Floor and ceiling functions

Floor and ceiling functions

Floor function

Ceiling function

Inmathematics,thefloor function(orgreatest integer function) is thefunctionthat takes as input areal numberx,and gives as output the greatestintegerless than or equal tox,denoted⌊x⌋orfloor(x).Similarly, theceiling functionmapsxto the smallest integer greater than or equal tox,denoted⌈x⌉orceil(x).

For example, for floor:⌊2.4⌋ = 2,⌊−2.4⌋ = −3,and for ceiling:⌈2.4⌉ = 3,and⌈−2.4⌉ = −2.

Historically, the floor ofxhas been–and still is–called theintegral part,integer part,orentierofx,often denoted[x](as well as a variety of other notations).However, the same term,integer part,is also used fortruncationtowards zero, which differs from the floor function for negative numbers.

Fornan integer,⌊n⌋ = ⌈n⌉ = [n] =n.

Althoughfloor(x+1)andceil(x)produce graphs that appear exactly alike, they are not the same when the value of x is an exact integer. For example, whenx=2.0001;⌊2.0001+1⌋ = ⌈2.0001⌉ = 3.However, ifx=2, then⌊2+1⌋ = 3,while⌈2⌉ = 2.

Examples

x	Floor⌊x⌋	Ceiling⌈x⌉	Fractional part{x}
2	2	2	0
2.0001	2	3	0.0001
2.4	2	3	0.4
2.9	2	3	0.9
2.999	2	3	0.999
−2.7	−3	−2	0.3
−2	−2	−2	0

Notation

Theintegral partorinteger partof a number (partie entièrein the original) was first defined in 1798 byAdrien-Marie Legendrein his proof of theLegendre's formula.

Carl Friedrich Gaussintroduced the square bracket notation[x]in his third proof ofquadratic reciprocity(1808).This remained the standardin mathematics untilKenneth E. Iversonintroduced, in his 1962 bookA Programming Language,the names "floor" and "ceiling" and the corresponding notations⌊x⌋and⌈x⌉.(Iverson used square brackets for a different purpose, theIverson bracketnotation.) Both notations are now used in mathematics, although Iverson's notation will be followed in this article.

In some sources, boldface or double brackets⟦x⟧are used for floor, and reversed brackets⟧x⟦or]x[for ceiling.

Thefractional partis thesawtooth function,denoted by{x}for realxand defined by the formula

{x} =x− ⌊x⌋

For allx,

0 ≤ {x} < 1.

These characters are provided in Unicode:

• U+2308⌈LEFT CEILING(&lceil;, &LeftCeiling;)
• U+2309⌉RIGHT CEILING(&rceil;, &RightCeiling;)
• U+230A⌊LEFT FLOOR(&LeftFloor;, &lfloor;)
• U+230B⌋RIGHT FLOOR(&rfloor;, &RightFloor;)

In theLaTeXtypesetting system, these symbols can be specified with the\lceil, \rceil, \lfloor,and\rfloorcommands in math mode, and extended in size using\left\lceil, \right\rceil, \left\lfloor,and\right\rflooras needed.

Definition and properties

Given real numbersxandy,integersmandnand the set ofintegers$\mathbb {Z}$,floor and ceiling may be defined by the equations

$\lfloor x\rfloor =\max\{m\in \mathbb {Z} \mid m\leq x\},$

$\lceil x\rceil =\min\{n\in \mathbb {Z} \mid n\geq x\}.$

Since there is exactly one integer in ahalf-open intervalof length one, for any real numberx,there are unique integersmandnsatisfying the equation

$x-1<m\leq x\leq n<x+1.$

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

Equivalences

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

${\begin{alignedat}{3}\lfloor x\rfloor &=m\ \ &&{\mbox{ if and only if }}&m&\leq x<m+1,\\\lceil x\rceil &=n&&{\mbox{ if and only if }}&\ \ n-1&<x\leq n,\\\lfloor x\rfloor &=m&&{\mbox{ if and only if }}&x-1&<m\leq x,\\\lceil x\rceil &=n&&{\mbox{ if and only if }}&x&\leq n<x+1.\end{alignedat}}$

In the language oforder theory,the floor function is aresiduated mapping,that is, part of aGalois connection:it is the upper adjoint of the function that embeds the integers into the reals.

