Math 3210-3
HW 1
Solutions
Logic
1. Write the negation of each statement.
(a) G is an abelian group.
(b) The set of prime numbers is finite.
(c) Jack and Jill are on the hill.
(d) Three is odd, or four is prime.
(e) If today is not stormy, then I am riding my bike.
(f) If f is bounded and linear, then f is continuous.
Solution 1
(a) G is not an abelian group.
(b) The set of prime numbers is not finite, or the set of prime numbers is infinite.
(c) Jack is not on the hill or Jill is not on the hill.
(d) Three is not odd, and four is not prime, or three is even, and four is composite.
(e) If today is not stormy, then I am not riding my bike.
(f) If f is bounded and linear, then f is not continuous.
2. Construct a truth table for each statement.
(a) p =⇒∼ q
(b) ∼ p ∨ q
(c) p ∧ ∼ p
(d) [∼ q ∧ (p =⇒ q)] =⇒∼ p
(e) [p ∧ (p =⇒ q)] =⇒ q
(f) [p =⇒ (q ∨ ∼ q)] ⇔∼ p
q
2T
F
T
F
∼q
F
T
F
T
p =⇒∼ q
F
T
T
T
(b)
p
T
T
F
F
q
T
F
T
F
∼p
F
F
T
T
∼p ∨q
T
F
T
T
(c)
p
T
F
∼p
F
T
(d)
p
T
T
F
F
q
T
F
T
F
p
T
Solution
(a) T
F
F
p∧ ∼ p
F
F
∼q
F
T
F
T
p =⇒ q
T
F
T
T
∼ q ∧ (p =⇒ q)
F
F
F
T
∼p
F
F
T
T
[∼ q ∧ (p =⇒ q)] =⇒∼ p
T
T
T
T
(e)
p
T
T
F
F
q
T
F
T
F
p =⇒ q
T
F
T
T
p ∧ (p =⇒ q)
T
F
F
F
(f)
p
T
T
F
F
q
T
F
T
F
q∨ ∼ q
T
T
T
T
p =⇒ (q ∨ ∼ q)
T
T
T
T
[p ∧ (p =⇒ q)] =⇒ q
T
T
T
T
[p =⇒ (q ∨ ∼ q)] ⇔∼ p
F
F
T
T
3. Indicate whether each statement is true or false.
(a) 5 is prime and 3 is even.
(b) 5 is prime or 3 is even.
(c) 1 > 3 or 6 is prime.
(d) If 2 is prime, then 2 + 2 = 3.
(e) If π is rational, then 6 is prime.
(f) If 2 > 5 and 3 is prime, then 32 = 9.
(g) If 12 < 5 only if 3 is prime, then 4 is odd.
Solution 3
(a) False.
(b) True.
(c) False.
(d) False.
(e) True.
(f) True.
(g) False.
Quantifiers
4. Write the negation of each statement.
(a) Some apples are blue.
(b) All dogs have four legs.
(c) ∃ x > 1 3 f (x) = 3.
(d) ∀ x ∈ A, ∃ y ∈ B 3 x < y < 1.
(e) ∀ x ∃ y 3 ∀ z, x + y + z ≤ xyz.
Solution 4
(a) All apples are not blue.
(b) There exists a dog that does not have four legs.
(c) ∀ x > 1, f (x) 6= 3.
(d) ∃ x ∈ A 3 ∀ y ∈ B, y ≤ x or, y ≥ 1.
(e) ∃ x 3 ∀ y ∃ z 3 x + y + z > xyz.
5. Determine the truth value of each statement, assuming that x, y, and z are real numbers.
(a) ∃ x 3 ∀ y ∃ z 3 x + y = z.
(b) ∃ x 3 ∀ y and ∀ z, x + y = z.
(c) ∀ x and ∀ y, ∃ z 3 xz = y.
(d) ∃ x 3 ∀ y and ∀ z, z > y =⇒ z > x + y.
(e) ∀ x, ∃ y and ∃ z 3 z > y =⇒ z > x + y.
Solution 5
(a) True.
(b) False.
(c) False.
(d) True.
(e) True.
The following two problems define certain properties of functions. You are to do two things:
(a) rewrite the defining condition in logical symbolism using ∀, ∃, 3, and =⇒ or ⇐, as appropriate;
and
(b) write the negation of part (a) using the same symbolism.
6. A function f is periodic iff there exists a k > 0 such that, for every x, f (x + k) = f (x).
Solution 6
(a) ∃ k > 0 3 ∀ x, f (x + k) = f (x).
(b) ∀ k > 0 ∃ x 3 f (x + k) 6= f (x).
7. A function f is strictly decreasing iff for every x and for every y, if x < y, then f (x) > f (y).
Solution 7
(a) ∀ x and ∀ y, x < y =⇒ f (x) > f (y).
(b) ∃ x and ∃ y 3 x < y =⇒ f (x) ≤ f (y).