Question

Show that 3√7 is an irrational number.​

Answer 1

Answer:

let us suppose that 3√7 is a rational number

so 3√7= a/b which are coprime number

√7=a/3b

√7=intergers/3×integers

√7 is a rational number

which contradicts our Statment so 3√7 is a irrational number

Answer 2

Step-by-step explanation:

$let \: us \: assume \: that \: 3 \sqrt{7} \: \: is \: not \: irrational \: \\ therefore \: 3 \sqrt{7} \: \: rational \: . \\ now \\ \: 3 \sqrt{7} = \frac{a}{b} \: where \: b \: is \: not \: equal \: to \: zero \: \: and \: a \: and \: b \: are \: co \: prime \: \\ 3 \sqrt{7} = \frac{a}{b} \: \\ \sqrt{7} = \frac{a}{b} - 3 \\ \sqrt{7} = \frac{a - 3b}{3b} \: \\ here \: \: a \: \: b \: and \: 3 \: are \: integers \: . \\ therefore \: \sqrt{7} \: \: is \: \:rational. \\ but \: \: we \: know \: that \: \\ \sqrt{7} \: is \: irrational \: . \\ this \: contradiction \: arisen \: because \: of \: \\ our \: wrong \: assumption \: that \: 3 \sqrt{7} \: is \: rational .\: \\ therefore \: 3 \sqrt{7} \: \: is \: \: irrational \: \\ hence \: proved \:$