If $$z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$$ find $\lfloor z \rfloor$.
I don't rely know how to do this, but I was thinking about multiplying the denominator by it's conjugate, but idk.
Assuming $\{\cdot \}$ is fractional part, $\{\sqrt{3}\} = \sqrt{3}-1$ and $\{\sqrt{2}\} = \sqrt{2}-1$, so
$$ z = \frac{(\sqrt{3}-1)^2 - 2 (\sqrt{2}-1)^2}{\sqrt{3} + 1 - 2 \sqrt{2}}$$ Expand out the numerator and you should recognize that it is a certain integer times the denominator.
