Mathematical Induction Proofs, Assignments of Mathematical Methods for Numerical Analysis and Optimization

Download Mathematical Induction Proofs and more Assignments Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Assignment #1 — Solutions Math 315. III.4 For each n ∈ IN consider the statement Pn : 7 + 11 + 15 + · · · + (3 + 4n) = 2n2 + 5n. We claim that Pn is true for all n ∈ IN. Use induction. (B) 7 = 2 · 12 + 5, so P1 is true. (IS) Let n ∈ IN and assume that Pn is true. Then 7 + 11 + 15 + · · ·+ (3 + 4n) + (3 + 4(n + 1)) (IH)= 2n2 + 5n + (3 + 4(n + 1)) = 2n2 + 9n + 7 = (2n2 + 4n + 2) + (5n + 5) = 2(n + 1)2 + 5(n + 1) So by induction Pn is true for all n ∈ IN. III.7 Given that s1 = 1 and sn+1 = sn + 3n for all n ∈ IN, we claim that Pn : sn = (3n − 1)/2 for all n ∈ IN. We use induction. (B) s1 = 1 = (31 − 1)/2, so P1 is true. (IS) Let n ∈ IN and assume that sn = (3n − 1)/2. Then, first using recursion, we have sn+1 = sn + 3n (IH) = (3n − 1)/2 + 3n = (3n − 1 + 2 · 3n)/2 = (3 · 3n − 1)/2 = (3n+1 − 1)/2 So, by induction, Pn is true for all n ∈ IN. S1. Use Mathematical Induction to prove that 5n − 2n is divisible by 3 for all n ∈ IN. Let Pn be the statement that 5n − 2n = 3m for some m ∈ IN. Using induction we prove that Pn is true for all n ∈ IN. (B) 51 − 21 = 3 = 3 · 1, so P1 is true. (IS) Let n ∈ IN and assume that 5n − 2n = 3m for some m ∈ IN. Then, we have 5n+1 − 2n+1 = 5n+1 − 5(2n) + 5(2n)− 2n+1 = 5(5n − 2n) + 2n(5− 2) (IH) = 5(3m) + 2n · 3 = 3(5m + 2n) so Pn+1 is true and the proof is complete by induction. S2. Prove that each of the following real numbers is irrational: (a) r = 3 √ 5; Assume that r = 3 √ 5 is rational, so r = m/n with m,n ∈ IN relatively prime. Then r3 = 5 or m3 = 5n3, so that m = 5k for some k ∈ IN. Thus, m3 = 53k3 = 5n3 and n3 = 52k3. But then n = 5h for some h ∈ IN which is contrary to the fact that m and n are relatively prime. Therefore, our original assumption must be wrong and so r is not rational. [Alternate Proof.] By Theorem 2.2 if r = m/n (in lowest terms) is a rational root of x3 − 5 = 0, then m is a factor of 5 and n is a factor of 1; that is, m = ±1 or ±5 and n = ±1. But, trivially, none of ±1 or ±5 is a root of x3 − 5 = 0, so no rational number is 3 √ 5.