Question

21
Prove that
$\sqrt{3}$
is an irrational number.​

Answer 1

Step-by-step explanation:

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Answer 2

$let \: \sqrt{3} \: is \: a \: rational \: number \: and \: equal \: to \: \frac{p}{q} (q \: is \: not \: equal \: 0) \\ \sqrt{3} = \frac{p}{q} \\ s.b.s \\ 3 = \frac{ {p}^{2} }{ {q}^{2} } \\ {q}^{2} = \frac{ {p}^{2} }{3} ( \: it \: means \: p \: has \: two \: factors \: 3 \: \: and \: a \\ p = 3a \\ {q}^{2} = \frac{ {(3a)}^{2} }{3} \\ {q}^{2} = \frac{ {9a}^{2} }{3} \\ \frac{ {q}^{2} }{3} = {a}^{2} (q \: has \: two \: factors \: 3 \: and \: b) \\ here \: we\: find \: a \: common \: factor \: 3 \: in \: the \: factors \: of \: p \: and \: q \: which \: is \: not \: possible \: so \: our \: assumption \: is \: wrong \: \\ so \: \sqrt{3} is \: irrational$