Question

prove that √2 + √5 is irrational​

Answer 1

Answer:

By a theorem,

We know that the sum of two irrational numbers will irrational.

So here root 2 and root 5 are irrational.

So, It is irrational...

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Answer 2

AnswEr :

$\normalsize\sf\ Let \: \red{\sqrt{2} + \sqrt{5}} \: be \: a \: rational \: number \\ \\ \normalsize\sf\ As, \: we \: know \: rational \: number \: is \: in \: form \: of \: \green{\frac{p}{q} }$

$\normalsize\ : \implies\sf\red{\sqrt{2} + \sqrt{5} } \: = \: \green{\frac{p}{q} }$

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

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

$\normalsize\ : \implies\sf\ [\sqrt{2} + \sqrt{5}]^2 \: = \: [\frac{p}{q} ]^2 \\ \\ \normalsize\ : \implies\sf\ [2 + 5 + 2\sqrt{5} ] \: = \: \frac{p^2}{q^2} \\ \\ \normalsize\ : \implies\sf\ 7 + 2\sqrt{10} \: = \: \frac{p^2}{q^2} \\ \\ \normalsize\ : \implies\sf\ 2\sqrt{10} \: = \: \frac{p^2}{q^2} - 7 \\ \\ \normalsize\ : \implies\sf\ 2\sqrt{10} \: = \: \frac{p^2 - 7q^2}{q^2} \\ \\ \normalsize\ : \implies\sf\sqrt{10} \: = \: \frac{p^2 - 7q^2}{2q^2}$

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

$\normalsize\ : \implies\sf\frac{p^2 - 7q^2}{2q^2} = Rational \: number ≠ \sqrt{10}$

$\normalsize\sf\sqrt{10} \: is \: equal \: to \: a \: rational \; number \: but \: it \: contradicts$

$\normalsize\sf\ that \: \red{\sqrt{10} } \: is \: \red{irrational }$

$\normalsize\sf\ This \: shows \: that \; our \: assumption \: is \: wrong$

$\therefore\sf\ Hence, \: \sqrt{2} + \sqrt{5} \: is \: an \: irrational \; number$