Matthijs J. Warrens
On Irrational and Transcendental Numbers
Bachelor thesis, 9 augustus 2012
Supervisor: Dr. Jan-Hendrik Evertse
Mathematical Institute, Leiden University
1
Contents
1 Introduction
1.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 The Riemann zeta function . . . . . . . . . . . . . . . . . . . .
1.3 Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
4
4
2 Irrational numbers
2.1 A theorem for the numbers e and π . . . . . . . . . . . . . . . .
2.2 Auxiliary results for the irrationality of ζ(3) . . . . . . . . . . .
2.3 The irrationality of ζ(3) . . . . . . . . . . . . . . . . . . . . . .
5
5
7
11
3 The
3.1
3.2
3.3
3.4
3.5
Hermite-Lindemann approach
Hermite’s identity . . . . . . . . . . . . . . . . . . . . .
The number e . . . . . . . . . . . . . . . . . . . . . . . .
Algebraic integers and the house of an algebraic number
The number π . . . . . . . . . . . . . . . . . . . . . . . .
The Lindemann-Weierstrass theorem . . . . . . . . . . .
14
14
14
16
17
19
4 The
4.1
4.2
4.3
Gel’fond-Schneider theorem
23
Some auxiliary results . . . . . . . . . . . . . . . . . . . . . . . 23
The case α, β ∈ R, α > 0 . . . . . . . . . . . . . . . . . . . . . . 25
The general case . . . . . . . . . . . . . . . . . . . . . . . . . . 29
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1
1.1
Introduction
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
numbers that are fractions, ratios of integers. Examples are 12 and 22
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! = 0.1100010... was the
10
fractions. The so-called Liouville constant ∞
n=1
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 , π e or π 2 are irrational.
1.2
The Riemann zeta function
The Riemann zeta function is defined as
∞
X
1
1
1
1
= s + s + s + ···
ζ(s) =
s
n
1
2
3
n=1
for a complex number s with Re s > 1. For any positive even integer 2n we
have the expression
B2n (2π)2n
,
ζ(2n) = (−1)n+1
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 Apéry showed 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
2.1
Irrational numbers
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
of anti-derivatives that are integer-valued at 0 and c and satisfy f10 = f and
fi0 = 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.
Since p0 (0) = m, p0 (c) = −m and p0 (x) = −2n are all integers, we have
p0 (x) ∈ Pc . Next, let k ∈ Z≥1 and let
pk (x) =
xk (m − nx)k
.
k!
Using induction on k we will show that pk (x) ∈ Pc . For p1 (x) = x(m − nx)
we have p1 (0) = p1 (c) = 0 and p01 (x) = p0 (x). Hence, p1 (x) ∈ Pc . Next,
suppose p` (x) ∈ Pc and consider
p`+1 (x) =
x`+1 (m − nx)`+1
.
(` + 1)!
We have p`+1 (0) = p`+1 (c) = 0. Furthermore, using the chain rule we have
p0`+1 =
x` (m − nx)`
(m − 2nx) = p` (x)p0 (x).
`!
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
on [0, c], either in an endpoint or in the interior (0, c) where p0k (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
x = c, we must have p0 (x) = 0, or x = m/2n, in order to have p0k (x) = 0. At
x = m/2n we have
2 k
m
m
4n
pk
=
.
2n
k!
Replacing both f (x) and pk (x) by their maxima, we obtain
Z
M
c
f (x)pk (x)dx ≤
0
m2
4n
k!
k
Z
Mc
c
dx =
0
m2
4n
k!
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
f (x)pk (x)dx = f1 (x)pk (x)
−
f1 (x)p0k (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,
Rc
we obtain a sum of integers. Hence, the integral 0 f (x)pk (x)dx is an integer
for all k.
Rc
Since 0 f (x)pk (x)dx is an integer, since f (x) is positive on (0, c), and since
pk (x)Ris positive at c/2 and equal to zero only at 0 and c for all k, it follows
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
ζ(3) =
∞
X
1
1
1
=1+ +
+ ···
3
n
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 k−j
k−j
x
= j!
x ,
(k − j)!
j
in which
k
j
is an integer. Lemma 6. Let > 0. Then there is an N such that if n ≥ N , then
dn := lcm(1, 2, . . . , n) < e(1+)n .
