29
13. Proof. Suppose that a ≤ 2, b ≤ 1 and c ≤ 0. Thus c = 0 and a + 2b + 3c = a + 2b ≤ 2 + 2 · 1 =
4 < 5.
14. The preceding proof verifies that 5n + 7 is even if and only if n is odd. This is not the stated
result.
Exercises for Section 3.4. Proof by Cases
1. Proof. Let n be an integer. We consider two cases.
Case 1. n is even. Then n = 2a for some integer a. Therefore,
n2 − 3n + 5 = (2a)2 − 3(2a) + 5 = 4a2 − 6a + 5 = 2(2a2 − 3a + 2) + 1.
Since 2a2 − 3a + 2 is an integer, n2 − 3n + 5 is odd.
Case 2. n is odd. Then n = 2b + 1 for some integer b. Therefore,
n2 − 3n + 5
= (2b + 1)2 − 3(2b + 1) + 5 = (4b2 + 4b + 1) − 6b − 3 + 5
= 4b2 − 2b + 3 = 2(2b2 − b + 1) + 1.
Since 2b2 − b + 1 is an integer, n2 − 3n + 5 is odd.
2. Proof. Let n be an integer. We consider two cases.
Case 1. n is even. Then n = 2a for some integer a. Therefore,
n3 − n = (2a)3 − (2a) = 8a3 − 2a = 2(4a3 − a).
Since 4a3 − a is an integer, n3 − n is odd.
Case 2. n is odd. Then n = 2b + 1 for some integer b. Therefore,
n3 − n
= (2b + 1)3 − (2b + 1) = (8b3 + 12b2 + 6b + 1) − 2b − 1
= 8b3 + 12b2 + 4b = 2(4b3 + 6b2 + 2b).
Since 4b3 + 6b2 + 2b is an integer, n3 − n is even.
3. Proof. Assume that n2 + n = n(n + 1) = 0. Then n = 0 and n = −1. We consider these two
cases.
Case 1. n = 0. Then
20 + 30
1+1
2n + 3n
=
=
= 2,
n
12
120
1
which is even.
Case 2. n = −1. Then
2n + 3n
2−1 + 3−1
=
=
n
12
12−1
which is even.
1
2
+
1
12
1
3
= 12
!
1 1
+
2 3
"
= 6 + 4 = 10,
30
4. Proof. Let n be an integer. We consider two cases.
Case 1. n is even. Then n = 2a for some integer a. Therefore,
3n + 1
5n + 2
=
=
3(2a) + 1 = 6a + 1 = 2(3a) + 1 and
5(2a) + 2 = 10a + 2 = 2(5a + 1).
Since 3a and 5a + 1 are integers, 3n + 1 is odd and 5n + 2 is even.
Case 2. n is odd. Then n = 2b + 1 for some integer b. Therefore,
3n + 1
= 3(2b + 1) + 1 = 6b + 4 = 2(3b + 2) and
5n + 2
= 5(2b + 1) + 2 = 10b + 7 = 2(5b + 3) + 1.
Since 2b + 1 and 5b + 3 are integers, 3n + 1 is even and 5n + 2 is odd.
5. (a) Proof. Assume that m and n are of the same parity. We consider two cases.
Case 1. m and n are even. Therefore, m = 2a and n = 2b for integers a and b. So
m + n = (2a) + (2b) = 2(a + b).
Since a + b is an integer, m + n is even.
Case 2. m and n are odd. Therefore, m = 2a + 1 and n = 2b + 1 for integers a and b. So
m + n = (2a + 1) + (2b + 1) = 2(a + b + 1).
Since a + b + 1 is an integer, m + n is even.
For the converse, assume that m and n are of opposite parity, say m is even and n is
odd. Then m = 2x and n = 2y + 1 for integers x and y. Thus
m + n = 2x + (2y + 1) = 2(x + y) + 1.
Since x + y is an integer, m + n is odd.
(b) Proof. First, observe that there are six pairs of integers of S, namely, {a, b}, {a, c}, {a, d},
{b, c}, {b, d} and {c, d}. Thus m + n = 6, which is even. By (a), m and n are of the same
parity.
6. Proof. First, we show that if m and n are of the same parity, then 3m − n is even. Assume
that m and n are of the same parity. We consider two cases.
Case 1. m and n are even. Therefore, m = 2a and n = 2b for integers a and b. So
3m − n = 3(2a) − (2b) = 6a − 2b = 2(3a − b).
Since 3a − b is an integer, 3m − n is even.
Case 2. m and n are odd. Then m = 2a + 1 and n = 2b + 1, where a, b ∈ Z. Hence
3m − n
= 3(2a + 1) − (2b + 1) = 6a + 3 − 2b − 1
= 6a − 2b + 2 = 2(3a − b + 1).
