How to prove that $\sqrt 3$ is an irrational number? 
Possible Duplicate:
$a^{1/2}$ is either an integer or an irrational number 

I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?

If you follow through the usual proof for $\sqrt{2}$ substituting $3$ for $2$, it goes through just fine.  Let $\sqrt{3}=\frac{p}{q}, p,q $ relatively prime.  $3=\frac{p^2}{q^2}$, so $3$ divides $p$ and so on.
