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An Invitation to Modern Number TheoryCountable, Uncountable, Algebraic and

Transcendentals

Steven J. Miller andRamin Takloo-Bighash

October 23, 2003


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Contents

1 Algebraic and Transcendental Numbers 2

1.1 Definitions and Cardinalities of Sets . . . . . . . . . . . . . . . . 31.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.2 Countable Sets . . . . . . . . . . . . . . . . . . . . . . . 41.1.3 Algebraic Numbers . . . . . . . . . . . . . . . . . . . . . 51.1.4 Transcendental Numbers . . . . . . . . . . . . . . . . . . 71.1.5 Continuum Hypothesis . . . . . . . . . . . . . . . . . . . 8

1.2 Properties ofe . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.1 e is Transcendental . . . . . . . . . . . . . . . . . . . . . 10

1.3 Properties of Algebraic Numbers . . . . . . . . . . . . . . . . . . 151.3.1 Symmetric Polynomials . . . . . . . . . . . . . . . . . . 151.3.2 Needed Lemmas . . . . . . . . . . . . . . . . . . . . . . 161.3.3 Proof of Theorem 1.3.3 . . . . . . . . . . . . . . . . . . . 181.3.4 Theory for More Variables . . . . . . . . . . . . . . . . . 191.3.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Liouvilles Theorem Constructing Transcendentals 22

2.1 Review of Approximating by Rationals . . . . . . . . . . . . . . 222.2 Liouvilles Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 232.3 Constructing Transcendental Numbers . . . . . . . . . . . . . . . 26

2.3.1

m 10

m! . . . . . . . . . . . . . . . . . . . . . . . . . 262.3.2 [101!, 102!, . . . ] . . . . . . . . . . . . . . . . . . . . . . . 27

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Chapter 1

Algebraic and Transcendental

Numbers

Definition 1.0.1 (Algebraic Number). C is an algebraic number if it is aroot of a polynomial with finite degree and integer coefficients.

Definition 1.0.2 (Transcendental Number). C is a transcendental numberif it is not algebraic.

Thus, a transcendental number is a number that does not satisfy any poly-nomial equation with integer coefficients. Fortunately primitive man must havethought that every number is algebraic otherwise the development of mathematicswould have suffered greatly. But transcendental numbers do exist. The mere exis-tence of such numbers was a puzzling problem for hundreds of years. Rememberthat back in the Pythagorean era the existence of irrational numbers was quite adevastating event. The existence of transcendental numbers, however, must havebrought a sense of relief to the mathematical psyche. For one, the transcendenceof a certain number, , settled the long-standing problem of proving the impossi-bility of squaring a circle. Also, it showed that the theory of equations is simplynot enough, and hence it opened the door for the development of other branches ofmathematics. The purpose of this chapter is to prove the existence of transcenden-

tal numbers. While it is possible to write down explicit examples of transcenden-tal numbers (e, , etc!), we prefer to show the existence using a different method.Here we will use Cantors ingenious counting argument. The basic idea is to showthat there are a lot more real numbers than there are algebraic numbers. This willthen show that there must be a left-over set, entirely consisting of transcendentalnumbers. We will see from the proof, that are a lot more transcendental numbers

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than there are algebraic ones; in fact, if one chooses a random number, the chance

of it being transcendental is effectively one hundred percent!

1.1 Definitions and Cardinalities of Sets

1.1.1 Definitions

A function f : A B is one-to-one (or injective) iff(x) = f(y) implies x = y;f is onto (or surjective) if given any b B, a A with f(a) = b. A bijectionis a one-to-one and onto function.

We say two sets A and B have the same cardinality (ie, are the same size) if

there is a bijection f : A B. We denote the common cardinality by |A| = |B|.IfA has finitely many elements (say n elements), A is finite and |A| = n < .Exercise 1.1.1. Show two finite sets have the same cardinality if and only if they

have the same number of elements.

Exercise 1.1.2. Iff is a bijection from A to B, prove there is a bijection g = f1

from B to A.

