Week 4-5: Binary Relations

1 Binary Relations Week 4-5: Binary Relations The concept of relation is common in daily life and seems intuitively clear. For instance, let X be the set of all living human females and Y the set of all living human males. The wife-husband relation R can be thought as a relation from X to Y. For a lady x X and a gentleman y Y, we say that x is related to y by R if x is the wife of y, written as xry. To describe the relation R, we may list the collection of all ordered pairs (x, y) such that x is related to y by R. The collection of all such related ordered pairs is simply a subset of the product set X Y. This motivates the following definition of relations. Definition 1.1. Let X and Y be nonempty sets. A binary relation (or just relation) from X to Y is a subset R X Y. If (x, y) R, we say that x is related to y by R, denoted xry. If (x, y) R, we say that x is not related to y, denoted x Ry. For each element x X, we denote by R(x) the subset of elements of Y that are related to x, that is, R(x) = {y Y : xry} = {y Y : (x, y) R}. For each subset A X, we define R(A) = {y Y : x A such that xry} = x A R(x). When X = Y, we say that R is a binary relation on X. Since binary relations from X to Y are subsets of X Y, we can define intersection, union, and complement for binary relations. The complementary relation of a binary relation R X Y is the binary relation R X Y defined by x Ry (x, y) R. 1

The inverse relation of R is the binary relation R 1 Y X defined by yr 1 x (x, y) R. Example 1.1. Consider a family A with five children, Amy, Bob, Charlie, Debbie, and Eric. We abbreviate the names to their first letters so that A = {a, b, c, d, e}. (a) The brother-sister relation R bs is the set R bs = {(b, a), (b, d), (c, a), (c, d), (e, a), (e, d)}. (b) The sister-brother relation R sb is the set R sb = {(a, b), (a, c), (a, e), (d, b), (d, c), (d, e)}. (c) The brother relation R b is the set {(b, b), (b, c), (b, e), (c, b), (c, c), (c, e), (e, b), (e, c), (e, e)}. (d) The sister relation R s is the set {(a, a), (a, d), (d, a), (d, d)}. The brother-sister relation R bs is the inverse of the sister-brother relation R sb, i.e., R bs = R 1 sb. The brother or sister relation is the union of the brother relation and the sister relation, i.e., R b R s. The complementary relation of the brother or sister relation is the brother-sister or sister-brother relation, i.e., R b R s = R bs R sb. Example 1.2. (a) The graph of equation 3 + y2 2 2 = 1 2 is a binary relation on R. The graph is an ellipse. x 2 2

(b) The relation less than, denoted by <, is a binary relation on R defined by a < b a is less than b. As a subset of R 2 = R R, the relation is given by the set {(a, b) R 2 : a is less than b}. (c) The relation greater than or equal to is a binary relation on R defined by a b a is greater than or equal to b. As a subset of R 2, the relation is given by the set {(a, b) R 2 : a is greater than or equal to b}. (d) The divisibility relation about integers, defined by a b a divides b, is a binary relation on the set Z of integers. As a subset of Z 2, the relation is given by {(a, b) Z 2 : a is a factor of b}. Example 1.3. Any function f : X Y can be viewed as a binary relation from X to Y. The binary relation is just its graph G(f) = {(x, f(x)) : x X} X Y. Exercise 1. Let R X Y be a binary relation from X to Y. Let A, B X be subsets. (a) If A B, then R(A) R(B). (b) R(A B) = R(A) R(B). (c) R(A B) R(A) R(B). Proof. (a) For any y R(A), there is an x A such that xry. Since A B, then x B. Thus y R(B). This means that R(A) R(B). 3

(b) For any y R(A B), there is an x A B such that xry. If x A, then y R(A). If x B, then y R(B). In either case, y R(A) R(B). Thus R(A B) R(A) R(B). On the other hand, it follows from (a) that R(A) R(A B) and R(B) R(A B). Thus R(A) R(B) R(A B). (c) It follows from (a) that R(A B) R(A) and R(A B) R(B). Thus R(A B) R(A) R(B). Proposition 1.2. Let R 1, R 2 X Y be relations from X to Y. If R 1 (x) = R 2 (x) for all x X, then R 1 = R 2. Proof. If xr 1 y, then y R 1 (x). Since R 1 (x) = R 2 (x), we have y R 2 (x). Thus xr 2 y. A similar argument shows that if xr 2 y then xr 1 y. Therefore R 1 = R 2. 2 Representation of Relations Binary relations are the most important relations among all relations. Ternary relations, quaternary relations, and multi-factor relations can be studied by binary relations. There are two ways to represent a binary relation, one by a directed graph and the other by a matrix. Let R be a binary relation on a finite set V = {v 1, v 2,..., v n }. We may describe the relation R by drawing a directed graph as follows: For each element v i V, we draw a solid dot and name it by v i ; the dot is called a vertex. For two vertices v i and v j, if v i Rv j, we draw an arrow from v i to v j, called a directed edge. When v i = v j, the directed edge becomes a directed loop. The resulted graph is a directed graph, called the digraph of R, and is denoted by D(R). Sometimes the directed edges of a digraph may have to cross each other when drawing the digraph on a plane. However, the 4

