Theorem: Liouville's Theorem (Number Theory)

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:
{{begin-eqn}}
{{eqn | l = \map P \alpha
      | o = \ne
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {\map P \alpha}
      | o = \ge
      | r = \frac 1 {q^n}
      | c = as $\map P \alpha$ is a multiple of $\dfrac 1 {q^n}$
}}
{{eqn | ll= \leadsto
      | l = \size {\map P x - \map P \alpha}
      | o = \ge
      | r = \frac 1 {q^n}
      | c = because $\map P x = 0$
}}
{{end-eqn}}
Suppose:
:$\ds \map P x = \sum_{k \mathop = 0}^n a_k x^k$
Then:
{{begin-eqn}}
{{eqn | l = \map P x - \map P \alpha
      | r = \sum_{k \mathop = 0}^n a_k x^k - \sum_{k \mathop = 0}^n a_k \alpha^k
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^n a_k \paren {x^k - \alpha^k}
      | c =
}}
{{eqn | r = \sum_{k \mathop = 1}^n a_k \paren {x^k - \alpha^k}
      | c = $x^0 - \alpha^0 = 0$
}}
{{eqn | r = \sum_{k \mathop = 1}^n a_k \paren {x - \alpha} \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i
      | c = Difference of Two Powers
}}
{{eqn | r = \paren {x - \alpha} \sum_{k \mathop = 1}^n a_k \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i
      | c = 
}}
{{end-eqn}}
'''Case 1''': Let $\size {x - \alpha} \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \size \alpha - \size x
      | o = \le
      | r = \size {x - \alpha}
      | c = Reverse Triangle Inequality
}}
{{eqn | ll= \leadsto
      | l = \size \alpha - \size x
      | o = \le
      | r = 1
      | c = Transitive
}}
{{eqn | ll= \leadsto
      | l = \size \alpha
      | o = \le
      | r = \size x + 1
      | c = rearranging
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \size {\map P x - \map P \alpha}
      | o = \le
      | r = \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k} \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 - i} \alpha^i}
      | c = Triangle Inequality
}}
{{eqn | o = \le
      | r = \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k} \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 - i} \paren {\size x + 1}^i}
      | c = substituting $\size x + 1$ for $\alpha$
}}
{{eqn | o = \le
      | r = \size {x - \alpha}\sum_{k \mathop = 1}^n \size {a_k} \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1} \paren {\frac {\size x + 1} x}^i}
      | c = moving $ x^{-i}$ over
}}
{{eqn | o = \le
      | r = \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k x^{k - 1} } \sum_{i \mathop = 0}^{k - 1} \size {\paren {1 + \frac 1 x}^i}
      | c = moving $x^{k - 1}$ out of the sum
}}
{{eqn | o = \le
      | r = \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k x^{k - 1} } \frac {\paren {1 + \frac 1 x}^k - 1} {\paren {1 + \frac 1 x} - 1}
      | c = Sum of Geometric Sequence
}}
{{eqn | o = \le
      | r = \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k x^k} \paren {\paren {1 + \frac 1 x}^k - 1}
      | c = 
}}
{{eqn | o = \le
      | r = \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k} \paren {\paren {\size x + 1}^k - \size x^k}
      | c = Distribute
}}
{{end-eqn}}
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 $\alpha$:
:$\size {x - \alpha} \ge \dfrac {\size {\map P x - \map P \alpha} } {c_x} \ge \dfrac 1 {c_x q^n}$
{{qed|lemma}}
'''Case 2''': Let $\size {x - \alpha} > 1$.
Then:
:$\size {x - \alpha} > 1 \ge \dfrac 1 {q^n}$
{{qed}}