\documentclass[全部作业]{subfiles}
\usepackage{mylatex}
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\fancyhead[C]{\mysignature}
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\begin{document}
\section{集合运算的性质}
\begin{enumerate}
    \item[9.] \ExplSyntaxOn
    \begin{zhongwen}
        $设A、B、C是集合，证明下列结论：$
    \end{zhongwen}
    \ExplSyntaxOff
    \begin{enumerate}
        \item[(3)] \ExplSyntaxOn
        \begin{zhongwen}
            $若A\cap B=\varnothing，则A-B=A$
        \end{zhongwen}
        \ExplSyntaxOff
        \begin{proof}
        \ExplSyntaxOn
        \begin{zhongwen}
            $$
            \begin{aligned}
            &\because A\cap B=\varnothing\\
            &\therefore \forall x \in A \implies x\notin B\\
            &\therefore \forall x \in A\implies x\in \bar{B}\\
            &\therefore A \subseteq \bar{B}\\
            &\therefore A\cap \bar{B}=A\\
            &\therefore A-B=A\\
            \end{aligned}
            $$
        \end{zhongwen}
        \ExplSyntaxOff
        \end{proof}
    \end{enumerate}
    \item[11.] \ExplSyntaxOn
    \begin{zhongwen}
        $指出下列集合等式成立的充分必要条件，其中A、B和C是集合：$
    \end{zhongwen}
    \ExplSyntaxOff
    \begin{enumerate}
        \item[(5)] $(A-B)\cap (A-C)=\varnothing$
        $$
        \begin{aligned}
        &(A-B)\cap (A-C)=A\cap \bar{B}\cap A \cap \bar{C}=A\cap \bar{B}\cap \bar{C}=A-B-C=\varnothing\\
        &\iff A-B-C=\varnothing\\
        \end{aligned}
        $$
    \end{enumerate}
    \item[13.] \ExplSyntaxOn
    \begin{zhongwen}
            $设A和B是集合，集合运算对称差\oplus 定义如下：A\oplus B=(A-B)\cup (B-A).\\证明下列恒等式，其中A、B和C是任意集合：$
    \end{zhongwen}
    \ExplSyntaxOff
    \begin{enumerate}
        \item[(5)] $A\oplus B=(A\cup B)-(A\cap B)$
        \begin{proof}
        \ExplSyntaxOn
        \begin{zhongwen}
            $$
            \begin{aligned}
            (A\cup B)-(A\cap B) & = (A\cup B)\cap \overline{A\cap B}  \\
            & = (A\cup B)\cap (\bar{A}\cup \bar{B})  \\
            & = ((A\cup B)\cap \bar{A})\cup ((A\cup B)\cap \bar{B})  \\
            & = (B\cap \bar{A})\cup (A\cap \bar{B})  \\
            & = (A-B)\cup (B-A)  \\
            & = A\oplus B \\
            \end{aligned}
            $$
        \end{zhongwen}
        \ExplSyntaxOff
        \end{proof}
    \end{enumerate}
    \item[21.] \textit{设A、B、C和D是任意集合.请证明：}
    \begin{enumerate}
        \item $(A\times C)\cap (B\times D)=(A\cap B)\times (C\cap  D)$
        \begin{proof}
        \ExplSyntaxOn
        \begin{zhongwen}
            $$
            \begin{aligned}
            \forall (x,y), \quad&(x,y)\in (A\times C)\cap (B\times D)\\
            \iff &(x,y)\in (A\times C)且(x,y)\in (B\times D)\\
            \iff &x\in A,y\in C且x\in B,y\in D\\
            \iff &x\in A\cap B且y\in C\cap D\\
            \iff &(x,y)\in (A\cap B)\times (C\cap D)\\
            \therefore \quad&(A\times C)\cap (B\times D)=(A\cap B)\times (C\cap D)\\
            \end{aligned}
            $$
        \end{zhongwen}
        \ExplSyntaxOff
        \end{proof}
    \end{enumerate}
    \item[22.] \textit{设$\{ A_\beta|\beta\in B \}$是以B为下标集的集合族.证明下列恒等式.}
    \begin{enumerate}
        \item $\overline{\displaystyle \bigcup_{\beta\in B} A_\beta}=\displaystyle \bigcap_{\beta\in B} \overline{A_\beta}$
        \begin{proof}
        % \ExplSyntaxOn
        \begin{zhongwen}
            $$
            \begin{aligned}
            \forall x, \quad&x\in \overline{\bigcup_{\beta\in B}A_\beta}\\
            \iff& x\notin \bigcup_{\beta\in B} A_\beta\\
            \iff&\lnot (\exists \beta\in B, x\in A_\beta)\\
            \iff&\forall \beta\in B, x\notin A_\beta\\
            \iff&\forall \beta\in B, x\in \overline{A_\beta}\\
            \iff&x\in \bigcap_{\beta\in B} \overline{A_\beta}\\
