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Math 415 Test 1 Solutions: Odd Numbers, Logic, Set Operations, Exams of Mathematics

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.

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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Math 415, Test 1, October 20, 2003
Answers
Instructions: Do problem 1and any three of probems 2,3,4,5. Please do your best, and
show all appropriate details in your solutions.
1. (20 pts) (a) State the contrapositive of “If nkis odd, then nis odd.”
Answer. “If nis even, then nkis even.”
(b) Let nbe an integer, and let kN, prove that nis odd if nkis odd.
Proof. We will prove the contrapositive statement given in (a). Let nbe even. Then n= 2m
for some integer m. Therefore, nk= (2m)k= 2(2k1mk) is an even integer.
(c) Let aand bbe odd integers, prove that their product ab is odd.
Proof. Since aand bbe two odd integers, we can write a= 2k+ 1 and b= 2l+ 1 for some
integers kand l. Now, ab = 4kl + 2k+ 2l+ 1 = 2(2kl +k+l) + 1 which is odd.
(d) Let nbe an odd integer. Use the Principle of Mathematical Induction to prove that nkis
odd for all kN.
Proof. Let P(k) be the statement nkis odd.” Because nis odd, this true when k= 1. Now
suppose P(k) is true for some kN. 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 kN. By the Principle of
Mathematical Induction, nkis odd for kN.
(e) Let nbe an integer, and let kN. Is it true that nkis odd if and only if nis odd?
Answer. Yes this is true. Part (b) shows that nkis odd only if nis odd, while part (d) showed
nkis odd if nis odd.
2. (10 pts) (a) Complete the truth table for ¬(PQ) ¬P ¬Q.
Answer. The truth table is as follows:
P Q ¬(PQ) ¬P ¬Q
T T F T F
T F T T T
F T T T T
F F T T T
1
pf3

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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.

  1. (20 pts) (a) State the contrapositive of “If nk^ is odd, then n is odd.”

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.

  1. (10 pts) (a) Complete the truth table for ¬(P ∧ Q) ←→ ¬P ∨ ¬Q.

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.

  1. (10 pts) Let A = { 1 , 2 , 3 , 4 }, B = { 2 , 3 , 5 } and C = { 1 , 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) =

  1. (10 pts) (a) Prove or disprove that A − (B − C) = (A − B) ∪ (A ∩ C).

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).

  1. (10 pts) Prove the following versions of De Morgan’s Laws.

(a)

i∈I

Ai

)c

i∈I

Aci