On Irrational and Transcendental Numbers

Home , Irrational number, Transcendental number

Matthijs J. Warrens

Bachelor thesis, 9 augustus 2012

Supervisor: Dr. Jan-Hendrik Evertse

Mathematical Institute, Leiden University

1 Contents

1 Introduction 3 1.1 History ...... 3 1.2 The Riemann zeta function ...... 4 1.3 Outline ...... 4

2 Irrational numbers 5 2.1 A theorem for the numbers e and π ...... 5 2.2 Auxiliary results for the irrationality of ζ(3) ...... 7 2.3 The irrationality of ζ(3) ...... 11

3 The Hermite-Lindemann approach 14 3.1 Hermite’s identity ...... 14 3.2 The number e ...... 14 3.3 Algebraic integers and the house of an algebraic number . . . . 16 3.4 The number π ...... 17 3.5 The Lindemann-Weierstrass theorem ...... 19

4 The Gel’fond-Schneider theorem 23 4.1 Some auxiliary results ...... 23 4.2 The case α, β ∈ R, α > 0...... 25 4.3 The general case ...... 29

2 1 Introduction

1.1 History

Number theory is the branch of mathematics that is devoted to the study of integers, subsets of the integers like the prime numbers, and objects made out of the integers. An example of the latter are the rational numbers, the 1 22 numbers that are fractions, ratios of integers. Examples are 2 and 7 . The irrational numbers are those numbers that cannot be represented by fractions of integers. An example√ of an irrational number that was already known in Ancient Greece is 2. The irrationality of e, the base of the natural logarithms, was established by Euler in 1744. The irrationality of π, the ratio of the circumference to the diameter of a circle, was established by Lambert in 1761 (see Baker 1975). A generalization of the integers are the algebraic numbers. A complex number α is called algebraic if there is a polynomial f(x) 6= 0 with integer coefficients such that f(α) = 0.√ If no such polynomial exists α is called transcendental. The numbers 2 and i are algebraic numbers since they are zeros of the polynomials x2 − 2 and x2 + 1 respectively. The first to prove the existence of transcendental numbers was Liouville in 1844, using continued P∞ −n! fractions. The so-called Liouville constant n=1 10 = 0.1100010... was the first decimal example of a transcendental number (see Burger, Tubbs 2004). In 1873 Hermite proved that e is transcendental. This was the first number to be proved transcendental without having been specifically constructed for the purpose. Building on Hermite’s result, Lindemann showed that π is transcendental in 1882. He thereby solved the ancient Greek problem of squaring the circle. The Greeks had sought to construct, with ruler and compass, a square with area equal to that of a given circle. If a unit length is prescribed this amounts to constructing two points in the plane at a distance √ π apart. In 1837 Wantzel showed that the constructible numbers are a √ subset of the algebraic numbers. Lindemann showed that π is however transcendental. For a historical overview, see Burger and Tubbs (2004), and Shidlovskii (1989). In 1874 Cantor showed that the set of algebraic numbers is countably infinite. This follows from the fact that the polynomials with integer coefficients form a countable set and that each polynomial has a finite number of zeros. In the same paper Cantor also showed that the set of real numbers is uncountably infinite. Since the algebraic numbers are countable while the real numbers are uncountable, it follows that most real numbers are in fact transcendental (see Dunham 1990). At the Second International Congress of Mathematicians in 1900, Hilbert posed a set of 23 problems “the study of which is likely to stimulate the further development of our science”. In the 7th of these problems he

3 conjectured that if α and β are algebraic numbers, α 6= 0, 1 and β irrational, then αβ is transcendental. In 1934 both Gel’fond and Schneider independently and using different methods obtained a proof of Hilbert’s conjecture.√ It follows from the Gel’fond-Schneider theorem that the numbers 2 2 and eπ are transcendental (see Shidlovskii 1989). Since irrational and transcendental numbers are defined by what they are not, it may be difficult, despite their abundance, to show that a specific number is irrational or transcendental. For example, although√ e and π are irrational, it is unknown whether e + π, e − π, eπ, 2e, πe or π 2 are irrational.

1.2 The Riemann zeta function

The Riemann zeta function is defined as ∞ X 1 1 1 1 ζ(s) = = + + + ··· ns 1s 2s 3s n=1 for a complex number s with Re s > 1. For any positive even integer 2n we have the expression B (2π)2n ζ(2n) = (−1)n+1 2n , 2(2n)! where B2n is the 2n-th Bernoulli number (see Abramowitz, Stegun 1970, chapter 23). The first few Bernoulli numbers are B0 = 1, B1 = −1/2, B2 = 1/6, B4 = −1/30, B6 = 1/42 and B8 = −1/30. The odd Bernoulli numbers B3,B5,... are zero. The expression for ζ(2n) is due to Euler (Dunham 1990). It is unknown whether there is such a simple expression for odd positive integers. Since π is a transcendental number, it follows from the above expression for even numbers that ζ(2n) is transcendental. In 1979 Apéryshowed that the number ζ(3) is irrational. It is unknown if ζ(3) is also transcendental. Furthermore, it is unknown whether ζ(5), ζ(7), ζ(9) and ζ(11) are all irrational, although Zudilin (2001) showed that at least one of them is irrational. Moreover, Rivoal (2000) showed that infinitely many of the numbers ζ(2n + 1), where 2n + 1 is an odd integer, are irrational.

1.3 Outline

In this bachelor thesis we consider the proofs of some results on irrational and transcendental numbers. The thesis is organized as follows. In Section 2 we consider the irrationality of e, π and ζ(3). In Section 3 we prove the transcendence of the numbers e and π. We also consider the Lindemann-Weierstrass theorem in this section. In Section 4 we discuss the Gel’fond-Schneider theorem.

4 2 Irrational numbers

2.1 A theorem for the numbers e and π

In this subsection all integers are rational integers. Let c ∈ R>0. Suppose f(x) is a function that is continuous on [0, c] and positive on (0, c). ∞ Furthermore, suppose there is associated with f an inﬁnite sequence {fi}i=1 0 of anti-derivatives that are integer-valued at 0 and c and satisfy f1 = f and 0 fi = fi−1 for i ≥ 2. Theorem 1 shows that if such a f exists for c, then the number c is irrational. The proof comes from Parks (1986). It is an extension of a simple proof by Niven (1947) that π is irrational.

Theorem 1. Let c and f(x) be as above. Then c is irrational.

Proof: Suppose c is rational. Then there are m, n ∈ Z such that c = m/n.

First, let Pc be the set of polynomials p(x) ∈ R[x] such that p(x) and all its derivatives are integer-valued at 0 and c. The set Pc is closed under addition. Furthermore, repeated application of the product rule shows that Pc is also closed under multiplication. Consider

p0(x) = m − 2nx.

0 Since p0(0) = m, p0(c) = −m and p (x) = −2n are all integers, we have p0(x) ∈ Pc. Next, let k ∈ Z≥1 and let

xk(m − nx)k p (x) = . k k!

Using induction on k we will show that pk(x) ∈ Pc. For p1(x) = x(m − nx) 0 we have p1(0) = p1(c) = 0 and p1(x) = p0(x). Hence, p1(x) ∈ Pc. Next, suppose p`(x) ∈ Pc and consider

x`+1(m − nx)`+1 p (x) = . `+1 (` + 1)!

