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The greatest common divisor of two numbers (not both equal to $0$) is defined in $\mathbb N^*$ as the greatest element of the set of common divisors of these two numbers, with respect to the partial order $\leq$.

However, in my book is shown the fact that this there is another equivalent way of defining the gcd(a,b), namely by choosing the greatest element of the set of common divisors, with respect to the partial order "divides": $\mid$.

If I am not mistaken, here $\mathbb N^*$ can be equipped either with the partial order $\leq$ or with the partial order $\mid$.

My question is: how is it that we switch from one partial order to another in the definition of the $gcd$?

I mean, more generally, when two partial orders are available, can we always use one instead of the other when defining objects such as greatest elements?

Thank you very much in advance

PS: I am not looking for a proof of the equivalent way of defining the $gcd$, I am wondering whether this can be generalized to any set that has two partial orders available.

niobium
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    Possibly helpful: https://math.stackexchange.com/q/4561330/42969 – Martin R Sep 09 '23 at 10:05
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    Not sure what you are asking. It's a theorem that the two definitions are equivalent...it's not obvious. It's obviously not true that, in any set with multiple partial orders, a "greatest element" must exist simultaneously for all of them. – lulu Sep 09 '23 at 10:49
  • @lulu Yes you are right, $(\mathbb N^* , \le)$ has the property that any nonempty subpart of it bounded from above, has a greatest element (cf. axiom of completeness). $(\mathbb N^* , \mid )$ doesn't. – niobium Sep 09 '23 at 11:27
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    Sure. consider any finite collection of natural numbers. Of course such a collection has a unique maximal element under $≤$ but most of them won't have a unique maximal element under divisibility. There is no reason to imagine that your finite collection should have an element divisible by all the others. This property does hold for the set of divisors of a fixed natural number, but that's a special property. – lulu Sep 09 '23 at 11:40

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For $a, b \in \mathbb{N}^*$, if $a \mid b$ then $a \le b$. So if a set $S \subset \mathbb{N}^*$ has a maximum $M$ under $\mid$ then $M$ is also the maximum under $\le$. Now take $S$ to be the set of common divisors of some $a$ and $b$. Since $\le$ is total, $S$ of course has a maximum under $\le$ but because it is of this special form it also has a maximum under $\mid$ (ie $(\mathbb{N}^*, \mid)$ has meets). By the above reasoning these maxima must coincide.

A similar fact applies more generally to any map of partial orders: if $f : (P, \preceq) \to (Q, \le)$ satisfies $f(a) \le f(b)$ for $a \preceq b$ (ie $f$ is an increasing function) and $S \subset P$ has a maximum $M$ then $f(M)$ is the maximum of $f(S)$. In the divisibility case you are looking at the map $\mathrm{id} : (\mathbb{N}^*, \mid) \to (\mathbb{N}^*, \le)$.

ronno
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