Exercise 1.1.3. Suppose A and B are two sets, and suppose we have two ontomaps f : A B andg : B A. Then show that |A| = |B|. NOT AS EASY ASIT SEEMS

Exercise 1.1.4. A setA is called infinite if there is a one-to-one map f : A Awhich is not onto. Using this definition, show that the sets N andZ are infinite

sets. In other words, prove that an infinite set has infinitely many elements.

Exercise 1.1.5. Show that the cardinality of the even integers is the same as the

cardinality of the integers.

Remark 1.1.6. The above example is surprising to many. MAYBE ADD RE-

MARK HERE ABOUT COUNTING INTEGERS UP TO X, AND LOOKING

AT LIMITS.

A is countable if there is a bijection between A and the integers Z. A is atmost countable ifA is either finite or countable.

Exercise 1.1.7. Let x,y,z be subsets of X (for example, X = Q,R,C,Rn , etcetera). Define R(x, y) to be true if |x| = |y| (the two sets have the same cardi-nality), and false otherwise. Prove R is an equivalence relation.

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1.1.2 Countable Sets

We show that several common sets are countable. Consider the set of whole num-bers W = {1, 2, 3, . . . }. Define f : W Z by f(2n) = n 1, f(2n + 1) =n 1. By inspection, we see f gives the desired bijection betweenW and Z.

Similarly, we can construct a bijection from N to Z, where N = {0, 1, 2, . . . }.Thus, we have proved

Lemma 1.1.8. To show a setS is countable, it is sufficient to find a bijection fromS to eitherW orN.

We need the intuitively plausible

Lemma 1.1.9. IfA B, then |A| |B|.Definition 1.1.10. If f : A C is a one-to-one function (not necessarily onto),then |A| |C|. Further, ifC A, then |A| = |C|.Exercise 1.1.11. Prove Lemmas 1.1.9 and 1.1.10.

IfA and B are sets, the cartesian product A B is {(a, b) : a A, b B}.Theorem 1.1.12. IfA andB are countable, so is A B andA B.Proof. We have bijections f : N

A and g : N

B. Thus, we can label the

elements ofA and B by

A = {a0, a1, a2, a3, . . . }B = {b0, b1, b2, b3, . . . }. (1.1)

Assume AB is empty. Define h : N AB by h(2n) = an and h(2n+1) =bn. We leave to the reader the case when A B is not empty.

To prove the second claim, consider the following function h : N A B:

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h(1) = (a0, b0)

h(2) = (a1, b0), h(3) = (a1, b1), h(4) = (a0, b1)

h(5) = (a2, b0), h(6) = (a2, b1), h(7) = (a2, b2), h(8) = (a1, b2), h(9) = (a0, b2)...

h(n2 + 1) = (an, b0), h(n2 + 2) = (an, bn1), . . . ,

h(n2 + n + 1) = (an, bn), h(n2 + n + 2) = (an1, bn), . . . ,

h((n + 1)2) = (a0, bn)... (1.2)

Basically, look at all pairs of integers in the first quadrant (including those onthe axes). Thus, we have pairs (ax, by). The above function h starts at (0, 0), andthen moves through the first quadrant, hitting each pair once and only once, bygoing up and over. Draw the picture!

Corollary 1.1.13. Let Ai be countable i N. Then for any n, A1 Anand A1 An are countable, where the last set is all n-tuples (a1, . . . , an),ai Ai. Further, i=0Ai is countable. If each Ai is at most countable, then

i=0Ai is at most countable.

Exercise 1.1.14. Prove Corollary 1.1.13. Hint: fori=0Ai, mimic the proof usedto show A B is countable.

As the natural numbers, integers and rationals are countable, by taking eachAi = N, Z or Q we immediately obtain

Corollary 1.1.15. Nn, Zn andQn are countable. Hint: proceed by induction. For

example write Qn+1 as Qn Q.Exercise 1.1.16. Prove that there are countably many rationals in the interval

[0, 1].

