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$$z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$$, find  $\lfloor z \rfloor.$

 Sep 25, 2021
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$z = -2 - \sqrt{3}$, so $\lfloor z \rfloor = \boxed{-4}$.

 Sep 25, 2021

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