Floor and ceiling functions

These formulas can be used to simplify expressions involving floors and ceilings.[10]

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

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.

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

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

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

The above are never true if n is not an integer; however, for every x and y, the following inequalities hold:

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

Both floor and ceiling functions are the monotonically non-decreasing function:

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

It is clear from the definitions that

${\displaystyle \lfloor x\rfloor \leq \lceil x\rceil ,}$   with equality if and only if x is an integer, i.e.

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

In fact, for integers n, both floor and ceiling functions are the identity:

${\displaystyle \lfloor n\rfloor =\lceil n\rceil =n.}$

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

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

and:

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

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

Negating the argument complements the fractional part:

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

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

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

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

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

due to the identity property for integers.

If m and n are integers and n ≠ 0,

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

If n is a positive integer[11]

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

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

If m is positive[12]

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

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

For m = 2 these imply

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

More generally,[13] for positive m (See Hermite's identity)

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

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

The following can be used to convert floors to ceilings and vice versa (m positive)[14]

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

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

For all m and n strictly positive integers:[15]

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

which, for positive and coprime m and n, reduces to

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

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

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

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

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

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

More generally, if m and n are positive,

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

This is sometimes called a reciprocity law.[16]

For positive integer n, and arbitrary real numbers m,x:[17]

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

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

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

${\displaystyle \lfloor x\rfloor }$  is upper semi-continuous and  ${\displaystyle \lceil x\rceil }$  and ${\displaystyle \{x\}}$  are lower semi-continuous.

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 uniformly convergent Fourier series expansions. The fractional part function has Fourier series expansion[18]

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

for x not 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: for y fixed and x a multiple of y the Fourier series given converges to y/2, rather than to x mod y = 0. At points of continuity the series converges to the true value.

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

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

for x not an integer.

For an integer x and a positive integer y, the modulo operation, denoted by x mod y, gives the value of the remainder when x is divided by y. This definition can be extended to real x and y, y ≠ 0, by the formula

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

Then it follows from the definition of floor function that this extended operation satisfies many natural properties. Notably, x mod y is always between 0 and y, i.e.,

if y is positive,

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

and if y is negative,

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

Gauss's third proof of quadratic reciprocity, as modified by Eisenstein, has two basic steps.[19][20]

Let p and q be distinct positive odd prime numbers, and let

${\displaystyle m={\frac {p-1}{2}},}$ ${\displaystyle n={\frac {q-1}{2}}.}$

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

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

and

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

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

${\displaystyle \left\lfloor {\frac {q}{p}}\right\rfloor +\left\lfloor {\frac {2q}{p}}\right\rfloor +\dots +\left\lfloor {\frac {mq}{p}}\right\rfloor +\left\lfloor {\frac {p}{q}}\right\rfloor +\left\lfloor {\frac {2p}{q}}\right\rfloor +\dots +\left\lfloor {\frac {np}{q}}\right\rfloor =mn.}$

Combining these formulas gives quadratic reciprocity in the form

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

There are formulas that use floor to express the quadratic character of small numbers mod odd primes p:[21]

${\displaystyle \left({\frac {2}{p}}\right)=(-1)^{\left\lfloor {\frac {p+1}{4}}\right\rfloor },}$

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

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

If tie-breaking is away from 0, then the rounding function is ${\displaystyle {\text{ri}}(x)=\operatorname {sgn}(x)\left\lfloor |x|+{\tfrac {1}{2}}\right\rfloor }$ (see sign function), and rounding towards even can be expressed with the more cumbersome ${\displaystyle \lfloor x\rceil =\left\lfloor x+{\tfrac {1}{2}}\right\rfloor +\left\lceil {\tfrac {2x-1}{4}}\right\rceil -\left\lfloor {\tfrac {2x-1}{4}}\right\rfloor -1}$, which is the above expression for rounding towards positive infinity ${\displaystyle {\text{rpi}}(x)}$ minus an integrality indicator for ${\displaystyle {\tfrac {2x-1}{4}}}$.

The number of digits in base b of a positive integer k is

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

The number of possible strings of arbitrary length that don't use any character twice is given by[22]

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

where:

• n > 0 is the number of letters in the alphabet (e.g., 26 in English)
• the falling factorial ${\displaystyle (n)_{k}=n(n-1)\cdots (n-k+1)}$ denotes the number of strings of length k that don't use any character twice.
• n! denotes the factorial of n
• e = 2.718… is Euler's number

For n = 26, this comes out to 1096259850353149530222034277.

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 a version of Legendre's formula[23]

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

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

The Beatty sequence shows how every positive irrational number gives rise to a partition of the natural numbers into two sequences via the floor function.[24]

There are formulas for Euler's constant γ = 0.57721 56649 ... that involve the floor and ceiling, e.g.[25]

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

${\displaystyle \gamma =\lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}\left(\left\lceil {\frac {n}{k}}\right\rceil -{\frac {n}{k}}\right),}$

and

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

The fractional part function also shows up in integral representations of the Riemann zeta function. It is straightforward to prove (using integration by parts)[26] that if ${\displaystyle \phi (x)}$ is any function with a continuous derivative in the closed interval [a, b],

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

Letting ${\displaystyle \phi (n)={n}^{-s}}$ for real part of s greater than 1 and letting a and b be integers, and letting b approach infinity gives

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

This formula is valid for all s with real part greater than −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.[27]

For s = σ + it in the critical strip 0 < σ < 1,

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

In 1947 van der Pol used this representation to construct an analogue computer for finding roots of the zeta function.[28]

The floor function appears in several formulas characterizing prime numbers. For example, since ${\displaystyle \left\lfloor {\frac {n}{m}}\right\rfloor -\left\lfloor {\frac {n-1}{m}}\right\rfloor }$ is equal to 1 if m divides n, and to 0 otherwise, it follows that a positive integer n is a prime if and only if[29]

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

One may also give formulas for producing the prime numbers. For example, let p_n be the n-th prime, and for any integer r > 1, define the real number α by the sum

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

Then[30]

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

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

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

are all prime.[31]

There is also a number ω = 1.9287800... with the property that

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

are all prime.[31]

Let π(x) be the number of primes less than or equal to x. It is a straightforward deduction from Wilson's theorem that[32]

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

Also, if n ≥ 2,[33]

${\displaystyle \pi (n)=\sum _{j=2}^{n}\left\lfloor {\frac {1}{\sum _{k=2}^{j}\left\lfloor \left\lfloor {\frac {j}{k}}\right\rfloor {\frac {k}{j}}\right\rfloor }}\right\rfloor .}$

None of the formulas in this section are of any practical use.[34][35]

Ramanujan submitted these problems to the Journal of the Indian Mathematical Society.[36]

If n is a positive integer, prove that

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

Some generalizations to the above floor function identities have been proven.[37]

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

Are there any positive integers k ≥ 6 such that[38]

${\displaystyle 3^{k}-2^{k}\left\lfloor \left({\tfrac {3}{2}}\right)^{k}\right\rfloor >2^{k}-\left\lfloor \left({\tfrac {3}{2}}\right)^{k}\right\rfloor -2}$ ?

Mahler[39] has proved there can only be a finite number of such k; none are known.