我們通常將\(y=[x]\)或\(y=\lfloor x \rfloor\)記作關於\(x\)的取整函數,也稱為高斯函數,其意義是不超過x的最大整數
\(\text{Lemma 0:}\)
\[\lfloor b \rfloor \le b<\lfloor b \rfloor+1 \]
\(\text{Lemma 0':}\)
\[\forall a,b,c\in N_+ ,\lfloor\lfloor\frac{a}{b}\rfloor/c\rfloor=\lfloor\frac{a}{bc}\rfloor \]
\(\texttt{Proof:}\)
\[a=\lfloor\frac{a}{b}\rfloor \times b+r_1=\lfloor\frac{a}{bc}\rfloor\times bc+r_2,r_1\in[0, b),r_2\in[0, bc),r_2-r_1\in(-bc,bc) \]
\[\lfloor\lfloor\frac{a}{b}\rfloor/c\rfloor=\lfloor \frac{a-r_1}{bc} \rfloor=\lfloor \lfloor\frac{a}{bc}\rfloor + \frac{r_2-r_1}{bc}\rfloor=\lfloor\frac{a}{bc}\rfloor \]
\(\text{Lemma 1:}\)
\[a\in Z,b\in R \]
\[a\le\lfloor b \rfloor \Leftrightarrow a\le b \]
\(\texttt{Proof:}\)
\[a\le\lfloor b \rfloor,\lfloor b \rfloor\le b \Rightarrow a\le b \]
\[a\le b \Rightarrow a<\lfloor b \rfloor+1 \Leftrightarrow a\le \lfloor b \rfloor \]
(整數的離散性:\(x,y\in Z,x<y\Leftrightarrow x\le y-1\))
\(\text{Lemma 2:}\)
\[x,y\in Z \]
\[x\le \lfloor \frac{n}{y} \rfloor\Leftrightarrow y\le\lfloor \frac{n}{x} \rfloor \]
\(\texttt{Proof:}\)
\[\text{By lemma1:}x\le\lfloor \frac{n}{y} \rfloor\Leftrightarrow x\le \frac{n}{y} \Leftrightarrow y\le\frac{n}{x}\Leftrightarrow y\le \lfloor \frac{n}{x} \rfloor \]
\(\text{Proposition 3:}\)
\[x,n\in Z \]
\[x\le\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor \]
\(\texttt{Proof:}\)
\[\text{By lemma2: }x\le\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor\Leftrightarrow\lfloor\frac{n}{x}\rfloor\le\lfloor\frac{n}{x}\rfloor \]
\(\text{Theorem 4:}\)
\[x\in Z,\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor=\lfloor\frac{n}{x}\rfloor \]
\(\texttt{Proof:}\)
\[\text{By prosition3: }\lfloor\frac{n}{x}\rfloor\le\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor--(1),x\le\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor \]
\[\Rightarrow\frac{n}{x}\ge\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\ge\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor \]
\[\text{By lamma1: }\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor\le\lfloor\frac{n}{x}\rfloor--(2) \]
\[(1)\text{ and }(2)\Rightarrow\lfloor\frac{n}{\lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor}\rfloor=\lfloor\frac{n}{x}\rfloor \]
\(\text{Corollary 5:}\)
\[y\in Z_+,\max\left\{x\in Z_+|\lfloor\frac{n}{x}\rfloor=\lfloor\frac{n}{y}\rfloor\right\}=\lfloor\frac{n}{\lfloor\frac{n}{y}\rfloor}\rfloor \]
\(\texttt{Proof:}\)
\[\forall x\in Z_+ ,\text{that } \lfloor\frac{n}{x}\rfloor=\lfloor\frac{n}{y}\rfloor \]
\[\text{By proposition3: }x \le \lfloor\frac{n}{\lfloor\frac{n}{x}\rfloor}\rfloor=\lfloor\frac{n}{\lfloor\frac{n}{y}\rfloor}\rfloor \]
\[\texttt{original author: 11Dimensions} \]