5-M131-Chapter8-Suha AlShaikh

M131-Chapter8 By Ms. Suha Al-Shaikh-Dammam

1

Binary relation

Let A, B be any two sets.

A binary relation R from A to B, written R:A↔B, is a subset of A×B.

The notation a R b or aRb means (a,b)R.

The notation a R b or aRb means (a,b) R.

If aRb we may say “a is related to b (by relation R)”, or “a relates to b (under

relation R)”.

:Example

Let R: A ↔ B

A = {1, 2, 3} represents students

B = {a, b} represents courses

A × B = { (1,a), (1,b), (2,a), (2,b), (3,a), (3,b)}

R = { (1,a), (1,b) } it means that student 1 registered in courses a and b

Roaster Notation: List of ordered pairsA.

Ex: Let A be the set {1,2,3,4}, which ordered pairs are in the relation R={(a,b)| a

divides b}?

R= {(1,1),(2,2),(3,3),(4,4),(1,3),(1,2),(1,4),(2,4)}

Set builder notation B.

Ex: R={(a,b)| a divides b}

Graph C.

R1 = {(1,1),(2,2),(3,3),(4,4),(1,3),(1,2),(1,4),(2,4)}

8.1 Relations and their properties

be represented by:Relations can


2

Table D.

R = {(1,1),(2,2),(3,3),(4,4),(1,3),(1,2),(1,4),(2,4)}

Relations on a Set

A (binary) relation from a set A to itself is called a relation on the set A.

of natural numbers. Nthe set on” relation defined as a relation <The “ :Example

let < : N↔N :≡ {(a,b) | a < b}

If (a,b)R then a < b means (a,b) <

ex: (1,2) <

* The identity relation IA on a set A is the set {(a,a)|aA}.

Example:

A = {1,2,3,4}

IA = {(1,1),(2,2),(3,3),(4,4)}

Example:

R1= {(a,b)| a ≤ b}

R2= {(a,b)| a =b or a=-b}

R3= {(a,b)| a+b ≤ 3}

Which of these relations contain each of the following pairs (1,1),(1,2),(1,-1)?

(1,1) is in R1,R2,R3

(1,2) is in R1,R3

(1,-1) is in R2,R3


3

Question:

How many relations are there on a set with n elements?

1. A relation on set A is a subset from A x A.

2. A has n elements so A x A has n2 elements.

3. Number of subsets for n2 elements is

2

2n

Thus there are 2

2n relations on a set with n elements.

Example

S ={a, b, c}

There are :

Relations

Reflexivity & Irreflexivity1.

A. a R for every element a,a)if ( eflexiveris Aon RA relation

: Consider the following relations on {1,2,3,4}Ex

1. R1={(1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4}

Not Reflexive

2. R2={(1,1),(2,1),(2,2),(3,3),(3,4),(4,4)}

Reflexive

3. R3 = {(a, b) | a ≤ b} Reflexive

R A, (a, a) a if for every element lexiveirref A relation R on A is

Note: “irreflexive” ≠ “not reflexive”

Example:

A= {1, 2}

R = {(1, 2), (2, 1), (1, 1)}

Not Reflexive because (2, 2) R

Not irrflexive because (1,1) R

51222 932

Properties of Relations


4

Example:

Not Reflexive and Not Irreflexive

:Example

2. Symmetry & Antisymmetry

:if symmetricis Aon RA binary relation *

(a,b)R ↔ (b,a)R. where a, b A

:if antisymmetricon A is RA binary relation *

(a,b)R → (b,a)R.

also: (a,b)R (b,a) R →(a=b)

:Example

Let A = {1,2,3}


5

Example:

Consider these relations on the set of integers:

R1={(a, b) | a=b}

symmetric , antisymmetric

R2={(a, b) | a>b}

not symmetric, antisymmetric

R3={(a, b) | a=b+1}

not symmetric, antisymmetric

Transitivity3.

A relation R is transitive iff (for all a,b,c)

(a,b)R (b,c)R → (a,c)R.

A= {1,2} :Examples

R1={(1,1),(1,2),(2,1),(2,2)} transitive

R2={(1,1),(1,2),(2,1)} not transitive, (2,2) R2

R3 ={(3,4)} transitive

Special cases

Empty set {}

irreflexive, transitive, symmetric, antisymmetric

U universal set.

