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

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

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