Math 3210-1
HW 1
Due Monday, June 11, 2007
Each problem is worth one point except for problems 2, 3 and 5. Problems 2, 3 and 5 are each worth 2
points: one point for the first half of the problem, and one point for the second half of the problem. There
are a total of 8 points possible in this assignment.
Logic
1. Write the negation of each statement.
(a)
(b)
(c)
(d)
(e)
(f)
G is an abelian group.
The set of prime numbers is finite.
Jack and Jill are on the hill.
Three is odd, or four is prime.
If today is not stormy, then I am riding my bike.
If f is bounded and linear, then f is continuous.
2. Construct a truth table for each statement.
(a)
(b)
(c)
(d)
(e)
(f)
p =⇒∼ q
∼p ∨q
p∧ ∼ p
[∼ q ∧ (p =⇒ q)] =⇒∼ p
[p ∧ (p =⇒ q)] =⇒ q
[p =⇒ (q ∨ ∼ q)] ⇔∼ p
3. Indicate whether each statement is true or false.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
5 is prime and 3 is even.
5 is prime or 3 is even.
1 > 3 or 6 is prime.
If 2 is prime, then 2 + 2 = 3.
If π is rational, then 6 is prime.
If 2 > 5 and 3 is prime, then 32 = 9.
If 12 < 5 only if 3 is prime, then 4 is odd.
Quantifiers
4. Write the negation of each statement.
(a)
(b)
(c)
(d)
(e)
Some apples are blue.
All dogs have four legs.
∃ x > 1 3 f (x) = 3.
∀ x ∈ A, ∃ y ∈ B 3 x < y < 1.
∀ x ∃ y 3 ∀ z, x + y + z ≤ xyz.
5. Determine the truth value of each statement, assuming that x, y, and z are real numbers.
(a)
(b)
(c)
(d)
(e)
∃ x 3 ∀ y ∃ z 3 x + y = z.
∃ x 3 ∀ y and ∀ z, x + y = z.
∀ x and ∀ y, ∃ z 3 xz = y.
∃ x 3 ∀ y and ∀ z, z > y =⇒ z > x + y.
∀ x, ∃ y and ∃ z 3 z > y =⇒ z > x + y.