Math 55 Homework 2 solutions

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1 Math 55 Homework solutions Section p q p q p q p q (p q) T T F F F T F T F F T T F T F T T F T F T F F T T T F T 8. a) Kwame will not take a job in industry and not go to graduate school. b) Yoshiko does not know Java or calculus. c) James is not young or not strong. d) Rita will not move to Oregon and Washington. 10. b) p q r p q q r p r (p q) (q r) ((p q) (q r)) (p r) T T T T T T T T T T F T F F F T T F T F T T F T T F F F T F F T F T T T T T T T F T F T F T F T F F T T T T T T F F F T T T T T c) p q p q p (p q) (p (p q)) q T T T T T T F F F T F T T F T F F T F T 14. Let s find a counterexample. If p F, q T, then (( p (p q)) q) F, so it is not a tautology. 4. If p F, then both are equivalent to T because F α T for any α. If p T, then both are equivalent to q r because T α α for any α. 30. If q r T, then the proposition is T. Otherwise, we have q = r F. In this case, one of p, p is F, so p q F or p r F. Therefore, (p q) ( p r) F, and the proposition is T. 3. Let s find a counterexample. If p r T, q F, then the first one is T, but the second one is F. 1

2 60. If α is unsatisfiable, then α F, so α T, i.e., α is a tautology. If α is tautology, then α T, so α F, i.e., α is unsatisfiable. 6. (a) If p T and q F, then the five propositions (p q r),... are all T, so it is satisfiable. Section a) There exists a student in your school who visited North Dakota. b) All students in your school visited North Dakota. c) It is not true that there exists a student in your school who visited North Dakota. d) There exists a student in your school who did not visit North Dakota. e) It is not true that all students in your school visited North Dakota. f) All students in your school did not visit North Dakota. 10. a) x(c(x) D(x) F (x)). b) x(c(x) D(x) F (x)). c) x(c(x) D(x) F (x)). d) x( C(x) D(x) F (x)). e) ( xc(x)) ( xd(x)) ( xf (x)) 16. a) If x =, then x =, so it is true. b) For all real numbers x, x 0. Then x 1, so it is false. c) For all real numbers x, x 0. Then x + 1, so it is true. d) If x = 0, then x = x, so it is false. 0. a) P ( 5) P ( 3) P ( 1) P (1) P (3) P (5). b) P ( 5) P ( 3) P ( 1) P (1) P (3) P (5). c) P ( 5) P ( 3) P ( 1) P (3) P (5). d) P (1) P (3) P (5) e) ( P ( 5) P ( 3) P ( 1) P (1) P (3) P (1)) P ( 5) P ( 3) P ( 1). 3. a) xp (x), x P (x), there exists a dog that does not have fleas. b) xp (x), x P (x), all horses can add. c) xp (x), x P (x), there exists a koala that cannot climb. d) x P (x), xp (x), there exists a monkey can speak French. e) x(p (x) Q(x)), x (P (x) Q(x)) = x P (x) Q(x), All pigs cannot swim or catch fish. 50. Let {0, 1} be the domain, P (x) = (x = 0), and Q(x) = (x = 1). Then xp (x) = xq(x) = F, so the first one is F. However, x(p (x) Q(x)) = T, so they are not logically equivalent. 60. a) P (x) Q(x).

3 b) x(r(x) Q(x)). c) x(r(x) P (x)). d) If b) holds, then there exists x such that R(x), Q(x) hold. By a), Q(x) implies P (x), so such an x satisfies R(x), P (x), i.e., c) holds. Section a) Randy Goldberg enrolled in CS 5. b) There exists a student in your school who is enrolled in Math 695. c) There exists a class being given at your school in which Carol Sitea is enrolled. d) There exists a student in your school who is enrolled Math and CS 5. e) There exists two different students in your school such that one is enrolled in the classes in which the other is enrolled. f) There exists two different students in your school such that they are enrolled in the same classes. 14. Let the domain of x be all students in this class. a) xp (x) where P (x) is that x can speack Hindi. b) xq(x) where Q(x) is that x plays some sport. c) x(r(x) S(x)) where R(x) is that x has visited Alaska and S(x) is that x has visited Hawaii. d) x yt (x, y) where the domain of y is all programing languages and T (x, y) is that x have learned y. e) x zu(x, z) where the domain of z is all departments in this school and U(x, z) is that x has taken every course offered by z. f) x x x ((x x ) V (x, x ) (((x x ) (x x )) V (x, x ))) where the domains of x, x are all students in this class and V (x, x ) is that x, x grew up in the same town. g) x x w((x x ) W (x, x, w)) where the domain of x is all students in this class, the domain of w is all chat groups, and W (x, x, w) is that x, x chatted in w. 0. a) x y(((x < 0) (y < 0)) (xy > 0)). b) x y(((x > 0) (y > 0) ( x+y > 0)). c) x y(((x < 0) (y < 0)) (x y < 0)). d) x y( x + y x + y ). 4. a) There exists a real number such that adding it to another real number does not change anything. b) The difference of a positive real number and a negative real number is positive. c) There exist two nonpositive real numbers whose difference is positive. d) The product of two nonzero real numbers is nonzero. 3

