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