Possible Duplicate:
How can you prove that the square root of two is irrational?
How to prove $\sqrt3$ is irrational using Fermat's infinite descent method?
Like says in Carl Benjamim Boyer's book.
Isnt the same prove to $\sqrt2$, in Boyer's book says something like this.
$\sqrt3=a1/b1$
$1/(\sqrt3-1)=(\sqrt3+1)/2$
$\sqrt3=(3b1-a1)/(a1-b1)$