chapter 8: relations. 8.1 relations and their properties binary relations: let a and b be any two...

Chapter 8: Relations

8.1 Relations and Their Properties

Binary relations:• Let A and B be any two sets.

• A binary relation R from A to B, written R : A B,

is a subset of the Cartesian product A×B.

• The notation a R b means (a, b) R.

• The notation a R b means (a, b) R.• If a R b we may say that a is related to b (by

relation R ), or a relates to b (under relation R ).

Example

Let R : A B, and 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)}.

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

Relations can be Represented by:

Let A be the set {1, 2, 3, 4} for which ordered pairs are in the relation

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

A- Roaster Notation: List of ordered pairs:

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

B- Set builder notation: R = {(a, b) : a divides b}

C- Graph:R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (1, 4), (2, 4)}

1 1

2 2

3 3

4 4

1 2 3 4OR

Relations can be Represented by:

Domain of R

D- Table:R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (1, 4), (2, 4)}

R 1 2 3 4

1 × × × ×

2 × ×

3 ×

4 ×

Relations can be Represented by:

Relations on a Set

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

e.g. The “<” relation defined as a relation on the set N of natural numbers:

let < : N N :≡ {(a, b) | a < b }If (a, b) R then a < b means (a, b) <

e.g. (1, 2) < .

Examples: Consider the relations on A = {-1, 0, 1, 2},

R1 = {(a, b) | a < b}R2 = {(a, b) | a = b or a = -b}

R3 = {(a, b) | 0 ≤ a + b ≤ 1}

R1 = {(-1,0), (-1,1), (-1,2), (0,1), (0,2), (1,2)}

R2 = {(-1,-1), (0,0), (1,1), (2,2), (-1,1), (1,-1)}R3 = {(-1,1), (1,-1), (-1,2), (2,-1), (0,0), (0,1), (1,0)}

Relations on a Set

Relations on a Set

• The identity relation IA on a set A is the set

{(a, a) | a A}.

e.g. If A = {1, 2, 3, 4},

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

Question

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

Answer:1. A relation on set A is a subset from A×A. 2. A has n elements so A×A has n2 elements.3. Number of subsets for n2 elements is , thus

there are relations on a set with n elements.

e.g. If S = {a, b, c}, there are relations.

2

2n2

2n

51222 932

Properties of Relations1. Reflexivity and Irreflexivity

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

e.g. Consider the following relations on {1, 2, 3, 4}R1 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 4), (4, 1), (4, 4)},

Not Reflexive.R2 = {(1, 1), (2, 1), (2, 2), (3, 3), (3, 4), (4, 4)},

Reflexive.R3 = {(a, b) | a ≤ b},

Reflexive.

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

Note: “irreflexive” ≠ “not reflexive”.

e.g. If A = {1, 2}, R = {(1, 2), (2, 1), (1, 1)} is not reflexive because (2, 2) R, not irrflexive because (1, 1) R.

Reflexivity and Irreflexivity

1 2

3 4Not Reflexive and Not Irreflexive

Example

1 2 1 2 3 4 3 4 Irreflexive Reflexive

Examples

2. Symmetry and Antisymmetry

• A binary relation R on A is symmetric if (a, b) R ↔ (b, a) R, where a, b A.

• A binary relation R on A is antisymmetric if (a, b) R → (b, a) R.

That is, if (a, b) R (b, a) R → a = b.

Examples

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.

Examples

Let A = {1, 2, 3}.

R1 = {(1, 2), (2, 2), (3, 1), (1, 3)} Not reflexive, not irreflexive, not symmetric, not antisymmetric

R2 = {(2, 2), (1, 3), (3, 2)} Not reflexive, not irreflexive, not symmetric, antisymmetric

R3 = {(1, 1), (2, 2), (3, 3)} Reflexive, not irreflexive, symmetric, antisymmetric

R4 = {(2, 3)} Not reflexive, irreflexive, not symmetric, antisymmetric

3. Transitivity

• A relation R is said to be transitive if and only if (for all a, b, c),

(a, b) R (b, c) R → (a, c) R.

e.g. Let A = {1, 2}.

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

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

R3 = {(3, 4)} is transitive.

Special Cases

Empty set { } Irreflexive, transitive, symmetric, antisymmetric.