Reflexive, transitive, symmetric

Combining Relations:

Let A={1,2,3} , B={1,2,3,4}

R1={(1,1),(2,2),(3,3)}

R2={(1,1),(1,2),(1,3),(1,4)}

R1 R2 = {(1,1),(2,2),(3,3), (1,2),(1,3),(1,4)}

R1 R2= {(1,1)}

R1- R2={(2,2),(3,3)}


6

:Composite Relations

If R1 a relation from a set A to a set B, and R2 is a relation from set B to set C, the

R2◦ R1 is a set from A to C.

If (a,c) is in R1 and (c,b) is in R2 then (a,b) is in R2◦ R1 and

: Ex

R is the relation from {1,2,3} to {1,2,3,4}

R={(1,1), (1,4),(2,3),(3,1),(3,4)}

S is the relation from {1,2,3,4} to {0,1,2}

S = {{1,0), (2,0),(3,1),(3,2),(4,1)}

S◦R = {(1,0),(1,1),(2,1),(2,2),(3,0),(3,1)}, a relation from {1,2,3} to {0,1,2}

Power

Let R be a relation on the set A.

the power Rn, n=1,2,3… are defined by

R1 =R and R

n = R

n-1 ◦ R

: Let R ={(1,1),(2,1),(3,2),(4,3)}. Ex

Find:

R2 = {(1,1),(2,1),(3,1),(4,2)}

R3 = R

2 ◦ R= {(1,1),(2,1),(3,1),(4,1)}

The Inverse Relation :

Let R be a Relation from a set A to a set B. The inverse relation from B to A denoted

by: R-1

= {(b,a)| (a,b) R}.

The Complementary Relation:

}),(|),{( RbabaR

A 1R B 2R C

CRR 12 A


7

Some special ways to represent binary relations:

*With a zero-one matrix.

*With a directed graph.

To represent a relation R by a matrix

MR = [mij], let mij = 1 if (ai,bj)R, else 0.

E.g., Joe likes Susan and Mary, Fred likes Mary, and Mark likes Sally.

The 0-1 matrix representation of that “Likes”relation:

: Example

A={1,2,3} , B={1,2} , R: A↔B such that:

R= {(2,1),(3,1),(3,2)} then the matrix for R is:

8.3 Representing Relations

One Matrices-Using Zero

1 00

010

0 1 1

Mark

Fred

Joe

SallyMarySusan

11

01

00

RM


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One Reflexive, Symmetric-Zero

.ymmetric and antisymmetric,sReflexive, irreflexive-non ,Reflexive :Terms

These relation characteristics are very easy to recognize by inspection of the zero-

one matrix.

all the elements on the main diagonal of1) The relation R is reflexive if : Remark

Are equal to 1 (note that is a square matrix)

(2) The relation R is symmetric if and only if jiij mm for all pairs of integers i

and j with i = 1,2,…..,n and j = 1,2,…….n

(3)The relation R is symmetric if and only if : t

RR MM )(

(4) The relation R is anti symmetric if and only if (a,b)R and (b,a) R → a=b. The

matrix of anti symmetric relation has the property that if

Example:

Is R reflexive, symmetric, antisymmetric?

Reflexive, symmetric, not antisymmetric

Operations

1) Union and the Intersection

The Boolean Operations join and meat can be used to find the matrices

representing the union and the intersection of two relations

Then:

110

111

011

RM

2121

2121

RRRR

RRRR

MMM

MMM

RM

RM

01 jiij mthenjim


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Example:

Suppose R1 and R2:relations on set A are represented by the matrices:

Then :

2) Composite

Suppose that R: A ↔ B, S: B ↔ C

(Boolean Product)

Example:

Let

Find the matrix of

3)Power

• R2 =R ◦ R = MR

[2]

• R3= R

2◦ R = MR

[3]

001

110

101

010

001

101

21 RR MandM

000

000

101

011

111

101

2121

2121

RRRR

RRRR

MMM

MMM

SRRS MMM

101

100

010

000

011

101

SR MandM

RS

000

110

111

RSM

][n

RRMM n


11

Example

Find the matrix that represent R2

where the matrix representing R is:

A directed graph or digraph G=(VG,EG) is a set VG of vertices (nodes) with a set

EGVG×VG of edges (arcs,links). Visually represented using dots for nodes, and

arrows for edges. Notice that a relation R:A↔B can be represented as a graph

GR=(VG=AB, EG=R).