We have p`+1(0) = p`+1(c) = 0. Furthermore, using the chain rule we have

x`(m − nx)` p0 = (m − 2nx) = p (x)p (x). `+1 `! ` 0

Since Pc is closed under multiplication, and since p0(x) and p`(x) are in Pc, it follows that p`+1(x) ∈ Pc. Next, since f(x) is continuous on [0, c], it attains a maximum on [0, c]. Let M denote this maximum. Furthermore, since pk(x) is a polynomial for all k it is continuous and diﬀerentiable on [0, c]. Hence, pk(x) attains a maximum 0 on [0, c], either in an endpoint or in the interior (0, c) where pk(x) = 0. The

5 derivative of pk(x) is pk−1(x)p0(x). Since pk−1(x) is only zero at x = 0 and 0 x = c, we must have p0(x) = 0, or x = m/2n, in order to have pk(x) = 0. At x = m/2n we have k m2 m 4n p = . k 2n k!

Replacing both f(x) and pk(x) by their maxima, we obtain

2 k 2 k c m c m Z M 4n Z Mc 4n f(x)pk(x)dx ≤ dx = . 0 k! 0 k! The expression on the right-hand side of the inequality goes to 0 when k → ∞. Hence, for suﬃciently large k we have the strict inequality Z c f(x)pk(x)dx < 1. 0

On the other hand, using integration by parts we obtain c Z c Z c 0 f(x)pk(x)dx = f1(x)pk(x) − f1(x)pk(x)dx. 0 0 x=0 The ﬁrst term on the right-hand side is an integer by hypothesis. By repeating integration by parts a number of times equal to the degree of p(x), repeatedly integrating the ‘f(x)’ part, while diﬀerentiating the ‘p(x)’ part, R c we obtain a sum of integers. Hence, the integral 0 f(x)pk(x)dx is an integer for all k. R c Since 0 f(x)pk(x)dx is an integer, since f(x) is positive on (0, c), and since pk(x) is positive at c/2 and equal to zero only at 0 and c for all k, it follows R c that 0 f(x)pk(x)dx is a positive integer, that is, Z c f(x)pk(x)dx ≥ 1, 0 for all k. Hence, we have a contradiction, and we conclude that c is irrational.

Corollary 2. π is irrational. Proof: π is a positive real number, and sin(x) is continuous on [0, π] and positive on (0, π). As a sequence of anti-derivatives of sin x we may take − cos x, − sin x, cos x, sin x, etc., which all have values from {−1, 0, 1} at x = 0 or x = π.

Corollary 3. Let a ∈ R>0, a 6= 1. If log a is rational, then a is irrational.

6 Proof: Since 1/a is rational if and only if a is rational, and log(1/a) = − log(a) is rational if and only if log a is rational, it suﬃces to prove the corollary for a > 1.

Suppose a is rational. Then there are m, n ∈ Z such that a = m/n. Since a > 1, we have log a > 0. Let c = log a and apply Theorem 1 with f(x) = nex. Then we may take the anti-derivatives of f all equal to f. We have f(0) = n and m f(c) = f log = m, n which are both integers. It follows from Theorem 1 that log a is irrational. This contradicts the hypothesis. Hence, we conclude that a is irrational.

Corollary 4. e is irrational. Proof: e is a real number, e 6= 1. Since log e = 1 is a rational number, it follows from Corollary 3 that e is irrational.

2.2 Auxiliary results for the irrationality of ζ(3)

In the next subsection we show that ∞ X 1 1 1 ζ(3) = = 1 + + + ··· n3 8 27 n=1 is irrational. We give a proof by Beukers (1979). We ﬁrst prove some lemmas.

Lemma 5. If f(x) ∈ Z[x], then for any j ∈ Z≥0 all the coefficients of the j-th derivative f (j)(x) are divisible by j!. Proof: Since differentiation is a linear operation, it suffices to prove the lemma for the polynomial xk for k > 0. The j-th derivative is 0 if j > k and if j ∈ {1, 2, . . . , k} then it is equal to

k! k xk−j = j! xk−j, (k − j)! j

k in which j is an integer.

Lemma 6. Let > 0. Then there is an N such that if n ≥ N, then

(1+)n dn := lcm(1, 2, . . . , n) < e .

r Proof: Let p be a positive prime number and r ∈ R>0. If p divides a number in the set {1, 2, . . . , n}, then pr ≤ n, and we have r ≤ log n/ log p. On the

7 other hand, p[log n/ log p] does divide one such number, namely itself. Thus,

Y [log n/ log p] dn = p . p≤n

Let π(n) be the prime-counting function that gives the number of primes less than or equal to n. The prime number theorem states that

π(n) log n lim = 1. n→∞ n Hence, for n suﬃciently large, we have   Y [log n/ log p] X π(n) log n (1+)n dn = p ≤ exp  log n = e < e . p≤n p≤n

Lemma 7. Let r, s ∈ Z>0. If r > s, then Z 1 Z 1 log(xy) − xrysdxdy (1) 0 0 1 − xy

3 is a rational number whose denominator when reduced divides dr. If r = s we have r ! Z 1 Z 1 log(xy) X 1 − xrysdxdy = 2 ζ(3) − . 1 − xy k3 0 0 k=1

Proof: Using the identity

∞ 1 X = xk = 1 + x + x2 + ··· , |x| < 1, 1 − x k=0 we obtain that (1) is equal to

∞ Z 1 Z 1 X − log(xy)xr+kys+kdxdy. (2) 0 0 k=0 Since x, y ∈ [0, 1], the series ∞ X xr+kys+k k=0 is convergent, and it follows that

Z 1 ∞ X r+k s+k log(xy)x y dx < ∞. 0 k=0

8 Hence, applying Fubini’s theorem we obtain that (2) is equal to

∞ ! Z 1 X Z 1 − log(xy)xr+kys+kdx dy. (3) 0 k=0 0 Let k ≥ 0. Using integrating by parts we obtain Z 1 Z 1 (log x)xr+kdx = lim (log x)xr+kdx 0 →0 1 xr+k+1 Z 1 xr+k

= lim log x − lim dx →0 r + k + 1 →0 r + k + 1 x= 1 xr+k = 0 − lim →0 (r + k + 1)2 x= −1 = . (r + k + 1)2

Using this identity and log(xy) = log x + log y in (3), we obtain that the expression in (3) and hence (1) is equal to

∞ X Z 1 ys+k log y ys+k − − dy. r + k + 1 (r + k + 1)2 k=0 0 Integrating next with respect to y we obtain, in a similar fashion,

∞ X 1 1 + . (4) (r + k + 1)(s + k + 1)2 (r + k + 1)2(s + k + 1) k=0 For r > s, we have r − s r − s + (r + k + 1)(s + k + 1)2 (r + k + 1)2(s + k + 1) r − s 1 1 = + (r + k + 1)(s + k + 1) s + k + 1 r + k + 1 1 1 1 1 = − + s + k + 1 r + k + 1 s + k + 1 r + k + 1 1 1 = − . (s + k + 1)2 (r + k + 1)2

Hence, if r > s, (4) and hence (1) can be written as

∞ ∞ 1 X 1 1 1 X 1 1 − = − r − s (s + k + 1)2 (r + k + 1)2 r − s (s + k)2 (r + k)2 k=0 k=1 r−s 1 X 1 = . r − s (s + k)2 k=1

9 The least common multiple of (r − s)(s + 1)2, (r − s)(s + 2)2,..., (r − s)r2 is 3 a divisor of dr, which completes the ﬁrst part of the lemma. Finally, if r = s (4) and hence (1) becomes

