User:Virginia-American/Sandbox/Floor and Ceiling



In mathematics and computer science, the floor and ceiling functions map a real number to the next smallest or next largest integer. More precisely, floor(x) is the largest integer not greater than x and ceiling(x) is the smallest integer not less than x.

This article also discusses two related functions, the fractional part function and the mod operator.

Synonyms and Notations
Gauss introduced the square bracket notation [x] for the floor function in his third proof of quadratic reciprocity (1808). This remained the standard in mathematics until Iverson introduced the names "floor" and "ceiling" and the corresponding notations $$\lfloor x \rfloor$$ and $$\lceil x \rceil$$ in his 1962 programming language APL. Both notations are now used in mathematics; this article follows Iverson.

The floor function is also called the greatest integer or entier (French for "integer") function, and the floor of a nonnegative x may be called the integral part or integral value of x. Computer languages (other than APL) commonly use ENTIER(x) (Algol), floor(x), or int(x) (C and C++).

The ceiling function is usually denoted by ceil(x) or ceiling(x) in non-APL computer languages; there is no common alternative to Iverson's $$\lceil x \rceil$$ in mathematics.

Fractional part
The fractional part function, denoted by $$\{x\}$$ for real x is defined by the formula
 * $$\{x\} = x -\lfloor x\rfloor.$$

For all x,
 * $$0\le\{x\}<1.\;$$

If x is positive, the floor of x is simply x with everything to the right of the decimal point replaced with 0, and the fractional part is x with everything to the left of the decimal point replaced with 0.

mod operator
The mod operator, denoted by x mod y for real x and y, y ≠ 0, is defined by the formula
 * $$x \mbox{ mod } y = x-y\left\lfloor \frac{x}{y}\right\rfloor.$$

x mod y is always between 0 and y; i.e.

if y is positive,
 * $$0 \le x \mbox{ mod } y y.$$

If x is an integer and y is a positive integer,
 * $$(x\mbox{ mod }y) \equiv x \pmod{y}.$$

Formulas
$$\mathbb{Z}$$ is the set of integers (positive, negative, and zero). In these formulas, x and y are real numbers and k, m, and n are integers.

Definitions
Floor and ceiling may be defined by the set equations


 * $$ \lfloor x \rfloor=\max\, \{n\in\mathbb{Z}\mid n\le x\}.$$


 * $$ \lceil x \rceil=\min\,\{n\in\mathbb{Z}\mid n\ge x\}.$$

Since there is exactly one integer in a half-open interval of length one, for any real x there are unique integers m and n such that


 * $$x-1<m\le x \le n <x+1.\;$$

Then $$\lfloor x \rfloor = m\;$$  and  $$\;\lceil x \rceil = n\;$$ may also be taken as the definition of floor and ceiling.

As stated above,
 * $$\{x\} = x -\lfloor x\rfloor,$$


 * $$x \mbox{ mod } y = x-y\left\lfloor \frac{x}{y}\right\rfloor.$$

Equivalences
These formulas can be used to simplify expressions involving floors and ceilings.



\begin{align} \lfloor x \rfloor = n &\;\;\mbox{ if and only if } &n &\le x < n+1,\\ \lceil x \rceil = n &\;\;\mbox{ if and only if } &n -1 &< x < n,\\

\lfloor x \rfloor = n &\;\;\mbox{ if and only if } &x-1 &< n \le x,\\ \lceil x \rceil = n &\;\;\mbox{ if and only if } &x &\le n < x+1. \end{align} $$

In the language of order theory, the floor function is a residuated mapping, that is, part of a Galois connection: it is the upper adjoint of the function that embeds the integers into the reals.



\begin{align} x<n &\;\;\mbox{ if and only if } &\lfloor x \rfloor &< n, \\ n<x &\;\;\mbox{ if and only if } &n &< \lceil x \rceil, \\ x\le n &\;\;\mbox{ if and only if } &\lceil x \rceil &\le n, \\ n\le x &\;\;\mbox{ if and only if } &n &\le \lfloor x \rfloor. \end{align} $$

These formulas show how adding integers to the arguments affects the functions.



\begin{align} \lfloor x+n \rfloor &= \lfloor x \rfloor+n,\\ \lceil x+n \rceil &= \lceil x \rceil+n,\\ \{ x+n \} &= \{ x \}. \end{align} $$

The above may or may not be true for if n is not an integer, but we do have:


 * $$\lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x + y \rfloor \leq \lfloor x \rfloor + \lfloor y \rfloor + 1.$$

Relations among the functions
It is clear from the definitions that
 * $$\lfloor x \rfloor \le \lceil x \rceil,$$  with equality if and only if x is an integer, i.e.
 * $$\lceil x \rceil - \lfloor x \rfloor = \begin{cases}

0&\mbox{ if } x\in \mathbb{Z}\\ 1&\mbox{ if } x\not\in \mathbb{Z} \end{cases}$$

Negating the argument switches floor and ceiling and changes the sign:
 * $$\lfloor x \rfloor +\lceil -x \rceil=0,$$  i.e.


