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

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Step 1: Let $\sqrt{5}$ be a rational number.
$	herefore \sqrt{5} = \frac{p}{q}$, where $q 
eq 0$ and let $p$ & $q$ be co-primes.
$5q^2 = p^2 \Rightarrow p^2$ is divisible by 5 $\Rightarrow p$ is divisible by 5
Step 2: $\Rightarrow p = 5a$, where '$a$' is some integer ... (i)
$25a^2 = 5q^2 \Rightarrow q^2 = 5a^2 \Rightarrow q^2$ is divisible by 5 $\Rightarrow q$ is divisible by 5
Step 3: $\Rightarrow q = 5b$, where '$b$' is some integer ... (ii)
(i) and (ii) leads to contradiction as '$p$' and '$q$' are co-primes.
$	herefore \sqrt{5}$ is an irrational number.

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