Liouville's Theorem (Number Theory)/Corollary

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:

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

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:

\(\ds \size {x - \dfrac p q}\)

\(<\)

\(\ds \frac 1 {q^{n + r} }\)

\(\ds \)

\(\le\)

\(\ds \frac 1 {2^r q^n}\)

\(\ds \)

\(\le\)

\(\ds \frac c {q^n}\)

which is a contradiction.

Thus $x$ is transcendental.

$\blacksquare$

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Liouville number
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Liouville number