4 30. a) y x P (x, y). b) x y P (x, y). c) y( Q(y) xr(x, y)). d) y( x R(x, y) x S(x, y)). e) y( x z T (x, y, z) x z U(x, y, z)). 40. a) x = 0 is a counterexample because there are no y such that 0 = 1 y. b) x = 100 is a counterexample because there are no y such that y < 0. c) x = 8, y = 4 is a counterexample. 46. a) Choose a real number y less than x. Then x > y, so the truth value of the statement is false. b) If x = 0, then for all integers y, x y, so the truth value of the statement is true. c) Choose a positive real number y less than x. Then x > y, so the truth value of the statement is false. Section Let p be the proposition George have eight legs and q the proposition George is a spider. The argument form is p q q p, and it is valid because (( p q) q) p ((q p) q) p T. Therefore, we can conclude that the conclusion is true if the premises are true. 4. a) Simplification b) Disjunctive syllogism c) Modus ponens d) Addition e) Hypothetical syllogism 8. If y is Manhattan, P (x) is that x is a man, and Q(x) is that x is an island, then the argument form is x(q(x) P (x)) P (y) Q(y). By the universal instantiation rule, Q(y) P (y), and by the modus tollens rule, this and P (y) implies Q(y). Therefore, the universal instantation and modulus tollens rules are used. 1. By Exercise 11, it suffices to show that if (p t) (r s), q (u t), u p, s, q are valid, then r is also valid. Under the assumptions, u t is valid because 4

5 q, q (u t) are valid (modus ponens), so u, t are valid (simplification). Then p is also valid because u p (modus ponens), so p t is valid (conjuction). By (p t) (r t), r t is valid (modus ponens), and then r is valid (resolution). 16. a) If Mia is enrolled in the university, then by the first premise, Mia has lived in a dormitory, which is contradict to the second premise. Therefore, Mia is not enrolled in the university, so the argument is correct. b) A car that is not convertible may be fun to drive. Therefore, the argument is incorrect. c) There may be movies that Quincy likes but is not an action movie. Therefore, the argument is incorrect. d) Hamilton is a lobsterman, which set at least a dozen traps, so the argument is correct. 18. The premise ss(s, Max) does not imply S(Max, Max) if the number of elements in domain of s is at least two is not correct because it is not simplification. 5 is also not correct because it is not simplification. 8. By universal instantiation, P (c) Q(c) and ( P (c) Q(c)) R(c) are valid for arbitrary c. Then the contrapositive of the second one is R(c) (P (c) Q(c)). By resolution, (P (c) Q(c)) and (P (c) Q(c)) imply P (c) P (c) = P (c), so by hypothetical syllogism, R(c) P (c) is valid. Therefore, by universal generalization, x( R(x) P (x)) is valid. 35. Assume that Superman exists. Then by the fifth premise, he is neither impotent nor malevolent. By the contrapositive of the second premise, he is able to prevent evil, and by the contrapositive of the third premise, he is willing to prevent evil. Therefore, by the first premise, he prevent evil, which is contradict to the fourth premise. Therefore, Superman does not exist, so the argument is valid. Section If x, y are odd integers, then by definition, x = m + 1, y = n + 1 for some integers x, y. Therefore, x + y = (m + n + 1), so x + y is an even integer by definition. 8. Let n be a perfect square m where m is an integer. If n + is also a perfect square, then n+ = l for some integer l. Then l m =, so (l m)(l+m) =. We have for cases: l m = 1; l + m =, l m = ; l + m = 1, l m = 1; l + m =, l m = ; l + m = 1, 5

6 because 1,, 1, are the only integers dividing. In all cases, (l m) + (l + m) = l is odd, which is a contradiction. Therefore, n + is not a perfect square. 10. Let x, y be rational numbers. By definition, x = p q, y = p q for some integers p, q, p, q such that q, q 0. Then xy = pp qq, and pp, qq are integers such that qq 0, so xy is also rational. 1. There is a counterexample. Consider 0 and. Then the first one is rational, and the second one is irrational, but the product of two numbers is 0, which is rational. Therefore, the proposition is false. 18. a) Assume that n is not even. Then n = m + 1 for some integer m, so 3n + = (3m + 1) + 1, which is not even. This proves the contrapositive of the proposition. b) Assume that 3n + is even and that n is not even. Then n = m + 1 for some integer m, so 3n + = (3m + 1) + 1, which is not even. This contradicts to the premise. 0. Let s prove it directly. Because 1 1, P (1) is true. 4. Assume that the conclusion is not true. Then for each month of the year, there are at most days chosen, so the number of chosen days is at most 1 = 4, which contradicts to the assumption that we chose 5 days. Therefore, at least three of any 5 days chosen must fall in the same month of the year. 6. If n is even, then n = m for some integer m, so 7n + 4 = (7m + ), which is also even. Conversely, if 7n + 4 is even, then 7n + 4 = l for some integer l, so n = (l 3n), which is also even. 30. If a < b, then a < a + b, so a < a+b. This shows that (i) implies (ii). Conversely, if a < a+b, then a < a + b, so a < b. This shows that (ii) implies (i). If a < b, then a + b < b, so a+b < b. This shows that (i) implies (iii). Conversely, if a+b < b, then a + b < b, so a < B. This shows that (iii) implies (i). Therefore, all three statements are equivalent. 34. The reasoning is incorrect. By the argument in the exercise, it is correct that (1) implies (5), but to find all the solutions, we should prove that (5) implies (1), which is not in the argument. 6