Question

Prove that √6+√2 is irrational no.

Answer 1

AnswEr :

$\normalsize\sf\ Let \: \red{\sqrt{6} + \sqrt{2} } \: be \: a \: rational \: number$

$\normalsize\sf\ As, \: we \: know \: every \: rational \: number \: is \: in \: form \: of \: \blue{\frac{p}{q} }$

$\normalsize\ : \implies\sf\red{\sqrt{2} + \sqrt{6} } = \blue{\frac{p}{q} }$

$\:\:\:\:\:\:\normalsize\sf\ [a \: and \: b \: are \: co- prime \: and \: b ≠ 0]$

$\:\:\:\:\underline{\dag\:\textsf{Squaring \: both \: sides}}$

$\normalsize\ : \implies\sf\ [\sqrt{6} + \sqrt{2}]^2 = [\frac{p}{q}]^2 \\ \\ \normalsize\ : \implies\sf\ [6 + 2 + 2\sqrt{12}] = \frac{p^2}{q^2} \\ \\ \normalsize\ : \implies\sf\ 8 + 2\sqrt1{12} = \frac{p^2}{q^2} \\ \\ \normalsize\ : \implies\sf\ 2\sqrt{12} = \frac{p^2}{q^2} - 8 \\ \\ \normalsize\ : \implies\sf\ 2\sqrt{12} = \frac{p^2 - 8q^2}{q^2} \\ \\ \normalsize\ : \implies\sf\ \sqrt{12} = \frac{p^2 - 8q^2}{2q^2}$

$\underline{\dag\:\textsf{Here, \: a \: and \: b \: are \: integers}}$

$\normalsize\ : \implies\sf\frac{p^2 - 8q^2}{2q^2} = Rational \: number ≠ \sqrt{12}$

$\normalsize\sf\ \sqrt{12} \: is \: equal \: to \: a \: rational \: number \: but \: it \: contradicts$

$\normalsize\sf\ that \: \red{\sqrt{12}} \: is \: \red{\sqrt{irrational}}$

$\normalsize\sf\ This \: show \: that \: our \: assumption \: is \: wrong$

$\normalsize\therefore\sf\ Hence, \: \sqrt{6} + \sqrt{2} \: is \: irrational$

Answer 2

Answer:

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