is it possible to find the closest rational number to an irrational number? given an irrational number is it possible to find the closest rational number to the irrational number? If so, how?
 A: 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 $x<r$).
A: 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.
