How to prove $\sqrt3$ is irrational?

Question

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.

$\sqrt3=a1/b1$

$1/(\sqrt3-1)=(\sqrt3+1)/2$

$\sqrt3=(3b1-a1)/(a1-b1)$

Answer 1

Because the greatest power of $3$ that divides $p^2$ must be even, whereas the greatest power of $3$ that divides $3q^2$ must be odd. So $(p/q)^2$ can't equal $3$.