intersection points of directed edges are not considered to be vertices of the digraph. The in-degree of a vertex v V is the number of vertices u such that urv, and is denoted by indeg (v). The out-degree of v is the number of vertices w such that vrw, and is denoted by outdeg (v). If R X Y is a relation from X to Y, we define outdeg (x) = R(x) for x X, indeg (y) = R 1 (y) for y Y. The digraphs of the brother-sister relation R bs and the brother or sister relation R b R s are demonstrated in the following. a 01 01 00 00 00 11 11 11 01 01 1000000000000000000000 111111111111111111111 00 0 0 11 1 1 e b d c a e b d 0 0 1 1 c Definition 2.1. Let R X Y be a binary relation from X to Y, where X = {x 1, x 2,...,x m }, Y = {y 1, y 2,..., y n }. The matrix of the relation R is an m n matrix M R = [a ij ], whose (i, j)-entry is given by { 1 if xi Ry a ij = j 0 if x i Ry j. The matrix M R is called the Boolean matrix of R. When X = Y, the matrix M R is a square matrix. 5

Let A = [a ij ] and B = [b ij ] be m n Boolean matrices. If a ij b ij for all (i, j)-entries, we write A B. The matrix of the brother-sister relation R bs on the set A = {a, b, c, d, e} is the square matrix 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 and the matrix of the brother or sister relation is the square matrix 1 0 0 1 0 0 1 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 1 0 1 Proposition 2.2. For any digraph D(R) of a binary relation R V V on V, indeg (v) = outdeg (v) = R. v V v V If R is a binary relation from X to Y, then indeg (y) = R. Proof. Trivial. x X outdeg (x) = y Y Let R be a relation on a set X. A path of length k from x to y is a finite sequence x 0, x 1,..., x k, beginning with x 0 = x and ending with x k = y, such that x 0 Rx 1, x 1 Rx 2,..., x k 1 Rx k. A path that begins and ends at the same vertex is called a cycle. For any fixed positive integer k, let R k X X denote the relation on X given by xr k y a path of length k from x to y. 6

Let R X X denote the relation on X given by xr y a path from x to y. The relation R is called the connectivity relation for R. Clearly, we have R = R R 2 R 3 = R k. The reachability relation of R is the binary relation R X X on X defined by xr y x = y or xr y. Obviously, R = I R R 2 R 3 = where I is the identity relation on X defined by xiy x = y. k=1 R k, We always assume that R 0 = I for any relation R on a set X. Example 2.1. Let X = {x 1,...,x n } and R = {(x i, x i+1 ) : i = 1,...,n 1}. Then R k = {(x i, x i+k ) : i = 1,..., n k}, 1 k n/2; R k =, k (n + 1)/2; R = {(x i, x j ) : i < j}. If R = {(x i, x i+1 ) : i = 1,...,n} with x n+1 = x 1, then R = X X = X 2. 3 Composition of Relations Definition 3.1. Let R X Y and S Y Z binary relations. The composition of R and S is a binary relation S R X Z from X to Z defined by k=0 x(s R)z y Y such that xry and ysz. When X = Y, the relation R is a binary relation on X. We have R k = R k 1 R, k 2. 7

