ON TRANSCENDENTAL NUMBERS

Murugaraj S/O Odiathevar

Supervisor: Associate Professor Victor Tan

Honours Thesis

Department of MathematicsNational University of Singapore

March 2011

Acknowledgements

I would like, above all, to express my heartfelt thanks to my supervisorAssociate Professor Victor Tan for his guidance and patience. Throughoutthe past year he has not only been motivating but also very supportive andunderstanding. This thesis would not have been possible if not for his in-sightful questions that always pointed me in the right direction and helpedme develop a much greater understanding of the subject. It is an honour tohave worked under him.

i

Summary

This thesis firstly serves as an introduction to transcendental number theoryand its related concepts. It explores, studies and analyses the ideas of manymathematicians over the ages who have contributed to this subject. Thetechniques and arguments are very interesting especially to see how highlyregarded numbers such as e, π are transcendental.Chapter 1, gives readers the basic concepts and definition needed to under-stand transcendental numbers and its opposite, the algebraic numbers witha bit of history on the topics.Chapter 2 explores the first ever transcendental numbers discovered and thethought processes behind the mathematicians who proved them. And withthe basic understanding, it describes simple methods for anyone to constructa transcendental number.In Chapter 3, a powerful theorem is stripped and analysed to show why num-bers such as e and π are transcendental. It also allows us to mix and matchdifferent numbers to get a wide range of transcendental numbers.Chapter 4 continues from Chapter 3 and analyses an even more powerfultheorem and explores an even a wider range of transcendental numbers.All of the numbers finally break into different classes in Chapter 5 and itinspects these classes giving the readers more amazing properties.

ii

Author’s Contribution

I have endeavored to present the intricate ideas behind the results in a clearand concise way for readers. The ideas and complexity of the arguments havebeen analysed in detail and many of them modified with additional conceptsand a different point of view for a better understanding. I have faithfullycitied the arguments back to the original authors.A few examples are Propositions 1.10, 1.14 and Theorems 1.11, 2.9, 3.12, 3.13,3.14 which can be attained through simple manipulation and observation.Theorem 2.7 which was stated without proof in [44] has been substantiatedby a simple proof. Also for section 2.2.2, I have modified and lessened thealgorithm given in [37]. Corollaries in section 3.2 are easily attainable. Manyconceptual steps have been included in the proofs of theorems 3.2 and 4.1.Theorem 5.9 from [2] have been modified to follow from Theorem 5.8. For therest of the theorems and proofs, I have tried to explain why each step holdsby citing them back to previous lemmas and theorems and by introducingvarious concepts which are not included in by the original authors. Theseare my modest contributions.

iii

Contents

1 Introduction 21.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Basic definitions and proofs . . . . . . . . . . . . . . . . . . . 3

1.2.1 Algebraic and Transcendental . . . . . . . . . . . . . . 31.2.2 Properties of algebraic numbers . . . . . . . . . . . . . 4

1.3 Continued Fractions . . . . . . . . . . . . . . . . . . . . . . . 10

2 Approximations of a real number 122.1 Liouville’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Constructing transcendental numbers . . . . . . . . . . . . . . 17

2.2.1 Other Liouville numbers . . . . . . . . . . . . . . . . . 172.2.2 Cantor’s semi-construction . . . . . . . . . . . . . . . . 18

3 Lindemann-Weierstrass Theorem 243.1 Proof of theorem . . . . . . . . . . . . . . . . . . . . . . . . . 243.2 Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.3 Algebraic independence . . . . . . . . . . . . . . . . . . . . . . 30

4 Linear forms in Logarithms 344.1 Proof of theorem . . . . . . . . . . . . . . . . . . . . . . . . . 344.2 Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5 Mahler’s Classification 515.1 Mahler’s Classification . . . . . . . . . . . . . . . . . . . . . . 515.2 A-numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.3 S, T, U -numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 56

A Useful results 66

1

Chapter 1

Introduction

Transcendental number theory is one the most baffling and intriguing top-ics in number theory. It is not widely known because of its obscurity andeven upon a third or fourth inspection the arguments seem impenetrable. Itseems paradoxical as well that essentially all the numbers are transcendentalwhile we only know a few. Meaning to say, if we close our eyes and put ourhand into a pot containing all the numbers, with probability 1, we will picka transcendental number. Yet it is so difficult to identify them. Further-more, proving that a number is transcendental can be a daunting task. Inthis project, I look at the proofs of the transcendence of different classes ofnumbers. They are indeed long, tedious, clever and thought provoking. Weshall try to explain the arguments in a more clear and concise way.We shall explore various methods of proofs of several mathematicians namely,Joseph Liouville, Georg Cantor, Ferdinand von Lindemann, Karl Weierstrass,Alan Baker and Kurt Mahler. With some of the theories, we will look to somemethods to construct transcendental numbers.

1.1 History

The study of the nature of numbers is one of the most ancient and fundamen-tal pursuits in all of mathematics. The history of transcendental numbersdates back to 1800 B.C. when the problem of squaring a circle arose. How-ever, at that time no one knew the existence of such numbers. LeonhardEuler(1707-1783) was the first mathematician to define transcendental num-bers in its modern sense, meaning the way it is defined now.In ancient times most mathematicians dealt with constructing numbers geo-metrically. Hence, the notion of constructible numbers arose; with the beliefthat all numbers were rational. Then, rational numbers broke through to

2

irrational numbers and mathematicians came up with methods to approx-imate them as close as possible to rational numbers. There were methodssuch as Euclids algorithm and continued fractions . While, in the study ofpolynomial equations, Euler conjectured that there exist numbers that can-not be roots of a polynomial equation with rational coefficients.Joseph Liouville(1809-1882) was the first mathematician to prove the exis-tence of such numbers, while Georg Cantor(1845-1918) used his theory ofcountability to show their existence. We will explore their results in Chapter2.Ferdinand von Lindemann(1852-1939) proved that π is transcendental andshowed the impossibility of squaring a circle. Karl Weierstrass(1815-1897)and Lindemann together proved a theorem which we will see in chapter 3that has many simple corollaries that transcends many numbers.In chapter 4, we explore some of Alan Baker’s method that transcends awhole new class of numbers. We will also look at some of Kurt Mahler’swork which shows different classes of transcendental numbers in chapter 5.

1.2 Basic definitions and proofs

1.2.1 Algebraic and Transcendental

In many areas of mathematics, to solve a problem, one is often brought to apoint where one has to solve a polynomial equation. We shall use the termQ[x] to denote the ring of polynomials with rational coefficients. When wetry to solve for the roots the natural way is to multiply through by the lowestcommon multiple of all the denominators so that all the coefficients will beintegers. Hence, without loss of generality one can switch from rationalcoefficients to integer coefficients freely when one is equating a polynomialto zero.

Definition 1.1. A number α ∈ C is an algebraic number of degree n if it isa root of an irreducible polynomial of degree n in Q[x]. We also say that αis algebraic over Q. This irreducible polynomial is referred to as the minimalpolynomial of α and denoted as fαQ(x).

We will assume that the leading coefficient of the minimal polynomial isstrictly greater than zero.

Definition 1.2. A number is transcendental if it is not algebraic.

3

All real transcendental numbers are irrational for any rational number pq

is a root of qx − p = 0. To prove a number is transcendenal it is usuallyfirst proven to be irrational then transcendental. However there are powerfultheorems that lead to transcendence right away. Even then, many numbersthat are conjectured to be transcendental have not even been proven to beirrational yet. An example is eπ.

Definition 1.3. A number α ∈ C is an algebraic integer of degree n if it isa root of an irreducible monic polynomial of degree n in Z[x].

Definition 1.4. The height, h of a polynomial is the maximum modulus ofits coefficients and the height of an algebraic number, hα is defined to be theheight of its irreducible primitive polynomial.

1.2.2 Properties of algebraic numbers

Properties relating the roots of the polynomial with its coefficients will be avery handy tool as well. Let α = α1, α2, . . . , αn, counting multiplicity; be allthe roots of P (x). And

P (x) = anxn + an−1x

n−1 + · · ·+ a0 = 0, where ai ∈ Z and an > 0, a0 6= 0

such that P (x) is irreducible over Q. a0 6= 0 otherwise it will contradict theirreducibility of P (x).

n∑i=1

αi =an−1

an∑i<j

αiαj = (−1)an−2

an...

α1 · · ·αn = (−1)na0

an

We can also look at the norm of an algebraic number.

Definition 1.5. The norm of a degree n algebraic number α is given by theproduct of all its conjugates. N (α) = αα2 · · ·αn.

By considering the properties above, one can see that the norm of analgebraic integer is an integer and the norm of an algebraic number is rational.We will need a lemma to proceed with our next property[5].

4

Lemma 1.6. The number θ ∈ C is an algebraic number if and only if thereexist w1, . . . , wn ∈ C not all zero such that

θwj =n∑i=1

aijwi for all 1 ≤ j ≤ n, aij ∈ Q (1.1)

Proof. (⇒) Suppose θ satisfies θn + an−1θn−1 + · · ·+ a0 = 0 where ai ∈ Q.

Let w1 = 1, w2 = θ, w3 = θ2, . . . , wn = θn−1 since w1 6= 0, not all wi is zero.Then

θw1 = θ = 0w1 + 1w2 + · · ·+ 0wnθw2 = θ2 = 0w1 + 0w2 + 1w3 + · · ·+ 0wn

......

......

...θwn−1 = θn−1 = 0w1 + 0w2 + · · ·+ 1wnθwn = θn = −a0w1 − a1w2 − · · · − an−1wn

Thus proven one direction(⇐) Suppose (1.1) holds. Then,

(θ − a11)w1 + (−a21)w2 + · · ·+ (−an1)wn = 0...

......

(−a1n)w1 + (−a2n)w2 + · · ·+ (θ − ann)wn = 0

If we put the above in matrix equation form, the determinant of the co-efficient matrix is ∣∣∣∣∣∣∣∣∣

θ − a11 −a21 · · · −an1

−a12 θ − a22 · · · −an2...

.... . .

...−a1n −a2n · · · θ − a22

∣∣∣∣∣∣∣∣∣ = 0.

Since not all wi are zero, the columns of the matrix are linearly depen-dent. Thus, giving us the zero determinant.The determinant of the above matrix shows that the characteristic equationf(x) = det(xI − A) = 0 of the matrix

A =

a11 · · · an1...

. . ....

a1n · · · ann

has solution θ and f(x) is a polynomial in Q[x] as all the aij ∈ Q. Hence θis an algebraic number.

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Theorem 1.7. The set of algebraic numbers and algebraic integers form afield and ring respctively.

Proof. Suppose θ, φ are algebraic numbers. Then, they each have a minimalpolynomial. Namely,

f θQ(θ) = θm + am−1θm−1 + · · ·+ a0

gφQ(φ) = φn + bn−1φn−1 + · · ·+ b0

Let w1, w2, . . . , wM ∈ C with M = mn be 1, θ, . . . , θm−1, φ, θφ, . . . , θm−1φ,φ2, θφ2, . . . , θm−1φ2, . . . , φn−1, θφn−1, . . . , θm−1φn−1 respectively where

θ(θrφs) =

θr+1φs if r < m− 1

(−am−1θm−1 − · · · − a0)φs if r = m− 1

for θrφs where 0 ≤ r < m, 0 ≤ s < n.

Hence, θwj =M∑i=1

cijwi and φwj =M∑i=1

dijwi where cij, dij ∈ Q.

It follows that

(θ + φ)wj =M∑i=1

(cij + dij)wi where cij + dij ∈ Q

By Lemma 1.6, θ + φ is algebraic.

Similarly, θφwj = θM∑i=1

dijwi =M∑i=1

dij(θwi) =M∑i=1

dij

M∑k=1

ckiwk =M∑k=1

(M∑i=1

ckidij)wk.

Since,M∑i=1

ckidij ∈ Q, we again have θφ is algebraic.

−θ is algebraic as it satisfies (−1)mf θQ(x)

For θ 6= 0,Suppose a0 = 0,m > 1. Then θ(θm−1 + · · · + a1) = 0 which contradicts theminimality of f θQ(x).Suppose a0 = 0,m = 1. Then a1θ = 0 which implies θ = 0. Again, a contra-diction.

For a0 6= 0, consider1

a0

(1

θ)mf(θ) = (

1

θ)m +

a1

a0

(1

θ)m−1 + · · ·+ am−1

a0

(1

θ) +

1

a0

= 0

So θ−1 is algebraic.

6

Similarly we can show that algebraic integers form a ring by using the samesteps except for the last one.

However, these nice properties are only for algebraic numbers[31] anddo not extend to transcendental numbers. Adding, subtracting, multiplyingor dividing a transcendental number by another transcendental number maynot be a transcendental number. The result might even be a rational numberinstead. However, if you apply the similar operations to an algebraic numberand a transcendental number you will get a new transcendental number.

Definition 1.8. The field of algebraic numbers is called the algebraic closureof Q and is denoted by Q.

We shall denote the field of algebraic numbers by Q.

Theorem 1.9. Algebraic numbers are countable.

Before we can prove this we need some idea of countability. Firstly, anyfinite set is countable. For infinite sets,

• A set S is countable if there exist a bijection from S to N.

• If A and B are 2 countable sets then A⋃B is countable.

• Let Ai be countable for all i ∈ N, thenn⋃i=1

Ai is countable. Moreover,

∞⋃i=1

Ai is countable.

Using this for Theorem 1.9,

Proof. For each algebraic number, it is the root of exactly one irreducibleprimitive polynomial in Z[x]. Hence we take P (x) = anx

n +an−1xn−1 + · · ·+

a0 = 0 such that gcd(an, . . . , a0) = 1 and an > 0.Let N = n+ |an|+ |an−1|+ · · ·+ |a0|.

For each N , there exist finite number of polynomials of which each admits afinite set of algebraic numbers that are its roots.Let SN be the set of algebraic numbers corresponding to each N , whereN = 2, 3, 4, . . .

Since each SN is finite, it is countable. If we consider the union∞⋃N=2

SN , it is

also countable.

7

This extends to the idea that there exist transcendental numbers as weknow that the set of real numbers is uncountable. The next few results areimportant to the understanding of algebraic numbers.

Proposition 1.10. Suppose α is an algebraic number and g(x) be any poly-nomial in Q[x]. Then the conjugates of the algebraic number g(α) is obtainedby substituting arbitary conjugates of α.

Proof. Let ξ = g(α) and since 0 = f ξQ(g(α)) = h(α) for some h(x).We have fαQ(x)|h(x) implying h(β) = 0 for any conjugate β of α.

Therefore f ξQ(g(β)) = 0 which gives us the result that g(β) is a conjugate ofg(α).

Theorem 1.11. For any algebraic number α, there exists a r ∈ N such thatrα is an algebraic integer.

Proof. Consider the minimal polynomial of α in Q[x]. We can multiplythe lowest common multiple of the denominators throughout to get f(x) =anx

n + · · ·+ a0 where all the ai ∈ Z and an > 0 else we can multiply a minussign throughout. So f(α) = anα

n + an−1αn−1 + · · ·+ a0 = 0.

Rewriting, we get(anα)n

an−1n

+an−1(anα)n−1

an−1n

+an−2an(anα)n−2

an−1n

+ · · ·+ a1an−2n (anα)

an−1n

+ a0 = 0.

And multiplying by an−1n , we get

(anα)n + an−1(anα)n−1 + an−2an(anα)n−2 + · · ·+ a1an−2n (anα) + a0a

n−1n = 0.

Let g(x) = an−1n f(x) = xn + an−1x

n−1 + an−2anxn−2 + · · ·+ a1a

n−2n x+ a0a

n−1n

which is a monic polynomial of degree n in Z[x] with anα as its root. There-fore, anα is an algebraic integer.

Hence all we need to do to get an algebraic integer from an algebraicnumber is multiply the leading coefficient of its minimal polynomial. Thenext result gives a bound to any algebraic number with respect to its height.

Lemma 1.12. If f(x) = anxn + · · · + a0 where f(x) ∈ C[x] is of height h,

and ana0 6= 0. Let α be a root of f(x). Then

|a0|1 + |a0|

< |α| < h+ |an||an|

and |anα| < 2h

8

Proof. For |α| ≤ 1, the right inequality is obvious.For |α| > 1,

|an| |α|n =∣∣an−1α

n−1 + · · ·+ a0

∣∣ ≤ h(|α|n−1 + |α|n−2 + · · ·+ 1)

= h|α|n − 1

|α| − 1

< h|α|n

|α| − 1.

Thus we have |α| < h+ |an||an|

.

Since1

αis a root of a0x

n + a1xn−1 + · · ·+ an = 0, we have

1

|α|<h+ |a0||a0|

and |α| > |a0|h+ |a0|

.

From the right inequality, we have |anα| < h+ |an| < 2h.

The next result gives us a lower bound of an algebraic number evaluatedat a polynomial other than its minimal polynomial.

Theorem 1.13. Given an algebraic number α of degree n and height h, and

a P (x) ∈ Z[x] of degree k and height H. If P (α) 6= 0, then for c =1

3n−1hn.,

|P (α)| ≥ ck

Hn−1.

Proof. Let fαQ(x) = bnxn + · · ·+ b0 where bi ∈ Z and bn > 0.