Proof: Let p be a positive prime number and r ∈ R>0 . If pr 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
p[log n/ log p] .
dn =
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
lim
n→∞
π(n) log n
= 1.
n
Hence, for n sufficiently large, we have


Y
X
p[log n/ log p] ≤ exp 
log n = eπ(n) log n < e(1+)n .
dn =
p≤n
p≤n
Lemma 7. Let r, s ∈ Z>0 . If r > s, then
Z 1Z 1
log(xy) r s
x y dxdy
−
1 − xy
0
0
(1)
is a rational number whose denominator when reduced divides d3r . If r = s we
have
!
Z 1Z 1
r
X
log(xy) r s
1
−
x y dxdy = 2 ζ(3) −
.
1 − xy
k3
0
0
k=1
Proof: Using the identity
∞
X
1
=
xk = 1 + x + x2 + · · · ,
1−x
|x| < 1,
k=0
we obtain that (1) is equal to
Z
−
∞
1Z 1X
0
log(xy)xr+k y s+k dxdy.
0 k=0
Since x, y ∈ [0, 1], the series
∞
X
xr+k y s+k
k=0
is convergent, and it follows that
Z
∞ 1X
0 k=0
log(xy)xr+k y s+k dx < ∞.
8
(2)
Hence, applying Fubini’s theorem we obtain that (2) is equal to
!
Z 1 X
∞ Z 1
log(xy)xr+k y s+k dx dy.
−
0
k=0
(3)
0
Let k ≥ 0. Using integrating by parts we obtain
Z 1
Z 1
r+k
(log x)x dx = lim
(log x)xr+k dx
→0
0
1
Z 1
xr+k+1 xr+k
= lim log x
− lim
dx
→0
→0 r + k + 1
r + k + 1
x=
1
r+k
x
= 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
−
∞ Z
X
k=0
0
1 s+k
y
y s+k
log y
−
r+k+1
(r + k + 1)2
dy.
Integrating next with respect to y we obtain, in a similar fashion,
∞ X
1
1
+
.
(r + k + 1)(s + k + 1)2 (r + k + 1)2 (s + k + 1)
(4)
k=0
For r > s, we have
r−s
r−s
+
2
(r + k + 1)(s + k + 1)
(r + k + 1)2 (s + k + 1)
1
r−s
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
=
−
.
2
(s + k + 1)
(r + k + 1)2
Hence, if r > s, (4) and hence (1) can be written as
∞
1 X
r−s
k=0
1
1
−
(s + k + 1)2 (r + k + 1)2
∞
1 X
=
r−s
k=1
=
9
1
r−s
r−s
X
k=1
1
1
−
(s + k)2 (r + k)2
1
.
(s + k)2
The least common multiple of (r − s)(s + 1)2 , (r − s)(s + 2)2 , . . . , (r − s)r2 is
a divisor of d3r , which completes the first part of the lemma.
Finally, if r = s (4) and hence (1) becomes
2
∞
X
k=0
∞
r
X
X 1
1
1
=
2
=
2
ζ(3)
−
(r + k + 1)3
(r + k)3
k3
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
2
f (u, v, w) = f u, v,
1−w
1 − (1 − uv)w
1−w
1−(1−uv)w
1−w
− uv) 1−(1−uv)w
1−
=
u, v,
1 − (1
1 − (1 − uv)w − (1 − w)
1 − (1 − uv)w − (1 − uv)(1 − w)
= (u, v, w) ,
= u, v,
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
1
0 =
= det  0
=
.
∂(u, v, w)
∂w
(1 − (1 − uv)w)2
∂y
∂x
∂z
∂w
∂w
∂w
10
!
Lemma 9. In the region D = {(u, v, w) : u, v, w ∈ (0, 1)}, the function
f (u, v, w) =
u(1 − u)v(1 − v)w(1 − 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
p
√
1√
u(1 − u)v(1 − v)w(1 − w)
√
=
u(1 − u) v(1 − v) w(1 − w).
√
2
2 1 − w uvw
√
Forpt ∈ [0, 1], the maximum of t(1 − t) occurs at t = 1/3 and the maximum
of t(1 − t) occurs at t = 1/2. Hence, we have
s 1
1
1
1
1
1
1 1
1−
·√
1−
·
1−
= .
f (u, v, w) ≤ · √
2
3
3
2
2
27
3
3
f (u, v, w) ≤
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
Pn (x) =
1 dn
(xn (1 − x)n ) .
n! dxn
The first three polynomials are
P1 (x) = 1 − 2x
P2 (x) = 1 − 6x + 6x2
P3 (x) = 1 − 12x + 30x2 − 20x3 .
