Question

11) prove that √2+√7 is an irrational
number​

Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: assume \: that \: \sqrt{2} + \sqrt{7} \: is \: not \: irrational.$

So,

$\sf :\implies\: \sqrt{2} + \sqrt{7} \: is \: rational.$

$\sf \: Let \: \sqrt{2} + \sqrt{7} = \dfrac{a}{b} - - - (1)$

$\{ \sf \: where \: a \: and \: b \: are \: integers \: such \: that \: b \ne \: 0 \: and \: \\ \sf \: a \: and \: b \: are \: coprime \: numbers \} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf :\longmapsto\: \sqrt{7} = \dfrac{a}{b} - \sqrt{2}$

On squaring both sides, we get

$\sf :\longmapsto\: {( \sqrt{7})}^{2} = { \bigg(\dfrac{a}{b} - \sqrt{2} \bigg)}^{2}$

$\rm :\longmapsto\:7 = \dfrac{ {a}^{2} }{ {b}^{2} } + 2 - 2 \sqrt{2}\dfrac{a}{b}$

$\rm :\longmapsto\:2 \sqrt{2} \: \dfrac{a}{b}= \dfrac{ {a}^{2} }{ {b}^{2} } + 2 - 7$

$\rm :\longmapsto\:2 \sqrt{2} \: \dfrac{a}{b}= \dfrac{ {a}^{2} }{ {b}^{2} } - 5$

$\rm :\longmapsto\:2 \sqrt{2} \: \dfrac{a}{b}= \dfrac{ {a}^{2} - {5b}^{2} }{ {b}^{2} }$

$\rm :\longmapsto\:\sqrt{2} \: = \dfrac{ {a}^{2} - {5b}^{2} }{2a {b}}$

$\red{ \rm \: As \: a \: and \: b \: are \: integers \implies \: \dfrac{ {a}^{2} - {5b}^{2} }{2a {b}} \: is \: rational}$

$\bf\implies \: \sqrt{2} \: is \: rational$

$\sf \: which \: is \: contradiction \: as \: \sqrt{2} \: is \: irrational.$

$\sf :\implies\:Our \: assumption \: is \: wrong$

$\sf :\implies\: \sqrt{2} + \sqrt{7} \: is \: irrational.$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Irrational number are those numbers whom decimal representation is non - terminating nor repeating. Basically square roots of non perfect square numbers are irrational.

For example :- √(2), √(3) are irrational numbers.

Rational number are those numbers whom decimal representation is terminating or non-terminating but repeating. The rational numbers are represented in the form a/b, where a and b are integers, b is non-zero.