P (α) is an algebraic number by properties of field and bnα is an algebraicinteger.For P (α) = akα

k + · · ·+ a0 where ai ∈ Z, we have

bknP (α) = akbknα

k + · · ·+ bkna0

= ak(bnα)k + ak−1bn(bnα)k−1 + · · ·+ bkna0.

Since algebraic integers form a ring, bknP (α) is an algebraic integer. Letα = α1, . . . , αn be conjugates and we see that∣∣N (bknP (α))

∣∣ =

∣∣∣∣∣bknP (α)n∏i=2

bknP (αi)

∣∣∣∣∣ ≥ 1

9

because the norm of an algebraic integer is an integer.

bknP (αi) ≤ bknH(1 + |αi|+ · · ·+ |αi|k) ≤ bknH(1 + |αi|)k

= H(bn + bn |αi|)k

< H(3h)k by Lemma 1.12.

And |P (α)| ≥ 1

bkn

n∏i=2

bkn |P (αi)|

>1

bknHn−1(3h)k(n−1)

≥ 1

Hn−1(3n−1hn)k.

So c =1

3n−1hn.

And of course we have a simple upper bound in general for any numberevaluated at any polynomial.

Proposition 1.14. Let ξ ∈ C and P (x) be a polynomial of degree n andheight h. Then |P (ξ)| ≤ ch where c = c(n, ξ).

Proof. For P (x) = anxn + an−1x

n−1 + · · ·+ a0, we have

|P (ξ)| = |anξn + an−1ξn−1 + · · ·+ a0| ≤ (ξn + · · ·+ 1)h = ch

1.3 Continued Fractions

Continued fraction is yet another elegant way of writing irrational numbers.It was introduced by Euclid in 300BC. A continued fraction of a number xis of the form,

x = a0 + 1a1+ 1

a2+1

a3+···

where a0 ∈ Z and the rest of ai ∈ Z+ are called partial quotients. Anotherneater expression for the above is

[a0; a1, a2, a3, . . .].

Here is a simple method to calculate the continued fraction of any real num-ber, X

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1. Write the integer part of X

2. Subtract the integer part from X

3. If the difference is 0, stop; else compute the reciprocal of the differenceand return to step one with this reciprocal as the new X

Here is an example. For the number π

3 3.1415 . . .− 3 = 0.1415 . . . 1/0.1415 . . . = 7.0625 . . .7 7.0625 . . .− 7 = 0.0625 . . . 1/0.0625 . . . = 15.9965 . . .15 15.9965 . . .− 15 = 0.9965 . . . 1/0.9965 . . . = 1.0034 . . .1 1.0034 . . .− 1 = 0.0034 . . . 1/0.0034 . . . = 292.63 . . .

Hence, π = [3; 7, 15, 1, 292, . . .].For any rational number the procedure will eventually stop. We can also ap-proximate any real number by considering its continued fraction. The morepartial quotients you take, the more accurate the approximation is. Takingn partial quotients, can be termed as the nth convergent of the real number.

These basic definitions and results will be used throughout the paper. Wewill also define new terms as we go on. Many essential lemmas and tedioussteps can be refered from the appendix.

As we move on from these modest preliminaries into establishing transcen-dence we are faced with an almost insurmountable obstacle. Firstly, tran-scendental numbers are defined by what it is not. Hence, the most obvioustechnique of proof to prove a number is transcendental is proof by contra-diction by showing that it is not algebraic. But where will this contradictionbe? What basic principles will it violate? Since any real number can beapproximated to by a rational number, perhaps that is a decent place tobegin. With approximations it is also possible to construct or semi-constructa transcendental number. We will see in the next chapter methods employedby Joseph Liouville and Georg Cantor.

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Chapter 2

Approximations of a realnumber

Joseph Liouville, a French mathematician, was the first one to prove the exis-tence of transcendental numbers[1]. He also gave an explicit example for one.Liouville discovered that there is a limit to the accuracy of approximationof algebraic numbers by rational numbers. So one can see that if a numberdoes not have any limit, then transcendence follows.Georg Cantor, who proved that the set of real numbers is uncountable us-ing his ingenious diagonal argument; also proved that algebraic numbers arecountable, thus transcendental numbers exist. Many people were not con-vinced by his proof as it did not give an explicit transcendental number.However, nowadays with the help of computers, we can semi-construct adecimal representation of transcendental numbers using his methods[37].

2.1 Liouville’s theorem

Theorem 2.1. Liouville’s TheoremIf α is a real algebraic number of degree n with n ≥ 1, then there exists aconstant c = c(α) such that ∣∣∣∣α− p

q

∣∣∣∣ > c

qn

for all p ∈ Z, q ∈ N, pq6= α.

With this theorem, we can see that any irrational algebraic number canonly be approximated to a rational number to a certain degree.

12

Proof. Let f(x) = anxn + · · · + a1x + a0 ∈ Z[x] be the minimal polynomial

of α = α1. Then f(x) = an(x− α)(x− α2) · · · (x− αn) where α2, . . . , αn areconjugates of α.

Let p ∈ Z, q ∈ N, if

∣∣∣∣α− p

q

∣∣∣∣ ≥ 1, then

∣∣∣∣α− p

q

∣∣∣∣ > 1

qn.

When

∣∣∣∣α− p

q

∣∣∣∣ < 1, thenp

q< α + 1.

As f(x) is irreducible, it has no rational roots and so f(pq) 6= 0.∣∣∣∣f(

p

q)

∣∣∣∣ =

∣∣∣∣an(p

q)n + · · ·+ a1(

p

q) + a0

∣∣∣∣=|anpn + an−1p

n−1q + · · ·+ a1pqn−1 + a0q

n|qn

≥ 1

qn∣∣∣∣αi − p

q

∣∣∣∣ =

∣∣∣∣αi − α + α− p

q

∣∣∣∣ ≤ |αi − α|+ ∣∣∣∣α− p

q

∣∣∣∣≤ max

1≤i≤n|αi − α|+ 1 = K.

And finally,∣∣∣∣α− p

q

∣∣∣∣ =

∣∣∣f(pq)∣∣∣

|an|∣∣∣α2 − p

q

∣∣∣ · · · ∣∣∣αn − pq

∣∣∣ ≥1qn

|an|Kn−1=

1|an|Kn−1

qn.

So our c = 1|an|Kn−1 .

Liouville’s theorem leads to a natural corollary.

Corollary 2.2. If α is a real algebraic number of degree n, then there existsa c = c(α) such that ∣∣∣∣α− p

q

∣∣∣∣ < c

qn(2.1)

has no solution in (p, q) ∈ Z× N, pq6= α.

From Liouville’s theorem, we use the same constant c.

The challenge naturally arose to refine Liouville’s theorem. The challengewas to look for the smallest possible value for v such that for a real irrationalalgebraic number α, given any ε > 0, there are only finitely many solutions

in (p, q) ∈ Z×N for

∣∣∣∣α− p

q

∣∣∣∣ < 1

qv+ε. Over the ages the bound has decreased

to 2. The details can be found in [17,34].

13

• Liouville’s Theorem. v = n

• Thue, 1909. v = n2

+ 1

• Sigel, 1921. v = 2√n

• Dyson, 1947. v =√

2n

• Roth, 1955. v = 2

Definition 2.3. A real number α is a Liouville number if for all v ∈ N,∣∣∣∣α− p

q

∣∣∣∣ < 1

qv

has infinite set of solutions for p ∈ Z, q ∈ N.

Theorem 2.4. If α is a Liouville number, then it is transcendental.

Proof. Suppose α is algebraic of degree n. Then by Liouville’s theorem thereexists a c = c(α) > 0 such that for all (p, q) ∈ Z× N,∣∣∣∣α− p

q

∣∣∣∣ > c

qn.

Let v = n+ 1, then by Definition 2.3 of a Liouville number,

∣∣∣∣α− p

q

∣∣∣∣ < 1

qn+1.

Choose ap

qsuch that

1

q< c. Then∣∣∣∣α− p

q

∣∣∣∣ < 1

qn+1<

c

qn

which contradicts Corollary 2.2.

Liouville numbers forms a whole new class of transcendental numbers.However not all transcendental numbers are Liouville numbers and we willsee this more explicitly in Chapter 5. Next we will look at Liouville’s proofof existence of transcendental numbers which was not established until 1844.

Corollary 2.5. The number α =∞∑k=1

1

ak!for all a ∈ N, a > 1, is a Liouville

number, and hence a transcendental number.

Proof. We express α as

α =n∑k=1

1

ak!+

∞∑k=n+1

1

ak!.

14

wheren∑k=1

1

ak!=p

qand hence the denominator, q = an!.

α− p

q=

∞∑k=n+1

1

ak!<

∞∑k=(n+1)!

1

ak=

a

a(n+1)!=

a

qn+1=

aq

qn

If α is algebraic of degree d, then by Liouville’s theorem,

∣∣∣∣α− p

q

∣∣∣∣ > c

qd

for some c.

On the other hand choose n ≥ d, n big enough such thata

q=

a

an!< c

then

∣∣∣∣α− p

q

∣∣∣∣ < aq

qn≤ c

qn≤ c

qd, a contradiction.

The first transcendental number discovered is a Liouville number witha = 10.

i.e.∞∑n=1

1

10n!= 0.110001000000000000000001000 . . .

It is also known as the Liouville’s constant. As we can see that the runsof zero in the decimal representation increase at an alarming rate. We cancompute the continued fraction of the Liouville’s constant by the methodmentioned in Chapter 1.

∞∑n=1

1

10n!= [0; 91, 11, 92, 1, 10, 9, 912, 1, 8, 10, 1, 99, 11, 9, 972, . . .]

where 9k means the there are k 9s in that particular partial quotient. Thepattern is at every (2n − 1)th position the term will have (n− 1)n! 9s.

We can write the definition of Liouville numbers in an alternate way takenfrom [2].

Definition 2.6. A real number α is a Liouville number if there exists a

sequence of distinct rational approximationspnqn

(n = 1, 2, 3, . . .) such that∣∣∣∣α− pnqn

∣∣∣∣ < 1

qwnnwhere lim sup

n→∞wn =∞

15

From this definition we can infer that the partial quotients in the contin-ued fraction of a Liouville number are unbounded[38]. So any transcendentalnumber with bounded partial quotients cannot be a Liouville number. Alsofrom the continued fraction expression of Liouville’s constant, we see that theincrease in the partial quotients is very fast. Thus, we can also see intuitivelythe difference between Liouville numbers and other numbers.

Theorem 2.7. For a function f : N −→ N where f(x) is strictly increasing,

α =∞∑k=1

1

10f(k)is Liouville if lim inf

k→∞

f(k + 1)

f(k)=∞.

Proof. We can considerpnqn

=n∑k=1

1

10f(k).Then qn = 10f(n) by taking a com-

mon denominator.∣∣∣∣α− pnqn

∣∣∣∣ =∞∑

k=n+1

1

10f(k)

<∞∑

k=f(n+1)

1

10kas there are more terms here

=10

9

1

10f(n+1)by sum of geometric series

<10

10f(n+1).

By the definition of lim infk→∞

f(k + 1)

f(k),

inf

x; infinitely many

f(k + 1)

f(k)< x

=∞

Let wk =f(k + 1)

f(k)− 1 for each k. Then lim sup

k→∞wk = ∞ and f(k + 1) =

wkf(k) + f(k)

Putting this into the above inequality, we get∣∣∣∣α− pnqn

∣∣∣∣ < 10

10f(n+1)=

10

10wnf(n)+f(n)

<1

10wnf(n)as the smallest f(n) = 1

16

Of course we can extend Liouville’s theorem to complex algebraic numbersas well.

Theorem 2.8. If α is an algebraic number of degree n, n ≥ 1, then thereexists a constant c = c(α) > 0 such that for all v ∈ N, the following inequalityholds for any algebraic number θ of degree k, k ≥ 1 and height Hθ, for θ 6= α

|α− θ| > ck

Hvθ

We can see that when k = 1, it is exactly Liouville’s theorem. Similarly,

if α ∈ C is such that for any v ∈ N, the inequality |α− θ| < 1

Hvθ

, has infinite

set of solutions in θ, then α is transcendental. However, we shall not go intothe complex case in this chapter. More details can be found in [1]

Now with the idea of approximation, we have a few methods to constructor semi-construct a real transcendental number.

2.2 Constructing transcendental numbers

2.2.1 Other Liouville numbers

From Corollary 2.5, we can replace a by any positive integer greater than 1to get other Liouville numbers[30].

From Theorem 2.7, we can replace f(n) = n! by other functions which satis-

fies lim infk→∞

f(k + 1)

f(k)=∞ to get an even wider range of Liouville numbers.

We can also take Liouville’s constant,∞∑k=1

1

10k!and replace the numerator

with any integer ak from 0 ≤ ak ≤ 9. Then

c =∞∑k=1

ak10k!

is again Liouville.

This gives us many possibilities for Liouville numbers. Also, with this andusing Cantor’s diagonal argument we can show that Liouville numbers areuncountable.

Theorem 2.9. The set of Liouville numbers in the interval [0, 1] is uncount-able.

17

Proof. We do not even need to consider the whole set of Liouville numbers.Just those that are in this form

c =∞∑k=1

ak10k!

(*)

Suppose they are countable, then there exist a bijection f between N andnumbers of the form (*).

f(1) = 0.a11a12000a130 · · · 0a140 · · · 0a1500 · · ·f(2) = 0.a21a22000a230 · · · 0a240 · · · 0a2500 · · ·f(3) = 0.a31a32000a330 · · · 0a340 · · · 0a3500 · · ·

...

Consider d = 0.b1b2000b30 · · · 0b40 · · · 0b50 · · · =∞∑k=1

bk10k!

, where bi 6= aii.

Then d is a Liouville number of the form (*) which is not an image of anyr ∈ N. This contradicts the definition of countability by bijectivity of f .

We will see in chapter 5 that even though there are so many types andforms of Liouville numbers, they are not much in the grand scheme of things.

2.2.2 Cantor’s semi-construction

In this section we will show that Cantor’s proof of the existence of transenden-tal numbers do give us a transendental number. However, we call it semi-construction because it is not in any explicit form but an approximation ifyou stop at any finite step. We start with two facts.Firstly, for any sequence of real numbers on any interval [α, β] one can findan η ∈ [α, β] that does not belong to the sequence and secondly all algebraicreals can be written as an infinite sequence[37].

Theorem 2.10. For any infinite sequence wk consisting of all algebraicreals in an interval [α, β] where each wk is distinct, there exists a subsequencewhich converges to a transcendental number η ∈ [α, β].

Proof. Suppose we have a sequence (wn)∞n=1 of all algebraic reals in the in-terval [α, β] where each wn is distinct.

1. Choose α1 = w1 and as we go down the sequence let β1 be the first wk1such that w1 < wk1 < β.

18

2. We have now a new interval [α1, β1] contained in [α, β].

3. As we go down the sequence, look for the first wk2 such that α1 <wk2 < β1 and call it α2.

4. Next look for the first wk3 such that wk2 < wk2 < β1 and call it β2 andwe have a new interval [α2, β2] contained in [α1, β1].

5. Keep on repeating step 3 and 4 by getting a smaller and smaller intervaleach time.

Case 1.Suppose this algorithm terminates and there is only finitely many such in-tervals. Then let [αN , βN ] be the last interval.There exists at most one wk ∈ [αN , βN ] else if we have more than one, wecan get another interval contained in [αN , βN ].Then, the rest of the numbers within [αN , βN ] are transcendental which isnot possible as we can find a rational number in this interval.

Case 2.The algorithm does not stop and we have

α∞ = limn→∞

αn and β∞ = limn→∞

βn.

This limit exists as (αn)∞n=1 is an increasing sequence bounded above and(βn)∞n=1 is decreasing and bounded below.Here, we have one of 2 cases. Either α∞ = β∞ or α∞ < β∞.For the latter, it is the same as case 1 as any wn < α∞ or wn > β∞. Therefore,all numbers within the interval [α∞, β∞] are transcendenal which is again notpossible.

So we are left with α∞ = β∞ = η and η is transcendenal as any wn < α∞and wn > β∞. Thus η /∈ (wn)∞n=1.

Now given any interval if we can get a sequence of all algebraic numbersin the interval, we can use this argument to obtain a transcendenal number.Hence, we first need a method to generate this sequence of algebraic num-bers. Below is a proposed method.

Nested Intervals AlgorithmConsider polynomials of the form P (x) = anx

n + an−1xn−1 + · · ·+ a0 in Z[x],

where gcd(an, . . . , a0) = 1 and let N = n+ |an|+ |an−1|+ · · ·+ |a0|.

19

Then we can generate polynomials for each N and eliminate all the con-stant polynomials:N = 2 xN = 3 x+ 1, x− 1, 2x, x2

N = 4 2x+ 1, 2x− 1, x+ 2, x− 2, x2 + 1, x2 − 1, 3x, x3

......

We can simplify the process if we only generate polynomials P (x) such thatP (x) and P ′(x) does not have any common roots to eliminate repeated roots.

For each N after we have generated our polynomials, select the roots ofthe polynomials that fall in the desired interval. We do not take roots thatwe have selected before. We go horizontally for each N and then verticallyto get our sequence (wn)∞n=1 of algebraic numbers in the interval. We cando this because for each N there is only finitely many such polynomials andwe can exhaust all the algebraic reals in the interval. With this sequence wecan use this procedure in the proof of Theorem 2.10 to get a transcendentalnumber.

