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As I was graphing functions in Desmos graphing calculator, I typed in the function $$\lceil{x-\lfloor{x}\rfloor}\rceil$$

which, after some reasoning, unsurprisingly generates the values $0$ or $1$. My question is, can you - with any given amount of floor and ceiling functions - get a the values $0,1$ and $2$? In general, can you prove that with any amount of floor and ceiling functions you can obtain only whole numbers from $0$ to $n$.

Note???

Modular/remainder functions aren't allowed. I'm still in high school so I would appreciate an informal way of either proving the existence of such a function or actually showing a function that can cycle from $0$ to $n$.

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    $\begingroup$ "Modular/remainder functions aren't allowed." — $\ $ $x-\lfloor x\rfloor$ is exactly a remainder function: It's the remainder for division by $1$. $\endgroup$
    – celtschk
    Aug 30, 2016 at 7:38

3 Answers 3

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The function $f(x)$ that cycles through $0,1,...,n-1$ is $f(x) = x \bmod n$. So the question is basically how to write it without using $\operatorname{mod}$.

The largest multiple of $n$ lower than or equal to $x$ is $n \left\lfloor \frac{x}{n} \right\rfloor$ and $\operatorname{mod}$ is the difference between $x$ and this largest multiple of $n$. So, in the end:

$$ f(x) = x - n \left\lfloor \frac{x}{n} \right\rfloor $$

is equivalent to the $\operatorname{mod}$ function and will loop through $0,1,...,n-1$ for sequential $x \in \mathbb{N}$.


[ EDIT ] Also, for an example of a function that loops through the same values, but takes only integer values between them:

$$ \lfloor f(x) \rfloor = \left\lfloor{x}\right\rfloor - n \left\lfloor\frac{x}{n}\right\rfloor $$

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    $\begingroup$ You can write a congruency with parentheses using the LaTeX \pmod operator: f(x) \equiv x \pmod n renders as: $f(x) \equiv x \pmod n$. For a single $mod$ to appear in upright font declare it explicitly as an operator name: \operatorname{mod} → $\operatorname{mod}$. $\endgroup$
    – CiaPan
    Aug 30, 2016 at 15:59
  • $\begingroup$ @CiaPan Thanks for the tips. Edited to fix the typesetting. $\endgroup$
    – dxiv
    Aug 30, 2016 at 16:27
  • $\begingroup$ @CiaPan: is it correct to write $f(x) = x \pmod n$ ? With $n>0$, I use $a\equiv b\pmod n$ to mean that $n$ divides $b-a$; and $a=b\bmod n$ to additionally mean that $0\le a<n$. $\endgroup$
    – fgrieu
    Aug 30, 2016 at 16:39
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    $\begingroup$ @dxiv: my problem is with $x\pmod n$; unless I err, my math teacher (France, 1978-1982) thought that the remainder of the division of $x$ by $n$ (as meant here) is noted $x\bmod n$, not $x\pmod n$; that $y\equiv x\pmod n$ means $y-x$ is divisible by $n$; and that $x\pmod n$, if used at all, would be the infinite set of $y$ with $y\equiv x\pmod n$. $\endgroup$
    – fgrieu
    Aug 30, 2016 at 19:25
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    $\begingroup$ @fgrieu is correct here. It is a common mistake to use "$x \pmod{n}$" instead of the intended "$x \bmod n$". The accepted answer to the very question you linked uses the right notation. Don't assume that the asker was using the correct notation. $\endgroup$
    – user21820
    Aug 31, 2016 at 2:38
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Sure. $x-\lfloor x \rfloor$ is the fractional part of $x$, so it can take any value in the range $[0,1)$.

So $2(x-\lfloor x \rfloor)$ can take any value in the range $[0,2)$, which means $\lceil 2(x-\lfloor x \rfloor) \rceil$ can be any of $0$, $1$, or $2$.

Similarly $n(x-\lfloor x \rfloor)$ can take any value in the range $[0,n)$, which means $\lceil n(x-\lfloor x \rfloor) \rceil$ can take on any integer value between $0$ and $n$ inclusive.

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  • $\begingroup$ This doesn't necessarily give $x \in \mathbb{N}$ which (is not entirely clear in the question, but) seems to be required. $\endgroup$
    – dxiv
    Aug 30, 2016 at 3:25
  • $\begingroup$ @dxiv When does this occur? $\endgroup$
    – Ian L
    Aug 30, 2016 at 3:29
  • $\begingroup$ @IanLimarta For example $\lceil 2(x - \lfloor x \rfloor) \rceil = 0$ for all $x \in \mathbb{N}$. Please edit the original question and clarify what you mean by "achieve the set of values $\{x\space|\space\space{x\in{\mathbb{Z_{\geq0}\}}}}$". $\endgroup$
    – dxiv
    Aug 30, 2016 at 3:34
  • $\begingroup$ @dxiv Edited. I'm sorry but I meant whole numbers. I should have just put that instead. $\endgroup$
    – Ian L
    Aug 30, 2016 at 3:45
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$\def\floor#1{\lfloor#1\rfloor}$Take any positive integer $n$ and real $x$.

$\floor{x-n\floor{\frac{x}{n}}} = \floor{x} \bmod n$. [This cycles through $0$ to $n-1$ with one unit for each value.]

$\floor{n(x-\floor{x})} = \floor{nx} \bmod n$. [This cycles through $0$ to $n-1$ with period length $1$.]

The easiest way to see why is to notice that $n \floor{\frac{x}{n}}$ is the nearest multiple of $n$ no greater than $x$, so subtracting that from $x$ yields the remainder of $x$ modulo $n$. If you then perform a further floor, you will get the integer part of the remainder, as desired. You can first scale $x$ by a constant before the whole process, to change the period length. This is how we can get the second expression above.

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