Math220, Homework 3
Due in class January 28 or or 29 (depending on your section)
1. (Modified Question 2.2, p. 65) Consider the sets A, B, C, and D below. First,
describe the sets A and D using a complete sentence (”A is the set of numbers which...”).
Then determine which of the following statements are true: Give an explanation for each
false statement.
A = {1, 4, 7, 10, 13, 16, . . . } B = {x ∈ Z : x is odd}
C = {x ∈ Z : x is prime and x =
6 2} D = {1, 2, 3, 5, 8, 13, 21, 34, 55, . . . }.
(a)
(b)
(c)
(d)
25 ∈ A
22 ∈ A ∪ D
C⊆B
∅∈B∪D
Solution: In the language that we learned recently, A is the set of numbers that are
congruent to 1 mod 3. In the homework, you could say “A is the set of numbers
that give remainder 1 when divided by 3”. You could also write A = {n ∈ N : ∃k ∈
N s.t. n = 3k + 1}.
The set D is the set of so-called Fibonacci numbers; they are defined by a recursive
formula f0 = f1 = 1, fn+1 = fn + fn−1 for all n.
(a) 25 ∈ A. True, since 25 − 1 = 24 is divisible by 3.
(b) 22 ∈ A ∪ D. False since 22 ∈
/ D.
(c) C ⊆ B. True: all the prime numbers except 2 are odd.
(d) ∅ ∈ B ∪ D. False: ∅ is not an element of B or D!
2. (Modified Question 2.6, p. 65) For the open sentence P (A) : A ⊆ {1, 2, 3} over the
domain S = P({1, 2, 4}), determine:
(a) all A ∈ S for which P (A) is true
(b) all A ∈ S for which P (A) is false
(c) Let W be the set of all A ∈ S for which A ∩ {1, 2, 3} = ∅. List all the elements of
W . Then find the intersection of W with the set of all A ∈ S for which P (A) is
true.
Solution: Since A ∈ P ({1, 2, 4}), A is a subset of {1, 2, 4}.
• We need A to be a subset of {1, 2, 3}, so we need A = ∅, {1} , {2} or {1, 2}.
• We need A to not be a subset of {1, 2, 3}, so we need 4 ∈ A. Thus A =
{4} , {1, 4} , {2, 4} or {1, 2, 4}.
• A is in W if and only if A is disjoint from {1, 2, 3}, so W = {∅, {4}}. Then
W ∩ {A : P (A)is true} = {∅}.
Page 1 of 6
Math220, Homework 3
Due in class January 28 or or 29 (depending on your section)
3. (Question 2.14, p. 66) State the negation of each of the following statements:
(a) At least two of my library books are overdue.
(b) One of my two friends misplaced his homework assignment.
(c) No one expected that to happen.
(d) It’s not often that my instructor teaches that course.
(e) It’s surprising that two students received the same exam score.
Solution:
(a) At most one of my library books is overdue.
(b) Here the negation depends on how you interpret the original statement: if you
interpret it as saying that one of your friends misplaced his assignment but not
saying anything about the other friend (that is, implying that at least one of
your friends misplaced their assignment), then the negation would be: “None of
my friends misplaced his homework assignment”. However, if you interpret the
original statement as saying that exactly one of your two friends misplaced his
homework assignment, then the negation would be “None of my friends misplaced
his homework assignment or both of them did”. You see that language is always
ambiguous! In mathematical writing, if you meant the first case, you would have
used the existential quantifier; if you meant the second case, you would have to
use the symbol ∃!, which indicates that the object in question (in this case, the
friend) exists and is unique.
(c) Someone expected that to happen.
(d) My instructor teaches that course often.
(e) It’s not surprising that two students received the same exam score.
4. (Question 2.16, p. 66) For the sets A = {1, 2, . . . , 10} and B = {2, 4, 6, 9, 12, 25},
consider the statements
P : A ⊆ B. and Q : |A − B| = 6.