∞ ∞ r ! X 1 X 1 X 1 2 = 2 = 2 ζ(3) − . (r + k + 1)3 (r + k)3 k3 k=0 k=1 k=1

Lemma 8. Let D = {(u, v, w): u, v, w ∈ (0, 1)}. Then the function f given by 1 − w f(u, v, w) = u, v, 1 − (1 − uv)w is a bijection from D to D. Furthermore, its Jacobian determinant is

∂f(u, v, w) −uv = . ∂(u, v, w) (1 − (1 − uv)w)2

Proof: Note that f is defined on D. We first show that f(D) ⊂ D. Let (u, v, w) ∈ D. Since 0 < 1 − uv < 1, we have 0 < 1 − w < 1 − (1 − uv)w < 1, or 1 − w 0 < < 1, 1 − (1 − uv)w and hence f(u, v, w) ∈ D, and it follows that f is well-defined. Next, let f 2 = f ◦ f denote the two times iteration of f. We have

1−w ! 1 − w 1 − f 2(u, v, w) = f u, v, = u, v, 1−(1−uv)w 1 − (1 − uv)w 1−w 1 − (1 − uv) 1−(1−uv)w 1 − (1 − uv)w − (1 − w) = u, v, 1 − (1 − uv)w − (1 − uv)(1 − w) = (u, v, w) , that is, f is self-inverse. In particular, f is bijective. Finally, if we denote f(u, v, w) = (x, y, z), then we have

∂z −uv = , ∂w (1 − (1 − uv)w)2 and the Jacobian determinant equals

 1 0 0  ∂(x, y, z) ∂z −uv = det  0 1 0  = = 2 . ∂(u, v, w) ∂x ∂y ∂z ∂w (1 − (1 − uv)w) ∂w ∂w ∂w

Lemma 9. In the region D = {(u, v, w): u, v, w ∈ (0, 1)}, the function u(1 − u)v(1 − v)w(1 − w) f(u, v, w) = 1 − (1 − uv)w is bounded from above by 1/27. Proof: Let (u, v, w) ∈ D. Using the arithmetic-geometric means inequality we obtain the inequality √ √ 1 − (1 − uv)w = (1 − w) + uvw ≥ 2 1 − w uvw.

Hence, we have

u(1 − u)v(1 − v)w(1 − w) 1√ √ p f(u, v, w) ≤ √ √ = u(1 − u) v(1 − v) w(1 − w). 2 1 − w uvw 2 √ For t ∈ [0, 1], the maximum of t(1 − t) occurs at t = 1/3 and the maximum of pt(1 − t) occurs at t = 1/2. Hence, we have s 1 1 1 1 1 1 1 1 f(u, v, w) ≤ · √ 1 − · √ 1 − · 1 − = . 2 3 3 3 3 2 2 27

2.3 The irrationality of ζ(3)

Theorem 10. The number ζ(3) is irrational. Proof: The n-th shifted Legendre polynomial is given by 1 dn P (x) = (xn(1 − x)n) . n n! dxn The ﬁrst three polynomials are

P1(x) = 1 − 2x 2 P2(x) = 1 − 6x + 6x 2 3 P3(x) = 1 − 12x + 30x − 20x .

Consider the double integral

Z 1 Z 1 log(xy) − Pn(x)Pn(y)dxdy. 0 0 1 − xy

It follows from Lemma 5 that Pn(x) ∈ Z[x]. Since Pn(x) is of degree n, the i j quantity Pn(x)Pn(y) is a sum of terms of the form aijx y where

11 i, j ∈ {0, 1, . . . , n}, and aij ∈ Z. Since aii is a square for each i, we have aii > 0 for each i. Note that the double integral can be written as a sum of double integrals of the form in Lemma 7. It follows from Lemma 7 that the 3 double integral is a sum of rational numbers whose denominators divide dn plus a positive integer multiple of ζ(3). Hence, there exists integers An and 3 Bn > 0 such that the double integral equals (An + Bnζ(3))/dn. Next, we ﬁnd a second expression for the double integral. Since 1 log(xy) log(1 − (1 − xy)z) Z 1 1

− = − = dz, 1 − xy 1 − xy 0 1 − (1 − xy)z z=0 the double integral becomes Z 1 Z 1 Z 1 P (x)P (y) n n dxdydz. (5) 0 0 0 1 − (1 − xy)z For k ∈ {0, 1, . . . , n − 1} the multiple derivative (dk)/(dxk)(xn(1 − x)n) can be expressed as a sum of terms each having both x and 1 − x as a factor. Switching order of integration and integrating by parts repeatedly, the triple integral (5) becomes Z 1 Z 1 Z 1 dn n n 1 dxn (x (1 − x) ) Pn(y) dxdydz n! 0 0 0 1 − (1 − xy)z Z 1 Z 1 Z 1 n−1 1 1 d n n = Pn(y) d n−1 (x (1 − x) ) dydz n! 0 0 0 1 − (1 − xy)z dx Z 1 Z 1 Z 1 dn−1 n n 1 dxn−1 (x (1 − x) ) = Pn(y)yz 2 dxdydz n! 0 0 0 (1 − (1 − xy)z) Z 1 Z 1 Z 1 n n 1 n x (1 − x) = ··· = Pn(y)n!(yz) n+1 dxdydz n! 0 0 0 (1 − (1 − xy)z) Z 1 Z 1 Z 1 n n n n x y z (1 − x) Pn(y) = n+1 dxdydz. (6) 0 0 0 (1 − (1 − xy)z) Applying the transformation of Lemma 8 we have u = x, v = y, (1 − w)n zn = (1 − (1 − uv)w)n and 1 − w n+1 (uv)n+1 (1 − (1 − xy)z)n+1 = 1 − (1 − uv) = . 1 − (1 − uv)w (1 − (1 − uv)w)n+1 The triple integral then becomes Z 1 Z 1 Z 1 n n n n n+1 u v (1 − u) (1 − w) Pn(v) (1 − (1 − uv)w) uv n n+1 · 2 dudvdw 0 0 0 (1 − (1 − uv)w) (uv) (1 − (1 − uv)w) Z 1 Z 1 Z 1 P (v) = (1 − u)n(1 − w)n n dudvdw. 0 0 0 1 − (1 − uv)w

12 With the same arguments we used to show that the triple integral in (5) is equal to the integral in (6), but now with respect to v instead of x, we ﬁnally obtain the identity

Z 1 Z 1 log(xy) − Pn(x)Pn(y)dxdy 0 0 1 − xy Z 1 Z 1 Z 1 n n n n n n dudvdw = u (1 − u) v (1 − v) w (1 − w) n+1 . 0 0 0 (1 − (1 − uv)w)

Applying Lemma 9 and Lemma 7 (with r = s = 0) we obtain

Z 1 Z 1 log(xy) 1 n Z 1 Z 1 Z 1 dudvdw 0 < − Pn(x)Pn(y)dxdy ≤ 0 0 1 − xy 27 0 0 0 1 − (1 − uv)w 1 n Z 1 Z 1 log(uv) = − dudv 27 0 0 1 − uv 1 n = 2ζ(3) . 27

For a positive integer n and integers An and Bn we have n |An + Bnζ(3)| 1 0 < 3 < 2ζ(3) . dn 27 Assume now that ζ(3) = a/b for some integers a, b with b > 0. By Lemma 6, we have, for suﬃciently large n,

1 n 0 < |bA + aB | ≤ 2ζ(3) d3 b n n 27 n 1 n (2.8)3 n < 2ζ(3) (2.8)3nb = 2ζ(3) b < 2ζ(3)(0.9)nb. 27 27

Since bAn + aBn is an integer, we obtain a contradiction for suﬃciently large n. Hence, ζ(3) is irrational.