Example:

List the vertices and the edges.

Answer: Vertices are a,b,c, and d

Edges are (a,b), (b,b), (c,b), (a,d), (b,d), (d,b), and (c,a).

Example: Draw a diagraph of the relation

R={(1,1),(1,3),(2,1),(2,3),(2,4),(3,1),(3,2),(4,1)} on the set {1,2,3,4}

Solution:

010

111

110

001

110

010

2RR MthenM

1 00

010

0 1 1

Mark

Fred

Joe

SallyMarySusan

Note that : an edge from (a,a) represented using an arc from the

vertex a back to it self. Such an edge is called a loop

Using Directed Graphs


11

Digraph Reflexive, Symmetric

It is extremely easy to recognize the reflexive/irreflexive/ symmetric/antisymmetric

properties by graph inspection.

Example: Determine whether the relation for the following directed graph are

reflexive, symmetric , anti symmetric, and or transitive.

Solution: 1) because there are loops at every vertex R is reflexive.

(a,a), (b,b), (c,c) R

2)It is not symmetric because there are (a,b) but not (b,a).

3)R is not anti symmetric, because (b,c) and (c,b) in R.

4)R is not transitive because (a,b), and (b,c) belongs to S, but (a,c) doesn’t belong.


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Definition: For any property X, the “X closure” of a set A is defined as the “smallest”

superset of A that has the given property.

Example

The relation R={(1,1),(1,2),(2,1),(3,2)} on the set A ={1,2,3} is not reflexive.

How can you produce a reflexive relation containing R that is as small as possible?

Answer: By adding (2,2) and (3,3) so :

Reflexive closure of R is:

{(1,1),(1,2),(2,1),(3,2),(2,2),(3,3)}

Example

What is the reflexive closure of the relation:

R={(a ,b)| a < b} on the set of integers?

Answer:

The Reflexive Closure of R is:

{(a ,b)| a < b} {(a,a)| a Z} = {(a,b) | a ≤ b}

The reflexive closure of a relation R on A is obtained by adding (a,a) to R for each

aA not already in R

I.e., it is R IA

Closures of Relations


13

Example

The relation

{(1,1),(2,2),(1,2),(3,1),(2,3),(3,2)} on the set {1,2,3} is not symmetric.

How can we produce a symmetric relation that is as small as possible and contains R?

Answer: by adding (2,1) and (1,3) so theSymmetric Closure of R is:

{(1,1),(2,2),(1,2),(3,1),(2,3),(3,2),(2,1),(1,3)}

Example

What is the symmetric closure of the relation:

R={(a ,b)| a > b} on the set of positive integers?

Answer:

The Symmetric Closure of R is:

{(a ,b)| a >b} {(b, a)| a > b} = {(a, b) | a ≠ b}

Example

R={(1,1), (1,2), (2,1),(3,2)} on the set A= {1, 2, 3}

• R* = R R2 R

3

• R2 = R ◦ R = {(1,1),(1,2),(2,2),(3,1)}

• R3 = R

2o R = {(1,1),(1,2),(2,1),(2,2),(3,2)}

• R* = {(1,1),(1,2),(2,1),(3,2),(2,2),(3,1)}

The transitive closure or connectivity relation of R is obtained by repeatedly

adding (a, c) to R for each (a, b), (b, c) in R.

i.e., it is

Or in term of zero-one matrices:

MR*=MR MR[2]…………MR[n]

The symmetric closure of R is obtained by adding (b,a) to R for each (a,b) in R.

i.e., it is R R−1

Zn

nRR*


14

Example:

Find MR* for

Questions: 1) let R be the relation on the set {0,1,2,3} containing the pairs

(0,1),(1,1),(1,2),(2,0),(2,2) and (3,0). Find the

a) Reflexive closure of R

b) Symmetric closure of R

Answer: a) reflexive closure of R={(0,0),(0,1),(1,1),(1,2),(2,0),(2,2),(3,0),(3,3)}

b) Symmetric closure of R={(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2),(3,0)}

_____________________________________________________________________

2) Let R be the relation {(a,b)| a divides b} on the set of integers. What is the

symmetric closure of R?

