Page 350 - HOW TO PROVE IT: A Structured Approach, Second Edition

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336 Appendix 1: Solutions to Selected Exercises
14. P Q R P → (Q → R) ¬R → (P →¬Q)
F F F T T
F F T T T
F T F T T
F T T T T
T F F T T
T F T T T
T T F F F
T T T T T

Section 3.3
1. Suppose ∃x(P(x) → Q(x)). Then we can choose some x 0 such that
P(x 0 ) → Q(x 0 ). Now suppose that ∀xP(x). Then in particular, P(x 0 ),
and since P(x 0 ) → Q(x 0 ), it follows that Q(x 0 ). Since we have found a
particular value of x for which Q(x) holds, we can conclude that ∃xQ(x).
Thus ∀xP(x) →∃xQ(x).
3. Suppose that A ⊆ B \ C,but A and C are not disjoint. Then we can choose
some x such that x ∈ A and x ∈ C. Since x ∈ A and A ⊆ B \ C, it follows
that x ∈ B \ C, which means that x ∈ B and x /∈ C. But now we have both
x ∈ C and x /∈ C, which is a contradiction. Thus, if A ⊆ B \ C then A and
C are disjoint.
√
2
7. Suppose x > 2. Let y = (x + x − 4)/2, which is defined since
2
x − 4 > 0. Then
√ √
2
2
1 x + x − 4 2 2x + 2x x − 4
2
y + = + √ = √ = x.
y 2 x + x − 4 2(x + x − 4)
2
2
9. Suppose F is a family of sets and A ∈ F. Suppose x ∈∩F. Then by the
definition of ∩F, since x ∈∩F and A ∈ F, x ∈ A. But x was an arbitrary
element of ∩F, so it follows that ∩F ⊆ A.
12. Hint: Assume F ⊆ G and let x be an arbitrary element of ∪F. You must
prove that x ∈∪G, which means ∃A ∈ G(x ∈ A), so you should try to find
some A ∈ G such that x ∈ A. To do this, write out the givens in logical
notation. You will find that one of them is a universal statement, and one
is existential. Apply existential instantiation to the existential one.
14. Suppose x ∈∪ i∈I P (A i ). Then we can choose some i ∈ I such that x ∈
P (A i ), or in other words x ⊆ A i . Now let a be an arbitrary element of x.
Then a ∈ A i , and therefore a ∈∪ i∈I A i . Since a was an arbitrary element
of x, it follows that x ⊆ U i∈I A i , which means that x ∈ P (U i∈I A i ). Thus
∪ i∈I P (A i ) ⊆ P (∪ i∈I A i ).