Liouville's Theorem (Number Theory)/Corollary

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Corollary to Liouville's Theorem

Liouville numbers are transcendental.


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:

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

which is a contradiction.

Thus $x$ is transcendental.

$\blacksquare$


Sources