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Let $a$, $b$, $c \in \mathbb{N}$. $[a, b]$ denotes $\mathrm{lcm}(a, b)$ and $(a,b)$ denotes $\gcd(a, b)$

Show that

  1. $(a,[b,c]) = [(a,b),(a,c)]$.
  2. $[a,(b,c)] = ([a,b],[a,c])$.
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    as @Srivatsan Narayanan said, it would be nice to see a proof by definition. I donot know if every such properties that is something (equations) about gcd and lcm hold in ufd always hold in a [GCD](http://en.wikipedia.org/wiki/Gcd_domain) domain?2011-09-14

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HINT $\ $ Using the basic GCD laws (associative, commutative, distributive) and, furthermore, employing $\rm\:[x,y] = xy/(x,y)\:$ to eliminate LCMs, we obtain

$\begin{array}{lrll} &\rm(a,[b,c]) &=&\rm [(a,b),(a,c)]&\qquad\qquad\qquad\qquad\quad \\ \iff &\rm(a,bc/(b,c)) &=&\rm (a,b)(c,a)/(a,b,c)& \\ \iff &\rm(a,b,c)(ab,ac,bc) &=&\rm (a,b)(a,c)(b,c)& \end{array} $

true since both sides $\rm\: =\: (abc, baa,caa, abb,cbb, acc,bcc)\:$ (i.e. all trinomials except cubes), after expanding, by distributivity. The dual identity is proved similarly, yielding the same equality.

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    @Thi That's why I always stress the GCD laws when teaching. These GCD proofs are all quite trivial once one intuitively grasps that GCD arithmetic is essentially the same as integer arithmetic (modulo a couple GCD-specific laws). Such axiom-based proofs are more general than proofs depending on primes (UFDs) or the Bezout identity (PIDs), e.g. the same proof often works for gcds and ideals (which is abstracted by *divisor theory*), e.g. the [Freshman's Dream](http://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring/10416#10416) $\rm\ (A+B)^n\:=\:A^n + B^n$2011-09-14