Quiz 1 1. Let p and q be the following statements. p : Maxwell is a mathematics major. q : Maxwell is a chemistry major. (1) a. Write each of the following in symbolic form using logical connectives. i. Maxwell is a mathematics major or a chemistry major. p q ii. Maxwell is a mathematics major but not a chemistry major. p q (1) b. Write each of the following in words. i. p q Maxwell is either a mathematics major or a chemistry major but not both. ii. p q Maxwell is not a mathematics major or he is a chemistry major. If Maxwell is a mathematics major, then he is a chemistry major. (1) 2. Give the truth table for the following. [p ( p q)] q p q p q p q [p ( p q)] q T T F F T F T F F T F F F T T F T F F F T T T F Recall that [p (p q)] q [p ( p q)] q is modus ponens. So ([p ( p q)] q) (q q) which is a contradiction.
Quiz 2 1. Use a truth table to verify each of the following. (1) a. p q p q p q p p q p q T T F T T T F F F F F T T T T F F T T T (1) b. An implication is logically equivalent to its contrapositive. p q p q q p T T T T T F F F F T T T F F T T (1) 2. Give the truth table for (p r) (r q). p q r p r r q (p r) (r q) T T T T T T T T F F T T T F T T F T T F F F T T F T T T T T F T F T T T F F T T F T F F F T T T
Quiz 3 01-26-18 (1) 1. Suppose that x, ρ, and θ are distinct real numbers. Put them in numerical order given the following. a. If ρ is not the largest, then θ is the largest. b. If θ is not the smallest, then x is the largest. θ < ρ < x Note that by the first statement, either ρ or θ is the largest. If θ is not the smallest, then the second statement implies that x is the largest which contradicts the first statement. Therefore, θ is the smallest. Since θ is not the largest, then ρ is the largest (modus tollens) by the first statement. (1) 2. Suppose you encounter three inhabitants of the island of knights and knaves created by Smullyan. Knights always tell the truth and knaves always lie. If possible, determine the relative ages of the three inhabitants given that there is exactly one knave in the group and they make the following statements. A : I am the oldest. B : I am older than A. C : I am the oldest. Solution: Inhabitant C is the oldest, B is the next oldest, and A is the youngest. Since the statements made by A and C are contradictory, one of them is the knave. Therefore, B is a knight and is telling the truth when claiming to be older than A. This means that A is the knave and C is the other knight. Since the statements made by B and C are true, C is the oldest, B is the next oldest, and A is the youngest.
Quiz 4 (2) 1. Prove that the sum of two rational numbers is a rational number. You may use without proof that the sum or product of two integers is an integer. Proof: Suppose that r,s Q. Then there are p a,p b,q a,q b Z such that a = p a and b = q a p b. Then a+b = p a + p b = p a q b +p b q a which is the quotient of integers. q b q a q b q a q b (2) 2. Identify the error in this proof that 1 = 2. Proof: Let x = 1. Then we have x 2 x 2 = x 2 x 2 x(x x) = (x+x)(x x) (factor each side differently) x = x+x (divide each side by x 2 x 2 ) x = 2x (x+x = 2x) 1 = 2 1 (substitute x = 1) 1 = 2 (2 1 = 2) The erroneous step is dividing by x 2 x 2 = 0.
Quiz 5 (2) 1. Find integers p and q such that p q = 0.51 Note that 100x x = 51 99x = 51 x = 51 99 (2) 2. Prove that if 2 divides n, then 4 divides (5n+2) 2. Proof: Since 2 n, there exists k Z such that 2k = n. Then (5n+2) 2 = 25n 2 +20n+4 = 25(2k) 2 +40k +4 = 4(25k 2 +10k +1) which is divisible by 4. 3. Prove the converse of the statement in the previous problem. Direct Proof: Proof: Suppose that 4 divides (5n+2) 2 = 25n 2 +20n+4. Since 4 divides 20n+4, 4 must also divide 25n 2 +20n+4 (20n+4) = 25n 2. This implies that n 2 is even which implies that n is even. Therefore, 2 n. Contraposition Proof: Recall that the product of two odd integers is odd. Proof: Suppose that 2 does not divide n. Then n is odd which implies that 5n and 5n+2 are odd. Since 5n+2 is odd, (5n+2) 2 is odd which implies that it is not divisible by 4.
Quiz 6 (2) 1. Prove that 3 is irrational. (2) 2. Suppose that n N and p is a prime number. Prove that p n! if and only if p n.
