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Liouville's theorem (On diophantine approximation)
12-23-2010, 04:45 PM
Post: #1
Liouville's theorem (On diophantine approximation)
Theorem: If $\alpha$ is an irrational number which is the root of a polynomial $P$ of degree $n > 0$
with integer coefficients, then $\exists \;A \in \mathbb{R}, \; (A>0)$ such that $\displaystyle{\left|\alpha - \frac{p}{q}\right | > A/q^n \quad\forall p,q \in \mathbb{Z}, \; (q > 0)}$


Proof: Let $M = \displaystyle{\max_{[\alpha -1, \alpha +1]} |P'|}$ and take $A \in (0, \min(1,1/M,|\alpha - \alpha_1| , \cdots, |\alpha-\alpha_m|))$
where $\alpha_1, \cdots, \alpha_m$ are distinct roots of $P$ which differ from $\alpha$.

Assume the claim is not true: $\displaystyle{\left|\alpha - \frac{p}{q}\right| \le A/q^n < \min(1,1/M,|\alpha - \alpha_1| , \cdots, |\alpha-\alpha_m|)}$ ,
for some integers $p$ and $q \ (>0)$, then $\displaystyle{\frac{p}{q} \in [\alpha-1,\alpha+1]}$ and $\displaystyle{\frac{p}{q} \not\in \{ \alpha_1,\cdots,\alpha_m\}}$
and so $\displaystyle{\frac{p}{q}}$ is not a root of $P$ and there is no root between $\alpha$ and $\displaystyle{\frac{p}{q}}$.

By the mean value theorem, there is a $\theta$ between $\alpha$ and $p/q$ such that
$\displaystyle{0 \ne -P\left(\frac{p}{q}\right) = P(\alpha)-P\left(\frac{p}{q}\right) = \left(\alpha-\frac{p}{q}\right)P'(\theta)}$ hence
$\displaystyle{\left|\alpha - \frac{p}{q}\right| = \frac{|P(\frac{p}{q})|}{|P'(\theta)|}>A\left|P\left(\frac{p}{q}\right)\right|=A\left|\sum_{i=0}^n c_i p^i q^{-i}\right| = \frac{A|\sum_{i=0}^n c_i p^i q^{n-i}|}{q^n} \ge A/q^n \ge \left|\alpha - \frac{p}{q}\right|}$
A contradiction! and this completed the proof.
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