Question

prove that 2 root 3 + root 5 is an irrational number also check whether( 2 root 3+root 5) (2 root 3 -root 5 ) is rational or irrational

Answer 1

$let \: 2 \sqrt{3 } + \sqrt{5} be \: rational \: number \\ \\ thus \: it \: can \: be \: represented \: in \: \\ the \: form \: of \: \frac{p}{q}$

$here \: p \: and \: q \: are \: co \: - primes \: \\ (only \: one common \: factor)$

$and \: q \: i s \: not \: = 0$

$2 \sqrt{3} + \sqrt{5} = \frac{p}{q}$

$multiply \: \: by \: \frac{1}{2} both \: sides$

$\sqrt{3} + \sqrt{5} \times \frac{1}{2} = \frac{p}{2q} \\ \\ \frac{ \sqrt{5} }{2} = \frac{p}{2q} - \sqrt{3}$

$\frac{ \sqrt{5} }{2} = \frac{p - 2 \sqrt{3} q}{2q}$

$\sqrt{5} = \frac{p - 2 \sqrt{3}q }{q}$

$multiply \: by \: 2 \sqrt{3}both \: sides \\ \sqrt{5 } \times 2 \sqrt{3} = \frac{p - 2 \sqrt{3} q}{q} \times 2 \sqrt{3}$

$2 \sqrt{15 } = \frac{2 \sqrt{3p} }{q} - \frac{12q} {q}$

$\frac{12} \: \: = \frac{2 \sqrt{3}p }{q} - 2 \sqrt{15}$

$12 = \frac{2 \sqrt{3}p - 2 \sqrt{15}q }{q}$

$but \: we \: know \: that \: a \: rational \: number \\ \: is \: not \: equal \: to \: an \: irrational \: number \\ \\ this \: contradiction \: has \: arisen \\ \: due \: to \: our \: incorrect \: assumption \: \\ \\ thus \: our \: assumption \: is \: wrong \: \\ and \: 2 \sqrt{3} + \sqrt{5} is \: irrational$

Answer 2

Answer:

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