First course in abstract algebra with applications 3rd by joseph j rotman

For example, the statement “2n > n for all 3The contrapositive of an implication “ P implies Q” is the implication “not Q implies not P.” For example, the contrapositive of “If a series

Special Notation

Set Theory and Number Theory

natural numbers 1

integers 1

n r
binomial coefficient 18

bxc greatest integer in x 25

8d(x ) dth cyclotomic polynomial 29

φ (n) Euler φ-function 30

 rational numbers 34

 real numbers 34

 complex numbers 35

a | b a is a divisor of b 37

(a, b) gcd of a and b 37

[a, b] lcm of a and b 55

a ≡ b mod m a congruent to b mod m 57

X ⊆ Y X is subset of Y 81

X  Y X is proper subset of Y 81

 empty set 81

X × Y cartesian product 84

1X identity function on set X 84

|X| number of elements in finite set X 84

im f image of function f 85

f : a 7→ b f (a) = b 86

a ≡ b a is equivalent to b 96

[a] equivalence class of a 97

[a] congruence class of a 168

 m integers modulo m 168

δi j Kronecker delta 370

Group Theory S X symmetric group on set X 103

S n symmetric group on n letters 103

sgn(α) signum of permutation α 117

GL(n, k) general linear group 128

Isom( 2) group of isometries of the plane 136

O2( ) orthogonal group of the plane 136

D 2n dihedral group of order 2n 141

6(2, R) stochastic group 144

V four-group 145

H ≤ G H is subgroup of G 145

H < G H is proper subgroup of G 145

A n alternating group on n letters 147

a H coset 151

[G : H] index of H in G 153

SL(n, k) special linear group 154

G ∼ = H isomorphic 156

ker f kernel of f 160

H G H is normal subgroup of G 161

Z (G) center of group G 163

Q quaternion group of order 8 164

G/H quotient group 176

H × K direct product 183

G x stabilizer of x 194

(x ) orbit of x 194

C G(a) centralizer of a∈ G 195

GL(V ) all automorphisms of vector space V 381

H ⊕ K direct sum 473

Pn i=1S i sum of subgroups 477

Ln i=1S i direct sum 477

N G(H ) normalizer of H ≤ G 489

UT(n, k) unitriangular group 493

Commutative Rings and Linear Algebra I or I n identity matrix 128

[i] Gaussian integers .217

 (  ) ring of functions on  222

U (R) group of units in ring R 226

 (R) ring of functions on ring R 227

p,
q finite field having p, or q, elements 228

Frac(R) fraction field of domain R 231

R× nonzero elements in ring R 232

deg( f ) degree of polynomial f (x ) 233

k [x] polynomial ring over k 237

k(x ) field of rational functions over k 238

k [[x]] power series ring over k 240

R ∼ = S isomorphic 241

(a1, ,a n) ideal generated by a1, ,a n 246

(a) principal ideal 246

R × S direct product 249

a + I coset 290

R/I quotient ring 291

k(z) adjoining z to field k 297

A ◦ B Hadamard product 306

A ⊗ B Kronecker product .309

Matn(k) all n× n matrices over k 323

A T transpose 325

Row( A) row space of matrix A 329

Col( A) column space of matrix A 329

dim(V ) dimension of vector space 336

E / k field extension 341

[E : k] degree of field extension E/k 341

Homk(V , W ) all linear transformations V → W 367

Y [T ] X matrix of transformation T relative to bases X , Y 370

det( A) determinant 385

tr( A) trace 392

Supp(w) support of w∈ k n 408

Gal(E / k) Galois group of E / k 452

Var(I ) algebraic set of ideal I 540

Id(V ) ideal of algebraic set V 544√

I radical of ideal I 545

DEG(f ) multidegree of polynomial f (x1, ,x n) 559

Danny and Ella, whom I love very much

Special Notation i

Contents v

Preface to the Third Edition viii

Chapter 1 Number Theory 1

Section 1.1 Induction 1

Section 1.2 Binomial Coefficients 16

Section 1.3 Greatest Common Divisors 34

Section 1.4 The Fundamental Theorem of Arithmetic 53

Section 1.5 Congruences 57

Section 1.6 Dates and Days 72

Chapter 2 Groups I 80

Section 2.1 Some Set Theory 80

Functions 83

Equivalence Relations 96

Section 2.2 Permutations 103

Section 2.3 Groups 121

Symmetry 134

Section 2.4 Subgroups and Lagrange’s Theorem 144

Section 2.5 Homomorphisms 155

Section 2.6 Quotient Groups 168

Section 2.7 Group Actions 189

Section 2.8 Counting with Groups 205

Chapter 3 Commutative Rings I 214

Section 3.1 First Properties 214

Section 3.2 Fields 227

Section 3.3 Polynomials 232

Section 3.4 Homomorphisms 240

Section 3.5 Greatest Common Divisors 250

Euclidean Rings 265

Section 3.6 Unique Factorization 272

Section 3.7 Irreducibility 278

Section 3.8 Quotient Rings and Finite Fields 288

Section 3.9 Officers, Magic, Fertilizer, and Horizons 305

Officers 305

Magic 310

Fertilizer 314

Horizons 317

Chapter 4 Linear Algebra 321

Section 4.1 Vector Spaces 321

Gaussian Elimination 345

Section 4.2 Euclidean Constructions 354

Section 4.3 Linear Transformations 367

Section 4.4 Determinants 385

Section 4.5 Codes 400

Block Codes 400

Linear Codes 407

Chapter 5 Fields 430

Section 5.1 Classical Formulas 430

Vi`ete’s Cubic Formula 443

Section 5.2 Insolvability of the General Quintic 447

Formulas and Solvability by Radicals 458

Translation into Group Theory 460

Section 5.3 Epilog 469

Chapter 6 Groups II 473

Section 6.1 Finite Abelian Groups 473

Section 6.2 The Sylow Theorems 487

Section 6.3 Ornamental Symmetry 498

Chapter 7 Commutative Rings II 516

Section 7.1 Prime Ideals and Maximal Ideals 516

Section 7.2 Unique Factorization 522

Section 7.3 Noetherian Rings 532

Section 7.4 Varieties 538

Section 7.5 Gr¨obner Bases 556

Monomial Orders 557

Generalized Division Algorithm 564

Gr¨obner Bases 569

Appendix A Inequalities 581

Appendix B Pseudocodes 583

Hints for Selected Exercises 587

Bibliography 601

Index 604

Preface to the Third Edition

A First Course in Abstract Algebra introduces groups and commutative rings.

