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An inductionish problem.

  1. May 8, 2014 #1
    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.
     
  2. jcsd
  3. May 9, 2014 #2

    pasmith

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    Doesn't there exist a [itex]k \in \mathbb{N}[/itex] such that [itex]2^k \geq m[/itex]?
     
  4. May 9, 2014 #3
    That does make sense. That had crossed my mind briefly, but now that I've spent a little more time thinking about it I feel more comfortable. I'd been confusing myself since m doesn't have an upper limit, but since N isn't bounded above even if you pick an m higher than 2k, there's still going to be a 2k+1 that is bigger. I think I should be able use the second hypothesis to work my way back down to m.

    Thank you much!
     
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