Question

Prove that √3 is not a rational number.​

Answer 1

Answer:

First assume that root 3 is a rational number

apply the conditions

Then you will prove that root 3 is not a rational number

Answer 2

Answer:

$let \: us \: suppose \: that \: \sqrt{3} is \: rational \: number \: \\ \\ which \:means \: it \: can \: be \: wrtten \: in \: the \: form \: of \\ \\ \: \sqrt{3} = \frac{a}{b} \: where \: a \: and \: b \: are \:co prime \: and \: b \: is \: not = 0 \\ \\ so \\ \\ b \sqrt{3} = a \\ \\ {b}^{2} \times 3 = {a}^{2} \: (by \: squaring) \\ \\ 3 {b}^{2} = {a}^{2} \\ \\ {b}^{2} = \frac{ {a}^{2} }{3} \\ \\ which \: means \: 3 \: divides \: {a}^{2} \: \\ \\ and \: also \: 3 \: divides \: b \\ \\ because \: if \: prime \: number \: |p| \\ \\ divides \: {a}^{2} \: then \: it \: divides \: (a) \: also \\ \\ \: where \: is \: (a )\: postive \: integer .\\ \\ \: \: \: \: \: \: \: then \: \: it \: \: means \: 3 \: is \: factor \: of \: a \: \\ \\ which \: is \: contradict \: to \: our \: statement \\ \\ that \: a \: and \: b \: are \: coprime \: and \: have \\ \\ \: only \: 1 \: as \: factor. \\ \\ therefore \: \sqrt{3} \: is \: not \: rational$