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given an irrational number is it possible to find the closest rational number to the irrational number? If so, how?

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6  
No: if $x$ is any irrational number, and $q$ is any rational number, there is another rational number between $x$ and $q$. –  Brian M. Scott Dec 13 '12 at 9:45
    
On the other hand, Liouville's theorem on irrational algebraic numbers says there is an upper bound on how well you can approximate such numbers by rationals. –  Zhen Lin Dec 13 '12 at 10:12
    
The continued fraction of your irrational number will allow you to get fractions as near as you want from your irrational (in some way the bests possible). –  Raymond Manzoni Dec 13 '12 at 22:35

2 Answers 2

No. It is a fact that in any open interval $]a,b[ $ there exists a rational number.

Proof:

Assume WLOG that $a>0$. Let $n$ be a positive integer such that $\frac{1}{b-a}<n$. Now consider the subset of natural numbers $\{m\in N|a<\frac{m}{n}\}$. By the well ordering principle, we know that this set has a minimum $m_0$. Because of the way $m_0$ was chosen we know that: $$\frac{m_0-1}{n}\leq a<\frac{m_0}{n}$$ Thus: $a<\frac{m_0}{n}\leq a+1/n<a+b-a=b$

Let $x$ be irrational and $r$ be the closest rational number, now get a closer rational from the interval $]r,x[$ (or $]x,r[$ if $x<r$).

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That's what I thought. So is there then an interval on which there aren't any rational numbers around an irrational, and is it possible to find it? –  user1066113 Dec 13 '12 at 9:57
    
I edited my answer to include a proof of the first fact I used –  Amr Dec 13 '12 at 10:10

There is no closest one, as already pointed out. However, you can sometimes estimate how far a rational number is from an irrational one. Take $\sqrt{2}$ for example and a rational number $\frac{p}{q}$. Then

$$ \frac{p^2}{q^2} - 2 = \frac{p^2 - 2q^2}{q^2}. $$

Since the numerator is a non-zero integer this shows

$$ \left|\frac{p}{q} - \sqrt{2}\right| \cdot \left| \frac{p}{q} + \sqrt{2} \right| = \left|\frac{p^2}{q^2} - 2 \right| \geq \frac{1}{q^2} $$

and so if $\left| \frac{p}{q} - \sqrt{2} \right| \leq \varepsilon$ then

$$ \left|\frac{p}{q} - \sqrt{2}\right| \geq \frac{1}{q^2 ( 2\sqrt{2} + \varepsilon)}. $$

See here for a more general discussion.

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