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Prove that $\sqrt{5}$ is an irrational number.
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Sol. Let $\sqrt{5}$ be a rational number.
$\therefore \sqrt{5} = \frac{a}{b}$, where $a$, $b$ are coprime and $b \neq 0$.
$\Rightarrow a^2 = 5b^2 \Rightarrow a^2$ is divisible by $5$.
$\Rightarrow a$ is divisible by $5$. ----- (i)
Let $a = 5m$, where 'm' is any natural number.
$\Rightarrow b^2 = 5m^2 \Rightarrow b^2$ is divisible by $5$.
$\Rightarrow b$ is divisible by $5$. ----- (ii)
From (i) and (ii), $a$ and $b$ have common factors which is contrary to our assumption.
Hence, $\sqrt{5}$ is an irrational number.
$\therefore \sqrt{5} = \frac{a}{b}$, where $a$, $b$ are coprime and $b \neq 0$.
$\Rightarrow a^2 = 5b^2 \Rightarrow a^2$ is divisible by $5$.
$\Rightarrow a$ is divisible by $5$. ----- (i)
Let $a = 5m$, where 'm' is any natural number.
$\Rightarrow b^2 = 5m^2 \Rightarrow b^2$ is divisible by $5$.
$\Rightarrow b$ is divisible by $5$. ----- (ii)
From (i) and (ii), $a$ and $b$ have common factors which is contrary to our assumption.
Hence, $\sqrt{5}$ is an irrational number.