13 3 The Hermite-Lindemann approach

In this section we prove the transcendence of the numbers e and π. We also present the Lindemann-Weierstrass theorem. In our proof we follow Baker (1975) and Shidlovskii (1989). We ﬁrst prove a lemma.

3.1 Hermite’s identity

Lemma 11. Let f ∈ C[x] with deg f = m, u ∈ C, and let Z u I(u; f) = eu−tf(t)dt (7) 0 be the integral along the line segment from 0 to u. Then

m m X X I(u; f) = eu f (j)(0) − f (j)(u). (8) j=0 j=0

Proof: Using integration by parts we obtain the relation

u Z u u−t u−t 0 I(u; f) = −e f(t) + e f (t)dt t=0 0 Z u = euf(0) − f(u) + eu−tf 0(t)dt. 0

If we repeat this process m − 1 times we obtain identity (8).

Identity (8) is also called Hermite’s identity (Shidlovskii 1989).

3.2 The number e

The proof of Theorem 12 is a simpliﬁed version of the original proof by Hermite. This version can be found in Baker (1975) and Shidlovskii (1989).

Theorem 12. e is transcendental.

Proof: Suppose e is algebraic. Then there are a1, a2, . . . , an ∈ Z with a0 6= 0 such that n X k n ake = a0 + a1e + ··· + ane = 0. (9) k=0

Let p be a prime number with p > max {n, |a0|} and deﬁne

f(x) = xp−1(x − 1)p ··· (x − n)p. (10)

14 Using this f with deg f = m = (n + 1)p − 1 and I(u; f) in (7), we deﬁne the quantity

n X J = akI(k; f) = a0I(0; f) + a1I(1; f) + ··· + anI(n; f) k=0 We ﬁrst derive an algebraic lower bound for |J|. Since (9) holds, the contribution to the ﬁrst summand on the right-hand side of (8) to J is 0, and we have n m X X (j) J = − akf (k). k=0 j=0 The polynomial f(x) in (10) has 0 as a root of multiplicity p − 1 and 1, 2, . . . , n as roots of multiplicity p. Hence, we have

m m n X (j) X X (j) J = − a0f (k) + akf (k). (11) j=p−1 j=p k=1

Since f(x) in (10) can be written as

p−1 2 np f(x) = x (−1)(−2) ··· (−n) + b1x + b2x + ··· + bnx for some b1, . . . , bn ∈ Z, we have f p−1(0) = (p − 1)!(−1)np(n!)p.

Due to Lemma 5 each term on the right-hand side of (11) is divisible by p!, p−1 except for f (0) since p > n. Furthermore, since p > |a0|, it follows that J is an integer which is divisible by (p − 1)! but not by p. Hence, J is an integer with |J| ≥ (p − 1)!. Next, we derive an analytic upper bound for |J|. On the interval x ∈ [0, n] each of the factors x − k for k ∈ {0, 1, . . . , n} is bounded by n. Thus,

p |f(x)| = |xp−1(x − 1)p ··· (x − n)p| ≤ n(n+1)p−1 ≤ nn+1 , for x ∈ [0, r]. Moreover, we have

k k Z Z p |I(k; f)| ≤ |ek−tf(t)|dt ≤ dt ek max |f(t)| ≤ kek nn+1 0 0 t∈[0,k] for k ∈ {0, 1, . . . , n} and, using the triangle inequality,

n n X X k n+1p p |J| ≤ |ak||I(k; f)| ≤ |ak|ke n ≤ c1c2, k=0 k=0 for some constants c1 and c2 that are independent of p. Since we also have |J| ≥ (p − 1)!, we obtain a contradiction for suﬃciently large p. The contradiction proves the theorem.

15 3.3 Algebraic integers and the house of an algebraic number

Recall that a complex number α is called algebraic if there is a non-zero polynomial f with integer coefficients such that f(α) = 0. There is a unique polynomial Fα ∈ Z[x] such that Fα(α) = 0, Fα is irreducible in Q[x], the leading coefficient of Fα is positive, and the coefficients of Fα have greatest common divisor 1. This polynomial Fα is called the minimum polynomial of α. The other zeros in C of the minimum polynomial of α are called the conjugates of α. An algebraic number α is said to be an algebraic integer if its minimum polynomial has leading coefficient 1. The algebraic integers form a subring of C. If α is algebraic we have

n n−1 bnα + bn−1α + ··· + b1α + b0 = 0

n−1 for certain b0, . . . , bn ∈ Z with bn 6= 0. If we multiply this equation by bn we obtain

n n−1 n−2 n−1 (bnα) + bn−1(bnα) + ··· + bn b1(bnα) + bn b0 = 0.

Hence, if α is an algebraic number and bn is the leading coeﬃcient of its minimal polynomial, then bnα is an algebraic integer.

Let α1 ∈ C be an algebraic number and let αi for i ∈ {2, 3, . . . , n} denote the conjugates of α1 in C. The house of α1 denoted by α1 is deﬁned as

α1 = max {|α1|, |α2|,..., |αn|} .

The following lemma will be used in the proof of the Gel’fond Schneider theorem in Section 4.

Lemma 13. Let α1 ∈ C, α1 6= 0 be algebraic and deg α1 = n. Let T ∈ Z, T > 0 be such that T α1 is an algebraic integer. Then 1 |α | ≥ . 1 n n−1 T α1

Proof: Let αi for i ∈ {2, 3, . . . , n} denote the conjugates of α1. Since the numbers T αi for i ∈ {1, 2, . . . , n} are algebraic integers, the number n T α1T α2 ··· T αn = T α1 ··· αn is an algebraic integer. Since the minimum polynomial of T α1 is given by

(x − T α1)(x − T α2) ··· (x − T αn) ∈ Z[x],

From here on we make a distinction between rational integers, which are simply elements of Z, and algebraic integers.

3.4 The number π

The proof of Theorem 14 is a simpliﬁed version of the original proof by Lindemann. This version can be found in Baker (1975) and Shidlovskii (1989).

Theorem 14. π is transcendental.

Proof: Suppose π is algebraic. Then πi is also algebraic. Let α1 = πi with πi deg α1 = d, and let α2, . . . , αd be the conjugates of α1. Since 1 + e = 0, we obtain d Y (1 + eα` ) = (1 + eα1 ) ··· (1 + eαd ) = 0. `=1 If we expand this product, we obtain

d 1 1 Y X X (1 + eα` ) = ··· e1α1+···+dαd

`=1 1=0 d=0 The exponents inside the multiple sum include some which are non-zero, for example, 1 = 1 and 2 = ··· = d = 0, and also some which are zero, for example, 1 = ··· = d = 0. Call the exponents θ1, θ2, . . . , θ2d and let the ﬁrst n be the non-zero ones. We have n < 2d, and

2d − n + eθ1 + eθ2 + ··· + eθn = 0. (12)

It turns out that the numbers θ1, . . . , θn are the zeros of a polynomial g(x) ∈ Z[x] of degree n. We have the polynomial

1 1 Y Y h(x) = ··· (x − (1α1 + ··· + dαd))

