# Difference between revisions of "Rational approximation of famous numbers"

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The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of <math>\pi</math> but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that <math>x</math> cannot be approximated by rationals too well, one needs to find plenty of good but not too good rational approximations of <math>x</math>. | The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of <math>\pi</math> but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that <math>x</math> cannot be approximated by rationals too well, one needs to find plenty of good but not too good rational approximations of <math>x</math>. | ||

==Proof of the Main Theorem== | ==Proof of the Main Theorem== | ||

+ | Choose the least <math>n</math> such that <math>Q^{(\mu-1) n}\ge 2q</math>, i.e., that <math>Q^n\ge (2q)^{1/(\mu-1)}</math>. Note that for such choice of <math>n</math>, we have <math>Q^n< Q(2q)^{1/(\mu-1)}</math>. There are two possible cases: | ||

+ | |||

+ | '''Case 1:''' <math>\frac pq=\frac {P_n}{Q_n}</math>. Then <math>\left|x-\frac pq\right|=\left|x-\frac {P_n}{Q_n}\right|>Q^{-\beta n}>Q^{-\beta}\frac 1{(2q)^{\frac \beta{\mu-1}}}>\frac 1{q^M}</math> if <math>q</math> is large enough. | ||

+ | |||

+ | '''Case 2:''' <math>\frac pq\ne \frac {P_n}{Q_n}</math>. Then <math>\left|x-\frac pq\right|\ge | ||

+ | \left|\frac pq-\frac{P_n}{Q_n}\right|-\left|x-\frac {P_n}{Q_n}\right| | ||

+ | >\frac 1{qQ_n}-Q^{-\mu n}\ge Q^{-n}\left(\frac 1q-\frac 1{Q^{(\mu-1)n}}\right)\ge \frac 1{2qQ^n}\ge | ||

+ | Q^{-1}\frac 1{(2q)^{\frac \mu{\mu-1}}}>\frac 1{q^M}</math> if <math>q</math> is large enough (recall that <math>\mu<\beta</math>). |

## Revision as of 11:43, 26 June 2006

*This article is a stub. Help us out by expanding it.*

## Introduction

The Dirichlet's theorem shows that, for each irrational number , the inequality has infinitely many solutions. On the other hand, sometimes it is useful to know that cannot be approximated by rationals too well, or, more precisely, that is not a Liouvillian number, i.e., that for some power , the inequality holds for all sufficiently large denominators . So, how does one show that a number is not Liouvillian? The answer is given by the following

## Main theorem

Suppose that there exist , and a sequence of rational numbers such that for all , and . Then, for every , the inequality has only finitely many solutions.

The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that cannot be approximated by rationals too well, one needs to find plenty of good but not too good rational approximations of .

## Proof of the Main Theorem

Choose the least such that , i.e., that . Note that for such choice of , we have . There are two possible cases:

**Case 1:** . Then if is large enough.

**Case 2:** . Then if is large enough (recall that ).