Math 42, Discrete Mathematics

Transcription

1 c Fall 2018 last updated 10/10/2018 at 23:28:03 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 basics Denition A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set A. The notation a / A means that a is not an element of the set A. Remarks Unordered means that the order in which we list items does not matter. So, for example etc. {1, 2, 3, 4} = {4, 3, 2, 1} = {2, 1, 3, 4}, We haven't specied what objects are, or even what a collection is. This ne print is mathematically important. But it's way beyond the level of this course. c R. P.

3 descriptions There are a number of ways of describing sets. Here are the most common: Roster Method. List the elements of the set between braces{ and }separated by commas. This method is most useful when the set doesn't have very many elements. Example: A = {red, green, blue, yellow}. Or B = {2, granite, topology, Proust}. (Nothing says the elements of the set must be alike in any way.) Ellipsis Method. When the elements of the set follow an obvious pattern, we can list the rst few elements followed by an ellipsis (...). Example: E = {2, 4, 6, 8, 10,...}. c R. P.

4 descriptions Set Builder Notation. Examples B = {x Q x = 1/n, 0 n Z} = {x Q : x = 1/n, 0 n Z} C = {2n + 1 n Z} = {2n + 1 : n Z} Remarks Some bookslike oursuse a vertical bar ( ), while others use a colon (:). I will likely be inconsistent in this class, sometimes using one style, sometimes the other. Note the subtle dierence between the two cases above. In the B" case, we want all elements from a given set that satisfy the condition stated after the vertical bar. In the C" case, we want all items of the form given before the vertical bar. c R. P.

5 Set Builder Notation Remark Note that repetition of elements is irrelevant. We list only the distinct elements of the set. Example {n 2 n Z} = {0 2, 1 2, ( 1) 2, 2 2, ( 2) 2,...} = {0, 1, 1, 4, 4,...} = {0, 1, 4,...} (1) c R. P.

6 : standard notation Some important setsand their notation. N = {0, 1, 2, 3,...}, the set of natural numbers. Z = {..., 2, 1, 0, 1, 2,...}, the set of integers. Z + = {1, 2, 3,...}, the set of positive integers. Q = {p/q p, q Z, q 0}, the set of rational numbers. R, the set of real numbers. R + = {x R x > 0}, the set of positive real numbers. C = {x + i y x, y R}, the set of complex numbers. c R. P.

7 : standard notation Some important sets of real numbers and their notation: (a, b) = {x R a < x < b} [a, b) = {x R a x < b} (a, b] = {x R a < x b} [a, b] = {x R a x b} Remark (a, b) is called an open interval. [a, b] is called a closed interval. [a, b) and (a, b] are called half-open intervals. Warning Pay careful attention to which type of numbers you are considering, e.g., real numbers vs. integers. For example, [0, 1) = {x R 0 x < 1} contains innitely many numbers, while {x Z 0 x < 1} contains just one. c R. P.

8 Equality of Denition Two sets are equal if and only if they have the same elements. That is, A = B if and only if ( a A a B) ( b B b A) Example {2n n Z} = {m Z m is divisible by 2} (2) To establish the truth of (2), show that any element in the left-hand set is in the right-hand set and any element in the right-hand set is in the left-hand set. c R. P.

9 Two Special Denition The set that has no elements at all is called the empty set (or null set) and is denoted by or {}. There is only one empty set. Denition A set that contains just one element is called a singleton set. {Juanita}, {0}, { }, { } are all (distinct) singleton sets. c R. P.

10 Subsets Denition A set A is a subset of B if and only every element of A is also an element of B. We denote this by A B. Remarks To show that A B, show that if x A, then x B. To show that A Bi.e., that A is not a subset of Bnd an x A such that x / B. Example Z + Z Q R C. All of these inclusions are, of course, strict inclusions. c R. P.

11 Subsets Remarks A = B if and only if A B and B A. S for every set S. S S for every set S. If A B and A B, we call A a proper subset of B. Example: Is 2Q = {2q q Q} a proper subset of Q? c R. P.

12 Size of a Set Denition If a set S has exactly n distinct elements, with n a nonnegative integer, we say that S is a nite set of cardinality n. We denote that cardinality of S by S, so S = n here. Remark If S has innitely many elements, i.e., more than n elements for any nonnegative integer n, we call S innite. We won't use the expression S =, since it turns out that there are dierent sizes of innity. c R. P.

