离散数学习题七答案 　离散数学课后习题答案 (邱学绍).doc

⑸(p(q)((((q(( q((r))是5层公式，这是因为一层：p(q，(q，(r二层：q((r三层：(q(( q((r)四层：(((q(( q((r))2．解 ⑴A=(p(q)(q是2层公式。真值表如表2-1所示：表2-1p q 0 0 0 0 0 1 1 1 1 0 1 0 1 1 1 1 ⑵是3层公式。真值表如表2-2所示：表2-2p q 0 0 1 0 1 0 1 1 1 0 1 0 0 0 1 1 1 1 1 1 ⑶是3层公式。真值表如表2-3所示：表2-3p q r 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 1 0 0 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 1 0 1 0 0 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 1 ⑷是4层公式。真值表如表2-4所示：3．解 ⑴真值表如表2-5所示：表2-5p q 0 0 1 1 1 1 0 1 1 0 0 0 1 0 0 1 0 1 1 1 0 0 0 1所以其成真赋值为：00，10，11；其成假赋值为01。⑵真值表如表2-6所示：表2-6p q r 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 1 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 1 1 1 所以其成真赋值为：000，010，100，110，111；其成假赋值为001，011，101。

⑶真值表如表2-7所示，所以其成真赋值为：00，11；成假赋值为：01，10，。4．解 ⑴设，其真值表如表2-8所示：表2-8p q 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 0 1 故为重言式。⑵设A=(p(q)(((p(q)，其真值表如表2-9所示：表2-9p q p(q p(q ((p(q) A 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 1 1 0 0 故A=(p(q)(((p(q)为矛盾式。⑶设A=(p(q)(((p(q)，其真值表如表2-10所示：表2-10p q 0 0 1 0 1 0 0 1 1 1 1 1 1 0 0 1 0 0 1 1 0 0 1 0 故A=(p(q)(((p(q)为可满足式。⑷设，其真值表如表2-11所示：表2-11p q r 0 0 0 1 1 1 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 故为重言式。

习题1.31．解 ⑴真值表如表2-12所示：表2-12p q 0 0 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 1 0 1 1 0 0 0 1 0 由真值表可以看出和所在的列相应填入值相同，故等值。⑵真值表如表2-13所示：表2-13p q 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 1 1 1 1 0 1 0 1 由真值表可以看出和所在的列相应填入值相同，故等值。⑶真值表如表2-14所示：表2-14p q 0 0 1 1 1 1 1 0 1 1 0 1 1 1 1 0 0 1 0 1 0 1 1 0 0 1 0 0 由真值表可以看出(p和(p(q)((p((q)所在的列相应填入值相同，故等值。⑷真值表如表2-15所示：p q r q(r p((q(r) p(q (p(q)(r 0 0 0 1 1 0 1 0 0 1 1 1 0 1 0 1 0 0 1 0 1 0 1 1 1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 1 0 1 1 1 0 0 0 1 0 1 1 1 1 1 1 1 表2-15由真值表可以看出p((q(r)和(p(q)(r所在的列相应填入值相同，故等值。

2．证明 ⑴(p(q)(( ((p(q)( (p(q)(( p((q)( p( (q((q)( p。⑵(p(q)((q(p)(((p(q) (((q(p)(((p((q)(((p( p)(( q((q)((q( p)(( p(q)(((p((q)。⑶由⑵可得，((p(q)(((( p(q)(((p((q))((( p((q)((p(q)((q((p)(((p(q)((p(q。⑷p((q(r)(( p(((q( r)(( q(((p( r)( q(( p (r)。⑸⑹3．解 ⑴((p((q)((((p((q)(p(q⑵(((p((q)((( p((q)((p(q⑶((p((q)((((p((q)(((q(p))(((p((q)((((q(p)((p(q) (((p((q)( p(q。⑷同理可证(((p(q)( p(q。4．解 ⑴与习题2(2第4（4）相同。⑵真值表如表2-16所示：表2-16p q (p (q p(q (q ((p A 0 0 1 1 1 1 1 0 1 1 0 1 1 1 1 0 0 1 0 0 1 1 1 0 0 1 1 1 所以公式是重言式。⑶真值表如表2-17所示，所以公式是矛盾式。

表2-170 0 1 1 1 0 0 0 1 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 0 1 0 0 ⑷真值表如表2-18所示，所以公式是重言式。表2-180 0 0 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 0 1 1 0 0 1 1 0 0 0 0 1 1 0 1 0 0 1 1 1 0 1 0 1 1 1 1 1 1 1 ⑸真值表如表2-19所示，所以公式仅为可满足式。表2-190 0 1 0 1 1 0 1 1 1 0 1 1 0 0 1 0 0 1 1 0 1 0 0 ⑹真值表如表2-20所示，所以公式是重言式。表2-20p q r p(q r(q p(r (p(q)((r(q) (p(r)(q A 0 0 0 1 1 0 1 1 1 0 0 1 1 0 0 0 1 1 0 1 0 1 1 0 1 1 1 0 1 1 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 0 1 0 0 1 0 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 15．解 ⑴设p：他努力学习；q：他会通过考试。则命题符号化p(q。

其否定((p(q)( p((q。所以语句的否定：他学习很努力但没有通过考试。⑵设p：水温暖；q：他游泳。则命题符号化p(q。其否定((p(q)( p((q。所以语句的否定：当且仅当水不温暖时他游泳。⑶设p：天冷；q：他穿外套；r：他穿衬衫。则命题符号化p((q((r)其否定(( p((q((r)) ((((p((q((r))( p((( q((r) ( p(((q( r)所以语句的否定：天冷并且他外套或者穿衬衫。⑷设p：他学习；q：他将上清华大学；r：他将上北京大学。则命题符号化其否定所以语句的否定：他努力学习，但是没有上清华大学，也没有上北京大学。6．解 设p：张三说真话；q：李四说真话；r：王五说真话。则：p((q, q((r(((q(r), r(((p((q)为真,因此p(((p((q)((p((p((q)(((p((p(q))((p(q为真。因此，p为假，q为真，所以r为假。故张三说谎，李四说真话，王五说谎。7．解 设p：甲得冠军；q：乙得亚军；r：丙得亚军；s：丁得亚军。前提：p((q(r)，q((p，s((r，p结论：(s证明 p((q(r)为真，其前件p为真，所以q(r为真，又q((p为真，其后件(p为假，所以要求q为假，所以r为真。

又s((r为真，其后件(r为假，所以要求s为假，故(s为真。习题1.41．解 ⑴设p：明天下雨；q：后天下雨。命题符号化。⑵设p：明天我将去北京；q：明天我将去上海。命题符号化。2．解 ⑴⑵⑶3．证明 因为，{}是功能完备联结词集，所以，含有{}外的其他联结词的公式均可以转换为仅含{}中的联结词的公式。又因为即含有的公式均可以转换为仅含{}中的联结词的公式。因此，含{}外其他联结词的公式均可以转换为仅含{}中的联结词的公式。故{}是功能完备联结词集。4．证明 是极小功能完备集，因而只需证明中的每个联结词都可以用( 表示，就说明是功能完备集。只有一个联结词，自然是极小功能完备集。事实上，(p(((p(p)(p(p，p(q((((p(q)(((p(q)((p(q)((p(q)。对于证明是极小功能完备集，可类似证明。习题1.51．解 ⑴；⑵2．解 ⑴即为其析取范式。即为其合取范式。⑵即为其合取范式。(p((q(r)((p(((q(r)(((q((r))(((p(q(r)(((p((q((r) 即为其析取范式。⑶即为其合取范式。为其析取范式。⑷即为其析取范式和合取范式。3．解 ⑴即为其主合取范式。

其主析取范式为(3(p(q。⑵。故其主析取范式为((0,1,2,3)=((p((q)(((p(q)((p((q)((p(q)。⑶即为其主合取范式。其主析取范式为((2,4,5,6,7) ( ((p(q((r)((p((q((r)((p((q(r)((p(q((r)((p(q(r)。⑷即为其主合取范式。其主析取范式为。4．解 ⑴真值表如表2-21所示, 所以其极小项是p((q，极大项为p(q，p((q，(p((q。表2-21p q 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 其主析取范式是：p((q，主合取范式为：(p(q)(( p((q)(((p((q)。⑵真值表如表2-222所示, 所以其极小项是(p(q, p((q, p(q, 极大项为p(q。表2-22p q 0 0 0 1 0 0 0 1 1 1 0 1 1 0 1 0 1 1 1 1 1 1 0 1 其主析取范式是：((p(q)((p((q)((p(q)，主合取范式为：p(q。⑶真值表如表2-23所示，所以其极小项是(p(q(r,p((q((r, p((q(r, p(q((r,p(q(r,表2-23p q r 0 0 0 1 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 1 1 1 1 1 1 0 0 0 0 1 1 0 1 0 0 1 1 1 0 0 0 1 1 1 1 0 0 1 极大项为p(q(r，p(q((r，p((q(r。

其主析取范式是：((p(q(r)((p((q((r)((p((q(r)((p(q((r)((p(q(r)，主合取范式为：(p(q(r)((p(q((r)((p((q(r) 。⑷真值表如表2-24所示,所以其极小项为(p((q(r,(p(q(r,p((q((r,p((q(r,p(q(r,而极大项分为p(q(r，p((q(r，(p((q(r.主合取范式为(p(q(r)((p((q(r)(((p((q(r),主析取范式为((p((q(r)(((p(q(r)((p((q((r,)((p((q(r)((p(q(r)。表2-24p q r 0 0 0 1 0 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1 0 0 0 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 15．解 ⑴((p(q)(((((p((q))(((p(q)((p(q)( q( ((p(q)((p(q)，故⑴为可满足式。⑵故⑵为重言式。⑶((p((q(r))(((p(q)((p(r))(((p((q(r))((p((q(r))((p((q(r))((p((q(r))(((p((q(r))(((p((q(r))((p((q(r))(((p((q(r))((p((q(r))((p(((q(r)(((p(q(r)(((q((r)(0。

故⑶为矛盾式。⑷故仅为可满足式。6．证明 ⑴右边已经是主合取范式。而左边主合取范式已是(p((q，因此，((p( q)((p((q，证毕。⑵右边(p( q)((p((q)已经是主合取范式。p(p((q((q)( (p( q)((p((q)。因此，。⑶左边p((q(r)((p(((q(r)((p((q(r，而右边(((p(q)(r((p((q(r，因此，。习题1.61．解 设p：这里有演出；q：这里通行是困难的；r：他们按照指定时间到达。前提：p(q, r((q,r结论：(p证明①r P②r((q P③(q T①②假言推理④p(q P⑤(p T③④拒取式2．⑴证明①s P②s(p P③p T①②假言推理④p(q P⑤q T③④假言推理⑵证明①r P附加前提引入②r(q P③q T①②假言推理④p((q P⑤(p T③④拒取式⑥(p(s P⑦s T⑤⑥假言推理⑧r(s T①⑦CP⑶证明①p P否定结论引入②p(q P③q T①②假言推理④q(r P⑤r T③④假言推理⑥(r(s P⑦(r T⑥化简⑧r((r T⑤⑦合取⑷证明①p P附加前提引入②(p(q P③q ①②析取三段论④r((q P⑤(r ③④拒取式⑥p((r ①⑥CP⑸证明①p P附加前提引入②p((q(r) P③q(r T①②假言推理④q P附加前提引入⑤r T③④假言推理⑥(r(s)(t P⑦(r((s(t T⑥蕴涵等价式⑧(s(t T⑤⑦析取三段论⑨(h((s((t) P⑩(s(t (h T⑨假言易位⒒ h T⑧⑩假言推理⒓ q(h T④⒒CP13. p((q(h) T①⒓CP3．解 推理不正确。

在①到②化简时，只能对整个公式进行而不是子公式。4．解 正确。⑴P，⑵P附加前提引入；⑶T①②析取三段论；⑷P；⑸T③④假言推理；⑹P；⑺T⑤⑥假言推理；⑻T②⑦CP。5．解 设p：张三努力工作，q：李四高兴，r：王五高兴，s：刘六高兴前提：p((q(r)，q((p，s((r结论：p((s证明：①p P附加前提引入②p((q(r) P③q(r T①②假言推理④q((p P⑤(q T①④拒取式⑥r T③⑤析取三段论⑦s((r P⑧(s T⑥⑦拒取式⑨p((s T①⑧CP6．解 设：p：天下雪；q：马路结冰；r：汽车开得快；s：马路塞车。前提：p(q，q((r，(r(s，(s结论：(p证明①p(q P②q((r P③p((r ①②推理三段论④(r(s P⑤p(s ③④推理三段论⑥(s P⑦(p ⑤⑥拒取式复习题11．解 ⑴ 设p：3是偶数，q：中国人的母语是汉语。命题符号化。⑵ 设p：你抽烟，q：你很容易得病。命题符号化。⑶ 设p：今天是星期一，q：明天才是星期二。命题符号化。⑷ 设p：李春这个学期《离散数学》考了100分。q：李春这个学期《数据结构》考了100分。命题符号化。⑸ 设p：下雪路滑，q：他迟到了。

命题符号化。⑹ 设p：经一事，q：长一智。命题符号化。⑺ 设p：一朝被蛇咬，q：十年怕井绳。命题符号化。⑻ 设p：以物喜，q：以己悲。命题符号化。2. 解 命题中的“或”是不可兼或，因此，可以直接用“”符号化；根据联结词的性质及其之间的转换关系，可知命题“李春生于1979年或生于1980年”的本意是“李春生于1979年（但不能生于1980年）或生于1980年（但不能生于1979年）”，因此，也可以转化为“”对其进行符号化。3．解 设p：会拳击，q：李春会唱歌。命题符号化((p((q)(((p(q)。而((p((q)(((p(q)(((p((q)((((p(q)(((p((q)(p((q(p((q因此，会拳击并且李春不会唱歌。4．解 ⑴A的极小项对应于其真值表中的成真赋值0001，0110，1000，1001，1010，1100，1101，1111。成真赋值对应二进制数转化为十进制数就是A的极小项的下标。由此可得，A的极小项为：；；；；；；；。相应的，A的极大项对应于其真值表中的成假赋值，成假赋值对应二进制数转化为十进制数就是A的极大项的下标。由此可得，A的极大项为：；；；；；；；。

⑵由问题⑴得到了A的极小项和极大项，于是与A等值的主析取范式和主合取范式可以直接得到，分别为：；。⑶从A的主析取范式出发，进行等值演算化简，可得析取范式的最简形式：((p((q((r(s)(((p(q(r(s)((p((q((r((s)((p((q((r(s)((p((q(r((s)((p(q((r((s)((p(q((r(s)((p(q(r(s)(((p((q((r(s)((q(r(s)((p((q((r)((p((q(r((s)((p(q((r)((p((r)(((p((q((r(s)((q(r(s)((p((q(r((s)((p((r)(((q((r(s)((q(r(s)((p((q(r((s)((p((r)(((q((r(s)((q(r(s)((p((q((s)5． 证明 ⑴⑵⑶⑷6．解 ⑴公式的真值表如表2-27所示：表2-27p 0 0 1 1 1 1 1 0 1 1 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0从真值表可见，公式所在列的填入值有1也有0，故仅为可满足式。(p((q)(((q((p)(((p((q)((q((p)(((2,3)为其主合取范式，可见公式仅为可满足式。

⑵公式真值表如表2-28所示：表2-28p q r 0 0 0 0 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 p((p(q(r)((p(p(q(r(1(((0,1,2,3,4,5,6,7)从真值表可见，公式所在的列的填入值均为1，等值演算，以及求出的主析取范式均说明公式是重言式。⑶A=((p(q)((q(r))((p(r)真值表见习题2(2第4(4)题。((p(q)((q(r))((p(r)(((((p(q)(((q(r))(((p(r)((p((q)((q((r))((p(r(1.从真值表可见，公式所在的列的填入值均为1，由等值演算，以及求出的主析取范式均说明公式是重言式。7．⑴证明①p P附加前提引入②p((q(r) P③q(r T①②假言推理④q P附加前提引入⑤q((r(s) P⑥r(s T④⑤假言推理⑦q(s T③⑥假言三段论⑧p((q(s) T①⑦CP⑵证明①(w P②u(w P③(u T①②拒取式④(s(u P⑤(s T③④析取三段论⑥(r(s P⑦(r T⑤⑥析取三段论⑧(p(q)(r P⑨((p(q) T⑦⑧拒取式⑩(p((q) T⑨德(摩根律⑶证明①p P附加前提引入②p(q(r P③q(r T①②假言推理④q((p P⑤(q T①④拒取式⑥r T③⑤析取三段论⑦s((r P⑧(s T⑥⑦拒取式⑨p((s T①⑧CP8．解①p(r P②p T①化简③p(q P④q T②③假言推理⑤((q(s) P⑥(q((s T⑤德(摩根律⑦(q T⑥化简⑧(q(q T④⑦合取由⑧得到矛盾，可见p(q，((q(s)，p(r不能同时成立。

9．解 设p：小王曾经到过受害人的房间，q：小王11点以前离开，r：小王犯了谋杀罪，s：看门人看到小王。符号化：(((p((q)(r)(p((q(s)((s)(r。⑴形式构造推理证明前提：(p((q)(r，p，q(s，(s结论：r证明①(s P②q(s P③(q T①②拒取式④p P⑤p((q T③④合取⑥(p((q)(r P⑦ T⑤⑥假言推理⑵真值表技术：真值表如表2-30所示，设A=(((p((q)(r)(p((q(s)((s)。表2-29p q r s (q (s p((q (p((q)(r q(s A 0 0 0 0 1 1 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 0 1 1 1 1 0 0 0 1 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 0 0 0 1 0 1 0 0 1 1 0 1 0 0 0 1 1 0 1 1 1 0 0 1 0 1 0 1 1 1 1 1 0 0 0 1 1 1 由真值表可以看出：(((p((q)(r)(p((q(s)((s)(r，所以，(((p((q)(r)(p((q(s)((s)(r成立。

⑶等值演算方法(((p((q)(r)(p((q(s)((s)(r((((((p((q)(r)(p(((q(s)((s)(r((((((p((q)(r)(p(((q((s))(r((((((p((q)(r)((p((((q((s))(r(((p((q((r)(((p((q((s))(r(1。由此可以说明(((p((q)(r)(p((q(s)((s)(r为重言式，即(((p((q)(r)(p((q(s)((s)(r成立。10．解 逻辑学家手指A门问旁边的一名强盗（A）说：“这扇门是生门，他（指强盗B）将回答‘是’，他的回答对吗？”设：强盗A回答“对”；：强盗B回答“对”；：这扇门（A）是生门。因为，两个强盗一个总说真话，而另一个强盗一个总说假话，因此该问题符号化为：((p(q)(r。((p(q)(r((((p(q)((p((q))(r(((p(q(r)((p((q(r)逻辑学家的提问可知，r和q都为真，由公式可以看出这时(p为真，即p为假。所以，当被问强盗A回答“否”，则逻辑学家开启所指的门从容离去。当被问强盗A回答“对”，则逻辑学家开启另一扇门从容离去。第二章 谓词逻辑习题2.11．解 ⑴个体：离散数学；谓词：…是一门计算机基础课程。

⑵个体：田亮；谓词：…是一名优秀的跳水运动员。⑶个体：大学生；谓词：…要好好学习计算机课程；量词：所有。⑷个体：推理；谓词：…是能够由计算机来完成的；量词：一切。2． 解 ⑴设：x是舞蹈演员；a：小芳。命题符号化：。⑵设：x是一位有名的哲学家；a：苏格拉底。命题符号化：。⑶设：x作完了他的作业家；a：张三。命题符号化：。⑷设：x身体很好；a：我。命题符号化：。3．解 ⑴选取个体域为整数集合。设：x的平方是奇数；：x是奇数。命题符号化：。⑵选取个体域为所有国家的集合。设：x在南半球；：x在北半球。命题符号化：。⑶选取个体域为所有人的集合。设：x在中国居住；：x是中国人。命题符号化：⑷选取个体域为所有人的集合。设：x是艺术家；：x是导演；：x是演员。命题符号化：(x(M(x)(F(x)(G(x))。⑸选取个体域为所有猫的集合。设M(x)：x是好猫；F(x)：x捉耗子。命题符号化：(x(M(x)((x(F(x)(M(x))。4．解 ⑴①设：x喜欢开汽车；：x喜欢骑自行车。命题符号化：。②设：x喜欢开汽车；：x喜欢骑自行车；：x是人。命题符号化：。⑵①设：x必须学好数学。命题符号化：。②设：x必须学好数学；：x是学生。

命题符号化：。⑶①设：x的平方是质数；：x是质数。命题符号化： 。②同①。③设：x的平方是质数。命题符号化：。习题2.21．解 ⑴(x的辖域为P(x)(Q(x)，个体变元x是约束变元。⑵(x的辖域为P(x)((yQ(x,y)，(y的辖域为Q(x,y)，个体变元x是约束变元，个体变元y是约束变元。⑶(x的辖域为F(x,y)，其中个体变元x是约束变元，个体变元y是自由变元；的辖域为Q(x,y)，其中个体变元x是自由变元，个体变元y是约束变元。2．解 ⑴。⑵。3．解 ⑴((xP(x)((yQ(x))(F(s,z)；⑵(y(P(s,y)(((zQ(s,z)(R(s,y,v)))((x(rS(x,t,r)。离散数学习题七答案4．解 ⑴的真值分别为：0，1，0。⑵的真值分别为：0，1，0。⑶的真值分别为：1，1，1。⑷的真值分别为：0，0，0。5．解 ⑴0。 ⑵1。 ⑶0。6．解 ⑴设I为任意解释，其个体域为D，若(xP(x)为真，即((xP(x)为假，则((xP(x)((xP(x)为真；若(xP(x)为假，即((xP(x)为真，则就是说在个体域中不存在使得P(x)为真的个体，故(xP(x)为假，即((xP(x)((xP(x)为假。

