Math 42, Discrete Mathematics

Transcription

1 c Fall 2018 last updated 12/05/2018 at 15:47:21 For use by students in this class only; all rights reserved. Note: some prose & some tables are taken directly from Kenneth R. Rosen, and Its Applications, 8th Ed., the ocial text adopted for this course.

2 Binary from A to B Denition Let A and B be sets. A binary relation from A to B is a subset of A B Suppose R A B is a relation from A to B. If (a, b) R, we write arb and say that a is related to b by R. If (a, b) / R, we write a Rb. Remark A relation as dened above is called binary because it involves ordered pairs. We could talk about ternary relations involving ordered triples, quaternary relations involving ordered quadruples, n-ary relations involving ordered n-tuples, etc. Since we will restrict our investigation to the binary case, we will henceforth simply speak of relations," instead of binary relations."

3 Examples Examples 1. Suppose A = {students in this Math 42 class} and B = {countries in the world other than the US}. Let R 1 = {(a, b) a has visited b}. 2. If f : A B is a function, then R 2 = {(a, f(a)) a A} is a relation, called the graph of the function. Remarks Conversely, if R A B is a relation with the property that every element of A is the rst entry in precisely one ordered pair of R, then setting f(a) = b whenever (a, b) R denes a function f : A B. Note that R 1 illustrates the fact that a given a A may correspond under a relation to many elements b Bor none.

4 Examples 3. R 3 = A B. 4. R 4 = A B A B. For the most part, we will be interested in relations where B = A. Denition A relation on A is a subset of A A, i.e., a relation from A to A. Examples The relations 3 and 4 above are relations on A when B = A. 5. For any set A, R 5 = {(a, a) a A} is a relation on A. This relation, of course, is just equality, and we write a 1 = a 2 " instead of a 1 R 5 a 2."

5 Examples Examples 6. R 6 = {(a 1, a 2 ) a 1 a 2 }. We write a 1 a 2 " instead of a 1 R 6 a 2." For the following examples, let A = Z R 7 = {(a 1, a 2 ) a 1 a 2 }. We write a 1 a 2 " instead of a 1 R 7 a 2." 8. R 8 = {(a 1, a 2 ) a 1 > a 2 }. We write a 1 > a 2 " instead of a 1 R 8 a 2." Remark 7 and 8 can be dened for A any subset of R. 9. R 9 = {(a 1, a 2 ) a 1 divides a 2 }. We write a 1 a 2 " instead of a 1 R 9 a 2."

6 Examples Examples 10. If A = P(S), for some set S, we may dene the relation R 10 = {(X, Y) X Y}. We write X Y" instead of XR 10 Y." For the following examples, let A = {all triangles}. 11. R 11 = {(a 1, a 2 ) a 1 is congruent to a 2 }. We write a 1 = a2 " instead of a 1 R 11 a 2." 12. R 12 = {(a 1, a 2 ) a 1 is similar to a 2 }. We write a 1 a 2 " instead of a 1 R 12 a 2." Remark For technical reasons, it turns out that there is no set of all triangles." But at this point there's no harm in pretending that there is.

7 Examples For the following two examples, let A be the set of all human beings, living and dead. 13. R 13 = {(a 1, a 2 ) a 1 is Facebook friends with a 2 }. 14. R 14 = {(a 1, a 2 ) a 1 loves a 2 }. 15. Let A = {(p, q) p, q Z q 0} and dene (p 1, q 1 )R 15 (p 2, q 2 ) if and only if p 1 q 2 = p 2 q 1. Warning The set A is already a set of ordered pairs. So the relation R 15 is a set of ordered pairs of ordered pairs: ((p 1, q 1 ), (p 2, q 2 )). This relation is much more complicated than the others but it is nonetheless very important.

