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第 3 周起每周一交作业，作业成绩占总成绩的1 5% ； 平时不定期的进行小测验，占总成绩的 15% ； 期中考试成绩占总成绩的 20% ；期终考试成绩占总成绩的 50%

第 3 周起每周一交作业，作业成绩占总成绩的1 5% ； 平时不定期的进行小测验，占总成绩的 15% ； 期中考试成绩占总成绩的 20% ；期终考试成绩占总成绩的 50% zhym@fudan.edu.cn 张宓 13212010027@fudan.edu.cn BBS id:abchjsabc 软件楼 103 杨侃 10302010007@fudan.edu.cn 程义婷 11302010050@fudan.edu.cn BBS id:chengyiting 刘雨阳 13212010013@fudan.edu.cn ，软件楼 405

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第 3 周起每周一交作业，作业成绩占总成绩的1 5% ； 平时不定期的进行小测验，占总成绩的 15% ； 期中考试成绩占总成绩的 20% ；期终考试成绩占总成绩的 50%

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  1. 第3周起每周一交作业，作业成绩占总成绩的15%；第3周起每周一交作业，作业成绩占总成绩的15%； 平时不定期的进行小测验，占总成绩的15%； 期中考试成绩占总成绩的20%；期终考试成绩占总成绩的50% zhym@fudan.edu.cn 张宓 13212010027@fudan.edu.cn BBS id:abchjsabc 软件楼103 杨侃 10302010007@fudan.edu.cn 程义婷 11302010050@fudan.edu.cn BBS id:chengyiting 刘雨阳 13212010013@fudan.edu.cn，软件楼405 liy@fudan.edu.cn李弋

  2. 2.Composition • Definition 2.14: Let R1 be a relation from A to B, and R2 be a relation from B to C. The composition of R1 and R2, we write R2R1, is a relation from A to C, and is defined R2R1={(a,c)|there exist some bB so that (a,b)R1 and (b,c)R2, where aA and cC}. • (1)R1 is a relation from A to B, and R2 is a relation from B to C • (2)commutative law?  • R1={(a1,b1), (a2,b3), (a1,b2)} • R2={(b4,a1), (b4,c1), (b2,a2), (b3,c2)}

  3. Associative law? • For R1A×B, R2B×C, and R3C×D • R3(R2R1)=?(R3R2)R1 • subset of A×D • For any (a,d)R3(R2R1), (a,d)?(R3R2)R1, • Similarity, (R3R2)R1R3(R2R1) • Theorem 2.3：Let R1 be a relation from A to B, R2 be a relation from B to C, R3 be a relation from C to D. Then R3(R2R1)=(R3R2)R1(Associative law)

  4. Definition 2.15: LetR be a relation on A, and nN. The relation Rn is defined as follows. • (1)R0 ={(a,a)|aA}), we write IA. • (2)Rn+1=RRn. • Theorem 2.4: LetR be a relation on A, and m,nN. Then • (1)RmRn=Rm+n • (2)(Rm)n=Rmn

  5. A={a1,a2,,an},B={b1,b2,,bm} • R1 and R2 be relations from A to B. • MR1=(xij), MR2=(yij) • MR1∪R2=(xijyij) • MR1∩R2=(xijyij) •  0 1  0 1 • 0 0 1 0 0 0 • 1 1 1 10 1 • Example:A={2,3,4},B={1,3,5,7} • R1={(2,3),(2,5),(2,7),(3,5),(3,7),(4,5),(4,7)} • R2={(2,5),(3,3),(4,1),(4,7)} • Inverse relation R-1 of R : MR-1=MRT, MRT is the transpose of MR.

  6. A={a1,a2,,an},B={b1,b2,,bm}, C={c1,c2,,cr}, • R1 be a relations from A to B, MR1=(xij)mn, R2 be a relation from B to C, MR2=(yij)nr. The composition R2R1 of R1 and R2,

  7. Example：R={(a,b),(b,a),(a,c)},is not symmetric • + (c,a),R'={(a,b),(b,a),(a,c), (c,a)}，R'is symmetric. • Closure

  8. 2.5 Closures of Relations • 1.Introduction • Constructa new relation R‘,s.t. RR’, • particular property, • smallest relation • closure • Definition 2.17: Let R be a relation on a set A. R' is called the reflexive(symmetric, transitive) closure of R, we write r(R)(s(R),t(R) or R+), if there is a relation R' with reflexivity (symmetry, transitivity) containing R such that R' is a subset of every relation with reflexivity (symmetry, transitivity) containing R.

