Prove that √5 is an irrational number.

Let $\sqrt{5}$ be a rational number. $\therefore \sqrt{5} = \frac{p}{q}$, where $q \neq 0$ and let $p$ & $q$ be the coprimes ($\frac{1}{2}$ mark). $\Rightarrow 5q^2 = p^2 \Rightarrow p^2$ is divisible by 5 $\Rightarrow p$ is divisible by 5 (1 mark). Let $p = 5a$, where 'a' is some integer $\therefore 25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by 5 $\Rightarrow q$ is divisible by 5 (1 mark). $\therefore 5$ divides both $p$ & $q$. ① and ② leads to contradiction as $p$ and $q$ are coprimes. Hence, $\sqrt{5}$ is an irrational number ($\frac{1}{2}$ mark).