2

Every proof of this fact that I've seen relies on guessing a "formula" for the GCD first, such as "the smallest positive integer of the form $ax+by$" or $\frac{ab}{\text{lcm}(a,b)}$. Then one shows that the guess was indeed correct and proves the result. I don't find these proofs very intuitive and I would like to know if there's a simpler proof that doesn't involve guessing what the GCD looks like (this includes the fundamental theorem of arithmetic, which seems like overkill).

The proof should go like this:

The statement is trivially true for $1$ and $(a,b)$ itself. Let $(a,b)=d$. Suppose $\exists c$ such that $1<c<d$, $c \mid a$ and $c \mid b$ but $c \not \mid d$. Since $c<d$, we have $1 \le (c,d) < c$. Suppose $(c,d)=1$. Then $a=dk$ and $c \mid a$ imply $c \mid k$, hence $cd \mid a$. In the same way $cd \mid b$, a contradiction.

Now suppose $1<(c,d)<c$. Then $\frac{c}{(c,d)} > 1$. I would like to show that $\frac{cd}{(c,d)} \mid a$, but here I get stuck. Can it be done with my restrictions? If not, why?

EDIT:

So my original proof only used multiplicative properties of $\Bbb Z$, but I have learned that the very existence of the GCD requires additive properties as well. However, I've found a new proof that doesn't seem to use any additive properties (not even duality with LCM). I believe it is closer to what I was looking for. The reasoning behind this proof relies on additive properties of $\Bbb Z$, but they seem to disappear in my formal proof. What's going on here? How is this proof equivalent to other proofs?

Proof. Let $c$ be a common divisor of $a$ and $b$ ($a<b$) but $c \not \mid d$.

Since $c \not \mid d$, we can't have $a=d$, so $a=kd$ for some $k>1$. Also $a=tc$, for some $t>k$. We have $kd=tc \implies c=\frac kt d$. Observe that $k \not \mid t$, otherwise $d=\frac tk c \implies c \mid d$. Let $v=(k,t)$; then $1 \le v < k$. Of course $(\frac kv, \frac tv)=1$. Now, $$b=k'd=t'c=t' \frac kt d \implies k'=t' \frac kt= t' \frac{k/v}{t/v} \implies t/v \mid t'$$ But then $b= t' \frac kt d = t' \frac {k/v}{t/v} d=\frac {t'}{t/v} \left(\frac kv d \right)$. Also $a=kd=v \left( \frac kv d \right)$. This shows that $\frac kv d > d$ is a common divisor and completes the proof. $\square$

Note that $c \mid \frac kv d$ as well.

This proof is a formalization of the following hand-waving:

suppose that $a=4d=6c$. Then the respective times $d$ and $c$ are contained in any common multiple of $d$ and $c$ must always have a ratio of $2:3$. This means that there must be a factor of $2d$ (and therefore $3c$) in any common multiple. If, for example, $b=5d$, then $b=6c+d$. But $c \mid b$ and $c \mid 6c$ imply $c \mid d$. This is impossible, because $3c=2d \implies 3=2 \frac dc$, a contradiction. This situation arises every time there are two common divisors and neither divides the other.

Bill Dubuque
  • 272,048
The Footprint
  • 603
  • 2
    How do you define the gcd? –  Jun 16 '19 at 18:46
  • The greatest common divisor of $a$ and $b$, i.e. $(a,b)$ is a common divisor and if $c$ is another common divisor, then $c \le (a,b)$. – The Footprint Jun 16 '19 at 18:49
  • See this duplicate. – Dietrich Burde Jun 16 '19 at 19:01
  • @DietrichBurde doesn't Math Gems's answer assume that GCD=$ab$/LCM ? What if I don't know that? That's why I don't like proofs that guess the form of the GCD first – The Footprint Jun 16 '19 at 19:04
  • This is a fact, where we don't have to guess the gcd. Also, you cannot avoid using $gcd(a, b)$, since you are using it (see the title, $(a, b)$) for the question. Perhaps I don't understand the question well. I find Bill's proof very good, and I would think it is even good know (and to use) gcd=ab/lcm. – Dietrich Burde Jun 16 '19 at 19:06
  • I don't think it's trivial, it certainly is debatable. Also, the definition of GCD doesn't depend on divisibility by every common divisor. It is just the greatest common divisor; we don't know beforehand that every other common divisor divides it. – The Footprint Jun 16 '19 at 19:10
  • This is why Bill invested such much work to prove it, you are right. I just do not agree with you to avoid good and useful properties related to a gcd. Of course, it should not be an overkill. – Dietrich Burde Jun 16 '19 at 19:11
  • It just seems that it can be proved in the way I want it... Maybe it's impossible because my restrictions hide key properties of the integers, and therefore guessing the form of GCD first is necessary? Then the question would be: what properties are missing? – The Footprint Jun 16 '19 at 19:15

