I'm trying to evaluate whether or not $\gcd(p,q) = \gcd(-p,q)$ for non-zero integers $p$ and $q$.
I was wondering if it's possible to find $\gcd(-p,q)$.
If so, this statement should be true, correct?
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5If we define $\gcd(a,b)$, where $a$ and/or $b$ may be negative, in the usual way (greatest common divisor), then it is not hard to show directly from the definition that $\gcd(a,b)=\gcd(|a|,|b|)$. – André Nicolas Sep 11 '14 at 02:21
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2The prescription Andre gives is also consistent with Bezout's theorem, since if $ap+bq=gcd(p,q)$ then $(-a)(-p)+bq=gcd(p,q)$ as well. – Semiclassical Sep 11 '14 at 03:25
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1Check my animated answer. – Felix Marin Sep 11 '14 at 03:49
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1Oh! yes. The word "greatest" . – PleaseHelp Mar 01 '15 at 09:48
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1Easy answer. Sign doesn't matter $\gcd(\pm a, \pm b) = \gcd(a,b)$. If you feel weird about that just keep in mind if $n|m$ so $m = kn$ then $-n|m$ ans $m = (-k)(-n)$ and $n|-m$ as $-m = (-k)n$. – fleablood Aug 23 '16 at 06:36
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See my answer below, if there is no notion of "greatest" in your ring, then do as @fleablood says. Say that your ring "has $\gcd$'s" which means each $\gcd(a,b)$ is only unique up to mulplication by a unit. – Daniel Donnelly Apr 01 '19 at 20:52
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Does this answer your question? How to prove $\gcd(m, n) = \gcd(-m, n)?$ – Bill Dubuque Aug 20 '23 at 06:49
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Somebody has a good proof on proofwiki which makes sense to me. It shows that $$ \gcd\{a,b\}=\gcd\{|a|,b\}=\gcd\{a,|b|\}=\gcd\{|a|,|b|\} . $$
Caleb Stanford
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milanchandna
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Define the $\gcd$ like a categorial universal property. This avoids having to make use of an ordering $\leq$ on the ring in which you desire a $\gcd$.
Let $R$ be a commutative ring with $1$.
Define $\gcd(a,b)$ to be any element $d \in R$ such that $d \mid a, b$ and if $e \mid a,b$ then $e \mid d$.
Then when $R = \Bbb{Z}$, $\gcd(-n, m),$ for $n,m \gt 0$ always has two elements, but they are "isomorphic" meaning they're the same upto a unit factor: $d = (-1)d'$.
So simply define $\gcd$ as you do "product" in a category i.e. there can be many products of objects $A, B$ but they are all isomorphic.
Daniel Donnelly
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