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