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:


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

By the Difference of Two Powers:


 * $\displaystyle x^k - \alpha^k = \paren {x - \alpha} \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i$

Hence, we have:


 * $\displaystyle \map P x - \map P \alpha = \paren {x - \alpha} \sum_{k \mathop = 1}^n a_k \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i$

Let us suppose that $\size {x - \alpha} \le 1$.

By the Reverse Triangle Inequality:
 * $\size \alpha \le \size x + 1$

Hence by the Triangle Inequality:


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

where:


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

So for such $\alpha$:


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

If $\alpha$ is one of the $r_i$, then:
 * $\size {x - \alpha} \ge c_1 \ge \dfrac {c_1} {q^n}$

If $\size {x - \alpha} \ge 1$, then:
 * $\size {x - \alpha} \ge \dfrac 1 {q^n}$

Choose $c = \min \set {1, c_1, \dfrac 1 {c_x} }$.

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

for all $\dfrac p q$.

Also see

 * Definition:Liouville Number