8 Properties of on A Denitions Let R be a relation on A. R is called reexive if ara for all a A. That is, if a A, (a, a) R. R is called symmetric if a 1 Ra 2 implies a 2 Ra 1. That is, if (a 1, a 2 ) A A, ((a 1, a 2 ) R) ((a 2, a 1 ) R). R is called antisymmetric if a 1 Ra 2 and a 2 Ra 1 implies a 1 = a 2. R is called transitive if a 1 Ra 2 and a 2 Ra 3 implies a 1 Ra 3. That is, if (a 1, a 2, a 3 ) A A A, (((a 1, a 2 ) R) (a 2, a 3 ) R)) ((a 1, a 3 ) R)

9 Properties of Example Table 1: Properties of Example R S T A R3 Y Y Y R4 Y Y Y R5 Y Y Y Y R6 Y R7 Y Y Y R8 Y Y R9 Y Y Y R10 Y Y Y R11 Y Y Y R12 Y Y Y R13 Y R14 R15 Y Y Y

10 A type of relation on A that is very important in many branches of mathematics is an equivalence relation. Denition A relation R on a set A is called an equivalence relation on A if R is 1. reexive, 2. symmetric, & 3. transitive. Example Let A = Z and x a positive integer n. Dene a relation R 16 on A by R 16 = {(r, s) r s = nk for some k Z} = {(r, s) n divides r s evenly} (1)

11 Examples Claim R 16 is an equivalence relation on Z. Proof. Reexive If r Z, we must show that rr 16 r. But r r = 0 = n 0, which is what we need. Symmetric If rr 16 s, we must show sr 16 r. rr 16 s gives us that r s = nk for some k Z. But then s r = nk = n( k), and k Z. So we have what we need. Transitive If rr 16 s and sr 16 t, then r s = nk and s t = nl for some k, l Z. Then r t = (r s) + (s t) = nk + nl = n(k + l). And since (k + l) Z, we have rr 16 t, which is what we needed.

12 Examples Remarks If (r, s) R 16 i.e., rr 16 swe write r s (mod n) and say r is congruent to s modulo n." If n = 2, r s (mod 2) if and only if r and s have the same parity, i.e., are both odd or both even.

13 Examples Example Recall that when A = {(p, q) p, q Z q 0}, we dened (p 1, q 1 )R 15 (p 2, q 2 ) if and only if p 1 q 2 = p 2 q 1. Claim R 15 is an equivalence relation on Z (Z {0}). Proof. Reexive If (p, q) A, then pq = pq. So (p, q)r 15 (p, q), as needed. Symmetric If (p 1, q 1 )R 15 (p 2, q 2 ), then p 1 q 2 = p 2 q 1. But then p 2 q 1 = p 1 q 2. So (p 2, q 2 )R 15 (p 1, q 1 ), as desired.

14 Examples Example (cont'd) Proof. (cont'd) Transitive If (p 1, q 1 )R 15 (p 2, q 2 ) and (p 2, q 2 )R 15 (p 3, q 3 ), then p 1 q 2 = p 2 q 1, and p 2 q 3 = p 3 q 2. So p 1 q 3 q 2 = p 2 q 3 q 1 = p 2 q 1 q 3 = p 3 q 1 q 2. (2) But the extreme left and right ends of (2) both contain a factor of q 2 0. Dividing by q 2 gives p 1 q 3 = p 3 q 1, and hence the desired (p 1, q 1 )R 15 (p 3, q 3 ).

15 Denition Two elements a 1, a 2 A that are related by an equivalence relation Ri.e., that satisfy a 1 Ra 2 are called equivalent. The notation a 1 a 2 is often used to denote that a 1 and a 2 are equivalent with respect to a particular equivalence relation. (Although specic equivalence relations like congruence mod n, and, of course, equality, have their own symbols, and =.)

16 Classes Denition If is an equivalence relation on A, and if a A, then we dene [a], the equivalence class of a, by [a] = {b A b a}. (3) Remark In cases where more than one equivalence relation on A is under consideration, we write [a] R = {b A bra} to emphasize the particular equivalence relation R.