1=0 d=0

d with deg h = 2 . If we consider h(x) as a polynomial in α1, . . . , αd, then h(x) is symmetric in α1, . . . , αd. Since α1, . . . , αd are a complete set of conjugates, it follows from the theory of elementary symmetric functions that d h(x) ∈ Q[x]. The zeros of h(x) are θ1, . . . , θn, and 0 with multiplicity 2 − n. 2d−n Hence, the polynomial h(x)/x ∈ Q[x] of degree n has precisely the numbers θ1, . . . , θn as its zeros. If we let r be the least common denominator of the coeﬃcients of h(x)/x2d−n, then the polynomial r g(x) = d h(x) ∈ Z[x] x2 −n

17 has also precisely θ1, . . . , θn as its zeros. Next, let p be a prime number, let b be the leading coefficient of g(x), and define (n−1)p p−1 p np p−1 p p f(x) = b x g (x) = b x (x − θ1) ··· (x − θn) with deg f = m = (n + 1)p − 1. Furthermore, using I(u; f) in (7) we define

n X J = I(θk; f) = I(θ1; f) + I(θ2; f) + ··· + I(θn; f). k=1 We first derive an algebraic lower bound for |J|. Using (8) and (12) we can write J as m m n d X (j) X X (j) J = − 2 − n f (0) − f (θk). (13) j=p−1 j=p k=1 It turns out that the inner sum over k is a rational integer. Indeed, first note that since bα` for ` ∈ {1, 2, . . . , d} is an algebraic integer, bθk for k ∈ {1, 2, . . . , n} is also an algebraic integer. Furthermore, since g(x) ∈ Z[x] we have that f(x) ∈ Z[x]. Hence, since the sum over k is a symmetric polynomial in bθ1, . . . , bθn with coefficients in Z and thus a symmetric polynomial with rational integer coefficients in the 2d numbers b(1α1 + ··· + dαd), it follows from the theory of elementary symmetric functions that the sum over k is a rational integer. (j) Since f (θk) = 0 for j < p, it follows from Lemma 5 that the double sum in (13) is a rational integer divisible by p!. Furthermore, we have f (j)(0) = 0 for j < p − 1 and f (j)(0) is divisible by p! for j ≥ p due to Lemma 5. It follows from the theory of elementary symmetric functions that

(p−1) np np p f (0) = b (p − 1)!(−1) (θ1θ2 ··· θn) , is divisible by (p − 1)!. However, if p is suﬃciently large f (p−1)(0) is not divisible by p!. Hence, if p > 2d − n it follows that |J| ≥ (p − 1)!. p Similar to the proof of Theorem 12 we can derive that |J| ≤ c1c2 where c1 and c2 are constants that are independent of p. We get a contradiction, which completes the proof.

18 3.5 The Lindemann-Weierstrass theorem

Theorems 12 and 14 on the transcendence of e and π are special cases of a more general result which Lindemann sketched in 1882. The result was later rigorously demonstrated by Weierstrass in 1885 (see Baker 1975). The proof of Theorem 15 comes from Baker (1975).

Theorem 15. For any distinct numbers α1, . . . , αn ∈ Q, and non-zero α α αn numbers β1, . . . , βn ∈ Q, we have β1e 1 + β2e 2 + ··· + βne 6= 0. Proof: Suppose α1 α2 αn β1e + β2e + ··· + βne = 0. (14)

We can assume that the βi are rational integers. If this is not the case, we consider the product of all the expressions formed by substituting for one or more of the βj one of its conjugates. Suppose βj has degree mj, let its mj conjugates be denoted by βj(ij) for ij ∈ {1, 2, . . . , mj}, and put

n Y M = mj. j=1 The product is given by

m1 mn Y Y α1 αn ··· (β1(i1)e + ··· + βn(in)e )

i1=1 in=1

X j1α1+···+jnαn = β(j1, . . . , jn)e ,

j1,...,jn where the latter sum is taken over all tuples of non-negative integers (j1, . . . , jn) with j1 + ··· + jn = M and β(j1, . . . , jn) is a polynomial expression in b1(1), . . . , βn(mn) which has rational integer coeﬃcients and which is invariant under any permutation of (βi(1), . . . , βi(mi)) for i ∈ {1, 2, . . . , n}. Hence, all β(j1, . . . , jn) ∈ Q. Let γ1, . . . , γt be the distinct numbers among the j1α1 + ··· + jnαn. Then the product becomes

γ1 γt δ1e + ··· + δte , where each δi is the sum of some of the terms β(j1, . . . , jn). Hence, δ1, . . . , δt ∈ Q. To complete, we multiply the rational numbers by a common denominator.

We now show that at least one of the new coeﬃcients δj is non-zero. To this end, we deﬁne on C a lexicographic ordering ≺ such that ζ ≺ η if Re ζ < Re η or Re ζ = Re η and Im ζ < Im η. If ζ1, . . . ζr, η1, . . . ηr are complex numbers with ζ1 ≺ η1,..., ζr ≺ ηr, then it holds that ζ1 + ··· + ζr ≺ η1 + ··· + ηr. We assume without loss of generality that α1 ≺ · · · ≺ αn and γ1 ≺ · · · ≺ γt. Hence, we have γt = Mαn and j1α1 + ··· + jnαn < γt for m ···mn (j1, . . . , jn) 6= (0,...,M), and thus δt = (βn(1) ··· βn(mn)) 1 6= 0.

19 Next, we can assume that the set {α1, . . . , αn} is closed under conjugation, that is, it contains all conjugates of each element occurring in it, and moreover, for any two indices j and k such that αj and αk are conjugates, we have βj = βk. If this is not the case, let K be any ﬁnite normal extension of Q containing α1, . . . , αn, and let {σ1, . . . , σm} be the Galois group of K/Q. Then clearly, m Y σi(α1) σi(αn) (β1e + ··· + βne ) = 0. i=1 By expanding the product on the left-hand side, we get

n n X X ··· βi1 ··· βim exp(σ1(αi1 ) + ··· + σm(αim )) = 0. i1=1 im=1 By grouping together those terms for which the exponents

σ1(αi1 ) + ··· + σm(αim ) have equal values we obtain an identity of the form

γ1 γt δ1e + ··· + δte = 0, where γ1, . . . , γt are the distinct numbers among the exponents

σ1(αi1 ) + ··· + σm(αim ). Clearly, {γ1, . . . , γt} is closed under conjugation, and δj = δk whenever γj and γk are conjugate to one another.

It remains to show that at least one of the numbers δk is non-zero, and for this, we use the argument from above. For i ∈ {1, 2, . . . , m}, let ji be the index j for which σi(αj) is the largest among σi(α1), . . . , σi(αn) in the lexicographic ordering. This index ji is unique since α1, . . . , αn are distinct.

Then σ1(αj1 ) + ··· + σm(αjm ) = γk is in the lexicographic ordering larger than all other exponents σ1(αi1 ) + ··· + σm(αim ) and thus, the coeﬃcient

δk = βi1 ··· βim 6= 0. For the remainder of the proof we can now assume that

α1 α2 αn β1e + β2e + ··· + βne = 0, (15) where α1, . . . , αn are distinct and the βi are rational integers, and that there are integers 0 = n0 < n1 < ··· < nr such that αnt+1, . . . , αnt+1 is a complete set of conjugates for each t, and

βnt+1 = βnt+2 = ··· = βnt+1 .