Group theory was invented by E Galois in the early 1800s, when he used groups

to completely determine when the roots of polynomials can be found by formulasgeneralizing the quadratic formula Nowadays, groups are the precise way to dis-cuss various types of symmetry, both in geometry and elsewhere Besides intro-ducing Galois’ ideas, we also apply groups to some intricate counting problems

as well as to the classification of friezes in the plane Commutative rings providethe proper context in which to study number theory as well as many aspects ofthe theory of polynomials For example, generalizations of ideas such as greatestcommon divisor and modular arithmetic extend effortlessly to polynomial ringsover fields Applications include public access codes, finite fields, magic squares,Latin squares, and calendars We then consider vector spaces with scalars in ar-bitrary fields (not just the reals), and this study allows us to solve the classicalGreek problems concerning angle trisection, doubling the cube, squaring the

circle, and construction of regular n-gons Linear algebra over finite fields is

applied to codes, showing how one can accurately decode messages sent over anoisy channel (for example, photographs sent to Earth from Mars or from Sat-urn) Here, one sees finite fields being used in an essential way In Chapter 5,

we give the classical formulas for the roots of cubic and quartic polynomials,after which both groups and commutative rings together are used to prove Ga-lois’ theorem (polynomials whose roots are obtainable by such formulas havesolvable Galois groups) and Abel’s theorem (there is no generalization of theseformulas to polynomials of higher degree) This is only an introduction to Galoistheory; readers wishing to learn more of this beautiful subject will have to see

a more advanced text For those readers whose appetites have been whetted bythese results, the last two chapters investigate groups and rings further: we provethe basis theorem for finite abelian groups and the Sylow theorems, and we in-troduce the study of polynomials in several variables: varieties; Hilbert’s basis

theorem, the Nullstellensatz, and algorithmic methods associated with Gr¨obnerbases.

Let me mention some new features of this edition I have rewritten the text,adding more exercises, and trying to make the exposition more smooth The fol-lowing changes in format should make the book more convenient to use Everyexercise explicitly cited elsewhere in the text is marked by an asterisk; moreover,every citation gives the page number on which the cited exercise appears Hintsfor certain exercises are in a section at the end of the book so that readers mayconsider problems on their own before reading hints One numbering systemenumerates all lemmas, theorems, propositions, corollaries, and examples, sothat finding back references is easy There are several pages of Special Notation,giving page numbers where notation is introduced

Today, abstract algebra is viewed as a challenging course; many bright dents seem to have inordinate difficulty learning it Certainly, they must learn

stu-to think in a new way Axiomatic reasoning may be new stu-to some; others may

be more visually oriented Some students have never written proofs; others mayhave once done so, but their skills have atrophied from lack of use But none ofthese obstacles adequately explains the observed difficulties After all, the sameobstacles exist in beginning real analysis courses, but most students in thesecourses do learn the material, perhaps after some early struggling However, thedifficulty of standard algebra courses persists, whether groups are taught first,whether rings are taught first, or whether texts are changed I believe that a ma-jor contributing factor to the difficulty in learning abstract algebra is that bothgroups and rings are introduced in the first course; as soon as a student begins to

be comfortable with one topic, it is dropped to study the other Furthermore, ifone leaves group theory or commutative ring theory before significant applica-tions can be given, then students are left with the false impression that the theory

is either of no real value or, more likely, that it cannot be appreciated until somefuture indefinite time (Imagine a beginning analysis course in which both realand complex analysis are introduced in one semester.) If algebra is taught as

a one-year (two-semester) course, there is no longer any reason to crowd bothtopics into the first course, and a truer, more attractive, picture of algebra ispresented This option is more practical today than in the past, for the many ap-plications of abstract algebra have increased the numbers of interested students,many of whom are working in other disciplines

I have rewritten this text for two audiences This new edition can serve as atext for those who wish to continue teaching the currently popular arrangement

of introducing both groups and rings in the first semester As usual, one begins

by covering most of Chapter 1, after which one chooses selected parts of ters 2 and 3, depending on whether groups or commutative rings are taught first.Chapters 2 and 3 have been rewritten, and they are now essentially independent

Chap-of one another, so that this book may be used for either order Chap-of presentation.(As an aside, I disagree with the current received wisdom that doing groups first

is more efficient than doing rings first; for example, the present version of ter 3 is about the same length as its earlier versions.) There is ample material inthe book so that it can further serve as a text for a sequel course as well

Chap-Let me now address a second audience: those willing to try a new approach

My own ideas about teaching abstract algebra have changed, and I now think that

a two-semester course in which only one of groups or rings is taught in the firstsemester, is best I recommend a one-year course whose first semester coversnumber theory and commutative rings, and whose second semester covers lin-ear algebra and group theory In more detail, the first semester should treat theusual selection of arithmetic theorems in Chapter 1: division algorithm; gcd’s;euclidean algorithm; unique factorization; congruence; Chinese remainder the-orem Continue with Section 2.1: functions; inverse functions; equivalence re-lations, and then commutative rings in Chapter 3: fraction fields of domains;

generalizations of arithmetic theorems to polynomials; ideals; integers mod m;

isomorphism theorems; splitting fields, existence of finite fields, magic squares,orthogonal Latin squares One could instead continue on in Chapter 2, coveringgroup theory instead of commutative rings, but I think that doing commutativerings first is more user-friendly It is natural to pass from

to k [x], and one can

watch how the notion of ideal develops from a technique showing that gcd’s arelinear combinations into an important idea

For the second semester, I recommend beginning with portions of Chapter 4:linear algebra over arbitrary fields: invariance of dimension; ruler-compass con-structions; matrices and linear transformations; determinants over commutativerings Most of this material can be done quickly if the students have completed

an earlier linear algebra course treating vector spaces over

If time permits,one can read the section on codes, which culminates with a proof that Reed-Solomon codes can be decoded The remainder of the semester should discussgroups, as in Chapter 2: permutations; symmetries of planar figures; Lagrange’stheorem; isomorphism theorems; group actions; Burnside counting; and friezegroups, as in Chapter 6 If there is not ample time to cover codes and friezegroups, these sections are appropriate special projects for interested students Iprefer this organization and presentation, and I believe that it is an improvementover that of standard courses

