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Math 2602(G). Exam 1 Solution.
February 7, 2012.
1. (9 points) Fill in the values in the truth table.
p
q
(p ∨ q) ↔ (p ∧ q) p ⊕ (p → q) (p ∨ q) → q
T
T
T
F
T
T
F
F
T
F
F
T
F
T
T
F
F
T
T
T
2. (9 points) Determine the truth value of each of the following statements. Answer
“True” or “False”.
(a) False
If 0 = 1 or 2 + 2 = 4, then cats can fly.
(b) False
∃x∀y(x + y = 0) where the domain is the set of real numbers.
(c) True
∀x∃y(x 6= 0 → xy = 1) where the domain is the set of real numbers.
3. (7 points) Show that (p → r) ∧ (q → r) ≡ (p ∨ q → r) by developing a series of
logical equivalences using the De Morgan laws, the Distributive laws, and the following
equivalence (1)
(s → t) ≡ (¬ s ∨ t)
(1)
Label each equivalence used.
(p → r) ∧ (q → r) ≡ (¬p ∨ r) ∧ (¬q ∨ r)
≡ (¬p ∧ ¬q) ∨ r
≡ ¬(p ∨ q) ∨ r
≡p∨q →r
(1)
(Distributive)
(De Morgan)
(1)
4. (9 points) Use predicates, quantifiers, logical connectives, and mathematical operators
to express the following statements. Specify the domains.
(a) Every positive integer is a sum of squares of four integers.
∀x > 0 ∃a ∃b ∃c ∃d (x = a2 + b2 + c2 + d2 )
where the domain is the set of integers for every variable.
(b) The difference of two positive integers is not necessarily positive.
∃x > 0 ∃y > 0 (x − y ≤ 0)
where the domain is the set of integers for every variable.
(c) The product of two negative real numbers is positive.
∀x < 0 ∀y < 0 (xy > 0)
where the domain is the set of real numbers for every variable.
5. (6 points) Rewrite the following statements without using the negation operator ¬.
(a) ¬ ∀x ∃y (x + y = 1 ∨ xy = 1)
∃x ∀y (x + y 6= 1 ∧ xy 6= 1)
(b) ¬ ∃x ∀y ∀z (y 6= z → xy 6= xz)
∀x ∃y ∃z (y 6= z ∧ xy = xz)
6. (4 points) Are the following arguments correct? Answer “Yes” or “No”.
(a) Yes
Students in Math 2602 has taken Calculus. Jess has not taken Calculus.
Therefore Jess is not in Math 2602.
(b) No
All math majors take number theory. Jin is taking number theory. Therefore Jin is a math major.
7. (6 points) Let A = {x ∈ Z : x2 ≤ 4} and B = {1, 2, 3, 4}. List all the elements in the
following sets.
(a) A = {−2, −1, 0, 1, 2}
(b) A − B = {−2, −1, 0}
(c) (A − B) ∪ (B − A) = {−2, −1, 0, 3, 4}
8. (13 points) Prove that the product of a nonzero rational number and an irrational
number is irrational.
We will prove by contradiction.
Suppose there is a non-zero rational number r and an irrational number s such that
their product rs is rational. Then r = a/b for some non-zero integers a and b, and
rs = c/d for some integer c and non-zero integer d. Since r is non-zero, we have
s=
rs
cb
bc
=
=
.
r
da
ad
Note that we have a 6= 0 and d 6= 0. Since bc and ad are integers and ad 6= 0, it follows
that s is rational, contradicting the assumption that s is irrational.
(Remark: It is incorrect to claim that an irrational number must be the squareroot of
an integer.)
9. (12 points) Disprove the following statements by giving counterexamples.
(a) If a and b are rational numbers, then ab is also rational.
Let a = 2 and b = 1/2. Then ab =
lecture.
√
2 is irrational, as seen in the textbook and
(b) The product of two irrational numbers is irrational.
The number
√
2 is irrational, but the product of
√
2 and
√
(c) ∀x(x2 > x). The domain is the set of real numbers.
Let x = 1/2. Then x2 = 1/4 ≤ 1/2 = x.
(Remark: x = 0 or x = 1 are also easy counterexamples.)
2 is 2, which is rational.
Prove the following statements using only the definitions of even/odd integers.
10. (13 points) For any integer a, a2 + a + 2 is an even number.
Let a be an integer. We will consider two cases.
Case 1: a is even. Then a = 2k for some integer k. Then
a2 + a + 2 = (2k)2 + (2k) + 2 = 2(2k 2 + k + 1).
Since k is an integer, 2k 2 + k + 1 is an integer as well. Thus a2 + a + 2 is even.
Case 2: a is odd. Then a = 2ℓ + 1 for some integer ℓ. Then
a2 + a + 2 = (2ℓ + 1)2 + (2ℓ + 1) + 2 = 2(2ℓ2 + 3ℓ + 2).
Since ℓ is an integer, 2ℓ2 + 3ℓ + 2 is an integer as well. Thus a2 + a + 2 is even.
11. (12 points) For any integers a and b, if ab is even, then a is even or b is even.
We will prove by contraposition.
Suppose a and b are odd integers. Then a = 2k + 1 and b = 2ℓ + 1 for some integers k
and ℓ. Then
ab = (2k + 1)(2ℓ + 1) = 4kℓ + 2k + 2ℓ + 1 = 2(2kℓ + k + ℓ) + 1.
Since k and ℓ are integers, 2kℓ + k + ℓ is also an integer. Therefore ab is odd.