2 Answers2

2

This is not so much a direct answer to your question as an indication of how one of the standard approaches might naturally be motivated

Suppose $c|a$ and $c|b$ then $c|ha+kb$ for any integer choice of $h$ and $k$.

It is natural to constrain $c$ as much as possible, and we do this by taking the least positive value of $ha+kb$. Let's call this $f$, so we have $c|f$.

Now let's think about how this relates to $a$. We have $f\le a$ since $1a+0b=a$ and so we can divide $a$ by $f$ to get $a=mf+n$ with $0\le n\lt f\le a$. But $n=a-mf=(1-mh)a-mkb$ can't be a positive value, so must be zero. We therefore have $f|a$. Likewise $f|b$.

We now know that any common factor of $a$ and $b$ divides $f$, and also that $f$ is a common factor.

The tricky part of the proof, which you can do by uniqueness of prime factorisation as well, is to show that any common factor divides the highest common factor. Note that proving uniqueness of prime factorisation uses the additive properties of the integers and doesn't just depend on multiplicative properties.

So you will find that, at least implicit within your argument is an appeal to the additive properties of the integers.

This is quite a subtle point, and is the reason why the most efficient proofs are written the way they are. I agree they can seem a bit like magic, but they can also be motivated, as I have tried to illustrate.

Mark Bennet
  • 100,194
  • But doesn't the answer to this duplicate question only use multiplicative properties of the integers? – The Footprint Jun 16 '19 at 19:29
  • No you're right, it looks like it's indeed impossible to prove without additive properties. Math Gems' question assumes that the LCM divides every common multiple, which begs the question. – The Footprint Jun 16 '19 at 19:43
  • 2
    @TheFootprint No, it doesn't beg. Rather, it deduces it from the dual: that the least common multiple $\ell $ divides every common multiple $,m,,$ whose proof is trivial (else $,m\bmod \ell = m - q\ell$ would be a smaller common multiple). See here for more. – Bill Dubuque Jun 16 '19 at 19:51
  • @TheFootprint One of the assumptions which can get made is that $(a, b)$ exists. The duality proof is nice. If you consider positive integers only of the form $4n+1$ then unique factorisation into irreducibles fails - we have $9\times 77 =21\times 33$ and although we can see common factors they are not admissible. This kind of system can be used to show muliplication alone is insufficient. – Mark Bennet Jun 16 '19 at 19:52
  • @BillDubuque I agree that the dual proof is nice, but, for the record, it still depends on additive properties of the integers and not just multiplicative ones, as you illustrate in your comment. [The dual proof is "more intuitive" in that the additive part is somehow "closer to the surface"] – Mark Bennet Jun 16 '19 at 19:55
  • @BillDubuque thank you! It's weird to me that a property is easier to prove for the LCM than GCD, but this is probably at the heart of that duality. – The Footprint Jun 16 '19 at 19:59
  • @MarkBennet But are the integers of the form $4n+1$ a multiplicative system? I don't understand what you mean. – The Footprint Jun 16 '19 at 20:02
  • 1
    @TheFootprint Yes - $(4a+1)(4b+1)=16ab+4a+4b+1=4(4ab+a+b)+1$ and they have an identity in $1$ as well. – Mark Bennet Jun 16 '19 at 20:04
