Questions on The Sum and Difference of Floor Functions I'm trying to combine floor functions for a problem I'm working on.
Given that $m$ and $n$ are integers:


*

*Is there a way to write $\mathbb{floor}(\frac{m-1}{n})$ in terms of $\mathbb{floor}(\frac{m}{n})$?

*Is there a way to simplify $\mathbb{floor}(\frac{m}{n})-\mathbb{floor}(\frac{m-1}{n})$?

*Is there a way to simplify $\mathbb{floor}(\frac{m}{n})+\mathbb{floor}(\frac{m-1}{n})$?

*Is there a way to simplify $\mathbb{floor}(\frac{m}{n})+1$? ($\mathbb{floor}((\frac {m+1}n)$--Solved by J.W. Tanner)

*Is there a way to write $\mathbb{floor}^2(\frac{m}{n})$ to remove the exponent (similar to $\sin^2(x)=\frac 12-\frac12\cos(2x)$)?

 A: This can be useful when working with the floor function.
Consider the sequence $\frac1b, \frac2b, \frac3b, \ldots,$
and think about taking the floor of each term. When is 
$\operatorname{floor}\left(\frac{a+1}{b}\right) > \operatorname{floor}\left(\frac ab\right)$?
$$
\operatorname{floor}\left(\frac{m-1}{n}\right)=
\begin{cases}
\operatorname{floor}\left(\dfrac{m}{n}\right) - 1 & \text{if $n$ divides $m$,} \\
\operatorname{floor}\left(\dfrac{m}{n}\right) & \text{otherwise.}\end{cases}
$$
You can take this and apply it to parts $(2)$ and $(3)$.
Is the result "simpler" than before? I don't think so, but depending on what you're doing with those expressions, the substitution can be useful.
Notice that if you have multiple occurrences of
$\operatorname{floor}\left(\frac{m-1}{n}\right)$ in a single equation, you still only have two cases to be concerned with ($n$ divides $m$ or it does not), and you can replace all the occurrences of 
$\operatorname{floor}\left(\frac{m-1}{n}\right)$ simultaneously.
For $(4)$, there are other ways to write the same thing, but are they "simpler"?
The form you have in $(4)$ may be the most useful form for most purposes already.
For $(5),$
\begin{align}
\operatorname{floor}\left(\dfrac{m}{n}\right)
\operatorname{floor}\left(\dfrac{m+n}{n}\right)
&= \operatorname{floor}\left(\dfrac{m}{n}\right)
\left(\operatorname{floor}\left(\dfrac{m}{n}\right) + 1\right) \\
&= \left(\operatorname{floor}\left(\dfrac{m}{n}\right)\right)^2
+ \operatorname{floor}\left(\dfrac{m}{n}\right) \\
\end{align}
And that is why I have so many doubts about the usefulness of "simplifying" example $(4).$ It seems more useful to go in the other direction, from something else to $(4)$.
