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Given:

  • $Floor(x)=\lfloor x \rfloor$
  • $Ceiling(x)=\lceil x \rceil$

Where $x$ is a real number.

Is there any other (mathematical) way for defining $Floor(x)$ and $Ceiling(x)$?

Restrictions:

  • Do not use the Floor function in order to define the Ceiling function.
  • Do not use the Ceiling function in order to define the Floor function.
  • Do not use the Round function in order to define either one of them.

Please excuse the possible duplicate, as I haven't been able to do find this question anywhere...

Thanks

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3 Answers 3

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Use $\bmod$ (not necessarily $\diamond$ mods) $$ \lfloor x \rfloor = x - \bmod(x,1) $$ To get $\bmod(x,1)$, use $\frac12\Biggr(\frac{\log\left(\exp\left(2\pi i(x- \frac12)\right)\right)}{\pi i}+1\Biggr)$, since $\exp\left(2\pi i(x- \frac12)\right)$ has a period of $1$.

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  • $\begingroup$ And how do you define $mod$? $\endgroup$
    – barak manos
    Apr 2, 2014 at 15:29
  • $\begingroup$ @barakmanos what about $\log(\exp(i2\pi x)/(i2\pi))$? $\endgroup$
    – draks ...
    Apr 2, 2014 at 15:38
  • $\begingroup$ I don't know. Can you please elaborate on this in your answer? It actually looks very interesting, though I'd like to be sure that it gives the correct answer (and whether it is $Floor$ or $Ceiling$). Thanks $\endgroup$
    – barak manos
    Apr 2, 2014 at 15:45
  • $\begingroup$ So the result is a complex number... Do I take the $Real$ part then? $\endgroup$
    – barak manos
    Apr 2, 2014 at 15:55
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    $\begingroup$ Here is the graph: wolframalpha.com/share/… $\endgroup$
    – barak manos
    Apr 3, 2014 at 9:32
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Floor and ceiling functions are usually defined as

$$ \lfloor x \rfloor=\max\, \{m\in\mathbb{Z}\mid m\le x\} $$ and $$ \lceil x \rceil=\min\,\{n\in\mathbb{Z}\mid n\ge x\} $$ for $x\in\mathbb R$ (see Floor and ceiling functions for more details).

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$\lfloor x \rfloor = \max\{n \in \mathbb Z: n \le x\}$

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