Question

prove that root x + root Y is irrational where x and y are prime

Answer 1

Hope this answer helped u

Answer 2

Answer:

$\sqrt{x}+\sqrt{y} \: is \\ irrational.$

Step-by-step explanation:

Given x and y are prime.

Let us assume √x+√y is Rational.

$\sqrt{x}+\sqrt{y}=\frac{a}{b} \\where \: a,b \: are \: integers\:and \: b≠0$

$\sqrt{x} = \frac{a}{b}-\sqrt{y}$

/* On Squaring both sides of the equation, we get

$\implies \left(\sqrt{x}\right)^{2}=\left(\frac{a}{b}-\sqrt{y}\right)^{2}$

$\implies x = \left(\frac{a}{b}\right)^{2}-2\times \frac{a}{b}\times \sqrt{y} + \left(\sqrt{y}\right)^{2}$

$\implies x = \frac{a^{2}}{b^{2}}-\frac{2a\sqrt{y}}{b}+y$

$\implies \frac{2a\sqrt{y}}{b}=\frac{a^{2}}{b^{2}}+y-x\\=\frac{a^{2}+b^{2}(y-x)}{b^{2}}$

$\implies \sqrt{y}=\frac{a^{2}+b^{2}(y-x)}{b^{2}}\times \frac{b}{2a}$

$=\frac{a^{2}+b^{2}(y-x)}{2ab}$

$Since,a\:and \:b \: are \: integers, \\\frac{a^{2}+b^{2}(y-x)}{2ab}\: is \:rational,\\So,\:\sqrt{y}\: rational.$

$But ,this \: contradicts \: the\\fact \: that \: \sqrt{y}\:is\: irrational$

$Hence,\sqrt{x}+\sqrt{y} \: is \\ irrational.$

•••♪