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If I define $\gcd(a_1,...,a_n)$ be the greatest positive common divisor or $a_1,...,a_n$.

How can I show $\gcd(a_1,...,a_n)=\gcd(\gcd(a_1,...,a_{n-1}),a_n)$?

The equality is trivial if we have $\forall $ common divisor $c$ of $a_1,...,a_n$, $c\mid \gcd(a_1,...,a_n)$. But I think this is a consequence of the above proposition by extending the result from the case $n=2$.

It is easy to show $\gcd(\gcd(a_1,...,a_{n-1}),a_n)\le \gcd(a_1,...,a_n),$ but I got stuck showing the another side. Any suggestion will be appreciated.

Bill Dubuque
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xyz
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    GCD is not the least common divisor – Raffaele Sep 16 '20 at 12:14
  • edited thank you. @Raffaele – xyz Sep 16 '20 at 12:19
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    The way that I was taught number theory, if you have any two groups of positive integers [i.e. LHS denotes ${a_1, a_2, \cdots, a_m}$ and RHS denotes ${b_1, b_2, \cdots, b_n}$], then the way to prove that gcd(LHS) = gcd(RHS) is to prove that any divisor of LHS is also a divisor of RHS, and vice-versa. Assuming that you accept my methodology, the implementation with respect to the problem you posted should be straight forward. – user2661923 Sep 16 '20 at 12:35
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    Note that one of the intermediate results of number theory is (re my previous note) if $k = $ gcd{LHS} (for example), then not only is $k$ greater than any other common divisor of LHS, but any common divisor of LHS divides $k$. – user2661923 Sep 16 '20 at 12:39
  • how can you should $gcd(a_1,...,a_n)$ is a divisor of $gcd(a_1,...,a_{n-1})$ if without the result that $gcd(a,....,a_{n-1})$ divides any common divisor or $a_1,...,a_{n-1}$?@user2661923 – xyz Sep 16 '20 at 15:35
  • by the way, if anyone think this deserve a down-vote, please help point our why it does not meet the standard. – xyz Sep 16 '20 at 15:37
  • @xyz For more on the method hinted by user2661923 see this answer. – Bill Dubuque Sep 16 '20 at 18:13

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Hint: Prove more generally that $D(a_1,\dots,a_n)=D(a_1,\dots,a_{n-1}) \cap D(a_n)$, where $D(S)$ is the set of all positive common divisors of all elements in a set $S$. Then, by definition, $\gcd(S)=\max D(S)$. Furthermore, $D(S)=D(\gcd(S))$.

lhf
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  • To be pedantic, $D$ takes as input sets of natural numbers, not natural numbers. Writing $D(\gcd(S))$ looks a bit strange to me. On the other hand, I'm not the intended audience for this answer post. Presumably, writing $D({\gcd(S)})$ might confuse people who aren't entirely on top of this. – Arthur Sep 16 '20 at 12:26
  • @Arthur, a bit of notational abuse is healthy :-) Your comment is quite right of course. – lhf Sep 16 '20 at 13:38
  • As long as there is no reasonable ambiguity, I'm all for abuse (of notation). It is, however, also healthy to be aware of when it happens – Arthur Sep 16 '20 at 13:43
  • Why $D(S)=D(gcd(S))?$ i.e, why common divisors of $S$ is also a divisor of $gcd(S)?$. – xyz Sep 16 '20 at 15:52
  • By the above notation, I have $gcd(a_1,...,a_n)=max(D(a_1,...,a_{n-1})\cap D(a_n))$, But I got stuck here. – xyz Sep 16 '20 at 15:53
  • @xyz, because $\gcd(S)$ can be written as an integer linear combination of the elements of $S$. – lhf Sep 16 '20 at 15:56
  • how? I have only learnt bezout's identity for $n=2$. Is it a extension? – xyz Sep 16 '20 at 15:57
  • @xyz, exactly.. – lhf Sep 16 '20 at 16:36
  • @lhf but the extension require this property asked right? is there way to prove it without this property? – xyz Sep 17 '20 at 00:05
  • @xyz, you may have a point there :-) – lhf Sep 17 '20 at 00:33