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.
$\blacksquare$