Download CSci/Math2112 Assignment 1 Due May 22, 2015 The solutions to

CSci/Math2112
Assignment 1
Due May 22, 2015
The solutions to Assignment 1 are below. I might be showing more steps than necessary, this is to aid
your understanding.
(2)
1. (BoP 1.1 #24) Write {−4, −3, −2, −1, 0, 1, 2} in set-builder notation.
Solution: {n ∈ Z | −4 ≤ n ≤ 2}
2. (BoP 1.2 #2ae, 1.6 #2abefg) Write the following sets by listing their elements between braces.
(2)
(a) A × B and A × ∅, where A = {π, e, 0} and B = {0, 1}.
Solution: A × B = {{π, 0}, {π, 1}, {e, 0}, {e, 1}, {0, 0}, {0, 1}}
A×∅=∅
(5)
(b) A, B, A − A, A ∪ B, and A ∩ B where A = {0, 2, 4, 6, 8} and B = {1, 3, 5, 7} have universal set
U = {0, 1, 2, . . . , 8}.
Solution: A = {1, 3, 5, 7}(= B)
B = {0, 2, 4, 6, 8}(= A)
A − A = {0, 2, 4, 6, 8}
A∪B =∅
A∩B =∅
(2)
3. (BoP 1.7 #10) Draw a Venn diagram for (A − B) ∪ C.
Solution:
(2)
(2)
4. (BoP 1.8 #4) For each n ∈ N, let An = {−2n, 0, 2n}. Find
S
(a) i∈N Ai
S
Solution: i∈N Ai = {2n | n ∈ Z}
(b)
T
i∈N
Ai
Solution:
(3)
T
i∈N
Ai = {0}
5. (BoP 2.2 #8) Express “At least one of the numbers x and y equals 0.” in one of the forms P ∧ Q, P ∨ Q,
or ∼ P . Be sure to also state exactly what statements P and Q stand for.
Solution: P = The number x equals 0.
Q = The number y equals 0.
The original statement is then P ∨ Q.
6. (BoP 2.3 #2 and #4) Without changing their meanings, convert each of the following sentences into a
sentence having the form “If P , then Q.”
(1)
(a) For a function to be continuous, it is sufficient that it is differentiable.
Solution: If a function is differentiable, then it is continuous.
(1)
(b) A function is rational if it is a polynomial.
Solution: If a function is a polynomial, then it is rational.
(1)
7. (BoP 2.4 #4) Without changing the meaning, convert “If a ∈ Q then 5a ∈ Q, and if 5a ∈ Q then a ∈ Q.”
into a sentence having the form “P if and only if Q.”
Solution: We know that a ∈ Q if and only if 5a ∈ Q.
(2)
8. Give an example of a tautology which is not of the form P ∨ ∼ P .
9. (BoP 2.6 #2 and #4) Use truth tables to show that the following statements are logically equivalent.
(5)
(a) P ∨ (Q ∧ R) = (P ∨ Q) ∧ (P ∨ R)
Solution:
P Q R Q ∧ R P ∨ Q P ∨ R P ∨ (Q ∧ R) (P ∨ Q) ∧ (P ∨ R)
T T T
T
T
T
T
T
T T F
F
T
T
T
T
T F T
F
T
T
T
T
T F F
F
T
T
T
T
F T T
T
T
T
T
T
F T F
F
T
F
F
F
F F T
F
F
T
F
F
F F F
F
F
F
F
F
Since the last two columns have the same entries row-by-row, the two statements are logically
equivalent.
(5)
(b) ∼ (P ∨ Q) =∼ P ∧ ∼ Q
Solution:
P Q ∼ P ∼ Q P ∨ Q ∼ (P ∨ Q) ∼ P ∧ ∼ Q
T T
F
F
T
F
F
T F
F
T
T
F
F
F T
T
F
T
F
F
F F
T
T
F
T
T
Since the last two columns have the same entries row-by-row, the two statements are logically
equivalent.
(5) 10. (BoP 2.6 #12) Decide whether or not ∼ (P ⇒ Q) and P ∧ ∼ Q are logically equivalent.
Solution: We know that P ⇒ Q =∼ P ∨ Q. Thus (by the algebraic rules)
∼ (P ⇒ Q) =∼ (∼ P ∨ Q)
=∼ (∼ P )∧ ∼ Q
= P ∧ ∼ Q.
Using a truth table is also fine for this.
(4) 11. Show that (∼ P ∨ Q) ∧ (P ∧ (P ∧ Q)) and P ∧ Q are logically equivalent using the rules from LDM BF-6
(i.e. do not use a truth table).
Solution:
(∼ P ∨ Q) ∧ (P ∧ (P ∧ Q)) = (∼ P ∨ Q) ∧ ((P ∧ P ) ∧ Q)
= (∼ P ∨ Q) ∧ (P ∧ Q)
= (∼ P ∧ (P ∧ Q)) ∨ (Q ∧ (P ∧ Q))
= ((∼ P ∧ P ) ∧ Q) ∨ ((Q ∧ Q) ∧ P )
= (F ∧ Q) ∨ (P ∧ Q)
= F ∨ (P ∧ Q)
= (P ∧ Q)
(3) 12. What is the disjunctive normal form of the statement S (defined in the below truth table) in terms of
P , Q, and R?
P
T
T
T
T
F
F
F
F
Q
T
T
F
F
T
T
F
F
R
T
F
T
F
T
F
T
F
S
T
F
F
T
F
F
F
T
Solution: S = (P ∧ Q ∧ R) ∨ (P ∧ ∼ Q∧ ∼ R) ∨ (∼ P ∧ ∼ Q∧ ∼ R)
13. (BoP 2.10 #2, #8, and #10) Negate the following sentences.
√
(3)
(a) If x is prime, then x is not a rational number.
Solution: There exists an integer x such that x is prime and
(3)
√
x is a rational number.
(b) If x is a rational number and x 6= 0, then tan(x) is not a rational number.
Solution: There exists a rational number x such that x 6= 0 and tan(x) is a rational number.
(3)
(c) If f is a polynomial and its degree is greater than 2, then f 0 is not constant.
Solution: There exists a polynomial f such that its degree is greater than 2 and f 0 is a constant.