Discrete Structures Chapter 5 Relations Nurul Amelina Nasharuddin Multimedia Department

2 Objectives On completion of this topic, student should be able to: a.Determine the properties of relations – reflexive, symmetric, transitive, and antisymmetric b.Determine equivalence and partial order relations c.Represent relations using matrix and graph

Outline Properties of relations Matrix and graph representation of relations Equivalence relations Partial order relations 3

4 The most basic relation is “=” (e.g. x = y) Generally x R y  TRUE or FALSE ‒ R(x,y) is a more generic representation ‒ R is a binary relation between elements of some set A to some set B, where x  A and y  B Binary relations: x R y On sets x  X, y  Y R  X  Y Recall: Relations

5 Example “less than” relation from A={0,1,2} to B={1,2,3} Use traditional notation 0 < 1, 0 < 2, 0 < 3, 1 < 2, 1 < 3, 2 < 3 Or use set notation A  B={(0,1),(0,2),(0,3),(1,1),(1,2),(1,3),(2,1),(2,2), (2,3)} R={(0,1),(0,2),(0,3),(1,2),(1,3),(2,3)}  A  B Or use Arrow Diagrams

6 Reflexive: Let R be a binary relation on a set A Eg: Let A = {1,2,3,4} R 1 ={(1,1),(1,2),(2,2),(2,3),(3,3),(4,4)}  R 1 is reflexive R 2 = {(1,1),(2,2),(3,3)}  R 2 is not reflexive (why?) Properties of Relations

7 Symmetric: Let R be a binary relation on a set A Eg: Let A = {1,2,3} R 1 ={(1,2),(2,1),(1,3),(3,1)}  R 1 is symmetric R 2 = {(1,1),(2,2),(3,3),(2,3)}.  R 2 is not symmetric because (3,2)  R 2 Properties of Relations

8 Transitive: Let R be a binary relation on a set A Let A = {1,2,3,4} R = {(2,1),(3,1),(3,2),(4,1),(4,2),(4,3)} R is transitive because (3,2) & (2,1) → ( 3,1) (4,2) & (2,1) → (4,1) (4,3) & (3,1) → (4,1) (4,3) & (3,2) → (4,2) Properties of Relations

9 xRy or (x,y)  R for all x and y in A, 1.R is reflexive  for all x in A, (x,x)  R 2.R is symmetric  for all x and y in A, if (x,y)  R then (y,x)  R 3.R is transitive  for all x, y and z in A, if (x,y)  R and (y,z)  R then (x,z)  R

10 Example (1) Define a relation of A called R A = {2,3,4,5,6,7,8,9} R = {(4,4),(4,7),(7,4),(7,7),(2,2),(3,3),(3,6),(3,9), (6,6),(6,3),(6,9),(9,9),(9,3),(9,6)} Is R Reflexive? No Symmetric? Yes Transitive? Yes

11 Example (2) A = {0,1,2,3} R over A = {(0,0),(0,1),(0,3),(1,0), (1,1),(2,2),(3,0),(3,3)} Is R Reflexive? Yes. Symmetric? Yes. Transitive? No. (1,0),(0,3)  R but (1,3)  R

12 Proving Properties on Infinite Sets -“equal” relation (1) Define a relation R on R (the set of all real numbers): For all x, y  R, x R y ↔ x = y Is R reflexive? symmetric? transitive? R is reflexive iff  x  R, x R x. By definition of R, this means x = x, for all x  R. This is true, since every real number is equal to itself. Hence, R is reflexive

13 Proving Properties on Infinite Sets -“equal” relation (1) R is symmetric iff  x,y  R, if x R y then y R x. By definition of R, this means that  x,y  R, if x = y then y = x. So this is true; if one number is equal to a second, then the second is equal to the first. Hence, R is symmetric R is transitive iff  x,y  R, if x = y and y = z, then x = z. So this is true; if one real number equals to a second, and the second equals a third, then the first equals the third

14 Proving Properties on Infinite Sets -“less than” relation (1) Define a relation R on R (the set of all real numbers): For all x, y  R, x R y ↔ x < y Is R reflexive? symmetric? transitive?

15 Proving Properties on Infinite Sets -“less than” relation (2) R is reflexive iff  x  R, x R x. By definition of R, this means x < x, for all x  R. But this is false. Hence, R is not reflexive R is symmetric iff  x,y  R, if x R y then y R x. By definition of R. This means that  x,y  R, if x x. But this is false. Hence, R is not symmetric R is transitive iff  x,y,z  R, if x < y and y < z, then x < z. Hence, R is transitive

16 Properties of Congruence Modulo 3 (1) Define a relation R on Z: For all m,n  Z, m R n  3|(m – n) R is called congruence modulo 3 Is R reflexive? Is R symmetric? Is R transitive?