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

These formulas show how adding an integernto the arguments affects the functions:

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

The above are never true ifnis not an integer; however, for everyxandy,the following inequalities hold:

${\begin{aligned}\lfloor x\rfloor +\lfloor y\rfloor &\leq \lfloor x+y\rfloor \leq \lfloor x\rfloor +\lfloor y\rfloor +1,\\[3mu]\lceil x\rceil +\lceil y\rceil -1&\leq \lceil x+y\rceil \leq \lceil x\rceil +\lceil y\rceil.\end{aligned}}$

Monotonicity

Both floor and ceiling functions aremonotonically non-decreasing functions:

${\begin{aligned}x_{1}\leq x_{2}&\Rightarrow \lfloor x_{1}\rfloor \leq \lfloor x_{2}\rfloor,\\x_{1}\leq x_{2}&\Rightarrow \lceil x_{1}\rceil \leq \lceil x_{2}\rceil.\end{aligned}}$

Relations among the functions

It is clear from the definitions that

$\lfloor x\rfloor \leq \lceil x\rceil,$with equality if and only ifxis an integer, i.e.

$\lceil x\rceil -\lfloor x\rfloor ={\begin{cases}0&{\mbox{ if }}x\in \mathbb {Z} \\1&{\mbox{ if }}x\not \in \mathbb {Z} \end{cases}}$

In fact, for integersn,both floor and ceiling functions are theidentity:

$\lfloor n\rfloor =\lceil n\rceil =n.$

Negating the argument switches floor and ceiling and changes the sign:

${\begin{aligned}\lfloor x\rfloor +\lceil -x\rceil &=0\\-\lfloor x\rfloor &=\lceil -x\rceil \\-\lceil x\rceil &=\lfloor -x\rfloor \end{aligned}}$

and:

$\lfloor x\rfloor +\lfloor -x\rfloor ={\begin{cases}0&{\text{if }}x\in \mathbb {Z} \\-1&{\text{if }}x\not \in \mathbb {Z},\end{cases}}$

$\lceil x\rceil +\lceil -x\rceil ={\begin{cases}0&{\text{if }}x\in \mathbb {Z} \\1&{\text{if }}x\not \in \mathbb {Z}.\end{cases}}$

Negating the argument complements the fractional part:

$\{x\}+\{-x\}={\begin{cases}0&{\text{if }}x\in \mathbb {Z} \\1&{\text{if }}x\not \in \mathbb {Z}.\end{cases}}$

The floor, ceiling, and fractional part functions areidempotent:

${\begin{aligned}{\big \lfloor }\lfloor x\rfloor {\big \rfloor }&=\lfloor x\rfloor,\\{\big \lceil }\lceil x\rceil {\big \rceil }&=\lceil x\rceil,\\{\big \{}\{x\}{\big \}}&=\{x\}.\end{aligned}}$

The result of nested floor or ceiling functions is the innermost function:

${\begin{aligned}{\big \lfloor }\lceil x\rceil {\big \rfloor }&=\lceil x\rceil,\\{\big \lceil }\lfloor x\rfloor {\big \rceil }&=\lfloor x\rfloor \end{aligned}}$

due to the identity property for integers.

