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

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

\[\begin{aligned} & \therefore \sqrt{3}=\frac{\mathrm{p}}{\mathrm{q}}, \text { where } \mathrm{q} \neq 0 \text { and let } \mathrm{p} \& \mathrm{q} \text { be coprimes. } \\ & \Rightarrow 3 \mathrm{q}^{2}=\mathrm{p}^{2} \end{aligned} \]

\(\displaystyle \Rightarrow \mathrm{p}^{2}\) is divisible by 3.

\(\displaystyle \Rightarrow \mathrm{p}\) is divisible by 3. ----- ($\displaystyle 1$)

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

\[\begin{aligned} & \therefore 9 a^{2}=3 q^{2} \\ & \Rightarrow q^{2}=3 a^{2} \end{aligned} \]

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

\(\displaystyle \Rightarrow \mathrm{q}\) is divisible by $\displaystyle 3$ ----- ($\displaystyle 2$)

\(\displaystyle \therefore 3\) divides both p & q.

and ($\displaystyle 2$) leads to contradiction as p and q are coprimes.

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