Since the α1, . . . , αn and β1, . . . , βn are algebraic, we can choose a non-zero rational integer b such that bα1, . . . , bαn and bβ1, . . . , bβn are algebraic integers. Let p be a prime number and deﬁne for i ∈ {1, 2, . . . , n} the functions p np [(x − α1) ··· (x − αn)] fi(x) = b (x − αi)

20 with deg fi = m = np − 1. Using these fi(x) and I(u; f) in (7) we deﬁne for i ∈ {1, 2, . . . , n} the quantities

n X Ji = βkIi(αk; fi) = β1Ii(α1; fi) + ··· + βnIi(αn; fi) k=1

We first derive an algebraic lower bound for |J1 ··· Jn|. Using (8) and (15) we obtain m n X X (j) Ji = − βkfi (αk). j=0 k=1 (j) Using a modification of Lemma 5, we find that fi (αk) is p! times an algebraic integer unless j = p − 1 and k = i. In this particular case we have

n (p−1) np Y p fi (αi) = b (p − 1)! (αi − αk) . k=1,k6=j

(p−1) Hence, fi (αi) is an algebraic integer divisible by (p − 1)! but not by p! if p is suﬃciently large. It then follows that Ji is an algebraic integer that is divisible by (p − 1)!.

Next, we show that Ji 6= 0. For suﬃciently large p, the number Ji can be written as

m r−1 X X h (j) (j) i Ji = − βnt+1 fi (αnt+1) + ··· + fi (αnt+1 ) . j=0 t=0

Note that by construction, fi(x) can be written as a polynomial whose coeﬃcients are polynomials in the αi, with rational integer coeﬃcients independent of the αi. Thus, noting that the αi form a complete set of conjugates and using the fundamental theorem on symmetric polynomials as in the previous proof, we see that the product of the Ji is in fact a rational number. Since it is an algebraic integer, it is an integer. Thus, J1 ··· Jn is a rational integer, and it is divisible by ((p − 1)!)n. Thus, n |J1 ··· Jn| ≥ [(p − 1)!] . Finally, using the triangle inequality we have, for each i,

n X |Ji| ≤ |βk||Ii(αk; fi)|. k=1 Hence, similar to the proofs of Theorems 12 and 14 we can derive that p |J| ≤ c1c2 where c1 and c2 are constants that are independent of p. We get a contradiction, which completes the proof.

The transcendence of e and π follows directly from Theorem 15. We also have the following corollaries.

21 Corollary 16. If α 6= 0 is algebraic, then eα is transcendental. Proof: If eα = β is algebraic, then we have eα − βe0 = 0, which contradicts Theorem 15.

Corollary 17. If α 6= 0 is algebraic, then sin α and cos α are transcendental. Proof: We have

eiα − e−iα eiα + e−iα sin α = , and cos α = . 2i 2 If sin α = β is algebraic, then eiα − e−iα − 2iβe0 = 0, which contradicts Theorem 15.

Corollary 18. If α ∈ C\{0, 1} is algebraic, then log α is transcendental for every branch of the logarithm. Proof: If log α = β, then eβ = α. By Corollary 16, since α is algebraic, β must be transcendental.

22 4 The Gel’fond-Schneider theorem

In this section we prove the Gel’fond-Schneider theorem. We ﬁrst prove some analytic lemmas. Before presenting the lemmas we introduce the following notation.

Let w ∈ C, R ∈ R>0, and let

D(R, w) = {z ∈ C : |z − w| < R} and D(R, w) = {z ∈ C : |z − w| ≤ R} . If w = 0 we write D(R) and D(R). Furthermore, let the maximum of |f(z)| on D(R, w) be denoted by M(R, w, f). If w = 0 we write M(R, f). If f(z) is analytic on D(R) and continuous on D(R), then it follows from the maximum modulus principle that |f(z)| attains its maximum on |z| = R. If f(z) is analytic on D(R, w), then N(R, w, f) will be used to denote the number of zeros of f(z) in D(R, w).

4.1 Some auxiliary results

Lemma 19. Let a1(t), . . . , an(t) be non-zero polynomials in R[t] of degrees d1, . . . , dn respectively. Let w1, . . . , wn be pairwise distinct real numbers. Then n X wj t f(t) = aj(t)e j=1 Pn has at most n − 1 + j=1 dj real zeros. Proof: By multiplying through by e−wnt if necessary, we may suppose that Pn wn = 0 and wj 6= 0 for j ∈ {1, 2, . . . , n − 1}. Let E = n + j=1 dj. We proceed by induction on E.

If E = 1, then n = 1 and d1 = 0. In this case there are no zeros, that is, there are at most E − 1 = 0 zeros. Next, suppose the lemma holds for ` ∈ {2, 3,...,E − 1} and consider ` = E. We have the ﬁrst derivative

n−1 0 X 0 wj t 0 f (t) = aj(t) + wjaj(t) e + an(t). j=1

Since the wj are pairwise distinct, and since wj 6= 0 for j ∈ {1, 2, . . . , n − 1}, 0 aj(t) + wjaj(t) has exactly degree dj for j ∈ {1, 2, . . . , n − 1}. Furthermore, 0 since we may suppose that wn = 0, the derivative an(t) has degree dn − 1. It follows from the induction hypothesis that f 0(t) has at most Pn (n − 2) + j=1 dj real zeros.

23 Finally, let N denote the number of real zeros of f(t), and let b1 < b2 < . . . < bN denote these zeros. Since f(t) is continuous on the intervals [bi, bi+1] and diﬀerentiable on (bi, bi+1) for i ∈ {1, 2,...,N − 1}, it follows from Rolle’s theorem that f 0(t) has at least N − 1 real zeros. Hence, Pn Pn N − 1 ≤ (n − 2) + j=1 dj, or N ≤ (n − 1) + j=1 dj.

Lemma 20. Let r, R ∈ R with 1 ≤ r ≤ R. Let f1(z), f2(z), . . . , fm(z) be analytic in D(R) and continuous on D(R). Let y1, y2, . . . , ym ∈ C with |yi| ≤ r for i ∈ {1, 2, . . . , m}. Then the determinant   f1(y1) ··· fm(y1)  . .. .  ∆ = det  . . .  f1(ym) ··· fm(ym) satisﬁes the inequality −m(m−1)/2 m R Y |∆| ≤ m! M(R, f ). r j j=1 Proof: Consider the determinant   f1(y1z) ··· fm(y1z)  . .. .  h(z) = det(fj(yiz)) = det  . . .  . f1(ymz) ··· fm(ymz)

Since the yi satisfy |yi| ≤ r, the functions fj(yiz) are analytic in D(R/r) and continuous on D(R/r). Since it is a sum of products of the fj(yiz), the determinant h(z) itself is analytic in D(R/r) and continuous on D(R/r).

Next, let K = m(m − 1)/2. Since the fj(yiz) are analytic functions on D(R/r) they can be expanded into power series on D(R/r). It follows that

K−1 X k k K fj(yiz) = bk(j)yi z + z gij(z), k=0 where bk(j) ∈ C for each k and gij(z) is analytic in D(R/r) and continuous on D(R/r). Since the determinant is linear in each of its columns, we can view h(z) as zK times an analytic function on D(R/r) plus terms involving the factor

 n1 nm  y1 ··· y1 nj . . zn1+n2+···+nm det y = zn1+n2+···+nm det  . .. .  , i  . . .  n1 nm ym ··· ym where n1, n2, . . . , nm ∈ Z≥1 and nj ∈ {0, 1,...,K − 1}. The determinant in the last expression is zero if two of the nj are identical. Therefore, the non-zero terms of this form satisfy m(m − 1) n + n + ··· + n ≥ 0 + 1 + ··· + (m − 1) = = K. 1 2 m 2

24 Hence, we deduce that h(z) is divisible by zK . Finally, since h(z) is analytic in D(R/r) and continuous on D(R/r), and since h(z) is divisible by zK , it follows that h(z)/zK is analytic in D(R/r) and continuous on D(R/r). Since h(z)/zK is analytic in D(R/r) and continuous on D(R/r), it follows from the maximum modulus principle that h(z)/zK attains its maximum value on the boundary ∂D(R/r). Hence, for w ∈ D(R/r), we have the inequality h(w) R h(z) r K ≤ M , = M (R/r, h(z)) . wK r zK R

For |z| = R/r we have |yiz| ≤ R. The determinant of a m × m matrix is the sum of m! products, where each product consists of m entries, such that for each row and column only one entry is part of a product. For each row index j we have |fj(yiz)| ≤ M(R, fj) for i ∈ {1, 2, . . . , m}. Thus, m Y M(R/r, h(z)) ≤ m! M(R, fj). j=1 Since |∆| = h(1) and 1 ≤ R/r ≤ R we obtain

m r K r K Y |∆| ≤ M(R/r, h(z)) ≤ m! M(R, f ), R R j j=1 from which the desired inequality follows.