Quotients

Ifmandnare integers andn≠ 0,

$0\leq \left\{{\frac {m}{n}}\right\}\leq 1-{\frac {1}{|n|}}.$

Ifnis a positive integer

$\left\lfloor {\frac {x+m}{n}}\right\rfloor =\left\lfloor {\frac {\lfloor x\rfloor +m}{n}}\right\rfloor,$

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

Ifmis positive

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

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

Form= 2 these imply

$n=\left\lfloor {\frac {n{\vphantom {1}}}{2}}\right\rfloor +\left\lceil {\frac {n{\vphantom {1}}}{2}}\right\rceil.$

More generally,for positivem(SeeHermite's identity)

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

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

The following can be used to convert floors to ceilings and vice versa (mpositive)

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

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

For allmandnstrictly positive integers:

$\sum _{k=1}^{n-1}\left\lfloor {\frac {km}{n}}\right\rfloor ={\frac {(m-1)(n-1)+\gcd(m,n)-1}{2}},$

which, for positive andcoprimemandn,reduces to

$\sum _{k=1}^{n-1}\left\lfloor {\frac {km}{n}}\right\rfloor ={\tfrac {1}{2}}(m-1)(n-1),$

and similarly for the ceiling and fractional part functions (still for positive andcoprimemandn),

$\sum _{k=1}^{n-1}\left\lceil {\frac {km}{n}}\right\rceil ={\tfrac {1}{2}}(m+1)(n-1),$

$\sum _{k=1}^{n-1}\left\{{\frac {km}{n}}\right\}={\tfrac {1}{2}}(n-1).$

Since the right-hand side of the general case is symmetrical inmandn,this implies that

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

More generally, ifmandnare positive,

${\begin{aligned}&\left\lfloor {\frac {x{\vphantom {1}}}{n}}\right\rfloor +\left\lfloor {\frac {m+x}{n}}\right\rfloor +\left\lfloor {\frac {2m+x}{n}}\right\rfloor +\dots +\left\lfloor {\frac {(n-1)m+x}{n}}\right\rfloor \\[5mu]=&\left\lfloor {\frac {x{\vphantom {1}}}{m}}\right\rfloor +\left\lfloor {\frac {n+x}{m}}\right\rfloor +\left\lfloor {\frac {2n+x}{m}}\right\rfloor +\cdots +\left\lfloor {\frac {(m-1)n+x}{m}}\right\rfloor.\end{aligned}}$

This is sometimes called areciprocity law.

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

$m\leq \left\lfloor {\frac {x}{n}}\right\rfloor \iff n\leq \left\lfloor {\frac {x}{m}}\right\rfloor \iff n\leq {\frac {\lfloor x\rfloor }{m}}.$

Similarly,

$m\geq \left\lceil {\frac {x}{n}}\right\rceil \iff n\geq \left\lceil {\frac {x}{m}}\right\rceil \iff n\geq {\frac {\lceil x\rceil }{m}}.$

Indeed,

$m\leq \left\lfloor {\frac {x}{n}}\right\rfloor \implies m\leq {\frac {x}{n}}\implies n\leq {\frac {x}{m}}\implies n\leq \left\lfloor {\frac {x}{m}}\right\rfloor \implies \ldots \implies m\leq \left\lfloor {\frac {x}{n}}\right\rfloor,$

keeping in mind that${\textstyle \lfloor x/n\rfloor ={\bigl \lfloor }\lfloor x\rfloor /n{\bigr \rfloor }.}$ The second equivalence involving the ceiling function can be proved similarly.

Nested divisions

For positive integern,and arbitrary real numbersm,x:

$\left\lfloor {\frac {\lfloor x/m\rfloor }{n}}\right\rfloor =\left\lfloor {\frac {x}{mn}}\right\rfloor$

$\left\lceil {\frac {\lceil x/m\rceil }{n}}\right\rceil =\left\lceil {\frac {x}{mn}}\right\rceil.$

Continuity and series expansions

None of the functions discussed in this article arecontinuous,but all arepiecewise linear:the functions$\lfloor x\rfloor$,$\lceil x\rceil$,and$\{x\}$have discontinuities at the integers.

$\lfloor x\rfloor$isupper semi-continuousand$\lceil x\rceil$and$\{x\}$are lower semi-continuous.

