A misunderstanding in the uniqueness of $q,r$ in $a=bq+r$?

Question

I've seen a proof of the uniqueness of $q,r$ in $a=bq+r$ when $a,b$ are given. The proof is as follows:

Suppose $a=bq+r=bq'+r'$ with the obvious assumptions on $q,r$, then:

$$|r'-r|<b \quad \quad b(q-q')=r'-r$$

If $q\neq q'$, then $|q-q'|\geq 1$, such that:

$$|b|\leq |b||q-q'|=|r'-r|<|b|\tag{$\star$}$$

A contradiction, and hence $q=q'$ and $r=r'$.

I've explored this proof in the following way: I assumed $|q-q'|=1$, and hence:

$$|b|\leq |b|=|b|<|b|$$

Picking any other value for $|q-q'|$ yield the same problem. Then I supposed $|q-q'|=0$ which is the consequence of the conclusion of the proof. And it gives me:

$$|b|\leq 0<|b|$$

Isn't this also a contradiction? I may have slipped some bad reasoning there, but I don't see where. I doesn't contradict the previous assumptions, but I guess somehow $(\star)$ can only be used when we suppose $|q-q'|\geq 1$.

Answer 1

You only have $|b|\leq |b||q-q'|$ because you assumed $q\neq q'\implies |q-q'|\ge 1.$ You can't change your mind and then assume $q=q'.$ If you assume $q\ne q'\wedge q= q'$ you can prove anything.

Answer 2

If $|q-q'|<1$ then $|b|>|b||q-q'|$

So you get $|b|>|b||q-q'|=0=|r-r'|<|b|$ which is not in the least bit a contradiction.

Instead if $|q-q'|= 0$ then $q=q'$ and $|r-r'|=0$ and so $r=r'$.