I saw a proof of Liouville's theorem on ProofWiki. It goes like this:
Let $x$ be a Liouville number. Also suppose that it is an algebraic number.
Liouville numbers are irrational, so there exists $c > 0$ and $n \in \mathbb{N}_{>0}$ such that:
$$\left | x – \frac{p}{q} \right | \geq \frac{c}{q^{n}}$$
for every pair $p, q \in \mathbb{Z}$ with $q \neq 0$. Let $r \in \mathbb{N}_{>0}$ such that $2^{r} \geq \frac{1}{c}$. Since $x$ is a Liouville number, there exists $p, q \in \mathbb{Z}$ with $q > 1$ such that:
$$ \left | x – \frac{p}{q} \right | < \frac{1}{q^{n+r}} \leq \frac{1}{2^{r} q^{n}} \leq \frac{c}{q^{n}}$$
which is a contradiction. Thus $x$ is transcendental. $\blacksquare$
This proof has 2 problems: this is not a contradiction (case where $\left | x – \frac{p}{q}\right | = \frac{c}{q^{n}}$) and there is no use of that $x$ is algebraic. The first problem could be solved like this:
$$\left | x – \frac{p}{q} \right | = \frac{c}{q^{n}} \Rightarrow \left | x q^{n} – p q^{n-1} \right | = c$$
which is false for sufficiently small $c$ (because $x$, $p$, $q$ are not said to be dependent on $c$). But the second problem still is present. How this "proof" could be correct? Is it just plain wrong?
Best Answer
If $x$ is some algebraic number, then there is come $c>0$ and $n\in\mathbb N$ such that
$$ \left|x-\frac pq\right|\ge{c\over q^n}.\tag1 $$
for all coprime integers $p,q$ such that $x\ne p/q$.
However, it follows from the definition of Liouville number that for all $m\in\mathbb N$ there exists infinitely many pairs of coprime $p,q$ such that
$$ 0<\left|x-\frac pq\right|<{1\over q^m}. $$
If $m$ is large enough such that $q^{-m}<cq^{-n}$. Then, we see that (1) is violated, which indicates that $x$ cannot be algebraic.
Proof of (1)
Since $x$ is algebraic, it must be a root of some polynomial $P(t)$ with integer coefficients:
$$ P(t)=a_nt^n+a_{n-1}t^{n-1}+\dots+a_0. $$
Because $p$ has finitely many roots, we can find some $k>0$ such that $p(t)\ne0$ whenever $0<|t-x|<k$. Let $M$ be the maximum of $P'(t)$ in $[x-k,x+k]$, so it follows from the mean value theorem that $\exists c\in(x-k,x+k)$ such that
$$ |P(t)|=|t-x|\cdot|p'(c)|\le M|t-x|\tag2 $$
When $t=p/q$ and $0<|x-t|<k$. By the properties of integers, we know
$$ q^n|P(t)|=|a_np^n+a_{n-1}p^{n-1}q+\dots+a_0q^n|\ge1.\tag3 $$
Combining (2) and (3), we get
$$ \left|x-\frac pq\right|\ge{|P(p/q)\over M}\ge{M^{-1}\over q^n}. $$