Week 5: Binary Relations
|
|
- Linette Howard
- 7 years ago
- Views:
Transcription
1 1 Binary Relations Week 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
2 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
3 (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
4 (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
5 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 e b d c a e b d 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
6 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 and the matrix of the brother or sister relation is the square matrix 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 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
7 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
8 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 2 x 3 x 4 R y y y S z z z z z x 1 x 2 x 3 x 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
9 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
10 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
11 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. The 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
12 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
13 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
14 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
15 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
16 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
17 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.
18 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
19 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
20 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 M R =
21 By Warshall s algorithm, we have W 0 = W 1 = W 2 = W 3 = W 4 = W 5 = (3, 1), (1, 5) (no change) (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
22 The binary relation for the Boolean matrix W 5 is the transitive closure of R 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
23 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 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
24 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 M 1 =, M 2 =, M 3 = (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
25 Solution: M R 1 1 = M R1 R 1 1 = , M R2 R 3 = = M 2, M R 1 1 R 1 =, M R 2 = 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
26 Then MR 5 is a Boolean matrix all whose entries are 1. Thus M R 8 is the same as MR 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) = 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
27 (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
28 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
Chapter 3. Cartesian Products and Relations. 3.1 Cartesian Products
Chapter 3 Cartesian Products and Relations The material in this chapter is the first real encounter with abstraction. Relations are very general thing they are a special type of subset. After introducing
More informationLecture 16 : Relations and Functions DRAFT
CS/Math 240: Introduction to Discrete Mathematics 3/29/2011 Lecture 16 : Relations and Functions Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT In Lecture 3, we described a correspondence
More informationLecture 17 : Equivalence and Order Relations DRAFT
CS/Math 240: Introduction to Discrete Mathematics 3/31/2011 Lecture 17 : Equivalence and Order Relations Instructor: Dieter van Melkebeek Scribe: Dalibor Zelený DRAFT Last lecture we introduced the notion
More informationDiscrete Mathematics. Hans Cuypers. October 11, 2007
Hans Cuypers October 11, 2007 1 Contents 1. Relations 4 1.1. Binary relations................................ 4 1.2. Equivalence relations............................. 6 1.3. Relations and Directed Graphs.......................
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationMathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson
Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement
More information8 Divisibility and prime numbers
8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express
More informationMATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set.
MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set. Vector space A vector space is a set V equipped with two operations, addition V V (x,y) x + y V and scalar
More informationQuotient Rings and Field Extensions
Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.
More informationINTRODUCTORY SET THEORY
M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H-1088 Budapest, Múzeum krt. 6-8. CONTENTS 1. SETS Set, equal sets, subset,
More informationCartesian Products and Relations
Cartesian Products and Relations Definition (Cartesian product) If A and B are sets, the Cartesian product of A and B is the set A B = {(a, b) :(a A) and (b B)}. The following points are worth special
More informationNOTES ON LINEAR TRANSFORMATIONS
NOTES ON LINEAR TRANSFORMATIONS Definition 1. Let V and W be vector spaces. A function T : V W is a linear transformation from V to W if the following two properties hold. i T v + v = T v + T v for all
More informationCHAPTER 5. Number Theory. 1. Integers and Division. Discussion
CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a
More informationa 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.
Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given
More informationHomework until Test #2
MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such
More informationSystems of Linear Equations
Systems of Linear Equations Beifang Chen Systems of linear equations Linear systems A linear equation in variables x, x,, x n is an equation of the form a x + a x + + a n x n = b, where a, a,, a n and
More informationLemma 5.2. Let S be a set. (1) Let f and g be two permutations of S. Then the composition of f and g is a permutation of S.
Definition 51 Let S be a set bijection f : S S 5 Permutation groups A permutation of S is simply a Lemma 52 Let S be a set (1) Let f and g be two permutations of S Then the composition of f and g is a
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More informationMatrix Representations of Linear Transformations and Changes of Coordinates
Matrix Representations of Linear Transformations and Changes of Coordinates 01 Subspaces and Bases 011 Definitions A subspace V of R n is a subset of R n that contains the zero element and is closed under
More information= 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that
Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without
More informationEvery Positive Integer is the Sum of Four Squares! (and other exciting problems)
Every Positive Integer is the Sum of Four Squares! (and other exciting problems) Sophex University of Texas at Austin October 18th, 00 Matilde N. Lalín 1. Lagrange s Theorem Theorem 1 Every positive integer
More information26 Ideals and Quotient Rings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 26 Ideals and Quotient Rings In this section we develop some theory of rings that parallels the theory of groups discussed
More information1. Prove that the empty set is a subset of every set.
