$c\mid a,b\iff c\mid\gcd(a,b)$ [GCD Universal Property]

If $c$ is a common divisor of $a$ and $b$ then $c$ divides the greatest common divisor of $a$ and $b$. What can we use to prove this?

The general definition of a gcd $$G$$ is that it is a common divisor that is divisibly greatest, i.e. if $$d$$ is any common divisor then $$\,d\mid G,\,$$ so $$\, d\le G,\,$$ thus $$G$$ is a greatest common divisor. Combining both directions we obtain the following handy bidirectional form of the general definition of a gcd

$$g\,\text{ is a gcd of }\,a,b\,\text{ in }R\ \ \text{ if }\ \ \bbox[5px,border:1px solid #c00]{d\mid a,b\iff d\mid g}\ \text{ holds for all}\ \ d\in R\qquad\qquad\ \ \ \ \ \ \ \$$

Indeed putting $$\,d=g\,$$ in $$(\Leftarrow)$$ yields $$\,g\mid a,b,\,$$ so $$\,g\,$$ is a common divisor of $$\,a,b,\,$$ and necessarily divisibly greatest since direction $$(\Rightarrow)$$ shows every common divisor $$\,d\,$$ divides $$\,g.$$

Below is a proof of the "divisibly greatest" form of the gcd in $$\Bbb Z,\,$$ via Bezout.

Theorem $$\ \ \ \ d\mid a,b\iff d\mid (a,b)\ \ \$$ [GCD Universal Property]

$${\bf Proof}\ \ (\Rightarrow)\ \ \ d\mid a,b\,\Rightarrow\, d\mid (a,b) = i\,a+j\,b,\,$$ some $$\, i,j\in\Bbb Z,\,$$ by Bezout.

$$(\Leftarrow)\ \ \ \ d\mid (a,b)\mid a,b\,\Rightarrow\, d\mid a,b\$$ by transitivity of  "divides".

Note that the proof shows that a linear common divisor is always greatest, i.e. any common divisor $$\:d\:$$ of $$\:a,b\:$$ that is an integral linear combination of them $$\:d = i\,a + j\,b\:$$ is necessarily the greatest common divisor, since, as above, every common divisor $$\:c\:$$ divides $$\:d,\:$$ hence $$\:c\le d.\,$$ If we inline the linked Bezout proof here we obtain a direct proof by Euclidean descent (induction).

Clearly the above proof immediately extends to any number of gcd arguments, since the linked Bezout proof is $$n$$-ary, and the $$n$$-ary converse is clear by $$\,d\mid (a_1,\ldots,a_n)\mid a_1,\ldots,a_n,\,$$ hence

Theorem$$\:\!_{\color{#c00}n}$$ $$\ \ \ d\mid a_1,\cdots a_{\color{#c00}n}\iff d\mid (a_1,\cdots a_{\color{#c00}n})\ \ \$$ [$$\color{#c00}n$$-ary GCD Universal Property]

Remark $$\$$ Dually we have the universal property of LCM

Lemma $$\ \ \ a,b\mid m\iff [a,b]\mid m\ \ \$$ [LCM Universal Property]

Beware  In other UFD's (e.g. $$\,\Bbb Z[x]\,$$ and $$\,\Bbb Q[x,y]$$) there generally is not a linear representation (Bezout equation) for gcds but we can instead use prime factorizations to prove the above properties (then these properties boil down to the universal propery of min & max on the exponents of primes, e.g. see here).

Or we can prove the GCD Universal Property directly by induction on $$\,\color{#90f}{{\rm size} := a\!+\!b}.\,$$ It's clearly true if $$\,a\!=\!0\,$$ or $$\,b\!=\!0,\,$$ or if $$\,a\! =\! b\!:\ c\mid a,a\!\iff\! c\mid (a,a)=a.\,$$ Else $$\,a\!\neq\! b\!\neq\!0.\,$$ By symmetry, wlog $$\,a>b.\,$$ so $$\, c\mid a,b\!\iff\! \color{#0a0}{c\mid a\!-\!b,b}\!\iff\! c\mid(a\!-\!b,b)=(a,b)\,$$ since $$\,\color{#0a0}{\rm green}\,$$ instance has smaller $$\,\color{#90f}{{\rm size}} = (a\!-\!b)+b = a < \color{#90f}{a+b},\,$$ so $$\rm\color{}{induction}\,$$ applies.

• What is your definition for the greatest common divisor?
– quid
Dec 18, 2014 at 18:35
• @quid It works for both common gcd definitions, i.e. "greatest" means either wrt (absolute) value, or wrt divisibility, as in the universal definition $\ a\mid b,c\iff a\mid (b,c)\ \$ Dec 18, 2014 at 18:51
• Part of your answer is in fact wrong for the former (and for the former with parenthesis and the latter there is not really such a thing as the greatest common divisor, at least not if one want it to be an element).
– quid
Dec 18, 2014 at 18:54
• @quid I see nothing wrong. The former definition is intrinsic to Euclidean domains, and the latter to gcd domains. And the abuse of notation (uniqueness of the gcd up to unit factors) is ubiquitous. But this is very far from the question at hand. Dec 18, 2014 at 18:56
• @quid Ah, I see now. That comment was explicitly addressed to the OP, even though it was in the comments of your answer. Today has been so busy that now I don't recall why I placed the comment at that spot (maybe because I thought some readers of your answer might find the links helpful). Apologies for any confusion it may have caused. Dec 18, 2014 at 19:40

First of all, it could be a definition. Otherwise, you may use the Euclidean algorithm and some elementary properties of division, or simply the unique prime factorisation of natural numbers.

• "First, of all it could be a definition." Really? What definition might that be. Sep 26, 2012 at 10:32
• @user42362 E.g. the universal definition of gcd, viz. $\rm\:c\:|\:(a,b)\iff c\:|\:a,b\ \$ Sep 26, 2012 at 16:26

Use unique prime factorization: say $m=\displaystyle\prod_{i}{p_i}^{\alpha_i}$ and $n=\displaystyle\prod_i{p_i}^{\beta_i}$, then $\gcd(m,n)$ will contain the exponents $\min(\alpha_i,\beta_i)$.

Meanwhile, if $d$ has exponents $\delta_i$, then $d|m$ means exactly that each $\delta_i \le \alpha_i$. So, this is all a reformulation of the ($\delta_i\le\alpha_i$ and $\delta_i\le\beta_i$ then $\delta_i\le\min(\alpha_i,\beta_i)$) setting.

This can be shown based on what we are given by showing $a$ and $b$ in terms of a product of $c$. If $c$ is a common divisor of $a$ and $b$, that means a and b are each the product of at least one $c$ and another natural number: $a = c^np$ and $b=c^nq$, with $a,b,c,n,p,q \in \mathbb N$. There are then two possibilities:

1. Either $c$, or a power of $c$ $c^n$, is the GCD of $a$ and $b$. Trivially, $c$ divides any $c^n$ where $n>0$.
2. Neither $c$ nor any $c^n$ is the GCD of $a$ and $b$; in that case, $p$ and $q$ have their own GCD $m > 1$, and the GCD of $a$ and $b$ is $c^nm$; again, by inspection, we see that $c$ divides any $c^nm$ to produce $c^{n-1}m$.
• @Downvoter - anything to add or amend? Sep 26, 2012 at 17:31