Universal set U 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)}, then

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

R1 R2 = {(1, 1)}

R1 − R2 = {(2, 2), (3, 3)}

Composite Relations

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

e.g. 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)}.

Powers

• Let R be a relation on the set A. the power Rn, n = 1, 2, 3, … are defined by

R1 = R and Rn = Rn -1 ◦ R

e.g. Let R = {(1, 1), (2, 1), (3, 2), (4, 3)}. FindR2 =R ◦ R = {(1, 1), (2, 1), (3, 1), (4, 2)}

R3 = R2 ◦ R = {(1, 1), (2, 1), (3, 1), (4, 1)}

8.3 Representing Relations

• Some special ways to represent binary relations:– With a zero-one matrix.– With a directed graph.

Using Zero-One Matrices

• To represent a relation R by a matrix MR = [mij ], let mij = 1 if (ai , bj) R, otherwise 0.

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

• The 0-1 matrix representation of that relation:

1 000100 1 1

MarkFredJoe

SallyMarySusan

Let 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: 1 2

1

2

3

Example

100

110

RM

100

110

Zero-One Reflexive, SymmetricThe terms: Reflexive, non-reflexive, irreflexive,

symmetric and antisymmetric.– These relation characteristics are very easy to

recognize by inspection of the zero-one matrix.

1 0 1 01 0 1 0 1 0 1

1 0 1

1 0 0 0

Reflexive:all 1’s on diagonal

Irreflexive:all 0’s on diagonal

Symmetric:all identical

across diagonal

Antisymmetric:all 1’s are across

from 0’s

any-thing

any-thing

any-thing

any-thing anything

anything

Example

Is R reflexive, symmetric, antisymmetric?

Reflexive, symmetric, not antisymmetric

110111011

RM

Operations

1- Union and the Intersection The Boolean Operations join and meet can

be used to find the matrices representing the union and the intersection of two relations

2121

2121

RRRR

RRRR

MMM

MMM

ExampleSuppose R1 and R2 are relations on a set A which are

represented by the matrices:

001110101

and010001101

21 RR MM

.000000101

,011111101

2121

2121

RRRR

RRRR

MMM

MMM

2- CompositeSuppose that R : A ↔ B, S : B ↔ C (Boolean Product)

SRRS MMM

Operations

Example

Let

Find the matrix of ?

RS

000110111

SRRS MMM

101100010

and000011101

SR MM

3- Power][n

RRMM n

Operations

.,

.,

]3[

23

]2[

2

3

2

RR

RR

MMRRR

MMRRR

Example

• Find the matrix that represents R2 where the matrix representing R is:

.010111110

then001110010

If ]2[2

RRR MMM

Using Directed Graphs

A directed graph or digraph G = (VG , EG) is a set VG of vertices (nodes) with a set EG VG ×VG of edges (arcs or 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 = A B, EG = R).

1 000100 1 1

MarkFredJoe

SallyMarySusanMR

GR

JoeFred

Mark

SusanMarySally

Node set VG (black dots)

Edge set EG

(blue arrows)

Digraph Reflexive, Symmetric

It is extremely easy to recognize the reflexive, irreflexive, symmetric, antisymmetric properties by graph inspection.

Reflexive:Every node

has a self-loop

Irreflexive:No node

links to itself

Symmetric:Every link isbidirectional

Antisymmetric:

No link isbidirectional

Not symmetric, non-antisymmetric Non-reflexive, non-irreflexive

8.4 Closures of Relations

• For any property X, the “X closure” of a set A is defined as the “smallest” superset of A that has the given property.

• The reflexive closure of a relation R on A is obtained by adding (a, a) to R for each a A not already in R .

i.e. It is R IA .

Example 1

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 the reflexive closure

of R is:

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

Example 2

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}.

Symmetric Closure

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 .

Example 1

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 the symmetric closure

of R is:{(1, 1), (2, 2), (1, 2), (3, 1), (2, 3), (3, 2), (2, 1), (1, 3)}.

Example 2

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}.

