取整函数

由上下取整函數的定義，可見

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

等號當且僅當${\displaystyle x}$為整數，即

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

實際上，上取整與下取整函數作用於整數${\displaystyle n}$，效果等同恆等函數：

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

自變量加負號，相當於將上取整與下取整互換，外面再加負號，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：

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

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

至於小數部分${\displaystyle \{x\}=x-\lfloor x\rfloor }$，自變量取相反數會使小數部分變成關於1的「補數」：

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

上取整、下取整、小數部分皆為冪等函數，即函數疊代兩次的結果等於自身：

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

而多個上取整與下取整依次疊代的效果，相當於最內層一個：

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

因為外層取整函數實際衹作用在整數上，不帶來變化。

若${\displaystyle m}$和${\displaystyle n}$為正整數，且${\displaystyle n\neq 0}$，則

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

若${\displaystyle n}$為正整數，則[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${\displaystyle m}$為正數，則[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，對正整數${\displaystyle m}$，有埃爾米特恆等式：[5]

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

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

對於正整數${\displaystyle m}$，以下兩式可將上下取整函數互相轉化：[6]

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

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

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

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

作為特例，當${\displaystyle m}$和${\displaystyle n}$互質時，上式簡化為

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

此等式可以幾何方式證明。又由於右式關於${\displaystyle m}$、${\displaystyle n}$對稱，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，對正整數${\displaystyle m,n}$，有

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

上式算是一種「互反律」（）[7]，與§ 二次互反律有關。

高斯給出二次互反律的第三個證明，經艾森斯坦修改後，有以下兩個主要步驟。[8][9]

設${\displaystyle p}$、${\displaystyle q}$為互異奇質數，又設

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

首先，利用高斯引理，證明勒让德符号可表示為和式：

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

同樣

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

其後，採用幾何論證，證明

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

總結上述兩步，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。一些小整數模奇質數${\displaystyle p}$的二次特徵標，可以高斯符號表示，如：[10]

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

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

下取整函數出現於若干刻畫質數的公式之中。舉例，因為${\displaystyle \left\lfloor {\frac {n}{m}}\right\rfloor -\left\lfloor {\frac {n-1}{m}}\right\rfloor }$在${\displaystyle m}$整除${\displaystyle n}$時等於${\displaystyle 1}$，否則為${\displaystyle 0}$，所以正整數${\displaystyle n}$為質數当且仅当[11]

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

除表示質數的條件外，還可以寫出公式使其取值為質數。例如，記第${\displaystyle n}$個質數為${\displaystyle p_{n}}$，任選一個整數${\displaystyle r>1}$，然後定義實數${\displaystyle \alpha }$為

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

則衹用取整、冪、四則運算可以寫出質數公式：[12]

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

類似還有米尔斯常数${\displaystyle \theta =1.3064\ldots }$，使

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

皆為質數。[13]

若不疊代三次方函數，改為疊代以${\displaystyle 2}$為㡳的指數函數，亦有${\displaystyle \omega =1.9287800\ldots }$使

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

皆為質數。[13]

以質數計算函數${\displaystyle \pi (x)}$表示小於或等於${\displaystyle x}$的質數個數。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，當${\displaystyle n\geq 2}$時：[15]

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

本小節的公式未有任何實際用途。[16][17]