Question

Prove that
$\sqrt{3}$
is an irrational number.​

Answer 1

Answer:

$if \: \sqrt{3 } \: is \: irrational \: so \: it \: cn \: written \: in \: the \: form \: of \: \frac{a}{b} \\ \\ \sqrt{3 } = \frac{a}{b} \\ squring \: both \: side \\ 3 = \frac{ {a}^{2} }{ {b}^{2} } \\ {a}^{2} = 3{b}^{2} ....1\\ let \: a = 3c \: put \: in1 \\ 9 {c}^{2} = 3 {b}^{2} \\ hence \: both \: and \: b \: have \: common \: facter \: 3 \\ \\ but \: contraticts \: fact \: that \: a \: and \: b \: no \: common \: facter \: other \: than \: 1 \\ hence \: \sqrt{3} \: is \: irrational.$