However, this method is very slow. Fortunately, we have another algorithmthat leads to a transcendental faster and if we stop at any step, it will be anapproximation of our transcendental number η.

Diagonal Number AlgorithmFrom our sequence of (wn)∞n=1, consider the decimal representation of eachwn.w1 = 0.a11a12a13a14 · · ·w2 = 0.a21a22a23a24 · · ·w3 = 0.a31a32a33a34 · · ·...

Construct our diagonal number η = 0.b1b2b3b4 · · · such that

bi =

aii + 1 when 0 ≤ aii ≤ 8,

0 when aii = 9

We need to make sure our η is in the desired interval. For example, if ourinterval is [0.5, 0.6], then all our wn = 0.5 · · · . So we let the first decimal digitfor our η be 5 and do the diagonal argument starting with the 2nd decimaldigit onwards.And by the diagonal argument this η /∈ (wn)∞n=1. Thus it is transcendental.

Note that there is no particular enumeration for our sequence (wn)∞n=1. Thus,

20

by interchanging the terms in the sequence, we get a new sequence of alge-braic reals in the desired interval and by applying the diagonal argument, weattain another transcendental number in the interval. Of course, there areinfinitely many ways to interchange the terms in the sequence.

We can also do the following to get a transcendental number. It is thesame method as above but we use instead the binary representation of a realnumber instead. Let br(α) denote the rth decimal digit of the binary repre-sentation for a real number α.

Diagonal Number Algorithm(Binary)We express each element of our sequence (wn)∞n=1in its binary representation.Without loss of generality we consider only the interval (0, 1) and let bk(wn)denote the kth decimal digit of wn.

wn =∞∑k=1

bk(wn)

2kwhere bk(wn) ∈ 0, 1

For example, 13

= 0.333333 · · · = 0(12) + 1( 1

22) + 0( 1

23) + 1( 1

24) + · · ·

So b1(13) = 0, b2(1

3) = 1, . . . and the binary representation of 1

3is 0.010101 . . .

Note that algebraic numbers of the form m2n

have 2 binary representations.For example, 1

2can be either 0.1 or 0.01111 · · ·

Thus we have to modify our sequence such that whenever a number wnin the sequence have 2 binary representation, the next number wn+1 will bethe same wn with the alternate binary representation. This is so that wehave included all the binary representation of algebraic reals in the intervalin our sequence.

Again we apply the diagonal argument. So if the kth decimal digit for wk is1, we change it to 0 and if it is 0, we change it to 1 to get a binary represen-tation of our diagonal number η.

We will use this to show that as long as the sequence encompasses all thealgebraic reals in the interval our η is transcendental. Thus the ordering ofour sequence (wn)∞n=1 does not matter.

Theorem 2.11. A real number in (0, 1) is transcendental if and only if itsbinary representation is a diagonal number of a sequence that consist of allbinary representation of algebraic reals in (0, 1).

Proof. (⇐) By diagonal argument.(⇒) Let t ∈ (0, 1) be transcendental in its binary form. Let (wn)∞n=1 be the

21

sequence of all binary representation of algebraic reals in (0, 1).We shall show that t is given by the diagonal number of a sequence (an)which is a permutation of (wn).

For any α in its binary form, let α(r) denote the rth decimal digit of αWe start to look for a1 by going through (wn)∞n=1. We take the first wk suchthat wk(1) 6= t(1).There definitely exist such a wk as 1

2is an algebraic real in (wn) with 2 binary

representation. One with 0 as first decimal digit and the other with 1 as thefirst decimal digit. Thus we can even let one of the binary representation of12

as a1.

For each ai we start our search from w1 and go down the sequence. Inthis way assume that we have found a1, . . . , an−1. For an we search for thefirst unused wk such that wk(n) 6= t(n).If t(n) = 0, and if we cannot find any number in the sequence (wk) until we

reach one of1

2n+

1

2n+ifor i = 1, . . . , n, then we let an be one of these.

The binary representation of one of these n numbers have 1 in their nth

decimal place and by pigeon hole priniciple at least one will not have beenused.

1

2n+

1

2n+i= 0.0 · · · 0100 · · · 10 · · ·

where the 1 falls on the nth and n+ ith place.If t(n) = 1 then we can use 1

2n+ifor i = 1, · · · , n. Again by pigeonhole

principle, at least one of them is unused.

1

2n+i= 0.0 · · · 01 where the 1 falls on the n+ ith place only.

So there exist a sequence of algebraic numbers (an) in (0, 1) such that t is adiagonal number of (an).

Suppose some wn was not used. Let wk be the unused one with the smallestindex.So each wi(i < k) was used to define some aj. Let N be greater than all theindices of these aj. Thus, all w1, · · · , wk−1 are all among a1, · · · , aN−1

If k = 1, then we let N = 1.Since wk is the first unused term and by the way we have constructed (an),for wk to be unused, wk(n) = t(n) for all n ≥ N .This implies that t − wk is rational as all the binary digits are 0 after andincluding the N th decimal place.

22

Since t is transcendental this implies that wk must be transcendental whichcontradicts the fact that wk is algebraic.

We call the above processes semi-constructing a transcendental numberbecause we can never get a nice expression for a transcndental number likethe Liouville’s constant. There are other ways to construct or semi-constructa transcendental number. Continued fraction is an elegant way. This theorembelow is by Davenport and Roth[39].

Theorem 2.12. Suppose we have ξ = [a0; a1, a2, a3, . . .] and let qn be thedenominator of the nth convergent of ξ. If

lim supn→∞

(log log qn)(log n)12

n=∞,

then ξ is transcendental.

There are others theorems on the partial quotients and convergents of con-tinued fraction of transcendental numbers. They can be found in [10,40,41].The theorem below was proved by Kurt Mahler[19].

Theorem 2.13. For a f(x) ∈ Q[x] of degree at least 1 such that f(k) is apositive integer whenever k is, then the number represented by the infinitedecimal

0.f(1)f(2)f(3) . . .

is transcendental and not Liouville.

Using this we can get the transcendence of Champernowne’s number,0.1234567891011 . . .. There is also an alternate proof for the transcendenceof Champernowne’s number in [27].

Thus approximations are a great way to view the real line. How about forcomplex numbers? If a and b are transcendental, it does not necessarily meanthat a+ib is transcendental. However, we can express any complex number inthe form of reiθ where r, θ ∈ R[6]. Hence, a good place to start is to evaluatethe exponential function at different values of θ. In the next chapter, we willlook at a powerful theorem and some properties of the exponential function atdifferent values of θ and see how the transcendence of many desired numbersare attained.

23

Chapter 3

Lindemann-WeierstrassTheorem

Following from Liouville and Cantor’s arguments that transcendental num-ber exists and there are much more of them than algebraic numbers, manymathematicians set out to find some of these numbers and prove them. Thenext challenge: To prove a number is transcendental instead of constructingone.

Historically, the most interesting of all the numbers were π and e[1,2,4,8].Charles Hermite proved that e is transcendental. Later, by modifying themethod, Ferdinand Lindemann proved π is transcendental and he sketcheda proof of a more general theorem which includes the transcendence of thesetwo numbers and many others as corollaries. Karl Weierstrass rigorouslyproved this and it is now known as the Lindemann-Weierstrass Theorem[14].

3.1 Proof of theorem

We start out with the definition of linear independence of numbers.

Definition 3.1. The set of numbers t0, . . . , tn is linearly independent overalgebraic numbers if for any algebraic numbers β1, . . . , βn not all zero, wehave β0t0 + · · ·+ βntn 6= 0.

The theorem states that exponential functions of algebraic numbers arelinearly independent over the field of algebraic numbers.

Theorem 3.2. Lindemann-Weierstrass theoremFor any distinct algebraic numbers α0, . . . , αm and any non-zero algebraicnumbers a0, . . . , am, we have

24

a0eα0 + . . .+ ame

αm 6= 0.

The proof of this theorem incorporates some applications of Galois andfield theory. Before we go on with the proof we identify 2 lemmas.

Lemma 3.3. Let α0, . . . , αm, a0, . . . , am ∈ C such that the function A(t) =a0e

α0t + · · ·+ ameαmt satisfies A(1) = 0.

For n ∈ N, let f(t) and g(t) be polynomials such that

f(t) = (t − α0)n(t − α1)n+1 · · · (t − αm)n+1 and g(t) =1

n!

∑i≥n

f (i)(t). Then

there exists a c ∈ R+ dependent on α0, . . . , αm, a0, . . . , am such that,

|a0g(α0) + · · ·+ amg(αm)| ≤ cn+1

n!.

Proof. For this f(t), f (i)(α0) = f (i)(α1) = . . . · · · = f (i)(αm) = 0 for all0 ≤ i ≤ n− 1. So for 0 ≤ k ≤ m, let

F (αk) =

m(n+1)+n∑i=0

f (i)(αk) =

m(n+1)+n∑i=n

f (i)(αk) = n!g(αk)

because, for i > m(n+ 1) + n, f (i)(αk) = 0.

From Hermite’s identity(Lemma A.3), let x = αk to get

eαk∫ αk

0

e−tf(t) dt = F (0)eαk − F (αk) = n!g(0)eαk − n!g(αk)

Soeαk

n!

∫ αk

0

e−tf(t) dt = g(0)eαk − g(αk) (3.1)

Multiply (3.1) by ak for each k and sum them up:

m∑k=0

akg(αk) = − 1

n!

m∑k=0

(akeαk

∫ αk

0

e−tf(t) dt) (3.2)

becausem∑k=0

akg(0)eαk = g(0)[a0eα0 + · · ·+ ame

αm ] = 0 as A(1) = 0.

Let r = maxk|αk|

So on the interval |t| ≤ r, we also have |t− αk| ≤ r for all 0 ≤ k ≤ m.

25

Then max|t|≤r|f(t)| ≤ rm(n+1)+n

Let m1 = maxk|ak| and m2 = max

k|eαk |

Then∣∣∣∣eαk ∫ αk

0

e−tf(t) dt

∣∣∣∣ ≤ eαkrm(n+1)+n

∫ αk

0

e−t dt = eαkrm(n+1)+n[1− e−αk ]

≤ rm(n+1)+n[eαk − 1]

≤ rm(n+1)+n(m2 − 1) < rm(n+1)+n+1m2.

Then (3.2) gives

m∑k=0

akg(αk) ≤1

n!((rm+1)n+1)m2

m∑k=0

|ak|

<(rm+1)n+1

n!m2(m+ 1)m1

<(rm+1m2(m+ 1)m1)n+1

n!.

Lemma 3.4. Let α0, . . . , αm, a0, . . . , am be algebraic numbers such that A(t) =a0e

α0t + · · ·+ ameαmt is not identically zero.

If the taylor series of A(t) near t = 0 has rational coefficients, then A(1) 6= 0

Proof. We can find an irreducible polynomial in Q[x] withα0, . . . , αm, a0, . . . , am as its roots.Let r1, . . . , rq be the remaining conjugates of α0, . . . , αm, a0, . . . , am.

Let K = Q(α0, . . . , αm, a0, . . . , am, r1, . . . , rq) and [K : Q] = v. Then K\Qis galois and let σ1, . . . , σv denote all the automorphisms of K. For allγ ∈ K, let γ(j) = σj(γ) for j = 1, . . . , v.

Without loss of generality, we can assume the following:

• a0, . . . , am are algebraic integers contained in K, as we can multiply anatural number through out.

• We can assume that all αi are distinct. If they are similar, we cancombine them as aeαit + beαit = (a+ b)eαit.

• At least one of the aj 6= 0. We can assume a0 6= 0.

26

We shall denote ZK to be the ring of algebraic integers contained in K.Let d ∈ N such that dαj ∈ ZK for each j and let D = dm+1. Then

Dn+1f(t) ∈ ZK[t] and Dn+1g(t) ∈ ZK[t].

for f(t) and g(t) are defined in Lemma 3.3. Let I = D2(n+1)[a0g(α0) + · · ·+amg(αm)]. Then I ∈ ZK as ZK is a ring.

For h(t) =1

(n+ 1)!

∑i≥n+1

f (i)(t), we have J = D2(n+1)[a0h(α0)+· · ·+amh(αm)] ∈

ZK.

Recall the norm N (a) of an algebraic number a in Definition 1.5. Let nbe a large natural number such that n + 1 is coprime with d, N (a0) andN (d(α0 − αj)) for all 1 ≤ j ≤ m.We have

I = a0D2(n+1)g(α0) +D2(n+1)[a1g(α1) + · · ·+ amg(αm)] (3.3)

I = a0D2(n+1)

[1

n!

∑i≥n

f (i)(α0)

]+D2(n+1) [a1g(α1) + · · ·+ amg(αm)]

= a0D2(n+1)

[1

n!f (n)(α0) +

1

n!

∑i≥n+1

f (i)(α0)

]+D2(n+1) [a1g(α1) + · · ·+ amg(αm)]

= a0D2(n+1) 1

n!f (n)(α0) +

n+ 1

(n+ 1)!D2(n+1) 1

n!

∑i≥n+1

f (i)(α0)

+D2(n+1)[a1g(α1) + · · ·+ amg(αm)]

= a0D2(n+1) 1

n!f (n)(α0) + (n+ 1)D2(n+1)[a0h(α0)]

+D2(n+1)[a1g(α1) + · · ·+ amg(αm)].

From our choice of f(x), we can see that f (n)(αk) = 0 for all k = 1, . . . ,mand g(αk) = (n+ 1)h(αk). Then

I = a0D2(n+1) 1

n!f (n)(α0) + (n+ 1)D2(n+1)[a0h(α0) + · · ·+ amh(αm)]

= a0D2(n+1) 1

n!f (n)(α0) + (n+ 1)J

I = a0[d2(m+1)(α0 − α1) · · · (α0 − αm)]n+1 + (n+ 1)J (3.4)

27

In the computation above, note that f (n)(t) =∑

(t− α0)s0 · · · (t− αm)sm

where the summation is over all s0, s1 . . . , sm such that s0 + · · · + sm =m(n+ 1).We have f (n)(α0) = (α0 − α1)s1 · · · (α0 − αm)sm as s0 = 0,since s1 + · · ·+ sm = m(n+ 1) and we have si = n+ 1 for each i for if si wasany other value the (t − α0) term will still be there and evaluating it at α0

you will get 0.

From (3.4), we deduce that I 6= 0 as, by our choice of n, the first termis not divisible by n+ 1.

Consider Aj(t) = σj(A(t)) = a(j)0 eα

(j)0 t + · · ·+ a

(j)m eα

(j)m t for j = 1, . . . , v.

Note that by considering the Taylor series of eαit, we have, for i = 1, . . . ,m,

σj(aieαit) = a

(j)i

[1 + α

(j)i t+

(α(j)i t)2

2!+ . . .

]= a

(j)i eα

(j)i t.

If the Taylor series of A(t) has coefficients in Q, then σj(A(t)) = A(t) for allj = 1, . . . , v and

A(t) = A1(t) = · · · = Av(t).

Suppose A(1) = 0 and hence Aj(1) = 0. By lemma 3.3,∣∣∣a(j)0 gj(α

(j)0 ) + · · ·+ a(j)

m gj(α(j)m )∣∣∣ ≤ cn+1

2

n!where c2 > 0 does not depend on n

and gj(t) = σj(g(t)). Thus from (3.3),∣∣I(j)∣∣ ≤ (d2m+2c2)n+1

n!for all j = 1, . . . , v.

Since I ∈ ZK, so N (I) ∈ Z and N (I) 6= 0 as I 6= 0 we get,

1 ≤ |N (I)| =∣∣I(1) · · · I(v)

∣∣ ≤ cn+13

(n!)v−→ 0 as n −→∞

which is a contradiction.

Proof of Lindemann-Weierstrass theorem

Proof. Let A(t) = a0eα0t + · · ·+ ame

αmt, where aj are algebraic numbers notall zero, and αj are distinct algebraic numbers. Suppose A(1) = a0e

α0 + · · ·+ame

αm = 0. Consider

28

B(t) =v∏j=1

Aj(t) = b0eβ0t + · · ·+ bMe

βM t

which is a non zero function with bj, βj ∈ K

Since K is a field, the coefficients of the Taylor expansion of B(t) lies inK.

So we can write B(t) =∞∑n=0

γntn where γn ∈ K.

Let σ be any automorphism of K. Let τ be the permutation of indices1, . . . , v such that σ(σi) = στ(i). Then

σ(B(t)) =∞∑n=0

σ(γn)tn =v∏j=1

Aτ(j)(t) =v∏j=1

Aj(t) = B(t)

This implies σ(γn) = γn and hence γn ∈ Q for all nSince A(1) = 0, implies B(1) = 0 which contradicts Lemma 3.4.

3.2 Corollaries

We now show the use of Lindemann-Weierstrass theorem to prove manyclasses of transcendental numbers.

Corollary 3.5. e is transcendental.

Proof. Suppose e is algebraic over Q.Then bne

n + · · ·+ b1e+ b0 = 0 with b0 6= 0 and bk ∈ Q for all k = 0, . . . , n.This implies 1 = e0, e1, . . . , en are linearly dependent over Q.Now let m = n, αi = i, ai = bi in the Lindermann-Weierstrass theorem.Then bne

n + · · ·+ b1e+ b0 6= 0. This gives us a contradiction.

Corollary 3.6. π is transcendental.

