Homework #4
1. Consider the following variable propositions:
Let P (x) be the proposition “1 ≤ x ≤ 3”
Let Q(x) be the proposition “(∃k ∈ Z)(x = 2k)”
Let R(x) be the proposition “x2 = 4”
Recall that a statement is in positive form if the only negation symbols in the statement appear next
to substatements that do not contain quantifiers or connectives.
For each of the following statements, write the negation in a logically equivalent positive form. Then
decide which claim (the original or the negation) is true (no proof required).
a.) (∀x ∈ Z)(P (x) ⇒ Q(x))
negation: (∃x ∈ Z)(1 ≤ x ≤ 3 ∧ (∀k ∈ Z)(x 6= 2k))
The negation is true
b.) (∃x ∈ Z)(R(x) ∧ P (x))
negation: (∀x ∈ Z)(x2 6= 4 ∨ (x < 1 ∨ x > 3))
The original statement is true
c.) (∀x ∈ Z)(R(x) ⇒ P (x))
negation: (∃E ∈ Z)(x2 = 4 ∧ (x < 1 ∨ x > 3))
The negation statement is true
d.) (∀x ∈ Z)(∃y ∈ Z)(x 6= y ∧ P (x) ∧ Q(x))
negation: (∃x ∈ Z)(∀y ∈ Z)(x = y ∨ x < 1 ∨ x > 3 ∨ (∀k ∈ Z)(x 6= 2k))
The negation statement is true
2. Consider the following propositions, which assert that the rational numbers are dense, and the integers
are discrete, respectively:
(a) Strictly between any two distinct rational numbers lies a third rational number.
(∀m, n ∈ Q)[(m 6= n) =⇒ (∃q ∈ Q)(m < q < n ∨ n < q < m)]
(b) For every integer n, there is a strictly larger integer m, such that there are no integers strictly
between n and m.
(∀n ∈ Z)(∃m ∈ Z)[(m > n ∧ ((∀q ∈ Z)(q ≤ n ∨ q ≥ m))]
Write out these propositions symbolically, using only logical symbols and the sets Q and Z.
3. For every i ∈ N, define a set Ai ⊆ N such that the indexed family of sets {Ai : i ∈ N} satisfies all of
the following properties (recall that “(” means “is a strict subset of”)
Then, prove that the family you’ve defined satisfies each of these properties.
A1 = {x S
∈ N | where k ∈ Z and x = 2k}
Ai = A1 {x ∈ N | where k ∈ Z and x = 2k − 1, 1 ≤ x ≤ 2i − 3}
a.) (∀n ∈ N)(∃i ∈ N)(n ∈ Ai )
•
•
•
•
fix n ∈ N
for a given N ∈ N, if n ≤ 2N − 3, then n ∈ AN
Let i be large enough such that n ≤ 2i − 3 i.e. i = 5n
then n ∈ Ai ; since n was arbitrary, the claim is proved
b.) (∀i ∈ N)(∃n ∈ N)(n 6∈ Ai )
• fix i ∈ N
Observe that Ai contains no odd numbers greater than 2i − 3
• if n = 2i − 1, then n ∈
/ Ai
• since i was arbitrary, the claim is proved
c.) (∀i, m ∈ N)(∃n ∈ N)(n > m ∧ n ∈ Ai )
• fix i and m ∈ N
• let n = 2m
Observe: 2m ∈ Ai since 2m is even and Ai contains the set of positive, even natural numbers
• 2m > m
• since i and m were arbitrary, claim is proved
d.) (∃j ∈ N)(∀i ∈ N)(i 6= j ⇒ Aj ( Ai )
•
•
•
•
let j = 1
fix i ∈ N
assume i 6= j, that is i 6= 1
since i ∈ N, i > 1
S
Observe: Ai = A1 {1, 3, ..., 2i − 3}
• 1 ∈ Ai , and 1 ∈
/ A1 , but Ai contains all the elements of A1 , that is, the positive even numbers–
hence A1 ⊂ Ai
• since i was arbitrary, we have for every i: i 6= 1 =⇒ A1 ⊂ Ai
4. Use a chain of logical equivalences to prove the following propositions.
a.) Given a universal set U and sets A, B ⊆ U , it is the case that (A ∪ B) ∩ A = B − A.
