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I'm working through the Euclidian Algorithm to find the greatest common divisor of two integers $n,m$. However I'm stuck at a very trivial step before the algorithm is even presented:

$n,m\in \mathbb{N}, m\leq n\Longrightarrow n = \lfloor\frac{n}{m}\rfloor m + n \text{ mod } m$

we define $q:= \lfloor\frac{n}{m}\rfloor$ and $r:=n\text{ mod } m \Longrightarrow n = qm + r$ with $1\leq q \leq n$, $0\leq r<m$

I define the divisor set $D(n):= \left\{ d\in \mathbb{N}\text{ }| \text{ } \exists k\in\mathbb{N}: \frac{n}{d} = k \right\}$

Now I can easily prove that:

(1) $d\in D(m) \cap D(r) \Longrightarrow d\in D(n)$

(2) $d \in D(n) \cap D(m) \Longrightarrow d\in D(r)$

Here is the missing step: Why do (1) and (2) imply the following identity of the greatest common denominator $\mathrm{gcd}(m,n)$?

$\mathrm{gcd}(n,m) = \mathrm{gcd}(m,r)$

There's some thought or idea that I'm missing here that I cannot quite grasp.

Thank you!

Bill Dubuque
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Michel H
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1 Answers1

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Rewrite these two implications by keeping the fact that $d \in D(m)$ around:

  1. $d\in D(m) \cap D(r) \implies d\in D(n) \cap D(m)$
  2. $d \in D(n) \cap D(m) \implies d\in D(m) \cap D(r)$.

We conclude that there's a two way implication $d\in D(m) \cap D(r) \iff d\in D(n) \cap D(m)$. As a result, $D(m) \cap D(r) = D(n) \cap D(m)$: the set of common divisors of $m$ and $r$ is the same as the set of common divisors of $n$ and $m$. In particular, the greatest common divisor is the same in both cases.

Misha Lavrov
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  • Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 10 '22 at 17:37
  • @BillDubuque I disagree that this is a dupe. It is a more specific question about a particular step in the argument. – Misha Lavrov Apr 10 '22 at 17:37
  • which is already explained in the linked dupe (and many other answers). – Bill Dubuque Apr 10 '22 at 17:39
  • Right, because clearly the way to resolve this question is to say the words "$\mathbb Z$ forms a subring of its fraction field $\mathbb Q$" three times fast. – Misha Lavrov Apr 10 '22 at 17:40
  • Yes, you can indeed make wildly misleading statements by quoting out of context. Please try to be more reasomable. – Bill Dubuque Apr 10 '22 at 17:44
  • Closing things as dupes the way you are, it doesn't seem like you think context is important in the least. Please try to be more reasonable. – Misha Lavrov Apr 10 '22 at 17:46
  • The context does not differ here. – Bill Dubuque Apr 10 '22 at 17:47
  • Engaging with discussions with you about this seems fruitless, so I will stop doing this. When I make a mistake and answer a question that's legitimately a duplicate and you spot it, I admit it. When you make a mistake, you seem reluctant to give even the least bit of ground. – Misha Lavrov Apr 10 '22 at 17:48
  • Beautiful. Thank you Misha! – Michel H Apr 10 '22 at 17:53
  • There is no mistake. Your answer exactly duplicates the proof here that I linked to in the dupe. Be aware that many users have been suspended for answering many dupes (& psqs), including high rep users. – Bill Dubuque Apr 10 '22 at 17:55
  • Yeah, and last time your argument was "I admit that the duplicate doesn't actually have an answer to this question, but it's on my to-do list". – Misha Lavrov Apr 10 '22 at 17:57
  • Posting more false and misleading "quotes" is not a logical way to refute the above duplication claim. Those are your words - not mine. – Bill Dubuque Apr 10 '22 at 18:05