = 6 (g) log., 2 = t (h) log., 9 = -i

EXERCISES

1. Given that log 2 ing: (a) (b) (c) (d) (e)

log 4 log 8 log 0.5 = log t log 5 = log ¥ log 9

= 0.301 and log 3 = 0.477, (f) (g) (h) (i)

W

compute the follow-

log 6 log 12 log t log t log-7!'-

2. Using the data computed above, approximate the following to two decimal places:

70

REAL AND COMPLEX !:iUMBERS

= approx.log.yoo (b) log 11 - approx. log (c) log 13 = approx.log -¢'2200 (d) log 17 = approx.log V2oo, if log 29

(a) log 7

[CHAP. m

Vl25

=

1.4624

10. Change of base (FULL COUBSE). If the base of a system of logarithms is replaced by a new base, the logarithms in the new system are related very simply to those in the old system. Indeed, let a be the old base and b be the new base. Then a == b" where r == .10g" a. If e is any positive number and e == a!' == 11" where n == log,. e and m == log" e then e == b"' so that m == m. This gives the formula

10gb e == (log,. e)(log" a),

(28)

that is, the logarithm of e to the base b is the product of the logarithm of e to the base a Iby the logarithm of a to the base b. Since it is true that log" b == 1 no matter what the value of the number b > 1, we have loga b log" a == 1, . that is, the numbers 10gb a and lo~ b are reciproc$. .particular 1 ' loge (29) In 10 == 1 - ' In e == (log e) (In 10) == 1 - '

,

~e

In

-~e

for all positive numbers e. Actual computation gives the approximate formulas . log 6 == 0.4342945,

In 10 == 2.302585.

BXDCISBS

Use the value In 10 = 2.30 and the values given in Sec. 9 to compute the following: (a) In 2 (b) In 3

(c) In I) (d) In 90

(e) In 60 (j) In t

(g) In Tr .(ll) In n

11. Irrati.onality of real numbers. A real number is ea11ed an irrational number if it is not a rationaZ number. ·If

SEC.Il]

IRRATIONALITY OF REAL NUMBERS

*

71

*

n is any integer greater than 1 and p is a positive prime integer, the number is irrational. -For let alb == where a and b are integers. Then a == b( vp), a" == b"p. This was shown to be i.ID.possible in an illustrative example at the end of Chap. II. ' It is not a simple matter to show whether or not a given real number is rational, but we may at least develop a criterion for this property. We call a decimal a repeating decimal oj period n if there is a group of n digits which are repeated from a certain place on to give all of the infinite sequence of digits of the decimal. For example, 361.793842568425684256 . . • is a repeating decimal of period five if we assume that all digits not shown are obtained by repeating the group of digits 84256. We may now give a formal proof of the criterion. . Theorem 2. An infinite decimal i8 a rational number iJ and only iJ it i8 a repeating decimal. IJ n iB the period oj d then there i8 a finite decimal b BUCk that (30)

d

b

== 10" - 1"

If we multiply a repeating decimal of period n by 10, the result is the same sequence of digits but with the decimal point moved one place to the right. Then d is a repeating decimal of period n if and only if n is·the least integer such tbat the digits of lO"d from a certain place on are exactly the same as the correspondingly placed digits of d. Then b == lO"d - d == (10" - l)d is a finite decimal and we have formula (30). . Conversely, let us write a rational number d as a quotient cle of two integers c and e > o. Then d will be a repeating decimal if we can show that there exists an integer n such that (10" - l)d' is a finite decimal. We observe that the only remainders on division of integers by e are the e numbers 0, 1, 2, • . . ,e - 1 and therefore that the e + 1

72


[CHAP. III

numbers 1, 10, lOs, • • . , 108 cannot all have different remainders. Then'there are two integers g and h such that e ~ g > h ~ 0 and such that 1(}11 has the same remainder on division by e as does lOla. Then 1(}11 - lOla will be divisible bye. Write g - h = n for a positive integer n and have 1(}11 - 10" = 10"(10n - 1) = eq where q is an integer. Then (10n

_

l)d - 1QB(lOn - l)e _ qee _.!JE.. - b lO"e - lO"e - 10" - •

Since q and e are integers, the number b is a finite decimal. This proves that d is a repeating decimal. EXERCISES

1. Express the following fractions as repeating decimaJs by actual division and state the length of the smallest period for each: (a) (b)

t t

(0) (d)

t

(e)

-fr

(I> +r

fi

(g) (h)

-it it

2. Express the following repeating decimals as rational numbers in least terms: (a) (b) (0) (d)

1.714285714285 ••• 3.70370 ••• 0.076923076923 ••. 0.1212 •••

(e) 0.120120 ••• Ans.+h. (f) 0.132132 • . . Ans.&. (g) 0.10241024 ••• Ans.tHt. (h) 2.111211121 •.• Ans.-w#-.

12. The sigma and pi symbols. A simple notation for 8" of the :first n terms of a sequence ai, as . . . is

the sum.

8n =at+as+'"

+an

(read It at, plus as, plus and so on, plus an "). For n = 1 this sum reduces to its first term, i.e., 8t = at. Another and more compact notation for the same sum is one which involves the use of the capital Greek letter sigma. It is frequently used in mathematics and is

SEC.

(read "the sum for i equals 1 to n of a/'). n

i

k==

73

T'HE SIGMA AND PI SYMBOLS

121

~ is used for the sum of the first n

The symbol

+ 1 terms of a

0 n

sequence whose first term is ao, and the more general ~ lJi '''III

is the sum of terms beginning with the mt~ term and proceeding through consecutive integer subscripts to the nth 10

term. For example

k ~ = a5 + as + a7 + as + ag + alO. i=5

We shall use this symbol only for m < n. The product Pn of the first n terms of a sequence a1, a2 • • • is representable by (read "al times a2 times and; so on times an"). AI! above Pl = ale We also have a notation using the Greek letter pi, which is

(read "the product for i equals 1 to n of a/'). uct

n

lli

i=O

where m

The prod-

n ai = amam-rl . . . an

= aOa1 • • . an, and

i = m

< n. ORAL EXERCISES 't

1. Read off the meaning of

k a, in the following cases: i .. 1

(a) (b) (c) (d)

a, = i a, = 2i a, = 2i - 1 a, = 2i 1

(e) a, =

+ 3i + 1

(g) a, = C7 •• (h) a, '" box' (i) = ~_.x, W a, = bi-lx7 -,

a,

3

(k)

k j' i" a, = k iyi i

a. =

1

7

(l)

= 1

74


{CHAP.

m

I. Read off· the meaning· of the following sums and products: 6

:t

(a)

(it - 1)

(e)

, .. 1

·5

(b)

10

n (1 +~)' n (!, __ . ,+

(J)

, .. 1

·8

(c)

1 ) 1

i .. 1 10

(d) i

(1

• (1i - i +1) :t. 1

, .. 1

(g)

1)

:

ni~1

i ... 1 10

n 2,-1 2' + 1 •

i .. 1 4. •

n i3 -:

,3-1

~1 i - , + 1

(h)

1

,- 1

18. Infinite series and products. Every infinite sequence aI, az • • • of real numbers determines an accompanying sequence 81, 8z • • • where

(31)

.8.

• = i=l :t a; = a1 +

az + • • ~ +

an

is the sum of the first n terms of the original sequence. Such a sequence of ~ums is called an infinite 8eries or simply a serie8 and it is sometimes designated by the notation a1+ aZ+" • or by the notation

(32)

•

:ta; i-I

(read" the sum for i equals 1 to ihfinity of the ~"). A series (32) is said to oorwerge if the sequence of real numbers 81, 8z • • • defined by formula (31) is a convergent sequence. Then the real number 8 which is the limit of the sequence 81, 8. • • • is called the sum of the series and we write . .

We shall not discuss such nonconvergent series as 1+2+3+" . which diverge nor 1 - 1 + 1 - 1 + . . . which o8cillate~

SEC.

IS]

75

INFINITE SERIES 'AND PRODUCTS

Every infinite sequence a1, all • • • also defines an infinite sequence P 1, PI. • • of products n

(33)

P"

n ~. '-1

=

Then we call the sequence P 1, PI. • • an infinite product and designate it by alaI • •• or, better, by

.

H the sequence Ph PI •.• converges and its limit is the real number P, we write·

.

P= n~.

(34)

i .. 1

Infinite series and products are of importance for the computation of various mathematical constants. For example, the ratio 'Ir of the circumference of any circle to its diameter may be computed by the use of (35)

1 1 1(12 + 31) + 51(12 + 31) - ~(i7 + i7) + .

4 == 2 + 3 - 3

'Ir

so that 'Ir/4 ==

i

k

lID

1

3

5

3

5

~ in which ~ == (-I~~11/m(z-m

+ 3-)

and m = 2i - 1. ' The base e of N apierian logarithms may be computed by the use of (36)

1+ -3!1+ .. . . + -n!1+ . .'. •

e == 1 + 1 + -2!

To compute the natural logarithm of any positive real number a, we proceed as follows. H a > 1 the real number 1 1-(37)

a-I a+l

a

:x;.=--=--

1+! a

76

REAL AND COMPLEX NUMBER·S

[CHAP. In

is positive and less than one, we may compute our logarithm by the formula

(38)

In a = 2 ( x

+ "3X8 + 5"x" :- '7X + . .. ) . 7

If 0 < a < 1 then b = 1/a > 1 and In a = - In b. Evidently the formula loglo a, = In a ·loglo e may be used with formula (38) to compute commop logarithms. The positive nth root of a positive real number may be computed by the use of the so-called binomial series 1

The (r

+ mx + m(m2 ,

1) X2

+

+ l)st term of this series is m~ ~ ~~(m.:.:...---.!.l)_._....,...~(m_---.:...:...n----:+~l)

-

nl

xr

and it will converge and have (1 + x)m as its limit if -1 < x < 1. To compute the nth root of a positive real number a, we write a = bn ' + c = bn(l

+ x),

c x=bn O. Then the ordinary so-called long-diviBion prOce88 begins with the subtraction J(z) - aoZ-bo-1g(z).

This eliminates the term aoZ", and so yields a polynomial of degree at most n - 1. A finite number of such subtractions of multiples of g(z) reduces the degree' of the final difierence r(i) below that of g(z). Then we have formula (5) where q(z) begins with aobo-lz- and has the degree stated in the algorithm.

SEC.

5)

REDUCIBILITY OF POLYNOMIALS

95

To show that q(x) and rex) are unique we suppose that + t(x) where the degree of t(x) is less than m. Then by Theorem 1 the degree of

f(x) = 8(X) . g(x)

rex) - t(x) = [8(X) - q(x)]g(x)

is less than m.

This is not possible, by Theorem 2, unless For g(x) has degree m. Hence

8(X) - q(x) = O.

8(X)

=

q(x),

= 0, rex) = t(x). We note that when c is any number and g(x) = x - c the remainder polynomial rex) has degree zero and is a constant r,

rex) - t(x)

f(x)

(6)

= q(x)'. (x

- c)

+ r.

When g(x) is a quadratic polynomial x 2 remainder polynomial rex) is linear, (7)

f(x)

=

q(x)(x 2

+ ax + b,

the

+ ax + b) + ex + d

for constants c and d. The student will have had practice in the division process from his elementary algebra experience. If further drill is deemed desirable, it is best provided for by the exercises of Sec. 7 on the g.c.d. process. 6. Reducibility of polynomials. A polynomial f(x) is divisible by a polynomial g(x) if f(x) = g(x)q(x) where q(x) is a polynomial. When g(x) is a nonconstant polYnomial, this occurs only if the remainder rex) of the division algorithm is the zero polynomial. The equality f(x) = g(x)q(x) is called a factorization of f(x) with polynomial factors (or divisors) g(x) and q(x). Every polynomial has the trivial factqrizations (8)

f(x)

= a[(r~(x)]

for any number a ;;t: O. Thus every polynomial has all nonzero constants and all nonzero constant multiples of itself as trivial factor8.

96

PO LYNOMIALS

[CHAP. IV

If a polynomial has no factorizations except its trivial factorizations, we call it an irreducible (or a prime) polynomial. All linear polynomials are irreducible. When a polynomial does have nontrivial factorizations, we call it a reducible polynomial. For example Z2 -

3z

+ 2 ==

(z - 1) (z - 2)

is reducible. The polynomial Z2 - 2 == (z + '\1'2) (z ~ V2) and so is a reducible polynomial if we permit the use of irrational coefficients. However if we insist that only rational coefficients be employed this polynomial is irreducible. For otherwise we Inay use the result about leading coefficients in Theorem 2 to write Z2 - 2 == (z - a)(z - b) for rational numbers a and b. Then a2

-

2 == (a - a)(a - b) == 0,

a2 == 2. This has already been shown to be impossible. It should be clear now that the question of the reducibility or irreducibility of a polynomial depends upon the nature of the coefficients that Inay be used. We shall assume throughout the remainder of this chapter that the coefficients will be restricted to lie in some fixed field F of complex numbers. The most important cases are those where F is the field of all rational numbers or the field of all re8J. numbers or the field of all complex numbers. We shall thus develop results holding for any of these cases. S. The factorization theorems. A reducible polynomial J(z) has degree 11. '> 1 and a nontrivial factorization J(z) == g(z)q(z).

Neither of the nontrivial fagtors g(z) and q(z) is a constant and the degree of J(z) is the sum of the degrees of these factors. We phrase this result as follows: ·Lemma 1. The degree oj a nontrivial Jactor oj J(z) i8 le" than the degree oj J(z).

SEC.

6]

THE FACTORIZATION THEOREMS

97

If f(x) = g(x)q(x) and g(x) = h(x)s(x) are nontrivial factorizations then f(x) = h(x)s(x)q(x) is a nontrivial factorization. in which we have three factors. Lemma 1 implies a lowering of degrees. Such a process must terminate and can terminate only when we arrive at a factorization in which all factors are irreducible. This gives us the first of our factorization theorems. Theorem 4. Every reducible polynomial is a product f(x) = Pl(X)P2(X) ••• Pt(x) of a finite number of irreducible polynomials.

By Theorem 2, the leading coefficient of f(x) is the product of the leading coefficients of its factors. We may thus remove these coefficients and carry their product to the left so as to obtain the following: Theorem 6. Every reducible polynomial is a product (9) of its leading coefficient ao and a finite number t of irreducible polynomials Pi(X) each having leading coefficient unity. We shall present a proof later that there is essentially only one factorization of the kind in formula (9). Of course

the order of the factors is not unique. (x - l)(x - 2)(x

+ l)(x -

l)(x - 3)

= (x - 3)(x - 2)(x

For example

+ l)(x -

l)(x - 1).

However, both the irreducible polynomials Pi(X) with leading coefficient unity and the number of times they occur in formula (9) are unique. If an irreducible polynomial p(x) occurs exactly e times as a factor of f(x) in formula (9), we may group these factors together and write the product as p(X)8. Then p(X)8 will divide f(x) and p(X)8+1 will not. We shall call e the multiplicity of the irreducible factor p(x) of f(x). EXERCISES

1. In the following polynomials the irreducible' factors are linear. Write the polynomials as products of constants by powers of distinct

98

POLYNOMIALS

[CHAP. IV

linear factors with leading coefficient one and state the multiplicity of each factor. (a) (b) (c) (d) (II) (J)

+

+

(z - l}(zl - 33; 2}(Z' - z - 2}(z 2}4 (z - l}'(z l}l(z - l}(z - 2}4 (2z 1}'(33; 1}(2z 1}(4:r;1 - I) . z(z l}8zI(zl - l}'(z' zl) (:r;4 - 33;8 2z1}(Z4 - z8)(ZI - 2z) (z 1}8(z - l}l(zl - 1}4(z' - 9}~(Z' 4:r; 3)' (Zl - 2z - 3}(2:r;1

+ + + + + +

+ +

+ +

+ 4:r; -

6).

I. In the following polynomials the factors, as written, are powers of polynomials irreducible in the field of rational numbers. Give the multiplicity of all the irreducible factors with leading coefficient unity. (a) (b) (c) . (d) (II) (J)

+ 1}8(z - 1}1(2z1 + 2}(z + I) - 2:r; - 2}'(2z' - 4}8z + 2}(z. + 2}(ZI + 2)

(Zl Z8(ZI (Zl (Zl (33; (2z1

2}I(ZI - 2}1(2:r;. - 4)' 6}I(Z - 1}4(z - 2)' 4}(ZI 2}(4:r;1 8)&

+

+

+

7. The EucHdean g.c.d. proc~ss

(FULL COURSE).

A c0m.-

mon diviBor of two or more polynomials is a polynomial

factor of all of them. The g.c.d. of several polynomials, not all zero, is that common divisor of largest degree and leading coefficient unity. It can be shown that there is only one such polynomial. The g.c.d. of two polynomials J(:x;) and g(:x;) ~ 0 may be computed by a polynomial analogue of the process of Sec. 7 of Chap. II. We first prove the following: Lemma I. Let g(:x;) ~ 0 be a divi80r oj J(:x;). Then the g.e.d. oj J(:x;) and g(:x;) is a eomtant multiple oj g(:x;). For every common divisor of J(:x;) and g(:x;) divides g(:x;). Thus the degree of their g.e.d. cannot exceed that of g(:x;). However g(:x;) is a common divisor of J(:x;) and g(:x;) and has

the maximum degree possible. Lemma 3. Let g(:x;) ~ 0, a and b be any nonuro

+

c0n.-

stants, J(:x;) = q(:x;)g(:x;) r(:x;). Then the common divisor8 oj J(:x;) and g(:x;) eoincide with the common· diviBor8 oj ag(:x;) and br(:x;). Hence the g.e.do oj J(:x;) and g(:x;) is that oj ag(:x;) and br(:x;).

800.7]

THE EUCLIDEAN G.C.D. PROCESS

99

For, if c(x) divides ag(x) and br(x), it divides f(x) == q(x)g(x)

1- b-1br(x)

and ag(x). Conversely, if c(x) divides f(x) and g(x), it divides ag(x) and br(x) == b[J(x) - q(x)g(x)]. Lemmas 2 and 3 result in a process for computing the g.c.d. d(x) of f(x) and g(x) ~ O. If g(x) divides f€x), the product of g(x) by the inverse of its leading coefficient is d(x). Otherwise we use the division algorithm to :find the polynomial rex) of Lemma 3. (The constants a and b are used to avoid fractions in the case where we study polynomials with integral coefficients.) If rex) divides g(x), some constant multiple of it is d(x). If not we divide g(x) by rex) and get a new remainder of lower degree. The process terminates with a :final nonzero remainder which is, apart from a constant factor, the required g.c.d. Let hex), f2(X); ••• ,fe(x) be any nonzero polynomials and ~(x) be the g.c.d. of hex): . . . ,h(x). Then it may be shown that the g.c.d. di+l(X) of hex), ••• ,h(x), fi+l(x) is the g.c.d. of d,(x) andfi+l(x). Thus the division pro.cess above may be utilized to :find the g.c.d. of any number t of polynomials. . In the rather rare case where polynomials are given in factored form, their g.c.d. may be computed trivially. Indeed it is the product of all the prime polynomials occurring in the various factorizations but with exponents the least that occur for the individual primes in all the polynomials. For example, let hex) = (x - 1)8(X 1- 1)2(X2 1- 1)(2x - 4)4, hex) == (x - 1)6(X 1- 1)8(2x - 4)4, f8(X) == (2x - 4)3(X - 1)4(X2 1- 1)3(X 1- 1)2.

Then the least exponent for x - 1 is 3, the least for (x 1- 1)2 is 2, the least for X2 1- 1 is 0, the least for x - 2 is 3. Thus the g.c.d. is (x - 1)3(X 1-·1)2(X - 2)3. Thls procedure is of relatively little use and so we shall not dwell on it further.

100

POLYNOMIALS

[CJLU>.

IV

The l.c.m. of two nonzero polynomials f(:r;) and g(:r;) is that polynomial M(:r;) with leading coefficient unity which is divisible by both f(:r;) and g(:r;) and whose degree is the smaJIest of aJl nonzero polynomials divisible by both f(:r;) and g(:r;). It may be shown that, if ao, bo are the respective leading coefficients of f(:r;) and g(:r;) and 'if fl(:r;) is their g.c.d., then f(:r;)g(:r;) aobod(:r;)

is their l.c.m. M(:r;). The l.c.m. of a set of polynomials /l(x), .•. ,f,(:r;) may be computed by using the property that if M,(:r;) is the l.c.m. of /l(:r;), ••• , f.(:r;)· then the l.c.m. of /l(:r;), ••• , f'+I(:r;) is the l.c.m. of M,(:r;) and f,+I(:r;). The computation of the l.c.m. is not particularly important and we shaJl only illustrate it below. DIU8trative Examples

I. Find the g.c.d. of z' - 2:z:1 - 21; - 1 and Zl - 2:z:a - 2:z:1 - 3:z: - 2.

Solution Zl

-2 a + ZZ4 - 2:z:I + lz~.

.+2

- 2:z: - lzl - 2z' - 2:z:1 - 3:z: - 2 za+zz' +Zl zl-2z4 -2zl - z o _2:z:8 zl-2z 2z'-2z' -2z-2 _21;8 -2z 2z'-4:J:a -4:J:-2 -Zl -1 21;8 +2z

Hence the g.c.d. is Zl

+ 1.

Note that since z is not a. factor of

g=z'-2:z:'-2z-1

+

+

the g.c.d of g and Zl z is the same as the g.c.d. of g and Zl 1. So we could save one step in the process. However this follows only when we have the factorization theory of Bee. 9. II. Find the g.c.d. of Zl + 2:z:1 - Z - 2, Z8 - Sa: + 2, .:Z:B + z - 2. Remar1D~ It would seem simplest to find the ,g.c.d. of Zl - 3:z: 2 and Zl + z - 2 and then the g.c.d. of the result of this first step and Zl + 2:z:1 - Z - 2. .

+

SEC.

THE EUCLIDEAN G.C.D. PROCESS

7]

101

Solution _ _--=!Z2+3Z+2 . z2+z+2 1 + z - 2 Z8 z - 1 Z8 + 2Z2 - Z - 2 z - 1 Z8 Z8 - Z2 Z3 - Z2 Z8 --Z2 + Z ' 3z2 - Z 3z2 -3z Z2_ Z 2z-2 --2z-2 2z-2 2z-2

o

' - 3z + 2 + z - 2 - 43: + 4

0 Am. G.c.d. = z - 1.

ITI. Find the l.c.m. of the polynomials in Illustrative Example IT.

Solution

+

= Z8 z - 2 = (Z2 + z + 2)(z Z8 - 3z + 2 = (Z2 + z - 2)(z so that Ml = (Z2 + z + 2)(Z2 + z - 2)(z - 1). fa = Z8 + 2Z2 - Z - 2 = (Z2 + 3z + 2)(z - 1) . = (z Ml = (Zl z + 2)(z + 2)(z -

/I

+

- 1), 1), Now + 1)(z + 2)(z - 1), 1)2

and, by using either this factored form or the g.c.d. process, we find that the g.c.d. of Ml and!8 is (z + 2)(z - 1). Then

M = (Z2 + z + 2)(z + 2)(z - 1)2(z + 1). EXERCISES

1. Find the g.c.d. of each of the following pairs of polynomials: (a) Z8 - 3z2 + 4

z8_2:l: 2 _Z+2

(f) z, + z + 2 z'-z2-1

(b) Z8 - Z2 + 2 Z8 - Z2 z - 1

Am. 1. (g) z, - Z8 - 7z 2 + z + 6 Z8 + Zll - 4:1; - 4

(c) z, - 3z 2 + 2

(k) z8 - 2:/:6

+

Ana. Z2

Z4 - Z2 - 2 (d) Z4 + Z8 - Zll + z - 2

(~)

Z4 + Z8 - 3z 2 - Z + 2

+ 43:2 + 1 z, - 4:1;2 + 8z - 7

(e) z,

W

+ Zl + 2:/:2 -

+ 3z + 2.

Z- 1 Z8 + z& + Z4 - Z - 2 Am. Z4 - 1. Z8 - 2:/: + 4 Z8 - 2:/: 4 + 43:3 + Z2 - 2:1: + 2 Am. Z2 - 2z + 2. z4-2z 3 +z 2 -1 z& - &8 + 2:/:2 - 1

Am. 1.

-102

POLYNOMIALS

[CHAP. IV

2. Find the g.o.d. of each of the following sets of polynomials: (c) Z4 - 23;3 - 2z1 - 2z - 3

(a) Z4 - 1

+

Z8 + 3z2 + 3z + 1 z8-2z 2 -3z

Z8 2z' - z - 2 z4-2:I)8+2z2 -2z+1

.

Ans. z - 1.

. (b) Z4 - 2:1)8 - 5z2 + 6z

Am. 1. (d) z& + 2Z4 - Z8 - 5z2 - 6z - 3

Z4 - 7z2 + 6z z4+43;8+3z2 -43;-4

Z8 + Z2 + 3z + 3 z4+ z 3 - z -1

Am. z - 1.

Am. z

+ 1.

(6) Z6 - 11z2 - 9z - 2

z8-3z 2 +3z-2 Zl1 - 2z& - 121z3 - 198z2 - 100z - 24 (f) Z8 + Z1 - 3z& - 43;4 + z.8 + 3z - 3 :l)6_:I)4+Z2_2:I)+2 z8 + 2:1)1 - 3:1)6 - 7z4 - 2:1)8 + 3z + 6 (g) :1)1 + 6z8 + 20 Z8 - 3z8 - 10 :1)8 - 2:1)4 - Z8 - 43; - 6 . . (h) Zl - :1)8 - 3:1)1 + 3z8 - z& + 43;4 - 5z2 + 43; - 2 :1)4 + :1)8 - :1)1 + 2z - 3 :1)1-2:1)+1

8. Linear combinations (FULL COURSE). A linear combination of two polynomials f(x) and g(x) is an expression a(x)f(x)

1- b(x)g(x),

where a(x) and b(x) are also polynomials in x. Then we have the following: Lemma 4. Let u(x) and v(x) be linear combinatio'M of f(x) and g(x). Then any linear combination of u(x) and v(x) is also a linear combination of f(x) and g(x). For suppose that u == af + bg, v == cj + dg, where all the letters represent polynomials in x. Then 8'U + tv == saf + sbg + tcj + tdg == (sa + tc)f + (sb + td)g is a linear combination of f and g•. We may use the simple result to prove the following.: Theorem 6. The g.c.d. of two nonzero polynomials f(x) and g(x) is their only common divisor of the form (10)

d(x) == a(x)f(x)

+ b(x)g(:e).,

which has leading coejficient one. "

SEC.

8]

LINEAR COMBINATIONS

103

For the first step in our g.c.d. process expresses f(x) in the form f(x) = q(x)g(x) + rex). Then g(x) = 0 . f(x)

+ 1 . g(x)

and rex) = 1 . f(x) + [-q(x)]g(x) are both linear combinations of f(x) and g(x). Each step in our process will then begin with a dividend and divisor polynomial which are linear combinations of f(x) and g(x) and yield a remainder polynomial which is also a linear combination of f(x) and g(x). Thus d(x), which is a constant multiple of the final remainder, has the form (10). Every common divisor c(x) of f(x) and g(x) divides d(x) if formula (10) holds. n c(x) = A(x)f(x) + B(x)g(x), then d(x) will divide c(x). But c(x) and 'd(x) differ by a constant factor and are equal if they both have leading coefficient 1. The result just proved implies that ,the g.c.d. d(x) is unique. For any other common divisor of the same degree aEl d(x) would divide d(x) and would then differ from d(x) by a constant factor just as does c(x) in the proof above. The problem of actually determining the polynomials a(x) and b(x) of formula (10) is easily solved but we shall not emphasize it by a set of exercises. n exercises are wanted, the polynomials of the Exercise 1 of Sec. 7 may be employed. We illustrate the procedure below. . nlu.trative Example , Find the polynomials a(z), b(z) of formula (10) for g(z) = Z4 - 2z8 - 23: - 1, I(z) = z& - 2z' - 2z1 - 3:r; - 2.

Solution By Dlustrative Example I of Sec. 7 we have g = (z a = (z -

a

1 = (z + 2)g + 2r,

+

2)r 80 that 2r = 1 - (z 2)g, 2)r - g = (z - 2)trJ - (z + 2)g] - g = !(z - 2)/

- !(Zl - 4)g - g = i(z - 2)J + i(2 - Zl)g. To check we compute (z - 2)1 + (2 - Zl)g = :r;I - 2z' - 2z' - 8z1 - 2z - 2z& + 4:/:8 + 4:/:1 + 6:r; + 4 + 2z' - 4:/:' - 4:/: - 2 - Zl + 2z& + 2z1 + Zl =2z l +2-2d

104

POLYNOMIALS

and have shown that a(z) of formula (10).

= Hz -

2), b(z)

[CHAP. IV

= t(2 -

Z2)

are solutions

'

9. The unique factoriZation theorem (FULL COURSE). Two nonzero polynomials f(x) and g(x) are called relatively prime if their only common factors are constants. Then their g.c.d. is 1 and, by Theorem 6; there are polynomials a(x) and b(x) such that (11)

a(x)f(x)

+ b(x)g(x)

== 1.

Conversely when formula (11) holds, every common divisor of f(x) and g(x) divides 1 and is a constant. We have proved the following: Theorem 7. Two nonzero polynomia18 f(x) and g(x) are relatively prime if and only if there are polynomia18 a(x) and b(x) 8UCh that formula (11) hold8.

We then have the following: Theorem 8. Let f(x) and g(x) be relatively prime aJ1,d let f(x) divide the product g(x)h(x).

Then f(x) divide8 hex).

For we may use formula _(11) to obtain h(x) == [a(x)f(x)

+ b(x)g(x)]h(x)

==' a(x)h(x)f(x)

+ b(x)g(x)h(x).

