Discrete Mathematics

Chapter 7

Relations

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7.1 Relations and their properties.※The most direct way to express a

relationship between elements of two sets

is to use ordered pairs.

For this reason, sets of ordered pairs are

called binary relations.

Example 1.A : the set of students in your school.B : the set of courses.R = { (a, b) : aA, bB, a is enrolled in course b }

7.1.1

Def 1. Let A and B be sets. A binary relation

from A to B is a subset R of AB.

( Note AB = { (a,b) : aA and bB } )

Def 1’. We use the notation aRb to denote

that (a, b)R, and aRb to denote that

(a,b)R.

Moreover, a is said to be related to b

by R if aRb.

7.1.2

Example 3. Let A={0, 1, 2} and B={a, b},

then {(0,a),(0,b),(1,a),(2,b)} is a relation R

from A to B. This means, for instance, that

0Ra, but that 1Rb

0

1

2

a

b

A B

R

R AB = { (0,a) , (0,b) , (1,a) (1,b) , (2,a) , (2,b)}

R R

7.1.3

Example ( 上例 ) : A : 男生 , B : 女生 , R : 夫妻關系 A : 城市 , B : 州 , 省 R : 屬於 (Example

2)

Note. Relations vs. Functions

A relation can be used to express a 1-to-many

relationship between the elements of the sets

A and B.

( function 不可一對多，只可多對一 )

Def 2. A relation on the set A is a subset of A A ( i.e., a relation from A to A ).

7.1.4

Example 4. Let A be the set {1, 2, 3, 4}. Which ordered pairs are in the relation R = { (a, b)| a divides b }?

Sol :

1

2

3

4

12

3

4

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

7.1.5

Example 5. Consider the following relations on Z.R1 = { (a, b) | a b }

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

R3 = { (a, b) | a = b or a = b }

R4 = { (a, b) | a = b }

R5 = { (a, b) | a = b+1 }

R6 = { (a, b) | a + b 3 }

Which of these relationscontain each of the pairs(1,1), (1,2), (2,1), (1,1),and (2,2)?

Sol : (1,1) (1,2) (2,1) (1,1) (2,2)

R1

R2

R3

R4

R5

R6

● ● ●

● ●

● ● ●

● ● ● ●

●

● ●

7.1.6

Def 3. A relation R on a set A is called reflexive

if (a,a)R for every aA.Example 7. Consider the following relations on

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

R3 = { (1,1), (1,2), (1,4), (2,1), (2,2), (3,3), (4,1), (4,4) }

R4 = { (2,1), (3,1), (3,2), (4,1), (4,2), (4,3) }

which of them are reflexive ?

反身性

Sol : R3

7.1.7

Example 8. Which of the relations from

Example 5 are reflexive ?

Sol :

R1, R3 and R4 7.1.8

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

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

R3 = { (a, b) | a = b or a = b }

R4 = { (a, b) | a = b }

R5 = { (a, b) | a = b+1 }

R6 = { (a, b) | a + b 3 }

Def 4.

(1) A relation R on a set A is called symmetric

if for a, bA, (a, b)R (b, a)R.

(2) A relation R on a set A is called antisymmetric ( 反對稱 ) if for a, bA,

(a, b)R and (b, a)R a = b.

即若 a≠b 且 (a,b)R (b, a)R

7.1.9

Example 10. Which of the relations from Example 7 are symmetric or antisymmetric ? R2 = { (1,1), (1,2), (2,1) }

R3 = { (1,1), (1,2), (1,4), (2,1), (2,2), (3,3), (4,1), (4,4) }

R4 = { (2,1), (3,1), (3,2), (4,1), (4,2), (4,3) }

Sol : R2, R3 are symmetricR4 are antisymmetric.

7.1.10

Def 5. A relation R on a set A is called transitive( 遞移 ) if for a, b, c A, (a, b)R and (b, c)R (a, c)R.

7.1.11

Example 13. Which of the relations in Example 7 are transitive ? R2 = { (1,1), (1,2), (2,1) }

R3 = { (1,1), (1,2), (1,4), (2,1), (2,2), (3,3), (4,1), (4,4) }

R4 = { (2,1), (3,1), (3,2), (4,1), (4,2), (4,3) }

Sol : R2 is not transitive since (2,1) R2 and (1,2) R2 but (2,2) R2. R3 is not transitive since (2,1) R3 and (1,4) R3 but (2,4) R3. R4 is transitive.

