Question

Find the smallest and greatest among $\sqrt{7}-\sqrt{5}$ and $\sqrt{8}-\sqrt{6}$

Answer 1

R.F of $\sqrt{7}-\sqrt{5}$ is $\sqrt{7}+\sqrt{5}$

$\therefore \sqrt{7}-\sqrt{5}=\sqrt{7}-\sqrt{5}\times \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}$

$=\frac{7-5}{\sqrt{7}+\sqrt{5}}$

$=\frac{2}{\sqrt{7}+\sqrt{5}}$

R.F of $\sqrt{8}-\sqrt{6}$ is $\sqrt{8}+\sqrt{6}$

$\therefore \sqrt{8}-\sqrt{6}=\sqrt{8}-\sqrt{6}\times \frac{\sqrt{8}+\sqrt{6}}{\sqrt{8}+\sqrt{6}}$

$=\frac{8-6}{\sqrt{8}+\sqrt{6}}$

$=\frac{2}{\sqrt{8}+\sqrt{6}}$

$\sqrt{7}+\sqrt{5}<\sqrt{8}+\sqrt{6}$

$\Rightarrow \frac{1}{\sqrt{7}+\sqrt{5}}>\frac{1}{\sqrt{8}+\sqrt{6}}$

$\Rightarrow \frac{2}{\sqrt{7}+\sqrt{5}}>\frac{2}{\sqrt{8}+\sqrt{6}}$

$\Rightarrow \sqrt{7}-\sqrt{5}>\sqrt{8}-\sqrt{6}$

$\therefore \sqrt{8}-\sqrt{6}$ is smaller than $\sqrt{7}-\sqrt{5}$