Quiz 7 (2) 1. Prove that 3 is irrational. Proof: Toward a contradiction, suppose that 3 is rational. Then there are integers p and q such that 3 = p and p and q are relatively prime. Then q p q = 3 p 2 q 2 = 3 p 2 = 3q 2. Since 3 p 2 and 3 is prime, 3 p. So there exists a natural number k such that p = 3k. So we have that p 2 = 3q 2 (3k) 2 = 3q 2 9k 2 = 3q 2 3k 2 = q 2. Then 3 q 2 which implies that 3 q since 3 is prime. This is a contradiction since p and q are relatively prime. Therefore, 3 is irrational. (2) 2. Suppose that n N and p is a prime number. Prove that p n! if and only if p n. Proof: First, suppose that p divides n! = n(n 1)(n 2) 3 2 1. Since p is prime, p divides one of n,(n 1),(n 2),...,3,2 which implies that p n. Now suppose that p n. Then p is one of n,(n 1),(n 2),...,3,2 which implies that n! = n(n 1)(n 2) 3 2 1 is a multiple of p.
Quiz 8 (1) 1. Let A = [4,7] and B = [5, ). Express the following sets using interval notation. a. A\B = [4,5) b. B \A = (7, ). (1) 2. Use Venn diagrams to illustrate the following. A (B C) = (A B) (A C) (2) 3. Suppose that A, B, and C are sets. Prove that A (B C) = (A B) (A C). Proof: First, suppose that x [A (B C)]. Then x A and x (B C). Since x (B C), x B or x C. If x B, then x A B and if x C, then x A C. Since x A B or x A C, x [(A B) (A C)]. Therefore, A (B C) (A B) (A C). Now suppose that x [(A B) (A C)]. Then x (A B) or x (A C). If x (A B), then x A and x B B C which means that x [A (B C)]. On the other hand, if x (A C), then x A and x C B C which means that x [A (B C)]. So in any case, x [A (B C)] which implies that (A B) (A C). A (B C). Since A (B C) (A B) (A C) and (A B) (A C) A (B C), A (B C) = (A B) (A C).
Quiz 9 (2) 1. Give an example of a set X and a set A such that A is both a subset of X and an element of X. Let X = {x,{x}} and A = {x}. Since x X, A = {x} X. Also, note that A = {x} X. (2) 2. Suppose that X is a set and S = {x X : x / x}. Prove that if S X, then S S if and only if S / S. Proof: First, suppose that S S. Then S is an element of the set of elements of X that are not elements of themselves. Therefore, S is not an element of itself. Now suppose that S / S. Then S is not an element of itself. This implies that S is an element of the set of elements that are not elements of themselves. Therefore, S S.
Quiz 10 1. Recall that N = {1,2,3,...} Consider the function f : (0, ) R defined by f(x) = lnx. Find the following sets. (1) a. f(n) = {ln1,ln2,ln3,ln4,...} (1) b. f -1 (N) = {e,e 2,e 3 e 4,...} (2) 2. Suppose that f : X Y, A,B X. Prove that f(a B) f(a) f(b). Below are two proofs. Proof: Suppose that y f(a B). Then there exists x A B such that f(x) = y. Since x A B, x A and hence y = f(x) f(a). Also, since x A B, x B and hence y = f(x) f(b). Since y f(a) and y f(b), y f(a) f(b). Proof: Since A B A, f(a B) f(a). Also, since A B B, f(a B) f(b). Since f(a B) f(a) and f(a B) f(b), f(a B) f(a) f(b).
Quiz 11 (2) 1. Define a relation R on P(N) by A R B if and only if A B. a. Is the relation R reflexive? Yes. Note that every set is a subset of itself. b. Is the relation R symmetric? No. Note that {1} N while N {1}. c. Is the relation R antisymmetric? Yes. If A B and B A, then A = B. d. Is the relation R transitive? Yes. If A B and B C, then A C. (1) 2. Using words and math symbols, describe the relation on R pictured below. For all x,y R, x is related to y if and only if x = y.
Quiz 12 (2) 1. Let A={1,2,3,4,5}. Give anexample of a relation R onasuch that R is reflexive, antisymmetric, transitive, and not symmetric. For all x,y A, (x,y) R if and only if x y.
Quiz 13 (3) 1. Let A = {1,2,3,4,5} and suppose that a relation R on A is reflexive, symmetric, and antisymmetric. List the elements of R. Justify your answer. R = {(1,1),(2,2),(3,3),(4,4),(5,5)} Proof: Since R is reflexive, all elements of A are related to themselves. Now suppose that x,y A such that x R y. Since R is symmetric, y R x. This implies that x = y since R is antisymmetric. Total Points: 44