By hypothesis, g(x)h(x) == q(x)f(x), hex)

== [a(x)h(x)

+ b(x)q(x)]f(x)

as desired. Theorem 9. Let f(x) be irreduc?'ole.

Then f(x) and g(x) ' are either relatively prime or f(x) divide8 g(x). For the g.c.d., of f(x) and g(x) divides f(x). If it is 1,

the polynomials are relatively prime. Otherwise it is a multiple of f(x) and divides g(x), so doesf(x). Theorem 10. An irreduc?'ole polynomial divide8 a product of polynomia18 if and only if it divide8 one of the factor8. Theorem 8 implies that f(x) divides a product only if f(x) and at least one of the factors of the product are not relatively prime. By Theorem 9, if f(x) is irreducible it

divides this product.

Be. 10] POLYNOMIALS IN SEVERAL SYMB OLS 105

We may now use Theorem 10 to prove that the factor:ization in Theorem 5 is unique. The result is trivial for polynomials of degree 1, and we assume that it is true for po1ynomials of degree less than the degree n of fez) = aOP1(z), . . . ,p,(z) = aoql(z), ••• ,q.(z). Here the polynomials Pl(Z), .•• ,p,(z), q,(z), ••• ,q.(z) are irreducible and have leading coefficient 1. By Theorem 10, Pl(Z) divides one of the q,(z) and, since both are nonconstant irreducible polynomials with leading coefficient 1, they must be equal. This proves that the factorization formula (9) of Theorem 5 is unique apart from the order in which we write the factors. 10. Polynomials in several symbols. A polynomial in z and y is an algebraic expressionf(z, y) (read "f of z and y") obtained by formally applying a finite number of integral operations to z, y and constants. It may be regarded as a polynomial in z and so written uniquely in the form (12)

fez, y) = ao(y):r

+ al(y):r-1 + . . . + a".(y).

.

The coefficients a.(y) will, of course, not be constants but polynomials in y, and m is the degree in z of fez, y); ao(Y) is not the zero polynomial unless fez, y) is the zero polynomial in z and y. The polynomialf(z, y) may also be written in the form (13)

fez, y) = bo(z)y'

+ b1(z)y-l + . . . + b,(z).

The coefficients bo(z) , ••• , b,(z) are then unique polynomials in z, t is the degree in y of fez, y), bo(z) is not zero unless fez, y) is the zero polynomial. Both formulas (12) and (13) ~xpress fez, y) as a sum of constant multiples of power products ziyi. The two expressions differ only in the manner in which we have ordered and grouped the terms. If we igno~ these groupings, we may define the degree of ziyi in both z and y to be i + j and the degree of fez, y) in z and y to be the largest degree of

106

POLYNOMIALS

[CHAP. IV

any of its terms with cOlU!tant coefficient not zero. example, I(x, y) = 3x4 + 5X8y8 + 6x2(y6 + 2y4) + y8

For

I

is'a polynomial of degree four in x, six in' y, seven iIi x a~d y. It can be written also as y6 + 6xly6 + 12x2y4 + 5x8y3 3x4 or as 6x8y6 (5X 8y8 + 12xly4 + y6) + 3x4, the laSt expression being an arrangement according to descending degrees in x and y. If all the terms of \ a polynomial have the same degree in all the symbols x, y, z •.•• , we call the polynomial a homogeneou8 polynomial. For example, 3x2 + 2Xy - 5y2 is homogeneou8 01 degree two in x and y. Every polynomial I(x, y) of degree n can be written as a sum

+

+

(14) I(x, y)

= lo(x, y)

+ /1 (x, y) + ... + In(x, y)

of homogeneous polynomials/i(x, y) of degree n "'7'" i. The leading polynomial lo(x, y) has degree n and is not zero unless I(x, y) = o. An example of such an expression is I(x, y) == (x6y

.

+ 3X8y4 + 6xly6) + (2x4y2 + 3xy6)

in which lo(x, y) /1(x, y) 12(X, y) h(x, y) /&(x, y) 16(X, y)

.

+ (2xly -

+ +

y8)

+ (3x 2 -

xy)

+

== x6y 3X8y4 6x 2y6, == 2x4y2 3xy6, == 18(X, y) = 0,

= 2xly -

y8, 2 = 3x - xy, == 17(X, y) = o.

Note that in our example we have written each homogeneous polynomial as a polynomial in x with coefficients constant multiples· of powers of y. Homogeneoq,s polynomials are sometimes called lorms. . The general polynomial I(x, y) of degree two is Qf considerable importance in other branches of elementary mathematics. It may be written in the fo~ (15) Ax2 + BX'Jj + Cy2 + Dx Ey F

+ +

SEC.

10]

POLYNOMIALS IN SEVERAL SYMBOLS

107

for constants A, B, 0, D, E, F which are usually real numbers. The de~tions above can be extended to polynomials in any number of symbols. When the number of symbols is large, it is customary to use Xl, X2, • • • ,Xn rather than X, y, s, t • .•. We shall be interested later principally in linear functions '

and in quadratic forms allXl 2

+ a22Z2 2 + . . . + a.nXn2 + 2a12XIX2 + + 2a-InXn-IXn. ORAL EXBRCISES

1. Give the degree in z, in y, and in z and y together of the following polynomials: (a)· &;Iy + &;'111 - 2zy8 - 3y' (b) -2:/:1 zlyl - Zl &;'y zill 5 (c) -zly! zy' &;& 2z'y (d) -2:/:iI' + 6z'1/' + &'yB - 2:/:11 (8) 4:/:1 3zy - 2y1 5y - 3z 1 2. Express each of the polynomials above in each of the forms (12), (13), (14). 3. Which of the polynomials above are forms? 4. Give the values of A, B, 0, D, E, F of formula (15) for each of the polynomials: . (a) &;! - 2zy 4y1 &; - 2y1 1 (b) y' 3y 1 2:/:1 - 4:/: 3 (c) (z + 3y)(z - 4y) + 2 (d) (2z + y)(2:/: - y) - (&; + 2y)(&; - 2y) (8) 4(z 3y)1 - 5(z 3y)(3:/l - y) 2(z - By)1 (J) 2(4:/: 3y)! - (4:/: 3y)(3:/l - 4y) (&; - 4y)1 2(3z - 4y) - (z 3y) 1 6. What are the degrees of the following polynomials? (a) Z8 3:/l1y 3s2 - 4ta ",a (b) (z 3y)(z - 21)' ",10 (c) (z + yl)1 - t B • \ (d) (Zl til)' - (z + 11 + S)I

+ +

+

+

+ + +

+

+

+ + + + + + +

+ +

+

+

+ +

+

+

+

+

+ + +

+

+

+

108

[ClLU'.

POLYNOMIALS

IV

6. Verify the following formulas by direct multiplication or use of earlier formulas in the list: (a) Zl - yl = (z y)(z - y) (b) Z8 - 11' = (z - Y)(ZI + zy + yl) . (c) Zl y' = (z Y)(ZI - zy yl)

+ + + + (d) z' - 'Y' = (Zl - yl)(ZI + yl) = (z - y)(z + Y)(ZI + yl) (6) Z8 + 11 = (Zl + yl)(Z' - zlyl + y') of n.

For let K be the set of all positive integers for which formula (6) holds. Since f(l) == al the integer 1 is in K. If Ie is in K thenf(le) == al + ... + a. and f(le)

+ a.....l

Replace n by Ie . f(1e and so Ie

== al

+ . . . + a. + a.....l.

+ 1 in formula (5) to obtain

+ 1) == f(le) + a.....l == al + ... + tlJl.f-l +1

is in K. The principle of mathematicol induction states that K is the set of all positive integers, that is, formula (6) is true for all values of n. .;

. lUustrative EumjJlu I. Prove that 1 2 n - n(n 1)/2. . ..R6fnmo1c: We are trying to prove that the sum of the first n integers is eq~a1 to n(n 1)/2,'

+ + ... +

+

+

SoZution In this sequence a. - n and J(n) = n(n J(I) ... 1 . 2/2 -.1 ... 1.11. .Also

+

1) n(n 1) (n - l)n J( ). J( n -. n == 2 2.

==

+ 1)/2.

n(n

Then

1.11

== 1,

+ 1 2- n -+ 1) 2ft

-2==n==a..,

and our result follows from the summation theorem. II. Prove that if r " 1 then 1 r == (,... - 1)/(r - 1).

+ + ... + ,...-1

In this sequence a.. ..

r·- 1,

SoZution J(n)

1.11 -

= (,... -

r 1- 1

1)/(r - 1). Then

1,

Also (,... - 1) - (,...-1 - 1) r.- 1(r - 1) J(n) - J(n - 1) = r _ 1 .. r _ 1 == a..

andJ(I) ... (r - 1)/(r - 1) - 1 ... ..

...

1.11.

as desired.

m. Prove that 1 • 3 + 2 • 4: + . . . + n(n + 2) .-

.. in(n

+ 1)(2n + 7).

121

SUMMATION FORMULAS

BEc.2]

Solution HereJ(1) -= 1(2·9) =- 3 ... al and J(n) - J(n - 1) =- i[n(n + 1)(2n + 7) - (n --1)n(2n + 5)] = 1n(2n1 + 9ft + 7 - 2n1 + 5) =- ..",(6n + 12) ... n(n + 2)

an

= a-

as desired. EXERCISBS 1. Prove the following summation formulas by the use of the summation theorem: -

+ + ...+ + + ... +

(a) 1 3 (2n - 1) = n l (b) 1 5 (4n - 3) = n(2n - 1) (c) l' + 21 + . . . + nl = in(n + 1)(2n + 1) (d) 11 + 31 + ... + (2n - 1)1 = In(4n1 - 1) (6) 21 + 41 + ... + (2n)l = fn(n + 1)(2n + 1)

(J) l' + 2' + ... + n' = [n(n:- 1)T

= nl(2n1 - 1) 1 1 1 n (h) 1. 2 + N + ... + nCn + 1) = n + 1 1 1 1 n (,") 1· 3 + 3 . 5 + . . . + (2n - 1)(2n + 1) =- 2n + 1 (J1 1· 2 . 3 + 2 . 3 . 4 + • . . + n(n + 1)(n + 2) (g) l' + 3' + . . . + (2n - 1)'

n(n

+ 1)(n + 2)(n + 3) 4

(k) 2·4 + 3 . 5 + . . . + (1

n

-t n)(3 + n) = 6 (2n1 +

(Z) 2· 10 + 3 . 11 + . . . + (1 + n)(9 + n) =

15ft + 31)

n

6 (2nl +

33n + 85)

(m) (6 + 1)(J + 1) + (6 + 2)(1 + 2) + . . . + (6 + n'(J + n) = "6n (n - 1)(2n - 1) + n(n 2- 1) (6 + 1 + 2) + n(6 + 1)([ + 1) (n) 11 .2+21 .3+ ... +nl (n+1) =-nn(n+1)(n+2)(3n+1) (0) 1· 4 + 4 . 7 + ... + (3n - 2)(3n + 1) = n(3nl + 3n - 2) 1 2 n (P) 3. 3 . 5 + 4 . 5 . 6 + ... + (n + 2)(n + 3)(n + 4) n(n 1) = 6(n 3)(n 4)

+

(q) (r)

(8) (t)

1 1 1 n 1· 4 + 4·7 + ..• + (3n - 2)(3n + 1) = 3n + 1 1· 2 + 2 . 21 + . . . + n l = (n - 1)2..+1 + 2 ! + t + .... + =13 + 31 + • • • + 3" = 1(3..+1 - 3)

m..

m..

+

+

122 '

IDENTITIES AND APPLICATIONS

[ClL\P.v

I. Prove (a) in Exercise 1 by subtraction of two formulas derived . from IDustrative Example L 8. Prove (d) in Exercise 1 by subtraction of two formulas derived from (0). '" Derive the formula. 1/;"

+ 1/;"-'1/ + . . . + :1;"-'11' + . . . + 1/.

0=

1/;,,+1 - 1/,,+1 ~_L..1/;-1/

by replaCing r by 1/;/1/ in mustrative Example n. .. I. Derive a. pairof corresponding formulas in the ca.ses where n 9r 2m 1 by replacing r by -1/;/1/.

+

= 2m

S. The method of unc:letermined coe:fJlcjents (FUI40 COt:llJSE). The student may- become Qurious as to how the l!Q.IIUDf'tion formulas he has proved were -obtained. We sh8.u .,atisfy this curiosity in the case where a. is a polynomial in n and shall thereby introduce the so-called method oj undet8rmined coejficientB. Let a. = g(n) be a polynomial of degree m. Then the exercises of Sec. 2 suggest that

+ cln'" + · .. + c......l is a polynomial of degree m + 1 whose coefficients are to be J(n) = conm+l

determined. By the summation theorem the expression (7)

co[nm+l - (n - l)m+l]

+' Cl[~'" -

(n - 1)"']

+ .. · + c".[n -

(n - 1)] - g(n)

must be identically zero and we must also have (8)

J(l)

= Co + . . . + c......l = gel).

The difference has degree at most m. By using the binomial theorem we obtain the term eo(m + l)n'" and may select Co so that the degree of the expression in formula (7) is at m9st m - 1. We then use this value of Co and select Cl in clmn--1 so that the degre'e of the expression in formula (7) is at most m - 2. The sequence of determinations terminates with the determination of c". so that the constant term of the expression in formula (7) is zero. We then select c......l = g(l) - (co + . . . c".) and have the desired summation formula.

+

BlBc.3]

123

UNDETERMINED COEFFICIENTS

It is really only necessary to use the process above to obtain formUlas for the sums of powers '

+ ~ + ... + nil. Indeed, if a,. == bon'" + b n-- + . . . + b"" then al + all + ... + am == bo(I'" + 2'" + ... + 11,"') + b1(1--1 + 2--1 + . . . + 11,--1) + . . . + b1(1 + 2 + . . . + 11,) 111

1

1

nb", and we may substitute the polynomial expressions for 111 + ~ + ... + nil, taken where ~ == 1, ••• , m, if they have already been obtained. nlustrative Example, I. Find a formula for 1· 5 + 2 . 9 + ••. + n(4:1& + 1) by the method of undetermined coefficients.

Solution Here a. == 4:1&1 + n and we wish a polynomial such that a + b + c + d = 5 and a[Z8 - (z - 1)']

+ b[Z' -

(z -'1)2]

+c=

tJII:I

+

ba;1

+

+ z. 1)1 ... raz -

c:J)

+ d

4z1

Then Zl - (z - 1)' == 3z1 - 3z + 1, Zl - (z 3a == 4:, ~ == t, -3(t) + 2b .. 1, 2b -= 5, b == t, t that c ... ¥ - t .. t. Finally, .

1 and 0 so

t +c =

a+b+c+d=t+¥+t+d=d+5=~

d == O. Hence!(n) ... In(Snt + 100 + 7) == in(n + I)(Sn + 7). II. Find !(n) in mustrative Example I above by the use of the formulaa for 1 + ... + n and 11 + ... + nt.

a. ==

4:1&'

Solution

+n and so

!(n) == 4(1 + ••• + nl) + (1 + ••• + n) ...

I + n(n:- 1) == n(n

t 1)

[(Sn

III. Find a formula for l' + 2' +

+ 4:) +

64: n(n +

1)(2n + 1)

3] == n(n + I~Sn + 7).

. . ~ + n'.

Solution We wish a[zl - (z - I)&]

+ b[z' -

(z - 1)4] + C[ZI - (z - 1)1] + d[Z' - (z - I),] + e[z - (z - 1)] - z'.

124

IDENTITIES AND ,APPLICATIONS

[ClU.P.v

,

Thus a(5:r;' - lOs-

Hence a ==

+ 10s1 -

1,

+ 1) + b(~a - &;1 + ~ - 1) + O(SsI - Ss + 1) + d(2:.; - 1) + e - :.;'. -2 + 4b == 0 and b ... ~; 2 - 3 + 30 = 0 and c ... ~,

-1 e

=

~

111

+ ~ - 1 + 2d == 0 and d == 0, 5 - 2 + 3 + e ... 0,

15 - 10 - 6 1 • 30 == - 30· Finally

a +b+0

+ d + e +1== 6 + 15

t,

10 - 1

+1,1" - :~.

We have

shown that l'

+ . . . + n' == anI + 100' +30IOn' -

n - 27

.

BDRCISBS 1. Use the method of undetermined coefficients to derive for al + ... + a.. in the following cases: (a) (b)

(0) (d) (e)

a.. - (n - l)(n' + 2)

(J)

a.. = (2n + 1)(2n - 3) . a.. ... n(3n - 1) a.. = 2n1 - nl a.. = (nl + l)(n - 1)

(g) (n) (i)

&

formula

a.. == n(n + l~(n + 2) a.. ... nl + 3n - 1 a.. - nl + 2n - 1 a.. ... n' - n

W a..

== nl

2. Derive the formulas in (a)-($) of Exercise 1 by the use of previously determined formulas for sums of powers.

4. Arithmetic progressions. Our sum.pl&tion formulas may be applied to obtain the sum of n terms of two kinds of special sequence called progre8BiO'fUl. The first of these is the arithmetic progresBion in which the difference d = ~l

(9)

-

a.

oj any two consecutive t6Nn8 is a jig;ed constant. We shall call d the common difference in the' progression. Let us now obtain a formula for the general term of an arithmetic progression. H al is the first term, the second term is all == al + d, the third term is as = al + 2d, and so forth. Thus it follows that the general term is (10)

CJ,a

== al

+ (n -

l)d.

no. 4]

ARITHMETIC PROGRESSIONS

125

The sum of n terms of an arithmetic progression will be denoted by B.. Since each term has a summand ai, we have B. == nal + [0 + 1 + 2 + ... + (n - l)]d. By using the illustrative Example I of Sec. 2 with n replaced by n - 1, we have . n(n - 1) n (11) B. == 001 + 2 d == 2 [2a l + (n - 1)d]. Since an == al yield

+ (n -

l)d, this formula may be modified to

(12) The finite sequence al; ..• , an may be thought of as having (al + an)/2 as an average term and formula (12) states that the sum is the number of terms multiplied by the average term. An arithmetic progression may be used to define other arithmetic progressions with the same common difference but with a different starting place. For example, if ai, at . • • is an arithmetic progression, so is bl , bs . . • where bl == ali, bt == as, . . . , b. == a.....4 • • • • We may also extend an arithmetic progression by adjoining the terms al + (n - l)d in which n takes on negative integral values. These are the terms al - d, as - 2d • • • and form an arithmetic progression by themselves with common difference - d. In the study of finite arithmetic progressions of n terms the number n is the number of term8 and it is usual to call al the fir8t term and a. the laat term. In cases where we wish to compute B., but a. and d are given, it is simplest to reverse the progression. Then it will become a progr.ession bl , • • • , b. in which bl == a. and the common difference is --d. Since b, == ~l we have

B. ==

al

+ . . . + a. == bl + . . . + b., _ n(n - 1) d 8 .-~ 2 .

126

IDE NTITIES AND APPLICATIONS

[CRAP. V

If two numbers a and b and a positive integer m are given, we can construct an arithmetic progression of n = m + 2 terms in which a = al is the first term and b = llm+. is th~ last term. The m intermediate terms are called a set of m arithmetic means between a and b. The common difference is

b-a

b-a

n-1

m+l

d=--=--,

(13)

and the means are (14)

b-a b-a a + m + l' a + 2 m + l'

When m

=

b-a

,a+m m +l"

1, the single mean is called the arithmetic mean

of a and b, its value is a

t b., It is the average of a and b'.

ntustrative Examples I. Find d and SlO if al alO

= al +

= -8 and alO = 19.

Solution 9d = -8 + 9d = 19, 9d = 27, d

SlO = ¥(al

= 3.

Also

+ a..) = 5 . 11 = 55.

II. Find'd and a8 if al == 15, SlO = 60. Solution SlO = ¥(2 . 15 + 9d) = 5(30 + 9d) = 60. 'Hence 9d = -18, d = -2. Then a. = 15+ 7d = 1. III. Find al, alO, SlO if al = -3 and au = 17.

9d + 30 = 12.

Solution Write a& = bl and au = bu. Then bu = bl lOd = -3 lOd = 17, d = 2. This is the same common difference as in (it, a2, • • • , and 80 al = al 4d = -3 = al + 8, al = -11 9d ='7, SlO = ¥( ..... 11 + 7) = -20. IV. Insert six arithmetic means between -5 and 23.

+ +

+

+

Solution al = -5, a. = 23 80 that 23 = -5 + 7d and 'd = 4. The means are -1,3,7, 11, 15, 19. V. In an arithmetic pro~on SlO = 115 and d == 3. Find alO.

ARITHMETIC PROGRESSIONS

127

Bolution

a

The reversed :progression has alO .. bit = -3, and the same sum + 9d) = 5(2bl - 27). "Then 23 = 2bl - 27, alD = bl ... 25.

BlO - 115 - 5(2b1

BDRCISBS 1. In an arithmetic progression find the following: (a)

B., as if al = 3, a = -4

Ana. B. B11, as if al = 5, a ... 2 (e) Bu, al' if al = -3", a - 3 Ana. B" (d) Bu , au if al .. -6, as = 6

0=

-88, as ... -17.

(b)

= 231,

au ... 36.

(e) Be, a, if /J7 ... 13, au ... 28

Ana. B, = 63, a. - 1.

M B., au if at = 12, as = -12 (g) a, as if Ba ... 108, al == 24

Ana. a ... -3, as ... 3. a, a7 if Bl = 25, al ... 20 ('1 a, al if B• ... 90, at = 14 Ana. a ... 1, al - 6. W a, au if B• ... -108, a, ... -15 (1:) al, at if B• ... 117, a = 6 Ana. al = -11, at ... 37. (I) al, au if B1I =- 138, a = 5 (h)

O

(m) n, al if B.. = 153, tl - 2, al integral

I. Insert m arithmetic means between a and b in the following cases: (a) m .. 6, a = 3, b ... -11 Ana. 1, -1, -3, -5, -7, -9. " (b) m = 5, a ... -2, b = 4 (e) m = 7, a .. 6, b = 10 A"". 6t, 7, 7t, 8, st, 9, 9t. (d) m = 11, a = 18, b = 14 (e) m ... 8, a = 10, b - -17 , Ana. 7, 4,1, -2, -5, -8; -11, -14. m ... 9, a ... "-9, b = -7 (U) m = 3, a ... -6, b ... 6 Ana. -3, 0, 3. (1&) m = 14, a = 0, b ... -5 ('1 m = 5, a = 16, b = -8 Ana. 12, 8, 4, 0, -4. W m=8,a-200,b=20

(n

.

128

InlllNTITUIS AND APPLICATIONS

.[CIIAP. V

6. Geometric progressions. A geometric progre8sion is a sequence aI, aI, • • • in which the rati(,)

r = ~ tli of any two consecutive terms is a constant. It is called the common ratio for the progression. Then tli.tl = rtli, al == ral, aa = ral· =. rIal, and we see that . (15)

(16)

By Illustrative Example II of Sec. 2 we see that the sum of n terms of the progression is given by

fA - 1 = antI - 1ale B. = al r - 1 r-

(17)

Just as finite arithmetic progressions may be reversed so may finite geometric progressions be reversed. This is done by writing bl = a. and Writing 11r for the common ratio instead of r. Thus, for the reversed progression, (llr)· - 1 1 - r" B. = a. (llr) _ 1 = a. '-1(1 - ·r)· A set of m nlimbers aI, • • • , a.n.t-l is called a set of m geometric meam between. a and b if a = aI, aI, • • • , tJn...rl, tJn...r1 = b form a geometric progression. Then ,.-+1

=£ a

and, if we restrict our attention to the case where a and b are positive real numbers, r is the positive (m + l)st root ef b/ a. The means are then ar, ar·, . • • , arm. nlustrative Examples I. Find al-and.S. in a·geometric pro~n if al == 2 and r == 1. Solution 1

a. == 2(1)' == 8'

1 at =- 16;

a,-al

-h-2

S. == 1 - 1 == 1 - 1

=2 (2 -

1) 31 31 16 == 2 16 ... S.

II. Find al and S. in a geometric progression if Ii., == 192 and r == 2.

GEOMETRIC PROGRESSIONS

SEc. 5]

0Ir

= a128 = 3 • 2~, al

129

Solution = 3 . 2& = 96 so that

= 3 and a.

S& = 96 - 3 = 93.

III. Find r in a geometric progression if al

=i

and S6"1= ¥-.

Solution S&

=i

e& ~ i) = 3i.

Evidently we cannot hope to solve the

equation r8 - 1 = 31(r - 1) except for integral values by substitution and we obtain r = 2 as a solution. IV. Find r and n in a geometric progression if al = 2, a.. = - 54,

.

&=-~

Solution -

a.. = 2r,,-1 ... -54, r,,-l == -27, r" =

-27r and

-40(r - 1) = al(r" - 1) = (-27r - 1)2.

Then -27r - 1 = -2Or + 20 and -7r = 21, r = 3, n = 4.

= -3, (-3),,-1 =

-27,

n- 1

EXERCISES

1. In a geometric progreBBion find (a) S., a, if al = 8, r = i Am. S, (b) S8, a& if al = 16, r = -i (c) Sa, a4 if al = t, r = 2 Am. S. (el) al, S& if aa = t, r = -! (6) a7, S, if a4 .. iT, r = -i Am. a7 (/) at if as = 42, a7 = 5,250 (g) as if au = 1,080, a14 = 233,280

(h) r, n if al = 2, a" = 8, Sn = 3 (1.) n if al = -1, r = 2, S" ... -63

= ¥, a. = t. = ¥, a4 = 1. = - vh, S, = - tn.

Am. as

= 5.

Am. n

=

6.

(3) r, n if al = 3, a" = 48, S,. = 93

2. Insert ~ geometric means between a and b in the following~: (a) m = 4, a = 1, b = 32 (b) m = 5, a = 2, b = iT (c) m = 3, a = 18, b = i

(el) m

= 7, a = ..;."

b

=3

(6) m = 5, a ... 2, b =

t

(J) ~ ... 2,(J==3,b ... 41

Am. 2, 4, 8, 16. Am. 6, 2, Am.

t.

V22, 1, 2 V21 1 ' 2' 2 V2'

lao:

IDENTITIES AND APPLICATIONS

[ClIAP.v

8. Harmonic progressions. A S'equen'ce (h, all • • • is said to form a harmonic progr68Bion if aU the numbers a. are not zero, and the sequence' of reciprucals 1 1

(18)

is an arithmetic progression. Then 'd = ~ _

(19)

a..rl

!

a.

==

1. _.! == (h all

al

- flll alai

for every i and

1.. == 1.. + (n _ 1) (a l tin

al

-

all) ==,all + (n -alaIl)(fll -

al ),

,

alaI

so that (20)

The terms of a hal'Illonic progression may be computed by the use of this formula, but it is prefera.ble to pass to the cottesponding arithmetic progression and make the computations directly by the Use of the theory of arithmetic progressions. See the illustrative examples below for this procedure. There is no formula for the sum-or n te1'mS of a. harmonic progression. However, harmonic m6an8 are defined in a fashion ~na1ogous to the definition for arithmetic means. Thus, if a and b are any two nonzero numbers, we may form a harmonic progression of n ==. m + 2 terms in which al == a, a....t-li == b,

a-b

d == (m + 1)00' The m arithmetic means between 11a and lib are then computed and their reciprocals are the m harmonic means between a and b. . lUustrtltiv. EZtmlpl.B I. In a harmonic progression find Ga, Gl0 if GI == -

t, Gil ... ir.

BEc.6]

HARMONIC PROGRESSIONS

131

Solution

The elements c, = -:-3, Cle = 17 form. the fifth and sixteenth terms of an arithmetic progression in which d = 2 by the IDustrative Example III of Sec. 4. Then in that example Cl = -11 and

=

Cs

-11

+ 2d = -7,

= -11 + 9d =

CIO

7.

Hence a, = -t, alO = t. II. Insert six harmonic means between -i and

n.

Solution We form. an arithmetic progression of eight terms with al == -5, as = 23. Then 23 == -5 + 7d and d ... 4, the arithmetic means are -1,3,7,11,15, 19. The required harmonic means are -1, i, t,

n.

~,n.

BDRCISBS

1. In a harmonic progression find

=-

(a) ai, ai, a, if al i, a, = t (b) a" a, if a., = -h, all = iT

(e) as, all if as = -h, as = (d) at, au if a. = 1, a, = 3 (e) al, a., if as = t, a, = h (f) au = if al = -h, a7 0= i (g) lJI8 if al = -h, as =

"*

Tr:

Am. -

i, 1,t.

Am. ae

= - i, all == -

n.

Am. --Ir.

S. Insert m harmonic means between a and b in the following cases: (a) . (b) (c) (d) (e)

m

= 2, a = 1, b ... 3

Am.

t, t .

m = 5, a = i, b = 1 m == 7, a = 1, b ... 2 Am. tt, tt, tt, tt, tt, tt, JoI. m == 4, a = i, b = -h m = 11, a = iT, b = t S. Describe each of the following progressions by the appropriate term "arithmetic," "geometric," "harmonic" or none of these: (a) 1, (b) 1, (e) 1, (d) i,

(e) (J) (g)

(1&) (t.')

i, t, t, T • . • i, t, T, n ... i, 0, - i, - 1 •••

t, n, n

.. .

i, t, -h, n .. . i, t, - t, i, 1, t, JoI, H .. . i, 1, t, i, t .. . 3, I, 1, i, i, i .. ·

*.. .

.

