Question

If p,q are prime positive integers , prove that √p+√q is an irrational number .​

Answer 1

you may go with the attachment

Answer 2

$\huge \mathfrak \red{ÄÑẞWÈR}$

$Suppose \: that  \sqrt{p} + \sqrt{q} \: is \: a \: rational \: number \: equal \: to \:   \frac{a}{b} , \: where \: a \: and \: b \: are \: integers \: having \: no \: common \: factor.$

$now \: \sqrt{p} + \sqrt{q} = \frac{a}{b}$

$\sqrt{p} = \frac{a}{b} -\sqrt{q} \: \: \: \: \: \ \red {(squaring \: both \: side)}$

$= (\sqrt{p})^{2} = ( \frac{a}{b} - \sqrt{q})^{2}$

$= p = \frac{ {a}^{2} }{ {b}^{2} } - 2 \: \: (\frac{a}{b}) \sqrt{q} + q$

$= 2( \frac{a}{b} ) \sqrt{q} = \frac{ {a}^{2} + {b}^{2}(q - p) }{ {b}^{2} }$

$= \sqrt{q} = \frac{ {a}^{2} + {b}^{2}(q - p) }{2ab}$

=√qis a rational number. (because sum of two rational numbers is always rational)

This is a contradiction as q√ is an irrational number.

Hence, √p+√q is an irrational number.

$\blue{i \: hope \: it \: helpfull \: for \: you}$