17 Classes Examples Example Recall that for a xed positive integer n, we dened the relation R 16 by R 16 = {(r, s) r s = nk for some k Z} = {(r, s) n divides r s evenly} Moreover, we proved that R 16 is an equivalence relation. So what are the equivalence classes for R 16? If m Z, we have [m] = {k Z k m mod n} = {k Z k m = nl for some l Z} = {k Z k = m + nl for some l Z} = {m, m ± n, m ± 2n,...} (4) (5)

18 Classes Examples Example Recall that when A = {(p, q) p, q Z q 0}, we dened (p 1, q 1 )R 15 (p 2, q 2 ) if and only if p 1 q 2 = p 2 q 1. Furthermore, we proved that R 15 is an equivalence relation. So what are the equivalence classes for R 15? If (p, q) A, we have [(p, q)] = {(r, s) A (r, s) (p, q)} (6) Case 1 (r, s) (0, q) if and only if r = 0. So [(0, q)] = {(0, s) s 0} (7)

19 Classes Examples Example (cont'd) Case 2 If p 0, (r, s) (p, q) rq = ps r p = s q = v Q {0} (8) so (r, s) = (vp, vq). Thus, [(p, q)] = {(r, s) (r, s) (p, q)} = {(vp, vq) v Q {0}} A (9)

20 Classes Examples Example (cont'd) For example, [(1, 2)] = {(1, 2), ( 1, 2), (2, 4), ( 2, 4), (3, 6),...} (10) Remark When we use the rational numbersnumbers of the form p/q where p, q Z and q 0we're actually dealing with the equivalence classes discussed in this example because we're using the fact that 1 2 = 1 2 = 2 4 = 2 4 = 3 6 = 3 6 =... That's why this rather complicated example is important.

21 Partitions Denition A partition of a set S is a collection of disjoint nonempty subsets of S whose union is all of S. In other words, P = {A i A i S i I} is a partition of S if and only if 1. A i i I; 2. A i A j = i j I; and 3. i I A i = S. Remark We call the A i in a partition mutually exclusive and exhaustive because they are mutually exclusive and together they exhaust all the possible elements of S.

22 Partitions Figure 1: A Partition of a Set

23 Partitions Partitions are intimately related to equivalence relations, as the following theorem shows. Theorem Let R be an equivalence relation on a set S. Then the equivalence classes of R form a partition of S. Conversely, given a partition {A i i I} of the set S, there is an equivalence relation R that has the sets A i, i I, as its equivalence classes.

24 Partitions Proof. " Assume that R is an equivalence relation on S and Let P = {[s] s S} be the set of its equivalence classes. We must show that P is a partition of S. 1. For each equivalence class [s] P, s [s] by reexivity, so [s]. 2. Given [s], [t] P, we must show that either [s] = [t] or [s] [t] =. Suppose u [s] [t], i.e., [s] [t]. Then u s and u t. Now if v [s], then v s. But symmetry of the relation gives that s u, and transitivity then implies that v u; transitivity again gives v t. Hence [s] [t]. A similar argument gives that [t] [s], and so [s] = [t].

25 Partitions Proof. (cont'd) 3. s S [s] S since [s] S s S. On the other hand, for every s S, s [s] s S [s]. Thus s S [s] = S, as needed. " If P = {A i i I} is a partition of S, we must show that P denes an equivalence relation R on S whose equivalence classes are the A i. Dene the relation R by s t if and only if s, t A i0 for some i 0 I, i.e., if and only if s and t are in the same piece of the partition. Reexive If s S, then since i I A i = S, we must have s A i0 for some i 0 I. But since s is in the same A i0 as itself, s s.

26 Partitions Proof. (cont'd) Symmetric If s t, then s, t A i0 for some i 0 I. But then t, s A i0, and so t s. Transitive If s t and t v, then s, t A i0 for some i 0 I, and t, v A j0 for some j 0 I. Then t A i0 A j0. This implies that A j0 = A i0, so s, v A i0, and thus s v, as needed.

27 Partitions Example Example Let A = R. And consider P = {[k + i/ , k + i/ ) k Z, i = 0,..., 99} (11) Claim P is a partition of A = R. So that means that P gives rise to an equivalence relation. Dene x y if and only if 100x = 100y This is the equivalence relation that gives rise to the partition P.

28 Partitions Example Example (cont'd) What's it all about? This is the equivalence relation we get when we round a number to the nearest hundredth. A similar argument yields the equivalence relation of rounding to any specied decimal place.