1.1.3 Algebraic Numbers

Consider a polynomial f(x) with rational coefficients. By multiplying by the leastcommon multiple of the denominators, we can clear the fractions. Thus, withoutloss of generality it is sufficient to consider polynomials with integer coefficients.

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The set ofalgebraic numbers,

A, is the set of all x

C such that there is a

polynomial of finite degree and integer coefficients (depending on x, of course!)such that f(x) = 0. The remaining complex numbers are the transcendentals.

The set ofalgebraic numbers of degree n, An, is the set of all x A suchthat

1. there exists a polynomial with integer coefficients of degree n such thatf(x) = 0

2. there is no polynomial g with integer coefficients and degree less than nwith g(x) = 0.

Thus, An is the subset of algebraic numbers x where for each x An, thedegree of the smallest polynomial f with integer coefficients and f(x) = 0 is n.

Exercise 1.1.17. Show the following are algebraic: any rational number, the

square-root of any rational number, the cube-root of any rational number, rpq

where r,p,q Q, i = 1,

3

2 5.Theorem 1.1.18. The algebraic numbers are countable.

Proof. If we show each An is at most countable, then as A = n=1An, by Corol-lary 1.1.13 A is at most countable.

Recall the Fundamental Theorem of Algebra (FTA): Let f(x) be a poly-nomial of degree n with complex coefficients. Then f(x) has n (not necessarilydistinct) roots. Of course, we will only need a weaker version, namely that theFundamental Theorem of Algebra holds for polynomials with integer coefficients.

Fix an n N. We now show An is at most countable. We can represent everyintegral polynomial f(x) = anxn + + a0 by an (n + 1)-tuple (a0, . . . , an).By Corollary 1.1.15, the set of all (n + 1)-tuples with integer coefficients (Zn+1)is countable. Thus, there is a bijection from N to Zn+1, and we can index each(n + 1)-tuple a Zn+1:

{a : a Zn+1

} =i=1

{i}, (1.3)where each i Zn+1.For each tuple i (or a Zn+1), there are n roots. Let Ri be the roots of the

integer polynomial associated to i. The roots in Ri need not be distinct, and theroots may solve an integer polynomial of smaller degree. For example, f(x) =

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(x2

1)4 is a degree 8 polynomial. It has two roots, x = 1 with multiplicity 4 and

x = 1 with multiplicity 4, and each root is a root of a degree 1 polynomial.Let Rn = {x C : x is a root of a degree n polynomial}. One can show that

Rn =i=1

Ri An. (1.4)

By Lemma 1.1.13, Rn is countable. Thus, by Lemma 1.1.9, as Rn is at mostcountable, An is at most countable.

Therefore, each An is at most countable, so by Corollary 1.1.13 A is at mostcountable. As A1 Q (given pq Q, consider qx p = 0), A1 is at leastcountable. As weve shown

A1 is at most countable, this implies

A1 is countable.

Thus, A is countable.

Exercise 1.1.19. Show the full force of the Fundamental Theorem of Algebra is

not needed in the above proof; namely, that it is enough that every polynomial

have finitely many roots.

Exercise 1.1.20. Prove Rn An.

1.1.4 Transcendental Numbers

A set is uncountable if there is no bijection between it and the rationals (or theintegers, or any countable set). The aim of this paragraph is to prove the followingfundamental theorem:

Theorem 1.1.21. The set of all real numbers is uncountable.

We first state and prove a lemma.

Lemma 1.1.22. LetSbe the set of all sequences (yi)iN with yi {0, 1}. Then Sis uncountable.

Proof. We proceed by contradiction. Suppose there is a bijection f : S N. It

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is clear that this is equivalent to giving a list of the elements of

S:

x1 = x11x12x13x14 x2 = x21x22x23x24 x3 = x31x32x33x34

...

xn = xn1xn2xn3xn4 xnn ... (1.5)

Define an element = (i)iN Sby i = xii, and another element =(1

i)iN. Now the element cannot be in the list; it is not xN because 1

xNN

=

xNN!

