Page 376 - HOW TO PROVE IT: A Structured Approach, Second Edition

P1: PIG/
0521861241apx01 CB996/Velleman October 20, 2005 2:56 0 521 86124 1 Char Count= 0

362 Appendix 1: Solutions to Selected Exercises
(b) Base case: n = 0. Then Q(n) is the statement ∀k < 0P(k), which is
vacuously true.
Induction step: Suppose Q(n) is true. This means that ∀k < nP(k)
is true, so by assumption, it follows that P(n) is true. Therefore ∀k <
n + 1(P(k)) is true, which means that Q(n + 1) is true.
√ √
3. (a) Suppose 6 is rational. Let S ={q ∈ Z |∃p ∈ Z (p/q = 6}.
+
+
Then S = ∅, so we can let q be the smallest element of S, and we can
√
2
2
choose a positive integer p such that p/q = 6. Therefore p = 6q ,
2
so p is even, and hence p is even. This means that p = 2 ¯ p, for some
2
2
2
2
2
integer ¯ p. Thus 4 ¯ p = 6q ,so2 ¯ p = 3q and therefore 3q is even.
2
It is easy to check that if q is odd then 3q is odd, so q must be even,
√
which means that q = 2 ¯ q for some integer ¯ q. But then 6 = ¯ p/ ¯ q and
¯ q < q, contradicting the fact that q is the smallest element of S.
√ √
(b) Suppose that 2 + 3 = p/q. Squaring both sides gives us 5 +
√ √
2
2
2
2
2
2 6 = p /q ,so 6 = (p − 5q )/(2q ), which contradicts part (a).
6. (a) We use ordinary induction on n.
Base case: n = 0. Both sides of the equation are equal to 0.
n
Induction step: Suppose that F i = F n+2 − 1. Then
i=0
n+1 n

F i = F i + F n+1 = (F n+2 − 1) + F n+1 = F n+3 − 1.
i=0 i=0
(b) We use ordinary induction on n.
Base case: n = 0. Both sides of the equation are equal to 0.
n 2
Induction step. Suppose that i=0 (F i ) = F n F n+1 . Then
n+1 n

2 2 2 2
(F i ) = (F i ) + (F n+1 ) = F n F n+1 + (F n+1 )
i=0 i=0
= F n+1 (F n + F n+1 ) = F n+1 F n+2 .
(c) We use ordinary induction on n.
Base case: n = 0. Both sides of the equation are equal to 1.
Induction step: Suppose that n F 2i+1 = F 2n+2 . Then
i=0
n+1 n

F 2i+1 = F 2i+1 + F 2n+3 = F 2n+2 + F 2n+3
i=0 i=0
= F 2n+4 = F 2(n+1)+2 .
(d) The formula is n F 2i = F 2n+1 − 1.
i=0
8. (a) (→) Suppose a 0 , a 1 , a 2 ,... is a Gibonacci sequence. Then in particular
2
a 2 = a 0 + a 1 , which means c = 1 + c. Solving this quadratic equa-
√
tion by the quadratic formula leads to the conclusion c = (1 ± 5)/2.
√ √
(←) Suppose either c = (1 + 5)/2or c = (1 − 5)/2. Then
2
n
c = 1 + c, and therefore for every n ≥ 2, a n = c = c n−2 2
c =
c n−2 (1 + c) = c n−2 + c n−1 = a n−2 + a n−1 .