29 Partitions Example Example Suppose f : X Y is a function. Dene a relation on X by x 1 x 2 if and only if f(x 1 ) = f(x 2 ). Show that is an equivalence relation. Proof. Reexive For each x X, f(x) = f(x) since = is reexive. Thus x x. Symmetric If x 1 x 2, then f(x 1 ) = f(x 2 ). But = is symmetric, so f(x 2 ) = f(x 1 ), and thus x 2 x 1. Transitive x 1 x 2 and x 2 x 3, then f(x 1 ) = f(x 2 ) and f(x 2 ) = f(x 3 ). But then f(x 1 ) = f(x 3 ) by the transitivity of =, and so x 1 x 3, as needed.

30 Partitions Example Example (cont'd) Remark Note that the same resultwith the same proofwould hold if were an equivalence relation on Y and we dened on X by x 1 x 2 if and only if f(x 1 ) f(x 2 ) Suppose that f : R R is given by f(x) = sin x and that we dene x 1 x 2 if and only if f(x 1 ) = f(x 2 ). What is [π/2], i.e., the equivalence class represented by π/2? [π/2] = {x R sin x = sin π/2 = 1} = {π/2 + k2π k Z}. (12) Note that x 2 x 1 if and only if x 2 x 1 = k 2π for some k Z. Does this look familiar?

31 Partitions Example Let A = R and dene on A by where k Z. a b if and only if a b = k 2π, (13) Remarks The relation dened in (13) is an equivalence relation. The proof is similar to the proof that congruence modulo n is an equivalence relation. The relation is the same relation as the one dened by the sine function above.

32 Matrices Denition A matrix is a rectangular array of numbers. A matrix with m rows and n columns is called an m n matrixpronounced m by n matrix. The plural of matrix is matrices. There's no such thing as a matrice. A matrix with the same number of rows as columns is called square. Two matrices are equal if they have the same number of rows and the same number of columns and the corresponding entries in every position are equal.

33 Matrices A generic matrix looks like a 11 a 12 a 1n a 21 a 22 a 2n A =..., (14) a m1 a m2 a mn or A = [a ij ], for short. We call a ij the (i, j)th entry or element of A. The left-hand subscript gives the row number, while the right-hand subscript gives the column number. Thus a ij gives the element that occupies the intersection of the i-th row and the j-th column. Remark Matrices are ubiquitous in science, mathematics, and even business, but we will be interested only in binary matrices, matrices whose entries are only 0's and 1's.

34 and Matrices Given a relation R on the set A = {a 1,..., a n }, we build a binary matrix M = [m ij ] as follows: m ij = { 1 if a i Ra j 0 if a i Ra j. (15)

35 and Matrices Example Let A = {2, 3, 4, 5, 6} and suppose R is dened by arb if and only if a b mod 3. Then R = {(2, 2), (2, 5), (3, 3), (3, 6), (4, 4), (5, 2), (5, 5), (6, 3), (6, 6)}. The matrix M associated to R is M = (16)

36 and Matrices Could we detect Reexivity in R by looking at M? How? Note that for Reexivity, we need ara for all a A. But since the elements of A are all numbered, that means we need a i Ra i for all i = 1,..., n. And that will be true exactly when m ii = 1 for all i = 1,..., n, i.e., when all the diagonal entries of M are 1's. Could we detect Symmetry in R by looking at M? How? For Symmetry, we need a j Ra i whenever we have a i Ra j, i.e., we need m ji = 1 whenever m ij = 1. On the other hand, if m ij = 0 and m ji = 1, we would have a i Ra j and a j Ra i, contradicting symmetry. So we conclude that R will be symmetric if and only if m ji = m ij for all i, j = 1,..., n. A square matrix M is called symmetric if m ji = m ij for all i, j = 1,..., n. So R will be symmetric exactly when its corresponding matrix M is symmetric.

37 Boolean Product of Matrices Denition Let S = {0, 1}, we dene two operations on S as follows: { 1 if a 1 = a 2 = 1 a 1 a 2 = 0 otherwise, a 1 a 2 = { 0 if a 1 = a 2 = 0 1 otherwise. (17) (18) Remark Note that if we interpret 1 as T and 0 as F, these are precisely the operations of disjunction ( ) and conjunction ( ) we studied in Chapter 1.