Remark. Given relations R X Y and S Y Z, the composition S R of R and S is backward. However, some people use the notation R S instead of our notation S R. But this usage is inconsistent with the composition of functions. To avoid confusion and for aesthetic reason, we write S R as RS = {(x, z) X Z : y Y, xry, ysz}. Example 3.1. Let R X Y, S Y Z, where X = {x 1, x 2, x 3, x 4 }, Y = {y 1, y 2, y 3 }, Z = {z 1, z 2, z 3, z 4, x 5 }. x 1 x x x 2 3 4 R y y y 1 2 3 S z z z z z 1 2 3 4 5 x 1 x x x 2 3 4 RS Example 3.2. For the brother-sister relation, sister-brother relation, brother relation, and sister relation on A = {a, b, c, d, e}, we have R bs R sb = R b, R sb R bs = R s, R bs R s = R bs, R bs R bs =, R b R b = R b, R b R s =. Let X 1, X 2,..., X n, X n+1 be nonempty sets. Given relations R i X i X i+1, 1 i n. We define a relation R 1 R 2 R n X 1 X n+1 from X 1 to X n+1 by xr 1 R 2 R n y, if and only if there exists a sequence x 1, x 2,..., x n, x n+1 with x 1 = x, x n+1 = y such that x 1 R 1 x 2, x 2 R 2 x 3,..., x n R n x n+1. 8 z 1 z 2 z 3 z 4 z 5

Theorem 3.2. Given relations R 1 X 1 X 2, R 2 X 2 X 3, R 3 X 3 X 4. We have R 1 R 2 R 3 = R 1 (R 2 R 3 ) = (R 1 R 2 )R 3. as relations from X 1 to X 4. Proof. For x X 1, y X 4, we have xr 1 (R 2 R 3 )y x 2 X 2, xr 1 x 2, x 2 R 2 R 3 y x 2 X 2, xr 1 x 2 ; x 3 X 3, x 2 R 2 x 3, x 3 R 3 y x 2 X 2, x 3 X 3, xr 1 x 2, x 2 R 2 x 3, x 3 R 3 y xr 1 R 2 R 3 y. Similarly, x(r 1 R 2 )R 3 y xr 1 R 2 R 3 y. Proposition 3.3. Let R i X Y be relations, i = 1, 2. (a) If R W X, then R(R 1 R 2 ) = RR 1 RR 2. (b) If S Y Z, then (R 1 R 2 )S = R 1 S R 2 S. Proof. (a) For any wr(r 1 R 2 )y, x X such that wrx and x(r 1 R 2 )y. Then xr 1 y or xr 2 y. Thus wrr 1 y or wrr 2 y. Namely, w(rr 1 RR 2 )y. Conversely, for any w(rr 1 RR 2 )y, we have wrr 1 y or wrr 2 y. Then x X such that wrx and xr 1 y, or, wrx and xr 2 y. This means that wrx and either xr 1 y or xr 2 y. Thus wrx and x(r 1 R 2 )y. Namely, wr(r 1 R 2 )y. The proof for (b) is similar. Exercise 2. Let R i X Y be relations, i = 1, 2,.... (a) If R W X, then R ( i=1 R i) = i=1 RR i. (b) If S Y Z, then ( i=1 R i)s = i=1 R is. 9

For the convenience of representing composition of relations, we introduce the Boolean operations and on real numbers. For a, b R, define a b = min{a, b}, a b = max{a, b}. Exercise 3. For a, b, c R, a (b c) = (a b) (a c), a (b c) = (a b) (a c). Proof. We only prove the first formula. The second one is similar. Case 1: b c. If a c, then the left side is a (b c) = a c = c. The right side is (a b) (a c) = b c = c. If b a c, then the left side is a (b c) = a c = a. The right side is (a b) (a c) = b a = a. If a b c, then the left side is a (b c) = a c = a. The right side is (a b) (a c) = a a = a. Case 2: b c. If a c, then a (b c) = a c = a and (a b) (a c) = a a = a. If b a c, then a (b c) = a c = a and (a b) (a c) = a c = a. If a b, then a (b c) = a b = b and (a b) (a c) = b c = b. Sometimes it is more convenient to write the Boolean operations as a b = min{a, b}, a b = max{a, b}. For real numbers a 1, a 2,..., a n, we define n n a i = a i = max{a 1, a 2,..., a n }. i=1 i=1 For an m n matrix A = [a ij ] and an n p matrix B = [b jk ], the Boolean multiplication of A and B is an m p matrix A B = [c ik ], whose (i, k)-entry is defined by n n c ik = (a ij b jk ) = (a ij b jk ). j=1 Theorem 3.4. Let R X Y, S Y Z be relations, where X = {x 1,...,x m }, Y = {y 1,...,y n }, Z = {z 1,..., z p }. 10 j=1