Proof. Suppose π is algebraic over Q. Then πi and 2πi are algebraic.Let α0 = πi, α1 = 2πi, a0 = a1 = 1 in the Lindermann-Weierstrass theorem.Then 0 6= a0e

α0 + a1eα1 = eπi + e2πi = −1 + 1 = 0 which gives us a contra-

diction.

Corollary 3.7. If α is a non-zero algebraic number, then eα is transcenden-tal.

29

Proof. Suppose eα is algebraic over Q.Let α0 = 0, α1 = α, a0 = eα, a1 = −1 in the Lindermann-Weierstrasstheorem.Then 0 6= a0e

α0 + a1eα1 = eα − eα = 0, contradicting the Lindemann-

Weierstrass theorem.

Corollary 3.8. If α > 0, α 6= 1 is an algebraic number, then logα is tran-scendental.

Proof. Suppose logα is algebraic over Q.Let β = logα. Then α = eβ is transcendental, contradicting our assumptionthat α is algebraic.

Corollary 3.9. If α 6= 0 is an algebraic number, then cosα and sinα aretranscendental.

Proof. If α is algebraic then iα is algebraic, which implies that eiα is tran-scendental.Using eiα = cosα + i sinα,

if cosα is algebraic then sinα =√

1− cos2 α is algebraic and vice versa.Thus, cosα + i sinα is algebraic as algebraic numbers form a field. Thiscontradicts Corollary 3.7.

Corollary 3.10. If α is an algebraic number, then tanα, cotα, secα, cscα,cos−1 α, sin−1 α, tan−1 α are all transcendental.

Proof. We shall only show the proof for tanα and sin−1 α. The rest can bedone similarly.

Using sec2 α = tan2 α + 1, we have cosα =√

1tan2 α+1

.

If tanα is algebraic then√

1tan2 α+1

is algebraic. This contradicts Corollary

3.9.If y = sin−1 α is algebraic, then sin y = α is transcendental. This is again acontradiction.

3.3 Algebraic independence

The notion of algebraic independence is very crucial to the study of tran-scendental number theory for if two numbers are algebraically independentover Q, they are not only transcendental but we can use them to attain manymore transcendental numbers[4]. Thus, in some sense it is a stronger asser-tion than transcendence. This notion can be generalised to any field.

30

Definition 3.11. The set of numbers t1, . . . , tn is algebraically independentover a field K if f(t1, . . . , tn) 6= 0 for all nonzero polynomials f(x1, . . . , xn) ∈K[x1, . . . , xn].

Here, however, we restrict our field to Q and Q. We shall look at algebraicindependence to see some simple results that lead us to an even wider classof transcendental numbers.

Theorem 3.12. If t1, . . . , tn are algebraically independent over Q, then t1, . . . , tnare algebraically independent over Q.

Proof. Suppose there exist P = P (x1, . . . , xn) ∈ Q[x1, . . . , xn] such thatP (t1, . . . , tn) = 0.Consider Pr = Pr(x1, . . . , xn) by the different combination of the conjugates

of each coefficient. Meaning to say, suppose P =∑

ai1,...,inxi11 · · ·xinn ,

Let Pr be P with different combinations of the conjugates of ai1,...,in . Mul-tiply all the Pr together to get R(x1, . . . , xn) ∈ Q[x1, . . . , xn] such thatR(t1, . . . , tn) = 0 which contradicts the algebraic independence of t1, . . . , tn.

Here is a simple example. Suppose P (x, y) =√

2x+√

3y, then

R(x, y) =[√

2x+√

3y] [√

2x−√

3y] [−√

2x+√

3y] [−√

2x−√

3y]

= 4x4 − 12x2y2 + 9y4

Theorem 3.13. If t1, . . . , tn are algebraically independent complex numbersover Q, then for all non-constant polynomial f(x1, . . . , xn) ∈ Q[x1, . . . , xn],f(t1, . . . , tn) is transcendental.

Proof. Suppose f(t1, . . . , tn) is algebraic, considerg(x1, . . . , xn) = f(x1, . . . , xn)− f(t1, . . . , tn) ∈ Q [x1, . . . , xn].We have g(x1, . . . , xn) is non-zero such that g(t1, . . . , tn) = 0, which contra-dicts the algebraic independence over Q by Theorem 3.12.

These theorems though are simple; they let us choose freely our polyno-mials. Furthermore, it does not restrict our choice to rational coefficientsbut algebraic ones. Now, if we evaluate any polynomial in Q[x1, . . . , xn] atalgebraically independent numbers, we can attain many more transcendentalnumbers.

Theorem 3.14. If ξ1, . . . , ξn with n ≥ 0 are algebraic numbers linearly in-dependent over Q, then eξ1 , . . . , eξn are algebraically independent over Q.

31

We shall show that this statement is equivalent to the Lindemann-Weierstrasstheorem.

Proof. Assuming Lindemann-Weierstrass theorem and suppose eξ1 , . . . , eξn

are algebraically dependent over Q.There exist P (x1, . . . , xn) ∈ Q[x1, . . . , xn] such that

P (eξ1 , . . . , eξn) =∑

(k1,...,kn)

ck1···knek1ξ1+···+knξm = 0

where each k1ξ1 + · · ·+ knξm are distinct as ξ1, . . . , ξn are Q-linear indepen-dent; contradicting the Lindemann-Weierstrass theorem.

Now for the other direction. Assume Theorem 3.14 holds,Let β1, . . . , βn denote the maximum set of Q-linearly independent elementsamong α0, . . . , αm. For the rest of the αi,

αi = ci,1β1 + · · ·+ ci,nβn for 0 ≤ i ≤ m and ci,j ∈ Q.

Let q ∈ N be the least common denominator of the ci,j and setβjq

= ξj andqci,j = bi,jWe have ξ1, . . . , ξn are linearly independent over Q

Consider Q(x1, . . . , xn) =m∑i=0

aixbi,11 · · ·xbi,nn ∈ Q[x].

Then Q(x1, . . . , xn) is not a zero function as each (bi,1, . . . , bi,n) − tuple isdistinct. Since eξ1 , . . . , eξn are algebraically independent over Q, they arealgebraically independent over Q by Theorem 3.12. So

Q(eξ1 , . . . , eξn) =m∑i=0

aiebi,1ξ1+···+bi,nξn

= a0eα0 + · · ·+ ame

αm 6= 0

By these statements we have the transcendence of numbers such as e√

2 +√3e√

5. The consequences of these theorems are also not restricted to realnumbers. For example, from the Lindemann-Weierstrass theorem, ei is tran-scendental. Using the theorems from this section, ei+ ie

√2 is transcendental.

It has been proven also that π and eπ are algebraically independent[15]. Wewill see that eπ is transcendental in the next chapter. There are many moretheorems and criterions regarding algebraic independence which can be foundin [15]. In this chapter, we shall look at one more simple consequence.

32

A single variable algebraic function, intuitively, are functions which are poly-nomials with algebraic coefficients like we have seen previously, functionswhich are nth roots of these polynomials and non-constant functions whichare addition, subtraction, multiplication or division of such functions.

Theorem 3.15. For any non-constant single variable algebraic functionf(x), f(t) is transcendental for any transcendental t.

Proof. Let t be a transcendental number, thus t is algebraically independentover QBy Theorem 3.13, we have for any P (x) ∈ Q[x], P (t) is transcendental.

Suppose the nth root of P (t) is algebraic. Then P (t) = [ n√P (t)]n is alge-

braic as algebraic numbers form a field, which contradicts the fact that P (t)is transcendental.

For addition, subtraction or multiplication the proof is similar to that ofTheorem 3.13. If P (x) is an nth root of a polynomial, take [P (x)]n so that itis in Q[x]. So without loss of generality we can take P (x), Q(x) ∈ Q[x]Suppose, P (t) +Q(t) = α where α is algebraic.Consider G(x) = P (x)+Q(x)−α which is a polynomial in Q[x] that vanishesat t. This contradicts Theorem 3.12.

For division, suppose P (t)Q(t)

= α where α is algebraic, then P (x)−αQ(x) ∈ Q[x]vanishes at t, contradicting Theorem 3.12.

Now using this theorem, we can show that numbers such as√

2(e2 +√e)2

andπ5 −

√π

4π2are transcendental.

There are still many open questions in algebraic number theory, one of whichis the algebraic independence of e and π. They are conjectured to be alge-braically independent.

So the Lindemann-Weierstrass theorem is a general theorem which givesus the transcendence of many numbers. We can also prove the transcendenceof each individual number by contradiction. We identify yet another methodused in transcendental number theory in the next chapter and explore aneven wider range of transcendental numbers.

33

Chapter 4

Linear forms in Logarithms

Since transcendental numbers are defined by what it is not rather than whatit is, the usual method to prove a number transcendental will be by contradic-tion. The main method is to first assume that the number is algebraic. Thenuse the coefficients and conjugates of its minimal polynomial to constructan auxiliary function which will have contradictory properties. We highlightthis main idea in this chapter by showing a proof of yet another powerfultheorem proved by Alan Baker[2,23] of Cambridge in 1970. Mainly the theo-rem is a generalisation of the Gelfond-Schneider theorem[13,16] which statesthat αβ is transcendental for algebraic α 6= 0, 1 and algebraic β /∈ Q. We willsee an explicit example of an auxiliary function in Lemma 4.2. The proofis long and tedious. All the important facts have been added for an easierunderstanding. Some of the tedious details are in the appendix.

4.1 Proof of theorem

Theorem 4.1. If α1, . . . , αn are algebraic numbers such that logα1,. . ., logαnare linearly independent over Q, then 1, logα1, . . . , logαn are linearly inde-pendent over Q.

This is the non-homogeneous version of Baker’s theorem. The homoge-nous version is the same as above but without the 1 in the result[21,22].

Suppose there exist algebraic numbers β0, . . . , βn not all zero, such that

β0 + β1 logα1 + · · ·+ βn logαn = 0. (4.1)

At least one of the β1, . . . , βn 6= 0. Let βn 6= 0 and divide throughout by−βn.We get − β0

βn− β1

βnlogα1 − · · · − logαn = 0; which is in the same form as

34

(4.1). So without loss of generality we can let βn = −1 and restart from thebeginning to get

logαn = β0 + β1 logα1 + · · ·+ βn−1 logαn−1

and αn = eβ0αβ11 · · ·αβn−1

n−1

Let c, c1, c2, . . . be constants greater than 1 depending on α1, . . . , αn, β0, . . . , βn.Let d be the maximum of the degrees of α1, . . . , αn, β0, . . . , βn−1.Let a1, . . . , an, b0, . . . , bn be the leading coefficients of their respective mini-mal polynomials and let h be a sufficently large positive number.

Also define fm0,...,mn−1(z0, . . . , zn−1) = (∂

∂z0

)m0 . . . (∂

∂zn−1

)mn−1f(z0, . . . , zn−1)

where f is an integral function and m0, . . . ,mn−1 are non-negative integers.

Lemma 4.2. There are integers p(λ0, . . . , λn) not all zero, with |p(λ0, . . . , λn)| ≤eh

3such that the function

Φ(z0, . . . , zn−1) =L∑

λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)zλ00 eλnβ0z0αγ1z11 · · ·αγn−1zn−1

n−1

where γr = λr + λnβr (1 ≤ r < n) and L = bh2− 14n c, satisfies

Φm0,...,mn−1(l, . . . , l) = 0 for l ∈ Z with 1 ≤ l ≤ h and m0, . . . ,mn−1 ∈N⋃0. And m0 + · · ·+mn−1 ≤ h2.

Proof. First of all we have to find the function Φm0,...,mn−1(z0, . . . , zn−1) byfinding the partial derivatives of each zr, mr times. For z0,

dm0

dzm00

zλ00 eλnβ0z0 = q(λ0, λn, z0)eλnβ0z0

where

q(λ0, λn, z0) =

[m0∑µ0=0

(m0

µ0

)λ0(λ0 − 1) · · · (λ0 − µ0 + 1)(λnβ0)m0−µ0zλ0−µ00

].

For zr with 1 ≤ r ≤ n− 1,

dmr

dzmrrαγrzrr = γmrr (logαr)

mrαγrzrr .

The details of the differentiation can be found in Appendix as Lemma A.6.

Let Ψ = Φm0,...,mn−1(z0, . . . , zn−1). Then

Ψ =L∑

λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)q(λ0, λn, z0)αλ1z11 · · ·αλn−1zn−1

n−1

γm11 · · · γ

mn−1

n−1 (logα1)m1 · · · (logαn−1)mn−1(αβ1z11 · · ·αβn−1zn−1

n−1 )λneλnβ0z0

35

Thus, by substituting the values of z0, z1, . . . , zn−1 by l we attain

Ψ(l, . . . , l) =L∑

λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)q(λ0, λn, l)αλ1l1 · · ·α

λn−1ln−1 γm1

1 · · · γmn−1

n−1

(logα1)m1 · · · (logαn−1)mn−1(αβ11 · · ·αβn−1

n−1 eβ0)λnl

=L∑

λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)q(λ0, λn, l)αλ1l1 · · ·α

λn−1ln−1 γm1

1 · · · γmn−1

n−1

(logα1)m1 · · · (logαn−1)mn−1αλnln .

Now we equate Ψ = 0 to get

L∑λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)q(λ0, λn, l)αλ1l1 · · ·αλnln γm1

1 · · · γmn−1

n−1 = 0.

Let P ′ = (a1 · · · an)Llbm00 · · · b

mn−1

n−1 and multiply the above equation by P ′.

Also, by expanding γmrr = (λr + λnβr)mr =

mr∑µr=0

(mr

µr

)λmr−µrr (λnβr)

µr ,

and using Proposition A.4, we have

(arαr)j =

d−1∑s=0

a(j)rs α

sr (brβr)

j =d−1∑t=0

b(j)rt β

tr

where a(j)rs , b

(j)rt ∈ Z with absolute values at most cj1. We get

d−1∑s1=0

· · ·d−1∑sn=0

d−1∑t0=0

· · ·d−1∑

tn−1=0

A(s, t)αs11 · · ·αsnn βt00 · · · β

tn−1

n−1 = 0

where A(s, t) = A(s1, . . . , sn, t0, . . . , tn−1). Then

A(s, t) =L∑

λ0=0

· · ·L∑

λn=0

m0∑µ0=0

· · ·mn−1∑µn−1=0

p(λ0, . . . , λn)q′q′′q′′′ and

q′ =n∏r=1

a(L−λr)lr a(λrl)

r,sr

q′′ =n−1∏r=1

[(mr

µr

)(brλr)

mr−µrλµrn b(µr)r,tr

]q′′′ =

(m0

µ0

)λ0(λ0 − 1) · · · (λ0 − µ0 + 1)λm0−µ0

n bµ00 lλ0−µ0b

(m0−µ0)0,t0

36

with reference to Appendix, Lemma A.6 for the expansion.It now suffices to solve for A(s, t) = 0. We can find an upper bound for eachq′, q′′, q′′′. (Lemma A.6)

|q′| ≤ cLh2 |q′′| ≤n−1∏r=1

(c3L)mr |q′′′| ≤ (c3L)m0hL∣∣∣∣∣m0∑µ0=0

· · ·mn−1∑µn−1=0

q′q′′q′′′

∣∣∣∣∣ ≤ (2c3L)h2cLh4 = U .

Since each l has h choices and each mi has at most h2 + 1 choices from1 ≤ l ≤ h and m0 + · · ·+mn−1 ≤ h2 with mi ∈ N

⋃0, we have h(h2 + 1)n

distinct (n+ 1)-tuples of l,m0, . . . ,mn−1.

Now A(s, t) = A(s1, . . . , sn, t0, . . . , tn−1) has 2n-tuples of si and tj with dpossibilities each.

We have M ≤ d2nh(h2 + 1)n equations A(s, t) = 0 and N = (L + 1)n+1

unknowns p(λ0, . . . , λn) as 0 ≤ λr ≤ L. So we see that

2M ≤ d2nh(h2 + 1)n < h2n+ 32

for a large enough h; as the highest exponent of h in the second term is2n+ 1. And

h2n+ 32 ≤ h2nh

74− 1

4n = h2n+2− 14− 1

4n ≤ (h2− 14n )n+1 < N .

The first inequality holds because n ≥ 1 and the second inequality rises fromL ≤ h2− 1

4n < L+ 1.

By Lemma A.5 in the appendix, since there are more unknowns than equa-tions, there exists p(λ0, . . . , λn) such that A(s, t) = 0. And for a large enoughh,

|p(λ0, . . . , λn)| ≤ (NU)M

N−M ≤ NU ≤ h2n+2(2c3L)h2cLh4 ≤ eh

3.

The notations in the above Lemma will follow through out the proof.

Lemma 4.3. Let m0, . . . ,mn−1 be any non-negative integers with m0 + · · ·+mn−1 ≤ h2 and f(z) = Φm0,...,mn−1(z, . . . , z). Then for any z, we have

|f(z)| ≤ ch3+L|z|5 and for any l ∈ Z+, either f(l) = 0 or |f(l)| > c−h

3−Ll6 .

