TITLE: Can floor functions be used to "cycle" between values?
QUESTION [7 upvotes]: 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 $n$.

REPLY [6 votes]: $\def\floor#1{\lfloor#1\rfloor}$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.