Giving the etymology of mathematical terms is rarely done Let me explain,with an analogy, why I have included derivations of many terms There are manyvariations of standard poker games and, in my poker group, the dealer announcesthe game of his choice by naming it Now some names are better than others.For example, “Little Red” is a game in which one’s smallest red card is wild; this

is a good name because it reminds the players of its distinctive feature On the

other hand, “Aggravation” is not such a good name, for though it is, indeed, gestive, the name does not distinguish this particular game from several others.Most terms in mathematics have been well chosen; there are more red names than

sug-aggravating ones An example of a good name is even permutation, for a

per-mutation is even if it is a product of an even number of transpositions Another

example of a good term is the parallelogram law describing vector addition But

many good names, clear when they were chosen, are now obscure because theirroots are either in another language or in another discipline The trigonomet-

ric terms tangent and secant are good names for those knowing some Latin, but

they are obscure otherwise (see a discussion of their etymology on page 31) The

term mathematics is obscure only because most of us do not know that it comes from the classical Greek word meaning “to learn.” The term corollary is doubly

obscure; it comes from the Latin word meaning “flower,” but why should sometheorems be called flowers? A plausible explanation is that it was common, inancient Rome, to give flowers as gifts, and so a corollary is a gift bequeathed by

a theorem The term theorem comes from the Greek word meaning “to watch”

or “to contemplate” (theatre has the same root); it was used by Euclid with its present meaning The term lemma comes from the Greek word meaning “taken”

or “received;” it is a statement that is taken for granted (for it has already beenproved) in the course of proving a theorem I believe that etymology of terms

is worthwhile (and interesting!), for it often aids understanding by removing necessary obscurity

un-In addition to thanking again those who helped me with the first two editions,

it is a pleasure to thank George Bergman and Chris Heil for their valuable ments on the second edition I also thank Iwan Duursma, Robert Friedman, Blair

com-F Goodlin, Dieter Koller, Fatma Irem Koprulu, J Peter May, Leon McCulloh,Arnold Miller, Brent B Solie, and John Wetzel

Joseph J Rotman

inductive reasoning is the assertion that a freqently observed phenomenon will

always occur Thus, one says that the Sun will rise tomorrow morning because,from the dawn of time, the Sun has risen every morning This is not a legitimatekind of proof in mathematics, for even though a phenomenon has been observedmany times, it need not occur forever However, inductive reasoning is still valu-able in mathematics, as it is in natural science, because seeing patterns in dataoften helps in guessing what may be true in general

On the other hand, a reasonable guess may not be correct For example, what

is the maximum number of regions into which

3 (3-dimensional space) can be

divided by n planes? Two nonparallel planes can divide

3 into 4 regions, andthree planes can divide

3 into 8 regions (octants) For smaller n, we note that

a single plane divides

3 into 2 regions, while if n = 0, then

3 is not divided

at all: there is 1 region For n = 0, 1, 2, 3, the maximum number of regions is

thus 1, 2, 4, 8, and it is natural to guess that n planes can be chosen to divide

3into 2nregions But it turns out that any four chosen planes can divide

3 into atmost 15 regions!

Before proceeding further, let us make sure that we agree on the meaning of

some standard terms An integer is one of the numbers 0, 1,−1, 2, −2, 3, ;the set of all the integers is denoted by

(from the German Zahl meaning ber):

Definition. An integer d is a divisor of an integer n if n = da for some

inte-ger a An inteinte-ger n is called prime1if n ≥ 2 and its only divisors are ±1 and

±n; an integer n is called composite if it is not prime.

If a positive integer n is composite, then it has a factorization n = ab, where

a < n and b < n are positive integers; the inequalities are present to eliminate the uninteresting factorization n = n × 1 The first few primes are 2, 3, 5, 7,

11, 13, 17, 19, 23, 29, 31, 37, 41, ; that this sequence never ends is proved inCorollary 1.30

Consider the assertion that

It is tedious, but not very difficult, to show that every one of these numbers is

prime (see Proposition 1.3) Inductive reasoning predicts that all the numbers of the form f (n) are prime But the next number, f (41)= 1681, is not prime, for

f (41)= 412− 41 + 41 = 412, which is obviously composite Thus, inductivereasoning is not appropriate for mathematical proofs

Here is an even more spectacular example (which I first saw in an article by

W Sierpinski) Recall that perfect squares are numbers of the form n2, where n

is an integer; the first few perfect squares are 0, 1, 4, 9, 16, 25, 36, For each

n≥ 1, consider the statement

S(n) : 991n2+ 1 is not a perfect square

The nth statement, S(n), is true for many n; in fact, the smallest number n for which S(n) is false is

n= 12, 055, 735, 790, 331, 359, 447, 442, 538, 767

≈ 1.2 × 1028

The equation m2 = 991n2+ 1 is an example of Pell’s equation—an equation

of the form m2 = pn2+ 1, where p is prime—and there is a way of

calcu-lating all possible solutions of it An even larger example involves the prime

1 One reason the number 1 is not called a prime is that many theorems involving primes would otherwise be more complicated to state.

p = 1,000,099; the smallest n for which 1,000,099n2+ 1 is a perfect squarehas 1116 digits The most generous estimate of the age of the Earth is 10 billion(10,000,000,000) years, or 3.65× 1012days, a number insignificant when com-pared to 1.2× 1028, let alone 101115 If, starting from the Earth’s very first day,

one verified statement S(n) on the nth day, then there would be today as much

evidence of the general truth of these statements as there is that the Sun will rise

tomorrow morning And yet some of the statements S(n) are false!

As a final example, let us consider the following statement, known as bach’s conjecture: every even number m≥ 4 is a sum of two primes No one hasever found a counterexample to Goldbach’s conjecture, but neither has anyoneever proved it At present, the conjecture has been verified for all even numbers

Gold-m < 1013, and it has been proved by J.-R Chen that every sufficiently large even

number m can be written as p + q, where p is prime and q is “almost” a prime; that is, q is either a prime or a product of two primes Even with all of this pos-

itive evidence, however, no mathematician will say that Goldbach’s conjecture

must, therefore, be true for all even m.

We have seen what (mathematical) induction is not; let us now discuss whatinduction is Our discussion is based on the following property of the set of

natural numbers (usually called the Well Ordering Principle).

Least Integer Axiom. There is a smallest integer in every nonempty2subset

C of the natural numbers

Although this axiom cannot be proved (it arises in analyzing what integersare), it is certainly plausible Consider the following procedure: check whether

0 belongs to C; if it does, then 0 is the smallest integer in C Otherwise, check whether 1 belongs to C; if it does, then 1 is the smallest integer in C; if not, check 2 Continue this procedure until one bumps into C; this will occur eventu- ally because C is nonempty.

Proposition 1.1 (Least Criminal). Let k be a natural number, and let S(k), S(k + 1), , S(n), be a list of statements If some of these statements are false, then there is a first false statement.

Proof Let C be the set of all those natural numbers n ≥ k for which S(n) is false; by hypothesis, C is a nonempty subset of The Least Integer Axiom

provides a smallest integer m in C, and S(m) is the first false statement. •This seemingly innocuous proposition is useful

Theorem 1.2. Every integer n ≥ 2 is either a prime or a product of primes.

