取整函数的性质

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

\(\text{Lemma 0:}\)

\[\lfloor b \rfloor \le b<\lfloor b \rfloor+1 \]

\(\text{Lemma 0':}\)

\[\forall a,b,c\in N_+ ,\lfloor\lfloor\frac{a}{b}\rfloor/c\rfloor=\lfloor\frac{a}{bc}\rfloor \]

\(\texttt{Proof:}\)

\[a=\lfloor\frac{a}{b}\rfloor \times b+r_1=\lfloor\frac{a}{bc}\rfloor\times bc+r_2,r_1\in[0, b),r_2\in[0, bc),r_2-r_1\in(-bc,bc) \]

\[\lfloor\lfloor\frac{a}{b}\rfloor/c\rfloor=\lfloor \frac{a-r_1}{bc} \rfloor=\lfloor \lfloor\frac{a}{bc}\rfloor + \frac{r_2-r_1}{bc}\rfloor=\lfloor\frac{a}{bc}\rfloor \]

\(\text{Lemma 1:}\)

\[a\in Z,b\in R \]

\[a\le\lfloor b \rfloor \Leftrightarrow a\le b \]

\(\texttt{Proof:}\)

\[a\le\lfloor b \rfloor,\lfloor b \rfloor\le b \Rightarrow a\le b \]

\[a\le b \Rightarrow a<\lfloor b \rfloor+1 \Leftrightarrow a\le \lfloor b \rfloor \]

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

\(\text{Lemma 2:}\)

\[x,y\in Z \]

\[x\le \lfloor \frac{n}{y} \rfloor\Leftrightarrow y\le\lfloor \frac{n}{x} \rfloor \]

\(\texttt{Proof:}\)

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

\(\text{Proposition 3:}\)

\[x,n\in Z \]

\[x\le\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor \]

\(\texttt{Proof:}\)

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

\(\text{Theorem 4:}\)

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

\(\texttt{Proof:}\)

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

\[\Rightarrow\frac{n}{x}\ge\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\ge\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor \]

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

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

\(\text{Corollary 5:}\)

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

\(\texttt{Proof:}\)

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

\[\text{By proposition3: }x \le \lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor=\lfloor\frac{n}{\lfloor\frac{n}{y}\rfloor}\rfloor \]

\[\texttt{original author: 11Dimensions} \]

posted @ 2020-07-05 21:12  hkr04  阅读(1751)  评论(0编辑  收藏  举报