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.:D

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- Jul 29th 2007, 02:59 AMDivideBy0Divisibility
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.:D - Jul 29th 2007, 04:59 AMPlato
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?

- Jul 29th 2007, 05:38 AMThePerfectHacker
- Jul 29th 2007, 07:40 AMDivideBy0
- Jul 29th 2007, 07:58 AMtopsquark
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