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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 ...

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    $\begingroup$ Rationalize the denominator $\endgroup$ Dec 30, 2015 at 14:32
  • $\begingroup$ Note $\{\sqrt{2}\}=\sqrt{2}-1$ $\endgroup$
    – Empy2
    Dec 30, 2015 at 14:32
  • $\begingroup$ Oops,sorry let me edit ! $\endgroup$
    – Mr. Y
    Dec 30, 2015 at 14:33
  • $\begingroup$ The best was to not rationalize D: $\endgroup$
    – Mr. Y
    Dec 30, 2015 at 14:46

1 Answer 1

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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*}

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    $\begingroup$ Me dumb, I didn't see that ,neat ! (thank you) $\endgroup$
    – Mr. Y
    Dec 30, 2015 at 14:45

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