 * $$\lfloor x \rfloor + \lfloor -x \rfloor = \begin{cases}

0&\mbox{ if } x\in \mathbb{Z}\\ -1&\mbox{ if } x\not\in \mathbb{Z}, \end{cases}$$


 * $$\lceil x \rceil + \lceil -x \rceil = \begin{cases}

0&\mbox{ if } x\in \mathbb{Z}\\ 1&\mbox{ if } x\not\in \mathbb{Z}. \end{cases}$$

Negating the argument complements the fractional part:


 * $$\{ x \} + \{ -x \} = \begin{cases}

0&\mbox{ if } x\in \mathbb{Z}\\ 1&\mbox{ if } x\not\in \mathbb{Z}. \end{cases}$$

The floor, ceiling, and fractional part functions are idempotent:


 * $$\begin{align}\lfloor\lfloor x\rfloor\rfloor&=\lfloor x\rfloor,\\

\lceil\lceil x\rceil\rceil&=\lceil x\rceil,\\ \{\{ x\}\}&=\{ x\}.\end{align}$$

In fact, since for integers n  $$\lfloor n \rfloor = \lceil n \rceil = n,$$
 * $$\begin{align}\lfloor\lceil x\rceil\rfloor&=\lceil x\rceil,\\

\lceil\lfloor x\rfloor\rceil&=\lfloor x\rfloor.\end{align}$$

For fixed y, x mod y is idempotent:
 * $$(x \mbox{ mod } y) \mbox{ mod } y = x\mbox{ mod } y.\;$$

Also, from the definitions,
 * $$\{x\}= x\mbox{ mod } 1.\;$$

Quotients
If n ≠ 0,
 * $$0 \le \left \{\frac{m}{n} \right\} \le 1-\frac{1}{|n|}.$$

If n is positive
 * $$\left\lfloor\frac{x+m}{n}\right\rfloor = \left\lfloor\frac{\lfloor x\rfloor +m}{n}\right\rfloor,

$$


 * $$\left\lceil\frac{x+m}{n}\right\rceil = \left\lceil\frac{\lceil x\rceil +m}{n}\right\rceil.

$$

If m is positive


 * $$n=\left\lceil\frac{n}{m}\right\rceil + \left\lceil\frac{n-1}{m}\right\rceil +\dots+\left\lceil\frac{n-m+1}{m}\right\rceil,

$$


 * $$n=\left\lfloor\frac{n}{m}\right\rfloor + \left\lfloor\frac{n+1}{m}\right\rfloor +\dots+\left\lfloor\frac{n+m-1}{m}\right\rfloor.

$$

For m = 2 these imply


 * $$\lfloor k / 2 \rfloor + \lceil k / 2 \rceil = k$$.

More generally, for positive m


 * $$\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,

$$


 * $$\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.

$$

These can be used to convert floors to ceilings and vice-versa


 * $$\left\lceil \frac{n}{m} \right\rceil = \left\lfloor \frac{n - 1}{m} \right\rfloor + 1, $$


 * $$\left\lceil \frac{n}{m} \right\rceil = \left\lfloor \frac{n+m-1}{m} \right\rfloor.$$

If m and n are positive and coprime, then
 * $$\sum_{i=1}^{n-1} \lfloor im / n \rfloor = (m - 1) (n - 1) / 2.$$

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


 * $$\left\lfloor \frac{m}{n} \right \rfloor + \left\lfloor \frac{2m}{n} \right \rfloor + \dots + \left\lfloor \frac{(n-1)m}{n} \right \rfloor =

\left\lfloor \frac{n}{m} \right \rfloor + \left\lfloor \frac{2n}{m} \right \rfloor + \dots + \left\lfloor \frac{(m-1)n}{m} \right \rfloor. $$

More generally, if m and n are positive,


 * $$\begin{align}

&\left\lfloor \frac{x}{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\\= &\left\lfloor \frac{x}{m} \right \rfloor + \left\lfloor \frac{n+x}{m} \right \rfloor + \left\lfloor \frac{2n+x}{m} \right \rfloor + \dots + \left\lfloor \frac{(m-1)n+x}{m} \right \rfloor. \end{align} $$

This is sometimes called a reciprocity law.