因此((xP(x)((xP(x)仅为可满足式。⑵设I为任意解释，其个体域为D，若((xA(x)为假,则(xA(x)为真，就是说对于个体域中任意一个个体A(x)均为真，那么(A(x)必为假，所以(x((A(x))必为假；若((xA(x)为真，即(xA(x)为假，则就是说对于个体域中至少存在一个体使A(x)均为假，那么对于个体域中至少存在一个个体使(A(x)为真，所以(x((A(x))必为真，总之((xA(x)((x((A(x))对于个体域中任意一个个体必为真，即其为逻辑有效式。⑶设I为任意解释，其个体域为D，若(x(P(x)(Q(x))为真，即是说在个体域中至少存在一个个体使得P(x)和Q(x)同时为真，此时(x(P(x)((Q(x))可真可假，所以，(x(P(x)(Q(x))((x(P(x)((Q(x))可真可假。因此，仅为可满足式。习题2.31．解 ⑴(((xA(x)((xB(x))(((((xA(x)((xB(x))((xA(x)(((xB(x))((xA(x)((x(B(x))⑵(((xA(x)((B(x)(((xC(x)((x(A(x)((B(x)((x(C(x)⑶(((xA(x)((x(B(x))(((x(A(x)((xB(x))((x(C(x)。

2．证明 ⑴(x(P(x)(Q(x))(((xP(x)((xQ(x))(((x((P(x)(Q(x))((((xP(x)((xQ(x))((x(P(x)((Q(x))((x(P(x)((xQ(x)(((P(1)((Q(1))(((P(2)((Q(2))(((P(3)((Q(3)))(((P(1)((P(2)((P(3))(Q(1)(Q(2)(Q(3)(((P(1)(Q(1))(((P(2)(Q(2))(((P(3)(Q(3)))(((P(1)((P(2)((P(3))((P(1)(P(2)(P(3))((Q(1)(Q(2)(Q(3))(((P(1)(P(2)(P(3))(1。故(x(P(x)(Q(x))(((xP(x)((xQ(x))为逻辑有效式。⑵((xP(x)((xQ(x))((x(P(x)(Q(x))(((((xP(x)((xQ(x))((x((P(x)(Q(x))(((xP(x)((x(Q(x))((x((P(x)(Q(x))(((P(1)(P(2)(P(3))(((Q(1)((Q(2)((Q(3))((((P(1)(Q(1))((P(2)(Q(2))((P(2)(Q(2)))([(P(1)((Q(1))((P(2)((Q(2))((P(3)((Q(3))((P(1)((Q(2))(]((((P(1)(Q(1))(((P(2)((Q(2))(((P(1)((Q(1)))((((P(1)(Q(1))((P(1)((Q(1))((P(2)((Q(2))((P(3)((Q(3))((P(1)((Q(2))()((((P(2)((Q(2))((P(1)((Q(1))((P(2)((Q(2))((P(3)((Q(3))((P(1)((Q(2))()((((P(1)((Q(1))((P(1)((Q(1))((P(2)((Q(2))((P(3)((Q(3))((P(1)((Q(2))()(1(1(1。

⑶习题2.41．解 ⑴④错。在一个逻辑推理过程中，若同时用到ES和US，并且选用代替的个体变元相同时应先用ES，再用US。⑵②错，在用UG规则时，引入的个体变元在原来的公式中不能自由出现过。③错。⑶④错，在用两次ES规则时，引入的个体常元不能是一样的。2．⑴证明①(x(Q(x) P②(Q(y) T①US③(x((P(x)(Q(x)) P④(P(y)(Q(y) T③US⑤P(y) T②④拒取式⑥(xP(x) T⑤EG⑵证明①(xP(x) P附加前提引入②P(c) T①US③(x(P(x)(Q(x)) P④P(c)(Q(c) T③US⑤Q(c) T②④假言推理⑥(xQ(x) T⑤UG⑦(xP(x)((xQ(x) T①⑥CP⑶证明①((x(P(x)(Q(x)) P②(x((P(x)((Q(x)) T①量词否定，德(摩根律③(P(c)((Q(c) T②ES④(xQ(x) P否定结论引入⑤Q(c) T④US⑥(Q(c) T③化简⑦Q(c)((Q(c) T⑤⑥合取由⑦得到矛盾，由间接证明原理，原命题得证明。3． 解 ⑴设M(x)：x是鸟；N(x)：x是猴子，F(x)：x会飞。前提：(x(M(x)(F(x))，(x(N(x)((F(x))结论：(x(N(x)((M(x))证明①(x(N(x)((F(x)) P②N(y)((F(y) T①US③(x(M(x)(F(x)) P④M(y)(F(y) T③US⑤(F(y)((M(y) T④假言易位⑥N(y)((M(y) T②⑤假言三段论⑦(x(N(x)((M(x)) T⑥UG⑵设M(x)：是学生；N(x)：是教师；F(x)：是骗子；R(x, y)：相信。

前提：(x(M(x)((y(N(y)(R(x, y)))，(x(M(x)((y(F(y)((R(x, y)))结论：(x(N(x)((F(x))证明①(x(M(x)((y(N(y)(R(x, y))) P②M(c)((y(N(y)(R(c, y)) T①ES③M(c) T②化简④(x(M(x)((y(F(y)((R(x, y))) P⑤M(c)((y(F(y)((R(c, y)) T④US⑥(y(F(y)((R(c, y)) T③⑤假言推理⑦F(d)((R(c, d) T⑥US⑧(y(N(y)(R(c, y)) T②化简⑨N(d)(R(c, d) T⑧US⑩R(c, d)((F(d) T⑦假言易位⑾N(d)((F(d) T⑨⑩假言三段论⑿(x(N(x)((F(x)) T⑾UG⑶设M(x)：是学术会成员；N(x)：是工人；R(x)：是专家；Q(x)：是青年人。前提：(x(M(x)((N(x)(R(x)))，(x(M(x)(Q(x))结论：(x(M(x)(Q(x)(R(x))证明①(x(M(x)(Q(x)) P②M(c)(Q(c) T①ES③(x(M(x)((N(x)(R(x))) P④M(c)((N(c)(R(c)) T③US⑤M(c) T②化简⑥N(c)(R(c) T④⑤假言推理⑦R(c) T⑥化简⑧M(c)(Q(c)(R(c) T②⑦合取⑨(x(M(x)(Q(x)(R(x)) T⑧EG复习题21．解 ⑴设个体域是整数集合I，F(x)：x是最大的整数，命题符号化为((xF(x)。

⑵设M(x)：x是学生，F(x)：x要好好学习。命题符号化(x(M(x)(F(x))。⑶设M(x)：x是液体，F(x)：x能溶于水。命题符号化((x(M(x)(F(x))。⑷设M(x)：x是人，F(x, y)：x与y一样高。命题符号化((x((M(x)(M(y))(F(x, y))。⑸设M(x)：x是数，F(x)：x是实数，G(x)：x是复数。命题符号化(x(M(x)((F(x)(G(x)))。⑹设M(x)：x是数，F(x)：x是奇数，G(x)：x是偶数，H(x)：x是2。命题符号化(x(M(x)(((F(x)(G(x))(H(x)))。⑺设M(x)：x不是地球，F(x)：x上有人，c：金星。命题符号化(x(M(x)(F(x))(F(c)。2．解 ⑴(x(A(x)(B(x))((A(1)(B(1))((A(2)(B(2))((A(3)(B(3))(0(0(0(0。⑵(x(A(x)(((x(2))((A(1)(((1(2))((A(2)(((2(2))((A(3)(((3(2))(1(0(1(0。3．解 ⑴(x的辖域P(x)(Q(y)，其中x是约束变元，y是自由变元；(y的辖域M(x,y)，其中x是自由变元，y是约束变元。

⑵(x的辖域P(x)，(x的辖域M(x)，其中x在两个量词的不同辖域中都是约束变元，y是自由变元。⑶(x的辖域P(x,y)，其中x是约束变元，y是自由变元；(y的辖域Q(y)，其中y是约束变元。⑷(x的辖域(yP(x,y)，(y的辖域P(x,y)，整个公式中x是约束变元，y约束变元1次，自由变元1次。4．解 (!xP(x)((x(P(x)((y(P(y)((y=x)))。5．解 ⑴(xP(x)((xQ(x)(P(a)(P(b)(P(c)(Q(a)(Q(b)(Q(c)⑵(x(P(x)((xQ(x))((P(a)((Q(a)(Q(b)(Q(c)))((P(b)((Q(a)(Q(b)(Q(c)))((P(c)((Q(a)(Q(b)(Q(c)))⑶(x(yR(x,y)((yR(a,y)((yR(b,y)((yR(c,y)((R(a,a)(R(a,b)(R(a,c))((R(b,a)(R(b,b)(R(b,c))((R(c,a)(R(c,b)(R(c,c))6．解 ⑴设个体域为D={a，b}，令P(a)=1；P(b)=0；Q(a)=0；Q(b)=1。则(xP(x)为假，(xQ(x)为假，从而(xP(x)((xQ(x)为真。

由于P(a)(Q(a)为假，所以(x(P(x)( Q(x))也为假，此时公式为假。因此，公式不是逻辑有效式。⑵ 设D={a}，若R(a)=1，P(a)=0，Q(a)=1，则(x(P(x)(Q(x))为假，而(x((P(x)(R(x))((Q(x)(R(x)))为真，因此原公式为假。因此，公式不是逻辑有效式。⑶设个体域D={a，b}，Q(a)=Q(b)=0，取P(a)=1，P(b)=0。则(x(P(x)(Q(y))为真，而((xP(x)(Q(y))为假。因此，原公式不是逻辑有效式。⑷(x(y(P(x)(Q(y))((x(y((P(x)(Q(y))((x((P(x)((yQ(y))((x(P(x)((y Q(y)(((xP(x)((yQ(y)((xP(x)((yQ(y)因此，原公式为逻辑有效式。7．⑴证明 (z(A(x)(B(x))((x((A(x)(B(x))((x(A(x)((x B(x))(((xA(x)((x B(x))((xA(x) ((x B(x))⑵证明 (xP(x)((yQ(y)(((xP(x)((yQ(y)((x(P(x)((yQ(y)((x(y((P(x)( Q(y))((x(y(P(x)(Q(y))9．⑴证明①(xF(x) P②F(c) T①ES③(xF(x)((y((F(y)(G(y))(R(y)) P④F(c)((y((F(y)(G(y))(R(y)) T③US⑤(y((F(y)(G(y))(R(y)) T②④假言推理⑥(F(c)(G(c))(R(c) T⑤US⑦F(c)(G(c) T⑥附加⑧R(c) T⑤⑦假言推理⑨F(c)(R(c) T⑥⑧合取⑩(x(F(x)(R(x)) T⑨EG⑵证明①(x(C(x)((B(x)) P②C(y)((B(y) T②US③(x(A(x)(B(x)) P④A(y)(B(y) T③US⑤(B(y)((A(y) T④假言易位⑥C(y)((A(y) T②⑤假言三段论⑦(x(C(x)((A(x)) T⑧UG(x(H(x))(A(x))((x(y((H(y)(N(x, y))((y(A(y)(N(x, y))⑶证明①(x(y((H(y)(N(x, y)) P附加前提引入②(y((H(y)(N(x, y)) T①US③H(v)(N(x, v) T②US④(x(H(x))(A(x)) P⑤H(v)(A(v) T④US⑥H(v) T③化简⑥A(v) T⑤⑥假言推理⑦N(x, v) T③化简⑧A(v)(N(x, v) T⑥⑦合取⑨(y(A(y)(N(x, y)) T⑧EG⑩(x(y((H(y)(N(x, y))((y(A(y)(N(x, y)) T①⑨CP10．解 ⑴设M(x)：x是航海家，F(x)：x教育自己的孩子成为航海家，G(x)：x教育他的孩子去做飞行家。

前提：(x(M(x)(F(x))，(x(G(x)((F(x))，(xG(x)结论：(x(M(x)证明①(xG(x) P②G(c) T①ES③(x(G(x)((F(x)) P④G(c)((F(c) T③US⑤(F(c) T②④假言推理⑥(x(M(x)(F(x)) P⑦M(c)(F(c) T⑥US⑧(M(c) T⑤⑦拒取式⑨(x(M(x) T⑨UG⑵设M(x)：x是哺乳动物，N(x)：x是脊椎动物，F(x)：x是胎生动物。前提：(x(M(x)(N(x))，(x(M(x)((F(x)).结论：(x(N(x)((F(x))证明①(x(M(x)((F(x)) P②M(c)((F(c) T①ES③(x(M(x)(N(x)) P④M(c)(N(c) T③US⑤M(c) T②化简⑥N(c) T④⑤假言推理⑦(F(c) T②化简⑧N(c)((F(c) T⑥⑦合取⑨(x(N(x)((F(x)) T⑧EG⑶设F(x)：x是有理数，G(x)：x是无理数，M(x)：x是实数，N(x)：x是虚数。前提：(x(F(x)(M(x))，(x(G(x)(M(x))，(x(N(x)((M(x))结论：(x(N(x)(((F(x) (G(x)))证明①(x(N(x)((M(x)) P②N(y)((M(y) T①US③(x(F(x)(M(x)) P④F(y)(M(y) T③US⑤(M(y)((F(y) T②④假言易位⑥N(y)((F(y) T②⑤假言三段论⑦(x(G(x)(M(x)) P⑧G(y)(M(y) T⑦US⑨(M(y)((G(y) T⑦假言易位⑩N(y)((G(y) T②⑨假言三段论⑾(N(y)((F(y))((N(y)((G(y)) T⑥⑩合取⑿N(y)(（(F(y)((G(y)） T⑾蕴涵等价式，分配律⒀(x(N(x)(((F(x) (G(x))) T⑿UG第三章 基础知识习题3.11．解 ⑴：A=；⑵：B={a, e, i,m, n, o, r, t, u}；⑶：C={-3,2}。

2．解 ⑴ A={x(1( x (79, x (N}；⑵ B={x( x=2k+1, k(N}；⑶ C={x( x=5n, n(I}。3．解 ⑴：1，2，3，4，6，9，12，18，36；⑵：a，b；⑶：1，，。习题3.21．解 互不相同。⑴是不包含任何元素的空集，⑵是以空集(为元素的单元素集合，⑶是以0为元素的单元素集合，但和⑵的集合中的元素不同。2．证明 若，则；反之，若，则 ，，因此，。3．解 ⑴设，则；⑵设，则；⑶设，则；⑷设，则。4．解 ⑴M(T；⑵N(P；⑶P(T=( 。5．解 由题意可得：；；；。⑴A((B((C(D)) = A(B(C(D ={0,1,2,3,4,5,6,7,8,9,12,15,16,18,21,24,27,30,32,64}；⑵A((B((C(D))=(；⑶因为，A(C={0,1,2,3,6,7,8,9,12,15,18,21,24,27,30},所以，B- A(C={4,5}；⑷，=；6．解 ⑴、⑵的文氏图如图1-1所示，图中阴影部分表示所求集合。7．解 ⑴所求集合的集合成员表如表1-1所示。表1-1A B 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 1 1 0 1 ⑵所求集合的集合成员表如表1-2所示。

表1-2A B C 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 1 1 1 1 1 0 0 1 0 0 1 0 1 1 1 0 1 1 0 1 0 0 1 1 1 1 1 08．证明 ⑴===⑵===⑶A- (B(C)==9．证明 ⑴⑵。 因为，，则，所以，，因此，。⑵⑶。 ，。⑶⑴。因为，，所以，B=( (B=因此。习题3.31．解 ⑴(((=0； ⑵({(}(=1； ⑶({1,2,{3,{2,1}}}(3； ⑷({1,2,1}(=2。2．解 ⑴①8，②8，③8，④10，⑤3，⑥6，⑦5，⑧12。⑵因为，=-，所以=-=60-25-26-26+9+11+8-8=3⑶=52-11-9-8+3(2=303．解 设A={x(x( N,1(x(100, x能被5整除}，B={x(x( N,1(x(100, x能被4整除}，C={x(x( N,1(x(100, x能被6整除}，则，，因此，=20-1=19。习题3.41．解 ⑴20=2(7+6；⑵58=2(27+4；⑶3=0(8+3；⑷57=3(19+0。2．解 ⑴因为,。所以，14和35的最大公因数为7，即(14,35)=7，且由以上两式可推得7=1(35-2(14。

⑵因为58=1(34+24,34=1(24+10,24=2(10+4,10=2(4+2,4=2(2。所以34和58的最大公因数为2，即(34,58)=2，且由以上各式可推得2=12(34-7(58。⑶因为，，。所以，180和252的最大公因数为36，即(180,252)=36，且由以上各式可推得36=3(180-2(252。⑷因为，，，。所以128和77互素，即(128,77)=1，且由以上各式可推得1=5(77-3(128。3．解 ⑴因为，，所以(108,72)=36，因此LCM(108,72)=108(72/36=216。⑵因为，，，所以(245,175)=35，因此LCM(245,175)=245(175/35=1225。⑶因为，，，所以(252,180)=36，因此LCM(252,180)=252(180/36=1260。⑷因为，，，，，所以(64,81)=1，因此LCM(64,81)=64(81=5184。4．证明 ⑴设，则存在整数、，使得 ，因为是、的最大公因数，所以，、。因此，①另一方面，因为是、的最大公因数，所以、，因此，与互素，否则与有大于1的公因数，则d2(c，d2(d1，又，由整除的传递性，所以，。

因此，与有大于1的因数，这与(a,c)=1矛盾。由于与互素，且，由定理6，则，又，因此②由于、都是正整数，由①②可得。⑵因为b,c都和a互素，则(a,b)=1，(a,c)=1，由⑴则有(a, bc)=1，即bc也和a互素。5．证明 设(a,b)=d，则d(a，d(b。因为c>0，所以，cd(ac，cd(bc。另一方面，因为(a,b)=d，所以存在s,t(I,使得sa+tb=d，此式两边同乘c得sac+tbc=cd因此，对于任何的正整数e, 若e(ac，e(bc，则e(cd。又因为cd>0，故cd是ac和bc的最大公因数，即(ac,bc)= cd= c((a,b)。6.证明 因为a(m，所以存在整数，使得又b(m，即，而、互素，由定理6，则有。因此，存在整数q1，使得，代入既可得即abm。7.证明 设（）,由，，则。同理，由，，则。即为、的公倍数，又是、的最小公倍数，因此，，即，所以mk。复习题31. 解 ⑴S5={3，5},,所以，X中不含元素3,5。故。⑵因为, 所以,因此或。⑶因为，X(S2=S1，所以，S1-S2(X，因此，X=S3或S1。

⑷S2={2，4，6，8},，所以X中不含元素2,4,6,8,所以X=S3或,但X(S4，故X=S5。⑸因为X(S1，所以，X=S2, S3, S4, S5，而X(S3=(，即X中无奇数，所以，X=S2。2.解 是可以作到的。如，，则.3.解 ⑴，则；⑵，则；⑶，则；⑷={{1，2}}，则。4.证明 对于任意的X(P(B)，则，又,，所以X(P(A)，由的任意性可知, .5.解 ⑴可构成个子集。⑵其中有个子集中元素个数为奇数。6.⑴证明 左边======右边,故原题得证.⑵证明==7.解 ⑴=；⑵=；⑶=；⑷=；⑸=。8. 证明 若，对于任意的x(B，⑴若x(A，则x(A(B，所以，x(A(B，则x(A(C，由可得x(A(C，而，则x(A(C，所以x(C；因此，B(C。⑵若x(A，则x(A(B，又x(B，所以，x(A(B，由且，可得，x(A(B，因此，x(A(C，但且，所以，x(C，因此，B(C。采用同样的方法可证。故。9.方法⑴证明 对于任意的x((A-B)((B-A)，则或，即有,或者；当时,且，当时,亦有且，故有。因此，同理可证综上知,方法⑵作集合成员表如表1-4所示可见与对应填入值相同故.表1-4A B A-B B-A 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 1 1 0 1 0 1 1 0 1 1 1 0 0 0 1 1 0 方法(3) 证明 =====10.解 设A、B、C分别表示选修法、德、俄语的学生的集合则，，， ，，，，=150。

=150+260+208+160-76-48-62+30=622⑵=76-30=46；⑶同理=62-30=32；⑷==260-76-48+30=166；⑸同理。11.解 |P(B)|=64；又；，则|A∩B|=6+3-8=1，所以|A-B|=|A|-|A∩B|=3-1=2；==8-2=6。12.证明 因为(a，b)=d，则存在整数、，使得又因为a，b不全为0，所以(a，b)=d，因此，、所以，、都是整数，由此可得所以，对于任意的正整数c，若，则c(1，因此。反之，因为d是a，b的正因数，所以、都是整数，又，则存在整数、，使得两边乘以得因此，对于任何的正整数c，若c(a，c(b，则c(d，故（a，b）=d。第四章 关系习题4.11.解 A(B={(a,1),(a,2),(b,1),(b,2),({a,b},1),({a,b},2)}，B(A={(1,a), (1,b), (1, {a,b}),(2, a), (2, b), (2, {a,b})}，B ( B={(1,2),(1,2),(2,1),(2,2)} ，B3={(1,1,1),(1,1,2),(1,2,1),(1,2,2), (2,1,1),(2,1,2),(2,2,1),(2,2,2)} 。

2.解 P(A)={, {x}, {y}, {x, y}}，因此，P(A)(A={<, x>, <, y>, <{x}, x>,<{x}, y>, <{y}, x>, <{y}, y>, <{x, y}, x>, <{x, y}, y>}。3.解 ⑴不正确。例如令A=，B={1，2}，C={x}，则A(B=A(C=⑵正确。因为，(<x, y>( (A-B)(C(x(A-B)∧yC(xA∧xB∧yC(xA∧yC∧xB∧yC(<x, y>A(C∧<x, y>B(C(<x, y>(A(C-B(C) 。所以，(A-B)(C((A(C-B(C) 。又因为，(<x, y>((A(C-B(C)(<x, y>(A(C∧<x, y>(B(C( (xA∧yC)∧[(x(B∧y(C)((x(B∧y(C)( (x(B∧y(C)]( xA∧(x(B∧y(C) ( (xA∧x(B)∧y(C( x(A-B)∧yC(<x, y>( (A-B)(C。

