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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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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% - PowerPoint PPT Presentation

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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% 张宓 BBS id:abchjsabc 软件楼 103 杨侃 程义婷 BBS id:chengyiting 刘雨阳 ,软件楼 405

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BBS id:abchjsabc 软件楼103



BBS id:chengyiting


  • 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)}
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)
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
  • 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.
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,
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
2 5 closures of relations
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.
  • 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
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
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\'\'
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\'\')
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.

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

2 6 equivalence relation
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?)

2.Equivalence classes


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},




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

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]。



the partition of Z =Z/R={[0],[1]}

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