取整函数的性质

我们通常将$y=[x]$或$y=\lfloor x \rfloor$记作关于$x$的取整函数,也称为高斯函数,其意义是不超过x的最大整数

$\text{Lemma 0:}$

$\lfloor b \rfloor \le b<\lfloor b \rfloor+1$

$\text{Lemma 1:}$

$a\in Z,b\in R$

$a\le\lfloor b \rfloor \Leftrightarrow a\le b$

$\texttt{Proof:}$

$a\le\lfloor b \rfloor,\lfloor b \rfloor\le b \Rightarrow a\le b$

$a\le b \Rightarrow a<\lfloor b \rfloor+1 \Leftrightarrow a\le \lfloor b \rfloor$

(整数的离散性:$x,y\in Z,x<y\Leftrightarrow x\le y-1$)

$\text{Lemma 2:}$

$x,y\in Z$

$x\le \lfloor \frac{n}{y} \rfloor\Leftrightarrow y\le\lfloor \frac{n}{x} \rfloor$

$\texttt{Proof:}$

$\text{By lemma1:}x\le\lfloor \frac{n}{y} \rfloor\Leftrightarrow x\le \frac{n}{y} \Leftrightarrow y\le\frac{n}{x}\Leftrightarrow y\le \lfloor \frac{n}{x} \rfloor$

$\text{Proposition 3:}$

$x,n\in Z$

$x\le\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor$

$\texttt{Proof:}$

$\text{By lemma2: }x\le\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor\Leftrightarrow\lfloor\frac{n}{x}\rfloor\le\lfloor\frac{n}{x}\rfloor$

$\text{Theorem 4:}$

$x\in Z,\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor=\lfloor\frac{n}{x}\rfloor$

$\texttt{Proof:}$

$\text{By prosition3: }\lfloor\frac{n}{x}\rfloor\le\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor--(1),x\le\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor$

$\Rightarrow\frac{n}{x}\ge\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\ge\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor$

$\text{By lamma1: }\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor\le\lfloor\frac{n}{x}\rfloor--(2)$

$(1)\text{ and }(2)\Rightarrow\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor=\lfloor\frac{n}{x}\rfloor$

$\text{Corollary 5:}$

$y\in Z_+,\max\left\{x\in Z_+|\lfloor\frac{n}{x}\rfloor=\lfloor\frac{n}{y}\rfloor\right\}=\lfloor\frac{n}{\lfloor\frac{n}{y}\rfloor}\rfloor$

$\texttt{Proof:}$

$\forall x\in Z_+ ,\text{that } \lfloor\frac{n}{x}\rfloor=\lfloor\frac{n}{y}\rfloor$

$\text{By proposition3: }x \le \lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor=\lfloor\frac{n}{\lfloor\frac{n}{i}\rfloor}\rfloor$