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Discrete Mathematics

Explore how binary relations represent relationships between elements of two sets using ordered pairs. Learn about properties like reflexive, symmetric, and antisymmetric relations. Dive into examples to grasp concepts easily.

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Discrete Mathematics

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  1. Discrete Mathematics Chapter 7 Relations

  2. 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, ais enrolled in courseb } 7.1.1

  3. 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AandbB } ) Def 1’.We use the notation aRb to denote that (a, b)R, and aRb to denote that (a,b)R. Moreover, ais said to be related tob by R if aRb. 7.1.2

  4. R R 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 toB. This means, for instance, that 0Ra, but that 1Rb R AB = { (0,a) , (0,b) , (1,a) (1,b) , (2,a) , (2,b)} A B 0 a 1 b 2 R 7.1.3

  5. 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

  6. 1 1 2 2 3 3 4 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 : R = { (1,1), (1,2), (1,3), (1,4), (2,2), (2,4), (3,3), (4,4) } 7.1.5

  7. ● ● ● ● ● ● ● ● ● ● ● ● ● ● Example 5. Consider the following relations on Z. Which of these relations contain each of the pairs (1,1), (1,2), (2,1), (1,-1), and (2,2)? 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 : ● 7.1.6

  8. 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

  9. 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 } Example 8. Which of the relations from Example 5 are reflexive ? Sol : R1, R3 and R4 7.1.8

  10. 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 Ais 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

  11. 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, R3are symmetric R4 are antisymmetric. 7.1.10

  12. 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

  13. 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

  14. 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

  15. sym. (b, a)R • 補充 : antisymmetric 跟 symmetric可並存 (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

  16. 1, if (ai,bj)R mij = 0, if (ai,bj)R 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 7.3.1

  17. B 1 2 1 A 2 3 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 : 0 0 0 1 1 1 7.3.2

  18. ※ Let A={a1, a2, …,an}.A relation R on A is reflexive iff (ai,ai)R,i. i.e., a2 … … an a1 a1 a2 ※ The relation R is symmetric iff (ai,aj)R  (aj,ai)R. This means mij = mji (即MR是對稱矩陣). 對角線上全為1 : : an 7.3.3

  19. ※ 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., ※ transitive 性質不易從矩陣判斷出來 7.3.4

  20. Example 3. Suppose that the relation R on a set is represented by the matrix Is R reflexive, symmetric, and / or antisymmetric ? Sol : reflexive, symmetric, not antisymmetric. 7.3.5

  21. 0 1 2 3 0 1 2 3 eg. Suppose that S={0,1,2,3} Let R be a relation containing (a, b) if a  b, where a  S andb  S. Is R reflexive, symmetric, antisymmetric ? Sol : ∴ R is reflexive and antisymmetric, not symmetric. 7.3.6

  22. 1 2 4 3 ※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, 4 edge(邊) : 11, 13, 21 23, 24, 31 32, 41 7.3.7

  23. ※ The relation R is reflexive iff for everyvertex,  x y x y (兩點間若有邊，必為一對不同方向的邊) (每個點上都有loop) ※ The relation R is symmetric iff ※ The relation R is antisymmetric iff 兩點間若有邊，必只有一條邊 7.3.8

  24. ※ 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

  25. a b S reflexive, not symmetric, not antisymmetric, not transitive (a→b, b→c, a→c) not reflexive, symmetric not antisymmetric not transitive (b→a, a→c, b→c) c d Example 10. Determine whether the relations R and S are reflexive, symmetric, antisymmetric, and / or transitive Sol : a R : b c Exercise : 1,13,26, 27,31 irreflexive(非反身性)的定義在 p.480 即 (a,a)R, aA 7.3.10

  26. 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

  27. 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 } Let Rs= R∪R-1={ (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

  28. 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

  29. 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

  30. 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 Rtof R ? Sol : ∴Rt= (1,2),(2,3),(3,4),(4,5) (1,3),(1,4),(1,5) (2,4),(2,5) (3,5) 3 1 5 2 4 Exercise : 1,9(改為找三種closure) 7.4.5

  31. 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 Yes.  (a,b)R  (b,a)R symmetric  (a,b)R,(b,c)R (a,c)R transitive 7.5.1

  32. ( a is congruent to b modulo m ) 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. Sol : Note that a≡b(mod m) iff m | (a-b). ∵ a≡a (mod m)(a, a)R  reflexive  If a≡b(mod m), then a-b=km, kZ  b-a≡(-k)m  b≡a (mod m)  symmetric  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

  33. 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

  34. 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

  35. Def. A partition (分割) of a set S is a collection of disjoint nonempty subsets Aiof S that have Sas their union. In other words, we have Ai ≠ , i, Ai∩Aj=  , when i≠j and ∪Ai = S. 7.5.6

  36. 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

  37. 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

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