Question

plz give solution of middle question...​

Answer 1

Step-by-step explanation:

$show \: that \: there \: is \: no. \: positive \: integer \: n \: for \: which \: \sqrt{n - 1} + \sqrt{n + 1} \: is \: rational$

$let \: is \: assume \: that \: for \: esome \: positive \: integer \: n \: is \: a \: rational \: no.$

$let \: \sqrt{n - 1} + \sqrt{n + 1} = \frac{a}{b} \: (1) \: a.b \: are \: tow \: positive \: integer$

$= \frac{1}{ \sqrt{n - 1} + \sqrt{n + 1} } = \frac{b}{a}$

$= \frac{ \sqrt{n - 1 } - \sqrt{n + 1} }{( { \sqrt{n - 1} })^{2} - ( { \sqrt{n + 1} })^{2} } = \frac{b}{a}$

$= \frac{ \sqrt{n - 1} - \sqrt{n + 1} } {( \sqrt{n - 1} + \sqrt{n + 1)} ( \sqrt{n - 1} - \sqrt{n - + 1}) } = \frac{b}{a}$

$= \frac{ \sqrt{n - 1} - \sqrt{n + 1} }{(n - 1) - (n + 1)} = \frac{b}{a}$

$= \frac{ \sqrt{n - 1} - \sqrt{n + 1} }{n - 1 - n - 1} = \frac{b}{a}$

$= \frac{ \sqrt{n - 1} - \sqrt{n + 1} }{ - 2} = \frac{b}{a}$

$= \sqrt{n + 1} - \sqrt{n - 1} = \frac{2b}{a} \: \: \: \: - (ii)$

$now \: by \: (i) + (ii)and \: (i) - (ii)$

$eq \: (i) + (ii)$

$= 2 \sqrt{n + 1} = \frac{a}{b} + \frac{2b}{a}$

$= 2 \sqrt{n + 1} = \frac{ {a}^{2} + 2 {b}^{2} }{ab}$

$= {n + 1} = \frac{ {a}^{2} + 2 {b}^{2} }{2ab}$

$eq \: (i) - (ii)$

$= 2 \sqrt{n - 1} = \frac{a}{b} - \frac{2b}{a}$

$= 2 \sqrt{n - 1} = \frac{ {a}^{2} - 2 {b}^{2} }{ab}$

$= \sqrt{n - 1} = \frac{ {a}^{2} - 2 {b}^{2} }{2ab}$

$since \: a.b \: and \: 2 \: are \: integers \: \frac{ {a}^{2} + 2 {b}^{2} }{2ab} and \: \frac{ {a}^{2} - 2 {b}^{2} }{2ab } \: must \:be \: rational \: numder$

$= \sqrt{n - 1} \: and \: \sqrt{n + 1} \: are \: laso \: rational$

= But they will be rational only when they are perfect squares

=Both n-1 and n+1 must be perfect squares

$i \: hope \: it \: helpfull \: for \: you$