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

Question

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?

Answer 1

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

$$g\,\text{ is a gcd of }\,a,b\,\text{ in }R\ \ \text{ if }\ \ \bbox[5px,border:2px solid #c40]{d\mid a,b\iff d\mid g\,}\ \text{ holds for all}\ \ d\in R\qquad\qquad\ \ \ \ \ \ \ \ $$

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 $\,g.$

gcds are unique only up to unit multiples, but we can normalize them in some domains, e.g. choosing them $\ge 0\,$ in $\,\Bbb Z,\,$ and monic for polynomials over a field - see here.

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". $\ \bf\small QED$

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 the common 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).

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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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $c^{n-1}m$.

Answer 5

(=>)

Assume c|a and c|b WTS there exists a k in Z s.t ck=gcd(a,b)

Then there exists q, p in Z s.t cq=a, cp=b

Then gcd(a,b)=gcd(cq,cp)=c*gcd(q,p)

Take k as gcd(q,p) then we’ve done.

(<=)

Assume c|gcd(a,b) WTS there exists q,p in Z s.t cq=a, cp=b

Since c|gcd(a,b) then there exists k in Z s.t ck=gcd(a,b)

By Bezout’s identity,

there exists u, v in Z s.t gcd(a,b)=au+bv gcd(a,b)=au+bv=ck k=a/c *u + b/c * v

Suppose c does not divide a, but c divides b

This leads to a contradiction that k is not in Z

The same contradiction holds when c divides a but does not divide b

So we must have both c divides a and c divides b as required.