Page 372 - HOW TO PROVE IT: A Structured Approach, Second Edition

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358 Appendix 1: Solutions to Selected Exercises
n
n
2 , but by part (c), it must fail for some list of length 2 . This is a
contradiction, so the inequality must always hold.
10. We proceed by induction on n.
Base case: n = 0. If A has 0 elements, then A = ∅,so P (A) ={∅},
0
which has 1 = 2 elements.
Induction step: Suppose that for every set A with n elements, P (A)
n
has 2 elements. Now suppose that A has n + 1 elements. Let a be any

element of A, and let A = A \{a}. Then A has n elements, so P (A )

n
has 2 elements. There are two kinds of subsets of A: those that contain a
as an element, and those that don’t. The subsets that don’t contain a are
n
just the subsets of A , and by inductive hypothesis there are 2 of these.

Those that do contain a are the sets of the form X ∪{a}, where X ∈ P (A ),
n
and there are also 2 of these, since by inductive hypothesis there are 2 n
possible choices for X. Thus the total number of elements of P (A)is
n
n
2 + 2 = 2 n+1 .
13. Base case: n = 1. One chord cuts the circle into two regions, and
2
(n + n + 2)/2 = 2.
Induction step: Suppose that when n chords are drawn, the circle is cut
2
into (n + n + 2)/2 regions. When another chord is drawn, it will intersect
each of the first n chords exactly once. Therefore it will pass through n + 1
regions, cutting each of those regions in two. (Each time it crosses one
of the first n chords, it passes from one region to another.) Therefore the
number of regions after the next chord is drawn is
2
2
2
n + n + 2 n + 3n + 4 (n + 1) + (n + 1) + 2
+ (n + 1) = = ,
2 2 2
as required.
Section 6.3
1. Hint: The formula is
n
1 n

= .
i(i + 1) n + 1
i=1
6. Base case: n = 1. Then
n
1 1
= 1 ≤ 1 = 2 − .
i 2 n
i=1
Induction step: Suppose that
n
1 1
≤ 2 − .
i 2 n
i=1