2Saying that C is nonempty merely means that there is at least one integer in C.

Proof Were this not so, there would be “criminals:” there are integers n ≥

2 which are neither primes nor products of primes; a least criminal m is the smallest such integer Since m is not a prime, it is composite; there is thus a factorization m = ab with 2 ≤ a < m and 2 ≤ b < m (since a is an integer,

1 < a implies 2 ≤ a) Since m is the least criminal, both a and b are “honest,”

i.e.,

a = pp0p00· · · and b = qq0q00· · · ,

where the factors p, p0,p00, and q, q0,q00 .are primes Therefore,

m = ab = pp0p00· · · qq0q00· · ·

is a product of (at least two) primes, which is a contradiction.3 •

Proposition 1.3. If m ≥ 2 is a positive integer which is not divisible by any prime p with p≤√m, then m is a prime.

Proof If m is not prime, then m = ab, where a < m and b < m are positive integers If a >√

m and b >√

m, then m = ab >√m√

m = m, a tion Therefore, we may assume that a ≤ √m By Theorem 1.2, a is either a prime or a product of primes, and any (prime) divisor p of a is also a divisor of

contradic-m Thus, if m is not prime, then it has a “small” prime divisor p; i.e., p ≤√m The contrapositive says that if m has no small prime divisor, then m is prime. •Proposition 1.3 can be used to show that 991 is a prime It suffices to check

whether 991 is divisible by some prime p with p ≤ √991 ≈ 31.48; if 991 isnot divisible by 2, 3, 5, , or 31, then it is prime There are 11 such primes,and one checks (by long division) that none of them is a divisor of 991 (Onecan check that 1,000,099 is a prime in the same way, but it is a longer enterprisebecause its square root is a bit over 1000.) It is also tedious, but not difficult, to

see that the numbers f (n) = n2− n + 41, for 1 ≤ n ≤ 40, are all prime.

Mathematical induction is a version of least criminal that is more convenient

to use The key idea is just this: Imagine a stairway to the sky If its bottom step

is white and if the next step above a white step is also white, then all the steps ofthe stairway must be white (One can trace this idea back to Levi ben Gershon

in 1321 There is an explicit description of induction, cited by Pascal, written

by Francesco Maurolico in 1557.) For example, the statement “2n > n for all

3The contrapositive of an implication “ P implies Q” is the implication “(not Q) implies

(not P).” For example, the contrapositive of “If a series P

a nconverges, then limn→∞a n=

0” is “If limn→∞a n 6= 0, then P

a n diverges.” If an implication is true, then so is its contrapositive; conversely, if the contrapositive is true, then so is the original implication The strategy of this proof is to prove the contrapositive of the original implication Although a statement and its contrapositive are logically equivalent, it is sometimes more convenient to

prove the contrapositive This method is also called indirect proof or proof by contradiction.

n ≥ 1” can be regarded as an infinite sequence of statements (a stairway to thesky):

21>1; 22>2; 23>3; 24>4; 25>5; · · ·

Certainly, 21 = 2 > 1 If 2100 > 100, then 2101 = 2 × 2100 > 2× 100 =

100+ 100 > 101 There is nothing magic about the exponent 100; the sameidea shows, having reached any stair, that we can climb up to the next one Thisargument will be formalized in Proposition 1.5

Theorem 1.4 (Mathematical Induction4). Given statements S(n), one for each natural number n, suppose that:

(i) Base Step : S(0) is true;

(ii) Inductive Step : if S(n) is true, then S(n + 1) is true.

Then S(n) is true for all natural numbers n.

Proof We must show that the collection C of all those natural numbers n for which the statement S(n) is false is empty.

If, on the contrary, C is nonempty, then there is a first false statement S(m) Since S(0) is true, by (i), we must have m ≥ 1 This implies that m − 1 ≥ 0, and

so there is a statement S(m − 1) [there is no statement S(−1)] As m is the least criminal, m − 1 must be honest; that is, S(m − 1) is true But now (ii) says that S(m) = S([m − 1] + 1) is true, and this is a contradiction We conclude that C

is empty and, hence, that all the statements S(n) are true. •

We now show how to use induction

Proposition 1.5. 2n >n for all integers n ≥ 0.

Proof The nth statement S(n) is

4Induction, having a Latin root meaning “to lead,” came to mean “prevailing upon to do

something” or “influencing.” This is an apt name here, for the nth statement influences the (n + 1)st one.

Inductive step If S(n) is true, then S(n + 1) is true; that is, using the inductive hypothesis S(n), we must prove

Induction is plausible in the same sense that the Least Integer Axiom is

plau-sible Suppose that a given list S(0), S(1), S(2), of statements has the erty that S(n + 1) is true whenever S(n) is true If, in addition, S(0) is true, then S(1) is true; the truth of S(1) now gives the truth of S(2); the truth of S(2) now gives the truth of S(3); and so forth Induction replaces the phrase and so forth by the inductive step which guarantees, for every n, that there is never an obstruction in the passage from any statement S(n) to the next one, S(n+ 1).Here are two comments before we give more illustrations of induction First,one must verify both the base step and the inductive step; verification of only

prop-one of them is inadequate For example, consider the statements S(n) : n2= n.

The base step is true, but one cannot prove the inductive step (of course, these

statements are false for all n > 1) Another example is given by the statements S(n) : n = n +1 It is easy to see that the inductive step is true: if n = n +1, then Proposition A.2 says that adding 1 to both sides gives n +1 = (n+1)+1 = n+2, which is the next statement, S(n+ 1) But the base step is false (of course, allthese statements are false)

Second, when first seeing induction, many people suspect that the inductive

step is circular reasoning: one is using S(n), and this is what one wants to prove!

A closer analysis shows that this is not at all what is happening The inductive

step, by itself, does not prove that S(n + 1) is true Rather, it says that if S(n)

is true, then S(n+ 1) is also true In other words, the inductive step proves that

the implication “If S(n) is true, then S(n+ 1) is true” is correct The truth ofthis implication is not the same thing as the truth of its conclusion For example,consider the two statements: “Your grade on every exam is 100%” and “Yourgrade in the course is A.” The implication “If all your exams are perfect, then youwill get the highest grade for the course” is true Unfortunately, this does not saythat it is inevitable that your grade in the course will be A Our discussion above

5 See Proposition A.2 in Appendix A, which gives the first properties of inequalities.

gives a mathematical example: the implication “If n = n +1, then n +1 = n +2”

is true, but the conclusion “n + 1 = n + 2” is false.

