Question

Prove That 7√5 Is Irrational

Answer 1

Let us assume that $7 \sqrt{5}$is a rational number

now

$7 \sqrt{5} = \frac{p}{q} \\ \\ [in \: \: which \: p \: and \: q \: are \: intgers \\ and \: q \: is \: not \: equal \: to \: 0 \\ also \: \: HCF(p \:,\: q) = 1]\\ \\ = > \sqrt{5} = \frac{p}{7q} \\ \\ now \: p ,\: q \: \: and \: 7 \: are \: \: integers \: \: and \\ \: \: q \: \: is \: \: not \: \: equal \:to \:0 \\ \\ so \\ \\ \frac{p}{7q} \: is \: rational \: number \\ \\ but \: \sqrt{5} \: is \: an \: irrational \: number \\ \\ \: so \: \\ \\ \sqrt{5} \: not \: \: equal \: \: to \: \frac{p}{7q} \:$

so

it arise due to our wrong assumption

therefore

$7 \sqrt{5} \:$is an irrational number

Hence Proved