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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 $$\begin{aligned}& 3 \dots (i) \\ & \Rightarrow p = 3a\end{aligned}$$, where '$a$' is some integer
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3 \Rightarrow q$ is divisible by $$\begin{aligned}& 3 \dots (ii) \\ & (i)\end{aligned}$$ 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 $$\begin{aligned}& 3 \dots (i) \\ & \Rightarrow p = 3a\end{aligned}$$, where '$a$' is some integer
$9a^2 = 3q^2 \Rightarrow q^2 = 3a^2 \Rightarrow q^2$ is divisible by $3 \Rightarrow q$ is divisible by $$\begin{aligned}& 3 \dots (ii) \\ & (i)\end{aligned}$$ and $(ii)$ leads to contradiction as 'p' and 'q' are coprime.
$\therefore \sqrt{3}$ is an irrational number.