Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$ for $0\ne a,b,c\in \Bbb{Z}$. I tried solving it with sets but I sense there are some details I am missing. I would truly appreciate your reference.


Hint $\qquad d\mid(a,b,c)\!\!\!\color{#c00}{\overset{\rm\ U}\iff} d\mid a,b,c\!\!\!\color{#c00}{\overset{\rm\ U}\iff} d\,\mid (a,b),c\!\!\!\color{#c00}{\overset{\rm\ U}\iff} d\mid ((a,b),c)$

This is the associative property of the GCD. In the same way, by induction, we can erase (or normalize) brackets in $n$-argument gcds, showing general associativity of the gcd, e.g. see here.

Remark $\ $ For completeness, below is a proof of the $\rm\color{#c00}{red\ arrows\ U}$

Lemma $\ \ d\mid a_1,\ldots,a_n\!\iff d\mid (a_1,\ldots,a_n)\ \ \ $ [GCD Universal Property]

${\bf Proof}\quad\ d\mid a_1,\ldots,a_n\Rightarrow\, d\mid (a_1,\ldots,a_n) = j_1 a_1\!+\ldots+j_n a_n\,$ for some $\, j_i\in\Bbb Z,\,$ by Bezout.

$\qquad\qquad\, d\mid (a_1,\ldots,a_n)\mid a_1,\ldots,a_n\,\Rightarrow\, d\mid a_1,\ldots,a_n\,$ by transitivity of $ $ "divides".

Dually we have the universal property of LCM

Lemma $\ \ a_1,\ldots,a_n\mid m\iff {\rm lcm}(a_1,\ldots,a_n)\mid m\ \ \ $ [LCM Universal Property]

These universal properties are the definitions of GCD & LCM in more general rings - where the Bezout identity need not hold true, e.g. in $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y]\,$ where the gcds $\,(x,2) = 1 = (x,y)\,$ cannot be written as linear combinations. Follow the links for further details.

  • $\begingroup$ How is it concluded (generally) that for evry $d|a,b \Rightarrow$ $d|\gcd(a,b)$? $\endgroup$ – Meitar Abarbanel Mar 14 '15 at 13:51
  • $\begingroup$ @Meitar See the added Remark. $\endgroup$ – Bill Dubuque Mar 14 '15 at 13:55
  • $\begingroup$ Oh... I wasn't taught Bezout theorem. I shall look for it. $\endgroup$ – Meitar Abarbanel Mar 14 '15 at 13:59
  • $\begingroup$ @Meitar I added a link to a simple, conceptual proof. $\endgroup$ – Bill Dubuque Mar 14 '15 at 14:01
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    $\begingroup$ @Elvin $\ a\mid b\ $ means $a$ divides $b,\,$ i.e. $\,an = b\,$ for some integer $\,n.\,$ This is widely used notation in number theory. $\endgroup$ – Bill Dubuque Sep 24 '19 at 13:48

The def of $d=\gcd(a,b)$ is $d|a$ and $d|b$ and if $f|a$ and $f|b$ then $f|d$.

Suppose $x=\gcd(a,b,c)$. Then $x|a$ and $x|b$ and $x|c$ so $x|\gcd(a,b)$ and $x|c$, so $x|\gcd(\gcd(a,b),c)$. Conversely if $x=\gcd(\gcd(a,b),c)$ then $x|\gcd(a,b)$ and $x|c$. So $x|a$ and $x|b$ and $x|c$, so $x|\gcd(a,b,c)$. Thus $\gcd(a,b,c) | \gcd(\gcd(a,b),c)$ and $\gcd(\gcd(a,b),c)| \gcd(a,b,c)$. Therefore $\gcd(\gcd(a,b),c) = \gcd(a,b,c)$

  • $\begingroup$ In the definition I was given, nothing is said about every divisor dividing the gcd. How is it shown? $\endgroup$ – Meitar Abarbanel Mar 14 '15 at 13:55
  • $\begingroup$ It doesn't need to be shown, it's part of the definition, so you can assume it once you know $d=\gcd(a,b)$ then automatically any other divisor of $a$ and $b$ must divide $d$. $\endgroup$ – Gregory Grant Mar 14 '15 at 15:30
  • $\begingroup$ Please post the definition you were given, so we can see how it differs from this one I gave. $\endgroup$ – Gregory Grant Mar 14 '15 at 15:30
  • $\begingroup$ "The greatest common divisor (gcd) of two or more integers, when at least one of them is not zero, is the largest positive integer that divides the numbers without a remainder." $\endgroup$ – Meitar Abarbanel Mar 14 '15 at 16:04

A different definition of gcd affords another proof.

For $p_i\in \mathbb P \land p_i\mid abc$, then $a=\prod p_i^{\alpha_i},\ b=\prod p_i^{\beta_i},\ c=\prod p_i^{\gamma_i} \Rightarrow \gcd(a,b,c)=\prod p_i^{\min(\alpha_i,\beta_i,\gamma_i)}$

Note that in some cases, an exponent $\alpha_i,\beta_i,\gamma_i$ will be $0$ when a particular $p_i$ is not a factor of one or another of $a,b,c$, but since $p_i\mid abc$, at least one of $\alpha_i,\beta_i,\gamma_i$ will be greater than $0$

Similarly, $\gcd(a,b)=\prod p_i^{\min(\alpha_i,\beta_i)}$

Hence, $\gcd(\gcd(a,b),c)=\prod p_i^{\min({\min(\alpha_i,\beta_i)},\gamma_i)}=\prod p_i^{\min(\alpha_i,\beta_i,\gamma_i)}=\gcd(a,b,c)$ QED


Since the minimum over a set is no greater than the minimum over a subset, $$ \begin{align} \gcd(\gcd(a,b),c) &={\min_{u,v}}^+\!\left({\min_{x,y}}^+(ax+by)u+cv\right)\\ &\ge{\min_{x,y,z}}^+(ax+by+cz)\\ &=\gcd(a,b,c)\tag1 \end{align} $$ $\gcd(a,b)\mid a,b$ and so $\gcd(\gcd(a,b),c)\mid a,b,c$. Thus, by the maximality of the gcd, $$ \gcd(\gcd(a,b),c)\le\gcd(a,b,c)\tag2 $$ $(1)$ and $(2)$ imply that $$ \gcd(\gcd(a,b),c)=\gcd(a,b,c)\tag3 $$


Proof: Let $d=gcd(a,b,c)$ for positive integers $a$,$b$, and $c$. Then $d\mid a$, $d\mid b$, and $d\mid c$. Hence, $d\mid gcd(a,b)$ and $d\mid c$ and so $d\mid gcd(gcd(a,b),c)$. Thus, $d\mid f$ where $f=gcd(e,c)$ and $e=gcd(a,b)$. Since $f=gcd(e,c)$, then $f\mid e$ and $f\mid c$. Hence, $f\mid a$, $f\mid b$, and $f\mid c$, so $f\mid gcd(a,b,c)$ which means $f\mid d$. Since $d\mid f$ and $f\mid d$, $d=f$.


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