i need to prove that if $n$ is a non zero integer, and $m > 0$ and $m \mid n$ ($m$ divides $n$), then $m \le |n|$.
I feel like i can do it by a combination of proof by contradiction and cases (ie assuming that the conclusion was false, then taking the cases of $m > n$ and $m \le n$ or something like that).
i haven't made a lot of headway, i'm not really sure how i can start other then that, though i get the feeling that saying if $n > m$, then it couldn't be a factor of $m$ is a decent place to start.
cany anybody offer any tips or advice on what path i can take to prove this?
thanks for your help!
This is a situation where you're trying to prove (from the axioms, essentially) something that we've taken for granted in practice. In such situations, you usually have to go look at the axioms and see what tools you have to work with. Axioms comparing numbers to other numbers are rare, but axioms comparing numbers to $0$ are easier to find, for example....
In this particular problem, when $n$ is positive, you can probably see how to write $n-m = (d-1)m$ for some positive integer $d$; can you compare this quantity to $0$? And then if $n$ is negative, can you derive the desired statement from the corresponding (already proved) statement for $|n|$?