Mod operator 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 :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, :0 \le x \bmod y and if
y is negative, :0 \ge x \bmod y >y.
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 = \tfrac12(p - 1), n = \tfrac12(q - 1). First,
Gauss's lemma is used to show that the
Legendre symbols are given by :\begin{align} \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{align} 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: :\begin{align} \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{align}
Rounding For an arbitrary real number x,
rounding x to the nearest integer with
tie breaking towards positive infinity is given by :\text{rpi}(x)=\left\lfloor x+\tfrac{1}{2}\right\rfloor = \left\lceil \tfrac12\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 \tfrac12 \lceil 2x \rceil \right\rfloor. If tie-breaking is away from 0, then the rounding function is :\text{ri}(x) = \sgn(x)\left\lfloor|x|+\tfrac{1}{2}\right\rfloor (where \sgn is the
sign function), and
rounding towards even can be expressed with the more cumbersome :\lfloor x\rceil=\left\lfloor x+\tfrac{1}{2}\right\rfloor+\left\lceil\tfrac14(2x-1)\right\rceil-\left\lfloor\tfrac14(2x-1)\right\rfloor-1, which is the above expression for rounding towards positive infinity \text{rpi}(x) minus an
integrality indicator for \tfrac14(2x-1). Rounding a
real number x to the nearest integer value forms a very basic type of
quantizer – a
uniform one. A typical (
mid-tread) uniform quantizer with a quantization
step size equal 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 in
base b of a positive integer
k is :\lfloor \log_{b}{k} \rfloor + 1 = \lceil \log_{b}{(k+1)} \rceil .
Number of strings without repeated characters The number of possible
strings of arbitrary length that doesn't use any character twice is given by :(n)_0 + \cdots + (n)_n = \lfloor e n! \rfloor where: • > 0 is the number of letters in the alphabet (e.g., 26 in
English) • the
falling factorial (n)_k = n(n-1)\cdots(n-k+1) denotes the number of strings of length that don't use any character twice. • ! denotes the
factorial of • = 2.718... is
Euler's number For = 26, this comes out to 1096259850353149530222034277.
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 a version of
Legendre'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 n = \sum_{k}a_kp^k is the way of writing
n in base
p. This is a finite sum, since the floors are zero when
pk >
n.
Beatty sequence 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.
Euler's constant (γ) There are formulas for
Euler's constant γ = 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 - \cdots - \tfrac1{15}\right) + \cdots
Riemann zeta function (ζ) The fractional part function also shows up in integral representations of the
Riemann 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 Letting \varphi(n) = n^{-s} for
real part of
s greater than 1 and letting
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} + \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. For
s =
σ +
it in the critical strip 0 \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.
Formulas for prime numbers The floor function appears in several formulas characterizing prime numbers. For example, since \left\lfloor\frac{n}{m} \right\rfloor -\left\lfloor\frac{n-1}{m}\right\rfloor = \begin{cases} 1 &\text{if } m \text{ divides } n \\ 0 &\text{otherwise}, \end{cases} it follows that a positive integer
n is a prime
if and only if :\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
pn be the
n-th prime, and for any integer
r > 1, define the real number
α by 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 number
θ = 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 number
ω = 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. :\pi(n) = \sum_{j=2}^n\Biggl\lfloor\frac{(j-1)!+1}{j} - \left\lfloor\frac{(j-1)!}{j}\right\rfloor\Biggr\rfloor. Also, if
n ≥ 2, :\pi(n) = \sum_{j=2}^n \left\lfloor \frac{1} {\displaystyle\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.
Solved problems Ramanujan submitted these problems to the
Journal of the Indian Mathematical Society. If
n is a positive integer, prove that \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, \left\lfloor\tfrac12 + \sqrt{n+\tfrac12}\right\rfloor = \left\lfloor\tfrac12 + \sqrt{n+\tfrac14}\right\rfloor, \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 of
Waring's problem has led to an unsolved problem: Are there any positive integers
k ≥ 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 \ ?
Mahler has proved there can only be a finite number of such
k; none are known.{{cite journal ==Computer implementations==