In general, in elementary numbers theory when we prove properties we begin with natural numbers then I was wondering how you can extend the proof to $\mathbb{Z}$ clearly and properly.
For instance, we take an assertion and we have to prove it for integers by using the Euclidean division. We know that for natural numbers we have $a=bq+r$ with $0\le r < b$ and $(q,r)$ is the unique couple of natural numbers which works. But in $\mathbb{Z}$ how do we proceed ? Do we have to consider four cases or can we just write $a=bq+r$ with $0 \le \vert r\vert <\vert b \vert$ (I suppose $r \in \mathbb{Z}$) ?
Thnaks in advance !
In the example you give, the way to extend it to $\Bbb Z$ is to allow $a,q$ to be positive, negative, or zero, but to keep the restrictions on $b,r$.
One of the reasons for this is that:
$b\Bbb Z = \{k \in \Bbb Z: \exists q \in \Bbb Z \text{ with } k = qb\}$
is the same as:
$(-b)\Bbb Z = \{k \in \Bbb Z: \exists q' \in \Bbb Z \text{ with } k = q'(-b)\}$.
That is, the positive multiples of $b$ are the negative multiples of $-b$, and the negative multiples of $b$ are the positive multiples of $-b$.
So when considering "multiples of $b$", we get the same set whether we use $b$ or $-b$, so we may as well use the positive $b$ to avoid as much "sign confusion" as possible.
It is a convention to measure remainders "counting up"; while this is by no means the ONLY way, it has the advantage of letting us determine the remainder of any $a$ (positive, negative, or zero) when divided by a positive $b$ unambiguously.
If $a \leq -b$, we will, of course, have a negative $q$.