2024/009) Find $\sqrt{\sqrt{9} - \sqrt{8}}$

We have $8 = 2 * 4 = 2 * 2^2$

so    $\sqrt{8} = 2 \sqrt{2}$

now $\sqrt{9} = 3$

so   $\sqrt{\sqrt{9} - \sqrt{8}}$

=  $\sqrt{3 - 2\sqrt{2}}$

this is of the form  $\sqrt{n - 2\sqrt{n-1}}$ when n = 3

so  $\sqrt{3 - 2\sqrt{2}} = \sqrt{2 + 1 - 2 * \sqrt{2} * 1}$

$= \sqrt{(\sqrt{2})^2 + 1^2 - 2 * \sqrt{2} * 1}$

$= \sqrt{2} - 1$