13 The Power Set of S Denition Given a set S, the power set of S, denoted by P(S), is the set of all subsets of S. P( ) = { } P({a}) = {, {a}} P({a, b}) = {, {a}, {b}, {a, b}} P({a, b, c}) = {, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} Remark If S = n, then P(S) = 2 n. c R. P.

14 Denition Given sets A 1, A 2,..., A n, the ordered n-tuple (a 1, a 2,..., a n ) is an ordered string of elements such that a i A i for i = 1, 2,..., n. Remarks An ordered n-tuple is not a set. Order matters, as does repetition. (1, 2, 3, 3) (3, 3, 2, 1) (1, 2, 3) If n = 2, we say ordered pair." For n = 3, ordered triple." For n = 4, ordered quadruple'.... But for n = 21, we say ordered 21-tuple." Denition The Cartesian product of the sets A 1, A 2,..., A n, denoted by A 1 A 2 A n, is the set of all ordered n-tuples (a 1, a 2,..., a n ) such that a i A i for i = 1, 2,..., n. c R. P.

15 Cartesian ProductsExamples Example Suppose A = {red, green, blue} and B = {0, 1}. Then A B = {(red, 0), (red, 1), (green, 0), (green, 1), (blue, 0), (blue, 1)} (3) B A = {(0, red), (1, red), (0, green), (1, green), (0, blue), (1, blue)} (4) As we see by comparing (3) with (4), it's not generally true that A B = B A. c R. P.

16 Cartesian ProductsExamples Example Suppose A = {a, b} and B = {0, 1}, and C = {α, β}. Then A B C = {(a, 0, α), (a, 0, β), (a, 1, α), (a, 1, β) (5) (b, 0, α), (b, 0, β), (b, 1, α), (b, 1, β)} but (A B) C = {((a, 0), α), ((a, 0), β), ((a, 1), α), ((a, 1), β) ((b, 0), α), ((b, 0), β), ((b, 1), α), ((b, 1), β)} (6) Again, by comparing (5) with (6), we see that A B C (A B) C c R. P.

17 Relations Denition A subset R A B is called a relation from A to B. A subset R A A is called a relation on A. If (a, b) R we write arb; if (a, b) / R we write a Rb. Examples 1. Let R 1 = {(n, m) n divides m evenly} Z + Z +. Then 2R 1 6, but 3 R 1 5. We could call this relation instead of R 1. So 2 6, but Given a set X, dene a relation R 2 on its powerset P(X) by R 2 = {(A, B) B A} P(X) P(X). Then {1, 2}R 2 {2}, but {1} R 2 {1, 2}. We could call this relation instead of R 2. So {1, 2} {2}, but {1} {1, 2}. c R. P.

18 Storing Subsets in a Computer Suppose we have an n-element set S = {a 1, a 2,..., a n } and we want to store some of its subsets eciently in a computer. As the book points out, we can completely describe a subset by simply giving a string of 0's and 1's as follows: Step 1 First list the elements in the set in a specic order: a 1, a 2,..., a n. Step 2 For each subset A S, generate k 1, k 2,..., k n an n-digit string of 0's and 1's such that k i = 1 if a i A and k i = 0 if a i / A. Example Suppose S = {red, green, blue}. Then let a 1 = red, a 2 = green, a 3 = blue. So the subset A = {red, blue} would be stored as 101, while the subset B = {blue, green} would be stored as 011. (Note that the order in which we list items in B doesn't aect the way in which it's stored digitally.) c R. P.

19 Cardinality of P(S) From the preceding discussion, we can get a proof of the conjecture we made earlier that if S = n, then P(S) = 2 n. Fact If A 1, A 2,..., A n are all nite sets, then A 1 A 2... A n = A 1 A 2 A n. So if A i = A for all i and A = m, then A A = m n. In the example under consideration, we have A i = {0, 1} i, so the number of subsets of Swhich is the number of n-digit strings of 0's and 1'sis just A 1 A 2... A n = 2 n. c R. P.