所以，(A(C-B(C)((A-B)(C。综上所述，(A–B)(C=(A(C–B(C)。⑶正确。例如A=，A(A=，显然AA(A。但是当A≠时，因为A(A是由集合A中的两个元素的序偶组成的集合，所以AA(A一般不成立。4.证明 (1)(<x, y>A((B∪C)((xA∧y(B∪C))((xA∧(yB∨yC))((xA∧yB)∨(xA∧yC)((<x, y>A(B) ∨(<x, y>A(C)(<x, y>((A(B) ∪(A(C))因此A((B∪C)=(A(B)∪(A(C)。同理可证(2)。5.证明 ⑴(<x, y>(A(C)∪(B(D)(<x, y>(A(C)∨<x, y>(B(D)(<x, y>(A(C)∨<x, y>(B(D)((xA∧yC)∨(xB∧yD)((xA∨xB)∧(xA∨yD)∧(yC∨xB)∧(yC∨yD)((xA∨xB)∧(yC∨yD)(x(A∪B)∧y(C∪D)(<x, y>(A∪B)((C∪D)。因此，(A(C)∪(B(D)((A∪B)((C∪D)。

⑵设(xA，若<x, x>A(A，而A(A=B(B，所以，<x, x>B(B，即xB，因此AB；同理可证BA；故A=B。习题4.21.解 因为，A(B={(1, a), (2, a) ,(3, a)}，所以，R1=、R2={<1, a>}、R3={<2, a>}、R4={<3, a>}、R5={<1, a>, <2, a>}、R6={<1, a>, <3, a>}、R7={<2, a>, <3, a>}、R8={<1, a>, <2, a>, <3, a>}。2.解 L={<1, 1>, <2, 1>, <2, 2>, <3, 1>, <3, 2>, <3, 3>, <4, 1>, <4, 2>, <4, 3>, <4, 4>, <8, 1>, <8, 2>, <8, 3>, <8, 4>, <8, 8>, <9, 1>, <9, 2>, <9, 3>, <9, 4>, <9, 8>, <9, 9>}；D={<1, 1>, <1, 2>, <1, 3>, <1, 4>, <1, 8>, <1, 9>, <2, 2>, <2, 4>, <2, 8>, <3, 3>, <3, 9>, <4, 4>, <4, 8>, <8, 8>, <9, 9>}L∩D={<1, 1>, <2, 2>, <3, 3>, <4, 4>, <8, 8>, <9, 9>}L∪D={<1, 1>, <1, 2>, <1, 3>, <1, 4>, <1, 8>, <1, 9>, <2, 1>, <2, 2>, <2, 4>, <2, 8>, <3, 1>, <3, 2>, <3, 3>, <3, 9>, <4, 1>, <4, 2>, <4, 3>, <4, 4>,<4, 8>, <8, 1>, <8, 2>, <8, 3>, <8, 4>, <8, 8>, <9, 1>, <9, 2>, <9, 3>, <9, 4>, <9, 8>, <9, 9>}L-D={<2, 1>, <3, 1>, <3, 2>, <4, 1>, <4, 2>, <4, 3>, <8, 1>, <8, 2>, <8, 3>, <8, 4>, <9, 1>, <9, 2>, <9, 3>, <9, 4>, <9, 8>}3.解 X∪Y={a, b, c, d}，则X∪Y上的全域关系U、恒等关系I分别为U={<a, a>, <a, b>, <a, c>, <a, d>, <b, a>, <b, b>, <b, c>, <b, d>, <c, a>, <c, b>, <c, c>, <c, d>, <d, a>, <d, b>, <d, c>, <d, d>}；I={<a, a>, <b, b>, <c, c>, <d, d>}。

4.解 P∩Q={<1, 2>, <2, 3>}；P∪Q={<1, 2>, <1, 4>, <2, 3>, <4, 4>, <4, 2>}；domP={1, 2, 4}，ranP={2, 3, 4}；domQ={1, 2, 4}，ranQ={2, 3}；dom(P∪Q)={1, 2, 4}，ran(P∪Q)={2, 3, 4}。5.证明 (1)对于任意元素xA，xdom(R1∪R2)((y(<x, y>R1∪R2)((y(<x, y>R1∨<x, y>R2)(xdom(R1)∨xdom(R2)(x(dom(R1)∪dom(R2))故dom(R1∪R2)=domR1∪domR2(2)对于任意元素yByran(R1∩R2)((x(<x, y>R1∩R2)((x(<x, y>R1∧<x, y>R2)((x(<x, y>R1)∧(x (<x, y>R2) (yranR1∧yranR2(y(ranR1∩ranR2)。

故ran(R1∩R2)(ranR1∩ranR26.解 R1={<1, 1>, <1, 2>, <2, 1>, <2, 2>}；R2={<2, 3>, <4, 1>}；R3={<0, 0>, <0, 1>, <0, 2>, <0, 4>,<0, 6>,<0, 8>, <1, 1>, <2, 2>, <4, 4>,<6, 6>,<8, 8>}；R4={<0,1>,<1,1>,<1,2>,<1,3>,<2, 1>,<2, 3>, <4, 1>,<4, 3>, <6, 1>,<8, 1>,<8, 3>}，，，。关系图如下：习题4.31.解 从关系复合的定义来求：R1·R2={<1, 4>, <1, 3>, <3, 2>, <4, 2>}；R2·R1={<1, 3>, <2, 3>}；R12=R1·R1={<1, 1>, <1, 2>, <3, 3>, <4, 3>}；R22=R2·R2={<2, 2>, <3, 3>, <3, 4>}。

2.解 dom(R1·R2)A；ran(R1·R2)C。3.解 (1)利用定义求解，R1={<1, 1>, <1, 2>,<1, 3>,<2, 1>,<2, 2>,<2, 3>,<3, 1>,<3, 3>}；R2={<3, 1>}；R1·R2={<1, 1>, <2, 1>,<3, 1>}；R1·R2·R1={<1, 1>,<1, 2>,<1, 3>,<2, 1>,<2, 2>,<2, 3>,<3, 1>, <3, 2>, <3, 3> }；R1-1={<1, 1>,<1, 2>,<1, 3>,<2, 1>,<2, 2>,<3, 1>, <3, 2>, <3, 3>}；(2)利用矩阵求解，；；；；。关系图如图4-2：4.解 R={<1, 2>, <2, 3>, <3, 1>, <4, 5>, <5, 4>}，R2={<1, 3>, <2, 1>, <3, 2>, <4, 4>, <5, 5>,}，R3={<1, 1>, <2, 2>, <3, 3>, <4, 5>, <5, 4>}，R4={<1, 2>, <2, 3>, <3, 1>, <4, 4>, <5, 5>}，R5={<1, 3>, <2, 1>, <3, 2>, <4, 5>, <5, 4>}，R6={<1, 1>, <2, 2>, <3, 3>, <4, 4>, <5,5>}，R7={<1, 2>, <2, 3>, <3, 1>, <4, 5>, <5, 4>}= R1。

所以，m=1、n=7可使得R1=R7。5.解=，因此，，一般地，。6.证明 ⑴对于任意的<a,b>((R-S)-1，则<b,a>(R-S，所以，<b,a>(R，但<b,a>(S，因此，再由逆关系的定义有，<a,b>(R-1，而<a,b>(S-1，故<a,b>(R-1-S-1。即(R-S)-1(R-1-S-1。同理可证R-1-S-1((R-S)-1。综合可得(R-S)-1=R-1-S-1。⑵由集合的笛卡尔积和逆关系的定义，因为，对于任意的a(A,b(B都有，<b, a>(B×A，则<a,b>(A×B，所以，<b, a>((A×B)-1，即B×A((A×B)-1，另一方面，因为，对于任意的a(A,b(B都有，<b,a>((A×B)-1，所以，则<a,b>(A×B，所以，<b, a>(B×A，即(A×B)-1(B×A。综合上述有，(A×B)-1=B×A。⑶用反证法证明-1=，假设存在a(A,b(B使得，<b,a>(-1，则<a,b>(，这与是空集矛盾。

⑷先证明R(S(R-1(S-1。对于任意的a(A,b(B，若<b,a>(R-1，由逆关系的定义则<a,b>(R，而R(S，所以，<a,b>(S，因此，<b,a>(S-1，即R-1(S-1。再证明R-1(S-1(R(S。若<a,b>(R，则<b,a>(R-1，而R-1(S-1，所以，<b,a>( S-1，因此，<a,b>(S，即R(S。习题4.41.解 如表4-2关系 自反的 反自反的 对称的 反对称的 传递的 R1 N N Y N Y R2 N Y Y N N R3 N Y N Y N2.解 ⑴R具有自反性，因为对于任意的xI，x(x-1)=x(x-1)，即<x, x>R，所以R具有自反性；⑵因为R具有自反性，所以R不具有反自反性；⑶R具有对称性，因为对于任意的x, yI ，若<x, y>R，则x(x-1)=y(y-1)，所以y(y-1)=x(x-1)，即<y, x>R，因此R具有对称性；⑷显然R不具有反对称性；⑸R具有传递性，设<x, y>R，<y, z>R，即x(x-1)=y(y-1)，y(y-1)=z(z-1)，显然有x(x-1) =z(z-1)，即<x, z>R，所以R具有传递性。

3.解 (1)R={<0, 0>, <0, 1>, <0, 2>, <0, 4>, <1, 0>,<1, 1>, <1, 2>, <1, 4>, <2, 0>, <2, 1>, <2, 2>,<4, 0>, <4, 1>}，关系图如图4-3。⑵R具有对称性。4.解 若A上关系R是反对称的，则R∩R-1关系矩阵MR∩R-1中最多只有主对角线上有非零值，即最多只有|A|个非零值。5.解 答案可以有很多组，下面给出一组答案如下。(1)R={<1, 1>}；(2)R={<1, 2>, <2, 1>, <3, 1>}；(3)R={<2, 2>, <3, 3>}；(4)R={<1, 2>, <2, 3>, <1, 3>}。6.证明 因为，R1，R2为集合A上的自反关系，则对于任意的x(A，有 <x, x>(R1，且<x, x>(R2，因此，<x, x>R1·R2，故R1·R2是A上的自反关系。

习题4.51.解 r(R1)=R1∪IA={<a, a>, <a, b>, <b, b>, <b, c>,<c, c>}；s(R1)=R1∪={<a, b>, <b, a>, <b, c>, <c, b>}；t(R1)={<a, b>, <a, c>, <b, c>}；r(R2)=R2∪IA={<a, a>, <a, b>,<b, b>,<b, c>, <c, a>, <c, c> }s(R2)=R2∪={<a, a>, <a, b>, <a, c>,<b, a>,<b, c>, <c, a>, <c, b>}t(R2)={<a, a>,<a, b>,<a, c>,<b, a>,<b, b>,<b, c>, <c, a>, <c, b>, <c, c>}关系矩阵如下：，，，，，，，。

关系图如图4-4：2.解 ，，则；；，，；；。3.证明 ⑴因为，R1R2，所以，IA∪R1IA∪R2，即r(R1)(r(R2)。⑵因为，R1R2，由习题4.3第6⑷题可得，所以，，即s(R1)(s(R2)。⑶因为，R1R2，所以，，则，即t(R1)(t(R2)。4.证明 ⑴由自反闭包的定义又r(R1)= IA∪R1(IA∪R1∪R2= r(R1∪R2)，同理有r(R2)= IA∪R2(IA∪R1∪R2= r(R1∪R2)，所以有r(R1)∪r(R2) ( r(R1∪R2)；另一方面，，下面分两种情况来证明。①如果a=b，则<a,b>(IA，所以；②如果a≠b，则，因此，，若，则，所以，，同理，若，则，所以，。故r(R1∪R2) ( r(R1)∪r(R2)。综上所述，r(R1∪R2)=r(R1)∪r(R2)。⑵因为R1R1∪R2，且R2R1∪R2，所以由第3题有s(R1)s(R1∪R2) ，且s(R2)s(R1∪R2)，因此，s(R1)∪s(R2)s(R1∪R2)；另一方面，，则或，①若，则或，所以，或，因此，；②若，则，所以，或，因此，或，所以，或，因此，。故s(R1∪R2)( s(R1)∪s(R2)。

综上所述，s(R1∪R2)=s(R1)∪s(R2)。⑶对于任取的<a, b>( t(R1)∪t(R2)，则<a, b>( t(R1)或<a, b>(t(R2)。若<a, b>( t(R1)，则存在正整数m使得，因此，所以，<a, b>( t(R1∪R2)；同理若<a, b>(t(R2)，可证<a, b>( t(R1∪R2)。综上所述，t(R1)∪t(R2)t(R1∪R2)。5.解 R={<x, y>}是集合A={x, y}上的二元关系，则st(R)={<x, y>, <y, x>}，而ts(R)={<x, y>, <y, x>, <x, x>, <y, y>}。因此st(R)(ts(R)。6.证明 rt(R)=r(R∪R2∪…∪Rn∪…)= IA∪R∪R2∪…∪Rn∪…=( IA∪R)∪(IA∪R2)∪…∪(IA∪Rn)∪…(( IA∪R)∪(IA∪R)2∪…∪(IA∪R)n∪…=tr(R)。反之，对于任意的<a, b>(tr(R)，则存在正整数n，使得<a, b>((IA∪R)n，即<a, b>(IA∪R∪R2∪…∪Rn，亦即<a, b>( rt(R)，所以，tr(R)(rt(R)。

综上所述，tr(R)=rt(R)。习题4.61.解 A上共有15个等价关系。由等价关系和划分是一一对应关系，而A共有如下的15个划分：(1={{1,2,3,4}}，(2={{1},{1,2,3,4}}，(3={{2},{1,3,4}}，(4={{3},{1,2,4}}，(5={{4},{1,2,3}}，(6={{1,2},{3,4}}，(7={{1,3},{2, 4}}，(8={{1,4}, {2,3}}，(9={{1,2},{3},{4}}，(10={{1,3},{2},{4}}，(11={{1,4},{2},{3}}，(12={{2,3},{1}, {4}}，(13={{2, 4},{1},{3}}，(14={{3,4},{1},{2}}，(15={{1},{2},{3},{4}}。与每个划分对应的等价关系由读者自行给出。2.证明 要证明R是A上的等价关系，只需证明R是A上的自反关系。事实上，因为，(aA，总存在一个元素bA，使<a, b>R，而R是集合A中的对称关系，所以<b, a>R，又因为R是A上的传递关系，有<a, b>R且<b, a>R时，可得<a, a>R，即R是集合A中的自反关系。

故R是等价关系。3.解 ⑴R={a, d, e}({a, d, e}∪{b, c}({b, c}={<a, a>, <a, d>, <a, e>,<d, a>, <d, d>, <d, e>,<e, a>, <e, d>, <e, e>}∪{<b, b>, <b, c>, <c, b>, <c, c>}⑵关系图如图4-5。4.解 ⑴不是等价关系。例如设A={1,2,3}，R1={<1,1>,<1,2>,<2,1>,<2,2>,<3,3>}，而显然A2-R1={<1,3>,<2,3>,<3,1>,<3,2>}不是等价关系；⑵不是等价关系。例如A={1,2,3}，R1={<1,1>,<1,2>,<2,1>,<2,2>,<3,3>}，R2={<1,1>,<2,2>,<2,3>,<3,2>,<3,3>}，而显然R1-R2={<1,1>,<1,2>}不是等价关系；⑶不是等价关系，令A={1, 2, 3}，R1={<1, 1>, <2, 2>, <3, 3>, <1, 2>, <2, 1>, <2, 3>, <3, 2>, <1, 3>, <3, 1>}，R2={<1, 1>, <2, 2>, <3, 3>, <1, 2>, <2, 1>}R1-R2={<2, 3>, <3, 2>, <1, 3>, <3, 1>}，r(R1-R2)=(R1-R2)∪IA={<1, 1>, <2, 2>, <3, 3>, <2, 3>, <3, 2>, <1, 3>, <3, 1>}，因为<1, 3>R1-R2，<3, 2>R1-R2，但<1, 2>R1-R2，显然R1-R2不具有传递性，所以R1-R2不是等价关系。

(4)不是等价关系，令A={1, 2, 3}，R1={<1, 1>, <2, 2>, <3, 3>, <2, 3>, <3, 2>}，R2={<1, 1>, <2, 2>, <3, 3>, <1, 2>, <2, 1>}，R1·R2={<1, 1>, <1, 2>, <2, 2>, <2, 1>, <3, 3>, <2, 3>, <3, 1>, <3, 2>}，因为<3, 1>R1·R2，但<1, 3>R1·R2，所以R1·R2不是等价关系。(5)不是等价关系，令A={1, 2, 3}，R1={<1, 1>, <2, 2>, <3, 3>, <1, 2>, <2, 1>}，R2={<1, 1>, <2, 2>, <3, 3>, <2, 3>, <3, 2>}R1∪R2={<1, 1>, <2, 2>, <3, 3>, <1, 2>, <2, 1>, <2, 3>, <3, 2>}，因为<1, 2>R1，<2, 3>R2，但<1, 3>R1·R2，显然R1∪R2不具有传递性，所以R1∪R2不是等价关系。

5.解 ⑴若(1=(2，则(1∪(2是集合A的划分，其它情况(1∪(2不是集合A的划分；例如设A={1,2,3}，(1={{1},{2,3}},(2={{1,2},{3}}，而(1∪(2={{1},{1,2},{2,3},{3}}不是划分。⑵若(1=(2，则(1∩(2是集合A的划分，其它情况(1∩(2不是集合A的划分，例如设A={1,2,3}，(1={{1},{2,3}},(2={{1},{2},{3}}，而(1∪(2={{1}}不是集合A的划分；⑶若(1∩(2=，则(1-(2是集合A的划分，其它情况(1-(2不是集合A的划分，例如设A={1,2,3}，(1={{1},{2,3}},(2={{1},{2},{3}}，而(1-(2={{2,3}}不是集合A的划分；⑷因为((1∩((2―(1))∪(1=(1，所以((1∩((2―(1))∪(1是集合A的划分。6.证明 (1)①对任意的xI，x2=x2，即<x, x >R，所以R是自反的；②对任意的<x, y>R，即x2=y2，则y2=x2，即<y, x>R，所以R是对称的；③对任意的<x, y>R，<y, z>R,即x2=y2，y2=z2，显然x2=z2，亦即<x, z>R，所以R是传递的；综合①、②、③可知R是等价关系。

(2)R的等价类可以分为{[0], [1], [2], …},其中 [0]={0}，[1]=[-1]={1, -1}，[2]=[-2]={2, -2}，…，[n]=[-n]={n, -n}，…。习题4.71.证明 (1)自反性。因为对任意的xX，<x, x>IX，所以<x, x>IX∪R∪R-1=R，即R是自反的；(2)对称性。对于任意的<x, y>IX∪R∪R-1，且x(y，显然<x, y>R∪R-1，则<x, y>R，或<x, y>R-1，因此，<x, y>R-1，或<x, y>R，从而有<y, x>R∪R-1，所以<y, x>IX∪R∪R-1=R，即R是对称的；综上所述，R是X上的相容关系。2.证明 (1)对任意的xA，有<x, x>R，所以R是自反的；(2) 设任意的<x, y>R，且x(y，都有<y, x>R，即R是对称的；综上所述，R是A上的相容关系。R的关系矩阵如下，关系图如图4-6：，最大相容类为：{1,2,3},{2,3,4},{2,4,5},{6}。

习题4.81.解 从R的定义中，显见R具有自反性、反对称性、传递性，所以R是A上的偏序关系，即<A, R>是偏序集。COVA={<1, 2>, <1, 3>, <2, 4>, <3, 5, >, <4, 5>}2.解 (1)从哈斯图可以看出，dRa，，，，aRa，eRa(2)R的关系图如图4-9：(3)A的最大元为a，无最小元，极大元为a，极小元为d，e；(4)出集合A及其子集B1={c, d, e}，B2={b, c,d}，B3={a, c, d, e}的上界，下界，上确界，下确界如表1。上界 下界 上确界 下确界 A={a, b, c, d, e} a 无 a 无 B1={c, d, e} c, a 无 c 无 B2={b, c, d} a d a d B3={a, c, d, e} a 无 a 无 表13.填写表2，区分偏序集<A，≤>的子集B上的最大(小)元，极大(小)元，上(下)界，上(下)确界。b是B的… 定 义 b(B否 存在性 唯一性 最大元素 b(B且b大于B中其它每个元素 是 不一定存在 存在则唯一 最小元素 b(B且b小于B中其它每个元素 是 不一定存在 存在则唯一 极大元素 b(B且b不小于B中其它每个元素 是 不一定存在 不一定唯一 极小元素 b(B且b不大于B中其它每个元素 是 不一定存在 不一定唯一 上界 b(A且b大于B中每个元素 不一定 不一定存在 不一定唯一 下界 b(A且b小于B中每个元素 不一定 不一定存在 不一定唯一 上确界 B的上界中的最小者 不一定 不一定存在 存在则唯一 下确界 B的下界中的最大者 不一定 不一定存在 存在则唯一4．解 ⑴是偏序集，不是其它集合；⑵是拟序集，不是其它集合；⑶是偏序集、全序集、良序集，不是拟序集；⑷是偏序集、全序集、良序集，不是拟序集；复习题四1．证明 对于任意的bB，因为A非空，所以存在aA，且<a, b>A(B，因为A(B=A(C，所以<a, b>A(C，从而bC，因此BC。

同理可证CB。故B=C。2．证明 利用集合相等的定义去证，即分别证明IA·RR，RIA·R；⑴对于任意的<x, y>IA·R，必存在z，满足<x, z>IA，<z, y>R，而<x, z>IA(x=z，所以<z, y>R，即IA·RR；⑵对于任意的<x, y>R，因为<x, x>IA，所以<x, y>IA·R，即RIA·R；综上所述，IA·R=R。4．解 ①R不具有自反性，因为 (A)，但(=，所以<, >R，故R不具有自反性。②R不具有反自反性，因为{1}(P(A)且{1}∩{1}={1}≠，所以<{1}, {1}>R，故R不具有反自反性。③R具有对称性，(x,y(P (A)，若<x, y>R，则x∩y≠，所以y∩x≠，因此<y, x>R，故R具有对称性。④R不具有反对称性，设x={1, 2}，y={1, 3}，则x∩y=y∩x={1}≠，由R的定义易知，<x, y>R且<y, x>R，但x≠y，故R不具有反对称性。