Let M R, M S, M RS be matrices of R, S, RS respectively. Then M RS = M R M S. Proof. We write M R = [a ij ], M S = [b jk ], and M R M S = [c ik ], M RS = [d ik ]. It suffices to show that c ik = d ik for any (i, k)-entry. Case I: c ik = 1. Since c ik = n j=1 (a ij b jk ) = 1, there exists j 0 such that a ij0 b j0 k = 1. Then a ij0 = b j0 k = 1. In other words, x i Ry j0 and y j0 Sz k. Thus x i RSz k by definition of composition. Therefore d ik = 1 by definition of Boolean matrix of RS. Case II: c ik = 0. Since c ik = n j=1 (a ij b jk ) = 0, we have a ij b jk = 0 for all j. Then there is no j such that a ij = 1 and b jk = 1. In other words, there is no y j Y such that both x i Ry j and y j Sz k. Thus x i is not related to z k by definition of RS. Therefore d ik = 0. 4 Special Relations We are interested in some special relations satisfying certain properties. For instance, the less than relation on the set of real numbers satisfies the socalled transitive property: if a < b and b < c, then a < c. Definition 4.1. A binary relation R on a set X is said to be (a) reflexive if xrx for all x in X; (b) symmetric if xry implies yrx; (c) transitive if xry and yrz imply xrz. A relation R is called an equivalence relation if it is reflexive, symmetric, and transitive. And in this case, if xry, we say that x and y are equivalent. The relation I X = {(x, x) x X} is called the identity relation. The relation X 2 is called the complete relation. 11

Example 4.1. Many family relations are binary relations on the set of human beings. a) The strict brother relation R b : xr b y x and y are both males and have the same parents. (symmetric and transitive) b) The strict sister relation R s : xr s y x and y are both females and have the same parents. (symmetric and transitive) c) The strict brother-sister relation R bs : xr bs y x is male, y is female, x and y have the same parents. d) The strict sister-brother relation R sb : xr sb x is female, y is male, and x and y have the same parents. e) The generalized brother relation R b : xr by x and y are both males and have the same father or the same mother. (symmetric, not transitive) f) The generalized sister relation R s: xr sy x and y are both females and have the same father or the same mother. (symmetric, not transitive) g) The relation R: xry x and y have the same parents. (reflexive, symmetric, and transitive; equivalence relation) h) The relation R : xr y x and y have the same father or the same mother. (reflexive and symmetric) Example 4.2. (a) The less than relation < on the set of real numbers is a transitive relation. (b) The less than or equal to relation on the set of real numbers is a reflexive and transitive relation. (c) The divisibility relation on the set of positive integers is a reflexive and transitive relation. (d) Given a positive integer n. The congruence modulo n is a relation n on Z defined by a n b b a is a multiple of n. 12

The standard notation for a n b is a b mod n. The relation n is an equivalence relation on Z. Theorem 4.2. Let R be a relation on a set X. Then (a) R is reflexive I R all diagonal entries of M R are 1. (b) R is symmetric R = R 1 M R is a symmetric matrix. (c) R is transitive R 2 R M 2 R M R. Proof. (a) and (b) are trivial. (c) R is transitive R 2 R. For any (x, y) R 2, there exists z X such that (x, z) R, (z, y) R. Since R is transitive, then (x, y) R. Thus R 2 R. R 2 R R is transitive. For (x, z) R and (z, y) R, we have (x, y) R 2 R. Then (x, y) R. Thus R is transitive. Note that for any relations R and S on X, we have R S M R M S. Since M R is the matrix of R, then M 2 R = M RM R = M RR = M R 2 is the matrix of R 2. Thus R 2 R M 2 R M R. 5 Equivalence Relations and Partitions The most important binary relations are equivalence relations. We will see that an equivalence relation on a set X will partition X into disjoint equivalence classes. Example 5.1. Consider the congruence relation 3 on Z. For each a Z, define [a] = {b Z : a 3 b} = {b Z : a b mod 3}. It is clear that Z is partitioned into three disjoint subsets [0] = {0, ±3, ±6, ±9,...} = {3k : k Z}, [1] = {1, 1 ± 3, 1 ± 6, 1 ± 9,...} = {3k + 1 : k Z}, [2] = {2, 2 ± 3, 2 ± 6, 2 ± 9,...} = {3k + 2 : k Z}. 13