37

Proof. Let P = (logα1)m1 · · · (logαn−1)mn−1 . We find a bound for the fol-lowing.

|q(λ0, λn, z)| =

∣∣∣∣∣m0∑µ0=0

(m0

µ0

)λ0(λ0 − 1) · · · (λ0 − µ0 + 1)(λnβ0)m0−µ0zλ0−µ00

∣∣∣∣∣≤

m0∑µ0=0

(m0

µ0

)Lµ0 |λnβ0|m0−µ0 |z|λ0−µ0

≤m0∑µ0=0

(m0

µ0

)(L |λnβ0|)m0 |z|L

≤ 2m0(c7L)m0 |z|L

≤ (2c7L)m0 |z|L by λi ≤ L and µj ≤ m0

and ∣∣∣αλ1z1 · · ·αλn−1zn−1

∣∣∣ ≤ (|α1|λ1 · · · |αn|λn)|z|

≤ (|α1|L · · · |αn|L)|z|

= (|α1| · · · |αn|)L|z| ≤ cL|z|8

and∣∣Pγm11 · · · γ

mn−1

n−1

∣∣ ≤ |P (λ1 + λnβ1)m1 · · · (λn−1 + λnβn−1)mn−1|≤ |logα1(λ1 + λnβ1)|m1 · · · |logαn−1(λn−1 + λnβn−1)|mn−1

≤ Lm1(|1 + β1| logα1)m1 · · ·Lmn−1(|1 + βn−1| logαn−1)mn−1

≤ Lm1+···+mn−1cm1+···+mn−1

9 where c9 = max1≤i<n

|1 + βi| logαi.

Note that the number of terms in the sum of f(z) ≤ h2n+2 because L ≤ h2

and there are n+ 1 sums. Thus, from

f(z) = P

L∑λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)q(λ0, λn, z)αλ1z1 · · ·αλn−1z

n−1 γm11 · · · γ

mn−1

n−1

we have

|f(z)| ≤ h2n+2(eh3

)(2c7L)m0 |z|L cL|z|8 (Lc9)m1+···+mn−1

≤ h2n+2(eh3

)(2c7L)m0 |z|L cL|z|8 (Lc9)h2

≤ h2n+2(eh3

)(2c7L)h2 |z|h

2

cL|z|8 (Lc9)h

2

≤ ch3+L|z|5

38

If f(l) 6= 0, let f ∗ = P ′

Pf(l) where P ′ = (a1 · · · an)Llbm0

0 · · · bmn−1

n−1 . So

f ∗ = (a1 · · · an)Llbm00 · · · b

mn−1

n−1

L∑λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)q(λ0, λn, l)

αλ1l1 · · ·αλn−1ln−1 γm1

1 · · · γmn−1

n−1 .

By Theorem 1.11, aiαi and bjβj are algebraic integers. Thus f ∗ is an algebraicinteger with degree at most d2n. And any conjugate of f ∗ is obtained bysubsituting arbitary conjugates of αi, βj.(Proposition 1.10) And hence,

|any conjugate of f ∗| ≤∣∣P ′P

∣∣ ch3+Ll5 ≤ ch

3+Ll10 as

∣∣P ′P

∣∣ can be absorbed.

If f(l) 6= 0 then f ∗ 6= 0. So for the norm of f ∗,

1 ≤ |N (f ∗)| ≤ c(h3+Ll)d2n10 . (4.1)

Hence, |f ∗| ≥ c−(h3+Ll)d2n

10 . For if not, suppose |f ∗| < c−(h3+Ll)d2n

10 . Then

c(h3+Ll)d2n

10 <|N (f ∗)||f ∗|

< c−(h3+Ll)d2n−1

10 which contradicts (4.1). Therefore we

have ∣∣∣∣P ′P f(l)

∣∣∣∣ ≥ c−(h3+Ll)d2n

10 implying that |f(l)| ≥ c−(h3+Ll)6 .

Lemma 4.4. Let J be any integer 0 ≤ J ≤ (8n)2. Then for f(z) defined in

Lemma 4.3, f(l) = 0 for all l ∈ Z, with 1 ≤ l ≤ h1+ J8n and m0 + · · ·+mn−1 ≤

h2

2J .

Proof. By mathematical induction. For J = 0, it is exactly Lemma 4.2

So we assume true for J = 0, 1, . . . K, with 0 ≤ K < (8n)2. Let f(l) be

defined as in Lemma 4.3 and RJ = bh1+ Jrn c and SJ = bh2

2Jc. For J = K+ 1 it

suffices to show that for all l ∈ Z, with RK < l ≤ RK+1 and m0+· · ·+mn−1 ≤SK+1, we have f(l) = 0.

fm(r) = (∂

∂z0

+ · · ·+ ∂

∂zn−1

)mΦm0,...,mn−1(z0, . . . , zn−1) |z0=z1=···=zn−1=r

=∑

m!(j0! · · · jn−1!)−1Φm0+j0,...,mn−1+jn−1(r, . . . , r) = 0

where the sum is over all non-negative integers j0, . . . , jn−1 with jo+· · · jn−1 =m. And we have m0 + · · ·+mn−1 + jo + · · · jn−1 ≤ SK+1 +SK+1 ≤ SK which

39

implies each Φm0+j0,...,mn−1+jn−1(r, . . . , r) = 0. Hence, by inductive hypothe-sis, fm(r) = 0 for all r,m ∈ Z with 1 ≤ r ≤ RK and 0 ≤ m ≤ SK+1.

Let F (z) = (z − 1) · · · (z −RK)SK+1 .

So, f(z)F (z)

is regular within and on the circle, C with center at origin and

radius R = RK+1h18n as we have eliminated all the repeated roots of f(z).

By Maximum modulus principle(Theorem A.8), we express the constant inthe form θ

Θ. So ∣∣∣∣ f(l)

F (l)

∣∣∣∣ ≤ θ

Θ

where |f(z)| ≤ θ and Θ ≤ |F (z)| are the respective bounds, with z on C.And by considering difference between the nearest point on the circle to RK ,

R−RK = h18nRK+1 −RK =

[h1+ k

8n+ 1

8n

]h

18n −

[h1+ k

8n

]>

1

2

[h1+ k

8n+ 1

8n

]h

18n =

1

2R

We get Θ ≥ (12R)RKSK+1 .

By Lemma 4.3, θ ≤ ch3+LR

5 and we also have |F (l)| ≤ RRKSK+1

K+1 .

Again by Lemma 4.3, either f(l) = 0 or |f(l)| > c−h3−LR

6 .

If |f(l)| > c−h3−LR

6 then by θ |F (l)| ≥ Θ |f(l)|, we have

(c5c6)h3+LR ≥ (

1

2h1/(8n))RKSK+1 . (4.2)

By looking at the powers and since K < (8n)2,

LR ≤ (h2− 14n )(h1+K+1

8n )h18n = h3+ K

8n =h2

2

h1+ k8n

24

≤ 222K+1SK+1RK

= 2K+3SK+1RK .

Thus (4.2) becomes

(12h1/(8n))RKSK+1 ≤ ((c5c6)2K+4

)SK+1RK .

This cannot hold for a sufficiently large h. Hence, f(l) = 0 by induction.

40

Lemma 4.5. By writing φ(z) = Φ(z, . . . , z), we have |φj(0)| < e−h8n

for0 ≤ j ≤ h8n.

Proof. From Lemma 4.4, we see that f(l) = 0 for all l ∈ Z with 1 ≤ l ≤ X

and m0 + · · ·+mn−1 ≤ Y where X = h8n and Y =

(h2

2(8n)2

).

Likewise, φm(r) = 0 for 1 ≤ r ≤ X and 0 ≤ m ≤ Y . And for

E(z) = (z − 1) · · · (z −X)Y

φ(z)

E(z)is regular within and on the circle, Γ with center origin and radius

R = Xh18n . Using Maximum modulus principle, for any w with |w| < X,

|φ(w)| ≤ ξΞ−1 |E(w)|

where |φ(w)| ≤ ξ and Ξ ≤ |E(w)| are the respective upper and lower boundsfor the 2 functions with z on Γ like in Lemma 4.4.

After considering a smaller circle with radius X, for |w| < X we can seethat each (z − r) ≤ 2X. Thus we get |E(w)| ≤ (2X)XY .

Also, |Ξ| ≥ (12R)XY as each (z − r) ≥ 1

2R for h big enough.

By lemma 4.3, ξ ≤ ch3+LR

5 . Hence we get,

|φ(w)| ≤ ξΞ−1 |E(w)| = ch3+LR

5 (1

2h

18n )−XY . (4.3)

And in the same manner as Lemma 4.4,

LR ≤ h2− 14nXh

18n ≤ h2− 1

8n+8n

≤ h8n+2 ≤ 2(8n)2+1XY.

Putting this into the powers of (4.3) and compressing h3 into XY ash3 < XY , for some constants c and k,we get

|φ(w)| ≤ (c−k)−XY (12h

18n )−XY = ( c

−k

2h

18n )−XY

Hence for a large enough h, (c−k

2h

18n )−XY < e−XY

By Cauchy’s formulae(Theorem A.10),

41

φj(0) =j!

2πi

∫Λ

φ(w)

wj+1dw

where Λ is the positively oriented circle |w| = 1 we arrive at∣∣∣∣ j!2πi

∫Λ

φ(w)

wj+1dw

∣∣∣∣ ≤ ∣∣∣∣ j!2πi

∫Λ

1

wj+1dw e−XY

∣∣∣∣=

∣∣∣∣ j!2πi2πie−XY

∣∣∣∣ by Cauchy’s residue theorem

= j!e−XY < jje−XY .

And

jje−XY < XXe−XY = (Xe−Y )X < e−X .

This holds because Xe−Y < e−1 as e−Y is very small compared to X for a hbig enough.

Lemma 4.6. For any integers t1, . . . , tn not all 0, with absolute values atmost T , we have

|t1 logα1 + · · ·+ tn logαn| > c−T11 (4.4)

Proof. Let aj be the leading coefficient of αj or 1αj

(1 ≤ j ≤ n) according to

when tj ≥ 0 or tj < 0 respectively.

Consider ω = a|t1|1 · · · a

|tn|n (αt11 · · ·αtnn − 1)

which is an algebraic integer with degree at most dn. And we obtain anyconjugate of ω by substituting arbitary conjugates of αi. We can say that

|any conjugate of ω| ≤ (|bβ|n)T = cT12.

for b = maxa1, . . . , an, 1 and β = max|α1| , . . . , |αn| , 1

If ω = 0, then αt11 · · ·αtnn = 1and, log(αt11 · · ·αtnn ) = 2πikand finally Ω = t1 logα1 + · · ·+ tn logαn = 2πik.

And Ω 6= 0 i.e. k 6= 0 as logα1, . . . , logαn are linearly independent overQ otherwise all the ti are zero. So |Ω| > c−T2 for some c2.

If ω 6= 0, then 1 ≤ |N (ω)| ≤ cTdn

12 and so

1

|ω|≤∣∣∣∣N (ω)

ω

∣∣∣∣ ≤ cTdn−1

12 ≤ cTdn

12 .

42

Thus, |ω| ≥ c−Tdn

12 .

Using |ez − 1| ≤ |z| e|z| which an easy consequence of taylor series. Letz = Ω and we attain ∣∣et1 logα1+···+tn logαn − 1

∣∣ ≤ |Ω| e|Ω|.From which

∣∣αt11 · · ·αtnn − 1∣∣ ≤ |Ω| e|Ω|

and so c−Tdn

12 ≤ |ω| ≤ |Ω| e|Ω|cT13.

Where cT13 = a|t1|1 · · · a

|tn|n and with some manipulation, we see that

|Ω| ≥ e−|Ω|(cdn

12c13)−T ≥(e|logα1+··· logαn|cd

n

12c13

)−T.

Hence |Ω| ≥ c−T11 .

Lemma 4.7. Let R, S ∈ Z+ and σ0, . . . , σR−1 ∈ C be distinct. Let ρ =min

0≤i<j<R1, |σi − σj| and σ = max1, |σ0| , . . . , |σR−1|. Then for all r, s with

0 ≤ r < R, 0 ≤ s < S, there exists wj ∈ C with |wj| ≤ (8σρ

)RS such that thepolynomial

W (z) =RS−1∑j=0

wjzj

satisfies Wk(σi) = 0 for all 0 ≤ i < R, 0 ≤ k < S other than i = r, k = s.And Ws(σr) = 1

Proof. We shall show that the polynomial below satisfies all the requiredproperties.

W (z) =

(−1

s!

)1

2πi

∫Cr

(ζ − σr)sU(z)

(ζ − z)U(ζ)dζ

where U(z) = [(z − σ0) · · · (z − σR−1)]S and Cr denotes a positively orientedcircle with center σr and radius less than min

z 6=σrρ, |z − σr|.

Note that the absolute value of the integrand multiplied by |ζ| goes to 0as |ζ| nears ∞. ∣∣∣∣(ζ − σr)sU(z) |ζ|

(ζ − z)U(ζ)

∣∣∣∣ −→ 0 (*)

as

∣∣∣∣(ζ − σr)sU(z) |ζ|(ζ − z)U(ζ)

∣∣∣∣ =

∣∣∣∣ U(z) |ζ|(ζ − z)(ζ − σ0)S · · · (ζ − σr)S−s · · · (ζ − σR−1)S

∣∣∣∣

43

For convenience, denote(ζ − σr)sU(z)

(ζ − z)U(ζ)by g(ζ).

Consider a positively oriented circle centered at origin and radius T withT big enough so that the circle contains all σis. Then by Cauchy’s theo-rem(Theorem A.9),∫

Cr

g(ζ) dζ =

∫CT

g(ζ) dζ −R−1∑

j=0,j 6=r

∫Cj

g(ζ) dζ −∫Cz

g(ζ) dζ

Also note that g(ζ) is analytic everywhere outside of CT . So by computingthe residue at infinity(Theorem A.11),∫

CT

g(ζ) dζ = 2πiResζ=∞

g(ζ) = −2πiResζ=0

1

ζ2g(

1

ζ)

= −∫CT

1ζg(1

ζ)

ζdζ

= −∫CT

h(ζ)

(ζ − 0)dζ where h(ζ) =

1

ζg(

1

ζ)

= g(0) = 0.

The above equalities are consequences of Cauchy’s residue theorem and (*).Computing h(ζ) as |ζ| nears 0 is same as computing ζg(ζ) as ζ tends to ∞.

By Cauchy’s integral formulae(Theorem A.10),

∫Cz

g(ζ) dζ = 2πi(z − σr)sU(z)

U(z).

Giving us

W (z) =(z − σr)s

s!+U(z)

s!

1

2πi

R−1∑j=0,j 6=r

∫Cj

(ζ − σr)s

(ζ − z)U(z)dζ

where Cj denotes a positively oriented circle with center σj and radius lessthan min

z 6=σjρ, |z − σj|.

For i = r, notice that U(z) = 0 at z = σr of order S,implying US(σr) = 0.Hence, for 0 ≤ k < s, we see that Wk(σr) = 0.For k = s, Ws(σr) = s!

s!+ 0 = 1 and for s < k < S, Wk(σr) = 0.

For the next part of the proof, using Cauchy’s integral formulae we havean alternate expression for W (z).

W (z) =−1

s!t!

[dt

dζt(ζ − σr)SU(z)

(ζ − z)U(ζ)

]ζ=σr

44

where t = S − s− 1. And by opening it,

W (z) =(−1)S−s−1

s!U(z)

∑j0≥0

· · ·∑

jR−1≥0

v(j0, . . . , jR−1)(σr − z)−jr−1. (4.5)

Where v(j0, . . . , jR−1) =R−1∏i=0i 6=r

(S + ji − 1

ji

)(σr−σi)−S−ji and j0+· · ·+jR−1 = t.

Since, jr ≤ t, we get jr + 1 ≤ S − s ≤ S and so 1 ≤ jr + 1 ≤ S.

The degree of U(z) is RS and it has a factor of (z − σr)S and since −S ≤−jr − 1 ≤ −1, W (z) is of degree at most RS − 1.

Since W (z) = 0 at z = σi(i 6= r) of order S, we have Wj(σi) = 0 for all0 ≤ j < S.

Since,

∣∣∣∣(S + ji − 1

ji

)(σr − σi)−S−ji

∣∣∣∣ ≤ 2S+ji−1ρ−S−ji , we have

|v(j0, . . . , jR−1)| ≤ 2S(R−1)2j0+···+jR−1−jr2−(R−1)ρ−S(R−1)ρ−j0−···−jR−1+jr

≤(

2

ρ

)(R−1)S+j0+···+jR−1−jr (1

2

)R−1

≤(

2

ρ

)(R−1)S+j0+···+jR−1

≤(

2

ρ

)RS.

The last inequality follows from the fact that j0 + · · ·+ jR−1 = t ≤ S.We can see that the coefficients of U(z)(σr − z)−j−1 has absolute values atmost (σ + 1)RS by looking U(z) alone.Number of terms in the sum of (4.5) is less than or equals to SR as there areR sums and each j0 ≤ S.

Thus, coefficients of W (z) has absolute value less than or equals to SR(σ +1)RS(2

ρ)RS.

And so SR(σ + 1)RS(

2

ρ

)RS≤ (2S)R(2σ)RS

(2

ρ

)RS=

(8σ

ρ

)RS.

For the main theorem,Proof of Theorem 4.1

45

Proof. Let S = L + 1, R = Sn = (L + 1)n and then any i ∈ Z, with0 ≤ i < RS = (L+1)n+1 can be expressed uniquely as i = λ0+λ1S+· · ·+λnSnwhere λr ∈ Z, with 0 ≤ λr ≤ L.

