I'm trying to understand a proof of the uniqueness of the quotient in the division algorithm, but the proof almost seems rather trivial to me, so it must be that I'm not understanding it. I've rewritten it somewhat.
For reference, the notation $\mathbb{N}_0$ is used to represent the set $\mathbb{N}\cup \{0\}$. This may be unique to this textbook.
Theorem: If $a = bq + r$ and $a = bq' + r$, for $a, b, q, q', r \in \mathbb{N}_0$ and $0 \leq r < b$, then $q = q'$.
Proof: This seems to follow from simple algebra. Since this chapter precedes establishing the rationals as an equivalence relation on $\mathbb{Z} \times \mathbb{Z}$, I assume that I cannot use division. But I believe all of the properties of the integers apply.
Equating the two expressions for $a$ gives us \begin{equation} bq + r = bq' + r. \end{equation} The cancellation law, since $\mathbb{N}_0 \subset \mathbb{Z}$, applies, so $bq = bq'$, which implies that $bq - bq' = b(q - q') = 0$, so either $b = 0$ or $q - q' = 0$. Since $0 \leq r < b$, and thus $b > 0$, meaning that $b \neq 0$, so $q - q' = 0$, and thus $q = q'$.
How does this look? Thanks in advance.
It's almost as easy to show, under the assumption $b > 0$, that
$bq + r = bq' + r' \Longrightarrow r = r', \; q = q'; \tag 1$
for if $r \ne r'$, we may without loss of generality assume
$r' > r; \tag 2$
then
$b(q - q') = r' - r > 0; \tag 3$
also, since
$0 \le r, r' < b, \tag 4$
we have
$r' - r < b \tag 5$
as well. Now (3) implies
$b > 0, \; q - q' > 0, \tag 6$
which in turn implies, also via (3), since $r, r' < b$,
$r' - r = b(q - q') > r' \ge r' - r, \tag 7$
a contradiction. Thus
$r' = r, \tag 8$
whence
$bq = bq', \tag 9$
whence by cancellation,
$q = q'. \tag{10}$