Can floor functions be used to "cycle" between values? 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 $n$.
 A: 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.
A: 
$\def\floor#1{\lfloor#1\rfloor}$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.
A: 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:

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