Editing Floor and ceiling functions (section)

===Quotients===
If ''m'' and ''n'' are integers and ''n'' ≠  0,
:<math>0 \le \left\{ \frac{m}{n} \right\} \le 1-\frac{1}{|n|}.</math>

If ''n'' is positive<ref>Graham, Knuth, & Patashnik, p. 73</ref>
:<math>\left\lfloor\frac{x+m}{n}\right\rfloor = \left\lfloor\frac{\lfloor x\rfloor +m}{n}\right\rfloor,
</math>

:<math>\left\lceil\frac{x+m}{n}\right\rceil = \left\lceil\frac{\lceil x\rceil +m}{n}\right\rceil.
</math>

If ''m'' is positive<ref>Graham, Knuth, & Patashnik, p. 85</ref>

:<math>n=\left\lceil\frac{n\vphantom1}{m}\right\rceil + \left\lceil\frac{n-1}{m}\right\rceil +\dots+\left\lceil\frac{n-m+1}{m}\right\rceil,
</math>

:<math>n=\left\lfloor\frac{n\vphantom1}{m}\right\rfloor + \left\lfloor\frac{n+1}{m}\right\rfloor +\dots+\left\lfloor\frac{n+m-1}{m}\right\rfloor.
</math>

For ''m'' = 2 these imply

:<math>n= \left\lfloor \frac{n\vphantom1}{2}\right \rfloor + \left\lceil\frac{n\vphantom1}{2}\right \rceil.</math>

More generally,<ref>Graham, Knuth, & Patashnik, p. 85 and Ex. 3.15</ref> for positive ''m'' (See [[Hermite's identity]])

:<math>\lceil mx \rceil =\left\lceil x\right\rceil  + \left\lceil x-\frac{1}{m}\right\rceil +\dots+\left\lceil x-\frac{m-1}{m}\right\rceil,
</math>

:<math>\lfloor mx \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{m}\right\rfloor +\dots+\left\lfloor x+\frac{m-1}{m}\right\rfloor.
</math>

The following can be used to convert floors to ceilings and vice versa (with ''m'' being positive)<ref>Graham, Knuth, & Patashnik, Ex. 3.12</ref>

:<math>\left\lceil \frac{n\vphantom1}{m} \right\rceil = \left\lfloor \frac{n+m-1}{m} \right\rfloor = \left\lfloor \frac{n - 1}{m} \right\rfloor + 1, </math>

:<math>\left\lfloor \frac{n\vphantom1}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil = \left\lceil \frac{n + 1}{m} \right\rceil - 1, </math>

For all ''m'' and ''n'' strictly positive integers:<ref>Graham, Knuth, & Patashnik, p. 94.</ref>

:<math>\sum_{k = 1}^{n - 1} \left\lfloor \frac{k m}{n} \right\rfloor = \frac{(m - 1)(n - 1)+\gcd(m,n)-1}2,</math>

which, for positive and [[coprime]]  ''m'' and ''n'', reduces to

:<math>\sum_{k=1}^{n-1} \left\lfloor \frac{km}{n} \right\rfloor = \tfrac{1}{2}(m - 1)(n - 1) ,</math>

and similarly for the ceiling and fractional part functions (still for positive and [[coprime]]  ''m'' and ''n''),

:<math>\sum_{k=1}^{n-1} \left\lceil \frac{km}{n} \right\rceil = \tfrac{1}{2}(m + 1)(n - 1),</math>

:<math>\sum_{k=1}^{n-1} \left\{ \frac{km}{n} \right\} = \tfrac{1}{2}(n - 1).</math>


Since the right-hand side of the general case is symmetrical in ''m'' and ''n'', this implies that

:<math>\left\lfloor \frac{m\vphantom1}{n} \right \rfloor + \left\lfloor \frac{2m}{n} \right \rfloor + \dots + \left\lfloor \frac{(n-1)m}{n} \right \rfloor =
\left\lfloor \frac{n\vphantom1}{m} \right \rfloor + \left\lfloor \frac{2n}{m} \right \rfloor + \dots + \left\lfloor \frac{(m-1)n}{m} \right \rfloor.
</math>

More generally, if ''m'' and ''n'' are positive,

:<math>\begin{align}
&\left\lfloor \frac{x\vphantom1}{n} \right \rfloor +
\left\lfloor \frac{m+x}{n} \right \rfloor +
\left\lfloor \frac{2m+x}{n} \right \rfloor +
\dots +
\left\lfloor \frac{(n-1)m+x}{n} \right \rfloor\\[5mu]
=
&\left\lfloor \frac{x\vphantom1}{m} \right \rfloor +
\left\lfloor \frac{n+x}{m} \right \rfloor +
\left\lfloor \frac{2n+x}{m} \right \rfloor +
\cdots +
\left\lfloor \frac{(m-1)n+x}{m} \right \rfloor.
\end{align}
</math>

This is sometimes called a [[#Quadratic reciprocity|reciprocity law]].<ref>Graham, Knuth, & Patashnik, p. 94</ref>

Division by positive integers gives rise to an interesting and sometimes useful property. Assuming <math>m,n >0</math>,

:<math> m \leq \left\lfloor \frac{x}{n} \right \rfloor \iff n \leq  \left\lfloor \frac{x}{m} \right \rfloor \iff n \leq  \frac{ \lfloor x \rfloor }{m}. </math>

Similarly,

:<math> m \geq \left\lceil \frac{x}{n} \right \rceil \iff n \geq  \left\lceil \frac{x}{m} \right \rceil \iff n \geq  \frac{ \lceil x \rceil }{m}. </math>

Indeed,

:<math> m \leq \left\lfloor \frac{x}{n} \right \rfloor \implies m \leq \frac{x}{n} \implies n \leq \frac{x}{m} 
\implies n \leq \left \lfloor \frac{x}{m}\right \rfloor \implies \ldots \implies m \leq \left\lfloor \frac{x}{n} \right \rfloor,</math> 
keeping in mind that <math display=inline> \left\lfloor \frac{x}{n} \right\rfloor = \left\lfloor \frac{\lfloor x \rfloor}{n} \right\rfloor.</math>
The second equivalence involving the ceiling function can be proved similarly.