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Take $60$ and $100$. Their respective divisors lists are:

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

1, 2, 4, 5, 10, 20, 25, 50, 100

The common divisors are:

1, 2, 4, 5, 10, 20

According to these answers : https://math.stackexchange.com/a/495125/1021982 & https://math.stackexchange.com/a/495127/1021982, the $gcd$ of two numbers is the greatest common factor (greatest in terms of divisibility).

My question is how can we compare the two common factors (in terms of divisibility) $4$ and $5$ when finding the $gcd(60,100)$

$4$ and $5$ are not comparable in terms of divisibility.

niobium
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    Not following. What's the question? In the usual divisibility partial order, $4,5$ are not comparable. Is that what you are asking? – lulu Oct 25 '22 at 15:22
  • Yes they are not comparable. However we have to compare them in order to find the $gcd$ – niobium Oct 25 '22 at 15:23
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    Sometimes you can't. Divisibility is a partial ordering which is not total. Still, there could be a notion of greatest and least, lower bound and upper bound, maximal element and minimal element etc. Start with Wikipedia. –  Oct 25 '22 at 15:24
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    Why? A maximal element with respect to a partial order would just be one which is greater than all the other elements in your collection. Certainly, $20$ is greater than either $4$ or $5$ with respect to that (partial) order. – lulu Oct 25 '22 at 15:25
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    You only need to compare them to $20$. $20$ is an upper bound for both of them (and actually for all divisors) - that qualifies it as the greatest common divisor. A sport analogy: people often argue who is the GOAT (say tennis), who had more Grand Slams, or who won more times on grass, or on hard surface etc. - here we have one who beats everyone else in every aspect. –  Oct 25 '22 at 15:25
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    "Yes they are not comparable. However we have to compare them in order to find the gcd" No you don't. Only need to find the largest division. As $5\mid 10$ we have found a divisor that is "bigger" than $5$ and we do not have to compare $5$ to anything else. And as $4\mid 20$ we do not have to compare it to anything else. ANd then $10\mid 20$ we can throw away the $10$ as well as the $5$ earlier. .... I suppose to be more technical..... – fleablood Oct 25 '22 at 15:53
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    Divisibility evokes a partial order and not a total order as Stinking Bishop mentioned. A total order has the condition that either $a<b;a=b$ or $b< a$ that exactly one must be true for every $a,b$. A partial order does not have that requirement. $4\not\mid 5$ and $5\not \mid 4$. But both have transitivity. If $a<b; b<c$ then $a|c$. And if $5\mid 10$ and $10\mid 20$ then $5\mid 20$. In this case you do not have to compare $4$ to $5$. You just have to compare $4$ to $20$ and $5$ to $10$ to $20$. Now you might be wondering how do we know there will always be a biggest? We dont – fleablood Oct 25 '22 at 15:59
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    If we had a list of numbers $1,2,3, 4,5, 6, 8, 10, 12$ and were asked which was the biggest there is no answer. We has $1\mid 2,2\mid 4, 4\mid 8$ and $8$ is the biggest of that chain and we have $1\mid 2, 2\mid 6, 6\mid 12$ and that's the biggest of that chain and $1\mid 5\mid 10$ and $10$ is the biggest of that but we can not compare $8$ to $10$ to $12$ so there is no biggest. So you might be asking how do we know that the list of common divisors will have a biggest. Well, that is not obvious and is a subtle result that natural numbers have unique factorizations. But it is true they will. – fleablood Oct 25 '22 at 16:05

2 Answers2

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One misconception is that the factors are written out in order of magnitude size when we are trying to compare divisibility order, not size order. It might be confusing to do what I am about to do (which is why most text don't) but let's try listing the strings in order of divisibility

The factors of 60 can be in the following "threads":

$1, 2, 4, 12, 60$
$1, 2, 6, 12, 60$
$1, 2, 10, 20, 60$
$1, 2, 10, 30, 60$
$1, 3, 6, 12, 60$
$1, 3, 15, 30, 60$
$1, 5, 10, 20, 60$
$1, 5, 10, 30, 60$
$1, 5, 15, 60$.