Consider the double integral
Z 1Z 1
−
0
0
log(xy)
Pn (x)Pn (y)dxdy.
1 − xy
It follows from Lemma 5 that Pn (x) ∈ Z[x]. Since Pn (x) is of degree n, the
quantity Pn (x)Pn (y) is a sum of terms of the form aij xi y j 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
double integral is a sum of rational numbers whose denominators divide d3n
plus a positive integer multiple of ζ(3). Hence, there exists integers An and
Bn > 0 such that the double integral equals (An + Bn ζ(3))/d3n .
Next, we find a second expression for the double integral. Since
1
Z 1
log(xy)
log(1 − (1 − xy)z) 1
−
=−
dz,
=
1 − xy
1 − xy
1
−
(1
− xy)z
0
z=0
the double integral becomes
Z 1Z 1Z
0
0
1
0
Pn (x)Pn (y)
dxdydz.
1 − (1 − xy)z
(5)
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 Z Z
dn
n
n
1 1 1 1
dxn (x (1 − x) )
Pn (y)
dxdydz
n! 0 0 0
1 − (1 − xy)z
n−1
Z Z Z
1 1 1 1
1
d
n
n
=
Pn (y)
d
(x (1 − x) ) dydz
n! 0 0 0
1 − (1 − xy)z
dxn−1
Z Z Z
dn−1
n
n
1 1 1 1
n−1 (x (1 − x) )
Pn (y)yz dx
=
dxdydz
n! 0 0 0
(1 − (1 − xy)z)2
Z Z Z
xn (1 − x)n
1 1 1 1
Pn (y)n!(yz)n
= ··· =
dxdydz
n! 0 0 0
(1 − (1 − xy)z)n+1
Z 1Z 1Z 1 n n n
x y z (1 − x)n Pn (y)
=
dxdydz.
(6)
(1 − (1 − xy)z)n+1
0
0
0
Applying the transformation of Lemma 8 we have u = x, v = y,
zn =
(1 − w)n
(1 − (1 − uv)w)n
and
n+1
(1 − (1 − xy)z)
=
1−w
1 − (1 − uv)
1 − (1 − uv)w
n+1
=
(uv)n+1
.
(1 − (1 − uv)w)n+1
The triple integral then becomes
Z 1Z 1Z 1 n n
u v (1 − u)n (1 − w)n Pn (v) (1 − (1 − uv)w)n+1
uv
·
dudvdw
n
n+1
(1 − (1 − uv)w) (uv)
(1 − (1 − uv)w)2
0
0
0
Z 1Z 1Z 1
Pn (v)
dudvdw.
=
(1 − u)n (1 − w)n
1 − (1 − uv)w
0
0
0
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 1Z 1
log(xy)
−
Pn (x)Pn (y)dxdy
1 − xy
0
0
Z 1Z 1Z 1
dudvdw
.
=
un (1 − u)n v n (1 − v)n wn (1 − w)n
(1 − (1 − uv)w)n+1
0
0
0
Applying Lemma 9 and Lemma 7 (with r = s = 0) we obtain
n Z 1 Z 1 Z 1
Z 1Z 1
log(xy)
1
dudvdw
0<
−
Pn (x)Pn (y)dxdy ≤
1
−
xy
27
1
−
(1 − uv)w
0
0
0
0
0
n Z 1 Z 1
1
log(uv)
=
−
dudv
27
1 − uv
0
0
n
1
.
= 2ζ(3)
27
For a positive integer n and integers An and Bn we have
n
|An + Bn ζ(3)|
1
0<
< 2ζ(3)
.
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,
n
1
0 < |bAn + aBn | ≤ 2ζ(3)
d3n b
27
n
n
1
(2.8)3
< 2ζ(3)
(2.8)3n b = 2ζ(3)
b < 2ζ(3)(0.9)n b.
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
eu−t f (t)dt
I(u; f ) =
(7)
0
be the integral along the line segment from 0 to u. Then
u
I(u; f ) = e
m
X
f
(j)
j=0
(0) −
m
X
f (j) (u).
