Discrete StructuresChapter 5Relations

Nurul Amelina NasharuddinMultimedia Department

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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

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• 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 yOn sets xX, yY R X Y

Recall: Relations

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Example

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

• Use traditional notation0 < 1, 0 < 2, 0 < 3, 1 < 2, 1 < 3, 2 < 3

• Or use set notationAB={(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

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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?)

Properties of Relations

A, xRxx R Reflexiveis

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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

Properties of Relations

yRxA, xRyyx R , Symmetricis

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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

xRzyRzA, xRyzyx R ,, Transitiveis

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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

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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 RReflexive? NoSymmetric? YesTransitive? Yes

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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 RReflexive? Yes.Symmetric? Yes.Transitive? No. (1,0),(0,3) R but (1,3) R

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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

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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

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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?

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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

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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?

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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 kn – m = - (m – n) = 3(-k)Hence 3|(n – m). Therefore R is symmetric

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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 rn – 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

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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)}

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110

321

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RM

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Example• Example:

A = {1,2,3,4}, B = {w,x,y,z} R = {(1,x),(2,x),(3,y),(3,z)}

00004

11003

00102

00101

zyxw

M R = zero-one matrix

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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)}

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Union, Intersection, Difference and Composition of Relations

• R: AB and S: AB

• R: AB and S: BC

}),(),(|),{( SyxRyxBAyxSR }),(),(|),{( SyxRyxBAyxSR }),(),(|),{( SyxRyxBAyxSR }),(),(|),{( RyxSyxBAyxRS

}),(),(,|),{( ScbRbaBbCAcaRS

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Composition of Relations

• 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)} andR2: BC = {(w,5),(x,6)}

Therefore, R2 o R1 = {(1,6),(2,6)}

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Equivalence Relations

• Any binary relation that is:ReflexiveSymmetricTransitive

• Eg: A = {1,2,3}, R = {(1,1),(2,2),(2,3),(3,2),(3,3)}R is reflexive, symmetric and transitiveTherefore, R is an equivalence relation

• Recall example earlier: Congruence modulo 3 is an equivalence

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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 antisymmetricR2 = {(0,0),(0,1),(0,2),(1,1),(1,2)}. R2 is antisymmetric

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Example (1)

• Let A = {1,2,3}, R = {(1,2),(2,1),(2,3)}

R is not symmetric, (3,2) RR 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

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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 = ba R1 b means a|b → b = k1ab R1 a means b|a → a = k2bIt follows that, b = k1a = k1(k2b) =

(k1k2)bThus, k1 = k2 = 1. Hence a = b

R1 antisymmetric

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Example (2)

• Let R2 be the divides relation on Z

• Is R2 antisymmetric? Prove or give counterexample.

Let a = 2, b = -2Hence, a|b (-2 = -1(2)) and b|a (2 = -1(-2)) but a b

R2 is not antisymmetric

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Partial Order Relations

• Any binary relation that is:ReflexiveAntisymmetricTransitive

• 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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Quiz 5

• Exercise 10.2 (No. 1, 2, 16)• Exercise 10.5 (No. 2, 5)

Send your answers in the next class!