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Math Help - Divisibility

  1. #1
    Senior Member DivideBy0's Avatar
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    Divisibility

    Prove that if q does not divide p, then q^n does not divide p^n.

    This isn't actually homework or anything, just a hypothesis I observed.
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  2. #2
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    Assuming that p \& q are positive integers, any prime factor of q^n must also be a prime factor of q. Can you use that to finish your question?
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  3. #3
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    Quote Originally Posted by DivideBy0 View Post
    Prove that if q does not divide p, then q^n does not divide p^n.

    This isn't actually homework or anything, just a hypothesis I observed.
    Here is a little bad proof, but it makes sense.

    \frac{p^n}{q^n} = \left(\frac{p}{q}\right)^n

    Now, \frac{p}{q} is not a whole number. So raised to any number it is not a whole number.
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  4. #4
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is a little bad proof, but it makes sense.

    \frac{p^n}{q^n} = \left(\frac{p}{q}\right)^n

    Now, \frac{p}{q} is not a whole number. So raised to any number it is not a whole number.
    Hahaha, awesome!
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    Hahaha, awesome!
    Ahhhh... But what TPH didn't tell you is that you have to be careful of the following trap:
    \sqrt{2} is not a whole number.

    But (\sqrt{2})^2 = 2 is.

    What you would need to show is something like that \sqrt{2} cannot be expressed as a rational number \frac{p}{q}.

    -Dan
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