Download 36 it follows that x4 − x2 + 2 ̸= 0. 11. Proof. Consider the number

36
it follows that x4 − x2 + 2 ̸= 0.
!
√
√
11. Proof. Consider the number 3
these two possibilities separately.
Case 1.
2
. This number is either rational or irrational. We consider
√
√
√ √2
3 is rational. Letting a = 3 and b = 2 verifies that the statement is true.
√ √2
3 is irrational.
In this case, consider the number obtained by raising the√ (irra√
√ 2
√ √2
b
tional)
√ number 3 to the (irrational) power 2; that is, consider a , where a = 3 and
b = 2. Observe that
! √ "√2
√ √2·√2 √ 2
√ 2
3
= 3
= 3 = 3,
ab =
Case 2.
which is rational.
12. The statement is true. Proof. Let b ∈ Z. Now let a = |b| + 1. Thus a ∈ N and |a − |b|| =
|(|b| + 1) − |b|| = 1.
13. (a) The statement is true. Proof. Let a = 3 and b = 32 . Then
# $
(a − 1)(b − 1) = 2 12 = 1.
(b) The statement is true. Proof. Let a = 3 and b = 32 . Then
1
1
1 1
1 2
+ = + 3 = + = 1.
a b
3
3
3
2
Proof Analysis.
Observe that if a and b are two (distinct) rational numbers that
satisfy a1 + 1b = 1, then a+b
ab = 1 and so a + b = ab. Thus ab − a − b = 0, which is
equivalent to ab − a − b + 1 = 1 and so (a − 1)(b − 1) = 1. Therefore, two distinct rational
numbers a and b satisfy
(a − 1)(b − 1) = 1
if and only if a and b satisfy
1 1
+ =1
a b
if and only if a and b satisfy
a + b = ab. !
14. The statement is true. Proof.
Let a = 6 and b = 2. Then
a
b
+
3b
a
= 3 + 1 = 4.
15. The statement is false. Let A = {1}, which is nonempty, and let B be an arbitrary set. Since
1 ∈ A ∪ B, it follows that A ∪ B ̸= ∅ for every set B.
16. The statement is true. Proof. Let A be a proper subset of S and let B = S − A. Then
B ̸= ∅, A ∪ B = S and A ∩ B = ∅.
Exercises for Section 3.7. Proof by Contradiction
1. Proof. Assume, to the contrary, that there is a largest negative rational number r. Then
a
. Since s is a negative rational number and s > r,
r = − ab , where a, b ∈ N. Let s = 2r = − 2b
this is a contradiction.
37
2. Proof. Assume, to the contrary, that 100 can be expressed as the sum of three odd integers,
say x, y and z. Then x = 2a + 1, y = 2b + 1 and z = 2c + 1 for integers a, b and c. Thus
100 = x + y + z = (2a + 1) + (2b + 1) + (2c + 1) = 2(a + b + c + 1) + 1.
Since a + b + c + 1 is an integer, 100 is odd, which is a contradiction.
3. Proof. Assume, to the contrary, that 101 can be expressed as the sum of two even integers,
say x and y. Then x = 2a and y = 2b for integers a and b. Furthermore,
101 = x + y = 2a + 2b = 2(a + b).
Since a + b is an integer, 101 is even, which is a contradiction.
4. Proof. Assume, to the contrary, that there exists an odd integer n that can be expressed as
the sum of three even integers x, y and z. Then x = 2a, y = 2b and z = 2c for integers a, b
and c. Thus
n = x + y + z = (2a) + (2b) + (2c) = 2(a + b + c).
Since a + b + c is an integer, n is even, which is a contradiction.
5. Proof. Assume, to the contrary, that there exist a rational number a and an irrational number
b such that a − b is rational. Since a and a − b are rational, a = p/q and a − b = x/y, where
p, q, x, y ∈ Z and q, y ̸= 0. Therefore,
x
p
= a − b = − b;
y
q
so
py − xq
p x
− =
.
q
y
qy
Since py − xq and qy are integers and qy ̸= 0, it follows that b is a rational number. This is a
contradiction.
√
√
6. (a) Solution. Let a = 0 and let b = 2. Then
√ a is a rational number, 2 is irrational and
ab = 0 is rational. Hence a = 0 and b = 2 form a counterexample.