For example, if λr = L for all r, L(1 +S + · · ·+Sn) = L(Sn+1−1)S−1

= RS− 1.

For each i, let

vi = λ0, Pi = p(λ0, . . . , λn), andψi = λ1 logα1 + · · ·+ λn logαn

Recall φ(z) from Lemma 4.5, and from Lemma 4.2,

φ(z) =L∑

λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)zλ0eλnβ0zαγ1z1 · · ·αγn−1zn−1

=RS−1∑i=0

Pizvieλnβ0zeγ1z logα1+···+γn−1z logαn−1

=RS−1∑i=0

Pizvieλnβ0zez[λ1 logα1+···+λn−1 logαn−1+λn(β1 logα1+···+βn−1 logαn−1)]

=RS−1∑i=0

Pizviez[λ1 logα1+···+λn−1 logαn−1+λn(β0+β1 logα1+···+βn−1 logαn−1)]

=RS−1∑i=0

Pizviez[λ1 logα1+···+λn−1 logαn−1+λn logαn]

=RS−1∑i=0

Pizvieψiz.

From Lemma 4.6, any two ψi with distinct λ1, . . . , λn, differs by at leastc−L11 because |ψi − ψj| = |ψk| > c−L11 as the difference is also in the same form(4.4).

Exactly R of the ψi are distinct because firstly logαis are linearly inde-pendent over Q by hypothesis, so we can compare the coefficients of logαisand secondly 0 ≤ λr ≤ L gives (L+ 1) choices and there are n of them. So,(L+ 1)n = R.

We can enumerate these distinct ψi by σ0, . . . , σR−1 and for σ, ρ defined asin Lemma 4.7, we get the following results.Firstly, σ ≤ c14L, as |ψi| ≤ L(logα1 + · · ·+ logαn) andsecondly, ρ ≥ c−L15 because distinct ψis differs by at least c−L11 .

46

Since not all Pt = 0 from the hypothesis of Lemma 4.2, we can let Pt 6= 0for some suffix t. Also let s = vt and r be the suffix for which ψt = σr. Andthen for the W (z) given in Lemma 4.7, we can write

Pt =RS−1∑i=0

PiWvi(ψi) as Wvi(ψi) =

1 when i = t

0 when i 6= t.

By Leibnitz’s Theorem, we have

Wvi(ψi) =RS−1∑j=0

j(j − 1) · · · (j − vi + 1)wjψj−vii

=RS−1∑j=0

wj

[dj

dzj(zvieψiz)

]z=0

Thus,

Pt =RS−1∑i=0

PiWvi(ψi)

=RS−1∑i=0

Pi

RS−1∑j=0

wj

[dj

dzj(zvieψiz)

]z=0

=RS−1∑j=0

wjφj(0)

And from Lemma 4.5,

|φj(0)| < e−h8n

for 0 ≤ j ≤ RS.

From Lemma 4.7, we have

|wj| ≤(

8σ

ρ

)RS≤ (8c14Lc

L15)RS ≤ ch

2n+4

16

and |Pt| ≥ 1 as Pt ∈ Z.

1 ≤ |Pt| ≤RS−1∑j=0

ch2n+4

16 e−h8n ≤ RS Ch2n+4−h8n

17 −→ 0 for h large.

which is a contradiction.

This is a very powerful theorem that gives transcendence of many numbersright away as we shall see in the next section.

47

4.2 Corollaries

Corollary 4.8. Any non-vanishing linear combination of logarithms of al-gebraic numbers with algebraic coefficents is transcendental.

Proof. We want to show that for all algebraic α1, . . . , αn, β0, . . . , βn with β0 6=0

β1 logα1 + · · ·+ βn logαn 6= β0. (4.6)

If logα1, . . . , logαn are linearly independent over Q then the above statementfollows from the Theorem 4.1.

If they are linearly dependent, we proceed by induction.(4.6) trivially holds for n = 0. So we assume it is valid for some n < m. Andfor n = m, suppose ρ1, ρ2, . . . , ρm ∈ Q, with not all ρi = 0. So let ρr 6= 0such that

ρ1 logα1 + · · ·+ ρm logαm = 0βr(ρ1 logα1 + · · ·+ ρm logαm) = 0

So,

ρr(β0 + β1 logα1 + · · ·+ βm logαm)− βr(ρ1 logα1 + · · ·+ ρm logαm)= ρrβ0 + (ρrβ1 − ρ1βr) logα1 + · · ·+ (ρrβm − ρmβr) logαm

ρr(β0 + β1 logα1 + · · ·+ βm logαm)= ρrβ0 + (ρrβ1 − ρ1βr) logα1 + · · ·+ (ρrβm − ρmβr) logαm.

On the right hand side, only ρrβr − ρrβr = 0. Hence it looks like,

ρrβ0 + (ρrβ1 − ρ1βr) logα1 + · · ·+ (ρrβr−1 − ρr−1βr) logαr−1 + (ρrβr+1 −ρr+1βr) logαr+1 + · · ·+ (ρrβm − ρmβr) logαm

= β′0 + β′1 logα1 + · · ·+ β′r−1 logαr−1 + β′r+1 logαr+1 + · · ·+ β′m logαm 6= 0

where β′0 = ρrβ0 and β′i = ρrβi− ρiβr. It is not zero by inductive hypothesis,as there are only m− 1 terms.And so, β0 + β1 logα1 + · · ·+ βn logαn 6= 0.

Corollary 4.9. eβ0αβ11 · · ·αβnn is transcendental for any non-zero algebraicnumbers α1, . . . , αn, β0, . . . , βn.

Proof. Suppose αn+1 = eβ0αβ11 · · ·αβnn is algebraic. Then

logαn+1 = β0 + β1 logα1 + · · ·+ βn logαn−β0 = β1 logα1 + · · ·+ βn logαn − logαn+1

would be algebraic which contradicts Corollary 4.8 as β0 6= 0.

48

Corollary 4.10. αβ11 · · ·αβnn is transcendental for any algebraic numbersα1, . . . , αn other than 0 or 1 and any algebraic numbers β1, . . . , βn with 1, β1, . . . , βnlinearly independent over Q.

Proof. Suppose αβ11 · · ·αβnn is algebraic, let αβ11 · · ·αβnn = αn+1.Then β1 logα1 + · · · + βn logαn = logαn+1. And letting βn+1 = −1 we canwrite it as

β1 logα1 + · · ·+ βn logαn − logαn+1 = 0 (4.7)

We want to show that the above expression is not zero for a contradiction.Again, we proceed by induction.It is trivially true for m = 1 as β1 logα1 6= 0 as β1, α1 6= 0.We again only consider the case where the logα1, . . . , logαn+1 are linearlydependent over Q otherwise the contradiction comes from the main theoremby letting β0 = 0.

Assume the above expression is not zero for m < n+1 and we want to show itis the same for m = n+ 1 and like Corollary 4.8, suppose ρ1, ρ2, . . . , ρm ∈ Q,with not all ρi = 0. So we can let ρr 6= 0 such that

ρ1 logα1 + · · ·+ ρm logαm = 0βr(ρ1 logα1 + · · ·+ ρm logαm) = 0.

Proceeding by the same method as Corollary 4.1, and by allowing β0 = 0 weget,

β′1 logα1 + · · ·+ β′r−1 logαr−1 + β′r+1 logαr+1 + · · ·+ β′m logαm 6= 0

by the inductive hyposthesis. So for m = n + 1, (4.7) is not zero and thecontradiction follows by induction.

Also from this theorem, we can deduce that

Example 4.11. eπ, π + logα for any algebraic α 6= 0 and eαπ+β for anyalgebraic α, β with β 6= 0 are transcendental.

Proof. We can express eπ in another way. eπ = (−1)(−i) which is transcen-dental by Corollary 4.10.

Suppose γ = π + logα is algebraic, then

eγ = eπα = (−1)(−i)αα = eγ(−1)(i)

49

which contradicts Corollary 4.9.

Suppose γ = eαπ+β is algebraic, then

γ = eαπ+β = eβ(−1)(−iα)

contradicting Corollary 4.9 again.

This theorem gives us the transcendence of an even larger class of tran-scendentals. Some examples are the Gelfond-Schneider constant 2

√2, Gel-

fond’s constant eπ and many others such as√

23√5,√

7 log 3√

2 − 4√

3 log√

2.And of course using Theorem 3.13 and Theorem 3.15, we can attain moretranscendentals.

So far the Lindemann-Weierstrass theorem and Baker’s theorem are pow-erful tools to prove a number is transcendental. Of course we can proveeach of the numbers attained from the corollaries of these theorems to betranscendental individually. Each of the proofs will have a unique auxiliaryfunction that can establish a contradiction. For the auxiliary functions of eand π, we refer readers to [1,2]. And the contradiction arises in the similarway to the proof of Theorem 4.1.: We attain an integer that is in between 0and 1. And we can also employ this method to many other numbers that areconjectured to be transcendental. The first challenge comes in constructingan auxiliary function that will work. And second to arrive at the mentionedcontradiction.In the next chapter we explore Kurt Mahler’s method to show transcen-dence by showing algebraic independence, an even stronger property thantranscendence as we have seen in chapter 3.

50

Chapter 5

Mahler’s Classification

In this chapter we will explore an even stronger assertion than merely tran-scendence. Algebraic independence of numbers is defined as in Chapter 3. Iftwo numbers are algebraically independent, then they are not only transcen-dental but they are also transcendental over one another.

By definition, a transcendental number does not satisfy any polynomialP (x) ∈ Z[x]. Kurt Mahler asked ”What is the smallest value of |P (ξ)|for a transcendental ξ with respect to some height and degree?” He proposeda classification[11,18] which lies central to algebraic independence and tran-scendental numbers.

5.1 Mahler’s Classification

In this chapter we will only look at polynomials with integer coefficients.

Given a complex number ξ and h, n ∈ Z, and P (x) is of degree at mostn and height at most h, Let

Ω(ξ, n, h) = min|P (ξ)| : P (ξ) 6= 0

and define w(n, h) as

Ω(ξ, n, h) = h−nw(n,h).

And also define the following,

wn = lim suph−→∞

w(n, h) w = lim supn−→∞

wn.

51

Definition 5.1. Mahler’s ClassificationWith the above definition, Mahler classified all numbers as follows:

A-number w = 0, v =∞S-number 0 < w <∞, v =∞T-number w =∞, v =∞U-number w =∞, v <∞.

Where v is the least positive integer n for which wn = ∞; and write v = ∞if wn <∞ for all n.

This classification does encompass all the possibilites. If we look at thedefinition of w(n, h),

w(n, h) =− log |Ω(ξ, n, h)|

n log h.

Which is always non-negative because we have Ω(ξ, n, h) = 1 for n = 0 andh = 1. So for any other degree and height, Ω(ξ, n, h) < 1. Thus, w ≥ 0.

5.2 A-numbers

Theorem 5.2. α is an A-number if and only if it is algebraic.

Proof. (⇒)Suppose ξ is transcendental. We shall show 2 cases, when ξ is realand when it is complex. Though the latter is enough, we wish to highlight acertain property of w when ξ ∈ R and when ξ ∈ C\R.

Suppose ξ ∈ R. Let Q(x) denote non-zero polynomials of degree at mostn and with integer coefficients between 0 and h.Let Qξ be the set of numbers evaluated at each of these Q(x).For Q(x) = anx

n + an−1xn−1 + · · ·+ a0, there are h+ 1 possibilities for each

ai and there are n+ 1 a′is. We subtract one for the case where all ai = 0.Thus this set has (h+ 1)n+1 − 1 elements.

Suppose Q(ξ) ∈ Qξ. Then by Proposition 1.14, |Q(ξ)| ≤ ch.So any element in Qξ has an absolute value less than ch where c = c(ξ, n).Divide the interval [−ch, ch] into hn+1 disjoint sub-intervals each of length2ch

hn+1= 2ch−n.

By Pigeon-hole principle, there will be 2 distinct numbers Q1(ξ) and Q2(ξ)

52

in the same interval.Let P (x) = Q1(x)−Q2(x) which satisfies |P (ξ)| ≤ 2ch−n. Then

wn = lim suph−→∞

w(n, h) = lim suph−→∞

− log |Ω(ξ, n, h)|n log h

≥ lim suph−→∞

− log |P (ξ)|n log h

≥ lim suph−→∞

− log 2ch−n

n log h

= −(

lim suph−→∞

log 2c

n log h

)+ 1 ≥ 1.

Hence, for any real transcendental number w ≥ 1.

Similarly, if ξ ∈ C, we want at most (h + 1)n+1 squares in the intervals[−ch, ch] on both the real and imaginary axes.

Area of all the squares = (2ch)(2ch) = 4c2h2

Area of each square =4c2h2

(h+ 1)n+1= 4c2h2(h+ 1)−(n+1)

Length of each square = 2ch(h+ 1)−12

(n+1)

< 2chh−12

(n+1) = 2ch−12

(n−1)

We can use a larger length for the squares, so let the length of each squarebe 2ch−

12

(n−1).Thus we obtain lesser squares with larger area in the interval.

By pigeon-hole priniciple, there will be 2 distinct numbers Q1(ξ) and Q2(ξ)in the same square. Then by pythagoras theorem,

|P (ξ)| ≤√

2(2ch−12

(n−1))2

53

as we look at the farthest distance of 2 points in a square. Now

wn = lim suph−→∞

w(n, h) = lim suph−→∞

− log |Ω(ξ, n, h)|n log h

≥ lim suph−→∞

− log |P (ξ)|n log h

≥ lim suph−→∞

− log

√2(2ch−

12

(n−1))2

n log h

= −

(lim suph−→∞

log 2√

2c

n log h

)+

1

2(1− 1

n) ≥ 1

2(1− 1

n).

Thus, for a complex transcendental, w ≥ 12.

Therefore, any transcendental number is not an A-number. By looking atthe contrapositive, every A-number is algebraic.

(⇐) Now to show that every algebraic number is an A-number.Suppose ξ is algebraic of degree m and height hξ. The smallest possiblevalue is zero given by the minimal polynomial of ξ. However, we want aΩ(x, n, h) ∈ Z[x] that will give us a smallest positive value.

For such a Ω(x, n, h), Ω(ξ, n, h) would admit an algebraic number of degreeat most m and height at most chξ where c = c(n, ξ) as algebraic numbersform a field.Since Ω(ξ, n, h) 6= 0,by Theorem 1.13,

|Ω(ξ, n, h)| > c′

hmwhere c′ = c′(m, c)

wn = lim suph−→∞

w(n, h) = lim suph−→∞

− log |Ω(ξ, n, h)|n log h

< lim suph−→∞

− log c′h−m

n log h

= −(

lim suph−→∞

log c′

n log h

)+m

n

=m

n.

Thus, w = lim supn−→∞

m

n= 0.

54

So A-numbers are precisely the algebraic numbers and since algebraicnumbers are countable, in the grand scheme of things, the next result is notsurprising.

Recall that a set Z ⊆ C is said to have measure zero if given any ε > 0,there exist a collection of balls Br each having a positive radius r such that

Z ⊂⋃

Br and Area(⋃

Br

)< ε.

So in terms of Lebesgue planar measure,

Theorem 5.3. A-numbers form a set of measure zero.

Proof. For each n, h ∈ Z+, let

An,h = α ∈ C : α algebraic , deg(α) ≤ n, height(α) ≤ h.

Thus, for any α ∈ Q, α ∈ An,h for some n, h.For any fixed n, h, any α ∈ An,h is contained in the open ball B(α, h−3n).The number of such balls which is the same as the number of α ∈ An,h. Forpolynomials with degree at most n and height at most h.

For P (x) = anxn + · · · + ao, we have −h ≤ ai ≤ h gives 2h + 1 possibil-

ities and there are n+ 1 of them and each polynomial has at most n roots.There are at most n(2h+ 1)n+1 such αs.

Define

U(n, h) =⋃

α∈An,h

B(α, h−3n).

For anyH ≥ h, if α ∈ An,h, automatically α ∈ U(n,H), since α ∈ B(α,H−3n).

Thus, for any H∗ ∈ Z+, all algebraic numbers, α ∈∞⋃

h=H∗

∞⋃n=2

U(n, h). Now

Area (U(n, h)) ≤∑

α∈An,h

Area(B(α, h−3n)) ≤ 3n+1nh2nArea(B(α, h−3n))

≤ 3n+1nh2nπ

h6n

≤ (3n+1nπ

h2n)

1

h2n≤ (

3n+1nπ

h2n)

1

h2.

A small remark before we proceed. For n ≥ 2, h ≥ 9, we have3n+1nπ

h2n<

1

n2.

55

Since n3π ≤ 3n+1, for n ≥ 2, we have

3n+1n3π ≤ 32n+2 = 9n+1

≤ 92n for n ≥ 2

≤ h2n for h ≥ 9.

So Area (U(n, h)) ≤ 1

n2h2

Area

(∞⋃n=2

U(n, h)

)≤

∞∑n=1

1

n2h2=π2

6

1

h2.

given ε > 0, choose H∗ > 9 large enough such that∞∑

h=H∗

1

h2<

6ε

π2.Then

Area

(∞⋃

h=H∗

∞⋃n=2

U(n, h)

)≤

∞∑h=H∗

π2

6

1

h2< ε

Hence, A-numbers form a set of measure 0.

