How to prove the division algorithm algorithmically?

Question

I did not get any answers for the following questions:

1. Which space do natural square-symmetric matrices belong to?
2. Sum preserving transformations of square symmetric matrices with natural elements

Therefore, I decided to follow a more fundamental approach to learning (abstract) algebra on my own. I am currently in the early stages and this question may seem too simple.

Theorem

Prove that $\forall a,b \in \mathbb{Z}\ \exists! q,r \in \mathbb{Z}$ such that $a = bq +r$ where $0\leq r < a$

A Beginning

My approach to this problem was a little more algorithmic than the one offered in the textbook:

Consider a set $R = \{a-bq\}\ \forall q\in\mathbb{Z}$.

For simplicity, I started with $a\geq 0$, $b>0$, and $a\geq b$.

Considering a sequence $Q = \{q | q = 0,1,2,3,\cdots\}$.

This generates a sequence $R'=\{r'\}$ such that, algorithmically, and speaking intuitively, $r'\in \mathbb{N}$ minimizes at the right value of $q$.

I am having difficulty proving this rigorously. Some help here.

P.S. I can totally understand the approach given in the textbook following the well-ordering principle, but I want to prove it using my approach and extend it to negative integers as well. 

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Edit

I am going to outline the proof here and will be happy to shift this to an answer if the question is reopened. Let me first state the well-ordering principle.

The Well-ordering Principle: Every set $S=\{s|s\in\mathbb{Z}^+\}$ contains a smallest member.

Proof Using Set Theory

Consider a set $S = \{ a-bk|k\in\mathbb{Z}, a-bk \geq 0 \}$

Case 1: $0 \in S$

$b|a$ and $r=0$. This is the trivial case.

Case 2: $0 \notin S$

$S$ is non-empty because if $a>0$ then $a-b\cdot 0 \in S$ and if $a<0$ then $a-b(2a)\in S$. Now, according to the well-ordering principle, $S$ contains a smallest member $k$ which we are going to call $q$ and $r = a-bq$. $r>0$ by definition. To prove $r<b$, we use proof by contradiction:

If $r \geq b$, then $a-b(q+1)= a-bq-b = r-b \geq 0$, so that $a-b(q+1)\in S$. But $a-b(q+1)<r$, and $r$ is the smallest member of $S$. Therefore, $r<b$

To prove the uniqueness of $q$ and $r$, assume $q'$ and $r'$ satisfy the given conditions.

$a=bq+r, 0\leq r < b$ and $a=bq'+r', 0\leq r' < b$. On substracting the two equations, we obtain $b(q-q')=r'-r$ and $b|(r'-r)$. However, $0 \leq r'-r < r' < b$ which implies $r'-r=0$ and $q'=q$, since positive $b$ is dividing a smaller positive number, which is only possible if the smaller positive number is $0$.

This proof is perfectly logical and, if you ask me, quite elegant as well. I am looking for a more algorithmic proof to this problem.