Solutions for exercises in chapter 1 E1.1 Prove by induction that for ...

(ii): Fix m and n. We prove by induction on p that for all p > n we have A(m, n) < A(m, p).
This is true for p = n + 1 by (i). Now assume it for p, where p > n. Then
A(m, n) < A(m, p) (by assumption) < A(m, p + 1) (by (i))
(iii): Induction on n. For n = 0, A(1, 0) = A(0, 1) = 2. Assume that A(1, n) = n+2. Then
A(1, n + 1) = A(0, A(1, n)) = A(1, n) + 1 = n + 2 + 1 = n + 1 + 2, finishing the induction.
(iv): Induction on n. For n = 0, A(m + 1, 0) = A(m, 1), as desired. Now assume that
A(m, n + 1) ≤ A(m + 1, n). Then A(m + 1, n) ≥ A(m, n + 1) ≥ n + 2 by the fact proved
in the notes, and so by (ii) we get
A(m + 1, n + 1) = A(m, A(m + 1, n)) ≥ A(m, n + 2),
finishing the induction.
(v): by (i) and (iv).
(vi): by induction on n. n = 0: A(2, 0) = A(1, 1) = 3, as desired. Now assume that
A(2, n) = 2n+3. Then A(2, n+1) = A(1, A(2, n)) = A(2, n)+2 = 2n+3+2 = 2(n+1)+3,
finishing the inductive proof.
E1.7 The Fibonacci sequence is defined by f1 = f2 = 1 and fn = fn−1 + fn−2 for all
n ≥ 3. Prove that for any positive integer n, f4n is divisible by 3.
Induction on n. The case n = 1:
f4 = f3 + f2 = f2 + f1 + f2 = 3,
as desired. Now suppose that n is a positive integer and f4n is divisible by 3. Then
f 4(n+1) = f4n+4 = f4n+3 + f4n+2 = f4n+2 + f4n+1 + f4n+2
= 2f4n+2 + f4n+1 = 2(f4n+1 + f4n) + f4n+1 = 3f4n+1 + 2f4n;
since f4n is divisible by 3, so is f 4(n+1), finishing the inductive proof.
E1.8 For the Fibonacci sequence, prove that f1+f2+· · ·+fn = fn+2 −1 for every positive
integer n.
Induction on n. The case n = 1: f3 − 1 = f2 + f1 − 1 = 1 = f1. Now assume that
f1 + f2 + · · · + fn = fn+2 − 1. Then
completing the inductive proof.
f1 + f2 + · · · + fn + fn+1 = fn+2 − 1 + fn+1 = fn+3 − 1,
E1.9 For the Fibonacci sequence, prove that f2 + f4 + · · · + f2n = f2n+1 − 1 for every
positive integer n.
Induction on n: For n = 1, f3 −1 = f2 +f1 − 1 = 1 = f2. Now assume that n is a positive
integer and f2 + f4 + · · · + f2n = f2n+1 − 1. Then
f2 + f4 + · · · + f2n + f 2(n+1) = f2n+1 − 1 + f2n+2 = f2n+3 − 1 = f 2(n+1)+1 − 1,
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