Transitive Closure

• 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:

Zn

nRR*

.][]2[* nRRRR MMMM

Example 1

e.g. R = {(1,1), (1, 2), (2,1), (3, 2)} on the set

A = {1, 2, 3}

• R* = R R2 R3

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

• R3 = R2 o R = {(1,1), (1, 2), (2,1), (2, 2), (3, 2)}

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

Example 2

• Find MR* for

111010111

011010101

]3[]2[* RRRR

R

MMMM

M

8.5 Equivalence Relations

An equivalence relation on a set A is simply any binary relation on A that is reflexive, symmetric, and transitive.

Example1: Show that a relation on the set of real numbersR = {(a, b) | a – b is an integer}.

is an equivalence relation.

R is reflexive: Since a – a = 0 is an integer a R a for all a.R is symmetric: If a R b then whenever a – b is an integer

b – a is an integer b R a.R is transitive: If a R b and b R c then a – b and b – c are

integers, therefore, a – b + b – c = a – c is also an integer, which means a R c.

Example 2: Let R : Z Z. Show that R = {(a, b) | a ≡ b (mod m)}, m > 1

is an equivalence relation.

Answer: a ≡ b (mod m) if and only if 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 m divides a – b, therefore m divides –(a – b) = b – a 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, therefore, m divides their sum (a – b) + (b – c) = a – c , thus a ≡ c (mod m).

Equivalence Relations

Example 3: Let R be a relation on Z defined as follows: a R b if and only if 5 | (a2 – b2).

Show that R is an equivalence relation. R is reflexive: a2 – a2 = 0 = 0×5 5 | (a2 – a2) a R a.

R is symmetric: If a R b 5 | (a2 – b2) a2 – b2 = m×5, m Z b2 – a2 = (-m)×5 5 | (b2 – a2) b R a.

R is transitive: Suppose that a R b 5 | (a2 – b2) a2 – b2 = m×5, m Z and b R c 5 | (b2 – c2) b2 – c2 = k×5, k Z . Now (a2 – b2) + (b2 – c2) = (a2 – c2) =(m + k)×5 5 | (a2 – c2) a R c.

Equivalence Relations

Equivalence Classes

• 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.

• The equivalence class of a with respect to R is denoted by [a]R . When only one relation is under consideration, we can delete the subscript R and write [a] for this equivalence class.

[a]R = {s | (a, s) R }.

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 a ≡ 0 (mod 4). The integers in this class are those divisible by 4. Hence, the equivalence class of 0 for this relation is

[0] = { . . . , -8, -4, 0, 4, 8 , . . . } . The equivalence class of 1 contains all the integers a

such that a ≡ 1 (mod 4). The integers in this class are those that have a remainder of 1 when divided by 4. Hence, the equivalence class of 1 for this relation is

[1] = { . . . , -7, -3, 1, 5, 9, . . . } .

Equivalence Classes

Equivalence Classes and Partitions

• A partition of a set S is a collection of disjoint nonempty subsets of S that have S as their union. In other words, the collection of subsets Ai , i I (where I is an index set) forms a partition of S if and only if

Example: What are the sets in the partition of the integers arising from congruence modulo 4?

Solution: There are four congruence classes, corresponding to [0]4, [1]4, [2]4, and [3]4 . They are the sets

[0]4 = { . . . , -8, -4, 0, 4, 8, . . . } ,

[1]4 = { . . . , -7, -3, 1, 5, 9, . . . } ,

[2]4 = { . . . , -6, -2, 2, 6, 10, . . . } ,

[3]4 = { . . . , -5, -1, 3, 7, 11, . . . } .

These congruence classes are disjoint, and every integer is in exactly one of them. In other words, these congruence classes form a partition.

Equivalence Classes and Partitions

8.6 Partial Orderings

• A relation R on a set S is called a partial ordering or partial order if it is reflexive, antisymmetric, and transitive.

• A set S together with a partial ordering R is called a partially ordered set, or poset, and is denoted by (S, R). Members of S are called elements of the poset.

Example 1

Show that the "greater than or equal" relation ( ) is a partial ordering on the set of integers.Solution:•a ≥ a for every integer a, ≥ is reflexive.•If a ≥ b and b ≥ a, then a = b. Hence, ≥ is antisymmetric. •Finally, ≥ is transitive because a ≥ b and b ≥ c imply that a ≥ c.It follows that ≥ is a partial ordering on the set of integers and (Z, ≥) is a poset.

