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.CompositionDefinition 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 R2R1, is a

relation from A to C, and is defined R2R1={(a,c)|

there exist some bB so that (a,b)R1 and

(b,c)R2, where aA and cC}.(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, R2B×C, and R3C×DR3(R2R1)=?(R3R2)R1

subset of A×DFor any (a,d)R3(R2R1), (a,d)?(R3R2)R1,Similarity, (R3R2)R1R3(R2R1) 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(R2R1)=(R3R2)R1(Associative law)

Definition 2.15: Let R be a relation on A, and nN. The relation Rn is defined as follows.

(1)R0 ={(a,a)|aA}), we write IA.(2)Rn+1=RRn.Theorem 2.4: Let R be a relation on A, and

m,nN. Then (1)Rm

Rn=Rm+n

(2)(Rm)n=Rmn

A={a1,a2,,an},B={b1,b2,,bm}

R1 and R2 be relations from A to B.

MR1=(xij), MR2=(yij)

MR1 R2∪ =(xijyij)

MR1∩R2=(xijyij) 0 1 0 10 0 1 0 0 0

1 1 1 1 0 1Example: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, MR

T 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)mn, R2 be

a relation from B to C, MR2=(yij)nr. The

composition R2R1 of R1 and R2,

rmkjik

n

kRR yxM

))((112

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 Relations1.IntroductionConstruct a new relation R‘, s.t. RR’, particular property, smallest relationclosureDefinition 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.

Condition:1)R' is reflexivity(symmetry, transitivity)2)RR'3)For any reflexive(symmetric, transitive)

relation R", If RR", then R'R" Example ： If R is symmetric, s(R)=? If R is symmetric ， then s(R)=RContrariwise, If s(R)=R ， then R is symmetricR is symmetric if only if s(R)=RTheorem 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 R1R2. Then

(1)r(R1)r(R2) ；(2)s(R1)s(R2) ；(3)t(R1)t(R2) 。Proof: (3)R1R2t(R1)t(R2)

Because R1R2 ， R1t(R2)

t(R2) :transitivity

Example:Let A={1,2,3},R={(1,2),(1,3)}. Then

2.Computing closuresTheorem 2.7: Let R be a relation on a set A,

and IA be identity(diagonal) relation. Then r(R)=R I∪ A(IA={(a,a)|aA})

Proof ： Let R'=R I∪ A.

Definition of closure (1)For any aA, (a,a)?R'. (2) R?R'. (3)Suppose that R'' is reflexive and

RR'' ， 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

RR'' ， 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 AThe symmetric closure of empty relation

on A, is an empty relation.

Theorem 2.9: Let R be a relation on A. Then

1

)(i

iRRt

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

n

i

iRRt1

)(

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

2.6 Equivalence Relation

1.Equivalence relationDefinition 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)|ab mod m}

Proof: (1)reflexive (for any aZ ， aRa?)(2)symmetric (for any aRb ， bRa?)(3)transitive (for aRb ， bRc ， aRc?)

2.Equivalence classespartitionDefinition 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

Example ： congruence modulo 2 is an equivalence relation.

For any xZ, or x=0 mod 2,or x=1 mod 2, i.e or xE ,or xO.

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] 。

[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]}

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

36