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 $c_0=c_0(\alpha ,ϵ)>0$ such that for all rational numbers $p/q$, the inequality

  $$\left|\alpha -\frac{p}{q}\right|>c_0{q}^{-1-ϵ-n/2}$$

  holds.

• Siegel: If $\alpha$ is an algebraic number of degree $n\ge 2$, then there is a constant $c_1=c_1(\alpha ,ϵ)>0$ such that for all rational numbers $p/q$, the inequality

  $$\left|\alpha -\frac{p}{q}\right|>c_1{q}^{-\lambda },\lambda =\underset{t=1,\mathrm{\dots },n}{\min }\left(\frac{n}{t+1}+t\right)+ϵ$$

  holds.

• Dyson: If $\alpha$ is an algebraic number of degree $n>3$, then there is a constant $c_2=c_2(\alpha ,ϵ)>0$ such that for all rational numbers $p/q$ with $q>c_2$, the inequality

  $$\left|\alpha -\frac{p}{q}\right|>q^{-\sqrt{2n}-ϵ}$$

  holds.

• Roth: If $\alpha$ is an irrational algebraic number and $ϵ>0$, then there is a constant $c_3=c_3(\alpha ,ϵ)>0$ such that for all rational numbers $p/q$, the inequality

  $$\left|\alpha -\frac{p}{q}\right|>c_3{q}^{-2-ϵ}$$

  holds.

Title	Liouville approximation theorem
Canonical name	LiouvilleApproximationTheorem
Date of creation	2013-03-22 11:45:45
Last modified on	2013-03-22 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 46-01
Classification	msc 46N40
Related topic	ExampleOfTranscendentalNumber