Page 356 - HOW TO PROVE IT: A Structured Approach, Second Edition

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342 Appendix 1: Solutions to Selected Exercises
Chapter 4

Section 4.1
1. (a) {(x, y) ∈ P × P | x is a parent of y}= {(George H. W. Bush, George
W. Bush), (Goldie Hawn, Kate Hudson), ...}.
(b) {(x, y) ∈ C × U | there is someone who lives in x and attends y}.If
you are a university student, then let x be the city you live in, and let
y be the university you attend; (x, y) will then be an element of this
truth set.
4. A × (B ∩ C) = (A × B) ∩ (A × C) ={(1, 4), (2, 4), (3, 4)},
A × (B ∪ C) = (A × B) ∪ (A × C) ={(1, 1), (2, 1), (3, 1), (1, 3), (2, 3),
(3, 3), (1, 4), (2, 4), (3, 4)},
(A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D) = ∅,
(A × B) ∪ (C × D) ={(1, 1), (2, 1), (3, 1), (1, 4), (2, 4), (3, 4), (3, 5),
(4, 5)},
(A ∪ C) × (B ∪ D) ={(1, 1), (2, 1), (3, 1), (4, 1), (1, 4), (2, 4), (3, 4),
(4, 4), (1, 5), (2, 5), (3, 5), (4, 5)}.
6. The cases are not exhaustive.
8. True.
12. The theorem is incorrect. Counterexample: A ={1}, B = C = D = ∅.
Notice that A ⊆ C. Where is the mistake in the proof that A ⊆ C?

Section 4.2
1. (a) Domain ={p ∈ P | p has a living child}; Range ={p ∈ P | p has a
living parent}.
(b) Domain = R; Range = R .
+
4. (a) {(1, 4), (1, 5), (1, 6), (2, 4), (3, 6)}.
(b) {(4, 4), (5, 5), (5, 6), (6, 5), (6, 6)}.
7. E ◦ E ⊆ F.
10. We prove the contrapositives of both directions.
(→) Suppose Ran(R) and Dom(S) are not disjoint. Then we can choose
someb ∈Ran(R) ∩Dom(S).Sinceb ∈Ran(R),wecanchoosesomea ∈ A
such that (a, b) ∈ R. Similarly, since b ∈ Dom(S), we can choose some
c ∈ C such that (b, c) ∈ S. But then (a, c) ∈ S ◦ R,so S ◦ R = ∅.
(←) Suppose S ◦ R = ∅. Then we can choose some (a, c) ∈ S ◦ R.
By definition of S ◦ R, this means that we can choose some b ∈ B such
that (a, b) ∈ R and (b, c) ∈ S. But then b ∈ Ran(R) and b ∈ Dom(S), so
Ran(R) and Dom(S) are not disjoint.