  • @MarkBennet so to sum up, the very existence of the GCD depends on the integers being a ring? – The Footprint Jun 16 '19 at 20:11
  • 2
    @TheFootprint Unlike common divisors, the common multiples form an ideal, and that's the structure needed to enable the Euclidean descent at the heart of the matter (closed under subtraction $\Rightarrow$ closed under mod $\Rightarrow$ closed under gcd). Indeed the least element $\ne 0$ of the ideal is the gcd of all elements, so the lcm is the gcd of all common multiples. Dually, the gcd is the lcm of all common divisors. – Bill Dubuque Jun 16 '19 at 20:11
  • @BillDubuque This goes way beyond the algebra I know, but it's perhaps comforting to know that it's okay if I don't fully understand these things yet, thank you. – The Footprint Jun 16 '19 at 20:14
  • 2
    @TheFootprint The proofs in $\Bbb Z$ all essentially use (Euclidean) Division with (smaller) Remainder (though some hide it by compiling into arithmetical assembly language). The same Euclidean proofs work in any domain enjoying such a division algorithm (Euclidean domains), e.g. univariate polynomials over a field. In domains lacking such division we define $(a,b)$ to be a gcd of $a,b$ if $,c\mid a,b\iff c\mid (a,b),$ and $[a,b]$ is an lcm of $,a,b,$ if $, a,b\mid m\iff [a,b]\mid m$ – Bill Dubuque Jun 16 '19 at 20:20
  • @BillDubuque Extra insight as usual. – Mark Bennet Jun 16 '19 at 20:23
  • @BillDubuque and Mark, sorry to bother you again, but would you be so kind as to take a look at my new edit? – The Footprint Jun 17 '19 at 15:40
  • 1
    @TheFootprint Duality yields ascent on common divisors: if $,c,d,$ are common divisors of $,a,b,$ then $,{\rm lcm}(c,d) = cd/(c,d),$ is also a common divisor, and a greater one if $,c,d,$ are incomparable, i.e. $,c\nmid d,\ d\nmid c.,$ Is that equivalent to what you do in your edit? (I haven't had a chance to read it) – Bill Dubuque Jun 17 '19 at 15:55
  • @TheFootprint Dually the gcd yields descent on common multiples: if $,m,n,$ are common multiples of $,a,b,$ then $,(m,n),$ is also a common multiple, and a smaller one if $,m,n,$ are incomparable. – Bill Dubuque Jun 17 '19 at 16:01
  • @BillDubuque I have used the cofactors of $c$ and $d$ in $a$ and $b$ respectively. I am sure you're right, but I am having a hard time seeing how this relates to duality. – The Footprint Jun 17 '19 at 16:04
  • 1
    @TheFootprint Said duality is in fact cofactor duality, e.g. see here or here. $\ \ $ – Bill Dubuque Jun 17 '19 at 16:09
  • @BillDubuque But that duality is between the common divisors of $a$ and $b$ and the common multiples of $a$ and $b$ which are $\le lcm(a,b)$. On the other hand, my duality is only between divisors of $a$ and $b$ separately. – The Footprint Jun 17 '19 at 16:13
  • 1
    @TheFootprint Your proof uses $\ d\mid e\mid a,b\ $ for $\ e = dk/v = a/(a/d,a/c).,$ Scaling that by $,cd,$ shows $e = cd/(c,d) = {\rm lcm}(c,d),$ so it is equivalent to the common divisor ascent in my 3rd last comment. – Bill Dubuque Jun 17 '19 at 17:14
  • 1
    @TheFootprint In fact the identity $\ {\rm lcm}(c,d) = a/(a/d,a/c)\ $ is precisely the identity in the answer I linked above in my 2nd last comment on cofactor duality. $\ \ $ – Bill Dubuque Jun 17 '19 at 17:28
  • 1
    @TheFootprint Duality rules these matters. Who cannot help but marvel at the simplicity and beauty of the prior identity (and its proof) when expressed using innate duality as $,\rm lcm' = gcd,,$ i.e.

