# Running Through Positive Divisors

• Feb 3rd 2010, 03:09 PM
KatyCar
Running 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.

• Feb 3rd 2010, 05:40 PM
Drexel28
Quote:

Originally Posted by KatyCar
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.

What does "run through" mean?
• Feb 3rd 2010, 07:01 PM
Bruno 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 PM
Drexel28
Quote:

Originally Posted by Bruno 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\$".

Isn't the question fairly trivial then...
• Feb 3rd 2010, 10:08 PM
KatyCar
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 AM
Bruno 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?