CHAPTER VI EQUATIONS

1. Conditional equations. Up to the present point in our study of polynomials we have written f(x) = 0 only in the case where f(x) is the zero polynomial. We have also written f(x) =:g(x) only in cases where f(x) and g(x) are equal polynomials in the sense of Sec. 1 of Chap. IV. We'shall henceforth use the notationsf(x) = 0, f(x) ='g(x) (read "f of x is identically equal to g of x") for the two properties above and shall give a new meaning to our earlier notation. If we replace x, by ,a complex number c in a polynomial f(x) whose coefficients are complex numbers, the resulting number: has been called the value of f(x) at x = c and has been designated by fCc). If fCc) is the complex number zero,- we call c a root or a zero of f(x). We shall devote this chapter to a study of the properties of the roots of polynomials, that is, to the roots of what we shall call conditional (polynomial) equations. A conditional equation (1) f(x) = 0 is not a statement but is a question. It asks: What are the complex numbers r such that fer) = O? It thus asks: What are the roots of f(x)? We shall call the answers to this question the roota (or 80lutions) of the (conditional) equation. An answer to formula (1) is, not a complete answer unless all roots are given. We shall say that we have 80lved the equation of formula (1) when we have found all its roots. If f(x) and g(x) are two polynomials in x, the conditional equation . (2)

f(x)

= g(x) 132

..

BEc.11

COND ITIO N AL EQUATIONS

133

is also a question. It asks: What are the complex numbers r such that fer) and g(r) are the same complex number? It should be clear that the answers to this question are precisely the same as the answers to the question. (3) fez) - g(z) == 0 in which the left member is the difference polynomial defined as in Sec. 2 of Chap. IV. Formula. (3) mayor may not be more convenient to solve than formula (2).' We call them equivalent equations and make th.e following general statement: Definition. Two equatiom are called equivalent if they have precisely the same roots.

Let us now observe that the equation fez) == g(z) is equivalent to the equations (4) aj(z) == ag(z), fez) + h(z) == g(z) + h(z) (a F 0) for any complex number a and any polynomial h(z). Thus, if we multiply (or divide) both sides of an equation by a nonzero number, we obtain an equivalent equation. H we add (or subtract) the same polynomial to (from) both sides of an equation, we obt~in an equivalent equation. The polynomial fez) is called the left member and the polynomial g(z) the right member of the equation fez) == g(z).

Both are sums of terms. .Any term of either member may be carried over as a term of the other, provided its sign is changed. The result will be an equivalent equation. This procedure is called the trampoBition process. It is the procedure by means of which formula (3) may be seen to arise from (2) and it is an immediate consequence of the second relation of formula (4). H a term (i.e., a summand) of f(z) coincides with a term of g(z), both terms may be cancelled, i.e., deleted. This is a consequence of formula, (4) or of the transposition process. For example, in the equation 9zS + 5z 3 - 7zS + 3z - 8 == 5z 3 + z - 2 + 2zs,

134

EQU ATIONS

[CHAP. VI

we cancel5x8 , add 9:& - 7X2 = 2:&, cancel 2x2, transpose to obtain 3x - x = - 2 + 8, 2x = 6, x = 3. 2. The degree of an equation. Every polynomial equation g(x) = hex) is equivalent to an equation f(x) = 0 where f(x) is a polynomial. Then we shall write (5) f(x) = aoXn + a l r1 + ... + an = 0, in which ao ~ O. The degree n of f(x) will be called the degree of the equation of formula (5) and the coefficients of f(x) will be called the coefficients of the equation. Then ao is the leading coefficient of the equation and an is the constant term. If we multiply formula (5) by a nonzero number a, we .obtain an equivalent equation af(x) = 0 with leading coefficient aao and other coefficients aQ,j. In case the coefficients are real this process is· usually employed to make ao positive. In case the coefficients are integerS the equation is usually divided by their g.c.d. We shall call the equation of formula (5) a linear equation if n = 1, a quadratic equation if n = 2, a cubic equation if n = 3, a quartic equation if n = 4, a quintic equatIon if n = 5. An equation of degree n may be properly referred to as an n-ic equation. "Note that an equation such as OX4

+ x + 2X2 8

3x

+ 5 = x + 2:& 8

4x

+6

is not a quartic, a cubic, or a quadratic equation but is a linear equation. For it is equivalent to x - 1 = O. ORAL EXERCISES

Give the degrees, the leading coefficients ao F 0, and the constant terms of the following equations: (a)· ~2 - 3~8 + 43; - 2 = 0 (b) ~8 - 3~2 + ~ - 1 = ~8 - 43;1 + 5~ - 1 (c) (~- 1)8(2~ + 1)2 = ~6 - 3~2 + ~ - 1

3. Linear equations. A linear equation is an equation = g(x) in which f(x) and g(x) are polynomials whose difference is a linear polynomial ax + b. Thusf(x) = g(x) is equivalent in this case to

f(x)

