Download COT 3100: Fall 2013 Exam 1, 50 min. PROBLEM 1

COT 3100: Fall 2013 Exam 1, 50 min. PROBLEM 1. SOLUTION:
There are 4 problems on this exam on 4 pages. Write your name on each page. Keep your
answers to each problem on its own page. You may write on the back, if needed.
(25 points) Let p be the statement “Prof. Davis is walking in the rain.” Let q be the statement
“Prof. Davis gets wet.” Suppose p → q is always true.
• (a) State the contrapositive as a logical proposition.
Solution: ¬q → ¬p
• (b) State the contrapositive in English.
Solution: If Prof Davis is not wet, then he isn’t walking in the rain.
(there are also other ways of saying it).
• (c) State the inverse ¬p → ¬q in English.
Solution: If Prof Davis does not walk in the rain, he does not get wet.
(there are also other ways of saying it).
• (d) If p → q is always true, does this imply the inverse is always true? (answer yes or no):
Solution: no
• (e) Prove your answer to (d). If your answer to (d) is “yes,” explain. If your answer to (d) is
“no,” determine the value(s) of p and q for which (p → q) is true yet the inverse is false.
Solution: If p is false but q is true, then (p → q) ≡ (F → T ) is true yet the inverse
(¬p → ¬q) ≡ (T → F ) is false.
COT 3100: Fall 2013 Exam 1, PROBLEM 2. SOLUTION:
(25 points)
• (part a) The following statements are false. Find all possible counter-examples, where the
domain is all real numbers.
1. ∀x(x2 6= x)
Solution: x = 1 and x = 0 are the only counter-examples
2. ∀x(|x| > 0)
Solution: x = 0 is the only counter-example.
• (part b) Determine the truth value of each of these statements if the domain of each variable
consists of all real numbers. Explain.
1. ∀x∃y(x = y 2 )
Solution: This is false, since no such y can exist if x < 0. Recall that the domain of
discourse is the real numbers, so y cannot be a complex number.
2. ∃x∀y((y 6= 0) → (xy = 1)) Solution: This is false. It attempts to say that there is an
x that is the reciprocal of all nonzero values y. Such a number x does not exist.
COT 3100: Fall 2013 Exam 1, PROBLEM 3. SOLUTION:
(25 points) Determine if the following argument is valid (show your work):
1.
2.
3.
4.
Premise 1: ¬a
Premise 2: s → a
Premise 3: (¬r ∨ ¬g) → (s ∧ d)
Conclusion: r
Solution: The argument is valid.
• Starting with the first premise, ¬a, and the 2nd, we can conclude that s is false.
• this means s ∧ d must be false (it does not matter what d is)
• since s ∧ d is false, we know (¬r ∨ ¬g) is false, from the contrapositive of premise 3.
• the DeMorgan equivalent of (¬r ∨ ¬g) is ¬(r ∧ g). Since this is false, (r ∧ g) is true.
• Since (r ∧ g) is true, both r and g are true.
• Thus, r is true.
COT 3100: Fall 2013 Exam 1, PROBLEM 4. SOLUTION:
(25 points) Use a direct proof to show that the product of two odd numbers is odd.
Solution: Consider two odd numbers m and n. Since they are odd, we can write m = 2k + 1
and n = 2t + 1 for some integers k and t. Multiplying,
mn = (2k + 1)(2t + 1) = 4kt + 2k + 2t + 1 = 2(2kt + k + t) + 1
Let s = (2kt + k + t). Since s is an integer, we have mn = 2s + 1 for some integer s, and thus mn
is odd.