  9. Condition: • 1)R' is reflexivity(symmetry, transitivity) • 2)RR' • 3)For any reflexive(symmetric, transitive) relation R", If RR", then R'R" • Example：If R is symmetric, s(R)=? • If R is symmetric，then s(R)=R • Contrariwise, If s(R)=R，then R is symmetric • R is symmetric if only if s(R)=R • Theorem 2.5: Let R be a relation on a set A. Then • (1)R is reflexive if only if r(R)=R • (2)R is symmetric if only if s(R)=R • (3)R is transitive if only if t(R)=R

  10. Theorem 2.6: Let R1 and R2 be relations on A, and R1R2. Then • (1)r(R1)r(R2)； • (2)s(R1)s(R2)； • (3)t(R1)t(R2)。 • Proof: (3)R1R2t(R1)t(R2) • Because R1R2， R1t(R2) • t(R2) :transitivity

  11. Example:Let A={1,2,3},R={(1,2),(1,3)}. Then • 2.Computing closures • Theorem 2.7: Let R be a relation on a set A, and IA be identity(diagonal) relation. Then r(R)=R∪IA(IA={(a,a)|aA}) • Proof：Let R'=R∪IA. • Definition of closure • (1)For any aA, (a,a)?R'. • (2) R?R'. • (3)Suppose that R'' is reflexive and RR''，R'?R''

  12. Theorem 2.8：Let R be a relation on a set A. Then s(R)=R∪R-1. • Proof：Let R'=R∪R-1 • Definition of closure • (1) R', symmetric? • (2) R?R'. • (3)Suppose that R'' is symmetric and RR''，R'?R'')

  13. Example ：symmetric closure of “<” on the set of integers,is“≠” • <,>， • Let A is no empty set. • The reflexive closure of empty relation on A is the identity relation on A • The symmetric closure of empty relation on A, is an empty relation.

  14. Theorem 2.9: Let R be a relation on A. Then Theorem 2.10：Let A be a set with |A|=n, and let R be a relation on A. Then

  15. Example：A={a,b,c,d},R={(a,b),(b,a), (b,c),(c,d)},t(R)=?

  16. 2.6 Equivalence Relation 1.Equivalence relation Definition 2.18: A relation R on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive. Example: Let m be a positive integer with m>1. Show that congruence modulo m is an equivalence relation. R={(a,b)|ab mod m} Proof: (1)reflexive (for any aZ，aRa?) (2)symmetric (for any aRb， bRa?) (3)transitive (for aRb，bRc，aRc?)

  17. 2.Equivalence classes partition Definition 2.19: A partition or quotient set of a nonempty set A is a collection  of nonempty subsets of A such that (1)Each element of A belongs to one of the sets in . (2)If Ai and Aj are distinct elements of , then Ai∩Aj=. The sets in  are called the bocks or cells of the partition. Example: Let A={a,b,c}, P={{a,b},{c}},S={{a},{b},{c}},T={{a,b,c}}, U={{a},{c}},V={{a,b},{b,c}},W={{a,b},{a,c},{c}}, infinite

  18. Example：congruence modulo 2 is an equivalence relation. For any xZ, or x=0 mod 2,or x=1 mod 2, i.e or xE ,or xO. And E∩O= E and O， {E, O} is a partition of Z

  19. Definition 2.20: Let R be an equivalence relation on a set A. The set of all element that are related to an element a of A is called the equivalence class of a. The equivalence class of a with respect to R is denoted by [a]R, When only one relation is under consideration, we will delete the subscript R and write [a] for this equivalence class. Example：equivalence classes of congruence modulo 2 are [0] and [1]。 [0]={…,-4,-2,0,2,4,…}=[2]=[4]=[-2]=[-4]=… [1]={…,-3,-1,1,3,…}=[3]=[-1]=[-3]=… the partition of Z =Z/R={[0],[1]}

  20. Exercise:P146 32,34 • P151 1,2,13, 17, 23,24 • P167 15,16,22,24,26,27,28,29,32, 36

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