00

ax+b=~

~~~

135

LINEAR EQUATIONS

SJDC.3]

as well as to az = -b, and to -b a

z = --.

(7)

It follows that a linear equation has a unique solution and that it is given by formula (7). The standard technique for solving the most complicated of linear equations is to transpose all term8 involving z as a factor to the left member of the equation, and all constants to the right member. The equation may then be put in the form cz = d,

(8)

where d is a constant, and will be a linear equation only when c is a nonzero constant. The unique solution will then be z = die. nlustrative ExamPle Solve the equation

aySz

+ yl -

2ay

+ 2tJ:1l =

a'IP - 4ay

+ 2a + tJ:Il + 2ayl + y -

1

for z.

Remark: We give this example as an illustration of the transposition procedure. Exercises of this type may be found in texts on elementary algebra. They have little value and will not be given here. Solution

We transpose all z terms to the left and all terms not involving z to the right so as to obtain

(ay'

+ 2a -

ay - a)z

= _yl + 2ay -

4ay

+ 2a + 2ayl + 11 -

1,

and the solution is Z ..

+ (2a -

yl(2a - 1) - y(2a - 1) a(yl - y 1)

+

1)

2a - 1 .. -a-·

I

ORAL BDRCISBS Drill until you can solve tJ:te following equations very rapidly: (a) 2z (b) 3z

+3 = 0 +4 = 0

(0) 2z - 4 = 0 (d) 2z - 3 - z - 4

(e) 3z - 4 = 5z - 2 (J) Szl 4z - 3 - Szl 2z - 11 (g) (z 1)(2z 3) = 2z' 4z I) (h) 2az 3a = 3z 2a1

+ + +

+

+

+

+ +

136

EQU ATIONS'

[CBAP. VI

4. Quadratic equations. An equation is a quadratic equation if it is equivalent to an equation (9)

f(x)

a:cl

EE

+ bx + c =

(a ~ 0),

0

where we use the notation a, b, c for the coefficients in this simple case rather than ao, al, al. We solve quadratic equations by the use of the polynomial identity (x

+ 1/)1 =

Xl

+ 2zy + 1/1

which becomes (10)

when we replace 1/ by t/2. Note the following principle: COMPLETING THE SQUARE. The e:tpre8Bion Xl + t:c i8 macle into a 8quare if we add to it the 8quare of half the coeiftcient of x. The re8ult of thi8 completion i8 the 8quare of x plU8 hall thi8 coejficient. Any quadratic equation may now be solved by writing f(x)

EE a (XI + ~ X+ ~) == a [ (x +

:ar + ~ - !:2}

We compute

and use the identity Xl

-

_bl(2a)2 - 4ac] [(x + !.)I 2a

1/1

= (x + 1/)(x -

:::; (x + _b + Vb=-=I------:-4a-c') (x + _b _ 2a

1/) to see that

Vb l

-

2a.

2a

2a

4ac

-b - Vb 2

4ac).

Then (11)

:

where (12)

.. ' -!J +. :Vb 2 rl = ... :

2q

-

,

, .2a

-

4ac

BEC.4]

137

QUADRATIC EQUATIONS

Let us suppose that r is a root of the equation f(x) == O. Thenf(r) == a(r - rl)(r - r.) == O. But a product of com. plex numbers is not zero unless one factor is zero. Hence r == rl or r == rs. Conversely f(rl) == a(rl - rl) (rl - r.) == 0, f(r.) == a(r. - rl)(r. - r.l == O.

Hence rl and r. are the roots of our quadratic equation and are its only roots. • Formula (12) for the two roots of the equation of formula (9) is called the quadratic formula. It reduces the problem of finding the roots of a quadratic equation to that of finding the square root of a == b· - 400 and to rational processes. We shall calla the discriminant of the equation of formula (9). Note that A (13) (rl - r.)S == "2' a and so the equation of formula (9) has equal roots if and only if a == O. If a, b, c are all real, so is a. When a is positive, its square root VA' is a perfectly well-defined positive real number and rl and r. are distinct and real. When a is negative, we write

a == where

-A,

VA == VAi,

VA is positive and the roots -b -b VA. rl == 2a

+

2a ~,

VA.

r.==---~

2a

2a

are a pair of conjugate imaginary numbers. If a, b, c are complex numbers, the number a might be an imaginary number, and we have not yet discussed the meaning of YK. We shall do so in Seo. 7 of Chap. VIII. Let us note that -b c (14) rlrS == -, rl + r. == -a , a

138

EQUATIONS

[CJl4P. VI

a;nd that these relations may be computed by explicit computation from formula (12) or from the identity

a(a: - rl)(a: - rl) == a[a:1

(rl

-

+ 1-1)a: + rlr.]

+ ba: + c.

= azl

We shall close our study of quadratic equations by a proof of an irreducibility theorem. Theorem 1. A polynomial f(a:) IE azl ba: c with real coejficient8 i8 irreducibk in the field of all real numberB if and only if 11 = bl - 4ac < O. For we have shown how to factor f(a:) if 11 ~ 0 as a product a(a: - rl) (a: - r.) where rl and rl are real. H 11 < 0 and f(a:) is not irreducible, it has a factorization f(a:) = (ga: h)(Ba: t) where g, h, 8, t are real and Bg = a ~ O. Then rl = -h/g, r. = -t/B are real roots of f(a:) contrary to our proof that f(a:) ~ a pair of conjugate imaginary numbers as its only roots. Formulas for solving cubic and quartic equations exist but are not of great value. We shall not study"them here.

+ +

+

+

nlustrative ExamPles I. Solve the equation

23;1 -

4z

+ 8 = Szt + 23; -

5.

Solution

This equation is equivalent to :n;t + &: - 13 = 0 and a == 3, h == 6, 12 '13 = 12 '16 == 3· 4 '16 == 81 • 3. c = -13, ht - 4ac = 36 The roots are

+

-6 ±68 v'3 == -1 ±

~v'3.

II. Determine Ie so that the equation 9bt equal roots. "

-

60z

+ 61e + 1 == 0 has

Solution

The disoriminant is

hi - 4ac = 3600 :- 4 . 9Ji(61e The equation 6k1

+ Ie -

+ 1) == 36(100 -

6let

-

100 = 0 has as roots

Ie== -1±V1+2400= -1±49==4 -25.

12

12'

6

Ie).

SEC.

5]

REMAINDER AND FACTOR THEOREMS

139

+ +

The value k = 4 gives the equation 363;1 - 60z 25 = (6z - 5)2 = 0, and the value k = - ¥ gives the equation - l/zl - 60z - 24 == 0 which is equivalent to 25z1 402: 16 = (52: 4)2 == O.

+

+

ORAL EXERCISES 1•. What are the sums and the products of the roots of the following equations?

+ + + + v=a +

+

(e) (1 0)xl z 0 (f) iz 2 - Z = 0 (g) (i I)ZI - Z = 0 (n) V3 Zl - 9 = 0

(a) Z2 - 3z 27 == 0 (b) Z2 3z - 27 ... 0 (/:) 2z2 4z - 27 = 0

(d)

+ + 3z + 12z + 8 = 0 1

=0

2. Give the discriminant of each of the equations (a) to (d) of O~al

Exercise 1. EXERCISES 1. Give the radical form of the solution of all the equations of Oral Exercise 1. 2. Solve the following equations by the use of the quadratic formula.: (a) Zl

+ 2z -

=0 52 = 0

63

(b) Zl - 9z (/:) Z2

+ 6z -

40

=0

(d) Z2 - 63; - 40 = 0 (e) Zl z 4 =0 (f) Zl - Z - 24 == 0

+ +

(g) 3z1

+ 2z -

1

5

== Z2 - 4z + 1

5

(n)

Z2 - ;; - 84 == 0 \

(i)

!? + 2: -

21

+

=0

+

W (z 1)1(z - 1) = Z8 - 3z 8 (k) 2(z - 1)8 = (2z l)(z ,1)2 3z - 5 (l) 12:1:2 4z - 1 = 0 (m) Zl = 10z 75

+

+

+

+

+

3. Determine k so that each of the following equations has two equoJ. roots:

+9 = 0 +k +3 = 0 + 9k + 2 = 0

(a) 2kz2 - & (b) kz2 - 14:1: (/:) kz2 - 12kz

4. Find k if 2 is a root of k 2z 2 6. Find k if 1 is a root of 2kz2

+ + +

(d) kzl kz 1 ,.. 0 (e) (2k + I)ZI - 2(k (J) (41c - 8)Z2 4kz

+ 2)z + k = 0 + (k + 3) = 0

+ z - 9k = O. + k z - 15 = O. 2

5. The remainder and factor theorems. If c is a number, the remainder on division of a polynomial/(x) by the linear polynomial x - c is a number. Its value is given by the following: Remainder theorem. The remainder on diui8ion oll(x) by x - c is/(c).

140

EQUATIONS

[CHAP. VI

For the division algorithm states that

+ r.

f(z) = q(z) • (z - c)

Then f(a) == q(a) • (a - c) + r for any number a. In particular f(e) == q(e) • (e - c) + r = r. There is an almost automatic consequence of our definitions and the remainder theorem which we state as the following: Factor theorem. A number e i8 a root of f(z) if and only if z - e i8 a factor of f(z) • • For by the remainder theorem (15)

f(z) = (z - c) • q(z)

+ f(e)

for any e. The statement that e is a root of f(z) means thatf(e) = 0 and thus thatf(z) = (z - c) • q(z), z - e is a factor of f(z). The statement that z - e is a factor of f(z) means that f(z) == (z - c) • q(z) and we conclude that f(e) = 0 either from formula (15) and the uniqueness of the remainder in the division algorithm, or by computing f(e) == (e - c) • q(e) == O. nlustratille E%amples I. Compute the remainder on division of :1:71

-

7:&,8

+ 1 bY:I: + 1.

Solution

The remainder is/( -1) .. 1 - 7 + 1 == -5. II. Show that:t + 2 is a factor of:tl + 7:t' + &:.

+ 2:t -

Solution 1(-2) ... -32

+ 112 -

64 - 4 -12 "" O.

ORAL EXERCISES

1. Tell why:t - r is a factor of 1(:1:) in the following cases: (a) r = -1,/(:t) "" :t' +:t8 - 2:&1 + 2:t + 4 (b) r .. 1,/(:t) == :tl - 7:t' + 2:t8 - 2:&1 - 1O:t + 16 (c) r = -2,/(:t) .. :tl + 3:t1 + 5:t + 6 (el) r = 3,/(:t) = :t8 - :tl - 4:& - 6 (6) r .. 2,/(:t) .... :tl - :t - 6 (f) r = -4,/(:t) = :t8 + 4:&1 + :t + 4 (g) r .. 3,/(:t) = :t8 - 3:t1 + 2:t - 6

12.

SEC.

SNYTHETIC DIVISION

6)

141

2. Give the remainder on division of the following: (a) x 8 - 3x 2 - 3x - 3 by x + 1 (b) x 3 + x2 + 5z + 7 by x + 2 (c) x 6 + 6x 4 - 7x2 + 23: - 1 by x-I (d) x 3 - x 2 - 3x + 1 by x - 3 (6) x 3 + 4:/: 2 + 23; + 4 by x + 4 (f) x 3 -x-3byx-2

6. Synthetic division. f(x) = aoX'"

If c is any number and

+ aIx"'-1 + ... + an

is any polynomial, we may write

f(x)

= q(x)(x

- c)

+ b""

where q(x) = boX"'-1 + bIx"'-2 + ... + bn-l, b", = f(c). Then (x - c)q(x) = f(x) - b", = xq(x) - cq(x), so that

+ b", = f(x) + cq(x), that is, boX'" + bIx"'-1 + . . . + b", = (aoX'" + aIx"'-1 + ... + an) + CboX"'-1 + cb Ix"'-2 + . . . + Cbn-l.

xq(x)

Comparing coefficients we have

+ cbo, + Cbi-I,

bl = al b. = a.

Thus the coefficients bo, . . . ,bn-l of the quotient q(x) and the remainder b", may be computed by the tabular form

+ + a2 + + an-I + an ~ + cao + cbl + . . . + Cbn-2 + Cbn-l bo + b + b2 + . . . + bn-I + b",

ao

al

i

This procedure is a modification of the ordinary division process and is called 8ynthetic division. It is not applicable without the use of special devices for the division of a polynomial by other divisors than those of the form x-c. Synthetic division may be used to compute f(c) when direct substitution of c for x might prove tedious. For

142

EQUATIONS

[CHAP. VI

example, if f(z) == z, - 7z 8 + &;2 - 100z - 31, then the computation of f(8) == 8' - 7(8)8 + 5(8)2 - 800 - 31 is complicated. But the synthetic division 1 - 7 + 5 - 100 - 31~ + 8 + .8 + 104 + 32 1 t 1 + 13 + 4 + 1 gives f(8) == 1. II c is a root of f(z) , the computation of fCc) by synthetic division gives the value fCc) == 0 as well as the quotient q(z) in f(z) == (z - c)q(z) •. Then the remaining roots of f(z) are the roots of the equation q(z) == O. This equation of degree n - 1 is called the depre88ed equation. nlustrative Examples I. Compute the quotient and remainder on division of &:' + 142;8 - 22:1;1 - 50 by z + 6. Solution 3+14-22+ 0-501-6 - 18 + 24 - 12 + 72 3 - 4 + 2 - 12 + 22 The quotient is &:' - 4z1 + 2:e - 12 and the remainder is +22. II. Verify that 8 is a root of the equation z' - 6Oz1 - 29:e - 24. Solution 1+0-60-29-24~

+8+64+32+24 1+8+ 4+ 3+ 0 EXERCISES

1. Use synthetic division to compute the quotient and remainder on division of the following: (a) ~I - 3z1 + 2:e - 1 by z - 2 (b) z' - 142;i + 2:e' + 49:& - 36 by z + 2 . Am. q(z) = Zl - 16:&1 + 34z - 19, r = 2. (c) z, + Zl + z - 2 by z + 3 (d) Zl - Z8 - 32:e by z - 3 Am. q = z, + 3z8 + &1 + 24z + 4O,'r = 120. (e) &:8 + 2zI + 5:1: + 10 by z + 2 (J) ZT - 6:&1 + &a - 3z1 - 7z + 1 by z + 1 Am. q - ~I - ~I - 5:1:' + 5:1:8 + &:1 - 6:& - 1, r ... 2.

SEC.

143

LINEAR FACTORS

7]

(g) :1:4 + 10:1:3 + 22:1:2 - 7:1:

+

+ 5 by :r: + 4

(h) :1: 4 + :1:8 - 22:1:2 15:1: - 32 by :r: - 4 Am. q = :1:8 + 5:1:2 - 2z + 7, r = -4 (i) :r: 4 + 2z8 - 13:r:2 + 13:1: - 21 bY:l: + 3 W 2z4 + 9:1:8 + 14:r: + 8 by :I: l Am. q = 2:1:8 8:1:2 - 4:r: + 16, r = O. (k) 6:1:4 + 5:1:8 10:1: - 4 by :I: - i

+

+

+ + 5:1:8 +:1: + 17:1: - 6 bY:l: - i Am. q = 3:1: 8 + 6:1:2 + 3:r: + 18, r = O. (m) 3:1: 4 - 14:r:8 - 57:1: 2 + 65:1: - 56 by :I: - 7 (n) 3:1:4 + 40:1:8 + 85:1: 2 + 97:1: + 99 by :I: + 11 Am. q = 3:1:8 + 7:1:2 + 8:1: + 9, r = O. (0) :1: 8Iz8 - 2:r: 8 + 18:1:2 + 8:1: - 72 bY:l: - 9 (p) 18:1:6 + 20:1: 4 + 29:1:8 + 57:r:2 + 6:r: by :r: + t Am. q = 9(2z4 + 2:1:8 + 3:1:2 + 6:1:), r = O. (q) :1:8 - 3:1:2 + 16:r: - 44 bY:l: - -nr (r) :1:8 + 0.4:r: 2 - 0.18:1: + 0.33 by :I: - 0.2 Am. q(:I:) = :1:2 + 0.6:1: - 0.06, r =: 0.318. (8) :1:8 - 0.2z2 + O.~Iz - 0.424 by :r: - 0.9 2. Use synthetio division to show that (a) (:I: - 2)2 is a faotor of :1: 4 - 4:r: 8 + 5:1:2 - 4:r: + 4 (b) (:I: + 3)2 is a faotor of :1: 4 - 17:1:2 + 6:1: + 90 (c) (:I: + 1)8 is a faotor of 2z6 + 6:1:4 + 5:1:8 - :1:2 -=:.j:r: (d) (:I: - 5)2 is a faotor of :1: 8 - 75:1: + 250 (e) (:I: + 4)8 is a faotor of :1: 4 + 8:1: 8 - 128:r:. - 256 (l) 3:1:4

2

8 -

1

7. Factorization into linear factors. There is a theorem called the fundamental theorem 0/ algebra which states that every polynomial/(x) with complex number coefficients has a complex root.. The pI:oof of this theorem is a complicated part of graduate mathematics· and will not even be suggested here. The result of this theorem and the factor theorem is that every polynomial/ex) with complex number coefficients has a linear factor x - c where c is a complex number. We state this result: Theorem 2. The only polynomials that are irreducible in the field 0/ all complex numbers are the linear polynomials. We may now apply our theory of the factorization of polynomials as products of irreducible factors. Theorem 2 states that the factors are linear and, since any linear polynomial ax + b = a(x - r), where r = -bfa, we may

144

EQUATIONS

[OBAl'. VI

take the factors to be polynomials z - rio H r is any root of f(z), then f(z) = (z - r)q(z) and we thus have a factorization of f(z) in which z - r is one of the linear factors. However the (linear) factors z - ri of f(z) were proved unique in Sec. 9 of Chap. IV. Hence r is one of the rio By the factor theorem, if z - ri is a factor of f(z), then r, is a root of f(z). We combine these results: Theorem 8. Let f(z) = ao:en + alzft-l + . . . + an where ao, ai, • • • , an are comple:x; number8, n > 0, ao ~ O. Then f(z) = ao(z - r0 (z - rll) ••• (z - rn),

(16)

where rl, • . • ,rn are comple:x; number8 which are the rootB of f(z). Thefactor8 z - ri are unique apartfrom their poBition in the factorization above. H r is. a root of f(z) and z - r occurs exactly m times in the factorization (16) of f(z) then f(z) is divisible by (z - r)m and not by (z - r)m+l. Then we call r an m-tupk root or a root of multiplicity m of the polynomial f(z) and of th~quationf(z) = '0. We" ~all r a Bimpk root if m = 1, a doubk root if m = 2, and a triple root if m = 3. ORAL EDRCISES

Give ities:

the roots of each of the following equations with their multiplic-

(a) (~- 8)1(~

+ 2)~ ... 0

+ 1) = 0

(~+ 8)1(~ - 2)'(~ - 1)(~ (c) (~+ 2)'(~~ - 4)1(~ - 2)8 ...

(b)

0

+ ~ + 2)'(~ + 2) ... 0 (~I - 4)1(~8 + ~)a(~1 + ~ + 1)1 = 0

(d) (~I

(e) (f)

(~I

-

~

+ 2)(~1I' - 4)(~1 - &: + 4)(~1 -~) = 0 + 3~ - 1)'(~5 - ~)(~I + ~I) ... 0

(g) (~I - ~.

8. Expression of coeftlcients in terms of roots. The factored form of a (17)

f(z)

pol~omi.al

==:x;"

+ C1Z-l + ... + en

with leading coefficient unity is (18)

f(z)

E

(z - rl)(z - r2)

(z ;- rn),

SEC.

8)

COEFFICIENTS IN TERMS OF ROOTS

145

where T1, T2 • • • T" are the roots of f(x). The factored form may be multiplied out and the coefficients in formula (17) compared with those in (18). We shall not give the details of the argument and recommend that the student carry out the details for n = 3, 4. The results may be expressed by the form~as

+ T2 + ... + T" = -C1, T1T2 + T1T a + . . . + Tn-1T" = C2, • Ti + T1T2 • • • T....1Ti+1 + .. . + Tn-{ ....t>Tn-{....2) • • • T" T1

(19) T1T2'

=

(_1)iCi,

Thus the sum of the Toots of f(x) is the negative of the coefficient C1 of its' second highest poweT X"-l. The sum of all possible formally distinct products of two roots is (-1)2c2 .. The sum of all possible formally distinct products of i roots is (-1)'c.. There is only one product of n roots and it is equal to (-1)"c". We have already noted these relations in the case n = 2. For n = 3 they become T1T2

T1 + T2 + Ta = -C1, + T1Ta + T2Ta = C2, T1T2T a

=

-Ca,

and for n = 4 they become T1 + 1"2 + Ta + T4 = -C1, + T1TS" + T.1T4 + T2Ta + T2T4 + TaT4 = C2, T1T2Ta + T1T2T4 + T1TaT 4 + T2TaT 4 = -Ca, T1T2TaT4 = C4. T1T2

The relations just given depend upon the hypothesis that the leading coefficient of f(x) is 1. If it is not, we may write f(x)

=

ao3!'

+ a1x"-1 + ... + an = ao(x" + C1x"-1 " + ... + c"),,

146

EQUATIONS

where the coefficients

Ci

[CHAP. VI

are the quotients.

.

The roots of I(x) are the same as those of

xn' + CIXn-

1

+ . . . + Cn

and so we may obtain the coefficients Ci in terms of the roots of I(x). Then I(x) is determined by its roots if we prescribe ao. mustrative Examples I. Find the equation with leading coefficients unity whose roots are 2, -1

+ 0,

-1 -

O.

Solution Let rl = -1 + 0, rs = -1 - 0, r8 = 2. Then rl + r2 = -2, rlr2 = (-1)2 - (O)2 = 1 - 3 = -2, (rl + r2) + r8 = -2 + 2 = 0, rlrS

+ rlr8 + rsr8 = rlrS + (rl + r2)r8 = (rlrS)r8

-2

= (-2)2 = -4.

+ (-2)2 =

-6,

The answer is:r:8 - (O):r:s + (-6):r: - (-4) = :r: 8 - 6:r: + 4 = O. II. Find the equation with leading coefficient 1 whose roots are -3 + y'2, -3 - 0, 2 + i, 2 - i. Solution

If rl = -3 + y'2, r2 = -3 - y'2, r8 = 2 + i, r, = 2 -i, we have r1 + r2 = -6, rlrS = 7, r8 + r, = 4, rar, = 5. Then -C1 = (r1

+ rs) + (r8 + r,)

= -2,

= ri.rs + (r1 + rS)(r8 + r,) + r8r, = 7 -C8 = rlr2(r8 + r,) + (r1 + rs)rar, = 7·4 c, = (rlr2)(rar,) = 35. Cs

Am. :r:'

+ 2:1:8 -

24 + 5 = -12, 6' 5 = -2,

12:r:2

+ 2:r: + 35 = O.

ORAL EXERCISES

1. Give the sum and product of the roots of· the following equations: (a) 3~s - 2:r: + 1 = 0 (b) 2:1:8 - 4:1)2 + 5:r: - 6 (c) 3:1)8 4:r; - 18 = 0

.

(d) 2:1:' + 2:r:8 + 7:1) = 0 (e) (:1)8 - 2:r:s + 1)(:r: + 3)

=0 (J) (:1)8 - 4:r;2 + 3)(:r:S - 2) = 0 (g) (2:1:8 - 3:1) + 1)(:r:2 + 4:r; - 7), ,= 0 (h) (:r: 2 - 3:1) + 2)(:r:8 - 2:1: 2 + 5)(:r:3 - 3:r;S + 6) = 0

+

=0

BlIIC.

9)

147

ROOTS OF REAL POLYNOMIALS

I. The sum of two of the roots of the polynomial2:l:1 is 3. What is the third root? S.· The product of two of the roots of the equation 2a;8

-

7~1

- 5a; + 4

+ 11:' - 1111: - 10 .. 0

is 2. What is the third root? BDRCISBS

Use the relations between roots and coefficients to find the polynomialJ(II:) = II:" + Cl/P-l + ... + c.. whose roots are (a) -3, 1 '\1'=3, 1 - '\1'=3 (c) 5, -I V2, - 1 - v'2 (b) 4, -2 + V -5, -2 - v'::5 (d) -2, i + " -3/2, i - ,,-3/2 (e) 1 V3, 1 - V3, -2 v'=2, -2 - v'=2 (J) 1+2i,I-2i,2-3i,2+3i (g) 2i - 1, -2i - 1, v'2 - 1, - v'2 - 1 (1&) 0, 3i + 2, -3i + 2, -2 + V5, -2 - V5

+

+

+

+

9. The imaginary roots of real polynomials. If r is a complex root of a polynomial J(:c) with real coefficients, the expression :c - r is a factor of J(:c). When r is real and the degree of J(:c) is greater than 1, the polynomial J(:c) cannot be irreducible in the field of all real numbers. Suppose next that r is imaginary and so write r =

where 8 and't are real, t

~

8

+ ti,

0, i" == -1. Then

'=8-ti is the conjugate imaginary, and (20)

g(:c) = (z - r)(:c - ,) = :c" - 28:&

has real coefficients. write

+ 8" + t"

We apply the division algorithm to

J(:c) = q(:c)g(:c)

+ a:c + b.

Since J(:c) and g(:c) have real coefficients, so does the remainder a:c + b. But J(r) = q(r)g(r)

+ ar + b' == ar + b =

0

148

EQUATIONS

[CHAP. VI

since fer) = g(r) = O. Hence either a = b =: 0 and g(x) divides f(x), or a ~ 0, r = -b/a is real. This is contrary to hypothesis and we have proved the following: Lemma. Let r be an imaginary root of f(:&). Then, i8 an imaginary root of f(x) , (x - r)(x - ,) i8 a factor of f(x). As a consequence of the lemma we see that every real polynomial (i.e., polynomial with real coefficients) has a real linear or a real quadratic factor. Using Theorem 1 we may state the following: Theorem 4. A real polynomial is irredUcible if and only if it is linear or is a quadratic axil + bx + c with bll - 4ac < O. We now apply the factorization theorem of Sec. 9 of Chap. IV with Theorem 4 and obtain the following: Theorem 6. Every real polynomial is ewpresBible uniquely, apart from arrange~ and constant factor8, as a product of real linear and real irreducible quadratic factors. The roots of the real linear factor8 are the real roots of f(x), and the roots of the quadratic factor8 are pairs of conjugate imaginary roots of f(x). . . H r is an imaginary root of multiplicity m of f(x), the corresponding polynomial (x - r)(x - ,) must occur exactly m times in the factorization of f(x). Then,. has multiplicity m. Theorem 6. Let a and b ~ 0 be real and a + bi be a root of multiplicity m of a real polynomial f(x). Then a - bi is a root of multiplicity m of f(x). ORAL BDllCISBS

1. The number 3 + 2i is a. root of the equation Z8 -

9z1

+ 31z -

39 = O•.

What are its other roots? So If 2 - 3i is a. root of Zl + Zl + ba: + c = 0, what are its other roots and what is the value of c? . 3. Let 2 + i be a double root of Zl + az4 + bz8 + CZI + d:& + 25 -0. What are its other roots and what is the value of a? " Give the factored form of the equation of degree six whioh has

SlDc.l0] ..

149

MULTIPLE ROOTS BY DEIUVATIVES .

real coefficients, has 2 simple root.

v=s as

&

double root, and has 1

+i

as

&

10. Multiple roots by derivatives (FULL COURSE). We may compute the derivative polynomial J'(z) of J(z) and the g.c.d. d(z) of J(z) and J'(z). The roots of the factor d(z) of J(z) are roots of J(z) and Sec. 13 of Chap. IV implies that, if r is a root of multiplicity m of J(z), then it is a root of multiplicity m - 1 of d(z). When d(z) == 1, the polynomial J(z) has no multiple roots. Otherwise we may find the multiple roots of J(z) by solving the equation d(z) == O. A nonconstant polynomial J(z), which is irreducible in a field F of numbers containing its coefficients, factors with linear factors in the field of all complex numbers but has no multiple root. For f(z) is not the zero polynomial and has coefficients in F, so does d(z); d(z) divides f(z) and has degree at most m - 1. Then d(z) is not a constant multiple of J(z) and can divide J(z) only if it is a constant. Let us apply the process above to any quintic polynomial J(z) with a root of multiplicity m > 1. H J(z) has only one mUltiple root r and this is a double root, the polynomial d(z) == z - r, (z - rr' is a factor of J(z) == (z - r)lIq(z). Then the problem of solving the equation J(z) == 0 is reduced to the solution of the cubic q(z) == O. H J(z) has two double roots or a triple root, the polynomial d(z) is a quadratic and may be solved by the quadratic formula. It determines three (or four) roots ofJ(z) and the determination of .the rema,;njng two roots'is a trivial matter. When i(z) has a double and a triple root, or a root of multiplicity four or five, the polynomial d(z) has degree three or four and a multiple root. The equation d(z) == 0 niay be solved .by inspection or the derivative process, and' the solution of J(z) == 0 completed by dividing J(z) by d(z) or by comparing coefficients and roots. lUustrative Examples I. Find the roots of J(z) - z' - 242;1 + 64z - 48.

150

EQUA.TIONS

[CRAP. VI

Solution /'(S) .. 42;1 - 4&

+ 16). 24:1:1 + 64s 12:1:1 + 1& 12:1:1 + 4& -

+ 64 = 4(SI -

+ 41~+ 4 - 12:1: + 16~ j;. - 42;1 + 42; s' 42;1 - lOa: + 16 42;1 - lOa: + 16 .. .. Sl - 42; + 4

12:1:

'Sl - 42;

des)

.. (s - 2)1. I(s) ... (s - 2)I(S - r),

-12(SI - 42;

( -8)( -r) - -48,

48 48

+ 4)

r - -6.

Am. 2 is a triple root, -6 is a simple root. Ii. Find the roots of Sl'- 7S1 - 2:1:1 + 12:1: +~. Solution

/'(s) - 5s' - 21s1

-

"'-11

1,_+10

42; + 12.

18'-.... -"+11 + .. -_-.1,.. + "'-.-10 MIz'+ MIz'+I01\11- 1NOII-7OCb , .. -

, .. -1M

lO.III-lo.-1O lO.III-lo.-1O

- 10l\Il -10&81-

-... - .... -lillif.+'o

1188 18'-1"'-'-'+1111 -1'-'- . . +.+~ lW + IiIHII + 1188 - -2('1"+"'-.-10)

OIl' - 10l11J111- 1 .

... +8.+800 , - 1"-'+ 1"- + lI88 ... 1~-_-J)

des) _ Sl - S - 2 t - (s - 2)(s + 1). I(s) - (s - 2)I(S + l)l(s - r), (-2)11'( -r) ... 8, r - -2. Am. Double roots 2, -1. Simple root -2. BIlmCISBS

Each of the following equations has at least one multiple root. Use this property to solve them. (a) s' - 0a:1 - 8s - 3 - 0 (b) s' 2:J:I Sl - 12:1: 8 - 0 Am. I, I, -2 2i, -2, -2i.· (e) s' - 24:1:1 64s - 48 - 0 (d) s' 4:J:I - 16:1: - 16 - 0 Am. -2, -2, -2, 2. (e) s' ~ 3:1:1 ~ U 4- 0 Am. I, I, -2, -2. (J) s' - u l - 2:1:1 12:1: 9 == 0 (g) Sl lOs' 15s 6 - 0 , Am. -I, -I, -1, t ± l V -15. (A) Sl 8Os1 240s 192 = 0 (i) sa - 6:1:' 5s' 12:1:1 42; .. 0 W Sl - 10s1 - 20:1:1 - 15s - 4 - 0

+

+ +

+ +

+ + -

+

+ + + + + + + + + +

+

CHAPTER VU REAL ROOTS OF REAL EQUATIONS 1. Transformations of polynomials. The principal tools used in the study of the real roots of equations J(x) = 0 are certain transformations which replace a polynomial (1) J(x)!5 aoX" + alxtl-l + . . . -t an =' ao(x - rl)(x - rs) ••• (x - r,,) by a polynomial g(x) whose roots are related to those of J(x) in a prescribed manner. We shall first prove the' following: Theorem 1. Let g(x) = J(x + a) and rl, rs, • • • , r" be the roots oj J(x). Then rl - a, rs - a, •••. , r" - a are the root8 oj g(x).

For J(x

+ a) lEI'ao(x + a -

But x

+a-

r.

rl)(x

!5

E!

ao(x

+a -

r,,).

ao(x - Bl) • • • (x - B,,)

where Ii ='r. - a. The coefficients bo, bl ,

.

rs) ••• (x

= x - (r. - a), and

g(x)

g(x)

+a -

• •

• ,

btl of

+ a)" + al(x + a)tl-l + . . . + a......l(X + a) + an boX" + blztl-l + . . . + btl E!

may be computed by a process of repeated synthetic division. We observe that since J(x + a) = g(x) then g(x - a)

= J(x)

&

bo(x - a)"

+ bl(x -

a)"-l + ... + btl-l(x - a) + btl.

Then btl is the remainder on division ofJ(x) by x - a and bo(x - a)tl-l + ... + btl-s(x - a) bn-lls the quotient. If we divide this quotient by x - a, the remainder is bn-l

+

151

152

REAL ROOTS OF. DEAL EQUATIONS

[CJLU>. VII

+ ... +

and the new quotient is bo(z - a)ft-I btl-2. We use this second quotient to compute btl-I as a third remainder, and so succ~BSive synthetic divisions will yield the desired coefficients bn, btl-I, . . • , bl, bo = ao as a sequence of remainders. The reader will have realized from the start that bo = ao. Our second transformation is that which replacesf(z) by

g(z)

~ tt1(I) ,

E

(I -

tnao ~ - rl) r2) ••. rn) - ao(z - WI)(Z - tr2) • • . (z - wn ).

(r -

The general term of fez) is a¢'-1 and that of g(z) is tnlSi

G)"""""'

E

lSirZ"""""'.

We have proved the following: Theorem 2. Let fez) be given by formula (1), t be any

1l.O'Tl.I67'0 number, and g(z)

5!!!

aoZ"

+ ta1z"-1 + t alr 2 + ... + t-llln-IZ + tnan. l

Then, if rl, . . • , rn are the roots of fez), the roots of g(z) are trl, WI • • • trn. If an - 0, then z is a factor of some multiplicity m of f(z) , zero is a root of multiplicity m of fez), and we may remove the factor zm from fez). Let us then restrict our attention to polynomials for which an ;oS 0, so that no root r, of fez) is zero and every rr l exists. We define

g(z) = z"f (~) - anz"

+ lln-lrl + . . . + ao,

and see that

g(z)

i5

aoZ" (~ - rl) (~ - r2) ••• E

(~ -

rn)

ao(l - rlz)(l - rlZ) ••• (1 - rnZ).

Then, if 8, = l/r" we ~ve g(z) = aorlrl .•• rn(81 - Z)(81 - z) ••• (8n - z).

SEC.

1]

TRANSFORMATIONS O.F POLYNOMIALS

153

Since TIT! • • • T" = (-I)"an/ao, we have g(x) = an(x - SI) • • • (x - B,,)

and have proved the following: 1).eorem 3. LetJ(x) Eii ao:c" +

... + + ... +

an with aoan ~ O. Then the TOOts oj g(x) = an:J!' ao aTe the Teciprocals oj the Toots oj J(x). This last transformation is used less frequently than the

others. The sum Tl + T! + . . . + T" of the roots of the equationJ(x) = ao:c" + a1xtl-l + ... + an = Ois -al/aO' Diminish the roots by - adnao and obtain an equation g(x) E!! bo:J!' + b1Xtl-l + ... + btl = 0 whose roots are T, + al/naO' Then the sum of the roots of g(x) = 0 is zero and b1 = O. Such an equation is called a Teduced equation. The solution of any equation J(x) = 0 may be achieved if we can solve the related reduced equation obtained above. A reduced cubic equation may be divided by its leading coefficient and so put in the form. h(x) ;;s x' + px + q = O. The number.6. = _4p' - 27q! is called the discriminant of h(x) = 0, and it can be shown that h(x) = 0 has three real roots except when.6. < O. We shall not try to prove this result. . nlustrative Examples I. Increase the roots of :1:' + 83:8 + 2:1:1 - 143: + 2 by 2. Solution 1 + 8 + 2 - 14 + 21-2 - 2 - 12 + 20 - 12 6 - 10 + 6 - 10 - 2 - 8+36 4 - 18 + 42 -2 - 4 2 - 22

-2 The desired

equa.ti~n

1+0 is :1:' - 22:1:1

+ 42:1: -

10

= O.

154

REAL ROOTS OF REAL EQUATIONS

[CHAP.VII

II. Find a polynomial whose roots are related to the roots r of 27z8 + 542;2 + 30z + 5 by the formula 8 = '3r + 2.