4.2 The case α, β ∈ R, α > 0

We ﬁrst present a proof of the Gel’fond-Schneider theorem for α, β ∈ R and α > 0. The proof comes from course notes by Filaseta (2011). The proof is based on the method of interpolation determinants developed by Laurent (1994).

β Theorem 21. If α, β ∈ Q ∩ R with α > 0 and α 6= 1, and β∈ / Q, then α is transcendental. An equivalent formulation of Theorem 21 is the following. Assume that β α, β, α ∈ Q ∩ R and α > 0. Then β ∈ Q. Proof: Part of our arguments will be needed also in the proof of the general Gel’fond-Schneider theorem (Theorem 25), where the condition α, β ∈ R, α > 0 is not needed. It is only when we apply Lemma 19 above that we have β to assume α, β ∈ R, α > 0. For the moment we assume α, β, α ∈ Q with α 6= 0, 1, where αβ = eβ log α is any choice of the branch of the logarithm. When we are at the point to apply Lemma 19, we use the assumption β α, β, α ∈ R and deduce that β ∈ Q.

25 Let L0,L1,S ∈ Z≥2, L = (L0 + 1)(L1 + 1), and K ∈ R. K, L0, L1 and L will be increasing functions of S, and repeatedly we will choose S suﬃciently large so that K is suﬃciently large. The functions K, L0 and L1 and S must be chosen such that the inequalities

2 KL0 log S ≤ L, KL1S ≤ L and L ≤ (2S − 1) hold. This can be done by taking, for example, S large, L0 = bS log Sc, L1 = bS/ log Sc and K = log log S. If we combine the ﬁrst two inequalities we obtain 2 KL (L0 log S + L1S) ≤ 2L . (16)

Next, consider some arrangement

(2S−1)2 (s1(i), s2(i))i=1 of the pairs (s1, s2) ∈ {0,...,S − 1} × {0,...,S − 1}. Furthermore, let

L (u(j), v(j))j=1 be an arrangement of the pairs (u, v) ∈ {0,...,L0} × {0,...,L1}. We deﬁne the (2S − 1)2 × L matrix

v(j) u(j) s1(i)+s2(i)β M = (s1(i) + s2(i)β) α .

Let u(j) v(j)z u(j) v(j)z log α fj(z) = z α = z e for j ∈ {1, 2,...,L} be functions in the complex variable z, let

yi = s1(i) + s2(i)β for i ∈ {1, 2,...,L}, and consider the determinant ∆ = det(fj(yi)) of an arbitrary L × L submatrix (fj(yi)). We will show that all L × L submatrices of M have determinant zero. Under the assumption that ∆ 6= 0 we derive an analytic upper bound and an algebraic lower bound for log |∆|, from which we derive a contradiction. v(j)z We ﬁrst derive an upper bound. We have α = exp(v(j)z log α), and fj(z) represents an entire function for each j ∈ {1, 2,...,L}. For z1, z2 ∈ C we have

|ez1z2 | = eRe(z1z2) ≤ e|z1z2| = e|z1||z2|.

Hence, for any R ∈ R>0, we have u(j) v(j)z log α u(j) v(j)R| log α| M(R, fj) = M(R, z e ) ≤ R e . (17)

Taking the log on both sides of (17) we obtain the inequality

log M(R, fj) ≤ u(j) log R + v(j)R| log α| ≤ L0 log R + L1R| log α|. (18)

26 Next, applying Lemma 20 to ∆ with r = S(1 + |β|) and R = e2r we obtain the inequality L −L(L−1) Y |∆| ≤ e L! M(R, fj). (19) j=1 Taking the log on both sides of (19), and using inequality (18), we obtain

L X log |∆| ≤ −L(L − 1) + log L! + log M(R, fj) j=1 2 ≤ −L + L + L log L + L max {log M(R, fj)} 1≤j≤L 2 ≤ −L + L (1 + log L + L0 log R + L1R| log α|) or 2 log |∆| ≤ −L + c1L (L0 log S + L1S) (20) for some absolute constant c1 ∈ R independent of S. If we choose S such that K ≥ 4c1, inequality (16) becomes L2 c L (L log S + L S) ≤ . 1 0 1 2 Combining this inequality with (20) we obtain L2 log |∆| ≤ − , (21) 2 which speciﬁes an upper bound for log |∆|. Next, we derive a lower bound for log |∆| under the assumption that ∆ 6= 0. β Fix T ∈ Z>0 such that T α, T β and T α are algebraic integers. Then T L0+2L1S has the property that T L0+2L1S times any element of M, and hence T L0+2L1S times any element of the matrix describing ∆ is an algebraic L(L +2L S) β integer. Hence, T 0 1 ∆ is an algebraic integer in Q α, β, α . Thus T L(L0+2L1S)∆ is a zero of a monic polynomial of degree N, where N is at most the product of the degrees of the minimal polynomials of α, β and αβ. The house ∆ is the maximum of the absolute values of ∆ and its conjugates. We have the upper bound

L L L1LS ∆ ≤ L!SL0L 1 + β 0 (1 + α )L1LS 1 + αβ .

In the latter inequality we have taken into consideration that β , α and αβ may be smaller than 1. If ∆ 6= 0, it follows from Lemma 13 that

1−N −N |∆| ≥ T −NL(L0+2L1S) ∆ ≥ T −NL(L0+2L1S) ∆ . Combining these two inequalities we obtain

−NL L −NL1LS |∆| ≥ T −NL(L0+2L1S)(L!)−N S−NL0L 1 + β 0 (1 + α )−NL1LS 1 + αβ .

27 Since N log L! ≤ NL log L, we obtain, after taking the log on both sides, the inequality log |∆| ≥ − NL(L0 + 2L1S) log T − NL log L − NL0L log S β − NL0L log 1 + β − NL1LS log (1 + α ) − NL1LS log 1 + α . Since N, T , log 1 + β , log (1 + α ) and log 1 + αβ are constants that only depend on α and β, there is an absolute constant c2 ∈ R independent of S for which

log |∆| ≥ −c2L (L0 + log L + L0 log S + L1S) . If we choose S suﬃciently large we obtain the inequality

log |∆| ≥ −c3L(L0 log S + L1S), (22) for some absolute constant c3 ∈ R independent of c. Furthermore, if we choose S such that K ≥ 6c3, inequality (16) becomes L2 c L (L log S + L S) ≤ . 3 0 1 3 Combining this inequality with (22) we obtain L2 log |∆| ≥ − , (23) 3 which speciﬁes a lower bound for log |∆|. We now get a contradiction between the upper bound in (21) and the lower bound in (23). Since ∆ was an arbitrary submatrix, this shows indeed that all L × L submatrices of M have determinant zero.

