取整函数

由上下取整函数的定义，可见

${\displaystyle \lfloor x\rfloor \leq \lceil x\rceil ,}$

等号当且仅当${\displaystyle x}$为整数，即

${\displaystyle \lceil x\rceil -\lfloor x\rfloor ={\begin{cases}0,&{\text{ 若 }}\ x\in \mathbb {Z} ,\\1,&{\text{ 若 }}\ x\not \in \mathbb {Z} .\end{cases}}}$

实际上，上取整与下取整函数作用于整数${\displaystyle n}$，效果等同恒等函数：

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

自变量加负号，相当于将上取整与下取整互换，外面再加负号，即：

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

且：

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

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

至于小数部分${\displaystyle \{x\}=x-\lfloor x\rfloor }$，自变量取相反数会使小数部分变成关于1的「补数」：

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

上取整、下取整、小数部分皆为幂等函数，即函数叠代两次的结果等于自身：

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

而多个上取整与下取整依次叠代的效果，相当于最内层一个：

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

因为外层取整函数实际衹作用在整数上，不带来变化。

若${\displaystyle m}$和${\displaystyle n}$为正整数，且${\displaystyle n\neq 0}$，则

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

若${\displaystyle n}$为正整数，则[3]

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

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

若${\displaystyle m}$为正数，则[4]

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

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

代${\displaystyle m=2}$，上式推出：

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

更一般地，对正整数${\displaystyle m}$，有埃尔米特恒等式：[5]

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

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

对于正整数${\displaystyle m}$，以下两式可将上下取整函数互相转化：[6]

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

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

对任意正整数${\displaystyle m}$和${\displaystyle n}$，有：[7]

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

作为特例，当${\displaystyle m}$和${\displaystyle n}$互质时，上式简化为

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

此等式可以几何方式证明。又由于右式关于${\displaystyle m}$、${\displaystyle n}$对称，可得

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

更一般地，对正整数${\displaystyle m,n}$，有

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

上式算是一种「互反律」（）[7]，与§ 二次互反律有关。

高斯给出二次互反律的第三个证明，经艾森斯坦修改后，有以下两个主要步骤。[8][9]

设${\displaystyle p}$、${\displaystyle q}$为互异奇质数，又设

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

首先，利用高斯引理，证明勒让德符号可表示为和式：

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

同样

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

其后，采用几何论证，证明

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

总结上述两步，得

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

此即二次互反律。一些小整数模奇质数${\displaystyle p}$的二次特征标，可以高斯符号表示，如：[10]

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

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

下取整函数出现于若干刻画质数的公式之中。举例，因为${\displaystyle \left\lfloor {\frac {n}{m}}\right\rfloor -\left\lfloor {\frac {n-1}{m}}\right\rfloor }$在${\displaystyle m}$整除${\displaystyle n}$时等于${\displaystyle 1}$，否则为${\displaystyle 0}$，所以正整数${\displaystyle n}$为质数当且仅当[11]

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

除表示质数的条件外，还可以写出公式使其取值为质数。例如，记第${\displaystyle n}$个质数为${\displaystyle p_{n}}$，任选一个整数${\displaystyle r>1}$，然后定义实数${\displaystyle \alpha }$为

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

则衹用取整、幂、四则运算可以写出质数公式：[12]

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

类似还有米尔斯常数${\displaystyle \theta =1.3064\ldots }$，使

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

皆为质数。[13]

若不叠代三次方函数，改为叠代以${\displaystyle 2}$为㡳的指数函数，亦有${\displaystyle \omega =1.9287800\ldots }$使

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

皆为质数。[13]

以质数计算函数${\displaystyle \pi (x)}$表示小于或等于${\displaystyle x}$的质数个数。由威尔逊定理，可知[14]

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

又或者，当${\displaystyle n\geq 2}$时：[15]

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

本小节的公式未有任何实际用途。[16][17]