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I thought that lemma (*) was $\gcd(a,b) = \gcd(b,r)$ where $a=bq + r, \, 0≤r <b$, and it worked when you divide $a$ by $b$ to get the quotient $q$ and a remainder $0≤r<b$.

But I don't think any of those conditions are even necessary. That is, for every integers $a,b,c,d$ such that $a=bc+d$ where $a$ and $b$ are not both zero (at the same time). Then $\gcd(a,b)= \gcd(b,d)$.

Proof: Let $\gcd(a,b)= g_1$ and $\gcd(b,d)=g_2$ so $g_1 \mid a,b \Rightarrow g_1 \mid a-bc=d$ Hence $g_1$ is a common divisor of $b$ and $d$ so $g_1≤g_2$. Similarly $g_2 \mid b,d \Rightarrow g_2 \mid bc+d = a$ so $g_2 ≤ g_1$. And hence proved.

Is this prove correct? If so, then why is it taught like (*). Is there a special reason for it? What am I missing?

amWhy
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William
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1 Answers1

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Your argument is correct. But without the inequality on the remainder it's not useful for proving that Euclid's algorithm helps you calculate the greatest common divisor.

See, e.g.:

Why does the Euclidean algorithm for finding GCD work?,

which may help you with your old question at

Why do we eventually end up with $0$ in Euclidean Algorithm?,

amWhy
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Ethan Bolker
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  • Hmm, I'd think teaching the general case would allow for more flexibility like to get $\gcd(a,b)$, simply writing $a=bc +d$ for some integers $c,d$ would do just fine in "reducing" the numbers $a$ and $b$ to $b$ and $d$. Don't you think? The restrictions are unnecessary and distracting. – William Dec 11 '21 at 12:34
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    Unless you specify that the $d$ is smaller this isn't "reducing". That said, the more general assertion is sometimes useful. – Ethan Bolker Dec 11 '21 at 12:36
  • Thank you for the links, I'm actually aware of Euclidean "Algorithm" and how it works. I'm only revisiting some concepts of number theory with my added experience after all those years since I asked the question (3 years, I guess) and I'm seeing that I took so many things for granted. – William Dec 11 '21 at 12:38
  • "It's not useful for proving that Euclidean Algorithm..." Wait why not? Doesn't $a =bc +d$ cover the case $a=bq +r $ where $0≤r <b$? – William Dec 11 '21 at 12:43
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    Of course it covers that case. The extra generality isn't useful for the most common application. Sometimes it's useful for other things. – Ethan Bolker Dec 11 '21 at 14:39
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    @amWhy I don't think this is a duplicate. The OP is asking whether a small generalization in one step in the Euclidean algorithm is true, and why that step isn't proved in the general form. That's what I tried to address. Perhaps the question is uninteresting (to others) or unimportant but it's not a duplicate. – Ethan Bolker Dec 14 '21 at 19:56
  • I wasn't the user first indicating it was a duplicate. And I question your answering of this when it should have been a comment: a comment plus link only. I have retracted my close vote that the question is a duplicate, given your argument. – amWhy Dec 14 '21 at 20:09
  • You convinced me. And forgive my trivial edits to your post, because you do point out the importance of the range of r; I edited solely to undownvote. Thanks for the comment to me! – amWhy Dec 14 '21 at 20:16
  • @amWhy You're welcome. Thanks for all the work you do on this site. – Ethan Bolker Dec 14 '21 at 20:30
  • And the same thanks to you! :-) – amWhy Dec 14 '21 at 20:32