Question

show that 4root11 is an irrational

Answer 1

Here is your answer.

Answer 2

$let \: us \: assume \: to \: the \: contrary \: that \: 4 \sqrt{11} is \: a \: rational \:number \\ 4 \sqrt{11} = \frac{a}{b} \\ \sqrt{11} = \frac{a}{4b} \\ \frac{a}{4b} is \: a \: rational \: number \\ it \: gives \: \sqrt{11} \: is \: a \: rational \\ but \: it \: contradicts \: the \: fact \: that \: \sqrt{11} is \: an \: irrational \: number \\ this \: assumption \: has \: arisen \: due \: to \: our \: incorrect \: assumption \: that \: 4 \sqrt{11} \: is \: rational \: number \\ hence4 \sqrt{11} is \: an \: irrational \: number$