If \(z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }\)
, find \(\lfloor z \rfloor\)
We can simplify this as :
z = 3 - 2(2) - 1
_______ ≈ ______ ≈ 0.9121
√3 - √8 -1.096
So....the floor of this is 0