Solution A polynomial whose roots are 3r is 27z. + 3 . 543;1 + 9 • 30z + 27 • 5 .. 27(Z8 + &:2 + 10z + 5). Diminish the roots by -2, using the synthetic division process.

1 + 6 + 10 + 51-2 - 2 - 8-4 4+ 2+1 -2- 4 2- 2 -2 1+0

Ana. Z8 - 2z

+ 1.

III. Find an equation whose roots are related to the roots r of 4z· - 38z2 + 120z - 127 = 0

by the formula

8

= 2(r - 3).

Solution 4 - 38 + 12 -26+ 12 -

-14+

120 - 12713 78 12642- 1 42

+

0

12 4- 2

An equation with roots r - 3 is 43;1 - 2z1 - 1 .. 0, and we multiply the roots by 2 to obtain.4z8 - 43;1 - 8 .. O. Ana. Zl - Zl - 2 =0. IV. Determine whether or not the roots of Z8 - 2z1 - 2 = 0 are all real.

Solution Here ao = 1, al = 2, and -al/3ao = f. To avoid fractions we multiply the roots by 3 and so obtain Z8 - 6zll - 54. Diminish the roots by 2. 1- 6+ 0 - 54~ +2 - 8 -16 - 4 - 8-70 2- 4 -2-12. +2

1+0

SEC.

1]

TRANSFORMATIONS OF POLYNOMIALS

The reduced cubic is Zl

-

155

12z - 70 = 0 and

A = -4( -12)' - 27(70)1

= 27(256 -

4,9(0)

is negative. The equation has only one real root.. ORAL BXERCISES

1. Use Theorem 2 to give an equation whose roots are the negatives of the roots of

+

+

(a) 3z4 4z1 - 2z1 7z - 1 (b) 2Z1 - Szl + 7z1 + 2z - 8 (c) Zl - Szl - 9 (g) Zl 4z' - Sz - 1 (h) 7:1:' Sz' - 9z1 - 10:1:1 - 2z

+ +

(d) Zl - 7z8 - Szl - 1 (e) Z8 - 9z' 10:1: (f) z, 6z1 - 7z'

+

+

- 1

2. Give polynomials whose roots are t times the roots of the following polynomials for the given values of t: (a) 32;1 - 2z1 4z - 5;.1 = 2, 3 (b) 2z' &1 - iz 1; t .. -2, 2 (c) 3z' - &1 + ISz - 54; t = i, i (d) z, - Szl - 24z - 64; t = i, i 3. Give!J. for each of tbe following equations and state whether or not the roots are all real:

+

(a) (b)

Zl Zl -

+

+

Z - 1 = 0 (c) 2z - 1 = 0 (d)

2 =0 + 2z + 1 = 0

Zl -

(e)

Zl -

Z8

(f)

Zl -

4:/: - 2 - 0 3z - 1 = 0

" Give polynomials whose roots are the reciprocals of the nonzero roots of the polynomials of Oral Exercise 1. EXERCISES

1. Diminish the roots of the following polynomials by the given n~bera:

+ 4z - 7, a = 1 Am. 3z1 + '32;1 + z - 6. 2z' + 3z1 - 5z + 8, a = 3 Zl -·1& + 17, a = 4 Am. Zl + 12z1 + 32z + 17. Zl - 32;1 + 2z - 4, a = -5 z, - 4:/:1 - 2z + 1, a = 2 Am. z' + Szl + 20:1:1 + I4:/:- 3.

(a) 32;1 - &1

"(b) (c) (d) (e)

(f)

Zl

+ 32;1 -

Z -

1, a = 3

S. Increase the roots by a in each case of Exercise 1. " 3. Find a polynomial whose roots B are related to the roots r of each of the following polynomials by the corresponding formula:

156 (a)

(b) (c) (d) (6)

(j) (0) (1&) (i)


[CHAP.VII

+ 56:1; - 40, 8 = it' - 2 Am. Zl + 2z + 7. Z8 - 6:1;1 + 831 + 16, 8 = l(r - 2) 4z1 - Z + 1, 8 == 2r + 1 Am. Zl - 3z1 + 2z + 2. 1 1 729:1: + 243z + 9:1; - 1, 8 = 3r + i 9:1;1 - 9:1;1 + 4z, 8 = 3(r - i) Am. Zl + z + 2. 8zI - 36:1;1 + 54z - 35, 8 == 2r - 3 Zl - 16z1 + 64z + 48, 8 = it' - 6 Am. z· + l(l:/;I + 28z + 30. 8zI - 12z1 - 1831 - 23, 8 -= 4r + 2 Z8 -

12z1

Zl -

9:1;1

(J) z·

+ 45z -

54,

8 ==

+ 21z1 + 14k + 54,

ir - 1

Am. Zl

8

= i(r

+ 6)

+ 2z + 1

" Check your answer g(z) = 0 in each case of Exercise 3 by solving for r in terms of 8 and applying the corresponding transformations to g(z) to obtain the original/(z). 5. Apply transformations that will repJace each of the following • equations by a reduced equation: 6:1;1 + 2 ... 0 3z1 + 4 == 0 (c) 2z' - 12z1 + 5z - 9°= 0 (d) Z4 + 8zI - 3z - 1 = 0 (6) 3z4 + 12z' + 7z1 + 2z 4

(a) (b)

Z8 -

Zl -

Am. Zl - 3z + 2 ... O. Am.

+ = O.

Z4 -

243;1

+ 61z -

43

= O.

'

6. Determine whether the roots of the following equations are or are not all real: (a),3z' -

Zl

+ 1 ... 0 == 0

(b)

Z8 -

Zl -

2

(c)

Zl -

2z1

+1

(d)

Zl -

6:1;1

+1 =

0=

0 0

(6) z· - 4z1

+ 3z -

1 == 0 Am. Not all real.

+ 6z 2z + 2 - 0 Am. All real. (0) 2z1 + 6z1 - 3z - 4 ... 0 Am. All real. (1&) 2z' + Zl - Z - 1 - 0 (J)

Zl

1 -

2. Integral roots. The study of the roots of equations with rational coefficients is equivalent to the study of the roots of equations with integral coefficients. For if we multiply an equation with ra.tional coefficients by the l.c.m. of the denominators of these coefficients, we obtain an equiva.lent equation with integral coefficients. Let us first study 8. process for finding the integral roots of such equations.

SEC.

157

INTEGRAL ROOTS

2]

If the coefficients of f(x) 5 aoXn + a1x"'-1 + + an are ordinary integers and r is an integral root of f(x) = 0, then fer) = br + an = 0 where b = aorn- 1 + ... + a-l is an integer. Then an = (-b)r is divisible by r. Since the integer an has only a finite number of divisors, it is possible to determine all integral roots of f(x) = 0 by listing these divisors d and by computing fed). We may, however, improve this procedure as follows. Let a be any integer and g(x) 5 f(x + a). By the process of Sec. 1 the coefficients

bo = ao,

bn

= g(O) = f(a)

are obtainable by addition and "multiplication of integers. Then they are integers. The roots Si of g(x) = 0 are related to the roots Ti of f(x) = 0 by the relation Si = ri - a and a is an integer. Then Si is an integer if and only if ri is an integer. It follows that r can be an integral root of f(x) = 0 only if S = r - a divides f(a). We have proved the following:

Theorem 4. The integral roots of a polynomial equation f(x) = 0 with integral coefficients are integers r such that r - a divide~ f(a) for every integer a. It is usually undesirable to utilize more than a few

values of a in applying Theorem 4 to sift out the integral roots of f(x). A systematic procedure might begin with an, the computation of f(O) = an, f(l) = ao + al + . . . and f( -1) equal to the sum of all of the coefficients of the even powers of x minus the sum of the coefficients of all the odd powers. We are assuming that f(O) ;;z!! o. If f(l) = 0, we ~vide f(x) by x - I to obtain a quotient equation cf>(x) = 0 of degree n - 1 which has already been called the depressed equation. Then we begin over again and study the integral roots of this new equation. We proceed similarly if f( -1) = O. Let us then assume that f(O), f(I), !( -1) are all not zero. Factor these integers. The three factorizations

-+

158


[CHAP. VII

will indicate which integer has fewest divisors and we list the divisors of this integer. Suppose first that we have listed the divisors d of !(O) omitting the values d == 1, d == .-1 which have already been tested. Then a particular d will not be a root of !(z) == 0 unless d - 1 divides !(1) and d + 1 divides !( -1). These divisibility conditions will sift out most divisors d that are not roots and we complete our solution by computing !(d) for the rema.ining values. . Our second possibility is that we have listed the divisors e of!(I). Then any particular e will arise from an integral root d of !(z) = 0 only if e == d - 1, d == e + 1. We omit d == 0, -1 and so e == -1, -2 from our divisor list in this case, and we see that e + 1 is an integral root of !(z) == 0 only if e + 1 divides !(O) and e + 2 == d + 1 divides !( -1). . Our final possibility is that of a list of divisors c of !( -1). Since c == d + 1 and d " 0, 1 we omit the values c == 1, 2 from this list. Then d = c - 1 will be a root of !(z) == 0 only if c - 1 divides !(O) and c - 2 == d - 1 divides !(1). We shall illustrate the procedure just described in the examples given below. However we shall delay the exercises until after the next section in which the results are given additional significance. mutTGtive E%Gmples I. Determine the integral roots of /(~) 51 ~,

-

49~1

+ &: + 56 .. O.

Solution Weseethat/(O) division

= 56,/(1) -

16,/(-1) .. O. Carryoutthesynthetic

1 + 0 - 49 + 8 + 561-1 - 1 + 1 + 48 - 561-1-48+56+ 0 Hence /(~) E (~ + 1)4>(~) where 4>(~) E ~. - ~. - 4&: + 56. Then 4>(0) ... 56 .. 8·7, 4>(1) = 8, 4>(-1) = 102 ... 2·3 ·17. The divisors 6 of 4>(1) which we list are 1, 2, 4, -4, 8, :-8. The eorrespondirig

SEC.

3]

159

RATION AL ROOTS

values of a = 6 + 1 are 2, 3, 5, -3, 9, -7 and we retain only 2,-7. Then a + 1 = 3, -6 are both retained. By the synthetic divisions 1-1 - 48 + 56~ 2+ 2-92 1+1-46-36

1 - 1 - 48 + 561-7 -7+56-561-8+ 8+ 0

we see that the integral roots of J(z) are -1, -7. Note that we may now compute the remaining roots of J(z). They are 4 + 2 0,

4-2'\1'2. II. Determine the integral roots ofJ(z)

E

Z4 -

3z1

-

63z - 72

=

o.

Solution WecomputeJ(O) = -72,J(I) = -137,J(-I) = -11. Thedivisors of J( -1) which we list are c = -1, 11, -11 and a = c - 1 = -2, 10, -12. We delete 10 and see that a-I = -3, -13 do not divide J(I). Thus J(z) has no integral roots.

3. Rational roots. A rational root r ~ 0 of an equation (1) with integral coefficients may be expressed as a fraction

r=l!. q.

where p and q are nonzero integers whose g.c.d. is unity; q may be taken to be positive. By Theorem 2 the integer p is a root of the equation aoXn + qalXn-1 + . . . + qnan = 0 and, by Sec. 2, p divides gnan. Since p and q are relatively prime, so are p and qn. By Lemma 7 in Sec. 8 of Chap. II we conclude that p divides an. The number q/ p is a root of anxn + . ... + ao = 0 and q is a root of anxn + PCln.-1Xn- 1 + . . . + pnao. Then the positive integer q divides pnao and by the lemma referred to above we see that q divides ao. We have proved the following: Theorem 5. Let p/ q be a nonzero rational root, in least terms, oj the equation (1) with integer coefficients. Then the integer p divides an and the positive integer q divides ao. In the case where ao = 1 the positive integer q divides 1 and q = 1. This yields the ·following:

160


[CBAP.vn

Theorem 8. Every rational root oj an equation with inte-

-ural coejficientIJ and leading C06jficient 1 is an integer. Theorem 5 may be used to find the rational roots of We list the divisors p of CIn, the divisors q of ao, the corresponding quotients plq and then use synthetic division to compute J(plq). However it is simpler to select a number t such that the equation

J(z) == O.

(2)

has integral coefficients. The rational roots of this new equation are integers u and may be determined by the method of Sec. 2. The rational roots of J(z) are then the numbers ult. Observe that a value of t which will suffice is t == ao. However it is preferable to use the least positive integer t for which formula (2) has integral coefficients. lUustrative Examples I. Find the rational roots of 1(1e) 55 &;8

+ 4z1 + 5z -

6 = O.

Solution The positive divisors of 6 are 1, 2, 3, 6 and those of 3 are 1, 3 so that the possible rational roots are 1, -1,2, -2, 3, -3, 6, - 6, i, - i, i, - i. Apply Theorem 2 with t = 3 to C?btain an equation g(1/) a i(3y8 + 121/1 + 45y - 6 . 27) 1/8 + 4y1l + 151/ - 54 = O. We multiply our previous possibilities by 3 to obtain 3, -3,6, -6,9, -9, 18, -18, 1, -1, 2, -2 as possible rational roots of g(1/). Note that all numbers divide 54 but some divisors of a.. == 54 do not occur in this list. We compute g(l) ,;.. -34, g( -1) = -66, omit d = 1, -1, and see that d - 1 divides 34 only for d = 3, 18, 2. Then d + 1 divides 66 only if d == 2. By synthetic division g(1/) = (yi + 6y + 27)(1/ - 2) and the only rational root of 1(1e) is f. Its other two roots are -1 + ~ ..-2,. ...., 1 - v'=2, since the polynomial whose roots are i those of 1/1 + 6y + 27 is leI + 24; + 3 -= (Ie + 1)1 + 2. n. Find the rational rOots of 1(1e) = 48z8 + 1&1 - 31e - 1 .. 0.

=

Solution Since a.. = 1, the rational roots of this equation are the reciprocals of the integral roots of g(1/) = 1/8 + 31/1 -.l6y - 48 = O. Now g(l) ... -60, g( -1) = -30. The divisors h of -30 .. -3' 5 . 2 are 1, -1,2, -2,3, -3,6, -6,5, - 5, 10, -10,15, -15,30, -30. Then d - h - 1 divides g(O) only if d = -2, -3, 2, -4, 4, -6, -16, and

161

RATION AL ROOTS

SEC. 3)

d - 1 divide.s -60 only if d = -2, -3,2, -4,4. We now compute = 8 + 12 - 32 - 48 ~ 0, g( -2) = -8 + 12 + 32 - 48 ~ 0,

g(2)

1 + 3 - 16 - 481-3 -3+ 0+48 0- 16 0

+

=0

and see that g(y) = (y2 - 16)(y -.\- 3). The roots of J(:c)

- i,l, -1. III. Determine the rational roots of :c 3

-

48:c

are

+ 64 = o.

Solution

+

We employ Theorem 2 with t = 1 to obtain g(y) = yS - 3y 1 = o. Evidently g(l) ~ 0, g( -1) ~ 0 and the equation has no rational roots. EXERCISES

1. Determine the integral roots of the following polynomials: 15:&2 + 54:& - 54. 8:&2 + 9:c - 36 19:c2 + 82:& + 60 17:c2 + 78:& - 56 4:&2 - 95:c + 198 :c 2 - 64:& - 72 2:&8 + 39:&2 - 188:& - 176 :c8 - 14:&2 + 9:& + 324 :c8 + 15:&2 + 84:c + 196 3:&3 + 2O:c2 - 131:c - 132 :c4 + :c 8 - 3O:c2 - 1O:c + 200 2:c4 - 47:c 2 - 42:c - 135 :c4 - 2:c 8 - 6x 2 - 2:& - 24 :c 8 2:c 2 - 15:c - 52 :c4 + 15:&8 - 2:c 2 - 34:& - 60 :c 8 - 9:&2 - 24:& + 216 :c4 + 4:&8 - 75:&2 - 324:& - 486 :c5 - 24:&4 149:c3 - 352:&2 + 294:& - 608

(a) :c 8

(b) (c) (d) (e) (J)

(g) (h) (i)

(;) (Te) (l) (m)

(n) (0) (p) (q) (r)

-

2:c8 :c 8 + :c 8 :c 8 :c 8 +

+

+

Am. 3. Am. -5. Am. 2, 11, -9. Am. None. Am. -7. Am. 4, -5. Am. 4. Am. 2, -15. Am. 9, -9.

s. What can be said about the rational roots of the polynomials above in all cases except (g), (i), and (l)? 3. Find all rational roots of the following:

+

X 2= 0 6z8 - 9:c2 + 2:& - 3 = 0 2O:c3 + 11:c 2 - 8:& + 1 = o· :c 3 + 3:&2 - 2:c + 15 = 0 3:c 8 - 37:c 2 93:c - 27 = 0 203:c 8 + 351:c2 + 39:& - 1 = 0

(a) 2z8 - 4:&2 -

(b) (c) (d) (e) (J)

+

Am. 2. Am. -1,1, i. Am.

i, 3, 9.

162

REAL ROOTS OF REAL EQUAT,IONS

(g) 2:1;' - z

+

+5=

Am. ,None.

0

+

(/I) 15:1;1 7z1 - 7z 1 = 0 (1.1 2:!;t + 16z' + 19z1 + 001: + 27

(J1

(I:)

(I) (m) (ft,)

[CHAP.VII

27z1 + 2z - 3 .. 0 9z8 + 9z1 - Z - 6 .. 0 6zI + 21z· + 26z - 13 - 0 4r - u 8 - 3z1 - Z - 6 - 0 2:1;8 - 15:1;1 + 42z 44 .. 0

=0

Am. None.

54z8 -

Am. f. Am. -1.

+

4. Upper and lower bounds. Let a and b be real numbers and a be less than b. Then we, designate the set of all real numbers z between a and b by a ~ z ~ b. If we omit a from this set, we designate the result by a 6,

+

Also /(-z) ==:c4 - 2z3 11Z2 - Z - 81 and k = 1, = 82. We examine the fractions 2, and -L2 = 1 ~f. = 1.;-. Hence - 1lf < r < 4.

and so U

= 4.

G = 81, so that -L1 = 1 + 81

-tt = ¥

=

+

ORAL EXERCISES

1. What can be said immediately about the real roots of (a) (b) (c) (d)

+ + +

+ 2:1; + 1 = 0 + 2Z2 = 0 + 2Z2 - 4:1; + 3 =

z' 3z2 Zl 3z3 z, - 3z3 Z6 2z' -

3z8 - 631 = 0

0

(e) (J) (g) (h)

-Z3 - 6z 2 - 53l - 4 = 0 -z' 2:1;3 - Z2 = 0 -Z2 2z - 1 = 0 Z3 - 3z2 3z - 1 = 0

+ +

+

2. What are the numbers k and G of the two methods for the following equations? (a) 2z3 - 4b2 320z - 800 = 0 (b) 4:1;6 - Sz' 20:1:8 100z2 - 73:1: 5 = 0 (c) z' - 36:1;2 49z 50 = 0 (d) 2:1;6 - 6:1;2 - 7z 14 = 0 (e) 4:1;8 6z7 - 18z' - 6z3 3z - 20 = 0 (I) Z7 4:1;6 4:1;' - 9z - 64 = 0 (g) z' 4:1;8 6z2 - 22z - 27 = 0 (h) 3z' - 12z2 15:1: - 8 = 0 (~) Z8 - 16z 6 24:1:' - 25:1: 2 3z 2 = 0 S. What are the fractions we compute in (e) of Oral Exercise 2? 4. What are the upper bounds in (e)?

+ + + + + + + + + + + + +

+

+

+ +

EXERCISES

1. Compute best upper and lower bounds for the polynomials of Oral Exercise 2. 2. Compute best bounds for the polynomials of the exercises of Sec.3.

6. Proof of the method of radicals (FULL roots of J(x) coincide with the roots of F(x) ==

COURSE).

..!. f(x) == x" + C1Xn - 1 + . . . + en. ao -

The

160

~EAL

ROOTS OF REAL EQUATIO'NS

We define k and G as in Sec. 4 and oDserve that first negative coefficient of F(x). Also define

!CHAP.VtI

Ck

is the

G d =-.

ao

Then -d is a negative coefficient of.F(x) such that d ~ 1c.;1 for every negative coefficient c.; of F(x). We let U = 1 + {Id and suppose that b ~ U. Then b - 1 ~ {Id and (b - 1)k ~ d~' But U > 1 and so b > 1, bk-l ~ (b - 1)k-l. Then bk-l(b - 1) ~ (b - 1)k ~ d and the number

The terms in F(b) with coefficients Cl, C2, • • • negative and so F(b) ~ bn + ckbn-k + .. every coefficient c.; ~ - d and so F(b)

~ 8,

8 =

bn - d(bn- k + bn-k-l

,CTc-l

are not But

+ en.

+ ... + 1).

However 8(b - 1) = bn(b - 1) - d(bn-k+l - 1) = bn+l - b'" - dbn-k+l + d = bn-k+l(bk - b'-l - d) + d = ebn-k+l + d

> 0.

Thus 8(b - 1) > 0, 8 > 0, F(b) > O. Since ao > 0, the number !(b) > 0, for every b ~ U, U is an upper bound for the real roots of lex) = 0. We have given the proof above mainly as a sample of a type of argument in the study of polynomials with real coefficients. We shall not give a proof here of the method of fractions. ~t may be found in L. E. Dickson's "First Course in the Theory of Equations." 6. Isolation of the real roots of a real equation. As we observed in Chap. VI, any polynomial lex) of degree n with real coefficients is expressible as a product

SEC.

6]

ISOLATION OF THE REAL ROOTS

167

where rl, ••• , rt are the real roots of f(x) = 0 and q,(x) is a product of factors of the type X2 - 2dx

+ d2 + e2,

for real numbers d and e such that e ~ O. The roots of q,(x) are the imaginary roots of f(x) = 0, the degree n - t of q,(x) is an even integer, and (_I)n-t = I, (-I)" = (-I)t. If c is any real number, then c2 - 200

+ d2 + e2 = (c -

d) 2

+ e2 > O.

Hence if c is real, the value q,(c) is a positive real number. If c > r, for i = I, • • • , t the factors c - r, are all positive and f(c) has the same sign as ao. If c < ri for i = I, ••• , t the numbers c - r, are all negative and f(c) .has the same sign as (-I)tao = (-I)"ao. We state this result. Theorem 7. Let U be any upper bound and L be any lower bound to the real roots of f(x) == aoX" + ... + a". Then f(c) has the 8ame sign as ao for every c ~ U and f(c) has the 8ame sign as (-I)"ao for every c ~ L. . We also prove the following: Theorem 8. Let a < b where a and b are real number8 and f(a) ~ 0, feb) ~ O. Then there is an even or an odd number of real roots between a and b according as the signs of J(a) and f(b) are the 8ame or opposite. For the sign of f(b)/f(a) is the same as the sign of b -- rl b - r2 ... b --rt Q= . a - rl a - r2 a - rt

If r, > b > a, then b - ri and a - r, are both negative so that their quotient is positive. If b > a > rj, then b - ri and a - ri are both positive and their quotient is positive. But if b > rk > a, the quotient (b - rk)/(a - rk) is negative. Thus the sign of Q is (_1)8 where 8 is the number of real roots rk between a and b. It follows that 8 is even or odd according as the sign of f(b) and f( a) are the same or opposite.

168

REAL· ROOTS OF REAL EQUATIONS

[CHAP. VB

The length 01 the interval between a and b is Ib - al. Evel'Y real root r of I(z) lies in some interval a < r < b such that b - a :i! 1, and such tha~ no root of I(z) distinct from r is in this same interval. When we find such an interval, we shall say that we have i80lated r. H m is any integer such that I(m) and I(m 1) have opposite signs, there is at least one real root of I(z) in the interval m < z < m + 1. Thus a procedure for isolating all real roots of I(z) is that of a computation of 1(0), 1(1), I( -1), 1(2), I( -2), etc. It is effective if no interval m of h(x) = O. If d has been correctly determined, the equation h(x) = 0 will have a root t between 0 and itr. The powers t2, t3, ... , tn are small real numbers and t will normally lie between two numbers

-nr

h=

Cn -, Cn-l

where

Moreover the difference It2 - til will be small and we shall be able to determine a digit 6 such that

r = a

+ d/1O + 6/100

to two decimal places. The process of diminishing roots is continued and it will be true in all cases that, after several steps, our process of squeezing the root between the roots of two linear equa-

176

REAL ROOTS OF REAL EQU .A.TIQNS

[CHAP. VII

tions will become valid and yield several digits instead of one. We illustrate this fact below. nlustrative Examples .1. Compute the positive real root of :1:3 transformations.

-

63; - 1 = 0, making three

Solution The equation has one positive root and no rational root. Also /(2) = "':'5,/(3) = 8 so that 2 < r < 3. 1 + 0 - 6 - 11~ 2+4-4 2-2-5 2+8 4+6 2

~ + 63;2 + 6:1: - 5 682 +68-5=0 -6 + "';f":36""'"+""'--;;-'12=0 -6 + VI56 8 = 12 = 12 12 < VI56 < 13, O.d = 0.5

g(:I:) =

1+6 1 + 6.0 + 6.00 - 5.000~ 0.5 + 3.25 + 4.625 6.5 + 9.25 -0.375 11,(:1:) = :1:8 + 7.5:1: 2 + 12.75:1: - 0.375 0.5 + 3.50 0.375 002 . I t = 12.75 = • 9 approXlmate y 7.0 + 12.75 0.5 . 1 + 7.5 1 + 7.50 + 12.7500 - '0.37500010.02 0.02 + 0.1504 + 0.258008 7.52 + 12.9004 - 0.116992 0.02 + 0.1508 7.54 + 13.0512 0.02 1 + 7.56 k(:I:) =;0 :1:3 + 7.563;2 + 13.0512:1: - 0.116992 0.116992 'Ul = 13.0512 = 0.0089+ _ 0.116992 - (0.01)8 - 7.56(0.01)2 OOOB 'U2 13.0512 = . 9+ The correct value 'U definitely lies between 'Ul and 'U2 in this case and so . II. Find the negative root of :1:8 - 3:1:2 - 4:Il + 11 =- O.

r =,2.5289, approximately.

Solution -/( -:I:) == :1:8 + 3:1: 2 - 4:Il - 11 == F(:I:) and F(l) Hence 1 < -r < 2. - . -.-

= -11, F(2) = 1.

SEC.

1 + 3 - 4 - 111!. 1 +4+ 0 4+0-11 1+5 5+5 1

g(z)

= Z8 + 6z1 +

1

" =

+ 6.0 + 5.00 0.9 + 6.21 + 6.9 + 11.21 0.9 + 7.02 7.8 + 18.23

5z - 11

-5 + 12

~5 + V25 + 240

= 11+, d = 9

+ V25 + 264 = 12

11.000/0.9 10.089 0.911

v28l) = 1

12

= -5

8

1+6 1

177

HORNER'S :METHOD

9]

12

h(z) = Zl + 8.7z' + 18.23z - 0.911 t _ 0.911 - 0 049+ - 18.23 - • e= 4

0.9 1 +8.7 1 + 8.70 + 18.2300 - 0.91100010.04 0.04 + 0.3496 - 0.743184 8.74 + 18.5796 - 0.167816 k(z) = Zl + 8.82z1 + 18.9308z - 0.167816 0.04 + 0.3512 8.78 + 18.9308 0.167816 00088 1£1 = 18.9308 = . 0.04 1 8.82 0.166933 _ 0 0088 1£, == 18.9308 . -r = 1.9488, r = - (1.9488) a.pproximately

+

EXERCISES

1. Use Horner's method to find all the real roots of the following equations. Make at most three transformations and obtain/our decimal places in each case.

(a) Zl .. 2 (b) Zl = 3 (c) Zl = 7 (d) Z8 - 5z - 1 ... 0 (e) Z8 + Zl - 2z - 1 (J) Zl + 3z' - 1 == 0 (g) Z8 + 3z - 1 = 0

(h) Z8 - 2z - 6 = 0

=

(i) 2z8 - 6z1 - 1 = 0 (J) Z8 + 9z - 6 ... 0 (1:) Z8 + 3z + 6 0 0 (l) Z8 + 9z + 2 = 0 (m) Z8 - Zl - 3 = 0 (tt) Z8 - Zl - 5 = 0

=

Am. 2.1799 Am. 3.0536 Am. 0.6378 Am. -1.2879. Am. -0.2208. Am. 1.8637 Am. 2.1154

Z. Use Horner's method to find all the real roots of the following equations, making two transformations: (a) Z4 - 3z + 1 (b) z, - 4z 2

=0

+ =0

(c) z' + Z8 - 2 = 0 (d) z, + 2z1 - Z + 1

=0

(e) Z4 (J) z, (g) Z4 (h) z, -

+

Z8 - 3z - 1 6z - 1 == 0 Zl - 1 = 0 7z' - 2 .. 0

=0

10. Other methods (FULL COURSE). The derivative of I'(x) is a polyp.omial f"(x). Let us suppose that we have isolated a simple real root r of f(x) in an interval a ~ x ~ b and that the sign of f"(x) is the same throughout the interval. Then one of- f(a)- and- feb) will have the same sign as f"(x) in this interval. Let us call this number g. It can be proved that f(g) c = g - f'(g) is closer to r than is g. Mter we compute c, we make a second such computation and determine a number f(c) Cl = C - f'(c)' which will be closer to r than is c. The method may be continued as far as desired. This procedure for approximating r is known as Newton' 8 method.

FIG. 9.

Another method for computing r is called the method of interpolation. The theory of similar triangles ~nd Fig. 9 may be used to show that d - af(b) - bf(a). - feb) - f(a)

Moreover in the case described for Newton's method dis closer to r than is either a or b and, indeed, if we compute c as above it can be shown that r is between c and d. The method of interpolation involves the least theory of our three methods for computing roots, but the computations themselvEls are not simple. We recommend that no' exer-

BEC.10]

179

OTHER METHODS

cises on this and Newton's method be assigned. If exercises are desired, those of Sec. 8 may be used. mustrative Examples I. Use Newton's method to approximate the positive root of :1:8 - 63; - 1 = O. Solutiun The root lies between 2 and 3. AlsoJ'(:I:) = 3:1:1 - 6,

>0 = 8; so g = 3. Then

/"(:1:) = 6:1: for 2 ;:ii

:I:

;:ii 3. C

=

But/(2) = -5,/(3) 1(3) 8 . 3 - 1'(3) = 3 - 21 = 2.7, approxuna.tely.

Also Cl

= 2.7 -

1(2.7) 2 7 2.483 2 55 • tel 1'(2.7) == • - 15.87 = . , appro:nm.a y.

Solution 2 In this solution we shall diminish the roots in each case but use the Newton method of approximation.

+0 -

6 - 11~ g(:I:) = :1:8 + 9:1:1 + 213: + 8 ' 3 + 9+9 g'(:I:) = 3:1:1 18:1: 21 3+ 3+8 g(O) -8 3 + 18 C = 0 - 0'(0) = 2f = -0.3 6+21 3 1+9 9.0 + 21.00 + 8.0001-0.3 k(:I:) = :1:3 + 8.1z1 + 15.873: + 2.483! 0.3 - 2.61 - 5.517 k(O) -2.483 8.7 + 18.39 + 2.483 - k'(O) = """I5.87 == -0.15 0.3 - 2.52 8.4 + 15.87 0.3 8.1 Am. r = 3 - 0.3 - 0.15 = 2.55.· 1

+

1+ 1+

:1: 3

+

II. Use the. method of interpolation to find the positive root of - 63; - 1 = O.

Solution /(2)

= -5,/(3) =

8, and so

d _ 3/(2) - ~(3) _ -15 - 16 = 31 = 237 - 1(2) - 1(3) -13 13 .

180


[CHAP.VII

In the next step we must compute 3/(2.37) - 2.37/(3) 1(2.37) - 1(3) . Such a oomputation may be conveniently made only with a computing machine. If instead we diminish the roots by 2 to obtain g(z) =

then g(I) = 8,

a=

Z8

+ 6z + 6z 2

5,

0, b = 1, and

d

= g(O)

g(O) _ g(I)

=

-5

-13

= 0.37

+

is neceBBarily the same value that we obtained above. The value 0.3 and even the value 0.4 are not so good as the 0.5 obtained by Horner's m~thod and so we shall not continue our solution directly. If we diminish the roots by 0.5 instead, to obtain h(z) =

Z8

+ 7.5z + 12.75:1: 2

C).375,

the next approximation according to the method of interpolation is

O.lh(O) 0.0375 _ 0 027 h(O) - h(I) -= 1.351 - . , which is, in this example, also Dot so good as the value obtained at the corresponding stage in Homer's method. '

11. The function concept. We shall close the present chapter with a discussion of the mathematical notation called the function concept. The germ of this idea has already been used here in our process of isolating the roots of an equation f(x) = ,0 by the q.se of a table consisting of values x == a and corresponding values f(a). We define a variable to be a symbol together with a set of objects which we call its range. If the range consists of a single object, we call the symbol a constant. The function concept involves two variables and a correspondence. We make the following: Definition. Let x and y, be two variable8 which are 80 related that to every element of the x range there corre8ponds one or more elements of the y range. Then y is called a function of x.

SliIC.

12]

OPERATIONS ON FUNCTIONS

lSl

The part played by the variable y in this relationship is different from that played by x. It is customary then to refer to x as the independent variable and to y as the depend-

ent variable. If it is true that to each element of the x range there corresponds exactly one element of the y range, we say that y is a Bingle-valued Junction of x. In all other cases we call y a multiple-valued or many-valued function of x. Most of the functions which the student will study :in plane a.naJ.ytic geometry are many valued, and some of the functions of plane trigonometry are infinitely many valued. We shall indicate the fact that y is a function of x by writing y = J(x). Of course we use other letters than J or, for that matter, than x and y. If we replace x in a single-valued function 'U = J(x) by an element a from the x range, the corresponding element of the y range is designated by J(a) and is called the value oj J(x) Jor x ....·a. This ~ the source of the language we have already used for the values J(a) of polynomia~ J(x) at x = a. Our definition of a function may be extended to functions s = J(x, y) of two independent variables x and 'U or, more generally, to functions of n variables. We let Xl, X2, ••.., x" be n independent variables and y be a variable: Suppose that every sequence aI, aI, • • • , an of elements al of the Xl range, al of the XI range, . • • , an of the x. range corresponds to one or more elements J(al, ••• , an) of the y range. Then we call the dependent variable y a Junction oj Xl, X2, • • • , x. and call y a Bingle-valued Junction oj Xl, X2, •.• , X" if every J(al, a2, ••. , an) is unique. The concept whose description we have given is of vital importance in all mathematics. We recommend that the student and the instructor construct illustrations of this concept from life as well as from the mathematics of this text. 12. Operations. on functions. The symbols n, a, b, c, i, j, . . . of Chap. I were varia.bles whose common range

182

REAL ROOTS O~ REAL EQUATIONS

[ClU.P. VD

was the set of all natural numbers. All the variables of Cha.p. II had the set of all integers as their common range and the ranges in Chap. III were other number systems. The range of a variable need not be a number system at ail. However most of the variables emphasized in elementary mathematics have as their ranges subsets of the complex number system. We shall restrict our attention to such variables. H'Y - f(:I:.) and s = g(z) are two functions of the same variable z, their sum 'Y + S' = h(z) = f(z) + g(z) is that function whose values h(a) are defined by h,(a) = f(a)