Proof of the theorem. Consider all those numbers in the interval [0, 1] whose dec-imal expansion consists entirely of numbers 0, 1. Clearly, there is a bijection be-tween this subset ofR and the set S. We have established that Sis uncountable.ConsequentlyR has an uncountable subset. This gives the theorem.

The above proof is due to Cantor (1873 1874), and is known as CantorsDiagonalization Argument. Note Cantors proof shows that most numbers aretranscendental, though it doesnt tell us which numbers are transcendental. Wecan easily show many numbers (such as 3 + 2 357) are algebraic. What ofother numbers, such as and e?

Lambert (1761), Legendre (1794), Hermite (1873) and others proved irra-tional. In 1882 Lindemann proved transcendental.

What about e? Euler (1737) proved that e and e2 are irrational, Liouville(1844) proved e is not an algebraic number of degree 2, and Hermite (1873) provede is transcendental.

Liouville (1851) gave a construction for an infinite (in fact uncountable) familyof transcendental numbers; we will discuss his construction later.

1.1.5 Continuum HypothesisWe have shown that there are more transcendental numbers than algebraic num-bers. Does there exist a subset of[0, 1] which is strictly larger than the rationals,yet strictly smaller than the transcendentals?

Cantors Continuum Hypothesis says that there are no subsets of intermediatesize.

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The standard axioms of set theory are known as the Zermelo-Fraenkel axioms

(note to the expert: often the Axiom of Choice is assumed, and we talk of ZF +Choice).

Kurt Gdel showed that if the standard axioms of set theory are consistent, sotoo are the resulting axioms where the Continuum Hypothesis is assumed true;Paul Cohen showed that the same is true if the negation of the Continuum Hy-pothesis is assumed.

These two results imply that the Continuum Hypothesis is independent of theother standard assumptions of set theory!

1.2 Properties ofeIn this section, we study some of the basic properties of the number e. One of themany ways to define the number e, the base of the natural logarithms, is to writeit as the sum of the following infinite series:

e =n=1

1

n!(1.6)

Now, let us denote the partial sums of the above series by

sm =

mn=1

1

n! (1.7)

Hence e is the limit of the convergent sequence sm. This idea will be the maintool in analyzing the nature ofe.

Theorem 1.2.1 (Euler, 1737). The numbere is irrational.

Proof. Assume that e Q. Then we can write e = pq

, where p, q are positiveintegers.

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Now,

e sm =

n=m+1

1

n!

=1

(m + 1)!

1 +

1

m + 1+

1

(m + 1)(m + 2)+ ...

|a0| and p > n. Let f equal

f(X) =1

(p 1)! Xp1(1 X)p(2 X)p (n X)p

=1

(p 1)!

(n!)pXp1 + higher order terms

. (1.20)

Though it plays no role in the proof, we note that the degree of f is

deg(f) := r = (n + 1)p 1. (1.21)In what follows we make a number of remarks which will help us finish the

proof. By pZ we mean the set of integer multiples ofp.

Remark 1.2.3. Fori p,f(i)(j) pZ, j N.Proof. Differentiate Equation 1.20 i p times. The only terms which survivebring down at least a p!. As each term off(x) is an integer over (p 1)!, we seethat all surviving terms are multiplied by p.

Remark 1.2.4. For0 i < p, f(i)

(j) = 0, j = 1, 2,..,n.

Proof. The multiplicity of a root of a polynomial gives the order of vanishing ofthe polynomial at that particular root. As j = 1, 2, . . . , n are roots of multiplicityp, differentiating f(x) less than p times yields a polynomial which still vanishesat these j.