Remark. The Least Integer Axiom is enjoyed not only by , but also by any

of its nonempty subsets Q (indeed, the proof of Proposition 1.1 uses the fact that the axiom holds for Q = {n in : n ≥ 2}) In terms of induction, this says that the base step can occur at any natural number k, not necessarily at k = 0

The conclusion, then, is that the statements S(n) are true for all n ≥ k The Least Integer Axiom is also enjoyed by the larger set Q m = {n in

: n ≥ m}, where m is any, possibly negative, integer If C is a nonempty subset of Q m and

if C ∩ {m, m + 1, , −1}6is nonempty, then this finite set contains a smallest

integer, which is the smallest integer in C If C ∩ {m, m + 1, , −1} is empty, then C is actually a nonempty subset of , and the original axiom gives a smallest number in C In terms of induction, this says that the base step can occur at any, possibly negative, integer k [assuming, of course, that there is a kth statement S(k)] For example, if one has statements S( −1), S(0), S(1), , then the base step can occur at n = −1; the conclusion in this case is that the statements S(n) are true for all n≥ −1

Here is an induction with base step occurring at n= 1

Proposition 1.6. 1+ 2 + · · · + n = 12n(n + 1) for every integer n ≥ 1 Proof The proof is by induction on n≥ 1

Base step If n= 1, then the left side is 1 and the right side is121(1+1) = 1,

as desired

Inductive step It is always a good idea to write the (n+ 1)st statement

S(n+ 1) so one can see what has to be proved Here, we must prove

6If C and D are subsets of a set X , then their intersection, denoted by C ∩ D, is the subset

consisting of all those x in X lying in both C and D.

volunteered that the answer was 5050 He let s denote the sum of all the numbers from 1 to 100; s = 1 + 2 + · · ·+ 99 + 100 Of course, s = 100 + 99 + · · ·+ 2 + 1.

Arrange these nicely:

s= 1+ 2 + · · · + 99 + 100

s= 100 + 99 + · · · + 2 + 1and add:

2s= 101 + 101 + · · · + 101 + 101,

the sum 101 occurring 100 times We now solve: s = 12(100× 101) = 5050

This argument is valid for any number n in place of 100 (and it does not use

induction) Not only does this give a new proof of Proposition 1.6, it also showshow the formula could have been discovered.7

It is not always the case, in an inductive proof, that the base step is verysimple In fact, all possibilities can occur: both steps can be easy; both can bedifficult; one is harder than the other

Proposition 1.7. If we assume ( f g)0= f0g + f g0, the product rule for tives, then

deriva-(xn)0= nx n−1 for all integers n≥ 1

Proof We proceed by induction on n≥ 1

Base step If n = 1, then we ask whether (x)0 = x0 ≡ 1, the constantfunction identically equal to 1 By definition,

Inductive step We must prove that (x n+1)0 = (n + 1)x n It is permissible

to use the inductive hypothesis, (x n)0= nx n−1, as well as (x )0≡ 1, for the base

7 Actually, this formula goes back at least a thousand years (see Exercise 1.10 on page 13) Alhazen (Ibn al-Haytham) (965-1039), found a geometric way to add

1k+ 2k + · · · + n k

for any fixed integer k≥ 1 [see Exercise 1.11 on page 13].

step has already been proved Since x n+1= x n x , the product rule gives

(xn+1)0 = (x n x )0= (x n)0x + x n(x )0

= (nx n−1)x+ x n1= (n + 1)x n

We conclude that (x n)0 = nx n−1is true for all n≥ 1 •

Here is an example of an induction whose base step occurs at n = 5 sider the statements

Con-S(n): 2n>n2

This is not true for small values of n: if n = 2 or 4, then there is equality, not

inequality; if n = 3, the left side, 8, is smaller than the right side, 9 However,

S(5) is true, for 32 > 25.

Proposition 1.8. 2n >n2is true for all integers n ≥ 5.

Proof We have just checked the base step S(5) In proving

S(n+ 1) : 2n+1> (n+ 1)2,

we are allowed to assume that n ≥ 5 (actually, we will need only n ≥ 3 to prove

the inductive step) as well as the inductive hypothesis Multiply both sides of

Definition. The predecessors of a natural number n ≥ 1 are the natural

num-bers k with k < n, namely, 0, 1, 2, , n− 1 (0 has no predecessor)

Theorem 1.9 (Second Form of Induction). Let S(n) be a family of ments, one for each natural number n, and suppose that:

state-(i) S(0) is true;

(ii) if S(k) is true for all predecessors k of n, then S(n) is itself true.

Then S(n) is true for all natural numbers n.

Proof It suffices to show that there are no integers n for which S(n) is false; that is, the collection C of all positive integers n for which S(n) is false is empty.

If, on the contrary, C is nonempty, then there is a least criminal m: there is

a first false statement S(m) Since S(0) is true, by (i), we must have m ≥ 1 As

m is the least criminal, k must be honest for all k < m; in other words, S(k)

is true for all the predecessors of m Then, by (ii), S(m) is true, and this is a contradiction We conclude that C is empty and, hence, that all the statements S(n) are true. •

The second form of induction can be used to give a second proof of rem 1.2 As with the first form, the base step need not occur at 0

Theo-Theorem 1.10 (= Theo-Theorem 1.2). Every integer n ≥ 2 is either a prime or a product of primes.

Proof.8 Base step The statement is true when n= 2 because 2 is a prime

Inductive step If n ≥ 2 is a prime, we are done Otherwise, n = ab, where

2≤ a < n and 2 ≤ b < n As a and b are predecessors of n, each of them is

either prime or a product of primes:

a = pp0p00· · · and b = qq0q00· · · ,

and so n = pp0p00· · · qq0q00· · · is a product of (at least two) primes •The reason why the second form of induction is more convenient here is that

it is more natural to use S(a) and S(b) than to use S(n− 1); indeed, it is not at

all clear how to use S(n− 1)

Here is a notational remark We can rephrase the inductive step in the first

form of induction: if S(n − 1) is true, then S(n) is true (we are still saying

that if a statement is true, then so is the next statement) With this rephrasing,

we can now compare the inductive steps of the two forms of induction Each

wants to prove S(n): the inductive hypothesis of the first form is S(n− 1); theinductive hypothesis of the second form is any or all of the preceding statements

S(0), S(1), , S(n− 1) Thus, the second form appears to have a strongerinductive hypothesis In fact, Exercise 1.21 on page 15 asks you to prove thatboth forms of mathematical induction are equivalent

The next result says that one can always factor out a largest power of 2 fromany integer

Proposition 1.11. Every integer n ≥ 1 has a unique factorization n = 2 k m, where k ≥ 0 and m ≥ 1 is odd.