Since none of the functions discussed in this article are continuous, none of them have apower seriesexpansion. Since floor and ceiling are not periodic, they do not have uniformly convergentFourier seriesexpansions. The fractional part function has Fourier series expansion

$\{x\}={\frac {1}{2}}-{\frac {1}{\pi }}\sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}$

forxnot an integer.

At points of discontinuity, a Fourier series converges to a value that is the average of its limits on the left and the right, unlike the floor, ceiling and fractional part functions: foryfixed andxa multiple ofythe Fourier series given converges toy/2, rather than toxmody= 0. At points of continuity the series converges to the true value.

Using the formula floor(x) = x − {x} gives

$\lfloor x\rfloor =x-{\frac {1}{2}}+{\frac {1}{\pi }}\sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}$

forxnot an integer.

Applications

Mod operator

For an integerxand a positive integery,themodulo operation,denoted byxmody,gives the value of the remainder whenxis divided byy.This definition can be extended to realxandy,y≠ 0, by the formula

$x{\bmod {y}}=x-y\left\lfloor {\frac {x}{y}}\right\rfloor.$

Then it follows from the definition of floor function that this extended operation satisfies many natural properties. Notably,xmodyis always between 0 andy,i.e.,

ifyis positive,

$0\leq x{\bmod {y}}<y,$

and ifyis negative,

$0\geq x{\bmod {y}}>y.$

Quadratic reciprocity

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

Letpandqbe distinct positive odd prime numbers, and let$m={\tfrac {1}{2}}(p-1),$$n={\tfrac {1}{2}}(q-1).$

First,Gauss's lemmais used to show that theLegendre symbolsare given by

${\begin{aligned}\left({\frac {q}{p}}\right)&=(-1)^{\left\lfloor {\frac {q}{p}}\right\rfloor +\left\lfloor {\frac {2q}{p}}\right\rfloor +\dots +\left\lfloor {\frac {mq}{p}}\right\rfloor },\\[5mu]\left({\frac {p}{q}}\right)&=(-1)^{\left\lfloor {\frac {p}{q}}\right\rfloor +\left\lfloor {\frac {2p}{q}}\right\rfloor +\dots +\left\lfloor {\frac {np}{q}}\right\rfloor }.\end{aligned}}$

The second step is to use ageometricargument 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 primesp:

${\begin{aligned}\left({\frac {2}{p}}\right)&=(-1)^{\left\lfloor {\frac {p+1}{4}}\right\rfloor },\\[5mu]\left({\frac {3}{p}}\right)&=(-1)^{\left\lfloor {\frac {p+1}{6}}\right\rfloor }.\end{aligned}}$

Rounding

For an arbitrary real number$x$,rounding$x$to the nearest integer withtie breakingtowards positive infinity is given by${\text{rpi}}(x)=\left\lfloor x+{\tfrac {1}{2}}\right\rfloor =\left\lceil {\tfrac {1}{2}}\lfloor 2x\rfloor \right\rceil$;rounding towards negative infinity is given as${\text{rni}}(x)=\left\lceil x-{\tfrac {1}{2}}\right\rceil =\left\lfloor {\tfrac {1}{2}}\lceil 2x\rceil \right\rfloor$.

If tie-breaking is away from 0, then the rounding function is${\text{ri}}(x)=\operatorname {sgn}(x)\left\lfloor |x|+{\tfrac {1}{2}}\right\rfloor$(seesign function), androunding towards evencan be expressed with the more cumbersome$\lfloor x\rceil =\left\lfloor x+{\tfrac {1}{2}}\right\rfloor +{\bigl \lceil }{\tfrac {1}{4}}(2x-1){\bigr \rceil }-{\bigl \lfloor }{\tfrac {1}{4}}(2x-1){\bigr \rfloor }-1$,which is the above expression for rounding towards positive infinity${\text{rpi}}(x)$minus anintegralityindicatorfor${\tfrac {1}{4}}(2x-1)$.

