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

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Step 1: Let $\sqrt{3} = \frac{p}{q}$, where $p$ and $q$ are co-prime integers, $q 
eq 0$
Step 2: $3q^2 = p^2 \Rightarrow p^2$ is divisible by 3 $\Rightarrow p$ is divisible by 3. Let $p = 3a$
Step 3: $9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q$ is divisible by 3
Step 4: This contradicts that $p$ and $q$ are co-prime. $	herefore \sqrt{3}$ is irrational.

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