(8)
j=0
Proof: Using integration by parts we obtain the relation
u
Z u
u−t
I(u; f ) = −e f (t)
+
eu−t f 0 (t)dt
0
t=0
Z u
u
= e f (0) − f (u) +
eu−t f 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
ak ek = a0 + a1 e + · · · + an en = 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 .
14
(10)
Using this f with deg f = m = (n + 1)p − 1 and I(u; f ) in (7), we define the
quantity
J=
n
X
ak I(k; f ) = a0 I(0; f ) + a1 I(1; f ) + · · · + an I(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 X
m
X
J =−
ak f (j) (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
J =−
m
X
a0 f
(j)
(k) +
j=p−1
m X
n
X
ak f (j) (k).
(11)
j=p k=1
Since f (x) in (10) can be written as
f (x) = xp−1 (−1)(−2) · · · (−n) + b1 x + b2 x2 + · · · + bn xn
p
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!,
except for f p−1 (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
Z
Z k
k−t
|I(k; f )| ≤
|e f (t)|dt ≤
0
k
p
dt ek max |f (t)| ≤ kek nn+1
0
t∈[0,k]
for k ∈ {0, 1, . . . , n} and, using the triangle inequality,
|J| ≤
n
X
k=0
|ak ||I(k; f )| ≤
n
X
|ak |kek nn+1
p
≤ c1 cp2 ,
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
bn αn + bn−1 αn−1 + · · · + b1 α + b0 = 0
for certain b0 , . . . , bn ∈ Z with bn 6= 0. If we multiply this equation by bn−1
n
we obtain
n−1
(bn α)n + bn−1 (bn α)n−1 + · · · + bn−2
n b1 (bn α) + bn b0 = 0.
Hence, if α is an algebraic number and bn is the leading coefficient of its
minimal polynomial, then bn α 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
|α1 | ≥
Tn
1
.
α1 n−1
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
T α1 T α2 · · · T αn = T n 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],
it follows that T n α1 · · · αn ∈ Z, and thus that |T n α1 · · · αn | ≥ 1. Hence,
|α1 | ≥
|T n α1 · · · αn |
1
|α1 · · · αn |
=
≥ n
.
n−1
n−1
n
α1
T α1
T α1 n−1
16
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
deg α1 = d, and let α2 , . . . , αd be the conjugates of α1 . Since 1 + eπi = 0, we
obtain
d
Y
(1 + eα` ) = (1 + eα1 ) · · · (1 + eαd ) = 0.
`=1
If we expand this product, we obtain
d
Y
(1 + eα` ) =
1
X
···
1 =0
`=1
1
X
e1 α1 +···+d αd
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
h(x) =
1
Y
1 =0
···
1
Y
(x − (1 α1 + · · · + d αd ))
d =0
with deg h = 2d . 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
h(x) ∈ Q[x]. The zeros of h(x) are θ1 , . . . , θn , and 0 with multiplicity 2d − n.
d
Hence, the polynomial h(x)/x2 −n ∈ Q[x] of degree n has precisely the
numbers θ1 , . . . , θn as its zeros. If we let r be the least common denominator
d
of the coefficients of h(x)/x2 −n , then the polynomial
g(x) =
r
x2d −n
17
h(x) ∈ Z[x]
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
f (x) = b(n−1)p xp−1 g p (x) = bnp xp−1 (x − θ1 )p · · · (x − θn )p
with deg f = m = (n + 1)p − 1. Furthermore, using I(u; f ) in (7) we define
J=
n
X
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 X
n
X
X
J = − 2d − n
f (j) (0) −
f (j) (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(1 α1 + · · · + d αd ), it follows from the theory of elementary symmetric
functions that the sum over k is a rational integer.
Since f (j) (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
f (p−1) (0) = bnp (p − 1)!(−1)np (θ1 θ2 · · · θn )p ,
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)!.
Similar to the proof of Theorem 12 we can derive that |J| ≤ c1 cp2 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
numbers β1 , . . . , βn ∈ Q, we have β1 eα1 + β2 eα2 + · · · + βn eαn 6= 0.
