Thread: Divisibility

1. 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.

2. 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?

3. Originally Posted by DivideBy0
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.

4. Originally Posted by ThePerfectHacker
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!

5. 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:
$\sqrt{2}$ 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