1.9. Transcendental numbers.

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 A complex number is said to be algebraic or transcendental

according as it is algebraic or transcendental over Q. First some history:

1844: Liouville showed that certain numbers (now called Liouville numbers) are transcendental.

1873: Hermite showed that e is transcendental.

1873: Cantor showed that the set of algebraic numbers is countable, but that R is not

countable. [Thus almost all numbers are transcendental, but it is usually very difficult to

prove that a particular number is transcendental.]

1882: Lindemann showed that π is transcendental.

1934: Gelfond-Schneider showed that if α and β are algebraic, α _= 0, 1, and β /∈ Q, then

αβ is transcendental. (This was one of Hilbert’s famous problems)

1994: Euler’s constant

γ = lim

n→∞

(

_n

k=1

1/k − log n)

has not yet been proven to be transcendental.

1994: The numbers e + π and e − π are surely transcendental, but they have not even

been proved to be irrational!

Proposition 1.22. The set of algebraic numbers is countable.

Proof. Define the height h(r) of a rational number to be max(|m|, |n|), where r = m/n

is the expression of r in its lowest terms. There are only finitely many rational numbers

with height less than a fixed number N. Let A(N) be the set of algebraic numbers whose

minimum equation over Q is of degree ≤ N and has coefficients of height < N. Then A(N)

is finite for each N. Count the elements of A(10);t hen count the elements of A(100);t hen

count the elements of A(1000), and so on.

A typical Liouville number is

_∞

n=0

1

10n!—in its decimal expansion there are increasingly

long strings of zeros. We prove that the analogue of this number in base 2 is transcendental.

FIELDS AND GALOIS THEORY 9

Theorem 1.23. The number α =

_ 1

2n! is transcendental.

Proof. Suppose not, and let

f(X) = Xd + a1Xd−1 + ・ ・ ・ + ad, ai ∈ Q,

be the minimum polynomial of α over Q. Thus [Q[α] : Q] = d. Let

f(X) =

_d

i=1

(X − αi), αi ∈ C, α1 = α,

and choose a nonzero integer D such that Df(X) ∈ Z[X]. Let ΣN =

_N

n=0

1

2n!, so that

ΣN → α as N →∞, and let xN = f(ΣN ).

Because f(X) is irreducible in Q[X], it has no rational root, except possibly α;but ΣN _= α,

and so xN _= 0. (In fact α is obviously nonrational because its expansion to base 2 is not

periodic.)

Clearly xN ∈ Q;in fact (2N!)dDxN ∈ Z, and so

|(2N!)dDxN| ≥ 1.

On the other hand,

|xN| =

_

|ΣN − αi| ≤ |α1 − ΣN|(M +ΣN)d−1, where M = max

i_=1

|αi|,

and

|α1 − ΣN| =

_∞

n=N+1

1

2n!

≤ 2

2(N+1)!

Hence

|(2N!)dDxN| ≤ 2 ・ 2d·N!D

2(N+1)!

・ (M +ΣN)d−1 →0 asN →∞

because 2d·N!

2(N+1)! =

_

2d

2N+1

            N!

→ 0. We have a contradiction.

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