Rounding areal number$x$to the nearest integer value forms a very basic type ofquantizer– auniformone. A typical (mid-tread) uniform quantizer with a quantizationstep sizeequal to some value$\Delta$can be expressed as

$Q(x)=\Delta \cdot \left\lfloor {\frac {x}{\Delta }}+{\frac {1}{2}}\right\rfloor$,

Number of digits

The number of digits inbasebof a positive integerkis

$\lfloor \log _{b}{k}\rfloor +1=\lceil \log _{b}{(k+1)}\rceil.$

Number of strings without repeated characters

The number of possiblestringsof arbitrary length that doesn't use any character twice is given by

$(n)_{0}+\cdots +(n)_{n}=\lfloor en!\rfloor$

where:

• n> 0 is the number of letters in the alphabet (e.g., 26 inEnglish)
• thefalling factorial$(n)_{k}=n(n-1)\cdots (n-k+1)$denotes the number of strings of lengthkthat don't use any character twice.
• n!denotes thefactorialofn
• e= 2.718... isEuler's number

Forn= 26, this comes out to 1096259850353149530222034277.

Factors of factorials

Letnbe a positive integer andpa positive prime number. The exponent of the highest power ofpthat dividesn!is given by a version ofLegendre's formula

$\left\lfloor {\frac {n}{p}}\right\rfloor +\left\lfloor {\frac {n}{p^{2}}}\right\rfloor +\left\lfloor {\frac {n}{p^{3}}}\right\rfloor +\dots ={\frac {n-\sum _{k}a_{k}}{p-1}}$

where${\textstyle n=\sum _{k}a_{k}p^{k}}$is the way of writingnin basep.This is a finite sum, since the floors are zero whenp^k>n.

Beatty sequence

TheBeatty sequenceshows how every positiveirrational numbergives rise to a partition of thenatural numbersinto two sequences via the floor function.

Euler's constant (γ)

There are formulas forEuler's constantγ = 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

$\gamma =\sum _{k=2}^{\infty }(-1)^{k}{\frac {\left\lfloor \log _{2}k\right\rfloor }{k}}={\tfrac {1}{2}}-{\tfrac {1}{3}}+2\left({\tfrac {1}{4}}-{\tfrac {1}{5}}+{\tfrac {1}{6}}-{\tfrac {1}{7}}\right)+3\left({\tfrac {1}{8}}-\cdots -{\tfrac {1}{15}}\right)+\cdots$

Riemann zeta function (ζ)

The fractional part function also shows up in integral representations of theRiemann zeta function.It is straightforward to prove (using integration by parts)that if$\varphi (x)$is any function with a continuous derivative in the closed interval [a,b],

$\sum _{a<n\leq b}\varphi (n)=\int _{a}^{b}\varphi (x)\,dx+\int _{a}^{b}\left(\{x\}-{\tfrac {1}{2}}\right)\varphi '(x)\,dx+\left(\{a\}-{\tfrac {1}{2}}\right)\varphi (a)-\left(\{b\}-{\tfrac {1}{2}}\right)\varphi (b).$

Letting$\varphi (n)=n^{-s}$forreal partofsgreater than 1 and lettingaandbbe integers, and lettingbapproach infinity gives

$\zeta (s)=s\int _{1}^{\infty }{\frac {{\frac {1}{2}}-\{x\}}{x^{s+1}}}\,dx+{\frac {1}{s-1}}+{\frac {1}{2}}.$

This formula is valid for allswith real part greater than −1, (excepts= 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.

Fors=σ+itin the critical strip 0 <σ< 1,

$\zeta (s)=s\int _{-\infty }^{\infty }e^{-\sigma \omega }(\lfloor e^{\omega }\rfloor -e^{\omega })e^{-it\omega }\,d\omega.$

In 1947van der Polused this representation to construct an analogue computer for finding roots of the zeta function.