20 Union, Intersection, Dierence, Complement Denitions Let A and B be sets. The union of the sets A and B, denoted by A B, is the set that contains those elements that are either in A or in B, or in both. The intersection of the sets A and B, denoted by A B, is the set containing those elements that are in both A and B. The dierence of A and B, denoted by A B, is the set containing those elements of A that are not in B. The dierence of A and B is also called the complement of B with respect to A. If U is the universal set, U A, the complement of A with respect to U, is called simply the complement of A. It is denoted by Ā. c R. P.

21 Denition Two sets A and B are called disjoint if A B =. Fact If A and B are disjoint, then A B = A + B. (7) Remark Since A = (A B) (A B) and since this is a disjoint union, i.e., the union of disjoint sets, then Similarly, B = B A + B A. A = A B + A B. (8) c R. P.

22 Cardinality of a Union We now note that any union A B can actually be written as a disjoint union: A B = (A B) B. (9) Now we can apply (7) to (9) to get A B = A B + B = ( A A B ) + B = A + B A B. (10) c R. P.

23 Set IdentitiesExample Fact If A, B, and C are sets, then Proof. A (B C) = (A B) (A C) (11) If x A (B C), then x A and x B C. So x A and either x B or x C. That means that either (x A and x B) or (x A and x C). So x (A B) or x (A C), and hence x (A B) (A C). Thus the left-hand set in (11) is contained in the right-hand set. Now if y (A B) (A C), then either y (A B) or y (A C). In the former case, y A (B C) since B (B C). In the latter case, y A (B C) since C (B C). Thus in either case y A (B C), as desired. c R. P.

24 Set IdentitiesExample Fact If A and B are sets, then (A B) = Ā B. (12) Proof. Let x (A B). Then x / (A B). This implies that x / A and x / B, and so x Ā and x B. Thus x (Ā B). So the left-hand set in (12) is contained in the right-hand set. Now if y (Ā B), then y Ā and y B. So y / A and y / B, and therefore y / (A B). Hence y (A B), as desired. c R. P.

25 Denition of a Function Denition A function f : X Y consists of 1. A set X of inputs of the function; X is called the domain of the function. 2. A set Y of potential outputs of the function; Y is called the codomain of the function. 3. A rule that associates to each element x of the domain a unique element f(x) of the codomain. Examples 1. f 1 : R R given by f 1 (x) = x f 2 : R + R given by f 2 (x) = x f 3 : R + R + given by f 3 (x) = x f 4 : Z + Z + given by f 4 (x) = x f 5 : R + R + given by f 5 (x) = x 3. c R. P.

26 Denition Two functions f and g are called equal if 1. domain(f)=domain(g) 2. codomain(f)=codomain(g) 3. f(x) = g(x) for all x domain(f) = domain(g). Remark None of the functions f 1,..., f 5 above is equal to any other. c R. P.

27 Images & Preimages Denitions Given a function f : X Y If f(x) = y, we call y the image of x under f. If f(x) = y, we call x a preimage of y (under f). If A X, we dene f(a), the image of the subset A, by f(a) = {f(x) x A} f(x), the set of all outputs of f, is called the range of the function. Note that we say the image of x but a preimage of y. That's because images are unique but preimages might not be. c R. P.

28 Preimages of Denition Given a function f : X Y and B Y, dene f 1 (B), the inverse image (or preimage) of B by f 1 (B) = {x X : f(x) B}. Warning The notation f 1 (B)" does not in any way imply the existence of an inverse for the function f. Just like a red herring is neither red nor a herring, the f 1 in f 1 (B) is just a part of the notation. In particular, f 1 (B) is dened for any function f, whether or not it is invertible. c R. P.

29 Preimages of Fact Given a function f : X Y and B 1, B 2 Y, then f 1 (B 1 B 2 ) = f 1 (B 1 ) f 1 (B 2 ). (13) Proof. x f 1 (B 1 B 2 ) f(x) B 1 B 2 f(x) B 1 f(x) B 2 x f 1 (B 1 ) x f 1 (B 2 ) x f 1 (B 1 ) f 1 (B 2 ) (14) On an exam, don't use. Instead, rst show f 1 (B 1 B 2 ) f 1 (B 1 ) f 1 (B 2 ) and then show f 1 (B 1 ) f 1 (B 2 ) f 1 (B 1 B 2 ). c R. P.