Proof: Suppose
β1 eα1 + β2 eα2 + · · · + βn eαn = 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
M=
n
Y
mj .
j=1
The product is given by
m1
Y
i1 =1
···
mn
Y
(β1 (i1 )eα1 + · · · + βn (in )eαn )
in =1
=
X
β(j1 , . . . , jn )ej1 α1 +···+jn αn ,
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 j1 α1 + · · · + jn αn . Then the product becomes
δ1 eγ1 + · · · + δt eγt ,
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 = M αn and j1 α1 + · · · + jn αn < γt for
(j1 , . . . , jn ) 6= (0, . . . , M ), and thus δt = (βn (1) · · · βn (mn ))m1 ···mn 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
(β1 eσi (α1 ) + · · · + βn eσi (αn ) ) = 0.
i=1
By expanding the product on the left-hand side, we get
n
X
i1 =1
···
n
X
βi1 · · · βim exp(σ1 (αi1 ) + · · · + σm (αim )) = 0.
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
δ1 eγ1 + · · · + δt eγt = 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
β1 eα1 + β2 eα2 + · · · + βn eαn = 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
[(x − α1 ) · · · (x − αn )]p
fi (x) = bnp
(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
Ji =
n
X
βk Ii (αk ; fi ) = β1 Ii (α1 ; fi ) + · · · + βn Ii (αn ; fi )
k=1
We first derive an algebraic lower bound for |J1 · · · Jn |. Using (8) and (15)
we obtain
m X
n
X
(j)
Ji = −
βk fi (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
(p−1)
fi
(αi ) = bnp (p − 1)!
n
Y
(αi − αk )p .
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
Ji = −
m X
r−1
X
h
i
(j)
(j)
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,
|J1 · · · Jn | ≥ [(p − 1)!]n .
Finally, using the triangle inequality we have, for each i,
|Ji | ≤
n
X
|βk ||Ii (αk ; fi )|.
k=1
Hence, similar to the proofs of Theorems 12 and 14 we can derive that
|J| ≤ c1 cp2 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
sin α =
eiα − e−iα
,
2i
and
cos α =
eiα + e−iα
.
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
f (t) =
aj (t)ewj t
j=1
has at most n − 1 +
Pn
j=1 dj
real zeros.
Proof: By multiplying through by e−wn t if necessary, we may
P suppose that
wn = 0 and wj 6= 0 for j ∈ {1, 2, . . . , n − 1}. Let E = n + nj=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
f 0 (t) =
n−1
X
a0j (t) + wj aj (t) ewj t + a0n (t).
j=1
Since the wj are pairwise distinct, and since wj 6= 0 for j ∈ {1, 2, . . . , n − 1},
a0j (t) + wj aj (t) has exactly degree dj for j ∈ {1, 2, . . . , n − 1}. Furthermore,
since we may suppose that wn = 0, the derivative a0n (t) has degree dn − 1. It
0
follows from
Pnthe induction hypothesis that f (t) has at most
(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’sPtheorem that f 0 (t) has at P
least N − 1 real zeros. Hence,
N − 1 ≤ (n − 2) + nj=1 dj , or N ≤ (n − 1) + nj=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
Y
R
m!
M (R, fj ).
r
j=1
Proof: Consider the determinant


fm (y1 z)

..
.
.
f1 (y1 z) · · ·

..
..
h(z) = det(fj (yi z)) = det 
.
.
f1 (ym z) · · ·
fm (ym z)
Since the yi satisfy |yi | ≤ r, the functions fj (yi z) are analytic in D(R/r) and
continuous on D(R/r). Since it is a sum of products of the fj (yi z), 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 (yi z) are analytic functions on
D(R/r) they can be expanded into power series on D(R/r). It follows that
fj (yi z) =
K−1
X
bk (j)yik z k + z K 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 z K times an analytic function on D(R/r) plus terms involving
the factor
 n1

y1 · · · y1nm
n 
..  ,
..
z n1 +n2 +···+nm det yi j = z n1 +n2 +···+nm det  ...
.
. 
n1
ym
···
nm
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
n1 + n2 + · · · + nm ≥ 0 + 1 + · · · + (m − 1) =
24
m(m − 1)
= K.
2
Hence, we deduce that h(z) is divisible by z K .