Formulas for prime numbers

The floor function appears in several formulas characterizing prime numbers. For example, since${\textstyle {\bigl \lfloor }{\frac {n}{m}}{\bigr \rfloor }-{\bigl \lfloor }{\frac {n-1}{m}}{\bigr \rfloor }}$is equal to 1 ifmdividesn,and to 0 otherwise, it follows that a positive integernis a primeif and only if

$\sum _{m=1}^{\infty }\left({\biggl \lfloor }{\frac {n}{m}}{\biggr \rfloor }-{\biggl \lfloor }{\frac {n-1}{m}}{\biggr \rfloor }\right)=2.$

One may also give formulas for producing the prime numbers. For example, letp_nbe then-th prime, and for any integerr> 1, define the real numberαby the sum

$\alpha =\sum _{m=1}^{\infty }p_{m}r^{-m^{2}}.$

Then

$p_{n}=\left\lfloor r^{n^{2}}\alpha \right\rfloor -r^{2n-1}\left\lfloor r^{(n-1)^{2}}\alpha \right\rfloor.$

A similar result is that there is a numberθ= 1.3064... (Mills' constant) with the property that

$\left\lfloor \theta ^{3}\right\rfloor,\left\lfloor \theta ^{9}\right\rfloor,\left\lfloor \theta ^{27}\right\rfloor,\dots$

are all prime.

There is also a numberω= 1.9287800... with the property that

$\left\lfloor 2^{\omega }\right\rfloor,\left\lfloor 2^{2^{\omega }}\right\rfloor,\left\lfloor 2^{2^{2^{\omega }}}\right\rfloor,\dots$

are all prime.

Letπ(x) be the number of primes less than or equal tox.It is a straightforward deduction fromWilson's theoremthat

$\pi (n)=\sum _{j=2}^{n}{\Biggl \lfloor }{\frac {(j-1)!+1}{j}}-\left\lfloor {\frac {(j-1)!}{j}}\right\rfloor {\Biggr \rfloor }.$

Also, ifn≥ 2,

$\pi (n)=\sum _{j=2}^{n}{\Biggl \lfloor }1\,{\bigg /}\ {\sum _{k=2}^{j}\left\lfloor \left\lfloor {\frac {j}{k}}\right\rfloor {\frac {k}{j}}\right\rfloor }{\Biggr \rfloor }.$

None of the formulas in this section are of any practical use.

Solved problems

Ramanujansubmitted these problems to theJournal of the Indian Mathematical Society.

Ifnis a positive integer, prove that

1. $\left\lfloor {\tfrac {n}{3}}\right\rfloor +\left\lfloor {\tfrac {n+2}{6}}\right\rfloor +\left\lfloor {\tfrac {n+4}{6}}\right\rfloor =\left\lfloor {\tfrac {n}{2}}\right\rfloor +\left\lfloor {\tfrac {n+3}{6}}\right\rfloor,$
2. $\left\lfloor {\tfrac {1}{2}}+{\sqrt {n+{\tfrac {1}{2}}}}\right\rfloor =\left\lfloor {\tfrac {1}{2}}+{\sqrt {n+{\tfrac {1}{4}}}}\right\rfloor,$
3. $\left\lfloor {\sqrt {n}}+{\sqrt {n+1}}\right\rfloor =\left\lfloor {\sqrt {4n+2}}\right\rfloor.$

Some generalizations to the above floor function identities have been proven.

Unsolved problem

The study ofWaring's problemhas led to an unsolved problem:

Are there any positive integersk≥ 6 such that

$3^{k}-2^{k}{\Bigl \lfloor }{\bigl (}{\tfrac {3}{2}}{\bigr )}^{k}{\Bigr \rfloor }>2^{k}-{\Bigl \lfloor }{\bigl (}{\tfrac {3}{2}}{\bigr )}^{k}{\Bigr \rfloor }-2\?$

Mahlerhas proved there can only be a finite number of suchk;none are known.