Continuity
None of the functions discussed in this article are continuous, but all are piecewise linear. $$\lfloor x \rfloor$$ and $$\lceil x \rceil$$ are piecewise constant functions, with discontinuites at the integers. $$\{ x\}$$ also has discontinuites at the integers, and  $$x \mbox{ mod }y$$ as a function of x for fixed y is discontinuous at multiples of y.

$$\lfloor x \rfloor$$ is upper semi-continuous and  $$\lceil x \rceil$$  and $$\{ x\}\;$$  are lower semi-continuous. x mod y is lower semicontinuous for positive y and upper semi-continuous for negative y.

Series expansions
Since none of the functions discussed in this article are continuous, none of them have a power series expansion. Since floor and ceiling are not periodic, they do not have Fourier expnsions.

x mod y for fixed y has the Fourier series expansion


 * $$x \mbox{ mod } y = \frac{y}{2} - \frac{y}{\pi} \sum_{k=1}^\infty

\frac{\sin(\frac{2 \pi k x}{y})} {k}. $$

in particular {x} = x mod 1 is given by


 * $$\{x\}= \frac{1}{2} - \frac{1}{\pi} \sum_{k=1}^\infty

\frac{\sin(2 \pi k x)} {k}. $$

At points of discontinuity, a Fourier series converges to a value that is the average of its limits on the left and the right. For x mod y, y fixed, the Fourier series converges to y/2 at multiples of y. At points of continuity the series converges to the true value.

Using the formula {x} = x &minus; floor(x), floor(x) = x &minus; {x} gives


 * $$\lfloor x\rfloor = x - \frac{1}{2} + \frac{1}{\pi} \sum_{k=1}^\infty \frac{\sin(2 \pi k x)}{k}.$$

Quadratic reciprocity
Gauss's third proof of quadratic reciprocity, as modified by Eisenstein, has two basic steps.

Let p and q be distinct positive odd prime numbers, and let
 * $$m = \frac{p - 1}{2},\;\; n = \frac{q - 1}{2}.$$

First, Gauss's lemma is used to show that the Legendre symbols are given by


 * $$\left(\frac{q}{p}\right) = (-1)^{\left\lfloor\frac{q}{p}\right\rfloor +\left\lfloor\frac{2q}{p}\right\rfloor +\dots +\left\lfloor\frac{mq}{p}\right\rfloor }

$$

and
 * $$\left(\frac{p}{q}\right) = (-1)^{\left\lfloor\frac{p}{q}\right\rfloor +\left\lfloor\frac{2p}{q}\right\rfloor +\dots +\left\lfloor\frac{np}{q}\right\rfloor }.

$$

The second step is to use a geometric argument to show that


 * $$\left\lfloor\frac{q}{p}\right\rfloor +\left\lfloor\frac{2q}{p}\right\rfloor +\dots +\left\lfloor\frac{mq}{p}\right\rfloor

+\left\lfloor\frac{p}{q}\right\rfloor +\left\lfloor\frac{2p}{q}\right\rfloor +\dots +\left\lfloor\frac{np}{q}\right\rfloor

= mn. $$

Combining these formulas gives quadratic reciprocity in the form


 * $$\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{mn}=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}.$$

There are formulas that use floor to express the quadratic character of small numbers mod odd primes p:
 * $$\left(\frac{2}{p}\right) = (-1)^{\left\lfloor\frac{p+1}{4}\right\rfloor},$$


 * $$\left(\frac{3}{p}\right) = (-1)^{\left\lfloor\frac{p+1}{6}\right\rfloor}.$$

Rounding
The ordinary rounding of the positive number x to the nearest integer can be expressed as $$\lfloor x + 0.5\rfloor.$$

Number of digits
The number of digits in base b of a positive integer k is
 * $$\lfloor \log_{b}{k} \rfloor + 1.$$

Factors of factorials
Let n be a positive integer and p a positive prime number. The exponent of the highest power of p that divides n! is given by the formula


 * $$\left\lfloor\frac{n}{p}\right\rfloor + \left\lfloor\frac{n}{p^2}\right\rfloor + \left\lfloor\frac{n}{p^3}\right\rfloor + \dots

$$

Note that this is a finite sum, since the floors are zero when pk > n.

