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:


 * $\left \lvert{x - \dfrac p q}\right \rvert \ge \dfrac c {q^n}$

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 $\left \lvert{x - r_i}\right \rvert$.

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

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

Then:

Suppose:


 * $\displaystyle P \left({x}\right) = \sum_{k \mathop = 0}^n a_k x^k$

Since:


 * $\displaystyle x^k - \alpha^k = \left({x - \alpha}\right) \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i$

we have:


 * $\displaystyle P \left({x}\right) - P \left({\alpha}\right) = \left({x - \alpha}\right) \sum_{k \mathop = 1}^n a_k \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i$

Let us suppose that $\left \lvert{x - \alpha}\right \rvert \le 1$.

By the Reverse Triangle Inequality:
 * $\left \lvert{\alpha}\right \rvert \le \left \lvert{x}\right \rvert + 1$

Hence by the Triangle Inequality:


 * $\left \lvert{P \left({x}\right) - P \left({\alpha}\right)}\right \rvert \le \left \lvert{x - \alpha}\right \rvert c_x$

where:


 * $\displaystyle c_x = \sum_{k \mathop = 1}^n \left \lvert{a_k}\right \rvert k \left({\left \lvert{x}\right \rvert + 1}\right)^{k - 1}$

So for such $\alpha$:


 * $\left \lvert{x - \alpha}\right \rvert \ge \dfrac {\left \lvert{P \left({x}\right) - P \left({\alpha}\right)}\right \rvert} {c_{x}} \ge \dfrac 1 {c_{x} q^n}$

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

If $\left \lvert{x - \alpha}\right \rvert \ge 1$, then:L
 * $\left \lvert{x - \alpha}\right \rvert \ge \dfrac 1 {q^n}$

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

Then:
 * $\left \lvert{x - \dfrac p q}\right \rvert \ge \dfrac c {q^n}$

for all $\dfrac p q$.

Also see

 * Definition:Liouville Number