280, MATH Section 1 Foundations of Mathematics Spring 2009
Greatest Hits, volume 3
Ive selected one each from among the solutions to dierent problems for this weeks installment of MATH 280s Greatest Hits. This rst one is distinguished by its clarity and composition: its among the shortest of the proofs submitted, but astute choice of wording and phrasing leaves nothing out. Note that even though the second case isnt proven, that the author of this proof recognized the need for a second case is the crucial fact: as we mentioned in class the other day, the appeal to an analogous proof is perfectly ne if theres really nothing substantively dierent in the second proof. (2) Theorem: The dierence between two consecutive cubes is odd. Proof: If n Z, then we have two cases to consider: Case I: The larger consecutive cube is odd. Case II: The larger consecutive cube is even. Case I: (2n + 1)3 (2n)3 = 8n3 + 12n2 + 6n + 1 8n3 2(6n2 = + 3n) + 1 Let 6n2 + 3n = k Z, then (2n + 1)3 (2n)3 = 2k + 1, which is an odd number. Case II: Is analogous to Case I.
This weeks second proof is one of the nest responses to 5.b, the tricky committee problem in which one must be sure to prove both implications. That is, one must show both that if n is prime then (n) = n + 1, and that if (n) = n + 1, then n is prime. Lets see how our resident expert managed this: (n) = n + 1 if and only if n is prime Proof. Assume some natural number n is prime. That is to say that ns only divisors are n and 1...