Liouville's Theorem (Number Theory)/Corollary

Corollary to Liouville's Theorem
Liouville numbers are transcendental.

Proof
Let $x$ be a Liouville number.

$x$ is an algebraic number.

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


 * $\size {x - \dfrac p q} \ge \dfrac c {q^n}$

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:

which is a contradiction.

Thus $x$ is transcendental.