Prove that $\sqrt{5}$ is an irrational number.

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Step 1: Let $\sqrt{5}$ be a rational number. Then $\sqrt{5} = \frac{p}{q}$, where $q 
eq 0$ and $p$, $q$ are co-prime.
Step 2: $5q^2 = p^2 \Rightarrow p^2$ is divisible by $5 \Rightarrow p$ is divisible by $5$ ... (i). Let $p = 5a$.
Step 3: $25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q$ is divisible by $5$ ... (ii)
Step 4: (i) and (ii) lead to a contradiction as $p$ and $q$ are co-prime. $	herefore \sqrt{5}$ is an irrational number.

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