Since 3a − b + 1 is an integer, 3m − n is even.
Next, we verify the converse, that is, if 3m − n is even, then m and n are of the same parity.
Assume that m and n are of opposite parity. We consider two cases.
31
Case 1. m is even and n is odd. So m = 2a and n = 2b + 1 for integers a and b. Therefore,
3m − n
=
=
3(2a) − (2b + 1) = 6a − 2b − 1
6a − 2b − 2 + 1 = 2(3a − b − 1) + 1.
Since 3a − b − 1 is an integer, 3m − n is odd.
Case 2. m is odd and n is even. Thus m = 2a + 1 and n = 2b for integers a and b. Therefore,
3m − n = 3(2a + 1) − (2b) = 6a + 3 − 2b = 2(3a − b + 1) + 1.
Since 3a − b + 1 is an integer, 3m − n is odd.
7. Proof. We first show that if m and n are of opposite parity, then 7m + 3n is odd. We consider
two cases.
Case 1. m is odd and n is even. Therefore, m = 2a + 1 and n = 2b for integers a and b. So
7m + 3n = 7(2a + 1) + 3(2b) = 14a + 7 + 6b = 2(7a + 3b + 3) + 1.
Since 7a + 3b + 3 is an integer, 7m + 3n is odd.
Case 2. m is even and n is odd. Therefore, m = 2a and n = 2b + 1 for integers a and b. So
7m + 3n = 7(2a) + 3(2b + 1) = 14a + 6b + 3 = 2(7a + 3b + 1) + 1.
Since 7a + 3b + 1 is an integer, 7m + 3n is odd.
Next, we show that if 7m + 3n is odd, then m and n are of opposite parity. We use a proof
by contrapositive. Assume that m and n are of the same parity. We consider two cases.
Case 1. m and n are even. Therefore, m = 2a and n = 2b for integers a and b. So
7m + 3n = 7(2a) + 3(2b) = 14a + 6b = 2(7a + 3b).
Since 7a + 3b is an integer, 7m + 3n is even.
Case 2. m and n are odd. Then m = 2a + 1 and n = 2b + 1, where a, b ∈ Z. Hence
7m + 3n =
=
7(2a + 1) + 3(2b + 1) = 14a + 7 + 6b + 3
14a + 6b + 10 = 2(7a + 3b + 5).
Since 7a + 3b + 5 is an integer, 7m + 3n is even.
8. Proof. Assume that m and n are of opposite parity. We consider two cases.
Case 1. m is even and n is odd. Then m = 2k and n = 2ℓ + 1 for some integers k and ℓ. Then
2m + n
= 2(2k) + (2ℓ + 1) = 4k + 2ℓ + 1
= 2(2k + ℓ) + 1,
3m − 4n
= 3(2k) − 4(2ℓ + 1) = 6k − 8ℓ − 4
while
= 2(3k − 4ℓ − 2).
Since 2k + ℓ and 3k − 4ℓ − 2 are integers, 2m + n is odd and 3m − 4n is even and so 2m + n
and 3m − 4n are of opposite parity.
32
Case 2. m is odd and n is even. Then m = 2k + 1 and n = 2ℓ for some integers k and ℓ. Then
2m + n
= 2(2k + 1) + (2ℓ) = 4k + 2 + 2ℓ
= 2(2k + 1 + ℓ),
3m − 4n
= 3(2k + 1) − 4(2ℓ) = 6k + 3 − 8ℓ
= 2(3k − 4ℓ + 1) + 1.
while
Since 2k + 1 + ℓ and 3k − 4ℓ + 1 are integers, 2m + n is even and 3m − 4n is odd. Therefore,
2m + n and 3m − 4n are of opposite parity.
9. Proof. First, we show that if m and n are odd, then mn2 is odd. Let m and n be odd integers.
Thus m = 2a + 1 and n = 2b + 1 for integers a and b. Therefore,
mn2
=
(2a + 1)(2b + 1)2 = (2a + 1)(4b2 + 4b + 1)
=
8ab2 + 8ab + 2a + 4b2 + 4b + 1 = 2(4ab2 + 4ab + a + 2b2 + 2b) + 1.
Since 4ab2 + 4ab + a + 2b2 + 2b is an integer, mn2 is odd.
Next, we verify the converse, that is, if mn2 is odd, then m and n are odd. Assume that m
or n is even. We consider two cases.
Case 1. m is even. Then m = 2c for some integer c. Thus
mn2 = (2c)n2 = 2(cn2 ).
Since cn2 is an integer, mn2 is even.
Case 2. n is even. Then n = 2d for some integer d. Thus
mn2 = m(2d)2 = m(4d2 ) = 4md2 = 2(2md2 ).
Since 2md2 is an integer, mn2 is even.
