Liouville approximation theorem
Given $\alpha $, a real algebraic number^{} of degree $n\ne 1$, there is a constant $c=c(\alpha )>0$ such that for all rational numbers^{} $p/q,(p,q)=1$, the inequality
$$\left\alpha \frac{p}{q}\right>\frac{c(\alpha )}{{q}^{n}}$$ 
holds.
Many mathematicians have worked at strengthening this theorem:

•
Thue: If $\alpha $ is an algebraic number of degree $n\ge 3$, then there is a constant ${c}_{0}={c}_{0}(\alpha ,\u03f5)>0$ such that for all rational numbers $p/q$, the inequality
$$\left\alpha \frac{p}{q}\right>{c}_{0}{q}^{1\u03f5n/2}$$ holds.

•
Siegel: If $\alpha $ is an algebraic number of degree $n\ge 2$, then there is a constant ${c}_{1}={c}_{1}(\alpha ,\u03f5)>0$ such that for all rational numbers $p/q$, the inequality
$$\left\alpha \frac{p}{q}\right>{c}_{1}{q}^{\lambda},\lambda =\underset{t=1,\mathrm{\dots},n}{\mathrm{min}}\left(\frac{n}{t+1}+t\right)+\u03f5$$ holds.

•
Dyson: If $\alpha $ is an algebraic number of degree $n>3$, then there is a constant ${c}_{2}={c}_{2}(\alpha ,\u03f5)>0$ such that for all rational numbers $p/q$ with $q>{c}_{2}$, the inequality
$$\left\alpha \frac{p}{q}\right>{q}^{\sqrt{2n}\u03f5}$$ holds.

•
Roth: If $\alpha $ is an irrational algebraic number and $\u03f5>0$, then there is a constant ${c}_{3}={c}_{3}(\alpha ,\u03f5)>0$ such that for all rational numbers $p/q$, the inequality
$$\left\alpha \frac{p}{q}\right>{c}_{3}{q}^{2\u03f5}$$ holds.
Title  Liouville approximation theorem^{} 

Canonical name  LiouvilleApproximationTheorem 
Date of creation  20130322 11:45:45 
Last modified on  20130322 11:45:45 
Owner  KimJ (5) 
Last modified by  KimJ (5) 
Numerical id  13 
Author  KimJ (5) 
Entry type  Theorem 
Classification  msc 11J68 
Classification  msc 4601 
Classification  msc 46N40 
Related topic  ExampleOfTranscendentalNumber 