Question

5
3-√2
Rationalize the denominator ​

Answer 1

$let \: \sqrt{2} \: be \: rational \: no.$

$so \: \frac{p}{q} = \sqrt{2}$

$q \: not \: equal \: to \: 0$

$p \: and \: q \: should \: be \: coprime$

$sq. \: on \: both \: sides$

${p}^{2} = 2 {q}^{2}$

$i.e. 2 \: is \: the \: factor \: of \: p$

$let \: p \: be \: 2m$

${2m}^{2} = {2q}^{2}$

${4m}^{2} = {2q}^{2}$

${2m}^{2} = {q}^{2}$

$i.e. \: 2 \: is \: the \: factor \: of \: \: p$

$but \: p \: and \: q \: should \: be \: coprime$

$so \: our \: supposition\: is \: wrong \:$

$as \: we \: know \: that \:$

$i.r \: + r = i.r$

$so \: 3 - \sqrt{2} \: is \: an \: irrational \: \: no.$

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Answer 2

Please mark me Brainlist.