Liouville Numbers are Irrational

Proof

Let $x$ be a Liouville number.

Aiming for a contradiction, suppose $x$ were rational, that is:

$x = \dfrac a b$

with $a, b \in \Z$ and $b > 0$.

By definition of a Liouville number, for all $n \in \N$, there exist $p,q \in \Z$ (which may depend on $n$) with $q > 1$ such that:

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

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

Let $p$ and $q$ be any integers with $q > 1$.

We have:

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

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

If $\left \lvert a q - b p \right \rvert \ne 0$, then:

 $\displaystyle \left \lvert x - \frac p q \right \rvert$ $\ge$ $\displaystyle \frac 1 {b q}$ as $\left \lvert a q - b p \right \rvert$ is a positive integer $\displaystyle$ $>$ $\displaystyle \frac 1 {2^{n - 1} q}$ by our choice of $n$ $\displaystyle$ $\ge$ $\displaystyle \frac 1 {q^n}$ as $q > 1$ by definition

which would violate the second inequality.

Therefore, in any case, if $n$ is sufficiently large, there cannot exist integers $p$ and $q$ with $q > 1$ satisfying the two inequalities.

Thus $x$ is irrational.
$\blacksquare$