Moreover, for all k Z, [0] = [3k], [1] = [3k + 1], [2] = [3k + 2]. Theorem 5.1. Let be an equivalence relation on a set X. For each x of X, let [x] = {y X : x y}, the set of elements equivalent to x. Then (a) x [x] for any x X, (b) [x] = [y] if x y, (c) [x] [y] = if x y, (d) X = x X [x]. The subset [x] is called an equivalence class of, and x is called a representative of [x]. Proof. (a) It is trivial because is reflexive. (b) For any z [x], we have x z by definition of [x]. Since x y, we have y x by the symmetric property of. Then y x and x z imply that y z by transitivity of. Thus z [y] by definition of [y]; that is, [x] [y]. Since is symmetric, we have [y] [x]. Therefore [x] = [y]. (c) Suppose [x] [y] is not empty, say z [x] [y]. Then x z and y z. By symmetry of, we have z y. Thus x y by transitivity of, a contradiction. (d) This is obvious because x [x] for any x X. Definition 5.2. A partition of a nonempty set X is a collection of subsets of X such that (a) A i for all i; (b) A i A j = if i j; (c) X = j J A j. P = {A j : j J} 14

Each subset A j is called a block of the partition P. Theorem 5.3. Let P be a partition of a set X. Let R P denote the relation on X defined by xr P y a block A j P such that x, y A j. Then R P is an equivalence relation on X, called the equivalence relation induced by P. Proof. (a) For each x X, there exits one A j such that x A j. Then by definition of R P, xr P x. Hence R P is reflexive. (b) If xr P y, then there is one A j such that x, y A j. By definition of R P, yr P x. Thus R P is symmetric. (c) If xr P y and yr P z, then there exist A i and A j such that x, y A i and y, z A j. Since y A i A j and P is a partition, it forces A i = A j. Thus xr P z. Therefore R P is transitive. Given an equivalence relation R on a set X. The collection P R = {[x] : x X} of equivalence classes of R is a partition of X, called the quotient set of X modulo R. Let E(X) denote the set of all equivalence relations on X and P(X) the set of all partitions of X. Then we have functions f : E(X) P(X), f(r) = P R ; g : P(X) E(X), g(p) = R P. The functions f and g satisfy the following properties. Theorem 5.4. Let X be a nonempty set. Then for any equivalence relation R on X, and any partition P of X, we have (g f)(r) = R, (f g)(p) = P. In other words, f and g are inverse of each other. 15

Proof. Recall (g f)(r) = g(f(r)), (f g)(p) = f(g(p)). Then x[g(p R )]y A P R s.t. x, y A xry; A f(r P ) x X s.t. A = R P (x) A P. Thus g(f(r)) = R and f(g(p)) = P. Example 5.2. Let Z + be the set of positive integers. Define a relation on Z Z + by (a, b) (c, d) ad = bc. Is an equivalence relation? If Yes, what are the equivalence classes? Let R be a relation on a set X. The reflexive closure of R is the smallest reflexive relation r(r) on X that contains R; that is, a) R r(r), b) if R is a reflexive relation on X and R R, then r(r) R. The symmetric closure of R is the smallest symmetric relation s(r) on X such that R s(r); that is, a) R s(r), b) if R is a symmetric relation on X and R R, then s(r) R. The transitive closure of R is the smallest transitive relation t(r) on X such that R t(r); that is, a) R t(r), b) if R is a transitive relation on X and R R, then t(r) R. Obviously, the reflexive, symmetric, and transitive closures of R must be unique respectively. Theorem 5.5. Let R be a relation R on a set X. Then a) r(r) = R I; b) s(r) = R R 1 ; 16

c) t(r) = k=1 Rk. Proof. (a) and (b) are obvious. (c) Note that R ( ) R i R j = i=1 j=1 k=1 Rk and i,j=1 R i R j = i,j=1 R i+j = R k k=2 R k. This shows that k=1 Rk is a transitive relation, and R k=1 Rk. Since each transitive relation that contains R must contain R k for all integers k 1, we see that k=1 Rk is the transitive closure of R. Theorem 5.6. Let R be a relation on a set X with X = n 2. Then t(r) = R R 2 R n 1. In particular, if R is reflexive, then t(r) = R n 1. Proof. It is enough to show that for all k n, R k This is equivalent to showing that R k k 1 i=1 Ri for all k n. Let (x, y) R k. There exist elements x 1,...,x k 1 X such that n 1 i=1 R i. (x, x 1 ), (x 1, x 2 ),..., (x k 1, y) R. Since X = n 2 and k n, the following sequence x = x 0, x 1, x 2,..., x k 1, x k = y has k + 1 terms, which is at least n + 1. Then two of them must be equal, say, x p = x q with p < q. Thus q p 1 and k=1 (x 0, x 1 ),..., (x p 1, x p ), (x q, x q+1 ),..., (x k 1, x k ) R. Therefore (x, y) = (x 0, x k ) R k (q p) 17 k 1 i=1 R i.