5.3 S, T, U-numbers

Now, we can see that transcendental numbers have 3 different classes and ifnot for the following theorem the classification will not have much use.

Theorem 5.4. Algebraically dependent numbers belong to the same class.

Proof. Suppose ξ, η are algebraically dependent, then Q(ξ, η) = 0 for someQ(x, y) ∈ Z[x, y].Let Q(x, y) be of degree k in x and degree l in y.

If either ξ or η is algebraic then both of them are algebraic, hence theywill be A-numbers.

So assume both ξ and η are transcendental.All the zeros, ξ = ξ1, ξ2, . . . , ξk of Q(x, η) are transcendental. Else, if ξi isalgebraic for some i, it has a minimal polynomial f ξiQ (x). Since, ξi is a root of

both f ξiQ (x) and Q(x, η), we have f ξiQ (x)|Q(x, η) and so f ξiQ (x) divides all the

coefficients of Q(x, y) as a polynomial in y. So we can consider Q(x,y)

fξiQ (x)

instead.

From Ω(ξ, n, h), we take the polynomial Ω(x, n, h). We denote it as Ω(x)

56

in this proof for convenience and compute it at each of the other ξi’s. andlet J = Ω(ξ)Ω(ξ2) · · ·Ω(ξk)

|J | = |Ω(ξ)| |Ω(ξ2)| · · · |Ω(ξk)|= h−nw(n,h) |Ω(ξ2)| · · · |Ω(ξk)|≤ h−nw(n,h)c1h

k−1 = c1h−nw(n,h)+k−1

for each |Ω(ξi)| ≤ c(i)h for some constant c(i) by Proposition 1.14 and so c1

depends on ξ, η, n,Q(x, y). Observe that J is symmetric in ξ1, . . . , ξk. Byfundamental theorem on symmetric functions(Theorem A.1), J can be ex-pressed as a polynomial in elementary symmetric functions of degree at mostn and height at most c2h

k as there are k Ω(ξi)s and c2 depends ξ, η, n,Q(x, y).

J = S(σ1, . . . , σk).

For each elementary symmetric function,

σ1 =k∑i=1

ξi

σ2 =∑i<j

ξiξj

...σk = ξ1 · · · ξk

Express Q(x, η) as polynomial in Q(η)[x] with roots ξ, ξ2, . . . , ξk:

Q(x, η) = q0(η)xk + q1(η)xk−1 + · · ·+ qk(η)

By properties of coefficients of polynomials we have,

q1

q0

=k∑i=1

ξi

q2

q0

=∑i<j

ξiξj

...(−1)k

qkq0

= ξ1 · · · ξk.

Thus (−1)iqi(η)

q0(η)= σi.

So J can be expressed as a polynomial inqi(η)

q0(η). And since it is of de-

gree at most n, q0(η)nJ is a polynomial in η of degree at most ln and height

57

at most h′ = c3hk where c3 depends on ξ, η, n,Q(x, y).

We can define like in the previous section, Ω(η, n, h′), w′(n, h′), w′n, w′

and v′ for η in the same way.

Ω(η, n, h′) = h′−nlw′(nl,h′) ≤ |q0(η)nJ |

≤ |q0(η)n| c1h−nw(n,h)+k−1 ≤ ch−nw(n,h)+k−1

where c = c1q0(η)n. Then

c3hknlw′(nl,h′) = h′nlw

′(nl,h′) ≥ chnw(n,h)−k+1.

Thus, knlw′nl ≥ nwn − k + 1 and klw′ ≥ w.

Also by interchanging η and ξ we obtain lnkwnk ≥ nw′n− l+ 1 and klw ≥ w′

So if w is finite, then w′ is finite and vice versa. If w is infinite, then w′ isinfinite and vice versa. And v and v′ is either finite or infinite at the sametime. For example, suppose v is finite, then wn is infinite for some n. Thenwe can see that w′nl is infinite for the same n and vice versa. Likewise wesee that when v is infinite, v′ is infinite and vice versa. Thus, proving thetheorem.

Note that the converse of Theorem 5.4 does not hold in general. I.e. alge-braically independent numbers can still be in same classes. Recall, Liouvillenumbers in Chapter 2. Most of Mahler’s research was based on numberswhich are not Liouville. Kurt Mahler proved that e, π and the Champer-nowne’s number are not Liouville numbers by considering their continuedfractions[20]. From that we know that Liouville numbers are not the onlytranscendentals. There are many other types of U -numbers[25,26]. Here, weshall see that all Liouville numbers are U -numbers[3].

Theorem 5.5. Liouville numbers are U-numbers.

Proof. Let ξ be a Liouville number. By Definition 2.6 of Liouville number,

there exists infinite sequence ofpnqn

with qn ≥ 1, pn, qn ∈ Z such that∣∣∣∣ξ − pnqn

∣∣∣∣ < 1

qnn.

Consider Pn(x) = qnx− pn, then |Pn(ξ)| < 1

qn−1n

≤ 1 and

|pn| = |−ξqn + pn + ξqn| ≤ |ξqn − pn|+ |ξqn|≤ 1 + |ξqn|≤ (1 + |ξ|) |qn| .

58

Let hn be the height of Pn(x).

If hn = |qn|, then |Pn(ξ)| < 1

qn−1n

<(1 + |ξ|)n−1

hn−1n

.

If hn = |pn|, then |Pn(ξ)| <∣∣∣∣ 1

qn−1n

pn−1n

pn−1n

∣∣∣∣ =

∣∣∣∣pn−1n

qn−1n

1

hn−1n

∣∣∣∣ ≤ (1 + |ξ|)n−1

hn−1n

.

In either cases, we have |Pn(ξ)| < (1 + |ξ|)n−1

hn−1n

.

So Ω(ξ, 1, hn) = h−w(1,hn)n <

(1 + |ξ|)n−1

hn−1n

and

−w(1, hn) log hn <(n− 1) log [1 + |ξ|]− (n− 1) log hn

which gives (n− 1) <w(1, hn) log hn

log hn − log(1 + |ξ|).

As n tends to ∞ and we see that the right hand side tends to ∞.

Sincelog hn

log hn − log(1 + |ξ|)tends to 1, we deduce that w(1, hn) tends to ∞.

Since the set hn of possible heights of Pn(x) is a subset of h the set ofall heights, we have

w(1, h) tends to ∞ and so w1 =∞.

Hence ξ is a U -number.

We call ξ a U -number of degree 1 as v = 1 from the classification. Sim-ilarly, it can be shown that ξ

1n is a U -number of degree n [2]. Our next

theorem shows a certain property that only S-numbers have.

Theorem 5.6. Let ξ be transcendental. Then 0 < w < ∞ if and only ifthere exist a ρ ∈ R+ for all n ∈ Z+, there exist a c = c(ξ, n) for all h ∈ Z+

and P (x) ∈ Z[x] with degree at most n and height at most h such that

c

hρn< |P (ξ)|.

Proof. (⇒) 0 < w < ∞ means, there exist a ρ > 0 such that for all n,wn < ρ.This implies that for any fixed n, w(n, h) ≤ ρ for all but finitely many h. Sofor all but finitely many h,

h−ρn ≤ h−w(n,h)n = Ω(ξ, n, h) ≤ |P (ξ)|.

For the finitely many h, let h1 < h2 < · · · < hL and let c = m where

m = minΩ(ξ, n, hk) : k = 1, . . . , L

59

andm

hρnL< m ≤ Ω(ξ, n, hL) ≤ |P (ξ)|.

(⇐) Assume for some ρ,c

hρn< |P (ξ)| holds.

After taking log with base h,

logh c− ρn < −nw(n, h)

and w(n, h) < ρ− logh c

n.

So as n tends to ∞ we can see that w ≤ ρ.

Mahler showed that e is an S-number[9,11]. We will not show the proofof this here but we will state the result.

Theorem 5.7. If α is a non-zero algebraic number, then there exist a c =c(α) > 0 such that for any P (x) ∈ Z[x] of degree, n and height, h largecompared to its degree,

h−cn ≤ |P (eα)|

With this result, we can show that eα is an S-number for any non-zeroalgebraic α.Since h−cn ≤ Ω(eα, n, h) = h−nw(n,h),so cn ≥ nw(n, h) and hence w(n, h) ≤ c for all n and h.

Thus, 0 < w < c implying that eα is an S-number.

Mahler also showed that Champernowne’s number is an S-number[18]. Hence,

using Theorem 5.4 and Theorem 3.13, we get the transcendence of e+∞∑k=1

1

10k!

and many more such numbers.

Now the question arises, not only to search for the mysterious transcen-dental numbers but also to classify them. Till today, we do not know theclassification of π. It has been proven by Mahler that π is not a U -number.Thus, it is either an S-number or a T -number. And suppose it is a T -number,the question of algebraic independence of π and e will be answered.The above theorem gives a sufficient and necessary condition for S-numbers.But what of the illusive T and U numbers?

Intuitively, here is a remark from Theorem 5.6 and the classification.

60

If w = ∞ for some ξ, then the sequence wn∞n=1 contains an unboundedsubsequence.for any ρ > 0, in other words, there exists a subsequence wnk∞k=1 such thatwnk > ρ for all k, where n1 < n2 < n3 < · · · .This implies that w(nk, h) > ρ for infinitely many h or elsewnk = lim sup

h−→∞w(nk, h) < ρ.

So for any ρ and any nk satisfying wnk > ρ and each having infinitely manychoices for h related to nk, there exists a P (x) ∈ Z[x] of degree at most nkand height at most h such that

|P (ξ)| = Ω(ξ, nk, h) = h−nkw(nk,h) < h−ρnk

and this holds for each k.

This shows that T and U numbers can be approximated sufficiently wellcompared to S-numbers. And because of this, it follows that the next the-orem not only states that almost all numbers are transcendental, but thatthey are S-numbers as well; in terms of Lebesgue planar measure[7].

Before we proceed, we give a simple lemma.

Lemma 5.8. For P (x) of height h, degree n, ξ ∈ C. Let α1, . . . , αn be theroots of P (x). Then

|ξ − α| < 2n|P (ξ)||P ′(α)|

where |ξ − α| = min1≤i≤n

|ξ − αi|.

Proof. For P (x) = (x− α1) · · · (x− αn),

P ′(x) =n∑i=1

P (x)

(x− αi).

Without loss of generality, let α1 = α. Then for all other roots αi, i 6= 1,

|α− αi| ≤ |ξ − α|+ |ξ − αi|≤ |ξ − αi|+ |ξ − αi| = 2 |ξ − αi| .

So |P ′(α)| =n∏i=2

|α− αi| < 2nn∏i=2

|ξ − αi| = 2n|P (ξ)||ξ − α|

and thus

61

|ξ − α| < 2n |P (ξ)||P ′(α)|

.

For the next theorem we will use Vinogradov’s notation or , wherea b means |a| < bc for a constant c depending only on n.

Theorem 5.9. Almost all numbers are S-numbers.

This theorem states that even among the transcendental numbers whichare uncountable and even with uncountable U -numbers, we still have ”muchmore” S-numbers. So if you were given a big pot filled with all complexnumbers, and if you drew out a number, then with probability 1, you havedrawn a S-number.

Proof. We need to show that the set of U and T numbers form a set ofmeasure zero. Let P (x) ∈ Z[x] be irreducible of degree at most n and heightat most h.Let α1, . . . , αn be roots of P (x) and let α = α1 be the nearest root to ξ ∈ C.Then

P (x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0

= an(x− α1) · · · (x− αn)

and P ′(x) = annxn−1 + an−1(n− 1)xn−2 + · · ·+ a1.

Let αi = max1, |αi|

|P ′(αi)| =∣∣annαn−1

i + an−1(n− 1)αn−2i + · · ·+ a1

∣∣≤ h(n |αi|n−1 + (n− 1) |αi|n−2 + · · ·+ 1

≤ h(n+ n− 1 + · · ·+ 1)αin−1.

P ′(α) is an algebraic number as algebraic numbers form a field. Each conju-gate of P ′(α) is obtained by substituting distinct conjugates of α. Withoutloss of generality, we assume α2, . . . , αn be the distinct conjugates of of α.Then by using the above inequality,∣∣an−1

n N (P ′(α))∣∣ = |P ′(α)|

∣∣an−1n P ′(α2) · · ·P ′(αn)

∣∣ |P ′(α)|

∣∣an−1n hn−1

∣∣ (α2 · · · αn)n−1

= |P ′(α)|∣∣an−1n hn−1

∣∣ |αk1 · · ·αkm|n−1

= |P ′(α)|hn−1∣∣an−1n αk1 · · ·αkm

∣∣n−1.

62

where αkj ∈ α2, . . . , αn such that∣∣αkj ∣∣ > 1 and j ≥ 2.

Consider Q(x) = an(x − αk1) · · · (x − αkm) and let Qj(x) = (x − αkj) andQ1(x) = (anx− anαk1).Let Hj be the height of Qj(x) and hi be the height of (x− αi).

By applying result in Theorem A.14,anαk1 · · ·αkm ≤ H1 · · ·Hm h1 · · ·hn h. Thus,

|an−1n N (P ′(α))| |P ′(α)|h2n−2 |P ′(α)|h2n.

And as for the lower bound for |an−1n N (P ′(α))|, we observe that when we

have our P (x), we have an, α and its conjugates. Hence |an−1n N (P ′(α))| is

fully determined. We can consider c = minP (x)∣∣an−1n N (P ′(α))

∣∣ over all P (x)

of degree at most n and height at most h. c will then be independent of αand an.

Therefore, we can say1

|P ′(α)| h2n as 1 c. (5.1)

If ξ is a T or a U -number, there exist for some n infinitely many polynomialsP such that

|P (ξ)| < h−6n with degree(P ) ≤ n and height(P ) ≤ h. (5.2)

This comes from the remark of theorem 5.6 that there are infinitely manyheights h for each ξ.

By Lemma 5.8, (5.1) and (5.2) above,

|ξ − α| < 2n |P (ξ)|P ′(α)

h−4n.

Consider U(n, h) =⋃

α∈An,h

B(α, h−3n) and ξ ∈ U(n, h) as for each such ξ

there exist some polynomial of degree n and height h.

Since there are infinitely many choices for the heights of polynomials, forany H > h, there exist another polynomial satisfying the above inequality,thus ξ ∈ U(n,H). Thus, for any H∗ ∈ Z+, any T and U -number, ξ,

ξ ∈∞⋃

h=H∗

∞⋃n=2

U(n, h)

And the proof follows as Theorem 5.3.

63

We have seen examples of S and U numbers but what of T -numbers. Tilltoday, no explicit example of a T -number has been found but a mathemati-cian named Wolfgang M. Schmidt[28,29] gave a proof that T -numbers existusing an inductively constructive method. The difficulty arises if we look atthe remark of Theorem 5.6 and the classification. T -numbers are those ξ’sfor which we have infinitely many polynomials P (x) ∈ Z[x] such that |P (ξ)|can be made small but the degrees of these polynomials are unbounded. Soensuring that these degrees of the best polynomial approximations are un-bounded poses the difficulty. That is as much as we know of T -numbers.Before concluding, we like to highlight an important conjecture that standson the brink of solving hundreds of open questions in transcendental numbertheory. For a detailed definition on transcendence degree and transcendentalfield extensions we refer readers to [4].

Conjecture 5.10. Schanuel’s ConjectureLet α1, . . . , αn ∈ C be linearly independent over Q, then the extension field

Q (α1, . . . , αn, eα1 , . . . , eαn)

has a transcendence degree of at least n over Q.

This conjecture implies the Lindemann-Weierstrass theorem and Baker’slinear independence of logarithms over algebraic numbers. It also implies thealgebraic independence of π and e and many others. Many mathematiciansare still looking into this problem and they have found alternative expressionsfor this conjecture[32,36].

64

Conclusion

We have seen many different ways and forms a transcendental number cantake. Also different concepts from different areas of mathematics come to-gether in transcendental number theory. They are complex analysis, algebra,Galois Theory, measure theory, continued fractions, to name a few. Thus,if one were to explore transcendental numbers he or she needs to be able tobring different concepts together.

We have also identified a few methods to prove transcendence. Thereare other methods for example, using differential operators to prove tran-scendence[13]. Just like transcendental numbers are enigmatic species ofnumbers, the proofs can take any shape and come in many unexpected ways.Each new number proved to be transcendental is an achievement and thereare many out there that have yet to cross the eyes of a mathematician.

65

Appendix A

Useful results

In this section, we give a few useful lemmas and theorems used in the paper.These theorems are general and can be used in many contexts. Hence, theyare stated here in the appendix.

Results on polynomials

We start out with a well known result.

Theorem A.1. Fundamental theorem on symmetric polynomialsEvery symmetric polynomial P = P (x1, . . . , xn) in indeterminates x1, . . . , xncan be expressed as a polynomial Q = Q(σ1, . . . , σn) in the elementary sym-metric functions where Q is unique and the degree of Q with respect toσ1, . . . , σn is degree of P with respect to x1.

The results below are used in Chapter 3 in Lemma 3.1.

Lemma A.2. If g(x) ∈ Z[x], then for any k ∈ N, all of the coefficients ofthe kth derivative g(k)(x) are divisible by k!.

