(1) Question 3.18. Prove that 5x − 11 is even if and only if x is odd.
We have to prove both directions. We have done this before — the contrapositive helps a lot.
Proof. Assume x is odd, and so x = 2k + 1 for some k ∈ Z. Then 5x − 11 =
10k − 6 = 2(5k − 3). Since 5k − 3 is an integer, 5x − 11 is even.
We use the contrapositive to prove the second implication. Assume x is
even and so x = 2k for some k ∈ Z. Then 5x − 11 = 10k − 11 = 2(5k − 6) + 1.
Since 5k − 6 is an integer, 5x − 11 is odd.
(2) Prove that the product of two integers ab is odd if and only if both a and b
are odd.
Proof. We prove ⇐= directly. If a and b are both odd then a = 2k + 1 and
b = 2` + 1 for some k, ` ∈ Z. Thus ab = 4k` + 2k + 2` + 1 = 2(2k` + k + `) + 1
is odd.
To prove =⇒ , we deal with the contrapositive instead. If a and b are not
both odd, there are two possible cases (why?):
• If a is odd and b is even then a = 2k + 1 and b = 2` for some k, ` ∈ Z.
Therefore, ab = 2`(2k + 1) is even.
• If a is even and b is even then a = 2k and b = 2` for some k, ` ∈ Z.
Therefore, ab = 2k(2` + 1) is even.
(3) Prove that n3 is even only if n is even.
Proof. Seeking a contradiction, suppose n = 2k + 1 for some k ∈ Z. It then
follows that n3 = (2k + 1)3 = 8k 3 + 12k 2 + 6k + 1 = 2(4k 3 + 6k 2 + 3k) + 1 is
odd. This contradicts our assumption that n3 is even.
(4) Prove that for any sets A and B, A∆B = ∅ iff A = B.
Here recall that A∆B = (A − B) ∪ (B − A).
Proof. First, let us prove that A = B implies A∆B = ∅. We use direct
proof. Since A = B, we have A − B = ∅ and B − A = ∅. Then A∆B =
(A − B) ∪ (B − A) = ∅ ∪ ∅ = ∅.
Conversely, suppose A∆B = ∅. Since A∆B contains both the sets A − B
and B − A, we conclude that both A − B and B − A are empty. Since
1
2
B − A = ∅, we have B ⊆ A; since A − B = ∅, we have A ⊆ B. We conclude
that A = B.
(5) Consider the statement: For any sets A and B, (A ∪ B) − B = A. If it is
true, provide a proof. If it is false give a counter-example.
It is false. For example, take A = {1}, B = {1, 2}. Then A ∪ B = {1, 2},
and (A ∪ B) − B = ∅, but A 6= ∅.
(6) Question 4.56 Let A, B, C be sets. Prove that (A − B) ∪ (A − C) = A −
(B ∩ C).
Proof. To show ⊆, notice that x ∈ (A − B) ∪ (A − C) =⇒ x ∈ (A − B) ∨ x ∈
(A−C) =⇒ (x ∈ A∧x ∈
/ B)∨(x ∈ A∧x ∈
/ C). Since x ∈
/ B =⇒ x ∈
/ B ∩C
and x ∈
/ C =⇒ x ∈
/ B ∩ C we we find that x ∈ A and x ∈
/ B ∩ C , i.e.,
x ∈ A − (B ∩ C).
To show ⊇, notice that x ∈ A − (B ∩ C) =⇒ x ∈ A ∧ x ∈
/ (B ∩ C) =⇒
x ∈ A∧x ∈
/ B∧x ∈
/ C =⇒ x ∈ A ∧ x ∈
/ B =⇒ x ∈ (A − B) =⇒ x ∈
(A − B) ∪ (A − C).
(7) Question 4.58 Let A, B, C, D be sets. Prove that (A × B) ∩ (C × D) =
(A ∩ C) × (B ∩ D). Note: (A × B) ∪ (C × D) = (A ∪ C) × (B ∪ D) is false:
can you find a counterexample?
Proof. To see that the two sets are equal notice that (x, y) ∈ (A × B) ∩ (C ×
D) ⇐⇒ (x, y) ∈ (A × B) ∧ (x, y) ∈ (C × D) ⇐⇒ x ∈ A ∧ y ∈ B ∧ x ∈
C ∧ y ∈ D ⇐⇒ x ∈ A ∩ C ∧ y ∈ B ∩ D ⇐⇒ (x, y) ∈ (A ∩ C) × (B ∩ D). For a possible counterexample, let A := {1}, B := {2}, C := {3}, D := {4}.
Then (A × B) ∪ (C × D) = {(1, 2), (3, 4)} =
6 {(1, 2), (1, 4), (3, 2), (3, 4)} =
(A ∪ C) × (B ∪ D).
(8) Question 4.2. Let a, b ∈ Z, a, b 6= 0. Prove that if a|b and b|a, then a = b
or a = −b.
Proof. Assume that a|b and b|a. Thus b = ka for some k ∈ Z and a = `b for
some ` ∈ Z. But this implies b = ka = k`b. Since b 6= 0 we have k` = 1, or
k = 1/`. Since k, ` ∈ Z we must have either that k = ` = 1 or k = ` = −1.
It follows that a = b or a = −b.
(9) Question 4.4 Let x, y ∈ Z. Prove that if 3 - x and 3 - y then 3 | (x2 − y 2 ).
3
Proof. Assume that 3 - x and 3 - y. Hence x = 3k + i and y = 3` + j for
k, l ∈ Z and i, j ∈ {1, 2}. Then
x2 − y 2 = (3k + i)2 − (3` + j)2 = 3k 2 + 6ki + i2 − 3`2 − 6`j − j 2
= 3(k 2 + 2ki − `2 − 2`j) + (i2 − j 2 ).
Since (k 2 + 2ki − `2 − 2`j) ∈ Z, if we can show i2 − j 2 is a multiple of 3 then
we know that x2 − y 2 is divisible by 3. There are 4 cases
• i, j = 1 then i2 − j 2 = 0,
• i = 1, j = 2 then i2 − j 2 = −3,
• i = 2, j = 1 then i2 − j 2 = 3, and
• i = 2, j = 2 then i2 − j 2 = 0.
Hence no matter what the possible values of i, j we have that 3 | x2 − y 2