1. Prove that the empty set is a subset of every set. Basic Topology Written by Men-Gen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since
More informationTHE NUMBER OF GRAPHS AND A RANDOM GRAPH WITH A GIVEN DEGREE SEQUENCE. Alexander Barvinok
THE NUMBER OF GRAPHS AND A RANDOM GRAPH WITH A GIVEN DEGREE SEQUENCE Alexer Barvinok Papers are available at http://www.math.lsa.umich.edu/ barvinok/papers.html This is a joint work with J.A. Hartigan
More information. 0 1 10 2 100 11 1000 3 20 1 2 3 4 5 6 7 8 9
Introduction The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive integer We say d is a
More informationNotes on Algebraic Structures. Peter J. Cameron
Notes on Algebraic Structures Peter J. Cameron ii Preface These are the notes of the second-year course Algebraic Structures I at Queen Mary, University of London, as I taught it in the second semester
More informationCS 103X: Discrete Structures Homework Assignment 3 Solutions
CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering
More informationLecture 13 - Basic Number Theory.
Lecture 13 - Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are non-negative integers. We say that A divides B, denoted
More informationSCORE SETS IN ORIENTED GRAPHS
Applicable Analysis and Discrete Mathematics, 2 (2008), 107 113. Available electronically at http://pefmath.etf.bg.ac.yu SCORE SETS IN ORIENTED GRAPHS S. Pirzada, T. A. Naikoo The score of a vertex v in
More informationDiscrete Maths. Philippa Gardner. These lecture notes are based on previous notes by Iain Phillips.
Discrete Maths Philippa Gardner These lecture notes are based on previous notes by Iain Phillips. This short course introduces some basic concepts in discrete mathematics: sets, relations, and functions.
More informationNumber Theory. Proof. Suppose otherwise. Then there would be a finite number n of primes, which we may
Number Theory Divisibility and Primes Definition. If a and b are integers and there is some integer c such that a = b c, then we say that b divides a or is a factor or divisor of a and write b a. Definition
More informationGeometric Transformations
Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted
More informationSUM OF TWO SQUARES JAHNAVI BHASKAR
SUM OF TWO SQUARES JAHNAVI BHASKAR Abstract. I will investigate which numbers can be written as the sum of two squares and in how many ways, providing enough basic number theory so even the unacquainted
More informationCHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY
January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.
More informationMath 319 Problem Set #3 Solution 21 February 2002
Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod
More informationInner Product Spaces
Math 571 Inner Product Spaces 1. Preliminaries An inner product space is a vector space V along with a function, called an inner product which associates each pair of vectors u, v with a scalar u, v, and
More informationON SOME CLASSES OF GOOD QUOTIENT RELATIONS
Novi Sad J. Math. Vol. 32, No. 2, 2002, 131-140 131 ON SOME CLASSES OF GOOD QUOTIENT RELATIONS Ivica Bošnjak 1, Rozália Madarász 1 Abstract. The notion of a good quotient relation has been introduced as
More informationABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS
ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS John A. Beachy Northern Illinois University 2014 ii J.A.Beachy This is a supplement to Abstract Algebra, Third Edition by John A. Beachy and William D. Blair
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationTest1. Due Friday, March 13, 2015.
1 Abstract Algebra Professor M. Zuker Test1. Due Friday, March 13, 2015. 1. Euclidean algorithm and related. (a) Suppose that a and b are two positive integers and that gcd(a, b) = d. Find all solutions
More informationAbstract Algebra Theory and Applications. Thomas W. Judson Stephen F. Austin State University
Abstract Algebra Theory and Applications Thomas W. Judson Stephen F. Austin State University August 16, 2013 ii Copyright 1997-2013 by Thomas W. Judson. Permission is granted to copy, distribute and/or
More informationAlgorithmic Mathematics
Algorithmic Mathematics a web-book by Leonard Soicher & Franco Vivaldi This is the textbook for the course MAS202 Algorithmic Mathematics. This material is in a fluid state it is rapidly evolving and as
More informationThe last three chapters introduced three major proof techniques: direct,
CHAPTER 7 Proving Non-Conditional Statements The last three chapters introduced three major proof techniques: direct, contrapositive and contradiction. These three techniques are used to prove statements
More informationSolutions to In-Class Problems Week 4, Mon.