30 One-to-One Denition A function f : X Y is called one-to-one (written 1-1) if f(x 1 ) = f(x 2 ) implies x 1 = x 2. Equivalently, f is 1-1 if x 1 x 2 implies f(x 1 ) f(x 2 ). A one-to-one function is called an injection or injective. Symbolically, f is one-to-one if x 1 x 2 (f(x 1 ) = f(x 2 ) x 1 = x 2 ). (15) Example Referring to the examples above f 1 is not 1-1, but f 2,..., f 5 are all 1-1. c R. P.

31 One-to-One Denition Suppose we are given a function f : R R. The function f is called increasing if f(x) f(y) whenever x < y; we call f strictly increasing if f(x) < f(y) whenever x < y. Decreasing and strictly decreasing functions are dened similarly. Fact If f : R R is strictly increasing (resp., strictly decreasing) then f is 1-1. Proof. We need to show that if x 1 x 2 then f(x 1 ) f(x 2 ). But since we're dealing with real numbers, if x 1 x 2 then either x 1 < x 2 or x 2 < x 1. Assume that x 1 < x 2. Then f strictly increasing implies that f(x 1 ) < f(x 2 ), and so f(x 1 ) f(x 2 ). The proofs in the other cases proceed similarly. c R. P.

32 Onto Denition A function f : X Y is called onto if for each y Y, there exists an x X with f(x) = y. An onto function is called a surjection or surjective. Symbolically, f is onto if y Y x X f(x) = y. (16) Example Referring to the examples above f 3 and f 5 are onto, but the others are not. c R. P.

33 Refuting Injectivity We can use (15) to determine when a function fails to be one-to-one. f is not 1-1 if x 1 x 2 (f(x 1 ) = f(x 2 ) x 1 = x 2 ) x 1 x 2 (f(x 1 ) = f(x 2 ) x 1 = x 2 ) x 1 x 2 (f(x 1 ) f(x 2 ) x 1 = x 2 ) x 1 x 2 (f(x 1 ) = f(x 2 ) x 1 x 2 ) (17) That is, f is not 1-1 if there exist x 1 x 2 with f(x 1 ) = f(x 2 ). c R. P.

34 Refuting Surjectivity We can use (16) to determine when a function fails to be onto. f is not onto if y Y x X f(x) = y y Y x X (f(x) = y) y Y x X f(x) y (18) That is, f is not onto if there exists a y Y such that the equation f(x) = y has no solution. c R. P.

35 An Interesting Function Let us dene a function ϕ : Z + Z by ϕ(n) = { n 2 n 1 2 if if n is even n is odd Show that ϕ is 1-1. Show that ϕ is onto. Denition A function f : X Y is called a one-to-one correspondence (or a bijection) if it is both one-to-one and onto. So ϕ is a bijection between the positive integers and all of the integers. Thus there are just as many positive integers as there are integers altogether! c R. P.

36 Inverses of Suppose f : X Y is a bijection, i.e., a one-to-one and onto function. We can dene a new function g : Y X as follows: 1. The domain of g will be Y, the codomain of f; the codomain of g will be X, the domain of f. So all we need now to complete the denition of g is a rule that assigns to each element of Y a unique element of X. 2. So take y Y. Since f is onto, there exists an x X such that f(x) = y. But since f is one-to-one, for each y Y there is only one such x X. We dene g(y) = x, i.e., that x, the one such that y = f(x). Remark The function g dened as above is called an inverse of f. c R. P.

37 Compositions of functions Denition Suppose k : X Y and h : Y Z are two functions. We dene the function h k : X Z, the composition of the functions h and k, by (h k)(x) = h(k(x)) for x X. (19) Remark Note that the denition given by (19) makes sense because for each x X, k(x) Y, the domain of the function h. c R. P.

38 Inverses of Recall that, given a bijection f : X Y, we dened a function g : Y X such that g(y) = x, where f(x) = y. In other words, f(g(y)) = y (20) for all y Y. Denition For any set A, dene the identity function ι A : A A by setting ι A (a) = a, a A. Using this denition, we can rewrite (20) as f g = ι Y (21) c R. P.

39 Inverses of So what about g(f(x))? To nd the value of g at a certain input, we need the element of X that gets sent by f to that input. So what element in X gets sent to f(x)? x gets sent to f(x)! So for all x X. And, as above, we can interpret (22) as g(f(x)) = x (22) g f = ι X. (23) c R. P.