That is R k k 1 i=1 R i. If R is reflexive, then R k R k+1 for all k 1. Hence t(r) = R n 1. Proposition 5.7. Let R be a relation on a set X. Then I t(r R 1 ) is an equivalence relation. In particular, if R is reflexive and symmetric, then t(r) is an equivalence relation. Proof. Since I t(r R 1 ) is reflexive and transitive, we only need to show that I t(r R 1 ) is symmetric. Let (x, y) I t(r R 1 ). If x = y, then obviously (y, x) I t(r R 1 ). If x y, then (x, y) t(r R 1 ). Thus (x, y) (R R 1 ) k for some k 1. Hence there is a sequence such that Since R R 1 is symmetric, we have x = x 0, x 1,..., x k = y (x i, x i+1 ) R R 1, 0 i k 1. (x i+1, x i ) R R 1, 0 i k 1. This means that (y, x) (R R 1 ) k. Hence (y, x) I t(r R 1 ). Therefore I t(r R 1 ) is symmetric. In particular, if R is reflexive and symmetric, then obviously I t(r R 1 ) = t(r). This means that t(r) is reflexive and symmetric. Since t(r) is automatically transitive, so t(r) is an equivalence relation. 18

Let R be a relation on a set X. The reachability relation of R is a relation R on X defined by such that That is, R = I t(r). xr y x = y or finite x 1, x 2,..., x k (x, x 1 ), (x 1, x 2 ),..., (x k, y) R. Theorem 5.8. Let R be a relation on a set X. Let M and M be the Boolean matrices of R and R respectively. If X = n, then Moreover, if R is reflexive, then Proof. It follows from Theorem 5.6. M = I M M 2 M n 1. R k R k+1, k 1; M = M n 1. 6 Washall s Algorithm Let R be a relation on X = {x 1,..., x n }. Let y 0, y 1,..., y m be a path in R. The vertices y 1,..., y m 1 are called interior vertices of the path. For each k with 0 k n, we define the Boolean matrix W k = [w ij ], where w ij = 1 if there is a path in R from x i to x j whose interior vertices are contained in X k, otherwise w ij = 0, where X 0 = and X k = {x 1,..., x k } for k 1. Since the interior vertices of any path in R is obviously contained in the whole set X = {x 1,..., x n }, the (i, j)-entry of W n is equal to 1 if there is a path in R from x i to x j. Then W n is the matrix of the transitive closure t(r) of R, that is, W n = M t(r). 19

Clearly, W 0 = M R. We have a sequence of Boolean matrices M R = W 0, W 1, W 2,..., W n. The so-called Warshall s algorithm is to compute W k from W k 1, k 1. Let W k 1 = [s ij ] and W k = [t ij ]. If t ij = 1, there must be a path x i = y 0, y 1,..., y m = x j from x i to x j whose interior vertices y 1,...,y m 1 are contained in {x 1,...,x k }. We may assume that y 1,..., y m 1 are distinct. If x k is not an interior vertex of this path, that is, all interior vertices are contained in {x 1,..., x k 1 }, then s ij = 1. If x k is an interior vertex of the path, say x k = y p, then there two sub-paths x i = y 0, y 1,..., y p = x k, x k = y p, y p+1,..., y m = x j whose interior vertices y 1,..., y p 1, y p+1,..., y m 1 are contained in {x 1,..., x k 1 } obviously. It follows that s ik = 1, s kj = 1. We conclude that t ij = 1 { sij = 1 or s ik = 1, s kj = 1 for some k. Theorem 6.1 (Warshall s Algorithm for Transitive Closure). Working on the Boolean matrix W k 1 to produce W k. a) If the (i, j)-entry of W k 1 is 1, so is W k. Keep 1 there. b) If the (i, j)-entry of W k 1 is 0, then check the entries of W k 1 at (i, k) and (k, j). If both entries are 1, then change the (i, j)-entry in W k 1 to 1. Otherwise, keep 0 there. Example 6.1. Consider the relation R on A = {1, 2, 3, 4, 5} given by the Boolean matrix 0 0 0 0 1 0 1 1 0 0 M R = 1 0 0 0 0. 0 0 0 1 0 0 0 1 1 0 20

