Problem:
If $$z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$$ find $\lfloor z \rfloor$, where $\{x\} = x-\lfloor x \rfloor.$
This seems like a whole lot of floors! How would I do this?
Hint: substitute and simplify using the following:
$\{\sqrt{3}\} = \sqrt{3} - \lfloor \sqrt{3}\rfloor = \sqrt{3} - 1\,$, $\;\;\{\sqrt{2}\} = \sqrt{2} - \lfloor \sqrt{2}\rfloor = \sqrt{2} - 1\,$
$\{\sqrt{3}\}^2 = (\sqrt{3} - 1)^2 = 4 - 2 \sqrt{3}\,$, $\;\;\{\sqrt{2}\}^2 = (\sqrt{2} - 1)^2 = 3 - 2 \sqrt{2}\,$