7.1.12

Example 17. Let A = {1, 2, 3} and B = {1, 2, 3, 4}.

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

and R2 = {(1,1), (1,2), (1,3), (1,4)} can be

combined to obtain

R1 ∪ R2

R1 ∩ R2 = {(1,1)}

R1 － R2 = {(2,2), (3,3)}

R2 － R1 = {(1,2), (1,3), (1,4)}

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

Exercise : 1, 7, 437.1.13

symmetric difference, 即 (A B) – (A B)

補充 :

antisymmetric 跟 symmetric 可並存

(a, b)R, a≠b

sym. (b, a)R

antisym. (b,a)R

故若 R 中沒有 (a, b) with a≠b 即可同時滿足

eg. Let A = {1,2,3}, give a relation R on A s.t. R is both symmetric and antisymmetric, but not reflexive.Sol :

R = { (1,1),(2,2) } 7.1.14

7.3 Representing Relations by matrices and digraphs

Suppose that R is a relation from A={a1, a2, …, am}

to B = {b1, b2,…, bn }.

The relation R can be represented by the matrix

MR = [mij], where

mij =1, if (ai,bj)R

0, if (ai,bj)R

7.3.1

※ Matrices

Example 1. Suppose that A = {1,2,3} and B = {1,2} Let R = {(a, b) | a > b, aA, bB}. What is MR ?Sol :

0 0

1

1 1

0

11

01

00

RM

7.3.2

1 2

1

2

3

A

B

※ The relation R is symmetric iff (ai,aj)R (aj,ai)R. This means mij = mji ( 即 MR 是對稱矩陣 ).

1

1

1

RM

tRR MM )(

0

01

1

※ Let A={a1, a2, …,an}. A relation R on A is reflexive iff (ai,ai)R,i.

i.e., a1

…a2 …a1 an

:an

:

a2

對角線上全為 1

7.3.3

※ The relation R is antisymmetric iff (ai,aj)R and i j (aj,ai)R.

This means that if mij=1 with i≠j, then mji=0.

i.e.,

0

01

00

1

RM

7.3.4

※ transitive 性質不易從矩陣判斷出來

Example 3. Suppose that the relation R on a set is represented by the matrix

110

111

011

RM

Is R reflexive, symmetric, and / or antisymmetric ?

Sol :reflexive, symmetric, not antisymmetric.

7.3.5

eg. Suppose that S={0,1,2,3} Let R be a relation

containing (a, b) if a b, where a S and b S. Is R reflexive, symmetric, antisymmetric ?

Sol :

1000

1100

1110

1111

RM

00

1 2 3

1

2

3

∴ R is reflexive and antisymmetric, not symmetric.

7.3.6

※Representing Relations using Digraphs. (directed graphs)

Example 8. Show the directed graph (digraph)

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}.

Sol :

vertex( 點 ) : 1, 2, 3, 4edge( 邊 ) : 11, 13, 21 23, 24, 31 32, 41

1 2

4 37.3.7

※ The relation R is reflexive iff for every vertex,

( 每個點上都有 loop)

※ The relation R is symmetric iff

兩點間若有邊，必只有一條邊

7.3.8

※ The relation R is antisymmetric iff

x y

x y

( 兩點間若有邊，必為一對不同方向的邊 )

※ The relation R is transitive iff for a, b, c A, (a, b)R and (b, c)R (a, c)R.

This means:

a b

d c

a b

d c

( 只要點 x 有路徑走到點 y ， x 必定有邊直接連向 y)

7.3.9

Example 10. Determine whether the relations R and S are reflexive, symmetric, antisymmetric, and / or transitive

Sol :

R : a

b c

Exercise : 1,13,26, 27,31

irreflexive( 非反身性 ) 的定義在 p.480即 (a,a)R, aA

7.3.10

a b

c d

Sreflexive,not symmetric,not antisymmetric,not transitive (a→b, b→c, a→c)

not reflexive,symmetricnot antisymmetricnot transitive(b→a, a→c, b→c)

7.4 Closures of Relations

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

Q: How to construct a smallest reflexive relation Rr such that R Rr ?

7.4.1

※ Closures

Sol: Let Rr = R {(2,2), (3,3)}.

i. e., Rr = R {(a, a)| a A}.

