Page 148 - HOW TO PROVE IT: A Structured Approach, Second Edition

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134 Proofs
11. Consider the following putative theorem:
Theorem? Suppose m is an even integer and n is an odd integer. Then
2
2
n − m = n + m.
(a) What’s wrong with the following proof of the theorem?
Proof. Since m is even, we can choose some integer k such that
m = 2k. Similarly, since n is odd we have n = 2k + 1. Therefore
2
2
2
2
2
2
n − m = (2k + 1) − (2k) = 4k + 4k + 1 − 4k = 4k + 1

= (2k + 1) + (2k) = n + m.
(b) Is the theorem correct? Justify your answer with either a proof or a
counterexample.
∗
12. Prove that ∀x ∈ R[∃y ∈ R(x + y = xy) ↔ x = 1].
+
13. Prove that ∃z ∈ R∀x ∈ R [∃y ∈ R(y − x = y/x) ↔ x = z].
p d14. Suppose B is a set and F is a family of sets. Prove that ∪{A \ B | A ∈
F}⊆∪(F \ P (B)).
∗
15. Suppose F and G are nonempty families of sets and every element of F
is disjoint from some element of G. Prove that ∪F and ∩G are disjoint.
p d16. Prove that for any set A, A =∪P (A).
p ∗
d 17. Suppose F and G are families of sets.
(a) Prove that ∪(F ∩ G) ⊆ (∪F) ∩ (∪G).
(b) What’s wrong with the following proof that (∪F) ∩ (∪G) ⊆
∪(F ∩ G)?
Proof. Suppose x ∈ (∪F) ∩ (∪G). This means that x ∈∪F and
x ∈∪G,so ∃A ∈ F(x ∈ A) and ∃A ∈ G(x ∈ A). Thus, we can
choose a set A such that A ∈ F, A ∈ G, and x ∈ A. Since A ∈ F and
A ∈ G, A ∈ F ∩ G. Therefore ∃A ∈ F ∩ G(x ∈ A), so x ∈∪(F ∩
G). Since x was arbitrary, we can conclude that (∪F) ∩ (∪G) ⊆
∪(F ∩ G).

(c) Find an example of families of sets F and G for which ∪(F ∩ G) =
(∪F) ∩ (∪G).
p d18. Suppose F and G are families of sets. Prove that (∪F) ∩ (∪G) ⊆
∪(F ∩ G)iff ∀A ∈ F∀B ∈ G(A ∩ B ⊆∪(F ∩ G)).
p d19. Suppose F and G are families of sets. Prove that ∪F and ∪G are disjoint
iff for all A ∈ F and B ∈ G, A and B are disjoint.
p d20. Suppose F and G are families of sets.
(a) Prove that (∪F) \ (∪G) ⊆∪(F \ G).
(b) What’s wrong with the following proof that ∪(F \ G) ⊆ (∪F) \
(∪G)?