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 k∈N, 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(2k−1mk) 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 k∈N.
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 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, nkis odd for k∈N.
(e) Let nbe an integer, and let k∈N. 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 ¬(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
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