Proving the Division Algorithm using induction Let $n \in \mathbb{N}$. For every $m \in \mathbb{Z}$, there exist unique $q, r \in \mathbb{Z}$ such that $ m = qn+r$ and $0 \le r \le n-1$. We call $q$ the quotient and $r$ the remainder when dividing into $m$.
I'm having trouble proving this with induction. I believe the idea is to first prove for all $m \in \mathbb{Z} _{\ge 0}$ by induction, then prove for all $m \in \mathbb{Z}$ but $m \notin \mathbb{Z} _{\ge 0}$ using the induction proof on $-m$, since $-m \in \mathbb{N}$.
I'd appreciate any help, thank you.

Theorem : If $a,b∈ ℤ$ such that $ b>0$ then
$\exists!$ $q,r\in$ Z such that $a=bq+r$ , $0≤r<b$
Proof : Consider, $S=\big\{ a-nb≥0 | n\in Z \big\}$
First thing to prove that $S≠∅$
It is clear that $a-(-|a|)b ≥ 0 $
So $a-(-|a|)b\in S$
Hence $S≠∅$
By WOP , there exists $r=min(S)$
So, $r = a-qb$ for $q\in\mathbb{Z}$
Hence, $a = bq+r$
So Here Existence part completes here!!
Now we have to prove that $r<b$
Let's assume to the contrary that $r≥b$
So, $r-b≥0$
since $r=a-qb$
So $a-qb-b≥0$
$a-(q+1)b≥0$
So, $r-b\in$ S
Since $r-b<r$ ( because $b>0$ )
This contradicts the minimality of $r$
Hence $r < b$ 
This completes our proof !!!
I am sorry that I couldn't provide the uniqueness part of division algorithm
Hope it helps 
