3

My Professor presented the following theorem:

Theorem If $a \in \mathbb{Z}$ and $b \in \mathbb{Z}^+$, then there exist unique $q, r \in \mathbb{Z}$ such that

$$a = bq + r$$ and $$0 \le r < b$$

And then the following proof of uniqueness:

Suppose $q$ and $r$ are not unique, and so

$$a = qb + r = q'b + r',$$ with $$0 \le r \le r' < b$$

i.e. assume, WLOG, that $r' \ge r$.

If $r' > r$, then $0 < r' - r < b$, and the expressions for $a$ imply

$$0 \le (q - q')b = r' - r < b,$$

which is impossible.

Thus $r' = r$ and $qb + r = q' b + r$ implies $q = q'$, so the two expressions for $a$ are the same.

But this proof begins by assuming that $q$ and $r$ are not unique; specifically, that $r' \ge r$. If we get a contradiction during this, then doesn't that simply prove that $r' \not\ge r$? In that case, we still haven't ruled out whether $r' < r$. So, in the end, it seems like the proof doesn't necessarily prove that $r' = r$, as it claims?

I would greatly appreciate it if people could please take the time to clarify this.

Community
  • 1
The Pointer
  • 4,686
  • 2
    Since $r$ and $r'$ are real numbers they are ordered, and one of these three cases must be true: $r'<r$, or $r'=r$ or $r'>r$. If $r'>r$ we can relabel $r$ to be $r'$ and $r'$ to be $r$ to obtain $r'<r$ again. That's what the "WLOG" means -- that we only need to consider the two cases. Can you now see how the proof by contradiction works? – postmortes Aug 21 '18 at 06:33
  • @postmortes Yes, that makes sense. Thanks for the clarification! – The Pointer Aug 21 '18 at 06:34
  • How do we derive a contradiction in the second case: $r'<r$? – Vlad Mikheenko Oct 09 '24 at 06:46

1 Answers1

0

WLOG we could choose such that $r > r'$ as otherwise we could simply just swap the variables.

By property of $\mathbb{Z}$, $(q - q') b \: < \: b$ is only possible if $q - q' = 0$ thus $q = q'$. As $(q - q')b = 0b = 0 = r - r'$, thus again $r = r'$.

Ari Royce Hidayat
  • 1,111