Page 377 - 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

Appendix 1: Solutions to Selected Exercises 363
√
(b) It will be convenient to introduce the notation c 1 = (1 + 5)/2
√
n n
and c 2 = (1 − 5)/2. Then for any n ≥ 2, a n = sc + tc =
1
2
sc n−2 2 1 n−2 2 2 n−2 (1 + c 1 ) + tc n−2 (1 + c 2 ) = (sc 1 n−2 +
c + tc
c = sc
1
1
2
2
n−2 n−1 n−1
tc 2 ) + (sc 1 + tc 2 ) = a n−2 + a n−1 .
√
(c) Hint: Let s = (5a 0 + (2a 1 − a 0 ) 5)/10 and t = (5a 0 − (2a 1 − a 0 )
√
5)/10.
n
n
10. Hint: The formula is a n = 2 · 3 − 3 · 2 .
14. Let a be the larger of 5k and k(k + 1). Now suppose n > a, and by the
division algorithm choose q and r such that n = kq + r and 0 ≤ r < k.
Now note that if q ≤ 4 then n = kq + r ≤ 4k + r < 5k ≤ a, which is a
contradiction. Therefore q > 4, so q ≥ 5, and by Example 6.1.3 it follows
q 2 2
that 2 ≥ q . Similar reasoning shows that q ≥ k + 1, so q ≥ q(k + 1) =
2 k
k
n
q k
kq + q > kq + r = n. Therefore 2 ≥ 2 kq = (2 ) ≥ (q ) ≥ n .
18. We proceed by induction on j.
Base case: j = 1. Suppose that p 1 , q 1 , q 2 ,..., q k are prime num-
bers and p 1 = q 1 q 2 ··· q k . Since p 1 is prime, we must have k = 1 and
p 1 = q 1 .
Induction step: Suppose that j ≥ 1, and for all k ≥ 1 and all non-
decreasing sequences of primes p 1 , p 2 ,..., p j and q 1 , q 2 ,..., q k ,if
p 1 p 2 ··· p j = q 1 q 2 ··· q k then j = k and p i = q i for all i, 1 ≤ i ≤ j.Now
suppose p 1 , p 2 ,..., p j+1 and q 1 , q 2 ,..., q k are nondecreasing sequences
of primes and p 1 p 2 ··· p j+1 = q 1 q 2 ··· q k . Then k ≥ 2, since otherwise
we have p 1 p 2 ··· p j+1 = q 1 , contradicting the fact that q 1 is prime. Also,
p j+1 | (q 1 q 2 ··· q k ), so by exercise 17(b), p j+1 | q i for some i. But then
p j+1 = q i ≤ q k . Similar reasoning shows that q k ≤ p j+1 ,so p j+1 = q k .
Therefore p 1 p 2 ··· p j = q 1 q 2 ··· q k−1 , and by inductive hypothesis it fol-
lows that k − 1 = j and p i = q i for 1 ≤ i ≤ j.
20. Hint: The formula is a n = F n+2 /F n+1 .
Section 6.5
1. (a) To see that F = ∅, notice that B ⊆ A ⊆ A and A is closed under
f,so A ∈ F. It follows that ∩F is defined, and in fact by exercise 9
of Section 3.3, ∩F ⊆ A. According to the definition of F, for every
C ∈ F, B ⊆ C, so by exercise 10 of Section 3.3, B ⊆∩F. Thus, we
have B ⊆∩F ⊆ A. To see that ∩F is closed under f, suppose that
x ∈∩F. LetC ∈ F be arbitrary. Then x ∈ C and C is closed under f,so
f (x) ∈ C. Since C was arbitrary, this shows that ∀C ∈ F( f (x) ∈ C),
so f (x) ∈∩F. Finally, to see that F is smallest, suppose that B ⊆
C ⊆ A and C is closed under f. Then C ∈ F, and therefore, applying
exercise 9 of Section 3.3 again, ∩F ⊆ C.