Difference between revisions of "Rational approximation"

m (in trivial theorem, pq --> p/q)
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The main theme of this article is the question how well a given [[real number]] <math>x</math> can be approximated by [[rational number]]s. Of course, since the rationals are dense on the real line, we, surely, can make the difference between <math>x</math> and its rational approximation <math>\frac pq</math> as small as we wish. The problem is that, as we try to make <math>\frac pq</math> closer and closer to <math>x</math>, we may have to use larger and larger <math>p</math> and <math>q</math>. So, the reasonable question to ask here is how well can we approximate <math>x</math> by rationals with not too large denominators.
 
The main theme of this article is the question how well a given [[real number]] <math>x</math> can be approximated by [[rational number]]s. Of course, since the rationals are dense on the real line, we, surely, can make the difference between <math>x</math> and its rational approximation <math>\frac pq</math> as small as we wish. The problem is that, as we try to make <math>\frac pq</math> closer and closer to <math>x</math>, we may have to use larger and larger <math>p</math> and <math>q</math>. So, the reasonable question to ask here is how well can we approximate <math>x</math> by rationals with not too large denominators.
 
==Trivial theorem==
 
==Trivial theorem==
Every real number <math>x</math> can be approximated by a rational number <math>pq</math> with a given denominator <math>q\ge 1</math> with an error not exceeding <math>\frac 1{2q}</math>.  
+
Every real number <math>x</math> can be approximated by a rational number <math>\frac{p}{q}</math> with a given denominator <math>q\ge 1</math> with an error not exceeding <math>\frac 1{2q}</math>.  
 
==Proof==
 
==Proof==
 
Note that the closed interval <math>\left[qx-\frac12,qx+\frac12\right]</math> has length <math>1</math> and, therefore, contains at least one integer. Choosing <math>p</math> to be that integer, we immediately get the result.
 
Note that the closed interval <math>\left[qx-\frac12,qx+\frac12\right]</math> has length <math>1</math> and, therefore, contains at least one integer. Choosing <math>p</math> to be that integer, we immediately get the result.

Revision as of 16:52, 23 June 2006

Introduction

The main theme of this article is the question how well a given real number $x$ can be approximated by rational numbers. Of course, since the rationals are dense on the real line, we, surely, can make the difference between $x$ and its rational approximation $\frac pq$ as small as we wish. The problem is that, as we try to make $\frac pq$ closer and closer to $x$, we may have to use larger and larger $p$ and $q$. So, the reasonable question to ask here is how well can we approximate $x$ by rationals with not too large denominators.

Trivial theorem

Every real number $x$ can be approximated by a rational number $\frac{p}{q}$ with a given denominator $q\ge 1$ with an error not exceeding $\frac 1{2q}$.

Proof

Note that the closed interval $\left[qx-\frac12,qx+\frac12\right]$ has length $1$ and, therefore, contains at least one integer. Choosing $p$ to be that integer, we immediately get the result.


So, the interesting question is whether we can get a smaller error of approximation than $\frac 1q$. Surprisingly enough, it is possible, if not for all $q$, then, at least for some of them.

Dirichlet's theorem

Let $n\ge 1$ be any integer. Then there exists a rational number $\frac pq$ such that $1\le q\le n$ and $\left|x-\frac pq\right|<\frac 1{nq}$.

Proof of Dirichlet's theorem

Consider the fractional parts $\{1\cdot x\}, \{2\cdot x\},\dots, \{n\cdot x\}$. They all belong to the half-open interval $[0,1)$. Represent the interval $[0,1)$ as the union of $n$ subintervals $[0,1)=\left[0,\frac 1n\right)\cup\dots\cup\left[\frac{n-1}{n},1\right),$.

to be continued

Liouville Approximation Theorem

We can generalize Dirichlet's theorem as follows: If $\alpha$ is an algebraic number of degree $n$, then there are only finitely many rational numbers $\frac{p}{q}$ satisfying the following inequality: $\Bigg|\alpha-\frac{p}{q}\Bigg| \le\frac{1}{q^n}$. This gives us the following corollary: $\sum_{n=0}^\infty 10^{-n!}$ is a transcendental number, known as Liouville's constant.