Solution

Liouville approximation theorem states that if an irrational algebraic number

α

has degree

d \geq 2

, then some

C (α) > 0

satisfies

α - \frac{p}{q} > \frac{C ( α )}{q ^{d}}

for every rational

p / q

with

q > 0

The omitted

n = 0

term is rational and does not affect transcendence. Let

ξ = \sum_{n \geq 1} 1 0^{- n^{n}}, ξ_{N} = \sum_{n = 1}^{N} 1 0^{- n^{n}} = \frac{p _{N}}{q _{N}}, q_{N} = 1 0^{N^{N}} .

The decimal expansion has ones at the increasingly separated positions

n^{n}

and zeros elsewhere, so it is not eventually periodic and

ξ

is irrational. Moreover,

0 < ξ - ξ_{N} < 21 0^{- (N + 1)^{N + 1}} .

For every fixed

d

, the exponent

(N + 1)^{N + 1}

eventually exceeds

d N^{N}

by an arbitrarily large amount, so this upper bound is smaller than

C q_{N}^{- d}

. Liouville's theorem therefore rules out every finite algebraic degree, proving that

ξ

is transcendental.

There are only countably many integer polynomials and each has finitely many roots, so the algebraic numbers are countable. Since

R

is uncountable, its complement, the set of transcendental numbers, is uncountable.

Solved by gpt-5.6-sol high.

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