Hi there! This is my first post!

I am confused over what is means to run through the divisors. The question is:

Show that if n>0, then as d runs through the positive divisors of n, so does n/d.

Thanks for your help!

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- Feb 3rd 2010, 03:09 PMKatyCarRunning Through Positive Divisors
Hi there! This is my first post!

I am confused over what is means to run through the divisors. The question is:

Show that if n>0, then as d runs through the positive divisors of n, so does n/d.

Thanks for your help! - Feb 3rd 2010, 05:40 PMDrexel28
- Feb 3rd 2010, 07:01 PMBruno J.
It's an expression used in many number theory texts, which essentially means : "as $\displaystyle d$ takes on all possible values of divisors of $\displaystyle n$, so does $\displaystyle n/d$".

- Feb 3rd 2010, 07:02 PMDrexel28
- Feb 3rd 2010, 10:08 PMKatyCar
The question most likely has a very simple solution :)

Unfortunately I am not familiar with number theory as I have just decided to start studying it so I do not see the solution.

If someone wouldn't mind giving a little more of an explanation it would me most helpful. Thanks! - Feb 4th 2010, 09:28 AMBruno J.
Well consider the set of all pairs $\displaystyle (d, n/d)$ where $\displaystyle d$ is a divisor of $\displaystyle n$. What you want to show is that every divisor $\displaystyle d$ occurs not only as a first element of some pair (which it does by definition) but also as the second element of some pair. Hint : show that when $\displaystyle d$ is a divisor of $\displaystyle n$, $\displaystyle d'=n/d$ is also a divisor of $\displaystyle n$. Which pair has $\displaystyle d'$ as a first element?