By Warshall s algorithm, we have 0 0 0 0 1 0 1 1 0 0 W 0 = 1 0 0 0 0 W 1 = 0 0 0 1 0 0 0 1 1 0 W 2 = W 3 = W 4 = W 5 = 0 0 0 0 1 0 1 1 0 0 1 0 0 0 (1) 0 0 0 1 0 0 0 1 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0 0 0 1 (1) 1 1 0 (1) 1 0 0 0 1 0 0 0 1 0 (1) 0 1 1 (1) 0 0 0 0 1 1 1 1 0 1 1 0 0 0 1 0 0 0 1 0 1 0 1 1 1 (3, 1), (1, 5) (no change) (1) 0 (1) (1) 1 1 1 1 (1) 1 1 0 (1) (1) 1 0 0 0 1 0 1 0 1 1 1 (2, 3), (3, 1) (2, 3), (3, 5) (5, 3), (3, 1) (5, 3), (3, 5) (no change) (1, 5), (5, 1) (1, 5), (5, 3) (1, 5), (5, 4) (2, 5), (5, 4) (3, 5), (5, 3) (3, 5), (5, 4). 21

The binary relation for the Boolean matrix W 5 is the transitive closure of R. 5 1 4 2 3 Definition 6.2. A binary relation R on a set X is called a) asymmetric if xry implies y Rx; b) antisymmetric if xry and yrx imply x = y. Problem Set 4 1. Let R be a binary relation from X to Y, A, B X. (a) If A B, then R(A) R(B). (b) R(A B) = R(A) R(B). (c) R(A B) R(A) R(B). Proof. (a) For any y R(A), there is an a A such that (a, y) R. Obviously, a B. Thus b R(B). (b) Since R(A) R(A B), R(B) R(A B), we have R(A) R(B) R(A B). On the other hand, for any y R(A B), there is an x A B such that (x, y) R. Then either x A or x B. Thus y R(A) or y R(B); i.e., y R(A) R(B). Therefore R(A) R(B) R(A B). (c) It follows from (a) that R(A B) R(A) and R(A B) R(B). Hence R(A B) R(A B). 22

2. Let R 1 and R 2 be relations from X to Y. If R 1 (x) = R 2 (x) for all x X, then R 1 = R 2. Proof. For any (x, y) R 1, we have y R 1 (x). Since R 1 (x) = R 2 (x), then y R 2 (x). Thus (x, y) R 2. Similarly, for any (x, y) R 2, we have (x, y) R 2. Hence R 1 = R 2. 3. Let a, b, c R. Then a (b c) = (a b) (a c), a (b c) = (a b) (a c). Proof. Note that the cases b < c and b > c are equivalent. There are three essential cases to be verified. Case 1: a < b < c. We have Case 2: b < a < c. We have Case 3: b < c < a. We have a (b c) = a = (a b) (a c), a (b c) = b = (a b) (a c). a (b c) = a = (a b) (a c), a (b c) = a = (a b) (a c). a (b c) = c = (a b) (a c), a (b c) = a = (a b) (a c). 4. Let R i X Y be a family of relations from X to Y, i I. (a) If R W X, then R ( i I R ) i = i I RR i; (b) If S Y Z, then ( i I R ) i S = i I R is. 23

Proof. (a) Note that (w, y) R ( i I R i) x X s.t. (w, x) R and (x, y) i I R i; and (x, y) i I R i i 0 I s.t. (x, y) R i. Then (w, y) R ( i I R i) i0 I s.t. (w, y) RR i (w, y) i I RR i. (b) Note that (x, z) ( i I R i) S y Y s.t. (x, y) R and (x, y) i I R i; and (x, y) i I R i i 0 I s.t. (x, y) R i. Then (w, y) R ( i I R i) i0 I s.t. (w, y) RR i (w, y) i I RR i. 5. Let R i (1 i 3) be relations on A = {a, b, c, d, e} whose Boolean matrices are 0 1 1 0 1 1 0 0 1 0 M 1 =, M 2 =, 0 1 1 0 1 1 0 0 1 0 M 3 = 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1. (a) Draw the digraphs of the relations R 1, R 2, R 3. (b) Find the Boolean matrices for the relations and verify that R 1 1, R 2 R 3, R 1 R 1, R 1 R 1 1, R 1 1 R 1; R 1 R 1 1 = R 2, R 1 1 R 1 = R 3. (c) Verify that R 2 R 3 is an equivalence relation and find the quotient set A/(R 2 R 3 ). 24

