Find $\lfloor z \rfloor$

Question

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

What should I do? I can't split square roots into integers and their fractional part!

Answer 1

Hint: $\{\sqrt{3}\} = \sqrt{3}-1$ and $\{\sqrt{2}\} = \sqrt{2}-1$ because $\sqrt{3}$ and $\sqrt{2}$ lie between $1$ and $2$.

Comment: Since you have accepted, I complete the simplification. We have $$\frac{(\sqrt{3}-1)^2 - 2(\sqrt{2}-1)^2}{(\sqrt{3}-1)-2(\sqrt{2}-1)}$$ $$=\frac{(3-2\sqrt{3}+1)-2(2-2\sqrt{2}+1)}{\sqrt{3}-2\sqrt{2} +1}$$ $$=\frac{-2\sqrt{3} + 4\sqrt{2} -2}{\sqrt{3}-2\sqrt{2} +1}$$ $$= \frac{-2\cdot(\sqrt{3}-2\sqrt{2}+1)}{\sqrt{3}-2\sqrt{2} +1}$$ $$=-2\cdot\frac{\sqrt{3}-2\sqrt{2} +1}{\sqrt{3}-2\sqrt{2} +1}$$ $$=-2$$

So $z=-2$ and thus $\lfloor z \rfloor = \boxed{-2}$.

Answer 2

This is a case where approximation can lead to disasters: if we use 0.73 and 0.41 for the two fractional parts, we get, according to bc -l

(0.73^2-2*0.41^2)/(0.73-2*0.41)
-2.18555555555555555555

and this would lead to $-3$ as the result. Not better if we use five digits:

(0.73205^2-2*0.41421^2)/(0.73205-2*0.41421)
-2.00018007990038393690

However, if one uses rounding to ten decimal digits, the result would be -1.99999999953945978587 (truncation would still lead to a number less than $-2$).

This of course makes one suspect that the fraction's value is $-2$, which it indeed is. The algebraic proof is not difficult by considering that $\{\sqrt{3}\}=\sqrt{3}-1$ and $\{\sqrt{2}\}=\sqrt{2}-1$ (done in another answer).