Which of the following statements are true:
(a) P ∨ Q
(b) P ∨ ∼ Q
(c) P ∧ Q
(d) (∼ P ) ∧ Q
(e) (∼ P ) ∨ (∼ Q).
Page 2 of 6
Math220, Homework 3
Due in class January 28 or or 29 (depending on your section)
Solution: Note that P is False and Q is True. Thus:
(a) P ∨ Q is True.
(b) P ∨ ∼ Q is False.
(c) P ∧ Q is False.
(d) (∼ P ) ∧ Q is True.
(e) (∼ P ) ∨ (∼ Q) is True.
5. Prove the distributive law P ∧ (Q ∨ R) ≡ (P ∧ Q) ∨ (P ∧ R) by constructing the truth
table. (Here P , Q, and R are statements).
Solution:
P Q R
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F
Q∨R
T
T
T
F
T
T
T
F
P ∧ (Q ∨ R)
T
T
T
F
F
F
F
F
P ∧Q
T
T
F
F
F
F
F
F
P ∧R
T
F
T
F
F
F
F
F
(P ∧ Q) ∨ (P ∧ R)
T
T
T
F
F
F
F
F
Note that the column for P ∧(Q∨R) and the last column representing (P ∧Q)∨(P ∧R)
are identical, hence, these statements are logically equivalent.
6. (Question 2.20, p. 67) For statements P and Q, construct a truth table for
(P ⇒ Q) ⇒ (∼ P ).
Solution:
P Q P
T T
T F
F T
F F
⇒Q
T
F
T
T
∼P
F
F
T
T
(P ⇒ Q) ⇒ (∼ P )
F
T
T
T
7. (Question 2.24, p. 67) Two sets A and B are non-empty disjoint subsets of a set S. If
x ∈ S, which of the following are true? (If a general statement is false, give an example
when the statement does not hold, called ”counterexample”).
(a) It is possible that x ∈ A ∩ B.
Page 3 of 6
Math220, Homework 3
Due in class January 28 or or 29 (depending on your section)
(b) If x is an element of A, then x can’t be an element of B.
(c) If x is not an element of A, then x must be an element of B.
(d) It’s possible that x ∈
/ A and x ∈
/ B.
(e) For each nonempty set C, either x ∈ A ∩ C or x ∈ B ∩ C.
(f) There exists a nonempty set C, such that both x ∈ A ∪ C and x ∈ B ∪ C.
Solution:
(a) False: since A and B are disjoint, their intersection is empty by definition, so
x ∈ A ∩ B is not possible.
(b) True, again because A and B are disjoint.
(c) False. Counterexample: take S = {1, 2, 3}, let A = {1}, B = {2}. Then A and
B are disjoint non-empty subsets of S, and the condition is satisfied. Take x = 3.
It is not an element of A and not an element of B, so the statement is False.
(d) True: see the above example in (c).
(e) False: in the same example as in (c), take C = {3}. Then C is non-empty. Take
x = 1. Then x is neither in A ∩ C nor in B ∩ C.
(f) True: given x ∈ S, take C = {x}. Then by definition, x ∈ A ∪ C and x ∈ B ∪ C.
8. (Question 2.28, p. 68) Consider the statement (implication):
If Bill takes Sam to the concert, then Sam will take Bill to dinner.
Which of the following implies that this statement is true:
(a) Sam takes Bill to dinner only if Bill takes Sam to the concert.
(b) Either Bill doesn’t take Sam to the concert or Sam takes Bill to dinner.
(c) Bill takes Sam to the concert.
(d) Bill takes Sam to the concert and Sam takes Bill to dinner.
(e) Bill takes Sam to the concert and Sam doesn’t take Bill to dinner.
(f) The concert is canceled.
(g) Sam doesn’t attend the concert.
Solution: Let P be the statement “Bill takes Sam to the concert”, and Q be the
statement “Sam takes Bill to dinner”. Then the original statement is the implication P ⇒ Q. Now we need to analyze which statements on the list imply that this
implication is true.
