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The solutions to math 415 test 1, covering topics such as odd numbers, logic, and set operations. It includes proofs for statements about odd numbers, the completion of a truth table, and solutions to problems involving set operations. Students can use this document to check their answers or study for exams.
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Math 415, Test 1, October 20, 2003 Answers
Instructions: Do problem 1 and any three of probems 2,3,4,5. Please do your best, and show all appropriate details in your solutions.
Answer. “If n is even, then nk^ is even.”
(b) Let n be an integer, and let k ∈ N, prove that n is odd if nk^ is odd.
Proof. We will prove the contrapositive statement given in (a). Let n be even. Then n = 2m for some integer m. Therefore, nk^ = (2m)k^ = 2(2k−^1 mk) is an even integer.
(c) Let a and b be odd integers, prove that their product ab is odd.
Proof. Since a and b be two odd integers, we can write a = 2k + 1 and b = 2l + 1 for some integers k and l. Now, ab = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1 which is odd.
(d) Let n be an odd integer. Use the Principle of Mathematical Induction to prove that nk^ is odd for all k ∈ N.
Proof. Let P (k) be the statement “nk^ is odd.” Because n is odd, this true when k = 1. Now suppose P (k) is true for some k ∈ N. Then nk+1^ = (nk)(n) is a product of two odd integers. By part (c), nk+1^ is odd. Thus P (k) implies P (k + 1) for all k ∈ N. By the Principle of Mathematical Induction, nk^ is odd for k ∈ N.
(e) Let n be an integer, and let k ∈ N. Is it true that nk^ is odd if and only if n is odd?
Answer. Yes this is true. Part (b) shows that nk^ is odd only if n is odd, while part (d) showed nk^ is odd if n is odd.
Answer. The truth table is as follows:
P Q ¬ (P ∧ Q) ←→ ¬P ∨ ¬Q T T F T F T F T T T F T T T T F F T T T
(b) Is the statement in (a) a tautology? Explain.
Answer. Yes, both sides of the truth table in (a) agree for all truth values of P and Q.
(c) State the negation of x > 0 and y = 5.
Answer. x ≤ 0 or y 6 = 5.
(a) C − B = { 1 }.
(b) (A ∩ B) ∪ C = { 2 , 3 } ∪ { 1 , 5 } = { 1 , 2 , 3 , 5 }.
(c) A − (C − B) = A − { 1 } = { 2 , 3 , 4 }.
(d) P(A) =
Proof. This statement is true, as we prove below.
x ∈ A − (B − C) ⇐⇒ x ∈ A and x 6 ∈ B − C ⇐⇒ x ∈ A and ¬(x ∈ B and x 6 ∈ C) ⇐⇒ x ∈ A, and x 6 ∈ B or x ∈ C ⇐⇒ x ∈ A and x 6 ∈ B, or x ∈ A and x ∈ C ⇐⇒ x ∈ (A − B) ∪ (A ∩ C).
(b) Prove or disprove that P(A ∪ B) = P(A) ∪ P(B).
Proof. This statement is false, as we prove with the following counterexample. Let A = { 1 } and B = { 2 }. Then A ∪ B = { 1 , 2 } and so P(A ∪ B) =
{. While^ P(A)^ ∪ P(B) = ∅, { 1 }, { 2 }
. Consequently, P(A ∪ B) 6 = P(A) ∪ P(B).
(a)
i∈I
Ai
i∈I
Aci