⑤R不具有传递性，设x={1}，y={1, 2}，z={2}，则有x∩y={1}≠且y∩z={2}≠，所以<x, y>R且<y, z>R，但x∩z=，所以<x, z>R，故R不具有传递性。5．证明 设R是集合X上的一个自反关系，如果R是X上对称和传递的，则对于任意a,b,c(X，若有<a, b>R∧<a, c>R，则<b, a>R∧<a, c>R，故得<b, c>R反之，若<a, b>R，<a, c>R，必有<b, c>R，则对任意a,bX，若<a, b>R，而因为R是自反关系，所以<a, a>R)，故<b, a>R，即R是对称的。若<a, b>R∧<b, c>R，则<b, a>R∧<b, c>R，所以<a, c>R，即R是可传递的。6．证明 (1)因为R+=t(R)是传递的，所以，(R+)+=t(R+)=t(t(R))=t(R)=R+(2)因为R*=IA∪R+=IA∪(R∪R2∪R3∪…)R·R*=R·(IA∪(R∪R2∪R3∪…))=R·IA∪R·R∪R·R2∪…=R∪R2∪R3∪…=R+同理可证，R*·R=R+。

(3)(R*)*=(R*)+∪IA=t(R*)∪IA=R*∪IA(因为R*是传递的)=R*7．证明 (1)自反性。对于任意的(xA, (yB，若<x, y>A(B，因为R1和R2分别是A和B上的等价关系，所以有<x, x>R1，<y, y>R2，因此根据R3的定义有<<x, y>, <x, y>>R3，即R3是自反的；(2)对称性。设任意的<<x1, y1>, <x2, y2>>R3，根据R3的定义有<x1, x2>R1且<y1, y2>R2，而因为R1和R2都具有对称性，所以<x2, x1>R1且<y2, y1>R2，因此<<x2, y2>, <x1, y1>>R3，即R3是对称的；(3)传递性。设任意<<x1, y1>, <x2, y2>>R3，<<x2, y2>, <x2, y2>>R3，根据R3的定义有<x1, x2>R1且<y1, y2>R2，<x2, x3>R1且<y2, y3>R2，因为R1和R2都具有传递性，所以<x1, x3>R1且<y1, y3>R2，因此<<x1, y1>, <x3, y3>>R3，即R3是传递的。

综上所述，R3是A(B上的等价关系。8．解 (1) E=rts(R)。(2)要证明E=rts(R)是等价关系，只要证明ts(R)具有对称性即可，对任意的<x, y>(ts(R)，则存在正整数k使得<x, y>((s(R))k，存在z1, z2, ┅, zkA，满足<x, z1>(s(R), <x, z2>s(R), ┅, <x, zk>s(R)，因为s(R)是对称的，所以<z1, x>(s(R), <z2, x>s(R), ┅, <zk, x>(s(R)，因此<y, x>((s(R))k ，即<y, x>(ts(R)，亦即ts(R)是对称的。(3) 因为R={<1, 2>, <1, 3>, <4, 4>, <4, 5>}，则s(R)={<1, 2>, <1, 3>,<2,1> ,<3,1>,<4, 4>, <4, 5>,<5,4>}，ts(R)={<1,1>,<1, 2>, <1, 3>,<2,1>,<2,2>,<2,3>,<3,1>,<3,2>,<3,3>,<4, 4>, <4, 5>,<5,4>,<5,5>}，显然rts(R)= ts(R)是等价关系。

9．证明 (1)①必要性。若R是一偏序关系，则R是自反的、反对称的和传递的，。而又R是传递的，可得R=t(R)；又因为R是自反的，所以IAR，因而R=rt(R)。若R是偏序关系，则R-1也是偏序关系，所以IAR-1。所以IAR-1∩R。又<x, y>R-1∩R(<x, y>R-1∧<x, y>R(<y, x>R∧<x, y>R，由R的反对称性，所以x=y，即<x, y>IA。因而R-1∩RIA。故IA= R-1∩R②充分性。若R-1∩R=IA且R=t(R)，则R是自反的、传递的。下面证明R是反对称的，若<x, y>R，<y, x>R，则<x, y>R且<x, y>R-1，所以<x, y>R∩R-1=IA，即x=y。因而R是反对称的。故R是偏序关系。(2)①必要性。若R是拟序关系，则R是反自反的、反对称的和传递的，因为R是传递的，所以R=t(R)。下面再用反证法证明R-1∩R=；假设R∩R-1≠，则存在某一<x, y>R-1∩R(<x, y>R-1∧<x, y>R(<y, x>R∧<x, y>R，由R的传递性可知，<x, x>R，这与R的反自反性矛盾，所以R∩R-1=。

②充分性。若R=t(R)，可得R是传递的。若R不是反自反的，则存在某一xA，使得<x, x>R，所以<x, x>R-1，这与R-1∩R=矛盾，因此R是反对称的。故R是拟序关系。第五章 函数( I+,由f(x)= |x|+2给出，,|x|≥0,则f(x)=|x|+2≥2故它的值域为ranf=N-{0,1}.3．解 （1）f(A)=f({5})={<5,6>}, f-1(B)=f-1({<2，3>})={2};(2) f(A)=f({2,3})={5,7}, f-1(B)= f-1({1，3})={0,1};(3) f(A)=f((0,1))=(1/4,3/4), f-1(B)= f-1([1/4,1/2])=[0,1/2];(4) f(A)=f({0,1/2})={1,2/3}, f-1(B)= f-1({1/2})={1}.4．解 ∵|A|=3,|B|=2,∴|BA|=8.即A→B的函数有8个，具体如下：f1={<x,a>,<y,a>,<z,a>}, f2={<x,a>,<y,a>,<z,b>}, f3={<x,a>,<y,b>,<z,a>},f4={<x,a>,<y,b>,<z,b>}, f5={<x,b>,<y,a>,<z,a>}, f6={<x,b>,<y,a>,<z,b>},f7={<x,b>,<y,b>,<z,a>}, f8={<x,b>,<y,b>,<z,b>}.因此，BA={f1, f2, f3, f4, f5, f6, f7, f8}。

习 题5.21．解 （1）是单射但不是满射；（2）既不是单射也不是满射；（3）是满射；（4）是满射，单射，双射。2．解 （1）f：I(I, f(x)= x3;(2) ;(3) f：R(R, f(x)= x2+1;(4) f：I(I, f(x)= x+1.3．解 （1）f={<0,0>,<1,4>,<2,3>,<3,2>,<4,1>},如图显然可知，是双射。(2) f={<0,0>,<1,4>,<2,2>,<3,0>,<4,4>}，由图可知，f即不是单射也不是满射。4．解 f1={<a,0>,<b,0>}, f2={<a,0>,<b,1>}, f3={<a,1>,<b,0>}, f4={<a,1>,<b,1>}.其中f2，f3是双射，而f1，f4既不是满射也不是单射。5．证明 设任意n∈N,则至少<n,0>∈N(N则f(n,1)=n∈N，g(n,1)=n∈N故f,g是满射.但f,g都不是单射，如f(2,2)=4=f(3,1),g(3,4)=g(2,6)=12。

6．证明 （1）对于<x,y>,<u,v>∈R×R，设f(<x,y>)=f(<u,v>)，即<(x+y)/2, (x-y)/2>=<(u+v)/2, (u-v)/2>，亦即(x+y)/2=(u+v)/2，(x-y)/2=(u-v)/2，解得x=u,y=v，故<x,y>=<u,v>，因此，f是单射；(2)证f是满射，既任意<u,v>∈R×R,令f(<x,y>)=<u,v>，则有<(x+y)/2, (x-y)/2>=<u,v>,既有(x+y)/2=u，(x-y)/2= v只要取x=u+v，y=u-v,就可使上式成立，且因为<u,v>∈R×R，所以，<x,y>∈R×R。故f是满射。综合上述f是双射。7．解 ψA(1)= ψA(2)=1; ψA(3)= ψA(4)=0;ΨB(1) =1; ψB(2) =ψB(3)= ψB(4)=0;ΨC(1)=ψC(2) =ψC(3)= ψC(4)=0;ΨM(1)=ψM(2) =ψM(3)=ψM(4)=1。

8．证明ψA(B(x)=1, ψA(x)+ ψB(x)-ψA(x)ψB(x)=1+1-1(1=1，所以，ψA(B(x)= ψA(x)+ ψB(x)-ψA(x)ψB(x)；② x∈A，x(B，则x∈A∪B，因此，ψA(B(x)=1, ψA(x)+ ψB(x)-ψA(x)ψB(x)=1+0-1(0=1，所以，ψA(B(x)= ψA(x)+ ψB(x)-ψA(x)ψB(x)；③ x(A，x∈B，同②可证ψA(B(x)= ψA(x)+ ψB(x)-ψA(x)ψB(x)；④ x(A，x(B，则x(A∪B，因此，ψA(B(x)=0, ψA(x)+ ψB(x)-ψA(x)ψB(x)=0+0-0(0=0，所以，ψA(B(x)= ψA(x)+ ψB(x)-ψA(x)ψB(x)。综合①②③④，对所有ψA(B(x)= ψA(x)+ ψB(x)-ψA(x)ψB(x)。（2）若x∈A则x(A，所以，；若x(A则x∈，因此，。离散数学习题七答案（3）①x(A，则x(A-B，所以，ψA-B(x)=0，ψA(x)=0，故ψA(x)(1-ψB(x))=0((1-ψB(x))=0=ψA-B(x);②x∈A且x∈B，则x(A-B，所以，ψA-B(x)=0，ψA(x)=1，ψB(x)=1，故ψA(x)(1-ψB(x))=1((1-1)=0=ψA-B(x);③x∈A，x(B，则x∈A-B，所以，ψA-B(x)=1，ψA(x)=1，ψB(x)=0，故ψA(x)(1-ψB(x))=1((1-0)=1=ψA-B(x)。

习题5.31．解 对于任意b∈B, 因为f：A(B是双射，即f：A(B是满射，则存在a∈A使<a,b>∈f，由逆关系定义有<b,a>∈f-1；若<y,x>∈f-1且<y,x1>∈f-1，则又由逆关系定义得<x,y>∈f且<x1,y>∈f，又因为f：A(B是单射故x=x1。综上所述由函数定义知是B到A的函数。2．解 没有，因为f不是双射函数。若将函数f的定义域和值域分别改为[0,(]和[-1,1]，则f有逆函数。3．解 g·f(x)=g(f(x))=g(2x+5)=(2x+5)+7=2x+12；f·g(x)=f(g(x))=f(x+7)=2(x+7)+5=2x+19;f·f(x)=f(f(x))=f(2x+5)=2(2x+5)+5=4x+15;g·g(x)=g(g(x))==g(x+7)=(x+7)+7=x+14;f·k(x)=f(k(x))=f(x-4)=2(x-4)+5=2x-3;g·h(x)=g(h(x))=g(x/3)=x/3+7.4．解f(f={<a,b >,<b,a>,<c,a >}({<a,b >,<b,a>,<c,a>}={<a,a>,<b,b>,<c,b>};f(f(f= f((f(f)= {<a,b>,<b,a>,<c,a>}({<a,a>,<b,b>,<c,b>}={<a,b>,<b,a>,<c,a>}。

5．解 （1）g·f(x)=g(f(x))=g(x2-2)= (x2-2)+4= x2+2;f·g(x)=f(g(x))=f(x+4)=(x+4)2-2= x2+8x+14.(2) g·f(x)=x2+2,不是单射，也不是满射和双射；f·g(x)= x2+8x+14也不是单射，满射和双射。6．证明 （1）因为g·f：A→C是双射，则g·f：A→C是单射。假设a1,a2∈A且a1≠a2，f(a1)= f(a2)，而g：B→C是函数，则g·f(a1)= g·f(a2)，这与g·f：A→C是单射矛盾，故f是单射;(2) 因为g·f：A→C是双射，则g·f：A→C是满射。所以，对于任意c∈C,存在a∈A,使g·f(a)=c即g(f(a))=c,又因为f：A→B是函数，故存在b=f(a)∈B，因此，存在b∈B使得g(b))=c，故g是满射.7．证明 因为，f：A→B是双射，由定理2知f-1:B→A是双射，故f-1也存在逆函数(f-1)-1：A→B，故对任意a∈A，设f(a)=b，则f-1(b)=a，因此有(f-1)-1(a)=b，于是f(a)=(f-1)-1(a)，由a的任意性可知f=(f-1)-1。

习题5.41．证明 要证A≈N,只须证明存在N到A的双射函数即可。设f：N→A，f(n)=11n+3,对与任意n∈N，显然，f：N→A是单射。下面证明f：N→A是满射。事实上，任取a∈A，由A中元素的形式，则存在x∈N,使得a=11x+3,且f(x)=11x+3=a,故f是满射,即f是双射。综合上述，A≈N.2．解 A={2n|n∈N},B={2n+1|n∈N},C={3n|n∈N}，A,B,C这三个集合均是N的子集且都N等势。事实上，可以作如下的三个函数。f:N→A,f(n)=2n,n∈N;g:N→B,g(n)=2n+1,n∈Nh:N→C,h(n)=3n,n∈N容易证明这三个都是双射函数。3．证明 作函数f：[0,1]→[2,3]，f(x)=x+2,x∈[0,1]。因为，f是严格单调的函数，所以，f是单射，又任意x∈[2,3]，则x-2∈[0,1]，且f(x-2)=(x-2)+2=x，故f是满射，故[2,3]≈[0,1]4．解 （1）作恒等函数任意a∈，IA(a)=a，显然IA是A上的双射函数，故A≈A（2）若A≈B，则存在双射函数f：A→B，由5.3节定理1和定理2知f：A→B存在逆函数，且:B→A也是双射双射，故B≈A。

（3）因为，A≈B,B≈C，则存在双射函数f:A→B和g:B→C，则f和g的复合函数g·f：A→C也是双射（5.3节定理5）即A≈C。5．证明 因为A≈C,B≈D，则存在双射函数f:A→C和g:B→D。由此可定义函数h:A×B→C×D，对于任意<a,b>∈A×C，h(a,b)=<c,d>,其中c=f(a),d=g(b)。下面证明：h:A×B→C×D是单射。对<a1,b1>,<a2,b2>∈A×B.若h(a1,b1)=h(a2,b2)= <c,d>，即f(a1)=f(a2)=c,g(b1)=g(b2)=d,而f和g都是单射，所以有a1=a2，b1=b2，即<a1,b1>=<a2,b2>。再证明h:A×B→C×D是满射。事实上对任意的<c,d>∈C×D,则c∈C,d∈D,由于f,g都是满射，所以存在a∈A,b∈B使得f(a)=c,g(b)=d。即存在<a,b>∈A×B，使得h(a,b)= <c,d>，故h是A×B到C×D的满射。因此，h是A×B到C×D的双射，故A×B≈C×D。

复习题五1．解 （1）是函数，定义域是{1,2,3,4}，值域是{x, y, z}，非满射，也非单射；(2) 不是函数；(3) 是函数，定义域是{1,2,3,4}，值域是{x, y, z, w},是双射，故有逆函数，则逆函数是{<z,1>,<w,2>,<x,3>,<y,4>},则定义域{x, y, z, w}，值域{1,2,3,4}；(4) 不是函数；(5) 是函数，定义域是{1,2,3,4}，值域是{y},单射，非满射，更不是双射。2．解 因为S=(A((B(C))((B(C)，所以(S(x)=( A((B(C)(x)+(B(C(x)-(A((B(C)(B(C(x)=(A(x)((B(C(x)+(B(x)(( C(x)-(A(B(C(x)=(A(x)((B(x)+(C(x)-(B(C(x))+(B(x)(( C(x)-(A(x)((B(x)((C(x)=(A(x)(B(x)+ (A(x)(C(x)-(A(x)((B(x)((C(x)+(B(x)(( C(x)-(A(x)((B(x)((C(x)=(A(x)(B(x)+(A(x)(C(x)+(B(x)(( C(x)-2(A(x)((B(x)((C(x)。

3．解 使f是双射，需要满足的条件和(y时，f(x)(f(y)，而nx=mk+r1，ny=ml+r2，(r1(r2)这里f(x)(f(y)就是r1(r2，r2-r1(0，即m(r2-r1)而由上式可得r2-r1=n(y-x)+m(k-l)要m (( r2-r1)，需要m ( n(y-x) ，因此需要m和证明是到的满射(B，都存在a(A使得f(a)=b，因此，f-1({b})非空；对于任何的b1, b2(B，b1(b2，则f-1({b1})(f-1({b2})=(。若不然，，则存在a(f-1({b1})(f-1({b2})，即a(f-1({b1})且a(f-1({b2})，亦即有f(a)=b1，f(a)=b2，这与f是函数矛盾。下面证明。对于任何的b(B，显然有f-1({b})(A，所以①另一方面，由于f是到的(A，都有b(B使得f(a)=b，即a(f-1({b})，因此②由①②可得。综上所述，(={f-1({b})(b(B}是A的一个划分。证明(B，都有a(A使得g(a)=[a]，(A，若要g(a)=g(b)，即要[a]=[b]，由等价类的性质，即需要aRb。6．解 （1）假。例如，设A={a,b,c}, B={x,y}, C={1,2,3}, 尽管f={< x, 1>,< y, 2>}是单射f·g={<a, 1>,<b, 1>,<c, 2>}不是单射；是满射f·g={<a, 1>,<b, 1>,<c, 1>}不是满射；f·g={<a, 1>,<b, 2>,<c, 3>}是单射是单射；(A且a1(a2但g(a1)=g(a2)，(C是函数，f·g(a1)= f·g(a2)，这与f·g：A(C是单射矛盾。

故g是单射f·g是满射，(C,存在a(A，使得f·g(a)=c，即f(g(a))=c。又因为g:A(B是函数，所以有b=g(a)(B，使得f(b)=c，因此f是满射。（6）证明 因为，f·g是双射，f·g既是满射，是单射是满射，是单射。f是从到的一个函数，f(a)=f(a)，因此aRa，既R是自反的；对于任意的a,b(A，若aRb，则f(a)=f(b)，所以f(b)= f(a)，因此bRa，即R是对称的；对于任意的a,b,c(A，若aRb，bRc，则f(a)=f(b)，f(b)=f(c)，所以f(a)=f(c)，即aRc，因此R是传递的。⑵等价类是(B→B(A，f(a,b)=<b,a>，对任意<a,b>(A(B。则任意a1,a2(A,b1,b2(B,则由函数f:A(B→B(A得，f(a1,b1)=<b1,a1>,f(a2,b2)=<b2,a2>,若f(a1,b1)＝f(a2,b2)，则<b1,a1>＝<b2,a2>，因此，b1=b2, a1=a2，所以<a1,b1>=< a2,b2>，故f是单射。

若任意<b,a>∈B×A,这里b∈B,a∈A，根据定义可知：<b,a>的原象是<a,b>∈A×B,则满足满射函数的定义，故f是满射函数。9．解 对于任意的y(f(X(Y)，则存在x(X(Y使得y=f(x)，这时因为x(X(Y，则x(X且x(Y，因此，f(x)(f(X)，f(x)(f(Y)，所以，f(x)(f(X)(f(Y)，即f(X(Y)(f(X)(f(Y)。任取u(f(X)(f(Y)，则u(f(X)，u(f(Y)，因此，存在x(X且y(Y，使得u=f(x), u=f(y)，即f(x)=f(y)，而因为f:A(B是单射(X(Y，因此，f(x)(f(X(Y)，即f(X)(f(Y)(f(X(Y)。综上所述：f(X(Y)=f(X)(f(Y)。第六章习题6.11．解 图G的图形如图6-1(a)，图H的图形如图6-1(b).2．解 ⑴、⑵、⑶、⑸能构成无向图的度序列，其中⑴、⑵、⑶能构成无向简单图的度序列。3．解 由握手定理可知，图G中所有结点度数之和，所以G应该有8条边。4．解 由握手定理可知，图G中所有结点度数之和为24，三个4度接点的度数之和为12，则其余度数的和也是12，而其余结点的度均为3，所以3度接点应该有4个，因此，图G有7个结点。

5．解 图中。结点之间的对应关系为(: V(G1)(V(G2)：((g)=1, ((a)=8, ((h)=2, ((b)=7, ((i)=10, ((c)=4, ((j)=3, ((d)=9, ((f)=6, ((e)=5。6．解 。因为与的结点间存在一一对应关系，边的方向一致，其对应关系为：f: V(G1)(V(G2)，f(a)=2, f(b)=5, f(c)=1, f(d)=4, f(e)=3。习题6.21．解 ⑴ 从a到f的链有:abcf, abef, abcef, abecf, adef, adecf, adebcf, adecbef,adebcef，共 9条。⑵ 从a到f的路有：abcf,abef, abcef，abecf, adef,adecf,adebcf，共7条。⑶ 从a到f的短程线有: abcf, abef, adef, ，共3条。距离为3。⑷ 所有从a出发的圈有:abeda, abceda, abcfeda。2．⑴ 证明 设G中的两个奇度结点分别为u和v，若u与v不连通，即他们之间无通路，则G至少有两个连通分支。记一个连通分支G1，G2=G- G1，这时u、v分别属于G1 和G2，于是G的子图G1 和G2各含有一个奇度结点，这与握手定理的推断是矛盾的，因此u与v一定是连通的。

⑵ 解 若有向图G中只有两个奇度结点u和v，u与v不一定互相可达，也不一定一个可达另一个。例如：图G=<V, E> (其中V ={u, v, w}，E={( u, w),( v, w) })中，结点u、v的度数均为1，w度数为2，但u不可达v，v也不可达u。3．解 图6-5所示的4个图中，(a) 是强连通图,(a)、(b)、(d) 是单向连通图, (a)、(b)、(c)、(d) 是弱连通图。4．证 必要性。用反证法,假设e在某个圈C上,则G-e的连通分支与G的连通分支相同,这与e是割边矛盾.充分性。用反证法, 假设e=(u,v)不是割边,则在图G-e中u,v仍连通,即在图G-e中有u到v路P,则图G有圈P+e, 这与e不在任何圈上矛盾。5．证 必要性。设u是连通图G的割点，则G-u至少有两个连通分支，设G1，G2是它的两个不同的连通分支，则存在v(G1,w(G2，使得v到w的每一条路都经过u。充分性。若存在v，w∈V，使得v到w的每一条路都经过u，则u是图G的割点。若不然，则G-u中任何两个点都是连通的，即G-u连通，亦即图G中任何有别于点u的两个点v、w，都有一条路不经过u，这与题设矛盾。