38 Denition Given two n n binary matrices A and B, the Boolean product C = A B is the n n binary matrix whose (i, j)th entry c ij is given by: c ij = (a i1 b 1j ) (a i2 b 2j ) (a in b nj ) (19) (See p. 192, Rosen, 8th Ed., for this denition.) Remark We denote A A by A [2]. Denition Given two n n binary matrices N and M, we say N M if n ij m ij for all i, j = 1,..., n.

39 Binary Matrices and Transitivity Facts Suppose R 1 and R 2 are two relations on A = {a 1,..., a n }, and M R1 and M R2 are their binary matrices, then M R1 M R2 if and only if R 1 R 2. A relation R on A = {a 1,..., a n } is transitive, if and only if M [2] R M R.

40 with Digraphs Denition A directed graph, or digraph, consists of a set V of vertices (or nodes) together with a set E of ordered pairs of elements of V called edges (or arcs). The vertex a is called the initial vertex of the edge (a, b) and the vertex b is called the terminal vertex of this edge. Remark Note any relation R on a set A = {a 1,..., a n } immediately gives rise to a digraph: let V = A and let E = R. This works because R is already a set of ordered pairs of the elements of V = A.

41 Digraphs Figure 2: Digraph 1

42 with Digraphs Example The matrix for the relation R represented by the digraph in Figure 2 is M R = (20) Is R reexive? No, since not all the diagonal entries of M R are 1s: m 11 = 0 = m 22 Is R symmetric? No, since M R is not a symmetric matrix. is R transitive? Well, we need to compute M [2] R.

43 with Digraphs Example (cont'd) The matrix M [2] R in Figure 2 is for the relation represented by the digraph N R = M [2] R = (21) Since n 11 = 1 and m 11 = 0, R cannot be transitive. Why is n 11 = 1? Because, for example, m 13 m 31 = 1 1 = 1. This means that a 1 Ra 3 and a 3 Ra 1. Transitivity would then imply a 1 Ra 1. But m 11 = 0, so a 1 Ra 1, a contradiction.

44 with Digraphs Example (cont'd) Remark Note that n 14 = 0 while m 14 = 1. What's the signicance of this? The fact that m ij = 1 encodes the fact that we can get from a i to a j in one step, while n ij = 0 encodes the impossibility of getting from a i to a j in exactly two steps. So here we can get from a 1 to a 4 in one step but not in two steps.

45 Digraphs Figure 3: Digraph 2

46 with Digraphs Example The matrix for the relation R represented by the digraph in Figure 3 is M R = (22) Is R reexive? Yes, since all the diagonal entries of M R are 1s. Is R symmetric? Yes, since M R is a symmetric matrix. is R transitive? Well, we need to compute M [2] R.

47 with Digraphs Example (cont'd) The matrix M [2] R in Figure 3 is for the relation represented by the digraph N R = M [2] R = (23) Since n 34 = 1 and m 34 = 0, R cannot be transitive. Why is n 34 = 1? Because, for example, m 31 m 14 = 1 1 = 1. This means that a 3 Ra 1 and a 1 Ra 4. Transitivity would then imply a 3 Ra 4. But m 34 = 0, so a 3 Ra 4, a contradiction.

48 Example (cont'd) Now suppose we consider the relation represented by the digraph in Figure 3 but with A = {a, c, d, b}, i.e., with the elements of A written in a dierent order. The new matrix M R will be M R = (24) Notice that we can partition the matrix M R into blocks as follows: M R = (25)

49 Digraphs Figure 4: Digraph 3

50 with Digraphs Example The matrix for the relation R represented by the digraph in Figure 4 is M R = (26) Is R reexive? Yes, since all the diagonal entries of M R are 1s. Is R symmetric? No, since M R is not a symmetric matrix. is R antisymmetric? Well, we haven't looked at that yet.

51 with Digraphs Example (cont'd) Recall that R is antisymmetric if and only if a i Ra j and a j Ra i together imply that a i = a j, i.e., i = j. In terms of the matrix M R, this means that if i j and m ij = 1, then we must have m ji = 0. M R satises this condition, so R is antisymmetric. The matrix M [2] R for the relation represented by the digraph in Figure 4 is N R = M [2] R = = = M R (27) But since N R M R, R is transitive.