# Find $\lfloor z \rfloor$ given that $z=(\{\sqrt{3}\}^2-2\{\sqrt{2}\}^2)/(\{\sqrt{3}\}-2\{\sqrt{2}\})$

Let $\lfloor x \rfloor$ denote the greatest integer function, and $\{x\}=x-[x]$ the fractional part of $x$. If $$z=\cfrac{\{\sqrt{3}\}^2-2\{\sqrt{2}\}^2}{\{\sqrt{3}\}-2\{\sqrt{2}\}}$$ find $\lfloor z \rfloor$

My attempt

$$z=\cfrac{\{\sqrt{3}\}^2-2\{\sqrt{2}\}^2}{\{\sqrt{3}\}-2\{\sqrt{2}\}}=\cfrac{(\sqrt{3}-1)^2-2(\sqrt{2}-1)^2}{\sqrt{3}-1-2\cdot (\sqrt{2}-1)}=\cfrac{3+1-2(3 -2\sqrt{2})}{\sqrt{3}-1-2\cdot \sqrt{2}+2}$$

After rationalizing I have $$z=\cfrac{4\sqrt{2}+4\sqrt{6}-2-2\sqrt{3}}{-2}$$ (I hope I haven't made any careless mistake now )

Now I am stuck as I don't know how I should take $\lfloor z \rfloor$ as I don't have that if $x=\cfrac{a}{b}$ then $\lfloor x \rfloor =\cfrac{\lfloor a \rfloor}{\lfloor b \rfloor}$ .

I also think I haven't noticed a more straightforward way to do the problem ...

• Rationalize the denominator – lab bhattacharjee Dec 30 '15 at 14:32
• Note $\{\sqrt{2}\}=\sqrt{2}-1$ – Empy2 Dec 30 '15 at 14:32
• Oops,sorry let me edit ! – Mr. Y Dec 30 '15 at 14:33
• The best was to not rationalize D: – Mr. Y Dec 30 '15 at 14:46

## 1 Answer

We have

\begin{align*} z = \cfrac{(\sqrt{3}-1)^2-2(\sqrt{2}-1)^2}{\sqrt{3}-1-2\cdot (\sqrt{2}-1)} &= \frac{3+1-2\sqrt{3}-2(2+1-2\sqrt2)}{\sqrt{3}-2\sqrt2+1}\\ &= \frac{-2\sqrt3 + 4\sqrt2-2}{\sqrt{3}-2\sqrt2+1}\\ &= -2\frac{\sqrt{3}-2\sqrt2+1}{\sqrt{3}-2\sqrt2+1}\\ &= -2. \end{align*}

• Me dumb, I didn't see that ,neat ! (thank you) – Mr. Y Dec 30 '15 at 14:45