习题6.31．解 由图G的邻接矩阵作出其图如图6-6。.⑴ d（v1）=2 d（v2）= 3。⑵ 图G不是完全图。⑶因为,而,所以,从v1到v2长为3的路有4条。⑷从v1到v2长为3的4条路分别为：v1v2 v1v2，v1v2v3v2，v1v2v4 v2，v1v4v1v2。2．解 邻接矩阵为A的无向图G的图形如图6-7所示：3．解 ⑴ 图G的邻接矩阵为⑵ 为了求G中长度为3的通路的数目，就要计算A3，为此先计算A2。，因， 故G中长度为3的通路有20条。因为A3的主对角线上的元素的和为12，所以G中长度为3 的回路有12条。⑶ 因为,,所以G的可达性矩阵为因为P中的每个元素都是1，所以G是强连通的，当然也是单向连通的。4．解 图6-9、图6-10的关联矩阵分别为，，5．解 图6-9、图6-10的邻接矩阵分别为：，。习题6.41．解 图(a)既是Euler图又是Hamilton图，(b)是Euler图但不是Hamilton图，图(c)是Hamilton图但不是Euler图， (d) 既不是Euler图又不是Hamilton图。2．解 当n=2k+1（k∈N,k≠0）时，Kn是Euler图. 当n=2时, Kn仅存在Euler链而不存在Euler回路.3．解 图6-12中的图 (2)、(3)、(4)中有Hamilton圈，(1) 、(5) 中只有Hamilton路而没有Hamilton圈。

(1)中的路为abcdejhgig；(2)中的圈为afidejhcbga；(3)中的圈为agkfeicbhdja；(4)中的圈为abrfgcdihja；(5) 中的路为jabihfgkdec。4．证明 设，则有((G-V1)=9>7=(V1(，所以它不是Hamilton图。5．解 我们分别用a, b, c, d, e, f, g七个结点表示七个人，若两人能交谈（会讲同一种语言），就在代表它们的结点之间连一条无向边，所得无向图如图6-14(a)。此图存在一条Hamilton圈：abdfgeca，于是按图6-14(b)所示是顺序安排座位即可。习题6.51．解 如图6-16,求得v1到v8的最短路是v1v3v6v7v8,其总长度为13.2．解 首先观察图中的奇数度结点，此图中只有两个奇数度结点E和F。用标号法求E到F的最短路径，容易算出E到F的最短路径为EGF，其权为28。然后将最短路径上的边均重复依次（如图6-17(b)中虚线）。于是得欧拉图。求从D点出发的一条欧拉回路，如，其权为281。3．解 从a开始，可得哈密尔顿回路，如图（b），长度37，而最短哈密尔顿回路如（c）长度36。

复习题61．解 因为G中有6个3度结点，其度数和为18，由握手定理可知，所有结点度数的和为24，所以其余结点的度数和为6，而其度数均小于3，因此最大只能为2，则其余结点至少有3个，故G中至少有9个结点。2．解 因为n是奇数，所以Kn中每个结点均为偶度，因此G中每个奇度结点在补图中仍为奇度结点，故G的补图中有r个奇度结点。3． 证明 用结点分别表示6个人。若与彼此认识（）就在与之间连边于是构成无向简单图。在中，若与之间有边，说明，所代表的人彼此不认识。有了图论模型,要证明命题,就是要证明,要么在图中有三角形, 要么在图中有三角形。设在中v1与三个以上的结点相邻,不妨v1与v2,v3, v4相邻,如图6-19,这时若v2,v3, v4中有某两个相邻，则在中存在三角形，若v2,v3, v4中任何两个都不相邻，则在中v2,v3, v4构成三角形。若在中v1与三个以下的结点相邻，则可先在中作类似的证明。4．解 K4的所有非同构的子图如图6-20共18个，其中有11个是生成子图，生成子图中有6个是连通图。5．解 图1同构于图2。只要作映射f : u1→v1, u2→v2, u3→v3, u4→v4,u5→v5, u6→v6即可.6．解 首先要注意5阶自补图的边数应该是的边数的一半,而有10条边,所以5阶自补图应该有5条边;其次,不连通图的补图是连通的,因此, 自补图是连通的.由以上分析,应先画出的全部5条边的连通生成子图，再利用同构的性质来判断哪些是自补图。

的全部5条边的连通生成子图如图6-22。其中，(a)与(a)自身互为补图，(b)与(c)互为补图，(d)与(d)互为补图。从而可得(a)和(d)为自补图。7．解 图6-23中的路有：v1v2, v1v4, v1v2v4, v1v4v2, v1v2v5, v1v4v3, v1v4 v5, v1v2v4 v3, v1v2v4 v5, v1v2 v5 v4, v1v4v5v2, v1v4v2v5, v1v2v5v4v3, v2v4, v2v5, v2v1v4, v2v4v3, v2v4v5, v2v5v4, v2v1v4v3, v2v1v4v5, v2v5v4v3, v3v4, v3v4v5, v3v4v2v5, v3v4v1v2v5, v4v5, v4v2v5, v4 v1v2v5共29条。图6-23中的4个圈分别为：v3ev3，v1v2v4v1, v2v4v5v2, v1v2 v5 v4 v1。8．证明 用反证法。假设G不连通，则G至少有两个连通分支G1 、G2，设连通分支G1中有n1个结点，G2中有n2个结点，且n1+ n2≤n。分别从G1和G2中任取一个结点u和v，由于G是简单图，从而G1和G2也是简单图。

所以，deg(u)≤n1-1, deg(v)≤n2-1,故deg(u)+ deg(v)≤n1+n2-2≤n-2，这与G中每对结点度数之和均大于等于n-1相矛盾。因此，G是连通图。9．解 图6-24中的图D1,D2,D3就是符合题意的3个4阶有向简单图。10．解 图6-25的关联矩阵M和邻接矩阵A分别为如下的矩阵：11．解 ⑴其邻接矩阵为：(2)由 ，，可知从到长度为1，2，3，4的路径分别有1，1，2，3条。(3)中第（2，3）个元素为1，说明从v2到v3引出的边能共同终止于同一结点的只有一个，。中第（2，2）个元素为2，说明的引出度数为2。中第（2，3）个元素为0，说明没有结点引出的边同时终止于和。中第（2，2）个元素为3，说明的引如度数为3。⑷ ，⑸ 。所以，强连通子图的顶点集为：{v1},{v2, v3, v4}。12．解：不存在。因为G为欧拉图，因而G是连通图。若G中存在割边e={u, v}，则u，v分别属于G-e的两个连通分支G1与G2。设w为G1中的一个结点，可从w出发走一条欧拉回路C：从w开始，一旦行到u，沿割边到达v，则在G2中行遍后无法回到G1达到w，这于G是欧拉图矛盾。

故G中无割边。13．解 abjibcdlchdefghfihbka是图6-27中的一条Euler回路14．解 ABCDFCEFGAHGBH是图9中找出一条Euler路。15．解 在图6-29中⑴与⑷两图为哈密尔顿图，⑵，⑶为半哈密尔顿图。在⑴中，为一条哈密尔顿回路。在⑷中为一条哈密尔顿回路。在⑶中为一条哈密尔顿通路。在⑵中，为一条哈密尔顿通路。16．解 图6-30中的图(a) 、(b)、 (c) 、(d)分别是⑴、⑵、⑶、⑷题要求的实例。17．解 图6-31中的图(a) 、(b) 能够一笔画出，但图(c) 不能够一笔画出。图(a)的具体画法是：v1 v8 v9 v3 v10 v11 v5 v12 v7 v2 v9 v10 v4 v11 v12 v6 v7 v1。图(b)的具体画法是：1,2,3,4,5,6,7,2,8,9,10,11,12,13,8,14,15,16,17,18,19,14,17,11,5,20。第七章 特 殊 图 类与平凡图构成的非连通图中有4个结点3条边，但是它不是树。3．证明 必要性。因为G中有n个结点，边数m=n-1，又因为G是连通的，由本节定理1可知，G为树，因而G中无回路。

再证充分性。因为G中无回路，又因为边数m=n-1，由本节定理1，可知G为树，所以G是连通的。4．解 因 m=n-r,这里n=15,r=3,所以m=15-3=12，即G有12条边。5．解6个结点的所有不同构的树如图7-1所示。6．证明 由定理，在任意的树中，边数；所以，由握手定理得 ①⑴若没有树叶，则由于是连通图，所以中任一结点均有，从而 ②则与矛盾。若树叶，则其余个结点的度数于2，于是 ③从而、相矛盾。⑴，⑵得知T中至少有两片树叶。表示。，，，。其中T2，T5是图中的最小生成树。9．解 最小生成树T如图7-7所示，W（T）=18。习题7.21．解 不一定是。如图7-8就不是根树.2．解 五个结点可形成3棵非同构的无向树，如图7-9⑴，⑵，⑶所示。由⑴可生成2棵非同构的根树，如图7-9⑷，⑸所示。⑷为3元树，⑸为4元树。由⑵可生成4棵非同构的根树，如图7-9⑹，⑺，⑻，⑼所示。⑹为2元树，⑺为2元树，⑻为3元树，⑼为2元树。由⑶可生成3棵非同构的根树，如图7-9⑽，⑾，⑿所示。⑽为1元树，⑾，⑿为2元树。由此可知，五个结点共形成9棵非同构的根树。3．解 不是。根树中最长路径的端点一个是树根，另一个是树叶，因为根树的高等于最长路径的长度，应从树根开始。

4．证明 设完全二元树T有n0个叶结点，n2分支结点，则T的结点数为n =n0+n2，边数m=2n2，有握手定理可得：2n2=n0+n2-1，所以，n2=n0 -1，因此，n =n0+n2=2 n0-1。即二元树有奇数个结点。5．解 先根遍历：abdfgechi中根遍历：fdgbeahci后根遍历：fgdebhica6．解：习题7.31．解 图⑴是偶图，互补结点子集为：；图⑵是偶图，互补结点子集为：；图⑶不是偶图。2．证明 设G=<V, E>是一棵树，任选v0∈V，定义V的两个子集如下：, V2 = V – V1。现证明V1中任二结点之间无边存在。若存在一条边(u,v)∈E，u,v∈V1,由于树中任意两个结点之间仅存在惟一一条路，设v0到u的路为v0v1v2…vku，则v0v1v2…vku的长度为偶数，因(u,v)∈E，所以v0v1v2…vkuv是v0到v的一条通路，且该通路的长度k+2为奇数，从而v0v1v2…vkuv不是路，因此v必与某个vi(i=0, 1, 2,…, k)相同，从而vvi+1 vi+2…vkuv是G中的一个圈，这与G是树矛盾。同理可证，V2中任意两个结点之间无边存在。

故G中的每条边(u,v)∈E，必有u∈V1，v∈V2或u∈V2，v∈V1，因此G是具有互补结点子集V1和V2的偶图。3．解 将n位男士和n位女士分别用结点表示，若某位男士认识某位女士，则在代表他们的结点之间连一条线，得到一个偶图G，它的互补结点子集V1和V2分别表示n位男士和n位女士，由题可知，V1中的每个结点度数至少为2，而V2中的每个结点度数至多为2，从而它满足t条件（t=2），因此存在从V1到V2的匹配，故可将男士和女士分配为n 对，使得每对中的男士和女士彼此都认识。4．解 不能。用结点表示五位教师（V1）和五门课（V2），在教师和他熟悉的课程之间连一条线，得到一个偶图G，其中，V1中的孙、李、周三个结点只与数学、物理两个结点相邻接，故不满足相异性条件，因此不存在从V1到V2的匹配，故不能按题设要求的安排。习题7.41．证明 将图7-13所示的两个图改画为图7-14所示的两个图,可以看出图(1),(2)任何两边除在结点处相交外，无其它交叉点即可。因此, 图7-13所示的两个图都是平面图.2．解 图7-15⑴中有五个面，分别为F1:abea,F2:acea, F3:acda, F4:cdec,F5:abeda。

它们的秩分别为r(F1)=r(F2)=r(F3)=r(F4)=3, r(F5)=4。图7-15⑵有两个面，其中有限面为F1:acfedea,无限面F2: acbcfea。它们的秩r(F1)=r(F2)=6。3．证明 设该连通简单图的面数为r，由Euler公式可得，6-12+r=2，所以r=8，其8个面分别设为ri(i=1,2,┅,8)。因是简单图，故每个面至少由3条边围成。即。而由本届定理1知，。因此每个面只能由3条边围成。4．解 去掉图7-16中的两条边，并给结点表上名称的图7-17⑴，在将图7-17⑴改画图7-17⑵，而显然图7-17⑵与K5是同胚的，由库拉图斯基定理知，图7-16所示的图为非面图。5．证明 若G中无圈,则G为森林,结论显然成立,若G中有圈,假设G中有n个结点, m条边,并假设G的所有连通分支为G1, G2,… , Gk,每个Gi有ni个结点, mi条边,则有由于G是简单图,所以每个Gi也是简单图,由本节定理2的推论可知mi≤3ni-6, i=1, 2, …,k。从而有所以m≤3n-6。再用反证法证明，简单平面图G中至少有一个度数小于等于5的结点。如果G中每个结点的度数均大于等于6,由握手定理可知,因此,代入m≤3n-6得m≤m-6,这显然的矛盾的。

故G中至少有一个度数小于等于5的结点。6．证明 设G是连通平面图。假设G中每一结点v都有deg(v)≥5。因为2m≥5n,所以n≤2m/5。于是，m≤3n-6≤6m/5-6 ，即5m≤6m-30。因此，m≥30，与题设m<30矛盾。所以G中必有一个结点的度小于或等于4。复习题71．解 假设T有m条边，x个1度结点，则有：m=5+3+4+2+x-12m=5×2+3×3+4×4+2×5+x解得：x=19,m=32，即T有19个1度结点。2．证明 因为，图G是连通图，所以，由本节定理2知图G存在生成树T，而生成树T的边数n -1是不超过图G的边数m的，即m≥n -1。3．证明 设G的p个连通分支分别为<n1，m1>，<n2，m2>，┅，<np，mp>，由于森林的每个连通分支都是树，因此，有：m1= n1 -1，m2 = n2-1，┅，mp= np-1 ⑴又 m1 +m2+┅+ mp=m ; n1+ n2+…+ np=n ⑵由（1），（2）得：m=n-p。4．证明 因为，a是在T1中但不在T2中的一条边，所以，T2({a}有惟一的圈C，而T1是树，则圈C上一定有一边b不在T1中。

因此，(T2-{b})({a}= T2({a}-{b}是G的生成树。下面证明，(T1-{a})({b}= T1({b}-{a}也是G的生成树，事实上，因为T1是树，所以T1中的边a是T1的割边，因此T1去掉边a后可得两个连同分支，设为T11和T12。又b不在T1中，所以T1({b}有惟一的圈C1，5．解 设G中的k个连通分支为，设结点。在G中添加边，设所得新图为T，则T连通且无回路，因此T是树。所加边的条数k-1是使得G为树的最小数目。6．解 取图7-18(a)中最小权的边为e1=(d,e); 再在E-{ e1}取中权最小的边 e2=(d,a); 在E-{ e1, e2}中权最小的边有两条(a,e)和(b,c) 但若选(a,e)就会有圈,因此取e3=(b,c)；最后取e4=(b,c)，则得最小树如图7-18 (b)，最小生成树的权为W=4+5+6+7=22。7．解 题目就是求图7-19⑴的最小生成树问题。因此，图7-19⑴的最小生成树为图7-19⑵所示，即是所求的通讯线路图。其权即是最小总造价，其权为：。8．解 高为2的所有不同构的二元树有7棵，如图7-20所示。其中有2棵完全二元树，图7-20中的⑸和⑺所示，有1棵满二元树，图7-20中⑺所示。

9．证明 方法1：设T结点数为n，分支点数为i。根据完全二元树的定义，可知下面等式均成立：⑴⑵⑶由式⑴，⑵，⑶易知。方法2：在完全二元树中，除树叶外，每个结点的出度为2。除树根外，每个结点的入度为1。由握手定理知解得：。10．证明 设完全二元树T有i个分支点，t片树叶，由T为完全二元树，则由7.2节定理3有。又结点数，所以为T的树叶数。11．解 对于图7-21所示的二元树，三种遍历方法的结果如下：先根遍历：中根遍历：后根遍历：12．解 设图7-22⑴所示的树为，是完全二元树。在每个分支结点引出的两条边上分别标上0（左）和1（右），则得如图7-23⑴的树，将树根到每片树叶的通路上所标的数字构成的符号串组成集合，则为前缀码。设图7-22⑵中所示的树为，是二元树，但不是完全二元树。对于有一个儿子的分支结点引出的边可随便标上0或1；有两个儿子的分支点标法同，则得如图7-23⑵的树，所得前缀码为。13．解：其实，等于T的各分支点的权之和，即。⑵ 由T形成的二元前缀码为。14．解 应该用较短的符号串传输出现频率高的数字，因而可用100乘各数字出现的频率作为权，求最优二元树，然后用这样的二元树产生前缀码传输上面给定的数字。

具体做法如下：用100乘各频率得权 。将这些权由小到大排列得到4，5，6，10，10，15，20，30。⑴ 所求最优二元树如图7-25所示。⑵ 用所求的最优二元树产生二元前缀码如图7-25所示。带权为的树叶对应的符号串就为传输i的符号串。数字0,1,2,3,4,5,6,7对应的符号串分别为01，11，001，100，101，0001，00000，00001．⑶用这样的符号串传输按上述比例出现的数字最少。所以传输10000个上述比例出现的数字至少要用27400个二进制数字。15．证明 设二部图G的互补结点子集为V1、V2，则m=(V1(((V2(且(V1(+(V2(=n，我们知道两个数的和一定，只有当它们相等时积最大，即当(V1(=(V2(=n/2时，积(V1(((V2(最大为n2/4，亦即m( n2/4。16．解 令V1= { P1，P2，…，P7}，V2={ a1，a2，…，a10}，以V1和V2的元素作结点，若Pi是aj的合格工作岗位，则在P i和aj之间连一边，因此，可得二部图如图7-26。去掉图7-26边(P1,a4), (P1,a8), (P3,a7), (P1,a1) ,(P2,a2) ,(P5,a1) ,(P6,a2), (P6,a5)，则图7-26的子图如图7-27。

而图7-27满足t(t=1)条件，所以，存在V1到V2的匹配M={(P1,a9), (P2,a7), (P3,a6), (P4,a3) ,(P5,a10) ,(P6,a1) ,(P7,a2)}，因此，图7-26中也存在V1到V2的匹配M={(P1,a9), (P2,a7), (P3,a6), (P4,a3) ,(P5,a10) ,(P6,a1) ,(P7,a2)}。这样安排使得所有的人都有工作。17．解 以V1={ L1,L2,L3,L4,L5,L6}, V2={ G1,G2,G3,G4,G5,G6}作为互补结点集，若Li和Gj互相满意对方，则在Li和Gj之间连一边，这样得到一个二部图如图7-28，由图7-28可以看出，此图满足满足t(t=2)条件，所以，存在V1到V2的匹配，因此，可使得每一个青年男女都能够找到自己满意的对象，其中一个分配方法是M={(L1,G1), (L2,G3), (L3,G4), (L4,G2), (L5,G6), (L6,G5) }。第八章 代数系统习题8.11．解 ⑴是，⑵不是，⑶是，⑷不是。2．解 若﹡对是可分配的，则有任意a,b,c∈,均有a﹡(bc)=(a﹡b)(a﹡c)= abac =( ab( ac )= ab+c而a﹡(bc)=a﹡(b(c)= ab(c≠ab+c 故﹡对 是不可分配的。

3．解 ⑴对于任意A∈P(S), 因为A(S，所以，A(S=S，因此，S是关于(运算的零元；⑵对于任意A∈P(S), 因为A(S，所以，A(S= A，因此，S是关于(运算的零元单。4．解 ⑴①因为x*y=xy-2x-2y+6，则y*x=yx-2y-2x+6= x*y，满换律；②任意x,y,z∈R有x*(y*z)=x*(yz-2y-2 z +6)=x(yz-2y-2 z +6)-2x-2(yz-2y-2z+6)+6=xyz-2xy-2xz+6x-2x -2yz+4y+4z-12+6= xyz-2xy-2xz-2yz+4x+4y+4z-6.(x*y)*z=(xy-2x-2y+6) *z =(xy-2x-2y+6)z-2(xy-2x-2y+6)-2z+6=xyz-2xz-2yz+6z-2xy+4x+4y-2z-6=x*(y*z).故满足结合律。(2) ①设任意a∈R,存在e∈R,要e*a= ea-2e-2a+6=a ，由于a的任意性则e=3。因此e=3是其单位元;②设任意b∈R, z∈R，要有z*b= zb-2 z-2b+6= z ，由于b的任意性则z=2，因此z=2是其零元。（3）因为*是满换律，对于x∈R，要存在∈R，须有x*= x-2x-2+6= e=3, 当x(2时，。

即对于任意的x，当x(2时x都是可逆的，且。5．解 f1,f2,f3都满换律，f4满足等幂率，f2有单位元a，f1有零元a，f3有零元b。习题8.21．解 构成代数系统的运算有（2），（3），（4）。2．解习题8.31．证明 作函数f:{a,b,c}→{(, (, (}，f(a)= (,f(b)= (,f(c)= (.显然此映射是双射。由表8-2可知对于任意的x,y(A都有有f(xy)=f(x) of(y)，故<A, *>≌<B, o>。2．解 代数系统不可能同构。因为，由同构的性质，如果两个代数系统同构，则两个系统的单位元对应，零元对应，而这里，代数系统<R,(>的零元是0，而<R,+>没有零元。故代数系统不可能同构。复习题八1．解 ⑴有单位元e=<1,0>，因为，对于任意<a,b>∈S，均有<，且，，故<1,0>单位元⑵对于<a,b>∈S，要<a,b>有逆元，需要有<x,y>∈S使得，<a,b>(<x,y>=<x,y>(<a,b>=<1,0>事实上，即<1,0>=<a,b>(<x,y>=<ax,ay+b>，因此，ax=1，ay+b=0，当a(0时可解得，且又有。