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Remark 1.2.5. LetF be the polynomial associated to f. Then F(1), F(2), . . . , F (n)

pZ.Proof. Recall that F(j) = f(j) + f(j) + .. + f(r)(j). By the first remark, f(i)(j)is a multiple ofp for i p and any integer j. By the second remark, f(i)(j) = 0for 0 i < p and j = 1, 2, . . . , n. Thus, F(j) is a multiple ofp for these j.Remark 1.2.6. For0 i p 2, f(i)(0) = 0.Proof. Similar to the second remark, we note that f(i)(0) = 0 for 0 i < p 2,because 0 is a root off(x) of multiplicity p 1.Remark 1.2.7. F(0) is not a multiple ofp.

Proof. By the first remark, f(i)(0) is a multiple of p for i p; by the fourthremark, f(i)(0) = 0 for 0 i p 2. Since

F(0) = f(0)+f(0)+ +f(p2)(0)+f(p1)(0)+f(p)(0)+ +f(r)(0), (1.22)

to prove F(0) is a not multiple of p it is sufficient to prove f(p1)(0) is notmultiple ofp (as all other terms are multiples ofp).

However, from the Taylor Series expansion off in Equation 1.20, we see that

f(p1)(0) = (n!)p +

terms that are multiples ofp

. (1.23)

Since we chose p > n, n! is not divisible by p, proving the remark.

We resume the proof of the transcendence ofe.We also chose p such that a0 is not divisible by p. This fact plus the above

remarks imply first that

k akF(k) is an integer, and second that

nk=0

akF(k) a0F(0) 0 mod p. (1.24)

Thus,

k akF(k) is a non-zero integer.Let us recall equation 1.19:

nk=0

akF(k) = a11 + + ann. (1.25)

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We have already proved that the left hand side is a non-zero integer. We ana-

lyze the sum on the right hand side. We have

k = kekckf(ck) = kekckcp1k (1 ck)p (n ck)p

(p 1)1 . (1.26)

As 0 ck k n we obtain

|k| ekkp(1 2 n)p

(p 1)! en(n!n)p

(p 1)! 0 as p . (1.27)

Now recall that n is fixed, and so are the constants a0, . . . , an (they define thepolynomial equation supposedly satisfied by e), and the only thing that varies inour argument is the prime number p. Hence, by choosing a prime number p largeenough, we can make sure that all ks are uniformly small, in particular we canmake them small enough such that the following holds:

n

k=1

akk < 1 (1.28)To be more precise, we only have to choose p such that p > n, |a0| and:

en(n!n)p

(p 1)! 2(2C)4. Then (p1)! > (p1)(p2) (p p

2) (p

2)p2 . Substitute

and compare.

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1.3 Properties of Algebraic Numbers

Let and be two algebraic numbers. Is their sum algebraic? Is their product?What about a general linear combination? What if both are transcendental?

1.3.1 Symmetric Polynomials

Consider the set {1, . . . , n}. There are n! ways to permute the n elements (whereorder counts). Let denote one of these permutations. Thus, {(1), . . . , (n)} isthe same set as {1, . . . , n}, with the elements (possibly) in a different order. Letthe group of permutations of{1, . . . , n} be denoted Sn, where the product of two

permutations is given by composition.Definition 1.3.1 (Symmetric Functions). Let f : Rn C and let be anypermutation of the set{1, . . . , n}. Then f is symmetric if for any permutation Sn,

f(x1, x2, . . . , xn1, xn) = f(x(1), x(2), . . . , x(n1), x(n)). (1.30)

For n = 2, there are two possible permutations, and we have

f(x1, x2) = f(x2, x1). (1.31)For n = 3, there are six possible permutations, and we have

f(x1, x2, x3) = f(x1, x3, x2) = f(x2, x1, x3) = f(x2, x3, x1) = f(x3, x1, x2) = f(x3, x2, x1).(1.32)

Example 1.3.2. For example, f(x1, x2) = x1x2 orx1x2

+ x2x1

or x21 + x22 are sym-

metric, butf(x1, x2) = x21x2 is not.

If we have n = 2, we often denote the variables by x and y instead ofx1 and

x2.In two variables, there are two basic symmetric polynomials:

1. 1 = 1(x, y) = x + y.