Solution: M R 1 1 = M R1 R 1 1 = 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0, M R2 R 3 = 1 0 0 1 0 0 1 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 1 0 1 = M 2, M R 1 1 R 1 =, M R 2 = 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 6. Let R be a relation on Z defined by xry if x + y is an even integer. (a) Show that R is an equivalence relation on Z. (b) Find all equivalence classes of the relation R. = M 3. Proof. (a) Since x + x = 2x is even for any x, then xrx, so R is reflexive. If xry, then x + y is even. Of course, y + x is even, i.e., yrx. So R is symmetric. If xry and yrz, then x +y and y +z are even, so xrz. Thus R is reflexive. Therefore R is an equivalence relation. (b) Note that xry if and only if x and y are both odd or both even. Thus there are only two equivalence classes: the set of even integers, and the set of odd integers. 7. Let X = {1, 2,...,10} and let R be a relation on X such that arb if and only if a b 2. Determine whether R is an equivalence relation. Let M R be the matrix of R. Compute M 8 R. Solution: The following is the graph of the relation. 1 2 3 4 5 6 7 8 9 10 25

Then MR 5 is a Boolean matrix all whose entries are 1. Thus M8 R is the same as MR 5. 8. A relation R on a set X is called a preference relation if R is reflexive and transitive. Show that R R 1 is an equivalence relation. Proof. Obviously, R R 1 is reflexive. If x(r R 1 )y, then xry and xr 1 y, i.e., yr 1 x and yrx. Hence y(r R 1 )x. So R R 1 is symmetric. If x(r R 1 )y and y(r R 1 )z, then xry, yrz, yrx, zry, thus xrz and zrx, i.e., xrz and xr 1 z. Therefore x(r R 1 )z. So R R 1 is transitive. 9. Let n be a positive integer. The congruence relation of modulo n is an equivalence relation on Z. Let Z n denote the quotient set Z/. For any integer a Z, we define a function f a : Z n Z n, (a) Find the cardinality of the set f a (Z n ). f a ([x]) = [ax]. (b) Find all integers a such that f a is invertible. Solution: (a) Let d = gcd(a, n), a = kd, n = ld. Let x = ql + r. Then ax = kd(ql + r) = kdql + kdr = kqn + ar ar(mod n). For 1 r 1 < r 2 < l, we claim that ar 1 ar 2 (mod n). In fact, if ar 1 ar 2 (mod n), then n a(r 2 r 1 ) or equivalently l k(r 2 r 1 ). Since gcd(k, l) = 1, we have l (r 2 r 1 ). Thus r 1 = r 2, a contradiction. Therefore n f a (Z n ) = l = gcd(a, n). (b) Since Z n is finite, then f a is a bijection if and only if f a is onto. However, f a is onto if and only if f a (Z n ) = n, i.e., gcd(a, n) = 1. 10. For a positive integer n, let φ(n) be the number of positive integers x n such that gcd(x, n) = 1, called Euler s function. Let R be the relation on X = {1, 2,...,n} defined by xry x y, y n, gcd(x, y) = 1. 26

(a) Find the cardinality R 1 (y) for each y X. (b) Show that R = φ(x). x n (c) Prove R = n by showing that the function f : R X, defined by f(x, y) = xn y, is a bijection. Solution: (a) For each y X and y n, we have R 1 (y) = {x X x y, gcd(x, y) = 1} = φ(y). (b) It follows obviously that R = y X R 1 (y) = y 1, y n R 1 (y) = y n φ(y). (c) It is clear that the function f is well-defined. For x 1 Ry 1 and x 2 Ry 2, if = x 2n y 2, i.e., x 1 y 1 = x 2 y 2, then (x 1, y 1 ) and (x 2, y 2 ) are integral proportional, x 1 n y 1 say, (x 2, y 2 ) = c(x 1, y 1 ). Since gcd(x 1, y 1 ) = gcd(x 2, y 2 ) = 1, then c = 1. We thus have (x 2, y 2 ) = (x 1, y 1 ). On the other hand, for any z X, let d = gcd(z, n). Then ( z f d, n = d) (z/d) n = z. n/d This means that f is surjective. Thus f is a bijection. 11. Let X be a set of n elements. Show that the number of equivalence relations on X is n n ( 1) k (l k) n k!(l k)!. k=0 l=k (Hint: Each equivalence relation corresponds to a partition. Counting number of equivalence relations is the same as counting number of partitions.) Proof. Note that partitions of X with k parts are in one-to-one correspondent with surjective functions from X to {1, 2,...,k}. By the inclusion-exclusion 27

principle, the number of such surjective functions is k ( ) k ( 1) i (k i) n i k=1 i=0 Thus the answer is given by n k ( ) k k! ( 1) i (k i) n = i i=0 (Note that for k = 0, the sum k i=0 ( 1)i ( k i n ( 1) i i=0 n k=i ) (k i) n = 0.) (k i) n i!(k i)!. 28