Finally, since h(z) is analytic in D(R/r) and continuous on D(R/r), and
since h(z) is divisible by z K , it follows that h(z)/z K is analytic in D(R/r)
and continuous on D(R/r). Since h(z)/z K is analytic in D(R/r) and
continuous on D(R/r), it follows from the maximum modulus principle that
h(z)/z K attains its maximum value on the boundary ∂D(R/r). Hence, for
w ∈ D(R/r), we have the inequality
K
h(w) ≤ M R , h(z) = r
M (R/r, h(z)) .
wK r zK
R
For |z| = R/r we have |yi z| ≤ 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 (yi z)| ≤ M (R, fj ) for i ∈ {1, 2, . . . , m}. Thus,
M (R/r, h(z)) ≤ m!
m
Y
M (R, fj ).
j=1
Since |∆| = h(1) and 1 ≤ R/r ≤ R we obtain
|∆| ≤
r K
R
M (R/r, h(z)) ≤
r K
R
m!
m
Y
M (R, fj ),
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).
Theorem 21. If α, β ∈ Q ∩ R with α > 0 and α 6= 1, and β ∈
/ Q, then αβ is
transcendental.
An equivalent formulation of Theorem 21 is the following. Assume that
α, β, αβ ∈ 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
to assume α, β ∈ R, α > 0. For the moment we assume α, β, αβ ∈ 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
α, β, αβ ∈ 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
KL0 log S ≤ L,
KL1 S ≤ L and L ≤ (2S − 1)2
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
KL (L0 log S + L1 S) ≤ 2L2 .
(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
(u(j), v(j))L
j=1
be an arrangement of the pairs (u, v) ∈ {0, . . . , L0 } × {0, . . . , L1 }. We define
the (2S − 1)2 × L matrix
v(j) s1 (i)+s2 (i)β
u(j)
.
α
M = (s1 (i) + s2 (i)β)
Let
fj (z) = z u(j) αv(j)z = z u(j) ev(j)z log α
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.
We first derive an upper bound. We have αv(j)z = 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
|ez1 z2 | = eRe(z1 z2 ) ≤ e|z1 z2 | = e|z1 ||z2 | .
Hence, for any R ∈ R>0 , we have
M (R, fj ) = M (R, z u(j) ev(j)z log α ) ≤ Ru(j) ev(j)R| log α| .
(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 + L1 R| log α|.
26
(18)
Next, applying Lemma 20 to ∆ with r = S(1 + |β|) and R = e2 r we obtain
the inequality
L
Y
−L(L−1)
|∆| ≤ 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 + L1 R| log α|)
or
log |∆| ≤ −L2 + c1 L (L0 log S + L1 S)
(20)
for some absolute constant c1 ∈ R independent of S. If we choose S such
that K ≥ 4c1 , inequality (16) becomes
c1 L (L0 log S + L1 S) ≤
L2
.
2
Combining this inequality with (20) we obtain
log |∆| ≤ −
L2
,
2
(21)
which specifies an upper bound for log |∆|.
Next, we derive a lower bound for log |∆| under the assumption that ∆ 6= 0.
Fix T ∈ Z>0 such that T α, T β and T αβ are algebraic integers. Then
T L0 +2L1 S has the property that T L0 +2L1 S times any element of M, and
hence T L0 +2L1 S times any element of the matrix describing ∆ is an algebraic
integer. Hence, T L(L0 +2L1 S) ∆ is an algebraic integer in Q α, β, αβ . Thus
T L(L0 +2L1 S) ∆ 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
L1 LS
L0 L
∆ ≤ L!S L0 L 1 + β
.
(1 + α )L1 LS 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
|∆| ≥ T −N L(L0 +2L1 S) ∆
1−N
≥ T −N L(L0 +2L1 S) ∆
−N
.
Combining these two inequalities we obtain
|∆| ≥ T −N L(L0 +2L1 S) (L!)−N S −N L0 L 1 + β
27
−N L0 L
−N L1 LS
(1 + α )−N L1 LS 1 + αβ
.
Since N log L! ≤ N L log L, we obtain, after taking the log on both sides, the
inequality
log |∆| ≥ − N L(L0 + 2L1 S) log T − N L log L − N L0 L log S
− N L0 L log 1 + β − N L1 LS log (1 + α ) − N L1 LS log 1 + αβ .
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 |∆| ≥ −c2 L (L0 + log L + L0 log S + L1 S) .
If we choose S sufficiently large we obtain the inequality
log |∆| ≥ −c3 L(L0 log S + L1 S),
(22)
for some absolute constant c3 ∈ R independent of c. Furthermore, if we
choose S such that K ≥ 6c3 , inequality (16) becomes
c3 L (L0 log S + L1 S) ≤
L2
.
