Liouville's Theorem (Number Theory)

Theorem
Let $x$ be an irrational number that is algebraic of degree $n$.

Then there exists a constant $c > 0$ (which can depend on $x$) such that:


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

for every pair $p, q \in \Z$ with $q \ne 0$.

Proof
Let $r_1, r_2, \ldots, r_k$ be the rational roots of a polynomial $P$ of degree $n$ that has $x$ as a root.

Since $x$ is irrational, it does not equal any $r_i$.

Let $c_1 > 0$ be the minimum of $\size {x - r_i}$.

If there are no $r_i$, let $c_1 = 1$.

Now let $\alpha = \dfrac p q$ where $\alpha \notin \set {r_1, \ldots, r_k}$.

Then:

Suppose:


 * $\ds \map P x = \sum_{k \mathop = 0}^n a_k x^k$

Then:

Case 1: Let $\size {x - \alpha} \le 1$.

Then:

Therefore:

To summarize:
 * $\size {\map P x - \map P \alpha} \le \size {x - \alpha} c_x$

where:


 * $\ds c_x = \sum_{k \mathop = 1}^n \size {a_k} \paren {\paren {\size x + 1}^k - \size x^k}$

So for such $\alpha$:


 * $\size {x - \alpha} \ge \dfrac {\size {\map P x - \map P \alpha} } {c_x} \ge \dfrac 1 {c_x q^n}$

Case 2: Let $\size {x - \alpha} > 1$.

Then:
 * $\size {x - \alpha} > 1 \ge \dfrac 1 {q^n}$

Also see

 * Definition:Liouville Number