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Homework Help: Help with a Real Analysis Proof

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that 2^n + 3^n is a multiple of 5 for all odd n that exist in the set of natural numbers.

    2. Relevant equations



    3. The attempt at a solution

    Suppose the contrary perhaps and do a proof by contradiction? Perhaps induction?

    edit: done, thank you. please look at second proof :)
     
    Last edited: Feb 4, 2010
  2. jcsd
  3. Feb 4, 2010 #2
    Also, another proof:
    1. The problem statement, all variables and given/known data

    Prove that for each a that exists in real numbers and n that exists in natural numbers there exists a rational r such that abs(a-r) < 1/n

    2. Relevant equations



    3. The attempt at a solution

    I believe it has something to do with the theorem on density of rationals.
     
  4. Feb 4, 2010 #3
    i don't think the first question is a real analysis problem. try working in mod 5. whats 2^n mod 5 and 3^n mod 5 for odd n?
     
  5. Feb 4, 2010 #4
    yeah, i redid the first one using a proof by induction. thanks for the response though.
     
  6. Feb 4, 2010 #5

    HallsofIvy

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    Yes, saying that |a- r|= |r- a|< 1/n is the same as saying that -1/n< r- a< 1/n or that a- 1/n< r< a+ 1/n. Since there exist a rational number in any interval, there exist a rational number in the interval (a- 1/n, a+ 1/n).
     
  7. Feb 4, 2010 #6
    is that valid as a proof though, starting from the conclusion and then reaffirming it?
     
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