Prove that $\sqrt{3}$ is an irrational number.

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Step 1: Let $\sqrt{3}$ be a rational number. Then $\sqrt{3} = \frac{p}{q}$, where $p$ and $q$ are co-primes and $q 
eq 0$.
Step 2: $3q^2 = p^2 \Rightarrow p^2$ is divisible by $3 \Rightarrow p$ is divisible by $3$
Step 3: $\Rightarrow p = 3a$, where $a$ is some integer ... (i)
Step 4: $9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3 \Rightarrow q$ is divisible by $3$
$\Rightarrow q = 3b$, where $b$ is some integer ... (ii)
Step 5: (i) and (ii) lead to contradiction as $p$ and $q$ are co-primes.
Step 6: $	herefore \sqrt{3}$ is an irrational number.

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