40 Uniqueness of Inverses Fact If f : X Y has an inverse g : Y Xi.e., g satises (20) and (22)then g is unique. Proof. Suppose g 1 : Y X and g 2 : Y X are two functions satisfying (20) and (22) and suppose g 1 g 2. That means that there exists some y 0 Y with g 1 (y 0 ) g 2 (y 0 ). But now applying (20) to both g 1 and g 2, we get f(g 1 (y 0 )) = y 0 = f(g 2 (y 0 )). (24) But (24) and the assumption that g 1 (y 0 ) g 2 (y 0 ) contradict the assumption that f is one-to-one, a necessary condition for f's having an inverse. c R. P.

41 Finding an Inverse Example Given the function f(x) = x 3 x+1, nd its inverse if you can. If you can't, tell why not. Remark I said Given the function...," but I didn't say anything about the domain and codomain of f. In mathematics, especially when dealing with real functions, we often get lazy and give only the rule of assignment for the function, usually by giving a formula. In that case, we take the domain of the function to be the set of all real numbers for which the formula makes sense. In that case, what would the domain of this f be? domain(f) = {x R x 1} c R. P.

42 Finding an Inverse Example How do we nd an inverse for a function f? We write y = f(x) and try to solve for x in terms of y. y = x 3 x + 1 y(x + 1) = x 3 yx + y = x 3 y + 3 = x yx y + 3 = x(1 y) 3 + y 1 y = x (25) c R. P.

43 Finding an Inverse Example Claim g(y) = x = 3 + y 1 y Remarks is the inverse for f. Note that we haven't yet proven that this g is an inverse for f. We assumed that f had an inverse. Why is that justied? It isn't! The g that we found is what any inverse must look like, but that doesn't mean that g is actually an inverse. We've only discovered necessary conditions on an inverse; they may not be sucient. To prove that g is an inverse for f we must verify that (20) and (22) both hold, i.e., that f(g(y)) = y for all y Y and g(f(x)) = x for all x X. c R. P.

44 Finding an Inverse Example Remarks Once again, we're being lazy. We need to specify the domain Y and codomain X for g. And once again, we take the domain of g to consist of all those real numbers y for which the formula makes sense. So Y = domain(g) = {y y 1} = R {1}. But this time we take the codomain of g to be X = R { 1}. c R. P.

45 Finding an Inverse Example We rst verify (20): as desired. f(g(y)) = g(y) 3 g(y) y 1 y 3 = 3 + y 1 y + 1 = (3 + y) 3(1 y) (3 + y) + (1 y) = 4y 4 = y, (26) c R. P.

46 Finding an Inverse Example Now verify (22): as desired. g(f(x)) = 3 + f(x) 1 f(x) = 3 + x 3 x+1 1 x 3 x+1 3(x + 1) + (x 3) = (x + 1) (x 3) = 4x 4 = x, (27) c R. P.

47 Finding an Inverse Example Remarks Another way to see that g is an inverse for f is to note that all the steps in the derivation (25) except that last are reversible. And even the last step is reversible as long as y 1. The fact that the last step in the derivation is not reversible when y = 1 reveals that the original function f : R { 1} R is not actually onto: 1 / f(r { 1}). So in order to claim that f is invertiblehas an inversewe must modify its codomain so that f : (R { 1}) (R {1}) c R. P.

48 Finding an Inverse Example Find the inverse of f : [0, ) R if f(x) = x. y = x y 2 = x (28) But is g : R [0, ) with g(y) = y 2 really an inverse for f? Check (22): g(f(x)) = (f(x)) 2 = ( x) 2 = x (29) for all x [0, ). What about (20)? f(g(y)) = g(y) = y 2 = y y (30) if y < 0. So g is not an inverse for f. (f is not onto.) c R. P.

49 Finding an Inverse Example Remark If we were to modify the codomain of f, changing it to [0, ) from R, then g : [0, ) [0, ) given by g(y) = y 2 would be an inverse for f because both (20) and (22) hold for the new f and the new g. With the modied codomain, f : [0, ) [0, ) is now onto as well as one-to-one. Of course technically we changed f to a new function when we changed its codomaineven though the rule f(x) = x stays the same. But sometimes we use the same namefafter we make such a change. c R. P.