We will write “Yes” when the statement in question implies that P ⇒ Q is true, and
“No” otherwise.
Page 4 of 6
Math220, Homework 3
Due in class January 28 or or 29 (depending on your section)
(a) Sam takes Bill to dinner only if Bill takes Sam to the concert.
This is saying P is required for Q, which is, ∼ P ⇒ ∼ Q, which is the contrapositive of Q ⇒ P (and thus logically equivalent to Q ⇒ P ), which is the converse of
our statement and implies nothing about it. (That is, even if Sam is determined
not to take Bill to dinner unless Bill takes him to the concert, this says nothing
about whether Bill will actually take Sam to the concert). So here the answer is
”No”.
(b) Either Bill doesn’t take Sam to the concert or Sam takes Bill to dinner.
This is saying, ∼ P ∨ Q, which is equivalent to P ⇒ Q, so the answer is ”Yes”.
(c) Bill takes Sam to the concert.
This just says P is true without saying anything about Q, so this says nothing
about whether the implication P ⇒ Q is true, and so the answer is “No”.
(d) Bill takes Sam to the concert and Sam takes Bill to dinner.
This says, P is True and Q is True, which makes the implication True, so the
answer is “Yes”.
(e) Bill takes Sam to the concert and Sam doesn’t take Bill to dinner.
This says, P is True and Q is False, which makes the implication False, so the
answer is “No”.
(f) The concert is canceled.
This says P is False; regardless of Q then, the implication is True, so the answer
is“Yes”.
(g) Sam doesn’t attend the concert. Same as above: this says P is False; regardless
of Q then, the implication is True, so the answer is“Yes”.
9. (Question 2.30, p. 68) Consider open sentences P (n) : 5n + 3 is prime, and Q(n) :
7n + 1 is prime, both over the domain N.
(a) State P (n) ⇒ Q(n) in words.
(b) State P (2) ⇒ Q(2) in words. Is this statement true or false?
(c) State P (6) ⇒ Q(6) in words. Is this statement true or false?
Solution:
• If 5n + 3 is prime then 7n + 1 is prime.
• If 13 is prime then 15 is prime — T ⇒ F is false.
• If 33 is prime then 43 is prime — F ⇒ T is true.
Page 5 of 6
Math220, Homework 3
Due in class January 28 or or 29 (depending on your section)
10. Consider the statement: The fish are biting and there are no bugs, or the fish are not
biting and there are bugs, or it is winter. Write out the negation of the above in English.
You should simplify your answer as much as possible, being sure of course that it is
logically equivalent to the negation. Justify your answer.
Solution: P : the fish are biting; B: there
are bugs; W : it is winter.
So we want ¬ (P ∧¬B) ∨(¬P ∧B) ∨W . Using a double application of De Morgan’s
laws this is logically equivalent to
(¬(P ∧ ¬B)) ∧ (¬(¬P ∧ B)) ∧ ¬W
≡ ((¬P ) ∨ B) ∧ (P ∨ (¬B)) ∧ ¬W.
(1)
This answer is not bad but one can simplify further using the distributive laws. A
double application of the distributive laws given in the text (we did this version in
class in Sec. 201) shows that
(R ∨ S) ∧ (T ∨ U) ≡ (R ∧ T ) ∨ (R ∧ U) ∨ (S ∧ T ) ∨ (S ∧ U).
Use this in (4), noting that P ∧ ¬P and B ∧ ¬B are always F to see that (4) is
logically equivalent to
F ∨ (¬P ∧ ¬B) ∨ (B ∧ P ) ∨ F ∧ (¬W )
≡ (¬P ∧ ¬B) ∨ (B ∧ P ) ∧ (¬W )
≡ (¬P ∧ ¬B ∧ ¬W ) ∨ (P ∧ B ∧ ¬W ).
This reads: the fish are not biting and there are no bugs and it is not winter, or the
fish are biting and there are bugs and it is not winter.
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