Page 351 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 337
17. Hint: The last hypothesis means ∀A ∈ F ∀B ∈ G(A ⊆ B), so if in the
course of the proof you ever come across sets A ∈ F and B ∈ G, you can
conclude that A ⊆ B. Start the proof by letting x be arbitrary and assuming
x ∈∪F, and prove that x ∈∩G. To see where to go from there, write these
statements in logical symbols.
2
20. The sentence “Then for every real number x, x < 0” is incorrect. (Why?)
22. Based on the logical form of the statement to be proven, the proof should
have this outline:
Let x = ....
Let y be an arbitrary real number.
2
[Proof of xy = y − x goes here.]
2
Since y was arbitrary, ∀y ∈ R(xy = y − x).
2
Thus, ∃x ∈ R∀y ∈ R(xy = y − x).
This outline makes it clear that y should be introduced into the proof
after x. Therefore, x cannot be defined in terms of y, because y will not yet
have been introduced into the proof when x is being defined. But in the
given proof, x is defined in terms of y in the first sentence. (The mistake
has been disguised by the fact that the sentence “Let y be an arbitrary real
number” has been left out of the proof. If you try to add this sentence to
the proof, you will find that there is nowhere it could be added that would
lead to a correct proof of the incorrect theorem.)
25. Here is the beginning of the proof: Let x be an arbitrary real number. Let
y = 2x. Now let z be an arbitrary real number. Then....

Section 3.4

1. (→) Suppose ∀x(P(x) ∧ Q(x)). Let y be arbitrary. Then since ∀x(P(x) ∧
Q(x)), P(y) ∧ Q(y), and so in particular P(y). Since y was arbitrary,
this shows that ∀xP(x). A similar argument proves ∀xQ(x): for arbitrary
y, P(y) ∧ Q(y), and therefore Q(y). Thus, ∀xP(x) ∧∀xQ(x).
(←) Suppose ∀xP(x) ∧∀xQ(x). Let y be arbitrary. Then since ∀xP(x),
P(y), and similarly since ∀xQ(x), Q(y). Thus, P(y) ∧ Q(y), and since y
was arbitrary, it follows that ∀x(P(x) ∧ Q(x)).
4. Suppose that A ⊆ B and A ⊆ C. Since A ⊆ C, we can choose some a ∈ A
such that a /∈ C. Since a ∈ A and A ⊆ B, a ∈ B. Since a ∈ B and a /∈
C, B ⊆ C.
7. Let A and B be arbitrary sets. Let x be arbitrary, and suppose that
x ∈ P (A ∩ B). Then x ⊆ A ∩ B. Now let y be all arbitrary element of
x. Then since x ⊆ A ∩ B, y ∈ A ∩ B, and therefore y ∈ A. Since y was