電大離散數(shù)學(xué)(本)復(fù)習(xí)題（小抄參考）

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1、專業(yè)好文檔 離散數(shù)學(xué)（本）復(fù)習(xí)題 1．設(shè)A={1，2}，B={2，3，4}，求r(AB)，r(A)r(B)。 2．設(shè)A={a，b，c}，問(wèn)IA，EA是否具有自反性，反自反性，對(duì)稱性，反對(duì)稱性，傳遞性？ 3．R，S是集合A上的兩個(gè)關(guān)系。試證明下列等式： （1）(R?S)-1= S-1?R-1 （2）(R-1)-1= R （3）(R∪S)-1= R-1∪S-1 （4）(R∩S)-1= R-1∩S-1 4．設(shè)R是集合A上的關(guān)系，令 R+={(x, y)|xA，yA，并且存在n>0，使得xRny}， 則稱R+是R的傳遞閉包，證明：R+是包含R的最小具有傳遞性的關(guān)系。 5．

2、若非空集合上的非空關(guān)系R是反自反的，是對(duì)稱的，試證明R不是傳遞的。 6．A={1，2，3，4，5，6，7，8，9，10}，R為A上的整除關(guān)系，請(qǐng)給出A的Hasse圖，并求出所有的極大元素，極小元素，最大元素，最小元素。 7．設(shè)G是含有3個(gè)不同原子的命題公式，當(dāng)G是恒假公式的時(shí)候，G的主析取范式中有多少極小項(xiàng)，主合取范式中有多少極大項(xiàng)？ 8．有人說(shuō)：“等價(jià)關(guān)系中的反身性可以不要，因?yàn)榉瓷硇钥梢詮膶?duì)稱性和傳遞性推出：由對(duì)稱性，從a @ b可得b @ a，再由傳遞性得a @ a”。你的意見(jiàn)呢？ 9．若集合A上的關(guān)系R，S具有對(duì)稱性，證明：R?S具有對(duì)稱性的充要條件為R?S= S?R。 1

3、0．若R是等價(jià)關(guān)系，試證明R-1也是等價(jià)關(guān)系。 11．給P和Q指派真值1，給R和S指派真值0，求出下面命題的真值： a) (P(QR))((PQ)(RS)) b) ((PQ)R)(((PQ)R)S) c) ((PQ)R)((QP)(RS)) d) (P(Q(RP)))(QS) 12．指出下列公式哪些是恒真的哪些是恒假的： (1)P（P Q）Q (2)（P Q）（PQ） (3)（P Q） （QR）（P R ） (4)（P Q）（P QP Q） 13．設(shè)S={G1，…，Gn}是命題公式集合。試求出在不增加新原子的情況下從S出發(fā)演繹出的所有命題公式。 14．證明下面的等價(jià)式：

4、 (1) (P(QR))(QR)(PR)=R (2) P(QP)=P(PQ) (3) P(QR)=(PQ)(PR) (4) (PQ)(RQ)=(PR)Q 15．找出下面公式的Skolem范式： (1)("xP(x)$y"zQ(y，z))； (2)"x(E(x，0)($y(E(y，g(x))"z(E(z，g(x))E(y，z)))))。 16．G=(P，L)是有限圖，設(shè)P(G)，L(G)的元數(shù)分別為m，n。證明：n ，其中 表示m中取2的組合數(shù)。 17．設(shè)G是有限圖，M，A分別是G的關(guān)聯(lián)矩陣和相鄰矩陣，證明：MM’和A2的對(duì)角線上的元素是G中所有點(diǎn)的度。 18．設(shè)G為圖（可

5、能無(wú)限），無(wú)回路，但若任意外加一邊于G后就形成一回路，試證G必為樹(shù)。 19．試舉出一個(gè)連通的(即漠視為圖后是連通的)，但無(wú)根的有向圖。 20．設(shè)G是有向圖，其中含一有向路(e1，…，en)，其中fin(en)=init(e1)，證明：G不是有向樹(shù)。 21．設(shè)(I，+)為整數(shù)加群，（5I，+）為I的子群，請(qǐng)給出mI的所有陪集。 22．證明：若一個(gè)圖G的任意兩點(diǎn)度數(shù)之和n-1，n=|P(G)|，則該圖有Hamilton路。 23．給出一個(gè)具有5個(gè)點(diǎn)的邊數(shù)最多的非Hamilton圖。 24．給出代數(shù)格的定義。 25．設(shè)G為有向圖，若G具有有向樹(shù)定義中的1)和2)，并且沒(méi)有有向回路。問(wèn)：

