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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 let $p \& q$ be coprimes.
$\implies 3q^2 = p^2$
$\implies p^2$ is divisible by 3.
$\implies p$ is divisible by 3. ----- (1)
Let $p = 3a$, where 'a' is some integer
$\therefore 9a^2 = 3q^2$
$\implies q^2 = 3a^2$
$\implies q^2$ is divisible by 3
$\implies q$ is divisible by 3 ----- (2)
$\therefore 3$ divides both $p \& q$.
(1) and (2) leads to contradiction as $p$ and $q$ are coprimes.
Hence, $\sqrt{3}$ is an irrational number.
$\therefore \sqrt{3} = \frac{p}{q}$, where $q \neq 0$ and let $p \& q$ be coprimes.
$\implies 3q^2 = p^2$
$\implies p^2$ is divisible by 3.
$\implies p$ is divisible by 3. ----- (1)
Let $p = 3a$, where 'a' is some integer
$\therefore 9a^2 = 3q^2$
$\implies q^2 = 3a^2$
$\implies q^2$ is divisible by 3
$\implies q$ is divisible by 3 ----- (2)
$\therefore 3$ divides both $p \& q$.
(1) and (2) leads to contradiction as $p$ and $q$ are coprimes.
Hence, $\sqrt{3}$ is an irrational number.