Question

HEYA!!!!

$prove \: that \: \sqrt{2} + 2 \sqrt{3} \: \: is \: \: irrational$

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5 Points : D

Answer 1

Hey friend!!!

Here's ur answer...

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$let \: \sqrt{2} + 2 \sqrt{3} \: be \: a \: rational \: number \: \\ so \: \: \sqrt{2} + 2 \sqrt{3} = \frac{a}{b} \: \: where \: a \: and \: b \: are \: co \: primes \: i.e.they \: have \: 1 \: as \: their \: hcf \\ \\ so \: \sqrt{2 } = \frac{a}{b} - 2 \sqrt{3} \\ \\ now \: taking \: \sqrt{3 } = \frac{p}{q} \: where \: p \: and \: q \: are \: co \: primes \\ \\ squaring \: both \: sides \: \\ 3 = \frac{ {p}^{2} }{ {q}^{2} } \\ 3 {q}^{2} = {p}^{2} - - - - ( 1) \\ = > 3 \: is \: a \: factor \: of \: {p}^{2} \\ = > 3 \: is \: a \: factor \: of \: p - - - - (2) \\ \\ put \: p = 3c \: in \: (1) \\ \\ 3 {q}^{2} = ( {3c)}^{2} \\ 3 {q}^{2} = 9 {c}^{2} \\ {q}^{2} = 3 {c}^{2} \\ \\ = > 3 \: is \: a \: factor \: of \: {q}^{2} \\ = > 3 \: is \: a \: factor \: of \: q - - - - - - (3) \\ \\ \\ \\ from \: (2) \: and \: (3 ) \\ 3 \: is \: a \: common \: factor \: of \: p \: and \: q \: \\ but \: p \: and \: q \: were \: co \: primes \: \\ \\ it \: means \: that \: our \: assumption \: was \: wrong \: . \\ = > \sqrt{3} \: is \: irrational \: \\ \\ so \: \sqrt{2} + 2 \sqrt{3} \: is \: also \: an \: irrational \: number.$
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Answer 2

$here \: is \: your \: answer \\ \\ to \: prove \: \sqrt{2} + 2 \sqrt{3} \: is \: an \: irrational \: number \\ \\ let \: \sqrt{2} + 2 \sqrt{3} = m \: is \: an \: rational \: number \: \: (where \: m \: = \: N) \\ \\ \sqrt{2 } + 2 \sqrt{3} = m \\ 2 \sqrt{3} = m - \sqrt{2} \\ \\ take \: square \: on \: both \: side \\ \\ {(2 \sqrt{3}) }^{2} = {(m - \sqrt{2)} }^{2} \\ 12 = {m}^{2} - 2 \sqrt{2} + 2 \\ 2 \sqrt{2} = {m}^{2} + 2 - 12 \\ 2 \sqrt{2} = {m}^{2} - 10 \\ \sqrt{2} = \frac{ {m}^{2 } - 10 }{2} \\ \\ \\ but \: \sqrt{2} is \: an \: irrational \: number \\ \\ since \: rational \: is \: not \: equal \: to \: irrational \\ \\ hence \: our \: assumption \: was \: wrong \\ \\ \sqrt{2} + 2 \sqrt{3} is \: an \: irrational \: number$

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