故当a(0时，形式的元素都可逆，且。2．解 因为a*b=b*aa=b，则任意a∈A，而*是可结合的，则有a*(a*a)=(a*a)*a，因此a*a=a，即*满足等幂律.3．证明 假设f: Q→Q-{0}是从<Q，+>到<Q-{0}，(>的同构，则两个系统的单位元对应，即有f(0)=1。因为f是从Q到Q-{0}的满射，所以，对于任意一个素数p∈Q-{0}必存在某个x∈Q，使得f(x)=p，又由于f是一个同构，因此有p=f(x)=f((x-1)+1)=f(x-1)(f(1) ，而在Q-{0}中有无穷多个素数，因此，总可以找到一个素数p，使得x-1(0，则f(x-1)不是1，这与p是素数矛盾。证毕。4．证明 因为，，，所以，。5．证明 对n用数学归纳法。当n=1时，由幂的定义则(a*b)1= a*b =(a1)*(b1)，所以结论成立。假设n=k时结论成立，即(a*b)k=ak*bk，下面考察n=k+1时，(a*b)k+1=(a*b)k*(a*b)=( ak*bk ) *(a*b)= ( ak* a ) *( bk * b) =ak+1*bk+1。即n=k+1时，结论也成立。

由归纳法原理，对于任意的正整数n, 都有(a*b)n=an*bn。6．证明 任意n1,n2∈N，只有如下的三种情况：①n1,n2都能表示成2的幂的形式，②n1,n2都不能表示成2的幂的形式，③一个能表示成2的幂的形式，而另一个不能。下面就这三种情况分别考虑。①设存在k1,k2∈N，使得n1=2k1，n2=2k2，则n1×n2=2k1×2k2=2k1+k2 ∈N，且f(n1)= f(n2)=1，因此f(n1×n2)=f(2k1+k2)=1=f(n1)×f(n2) =f(2k1)×f(2k2);②n1,n2都不能表示成2的幂的形式，则n1×n2也不能表示成2的幂的形式，所以，f(n1)= f(n2)=0，因此f(n1×n2)=0=f(n1)×f(n2)。③不妨设存在k∈N，使得n1=2k，，而n2不能表示成2的幂的形式，则n1×n2也不能表示成2的幂的形式，所以，f(n1)= 1，f(n2)=0，因此，f(n1×n2)=0=f(n1)×f(n2)。综上所述，代数结构 与同态。第九章 特殊的代数系统习题9.11．解 ⑴是半群。显然，二元运算“”在N上是封闭的， 所以，是一个代数系统，另一方面，有，而，因此，，所以，运算“”满足结合律的，故是半群；⑵是半群。

显然，二元运算“”在N上是封闭的， 所以，是一个代数系统，另一方面，，有，而，则，所以，运算“”满足结合律，故是半群；⑶是半群。显然，二元运算“”在N上是封闭的， 所以，是一个代数系统，另一方面，，有，，即 ，所以，运算“”满足结合律，故是半群。⑷不是半群。虽然，二元运算“”在N上是封闭的，即是一个代数系统，但是对于5，3，6，因为，，而，即，所以，运算“”不满足结合律，故不是半群。2．解 ⑴正确。因为，运算显然封闭。⑵正确。，，即是，所以(满足结合律。故是半群。⑶，有，又有即存在单位元是0，故是独异点。3．解 都不能使<{a,b},(>构成独异点，因为没有一个函数存在单位元。而的单位元是a, <{a,b},>能构成独异点。4．解 ⑴是，因为M={2,3}关于min是封闭的，故<M ,min>是<S,min>的子代数；⑵<M ,min>是<S,min>的子半群；⑶不是，因为S的单位元是4，而4M,故<M ,min>不是<S,min>的子独异点。习题9.21．解 ⑴是，因为实数乘法满足结合律，存在单位元a0=1，任意元素a存在逆元素a-1；⑵是，因为有理数乘法满足结合律，存在单位元1，任意元素a存在逆元素a-1；（3）是，因为复数乘法满足结合律，存在单位元1,任意元素z的逆元素是z共轭复数；（4）是，因为多项式的加法满足结合律，多项式关于加法的单位元是0多项式，任意元素P(x)的逆元素是-P(x).（5）是, 因为向量的加法满足结合律，n维实向量关于向量的加法的单位元是n维零向量，任意的n维实向量(的逆元素是-(。

2．解 可以构成群。⑴因为，对于任意的，所以，运算满足结合律；，⑵关于运算有单位元2，这是因为对于任意的都有，且；⑶对于任意的a (I，若要a有逆元b，需要有a(b=b(a=2，即需要a+b-2= b+a-2=2，事实上只要b=a-4即可。因此，对于任意的a (I，a都可逆，且a的逆元是a-4。综上所述，由⑴，⑵，⑶得出结论I与运算能构成群。3．证明 因为对于任意的，所以a可逆,且，因此，<G,*>是群。要证明<G,*>是Abel群，只需证明运算满换律，事实上，因为，对于任意的，所以， 因此，由结合律则有，再由消去律得：。故<G,*>是Abel群。4．证明 当时，因为，=，所以，是方程的解。下面方程的解是唯一的。对于若解y，即，由于群中的任何元素都可逆，则对上式两边同时左乘a-1，并两边同时右乘a-1b-1则得，由结合律则有，。证毕。5．证明 设1是群G的单位元，若G中存在幂等元a，即因为群中的任何元素都可逆，因此，a也可逆，则有故单位元为G中惟一的幂等元。6．解 答案是A，因为存在同态映射f:R(R-{0},f(x)=ex，但不存在同构映射。

习题9.31．解 1，5，7，11为其生成元，任何与12互素的正整数都可作<N12,+12>的生成元。2．证明 设H是循环群G的子群，且G的生成元是a。若H={e}，则H是循环群。若H≠{e}，由于H非空，则必存在正整数m>0使am∈H。设m是使am∈H的最小的正整数，若对于任何的an∈H(n(N)，则由带余除法有n=mk+r,0≤r<m则有ar=an-mk=ana-mk=an(am)∈H，而因为m是使am∈H的最小的正整数，且0≤r<m，所以r=0。这样n=mk, an= amk= (am) ，再由an∈H的任意性知，H中的任意元素都是am的幂，故H=（am）,即循环群的任何子群都是循环群。习题9.41．证明 ①显然H(G；②证明运算*关于H的封闭性。任取a,b(H，对于任意的x ( G有，则，因此，；③设1是G中的单位元，因为对于任意的x ( G有故，；④任取a(H(G，对于任意的x(G，则由H的定义有, x*a=a*x ，由于群的元素都有逆元，因此a也有逆元。等式x*a=a*x两边同时左乘、并同时右乘a的逆元a-1则有，，即，亦即。综合①、②、③、④，<H,*> 是<G,*> 的子群。

2．解 群真子群有如下4个：<{1},(>，<{1,5},(>，<{1,7},(>，<{1,11},(>。习题9.51．解 ⑴设，则集合G={E, A, B, C,-E,-A,-B,-C }，G关于运算(的运算表如下。表2 G关于运算(的运算表( E A B C -E -A -B -C E E A B C -E -A -B -C A A -E -C B -A E C -B B B C -E -A -B -C E A C C -B A -E -C B -A E -E -E -A -B -C E A B C -A -A E C -B A -E -C B -B -B -C E A B C -E -A -C -C B -A E C -B A -E由表1可以看出G关于运算(是封闭的。而运算(是矩阵的乘法运算，因此满足结合律。由表1可以看出G关于运算(的单位元是E。由表1可以进一步看出关于运算(，G中的每一个元素都有逆元，E-1 =E, A-1 = -A, B-1 = -B, C-1 = -C, (-E)-1 = -E, (-A)-1 = A, (-B)-1 = B, (-C)-1 = C。

因此，<G,(>是一个群。⑵G的所有子群是：<{E},(>,<{E,-E},(>,<{E, A, -A, -E},(>,<{E, B, -B,-E },(>,<{E, C, -C,-E },(>。⑶证明 显然<{ E },(>，<{ E,-E },(>是正规子群，下面证明<{E, A, -A, -E },(>是正规子群。设H={E, A, -A, -E }，显然有EH=HE =(-E)H=H(-E) = AH=HA= (-A)H=H(-A)={E, A, -A, -E }。又BH={B, C, -C, -B}，HB={B, -C,C, -B}= BH，因此有H(-B)={B, -C,C, -B}=(-B)H。同理可得，CH=HC=H(-C) =(-C)H={ C,-B, B,-C }。综上所述，对于任意的a(G都有aH=Ha，即<H,(>是正规子群。同理可证，<{E, B, -B,-E },(>,<{E, C, -C,-E },(>也是正规子群。

2．解 5，{3，7，11}，{0，4，8}。习题9.61．解I[i]对于普通加法和乘法能构成环。这是因为：⑴显然I[i]对加法+是封闭的，而复数的加法是满换律和结合律的，<I[i], +>的单位元是0=0+0i，任意元素a+bi的逆元是-a-bi。所以，<I[i], +>是可交换群。⑵I[i]对复数的加法是封闭的，且复数的乘法是满足结合律的，即<I[i], (>是半群。⑶复数的乘法对复数的加法满足分配律。综合⑴⑵⑶，I[i]对于普通加法和乘法能否构成环。2．证明 ⑴首先证明是一个环。设对于任意的,，因为， a,b,c,d(I，所以，(a+c)，(b+d)，(ac+2bd)，(ad+bc)(I，因此，，=(R，故R对于加法和乘法都是封闭的。另一方面，实数的普通加法满足结合律和交换律，且关于加法单位元是0，每个元素都有逆元，就是相反数。实数的普通乘法也满足结合律。综合上述<R,+,( >是一个环。⑵因为，实数的普通乘法也满换律，且R关于乘法有单位元1，又对于,，若两者都不为零，则 ，即<R,+,( >无零因子。综合⑴⑵可知，<R,+,( >是一个整环。

3．证明 设F={a+bi( a, b(Q}，对于任意的a+bi, c+di(F，因为，a,b, c,d(Q，所以，a+c, b+d, ac-bd , ad +bc(Q，因此，(a+bi)+(c+di)= (a+c)+(b+d)i(F，且(a+bi)(c+di)=(ac-bd)+( ad+bc)i(F，即F关于加法和乘法运算都是封闭的；而普通的加法和乘法都满足结合律和交换律；又关于加法+，F中有单位元0=0+0i，且每个元素a+bi都有逆元-a-bi。故<F,+,(>是一个环。另一方面，<F,(>中有单位元1=1+0i；又对于任何的非零元a+bi，因为，a,b不全为零，所以，a2+b2(0，因此对于任何的非零元a+bi都有逆元。综上所述，<F,+,(>是一个域。复习题九1．解 ⑴<G,*> 是半群，但不是独异点，更不是群；⑵不是半群；⑶不是半群，⑷是群，其单位元是。2．证明 对于任意的有同时，，故，所以满足结合律，故是<R,*>半群；又存在，且故0是单位元。综上所述，<R,*>是独异点。3．证明 对于任意的，因为，，其中为的二元运算， 而是半群，又，所以，，即对运算是封闭的。

又因为是半群，所以，有结合律，即对于任意的有，故<S,*>满足结合律，综上所述，<S,*>是半群。4．证明 对于任意，由于<S,*>是半群，所以，<S,*>有结合律，a*c=c*a且b*c=c*b，则。5．证明 对于任意，若a,b是<S,*>的幂等元，则a(a=a，b(b=b，再由交换律和结合律则，故a*b也是<S,*>的幂等元。6．证明 对于任意的，有，所以，<S,*>是封闭的，又因为因此，<S,*>是半群；又取，所以，，故存在单位元<0,0>所以<S,*>是独异点，且即交换律成立，故<S,*>是可交换独异点。7．证明 因为，0×0=0∈{0}，所以，运算封闭，又结合律是显然的，故<{0},×>是子半群。但所以，<{0},×>不是子独异点。8．证明 则{0}，{1}，{0，1}，{0，2}，{0，1，2}，{0，1，2，3}都能与×4构成<N4,×4>的子半群，其中<{0},×4>是独异点，但不是<N4,×4>的子独异点。

9．证明 对于任意的a,b∈G，由题设则有，因此，由消去律可得，①同理，由可得，②再由①②两式可得，由消去律消去上式中的b3可得，。故<G,*>是可交换群。10．解 群<N5,+5>只有两个子群H1={0}和H2= N5，其中H1的左陪集和右陪集如下：0H1= H10={0}, 1H1= H11={1},2H1= H12={2},3H1= H13={3},4H1= H14={4}；H2的左陪集和右陪集如下：0H2= H20=1H2= H21=2H2= H22=3H2= H23=4H2= H24= N5。11．证明 任取aH,bH∈G/H，因为，<G,*>是可交换群，所以对于a,b∈G，有a*b= b*a又因为，<H,*>是<G,*>的正规子群，所以，任意的a∈G，都有aH=Ha由以上两式，则有，(aH) * (bH)= a(Hb) * H= a(bH) * H= (ab)(H*H)= (ab)H(bH) * (aH)= b (Ha) * H=b(aH) * H= (ba)(H*H) = (ba)H = (ab)H即(aH) * (bH)= (bH) * (aH)。

由于aH,bH的任意性，所以，商群<G/H,*>也是可交换群。12．证明 ⑴先证明f是双射。对任意的 若，即有a*x1*a-1= a*x2*a-1，由群的消去律可得x1=x2，所以，f是单射。又对于任意的y∈G，则有x=a-1*y*a∈G，使得f(x) =a*x*a-1= a*(a-1*y*a) *a-1=y。即f是满射，故f是双射。⑵再证f是同态映射。对于任意的有综合⑴⑵可知，f是<G,*>到其自身的同构映射。第十章 格和布尔代数习题10.11．解 ⑴不是，因为L中的元素对{2,3}没有最小上界；⑵是，因为L=｛1,2,3,4,6,9,12,18,36｝任何一对元素a,b，都有最小上界和最大下界；⑶是，与⑵同理；⑷不是，因为L中的元素对{6,7}没有最小上界不存在最小上界。2．证明 ⑴因为，a≤b,所以，a∨b=b；又因为，b≤c,所以，b∧c=b。故a∨b=b∧c；⑵因为，a≤b≤c,所以，a∧b=a,b∧c=b,而a∨b=b，因此，(a∧b)∨(b∧c)=b；又a∨b=b,b∨c=c,而b∧c=b, 因此，(a∨b)∧(b∨c)=b。即(a∧b)∨(b∧c)=(a∨b)∧(b∨c)。

习题10.21．解 由图1知：＜S1,≤＞不是＜L,≤＞的子格，这是因为，e∨f=gS1;＜S2,≤＞不是＜L,≤＞的子格， ∵e∧f=cS2;＜S3,≤＞是＜L,≤＞的子格.2．解 S24的包含5个元素的子格有如下的8个：S1={1,3,6,12,24}, S2={1,2,6,12,24}, S3={1,2,4,12,24}, S4={1,2,4,8,24},S5={1,2,3,6,12}, S6={1,2,4,6,12}, S7={2,4,6,12,24}, S7={2,4,8,12,24}.3．证明 因为，一条线上的任何两个元素都有（偏序）关系，所以，都有最大下界和最小上界，故它是格，又因为它是<L，∨,∧>的子集，即是<L，∨,∧>的子代数，故是子格。4．证明 由(10-4)有，a∧b≤a,由已知a≤c,由偏序关系的传递性有，a∧b≤c；同理 a∧b≤d。由(10-5)和以上两式有，a∧b≤c∧d.5．证明 因为由(10-4)有，a∧b≤a，因此，(a∧b)∨(c∧d)≤a∨(c∧d) ①由分配不等式有，a∨(c∧d)≤(a∨c)∧(a∨d) ②再由由(10-4)有，(a∨c)∧(a∨d) ≤a∨c ③由偏序关系的传递性和①②③则有，(a∧b)∨(c∧d)≤a∨c同理 (a∧b)∨(c∧d)≤b∨d因此有， (a∧b)∨(c∧d)≤(a∨c) ∧(b∨d)。

习题10.31．解 ⑴ 是，全上界是24，全下界是1；⑵1的补元是24；3的补元是8；8的补元是3，4、6没有补元。2．解 图3是两个格的哈斯图，其中图⑴是有补格但不是分配格的例子；图⑵是分配格但不是有补格的例子。3．证明 先证充分性。由已知条件知，对于任何的a,b,c∈L,有(a∨b)∧c≤a∨(b∧c)，因此和等幂律、交换律可得，(a∨b)∧c=((b∨a)∧c)∧c≤(b∨(a∧c))∧c=((a∧c)∨b)∧c≤(a∧c)∨(b∧c) ①又因为，(a∧c)≤(a∨b)∧c且(b∧c)≤(a∨b)∧c，所以， (a∧c)∨(b∧c)≤(a∨b)∧c ②由①②可得， (a∧c)∨(b∧c)=(a∨b)∧c再由交换律得到， c∧(a∨b)=(c∧a)∨(c∧b) ③由此式容易证明c∨(a∧b)=(c∨a)∧(c∨b) ④由③④可知它是分配格。再证必要性。因为<L,≤>是分配格，则(a∨b)∧c=(a∧c)∨(b∧c)≤a∨(b∧c)。4．证明 因为，；同理有，；又因为补元素是唯一的，故成立。习题10.41．解 是布尔代数，因为<A，≤>是有补分配格。2．证明 因为，<B，-,∨,∧>是布尔代数，所以，运算-，∨,∧在B上都是封闭的，因此，由运算(的定义可知，运算(在B上也是封闭的。

又运算∨,∧都满换律。因此，对于任意的a,b(B,==由其对称性可知(满换律。下面证明运算(满足结合律，对于任意的a,b,c(B由上式则有同理可得即，，亦即(满足结合律。下面再证0是关于(的单位元。事实上对于任意的a(B，。最后证明任意的a(B关于运算(都可逆，且其逆元就是a自身，事实上综上所述，是交换群。复习题十1．证明 显然，a,b∈B，所以，B非空。对于任意的x, y∈B，则a≤x≤b, a≤ y≤b，由格的保序性和等幂律则有，a≤x(y≤b, a≤x(y≤b即集合B对于运算(和(是封闭的。因此，<B，≤>是<L,≤>的子格。而子格也是格，故<B，≤>也是一个格。2．证明 因为，<L,∨,∧>是一个格，由格的分配不等式则得（(a∧b）(a∧c)）∧((a∧b)∨(b∧c))≥(a∧b)∨(a∧b∧c)=a∧b ①（a∧b）∨(a∧c)≤a∧(b∨c) ②（a∧b）∨(b∧c)≤b∧(a∨c) ③由②③和格的保序性可得，((a∧b)∨(a∧c))∧((a∧b)∨(b∧c))≤(a∧(b∨c))∧(b∧(a∨c)=a∧b∧(b∨c)∧(a∨c)=a∧b ④由①④和反对称性则有，((a∧b)∨(a∧c))∧((a∧b)∨(b∧c))= a∧b。

3．证明 因为<L,≤>是格，对任意a,b,c∈L,(a∧b)∨(b∧c)∨(c∧a) ≤ [((a∧b)∨b)∧((a∧b)∨c)]∨(c∧a)=[b∧((a∧b)∨c)]∨(c∧a) ≤[b∧(a∨c)∧(b∨c)]∨(c∧a)=[b∧(a∨c)]∨(c∧a) ≤(b∨(c∧a))∧((a∨c)∨(c∧a))=(b∨(c∧a))∧(a∨c) ≤ (b∨c)∧(b∨a)∧(a∨c)= (a∨b)∧(b∨c)∧(c∨a)。4．证明 因为有限格都是有界格，而有界格必存在最大元素和最小元素，故有限格一定有最大元素和最小元素。5．证明 因为，a≤b,所以，a∨b=b；因此有，a∨(b∧c)≤(a∨b)∧(a∨c)=b∧(a∨c)。6．证明 因为，h将运算∨传送到运算∪，将运算“-”传送到运算“＇x,x1,x2∈B1有：h(x1∨x2)=h(x1)∪h(x2) ①②所以，对于任意的a,b∈B1,而，因此有：。即h将运算∧传送到运算∩。7．证明 由习题10.4第2题可知<B,⊕>是一个交换群。由于，在布尔代数<B，-,∨,∧>中∧是可结合的且是可交换的，由(运算的定义可知，(是可结合的且是可交换的。

