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I'm watching a number theory lecture, and the lecturer proves the proposition that for any $a \in \mathbb{Z}$ and $b > 0$, there exist unique $q,r \in \mathbb{Z}$ with $0 \leq r < b$ such that $a = bq + r$. He begins the proof by letting $q$ be the largest integer less than or equal to $\frac{a}{b}$, and proceeds to prove existence in this manner.

I understand all of the steps of the proof, but is this strategy valid? There isn't really a valid "division" on the integers, except for what this proof is aiming to establish, so this seems somewhat circular. I've always begun this proof with considering the set of all non-negative $a - bx$ and finding its minimal element, which is the remainder. I did some research for whether this alternate approach was used but could not find it.

EDIT: Actually, I think it's wrong for me to call it circular. One can prove the existence of the floor function for the integers and I believe for the rationals as a result, which is what $q$ is in this case. It wasn't previously proved, but I can understand it being taken for granted.

Bill Dubuque
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Brad G.
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    I usually did a straightforward inductive proof. I also dislike leaving the world of integers for the rationals. Your proof — dividing by repeatedly — is nice foreshadowing of the Euclidean algorithm. – Ted Shifrin Jul 18 '23 at 21:02
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    We can eliminate rationals using $,q\le a/b\iff bq \le a,,$ but we still need induction to finish. Without further context it is impossible to know what the author intended. The common inductive proofs are already given in many prior answers, so please don't duplicate them in answers below. – Bill Dubuque Jul 18 '23 at 21:10
  • I think that proving $\lfloor \frac{a}{b} \rfloor$ is well-defined requires well-ordering of $\mathbb{Z}$. You might be able to find a counterexample in $\mathbb{Z}[\sqrt{2}]$. But I agree with Ted; it's “better” to give an integers-only proof of this integers-only fact. – Matthew Leingang Jul 18 '23 at 21:15
  • I think normally you would compute gcd via the Euclidean algorithm, then reduce the rational. However, there are several choices as to which operations are taken as simplest primitives, and accordingly different approaches for computing division. – Esa Pulkkinen Jul 18 '23 at 21:30
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    "I've always begun this proof with considering the set of all non-negative $a−bx$ and finding its minimal element, which is the remainder. I did some research for whether this alternate approach was used but could not find it." I believe your approach is actually fairly standard. For example, it is the first proof (!) in the book Elementary Number Theory by Springer. – Adam Rubinson Jul 18 '23 at 21:36
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    The proof you could not find is a duplicate of Strong inductive proof of division algorithm (quotient with remainder). The question about the video cannot be answered since essential context is missing from the question (and this context might not even exist in the video). – Bill Dubuque Jul 19 '23 at 00:21
  • @OP is the video publicly available? If so, a link and a timestamp would be helpful to get the context. – Matthew Leingang Jul 19 '23 at 17:52

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