Rr is a reflexive relation containing R that is as small as possible. It is called the reflexive closure of R.

Example 1. What is the reflexive closure of the relation R={(a,b) | a < b} on the set of integers ?

Sol : Rr = R ∪ { (a, a) | aZ }

= { (a, b) | a b, a, bZ }Example :

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

Sol : Let R1={ (a, b) | (a, b)R } Let Rs= R∪R1={ (1,1),(1,2),(2,1),(2,2),(2,3), (3,1),(1,3),(3,2) }Rs is the smallest symmetric relation containing R, called the symmetric closure of R. 7.4.2

Def : 1.(reflexive closure of R on A)

Rr=the smallest set containing R and is reflexive.

Rr=R {∪ (a, a) | aA , (a, a)R}

2.(symmetric closure of R on A)

Rs=the smallest set containing R and is symmetric

Rs=R {∪ (b, a) | (a, b)R & (b, a) R}

3.(transitive closure of R on A)

Rt=the smallest set containing R and is transitive.

Rt=R {∪ (a, c) | (a, b)R & (b, c)R, but (a, c) R}Note. 沒有 antisymmetric closure ，因若不是 antisymmetric ， 表示有 a≠b, 且 (a, b) 及 (b, a) 都 R ，此時加任何 pair 都不可能變成 antisymmetric. 7.4.3

Example 2. What is the symmetric closure of the relation R={(a, b) | a > b } on the set of positive integers ?

Sol :

R { (∪ b, a) | a > b }={ (a,b) | a b }

{ (a, b) | a < b }

||

7.4.4

Example. Let R be a relation on a set A, where A={1,2,3,4,5}, R={(1,2),(2,3),(3,4),(4,5)}. What is the transitive closure Rt of R ?Sol :

1

2

3

4

5

∴Rt = (1,2),(2,3),(3,4),(4,5) (1,3),(1,4),(1,5) (2,4),(2,5) (3,5)

Exercise : 1,9( 改為找三種 closure)7.4.5

7.5 Equivalence Relations ( 等價關係 )Def 1. A relation R on a set A is called an

equivalence relation if it is reflexive,

symmetric, and transitive.

Example 1. Let L(x) denote the length of the string x. Suppose that the relation R={(a,b) | L(a)=L(b), a,b are strings of English letters } Is R an equivalence relation ?Sol :

(a,a)R string a reflexive (a,b)R (b,a)R symmetric (a,b)R,(b,c)R (a,c)R transitive

Yes.

7.5.1

Example 4. (Congruence Modulo m)Let m Z and m > 1. Show that the relation

R={ (a,b) | a≡b (mod m) } is an equivalence relation on the set of integers.

Note that a≡b(mod m) iff m | (ab).∵ a≡a (mod m) (a, a)R reflexive If a≡b(mod m), then ab=km, kZ ba≡(k)m b≡a (mod m) symmetric If a≡b(mod m), b≡c(mod m) then ab=km, bc=lm ac=(k+l)m a≡c(mod m) transitive

∴ R is an equivalence relation.

Sol :

7.5.2

( a is congruent to b modulo m )

Def 2.

Let R be an equivalence relation on a set A.

The equivalence class of the element aA is

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

For any b[a]R , b is called a representative of this equivalence class.

Note:If (a, b)R, then [a]R=[b]R.

7.5.3

Example 6.

What are the equivalence class of 0 and 1

for congruence modulo 4 ?

Sol :

Let R={ (a,b) | a≡b (mod 4) }

Then [0]R = { s | (0,s)R }

= { …, 8, 4, 0, 4, 8, … }

[1]R = { t | (1,t)R } = { …,7, 3, 1, 5, 9,…}

7.5.4

Def.

A partition ( 分割 ) of a set S is a collection of disjoint nonempty subsets Ai of S that have S as their union.

In other words, we have

Ai ≠ , i,

Ai∩Aj = , when i≠j

and

∪Ai = S.

7.5.6

Example 7.

Suppose that S={ 1,2,3,4,5,6 }. The collection

of sets A1={1,2,3}, A2={ 4,5 }, and A3={ 6 } form a partition of S.

Thm 2. Let R be an equivalence relation on a set A.

Then the equivalence classes of R form a partition of A.

7.5.7

Example 9.

The congruence modulo 4 form a partition of the integers.

Sol :

[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, … }

Exercise : 3,17,19,237.5.8