elementary number theory - Prove that $sqrt 3$ is irrational

I have to prove that $\sqrt 3$ is irrational.

let us assume that $\sqrt 3$ is rational. This means for some distinct integers $p$ and $q$ having no common factor other than 1,

$$\frac{p}{q} = \sqrt3$$

$$\Rightarrow \frac{p^2}{q^2} = 3$$

$$\Rightarrow p^2 = 3 q^2$$

This means that 3 divides $p^2$. This means that 3 divides $p$ (because every factor must appear twice for the square to exist). So we have, $p = 3 r$ for some integer $r$. Extending the argument to $q$, we discover that they have a common factor of 3, which is a contradiction.

Is this proof correct?