Assignment #1 Sample Solutions

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1 CS 220/MATH 320 Applied Discrete Mathematics Fall 2018 Instructor: Marc Pomplun Assignment #1 Sample Solutions Question 1: Say it with Propositional Functions Let Takes(x, y) be the propositional function x takes course y, Teaches(x, y) be the propositional function x teaches course y, and Passes(x, y) be the propositional function x passes course y. The universe of discourse is the set of all living people and all courses (i.e., you do not have to check this in your expressions). Write each of the following propositions symbolically in one expression: a) Peter takes CS 220 and CS 410, but not CS 680. Takes(Peter, CS220) Takes(Peter, CS410) Takes(Peter, CS680) b) Bob passes every course that he takes except CS220. Passes(Bob, CS220) x [(x CS220 Takes(Bob, x)) Passes(Bob, x)] c) Francesca passes every course that is taught by Prof. Einstein. x (Teaches(Einstein, x) Passes(Francesca, x)) d) There is a course that both Julia and Peter took, but both of them failed it. x (Takes(Julia, x) Takes(Peter, x) Passes(Julia, x) Passes(Peter, x)) Question 2: Tautologies and Contradictions Find out for each of the following propositions whether it is a tautology, a contradiction, or neither (a contingency). Prove your answer. 1

2 a) [(p q) (q p)] (p q) p q p q q p (p q) (q p) p q [(p q) (q p)] (p q) T T T T T T T T F F T F F T F T T F F F T F F T T T T T It is a tautology! b) (p q r) [(q r) (p q)] p q r p q r q r p q (q r) (p q) (p q r) [(q r) (p q)] T T T T T T T T T T F T F T T T T F T T T F T T T F F T T F T T F T T T T T T T F T F T F T T T F F T T T T T T F F F F T T T T Another tautology! Question 3: Set Operations Let us take a look at the sets A = {x, y, z}, B = {1, 2}, C = {x, z}. List the elements of the following sets D, E, F, G, H, and I: a) D = (A B) (B C) = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)} b) E = 2 C 2 A = c) F = 2 (2B ) = 2 {, {1}, {2}, {1, 2}} = {, { }, {{1}}, {{2}}, {{1, 2}}, {, {1}}, {, {2}}, {, {1, 2}}, {{1}, {2}}, {{1}, {1, 2}}, {{2}, {1, 2}}, {, {1}, {2}}, {, {1}, {1, 2}}, {, {2}, {1, 2}}, {{1}, {2}, {1, 2}}, {, {1}, {2}, {1, 2}}} d) G = (A B C) (C B A) = {(x, 1, x), (z, 1, x), (x, 2, x), (z, 2, x), (x, 1, z), (z, 1, z), (x, 2, z), (z, 2, z)} e) H = {(a, b, c) a, b, c Β a b a c} = {(1, 2, 2), (2, 1, 1)} 2

3 f) I = {(a, b, c) a B b A c B a c} = {(1, x, 2), (2, x, 1), (1, y, 2), (2, y, 1), (1, z, 2), (2, z, 1), Question 4: Cardinality Are the following statements true for all sets A, B and C? Prove your answers. a) A B C = A - B - C + B A - C + C - A - B No. Counterexample: A = {1}, B = {1}, C = {1} gives us 1 = 0 b) A B C = A + B + C - A B - A C - B C + A B C Here we can use a variant of a membership table. Idea: If adding any element to one or more sets always changes the cardinalities in the same way on the left and the right, then no matter which items A, B, and C contain, the statement will always be true. So our table should tell us how the numbers on the left and on the right change when we add an element of each of the eight membership types. A B C A B C A + B + C A B A C B C A B C total The increase on the left side of the equation always matches the increase on the right side (shown as total in the rightmost column). Therefore, no matter what elements we add, the number on the left side will always increase by the same amount as the right side. If we start with empty sets A, B, and C, the equation is obviously true. By adding more elements, we can create any sets A, B, and C that we want, and now we know that the equation will still be true. Question 5: Functions Find out whether the following functions from R to R are injective, surjective, and/or bijective. a) f(z) = z 2 - z Not injective: f(0) = f(1) = 0 3

4 Not surjective: For example, there is no z such that f(z) = -10. Therefore, not bijective. b) f(z) = 3z 3-7 Injective, surjective, and bijective! c) f(z) = z sin z Not injective: For example, f(0) = f(π) = 0 Surjective, because with growing z we can reach any value for f(z). Not bijective. d) f(z) = z 2 /(z 2 + 1) Not injective. For example, f(1) = f(-1) = 0.5 Not surjective. For example, there is no z such that f(z) = -1 Question 6 (Bonus Question): Function Proofs Are the following statements true or false? In each case, prove your answer. a) There is a strictly decreasing function f from N to N with f(0) = 100. Since f(0) and f is strictly decreasing, f(1) must be less than or equal 99. Similarly, f(2) must be less than or equal 98, and so on, until f(101) must be less than or equal (-1). However, this means that f(101) is not a natural number, which contradicts the definition of f. Therefore, no such function exists. b) Let f(x) and g(x) be strictly increasing functions from R to R. Then (f + g)(x) is also strictly increasing. True. Proof: If f(x) is strictly increasing, then if I pick any two real numbers a and b with a < b, f(a) < f(b). If g(x) is strictly increasing, then for the same numbers a and b, g(a) < g(b). 4

5 Adding the two inequations gives us: f(a) + g(a) < f(b) + g(b) or (f + g)(a) < (f + g)(b) Since this is true for any a and b with a < b that we pick, it means that (f + g)(x) is also strictly increasing. c) Once again, let f(x) and g(x) be strictly increasing functions from R to R. Then (f g)(x) is also strictly increasing. False. Counterexample: f(x) = x, g(x) = x Then (f g)(x) = x 2. If we pick a = -20 and b = -10 then we find: f(a) = 400 and f(b) = 100 Therefore, (f g)(x) is not strictly increasing. 5