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Prove that $\sqrt{2}$ is an irrational number.
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Let $\sqrt{2}$ be a rational number.
$\therefore \sqrt{2} = \frac{p}{q}$, where $q \neq 0$ and let $p$ & $q$ be co-primes.
$2q^2 = p^2 \Rightarrow p^2$ is divisible by $2 \Rightarrow p$ is divisible by $2$ -----(i)
$\Rightarrow p = 2a$, where 'a' is some integer
$4a^2 = 2 q^2 \Rightarrow q^2 = 2 a^2 \Rightarrow q^2$ is divisible by $2 \Rightarrow q$ is divisible by $2$ -----(ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{2}$ is an irrational number.
$\therefore \sqrt{2} = \frac{p}{q}$, where $q \neq 0$ and let $p$ & $q$ be co-primes.
$2q^2 = p^2 \Rightarrow p^2$ is divisible by $2 \Rightarrow p$ is divisible by $2$ -----(i)
$\Rightarrow p = 2a$, where 'a' is some integer
$4a^2 = 2 q^2 \Rightarrow q^2 = 2 a^2 \Rightarrow q^2$ is divisible by $2 \Rightarrow q$ is divisible by $2$ -----(ii)
(i) and (ii) leads to contradiction as 'p' and 'q' are co-primes.
$\therefore \sqrt{2}$ is an irrational number.