Since ∆ = det(fj(yi)) = 0 for any sub-determinant ∆, it follows that the columns of the matrix (fj(yi)) are linearly dependent over R. Hence, there exists b1, b2, . . . , bL ∈ R, not all 0, such that L X 2 bjfj(yi) = 0, for i ∈ 1, 2,..., (2S − 1) . (24) j=1

u(j) v(j)yi Since fj(yi) = yi α , identity (24) is equal to

L X u(j) v(j)yi 2 bjyi α = 0, for i ∈ 1, 2,..., (2S − 1) . (25) j=1

If we consider identity (25) for all pairs (u, v) with u ∈ {0, 1,...,L0} and v ∈ {0, 1,...,L1}, we obtain

L1 L0 ! X X u vyi 2 b(L0+1)v+u+1yi α = 0, for i ∈ 1, 2,..., (2S − 1) . (26) v=0 u=0

28 Choosing

L0 X u av(t) = b(L0+1)v+u+1t , wv = v log α, and t = yi = s1(i) + s2(i)β, u=0 we can write the left-hand side of (26) as

L1 L0 ! L1 X X u vyi X wvt b(L0+1)v+u+1yi α = av(t)e . v=0 u=0 v=0

PL1 wvt Each of the L values of yi is a zero of v=0 av(t)e . Note that this sum consists of L1 + 1 polynomials, each of degree L0. We now at last use our assumption α, β ∈ R, α > 0, and apply Lemma 19. By that lemma, there are at most L0(L1 + 1) + (L1 + 1) − 1 = L − 1 2 distinct zeros. Since L − 1 < L ≤ (2S − 1) , two of the yi must be the same, and we have

0 0 0 0 2 s1(i) + s2(i)β = s1(i ) + s2(i )β for some i, i with 1 ≤ i < i ≤ (2S − 1) .

0 0 However, since the pairs (s1(i), s2(i)) and (s1(i ), s2(i )) are distinct, it follows that 0 s1(i ) − s1(i) β = 0 ∈ Q. s2(i) − s2(i )

4.3 The general case

In this subsection we consider the Gel’fond-Schneider theorem for the complex case. In the proof for the real case in the previous subsection, only in Lemma 19 we used the assumptions α, β ∈ R and α > 0. The idea is to replace Lemma 19 by Proposition 22 for complex numbers. The following result comes from Tijdeman (1971).

Proposition 22. Let a1(z), . . . , an(z) be non-zero polynomials in C[z] of degrees d1, . . . , dn respectively, let w1, . . . , wn be pairwise distinct complex numbers, let n X wkz f(z) = ak(z)e , k=1 and put n X E = n + dk, and m = max |wk|. k k=1

29 Furthermore, let R, s, t ∈ R>0, s > 1, and let y ∈ C. Then 1 st + s + t 1 N(R, y, f) ≤ (E − 1) log + (st + s + 2t)Rm + . (27) log s t s

We ﬁrst prove the following result of Tijdeman (1971, Lemma 1).

Lemma 23. Let R, s, t ∈ R>0, s > 1, and let f 6= 0 be analytic on D((st + s + t)R). Then

1 M((st + s + t)R, f) N(R, f) ≤ log . log s M(tR, f)

Proof: Let w ∈ D(tR) such that |f(w)| = M(tR, f). It then follows that

D(R) ⊂ D((1 + t)R, w) (28) and D((st + s)R, w) ⊂ D((st + s + t)R). (29) By Jensen’s formula (Greene, Krantz 2006, p. 279) we have

Z sR N(r, w, f) 1 iθ dr = log f w + sRe dθ − log |f(w)|. 0 r 2π We also have Z sR N(r, w, f) Z sR N(R, w, f) dr ≥ dr = N(R, w, f) log s. 0 r R r Combining the two previous formulas we obtain

1 |f| N(R, w, f) ≤ M sR, w, log . log s |f(w)|

This inequality, together with the inclusions (28) and (29), implies that

1 |f| N(R, f) ≤ N((1 + t)R, w, f) ≤ M sR(1 + t), w, log log s |f(w)| 1 M((st + s + t)R, f) ≤ log . log s M(tR, f)

Lemma 23 is used in the proof of Proposition 22. The following result from Balkema and Tijdeman (1973, Theorem 2) is also used in the proof of Proposition 22.

30 Lemma 24. Let a1(z), . . . , an(z) be non-zero polynomials in C[z] of degrees d1, . . . , dn respectively, let w1, . . . , wn be pairwise distinct complex numbers, let n X wkz f(z) = ak(z)e , k=1 and put n X E = n + dk, and m = max |wk|. k k=1 Furthermore, let R, γ ∈ R>0, γ > 1. Then γE − 1 M(γR, f) ≤ eRm(γ+1)M(R, f). γ − 1

We are now ready to present the proof of Proposition 22.

Proof of Proposition 22: Let γ ∈ R>1. Using Lemma 24 we obtain the inequality γE − 1 M(γtR, f) ≤ etRm(γ+1)M(tR, f). γ − 1 Taking γ = (st + s + t)/t we have

γE − 1 t st + s + tE 1 st + s + tE−1 ≤ ≤ 1 + . γ − 1 st + s t s t Combining the previous two inequalities we obtain 1 st + s + tE−1 M((st + s + t)R, f) ≤ 1 + e(st+s+2t)RmM(tR, f). s t Combining this inequality with Lemma 23 we obtain the desired inequality.

In the complex case, the Gel’fond-Schneider theorem holds for every branch of the complex logarithm. By replacing Lemma 19 with Proposition 22 in the proof of Theorem 21 we obtain the following result.

Theorem 25. Let α, β ∈ Q and α 6= 0, 1, and β∈ / Q. For any branch of log z we have that αβ = eβ log α is transcendental. Proof: Following the same set up and arguments as in the proof of Theorem 2 21, we ﬁnd at the end that each of the (2S − 1) values of yi is a zero of PL1 wvz f(z) = v=0 av(z)e . Let E, R and m be as deﬁned in Proposition 22, 1 and let 0 be the center of the disc D(R). Taking s = 5 and t = 5 in inequality (27), and using that log 31 32 < 2.2 and < 3.9, log 5 5 log 5

31 we obtain N(R, f) ≤ 3(E − 1) + 4Rm. (30)

The sum f(z) consists of L1 + 1 polynomials, each of degree L0. Hence,

E = L1 + 1 + L0(L1 + 1) = L.

The complex numbers are of the form yi = s1(i) + s2(i)β where s1, s2 ∈ Z with |s1|, |s2| < S. Hence, since we consider the disc with center 0, we have the upper bound R ≤ S 1 + β . Finally, we have

m = max |wv| = max |v log α| = L1| log α|. v∈{0,...,L1} v∈{0,...,L1} Using the values of E and m and the upper bound for R in inequality (30) we obtain that the number of zeros of f satisfy N(f) ≤ 3(L − 1) + 4SL1 1 + β | log α|.

Using here the specific definitions L0 = bS log Sc and L1 = bS/ log Sc, we obtain for sufficiently large S that S 4S2 N(f) ≤ 3 S2 + S log S + + 1 + β | log α| log S log S < 4S2 − 4S + 1 = (2S − 1)2.

Hence, at least two of the yi must be the same, which completes the proof.

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