# Liouville Numbers are Irrational

## Theorem

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:

- $0 < \size {x - \dfrac p q} < \dfrac 1 {q^n}$

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

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

We have:

- $\size {x - \dfrac p q} = \dfrac {\size {a q - b p} } {b q}$

If $\size {a q - b p} = 0$, this would violate the first inequality.

If $\size {a q - b p} \ne 0$, then:

\(\ds \size {x - \frac p q}\) | \(\ge\) | \(\ds \frac 1 {b q}\) | as $\size {a q - b p}$ is a positive integer | |||||||||||

\(\ds \) | \(>\) | \(\ds \frac 1 {2^{n - 1} q}\) | by our choice of $n$ | |||||||||||

\(\ds \) | \(\ge\) | \(\ds \frac 1 {q^n}\) | 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.

This is a contradiction.

Thus $x$ is irrational.

$\blacksquare$