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