17 Properties of Congruence Modulo 3 (2) R is reflexive iff for all m in Z, m R m. By definition of R, this means 3|m – m or 3|0. This is true since 0 = 0. 3, so R is reflexive R is symmetric iff for all m,n in Z, m R n then n R m. By definition of R, this means if 3|(m – n) then 3|(n – m). This is true m – n = 3k, for some integer k n – m = - (m – n) = 3(-k) Hence 3|(n – m). Therefore R is symmetric

18 Properties of Congruence Modulo 3 (2) R transitive iff for all m,n,p  Z, if m R n and n R p then m R p. By definition of R, if 3|(m – n) and 3|(n – p) then 3|(m – p). So by definition of divide m – n = 3r for some r n – p = 3s for some s It is crucial to observe that (m – n) + (n – p) = m – p (m – n) + (n – p) = m – p = 3r + 3s m – p = 3(r + s) Hence 3|(m – p). Therefore, R is transitive

19 Matrix Representation of a Relation M R = [m ij ] (where i=row, j=col)  m ij ={1 iff (i,j)  R and 0 iff (i,j)  R} Eg: R : {1,2,3}  {1,2} where x > y R = {(2,1),(3,1),(3,2)}

20 Example Example: A = {1,2,3,4}, B = {w,x,y,z} R = {(1,x),(2,x),(3,y),(3,z)} = zero-one matrix

21 Graph Representation of a Relation Let A = {1,2,3,4} and R is relation on A where R = {(1,1),(1,2),(2,3),(3,2),(3,3),(3,4),(4,2)}

22 Union, Intersection, Difference and Composition of Relations R: A  B and S: A  B R: A  B and S: B  C

23 Composition of Relations Example: A = {1,2,3,4}, B = {w,x,y,z}, C = {5,6,7}, R 1 : A  B = {(1,x),(2,x),(3,y),(3,z)} and R 2 : B  C = {(w,5),(x,6)} Therefore, R 2 o R 1 = {(1,6),(2,6)}

24 Equivalence Relations Any binary relation that is: Reflexive Symmetric Transitive Eg: A = {1,2,3}, R = {(1,1),(2,2),(2,3),(3,2),(3,3)} R is reflexive, symmetric and transitive Therefore, R is an equivalence relation Recall example earlier: Congruence modulo 3 is an equivalence

25 Antisymmetry Relations Let R be a relation on a set A. R is antisymmetric iff for all a and b in A, if a R b and b R a then a = b In other words, a relation is antisymmetric iff there are no pairs of distinct elements a and b with a related to b and b related to a Eg: Let A = {0,1,2} R 1 = {(0,2),(1,2),(2,0)}. R 1 is not antisymmetric R 2 = {(0,0),(0,1),(0,2),(1,1),(1,2)}. R 2 is antisymmetric

26 Example (1) Let A = {1,2,3}, R = {(1,2),(2,1),(2,3)} R is not symmetric, (3,2)  R R is also not antisymmetric because (1,2),(2,1)  R Let A = {1,2,3}, S = {(1,1),(2,2)} S is both symmetric and antisymmetric

27 Example (2) Let R 1 be the divides relation (a|b) on Z + Is R 1 antisymmetric? Prove or give counterexample If a R 1 b and b R 1 a, then a = b a R 1 b means a|b → b = k 1 a b R 1 a means b|a → a = k 2 b It follows that, b = k 1 a = k 1 (k 2 b) = (k 1 k 2 )b Thus, k 1 = k 2 = 1. Hence a = b  R 1 antisymmetric

28 Example (2) Let R 2 be the divides relation on Z Is R 2 antisymmetric? Prove or give counterexample. Let a = 2, b = -2 Hence, a|b (-2 = -1(2)) and b|a (2 = -1(-2)) but a  b  R 2 is not antisymmetric

29 Partial Order Relations Any binary relation that is: Reflexive Antisymmetric Transitive Partial Order Set (POSET) (S,R) = R is a partial order relation on set S Examples: (Z,  ) (Z +,|){note: | symbolizes divides}

30 Quiz 5 Exercise 10.2 (No. 1, 2, 16) Exercise 10.5 (No. 2, 5) Send your answers in the next class!