Page 349 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 335
6
the theorem tells us that 2 − 1 is not prime. This is correct, since
6
2 − 1 = 63 = 9 · 7.
(b) We can conclude that 32767 is not prime. This is correct, since
32767 = 151 · 217.
(c) The theorem tells us nothing; 11 is prime, so the hypotheses are not
satisfied.
4. Suppose 0 < a < b. Then b − a > 0. Multiplying both sides by the pos-
itive number b + a, we get (b + a) · (b − a) > (b + a) · 0, or in other
2
2
2
2
2
2
words b − a > 0. Since b − a > 0, it follows that a < b . Therefore
2
2
if 0 < a < b then a < b .
9. Hint: Add b to both sides of the inequality a < b.
11. We will prove the contrapositive. Suppose c ≤ d. Multiplying both sides of
this inequality by the positive number a, we get ac ≤ ad. Also, multiplying
both sides of the given inequality a < b by the positive number d gives us
ad < bd. Combining ac ≤ ad and ad < bd, we can conclude that ac <
bd. Thus, if ac ≥ bd then c > d.
2
14. Since x > 3 > 0, by the theorem in Example 3.1.2, x > 9. Also, multi-
plying both sides of the given inequality y < 2by −2 (and reversing
the direction of the inequality, since −2 is negative) we get −2y > −4.
2
Finally, adding the inequalities x > 9 and −2y > −4 gives us
2
x − 2y > 5.

Section 3.2

1. (a) Suppose P. Since P → Q, it follows that Q. But then, since Q → R,
we can conclude R. Thus, P → R.
(b) Suppose P. To prove that Q → R, we will prove the contrapositive,
so suppose ¬R. Since ¬R → (P →¬Q), it follows that P →¬Q,
and since we know P, we can conclude ¬Q. Thus, Q → R,so
P → (Q → R).
5. Suppose a ∈ A \ B. This means that a ∈ A and a /∈ B. Since a ∈ A and
a ∈ C, a ∈ A ∩ C. But then since A ∩ C ⊆ B, it follows that a ∈ B, and
this contradicts the fact that a /∈ B. Thus, a /∈ A \ B.
8. Hint: Assume a < 1/a < b < 1/b. Now prove that a < 0, and then use
this fact to prove that a < −1.
11. (a) The sentence “Then x = 3 and y = 8” is incorrect. (Why?)
(b) One counterexample is x = 3, y = 7.