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Showing a set if denumerable

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data
    vws7qu.png
    I got to proof that the statement is denumberable

    2. Relevant equations
    vws7qu.png

    3. The attempt at a solution
    My attempt was that 2^k != 3^k (is not) for all k in N (natural numbers)
     
  2. jcsd
  3. Dec 3, 2009 #2
    Re: Denumberable

    The set is obviously infinite. So now all you have to do is prove that a subset of a countable set is also countable.

    I think proving this in general will be easier than attacking the specific question you posted.
     
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