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Let $a$ and $b$ be two coprime integers and $q \in \mathbb{Q}^*$. If $qa,\ qb \in \mathbb{Z}$ then $q \in \mathbb{Z}.$

From which historical mathematician does this lemma come? Any references?

Maman
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    Euclid?${{{}}}$ – Andrés E. Caicedo Apr 24 '20 at 00:28
  • Denominators for $q$ are closed under subtraction so also under gcd, thus if $,a,b,$ are deominators for $q$ so too is $,\gcd(a,b) = 1,,$ i,e, $,q\in\Bbb Z.,$ Proofs like this became obvious once ideals were known. See here for further discussion of such denominator (and order) ideals. It is probably difficult to date the first time this precise statement is known (before ideal theory). – Bill Dubuque Apr 24 '20 at 00:43
  • Not all lemma's are historically traceible. I've never heard of this lemma but it fairly clear and easy to prove. I doubt there is any history to it. Euclid may or may not have proven it. – fleablood Apr 24 '20 at 00:43
  • @Maman FYI, in case you didn't know about it, you may wish to consider in the future if these types of questions may be better suited for the History of Science and Mathematics site. – John Omielan Apr 24 '20 at 00:43
  • fyi: it's easy to show that this is equivalent to Euclid's Lemma and closely related results. @fleablood Surprised to hear that you've never seen it since I explicitly highlight it frequently here when discussing denominator (and order) ideals (e.g. the link in my prior comment) Maybe I should use brighter colors! – Bill Dubuque Apr 24 '20 at 00:58
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    Well... I never heard of this lemma as stated as a specific result worth remembering and referring. If I ever needed the results I'd figure they'd be easy to derive or even to just dismiss as "a clear immediate corollary to Euclid's Lemma". There's lots of lemmas I never heard of. I've never heard of the Lemma that all positive integers may be written as $3^k*M$ where $M$ is an integer not divisible by $3$ and I dare say you've never heard it either (odd that I can quote it despite never having heard it....) – fleablood Apr 24 '20 at 02:53
  • @fleablood Not only have I heard it, but I've also proved it here at least a handful of times (e.g. the analog for the prime $2$ when discussing proofs of irrationality of $\sqrt 2$). – Bill Dubuque Apr 24 '20 at 16:44
  • @Gone Thanks for your suggestion ! – Maman Apr 24 '20 at 17:36
  • Special case $(a,b)= 1$ of the Lemma in the linked dupe. – Bill Dubuque May 19 '22 at 18:24

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Setting $q=r/s$, this is equivalent to: if $(a,b)=1$ and $s\mid ra$ and $s\mid rb$, then $s\mid r$.

I doubt there is a specific significant first appearance of this lemma, but indeed it could have been proved by Euclid: write $1=ax+by$ for integers $x,y$, and then $$ r=1r=(ax+by)r = (ra)x+(rb)y $$ is a multiple of $s$.

While this proof uses the "Bezout identity" $(a,b)=ax+by$, which was apparently due to Bachet in the early 17th century, the Euclidean algorithm leads naturally to that statement, and thus one could algorithmically compute $r/s$ and know from Euclidean principles that the answer will be an integer.

Greg Martin
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  • Almost everything in elementary divisibility theory is equivalent to results on gcds and lcms. But that doesn't mean they were all discovered at the same time. Many things that are easily seen to be equivalent nowadays were very difficult to see in older times due to lack of modern theory. – Bill Dubuque Apr 24 '20 at 00:46