8 The similarity of the proofs of Theorems 1.2 and 1.10 indicates that the second form of induction is merely a variation of least criminal.

Proof We use the second form of induction on n ≥ 1 to prove the existence of

k and m; the reader should see that it is more appropriate here than the first form Base step If n = 1, take k = 0 and m = 1.

Inductive step If n ≥ 1, then n is either odd or even If n is odd, then take

k = 0 and m = n If n is even, then n = 2b Because b < n, it is a predecessor

of n, and so the inductive hypothesis allows us to assume S(b) : b = 2`m, where

`≥ 0 and m is odd The desired factorization is n = 2b = 2`+1m.

The word unique means “exactly one.” We prove uniqueness by showing that if n = 2k m = 2t m0, where both k and t are nonnegative and both m and m0are odd, then k = t and m = m0 We may assume that k ≥ t If k > t, then

canceling 2t from both sides gives 2k −t m = m0 Since k − t > 0, the left side

is even while the right side is odd; this contradiction shows that k = t We may

thus cancel 2k from both sides, leaving m = m0 •

The ancient Greeks thought that a rectangular figure is most pleasing to the

eye if its edges a and b are in the proportion

Definition. The Fibonacci sequence F0,F1,F2, is defined as follows:

F0= 0, F1= 1, and F n = F n−1+ F n−2 for all integers n≥ 2.The Fibonacci sequence begins: 0, 1, 1, 2, 3, 5, 8, 13,

Theorem 1.12. If F n denotes the nth term of the Fibonacci sequence, then for all n ≥ 0,

F n = √1

5(αn− βn),

where α= 12(1+√5) and β= 12(1−√5)

Proof. We are going to use the second form of induction [the second form is

the appropriate induction here, for the equation F n = F n−1+ F n−2suggests that

proving S(n) will involve not only S(n − 1) but S(n − 2) as well].

Base step The formula is true for n = 0 : √1

h1

2(1+√5)−12(1−√5)i

=√1

5(√5)= 1 = F1

(We have mentioned both n = 0 and n = 1 because verifying the inductive hypothesis for F n requires our using the truth of the statements for both F n−1

and F n−2 For example, knowing only that F2= √1

5(α2− β2)is not enough to

prove that the formula for F3is true; one also needs the formula for F1.)

Inductive step If n ≥ 2, then

h(αn−1+ αn−2)− (βn−1+ βn−2)i

= √15

h

αn−2(α+ 1) − βn−2(β+ 1)i

= √15

It is curious that the integers F nare expressed in terms of the irrational ber√

num-5

Corollary 1.13. If α= 12

1+√5

, then F n > αn−2for all integers n ≥ 3.

Remark. If n = 2, then F2= 1 = α0, and so there is equality, not inequality

Proof Base step If n = 3, then F3= 2 > α, for α ≈ 1.618

Inductive step We must show that F n+1> αn−1 By the inductive esis,

hypoth-F n+1 = F n + F n−1> αn−2+ αn−3

= αn−3(α+ 1) = αn−3α2= αn−1 •

One can also use induction to give definitions For example, we can define

n factorial,9denoted by n!, by induction on n ≥ 0 Define 0! = 1, and if n! is

known, then define

(n+ 1)! = n!(n + 1).

One reason for defining 0! = 1 will be apparent in the next section

EXERCISES

1.1 Find a formula for 1+ 3 + 5 + · · · + (2n − 1), and use mathematical induction to

prove that your formula is correct (Inductive reasoning is used in mathematics tohelp guess what might be true Once a guess has been made, it must still be proved,perhaps using mathematical induction, perhaps by some other method.)

1.2 Find a formula for 1+Pn j=1j ! j , and use induction to prove that your formula is

1.4 Show, for all n≥ 1, that 10nleaves remainder 1 after dividing by 9

1.5 Prove that if 0≤ a ≤ b, then a n ≤ b n for all n≥ 0

1.6 Prove that 12+ 22+ · · · + n2= 16n(n + 1)(2n + 1) =13n3+12n2+16n.

1.7 Prove that 13+ 23+ · · · + n3= 14n4+12n3+14n2

1.8 Prove that 14+ 24+ · · · + n4= 15n5+12n4+13n3−301n.

1.9 (M Barr) There is a famous anecdote describing a hospital visit of G H Hardy

to Ramanujan Hardy mentioned that the number 1729 of the taxi he had taken tothe hospital was not an interesting number Ramanujan disagreed, saying that it

is the smallest positive integer that can be written as the sum of two cubes in twodifferent ways

(i) Prove that Ramanujan’s statement is true

(ii) Prove that Ramanujan’s statement is false

*1.10 Derive the formula for Pn

i=1i by computing the area (n+ 1)2 of a square with

sides of length n+ 1 using Figure 1.1

*1.11 (i) Derive the formula forPn

i=1i by computing the area n(n+ 1) of a

rect-angle with height n + 1 and base n, as pictured in Figure 1.2.

(ii) (Alhazen) For fixed k≥ 1, use Figure 1.3 to prove

9The term factor comes from the Latin “to make” or “to contribute”; the term factorial recalls that n! has many factors.

k+1 k+1

Figure 1.3 Alhazan’s Dissection

(iii) Given the formulaPn

i=1i= 12n(n+1), use part (ii) to derive the formulaforPn

i=1i2

1.12 (i) Prove that 2n>n3for all n≥ 10

(ii) Prove that 2n>n4for all n≥ 17

(iii) If k is a natural number, prove that 2 n >n k for all n ≥ k2+ 1

1.13 Around 1350, N Oresme was able to sum the seriesP∞

n=1n/2 nby dissecting the

region R in Figure 1.4 in two ways Let A n be the vertical rectangle with base21n

and height n, so that area( A n) = n/2 n , and let B n be horizontal rectangle withbase 21n +2n1+1 + · · · and height 1 Prove thatP∞n=1n/2 n = 2

*1.14 Let g1(x), , g n(x) be differentiable functions, and let f (x) be their product:

f (x) = g1(x) · · · g n(x) Prove, for all integers n≥ 2, that the derivative

Figure 1.4 Oresme’s Dissections

1.15 Prove, for every n ∈ , that (1 + x) n ≥ 1 + nx whenever x ∈
and 1+ x > 0.

1.16 Prove that every positive integer a has a unique factorization a = 3k m, where

k ≥ 0 and m is not a multiple of 3.

