Math, asked by yash305158, 10 months ago

prove that √6 + 2√3 is irrational number

Answers

Answered by sudhanshudhek76
2

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

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