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As you know, it's about (informally) proving that for d and n there are q and r that are unique. We start with n = qd+r.
Remainder is smaller than d. 0 ≤ r < d. Then we suppose that there are q' and r' so that qd+r = q'd+r'.

So far so good. But how does it follow here that r-r' = 0, for the reason that r-r' is a multiple of d' ? Is there not any other multiple between 0 and d that r-r' could be? How did the writer conclude this? How is it clear that there can be no other multiple of d other than 0?

(It's all natural numbers.) I also don't know why zero looks like letter 'o' in this text. I have also seen other questions dealing with this proof, but I could not find the answer to my specific problem, or rather, could not recognize it because it was not answered directly anywhere for me to easily understand, that is why I am opening a new question.

Christelle Augustin
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    There is no non-zero multiple of $d$ between $0$ and $d-1$: is you assume $r\geqslant r'$ then $r-r'$ lies between $0$ and $d-1$. – ancient mathematician Aug 22 '22 at 07:33
  • Thanks! I got really confused because I was thinking there would be multiples between d and 0. It's all clear now, what the written proof is expressing. It is uncomfortable for beginners like me. Thanks for the tolerance – Christelle Augustin Aug 22 '22 at 07:57

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So you have that $$qd + r = qd' + r' \Longleftrightarrow qd = qd' - (r - r').$$ You can assume without loss of generality that $r \geq r'$. By assumption $0\leq r < q, 0\leq r' < q$, hence $0\leq r - r' < q$.

Now, the equality $qd = qd' - (r - r')$ implies that $r - r'$ is a non-negative integer multiple of $q$, but the only integer multiple of $q$ in the interval $[0, q)$ is $0$. Thus $r - r' = 0$.

R. W. Prado
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Epsilon Away
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