Woosh, I probably missed a few. (That was actually more tedious than I thought it would be).

We can't compare $2$ with $3$ or $10$ with $12$ because the do not divide. But we don't have to.

A "total" order is one where any to values can be compared and one will be bigger and the other will be smaller. This is not a total order. But it is a partial order. And it does have transitivity so that if $a\mid b$ and $b\mid c$ we have $a\mid c$.

As such and these are all divisors of $60$ we will have $60\mid 60$ and $60 = 60\times 1$ that $60$ is the "biggest" and top of all the threads.

Now common divisors..... We can compare the common divisors and make the following threads:

$1, 2 , 4, 20$
$1, 2, 10, 20$
$1, 5, 10, 20$

In both cases $20$ is the biggest of all threads. We do not need to compare the incomparable. We do not need to compare $5$ to $2$ or $4$, nor do we have to compare $4$ to $5$ or $10$.

fleablood
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First of all, the "greatest" used in "greatest common divisor" refers to the usual $\le$ relation on the integers (and real numbers), so 4 and 5 are certainly comparable under the relevant relation. However, if we wanted to consider the divisibility relation from here on out:

The definition of the greatest common divisor is a common divisor that's greater than every other common divisor. Nothing about greatest common divisors requires that all common divisors can be compared; the only necessary fact is that every divisor can be compared, under the divisibility relation, to the greatest common divisor. (That fact in turn needs be derived from the definition, which utilizes the $\le$ relation, and properties of divisibility.)

A simpler version of this phenomenon avoids common divisors entirely: the set of divisors of a positive integer $n$ is only partially ordered under the divisibility relation, but $n$ itself is greater than (hence comparable to) every other divisor of $n$.

Greg Martin
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    "irst of all, the "greatest" used in "greatest common divisor" refers to the usual ≤ relation" Not in all texts. In nearly all advanced texts for "grown up mathematicians" it refers to greatest in terms of divisibility. Divisibility is simply more relevant and important. However it is a harder concept for the novice. Unnecessarily harder. It makes answering these question harder as we don't know which def the OPs are using (the are both equiv). Still this was a question about divisibility and comparing $4$ to $5$. – fleablood Oct 25 '22 at 16:27
  • @fleablood I am using the definition in terms of divisibility (i.e. not the one of Greg Martin). The one that makes possible to define $gcd(0,0)$ as $0$. – niobium Oct 25 '22 at 16:31
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    "The one that makes possible to define gcd(0,0) as 0" Actually no it doesnt. $0 = 15678\times 0$ and therefore $15678$ is a divisor of $0$. Likewise every integer is a divisor of $0$ and so the $\gcd(0,0) = $ the greatest integer which is.... undefined..... Oops, you meant divisibility wise.... Yes, by divisibility $0$ is the largest possible number in the univers. (However it is never a factor of anything but itself. So $\gcd(n, 0) = n$ always. – fleablood Oct 25 '22 at 16:37
  • @fleablood I take note of that By the way (I am curious) what was the "subtle result that natural numbers have unique factorizations" you were talking earling? Is that the the result that states that all natural numbers have a prime number factorization? Or is it something else... – niobium Oct 25 '22 at 16:55
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    "Is that the the result that states that all natural numbers have a prime number factorization? " Yes, it is the unique prime factorization. I don't want to go into details but it's fairly straightforward if we assume UPF. (And I really don't want to get into how to prove UPF) – fleablood Oct 25 '22 at 17:18
  • @fleablood As a grown-up mathematician, I disagree with your claim: It's quite common for the definition of "greatest common divisor" to be the phrase's literal meaning in terms of $\le$, and then it is an early theorem, not definition, that all common divisors divide the greatest common divisor (and is thus "greatest" relative to divisibility as well). The advantage of this practice is that it's immediately obvious that greatest common divisors exist, which is not obvious with the definition you claim is prevalent. – Greg Martin Oct 25 '22 at 20:11