1.17 Prove that F n<2n for all n ≥ 0, where F0,F1,F2, is the Fibonacci sequence

1.18 If F n denotes the nth term of the Fibonacci sequence, prove that

m

X

n=1

F n = F m+2− 1

1.19 Prove that 4n+1+ 52n−1is divisible by 21 for all n≥ 1

1.20 For any integer n ≥ 2, prove that there are n consecutive composite numbers.

Conclude that the gap between consecutive primes can be arbitrarily large

*1.21 Prove that the first and second forms of mathematical induction are equivalent; that

is, prove that Theorem 1.4 is true if and only if Theorem 1.9 is true

*1.22 (Double Induction) Let S(m, n) be a doubly indexed family of statements, one for

each m ≥ 0 and n ≥ 0 Suppose that

(i) S(0, 0) is true;

(ii) if S(m, 0) is true, then S(m+ 1, 0) is true;

(iii) if S(m, n) is true for all m, then S(m, n + 1) is true for all m.

Prove that S(m, n) is true for all m ≥ 0 and n ≥ 0.

1.23 Use double induction to prove that

(m+ 1)n>mn

for all m, n≥ 0

1.2 BINOMIALCOEFFICIENTS

What is the pattern of the coefficients in the formulas for the powers (1+ x) nofthe binomial 1+ x? The first few such formulas are:

(1+ x)0= 1(1+ x)1= 1 + 1x

(1+ x)2= 1 + 2x + 1x2

(1+ x)3= 1 + 3x + 3x2+ 1x3

(1+ x)4= 1 + 4x + 6x2+ 4x3+ 1x4

Figure 1.5, called Pascal’s triangle, after B Pascal (1623–1662), displays

an arrangement of the first few coefficients Figure 1.6, a picture from China in

The coefficients c r are called binomial coefficients.10 L Euler (1707–1783)

10Binomial, coming from the Latin bi, meaning “two,” and nomen, meaning “name” or

“term,” describes expressions of the form a + b Similarly, trinomial describes expressions of

the form a + b + c, and monomial describes expressions with a single term The word is used

here because the binomial coefficients arise when expanding powers of the binomial 1+ x.

The word polynomial is a hybrid, coming from the Greek poly meaning “many” and the Latin

nomen; polynomials are certain expressions having many terms.

Figure 1.6 Pascal’s Triangle, China, ca 1300

introduced the notation n r

for them; this symbol evolved into n r

, which isgenerally accepted nowadays:



n r



= coefficient c r of x rin (1+ x) n.Hence,



x r

The number n r

is pronounced “n choose r ” because it also arises in counting

problems, as we shall see later in this section

Observe, in Figure 1.5, that an inside number (i.e., not a 1 on the border)

of the (n + 1)th row can be computed by going up to the nth row and adding

the two neighboring numbers above it For example, the inside numbers in row 4

be computed from row 3 as follows: 4= 1 + 3, 6 = 3 + 3, and 4 = 3 + 1 Let

us prove that this observation always holds

Lemma 1.14. For all integers n ≥ 1 and all r with 0 < r < n + 1,



n r



Proof We must show, for all n≥ 1, that if

n r

 •

Proposition 1.15 (Pascal). For all n ≥ 0 and all r with 0 ≤ r ≤ n,



n r



r !(n − r)!.Proof We prove the proposition by induction on n ≥ 0

Base step.11 If n= 0, then

00



= 0!/0!0! = 1

Inductive step Assuming the formula for n r

for all r , we must prove



n r

Proof. The first equation is the definition of the binomial coefficients, and thesecond equation replaces n r

by the value given in Pascal’s theorem •

Corollary 1.17 (Binomial Theorem). For all real numbers a and b and for all integers n ≥ 1,

Proof The result is trivially true when a = 0 (if we agree that 00 = 1) If

a 6= 0, set x = b/a in Corollary 1.16, and observe that

n

= (a+ b)

n

a n Therefore,



a n −r b r •

Remark. The binomial theorem can be proved without first proving

Corol-lary 1.16; just prove the formula for (a + b) n by induction on n ≥ 0 We havechosen the proof above for clearer exposition

Here is a combinatorial interpretation of the binomial coefficients Given a

set X , an r-subset is a subset of X with exactly r elements If X has n elements,

denote the number of its r -subsets by

sequence of r of these letters with no repetitions For example, the 2-anagrams

on the alphabet a, b, c are

(note that aa, bb, cc are not on this list) How many r -anagrams are there on an alphabet with n letters? We count the number of such anagrams in two ways (1) There are n choices for the first letter; since no letter is repeated, there are only n − 1 choices for the second letter, only n − 2 choices for the third letter, and so forth Thus, the number of r -anagrams is

n(n − 1)(n − 2) · · · (n − [r − 1]) = n(n − 1)(n − 2) · · · (n − r + 1).

Note the special case n = r: the number of n-anagrams on n letters is n! (2) Here is a second way to count these anagrams First choose an r -subset

of the alphabet (consisting of r letters); there are [n, r] ways to do this, for this

is exactly what the symbol[n, r] means For each chosen r-subset, there are r! ways to arrange the r letters in it (this is the special case of our first count when

n = r) The number of r-anagrams is thus



This is why the binomial coefficient n r

is often pronounced as “n choose r ”

As an example, how many ways are there to choose 2 hats from a closetcontaining 14 different hats? (One of my friends does not like the phrasing ofthis question After all, one can choose 2 hats with one’s left hand, with one’sright hand, with one’s teeth, ; but I continue the evil tradition.) The answer is

was algebraic; that

is, as coefficients of polynomials which can be calculated by Pascal’s formula;

our second interpretation is combinatorial; that is, as n choose r Quite often,

each interpretation can be used to prove a desired result For example, here is a

combinatorial proof of Lemma 1.14 Let X be a set with n+ 1 elements, and

let us color one of its elements red and the other n elements blue Now n+1

r

is the number of r -subsets of X There are two possibilities for an r -subset Y : either it contains the red element or it is all blue If Y contains the red element, then Y consists of the red element and r− 1 blue elements, and so the number of

such Y is the same as the number of all blue (r− 1)-subsets, namely, r−1n

The

other possibility is that Y is all blue, and there are n r

such r -subsets Therefore,

We are now going to apply the binomial theorem to trigonometry, but we

begin by reviewing properties of the complex numbers Recall that the modulus

|z| of a complex number z = a + ib is defined to be

|z| =p

a2+ b2

If we identify a complex number z = a + ib with the point (a, b) in the plane,

then its modulus|z| is the distance from z to the origin It follows that every

complex number z of modulus 1 corresponds to a point on the unit circle, and

so it has coordinates (cos θ , sin θ ) for some angle θ (in the right triangle O P A

in Figure 1.7, we have cos θ = |O A|/|O P| = |O A|, because |O P| = 1, and

sin θ = |P A|/|O P| = |P A|).

