Question

prove that root 3 is irrational number​

Answer 1

Statement:

$\sf{prove \: that \: \sqrt{3} \: is \: an \: irrational \: number}$

Proof:

$\sf{ \: let \: us \: assume \: the \: opposite \: } \\ \\ \sf{ \sqrt{3} \: is \: an \: rational \: number} \\ \\ \sf{ \: as \: we \: know \: that \: rational \: no. \: \: should \: be \: in \: the \: form \: of \: \frac{p}{q}} \\ \sf{ \: and \: where \: p \: and \: q \: are \: co - prime} \\ \\ \sf{ \sqrt{3} = \frac{p}{q} \: (where \: p \: and \: q \: are \: co \: prime)} \\ \\ \sf{ \sqrt{3}q = p} \: \\ \\ \sf{ \: now \: squaring \: both \: the \: sides \: } \\ \sf{we \: get \: } \\ \sf{( \sqrt{3} {q})^{2} } = {p}^{2} \\ \sf{3 {q}^{2} = {p}^{2} }......(1) \\ \sf{So, \: if \: 3 \: is \: the \: factor \: of \: {p}^{2} \: } \\ \sf{then, \: 3 \: is \: also \: a \: factor \: of \: p ..... ( ii )} \\ \sf{ \implies \: Let\: p = 3m (where \: m \: is \: any \: integer )} \: \\ \\ \sf \bold{squaring \: both \: sides} \\ \sf \implies {p}^{2} \: = (3 {m})^{2} \\ \sf{ {p}^{2} = 9 {m}^{2} } \\ \\ \sf{putting \: the \: value \: of \: {p}^{2} \: in \: equation \: \: (1) } \\ \sf{3 {q}^{2} = {p}^{2} } \\ \sf{ {q}^{2} = 3 {m}^{2} } \\ \sf \bold{ so} \\ \sf{if \: 3 \: is \: factor \: of \: {q}^{2} \: then \: 3 \: is \: also \: a \: factor \: of \: q } \\ \sf{since } \: 3 \: is \: factor \: of \: p \: and \: q \: both \: \\ \sf{So, \: our \: assumption \: that \: p \: and \: q \: are \: co- prime \: is wrong \: } \\ \\ \sf \orange{ \: hence \: , \: \sqrt{3} \: is \: an \: irrational \: number}$