Prove that $\displaystyle \sqrt{2}$ is an irrational number.

Let \(\displaystyle \sqrt{2}\) be a rational number.

\[\begin{aligned} & \therefore \sqrt{2}=\frac{p}{q}, \text { where } q \neq 0 \text { and } p \& q \text { are coprime. } \\ & 2 q^{2}=p^{2} \Rightarrow p^{2} \text { is divisible by } 2 \Rightarrow p \text { is divisible by } 2 \text {-----(i) } \end{aligned} \]

Let \(\displaystyle \mathrm{p}=2 \mathrm{a}\), where 'a' is some integer

\[4 a^{2}=2 q^{2} \Rightarrow q^{2}=2 a^{2} \Rightarrow q^{2} \text { is divisible by } 2 \Rightarrow q \text { is divisible by } 2 \text {----- (ii) } \]

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

\(\displaystyle \therefore \sqrt{2}\) is an irrational number.