Figure 1.7 (a, b) = r(cos θ + i sin θ)

Proposition 1.18 (Polar Decomposition) Every complex number z has a

fac-torization

z = r(cos θ + i sin θ), where r = |z| ≥ 0 and 0 ≤ θ < 2π.

Proof If z = 0, then |z| = 0 and any choice of θ works If z = a+bi 6= 0, then

|z| 6= 0 Now z/|z| = a/|z| + ib/|z| has modulus 1, for (a/|z|)2+ (b/|z|)2 =(a2+ b2)/|z|2= 1 Therefore, there is an angle θ with

z

|z| = cos θ + i sin θ, and so z = |z|(cos θ + i sin θ) = r(cos θ + i sin θ) •

If z = a + ib = r(cos θ + i sin θ), then (r, θ) are the polar coordinates12of

z; this is the reason Proposition 1.18 is called the polar decomposition of z.

The trigonometric addition formulas for cos(θ + ψ) and sin(θ + ψ) have alovely translation in the language of complex numbers

12A pole is an axis about which rotation occurs For example, the axis of the Earth has endpoints the North and South Poles Here, we take the pole to be the z-axis (perpendicular

to the plane).

Proposition 1.19 (Addition Theorem). If

z = cos θ + i sin θ and w= cos ψ + i sin ψ,

then

zw = cos(θ + ψ) + i sin(θ + ψ).

Proof.

zw = (cos θ + i sin θ)(cos ψ + i sin ψ)

= (cos θ cos ψ − sin θ sin ψ) + i(sin θ cos ψ + cos θ sin ψ).

The trigonometric addition formulas show that

zw = cos(θ + ψ) + i sin(θ + ψ) •

The addition theorem gives a geometric interpretation of complex

multipli-cation: if z = r(cos θ + i sin θ) and w = s(cos ψ + i sin ψ), then

zw = rs[cos(θ + ψ) + i sin(θ + ψ)], and the polar coordinates of zw are

(r s, θ+ ψ)

Corollary 1.20. If z and w are complex numbers, then

|zw| = |z| |w|.

Proof If the polar decompositions of z and w are z = r(cos θ + i sin θ) and

w= s(cos ψ+i sin ψ), respectively, then we have just seen that |z| = r, |w| = s,

and|zw| = rs •

It follows from this corollary that if z and w lie on the unit circle, then their product zw also lies on the unit circle.

In 1707, A De Moivre (1667–1754) proved the following elegant result

Theorem 1.21 (De Moivre). For every real number x and every positive ger n,

inte-cos(nx ) + i sin(nx) = (cos x + i sin x) n

Proof We prove De Moivre’s theorem by induction on n ≥ 1 The base step

n= 1 is obviously true For the inductive step,

(cos x+ i sin x) n+1= (cos x + i sin x) n(cos x+ i sin x)

= [cos(nx) + i sin(nx)](cos x + i sin x)

Let us find the value of (cos 3◦+ i sin 3◦)40 By De Moivre’s theorem,

(cos 3◦+ i sin 3◦)40= cos 120◦+ i sin 120◦= −12+ i√23

Here are the double and triple angle formulas

Corollary 1.23.

(i) cos(2x )= cos2x− sin2x = 2 cos2x− 1

sin(2x ) = 2 sin x cos x.

(ii) cos(3x )= cos3x − 3 cos x sin2x= 4 cos3x − 3 cos x

sin(3x )= 3 cos2x sin x− sin3x = 3 sin x − 4 sin3x Proof.

(i)

cos(2x ) + i sin(2x) = (cos x + i sin x)2

= cos2x + 2i sin x cos x + i2sin2x

= cos2x− sin2x + i(2 sin x cos x).

Equating real and imaginary parts gives both double angle formulas

(ii) De Moivre’s theorem gives

cos(3x ) + i sin(3x) = (cos x + i sin x)3

= cos3x + 3i cos2x sin x + 3i2cos x sin2x + i3sin3x

= cos3x − 3 cos x sin2x + i(3 cos2x sin x− sin3x ) Equality of the real parts gives cos(3x ) = cos3x − 3 cos x sin2x ; the second formula for cos(3x ) follows by replacing sin2x by 1− cos2x Equality of the imaginary parts gives sin(3x )= 3 cos2x sin x− sin3x = 3 sin x − 4 sin3x ; the

second formula arises by replacing cos2x by 1− sin2x •

Corollary 1.23 will be generalized in Proposition 1.24 If f2(x )= 2x2− 1,then

cos(2x )= 2 cos2x − 1 = f2(cos x ),

and if f3(x )= 4x3− 3x, then

cos(3x )= 4 cos3x − 3 cos x = f3(cos x )

Proposition 1.24. For all n ≥ 1, there is a polynomial f n(x ) having all

coeffi-cients integers such that

cos(nx ) = f n(cos x )

Proof. By De Moivre’s theorem,

cos(nx ) + i sin(nx) = (cos x + i sin x) n

(cos x )n −r(i sin x )r

The real part of the left side, cos(nx ), must be equal to the real part of the right side Now i ris real if and only if13r is even, and so

(cos x )n −r(i sin x )r

(cos x )n −2ksin2k x

(bn/2c denotes the largest integer m with m ≤ n/2).14But sin2k x = (sin2x ) k =(1− cos2x ) k , which is a polynomial in cos x This completes the proof. •

It is not difficult to show that f n(x ) begins with 2n−1x n A sine version ofProposition 1.24 can be found in Exercise 1.31 on page 33

13The converse of an implication “If P is true, then Q is true” is the implication “If Q is

true, then P is true.” An implication may be true without its converse being true For example,

“If a = b, then a2 = b2.” The phrase if and only if means that both the statement and its

converse are true.

14bxc, called the floor of x or the greatest integer in x, is the largest integer m with m ≤ x.

For example, b3c = 3 and bπc = 3.

We are now going to present a beautiful formula discovered by Euler, but webegin by recalling some power series formulas from calculus to see how it arises.

For every real number x ,

converges for every complex number z; the complex exponential e zis defined

to be the sum of this series

Euler’s Theorem For all real numbers x ,

Therefore, e i x = cos x + i sin x •

It is said that Euler was especially pleased with the equation

eπi = −1;

indeed, this formula is inscribed on his tombstone

As a consequence of Euler’s theorem, the polar decomposition can be

rewrit-ten in exponential form: Every complex number z has a factorization

Proof. If ζ = cos(2π/n) + i sin(2π/n), then De Moivre’s theorem,