Are you serious? Just replicate the proof of the $\sqrt{2}$ posted by someone else in another one of these threads.

This will be a proof by contradiction. Assume that $\sqrt3$ is rational. Observe that $\sqrt3$ is a root of $x^2-3=0.$ If $\sqrt3$ is rational, then by the rational root theorem, we can find that $\sqrt3$ is either 1, -1, 3, or -3. However, we can easily compute that $1^2 = 1 \neq 3$, $(-1)^2 = 1 \neq 3$, $3^2 = 9 \neq 3$, and $(-3)^2 = 9 \neq 3$. Thus, we can conclude that our assumption is false and that $\sqrt3$ is irrational.

Suppose that $\sqrt{3}$ is rational. We have $\sqrt{3} = \frac{p}{q} \Leftrightarrow p^2=3q$ where $\gcd (p,q)=1$. Note that $3|p \Rightarrow p=3r \Rightarrow q^2=3r^2 \Rightarrow 3|q \Rightarrow q=3s.$ Thus, $\gcd (p,q) = \gcd (3r,3s) \geqslant 3 > 1$, a contradiction. $\blacksquare$