Prove Corollary (1.8).
Corollary (1.8). Given any integers $a$ and $b$ with $b ≠ 0$, there exist unique integers $q$ and $r$ such that
$$a = bq + r \quad \text{where} \quad 0 \le r < \lvert b \rvert $$
The Division Algorithm (1.7). Given any integers $a$ and $b$ where $b ≥ 1$, then there exist unique integers $q$ called the quotient and $r$ called the remainder such that
$$a = bq + r \quad \text{where} \quad 0 ≤ r < b$$
To prove Corollary 1.8, we will show it follows from the Division Algorithm 1.7.
Given $b \ne 0$, there are two cases to consider, $b>0$ and $b<0$.
Case $b>0$
The Division Algorithm requires $b \ge 1$, which is equivalent to $b>0$.
The Division Algorithm then asserts there exist unique integers $q$ and $r$ such that
$$a = bq + r \quad \text{where} \quad 0 ≤ r < b$$
This is equivalent to the Corollary because $b = \lvert b \rvert$.
So for the case $b>0$, the Corollary is equivalent to the Division Algorithm.
Case $b<0$
We'll apply the Division Algorithm to $a$ and $\lvert b \rvert$. It then says that given any integers $a$ and $ \lvert b \rvert$ where $\lvert b \rvert ≥ 1$, then there exist unique integers $q$ and $r$ such that
$$a = \lvert b \rvert q + r \quad \text{where} \quad 0 ≤ r < \lvert b \rvert$$
Since $b<0$,
$$ \lvert b \rvert q = (-b)q = b(-q)$$
So we have
$$a =b (-q) + r \quad \text{where} \quad 0 ≤ r < \lvert b \rvert$$
This is Corollary 1.8 but with the unique integer $(-q)$ as quotient.
We have shown Corollary 1.8 follows from Division Algorithm 1.7 by showing it for both cases $b>0$ and $b<0$.