Liouville Numbers are Irrational

Theorem
Liouville Numbers are irrational.

Proof
Let $x$ be a Liouville Number.

Suppose $x = \dfrac a b$ with $a,b \in \Z$ and $b > 0$.

Then:


 * $\left \lvert x - \dfrac p q \right \rvert = \dfrac {\left \lvert a q - b p \right \rvert} {b q} $

By definition of a Liouville Number, $x$ must satisfy:


 * $0 < \left \lvert x - \dfrac p q \right \rvert < \dfrac 1 {q^n}$

If $\left \lvert a q - b p \right \rvert = 0$, this would violate the first inequality.

Let $n$ be a positive integer with $2^{n - 1} > b$.

If $\left \lvert a q - b p \right \rvert \ge 1$, then:

which violates the second inequality.

This is a contradiction. Thus, $x$ is irrational.