Solution: Symm. Closure of R={(a,b)| a divides b or b divides A}

_____________________________________________________________________

3) Draw the directed graph of the reflexive closure and the symm. Closure of the

relation with the directed graph shown

Solution:

111

010

111

011

010

101

]3[]2[* RRRR

R

MMMM

M


15

An equivalence relation on a set A is simply any binary relation on A that is reflexive,

symmetric, and transitive.

Example:

A relation on the set of real numbers such that

R={(a, b) | a-b is an integer}

Answer:

R is reflexive since (a,a) is an integer where a-a=0

R is symmetric since a-b and b-a is an integer

R is transitive since for (a,b) then a-b integer and for (b,c) then b-c integer ,

Therefore( a-b )+( b-c) =a-c is also an integer so (a,c) in R

so R is an Equivalence Relation.

Example:

Let R: Z Z, show that R = {(a,b)| a ≡b(mod m)}, m>1} is an equivalence

relation.

Answer:

a ≡b(mod m) iff m divides a-b

*R is reflexive, a ≡a(mod m) , m divides a-a=0

*R is symmetric: suppose a ≡b(mod m) then a-b is divisible by m so that a-b = km

It follows: b-a = (-k)m

So that : b ≡a(mod m)

*R is transitive: Suppose

a ≡b (mod m) and b ≡c ( mod m)

Then: m divides both a-b and b-c

a - b=km and b - c=lm

a-c=(a-b)+(b-c)=km+lm=(k+l)m

Thus: a ≡c ( mod m) So R is Equivalence

Equivalence Relations


16

Read example 4 and 5 page 557

Example: show that “divides” relation on the set of positive integers is not an

equivalence relation.

Solution:

reflexive: since a|a for all positive integers →aRa

Symmetric: since 2|4 but 4×2 so R is not symmetric .

so R is not an equivalence relation

Example: let R be the relation on the set of real numbers such that x Ry if and only if

x and y are real numbers that differ by less than 1. That is |x-y|<1. Show that R is not

an equivalence relation.

Solution:

reflexive since |x-x|=0<1 where ever xR Ris reflexive.

Symmetric: suppose xRy→|x-y|<1 but

|x-y|=|y-x|<1→yRx so R is symmetric.

Suppose xRy and yRz

So R is not transitive.

Equivalence classes:

Definition: let R be an equivalence relation on a set A. The set of all elements that are

related to an element a of A is called the equivalence class of a (denoted by [a]R ).

Do example8.

222

2211

11

11

zxzx

zyyxaddzy

yx

zyandyx

Rsasa R ),(|

Rab

Remark: in other words, if R is an equivalence relation on a set A, the equivalence

class of the element a is

If , then b is called a representative of this equivalence class.


17

Example: what are the equivalence classes of 0 and 1 for congruence modulo 4?

Solution:

*The equivalence class of 0 contains all integers a such that )4(modoa . The

integers in this class are divisible by 4. so

*The equivalence class of 1 contains all integers a such that )4(modoa

The integers in this class are those that have a remainder 1 when divided by 4.

So

Remark: The congruence classes of an integer a modulo m is denoted by

Refer to last example. Find

,....8,4,0,4,8....,0 4

,....9,5,1,3,7....,1 4

,...2,,,,2......, mamaamamaa m

44 32 and

,...11,7,3,1,5....,3

,...10,6,2,2,6....,2

4

4


18

Definition: A relation R on a set S is called a partial ordering or partial order if it is

reflexive, anti symmetric and transitive.

A set S together with R is called a partially ordered set, or poset and is denoted by

(S,R), members of S are called elements of the poset.

Example1: (page566)

Show that the “greater than or equal” relation (≥)is the partial ordering on the set of

integers.

Solution:

1) reflexive: because a≥a for every integer a≥ is reflexive.

2) Anti symmetric: if a ≥ b and b ≥ a, then a=b. Hence ≥ is anti symmetric.

3) transitive: if a ≥ b and b ≥ c then a ≥ b ≥ c. Hence a ≥ c so ≥ is transitive. It follows

that ≥ is a partial ordering on the set of integers and (Z, ≥) is a poset.

