# Liouville's Theorem (Number Theory)/Corollary

## Proof

Let $x$ be a Liouville number.

Aiming for a contradiction, suppose $x$ is an algebraic number.

By Liouville Numbers are Irrational, then there exists $c > 0$ and $n \in \N_{>0}$ such that:

$\left \lvert{x - \dfrac p q}\right \rvert \ge \dfrac c {q^n}$

for every pair $p, q \in \Z$ with $q \ne 0$.

Let $r \in \N_{>0}$ such that $2^r \ge \dfrac 1 c$.

Since $x$ is a Liouville number, there exists $p, q \in \Z$ with $q > 1$ such that:

 $\ds \left \lvert{x - \dfrac p q}\right \rvert$ $<$ $\ds \frac 1 {q^{n + r} }$ $\ds$ $\le$ $\ds \frac 1 {2^r q^n}$ $\ds$ $\le$ $\ds \frac c {q^n}$

Thus $x$ is transcendental.
$\blacksquare$