,

+ g(a)

'

for every a of the z range.. The difierence f(z) - g(z) and the product f(z)g(z) are functions defined similarly. The' quotient J(z) g(z)

~ a funcUon'defined only for those numbers a of the ~ range for which g(a) is not zero. We observe now that every polynomial J(z) defines a func,tion 'Y = f(z) where the z range and the 'Y range are subsets of the complex number system. H g(z) is a polynomial and s = g(z) is the corresponding function, the fQrmal sum polynomial J(z) + g(z) defined in Chap. IV defines the sum function 'Y + s. Similarly 'Y - s and 'YS are defined, respectively,. by the difierence and product polynomials. The function defined by' their formal quotient is the quotient function mentioned above, and the term roJ,ional function, which we have already used for the formal quotient, has its origin in the function concept. , The zero function is the function all of whose values are .zero. H f(z) and g(z) are two functions having the same z and 'Y ranges, we say that J(z) and g(z) are identical and ~te ...... 1' ~.

,"

J(z)

!E!

g(z)

8EC.

13J

THE EQUATIONS OF A CURVE

183

(read "f of x is identically equal to g of x") if it is true that f(a) and g(a) are the same elements of the y range for every x of the x range. Then f(x) ;;e g(x) if and only if f(x) - g(x) is the zero function, that is, f(x) - g(x) O. This is the inspiration of our definition of polynomial equality, the reason why we have regarded as equal all polynomials with coefficients all zero, ~nd the motivation of our use of the notationf(x) g(x) for equal pol~omials. The functions defined by polynomials and rational functions are functions defined by an algebraic formula. We shall have other examples in our next section. In Chap. VIII we shall define the functions of trigonometry by a geometric procedure involving angular measurement and the use of a coordinate system. 13. The equations of a curve. If the ranges of an independent variable x and a dependent variable y = f(x) are subsets of the real number system, we call y a real function of a real variable x. Then there is a corresponding graph of the function which involves the use of a coordinate system in the plane and is the set of all points P with coordinates x = a and y == b such that b == f(a). The graph might be a line, a circle, or some other geometric object which belongs to the family of what the mathematician calls curves. Let a curve C be defined geometrically. For example, a circle is the curve of all points at a :fixed distance r from a • :fixed point called its center. Let us also suppose that we have a polynomial F(x, y) with real coefficients. Then we shall say that the equation

=

=

F(x, y)

== 0

is an equation of the curve C if 1. Every point on C has coordinates x == a, y == b, such that F(a, b) == O. . 2. Every real number pair (a, b) BUch that F(a, b) == 0 defines a point with coordinates x = a, y == b on C.

184

RlilAL ROOTS OF REAL "EQUATIO-XS

[CHAP.VII

Plane analytic geometry is concerned with the study of curves whose equations are polynomial equations F(z, '1/) = 0,

and we shall return briefly to this study in the final two sections of Chap. VIII. " An equation F(z, '1/) = 0 of a curv:e 0 defines a function '1/- =" f(z) whose graph is O. This function is defined by the correspondence in which each real number z = a corresponds to the real roots '1/ ~ b of the equation -

F(a, '1/) = 0

and is called an algebraic Junction of z. When the degree in '1/ of F(z, '1/) is greater than 1, an equation F(a, '1/) may have several real roots and then '1/ will be a multiple-valued function of z. For example, the function defined by :x;I + '1/" = 1 is a real function of a real variable z whose range is the interval -1 ~ z ~ 1. Each value z = a such that -1 < a < 1 corresponds to the two values v1 - a", - v1 - a l of '1/, and hence to two points on the circle.

CHAPTER

vm

VECTORS IN THE PLANE

(FULL COURSE)

1. Vectors. In the present chapter we shall tie up the algebraic theory of vectors and linear equations with trigonometry and analytic geometry. Our presentation should' provide a strong foundation for these related subjects as well as a condensation of material that is common to all of them. We shall regard this entire chapter as optional material. A vector in the plane is a real number pair (z, 'II). We may interpret it geometrically either as the point P in the

,

~----------=---~~~----~~----~---.

FIG. 10.

plane whose coordinates relative to a fixed rectangular Cartesian coordinate system are z and y, or as the directed line segment OP from the origin 0 to P. We may also interpret it algebraically as the complex number z + yi. We shall usually write P = (z, y) here and shall use the words vector and point interchangeably. H t is any real number, the scalar product t(z, y) of the number t and the vector (z, y) is the vector (tz, ty). We construct Fig. 10 in which OHP, OH 1P 1, OH,}'" are evidently similar right triangles. Then, if P is not the origin, and P = (z, y), PI = (Z1, Y1), P s - (zs, Ys), we use similar 185

186

VECTORS IN THE PLA.NE

(CIUP. VIII

triangle properties to obtain Zl Z

= YIY = tl ~ 0 ,

.Z. z

= y.Y = #~

~ O.

"l_

It follows that if (Z, y) ~ (0, 0) every point on the same line through 0 as (Z, y) is a vector t(z, y). Also t ~ 0 or ;:iii 0 according as t(z, y) is or is not on the same ray from 0 as (z, y). Conversely, every vector t(z, y) defines a point on the same line through 0 as (z, y). For if PI = (tz, ty) the legs of the riglit trl&ngJ.e OPIHI and OPH are proportional, the right tri.angles are similar, the points 0, P, and PI are collinear. The norm of a vector (z, y) is the nonnegative real number z· 'V.. The length of (z, y) is

+

r =

V z'l. + y.

ie;

O.

We shall call a vector of length 1 a unit vector. All unit vectors are points on a circle with center at the origin and radius 1. If P is any point except 0, there are precisely two unit vectors on the line through 0 and P. They are ."

Uo

-z, -y) = (r r ,.

and P == rU == -rUo. Thus if we select any unit vector on a line through 0 and U, all vectors P on the line have the form tU where t is a real number. Then tU is on the same ray from 0 through U if and only if t !l: 0, and the nonnegative real number is the length of P. ORAL EXERCISES

1. Give the norms and the lengths of each of the vectors (0, 0), (2, 0), (0, -2), (3,4), (1, 1), (5, -12), (6,8), (3, -3), (-6, -8), (-10, -24). I. Give a unit vector on the same ray as each nonzero vector of Oral Exercise 1. 8. Give a unit vector on the same line but not on the. same ~y as each nonzero vector of Oral Exercise 1.

187

ADDITION OF VECTORS

BEo.21

4. Express each nonzero vector of Oral Exercise 1 in the form r(u, fJ) where r > 0 and (u, fJ) is a. unit vector. 6. Which vectors in Oral Exercise 1 are collinear and which are on the same ra.y?

2. Addition of vectors. The

BUm

== (Xl, Y1) + (X", y,,) of two vectors P 1(X1, Y1) and P" == (x", y,,) is defined to be PI

+ P"

the vector (Xl + X'" Y1 + y,,). In Fig. 11 PI and P" are two given vectors. We complete a parallelogram by drawing a line through PI parallel to OP" and a line through P" 11

--------~~--~~~--~-----L-------M

FIG. 11.

parallel to OP 1• These lines intersect in a point Q, and we show below that Q == (Xl + X., Y1 + y,,) == PI + P". This graphic method of constructing the suni of two vectors is called the parallelogram law for the addition of vectors. In the case of the di.agram. we see that the right triangles OP1H 1 and P .QR are equal and so P~ ==

H.Ha == X'" OHa == OH" + 1lJfa == Xl + X'" HaQ == HaR + RQ == H,p" + RQ == Y.

+ Y1.

A complete derivation merely requires the construction of additional diagrams and will be omitted. The vector (0,0) has the property that (0,0)

+ (x, y)

== (x, y)

+ (0, 0)

== (x, y).

188

[CHAP. vm


We call (0,0) the zero vector. The vector (-z, -Y) - -1(z, y) has the property that (z, y) + (-z, -y) = (0,0)

.

and so is called the negative, of (z, y) and is designated by - (z, y). It is a vector of the same length and on the same line through 0 as (z, y). It is not on the same ray. The difference P'J - P l of two vecto~ is given by (z'J, Y'l.) - (Zl, Yl) -, (z'J - Zl, YI -1/1). In Fig. 11 08 is parallel to P lP I and the length of 08 is the length P lPI. Then PI = P l 8 so that 8 == P'I. - Pl. It follows that the undirected distance d between P l and PI is given by the diBtance formula

+

(1)

d =

V (ZI -

Zl)'J

+ (YI -

Yl)'J.

ORAL BDRCISBS 1. Compute the coordinates of P 1 PI a.nd PI - P1 for each of the following pairs of points PlJ p.:

+

(d) (1, 2), (-4, -10) (e) (0, -1), (5,11) (f) (3, -2), (5, -1)

(a) (2, -1), (5, 3) (b) (-1,0), (1,1) (c) (-1, -2), (2, 2)

I. Compute the length of .

P,p.

for each line segment

P,p. in Oral

Exercise 1.

3. Angular measurement. An angle (J may be defined to be a rotation in the plane of a ray about its end point. The ray then starts its rotation as what we shall call the initial Bide of (J and ends its rotation as what we shall call the terminal Bide of (J. Angles may be compared by selecting the same ray as the initial side for all angles. Let us call this ray the polar azUl and its end point the pole. In Fig. 12 we have made a counterclockwi8e rotation of revolutions (complete rotations). We shall always measure counterclockwise rotations by positive real numbers and clockwise rotations by negative real numbers. The zero angle is that in which no rotation takes place. It should be clear now that,

In

smc • .3]

ANGULAR MEASUREMENT

189

if we use the revolution as a unit of measure, we shall have set up a one-to-one correspondence between the real num, ber system and the set of all angles. The use of the revolution as a unit of measurement is inconvenient and it is customary to divide it into 360 equal parts called degree8. Thus 3600 == 1 revolution, and the angle ill Fig. 12 is 3900 • Our definition of angular measurement, using the revolution as a unit of measure, has used the property of plane geometry which states that an angle at t'M center of a circle iB meaBUrea by the arc it, BUbtendB on t'M circumference.

~m FIG. 12.

We may use this property a bit more explicitly in introducing a new unit of measurement. We define an angle of one radian to be that angle which BUbtendB an arc wh08e length iB one unit on the circumference of a circle who8e radiU8 i8 one unit. Since the circumference of this circle is 2r units of length we have the fundamental relations 1 r6fJolution == 27r radiam == 3600 between our units. We shall use the radian as our standard unit in the next sections. The angles in' the accompanying table occur fmquently in the exercises of trigonometry. RevolutioDS Degrees Radians 0'

~

t I i i t

1

0

0

30·

.../6 .../4 .../3

45 60 90

180

'IT/2 'IT

270

3w-/2

360

2r

190

VECTORS IN THE J»LANE

[CIIAP.vm

BDRCISB

Make a table like the accompanying one for the angles obtained by adding 11'/2, 11', 3r/2, -11'/2 to each of the first four angles of that table.

,. Polar coordinates and trigonometric functions. The representation of a vector as a directed line segment of a given length suggests the introduction of a new type of coordinate system in the plane. We select a unit of measurement as. well as a pole 0 and a p~lar axis as defined in Sec. 3. Then every real number pair [r,8] such that r ~ 0 will define a ray which is the terminal side of an angle 11

W FIG. IS.

of 8 radians whose initial side is the polar axis. There is precisely one point P on this ray at a distance of r units from.J) and we call [r, 8] a set of polar coordinate8 oj P. Conversely, let the point P be given. Then the distance r = OP is unique and we may select 8 to be any angle whose terminal side is the ray from 0 through P. If r = 0, the angle 8 is arbitrary. Otherwise any two determinationB of 8 diifer by an integral multiple 0/ 21r radianB. Polar coordinates may be related to rectangular coordinates by rrw.king the polar a:r:i8 coincide with the poBif:ive ~ a:ti8 8UCh that the pole i8 the origin. The connection between coordinates in the two systems then involves the introduction of what are called trigonometric Juncti0n8.

SEC.

4]

191

POLAR COORDINATES

In Fig. 13 we suppose :first that only 6 is given. Then the terminal side of 6 determines a unique unit vector U on it. The x coordinate u of U is called the coBine of 6, its y coordinate v is called the Bine of 6 and we write u == COB 6, v == Bin 6. Now cos 6 and sin 6 are uniquely determined real numbers, for every real number 6, and we have defined two trigonometric functions. They are real functions of a real variable 6 defined for all real values of 6. We observe that since U is a unit vector we have the relation sinl 6 + cosl 6 == 1, and thus that -1 ~ sin 6 ~ 1, -1 ~ cos 6 ~ 1. Convel'Bely, if g is any real number such that -1 ~ g ~ 1, the vectors (g, h) and (h, g) defined for h == gl are unit vectors and g is the sine of an angle, and the cosine of another angle. We now pass. on to our connection between rectangular and polar coordinates. If P is a point whose polar coordinates are [r,6], its rectangular coordinates (x, y) have the property tl].at r == vxl + yl and (x, y) == r(u, v) so that x == fU, Y == ro. This is the same as x == r cos 6, y == r sin 6. Conversely, if x and y are given and (x, y) ~ (0, 0), the numbers r and 6 are determined by the relations

vI -

x r

• 6 ==-, Y sm

cos 6 == -,

(2)

r

The ratios (3)

tan 6 .

sec 6

sin 6 y = - - =-, cos 6

x

1

r

=cos-6 =-, x

cos 6 cot 6 ==sin6 1

x

=1/

esc 6 = sin 6 ==

r

1i

192


[CRAP. VIII

are called the tangent, the secant, the cotangent," and the cosecant of 8, respectively. Note that the first p~ of these functions are defined for all real numbers 8 eXcept odd integral multiples of 1r/2. For at the latter values cos 8 = 0, sin 8 = ± 1. The second pair are defined for all real numbers 8 except integral multiples of 1r since at the latter values sin 8 = 0, cos 8 = ± 1. Two angles are said to be coterminal if they have the same initial and terminal sides. Then they differ by integral multiple~ of 21r radians. It should be evident that such angles have the same trigonometric functions, i.e., f(8 + 2n1r) f(8) for all integers n, all real numbers 8, andf(8) interpreted as sin 8, cos 8, tan 8, cot 8, sec 8, or csc 8. We shall close our discussion of trigonometric functions with a further reference to Fig. 13. We let U = (u, v) be any unit vector so that U is on the terminal side of an angle 8, and let W = (uo, vo) be on the terminal side of cJ> = 8 + 1r/2. Drop a perpendicular from U to the z axis and let its foot be H. Similarly let the foot of the perpendicular from'W to the z axis be K. Then the right triangles" UOH and KOW have the same" l-ength 1 as hypotenuse and the sum of their angles at 0 is 1r/2. They must then be equal triangles and Uo = ± v, Vo = ± u. Since the terminal side of cJ> is in the quadrant following the terminal side of 8, we see that Uo and v have opposite signs, Vo and u have the same sign. Let the student draw three diagrams and verify this for all possible positions of the terminal side of 8. It follows that Uo = -v, Vo = u, and that

=

(4)

cos

(8 + ~) = - sin 8

sin

(8 +~) = cos 8

for all real numbers 8. EXERCISES

1. Use the properties of right triangles to construct the following table:

SEC.

51

193

ROTATION OF AXES

"

sin"

0

0

! 6

1

tan "

cot"

sec "

0

...

1

"2

a-

-va

"2

"2

1

1

V2

"2

1

-va a-

2

1

0

-va .

V2

T

i

-va

T

3 T

2

cos" 1

-va

-va

V2

"

0

csc "

-va

f

"

2

V2 f

.

-va 1

I. Extend the table above to angles 8 + 71'/2, 6 + 71', 8 + 37r/2, 8 - 71'/2 where 8 = 0,71'/2,71'/4,71'/3. 3. Determine the values of those trigonometric functions whose values are not explicitly given, if (a) (b) (c) (d) (6)

sin 8 = t, 0 < 8 < 71'/2 (J) sec 8 = ¥, 37r/2 < 8 < 2r sin 8 = t,7I'/2 < 8 < 71' (g) esc 8 = ti,7I'/2 < 8 < 71' tan 8 =- 1,71' < 8 < 3r/2 (h) sin 8 = i,O < 8 < 71'/2 cot 8 = - -h, 3r/2 < 8 < 21r (i) cos 8 = i,O < 8 < 71'/2 cot 8 = 1, sin 8 < 0 (3) sin 8 = i, cos 8 < 0

6. Rotation of axes. If we rotate the coordinate axes through an angle 8 as in Fig. 14, the. unit vector (1, 0) is rotated into a unit vector (u, v) where u = cos 8, v = sin 8. By formula (4) the unit vector (0, 1) is rotated into (-v, u). Let P = (x, y) be any point in the plane and regard the rotated axes as defining a new coordinate system in which OH = x' is the new ab8ci88a of P and OK = y' is the new ordinate of P. Then the (x, y) coordinates 'of H are x'u, x'v, and those of K are -y'v, y'u. But OHPK is a rectangle and the parallelogram law implies that P

= H + K = (x'u, x'v) + (-y'v,

y'u).

This yields the formulas (5)

{

X

y

= x'u - y/v, = x'v + y/U.

Formulas (5) are called the formulas for a rotation ofaxe8. They express the (x, y) coordinates of any point in the

194

[0lLU'. VIII


plane in terms of its. (:e', y') coordinates and the (:e, y) coordinates of the unit point on the :e' axis. Every point P = (:e, y) defines a complex number:e yi and formulas (5) are· equivalent to the complex number equation

+

(6)

x

+ yi =

(x'

+ y'i)(u + vi).

'I'

------~~~~--~~------~---.

FIa.14.

+

+

+

For formula (6) is :e 'Vi = :e'u - y'o :e'm y'ui. Since (u tn)(u - tn) = u· vi ,;,.. 1, formula (6) is also equiv. alent to

+

(7)

x'

+

+ y'i =

(x

+ yi)(u -

and hence to (8)

{

X' =

l'

xu

vi),

+ yv, + yu.

= -xv

These final formulas are called the 8oZtJed!qrm of the rotation formulas (5). ORAL BXEltCISBS

1. Give formulas (5) "and (8) with u replaced by cos (J and 11 by sin (J. I.' Give formulas (5) and (8) for (J = '11'/6, '11'/4, '11'/3, '11'/2. 8. The coordinates of a. vector on the positive :e' axis are (3,4). What are the equations (5)1

SEC.

6]

195

ADDITION FORMULAS

4. The coordinates of a vector on the positive y' axis are (3,4). What are the equations (5)1 6. Let the equations of a rotation be a;

3y' = &Il' -5 4y'' Y = 42;' + 5 •

Give the (a;', y') coordinates of P if its (a;, y) coordin8.tes are (a) (1, -1) (b) (0, 1)

(e) (-1,0) (d) (2,3)

(e) (3,4) (f) (-4, -3)

8. Let the number pairs given in Oral Exercise 5 be the (a;',1/) coordinates of P. Give the (a;, y) coordinates.

. 6. Addition formulas and inner products. Every angle 8 determines a rotation of axes and a unit vector (cos 8, sin 8) on the positive :c' axis. Let the initial side of a second angle q, be the positive :c' axis and ('Us, vs) be the unit vector on the terminal side of q,. Then the angle 8 + q, has the positive :c axis as initial side and the terminal side q,), of q, as terminal side. It follows that 'Us = cos (8 Vs = sin (8 q,), and that 'Us' = cos q" v,,' = sin q,. We use formulas (5) to obtain

+

+

(9)

{

COS (8 + q,) = cos 8 cos q, - sin 8 sin q, sin (8 + q,) = sin 8 cos q, + cos 8 sin q,.

These formulas are called the addition JfYI'm'Ula8 of trigonometry. The study of them and of their consequences makes up too large a part of trigonometry to be;' considered here. Any two nonzero vectors PI '7 (:C1' 'U1) and P s = (:Cs, 'Us) define angles 8 and q, as above, where the initial side of 8 is the positive :c axis, the terminal side of 8 and the initial side of q, coincide with the ray from 0 through P, and the terminal side of q, is the ray from 0 through P s. Then we shall call q, the angle between the vectors PI and P s. Now (:Cl, 'U1) = r1(cos 8, sin 8), (:Cs, 'Us) = rs('Us, vs) and cos q, = 'Us', We use formulas (8) with


[CHAP. VIlI

and. obtain (10)

XIX2 + YIY2 , 2 . VXi Yl 2 V X22 + Y2 2 • ,/;. YIX2 - XIYl SID¥,= • : VXl2 + Yl 2 VX22.+ Y2 2 COS

A.

~

=

+

T4e inner product of any two vectors (XII YI) and (X2, Y2) is : defined to be the real number XIX2 YIY2' Then

+

formula (10) states that the cosine of the angle between two nonzero vector8 i8 their inner product divided by the product of their lengths. If cos q, = 0 then q, = 7r' /2, or 37r'/2, and the ray through PI is perpendicular to that through P 2 • In this case we say that our vectors are orthogonal. Since rl > 0, r2 > 0, we s~ that two nonzero vectors (Xl, YI) and (X2' Y2) are orthog.onal if and only if their inner product XIX.

+ YIY2

o.

=

We observe finally that if (x, y) is any nonzero vector the vector (-y, x) is orthogonal to (x, y). Then the vectors

are orthogonal unit vectors which may be used, respectively, as the x' and y' unit points after a rotation of axes . such that (x, y) is on the positive x' axis. ORAL EXERCISES

1. Compute the inner products of the following pairs of vectors: (a) (1, 1), (2, -1) (b) (3, -1~, (1,3) (c) (-2, 1), (2, 1)

(d) (3, -4), (1, 2) (e) (-3,4), (7,12) (f) (1, ~ 1), (7, 12)

(g)

(va, -1), (1, -1)

(h)

(V2, - V2),

(i) (-1,

(1,0)

va), ,0, 1)

2. Give' the cosine of the angle between each pair of, vectors in Oral . Exercise 1. 3. Give a. unit vector orthogonal to P and a. unit vector on the same ray asP for each of the vectors P = (1, -1),P = (3,4),P = (-7,12), P = (-1,.2), P = (-1, - Va), P = (1).

va,

SEC.

THE BINOMIAL EQUATION

7]

197

7. The binomial equation. Every nonzero complex number :x: + yi corresponds to a vector (:x:, y) == r(u, v), where u == :x:/r == cos 8 and v == y/r == sin 8. Then :x: + yi == r(cos 8 + i sin 8). This is called the polar Jorm of a complex number. We shall write

e"

(12)

== cos 8

+i

sin 8

for the complex unit cos.8 + i sin 8 and call 8 the amplitude of :x: + yi. If :x: + yi == re" and :X:o + Yoi == 8e" then (:x: + yi)(:X:o + Yoi') == r8e"e~. But formula (6) implies that (cos 8 + i sin 8)(cos q,

+ i sin q,)

+ 4» + i sin (8 + q,).

== cos (8

This also follows by direct.multiplication from formula (9). It may be written as (13) and we see that (:x: + yi)(:X:o + y~) == r8e'(I#). We have multiplied the absolute values r of :x: + yi and 8 of :X:o + Yoi and have added their amplitudes 8 and q,. Formula (13) may be generalized to a product of any number of factol"! and states that the amplitude oj a product oj comple:x: number8 i8 the BUm oj the amplitude8 oj itB Jactor8. Th~n if n is any positive integer, we have

(re")" == r"e'''·.

(14)

This result is known as De M oivre' 8 theorem. If c == a + bi is any nonzero complex number and n is a positive integer, the equation :x:" == c is called a binomial equation. If we write c == rete where r == via' + b', the nnumbers

(15)

:X:i ==

V; e"',

8i == 8

+ (j n-

1)2'R' (j ==

.

all have the property that :X:i" == re'"'' == re£I+'''(i-1») == rete == c.

i . . . n)

198

. VECTORS IN THE PLANE

[CRAP. VIII

They are then all roots of the binomial equation :&" == c. For the angles 8 + 21r(j - 1) are all coterminal. By formula (13) . (16)

:&/

==

.yr eUl"ti-l,

t == e'"ftl" (j == 1 . . . 11.).

The complex unit t has amplitude 2r/11. and we see that each is obtained by drawing a circle whose radius is vr, locating the point· on this circle whose polar coordinates are [{IT,8/n], and then dividing the circumference of the circle into 11. equal parts by division points the first of which is [Vr, 8/11.]. It follows that the 11. numbers of formula (16) are all distinct and are the 11. roots of the equation :&~ == c. We may then pass on to the question as to what we ihall mean by the 81/rnbol where 11. > 1 iB an imeger and c i8 any compl6:& number. Let us define this symbol to pe that one of our numbers in formula {16) for which 8/ is the least nonnegative angle. Of course other definitions are possible. But we insist that they be such that, if c is a positive real number, and so 8 == 0, then shall be positive. Our definition satisfies this condition S;ll).ce ,8/11. == '0 if 8 == O. Now it is not true that == VCd for every c and d. Indeed, if 11. == 2 and c:= d == -1, the only possible /I , and in either definitions for v'='i are r l l or e'8r/2 == case (v'='i)(v'"=i) == -1 whereas

:&/

.yc

vc

vc.ytl

-e-

v( -1)( -1) == vI == 1. EXBRCISES

1. Use the values of the trigonometric functions to find r and 8 for the following complex numbers c: (a) c - 8 (b) c = -5

(Il) c

(c) c = 3i

( :'I

(d) c =

(t.") 3,

-7i

+

(8) c = 1 i (f) c = -2 2i (g) c = 3 --3i

+

=
0 if and only if P is on the ray from 0 through PI - Pl. Let us make Pl the zero point of a real number system on the.line'L, the direction from P l to p. the positive direction, and the distance from P 1 to PI the unit ·distance. Then it should be evident that t is the coordinate of P in this coordinate system on L. Also

P - P 1 = •(z - Zl,1/ - 1/1) = t(z. - Zl,1/1 - 1/1),

that is, (20)

{:1: = Zl 1/ = 1/1

+ t(ZI + t(1/1 -

:1:1), 1/1).

S:::C.

8J

EQU ATIONS OF LINES

201

Formulas (20) express the coordinates (z, y) of any point P on L in terms of Zl, Yl, Zll, Yll and the ratio t of the length P lP to PIP lI' They are called a pair oj parametric equatiOnB of the line L, and t is called the parameter. Note that if o ::!i t ::!i 1 the point P lies between PI and P 2• The midpoint of PIPII is the point where t = t so that its coordinates are Zl + (21)

Yl

21 (Z2

) Zl+Z2 - Zl = 2 '

+ 21 (Y2 - . Yl)

=

Yl + Yll

2

.

Formula (17) and its equivalent forms (18), (19) are referred to below as nonparametric Jorms of line equations, and we close by observing that formula (19) is obtainable from (20) by writing (Z -

zJ (Y2

- Yl) = t(Z2 - Zl) (Yll - Yl) = (y - Yl)(ZlI - Zl). BXERCISBS

1. Give nonparametric equations (a) (5,6) and (4, -2) (b) (5,3) and (5,2) (c) (5, -3) and (7, -3) (d) (-6, 2) and (-6, -3) (e) (4,1) and (6, 1) (J) (5, -3) and (6, 2)

of the lines through

VI) (:-9, -5) and (-6, -1) (k) (8, -2) and (-1, -8)

(i) (9, 7) and (6, 3) (J) (2, 0) and (0, 6) (k) (-1,0) and (0, -5) (l) (-5, 0) and- (0, 3)

I. Give parametric equations of the lines of Exercise 1. 8. Give formulas (17) of the lines L through PI = (Zl, 'U1) and orthogonal to Q = .(a, b) in each of the cases of Exercise 1 where we take PI as the first point and Q as the second point. 4. Replace each equation of Exercise 3 by an equivalent equation in which (a, b) is replaced by a unit vector (v, fI) = (air, blr). Ii. Find the coordinates of the mid-point of each line segment " ' where PI is the first given point in each part of Exercise 1 and PI is the second point. 6. Find the coordinates of a point P for PI, PI as in Exercise 5 such that 'FJiIPJ'1 = t where (a) t -= ti (b) t = ti (c) t = ti (d) t = - t. 'I. Put'U = 0 in formula (19) and solve for z to obtain the formula we used in Chap. VII to solve an equation by the method of interpolaiion.:

202


9. Conic sections. A quadratic equation (22) F(:c, y) = a:cl + 2b:cy + cyl + d:J; + ey

[CHAP.vm

+ 9 =0

defined for real numbers a, b, c, d, e, 9 such that a, b, c are not all zero is said to define a conic 86ction. If we replace (:c, y) by :c'u - y'v and :c'v + y'u, respectively, as in formula (5), we obtain a new equation (23)

~(:c,

y)

= a:c'~ + 2{J:c'y' + 'Yy'l + 8:c' + fI/I' + r = 0,

in which a, {J, 'Y, 8, E, r are real n.umbers. This new equation is the equation of the same curve with respect to the (:c', y,)-coordinate system. We shall show that u = cos (J and v = sin (J may be ·selected so that fJ = o. The second-degree terms of F(,;, y) make up a quadratic form Q(:c, y)" = a:cl

.(24)

+ 2b:cy + cyl,

and we call a + c the #:race of Q(:c, y) and ac - bl its determinant. These numbers determine an equation ~~

~-~+~+oo-~=~-~~-~=~

called the characteri8tic equation of Q(:c, y). The roots a and 'Y of this· equation are called the characteri8tic roo't8 of Q(x, y) and we already have seen that (26)

.

a"(

= 00 -

bl,

a

+ 'Y = a + c.

The discrimjnant of the quadratic equation. (25) is (a

+ C)I -

4(00 - bl) = (a -

C)I

+ 4bl

and is positive. Hence a and'Y are real and distinct. We nowprovethefollo~:

Theorem. Let Q(:c, y) = a:cl + 2b:cy + cyl where b ;o!! 0, and let a, 'Y be· the characteri8tic roo't8 oj Q(x, y). Then the rotation oj a:ce8 defined by a-a b v = -vr.b===.=;:+=G7"'a==a~)I (27) vbl + (a -

,,=

replaC68 Q(:c, y) byax'l

a/

+ 'Yy'l.

SlUe. 9]

.

203

CONIC SECTIONS

For the first of the equations (28)

+ bv == au -'YV + cv == av

au - av bu

+ bu ==

- bv

+ C'U

== 'YU

is satisfied if formula (27) holds and (27) is indeed derivable from the first equation if we use the equivalent form a-a

(29)

v == -b-u,

u2

+ vi == 1.

The second equation is equivalent, in view of formula' (29), to (a - 'Y)(a - a)u = b2u, which is true since, by formula (26), we have b2

+ (a -

a) (a - 'Y) = bll

+

all -

{a

+ c)a + (ae -

b2)

== O.