Computer implementations

Int function from floating-point conversion inC

In most programming languages, the simplest method to convert a floating point number to an integer does not do floor or ceiling, but truncation. The reason for this is historical, as the first machines usedones' complementand truncation was simpler to implement (floor is simpler intwo's complement).FORTRANwas defined to require this behavior and thus almost all processors implement conversion this way. Some consider this to be an unfortunate historical design decision that has led to bugs handling negative offsets and graphics on the negative side of the origin.

Anarithmetic right-shiftof a signed integer$x$by$n$is the same as$\left\lfloor x/2^{n}\right\rfloor$.Division by a power of 2 is often written as a right-shift, not for optimization as might be assumed, but because the floor of negative results is required. Assuming such shifts are "premature optimization" and replacing them with division can break software.

Many programming languages (includingC,C++,C#,Java,PHP,R,andPython) provide standard functions for floor and ceiling, usually calledfloorandceil,or less commonlyceiling.The languageAPLuses⌊xfor floor. TheJ Programming Language,a follow-on to APL that is designed to use standard keyboard symbols, uses<.for floor and>.for ceiling. ALGOLusesentierfor floor.

InMicrosoft Excelthe funtionINTrounds down rather than toward zero,whileFLOORrounds toward zero, the opposite of what "int" and "floor" do in other languages. Since 2010FLOORhas been changed to error if the number is negative.TheOpenDocumentfile format, as used byOpenOffice.org,Libreofficeand others,INTandFLOORboth do floor, andFLOORhas a third argument to reproduce Excel's earlier behavior.

See also

• Bracket (mathematics)
• Integer-valued function
• Step function
• Modulo operation

References

• J.W.S. Cassels (1957),An introduction to Diophantine approximation,Cambridge Tracts in Mathematics and Mathematical Physics, vol. 45,Cambridge University Press
• Crandall, Richard; Pomerance, Carl (2001),Prime Numbers: A Computational Perspective,New York:Springer,ISBN0-387-94777-9
• Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren (1994),Concrete Mathematics,Reading Ma.: Addison-Wesley,ISBN0-201-55802-5
• Hardy, G. H.; Wright, E. M. (1980),An Introduction to the Theory of Numbers (Fifth edition),Oxford:Oxford University Press,ISBN978-0-19-853171-5
• Nicholas J. Higham,Handbook of writing for the mathematical sciences,SIAM.ISBN0-89871-420-6,p. 25
• ISO/IEC.ISO/IEC 9899::1999(E): Programming languages — C(2nd ed), 1999; Section 6.3.1.4, p. 43.
• Iverson, Kenneth E. (1962),A Programming Language,Wiley
• Lemmermeyer, Franz (2000),Reciprocity Laws: from Euler to Eisenstein,Berlin:Springer,ISBN3-540-66957-4
• Ramanujan, Srinivasa (2000),Collected Papers,Providence RI: AMS / Chelsea,ISBN978-0-8218-2076-6
• Ribenboim, Paulo (1996),The New Book of Prime Number Records,New York: Springer,ISBN0-387-94457-5
• Michael Sullivan.Precalculus,8th edition, p. 86
• Titchmarsh, Edward Charles; Heath-Brown, David Rodney ( "Roger" ) (1986),The Theory of the Riemann Zeta-function(2nd ed.), Oxford: Oxford U. P.,ISBN0-19-853369-1

External links

• "Floor function",Encyclopedia of Mathematics,EMS Press,2001 [1994]
• Štefan Porubský,"Integer rounding functions",Interactive Information Portal for Algorithmic Mathematics,Institute of Computer Science of the Czech Academy of Sciences, Prague, Czech Republic, retrieved 24 October 2008
• Weisstein, Eric W."Floor Function".MathWorld.
• Weisstein, Eric W."Ceiling Function".MathWorld.