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

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Step 1: Let $\sqrt{5}$ be a rational number. $\sqrt{5} = \frac{a}{b}$, $b 
eq 0$, HCF$(a, b) = 1$
Step 2: $\Rightarrow 5 = \frac{a^2}{b^2} \Rightarrow a^2 = 5b^2$. $	herefore 5$ divides $a$
Step 3: Put $a = 5c$ (for some integer $c$). $\Rightarrow 25c^2 = 5b^2 \Rightarrow b^2 = 5c^2$. Then we get, 5 divides $b$
Step 4: Contradiction arises as HCF$(a, b) = 1$. $	herefore$ Our assumption is wrong. $	herefore \sqrt{5}$ is irrational number

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