Question

prove that root 2 is irrational​

Answer 1

Answer:

$let \: us \: suppose \: that \: \sqrt{2} is \: rational \: number \\ \\ which \: means \: it \: can \: be \: written \: in \: the \: form \: \\ \\ \sqrt{2} = \frac{a}{b} \: where \: a \: and \: b \: are \: coprime \: and \\ \\ \: b \: is \: not \: equal \: to \: 0 \\ \\ so \\ \\ b \sqrt{2} = a \\ \\ {b}^{2} \times 2 = {a}^{2} \: \: (squaring \: both \: side) \\ \\ 2 {b}^{2} = {a}^{2} \\ \\ {b}^{2} = \frac{ {a}^{2} }{2} \\ \\ which \: means \: 2 \: divides \: {a}^{2} \: \\ \\ and \: also \: 2 \: divides \: b \\ \\ because\: if \: a \: prime \: number \\ \\ dvides \: {a}^{2} \: then \: it \: divides \: a \: also \: \\ \\ where \: is \: a \: postive \: integer \\ \\ \: \: then \: it \: means \: 2 \: is \: factor \: of \\ \\ \: a \: which \: is \: contradict \: to \: our \: \\ \\ statement \: that \: a \: and \: b \: are \\ \\ \: coprime \: and \: have \: only \: 1as \: factor. \\ \\ therefore \: \sqrt{2} \: is \: irrational \: number.$

Answer 2

Hey your answer.

Hope it helps...

❤#thank you#❤