The third equation results when we use formula (29) to write its equivalent form (a - c)(a - a)u == b2u and thus [a 2 - (a - c)a + (ae - b2)]u == O. For a is a root of the equation of formula (25). Finally the fourth equation is a consequence of the third equation when we replace u by (a - a/b)v and so have [(c - 'Y)(a - 'Y) - b2]v

== ['Y 2 - (a

+ ch + ae -

b2]v == O.

The first and second equations of formula (28) may be multiplied respectively by u and -v and added to yield au2 + 'YV2 = a. Similarly we may use the third and fourth equations to obtain av2 + 'Yu2 == c. We multiply the first equation by v, the second by u and add to obtain (a - 'Y )uv == b.

Then ax'2

+ 'Y1I'2 E!!

+

55 a(zu + YV)2 + 'Y( -zv you)!! z2(au 2 + 'YV2) 2zy(a - 'Y)uv + y2(av 2 + 'Yu 2) Ei az2 + 2bzy cy2.

+

This proves our theorem.

+

204

(OIW'. vm


nlustrative Examples I. Remove the :cy .term from the equation 2z2 a rotation of axes.

Solution

-

4:cy

+ 5112 -

3 by

+

Here a = 2, b = -2, 0 = 5 so that a 0 = 7, ac - b2 ... 6 and the cha.ra.cteristic equation is al - 7a + 6 == (8 - 6)(8 - 1). We take a = 1, 'Y = 6 and the (z'y') equation is zl2 6y'2 = 3.

+

Also au

+ bv = au becomes 2u r(u,

II)

= r(2, 1),

u

211 =

so that u = 211 and

U

= 2/v'5,

" = l/VS.

The equations of rotation are

z, + 2y~

2z' - y' Z=

v'5'

v'5'

11 =:

n. Remove the:cy term from 9z2 + 2Uy + 16yl - 5z + lOtI + 3 by a rotation of axes.

=

0

Solution Here a = 9, b ... 12, 0 = 16 so that we solve

a2 Thus a

25a + (9 • 16 - 144) = al

-

258 = O.

= 0, 25. Take a ... 0 and so solve 9u + 1211 = 0, 3u + 411

= t, II = 511 = -3z'

U

-

t.

The equations for rotation are 5z = 43;' and

+ 411', -5z + lOtI = -43;' -

3y' - &'

= 0,

+ 3y',

+ Sri' = -10z' + 511'.

The rotation of axes replaces our equation by

2511'2 - 10z' + 5y' + 3 = O. EXERCISES

Remove the zy term from each of the following equations by a rota.tion of axes : (a) 4zy - 16 = 0 Am. Z'2 - V2 = 8; u = II = 1/0. (b) 8z2 4zy - 24y2 = 5 (0) Z2 2:cy y2 - 3 = 0 Am. 2:&'2 = 3, u = II .;,,'1/V2. (d) 1&2 - 24zy 9y2 = 17 (e) Z2 - 1O:cy + yl = 5

+ + +

+

Am. 12:&'2 - 10t1'2

=

5, u

=

-5/v'I46, II = 11/-vm.

BEC.9]

CONIC SECTIONS

205

(f) ~:l:1 + 120:1:1/ + 144y2 = 1 (g) 43;2 + 43;y + 4y2 = 12 Am. 3:1:'2 - y'2 = 6, U = II = 1/0. (h) 53;1 - 20:1:1/ + 2Oy2 = 4 (i) :1:1 + 12:1:1/ + y2 = 8 Am. 7:1:'2 - 5y'2 = 8, U = II = 1/0. W 25:1:2 +30:l:1/+9y2+20:1:+6y+1=0 (k) :1:1 - 4:1:11 + 4y2 + 6:1: + 3y - 9 = 0 Am. 5:1:'2 - 3 V5y' = 9, U = -l/V5, II = 2/V5. (l) Z7:1:2 - 36:1:1/ + 12y2 - 20:1: + 9y = 16 (m) :1:1 + :1:1/ + yS + 3y - 4 = 0 Am. 3:1:'2 - y'2 - 3 V2(a:' - y') = 8, U = II = 1/0. (n) 23;2 12:1:1/ 18y2 = 0 (0) 53;2 - 6:1:1/ + 5y1 - 10:1: - lOy - 15 = 0 Am. 2:1:'1 + By'2 + 5~' = 15, U = II = 1/0. (p) 43;2 + 12:1:Y + 9y2 - 2:1: - 3y + 1 = 0 (q) 4:1:11 - 3y2 + 6:1: + 4y = 5 (r) 11:1:2 - 243;y + 4y2 - 6:1: - 8y = 20 (8) 43;1 + 12:1:Y + 2Oy2 + 10:1: - 8y = 6 (t) 53;2 + 6:1:1/ - 3yl + 6:1: + 2y = 1 (U) 53;2 -,12:1:1/ + lOy! +:1: - Y = 4 (II) 53;2 + 8:1:1/ + 5y2 + 2:1: + 3y = 5 (w) 7:1:1 + 8:J:y - 8y2 + 9:1: - 6y = 11 (:I:) 43;1 + 24:1:11 - 3y2 + 8:1: + 8y = 10 (y) 7:1: 2 + 6:1:Y - yl - 76:1: - 60y - 100 =- 0 (z) 853;2 + 96:1:1/ + 45y1 + 1,0643; + 7441/ - 1,308 = 0

+

+

CHAPTER IX MATRICES, DETERMINANTS, AND LINEAR SYSTEMS 1. Systems of equations. Let f(3h, ••• ; Zn) be a complex-number valued. function of 11, complex variables Zl, . • • ,Zn. Then we have said that to each complexnumber sequence CI, • • • , Cn there corresponds a value'of our function which we designate by f(CI, ••• , c,~) and call the value of f(ZI, • • • , :en) at Zl = CI, Zi = Ci, • • • , Zn == Cn. We shall call (CI' • • • , en) a 'solution of the conditional equation f(ZI, • • • , Zn) ='0 if the value f( CI, • • • ,en) is the number zero. The set of conditional equations (1)

fl(ZI, . • •

,:en) == 0,

fi(ZI,···,

,.

:en) == 0,

fm(ZI,"', Zn) == 0,

defined by m functions of n variables, is called a system of equations. A system of equations (1) asks: What are all number sequences (CI' • • • , en) which are simultaneously solutions of all m of the equations? If the functions /1, fi, ••• , fm of formula (1) are the functions defined by linear polynomials in Zl, • • • , Zn, we call formula (1) a linear system. We shall concentrate our study of systems of equations in this text on linear systems. The theory of determinants is a tool for the solution of linear systems and will be presented. in this chapter. The theory of matrices, of which the theory of determinants is a part, also arises from the study of linear systems, and we shall present some of the elements of that theory here. ' We begin our presentation with a generalization of the vector concept. An n-dimensional vector is a sequence 206

SlIIC.

1]

207

SYSTEMS OF EQUATIONS

a == (ai, . . . , a.) of n numbers which are called the element8 oj a. A vector may be interpreted as the sequence of coordinates of a point in an n-dimensional geometric space, as the coefficient sequence of a linear equation alXl + azZlI + . . . + a,.z. == k, or as a value sequence Xl == ai, . . . , z. == a. of n variables. Our algebraic study will not use any of these interpretations but will treat the vector purely as a sequence. "The BUm a + ~ of two vectors a == (ai, . . . ", a.) and ~ == (b l , • • • ,b,,) is the vector

a. + b,,), obtained by adding corresponding elements. The sero vector is the vector (0, 0, .•• , 0) and will be designated simply by the symbol o. It has the property (2)

a

+~

== (al

+ bl, all + b

l,

••• ,

a+O==O+a==a for every vector a. The negative -:-a of a == (ai, . • • ,a.) is that vector with the property that a + (-a) == o. Thus -a == (-ai, . . . ,-a.). The difference a - ~ is defined to be a + (-~) and is (al - bl, • • • , a. - b,,). Addition of vectors may be shown to be commutative and associative. We leave the amplification of the meaning of this statement and its verification to the student. The 8caZar product da of a vector a == (ai, . . . , a.) by a number d is the vector da == (dal , • • • ,da.). We call a sum dial + dsal + . . . + dmam, of products ~ of numbers tis by vectors ai, a linear combination of ai, • • • ,

am· OlU.L EDR.CISES

1. Leta = (3, (a) a

-5,4,2),~

+~

(6) a - ~

2. Let a = (2, -3, 1), the linear combina.tions (a) a +~ +'Y (6) 2a - ~ ~ 'Y

= (-6,2, -3, -1).

+ 2fJ

(c) Sa (d) 2a ~

(6) 4a (f) a -

s.B

= (1,2, -2),

(c) 3a -"2fJ (d) 2a - s.B

Compute

+ 'Y + 'Y

+~

4fJ

'Y - (3, -1',4). Compute (6) 2(a +~) - 3-y (f) 3(2a 4-.y

+ m-

208

IeHAl'. IX

LiNEAR SYSTEMS

2. Rectangular matrices. A rectangular array of number8 is called a matrix.

The array

-5 -1 6 -2

1 2

o

"""

I'

1 1

3

-3

of the coefficients of Xl, Xs, X3, X4, Xli in the following system of four linear equations in five variables

'1-ax1 +,+ -

X2 2X2

5X3 -

X4

+ 6x

1i

=1

+ 6X8 - 2X4 =3 2x1 + 'X8 + X4 + 6x = -2 4x1 + 3X2 + X3 - 3X4 + 2x1i = 7 Xl

1i

is called a 4 by 5 matrix. The horizontal lines of numbers in a inatrix are vectors called its row~. The vertical lines also form finite sequences called its columna. In the ,example above (2

o

1

1

6)

is the third row of A and

is its fourth column. It is customary to call the rows of A row vector8 and the columns of A column vector8. We shall later add rows (columns) of A to other rows (columns) and shall multiply rows (columns) by numbers. 'We note that there is no need to separate the sequence elements in a column vector by commas. It is customary to omit commas also in the rows of a nmtrix and we shall henc~forth do so in all our vectors. A matrix which has m rows and n columns is called an m by n matrix. The vectors which are rows are 1 by n

SEC.

2]

RECTANGULAR MATRICES

209

matrices, and those which are its columns are m by 1 matrices. The numbers which appear in a matrix are called its elements. If all the elements of a matrix A are zero, we shall call it a zero matri:i; and shall write.A. == 0 regardle88 oj the number oj rOW8 and coZumm in A. We shall also say that two matrices .A. and B are equal, and write A == B; if they have the same size m by 11. and are such that correspondingly placed elements are equal. The position of an element of a matrix is at least as important as its value. It may be described by a statement as to the ~w and column in which the element appears. For example, in the matrix above the number 3 is in the fourth row and second column. It is convenient then to introduce a notation for the general element of any matrix. This is the symbol D.ii·

The subscript i is the row 8Ub8cMpt.· Its value is the label of the row' of our matrix A in which D.ii appears. The subscript j is the coZumn 8Ub8cMpt. Its value is the label of tb.e column in which a.j appears. When A is an m by 11. matrix the values of i range from 1 to m and those of j from 1 to n. In the matrix above au == 3, au == a" == -3, alii = aaa = au == au == 1, au == -1, au == -5, au == 6, etc. The third row of the matrix consists of all elements with first subscript 3 and they are all, aal, aaa, au, alii in our case. Let the reader give their values as well.as the notations and values of the elements in the fourth column. . We may now use the notation (3)

210

LINEAR

SYSTEMS

[CHAP. IX

for the arbitrary m by 11, matrix A. The ~otation (4) A· = (o,;i) '(i = 1, . . . , m; j- = 1, . . . ,

11,)

will also be used here. ORAL EXERCISES

1. The matrix at the beginning of this section is a 4 by 5 matrix. Make the corresponding statements about the following matrices. (aj

(_~

(.j

(J) (b) (1 (c)

3-5

(3 2) -1 3

(d)

(7 4 3 0 4 -2

6

2)

(_! ~ =v 7.

1

1

0

-1

/1

2

0

3 -4

1 0

2. Read off ~he elements au, au, au, alB, aas in those of the matriceS above 'Where theae elementa e:m8t. 3. Read off the first row a.nd the first column of each matrix in Oral Exercise 1. 4. Read. off the third row and the second column of those matrices where they exist. 6. What kind of matrix is a row of an m by n matrix? A column? An element?

S. Elementary transformations. We shall be concerned with certain processes for altering a matrix A which are called elementary tramJormation8 on A. They arise in the study of linear systems of equatiQD.S when the equations are interchanged, added, or multiplied by a number. There are three types of elementary transformations on the rows of a matrix and three co~esponding typ~s on its columns. The.first type is that of the interchange oj two rOW8 or two columns. By a finite sequence of such transformations we may obtain any desired permutation of the rows and columns of the matrix. The second type of transformation is that of the addition to a row (column) oj any numeriCal multiple oj another row (column). Our final type is that of multiplication oj a

SEC.

31

211

ELEMENTARY TRANSFORMATIONS

row (column) by any nonzero number. Observe that all our transformations are reversible. The result of applying a finite number of elementary transformations to an m by n matrix A is 'another m by n matrix B. We shall say that A and B are equivalent matrices and shall write A ,...., B. Observe that A :: B does not mean'that A and Bare equal. They are merely related in the manner we have described and the student should not try to read anything else into our definition. Illustrative Examples I. Find the matrix B which is the result of adding the second row plus twice the third row of

A

=

( 5 -4 4 2) -2 -1

0 2

-8 5 2-3

to;ts first row. Solution ,;' The new first row is (5

-4

4

2)

+ (-2 + (-2

-8

0

4

5) -6) = (1

4

0

0

1)

Thus

B =

(-~ -1

0 0 2

0

-8 2

-D

II. Subtract multiples of the columns of the matrix A from other columns and replace A by a matrix B ~ A which has all but one element in the first row zero. Solution Let the columns of A be represented by Cl, C2, Ca, C4. Form Cl - 2C4, C2 2C4, Ca - 2c 4 and then C4 - 2(Cl - 2C4) to obtain

+

A::::::: (-1251

1~ -1~8 -3~) ~ (-1~5 -41~ -1~8 -13 ,2~)'

-4

EXERCISES

1. Let Tl,

Ta, Ta

be the rows of A

=

(-~

2

-: -1

212

LINEAR SYSTEMS

[OBAl'. IX

Find the matrix B obtained by (a) Adding 2rs + r. to rl (b) Adding rl + rt to r. (c) Adding r. - 2rl to r. (d) Subtracting 3rl r. from r. (6) Subtracting r. - 2rl from rl I. Let Cl, CI, Ca, c,.be the columns of the matrix A above. Find the -matrix B obtained by (a) Adding CI Ca 2c, to Cl (b) Adding Cl c. c, to Cs (c) Adding 2cl Cs c, to Ca (d) Subtracting 2cl from c. (6) Subtracting 2cl Ca from c, (fJ Subtracting c, - 3Cl from c. a. Simplify the first row of each of the following matrices by the procedure used in Illustrative Example II.

+

+ + + + + + +

(a)

(6)

G G

-1

3 -3 4 -1 2 -1 4 -3 4 -5 1 2

(.) G

2

0

-1 1 -1

-V

D V

(d)

C5 -1

(.) (

:)

3 2

-2

1

-1

2

1

i are said to lie above the diagoruil of A and those with i < i to lie below the diagoruil. We shall say that ,A. has triangular form if either all elements below the diagonal of A, or all ele-

81110.

0]

213

SUB KATR I OlD S

ments above the diagona.I of A, are zero. the matrices

(j

0 4 2

0 0 3

~ G

G ~) (~ 0 3 1

1 4 0 0 1 0

For example,

4 5 3

~ 8.

V

have triangular form.. A matrix A is caJIed a Bquare matrix if it has exactly as many ro~ as columns, that is, A is an n by n matrix. We shaJI speak of such a matrix as an n-T0'W6d square matri:e or as a square matri:c oj order n. A square matriX in tria.ngula.r form. is caJIed a trianguZar matrix. If the nondiagonal elements of a square matrix A = (tlii) are aJI zero, we caJI A a diagonal matrix and write A = diag {d 1, • • • ,dn}. Here ~ = tli, is the ith diagonal element of A. A scalar matrix is a diagonal matrix with equal diagonal elements. For example, the first of the matrices

o 3

o

o 3

o

is a diagonal matrix which is not a scalar matrix while the second 'matrix is a scalar matrix. A scalar matrix whose diagonal elements are aJI1 is caJIed an identity matrix. It is customary to designate every identity matrix by the symbol I regardless of its order. 5. Submatrices. If we omit certain of the rows and columns of a rectangular matr:i.x A, the elements in the reIDa.;n;ng rows and columns also form. a matrix caJIed a submatri:e of A. For example, the omission of the first row and the second and fifth columns of the 4 by 5 matrix of Sec. 2 is the 3 by 3 submatrix

214

LINEAR SYSTEMS

[cauo.

6-2} 1

1

1

-3

IX

We shall be particularly interested in the result of deleting one row and one column of an n-rowed square matrix A. Each element D.ii of A determines the row -and column in which the element appears and we shall designate the submatrix obtamed by deleting this row and column by A'S. For example, in the 3 by 3 matrix above we have

Au

= (~

-~}

AI2

=

(-12 -2)1 .

ORAL BDRCISBS

1. Read off A 4I, A.a, As. in the following matrix: 1

-5 -1

2 4

2 0 3

-1

1

6 -2 1 1 1 -3 2 3

C -1

v·

I. In what row and column of Au does the element'as. of A appear? 8. In what row and column of Au do the elements au and a14 of A appear? ' 4r. In what row and column of A18 does all appear? '

8. Determinants. We shall associate with every square matrix A = (aai) a number which we shall call the determinant of .A and shall write IAI for this number. We shall also write IAI - laail as well as (5)

IAI-

au

au

au

a22

ani

'an2

altai

..... .

]'

and shall call this symbol an n-rowed determinant, or a determinant of order n. Matrices that are not, square do not have determinants• ..,

uc. 6)

215

. DETERMINANTS

The determinant of a 1 by 1 matrix A = (a) is defined to be the number a. We define

I: and au an an

al2 a21

au

:1

=

a18 au = au' la u au

aaa

ad - be,

aul - au' Ian aaa an

+ a18' Ian an

aul aas

au /. au

Then we have defined three-rowed determinants in terms of two-rowed determinants. We now proceed inductively. Let us suppose that (11, - I)-rowed determinants have been defined. Then every n-rowed square matrix A = (Q,ji) has elements Q,ji with corresponding (11, - I)-rowed square submatrices A#. We call the determinant IA~il the minor of Q,ji and the number

B# = (-I)C+i IA#1 the co/acInr of Q,ji' If we select any row of A, multiply the elements of this row by their cofactors, and add the resulting products we obtain the number

Q,jlBSl

+ Q,jIB,. + . . . + Q,j"Bi ".

Similarly, if we select any column, we obtain a sum aljBlj

+ a2iBli + . . . + aniB"i'

It is proved in the next section that all these numbers are equal. We 8hall call their common value the determinant 0/ A •. We have now given one expression of 114.1 for every row and one for every column of A. Each expression is called the e:tpamion of IA I according to the corresponding row or column. Each expansion is a sum of 11, terms. Every term is the product of· an element in a :fixed row by its

216

LINEAR SYSTEMS

(CHAP. IX

cofactor. The sign (-1)*"1, which we prefix to each minor /Ail/ to obtain the cofactor of 4ii, is plus if the sum of th~ row and column subscripts is even and is minus if this sum is odd. Thus the signs alternate and may be presented in a checker-board form. For n - 2, 3, 4 the checkerboards are

I~

+

-+1, /

+ +

+ + +

+

+ + +

,

+ + +

The following results are simple consequences of our definition. We shall not give their formal proofs. Theorem 1. Let all the element8 in a rO'lD or in a column oj A be sero. Then /A / - o. Theorem 2. The determinant oj a triangular rnatriz A i8 the product oj the diagonal elementB oj A. The determinant oj the n-r0'lD6d identity ~ is 1. Dlustratille ExamPles

I. What are the minors and cofactors of the elements of the second column of

IAI-

-1

2

_~

2 1 1

-5

: 2

-~4? 3

6

SoZtd.ion We use the technique of first deleting the second column of A simplify the eye work and so write the submatrix

to

DC.

217

DETERMINANTS

6]

The minors are then obtained by omitting the rows in turn and are 4 -1 -5 -1 4 = -5

al ...

a.

1 3, 2 -4 1, 2 "6 0 2 6 3 0

a2 =

-1 -1 -5

a. =

1-14 -I"

3 2 6 3 0 2

-4 3, 2 -41 l' 3

Since 1 + 2 is odd, the cofactors are -aI, at, -aI, a•. n. Give the terms in the expansion of the determinant above according to its third row.

Solution The row is (-1 matrix

1

3). If we delete it, we obtain the

2

Since 1 + 3 is even, the terms are 3

-4 1-1 3-4 1 - (1) 4 0" 1 6" 2 -5 6 2 1-1 2 -4 -1 " - + (2) 4"" 2 - 1 - (3) 4 ". ~5 1 2 -5"

o

2 2 1

3

O· 6

In. Complete the e~on of the determinant above ~cording to its third row. "" Solution We expand the first determinant according to its second row. The resuit is (":"2)(6 + 24) - (12 - 3) = -69. The second determirui.nt is expanded similarly to obtain -4(6

+ 24)

- (-6

+ 15) = -120 -

9

= -129.

Expand the third determinant according ~ its first row ~o obtain (-1)(4 -1) - 2(8 + 5) - 4(4 + 10) = -3 - 26.,... 56 == -:85. The last determinant is (-4)(12 - 3) + 2(-6 + 15) = -36 + 18 = -18. He~ce JAI = 69 + 129 - 170 + 54 = 82. "

218

[CHAP.

LINEAR SYSTEMS

]X

EXERCISES

1. Give the terms in the expansion of the following determinants according to the iP'st row: (a) 2

-1 1 3-1 6-2 -1 4

o

2 3

2

(b)

4

-1 3

o

(d) 1

4 8 2

3 4

0

(e)

2

-1 -1

3

-1

o

1 1

-3

o

-1 2

tn

,2

o

2

4 2 3 2 1

2 4 1 -1 4 000

o

1-1

1

2

-2 1 6-1

480 (c) 0

-2 3 1 -1 -1 2 4 -4

o

2-3

~I~ =~ -~ ~~I21 1

-1

2. Give the terms in the exPansion of each determinant of Exercise 1 according to the thi!d column. S. Compute the determinants of (c), (e), tn of Exercise 1, using a selection of a row 'or a column to make the computation 'as simple as possible. 4. Compute the following determinants:

(b) 1

o

(g)

A.m. -45.

2

-1

0

2 4 3

~I

A.m. -3.

1

1

-1

1

-1 4 -2 1 -4

5 5

0 1

1-2

2 1

0-1 1 ·-1 1 o 1-2

1

o

4

1 -1 2'

1 3 4 -1

1

-1

-7 2

o

2

I:

(d) 1

-3 2

tn

(c)

-1 2 -2 4 2-3

3

(e)

'-!I

-~

(x), f(x), g(x) are all polynomials in x and the degree of the numerator f(x) is less than the degree of the denominator g(x). Let us suppose now that it is possible to factor g(x) and so to write it as a product of powers of distinct irreducible factors. Then we may express p(x) as the sum of cJ>(x) and a number of what are called partial fraction8. The procedure used employs nothing but the solution of a system of linear equations and so is an application of the results of this chapter. If p(x) is an irreducible factor of g(x) and m is its multiplicity, there will be a corresponding sum of m partial fractions h1(x) p(x)

+ h2(X) + ... + h".(x) p(X)2

p(X)m

Each of the polynomials hi(x) is to be determined so that its degree is less than the degree of p(x), and the linear equations we solve are linear equation!!! in the coefficients of the polynomials hi(x). We shall not try to prove that solutions of these exist nor shall we give a general discus.sion of the process of finding the equations. The methods involved are adequately presented in the examples below. Note that in the case where our coefficient8 are permitted to be any real number8 the polynomials p(x) are either quadratic or linear, the polynomials hi(x) are correspondingly linear or constant. Illustrative Examples I. Express the rational function _2:1:4 + 15:1:2 - 20:1: + 13 (:I: - 1)3(:1: + 2)

as the sum of JVLrtial fractions.

240

LINEAR SYSTEMS

[CHAP. IX

Solution

By multiplication we find that the denominator is X4 - x 3 - 3x 2 + 5x - 2.

By division p(x1 == -2

+

-2x 3 9x 2 - lOx (x _ 1)3(x 2)

+

+

+9

,

and we use the theory above to write -2x 3 +9x 2 -10x+17 abc (x - 1)3(x 2) == (x - 1)3 (x - 1)2 x - I

+

+

for numbers a, b, c, d to be determined. -2x 3

+ 9x2 -

lOx

+

d

+x +2

Then

+ 9 = a(x + 2) + b(x -

+ + + d(x -

l)(x 2) 2) + c(x - 1)2(x

1)3.

This is to be a polynomial identity and so we equate coefficients of the powers x 3, X2, x, 1 of x on the two sides, using (x -1)3 = x 3- 3x 2+ 3x -1, (x - 1)2(x + 2) = x 3 - 3x + 2, (x - l)(x + 2) = X2 + X - 2, to obtain the linear system

+

c d = -2 b -3d= 9 a+ b-3c+3d= -10 2a - 2b 2c - d = 9.

+

The solutions are then a = 2, b = 0, c = 1, d = -3. Remark: We ~ay simplify our work by using values of x. vttlue x = 1 to obtain •

3a =

6,

We use the

a = 2,

the value x = -2 to obtain (-2)( -8)

+ 9(4) + 20 + 9 = 81 =

-27d,

and so have d = -3. The solution is completed by using the leading coefficient to obtain -2 ,;" c + d, c = 1, and the value x = 0 to obtain \J = 2a - 2b + 2c - d, 9 = 4 - 2b + 2 + 3 = 9 - 2b, b = O. 2 1 3 Ans. p(x) = -2 + (x - 1)3 + --1 x- - x- ,2'

II. Express p(x) =

x6

3 + 3x 2 - 4x +, 5x(x2++3x1)2(x3 _ 2)

as a sum of partial fractions.

4

5

SEC.

241

PARTIAL FRACTIONS

14J'

Solution

We write _ p(x)

=

ax + b (X2 j- 1)2

ex + d ex 2 + fx + g + X2 + 1 + x 2 ' 3 -

and obtain x6

+ 5x + 3x + 3X2 - 4x - 5 "'" (ax + b)(X3 - 2) + (ex + d)(x 2 + l)(x 2) + (ex2 + fx + O)(X2 + 1)2. 4

3

3 -

However, the factorization we are using permits rational numbers only as coefficients and so a, b, e, d, e, f, g are to be rational numbers. We put x 2 = -1, x = i, and obtain -1

+ 5-

3 - 5 - 3i - 4i = -4

+ 7i =

(ai

+ b)( -i -

2)

= a - 2b - (b Then a - 2b = -4, 2a then become

+ 3x + 3x 2 + 1 Then ci + d = 0, e = x6

4

+b= 7

+ 2a)i.

and a = 2, b = 3. Our equations

+ d)(x 2 + 1) + (ex2 + fx + g)(x 2 + 1j2. d = 0, ex 2 + fx + g == X2 + 1. 2x + 3 X2 + 1 \, "'" (ex

Ans. p(x) "'" (x2

+ 1)2 + x3 _

EXERCISE

Express the following as sums of partial fractions: 3(x 2

+ x)

(a) (x - 2)(x + 1)2 x 2 -.:x (b) (x + 1)(x2 + 1) 2X2 + 8x - 8 (e) (x + 2)(X2 + 4)

(d) x3

Ans. x

1

4x

Ans. X2

4x - 2 X2 - 2x 12 + 6x 2 (g) x 3 + 4X2 + 3x (f) x 3

x

~ 4x

+1 x3 + 8

2'

2

+4 -

17x

X4

(e)

2

+1+x -

+ 2' + 28 17 + 4) + 12(x + 2)'

Ans. x - 12(x2 _ 2x

-

(h) 5x 2

x3

(i) 43:3

-

-

4 'x

Ans - -

9 x+1

11 --+ x+3 -_.

3 x

+ ~X2 + 3x

Ans.

131 2x + 1 + 2x + 3'

x-

2'

LINEAR SYSTJUUiI

&;1 + 21;1 + 1 4m1 -:r: 21;4_:r:1 _&;1+21;+4 (Ie) (:r:1 - 4)(21; - 1) (3)

A

1

1.

1

Am. :r: + :r: _ 2 + :r: + 2 - 21; - 1· 24m1 - IO:r: + 5 (Z) (21; + 1)(21; - 1)1 4m :r: :r:1 - 3:r: (m) (:r:1 + 1)1 Am. - (:r:1 + 1)2 + ~I + 1· :r:1 - l (n) :r;I + 9:r: :r: 1 + 1 1& + 1 8:r: (0) (:r:1 + 4)1 Am. :r: + (:r:1 + 4)1 - :';1 + 4· 21;'-3:r:+2 (p) (:r: + 2)(2:r;' + 2:r; + 2)' (:r: - 3)1 -IO:r: 4 1 (g) (:r:1 + 4m + 5)1 Am. (:r:1 + 4m + 5)1 + :r:1 + 4m + 5· :r:1 + &; (r) (:r: + 2)(:r:' + 4) :r: 4 -3 :r:+1 1 (,), (:r: .....: I)(:r:1 + 1) Am. :r: + 1 + :r:1 + 1 - :r: - f • '3:r;8 - 3:r: 1 :r:4 + 3:r:1 (t) . :r: -10 21;-3 2 (u) :r:8 - 2:r;1 + &; Am. :r:1 - 2:r: + 5 -

+

+

z·

CHAPTER X MATRICES AND QUADRATIC FORMS (FULL COURSE)

1. Inner produCts. The theory of matrices is usually presented in more advanced courses. It! techniques are as simple as many techniques already presented here and are of sufficient importance to make an exposition of them very much worth while. We shaJI give such an exposition in this chapter without proofs of the principal theorems and shall regard the entire chapter as optional. We call a vector a = (ai, . . . , an) an rwlime'n8ional vector and its elements ai, . . . , an the coordinate8 of a. Then we may interpret a as representing a point in an n-dimensional geometric space whose origin is the uro vector (0, 0, . • • ,0). The inner product of a and another n-dimensional vector fJ - (b l , • • • , bn) is the number

(1)

otfJ'

= albl

+ aib! + . . . + anbn•

Our definition is such that otfJ' = fJa'. the inner pro!iuct (2)

aa' =

all

The norm of a is

+ ... + an!,

and we call a a unit vector if ota' - 1. Note that the vector E" whose ith- coordinate is 1 and whose other coordinates are zero, is a unit vector. Two vectors a and fJ are said to be orthogonal ~ otfJ' = o. We call a set ai, .. '. , ae of t distinct vectors a set of pairwise orthogonal vectors if 04.01/ = 0 for every i p6 j. We shall call a vector a with real coordinates a real vector. A real nonzero vector a has a positive nonn ota', and we c~ll the positive square root r = v'"ii.i2 the length of 243

244

MATRICES AND QUADRATIC FORMS

[CBAP.x

, /

Then the scalar product of the vector a by the number

a.

l/r is the unit vector (3)

Hence a ;:::: rp., that is, every real nonz~ro vector is the scalar product of its length r and a unit vector p.. If p. == (th, . . . , 'Un) and 71 == (VI, • • • , Vn ) are unit vectors, their inner product satisfies the relation (4)

-1

~ P.7I' ~

1.

,We shall not try to prove this but observe that it implies that P.7I' is the cosine of an angle 8 between 0 and '1r radians. We shall call 8 the angle between the unit vectors p. and 71. Then if a == rp. and {3 == 871 are any two nonzero ,vectors, we define the angle 8 between a and {3 to be that between the unit vector p. and 71. This yields the formula (5)

cos 8 ==

a{3'

.yaa; -v'ifPt' '

When a{3' == 0 the cosine of 8 is zero and 8 == '1r/2 radians. Then the vectors a and {3 may be thought of as being perpendicular and this is our reason for calling a 'and. {3 orthogonal if a{3' == O. nlustrative Examples I. Compute afJl where a = (1

-2

3

4)

and

(3 = (2

1

-3

2).

Solution

a(31 = (1 . 2)

+ (-2' 1) + (3' -3) + (4' 2) =

-1.

Remark: In cases where the coordinates of a and fJ are decimals it is desirable to superimpose the vectors to facilitate computation. For our simple case this would be a

,

= (1 -2 3 4) (3 = (2 1 -3 2) afJl = ""=2,----2=----"""'9-+:--:S"'"= -1.

II. Write a = (2

-6

IS

6) = rp. where p. is a unit vector.

SEC.

21

MATRIX MULTIPLICATION

SoZttI,ion

aa'=4+36+324+36=400 and -3 9 (1 a - 20 10 10 10

,. - 20,

:0)·

ORAL BDRCISBS

1. (a) Give the normS of the vectors (2), (3 0 2), (2 -1 3), -1 4 2), (4 1 3 -1), (t 1 - it). (b) Give the lengths of the vectors above. (c) Express each vector above as the product of its length by a unit vector. 2. Show that the following vectors are pairwJse orthogonal: (a) (1 -1 2), (1 3 1), (-7 1 4) (b) (1 1 1), (1 -1 0), (1 1 -2) (c) (1 2 3), (1 -5 3), (3 0 -1) (d) (2 1 0), (-1 2 1), (1 -2 5) (3

2. Matrix multiplication. The product AB of two matrices A and B i8 not defl,ned unless the nuniber of columri8 in the left-hand factor A is equal to the number of rOW8 in the right-hand factor B. . We define the product AB of an m by n matrix A == (~i)

(i == 1, .•• ,mj j ==

1, ...

,n), \

by an n by t matrix B

=

(bi ,,)

(j == 1, •.. ,

nj

Ie· == 1, ••• ,t),

; to be the m by t matrix (6)

a == AB

== (c.-.)

(i == 1, • • • ,mj Ie == 1, ••• , t),

in which (7)

c.-.

== ~lbl"

+ ~sbS1l + . . · + llMab"".

Thus every element of.the product matrix AB is the sum of the n products of the n elements in a row of A by the correspondingly placed elements in a column, of B. Our definition is called the row by column rule for matrix multiplication.

246


[CBAP.x

We note thatifa == (ala2 ••• an) and,8 == (b l b2 ••• bn ) are 1 by n matrices the matrix product 0

b1

al

bn

an

