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

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

\[\begin{aligned} & \therefore \sqrt{\mathbf{3}}=\frac{\mathbf{p}}{\mathbf{q}}, \text { where } \mathrm{q} \neq 0 \text { and let } \mathrm{p} \& \mathrm{q} \text { be co-primes. } \\ & 3 \mathrm{q}^{2}=\mathrm{p}^{2} \Rightarrow \mathrm{p}^{2} \text { is divisible by } 3 \Rightarrow \mathrm{p} \text { is divisible by } 3 \end{aligned} \]

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

\(\displaystyle 9 \mathrm{a}^{2}=3 \mathrm{q}^{2} \Rightarrow \mathrm{q}^{2}=3 \mathrm{a}^{2} \Rightarrow \mathrm{q}^{2}\) is divisible by \(\displaystyle 3 \Rightarrow \mathrm{q}\) is divisible by $\displaystyle 3$

\(\displaystyle \Rightarrow \mathrm{q}=3 \mathrm{~b}\), where 'b' is some integer

and (ii) leads to contradiction as 'p' and 'q' are co-primes.

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