由(运算的定义可知可进一步看出，关于(运算的单位元是布尔代数<B，-,∨,∧>的全上界1。事实上，对于任意的a(B,有因此，要证明<B,⊕, (>是一个含幺交换环，只需证明(对⊕满足分配律。事实上，对于任意的a,b,c(B,即综上所述，<B,⊕, (>是一个含幺交换环。Manufacturing transformation and development planning1 comprehensive strength of the manufacturing industry to upgradeRelying on good traffic conditions and location advantages of the industrial base, the manufacturing industry comprehensive strength steadily, the total size of the total, technical level, economic benefits and so on in the country has become the cornerstone of the precedent, the city's economic and social development and an important support. "12th Five-Year" period, the total industrial added value, the average annual growth rate of dada%, the amount of the total amount of investment of the city's industrial enterprises with an average annual growth rate%, above scale industrial enterprises realized a total profit of the average annual growth rate of%.xx years, the city's above scale industrial output mode billion yuan output value of over 80 billion yuan of industrial dada to 8, above scale industrial added value billion yuan, industrial growth. Added value accounted for GDDP%, the resident population per capita GDP reached million, ranked second in XXX Province, the national large and medium-sized cities in the forefront.2 continuous optimization of industrial structureThe benefit oriented, high end orientation and the direction of intensive development, gradually increase efforts to promote the transformation of industrial structure adjustment and optimization of industry has made remarkable achievements. As to XX, machinery, textile, petrochemical, metallurgy and electronics and other 55 pillar industries above scale industrial output value at current prices billion yuan, accounting for the proportion of for the high-tech industry output value accounted for%. The proportion of above scale industrial output value reached%, one percentage point increase over 2010. Sensor network innovation shows the construction of the demonstration zone to promote the smooth, emerging industry gradually expand the scale of production is expected to total output average growth rate, new energy, new materials, energy saving and environmental protection, Internet of things, the rapid development of microelectronics and other emerging industries, has become an important field of XX industrial transformation and upgrading.3 innovation capacity significantly improvedSeize the new round of the wave of technological innovation brings plenty of historical opportunity, in-depth implementation of innovation driven development strategy for the manufacturing industry's innovation and transformation and upgrading provides a strong intellectual support for the.Xx year, R & D expenditure accounting for the main business service income ratio reached%, among the province's first patent. The application amounted to 224197, the right amount of invention patents reached 5480, 000 people obviously patent has reached 25, are located in the forefront of the province of.Xx has a provincial engineering technology research centers heart 506, country, provincial high technology research key laboratory 10, International cooperation base, 9 national, provincial, foreign R & D center 41, provincial and international technology transfer center 8, selected "research research institutions" key enterprises XX 84. Tsinghua University, South, Fudan University, Harbin Institute of technology, Huazhong University of science and technology, Northeastern University and other colleges and universities have been set up in the XXX Research Institute.4 green manufacturing results significantly"XXX water crisis", energy saving and environment protection increasing pressure, vigorously promote the green, low-carbon, circular ring production mode, adhere to energy-saving emission reduction, the implementation of energy control, improve the energy source utilization rate of.Xx years, the city unit of GDP energy consumption decreased 20%, increase the added value of industrial energy consumption per ton standard coal / million. "" 12th Five-Year "period, the cumulative unit of GDPP energy consumption decreased percent, energy consumption consumption value XX lowest decline of XX south, South first. The elimination of backward production capacity can work orderly, shut down 8823 enterprises, 852 enterprises for rectification of family, eliminate backward production capacity of cement 75 million 770 thousand Tons, dyeing 244 million 600 thousand meters, 100 thousand tons of chemical fiber, Leather 3 million standard sheets, paper million tons, 2 million 280 thousand tons of steel, 106 battery Wanqianqian KVA, 200 thousand tons of steel casting, resolve the production capacity of 3 million 370 thousand tons, cement production capacity of 710 thousand tons, completed ahead of schedule into "12th Five-Year" planning objectives.5 enterprise vitality is increasingThe emergence of a number of key enterprises in the industry a lot of industries, some have developed into the world class level leading enterprises and leading enterprises. The key point of enterprises in technological innovation, product development, market development, has played an important leading role for.Xx and other aspects of social responsibility, listed 15 companies around the "top 500 enterprises Chinese" listed 221 companies "Chinese manufacturing enterprise 500 strong", all year for nine consecutive years, ranking first in the province. The city's domestic and foreign listed enterprises amounted to 94, ranked in the prefecture level city in the forefront. The integration of military and civilian combined to accelerate, the number of the city's participation in scientific research and production of weapons equipment industry the advantages of private enterprises Year after year, through the confidential qualification certification enterprises ranked 22 in the province.Two 6 integration steadily intoAs a national integration of the two test area, has been to two of the depth of financial integration and enhance the core competitiveness of enterprises as an important starting point, to accelerate the transformation of development mode. The information of all business to achieve the city's more than 80% large-scale enterprises, a total of at or above the provincial level demonstration pilot enterprises, 250 enterprises, including 22 national. Fusion demonstration of 4 enterprises, the provincial demonstration integration of the two test, Park 8. Baidu and Alibaba, Pakistan, Chinese manufacturing network, China network library and so on enterprises to carry out in-depth cooperation, support enterprises with the development of electronic commerce, there have been 4 enterprises selected for the national e-commerce service integrated innovation pilot .In a comprehensive summary of the "12th Five-Year" achievement at the same time, should also see some problems still exist in the development of XX manufacturing industry, difficulties and contradictions, need to be solved in the "Shisansanwu" period. The main is: not many major industrial projects, lack of a group with strong power, high output level large projects the industrial economic growth. Lack of weak industrial products to the intermediate product, the added value is not high. The emerging industry is small, difficult to support the city's economic growth. Science and technology innovation ability is not strong, high dependence on foreign technology, high level talent shortage. Excess capacity in some industries, some enterprise production and management difficulties. These difficulties all Urgent need to be addressed in the future development of the exhibition.1 from the global perspective, a new round of scientific and technological revolution and the rise of industrial change, to bring significant strategic opportunities for the transformation of the XXX industryThe development mode of the industry will undergo profound changes. When a new round of technological revolution in the application of the new generation of information technology integration as the core feature, "Internet plus" has become an important direction of industrial development. The high-end, smart, green, as the service has become a new trend of industrial development; the Internet network to break the constraints of time, the world famous enterprises and enterprises to achieve global collaborative design and production; intelligent manufacturing technology in the design, development, application, manufacturing and other aspects of the daily trend of generalization deepening; based on fusion, and services, new formats and new model into a platform for the new form of industry.The industrial division of labor and competition face remodeling. Europe and other developed countries to implement the "re industrialization strategy, trying to consolidate its leading position; emerging economies have focused on the implementation of catch-up strategy, competition to seize the high-end manufacturing areas to accelerate the end of a space for one person. To developed countries" reverse transfer, emerging countries rely on information the comparative advantage of labor resources, to attract international investment led shunt. Significant changes in Global trade investment system, the United States actively promote the TPP and TTIIP as the core of the trade and investment heavy order reconstruction, the future of China's foreign trade and absorbing the international direct investment is facing greater pressure.At present, the development of the XX industry development level in the nation, with the foundation and conditions follow a new round of revolution of science and technology revolution and industry industry, to seize the great opportunities in grasping the concept of development, innovation, production mode and format to change the strain, the first action, to create new competitive advantages of industry competition. At the same time. Despite the change of international trade and investment will lead to industrial growth slowed to put XXX of export-oriented industries, jobs and social challenges, increase the risk of wind, but also bring down forced backward industries and low added value of industry chain, promote industrial restructuring and upgrading opportunities.2 from across the country, China's economic development has entered a new normal norm, put forward urgent requirements of development quality and efficiency of quality industry in XX CityThe industrial economy is facing multiple conversion speed, structure and power. China's economic development has entered a new norm, is to form a higher division of advanced, more complex, more reasonable structure can decrease the growth rate of stage evolution. It may lead some challenges to There's no telling war, a variety of factors, constraints tightening, road development the traditional expansion gets narrower. Must take the innovation as the new engine driven by economic development, speeding up the drive to convert from elements of the scale of investment driven development to the development of innovation driven development based principal.The supply and demand of the market environment has changed greatly. Different from the demand side, the grade and residents' demand structure has undergone significant changes, the basic medium has been close to the level of developed countries, the individuality has become the competitiveness of products to reflect the heavy, large quantity of products demand is gradually being customized by small batch of personality replace. See from the aspect of supply, supply of existing products can not meet the rapid escalation of market demand, the lack of effective supply, supply mode and become the deep level reason of industrial economic downturn. Through product innovation arousing potential consumer market, stimulate investment demand, increase effective supply, has It has become an important direction to boost China's industrial economy.Industrial transformation and upgrading by elements to resource constraints more urgent. I China's per capita fresh water, arable land, forest resources is only the world average level of 28%, 40%, 25%%, oil, iron ore, copper and other important mineral resources per capita reserves respectively for the world average level%. 117%, 17%, the traditional industry has been difficult to support resources and high energy consumption. The long-term accumulation of environmental conflicts are concentrated, serious pollution of the atmosphere pollution has become a national environmental and social problems, water pollution, soil pollution is more prominent, increasingly, the industry restructuring and development to put forward urgent Pray.To deepen the reform, formation and development of new power will be the "13th Five-Year" industrial development in China. The theme of the next five years, XX industry development must establish the new ideas of thinking, to adapt to and guide the new norm as macro logic of industrial development, breaking the current resource bottleneck was completed, speed, actively and steadily. The structural change, multiple motive etc... to make good use of the big deployment of national level of the market and institutional environment of product of positive influence, seize the opportunity to expand the space of domestic development, focus on structural optimization, attention to enhance the quality, focus on environmental improvement, fully activate the development of energy industry xx, Provide continuous power for industrial upgrading.3 from the city to see, XX course of economic development, natural endowment and location, determine the XX must take industrial strong city roadRaise the level of independent innovation opportunities of the industrial structure of.Xxx is two, three, and a "heavy chemical characteristics of industrial scale manufacturing characteristic is apparent, especially in the equipment manufacturing sector, in the province and even the whole country has comparative advantage. To firmly firmly grasp the national independent innovation demonstration in South XX area XX sensor network innovation demonstration zone construction in a major strategic opportunity, perfect the system of innovation incentive mechanism, increasing the field of manufacturing and innovation investment, enhance the innovation output, stimulated to stimulate innovation and vitality, enhance innovation and lead development ability.The critical period of building a modern industry to develop new heights. From now until 2020, XXX should strive to build a "new Qiang Fu makall" XX, a high level of comprehensive well-off well-off society. The manufacturing industry is still an important force to promote sustained and healthy economic and social development xxx not indispensable, is to promote the XX economy jumped to a higher level basic support. To accelerate industrial restructuring and upgrading, stronger advanced manufacturing industry, fostering the growth of strategic emerging industries, accelerate the development of modern service industry, push to promote the establishment of innovation ability, quality benefit, structure reasonable layout, sustainable development and international competitive production A new system of industry, revitalize the vibration XX industry glory on the new starting point.Full integration into the national strategy of regional development gold period.Xx geographical advantages are obvious, and the historical and cultural heritage, has strong strong industrial strength and potential development of..xx should be fully docking "The Belt and Road Road, the Yangtze River economic belt, the Yangtze River Delta region a major strategic integration area, strengthen the strong docking with Shanghai, enhance with XXX, XX and even the entire north Yangtze River Basin industrial collaboration, open economy, to a higher level of development, and actively promote international industrial cooperation, the formation of a higher level and wider domain opening pattern.The next five years, XX must be the quasi future trend direction, continue to play in the field of manufacturing characteristics and advantages, will revive the industry itself as XX wind is the center of economic development as task, promote the industrial structure from low-end to high-end forward end, to create new heights of development of modern industry.Comprehensive implementation of the party's eighteen and eighteen session of the 33, fourth, Fifth Plenary Session spirit, foster innovation, harmony, green, open, sharing development idea, firm not to shift industries of City Road, take the initiative to melt into the global industrial division system, comprehensive implementation of the "made in China" and 2025 "XX the platform for action, to the depth of information technology and industrialization integration as the main line, in order to improve the quality and efficiency of economic development benefits as the center, and actively develop new industries, new technologies, new formats and new models, and vigorously promote industrial intelligent, green color, services, high-end development, and to build in emerging industries for the pilot to guide. As the main body of modern manufacturing, modern service industry as the support of the new modern industrial system, the XX to create a domestic first-class, internationally influential manufacturing city, revitalize the vibration XX industry glory in the new starting point for the construction of "the economy is strong, rich people, beautiful environment, social civilization degree is high" the new XX solid solid strong industrial base.Accelerate the innovation development, the implementation of talent drive. Firmly adhere to the development on the basis of innovation, and vigorously promote the comprehensive innovation in science and technology innovation as the core of the nucleus, since the focus on enhancing the capability of independent innovation, get rid of institutional system obstacles, to maximize the liberation and shock to stimulate innovation and vitality, accelerate the formation of innovation as the main leading and support the economic system and mode of development. Strengthening of "talent is the first resource" the idea, focus on the development of needs and direction of innovation, efforts to consolidate the foundation of science and education, optimize the talent environment, to create a high level of innovation of new talents for sustained economic and social healthy development continued to provide strong support.Play to the market mechanism, the transformation of government function. Focus on the decisive role of the market in resource allocation and better play the role of government, strengthen the supply of supply side reform, improve the quality and efficiency of the supply system, co-ordinate the use of economic, legal, policy and means, strengthen the guide and regulate the industry, promote the optimization the upgrading of the industrial structure industry. The implementation of negative list management clear, to further eliminate the obstacles of scientific development system, continuous development and promotion to stimulate the vitality of market players.Grasp the direction of development, firmly adhere to guidance. Based on the basis of the existing industry system and development, accurately grasp the main direction of holding a new round of technological revolution and the industrial revolution leather, strengthen strategic planning and forward-looking deployment, increase investment in innovation, enhance innovation output, stimulate a new vitality, stronger in advanced manufacturing industry, fostering the growth of strategic emerging industries, accelerate the development of producer services, promote industrial system to the reasonable structure, development, agglomeration, clear competitive advantage in the direction of development.To promote energy conservation, improve the ecological benefits. Firmly establish the concept of green development, unswervingly walk the green development, development cycle, low carbon development, increase energy conservation and pollution remediation efforts in resource saving, environment friendly, basis to improve the efficiency rate of the ecosystem, improve the quality and efficiency of development, strengthen the sustainable development the ability to force.Optimization of space layout, and promote regional cooperation. With the layout of concentration, land intensive, industrial cluster, cluster development oriented, optimize industrial park space layout, cultivate advantageous industry clusters, the formation of industrial development is relatively concentrated, production integration of the city space layout good interaction between elements, through the integration of resources. Expand the market space, improve the regional influence and competitiveness.To 20220when the Terminal level when installed on the line number (Word) should be arranged from top to bottom. 6.4.4 cable core must be completely loose and straight, but not damage the insulation and core. Core bundle of the same plate vertically or horizontally arrangedManufacturing transformation and development planning1 comprehensive strength of the manufacturing industry to upgradeRelying on good traffic conditions and location advantages of the industrial base, the manufacturing industry comprehensive strength steadily, the total size of the total, technical level, economic benefits and so on in the country has become the cornerstone of the precedent, the city's economic and social development and an important support. "12th Five-Year" period, the total industrial added value, the average annual growth rate of dada%, the amount of the total amount of investment of the city's industrial enterprises with an average annual growth rate%, above scale industrial enterprises realized a total profit of the average annual growth rate of%.xx years, the city's above scale industrial output mode billion yuan output value of over 80 billion yuan of industrial dada to 8, above scale industrial added value billion yuan, industrial growth. Added value accounted for GDDP%, the resident population per capita GDP reached million, ranked second in XXX Province, the national large and medium-sized cities in the forefront.2 continuous optimization of industrial structureThe benefit oriented, high end orientation and the direction of intensive development, gradually increase efforts to promote the transformation of industrial structure adjustment and optimization of industry has made remarkable achievements. As to XX, machinery, textile, petrochemical, metallurgy and electronics and other 55 pillar industries above scale industrial output value at current prices billion yuan, accounting for the proportion of for the high-tech industry output value accounted for%. The proportion of above scale industrial output value reached%, one percentage point increase over 2010. Sensor network innovation shows the construction of the demonstration zone to promote the smooth, emerging industry gradually expand the scale of production is expected to total output average growth rate, new energy, new materials, energy saving and environmental protection, Internet of things, the rapid development of microelectronics and other emerging industries, has become an important field of XX industrial transformation and upgrading.3 innovation capacity significantly improvedSeize the new round of the wave of technological innovation brings plenty of historical opportunity, in-depth implementation of innovation driven development strategy for the manufacturing industry's innovation and transformation and upgrading provides a strong intellectual support for the.Xx year, R & D expenditure accounting for the main business service income ratio reached%, among the province's first patent. The application amounted to 224197, the right amount of invention patents reached 5480, 000 people obviously patent has reached 25, are located in the forefront of the province of.Xx has a provincial engineering technology research centers heart 506, country, provincial high technology research key laboratory 10, International cooperation base, 9 national, provincial, foreign R & D center 41, provincial and international technology transfer center 8, selected "research research institutions" key enterprises XX 84. Tsinghua University, South, Fudan University, Harbin Institute of technology, Huazhong University of science and technology, Northeastern University and other colleges and universities have been set up in the XXX Research Institute.4 green manufacturing results significantly"XXX water crisis", energy saving and environment protection increasing pressure, vigorously promote the green, low-carbon, circular ring production mode, adhere to energy-saving emission reduction, the implementation of energy control, improve the energy source utilization rate of.Xx years, the city unit of GDP energy consumption decreased 20%, increase the added value of industrial energy consumption per ton standard coal / million. "" 12th Five-Year "period, the cumulative unit of GDPP energy consumption decreased percent, energy consumption consumption value XX lowest decline of XX south, South first. The elimination of backward production capacity can work orderly, shut down 8823 enterprises, 852 enterprises for rectification of family, eliminate backward production capacity of cement 75 million 770 thousand Tons, dyeing 244 million 600 thousand meters, 100 thousand tons of chemical fiber, Leather 3 million standard sheets, paper million tons, 2 million 280 thousand tons of steel, 106 battery Wanqianqian KVA, 200 thousand tons of steel casting, resolve the production capacity of 3 million 370 thousand tons, cement production capacity of 710 thousand tons, completed ahead of schedule into "12th Five-Year" planning objectives.5 enterprise vitality is increasingThe emergence of a number of key enterprises in the industry a lot of industries, some have developed into the world class level leading enterprises and leading enterprises. The key point of enterprises in technological innovation, product development, market development, has played an important leading role for.Xx and other aspects of social responsibility, listed 15 companies around the "top 500 enterprises Chinese" listed 221 companies "Chinese manufacturing enterprise 500 strong", all year for nine consecutive years, ranking first in the province. The city's domestic and foreign listed enterprises amounted to 94, ranked in the prefecture level city in the forefront. The integration of military and civilian combined to accelerate, the number of the city's participation in scientific research and production of weapons equipment industry the advantages of private enterprises Year after year, through the confidential qualification certification enterprises ranked 22 in the province.Two 6 integration steadily intoAs a national integration of the two test area, has been to two of the depth of financial integration and enhance the core competitiveness of enterprises as an important starting point, to accelerate the transformation of development mode. The information of all business to achieve the city's more than 80% large-scale enterprises, a total of at or above the provincial level demonstration pilot enterprises, 250 enterprises, including 22 national. Fusion demonstration of 4 enterprises, the provincial demonstration integration of the two test, Park 8. Baidu and Alibaba, Pakistan, Chinese manufacturing network, China network library and so on enterprises to carry out in-depth cooperation, support enterprises with the development of electronic commerce, there have been 4 enterprises selected for the national e-commerce service integrated innovation pilot .In a comprehensive summary of the "12th Five-Year" achievement at the same time, should also see some problems still exist in the development of XX manufacturing industry, difficulties and contradictions, need to be solved in the "Shisansanwu" period. The main is: not many major industrial projects, lack of a group with strong power, high output level large projects the industrial economic growth. Lack of weak industrial products to the intermediate product, the added value is not high. The emerging industry is small, difficult to support the city's economic growth. Science and technology innovation ability is not strong, high dependence on foreign technology, high level talent shortage. Excess capacity in some industries, some enterprise production and management difficulties. These difficulties all Urgent need to be addressed in the future development of the exhibition.1 from the global perspective, a new round of scientific and technological revolution and the rise of industrial change, to bring significant strategic opportunities for the transformation of the XXX industryThe development mode of the industry will undergo profound changes. When a new round of technological revolution in the application of the new generation of information technology integration as the core feature, "Internet plus" has become an important direction of industrial development. The high-end, smart, green, as the service has become a new trend of industrial development; the Internet network to break the constraints of time, the world famous enterprises and enterprises to achieve global collaborative design and production; intelligent manufacturing technology in the design, development, application, manufacturing and other aspects of the daily trend of generalization deepening; based on fusion, and services, new formats and new model into a platform for the new form of industry.The industrial division of labor and competition face remodeling. Europe and other developed countries to implement the "re industrialization strategy, trying to consolidate its leading position; emerging economies have focused on the implementation of catch-up strategy, competition to seize the high-end manufacturing areas to accelerate the end of a space for one person. To developed countries" reverse transfer, emerging countries rely on information the comparative advantage of labor resources, to attract international investment led shunt. Significant changes in Global trade investment system, the United States actively promote the TPP and TTIIP as the core of the trade and investment heavy order reconstruction, the future of China's foreign trade and absorbing the international direct investment is facing greater pressure.At present, the development of the XX industry development level in the nation, with the foundation and conditions follow a new round of revolution of science and technology revolution and industry industry, to seize the great opportunities in grasping the concept of development, innovation, production mode and format to change the strain, the first action, to create new competitive advantages of industry competition. At the same time. Despite the change of international trade and investment will lead to industrial growth slowed to put XXX of export-oriented industries, jobs and social challenges, increase the risk of wind, but also bring down forced backward industries and low added value of industry chain, promote industrial restructuring and upgrading opportunities.2 from across the country, China's economic development has entered a new normal norm, put forward urgent requirements of development quality and efficiency of quality industry in XX CityThe industrial economy is facing multiple conversion speed, structure and power. China's economic development has entered a new norm, is to form a higher division of advanced, more complex, more reasonable structure can decrease the growth rate of stage evolution. It may lead some challenges to There's no telling war, a variety of factors, constraints tightening, road development the traditional expansion gets narrower. Must take the innovation as the new engine driven by economic development, speeding up the drive to convert from elements of the scale of investment driven development to the development of innovation driven development based principal.The supply and demand of the market environment has changed greatly. Different from the demand side, the grade and residents' demand structure has undergone significant changes, the basic medium has been close to the level of developed countries, the individuality has become the competitiveness of products to reflect the heavy, large quantity of products demand is gradually being customized by small batch of personality replace. See from the aspect of supply, supply of existing products can not meet the rapid escalation of market demand, the lack of effective supply, supply mode and become the deep level reason of industrial economic downturn. Through product innovation arousing potential consumer market, stimulate investment demand, increase effective supply, has It has become an important direction to boost China's industrial economy.Industrial transformation and upgrading by elements to resource constraints more urgent. I China's per capita fresh water, arable land, forest resources is only the world average level of 28%, 40%, 25%%, oil, iron ore, copper and other important mineral resources per capita reserves respectively for the world average level%. 