Example 2

Show that the divisibility relation “|“ is a partial ordering on the set of positive integers Z+.Solution:•a | a for every positive integer a, | is reflexive.•If a | b and b | a, then a = b, | is antisymmetric. •Finally, | is transitive because a | b and b | c imply that a | c.We see that (Z+ , |) is a poset.

The elements a and b of a poset (S, ) are called comparable if either a b or b a . When a and b are elements of S such that neither a b nor b a, a and b are called incomparable.

Example: In the poset (Z+, |), The integers 3 and 9 are

comparable because 3 | 9. The integers 5 and 7 are incomparable, because

5 | 7 and 7 | 5.

Definition

If (S, ) is a poset and every two 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 1: The poset (Z, ≤) is totally ordered, because a ≤ b or b ≤ a whenever a and b are integers.

Example 2: The poset (Z+, |) is not totally ordered because it contains elements that are incomparable, such as 5 and 7.

Definition

Hasse Diagrams

We can represent a partial ordering on a finite set using this procedure:

• Start with the directed graph for this relation. Because a partial ordering is reflexive, a loop is present at every vertex. Remove these loops.

• Remove all edges that must be in the partial ordering because of the presence of other edges and transitivity. For instance, if (a, b) and (b, c) are in the partial ordering, remove the edge (a, c), because it must be present also. Furthermore, if (c, d) is also in the partial ordering, remove the edge (a, d), because it must be present also.

• Finally, arrange each edge so that its initial vertex is below its terminal vertex (as it is drawn on paper). Remove all the arrows on the directed edges, because all edges point "upward" toward their terminal vertex. (The edges left correspond to pairs in the covering relation of the poset.

Example 1

Constructing the Hasse Diagram for ({ 1, 2, 3, 4 }, ≤).

Example 1

Let A = {a, b, c}. Constructing the Hasse Diagram for (P(A), ).

P(A) = {, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

{a} {c}{b}

{a, b}{a, c} {b, c}

{a, b, c}

Maximal and Minimal Elements

• Elements of posets that have certain extremal properties are important for many applications.

• An element of a poset is called maximal if it is not less than any element of the poset. That is, a is maximal in the poset (S, ) if there is no b S such that a b. Similarly, an element of a poset is called minimal if it is not greater than any element of the poset. That is, a is minimal if there is no element b S such that b a. Maximal and minimal elements are easy to spot using a Hasse diagram. They are the "top" and "bottom" elements in the diagram.

Which elements of the Poset ({2, 4, 5 ,10, 12 , 20, 25 } , | ) are maximal, and which are minimal?

Solution: Maximal elements are 12, 20 and 25.Minimal elements are 2 and 5.

The Hasse Diagram of the Poset.This example shows that a poset can have more than

one maximal element and more than one minimal element.

Example

Greatest Element and Least Element

• We say that a is the greatest element of the poset (S, ) if b a for all b S.

The greatest element is unique when it exists.

• Also, a is siad the least element of the poset (S, ) if a b for all b S . The least element is unique when it exists.

Determine whether the posets represented by each of the Hasse diagrams in the Figure have a greatest element and/or a least element.

Answer: (a) No greatest element, a is the least element. (b) No greatest element, no least element. (c) d is the greatest element, no least element. (d) d is the greatest element, a is the least element.

Example

If A is a subset of the poset (S, ) and u is an element of S such that a u for all elements a A, then u is called an upper bound of A.

The element x is called the least upper bound of the subset A if x is an upper bound that is less than every other upper bound of A.

If l is an element of S such that I a for all elements a A, then I is called a lower bound of A.

The element y is called greatest lower bound of A if y is a lower bound of A and z y whenever z is a lower bound of A.

The greatest lower bound and least upper bound of a subset A are denoted by glb(A) and lub(A), respectively.

Example

Consider the poset ({2, 4, 5 ,10, 12 , 20, 25 } , | ). Find the lower and upper bounds of A = {2, 4}. Find the glb(A) and lub(A).

Solution: The lower bound of A is 2.The upper bound of A is 4,12 and 20.The glb(A) is 2.The lub(A) is 4.

The Hasse Diagram of the Poset.

Example