Massachusetts Institute of Technology 6.042J/18.062J, Fall 05: Mathematics for Computer Science September 26 Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld revised September 26, 2005, 1050 minutes Solutions
More informationZachary Monaco Georgia College Olympic Coloring: Go For The Gold
Zachary Monaco Georgia College Olympic Coloring: Go For The Gold Coloring the vertices or edges of a graph leads to a variety of interesting applications in graph theory These applications include various
More information1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]
1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not
More informationLecture 15 An Arithmetic Circuit Lowerbound and Flows in Graphs
CSE599s: Extremal Combinatorics November 21, 2011 Lecture 15 An Arithmetic Circuit Lowerbound and Flows in Graphs Lecturer: Anup Rao 1 An Arithmetic Circuit Lower Bound An arithmetic circuit is just like
More information8 Primes and Modular Arithmetic
8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.
More informationMath 55: Discrete Mathematics
Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 5, due Wednesday, February 22 5.1.4 Let P (n) be the statement that 1 3 + 2 3 + + n 3 = (n(n + 1)/2) 2 for the positive integer n. a) What
More informationDegree Hypergroupoids Associated with Hypergraphs
Filomat 8:1 (014), 119 19 DOI 10.98/FIL1401119F Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: http://www.pmf.ni.ac.rs/filomat Degree Hypergroupoids Associated
More informationCONTENTS 1. Peter Kahn. Spring 2007
CONTENTS 1 MATH 304: CONSTRUCTING THE REAL NUMBERS Peter Kahn Spring 2007 Contents 2 The Integers 1 2.1 The basic construction.......................... 1 2.2 Adding integers..............................
More informationMetric Spaces Joseph Muscat 2003 (Last revised May 2009)
1 Distance J Muscat 1 Metric Spaces Joseph Muscat 2003 (Last revised May 2009) (A revised and expanded version of these notes are now published by Springer.) 1 Distance A metric space can be thought of
More informationIntroduction to Topology
Introduction to Topology Tomoo Matsumura November 30, 2010 Contents 1 Topological spaces 3 1.1 Basis of a Topology......................................... 3 1.2 Comparing Topologies.......................................
More informationChapter 7: Products and quotients
Chapter 7: Products and quotients Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 7: Products
More informationThis asserts two sets are equal iff they have the same elements, that is, a set is determined by its elements.
3. Axioms of Set theory Before presenting the axioms of set theory, we first make a few basic comments about the relevant first order logic. We will give a somewhat more detailed discussion later, but
More informationJust the Factors, Ma am
1 Introduction Just the Factors, Ma am The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive
More informationBreaking The Code. Ryan Lowe. Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and
Breaking The Code Ryan Lowe Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and a minor in Applied Physics. As a sophomore, he took an independent study
More informationMATHEMATICAL INDUCTION. Mathematical Induction. This is a powerful method to prove properties of positive integers.
MATHEMATICAL INDUCTION MIGUEL A LERMA (Last updated: February 8, 003) Mathematical Induction This is a powerful method to prove properties of positive integers Principle of Mathematical Induction Let P
More informationMATH10212 Linear Algebra. Systems of Linear Equations. Definition. An n-dimensional vector is a row or a column of n numbers (or letters): a 1.
MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0-534-40596-7. Systems of Linear Equations Definition. An n-dimensional vector is a row or a column
More informationHandout #Ch7 San Skulrattanakulchai Gustavus Adolphus College Dec 6, 2010. Chapter 7: Digraphs
MCS-236: Graph Theory Handout #Ch7 San Skulrattanakulchai Gustavus Adolphus College Dec 6, 2010 Chapter 7: Digraphs Strong Digraphs Definitions. A digraph is an ordered pair (V, E), where V is the set
More informationNo: 10 04. Bilkent University. Monotonic Extension. Farhad Husseinov. Discussion Papers. Department of Economics
No: 10 04 Bilkent University Monotonic Extension Farhad Husseinov Discussion Papers Department of Economics The Discussion Papers of the Department of Economics are intended to make the initial results
More informationDiscrete Mathematics & Mathematical Reasoning Chapter 10: Graphs
Discrete Mathematics & Mathematical Reasoning Chapter 10: Graphs Kousha Etessami U. of Edinburgh, UK Kousha Etessami (U. of Edinburgh, UK) Discrete Mathematics (Chapter 6) 1 / 13 Overview Graphs and Graph
More information6.2 Permutations continued
6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of
More informationT ( a i x i ) = a i T (x i ).
