Question

proove that root p +root q is irrational

Answer 1

$\purple{ \huge \fbox{ \mathtt{ \: Solution : \: \: }}}$

Let us assume , to the contrary that , √p + √q is rational

Thus ,

$\sf \hookrightarrow \sqrt{p} + \sqrt{q} = \frac{a}{b}$

Squaring both sides , we get

$\sf \hookrightarrow {( \sqrt{p} + \sqrt{q})}^{2} = { (\frac{a}{b} )}^{2} \\ \\ \sf \hookrightarrow p + q + 2 \sqrt{pq} = { (\frac{a}{b} )}^{2} \\ \\ \sf \hookrightarrow \sqrt{pq} = \frac{1}{2} \bigg \{ { (\frac{a}{b} )}^{2} - p - q \bigg\}$

Since a and b are integers , we get $\frac{1}{2} \bigg \{ { (\frac{a}{b} )}^{2} - p - q \bigg\}$ is rational , and so $\sf \sqrt{pq}$ is rational

But this contradicts the fact that $\sf \sqrt{pq}$ is irrational

So , we conclude that √p + √q is irrational

________________

#answerwithquality #BAL