Proof. Since differentiation is a linear operator, it is enough to prove forg(x) = xs for s > 0.If k > s, then the kth derivative is zero. If 1 ≤ k ≤ s, then g(k)(x) =k!(sk

)xs−k

Lemma A.3. Hermite’s IdentityFor a degree m polynomial f(x) ∈ R[x], let

F (x) =m∑s=0

f (s)(x) = f(x) + f ′(x) + · · ·+ f (m)(x).

Then for any x ∈ C,

66

ex∫ x

0

e−tf(t)dt = F (0)ex − F (x).

Proof. By integration by parts,∫ x

0

e−tf(t)dt = f(0)− f(x)e−x +

∫ x

0

e−tf ′(t)dt.

Repeating the process for another m times, we get∫ x

0

e−tf(t)dt = F (0)− F (x)e−x

and the result follows.

Proof of Lemma 4.2

This is a general result used only once in the proof of Lemma 4.2.

Proposition A.4. If α is an algebraic number satisfying f(α) = A0αd +

A1αd−1 + · · ·+ Ad = 0 where Am ∈ Z, with |Am| ≤ A, then

(A0α)j = A(j)0 + A

(j)1 α + · · ·+ A

(j)d−1α

d−1 (*)

where A(j)m ∈ Z for all j ≥ 0 such that A

(j)m = A0A

(j−1)m−1 − Ad−mA

(j−1)d−1 for

j ≥ d with∣∣∣A(j)

m

∣∣∣ ≤ (2A)j, (0 ≤ m < d) and A(j−1)−1 = 0.

Proof. For 0 ≤ j < d, we get

(A0α)j = A0αd + A1α

d−1 + · · ·+ Ad + Aj0αj

= (−A1αd−1 − · · · − Ad) + A1α

d−1 + · · ·+ Ad + Aj0αj

= (Ad − Ad) + (Ad−1 − Ad−1)α + · · ·+ (Aj0 + Ad−j − Ad−j)αj + · · ·+ (A1 − A1)αd−1.

So A(j)m =

0 when m 6= j

Aj0 when m = j.

For j ≥ d, We follow by induction. Let j = d and we get

(A0α)d = Ad−10 (A0α

d) = Ad−10 (−A1α

d−1 − · · · − Ad)= −AdAd−1

0 − Ad−1Ad−10 α− · · · − A1A

d−10 αd−1.

67

Thus, A(d)m = −Ad−mAd−1

0 and∣∣∣A(d)

m

∣∣∣ =∣∣Ad−mAd−1

0

∣∣ ≤ Ad ≤ (2A)d

Suppose (*) is true for some j ≥ d. Then

(A0α)j+1 = (A0α)(A0α)j = (A0α)(A(j)0 + A

(j)1 α + · · ·+ A

(j)d−1α

d−1)

= A0A(j)0 α + · · ·+ A

(j)d−1(−A1α

d−1 − · · · − Ad)= −A(j)

d−1Ad + (A0A(j)0 − Ad−1A

(j)d−1)α + (A0A

(j)1 − Ad−2A

(j)d−1)α2 + · · ·

+ (A0A(j)d−2 − A1A

(j)d−1)αd−1

where, A(j+1)m = A0A

(j)m−1 − Ad−mA

(j)d−1 and

∣∣∣A(j+1)m

∣∣∣ ≤ A(2A)j + A(2A)j =

(2A)j+1. And so, the result holds by induction.

This next lemma states merely that when you have more unknowns thanequations, you will always have a solution. It is used in the proof of Lemma4.2.

Lemma A.5. Let w, v ∈ Z, with 0 < w < v. Let ajk(1 ≤ j ≤ w, 1 ≤ k ≤ v)denote integers such that |ajk| ≤ a for some a ≥ 1.Then there exists x1, . . . , xk ∈ Z not all zero with |xk| ≤ (av)

wv−w such that

v∑k=1

ajkxk = 0.

Proof. Let r = (av)wv−w and bxc denote the integral part of x.

Sincev − ww

> 0 and r > 1 we have av = rv−ww < (r + 1)

v−ww =

(r + 1)vw

r + 1.

And so avr + 1 ≤ av(r + 1) < (r + 1)vw as av ≥ 1 and from the above in-

equality.

Let yj = aj1x1 + · · ·+ ajvxv and −bj = sum of negative ajk and cj = sum ofpositive ajk.

For 0 ≤ xk ≤ r, we have −bjr ≤ yj ≤ cjr and bj + cj =v∑k=1

|ajk| ≤ av.

Which in turn leads to yj < (bj + cj)r + 1 ≤ avr + 1.

Thus, there are at most (avr + 1)w different w-tuples of y1, . . . yw andat most (r + 1)v different v-tuples of x1, . . . xv as 0 ≤ xk ≤ r.

Since (avr + 1)w < (r + 1)v, there exists 2 distinct sets of x1, . . . xvcorresponding to y1, . . . yw. Call it x′1, . . . x′v and x′′1, . . . x′′v and setxk = x′k − x′′k to get our solution.

68

The following is some of the details for Lemma 4.2 which was omitted inthe actual proof.

Lemma A.6. Details for proof of Lemma 4.2

For differentiating z0 we first observe a pattern.d

dz0

zλ00 eλnβ0z0 = (λ0zλ0−10 + λnβ0z

λ00 )eλnβ0z0

d2

dz20

zλ00 eλnβ0z0 = [λ0(λ0 − 1)zλ0−20 + 2λ0λnβ0z

λ0−10 + (λnβ0)2zλ00 ]eλnβ0z0

...dm0

dzm00

zλ00 eλnβ0z0 = q(λ0, λn, z0)eλnβ0z0 .

where

q(λ0, λn, z0) =

[m0∑µ0=0

(m0

µ0

)λ0(λ0 − 1) · · · (λ0 − µ0 + 1)(λnβ0)m0−µ0zλ0−µ00

]For the rest of the zr,d

dzrαγrzrr =

d

dzreγrzr logαr = γr(logαr)α

γrzrr .

Hence,dmr

dzmrrαγrzrr = γmrr (logαr)

mrαγrzrr .

Here are some of the omitted steps in the computation. From

L∑λ0=0

· · ·L∑

λn=0

p(λ0, . . . , λn)q(λ0, λn, l)αλ1l1 · · ·αλnln γm1

1 · · · γmn−1

n−1 = 0. (***)

Let P ′ = (a1 · · · an)Llbm00 · · · b

mn−1

n−1 and multiply (***) by P ′.

Also expand γmrr = (λr + λnβr)mr =

mr∑µr=0

(mr

µr

)λmr−µrr (λnβr)

µr ,

and by write (arαr)j =

d−1∑s=0

a(j)rs α

sr and (brβr)

j =d−1∑t=0

b(j)rt β

tr.

We display only the left hand side of (***). Denote p(λ0, . . . , λn) by p∗ andq(λ0, λn, l) by q∗ here only.

Also let Λ denoten−1∏r=1

mr∑µr=0

(mr

µr

)λmr−µrr (λnβr)

µr here only.

(a1 · · · an)Llbm00 · · · b

mn−1

n−1

L∑λ0=0

· · ·L∑

λn=0

p∗q∗αλ1l1 · · ·αλnln Λ

69

= bm00 · · · b

mn−1

n−1

L∑λ0=0

· · ·L∑

λn=0

p∗q∗n∏j=1

(ajαj)λj la

(L−λj)lj Λ

= bm00 · · · b

mn−1

n−1

L∑λ0=0

· · ·L∑

λn=0

p∗q∗n∏j=1

d−1∑sj=0

a(λj l)j,sj

αsjj a

(L−λj)lj Λ

= bm00 · · · b

mn−1

n−1

L∑λ0=0

· · ·L∑

λn=0

p∗q∗n∏j=1

d−1∑sj=0

a(λj l)j,sj

αsjj a

(L−λj)lj Λ

Then by computingn∏j=1

d−1∑sj=0

a(λj l)j,sj

αsjj , we get

d−1∑s1=0

· · ·d−1∑sn=0

n∏j=1

a(λj l)j,sj

αsjj . And

we do the same for βj.

The details of the inequalities left out in Lemma 4.2 are as follows.

|q′| ≤n∏r=1

a(L−λr)lr cλrl1 as

∣∣a(λrl)r,sr

∣∣ ≤ cλr1

=

(n∏r=1

a(L−λr)r cλr1

)l

≤

(n∏r=1

a(L−λr)r

)h

cLh1 as λr ≤ L, l ≤ h

≤

(n∏r=1

ar

)Lh

cLh1 =

(n∏r=1

arc1

)Lh

= cLh2 .

|q′′| ≤n−1∏r=1

2mr(brλr)mr−µrλµrn b

(µr)r,tr

≤n−1∏r=1

(2λnc1br)mrLmr−µr

≤n−1∏r=1

(2λnc1br)mrLmr =

n−1∏r=1

(c3L)mr .

70

|q′′′| ≤ 2m0λ0(λ0 − 1) · · · (λ0 − µ0 + 1)λm0−µ0n bµ00 l

λ0−µ0c(m0−µ0)1 as |b0,t0| ≤ c1

≤ 2m0λµ00 (λnc1)(m0−µ0)bµ00 lλ0−µ0

≤ (2b0λnc1)m0λµ00 hL as l ≤ h, λ0 ≤ L, µ0 ≤ m0

≤ cm03 Lm0hL = (c3L)m0 h

L.

And we use

(m0 + 1) · · · (mn−1 + 1) ≤ 2m02m1 · · · 2mn−1 = 2m0+m1+···+mn−1 ≤ 2h2

.

To show∣∣∣∣∣m0∑µ0=0

· · ·mn−1∑µn−1=0

q′q′′q′′′

∣∣∣∣∣ ≤∣∣∣∣∣m0∑µ0=0

· · ·mn−1∑µn−1=0

cLh2 (c3L)m1+···+mn−1(c3L)m0 hL

∣∣∣∣∣≤ cLh2 (c3L)m0+m1+···+mn−1(c3L)m0 h

L

m0∑µ0=0

· · ·mn−1∑µn−1=0

1

≤ cLh2 (c3L)m0+···+mn−1(c3L)m0 hL(m0 + 1) · · · (mn−1 + 1)

≤ cLh2 (2c3L)h2

hL

≤ (c2h)Lh(2c3L)h2

= (2c3L)h2

cLh4 .

Results from Complex analysis

Some results from Complex analysis[6] which are employed in Chapter 4.

Theorem A.7. If f is entire and bounded in C, then f(z) is constant.

Recall an entire function on C is a function that is analytic at each pointin the C plane. The term regular denotes analytic is used in Lemma 4.4.

Theorem A.8. Maximum modulus prinicpleLet D ⊆ C and f be an analytic function on D. If there exist a point z0 ∈ Dsuch that |f(zo)| ≥ |f(z)| for all z ∈ D, then f(z) has a constant valuethrough out D.

Theorem A.9. Cauchy’s residue theoremLet f(z) be analytic everywhere inside and on a simple closed contour Cdescribed in the positive sense, except for a finite number of singular pointssay z0, . . . , zn. And let Ck be the circle centered at zk small enough so that itdoes not contain any of the other singular points. Then We have

71

∫C

f(z)dz =n∑k=0

∫Ck

f(z)dz = 2πin∑k=0

Resz=zk

f(z).

Theorem A.10. Cauchy’s Integral formulaIf f is analytic everywhere inside and on a simple closed contour C in thepositive sense, then for any point z0 in the interior of C

f(z0) =1

2πi

∫C

f(z)

(z − z0)dz and f (n)(z0) =

n!

2πi

∫C

f(z)

(z − z0)n+1dz.

Theorem A.11. Residue at infinitySuppose f is analytic everywhere except for a finite number of singular pointscontained inside and a simple closed contour C described in the positive sense.Then, ∫

C

f(z)dz = −2πiResz=∞

f(z).

We also have the identity that

Resz=∞

f(z) = −Resz=0

[1

z2f(

1

z)

].

The residue at zk is computed by looking at the Laurent series representationof f(z) centered at zk.

Results involving Vinogradov’s notation

For the next few results constants we will use Vinogradov’s notation or, where a b means |a| < bc for a constant c depending only on the degreen of P (x). Where P (x) = anx

n + · · · + a0 ∈ C[x] of height h. And whenP (x) = P1(x) · · ·Pk(x) where Pi(x) ∈ Z[x], let

Pi(x) = a(i)ni x

ni + a(i)ni−1x

ni−1 + · · ·+ a(i)0 .

Where a(i)j ∈ Z and let h denote the height of P (x).

Lemma A.12. For some integer j, 0 ≤ j ≤ n, we have h |P (j)| h.

Proof. We can express P (x) in the form

P (x) =n∑j=0

P (j)A(x)

A′(j)(x− j)

where A(x) = x(x− 1) · · · (x− n) and A′(x) is the derivative of A(x).

Since A′(x) =n∑j=0

A(x)

(x− j), we have,

72

|A′(j)| = |j(j − 1) · · · (j − (j − 1))(j − (j + 1)) · · · (j − n)| ≥ 1

Since the roots ofA(x)

(x− j)are all integers from 1, . . . , n not including j, by

properties of the coefficients ofA(x)

(x− j), we can see that the coefficients of

A(x)

(x− j) 1.

Thus we have h ≤n∑j=0

|P (j)|ha, where ha is height ofA(x)

(x− j).

Let r be some j such that P (r) = max0≤j≤n

P (j). We then get h P (r).

We also have |P (j)| ≤n∑k=0

|ak| jk ≤ (n∑k=0

|ak|)jn ≤ h(n+ 1)jn.

Which gives us |P (j)| h

Lemma A.13. If there exist a number c, 0 < c ≤ 1 such that P (x) satisfies|an| > ch, then |α| 1 for any root α of P (x).

Proof. If α ≤ 1, then the inequality is satisfied trivially.Suppose α > 1, then anα = −an−1 − · · · − a0α

−n+1. Then

α <|an−1|+ · · ·+ |a0|

|an|≤ nh

ch=n

c.

Lemma A.14 is used in Theorem 5.9.

Lemma A.14. If P = P1 · · ·Pk where Pi is a polynomial with height hi, then

h1 · · ·hk h.

Proof. Choose an integer j that satisfies Appendix 15 for P (x). Denote Hi

as the height of Pi(x+ j). We can compute

Pi(x− j) = a(i)ni

(x− j)ni + a(i)ni−1(x− j)ni−1 + · · ·+ a

(i)0

= a(i)nixni +

(a(i)ni

(ni1

)(−j) + a

(i)ni−1

)xni−1

+

(a(i)ni

(ni2

)(−j)2 + a

(i)ni−1

(ni − 1

1

)(−j) + a

(i)ni−2

)xni−2 + · · ·

+ a(i)ni

(−j)ni + · · ·+ a(i)0 .

73

Where the constant term is P (−j). Also each coefficient is of the form belowand it can be bounded.∣∣∣∣∣

l∑k=0

a(i)n1−k

(nil

)(−j)l

∣∣∣∣∣ ≤l∑

k=0

h

(nil

)(j)l h

because l,(nil

), j are all bounded by a constant depending on n.

Thus we have height of Pi(x − j) height of Pi(x). Replacing x by x + j,we get height of Pi(x) height of Pi(x+ j),i.e. hi Hi.Let η be a zero of P (x+ j), and from mean value theorem, we obtain

h |P (j)| = |P (η + j)− P (j)| = |η| |P ′(ξ + j)|

where |ξ| ≤ |η|. Suppose |η| < 1 then

|P ′(x)| = |nanxn−1 + · · ·+ a1| ≤ nh(|x|n−1 + · · ·+ 1).

Since |η| < 1, then |ξ + j| ≤ |η| + j < 1 + n. We have |P ′(ξ + j)| ≤ nhcimplying |P ′(ξ + j)| h.

So we attain h ηh, implying that η 1. If η ≥ 1 naturally, η 1

Since the zeros of Pi(x+j) are included in P (x+j), we arrive at Pi(η+j) = 0for some i. And by computing

Pi(x+ j) = a(i)ni

(x+ j)ni + a(i)ni−1(x+ j)ni−1 + · · ·+ a

(i)0

= a(i)nixni +

(a(i)ni

(ni1

)(j) + a

(i)ni−1

)xni−1

+

(a(i)ni

(ni2

)(j)2 + a

(i)ni−1

(ni − 1

1

)(j) + a

(i)ni−2

)xni−2 + · · ·

+ a(i)ni

(j)ni + · · ·+ a(i)0

= b(i)nixni + b

(i)ni−1x

ni−1 + · · ·+ b(i)0 ,

we can see that b(i)0 = Pi(j).

Consider Qi(x+ j) = (x+ j)niPi(1

x+ j) = b

(i)0 x

ni + b(i)1 x

ni−1 + · · ·+ b(i)ni

.

And let α1, . . . , αni be the roots of Qi(x+j). By properties of its coefficients,

ni∑i=1

αi =b

(i)1

b(i)0

= σ1

∑i<j

αiαj = (−1)b

(i)2

b(i)0

= σ2

74

...

α1 · · ·αni = (−1)nib

(i)ni

b(i)0

= σni .

Since, Pi(j) h and Pi(j) is the leading coefficient of Qi(x+j), by TheoremA.13, each |αi| 1.

This implys that each σr 1, thus b(i)r = σrb

(i)0 Pi(j) for each r.

Therefore Hi Pi(j); and putting them all together, we get

h1 · · ·hk H1 · · ·Hk |P1(j) · · ·Pk(j)| = |P (j)| h.

75

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