# Divisibility

• Jul 29th 2007, 02:59 AM
DivideBy0
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.:D
• Jul 29th 2007, 04:59 AM
Plato
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 AM
ThePerfectHacker
Quote:

Originally Posted by DivideBy0
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

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.
• Jul 29th 2007, 07:40 AM
DivideBy0
Quote:

Originally Posted by ThePerfectHacker
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!
• Jul 29th 2007, 07:58 AM
topsquark
Quote:

Originally Posted by DivideBy0
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