Spectra (Beatty's theorem)
Beatty's theorem shows how every positive irrational number gives rise to a partition of the natural numbers into two sequences via the floor function.

Euler's constant γ
There are formulas for Euler's constant &gamma; = 0.57721 56649 ... that involve the floor and ceiling, e.g.


 * $$\gamma =\int_1^\infty\left({1\over\lfloor x\rfloor}-{1\over x}\right)\,dx,$$


 * $$ \gamma =     \lim_{n \to \infty} \frac{1}{n}\, \sum_{k=1}^n \left ( \left \lceil \frac{n}{k} \right \rceil - \frac{n}{k} \right ),

$$

and



\gamma = \sum_{k=2}^\infty (-1)^k \frac{ \left \lfloor \log_2 k \right \rfloor}{k} = \tfrac12-\tfrac13 + 2\left(\tfrac14 - \tfrac15 + \tfrac16 - \tfrac17\right) + 3\left(\tfrac18 - \dots - \tfrac1{15}\right) + \dots $$

Riemann ζ function
The fractional part function also shows up in integral representations of the Riemann zeta function. It is straightforward to prove (by integration by parts) that if &phi;(x) is any function with a continuous derivative in the closed interval [a, b],


 * $$\sum_{a<n\le b}\phi(n) =

\int_a^b\phi(x) dx + \int_a^b(\{x\}-\frac12)\phi'(x) dx + (\{a\}-\frac12)\phi(a) - (\{b\}-\frac12)\phi(b) $$

Letting &phi;(n) = n&minus;s for real part of s greater than 1 and a and b be integers, and letting b approach infinity gives


 * $$\zeta(s) = s\int_1^\infty\frac{\frac12-\{x\}}{x^{s+1}}\;dx + \frac{1}{s-1} + \frac12.

$$

This formula is valid for all s with real part greater than &minus;1, (except s = 1, where there is a pole) and combined with the Fourier expansion for {x} can be used to extend the zeta function to the entire complex plane and to prove its functional equation.

Formula for prime numbers
Let r > 1 be an integer, pn be the nth prime, and define


 * $$\alpha = \sum_{m=1}^\infty p_m r^{-m^2}.$$

Then


 * $$p_n = \lfloor r^{n^2}\alpha \rfloor - r^{2n-1}\lfloor r^{(n-1)^2}\alpha\rfloor.$$

Until a way is found to calculate &alpha; without using the primes explicitly, this formula is of no practical use.

Unsolved problem
The study of Waring's problem has led to an unsolved problem:

Are there any positive integers k such that


 * $$3^k-2^k\left\lfloor \left(\frac32\right)^k \right\rfloor > 2^k-\left\lfloor \left(\frac32\right)^k \right\rfloor -2$$

Mahler has proved there can only be a finite number of such k; none are known.

The operator in C
C and related programming languages convert floating point values into integers using the type casting syntax:. The fractional part is discarded (i.e., the value is truncated toward zero).

Spreadsheet software
Most spreadsheet programs support some form of a  function. Although the details differ between programs, most implementations support a second parameter—a multiple of which the given number is to be rounded to. As a typical example,   would round 2 up to the nearest multiple of 3, so this would return 3. The definition of what "round up" means, however, differs from program to program.

Microsoft Excel's  function does not follow the mathematical definition, but rather as with   operator in C, it is a mixture of the floor and ceiling function: for x ≥ 0 it returns ceiling(x), and for x < 0 it returns floor(x). This has followed through to the Office Open XML file format. For example,  returns -5. A mathematical ceiling function can be emulated in Excel by using the formula " " (please note that this is not a general rule, as it depends on Excel's  function, which behaves differently that most programming languages).

The OpenDocument file format, as used by OpenOffice.org and others, follows the mathematical definition of ceiling for its  function, with an optional parameter for Excel compatibility. For example,  returns -4.

Typesetting
The floor and ceiling function are usually typeset with left and right square brackets where the upper (for floor function) or lower (for ceiling function) horizontal bars are missing, and, e.g., in the LaTeX typesetting system these symbols can be specified with the \lfloor, \rfloor, \lceil and \rceil commands in math mode. Unicode contains codepoints for these symbols, at – : ⌈x⌉, ⌊x⌋.