Discrete Mathematics

Chapter 7 Relations 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) : a∈A, b∈B, 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 A×B. ( Note A×B = { (a,b) : a∈A and b∈B } )

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

A B R ⊆ A×B = { (0,a) , (0,b) , (1,a) (1,b) , (2,a) , (2,b)} 0 a ∈R ∈R 1 b 2

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 1 2 2 3 3 4 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 } Which of these relations

R3 = { (a, b) | a = b or a = −b } contain each of the pairs

R4 = { (a, b) | a = b } (1,1), (1,2), (2,1), (1,−1),

R5 = { (a, b) | a = b+1 } and (2,2)?

R6 = { (a, b) | a + b ≤ 3 } Sol : (1,1) (1,2) (2,1) (1,−1) (2,2)

R1 ●● ●

R2 ●●

R3 ●●●

R4 ●●

R5 ● ● ● R6 ● ● 7.1.6 反身性 Def 3. A relation R on a set A is called reflexive if (a,a)∈R for every a∈A. 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 ?

Sol :

R3 7.1.7 Example 8. Which of the relations from Example 5 are reflexive ?

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 }

Sol :

R1, R3 and R4 7.1.8 Def 4. (1) A relation R on a set A is called symmetric if for a, b∈A, (a, b)∈R ⇒ (b, a)∈R. (2) A relation R on a set A is called antisymmetric (反對稱) if for a, b∈A, (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 symmetric R4 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)} symmetric difference, 即 (A ∪ B) – (A ∩ B)

Exercise : 1, 7, 43 7.1.13 補充 : antisymmetric 跟 symmetric可並存 sym. ⇒ (b, a)∈R ∀(a, b)∈R, a≠b 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 ※ Matrices

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

1, if (ai,bj)∈R mij =

0, if (ai,bj)∉R

7.3.1 Example 1. Suppose that A = {1,2,3} and B = {1,2} Let R = {(a, b) | a > b, a∈A, b∈B}.

What is MR ? Sol : B 1 2 1 ⎡ 00⎤ ⎡ 00 ⎤ A ⎢ ⎥ ⎢ 1 0 ⎥ ∴M R = 01 2 ⎢ ⎥ ⎢ ⎥ ⎣⎢ 11 ⎦⎥ 3 ⎣⎢ 1 1 ⎦⎥

7.3.2 ※ Let A={a1, a2, …,an}. A relation R on A is reflexive iff (ai,ai)∈R,∀i. a … …a a i.e., 1 2 n a1 ⎡1 ⎤ ⎢ ⎥ a 1 2 ⎢ ⎥ ⎢ ⎥ M R = : O ⎢ ⎥ 對角線上全為1 : ⎢ O ⎥ a n ⎣⎢ 1⎦⎥

※ The relation R is symmetric iff (ai,aj)∈R ⇒ (aj,ai)∈R. This means mij = mji (即MR是對稱矩陣). ⎡ 1 ⎤ ⎢ ⎥ ⎢ ⎥ t M R = ⎢1 0⎥ = M R )( ⎢ ⎥ ⎢ ⎥ ⎣⎢ 0 ⎦⎥ 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., ⎡ 1 ⎤ ⎢ 00 ⎥ M = ⎢ ⎥ R ⎢ 01 ⎥ ⎢ ⎥ ⎣ 0 ⎦

※ transitive 性質不易從矩陣判斷出來

7.3.4 Example 3. Suppose that the relation R on a set is represented by the matrix ⎡ 011 ⎤ M = ⎢ 111 ⎥ R ⎢ ⎥ ⎣⎢ 110 ⎦⎥ 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 : 0 123 0 ⎡ 1111 ⎤ ⎢ ⎥ ∴ R is reflexive and 1 ⎢ 1110 ⎥ antisymmetric, M R = 2 ⎢ 1100 ⎥ not symmetric. ⎢ ⎥ 3 ⎣ 1000 ⎦

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 : 1 2 vertex(點) : 1, 2, 3, 4 edge(邊) : 1→1, 1→3, 2→1 2→3, 2→4, 3→1 3→2, 4→1 43 7.3.7 ※ The relation R is reflexive iff for every vertex, (每個點上都有loop)

※ The relation R is symmetric 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 antisymmetric iff

兩點間若有邊，必只有一條邊

7.3.8 ※ 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 a b

⇒

d c 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 a b Sol : reflexive, S R : a not symmetric, cd not antisymmetric, not transitive not reflexive, (a→b, b→c, a→c) symmetric not antisymmetric not transitive (b→a, a→c, b→c) bc

Exercise : 1,13,26, irreflexive(非反身性)的定義在 p.480 27,31 即 (a,a)∉R, ∀a∈A 7.3.10 7.4 Closures of Relations ※ Closures 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 ?

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.

7.4.1 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) | a∈Z } = { (a, b) | a ≤ b, a, b∈Z } 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 R−1={ (a, b) | (a, b)∈R } −1 Let Rs= R∪R ={ (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) | a∈A , (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 :

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

2 4 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 : c (a,a)∈R ∀string a ⇒ reflexive d (a,b)∈R ⇒ (b,a)∈R ⇒ symmetric Yes. e (a,b)∈R,(b,c)∈R ⇒ (a,c)∈R ⇒ transitive 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. ( a is congruent to b modulo m )

Sol : Note that a≣b(mod m) iff m | (a−b). ∵c a≡a (mod m) ⇒ (a, a)∈R ⇒ reflexive d If a≡b(mod m), then a−b=km, k∈Z ⇒ b−a≡(−k)m ⇒ b≡a (mod m) ⇒ symmetric e If a≡b(mod m), b≡c(mod m) then a−b=km, b−c=lm ⇒ a−c=(k+l)m ⇒ a≡c(mod m) ⇒ transitive

∴ R is an equivalence relation. 7.5.2 Def 2. Let R be an equivalence relation on a set A. The equivalence class of the element a∈A 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 : 3,17,19,23 7.5.8