is the inner product we defined in Sec. 1. This is the reason for the notation we used. We observed also that the element in the ith ..row and kth column oj our matriz product AB i8 the inner product oj tlJ,e ith row oj A by the kth column oj B.

Nothing we have said about the definition of the matrix product AB implies that when it is defined it is equal to BA. Indeed BA need not even be defined. However, even when ABo and BA are both matrices of the same size, they rp.ay be different. For example,

1) (0 o 1 ~)

o

The

matric~s

0\ _'/1

0) - \0

~}

(00 .. 01) -_10\0 0)1 .

whose elements

ar~

all zero are all called o. They have the property that OA == AO == 0 for any m by n matrix A and zero matrices of proper size. It can also ha.ppen -that A. ·and .B are nonzero inatrices and that AB == o. For example,

zero malrice8 aIid are represented by the symbol

The identity matrice8 are square matrices whose diagonal ele:m.ents are all 1 and whose nondiagonal elements are all zero. ' We represent all such matrices by I. Then

lA- = AI == A

0

0

SEC.

247

MATRIX MULTIPLICATION

2]

m

for all by n matrices A and a corresponding m-rowed I on the left, and n-rowed I on the right. We shall finally state the following results without proof: Theorem 1. Matrix multiplication i8 as80ciative, that i8, if AB.and BO are defined we have (AB)O = A(BO). Theorem 2. The transp08e of a product AlAI' .. At of matrice8 i8 the product At'At-I' ••• AI'AI' of the transpo8e8 of the factor8 in rever8e order. Theorem 3. The determinant of a product of 8f[Uare matrice8 i8 the product of the determinants of the factor8. Theorem 4. The rank of a product of matrices does not exceed the rank of any factor •. nlustrative Examples I. Compute

o=

(-:

_! _~) (_:

100 Cll Cll CIl CII

C8l

Cal Cn Cn

+

+

-~).

12

Sol'Ution

= (0, 2) (1 • 3) (2' -I), = (0' -1) (1 . 0) (2 . 2), = (-2' 2) (1 . 3) (1 . -I),

+ +

+

+ +

+ +( + +

= (-2' -1) (1'0) (1'2), = (2'2).+ (-4'3) + (-1' -I), = (2' -1) (-4 . 0) -1 . 2), = (1' 2) (0' 3) (0' -I), = (1 . -1) (0 . 0) (0 . 2).

+

II. Compute Y

x =

+ + =

(~)

-2

Ana. 0 = ( -71 2

AX where

y = (;)

A=

(-~ _~. =~)

Solution u=2z+2y+z tI = -z + 2y - 2z w=2z-y-2z. III. Compute AB and BA where

A =

(_=)

B = (4

2

3).

D 4

-4' -1

248'

MATRIOES AND QUADRATIO FORMS

Soltdion 2

3)

=(

: -12

BA = (4

2

[OHAP. x

2 3) 4.

6,

-6 -9

3) (_:) = (-1).

IV. Compute AB and BA where

Sol'l.lJ:lon

(0 1) -!) (-~ -2) 1 = 1 -2' -2) (-11 -4 "3) = (5 1) -2 -7"' 1 V. Compute AB and BA where B is the transpose of A and A =

( 2 2 1) -1

2

2

-1

-1'

-2

Solution

AA' =

(-~2 -1~ -2 -~)( -1~ -2 -~ -2 -~) = to \0 = (

~ ~~ _~) (_~

-1

-2

-2

2

i

~ _~).

o 9 o

~)

-1"-2

Hence AB = BA in this case. EXERCISES

1. Compute A', B', A' = (AB)' as well as B'A' for the products in Illustrative Examples I, II, IV. 2. Compute (AB)O and A(BO) where

A =

(~-A 0=

~),

(-~

"B "

1 1

o

= (~ -1

~ -~).

3

2

~

1

SJIC.

INVE'RBE OF A BQU ARE MATRIX

3)

249

3. Compute the following matrix products:

(a) (a b) (a -b) -b a b a (b)

-!be -(alb ~ ) (: c

(:00

l)

(c) (:

;)

(_~

~ _~ -~) (-1~

(d) (

-2 -4 (6)

-;)

(~

: 2

-2

5

-8

!) ( ~. =: 0

-1

~) (_~

(J) ( : -1

~

1

(g)

G-~){-~

(A)

(=~

(i)

fa

~

~

-!6-5_~)

~

2

:3)

D D

!)G-D ~) fo _~

~) (~

~ -~),

0 1 \~ 0 1 -1 0 2 4. The product of two diagonal matrices is a diagonal matrix. What are its diagonal elements?

\i

3. The inverse of a square matrix. A square matrix A is said to be nonsingular if its determinant is not zero. Then there exists a unique matrix A-I (read" A inverse ") called the inv6'f86.of A such that the product AA-l == A-IA == I

is the identity matrix of the same number of rows as A. By Theorem 3 we have III == lAllA-II. But III == 1 and so we have the following:' Theorem 8. The determinant oj A-I i8 the inv6'f86 oj the determinant oj A.

250

:MATRICES AND QUADR~TIC FOR:M&

[ClIAP.x

We also h~ve (AA-I), == (A-I)' A' == I' == I. This result is stated as a theorem. Theorem 6. The inver8e of the transp08e of a matriz A is the transp086 of A-I. Since (AB)(B"""IA-I) == A(BB"""I)A-I == AA-I == I, we have a special case of a result which we state without additional proof. Theorem 7. The inver8e of a product AlA • ... A, of nonsingtilar 8quare matrice8 is the product A,-l • . . A.-IAI-I

of their inver868 in r61J6r8e order. Two m by n matrices A and B are called equivalent if it i8 p08sible to carry A into' B by a finite 8equence of elementary transformations. Then we state the following: Theorem S. Two matrices A and B are equivalent if and only if there ~8t nonsingtilar matrice8 P and Q 8'UCh that B == PAQ. Indeed, let a sequence of row and column transformations CaITY an m by n matrix A into B. Apply all the row transformations tp the m-rowed identity matrix to obtainP and all the column transformations to the n-rowed identity matrix to obtain Q such that B == P AQ. The inverse of a two-rowed square matrix is given by the formulas: (8)

A == (:

t == ad - bc ~ 0,

A-' -

(~! I~.

The inverse of an n-rowed square matrix A with n > 2 may also be given by a formula involving the determinant ~ A and the cofactors of the elements of A, but it is preferable to compute A-I by solving a system of linear equations. We shall give this solution in our next section.

SlDC.4J

LINEAR TRANSFORMATIONS

251

ORAL EXERCISES

1. Give the inverses of the matrices

(-25 -2) l'

(-~

-D, (_!:5 ~_4:).

i. What is the inyerse of a diagonal matrix whose diagonal elements are all not zero?

4. Linear transformations. H A == (asi) is an n-rowed square matrix, we may form a corresponding linear system

+ a12XZ + . . . + alnXn == YI a21XI + azsZz + . . . + asnZn == Y2 allXI

(9)

This linear system is equivalent to the matrix equation AX == Y,

where

X==

(10)

y==

H IAI '" 0, so that A-I exists, we multiply AX == Yon the left by A-I to obtain X == A-IY. Thus if we solve' formula (9) for Xl, • • • ,:en we will obtain equations Xl

(11)

X2

== PllYI + PUY2 + == P21YI + P22YZ +

. . . + plnYn . . . + P2nYn

equiValent to the matrix equation X == PY, where

252 (12)

MATRICES AND QUADRATIC FORKS

P =

G:~"1 .. ~~ . :... ~) p""

•••

[CHAP.x

= A-1.

p""

This shows how to compute A-1 by solving a linear system. The equations of fomiulas (9) or (11) are called the equations of a noMingular linear (homogeneous) tramff1l'1Tl,Q,f,ion. If we interpret the n~bers :t1, • • • ,:en as the coordinates of a point then formula (9) expresses the coordinates 'V1, • • • , 'V". of thiS same point relative to a new coordinate system in terms of the old coordinates. The equations of formula (11) are called the 80lved form of the equations of formula (9). They express the old coordinates :t~, • • • ,:en in terms of the coordinates 'VI' • • • , 'V". Dlustrativ. E%am,z.

Compute A-I if

A -

~ -~ =~).

\i

0-1

8oh1tion The system of equations (9) becomes

211:-2y-,=u z+1/-28=" Z -,

=

tD.

E1imina1i!'l , to obtain· Z - 211 = u - w, -z + 1/ ... " - 2tD. Then - (z - 211) - 2( - Z + 1/) ... z ... -~u - w)- 2(" - 2tD) ... -u - 2v + 5w. Also1/-z+I1-2tD= -u-v+8tD,'-z-w= -u-2v+4tDj SO that

-1 -2

A-I... ( -1 -1 -1 -2

V·

We can check the answer by computing the product AA-l or A-IA. BDRCISES

Compute the inverses of the following matrices A and check your answers by computing AA-l 1. . .

=

SEC.

(a) (

(b)

5

2 0

4 1 0 0

~

0 0 2 1 1 0 2

-1

G

3 3

(_!

(g) (

-n

1

4 -4

V

3 -3 -4 7 3 -3

a

-3 7 -4

D

-D

G

4 -6 -1 -13) 0 1 -2 0 1 2 -5 -2 1 -5 1 0 1 1 1 ( 1 -2 1 -1 -4 -2 -3 1 1 1 -1 1 1 2 1 -2 1 0 -1 ( 2 -1 1 1 2 1 1 6 1 -2 ":'3 2 ( 2 -1 -4 3 -1 1 2 -1 0 -1 0 1 0 0 0 1 3 1 2 -1 1 5

W(~ (k)

D

-1

(Il)

(i)

a D

(d) (

tn

~)

1 2 -3 -2 1 4

(.) G (e)

253

SIMILAR MATRICES

5]

(ij

(..)

D -D G ~ -D V

(o)

(o)

n

!)

5. Similar matrices.

-)

-D -

If A == (o,;i) is any n-rowed square

matrix, we shall designate by A - xl the result of formally subtracting x from all the diagonal elements. Then (13)

A -xl

au - x . au

-

. a81

anI

au

alB

alta

au - x au

au

a2ta

ala - x

a8ta'

an2

an.

ann - x

. . . . . .....

254

MATRICES AND ~UADRATIC FORMS

[CHAP.x

The matrix of numbers ,obtained by replacing the v~riable x in formula (13) by a number d will be denoted by A - dl. The characteri8tic determinant of any square matrix A is .the polynomial in x with leading coefficient unity given by (14)

F(x) .= (-l)"IA - xII.

The equation F(x) = 0 is called the characteri8tic equaftio'(t of A, and the roots d 1, • • • , dn of this equation are called the characteri8tic root8 of A. They are a set consisting of all numbers di such that (15)

IA - dill =

o.

A s,uare matrix B is said to be similar to A if there exists a nonsingular matrix P such that B = P-IAP. We shall assume the following results without proof: Theorem 9., Similar matrice8 have the same characteri8tic determinant8 and therej01re the 8ame characteristic roota. Theorem 10. The characten8tic roota of a diagonal maf:!iix D are the diagonal elements of D. Theorem 11. Every real symmetric matrix i8 8imilar to a diagonal matrix. Theorem 12. Every matrix with di8tinct characteri8tic root8 i8 8imilar to a diagonal matrix. Theorem 10 implies that if a matrix A is similar to It diagonal matrix D the diagonal elements of D are the characteristic roots of A. The arrangement of these roots in the diagonal of D is actually arbitrary. Let us suppose' then that A is given and that we wish to determine whethe:r or not A is similar to a diagonal matrix p. We first determine the characteristic roots of A and hence a diagonal matrix D. We then seek a nonsingular matrixP such that (~6)

AP = PD.

If Pi is.the jth column of P, the jth column of PD is the s'calar mqltiple diPi . The jth column of AP is APi ~nd, formula (16) is equivalent to n linear homogeneous syste~s,

BBC.

5]

255

SIMILAR MATRICES

each of 11, eq1.1&tions in 11, variables. These systems are expressible in matrix form as (17) (A - diI)Pi = 0 (j = 1, . . . ,11,). It is alway, possible to solve these systems and determine P. When A is not similar to D no set of solutions of the equations of formula (17) will yield a nonsingular matrix p. mustrGtive EZGm;les I. Find the characteristic roots of

A

7 -6)

=(-!

4 2

8oltll.ion

-:e F(s) .. (-1)1-1

o

7

-6

O· -2

I

0

4- s 0 =-1 2 -s - 2 0

sI-4s+7

I,

4-s

o6

2

s+2

so that J(s) .. (s

+ 2)(SIl -

4s

= (s -

+

+ 7)

- 12 = sa - 2s1 - S 2 2)(SI - 1). Ana. ell = 1, ell = 2, ela = -1.

II. Show that the matrix of Illustrative Example I is similar to a diagonal matrix. We first solve the homogeneous 'system

-s +71/- & = 0 -s +31/ . = 0 2y - 3. = 0, whose matrix is A - I. Then s = 31/, 3s = 2y and the first column is a sca.la.r multiple ,of 9, 3, 2. The system with matrix A - 21 gives -s + 2y = 0, 21/ - 4s = 0 so that s = 2y - 4s and the second column may be taken to be 4, 2, 1. Finally, the system with matrix A - 41 gives -s + 411 = 0, 211 - • = 0 and so s = 411, • = 211. Hence we may take

IPI

-3

-4

= 1 3

2

-4

-3

III. Show that the matrix

A=G is not similar to a diagonal matrix.

1 .~).

o 1 "" O. o

256

MATRICES" AND QUADR'ATIC FORMS' [CHAP.x

Solution The characteristic determinant of A is (:I: - 1)2(:1; - 2) and d1 = ds = 1. Thus the first and second columns of P are obtained by solving the linear system with matrix A - I. The system is 'Y = 0"= s and the first two columns of P are necessarily proportional. Thus AP = .PD implies that IPI = 0, A and D are not simils.r. ORAL EXERCISES

Give the "chs.ra.cteristic determinants and roots of the following matrices: 0 (g) (~ (-~ 0 0 0 (6) ( 3 (b) (h) -4 6 0 4 1 3 (c) (~ t n (-4 3 o _ 4 0 -9 3 " 0 o 0 0 5 0 0 0

(a)

G

~)

G D'

~)

G ~)

i)

(~

D

°r~G

~)

D

EXERCISES

1. Compute the_characteristic roots of the following matrices:

(a):(_~ (b)

G

(.) G C~

1 0 5 1 0 0 0 0 1 0

!) o o

-4

2

-1

(d)

-V

-4 -6 1 -

-6 -6 (g) ( 9 -6 -6 8 2 -4 ,1 (h) ( ~ 0 tn

D

1

(c) (

3 -2

f 2 3

o

-2 0) 2-2 -2 1

2 -1 -4

D

Ana. -I, -I, -I, -1.

~) -!)

'D

Ana. I, 4, 16.

SEC.

6]

(i)

(!

010 -1

{il (

{ij

0 -2 4 9

~

257

QUADRATIC FORMS

0 0 1

-D. -1

1 0, 0 4 1 3 6 10 1 1 0 0

AM. -1, 2, 3.

0 1 0 '-6 0 0 0 1

D

~

,3

5

6 0 1

-V

AM. 1, -1,2, -3.

2. Find a nonzero matrix P such that AP = PD for each matrix of Exercise 1 where D is a diagonal matrix whose diagonal ele~ents are the chara.cteristic roots of A. State in each case whether or not A is similar to D. 1 3 3 1

AM. (6) P = (: : ·11

,

AM. (g) P

AM.

=

(1 2-2) 2 2

=: =~,i) (=~! -i : =r, V'

(~1 P = (-:1

AM. (0 P

=

2, A similar to D. -1

1 -2

1

A simila.r to D.

1

0

A similar to D.

0

6. Quadratic forms. An '1I/-O,ry iJua,dratic form. is a polynomial ,in n variables which is sum of terms all of degree two. The most general unary quadratic form. is ax l where a is a number and x a variable. The general binary quadratic form. may be represented as

I(x, y) = axl + 2bxy

+ C!l1

258

MATRIeE SAND QUAD RATIe FORMS

[CBAP.x

'where a, b, c are numbers. Any polynomial may be regarded as being a 1 by 1 matrix. Thenf(x, y) is expressible as the matrix product f(x, y) == (x

The general ternary quadratic form is f(x, y, z)

== ax! + by! + cz! + 2rxy + 28XZ + 2tyz.,

It may be expressed as the matrix product f(x~ y, z) == (x

r

y

b

t A quadratic form in Xl, X2, • • • , x A is a sum of terms for i, j = 1, .•. , n. The coefficient of Xi! is the number au. The total fJoefficient of-xiXi = XiXi is ~i aji. It is customary then to write this total coefficient as twice a' numbe~ which we now designate as ~i and so have ~i = aii· Then our quadratic form will be expres~ible as a matrix product f(X1, ••• , xA ) = X'AX. Here A = (~i) is an n-rowed symmetric matrix, that is aii = aii for every i and j. Also ~X'-Xi

+

so that X' is the transpose of X. We call the syn;unetric matrix A the matrix of the quadratic form f(X1, . • • , x A ) = X' AX and its rank the rank of f(X1, ••• , xA ). • nlustrative Examples I. Give the matrix of 3:1)2 + 23;y + 3yz + 51$2.

SEC.

7]

E Q U IV ALE NeE

0 F QUA D RAT I C FOR M S

259

Solution

We are using x

f(lfJ: Xl,

Y for X2, Z for

X3.

A=(! ~ II. Give the matrix of 2X2

Then

!}

+ 5xy + 6y2 + 3xz + yt + 5z2 + XZ + 2tz. Solution

f(x, y, z, t) = 2X2 + 2(l)yt 2zt and

+

+ 6y2 + 5z2 + Ot2 + 2(i)xy + 2(t)xz + Oxt + Oyz i

6

o i

t 0 5

V o :a

•

1

1

ORAL EXERCISES

Give the matrices of the following quadratic forms:

+

(a) 2X2 y2 - 4xy - 4yz (b) _2X2 4y2 6z 2 + 2xy 6xz 6yz (e) 4x 2 y2 - 8z 2 4xy - 4xz 8yz (d) 3x 2 - 3y2 - 5z 2 - 2xy - 4xz - 6yz (e) 3x 2 - y2 Z2 - xy xz + 3yz (f) 2X2 y2 - fuy - 3yz 6xz (g) 3xy - X2 - yz - 5xz (h) 4zx y2 zy Z2 3yx (i) x 2 - y2 + Z2 - 2xy 3xt 4yt (j) xy+xz+xt+3yz+5zt (k) 2xy 2xt - y2 yz - zx - t2 (l) X2 y2 Z2 t2 - 2xy + 2xz"- 2zt (m) (x - t)2 - (y - Z)2 + (2z - t)2 (n) 4(x y)2 - (2x ~ t)2 - x(y - z)

+

+

+

+

+

+ + + + + + + + + + + + + + + +

+

+

+ xt"

7. Equivalence of quadratic forms. ~et f(:lh, ••• -, xn) be a quadratic form so that f(Xl, . • . ,xn ) = X'AX where X is a column vector and A is a symmetric matrix. Write

I

(18)

y=

x = py, Yn

260

·MATRIOES AND QUADRATIO FORMS

[CRAP.x·

where P = (Pii) is a nonsingular matrix. Then J(:lh, •• • , z.)

= CPY)'.A.(PY)

= Y'(P'AP)Y ~ Y'BY = g(Yl, . • . ,Yta)

is a quadratic form in Yl, • . . , Yta with matrix (19) B = P'AP. Hence a linear tram/ormation with matrix P replaC68 a quadratic Jorm with matrix .A. by a quadratic Jorm 'With mat:ri:I: P'AP. We shall call two quadratic forms J(Xl, • • • , z.) and g(Xl, ••• , z.) equivalent if there exists a nonsingular linear traD.sformation (18) which replacesJ(xl, . . . ,xta ) by g(Yl, .•• ,Yta). We say that two' symmetric matrices .A. and B are CO'fI,(J'iouent if they are related as in formula (19). Then two quadratic JO'I'1Y/,8 are equivalent if and only iJ their matrice8 are congruent. It can be shown that .A. and B are congruent if and only if we can find a sequence of elementary transformations on the rows of .A. such ,that it, followed by the sequence of curresponding column transiQrmations, will carry .A. into B. We shall assume this as well as the following: Theorem 13. Every quadratic Jorm J ill equivalent to a diagonal quadraticJorm b1xl1 + ... + b,.xtal where.' bib. ••. br ~ 0, br+l -_.. . . = btJ ;", 0 and r i8 the rank oj the matrix oj J. In the study of quadratic forms with real coefficients we permit only linear transfo~tions with matrices of real elements. Let uS .then carry a real quadratic form J into an equivalent diagonal form b1xl 1 + ... + brXrl. Then the number of positive coefficients b1, • • • , br i8 the 8ame 1UJ matter what diagonal Jorm we obtain, and. we 'call this natural number the index of J. . It follows that two real quadratic JO'I'1Y/,8 are equivalent iJ and only iJ ·they have. the 8ame rank and index. ' . . A value of a real quadratic form is a real number J(Cl, ... , c,.).

SIIO.7]

EQUIVALENCE OF QUADRATIC FORMS

261

We call !(:Ih, '. . . , XII) a po8itive form if every value I(Cl, ••• , c,.) ~ 0, and call!(xl, . . . ~ x,.) a negative' form if -!(Xl, ••. , x,.) is positive. It can be shown that !(Xl, '. . . , XII) is po8itive i! and only i! its index i8 its rank. A quadratic form !(Xl, . . .., x,.) is said to be positive definite if !(Cl, . . . , c,.) > 0 unless Cl

= Ct = . . . = c,. = O.

This occurs when and only when n is the rank and index of !(Xl, . • • , x,.). When all the values of !(Xl, . . . , x,.) except !(O, 0, • • • ,0) are negative, we call I(Xl, • • • ,x,.) negative definite and this occurs when -!(Xl, . . . , x,.) is positive definite. Illu8trative Example

Compute the index of the quadratic form I(~,

1/,')

I!!!

43;1

+ 1/1 + 211 -

2zy

+ 10= -

4yz.

8oZw,Um The ma.trix of the quadratic form is

A

=

(-! =i

-0' .

We interchange the first and second rows and then the first and second columns to obtain

-~ -! -:). ( -2 5 2 Add the first row to the second and twice the first row to the third to obtain

(~

o

-! -!\. 3

-;J

Subtract the second row from the third to obtain (

- 001

-! -:). o

-5 '

262


[ClIAP.x

The general theory then states that the corresp,onding ,column' trans.. formatione will carry this matrix into the matrix

~ ~),

o

-5

which is congruent to A. Then the rank of j(x, y, s) is three, its index is two, and j(x, y, s) is equivalent to the diagonal quadratic form x 2 - Sy2' - 5s2• EXERCISES

Use elementary transformations involving rational numbers oruy to find diagonal quadratic forms to which each of the following quadratic forms are equivalent. Give the rank and index in each case. (a) 6x 2 + 14y2 + Z2 + 2xy - 2xs - 6ys . (b) 2X2 + 9y2 + 6s2 + 8xy + 2xs + 8ys (c) 2x 2 + Sy2 + 16s2 + 4xy - 8xs (d) 2xy - 4yz (j) 4xy - 4ys - 6xz W~-_+~-~-~ W2~+yz+~ (h) 5:£2 lSy2 S2 16xy - 2ys

+ + + + 6y2 - 2s t2 W 5x 2 - 2xs + 4xt + y2 (i) lSx 2

2 -

2yt + 8xy - 2st 4ys + 2yt -' 2s 2 - 2st - t2

8. Orthogonal matrices. A matrix P is called an orthogonal matrix if PF:' =, I. Then the transpose of P is the inverse of P. It should be clear that the rows of an orthogonal matrix are unit veators which are pairwise orthogonal. Since PP' = I implies that P'P = I the columns of an orthogonal matrix are also pairwise orthogonal unit vectors. ' If P is an orthogonal matrix and D is a diagonal matrix whose diagonal elements are dl , • • • , dft, the jth column of P D is dj times the jth' column of P. Then the length of the jth column of P D is dj and the colum,ns of P D are still pairwise orthogonal vectors. Hbwever the rows of P D' wi:ll ,usually no longer be pairwise orthogonal. Theorem 14. Let1al, ••• , at be t pairwise orthogonal n-dimensional real vectors. Then t ;;! n and there exist n - t vectors at+l, . • • , aft such that vectors aI, • • • , aft fJ,1'11 pairwise orthogonal.

BEc.8]

ORTHOGONAL MATRICES

263

Corollary. Let al, . . . , an be real vector8 selected as in Theorem 14. Then the matrix P, whose ith row is 1 (20) is an orthogonal matrix. Theorem 15. The product of two orthogonal matrice8 is an orthogonal matrix. Rlustrative Examples I. Find an orthogonal matrix P whose first row is a scalar multiple of(1 -1 2}. Solution The equation te - Y 2B = 0 has te = Y = 1 and z = 0 as a solu1 O) is a vector orthogonal to (1 -1 2). A tion. Thus (1 y z) orthogonal to both (1 -1 2) and (1 1 O) vector (te satisfies the -equations

+

te-y+2z=0 te+y=O. We see that y = -te, 2:e + 2z = 0, z = y = -te, and the rows of the matrix

~ -~

\i

~)

-1

-1

are pairwise orthogonal. If we replace the rows by unit vectors which are scalar multiples of them, we obtain an orthogonal matrix

p=

1

-1

2

v'6

V6

V6

1

1

v'2

v'2

1

0

-1

-1

va va va which is a solution of our problem. II. Find an orthogonal matrix P whose first two roW8 are scalar multiples Of (2 1 -2), (1 2 2). Solution

We solve the equations 2:e+y-2z=O te+2y+2z=O.

264

MATIUCES AND QUADRATIC FORMS

+

Then &; 31/ ... 0, Z = -1/, 2z II "'" 1 is a. solution. The ro~ of

Z

-=

Z

= 21 so tha.t Z

..

[CRAP,

2, 11

=

x

-2,

G-21. -2) 122 'I

are pairwise orthogonal, and a solution-of our problem is the orthogonal matrix '

(I I -I)I':

P=l 1 1 -I

i

:EXERCISES ' 1. Find an orthogonal matrix whose first row is a. scaJa.r multiple of (a) (1 (b) (2'

-1) 1)

(0) (-4

(d) (1 (e) (1

0)

(f) (1

-1 1 2

CJJ) (-3

1) 0) 1)

(h) (2

(,1

1 0

(-1

1) 1)

1

4)

i. Find an orthogonal matrix whose first two rows are scalar multiples, respectively, of the rows of the following: (a) (2

\0

(b) (1

\2

-1 -2) 2-1 1 -1) -1

1

(1 02 1) -1 1 (d) (1 2 3) -1...:1 1 (0)

(e)

(1

o

(f)(224) -2 0 1 CJJ) (1 1 -1 01) 1 -1 0 (h) (' 1 2 -1 01) . -1 1 1

(,1 (-1 -2 1

1

1 1

,

2\ II

2-2\ 1 1J

9. Orthogonal reduction of a quadratic form. An orthogonal liMar trarurjormation· is a linear transformation X == PY, where P is an orthogonal matrix. Then the solved form. of this transformation is Y = P'X, and so is obtained from X == PY simply by the interchange of rows with colmims in the coefficient matrix P. If Xl and X" are any two (column) vectors, their inner product is XI'X". The transformed vectors are Y1 == P'XI , Y" = P'X" and YlY" = X1'PP'X" == XI'X". Thus an orthogonal li'n6Gf 'rom/ormation pr886M188 tM inMr proa'1.1d8

SlilC.

9]

ORTHOGONAL REDUCTION'

265

of aU pair8 of f)ftctor8. It therefore preserves the length of a vector as well as the angle between two vectors. H P is an orthogonal matrix, we have PP' - I, IPP'I - 1 7 IPls and IPI = ± 1. We shall call an orthogonal transformation a rotation of tue8 if IPI = 1, and shall say that a quadratic formf(3h, ••• ,x,,) i8 equivalent under a rotation of (U68 to g(Xt, . . • , 3:,.) if· f(xt, ... ,3:,.)

5&

X'AX == g(Yt, ••• ,y,,) = Y'BY

where B - 'p'AP, PP' - I, IPI - 1. Then we state the following: Theorem 16. . Every real quadratic fqrm

f(xt, . . . ,x,,) - X' AX is equivalent under a rotation of (U68 to a diagonal quadratic fqrm dtxt S + ... + AAs, where d t , ••• ,d,. are the characteristic root8,of the real symmetric matrix A. The matrix P is determined as in Sec. 5 by the property that the jth column Pi of P is a unit vector such that (A - dJI)Pi = O. The property IPI = 1 may be obtained by changing the sign of all elements in a column of P if necessary. In the case n - 3 we shall write

f(x, y, z) = axS + bys + CZS + 2rxy + 28XZ + 2tyz and the equations 'of the rotation of axes will be

+ AsY' + AaZ' JJ.tX' + PosY' + JJ.aZ' 'IItX' + 'IIsy' + 'IIaZ'.

x - AtX' (21)

Y Z-

The solved form of these equations is then x' = AtX , y' = AsX z' - AaX

(22)

Here Xls

+ JJ.tY + 'IItZ . + JJ.sY + 'IIaZ + JJ.aY + 'IIaZ.

+ Pls + 'illS = Aa s + JJ.a s + 'IIz s =

Aa s

+ JJ.as + 'II. s = 1

MATRIC'ES AND QU,ADRATIC FO'~,;~ns

266

and we sdlve the systems (23) (A - d,I)

00 ~

0,

d,I)G~ ~ 0,

(A -

IA -

where d1, d2 , d3 are the roots of: F(x) = -

[OHAP.)f

xII =

o.

fllustrative Example Find the diagonal quadratic form to which we may carry lex, y, z)

== 3x2 + 4y2 -

+ 4yz

Z2 - 12xy - 8xz

by a rotation of axes, and give the equations of the rotation of axes. Solution The matrix of lex, y, z) is

A

=

(-!

-~ -~).

-4

2

-1

Its characteristic function is

, F(x)

== -

3 - x '-6 -4 -6 4- x 2 -4 2 -1 - x

-x - 9

-6

4

4 - x -2

-2x+2

'0

1+x

2

and

+ 9) - 4( -2x + 2)] + (1 + x)[(x + 9)(x - 4) + 6( -2x + 2)i == -2(10x + 10) + (x + 1){x2 - 7x - 24) == (:c + 1)(x2 - 7:c - 44) = (x + 1)(x + 4)(x - 11). Then!(x, y, z) will go into -X'2 - 4y'2 + l1z'2 under the required,rotation of axes. We solve the system of equations 4A1 - 61'1 - 4"1= 0; -6A1 + 51'1 + 2"1 = 0, -4A1 + 21'1 = 0 with matrix A + I to obtain 1'1 = 2A1, -6A1 + 101'1 + 2"1 = 0, "1 = -2A1' Thus our first column

F(x)

==

-2[2(x

is a unit vector which is a scalar multiple of 1, 2, -2. equations with matrix A + 41 is 7A2 - 61'2 - 4"2 = -6A2

+ 81'2 + 2"2 =

-4A2

and -oA2 + 101'2 = 0, A2 = 21'2, "2 = 3A2 - 41'2 second column is a multiple of 2, 1, 2. Finally --BAa - 6pa - 4"a

= -6},,8 -

7113

+ 2"8 9'

-4A3

The sys,tem,of

+ 21'2 + 3"2 = =

0

21'2, $0 that our

+ 21'3 -

12"8 = 0,

267

ORTHOGON AL REDUCTION

Sl!Io.9]

(-41\. - 31'. - 2".) + (-61\. - 71'. + 2".) = -101\8 - 101'. = 0 and 1\. = - 1'., 61'. - 71'. + 2"8 = 0, 1'. = 2"8. The determinant

2

1 2

-2

100 2 -3 6 = -27, -2 6-3

12=

2

-2

1

and so we must change the sign of the elements in one column. We change the signs of the elements in the last column. The length of each column vector is V1 + 4 + 4 = 3, and the matrix of the rotation of axes is

i i t -i' I -t

i

I -I

P=

The equations of the rotation of axes in matrix form are X = PY, where X and Y are column vectors, and so the equations become Z

=

z' + 2y' 3

+ 28'

'

2z'

y=

+ y' 3

2s' '

s =

-2z'

+32y' -

s'

.

We have used the rows of P as coefficients. The column denominators in this exercise are all the same but the denominators in different columns in other problems will uSually be different. The columns of P are the coefficients in the solved form

z'

=

z-2y-2s 3 '

,_2z-y-2s 3 '

Y -

s'

=

2z - 2y - s' 3 '

and it may be preferable to give the answers in this form. .Note, finally, that if p-1AP = D the sum oj the diagonal elements ot A is the sum oj the diagonal elements oj D. For the negative of this sum is the coefficient of 3:"-1 in their common characteristic equation. This result is useful as a check on answers. The matrix P is not unique if the characteristic roots of A are not all distinct. When these roots are all distinct, P is uniq~e apart from a change of sign of any two of its columns. EXD.CISBS Find a rotation of axes which will carry J(:e, y, s) into a diagonal 'quadratic form g(:e', y', s') in each of the following cases. Check the results in the cases where answers are not given by showing that 'AP = PD, where P is the matrix of the rotation of axes, A is the matrix 'of J(:e, y, s), and D is the matrix of g(:e', y', s'). (a)

J=

2:e 2 + y2

.~:e'

(b)

J=

-

- 4:ey - 4ys. Am. g = Z'2 - 2y'2 + 4z'2; + y - 28, 3y' = :e + 2y + 28, 3s' = 2:e - 2y + s. 2z2 + 4y2 + 6s + 2:ey + 6zz + 6ya.

=

2z

2

268 (co)

.M AT RIC E 8 A l'i D QUA D RAT I C FOR.M 8

[CHAP. X

J = 43:2 + y2 - Sz' + 4zy - 43:z +Syz.

Ana. g = 53:'2 + 2y'2 - 1Q~'2; -3: + 2y + z, v'3O z' = 3: ..:... .2y .5z. (d) '1 = 33: 2 - 3y2 - 5z 2 - 23:Y - 6zz - 6yz. (e) J = 33:2 -t y2 + 22 - 23:Y + 23:z - 211z. Ana. g = 43:'2 + Z'2; V63:' == 2x - y + z, v'2 y' = y + z, Z' = -3: - Y + z. (J) J = 23: 2 - y2 + 2Z2 - 63:Y - 6yz. • (g) = .53:2 + 5y2 + 3z2 - 23:Y + 23:Z + 2yz .. , Ana. g = 53:'2 + 6y'2 + 2Z'2; 3:' = 3: + y + z, v'2 y' = 3: - y, v'6 Z' = 3: + y - 2z. (h) J == 23: 2 - Z2 + 2:J:Y - 4yz. (t.') J = X2 + 23:Y - 2xz - 4yz. ' Ana. g = 3y'2 - 2Z'2; V6 x' = 23: - Y + z, y' = x + y - Z, v'2 Z' = Y + z. ' :(3') J = 5:r;2.+ 2y2 + 2Z2 + 23:y - 23:z - 4yz. ~k) I::; 2X2'+ 2y2 - Z2 + 8:J:Y - 43:z - 4yz. Ana. g. = -2X'2 - 2y'2 + 7Z'2;. 33:' = 2:1) - Y + 2z, 3y' = -3: + 2y + 2z, 3z' = 2x + 2y'';;'" z. (1) J = 2;2 + y2 2Z2 + 23:Y - 2yz. (m) J = 123: 2 - 6y2 - 4z2 - 123:y + 12yz. Ana. g = 143:'2 - 12y'2; '\I'9i x' = 9x - 3y - z, VI4 y' - x + 2y + 3z, V26z' = x + 4y - 3z. (n) J = X2 + y2 +. 9z 2 + 163:Y + Syz :- Su. ' (0) J = 43: 2 + 6y2 + 4z2 - 43:z. Ana. g = 2X'2 + 6y'2 + 6Z'2; v'2 x' = 3: + 2, y' = x + y - z, v'6 z' = - 3: + 2y + z. (p) 23: 2 + 3y2 + 2z2 + 2yz + 23:Y.

+

vA = 23: + y, v'6 y' = [1;'

va

r

va

va

+

va

10. Factoriz~tion of a positive symmetric_matrix. Every positive symmetric Ill{l.trix A has a factorization

A

=

FF',

where the number of columns in the rectangular matrix F is the rank: r of A. A method of determining F called the, diagonal method of factorization is presente

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