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Discrete Structures Chapter 5 Relations. Nurul Amelina Nasharuddin Multimedia Department. Objectives. On completion of this topic, student should be able to: Determine the properties of relations – reflexive, symmetric, transitive, and antisymmetric

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Discrete Structures Chapter 5 Relations

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Discrete structures chapter 5 relations

Discrete StructuresChapter 5

Relations

Nurul Amelina Nasharuddin

Multimedia Department


Discrete structures chapter 5 relations

Objectives

  • 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

Outline

  • Properties of relations

  • Matrix and graph representation of relations

  • Equivalence relations

  • Partial order relations


Discrete structures chapter 5 relations

Recall: 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 xA and yB

  • Binary relations: x R y

    On sets xX, yY R  X  Y


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

Properties of Relations

  • 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?)


Discrete structures chapter 5 relations

Properties of Relations

  • 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


Discrete structures chapter 5 relations

Properties of Relations

  • 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)


Discrete structures chapter 5 relations

Properties of Relations

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


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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?


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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?


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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)}


Discrete structures chapter 5 relations

Example

  • Example:

    • A = {1,2,3,4}, B = {w,x,y,z}

    • R = {(1,x),(2,x),(3,y),(3,z)}

= zero-one matrix


Discrete structures chapter 5 relations

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)}


Discrete structures chapter 5 relations

Union, Intersection, Difference and Composition of Relations

  • R: AB and S: AB

  • R: AB and S: BC


Discrete structures chapter 5 relations

Composition of Relations

  • Example: A = {1,2,3,4}, B = {w,x,y,z}, C = {5,6,7},

  • R1: AB = {(1,x),(2,x),(3,y),(3,z)} and

  • R2: BC = {(w,5),(x,6)}

  • Therefore, R2o R1 = {(1,6),(2,6)}


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

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


Discrete structures chapter 5 relations

Example (2)

  • 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


Discrete structures chapter 5 relations

Example (2)

  • 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


Discrete structures chapter 5 relations

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}


Discrete structures chapter 5 relations

Quiz 5

  • Exercise 10.2 (No. 1, 2, 16)

  • Exercise 10.5 (No. 2, 5)

Send your answers in the next class!


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