Let x be an arbitrary el’t in U, Then,
Proof. let x ∈ (A ∪ B) ∩ A
⇐⇒ x ∈ A ∪ B ∧ x ∈ A ,by def’n of ∩
⇐⇒ [(x ∈ A) ∨ (x ∈ B)] ∧ (x ∈ A) , by def’n of ∪
⇐⇒ [(x ∈ A) ∨ (x ∈ B)] ∧ (x ∈
/ A) by def’n of complement
⇐⇒ ((x ∈ A ∧ x ∈
/ A) ∨ (x ∈ B ∧ x ∈
/ A)) by distribution property
⇐⇒ (x ∈ B) ∧ (x ∈
/ A) since if P is false, P ∨ Q ⇐⇒ Q
⇐⇒ x ∈ B − A by def’n of Since x was arbitrary, the logical equivalences apply to all elements x ∈ (A∪B)∩A so (A∪B)∩
A=B−A
b.) For all sets A, B, and C, it is the case that A ∩ (B − C) = (A ∩ B) − (A ∩ C). Let x be an arbitrary
el’t in U , then
Proof. let x ∈ A ∩ (B − C)
⇐⇒ x ∈ A ∧ (x ∈ B − C) by def’n of ∩
⇐⇒ (x ∈ A) ∧ (x ∈ B) ∧ ¬(x ∈ C) by def’n of ⇐⇒ x ∈ A ∧ (x ∈ B ∧ x ∈
/ C) by DeMorgan’s laws and associative property
⇐⇒ (x ∈ A ∧ x ∈ B) ∧ (x ∈ A ∧ x ∈
/ C) by distribution
⇐⇒ (x ∈ A ∩ B) ∧ (x ∈
/ A ∩ C) by def’n of ∩
⇐⇒ x ∈ (A ∩ B) − (A ∩ C) by def’n of ∩ and −
Since x was arbitrary, it is the case that A ∩ (B − C) = (A ∩ B) − (A ∩ C).
(A possibly helpful hint: if P is a false statement, then P ∨ Q is logically equivalent to Q.)
5. Consider the following proposition:
For all integers n, n is an integer multiple of 3 if and only if n2 − 1 is not a multiple of 3.
a.) Write out this proposition symbolically, using only logical symbols and the set Z.
(∀n ∈ Z)((∃k ∈ N)(3k = n) ⇐⇒ (∀m ∈ N)(3m 6= n2 − 1))
b.) Prove the proposition. (You should be able to prove it using nothing more than the definition of
being a multiple of 3, and the fact that every integer has a remainder of 0, 1, or 2 when divided
by 3.)
Proof. (→) Let n be an element in N such that (∃k ∈ N)(n = 3k). If we square both sides of the
equation, we get n2 = 9k 2 . This means that n2 = 3(3k 2 ). Which is equal to 3M , where M = 3k 2 .
Since we were able to factor out a 3, this means that n2 is a multiple of 3– thus, n2 − 1 cannot be
a multiple of 3 because it will have a remainder of 1 or 2 when divided by 3. Hence, the forwards
implication claim is true.
Proof. (←) To prove this claim, we will use the contrapositive: (∀n ∈ Z)((∀k ∈ N)(3k 6= n) →
(∃l ∈ N)(3l = n2 − 1)). Given this claim, we know that n is an integer that is not a multiple of 3.
This means that (∃m ∈ N)(3m = n + 1 ∨ 3m = n + 2) since the remainder of a non-multiple of 3
when divided by 3 must be 1 or 2. If we consider the first case, 3m = n − 1, then n = 3m + 1.
Then n2 = (3k − 1)2 = 9k 2 + 6k + 1. We are concerned with n2 − 1, so if we subtract 1 from both
sides of the equation, we get 9m2 + 6m. We can factor a 3 out of this, which gives us 3M , where
M = 3m2 + 2m. This means that n2 − 1 must be a multiple of 3 when n = 3m + 1. Now we can
consider the second case in which n = 3m + 2. In this case, n2 = (3m + 2)2 = 9m2 + 12m + 4.
Again, we are concerned with n2 − 1, which gives us 9m2 + 12m + 3. We can factor 3 out of this to
get 3M where M = (3m2 + 4m + 1). This means that when n = 3k + 2, n2 + 1 is still a multiple
of 3. Since we proved the claim for both cases, we can say that if an integer n is not a multiple of
3, then n2 − 1 must be a multiple of 3. Thus, the contrapositive is true– if n2 − 1 is not a multiple
of 3, then n2 must be a multiple of 3.
Since we proved both directions of the claim, we can say that the bi-implication statement is true.