2. 2 = 2(x, y) = xy.

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Theorem 1.3.3. Let f(x, y) be a symmetric polynomial in two variables. Then

f(x, y) can be expressed in terms of1 and2.

For example,

x2 + y2 = 21(x, y) 22(x, y)yx4 + xy4 = 2(x, y)

31(x, y) 31(x, y)2(x, y)

. (1.33)

1.3.2 Needed Lemmas

Definition 1.3.4 (Zero Polynomial). A polynomial f is the zero polynomial if it

has no non-zero terms. In other words, iff is a function ofx1, . . . , xn, it containsno monomials Cxr11 xrnn .Lemma 1.3.5. Letf(x, y) be a polynomial with at least one non-zero term. Thenf(x, y) is not identically zero. Equivalently, if f(x, y) = 0 for all complex x andy, f(x, y) is the zero polynomial.

FROM REVIEWER: FOR ANY INFINITE FIELD K, AN ELEMENT

OF K(X1,...,XN) IS IDENTICALLY ZERO IF IT ONLY TAKES ZERO VAL-

UES. BY INDUCTION ITS ENOUGH TO PROVE FOR N=1, WHICH IS

OBVIOUS.

Proof. Without loss of generality, we can assume there is at least one term con-taining a power ofx, say xa. We collect terms where x has the same degree, andwrite f(x, y) as

f(x, y) =ni=0

gi(y)xi, (1.34)

where each gi(y) is a polynomial in y of finite degree (not necessarily the samedegree for each i). Clearly, ga(y) is not the zero polynomial (if it were, it wouldcontradict our assumption that we have a term xa).

By the Fundamental Theorem of Algebra, a polynomial in one variable withcomplex coefficients of degree m has at most m roots. Thus, there are only finitelymany y such that ga(y) = 0. Choose y0 such that ga(y0) = 0. Then

F(x) = f(x, y0) =ni=0

g(y0)xi (1.35)

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is not the zero polynomial, as ga(y0)

= 0. Therefore, F(x) is a finite polyno-

mial of degree at least a. By the Fundamental Theorem of Algebra, F(x) equalszero at most finitely often; therefore, F(x) is not identically zero, which impliesf(x, y) is not identically zero.

Lemma 1.3.6. Letf(x1, . . . , xn) have at least one non-zero term. Then f(x1, . . . , xn)is not the zero polynomial. Equivalently, iff(x1, . . . , xn) = 0 for all (x1, . . . , xn),then f is the zero polynomial.

The proof is by induction. The Fundamental Theorem of Algebra does thecase where we have just one variable; we did two variables above. The general

case is handled similarly. Briefly, we may assume without loss of generality thatthere is a term containing a power ofx1, say xa1. We may write

f(x1, . . . , xn) =Ni=0

gi(x2, . . . , xn)xi1. (1.36)

ga(x2, . . . , xn) is not the zero polynomial (otherwise we would not have an xa;by the inductive assumption (since ga is a function ofn1 instead ofn variables),there is a tuple (x20, . . . , xn0) such that ga(x20, . . . , xn0) = 0.

We form

F(x) = f(x1, x20, . . . , xn0) =Ni=0

gi(x20, . . . , xn0)xi. (1.37)

The rest of the proof is as before. 2

Lemma 1.3.7. If a symmetric polynomial contains a term Cxayb , then it mustcontain the term Cxbya.

Clearly, we only need to check when a = b.Assume f(x, y) is symmetric, so f(x, y) = f(y, x). Assume for some a and

b, Cxayb occurs in f(x, y) but Cxbya does not. Then Cxbya occurs in f(y, x) butCxayb does not.

Hence, if we look at f(x, y)f(y, x), we see the term CxaybCxbya occurs.Hence, f(x, y) f(y, x) is not the zero polynomial.

By Lemma 1.3.5, f(x, y) f(y, x) cannot be identically zero, which contra-dicts f(x, y) = f(y, x) (as we assumed f was symmetric). 2

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Remark 1.3.8. One could have defined symmetric polynomials slightly differently.