3
Combining this inequality with (22) we obtain
log |∆| ≥ −
L2
,
3
(23)
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
bj fj (yi ) = 0,
for i ∈ 1, 2, . . . , (2S − 1)2 .
(24)
j=1
u(j) v(j)yi
α
,
Since fj (yi ) = yi
L
X
identity (24) is equal to
u(j) v(j)yi
bj yi
α
= 0,
for i ∈ 1, 2, . . . , (2S − 1)2 .
(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
b(L0 +1)v+u+1 yi αvyi = 0, for i ∈ 1, 2, . . . , (2S − 1)2 .
(26)
v=0
u=0
28
Choosing
av (t) =
L0
X
b(L0 +1)v+u+1 tu ,
wv = v log α,
and t = yi = s1 (i) + s2 (i)β,
u=0
we can write the left-hand side of (26) as
!
L0
L1
L1
X
X
X
u
b(L0 +1)v+u+1 yi αvyi =
av (t)ewv t .
v=0
u=0
v=0
P 1
wv t . Note that this sum
Each of the L values of yi is a zero of L
v=0 av (t)e
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
distinct zeros. Since L − 1 < L ≤ (2S − 1)2 , two of the yi must be the same,
and we have
s1 (i) + s2 (i)β = s1 (i0 ) + s2 (i0 )β
for some i, i0 with 1 ≤ i < i0 ≤ (2S − 1)2 .
However, since the pairs (s1 (i), s2 (i)) and (s1 (i0 ), s2 (i0 )) are distinct, it
follows that
s1 (i0 ) − s1 (i)
β=
∈ Q.
s2 (i) − s2 (i0 )
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
f (z) =
ak (z)ewk z ,
k=1
and put
E =n+
n
X
dk ,
and
k=1
29
m = max |wk |.
k
Furthermore, let R, s, t ∈ R>0 , s > 1, and let y ∈ C. Then
st + s + t
1
1
(E − 1) log
.
N (R, y, f ) ≤
+ (st + s + 2t)Rm +
log s
t
s
(27)
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
N (R, f ) ≤
1
M ((st + s + t)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)
D((st + s)R, w) ⊂ D((st + s + t)R).
(29)
and
By Jensen’s formula (Greene, Krantz 2006, p. 279) we have
sR
Z
0
We also have
Z sR
0
N (r, w, f )
1
dr =
log f w + sReiθ dθ − log |f (w)|.
r
2π
N (r, w, f )
dr ≥
r
Z
sR
R
N (R, w, f )
dr = N (R, w, f ) log s.
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 |
M sR(1 + t), w, log
N (R, f ) ≤ N ((1 + t)R, w, f ) ≤
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
f (z) =
ak (z)ewk z ,
k=1
and put
E =n+
n
X
dk ,
and
k=1
m = max |wk |.
k
Furthermore, let R, γ ∈ R>0 , γ > 1. Then
M (γR, f ) ≤
γ E − 1 Rm(γ+1)
e
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
γ E − 1 tRm(γ+1)
M (γtR, f ) ≤
e
M (tR, f ).
γ−1
Taking γ = (st + s + t)/t we have
t
st + s + t E
1
st + s + t E−1
γE − 1
≤
≤ 1+
.
γ−1
st + s
t
s
t
Combining the previous two inequalities we obtain
st + s + t E−1 (st+s+2t)Rm
1
e
M (tR, f ).
M ((st + s + t)R, f ) ≤ 1 +
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
21, we find
that each of the (2S − 1)2 values of yi is a zero of
PL1at the end
w
z
f (z) = v=0 av (z)e v . Let E, R and m be as defined in Proposition 22,
and let 0 be the center of the disc D(R). Taking s = 5 and t = 15 in
inequality (27), and using that
log 31
< 2.2
log 5
and
31
32
< 3.9,
5 log 5
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
v∈{0,...,L1 }
|wv | =
max
v∈{0,...,L1 }
|v log α| = L1 | log α|.
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
4S 2
S
2
+
1 + β | log α|
N (f ) ≤ 3 S + S log S +
log S
log S
< 4S 2 − 4S + 1 = (2S − 1)2 .
Hence, at least two of the yi must be the same, which completes the proof. References
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33