Skip to main content

Which is larger ....

$\sqrt 7 - \sqrt 6$ Or $\sqrt 13 - \sqrt 12$

We have, $\sqrt 13 > \sqrt 7$ & $\sqrt 12 > \sqrt 6$

$\therefore \sqrt 13 + \sqrt 12 > \sqrt 7 + \sqrt 6 $

$\Rightarrow \frac {1}{\sqrt 13 + \sqrt 12} < \frac {1}{\sqrt 7 + \sqrt 6}$

$\Rightarrow \frac {\sqrt 13 - \sqrt 12}{(\sqrt 13 + \sqrt 12)(\sqrt 13 - \sqrt 12)} < \frac {\sqrt 7 - \sqrt 6}{(\sqrt 7 + \sqrt 6)(\sqrt 7 - \sqrt 6)}$

$\therefore \sqrt 13 - \sqrt 12 < \sqrt 7 - \sqrt 6 $