10. Proof. We first show that if m and n are both even, then mn and m + n are both even.
Assume that m and n are both even. Then m = 2a and n = 2b, where a, b ∈ Z. Hence
mn = (2a)(2b) = 2(2ab) and m + n = (2a) + (2b) = 2(a + b). Since 2ab and a + b are integers,
mn and m + n are both even.
Next, we verify the converse, that is, if mn and m + n are both even, then m and n are
both even. Assume that m and n are not both even. Then at least one of m and n is odd, say
m is odd and so m = 2a + 1 for some integer a. We consider two cases.
Case 1. n is even. Then n = 2b for some integer b. Thus
m + n = (2a + 1) + (2b) = 2(a + b) + 1.
Since a + b is an integer, m + n is odd.
Case 2. n is odd. Then n = 2b + 1 for some integer b. Thus
mn = (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1.
Since 2ab + a + b is an integer, mn is odd.
33
11. (a) Proof. Assume that |2n − 1| ≤ 5. Then −5 ≤ 2n − 1 ≤ 5. Therefore, −4 ≤ 2n ≤ 6.
Dividing by 2, we have −2 ≤ n ≤ 3.
(b) Proof. Assume that n > 3 or n < −2. We consider these two cases.
Case 1. n > 3. Then |2n − 1| = 2n − 1 > 5.
Case 2. n < −2. Then |2n − 1| = −(2n − 1) = −2n + 1 > 5.
12. Proof. Assume that m or n is even. We consider two cases.
Case 1. m is even. Then m = 2k for some integer k. Hence mn = (2k)n = 2(kn). Since kn is
an integer, mn is even.
Case 2. n is even. Then n = 2ℓ for some integer ℓ. Hence mn = m(2ℓ) = 2(mℓ). Since mℓ is
an integer, mn is even.
13. Proof. First, we show that (A − B) ∪ (A − C) ⊆ A − (B ∩ C). Let x ∈ (A − B) ∪ (A − C).
Then x ∈ A − B or x ∈ A − C. Assume, without loss of generality, that x ∈ A − B. Then
x ∈ A and x ∈
/ B. Since x ∈
/ B, it follows that x ∈
/ B ∩ C. Because x ∈ A and x ∈
/ B ∩ C, we
have x ∈ A − (B ∩ C). Hence (A − B) ∪ (A − C) ⊆ A − (B ∩ C).
Next, we show that A − (B ∩ C) ⊆ (A − B) ∪ (A − C). Let y ∈ A − (B ∩ C). Hence y ∈ A
and y ∈
/ B ∩ C. Since y ∈
/ B ∩ C, either y ∈
/ B or y ∈
/ C. Assume, without loss of generality,
that y ∈
/ B. Since y ∈ A, we have y ∈ A − B. Therefore, y ∈ (A − B) ∪ (A − C) and so
A − (B ∩ C) ⊆ (A − B) ∪ (A − C).
Therefore, (A − B) ∪ (A − C) = A − (B ∩ C).
14. Proof. We first show that A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C). Let x ∈ A ∩ (B ∪ C). Then x ∈ A
and x ∈ B ∪ C. Since x ∈ B ∪ C, it follows that x ∈ B or x ∈ C. We consider these two cases.
Case 1. x ∈ B. Since x ∈ A and x ∈ B, it follows that x ∈ A ∩ B. Because A ∩ B ⊆
(A ∩ B) ∪ (A ∩ C), we have x ∈ (A ∩ B) ∪ (A ∩ C).
Case 2. x ∈ C. Since x ∈ A and x ∈ C, it follows that x ∈ A ∩ C. Thus x ∈ (A ∩ B) ∪ (A ∩ C).
Therefore, A∩(B∪C) ⊆ (A∩B)∪(A∩C). Next, we show that (A∩B)∪(A∩C) ⊆ A∩(B∪C).
Let y ∈ (A ∩ B) ∪ (A ∩ C). Then y ∈ A ∩ B or y ∈ A ∩ C. There are two cases.
Case 1. y ∈ A ∩ B. Thus y ∈ A and y ∈ B. Since y ∈ B, it follows that y ∈ B ∪ C and so
y ∈ A ∩ (B ∪ C).
Case 2. y ∈ A ∩ C. Thus y ∈ A and y ∈ C. Since y ∈ C, it follows that y ∈ B ∪ C. Hence
y ∈ A ∩ (B ∪ C).
Therefore, (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C). Thus (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C).
15. Proof #1. First we show that (A−B)∪(B−A) ⊆ (A∪B)−(A∩B). Let x ∈ (A−B)∪(B−A).
Then x ∈ A − B or x ∈ B − A. We consider these two cases.
