How Can I Simplify An Inequality With A Floor Function? I'm trying to convert between pagination by starting index $i$ & length $l$ and pagination by page number $p$ & page size $s$. I've gotten far enough to know that: 
for given $i$ and $l$ such that $ i\gt l $:
$$\frac {i+l}{1+\lfloor \frac is \rfloor} \le s$$
How can I find the minimum $s$ for that inequality?
EDIT: Also,  $ \{i,l,p,s\} \in \Bbb N$
 A: $$i+l\le s-s\left\lfloor\frac is\right\rfloor=2s-i\bmod s.$$
Unfortunately, the function on the RHS is quite irregular. IMO exhaustive search is needed.
A: (Too long for a comment.) The following provides a lower bound for $\,s\,$, though not an actual minimum since it cannot always be attained.
The inequality is verified for $\,s=i\,$, so the minimum $\,s\,$ is $\,\le i\,$. Let $\,\dfrac{i}{k+1} \lt s \le \dfrac{i}{k}\,$ for some integer $\,k \ge 1\,$. Then $\,\left\lfloor \dfrac{i}{s}\right\rfloor = k\,$, and the inequality can be written as:
$$
i+l \le (k+1)s \tag{1}
$$
By the definition of $\,k\,$, the RHS is $\,(k+1)s \le \left(1+\dfrac{1}{k}\right)i\,$, so for the equality to hold this implies:
$$\require{cancel}
\bcancel{i}+l \le \left(\bcancel{1}+\dfrac{1}{k}\right)i \quad\implies\quad k \le \frac{i}{l} \quad\implies\quad k \le \left\lfloor \frac{i}{l} \right\rfloor \tag{2}
$$
Substituting back in $\,(1)\,$ gives the lower bound:
$$
s \ge \frac{i+l}{k+1} \ge \frac{i+l}{1 + \left\lfloor \frac{i}{l} \right\rfloor} \quad\implies\quad s \ge \left\lceil\frac{i+l}{1 + \left\lfloor \frac{i}{l} \right\rfloor}\right\rceil \tag{3}
$$
The lower-bound is not always attained, though. For example, with $\,i=5, l = 2\,$, $\,(3)\,$ gives $\,s \ge 3\,$, while the actual minimum is $\,s=4\,$.
