75 5.2 Proofs by Mathematical Induction 2k k−2 ≥ = +1+ 2 k−2 1 Σ every term in the summation is ≥ 1 2k 2k i=2k−1 +1 + 1 + 2k−1 · 2 k−2 1 = +1+ 2 2 k−1 = + 1. 2 1 there are 2k−1 terms in the summation 2k algebra algebra Thus P(k) is true too. 5.32 Assuming Theorem 5.9, we can finish the proof by observing the following. Let k = ⌊log n⌋, so that 2k ≤ n < 2k+1— that is, k ≤ log n < k + 1. Therefore, because Ha > Hb if and only if a > b, we know that H2k ≤ Hn < H2k+1 . (∗) By Theorem 5.9 and (∗), then, we know that k/2 + 1 ≤ H2k ≤ Hn < H2k+1 ≤ k + 2. Thus k/2 + 1 ≤ Hn < k + 2. By definition of k, we have that k ≤ log n < k + 1, which means that log n − 1 < k (and thus (log n − 1)/2 + 1 < k/2 + 1), and that log n ≥ k (and thus log n + 2 ≥ k + 2). Putting these facts together yields (log n − 1)/2 + 1 < k/2 + 1 ≤ Hn < k + 2 ≤ log n + 2. 5.33 Let x ≥ −1 be arbitrary. Let P(n) denote the claim that (1 + x)n ≥ 1 + nx. We’ll prove that P(n) holds for every integer n ≥ 1 by induction. base case (n = 1): P(1) follows because (1 + x)n and 1 + nx are both equal to 1 + x when n = 1. inductive case (n≥ 2): We assume the inductive hypothesis P(n − 1)—namely that (1 + x)n−1 ≥ 1 + (n − 1)x—and we must prove P(n). (1 + x)n = (1 + x) · (1 + x)n−1 ≥ (1 + x) · (1 + (n − 1)x) inductive hypothesis (Note that a ≤ b ⇒ ay ≤ by only when y ≥ 0, but because x ≥ −1 we know that 1 + x ≥ 0.) = 1 + (n − 1)x + x + (n − 1)x2 = 1 + nx + (n − 1)x2 algebra (multiplying out) algebra (combining like terms) ≥ 1 + nx because both n − 1 ≥ 0 (because n ≥ 2) and x2 ≥ 0 (because all squares are nonnegative; see Exercise 4.46), so (n − 1)x2 ≥ 0. 5.34 For n = 3, we have 2n = 8 and n! = 3 · 2 · 1 = 6. For n = 4, we have 2n = 16 and n! = 4 · 3 · 2 · 1 = 24. So we’ll prove that 2n ≤ n! for all integers n ≥ 4. For the base case (n = 4), we argued above that 2n = 16 and n! = 24. And, indeed, 24 ≥ 16. For the inductive case (n ≥ 5), we assume the inductive hypothesis 2n−1 ≤ (n − 1)! and we must prove 2n ≤ n!. But 2n = 2 · 2n−1 ≤ 2 · (n − 1)! ≤ n · (n − 1)! = n! inductive hypothesis n ≥ 5, so definitely n ≥ 2 5.35 Fix a particular integer b ≥ 1. Unlike most of these exercises, probably the hardest part here is identifying the integer k after which the inequality bn ≤ n! begins to hold. (Another way of stating it: unusually, the base case is probably harder to prove than the inductive case.) It turns out that k = 2b2 will work. Let P(n) be the claim that bn ≤ n!. Here’s a proof that P(n) holds for all n ≥ 2b2.