Namely, we could say a polynomial f(x, y) is symmetric if wheneverf contains aterm Cxayb, it contains a term Cxbya. For polynomials with variables in C, thetwo definitions are equivalent. Consider, however, the following:

f(x, y) = e2iy(x3+1)

3 + e2ix(y+1)3

3 , x, y Z. (1.38)Then f(x, y) = f(y, x) for all x, y Z, but the two terms look different.

1.3.3 Proof of Theorem 1.3.3

The following is due to Newton. We proceed by induction on the number of terms

of the symmetric polynomial f.By Lemma 1.3.7, if a symmetric polynomial contains a term Cxayb, then it

must contain the term Cxbya.Thus, the polynomial must contain Cxayb + Cxbya; if we subtract this expres-

sion from the original polynomial, the remaining polynomial will still be symmet-ric, and it will have fewer terms.

Thus, it is sufficient to prove that we can express xayb + xbya in terms of1(x, y) and 2(x, y).

Suppose a > b. Then

xayb + xbya = xbyb(xab + yab)

= 2(x, y)b(xab + yab). (1.39)

Thus, to complete the proof, it suffices to show

Lemma 1.3.9. Any polynomial xn + yn can be expressed in terms of1(x, y) and2(x, y).

We proceed by induction; the basis step n = 1 is clear. We also note that forn = 2, x2 + y2 = 1(x, y)2 22(x, y).

Look at

(X x)(X y) = X2 (x + y)X+ xy= X2 1(x, y)X+ 2(x, y). (1.40)

Setting X = x gives

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0 = x2 1(x, y)x + 2(x, y). (1.41)Therefore, we find

x2 = 1(x, y)x 2(x, y). (1.42)Multiplying by xn yields

xn+2 = 1(x, y)xn+1 2(x, y)xn

yn+2 = 1(x, y)yn+1 2(x, y)yn, (1.43)

where the second line follows from the symmetry of x and y (we can applysame type of argument with x replaced with y, as i(x, y) = i(y, x)). Addingthe two equations above yields

xn+2 + yn+2 = 1(x, y)

xn+1 + yn+1 2(x, y)(xn + yn). (1.44)

By induction, we are done. Note that it was important to verify for n = 1 andn = 2.

1.3.4 Theory for More Variables

There is a theory of symmetric polynomials in any number of variables.The basic symmetric functions in three variables are

1. 1 = x + y + z;

2. 2 = xy + xz + yz;

3. 3 = xyz;

in four variables, the basic symmetric functions are1. 1 = x + y + z + t;

2. 2 = xy + xz + xt + yz + yt + zt;

3. 3 = yzt + xzt + xyt + xyz;

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4. 4 = xyzt.

The main result, which we will not prove, is

Theorem 1.3.10 (Newton). For each n , there are n basic symmetric functions1, . . . , n such that any symmetric polynomial P can be expressed in terms of1through n. Furthermore, if P has rational coefficients, the expression of P interms in the i will have rational coefficients.

FROM REVIEWER: UTE PROOF OF THIS USING GALOIS THE-

ORY SEE E. ARTINS GALOIS THEORY

1.3.5 Applications

The formula

(X x)(X y) = X2 (x + y)X+ xy= X2 1(x, y)X+ 2(x, y), (1.45)

generalizes to

(X x1) (X xn) = Xn 1Xn1 + + (1)nnX0. (1.46)If we have any polynomial with coefficients in C, it factors over C into a

product of linear factors (the Fundamental Theorem of Algebra).Thus, if we take a polynomial with rational coefficients,

anXn + an1X

n1 + + a1X+ a0 = an(X 1) (X n),(1.47)

where the i C. A simple comparison combined with Newtons theorem

impliesLemma 1.3.11. Let 1, 2, . . . , n be the solutions of an algebraic equation ofdegree n with rational coefficients. Then every symmetric polynomial expressionwith rational coefficients in terms of the i will be a rational number.

