Question

prove that √6 + 2√3 is irrational number

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Answer 1

Answer:

$let \: 6+ 2 \sqrt{3} \: be \: rational \\ 6 + 2\sqrt{3} = \frac{p}{q} (p \: and \: q \: are \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: integer \: and \: \ \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: co - prime \: \\ q \geqslant 0) \\ \\ 2\sqrt{3} = p - 6q \\ \sqrt{3} = \frac{p - 6q}{2q} . \\ \\ since \: p \: and \: q \: are \: integer \ \\ : so \: \frac{p -3q}{2q} is \: rational \\ thus \: \sqrt{3} \: is \: rational \\ \\ but \: this \: contradicate \ \\ : say \: that \: \sqrt{3} \: is \: irrational \: \\ so \: our \: assumption \: is \: \\ incorrect. \\ \\ hence \: 6 + 2 \sqrt{3} is \: irrational$