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# Discrete Structures Chapter 5 Relations - PowerPoint PPT Presentation

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 StructuresChapter 5

Relations

Nurul Amelina Nasharuddin

Multimedia Department

• 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

• 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 xA and yB

• Binary relations: x R y

On sets xX, yY R  X  Y

• “less than” relation from A={0,1,2} to B={1,2,3}

• 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

• 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 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

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

• 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)}

• Example:

• A = {1,2,3,4}, B = {w,x,y,z}

• R = {(1,x),(2,x),(3,y),(3,z)}

= zero-one matrix

• 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)}

• R: AB and S: AB

• R: AB and S: BC

• 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)}

• 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}

• Exercise 10.2 (No. 1, 2, 16)

• Exercise 10.5 (No. 2, 5)