6、若G有限，G是否是有向樹(shù)？若G不是有限的，如何？ 26．設(shè) * 是集合S上的二元代數(shù)運(yùn)算，且滿足結(jié)合律，設(shè)x，y是S中任意元素，如果x * y = y * x，則x = y。試證明 * 滿足等冪律。 27．請(qǐng)給出一個(gè)布爾代數(shù)。 28．設(shè)R,S是A上的傳遞關(guān)系，證明或者反駁： (1) RS是傳遞關(guān)系； (2) RS是傳遞關(guān)系。 29．試用演繹法證明{PQ，QR，PM，M}共同蘊(yùn)涵R(PQ) 30. 求證G的任意多個(gè)子群的交集是G的子群。并且，G的任意多個(gè)正規(guī)子群的交集仍是G的正規(guī)子群。 31．設(shè)H是G的子群。N是G的正規(guī)子群。命HN為H的元素乘N的元素所得的所有元素的集合。求證H

7、N是G的子群。 32．設(shè)H是群G的一個(gè)有限非空子集，求證只要H中任意兩個(gè)元素的積仍在H內(nèi)，則H是G的子群。 33．求證循環(huán)群的子群仍是循環(huán)群。 34．求證若G的元數(shù)是一個(gè)質(zhì)數(shù)，則G必是循環(huán)群。 35．設(shè)K和H都是群G的子群，試證明：若HK是G的子群，則KH = HK。 36．什么是等價(jià)關(guān)系？ 37．如果A上的一個(gè)等價(jià)關(guān)系為R，如何求出一個(gè)等價(jià)類？ 38．給出命題公式PQ的真值表。 39．Skolem范式中的母式有什么特點(diǎn)？ 40．有根的有向圖，是否一定是強(qiáng)連通的？ 41．最優(yōu)樹(shù)是否一定唯一？ 42．什么是體？ 43．什么是代數(shù)格？ 44．半序子格與代數(shù)子格是什么關(guān)系？

8、 "If we dont do that it will go on and go on. We have to stop it; we need the courage to do it." His comments came hours after Fifa vice-president Jeffrey Webb - also in London for the FAs celebrations - said he wanted to meet Ivory Coast international Toure to discuss his complaint. CSKA gene

9、ral director Roman Babaev says the matter has been "exaggerated" by the Ivorian and the British media. Blatter, 77, said: "It has been decided by the Fifa congress that it is a nonsense for racism to be dealt with with fines. You can always find money from somebody to pay them. "It is a nonsense

10、to have matches played without spectators because it is against the spirit of football and against the visiting team. It is all nonsense. "We can do something better to fight racism and discrimination. "This is one of the villains we have today in our game. But it is only with harsh sanctions that

11、 racism and discrimination can be washed out of football." The (lack of) air up there Watch mCayman Islands-based Webb, the head of Fifas anti-racism taskforce, is in London for the Football Associations 150th anniversary celebrations and will attend Citys Premier League match at Chelsea on

12、 Sunday. "I am going to be at the match tomorrow and I have asked to meet Yaya Toure," he told BBC Sport. "For me its about how he felt and I would like to speak to him first to find out what his experience was." Uefa hasopened disciplinary proceedings against CSKAfor the "racist behaviour of the

13、ir fans" duringCitys 2-1 win. Michel Platini, president of European footballs governing body, has also ordered an immediate investigation into the referees actions. CSKA said they were "surprised and disappointed" by Toures complaint. In a statement the Russian side added: "We found no racist insu

14、lts from fans of CSKA." Baumgartner the disappointing news: Mission aborted. The supersonic descent could happen as early as Sunda. The weather plays an important role in this mission. Starting at the ground, conditions have to be very calm -- winds less than 2 mph, with no precipitation or humid

15、ity and limited cloud cover. The balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. It will climb higher than the tip of Mount Everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude o

16、f commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. As he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence. The balloon will slowly drift to the edge of space at 120,000 feet ( Then, I would assume, he will slowly step out ont

17、o something resembling an Olympic diving platform. Below, the Earth becomes the concrete bottom of a swimming pool that he wants to land on, but not too hard. Still, hell be traveling fast, so despite the distance, it will not be like diving into the deep end of a pool. It will be like he is diving

18、 into the shallow end. Skydiver preps for the big jump When he jumps, he is expected to reach the speed of sound -- 690 mph (1,110 kph) -- in less than 40 seconds. Like hitting the top of the water, he will begin to slow as he approaches the more dense air closer to Earth. But this will not be en

19、ough to stop him completely. If he goes too fast or spins out of control, he has a stabilization parachute that can be deployed to slow him down. His team hopes its not needed. Instead, he plans to deploy his 270-square-foot (25-square-meter) main chute at an altitude of around 5,000 feet (1,524 me

20、ters). In order to deploy this chute successfully, he will have to slow to 172 mph (277 kph). He will have a reserve parachute that will open automatically if he loses consciousness at mach speeds. Even if everything goes as planned, it wont. Baumgartner still will free fall at a speed that would cause you and me to pass out, and no parachute is guaranteed to work higher than 25,000 feet (7,620 meters). cause there 4