Chapter 2 Defn 1. (p. 65) Let V and W be vector spaces (over F ). We call a function T : V W a linear transformation form V to W if, for all x, y V and c F, we have (a) T (x + y) = T (x) + T (y) and (b)
More informationMidterm Practice Problems
6.042/8.062J Mathematics for Computer Science October 2, 200 Tom Leighton, Marten van Dijk, and Brooke Cowan Midterm Practice Problems Problem. [0 points] In problem set you showed that the nand operator
More informationLinear Algebra I. Ronald van Luijk, 2012
Linear Algebra I Ronald van Luijk, 2012 With many parts from Linear Algebra I by Michael Stoll, 2007 Contents 1. Vector spaces 3 1.1. Examples 3 1.2. Fields 4 1.3. The field of complex numbers. 6 1.4.
More informationApplications of Fermat s Little Theorem and Congruences
Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4
More informationGod created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886)
Chapter 2 Numbers God created the integers and the rest is the work of man. (Leopold Kronecker, in an after-dinner speech at a conference, Berlin, 1886) God created the integers and the rest is the work
More informationk, then n = p2α 1 1 pα k
Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square
More informationMath 312 Homework 1 Solutions
Math 31 Homework 1 Solutions Last modified: July 15, 01 This homework is due on Thursday, July 1th, 01 at 1:10pm Please turn it in during class, or in my mailbox in the main math office (next to 4W1) Please
More informationChapter 13: Basic ring theory
Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring
More informationLinear Maps. Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007)
MAT067 University of California, Davis Winter 2007 Linear Maps Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007) As we have discussed in the lecture on What is Linear Algebra? one of
More information3. SETS, FUNCTIONS & RELATIONS
3. SETS, FUNCTIONS & RELATIONS If I see the moon, then the moon sees me 'Cos seeing's symmetric as you can see. If I tell Aunt Maude and Maude tells the nation Then I've told the nation 'cos the gossiping
More information7 Gaussian Elimination and LU Factorization
7 Gaussian Elimination and LU Factorization In this final section on matrix factorization methods for solving Ax = b we want to take a closer look at Gaussian elimination (probably the best known method
More information7.1 RELATIONS AND THEIR PROPERTIES
Section 7.1 Relations and Their Properties 7.1.1 7.1 RELATIONS AND THEIR PROPERTIES def: A binary relation is a predicate on the cartesian product A B of two sets. Sometimes we say from A to B. CONCEPTUALIZATION
More information1 Sets and Set Notation.
LINEAR ALGEBRA MATH 27.6 SPRING 23 (COHEN) LECTURE NOTES Sets and Set Notation. Definition (Naive Definition of a Set). A set is any collection of objects, called the elements of that set. We will most
More information1 if 1 x 0 1 if 0 x 1
Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or
More informationElementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.
Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole
More informationSYSTEMS OF PYTHAGOREAN TRIPLES. Acknowledgements. I would like to thank Professor Laura Schueller for advising and guiding me
SYSTEMS OF PYTHAGOREAN TRIPLES CHRISTOPHER TOBIN-CAMPBELL Abstract. This paper explores systems of Pythagorean triples. It describes the generating formulas for primitive Pythagorean triples, determines
More information7 Relations and Functions
7 Relations and Functions In this section, we introduce the concept of relations and functions. Relations A relation R from a set A to a set B is a set of ordered pairs (a, b), where a is a member of A,
More informationINCIDENCE-BETWEENNESS GEOMETRY
INCIDENCE-BETWEENNESS GEOMETRY MATH 410, CSUSM. SPRING 2008. PROFESSOR AITKEN This document covers the geometry that can be developed with just the axioms related to incidence and betweenness. The full
More informationTriangle deletion. Ernie Croot. February 3, 2010
Triangle deletion Ernie Croot February 3, 2010 1 Introduction The purpose of this note is to give an intuitive outline of the triangle deletion theorem of Ruzsa and Szemerédi, which says that if G = (V,
More informationMATRIX ALGEBRA AND SYSTEMS OF EQUATIONS
MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a
More informationSo let us begin our quest to find the holy grail of real analysis.
1 Section 5.2 The Complete Ordered Field: Purpose of Section We present an axiomatic description of the real numbers as a complete ordered field. The axioms which describe the arithmetic of the real numbers
More informationDiscrete Mathematics Problems
Discrete Mathematics Problems William F. Klostermeyer School of Computing University of North Florida Jacksonville, FL 32224 E-mail: wkloster@unf.edu Contents 0 Preface 3 1 Logic 5 1.1 Basics...............................
