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Discrete Mathematics

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.

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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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