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

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(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, 4
completing the inductive proof. E1.10 For the Fibonacci sequence, prove that f 2 1 + f2 2 + · · · + f2 n = fnfn+1 for every positive integer n. Proof. Induction on n. For n = 1, f1f2 = 1 = f2 1 . Now assume that n is a positive integer and f2 1 + f2 2 + · · · + f2 n = fnfn+1. Then f 2 1 + f 2 2 + · · · + f 2 n + f 2 n+1 = fnfn+1 + f 2 n+1 = (fn + fn+1)fn+1 = fn+2fn+1, completing the inductive proof. E1.11 For 〈1, 3, 2〉 and 〈0, 4, 5〉, describe the domain of the function 〈1, 3, 2〉 ⌢ 〈0, 4, 5〉, and give its values for each element of the domain. This is the function h with domain 6 such that h(0) = 1, h(1) = 3, h(2) = 2, h(3) = 0, h(4) = 4, and h(5) = 5. E11.12 Let ϕ = 〈2, 3, 4〉. Express 〈1〉 ⌢ ϕ ⌢ 〈5〉 as a single sequence. 〈1〉 ⌢ ϕ ⌢ 〈5〉 = 〈1, 2, 3, 4, 5〉. E11.13 Use the recursion theorem (Theorem 1.11) to show that for each m ∈ ω there is a function gm with the following properties: (i) g1 is the function mapping R into R such that g1(a) = a for all a ∈ R. (ii) For each positive integer m, gm+1 is the function mapping m+1 R into R such that for any 〈a0, . . ., am ∈ R〉, gm+1(a0, . . ., am) = gm(a0, . . ., am−1) + am. [ m X is the set of all m-tuples of members of X, for any m ∈ ω and any set X. The value gm(a0, . . ., am−1) is denoted by � i
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