            \therefore \quad&\overline{\bigcup_{\beta\in B} A_\beta}=\bigcap_{\beta\in B} \overline{A_\beta}\\
            \end{aligned}
            $$
        \end{zhongwen}
        % \ExplSyntaxOff
        \end{proof}
    \end{enumerate}
    \item[24.] \textit{设集合族$\{A_n|n\in \mathbb{N}\}$，令$B_0=A_0$，$B_n=A_n-\displaystyle \bigcup_{k=0}^{n-1} A_k(n>0) $.证明：}
    \begin{enumerate}
        \item[(2)] $\displaystyle\bigcup_{ n\in \mathbb{N}} A_n =\displaystyle \bigcup_{n\in \mathbb{N}} B_n$
        \begin{proof}
        \begin{zhongwen}
            $$
            \begin{aligned}
            &\because  B_n=A_n-\displaystyle \bigcup_{k=0}^{n-1}A_k(n>0)\\
            &\therefore \forall n>0, B_n\in A_n\\
            &\forall x\in \bigcup_{n\in \mathbb{N}} B_n \iff\exists n_0\in \mathbb{N},x\in B_{n_0}\implies x\in A_{n_0}\\
            &\therefore \displaystyle \bigcup_{n\in \mathbb{N}}A_n\supseteq \bigcup_{n\in \mathbb{N}} B_n \\
            &\forall x\in \displaystyle \bigcup_{n\in \mathbb{N}} A_n, 则 \exists n\in \mathbb{N},x\in A_{n} \\
            &设C_x=\{ n|x\in A_n \} \\
            &设n_1=\min{C_x} \\
            &则x\in A_{n_1}且\forall n_2<n_1,x\notin A_{n_2} \\
            &\therefore x\notin \displaystyle \bigcup_{k=0}^{n_1-1}A_k \\
            &\therefore x\in A_{n_1}-\bigcup_{k=0}^{n_1-1}A_k \\
            &\therefore x\in B_{n_1} \\
            &\therefore x\in \displaystyle \bigcup_{n\in \mathbb{N}} B_n \\
            &\therefore \bigcup_{n\in \mathbb{N}}A_n \subseteq \bigcup_{n\in \mathbb{N}} B_n \\
            &\therefore \bigcup_{n\in \mathbb{N}} A_n =\bigcup_{n\in \mathbb{N}} B_n \\
            \end{aligned}
            $$
        \end{zhongwen}
        \end{proof}
    \end{enumerate}
\end{enumerate}
\section{有限集合的计数}
\begin{enumerate}
    \item[26.] \textit{以1开头或者以00结束的不同的字节（8位的二进制串）有多少个？}
    \begin{zhongwen}
        $$
        \begin{aligned}
        &设A为以1开头的不同的字节，B为以00结束的不同的字节。 \\
        &则\left\vert 以1开头或者以00结束的不同的字节 \right\vert =\left\vert A\cup B \right\vert =\left\vert A \right\vert +\left\vert B \right\vert -\left\vert A\cap B \right\vert =2^{7}+2^{6}-2^{5} = 160 \\
        \end{aligned}
        $$
    \end{zhongwen}
    \item[27.] \begin{zhongwen}
        $在1\sim 200的正整数中，$
    \end{zhongwen}
    \begin{enumerate}
        \item[(3)] \begin{zhongwen}
            $含因子3或5，但不同时含因子3和5的正整数共有多少个？$
        \end{zhongwen}
        \begin{zhongwen}
            $$
            \begin{aligned}
            &设A为1\sim 200中含因子3的正整数，B为1\sim 200中含因子5的正整数 \\
            &则\left\vert 在1\sim 200的正整数中 含因子3或5，但不同时含因子3和5的正整数 \right\vert  \\
            &= \left\vert A\oplus B \right\vert =\left\vert A \right\vert +\left\vert B \right\vert -2\left\vert A\cap B \right\vert =66+40-2*13 = 80 \\
            \end{aligned}
            $$
        \end{zhongwen}
        *用JavaScript再验证一下：
        \begin{minted}{javascript}
            > a =[]
            []
            > for (let i = 1; i<=200; i++){
            ... a.push(i);
            ... }
            >  a.reduce((out, cur) => {
            ... if ((cur%3==0 || cur%5==0) && cur%15!=0) out.push(cur);
            ... return out;
            ... }, Array()).length
            80
        \end{minted}
        \item[(5)] \begin{zhongwen}
            $与15互素的正整数（即与15之间的最大公因子为1的那些正整数）共有多少个？$
        \end{zhongwen}
        \begin{zhongwen}
            $$
            \begin{aligned}
            &A与B的定义和上题相同 \\
            &则\left\vert 在1\sim 200的正整数中与15互素的正整数 \right\vert  \\
            &=\left\vert 1\sim 200的正整数 \right\vert -\left\vert A\cup B \right\vert  \\
            &=200-(\left\vert A \right\vert +\left\vert B \right\vert -\left\vert A\cap B \right\vert )=200-(66+40-13) = 107 \\
            \end{aligned}
            $$
        \end{zhongwen}
    \end{enumerate}
\end{enumerate}
\end{document}