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§R1∪R2§R1∪R2 §R 1 ∩R 2 R1-R2R1-R2 2.4 Operations on Relations.

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Presentation on theme: "§R1∪R2§R1∪R2 §R 1 ∩R 2 R1-R2R1-R2 2.4 Operations on Relations."— Presentation transcript:

1 §R1∪R2§R1∪R2 §R 1 ∩R 2 R1-R2R1-R2 2.4 Operations on Relations

2 §1.Inverse relation §Definition 2.13: Let R be a relation from A to B. The inverse relation of R is a relation from B to A, we write R -1, defined by R -1 ={(b,a)|(a,b)  R} §Inverse relation of “ ” §R={(1,2),(2,3),(1,3)} §R -1 ={(2,1),(3,2),(3,1)}

3 §Theorem 2.1 : Let R,R 1, and R 2 be relation from A to B. Then §(1)(R -1 ) -1 =R; §(2)(R 1 ∪ R 2 ) -1 =R 1 -1 ∪ R 2 -1 ; §(3)(R 1 ∩R 2 ) -1 =R 1 -1 ∩R 2 -1 ; §(4)(A×B) -1 =B×A; §(5)  -1 =  ; (7)(R 1 -R 2 ) -1 =R 1 -1 -R 2 -1 (8)If R 1  R 2 then R 1 -1  R 2 -1

4 § (6),(8) § Proof:(6) (8)If R 1  R 2,then R 1 -1  R 2 -1 。 For any (b,a)  R 1 -1,

5 §Theorem 2.2 : Let R be a relation on A. Then R is symmetric if only if R=R -1. §Proof: (1)If R is symmetric, then R=R -1 。 §R  R -1 and R -1  R 。 §(2)If R=R -1, then R is symmetric §For any (a,b)  R, (b,a)  ?R

6 §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 )}

7 §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)

8 §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

9 §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 §0 0 1 0 0 0 §1 1 1 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.

10 §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,

11 §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

12 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.

13 §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

14 §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

15 §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''

16 §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) s(R), symmetric? §(2) R  ?R'. §(3)Suppose that R'' is symmetric and R  R'' , R'  ?R'')

17 §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.

18 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

19 §Example : A={a,b,c,d},R={(a,b),(b,a), (b,c),(c,d)}, t(R)=? §Equivalence Relation, P 136, i.e 4.5 §Partial order relations,P201-215,i.e 6.1,6.2

20 §Exercise: P135 32,34 §P139 1,2,12, 21,22, §P155 2,8,12,22,24,26,27,28,29,30


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