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Discrete Structures Chapter 5 RelationsPowerPoint Presentation

Discrete Structures Chapter 5 Relations

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

Outline

- Properties of relations
- Matrix and graph representation of relations
- Equivalence relations
- Partial order relations

- 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 xA and yB

- Binary relations: x R y
On sets xX, yY R X Y

- “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
- AB={(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)} AB
- Or use Arrow Diagrams

- Reflexive: Let R be a binary relation on a set A
- Eg: Let A = {1,2,3,4}
- R1 ={(1,1),(1,2),(2,2),(2,3),(3,3),(4,4)}
- R1 is reflexive
- R2 = {(1,1),(2,2),(3,3)}
- R2 is not reflexive (why?)

- Symmetric: Let R be a binary relation on a set A
- Eg: Let A = {1,2,3}
- R1 ={(1,2),(2,1),(1,3),(3,1)}
- R1 is symmetric
- R2 = {(1,1),(2,2),(3,3),(2,3)}.
- R2 is not symmetric because (3,2) R2

- 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)

xRy or (x,y) R for all x and y in A,

R is reflexive for all x in A, (x,x) R

R is symmetric for all x and y in A, if (x,y) R then (y,x) R

R is transitive for all x, y and z in A, if (x,y) R and (y,z) R then (x,z) R

- 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

- 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

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 iffx 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

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

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?

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 < y then y > 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

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?

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

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

Matrix Representation of a Relation

- MR = [mij] (where i=row, j=col)
- mij={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)}

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)}

Union, Intersection, Difference and Composition of Relations

- R: AB and S: AB
- R: AB and S: BC

- Example: A = {1,2,3,4}, B = {w,x,y,z}, C = {5,6,7},
- R1: AB = {(1,x),(2,x),(3,y),(3,z)} and
- R2: BC = {(w,5),(x,6)}
- Therefore, R2o R1 = {(1,6),(2,6)}

- 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

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}
- R1= {(0,2),(1,2),(2,0)}. R1 is not antisymmetric
- R2 = {(0,0),(0,1),(0,2),(1,1),(1,2)}. R2 is antisymmetric

- 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

- Let R1 be the divides relation (a|b) on Z+
- Is R1 antisymmetric? Prove or give counterexample
- If a R1 b and b R1 a, then a = b
- a R1 b means a|b → b = k1a
- b R1 a means b|a → a = k2b
- It follows that, b = k1a = k1(k2b) = (k1k2)b
- Thus, k1 = k2 = 1. Hence a = b
- R1 antisymmetric

- Let R2 be the divides relation on Z
- Is R2 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
- R2 is not antisymmetric

- 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}

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