Case 1. x ∈ A − B. So x ∈ A and x ∈
/ B. Since x ∈ A, it follows that x ∈ A ∪ B
and because x ∈
/ B, it follows that x ∈
/ A ∩ B. Hence x ∈ (A ∪ B) − (A ∩ B). Therefore,
(A − B) ∪ (B − A) ⊆ (A ∪ B) − (A ∩ B).
Case 2. x ∈ B − A. The proof in this case is similar to the proof in Case 1. So here too
(A − B) ∪ (B − A) ⊆ (A ∪ B) − (A ∩ B).
Next, we show (A ∪ B) − (A ∩ B) ⊆ (A − B) ∪ (B − A). Let x ∈ (A ∪ B) − (A ∩ B). Then
x ∈ A ∪ B and x ∈
/ A ∩ B. Since x ∈ A ∪ B, either x ∈ A or x ∈ B. We consider these two
cases.
Case 1. x ∈ A. Since x ∈
/ A ∩ B, it follows that x ∈
/ B. Hence x ∈ A − B. Therefore,
x ∈ (A − B) ∪ (B − A). Thus (A ∪ B) − (A ∩ B) ⊆ (A − B) ∪ (B − A).
34
Case 2. The proof in this case is similar to the proof in Case 1. Thus (A − B) ∪ (B − A) ⊆
(A ∪ B) − (A ∩ B) in this case as well.
Therefore, (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).
Proof #2. First we show that (A−B)∪(B−A) ⊆ (A∪B)−(A∩B). Let x ∈ (A−B)∪(B−A).
Then x ∈ A − B or x ∈ B − A. Assume, without loss of generality, that x ∈ A − B. So x ∈ A
and x ∈
/ B. Thus x ∈ A ∪ B and x ∈
/ A ∩ B. Consequently, x ∈ (A ∪ B) − (A ∩ B). Therefore,
(A − B) ∪ (B − A) ⊆ (A ∪ B) − (A ∩ B).
Next we show that (A ∪ B) − (A ∩ B) ⊆ (A − B) ∪ (B − A). Let x ∈ (A ∪ B) − (A ∩ B).
Then x ∈ A ∪ B and x ∈
/ A ∩ B. Since x ∈ A ∪ B, it follows that x ∈ A or x ∈ B. Without loss
of generality, let x ∈ A. Since x ∈
/ A ∩ B, it follows that x ∈
/ B. Therefore, x ∈ A − B and so
x ∈ (A − B) ∪ (B − A). Hence
(A ∪ B) − (A ∩ B) ⊆ (A − B) ∪ (B − A).
Therefore, (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).
16. Proof. First, observe that x2 − 4x + 3 = (x − 1)(x − 3). We consider three cases.
Case 1. x = 1 or x = 3. Then x is a solution to the equation x2 − 4x + 3 = 0 and so
x2 − 4x + 3 = 0.
Case 2. x = 0. Then x2 − 4x + 3 = 0 + 0 + 3 = 3.
Case 3. x = 4. Then x2 − 4x + 3 = 16 − 16 + 3 = 3.
Exercises for Section 3.5. Counterexamples
1. Solution. Since n = 2 is even and 3n + 2 = 3 · 2 + 2 = 8 is even, n = 2 is a counterexample. !
2. Solution. Since m = 2 and n = 0 are of the same parity and mn = 2 · 0 = 0 is even, m = 2
and n = 0 produce a counterexample.
!
3. Solution. The integers x = −1 and y = −2 form a counterexample since | − 1| = 1 < | − 2| = 2
but −1 > −2.
!
4. Solution. Since −1 < 0 < 1 but 02 − 0 − 2 < 0, it follows that x = 0 is a counterexample. !
5. Solution. The sets A = {1}, B = {1, 2}, C = {1, 3} form a counterexample since A ∩ B =
A ∩ C = {1} but B ̸= C.
!
6. Solution. The sets A = {1, 2}, B = {1} and C = {2} form a counterexample since A ∪ B =
A ∪ C = {1, 2} but B ̸= C.
!
7. Solution. The sets A = ∅ and B = {1} form a counterexample since A ∪ B ̸= ∅ but A = ∅. !
8. Solution. The sets A = {1} and B = {2} form a counterexample since A ∪ B ∈ P(A ∪ B) but
A∪B ∈
/ P(A) ∪ P(B).
!
9. Solution. The sets A = {1, 2, 3}, B = {1}, C = {2} form a counterexample since A∪(B−C) =
A but (A ∪ B) − (A ∪ C) = A − A = ∅.
!
10. Solution. Let n = 1. Then n2 = 1 and there is no m ∈ N such that n < m < n2 . Hence n = 1
is a counterexample.
!
11. Solution. Consider the positive integer 1. If 1 = a + b, where a, b ∈ Z, then 1 = a + b ≥
1 + 1 = 2, which is impossible.
!