More informationMatrix Algebra. Some Basic Matrix Laws. Before reading the text or the following notes glance at the following list of basic matrix algebra laws.
Matrix Algebra A. Doerr Before reading the text or the following notes glance at the following list of basic matrix algebra laws. Some Basic Matrix Laws Assume the orders of the matrices are such that
More informationMath 223 Abstract Algebra Lecture Notes
Math 223 Abstract Algebra Lecture Notes Steven Tschantz Spring 2001 (Apr. 23 version) Preamble These notes are intended to supplement the lectures and make up for the lack of a textbook for the course
More informationSection 4.2: The Division Algorithm and Greatest Common Divisors
Section 4.2: The Division Algorithm and Greatest Common Divisors The Division Algorithm The Division Algorithm is merely long division restated as an equation. For example, the division 29 r. 20 32 948
More informationS on n elements. A good way to think about permutations is the following. Consider the A = 1,2,3, 4 whose elements we permute with the P =
Section 6. 1 Section 6. Groups of Permutations: : The Symmetric Group Purpose of Section: To introduce the idea of a permutation and show how the set of all permutations of a set of n elements, equipped
More informationCOMBINATORIAL PROPERTIES OF THE HIGMAN-SIMS GRAPH. 1. Introduction
COMBINATORIAL PROPERTIES OF THE HIGMAN-SIMS GRAPH ZACHARY ABEL 1. Introduction In this survey we discuss properties of the Higman-Sims graph, which has 100 vertices, 1100 edges, and is 22 regular. In fact
More informationIntroduction to Modern Algebra
Introduction to Modern Algebra David Joyce Clark University Version 0.0.6, 3 Oct 2008 1 1 Copyright (C) 2008. ii I dedicate this book to my friend and colleague Arthur Chou. Arthur encouraged me to write
More informationMathematical Methods of Engineering Analysis
Mathematical Methods of Engineering Analysis Erhan Çinlar Robert J. Vanderbei February 2, 2000 Contents Sets and Functions 1 1 Sets................................... 1 Subsets.............................
More informationGraphs without proper subgraphs of minimum degree 3 and short cycles
Graphs without proper subgraphs of minimum degree 3 and short cycles Lothar Narins, Alexey Pokrovskiy, Tibor Szabó Department of Mathematics, Freie Universität, Berlin, Germany. August 22, 2014 Abstract
More informationON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS
ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for
More informationYou know from calculus that functions play a fundamental role in mathematics.
CHPTER 12 Functions You know from calculus that functions play a fundamental role in mathematics. You likely view a function as a kind of formula that describes a relationship between two (or more) quantities.
More informationMathematical Induction. Mary Barnes Sue Gordon
Mathematics Learning Centre Mathematical Induction Mary Barnes Sue Gordon c 1987 University of Sydney Contents 1 Mathematical Induction 1 1.1 Why do we need proof by induction?.... 1 1. What is proof by
More informationCardinality. The set of all finite strings over the alphabet of lowercase letters is countable. The set of real numbers R is an uncountable set.
Section 2.5 Cardinality (another) Definition: The cardinality of a set A is equal to the cardinality of a set B, denoted A = B, if and only if there is a bijection from A to B. If there is an injection
More informationChapter 17. Orthogonal Matrices and Symmetries of Space
Chapter 17. Orthogonal Matrices and Symmetries of Space Take a random matrix, say 1 3 A = 4 5 6, 7 8 9 and compare the lengths of e 1 and Ae 1. The vector e 1 has length 1, while Ae 1 = (1, 4, 7) has length
More informationCHAPTER 9. Integer Programming
CHAPTER 9 Integer Programming An integer linear program (ILP) is, by definition, a linear program with the additional constraint that all variables take integer values: (9.1) max c T x s t Ax b and x integral
More informationn 2 + 4n + 3. The answer in decimal form (for the Blitz): 0, 75. Solution. (n + 1)(n + 3) = n + 3 2 lim m 2 1
. Calculate the sum of the series Answer: 3 4. n 2 + 4n + 3. The answer in decimal form (for the Blitz):, 75. Solution. n 2 + 4n + 3 = (n + )(n + 3) = (n + 3) (n + ) = 2 (n + )(n + 3) ( 2 n + ) = m ( n
More information