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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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    The [continued fraction](http://en.wikipedia.org/wiki/$C$ontinued_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).2012-12-13

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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}. 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

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).

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    I edited my answer to include a proof of the first fact I used2012-12-13
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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.