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Prove that $\sqrt{3}$ is an irrational number.
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Let $\sqrt{3}$ be a rational number.
$\therefore \sqrt{3} = \frac{p}{q}$, where $q\neq 0$ and $p \& q$ are coprime.
$3q^2 = p^2 \Rightarrow p^2$ is divisible by $3$
$\Rightarrow p$ is divisible by $3----- (i)$
$\Rightarrow p = 3a$, where 'a' is a positive integer
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3$
$\Rightarrow q$ is divisible by $3 ----- (ii)$
(i) and (ii) leads to contradiction as 'p' and 'q' are coprime.
$\therefore \sqrt{3}$ is an irrational number.
$\therefore \sqrt{3} = \frac{p}{q}$, where $q\neq 0$ and $p \& q$ are coprime.
$3q^2 = p^2 \Rightarrow p^2$ is divisible by $3$
$\Rightarrow p$ is divisible by $3----- (i)$
$\Rightarrow p = 3a$, where 'a' is a positive integer
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3$
$\Rightarrow q$ is divisible by $3 ----- (ii)$
(i) and (ii) leads to contradiction as 'p' and 'q' are coprime.
$\therefore \sqrt{3}$ is an irrational number.