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Math Help - Proofs.

  1. #1
    Junior Member pearlyc's Avatar
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    Proofs.

    Prove that if n ∈ Z and log2 n is rational, then log2 n is an integer.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Maybe

    Quote Originally Posted by pearlyc View Post
    Prove that if n ∈ Z and log2 n is rational, then log2 n is an integer.
    I am misunderstanding or you are trying to prove the opposite...but ln(2) is transcendental therefore irrational...
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  3. #3
    Junior Member pearlyc's Avatar
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    Honestly, I really have no idea what the question is asking for! I do have a clue that it's related to proof by contradiction though. Anyone can help?
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  4. #4
    o_O
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    How can you prove something that's not true? Are you sure what you typed the question correctly?

    Prove that if n \in \mathbb{Z} and \log_{2} n is rational, then \log_{2} n is an integer.
    n = 5. \log_{2} 5 = 2.3219 .... Not an integer.
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  5. #5
    Junior Member pearlyc's Avatar
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    Hi! Yes, I have double checked the question and it is exactly what I typed. It is an assignment question. Gotta know how to do this!
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  6. #6
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    Then I don't know what we can do. Can't prove something that isn't true as I've shown for n = 5 .
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  7. #7
    Junior Member pearlyc's Avatar
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    Apparently, you are supposed to approach the question with 'Proof by Contradictions' but I'm not very sure of myself on how to apply proof by contradiction as I don't quite understand what is it about either.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    hmm

    Quote Originally Posted by pearlyc View Post
    Prove that if n ∈ Z and log2 n is rational, then log2 n is an integer.
    I would start with n=2^{\frac{p}{q}} then work from there
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  9. #9
    Junior Member pearlyc's Avatar
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    Yes, that's exactly how I started it as well.

    And I am stucked at,

    n^q = 2^p.

    I don't know where to go from here.
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  10. #10
    Moo
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    Hello,

    Yep, n^q=2^p

    n=a/b, as n is in Z.

    -> a^q=(2b)^p

    Now, we have to show that there exists k in Z such as p=kq

    Perhaps it can help..
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  11. #11
    Junior Member pearlyc's Avatar
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    Thanks, but I still don't know how to proceed from there. Sighhh. This is so difficult
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