Question

what is the perfect answer??​

Answer 1

$\normalsize\sf\ Let \: \red{\sqrt{2} + \sqrt{5}} \; be \: a \: rational \: number$

$\normalsize\sf\ As; \: we \: know \: Rational \: number \: is \: in \: form \: of \: \green{\frac{p}{q}}$

$\normalsize\dashrightarrow\sf\red{\sqrt{2} + \sqrt{5} } = \green{\frac{a}{b}}$

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

$\underline{\bigstar\:\textsf{Squaring \: both \: sides( L.H.S \& R.H.S):}}$

$\normalsize\ : \implies\sf\ \left[(\sqrt{2})^2 + (\sqrt{5})^2 + 2\sqrt{10} \right] = \left[\frac{p}{q} \right]^2$

$\normalsize\ : \implies\sf\ \left[2 + 5 + 2\sqrt{10} \right] = \left[\frac{p}{q} \right]^2$

$\normalsize\ : \implies\sf\ \left[7 + 2\sqrt{10} \right] = \left[\frac{p^2}{q^2} \right]$

$\normalsize\ : \implies\sf\ 2\sqrt{10} = \frac{a^2}{b^2} - 7$

$\normalsize\ : \implies\sf\ 2\sqrt{10} = \frac{a^2 - 7b^2}{b^2}$

$\normalsize\ : \implies\sf\ \sqrt{10} = \frac{a^2- 7b^2}{2b^2}$

$\normalsize\sf\therefore\ Here, \: a \: \& \: b \: are \: integers \: and \:\frac{a^2 - 7b^2}{2b^2} \: \\ \normalsize\sf\ is \: \blue{rational}. \: So, \: \red{\sqrt{10}} \: is \: also \: be \: rational$

$\normalsize\sf\ This \: contradicts \: the \: fact \: that \: \red{\sqrt{10}} \: is \: \blue{rational}$

$\normalsize\sf\ Hence, \: Our \: assumption \: is \: wrong \: \& \: \red{\sqrt{2} + \sqrt{5}} \: is \: \pink{irrational}$

$\large\maltese \: \: {\boxed{\sf \pink{Hence \: Prove \: !!}}}$