Question

Prove root 3 is irrational.

Answer 1

$\underline{\bf \: Correct \: Question:-}}}$

$\rm \: Prove \: that \: \sqrt{3} \: is \: irrational.$

$\underline{\bf \: Solution:-}}$

$\rm \: Let \: us \: assume, \: to \: the \: contrary ,\: that\\\rm \: \sqrt{3} \: is \: rational.$

$\rm \: So, \: we \: can \: find \: coprime \: integers \: a \: and \\\rm \: b \: (\neq \: 0) \: such \: that$

$\rm \: \sqrt{3} = \: \frac{a}{b} \: where \: a \: and \: b \: are \: co-prime \: i.e.,\\\rm \: their \: HCF \: is \: 1 . \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...[Eq. \: (1)]$

$\implies \: \rm \: \sqrt{3b} =a$

$\rm \: Squaring \: on \: both \: side, \: we \: get\\\rm \: 3b^{2} =a^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...[Eq. \: (2)]$

$\rm \: Therefore, \: 3 \: divides \: a^{2} .\: But \: 3 \: is \: prime$

$\rm \: Therefore, \: 3 \: divides \: a \: also$

$\rm \: \therefore \: a=3c \: for \: some \: integer \: c.$

$\rm \: Putting \: this \: value \: of \: a \: in \: Eq. \: (2),$

$\rm \: 3b^{2} =9c^{2}$

$\rm \: \implies \: Dividing \: by \: 3, \: b^{2} =3c^{2}$

$\rm \: This \: means \: that \: 3 \: divides \: b^{2} , \: and \: so \: 3 \\\rm \: divides \: b, \: also \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (\because \: 3 \: is \: prime)$

$\rm \: Therefore, \: a \: and \: b \: have \: 3 \: as \: a \: common\\\rm \: factor.$

But this contradicts ( 1 ) namely the fact

that a and be have no common factor other

than 1.

$\rm \: This \: contradiction \: arose \: because \: of \: our\\\rm \: incorrect \: assumption \: that \: \sqrt{3} \: is \: rational.$

$\rm \: So, \: we \: conclude \: that \: \sqrt{3} \: is \: irrational.$