    $$\overbrace{\frac{a}{{\rm lcm}(c,d)}}^{\Large {\rm lcm}(c,d)'} =,\overbrace{\gcd\left(\frac{a}c,\frac{a}d\right)}^{\Large \gcd,(c',,d')}\ \ \ {\rm for}\ \ \ x' := \dfrac{a}x $$

     – Bill Dubuque Jun 17 '19 at 18:11
  • @BillDubuque Indeed, it's quite enlightening! Thank you for using your time to explain this to people. One thing that still puzzles me is that, despite duality, we cannot prove that any common divisor divides GCD before proving that LCM divides any common multiple. Is this because the ideal structure of the multiples isn't preserved by duality, as you said before? – The Footprint Jun 17 '19 at 18:18
  • 1
    @TheFootprint As you noticed in your proof, we can in fact eliminate the lcm language and work directly with gcds. But this obfuscates the key innate arithmetical structure - that common multiples form an ideal, where Euclidean descent applies very simply. – Bill Dubuque Jun 17 '19 at 18:37
  • 1
    Similarly one can give "direct" proofs of FTA = existence and uniqueness of prime factorizations by eliminating higher-order concepts such as gcds and the division algorithm, by essentially compiling them into lower-level arithmetical assembly language. While that may be helpful for machines, it is opposite what is best for human comprehension. – Bill Dubuque Jun 17 '19 at 18:37
  • Comments are not for extended discussion; this conversation has been moved to chat. – robjohn Jun 17 '19 at 18:45
  • 1
    @BillDubuque: If there is content in comments that is of key importance to the answer, it would be better added to the answer. – robjohn Jun 17 '19 at 20:28
  • 1
    @robjohn I will simply collect them into an answer when I have time. As you must surely know, this site has a long history of tolerating on-topic mathematical elaborations in comments. So I was quite surprised to see everything deleted. Even more so considering that we have strongly disagreed on cloely related pedagogical matters in the past. Were I still a mod I would recuse myself from deleting your comments elaborating on some Bezout-based proof since that would be a major confict of interest imho. – Bill Dubuque Jun 17 '19 at 20:53
  • @BillDubuque For the record, I appreciated the care you took - I have found the element of interactivity here to be pedagogically significant more than a few times (understanding is important). – Mark Bennet Jun 17 '19 at 21:04
  • @Bill: If you would like your account deleted, please contact the SE team via https://math.stackexchange.com/contact, I'm sure that they would be able to assist. – Asaf Karagila Jun 17 '19 at 21:06
  • 1
    @robjohn Thanks for restoring the comments. Alas, it didn't update the above "Linked" questions list (bug?) so I will post the links again to achieve that link1, link2, link3, link4, – Bill Dubuque Jun 17 '19 at 22:24
2

It is easier to first show:

If $M$ is a common multiple of $a$ and $b$, then it is divisible by $m=\operatorname{lcm}(a,b).$

Proof: Apply the division algorithm: $M=mq+r$ with $0\leq r<m.$ But since $m,M$ are common multiples of $a,b$ then so is $r=M-mq<m.$ If $r\neq 0,$ then there would be a smaller common multiple than $m,$ which contradicts our definition $m=\operatorname{lcm}(a,b).$ So we must have $r=0,$ and hence $m\mid M.$

Now, if $d$ is a common divisor of $a$ and $b$ and $D=\gcd(a,b),$ then by the above, we must have that $D'=\operatorname{lcm}(d,D)$ is a divisor of $a$ and a divisor of $b.$ So it is a common divisor. If $d$ is not a divisor of $D$, then $\operatorname{lcm}(d,D)>D=\gcd(a,b),$ but that's not possible by the definition of $\gcd.$

Thomas Andrews
  • 177,126
  • 1
    I.e. common divisors are closed under LCM and this yields the sought ascent. This and the underlying duality is discussed at length in my comments on Mark's answer (and the posts linked there). – Bill Dubuque Jun 17 '19 at 18:16