Example: Show that |),( Z is a poset. Where Z+ is the set of positive integers,

and ( | is the devision)

Solution: 1) reflexive: since a|a for every positive integer a. | is reflexive.

2) Anti symmetric: if a|b and b|a then a=b. Hence | is anti symmetric.

3) transitive: Suppose a|b and b|c then b=ka and c=lb

C=lb=l(ka)=(lk)a→a|c so | is transitive

It follows that | is a partial ordering on the set of positive integers and and |),( Z

is a poset.

Example: show that the inclusion relation is a partial ordering on the power set of a

set S.

Solution:* reflexive because A A whenever A is a subset of S, is reflexive.

*Anti symmetric: because if A B and B A this implies that A=B.

*Transitive: suppose A B and B C then A B C→A C Hence is

transitive

Therefore is a partial ordering on the power set of a set S; and (p(s), ) is a poset.

8.6: Partial orderings


19

Example: page 567

Let R be the relation on the set of people such that xRy if x and y are people and x is

older than y. Show that R is not a partial ordering.

Solution: R is not reflexive because no person is older than himself or herself. That is

xRx for all people x it follows that R is not partial ordering.

Definition: page 567

The elements a and b of a poset (S, ) are called comparable if either a b or b a.

Example: In the poset |),( Z are the integers 3 and 9 comparable? Are 5 and 7

comparable?

Solution: because 3|9, the integers 3 and 9 are comparable. Because 5 | 7, the integers

5 and 7 are in comparable.

Definition: if is a poset and every 2 elements of S are comparable, S is called a

totally ordered or linearly ordered set, and is called a total order or a linear order.

A totally ordered set is also called a chain.

Example:page 568:

the poset (Z,≤) is totally ordered, because a≤b or b≤a whenever a and b are integers.

Example: the poset |),( Z is not totally ordered because it contains elements that are

in comparable, such as 5 and 7.

Lexicographic order: consider the 2 posets . The lexicographic

ordering ≤ on A1×A2 is defined by:

Remark: when every two elements in the set are comparable, the relation is called

a total ordering.

,S

2211 ,, AandA

22211111

21212121 ,,,,

baandbabothiforbaifeither

bbaaisthatbbthanlessisaa

Remark: The notation a is used to denote that (a,b)R in an arbitrary poset

(S,R).

is used to denote the relation in any poset


21

Example: determine whether (3,5) (4,8) whether (3,8) (4,5) and

whether (4,9) (4,11) in the poset where is the lexicographic ordering

constructed from the usual ≤ relation on Z.

Solution: because 3<4 it follows that (3,5) (4,8) and that (3,8) (4,5), we have

(4,9) (4,11), because the first entries of (4,9) and (4,11) are the same but 9<11.

Examplepage 569:

note that (1,2,3,5) (1,2,4,3), because the entries in the first 2 positions of these 4-

tuples agree, but in the third position the entry in the first 4-tuples,3 is less than 4 (

here the ordering on 4-tuples is the lexicographic ordering that comes from the usual

“less than or equals” relation on the set of integers).

Hasse diagrams: (finite set)

The Hasse Diagram for the partial ordering relation is obtained from the associated

diagrams by deleting all the loops and all the edges that occur from transitivity

1) Starts with the directed graphs for the given relation (partial ordering set)

2) Remove the loops (since it is reflexive)

3) Remove the edges that must be in-the partial ordering because of the presence

of other edges and transitivity. E.g if (a,b) and (b,c) are in the partial ordering

remove (a,c)

4) Finally arrange each edge so that its initial vertex is below its terminal vertex.

Example: Draw the Hasse Diagram representing the partial ordering {(a,b):a≤b} on

the set {1,2,3,4}

Solution: R{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}


21

Example: Draw the Hasse Diagram representing the partial ordering {(a,b):a|b} on

{1,2,3,4,6,8,12}

Answer:

R={(1,1),(1,2),(1,3),(1,4),(1,6),(1,12),(2,2),(2,4),(2,6),(2,8),(2,12),(3,3),(3,6),(3,12),

(4,4),(4,8),(4,12),(6,6),(6,12),(8,8)(12,12)}

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