Question

prove that 3√2 is irrational​

Answer 1

Answer:

$let \: u s \: assume \: that \: 3 \sqrt{2} \: is \: \: not \: irrational \\ therefore \: 3 \sqrt{2} is \: rational \: . \\ 3 \sqrt{2} = \frac{a}{b} \: where \: b \: not \: equal \: to \: 0 \: a \: and \: b \: are \: co - prime \\ 3 \sqrt{2} = \frac{a}{b} \\ \sqrt{2} = \frac{a}{3b} \\ here \: a \: b \: 3 \: are \: integers \: \\ therefore \: \frac{a}{3b} is \: a \: rational \: number. \\ therefore \: \sqrt{2} \: is \: rational .\: \\ w.k.t \\ \sqrt{2} \: is \: irrational \: . \\ this \: contradiction \: arisen \: because \: of \: our \: \\ wrong \: assumption \: that \: \sqrt{2} \: is \: rational \\ therefore \: \sqrt{2} is \: irrational \: . \\ therefore \: 3 \sqrt{2} \: is \: \: irrational \: . \\ hence \: proved \:$