1 RELATIONS Learning outcomes Students are able to: a. determine the properties of relations – reflexive, symmetric, transitive, and antisymmetric b. determine.

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RELATIONS Learning outcomes Students are 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.

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Contents Properties of relations Matrix and graph representation of

relations Equivalence relations Partial order relations

3April 21, 2023 Relations 3

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


Relations Binary relations: xRy

On sets xX yY R X Y 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 notationA 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)} Or use Arrow Diagrams


Formal Definition (Binary) relation from A to B

where xA, yB, (x,y)AB and R ABxRy (x,y)R

Finite example: A={1,2}, B={1,2,3} Infinite example: A = Z (set of

integers) and B = Z

aRb a-bZeven


Example Let A = {2,3}, B = {1,3,6}

Define a relation R from A to B such that:xRy x – y is odd

How could this explicitly be represented as tuples?– R = {(2,1),(2,3),(3,6)}

What if A and B were the set of all integers?– R = {(x,y)ZZ | kZ such that x – y = 2k +

1}


Properties of Relations Reflexive

Symmetric

Transitive

A, xRxx R Reflexiveis

yRxA, xRyyx R , Symmetricis

xRzyRzA, xRyzyx R ,, Transitiveis

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Example 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.

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Example 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) ε R2 .

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Example (transitive) 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)


Example 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)}

Draw the arrow diagram Is R

Reflexive? No. Symmetric? Yes Transitive? Yes


Example 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) ε R but (1,3) ε

R

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Question

?????

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Proving Properties on Infinite Sets -“less than” relation

Define a relation R on R (the set of all real numbers):

For all x, y ε R, x R y ↔ x < y Is R reflexive? symmetric? transitive? R is reflexive iff x ε R, x R x. By

definition of R, this means x < x, for all x ε R. But this is false. Hence, R is not reflexive.

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“less than” relation R is not symmetric. R is symmetric iff

x,y ε R, if x R y then y R x. By definition of R. this means that x,y ε R, if x < y then y > x. But this is false.

R is transitive.


Congruence Modulo 3 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

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 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).

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Properties of Congruence modulo 3

Is R transitive? It is necessary to show that For all m,n ε Z, if m R n and n R p then m R p. m – n = 3k for some k. n – p = 3l for some l. m – p = (m – n) + (n – p) = 3k + 3l = 3(k + l). Hence 3|(m – p). Therefore, R is transitive.


Exercise Define R(x,y) R: Z+ Z+ to be

{(x,y) Z+Z+ | x|y}

Prove whether or not this is:– Reflexive?– Symmetric?– Transitive?


Matrix Representation of a Relation

MR = [mij]

– mij={1 iff (i,j)R and 0 iff (i,j)R}

Example:– R : {1,2,3} {1,2} R = {(2,1),(3,1),

(3,2)}

100

110

321

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RM

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Example

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Graph Representation of a Relation


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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Compositions of Relations


Example - application Let ID = set of student IDs Let Course = set of courses offered Define relation Summer2007

{(x,y)IDCourse | student x is registered for course y}

Can we do this? Relational databases …

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DatabasesID Course

12345 Structure Discrete

45678 Java Programming

… …


Equivalence Relations Any binary relation that is:

– Reflexive– Symmetric– Transitive

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Antisymmetric relation 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.

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 symmetric.


Example Let R1 be the divides relation on Z+

Let R2 be the divides relation on Z

Is R1 antisymmetric? Prove or give counterexample.– a R1 b and b R1 a a = b

– True

Is R2 antisymmetric? Prove or give counterexample.– Counterexample (a = 2, b = -2)


Partial Order Relation R is a Partial Order Relation if and only if

– R is Reflexive, Antisymmetric and Transitive Partial Order Set (POSET)

(S,R) = R is a partial order relation on set S Examples

– (Z, ) – (Z+,|) {note: | symbolizes divides}– (S, ) {note: S indicates and set}

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End

Thank you

1 RELATIONS Learning outcomes Students are able to: a. determine the properties of relations – reflexive, symmetric, transitive, and antisymmetric b. determine.

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