I'm currently studying the division algorithm and was wondering how one could show this algorithm holds when $a \leq0$. Here is the entire question:
Show that if $a\in \mathbb{Z}$ with $a \leq 0$ and $b \in \mathbb{N}$, then $\exists!$ $q,r\in \mathbb{Z}$ s.t. $a = q \cdot b + r$, where $0 \leq r < b.$
Well, I want to show that there does exist an $r$ and a $q$, then show that they are unique. I already know the proof when $a>0$ (already discussed with my fellow peers), but I wouldn't imagine it to be much more different by having $a \leq0$. I also know that the Well-ordering Principle will have to come into play somehow, but how do I incorporate that into showing this result?
If $a=0$, then $a=0= 0\cdot b +0$ and the result is established.
Let $a < 0$. Assuming that the result holds for positive integers, there are unique integers $q$ and $r$ such that $|a| = qb + r$. Thus, $$ a =-|a| = -(qb + r) = (-q)b - r .$$ If $r=0$, then the result is established; otherwise, notice that $$ a =-(q+1)b +(b-r).$$ Since $0 < r < b$, it follows that $0< b-r < b$ and the result holds for this case as well.
The uniqueness argument is not case-dependent, so the result is established.