Page 375 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 361
(d) We let x and y be arbitrary and then prove the equation by induction
on n.
Base case: n = 0. Then both sides of the equation are equal to 1.
Induction step: We will make use of parts (a) and (b). Suppose that
n
n n n−k k
(x + y) = x y .
k
k=0
Then
(x + y) n+1 = (x + y)(x + y) n
n
n n−k k
= (x + y) x y (by inductive hypothesis)
k
k=0

n n n n−1 n n−2 2
= (x + y) x + x y + x y +· · ·
0 1 2

n n
+ y
n
n n+1 n n n n n n−1 2

= x + x y + x y + x y +· · ·
0 0 1 1
n n n n+1

+ xy + y
n n

n n n n n n−1 2
n+1
= x + + x y + + x y
0 1 1 2
n n n n+1

+··· + + xy + y
n − 1 n

n + 1 n+1 n + 1 n n + 1 n−1 2
= x + x y + x y
0 1 2
n + 1 n n + 1 n+1

+··· + xy + y
n n + 1
n+1
n + 1 n+1−k k
= x y .
k
k=0
20. Hint: Surprisingly, it is easier to prove that for all n ≥ 1, 0 < a n < 1/2.
Section 6.4
1. (a) (→) Suppose that ∀nQ(n). Let n be arbitrary. Then Q(n + 1) is true,
which means ∀k < n + 1(P(k)). In particular, since n < n + 1, P(n)
is true. Since n was arbitrary, this shows that ∀nP(n).
(←) Suppose that ∀nP(n). Then for any n, it is clearly true that
∀k < nP(k), which means that Q(n) is true.