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Thread: Divisibility

  1. #1
    Senior Member DivideBy0's Avatar
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    Divisibility

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

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

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

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

    Now, $\displaystyle \frac{p}{q}$ 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.

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

    Now, $\displaystyle \frac{p}{q}$ 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:
    $\displaystyle \sqrt{2}$ is not a whole number.

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

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

    -Dan
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