Prove that √ 5 is an irrational number.

Let \(\displaystyle \sqrt{\mathbf{5}}\) be a rational number. \[\begin{aligned} & \therefore \sqrt{\mathbf{5}}=\frac{\mathbf{p}}{\mathbf{q}}, \text { where } q \neq 0 \text { and let } p \& q \text { be co-primes. } \\ & 5 q^{2}=p^{2} \Longrightarrow p^{2} \text { is divisible by } 5 \Rightarrow p \text { is divisible by } 5 \\ & \Longrightarrow p=5 a, \text { where ' } a \text { ' is some integer } \\ & 25 a^{2}=5 q^{2} \Longrightarrow q^{2}=5 a^{2} \Longrightarrow q^{2} \text { is divisible by } 5 \Rightarrow q \text { is divisible by } 5 \\ & \Rightarrow q=5 b, \text { where ' } b \text { ' is some integer } \end{aligned} \] (i) and (ii) leads to contradiction as 'p' and 'q' are co-primes. \(\displaystyle \therefore \sqrt{\mathbf{5}}\) is an irrational number.