Page 369 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 355
n
n
Induction step: Suppose that n ≥ 1 and 0 < a < b . Multiplying
n
this inequality by the positive number a we get 0 < a n+1 < ab , and
n
multiplying the inequality a < b by the positive number b gives
n
us ab < b n+1 . Combining these inequalities, we can conclude that
0 < a n+1 < b n+1 .
√ √
(b) Hint: First note that n a and n b are both positive. (For n odd, this
follows from exercise 18. For n even, each of a and b has two nth
√ √
roots, one positive and one negative, but n a and n b are by definition
the positive roots.) Now use proof by contradiction, and apply part (a).
(c) Hint: The inequality to be proven can be rearranged to read a n+1 −
n
n
ab − ba + b n+1 > 0. Now factor the left side of this inequality.
(d) Hint: Use mathematical induction. For the base case, use the n = 1 case
of part (c). For the induction step, multiply both sides of the inductive
hypothesis by (a + b)/2 and then apply part (c).

Section 6.2
1. (a) We must prove that R is reflexive, transitive, and antisymmetric. For

the first, suppose x ∈ A . Since R is reflexive, (x, x) ∈ R,so(x, x) ∈

R ∩ (A × A ) = R . This shows that R is reflexive.

Next, suppose that (x, y) ∈ R and (y, z) ∈ R . Then (x, y) ∈
R, (y, z) ∈ R, and x, y, z ∈ A . Since R is transitive, (x, z) ∈ R,so

(x, z) ∈ R ∩ (A × A ) = R . Therefore R is transitive.

Finally, suppose that (x, y) ∈ R and (y, x) ∈ R . Then (x, y) ∈ R

and (y, x) ∈ R, so since R is antisymmetric, x = y. Thus R is anti-
symmetric.
(b) To see that T is reflexive, suppose x ∈ A.If x = a, then (x, x) =

(a, a) ∈{a}× A ⊆ T .If x = a, then x ∈ A , so since R is reflexive,
(x, x) ∈ R ⊆ T ⊆ T .

For transitivity, suppose that (x, y) ∈ T and (y, z) ∈ T .If x = a
then (x, z) = (a, z) ∈{a}× A ⊆ T . Now suppose x = a. Then
(x, y) /∈{a}× A, so since (x, y) ∈ T = T ∪ ({a}× A) we must have

(x, y) ∈ T . But T ⊆ A × A ,so y ∈ A and therefore y = a. Similar

reasoning now shows that (y, z) ∈ T . Since T is transitive, it follows

that (x, z) ∈ T ⊆ T .
To show that T is antisymmetric, suppose (x, y) ∈ T and (y, x) ∈ T .

If x = a then (y, x) /∈ T ,so(y, x) ∈{a}× A and therefore y = a =
x. Similarly, if y = a then x = y. Now suppose x = a and y = a. Then

as in the proof of transitivity it follows that (x, y) ∈ T and (y, x) ∈ T ,

so by antisymmetry of T , x = y.