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?

The proof is very similar to the irrationality of square root of two. 
Let $\sqrt{3} = \frac a b$, where a and b have no common factors besides $1$
As   $3b^2 = a^2$ so $a^2$ is a multiple of $3$, and hence $a$ should be a multiple of $3$. Let $a = 3k$, then $b^2 = 3k^2$, and $b$ must also be a multiple of three. You will arrive at a contradiction to the earlier assumption that $a$ and $b$ have no common factors.
