3
$\begingroup$

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 ...

$\endgroup$
  • 1
    $\begingroup$ Rationalize the denominator $\endgroup$ – lab bhattacharjee Dec 30 '15 at 14:32
  • $\begingroup$ Note $\{\sqrt{2}\}=\sqrt{2}-1$ $\endgroup$ – Empy2 Dec 30 '15 at 14:32
  • $\begingroup$ Oops,sorry let me edit ! $\endgroup$ – Mr. Y Dec 30 '15 at 14:33
  • $\begingroup$ The best was to not rationalize D: $\endgroup$ – Mr. Y Dec 30 '15 at 14:46
5
$\begingroup$

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*}

$\endgroup$
  • 1
    $\begingroup$ Me dumb, I didn't see that ,neat ! (thank you) $\endgroup$ – Mr. Y Dec 30 '15 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.