This lemma has a number of interesting consequences.

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Proposition 1.3.12. If and are two algebraic numbers, then + , , and

/ will be algebraic numbers.Proof. We will prove this for + ; the others are similar. Let 1, . . . , m(1, . . . , m, resp.) be all the roots of the algebraic equation satisfied by (,resp.). Consider the polynomial

1in

1jm

(x i j).

It is clear that (x)|P(x). So our result will follow if we can show that P(x)has rational coefficients. The coefficients ofP(x) are polynomials with integralcoefficients in terms of the i and the j . Also they are separately symmetric in

each set of the variables. The lemma then gives the result.Exercise 1.3.13. Complete the details of the proof.

Proposition 1.3.14. A number that satisfies an equation with algebraic coeffi-

cients (not necessarily rational) is algebraic.

Proof. Suppose our equation is

anxn + an1x

n1 + + a1x + a0 = 0,and suppose is a solution of this equation. We will find an equation with integralcoefficients that has as a root. For this we proceed as follows. Take a typicalcoefficient of the above equation, say ai. Since ai is algebraic it will be the solutionof an equation

b(i)mixmi + b(i)mi1x

mi1 + + b(i)1x1 + b(i)0 = 0,with b(i)js rational. Let c(i)1, c(i)2, . . . , c(i)mi be all the roots of this equation.Notice that by definition ai is one of the c(i)js. Next we consider the product

nk=0

1jkmk

c(n)jnx

n + c(n 1)jn1xn1 + + c(1)j1x + c(0)j0

.

This is an equation which has as a solution. Also it has rational coefficients!This proves the claim.

Exercise 1.3.15. Fill in the missing steps.

Remark 1.3.16. The last two propositions imply that the set of algebraic numbers

is an algebraically closed field.

REVIEWER: PROBABLY WANT TO MENTION RESULTS ON AL-

GEBRAIC INTEGERS

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Chapter 2

Liouvilles Theorem Constructing

Transcendentals

2.1 Review of Approximating by Rationals

Definition 2.1.1 (Approximated by rationals to order n). A real number x isapproximated by rationals to order n if there exist a constant k(x) (possibly de-pending on x) such that there are infinitely many rational p

qwith

x pq < k(x)qn . (2.1)Recall that Dirichlets Box Principle gives us, for Q,x pq

< 1q2 (2.2)for infinitely many fractions p

q. This was proved by choosing a large parameter

Q, and considering the Q + 1 fractionary parts {qx} [0, 1) for q {0, . . . , Q}.The box principle ensures us that there must be two different qs, say:

0 q1 < q2 Q (2.3)such that both {q1x} and {q2x} belong to the same interval [ aQ , a+1Q ), for some

0 a Q 1. Note that there are exactly Q such intervals partitioning [0, 1),and Q + 1 fractionary parts! Now, the length of such an interval is 1

Qso we get

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|{q2x} {q1x}| < 1Q. (2.4)There exist integers p1 and p2 such that

{q1x} = q1x p, {q2x} = q2x p. (2.5)Letting p = p2 p1 we find

|(q2 q1)x p| 1Q

(2.6)

Let q = q2

q1, so 1

q

Q, and the previous equation can be rewritten asx pq

< 1qQ 1q2 (2.7)Now, letting Q , we get an infinite collection of rational fractions p

q

satisfying the above equation. If this collection contains only finitely many dis-tinct fractions, then one of these fractions, say p0

q0, would occur for infinitely many

choices Qk ofQ, thus giving us:x p0

q0

0 and (pn, qn) = 1. Then {pnqn}n1 is a monotone increasingsequence converging to x. In particular, all these rational numbers are distinct.Not also that qn must divide 10n!, which implies

qn 10n!. (2.22)Using this, we get

0 < x pnqn

=m>n

1

10m!=

1

10(n+1)!

1 +

1

10n+2+

1

10(n+2)(n+3)+

countablealgtran

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