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

§2.Composition §Definition 2.14: Let R 1 be a relation from A to B, and R 2 be a relation from B to C. The composition of R 1 and R 2, we write R 2  R 1, is a relation from A to C, and is defined R 2  R 1 ={(a,c)|there exist some b  B so that (a,b)  R 1 and (b,c)  R 2, where a  A and c  C}. §(1)R 1 is a relation from A to B, and R 2 is a relation from B to C §(2)commutative law?  §R 1 ={(a 1,b 1 ), (a 2,b 3 ), (a 1,b 2 )} §R 2 ={(b 4,a 1 ), (b 4,c 1 ), (b 2,a 2 ), (b 3,c 2 )}

§Associative law? §For R 1  A×B, R 2  B×C, and R 3  C×D §R 3  (R 2  R 1 )=?(R 3  R 2 )  R 1 §subset of A×D §For any (a,d)  R 3  (R 2  R 1 ), (a,d)  ?(R 3  R 2 )  R 1, §Similarity, (R 3  R 2 )  R 1  R 3  (R 2  R 1 ) §Theorem 2.3 ： Let R 1 be a relation from A to B, R 2 be a relation from B to C, R 3 be a relation from C to D. Then R 3  (R 2  R 1 )=(R 3  R 2 )  R 1 (Associative law)

§Definition 2.15: Let R be a relation on A, and n  N. The relation R n is defined as follows. §(1)R 0 ={(a,a)|a  A}), we write I A.  (2)R n+1 =R  R n. §Theorem 2.4: Let R be a relation on A, and m,n  N. Then  (1)R m  R n =R m+n §(2)(R m ) n =R mn

§A={a 1,a 2, ,a n },B={b 1,b 2, ,b m } §R 1 and R 2 be relations from A to B. §M R1 =(x ij ), M R2 =(y ij ) §M R1 ∪ R2 =(x ij  y ij ) §M R1∩R2 =(x ij  y ij ) §  0 1  0 1 § § §Example:A={2,3,4},B={1,3,5,7} §R 1 ={(2,3),(2,5),(2,7),(3,5),(3,7),(4,5),(4,7)} §R 2 ={(2,5),(3,3),(4,1),(4,7)} §Inverse relation R -1 of R : M R -1 =M R T, M R T is the transpose of M R.

§A={a 1,a 2, ,a n },B={b 1,b 2, ,b m }, C={c 1,c 2, ,c r }, §R 1 be a relations from A to B, M R1 =(x ij ) m  n, R 2 be a relation from B to C, M R2 =(y ij ) n  r. The composition R 2  R 1 of R 1 and R 2,

§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 §1.Introduction §Construct a 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.

§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

§Theorem 2.6: Let R 1 and R 2 be relations on A, and R 1  R 2. Then §(1)r(R 1 )  r(R 2 ) ； §(2)s(R 1 )  s(R 2 ) ； §(3)t(R 1 )  t(R 2 ) 。 §Proof: (3)R 1  R 2  t(R 1 )  t(R 2 ) §Because R 1  R 2 ，  R 1  t(R 2 ) §t(R 2 ) :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 I A be identity(diagonal) relation. Then r(R)=R ∪ I A (I A ={(a,a)|a  A}) §Proof ： Let R'=R ∪ I A. §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

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

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 §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 A i and A j are distinct elements of , then A i ∩A j = . §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 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] 。 §[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 32,34 §P151 1,2,13, 17, 23,24 §P167 15,16,22,24,26,27,28,29,32, 36