Page 362 - HOW TO PROVE IT: A Structured Approach, Second Edition

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348 Appendix 1: Solutions to Selected Exercises
5. The assumption that is needed is that for every date d, someone was born
on the date d. What would go wrong if, say, just by chance, no one was
born on April 23? Where in the proof is this assumption used?
9. Since S is the equivalence relation determined by F, the proof of Theo-
rem 4.6.6 shows that A/S = F = A/R. The desired conclusion now fol-
lows from exercise 8.
12. Suppose a ≡ c (mod m) and b ≡ d (mod m). Then m | (a − c) and m |
(b − d). By exercise 18(a) of Section 3.3, it follows that m | (a − c +
b − d). But a − c + b − d = (a + b) − (c + d), so m | ((a + b) −
(c + d)), and therefore a + b ≡ c + d (mod m).
For the second half of the problem, you might find it useful to begin
with the equation ab − cd = (ab − ad) + (ad − cd).
15. By exercise 15(a) of Section 3.5, ∪(F ∪ G) = (∪F ) ∪ (∪G) = A ∪ B.To
see that F ∪ G is pairwise disjoint, suppose that X ∈ F ∪ G, Y ∈ F ∪ G,
and X ∩ Y = ∅.If X ∈ F and Y ∈ G then X ⊆ A and Y ⊆ B, and since
A and B are disjoint it follows that X and Y are disjoint, which is a con-
tradiction. Thus it cannot be the case that X ∈ F and Y ∈ G, and a similar
argument can be used to rule out the possibility that X ∈ G and Y ∈ F.
Thus, X and Y are either both elements of F or both elements of G.If
they are both in F, then since F is pairwise disjoint, X = Y. A similar
argument applies if they are both in G. Finally, we have ∀X ∈ F(X = ∅)
and ∀X ∈ G(X = ∅), and it follows by exercise 8 of Section 2.2 that
∀X ∈ F ∪ G(X = ∅).
19. (a) Here is the proof of transitivity: Suppose (x, y) ∈ T and (y, z) ∈ T .
Then since T = R ∩ S, (x, y) ∈ R and (y, z) ∈ R, so since R is tran-
sitive, (x, z) ∈ R. Similarly, (x, z) ∈ S,so(x, z) ∈ R ∩ S = T .
(b) Suppose x ∈ A. Then for all y ∈ A,
y ∈ [x] T iff (y, x) ∈ T iff (y, x) ∈ R ∧ (y, x) ∈ S
iff y ∈ [x] R ∧ y ∈ [x] S iff y ∈ [x] R ∩ [x] S .
(c) Suppose X ∈ A/T . Then since A/T is a partition, X = ∅. Also,
for some x ∈ A, X = [x] T = [x] R ∩ [x] S , so since [x] R ∈ A/R and
[x] S ∈ A/S, X ∈ (A/R) · (A/S).
Now suppose X ∈ (A/R) · (A/S). Then for some y and z in A, X =
[y] R ∩ [z] S . Also, X = ∅, so we can choose some x ∈ X. There-
fore x ∈ [y] R and x ∈ [z] S , and by part 2 of Lemma 4.6.5 it fol-
lows that [x] R = [y] R and [x] S = [z] S . Therefore X = [x] R ∩ [x] S =
[x] T ∈ A/T .
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21. F ⊗ F ={R × R , R × R , R × R , R × R , R ×{0}, R ×{0},
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{0}× R , {0}× R , {(0, 0)}}. In geometric terms these are the four
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