117%, 17%, the traditional industry has been difficult to support resources and high energy consumption. The long-term accumulation of environmental conflicts are concentrated, serious pollution of the atmosphere pollution has become a national environmental and social problems, water pollution, soil pollution is more prominent, increasingly, the industry restructuring and development to put forward urgent Pray.To deepen the reform, formation and development of new power will be the "13th Five-Year" industrial development in China. The theme of the next five years, XX industry development must establish the new ideas of thinking, to adapt to and guide the new norm as macro logic of industrial development, breaking the current resource bottleneck was completed, speed, actively and steadily. The structural change, multiple motive etc... to make good use of the big deployment of national level of the market and institutional environment of product of positive influence, seize the opportunity to expand the space of domestic development, focus on structural optimization, attention to enhance the quality, focus on environmental improvement, fully activate the development of energy industry xx, Provide continuous power for industrial upgrading.3 from the city to see, XX course of economic development, natural endowment and location, determine the XX must take industrial strong city roadRaise the level of independent innovation opportunities of the industrial structure of.Xxx is two, three, and a "heavy chemical characteristics of industrial scale manufacturing characteristic is apparent, especially in the equipment manufacturing sector, in the province and even the whole country has comparative advantage. To firmly firmly grasp the national independent innovation demonstration in South XX area XX sensor network innovation demonstration zone construction in a major strategic opportunity, perfect the system of innovation incentive mechanism, increasing the field of manufacturing and innovation investment, enhance the innovation output, stimulated to stimulate innovation and vitality, enhance innovation and lead development ability.The critical period of building a modern industry to develop new heights. From now until 2020, XXX should strive to build a "new Qiang Fu makall" XX, a high level of comprehensive well-off well-off society. The manufacturing industry is still an important force to promote sustained and healthy economic and social development xxx not indispensable, is to promote the XX economy jumped to a higher level basic support. To accelerate industrial restructuring and upgrading, stronger advanced manufacturing industry, fostering the growth of strategic emerging industries, accelerate the development of modern service industry, push to promote the establishment of innovation ability, quality benefit, structure reasonable layout, sustainable development and international competitive production A new system of industry, revitalize the vibration XX industry glory on the new starting point.Full integration into the national strategy of regional development gold period.Xx geographical advantages are obvious, and the historical and cultural heritage, has strong strong industrial strength and potential development of..xx should be fully docking "The Belt and Road Road, the Yangtze River economic belt, the Yangtze River Delta region a major strategic integration area, strengthen the strong docking with Shanghai, enhance with XXX, XX and even the entire north Yangtze River Basin industrial collaboration, open economy, to a higher level of development, and actively promote international industrial cooperation, the formation of a higher level and wider domain opening pattern.The next five years, XX must be the quasi future trend direction, continue to play in the field of manufacturing characteristics and advantages, will revive the industry itself as XX wind is the center of economic development as task, promote the industrial structure from low-end to high-end forward end, to create new heights of development of modern industry.Comprehensive implementation of the party's eighteen and eighteen session of the 33, fourth, Fifth Plenary Session spirit, foster innovation, harmony, green, open, sharing development idea, firm not to shift industries of City Road, take the initiative to melt into the global industrial division system, comprehensive implementation of the "made in China" and 2025 "XX the platform for action, to the depth of information technology and industrialization integration as the main line, in order to improve the quality and efficiency of economic development benefits as the center, and actively develop new industries, new technologies, new formats and new models, and vigorously promote industrial intelligent, green color, services, high-end development, and to build in emerging industries for the pilot to guide. As the main body of modern manufacturing, modern service industry as the support of the new modern industrial system, the XX to create a domestic first-class, internationally influential manufacturing city, revitalize the vibration XX industry glory in the new starting point for the construction of "the economy is strong, rich people, beautiful environment, social civilization degree is high" the new XX solid solid strong industrial base.Accelerate the innovation development, the implementation of talent drive. Firmly adhere to the development on the basis of innovation, and vigorously promote the comprehensive innovation in science and technology innovation as the core of the nucleus, since the focus on enhancing the capability of independent innovation, get rid of institutional system obstacles, to maximize the liberation and shock to stimulate innovation and vitality, accelerate the formation of innovation as the main leading and support the economic system and mode of development. Strengthening of "talent is the first resource" the idea, focus on the development of needs and direction of innovation, efforts to consolidate the foundation of science and education, optimize the talent environment, to create a high level of innovation of new talents for sustained economic and social healthy development continued to provide strong support.Play to the market mechanism, the transformation of government function. Focus on the decisive role of the market in resource allocation and better play the role of government, strengthen the supply of supply side reform, improve the quality and efficiency of the supply system, co-ordinate the use of economic, legal, policy and means, strengthen the guide and regulate the industry, promote the optimization the upgrading of the industrial structure industry. The implementation of negative list management clear, to further eliminate the obstacles of scientific development system, continuous development and promotion to stimulate the vitality of market players.Grasp the direction of development, firmly adhere to guidance. Based on the basis of the existing industry system and development, accurately grasp the main direction of holding a new round of technological revolution and the industrial revolution leather, strengthen strategic planning and forward-looking deployment, increase investment in innovation, enhance innovation output, stimulate a new vitality, stronger in advanced manufacturing industry, fostering the growth of strategic emerging industries, accelerate the development of producer services, promote industrial system to the reasonable structure, development, agglomeration, clear competitive advantage in the direction of development.To promote energy conservation, improve the ecological benefits. Firmly establish the concept of green development, unswervingly walk the green development, development cycle, low carbon development, increase energy conservation and pollution remediation efforts in resource saving, environment friendly, basis to improve the efficiency rate of the ecosystem, improve the quality and efficiency of development, strengthen the sustainable development the ability to force.Optimization of space layout, and promote regional cooperation. With the layout of concentration, land intensive, industrial cluster, cluster development oriented, optimize industrial park space layout, cultivate advantageous industry clusters, the formation of industrial development is relatively concentrated, production integration of the city space layout good interaction between elements, through the integration of resources. Expand the market space, improve the regional influence and competitiveness.To 20220when the Terminal level when installed on the line number (Word) should be arranged from top to bottom. 6.4.4 cable core must be completely loose and straight, but not damage the insulation and core. Core bundle of the same plate vertically or horizontally arranged1Manufacturing transformation and development planning1 comprehensive strength of the manufacturing industry to upgradeRelying on good traffic conditions and location advantages of the industrial base, the manufacturing industry comprehensive strength steadily, the total size of the total, technical level, economic benefits and so on in the country has become the cornerstone of the precedent, the city's economic and social development and an important support. "12th Five-Year" period, the total industrial added value, the average annual growth rate of dada%, the amount of the total amount of investment of the city's industrial enterprises with an average annual growth rate%, above scale industrial enterprises realized a total profit of the average annual growth rate of%.xx years, the city's above scale industrial output mode billion yuan output value of over 80 billion yuan of industrial dada to 8, above scale industrial added value billion yuan, industrial growth. Added value accounted for GDDP%, the resident population per capita GDP reached million, ranked second in XXX Province, the national large and medium-sized cities in the forefront.2 continuous optimization of industrial structureThe benefit oriented, high end orientation and the direction of intensive development, gradually increase efforts to promote the transformation of industrial structure adjustment and optimization of industry has made remarkable achievements. As to XX, machinery, textile, petrochemical, metallurgy and electronics and other 55 pillar industries above scale industrial output value at current prices billion yuan, accounting for the proportion of for the high-tech industry output value accounted for%. The proportion of above scale industrial output value reached%, one percentage point increase over 2010. Sensor network innovation shows the construction of the demonstration zone to promote the smooth, emerging industry gradually expand the scale of production is expected to total output average growth rate, new energy, new materials, energy saving and environmental protection, Internet of things, the rapid development of microelectronics and other emerging industries, has become an important field of XX industrial transformation and upgrading.3 innovation capacity significantly improvedSeize the new round of the wave of technological innovation brings plenty of historical opportunity, in-depth implementation of innovation driven development strategy for the manufacturing industry's innovation and transformation and upgrading provides a strong intellectual support for the.Xx year, R & D expenditure accounting for the main business service income ratio reached%, among the province's first patent. The application amounted to 224197, the right amount of invention patents reached 5480, 000 people obviously patent has reached 25, are located in the forefront of the province of.Xx has a provincial engineering technology research centers heart 506, country, provincial high technology research key laboratory 10, International cooperation base, 9 national, provincial, foreign R & D center 41, provincial and international technology transfer center 8, selected "research research institutions" key enterprises XX 84. Tsinghua University, South, Fudan University, Harbin Institute of technology, Huazhong University of science and technology, Northeastern University and other colleges and universities have been set up in the XXX Research Institute.4 green manufacturing results significantly"XXX water crisis", energy saving and environment protection increasing pressure, vigorously promote the green, low-carbon, circular ring production mode, adhere to energy-saving emission reduction, the implementation of energy control, improve the energy source utilization rate of.Xx years, the city unit of GDP energy consumption decreased 20%, increase the added value of industrial energy consumption per ton standard coal / million. "" 12th Five-Year "period, the cumulative unit of GDPP energy consumption decreased percent, energy consumption consumption value XX lowest decline of XX south, South first. The elimination of backward production capacity can work orderly, shut down 8823 enterprises, 852 enterprises for rectification of family, eliminate backward production capacity of cement 75 million 770 thousand Tons, dyeing 244 million 600 thousand meters, 100 thousand tons of chemical fiber, Leather 3 million standard sheets, paper million tons, 2 million 280 thousand tons of steel, 106 battery Wanqianqian KVA, 200 thousand tons of steel casting, resolve the production capacity of 3 million 370 thousand tons, cement production capacity of 710 thousand tons, completed ahead of schedule into "12th Five-Year" planning objectives.5 enterprise vitality is increasingThe emergence of a number of key enterprises in the industry a lot of industries, some have developed into the world class level leading enterprises and leading enterprises. The key point of enterprises in technological innovation, product development, market development, has played an important leading role for.Xx and other aspects of social responsibility, listed 15 companies around the "top 500 enterprises Chinese" listed 221 companies "Chinese manufacturing enterprise 500 strong", all year for nine consecutive years, ranking first in the province. The city's domestic and foreign listed enterprises amounted to 94, ranked in the prefecture level city in the forefront. The integration of military and civilian combined to accelerate, the number of the city's participation in scientific research and production of weapons equipment industry the advantages of private enterprises Year after year, through the confidential qualification certification enterprises ranked 22 in the province.Two 6 integration steadily intoAs a national integration of the two test area, has been to two of the depth of financial integration and enhance the core competitiveness of enterprises as an important starting point, to accelerate the transformation of development mode. The information of all business to achieve the city's more than 80% large-scale enterprises, a total of at or above the provincial level demonstration pilot enterprises, 250 enterprises, including 22 national. Fusion demonstration of 4 enterprises, the provincial demonstration integration of the two test, Park 8. Baidu and Alibaba, Pakistan, Chinese manufacturing network, China network library and so on enterprises to carry out in-depth cooperation, support enterprises with the development of electronic commerce, there have been 4 enterprises selected for the national e-commerce service integrated innovation pilot .In a comprehensive summary of the "12th Five-Year" achievement at the same time, should also see some problems still exist in the development of XX manufacturing industry, difficulties and contradictions, need to be solved in the "Shisansanwu" period. The main is: not many major industrial projects, lack of a group with strong power, high output level large projects the industrial economic growth. Lack of weak industrial products to the intermediate product, the added value is not high. The emerging industry is small, difficult to support the city's economic growth. Science and technology innovation ability is not strong, high dependence on foreign technology, high level talent shortage. Excess capacity in some industries, some enterprise production and management difficulties. These difficulties all Urgent need to be addressed in the future development of the exhibition.1 from the global perspective, a new round of scientific and technological revolution and the rise of industrial change, to bring significant strategic opportunities for the transformation of the XXX industryThe development mode of the industry will undergo profound changes. When a new round of technological revolution in the application of the new generation of information technology integration as the core feature, "Internet plus" has become an important direction of industrial development. The high-end, smart, green, as the service has become a new trend of industrial development; the Internet network to break the constraints of time, the world famous enterprises and enterprises to achieve global collaborative design and production; intelligent manufacturing technology in the design, development, application, manufacturing and other aspects of the daily trend of generalization deepening; based on fusion, and services, new formats and new model into a platform for the new form of industry.The industrial division of labor and competition face remodeling. Europe and other developed countries to implement the "re industrialization strategy, trying to consolidate its leading position; emerging economies have focused on the implementation of catch-up strategy, competition to seize the high-end manufacturing areas to accelerate the end of a space for one person. To developed countries" reverse transfer, emerging countries rely on information the comparative advantage of labor resources, to attract international investment led shunt. Significant changes in Global trade investment system, the United States actively promote the TPP and TTIIP as the core of the trade and investment heavy order reconstruction, the future of China's foreign trade and absorbing the international direct investment is facing greater pressure.At present, the development of the XX industry development level in the nation, with the foundation and conditions follow a new round of revolution of science and technology revolution and industry industry, to seize the great opportunities in grasping the concept of development, innovation, production mode and format to change the strain, the first action, to create new competitive advantages of industry competition. At the same time. Despite the change of international trade and investment will lead to industrial growth slowed to put XXX of export-oriented industries, jobs and social challenges, increase the risk of wind, but also bring down forced backward industries and low added value of industry chain, promote industrial restructuring and upgrading opportunities.2 from across the country, China's economic development has entered a new normal norm, put forward urgent requirements of development quality and efficiency of quality industry in XX CityThe industrial economy is facing multiple conversion speed, structure and power. China's economic development has entered a new norm, is to form a higher division of advanced, more complex, more reasonable structure can decrease the growth rate of stage evolution. It may lead some challenges to There's no telling war, a variety of factors, constraints tightening, road development the traditional expansion gets narrower. Must take the innovation as the new engine driven by economic development, speeding up the drive to convert from elements of the scale of investment driven development to the development of innovation driven development based principal.The supply and demand of the market environment has changed greatly. Different from the demand side, the grade and residents' demand structure has undergone significant changes, the basic medium has been close to the level of developed countries, the individuality has become the competitiveness of products to reflect the heavy, large quantity of products demand is gradually being customized by small batch of personality replace. See from the aspect of supply, supply of existing products can not meet the rapid escalation of market demand, the lack of effective supply, supply mode and become the deep level reason of industrial economic downturn. Through product innovation arousing potential consumer market, stimulate investment demand, increase effective supply, has It has become an important direction to boost China's industrial economy.Industrial transformation and upgrading by elements to resource constraints more urgent. I China's per capita fresh water, arable land, forest resources is only the world average level of 28%, 40%, 25%%, oil, iron ore, copper and other important mineral resources per capita reserves respectively for the world average level%. 117%, 17%, the traditional industry has been difficult to support resources and high energy consumption. The long-term accumulation of environmental conflicts are concentrated, serious pollution of the atmosphere pollution has become a national environmental and social problems, water pollution, soil pollution is more prominent, increasingly, the industry restructuring and development to put forward urgent Pray.To deepen the reform, formation and development of new power will be the "13th Five-Year" industrial development in China. The theme of the next five years, XX industry development must establish the new ideas of thinking, to adapt to and guide the new norm as macro logic of industrial development, breaking the current resource bottleneck was completed, speed, actively and steadily. The structural change, multiple motive etc... to make good use of the big deployment of national level of the market and institutional environment of product of positive influence, seize the opportunity to expand the space of domestic development, focus on structural optimization, attention to enhance the quality, focus on environmental improvement, fully activate the development of energy industry xx, Provide continuous power for industrial upgrading.3 from the city to see, XX course of economic development, natural endowment and location, determine the XX must take industrial strong city roadRaise the level of independent innovation opportunities of the industrial structure of.Xxx is two, three, and a "heavy chemical characteristics of industrial scale manufacturing characteristic is apparent, especially in the equipment manufacturing sector, in the province and even the whole country has comparative advantage. To firmly firmly grasp the national independent innovation demonstration in South XX area XX sensor network innovation demonstration zone construction in a major strategic opportunity, perfect the system of innovation incentive mechanism, increasing the field of manufacturing and innovation investment, enhance the innovation output, stimulated to stimulate innovation and vitality, enhance innovation and lead development ability.The critical period of building a modern industry to develop new heights. From now until 2020, XXX should strive to build a "new Qiang Fu makall" XX, a high level of comprehensive well-off well-off society. The manufacturing industry is still an important force to promote sustained and healthy economic and social development xxx not indispensable, is to promote the XX economy jumped to a higher level basic support. To accelerate industrial restructuring and upgrading, stronger advanced manufacturing industry, fostering the growth of strategic emerging industries, accelerate the development of modern service industry, push to promote the establishment of innovation ability, quality benefit, structure reasonable layout, sustainable development and international competitive production A new system of industry, revitalize the vibration XX industry glory on the new starting point.Full integration into the national strategy of regional development gold period.Xx geographical advantages are obvious, and the historical and cultural heritage, has strong strong industrial strength and potential development of..xx should be fully docking "The Belt and Road Road, the Yangtze River economic belt, the Yangtze River Delta region a major strategic integration area, strengthen the strong docking with Shanghai, enhance with XXX, XX and even the entire north Yangtze River Basin industrial collaboration, open economy, to a higher level of development, and actively promote international industrial cooperation, the formation of a higher level and wider domain opening pattern.The next five years, XX must be the quasi future trend direction, continue to play in the field of manufacturing characteristics and advantages, will revive the industry itself as XX wind is the center of economic development as task, promote the industrial structure from low-end to high-end forward end, to create new heights of development of modern industry.Comprehensive implementation of the party's eighteen and eighteen session of the 33, fourth, Fifth Plenary Session spirit, foster innovation, harmony, green, open, sharing development idea, firm not to shift industries of City Road, take the initiative to melt into the global industrial division system, comprehensive implementation of the "made in China" and 2025 "XX the platform for action, to the depth of information technology and industrialization integration as the main line, in order to improve the quality and efficiency of economic development benefits as the center, and actively develop new industries, new technologies, new formats and new models, and vigorously promote industrial intelligent, green color, services, high-end development, and to build in emerging industries for the pilot to guide. As the main body of modern manufacturing, modern service industry as the support of the new modern industrial system, the XX to create a domestic first-class, internationally influential manufacturing city, revitalize the vibration XX industry glory in the new starting point for the construction of "the economy is strong, rich people, beautiful environment, social civilization degree is high" the new XX solid solid strong industrial base.Accelerate the innovation development, the implementation of talent drive. Firmly adhere to the development on the basis of innovation, and vigorously promote the comprehensive innovation in science and technology innovation as the core of the nucleus, since the focus on enhancing the capability of independent innovation, get rid of institutional system obstacles, to maximize the liberation and shock to stimulate innovation and vitality, accelerate the formation of innovation as the main leading and support the economic system and mode of development. Strengthening of "talent is the first resource" the idea, focus on the development of needs and direction of innovation, efforts to consolidate the foundation of science and education, optimize the talent environment, to create a high level of innovation of new talents for sustained economic and social healthy development continued to provide strong support.Play to the market mechanism, the transformation of government function. Focus on the decisive role of the market in resource allocation and better play the role of government, strengthen the supply of supply side reform, improve the quality and efficiency of the supply system, co-ordinate the use of economic, legal, policy and means, strengthen the guide and regulate the industry, promote the optimization the upgrading of the industrial structure industry. The implementation of negative list management clear, to further eliminate the obstacles of scientific development system, continuous development and promotion to stimulate the vitality of market players.Grasp the direction of development, firmly adhere to guidance. Based on the basis of the existing industry system and development, accurately grasp the main direction of holding a new round of technological revolution and the industrial revolution leather, strengthen strategic planning and forward-looking deployment, increase investment in innovation, enhance innovation output, stimulate a new vitality, stronger in advanced manufacturing industry, fostering the growth of strategic emerging industries, accelerate the development of producer services, promote industrial system to the reasonable structure, development, agglomeration, clear competitive advantage in the direction of development.To promote energy conservation, improve the ecological benefits. Firmly establish the concept of green development, unswervingly walk the green development, development cycle, low carbon development, increase energy conservation and pollution remediation efforts in resource saving, environment friendly, basis to improve the efficiency rate of the ecosystem, improve the quality and efficiency of development, strengthen the sustainable development the ability to force.Optimization of space layout, and promote regional cooperation. With the layout of concentration, land intensive, industrial cluster, cluster development oriented, optimize industrial park space layout, cultivate advantageous industry clusters, the formation of industrial development is relatively concentrated, production integration of the city space layout good interaction between elements, through the integration of resources. Expand the market space, improve the regional influence and competitiveness.To 20220when the Terminal level when installed on the line number (Word) should be arranged from top to bottom. 6.4.4 cable core must be completely loose and straight, but not damage the insulation and core. Core bundle of the same plate vertically or horizontally arrangedNTFU②①③④⑤⑥⑦⑧图1-3（原教材图1-4）012468123R1R2012468123012468R3012468012468123R4图4-1123R2123R1·R2R1·R2·R1123R1123R1-1123图4-2表4-2102004图4-3a cR1R2r(R1)s(R1)t(R1)r(R2)s(R2)t(R2)图4-4ba cba cba cba cba cba cba cbabdec图4-5123456图4-624513图4-7abecd图4-8abcde图4-9表20123401234图：3（1）的A((f((B0123401234图：3（2）的A((f((B图6-1abdceabce图(a)dv4v1v2v3图6-6v5图6-8v1v2v3v4v4v2v1v5v3图6-7acbdefghijklnmpo图6-13abcdefg(a)abdfgec(b)v8v1v2v3图6-15v4v5v6v7842119511222418v1(0,0)v2(6,3)v3(2,1)v4(8,6)v5v6(7,3)v7(11,6)v8(13,7)8421195112224188621187151613152011图6-16B63540503019201082312ACDEFG63540503019201082312ACDEFGB(a)图6-17(b)5b10(c)67a85bced(a)66147a8514b(b)76614a5bcddcee图6-18图G图图6-19(a)(b)(c)(d)图6-23e图6-27abcldghijk图6-28ABCDEFGHabced(b)deabcdabc(c)dabc(d)图6-30图6-29图7-1⑵⑴⑶⑷⑸图7-3abcd12342图7-4abcdabcdabcdabcdabcdabcdabcdabcd⑴⑵⑶⑷⑸⑹⑺⑻图7-512367895104111236789510411图7-6图7-7图7-8⑵⑴⑶⑷⑸⑹⑺⑻⑽⑼⑾⑿图7-93758112235758111032575811151032558117151032558117213615103255811721(a)(b)(c)(d)(e)(f)图7-11图7-12aaabbbcccdddeeefffggg(1)(2)(3)abcde(1)eabcdf(2)图7-15图7-17abcdefg⑴abcdefg⑵图7-16ab(b)ecd7564a(a)becd16613785964图7-18图7-19abcdefg1204153816172823369⑴abcdefg1438239⑵⑵⑴⑷⑶⑸⑹⑺图7-20图7-21⑴⑵图7-22⑴010000001111111⑵000001111图7-23235⑴510235⑵510235⑶8157510235⑷8157图7-2425101015156945202030301006040⑴4312507610001⑵110100001图7-25P4P1P2P5P6P7a1a2a10a9a5a4a7a8a6a3图7-26P4P1P2P5P6P7a1a2a10a9a5a4a7a8a6a3图7-27L1L2L3L4L5L6图7-28G1G2G3G4G5G6表8-2abababb1b(a)表8-2ccc1bcccab(b)cabbbbbccccc表1abdcfeg图14121326图28241abc⑴0⑵1abc0图3

本文来自电脑杂谈，转载请注明本文网址：
http://www.pc-fly.com/a/jisuanjixue/article-35300-1.html