Step 1: Let us assume, to the contrary, that $\sqrt{3}$ is RATIONAL.
Then $$\sqrt{3} = \frac{a}{b}$$ where $a$ and $b$ are co-prime integers ($b \neq 0$).
Step 2: Squaring both sides :
$$3 = \frac{a^2}{b^2} \Rightarrow a^2 = 3b^2 \qquad \ldots (i)$$
Step 3: So $3$ divides $a^2$. Since $3$ is prime, $3$ divides $a$.
Step 4: Let $a = 3c$. Substituting in $(i)$ :
$$(3c)^2 = 3b^2 \Rightarrow 9c^2 = 3b^2 \Rightarrow b^2 = 3c^2$$
Step 5: So $3$ divides $b^2$, and hence $3$ divides $b$.
Step 6: Thus $3$ is a common factor of $a$ and $b$, contradicting that they are co-prime.
Step 7: Hence our assumption is wrong and $\sqrt{3}$ is an IRRATIONAL number. Hence proved.