!
b=
(b) Proof. Assume, to the contrary, that a is a nonzero rational number, b is an irrational
number and ab is rational. Since a and ab are nonzero rational numbers, a = p/q and
ab = x/y, where p, q, x, y ∈ Z and p, q, x, y ̸= 0. Therefore,
x
p
= ab = b
y
q
and so
qx
q x
· =
.
p y
py
Since qx and py are integers and py ̸= 0, it follows that b is a rational number. This is a
contradiction.
b=
7. Proof. Assume, to the contrary, that there is a smallest positive irrational number r. Let
s = r/2. Then 0 < s < r. We claim that s is irrational. Suppose that s is rational. Then
s = a/b, where a, b ∈ Z and b ̸= 0. Hence r = 2s = 2a/b and so r is rational, which is
impossible. Thus, as claimed, s is irrational. This is a contradiction.
8. Proof. Assume, to the contrary, that there exists an even integer n such that 7n + 9 is also
even. Since n is even, n = 2k for some integer k. Therefore,
7n + 9 = 7(2k) + 9 = 14k + 9 = 2(7k + 4) + 1.
Since 7k + 4 is an integer, 7n + 9 is odd. This contradicts our assumption that 7n + 9 is even.
38
9. Proof. Assume, to the contrary, that 3n + 14 is and n is also odd. Then n = 2k + 1 for some
integer k. Therefore,
3n + 14 = 3(2k + 1) + 14 = 6k + 17 = 2(3k + 8) + 1.
Since 3k + 8 is an integer, 3n + 14 is odd. This contradicts our assumption that 3n + 14 is
even.
√
√ √
10. Proof. Assume, to the contrary, that a and b are positive real numbers and a+ b = a + b.
Squaring both sides, we obtain
√
a + 2 ab + b = a + b
√
or 2 ab = 0. Thus 4ab = 0, which implies that a = 0 or b = 0. This contradicts our
assumption that a and b are positive real numbers.
√
√
11. Proof. Assume, to the contrary, that 3 is rational. Hence 3 = x/y, where x, y ∈ Z and
y ̸= 0. Furthermore, we may assume that x/y has been reduced to lowest terms. Since
√
3 = x/y, it follows that 3 = x2 /y 2 and so x2 = 3y 2 . Since y 2 is an integer, x = 3z for some
integer z. Hence (3z)2 = 3y 2 and so 9z 2 = 3y 2 . Thus y 2 = 3z 2 . Since z 2 is an integer, y = 3w
for some integer w. Since x = 3z and y = 3w for integers z and w, it follows that x/y is not
in lowest terms. This is a contradiction.
√
√
√
√
12. Proof. Assume, to the contrary, that a√= 2 + 3 is a rational
number. Then a − 2 = 3.
√
Squaring both sides, we obtain a2 − 2a 2 + 2 = 3 and so 2 = (1 − a2 )/2a, which is rational.
This is a contradiction.
√
√
13. Proof. Assume, to the contrary, that 6 is rational. Then 6 = a/b for nonzero integers a
and b. We can further assume that a/b has been reduced to lowest terms. Thus 6 = a2 /b2 ; so
a2 = 6b2 = 2(3b2 ). Because 3b2 is an integer, a2 is even. Thus a is even. So a = 2c, where
c ∈ Z. Thus (2c)2 = 6b2 and so 4c2 = 6b2 . Therefore, 3b2 = 2c2 . Because c2 is an integer,
3b2 is even. Thus either 3 is even or b2 is even. Since 3 is not even, b2 is even and so b is
even. However, since a and b are both even, each has 2 as a divisor, which contradicts our
assumption that a/b has been reduced to lowest terms.
14. Proof. Suppose that a + b, a + c and b + c are all rational and that at least one of a, b and
c is irrational. We may assume that a is irrational. Let a + b = q1 , a + c = q2 and b + c = q3 ,
where q1 , q2 , q3 are rational. Adding a + b = q1 and a + c = q2 , we obtain 2a + b + c = q1 + q2
and so 2a = q1 + q2 − (b + c) = q1 + q2 − q3 = q, which is rational. Thus a = q/2 is rational,
which is a contradiction.
√
√
√
15. √
Proof. Assume,
to the contrary, that two of a + b, a + c and b + c are equal, say
√
a + b = a + c. Squaring both sides, we get a + b = a + c and so b = c, which is a
contradiction.
Supplementary Exercises for Chapter 3
1. (a) For every positive even integer n, 2n−2 is an even integer.
(b) There exists a positive even integer n such that 2n−2 is an even integer.
(c) There exists a positive even integer n such that 2n−2 is not an even integer.
(d) For every positive even integer n, 2n−2 is not an even integer.