Page 354 - HOW TO PROVE IT: A Structured Approach, Second Edition

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340 Appendix 1: Solutions to Selected Exercises
that A B and C are disjoint. Therefore x ∈ B. Since we also know x ∈ C,
we have x ∈ B ∩ C. Since x was an arbitrary element of A ∩ C, this shows
that A ∩ C ⊆ B ∩ C. A similar argument shows that B ∩ C ⊆ A ∩ C.
(←) Suppose that A ∩ C = B ∩ C. Suppose that A B and C are not
disjoint. Then we can choose some x such that x ∈ A B and x ∈ C. Since
x ∈ A B, either x ∈ A \ B or x ∈ B \ A.
Case 1. x ∈ A \ B. Then x ∈ A and x /∈ B. Since we also know x ∈ C,
we can conclude that x ∈ A ∩ C but x /∈ B ∩ C. This contradicts the fact
that A ∩ C = B ∩ C.
Case 2. x ∈ B \ A. Similarly, this leads to a contradiction.
Thus we can conclude that A B and C are disjoint.
22. (a) Hint: Suppose x ∈ A \ C, and then break the proof into cases, depend-
ing on whether or not x ∈ B.
(b) Hint: Apply part (a).
23. (a) Suppose x ∈ (A ∪ B) C. Then either x ∈ (A ∪ B) \ C or x ∈ C \
(A ∪ B).
Case 1. x ∈ (A ∪ B) \ C. Then either x ∈ A or x ∈ B, and x /∈ C.
We now break case 1 into two subcases, depending on whether x ∈ A
or x ∈ B:
Case 1a. x ∈ A. Then x ∈ A \ C,so x ∈ A C,so x ∈ (A C) ∪
(B C).
Case 1b. x ∈ B. Similarly, x ∈ B C,so x ∈ (A C) ∪ (B C).
Case 2. x ∈ C \ (A ∪ B). Then x ∈ C, x /∈ A, and x /∈ B.It
follows that x ∈ A C and x ∈ B C, so certainly x ∈ (A C) ∪
(B C).
(b) Here is one example: A ={1}, B ={2}, C ={1, 2}.
26. The proof is incorrect, because it only establishes that either 0 < x or
x < 6, but what must be proven is that 0 < x and x < 6. However, it can
be fixed.
28. The proof is correct.
30. Hint: Here is a counterexample to the theorem: A ={1, 2}, B ={1},
C ={2}.

Section 3.6

2
1. Let x be an arbitrary real number. Let y = x/(x + 1). Then
3
x x + x x x 3 2 x 2
x − y = x − 2 = 2 − 2 = 2 = x 2 = x y.
x + 1 x + 1 x + 1 x + 1 x + 1