Prove that $\displaystyle \sqrt{3}$ is an irrational number.

Let \(\displaystyle \sqrt{3}\) be a rational number.

\(\displaystyle \therefore \sqrt{3}=\frac{\mathbf{p}}{\mathbf{q}}\), where \(\displaystyle \mathrm{q} \neq 0\) and p & q are coprime.

\[3 q^{2}=p^{2} \Rightarrow p^{2} \text { is divisible by } 3 \Rightarrow p \text { is divisible by } 3 \text {----- (i) } \]

Let \(\displaystyle \mathrm{p}=3 \mathrm{a}\), where 'a' is some integer

\[9 a^{2}=3 q^{2} \Rightarrow q^{2}=3 a^{2} \Rightarrow q^{2} \text { is divisible by } 3 \Rightarrow q \text { is divisible by } 3 \text {----- (ii) } \]

and (ii) leads to contradiction as 'p' and 'q' are coprime.

\(\displaystyle \therefore \sqrt{3}\) is an irrational number.