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

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