1. The problem statement, all variables and given/known data Let S is a subset of the natural numbers [denote S[itex]\subseteq[/itex]N] such that A: 2k[itex]\in[/itex]S [itex]\forall[/itex] k[itex]\in[/itex]N B: if k[itex]\in[/itex]s and k ≥ 2, then k-1[itex]\in[/itex]S Prove that S=N 2. Relevant equations 3. The attempt at a solution My first instinct was to approach it like one would prove the principle of mathematical induction, but I got to a point and it seems like it won't work that way. I'm not sure how else to approach the proof. Here's what I've got: Suppose S≠N. Then N\S [the compliment of N with respect to S, or the set of elements that are in N but not in S] is nonempty, and by the Well-Ordering Property of N, N\S has some least element m. 2[itex]\in[/itex]S by hypothesis, so m>2. This implies that m-1[itex]\in[/itex]N. Since m-1 < m, and m is the smallest element in N that is not in S, we conclude that m-1[itex]\in[/itex]S. This is where something seems wrong to me. I want to somehow get to m[itex]\in[/itex]S, so I have a proof by contradiction, but I'm not sure how I can do this with the given hypotheses.