Let $a$, $b$, $c \in \mathbb{N}$. $[a, b]$ denotes $\mathrm{lcm}(a, b)$ and $(a,b)$ denotes $\gcd(a, b)$
Show that
- $(a,[b,c]) = [(a,b),(a,c)]$.
- $[a,(b,c)] = ([a,b],[a,c])$.
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1$\begingroup$ Certainly prime factorization works, but it would be nice to see a proof that avoids it. $\endgroup$ – Srivatsan Sep 14 '11 at 14:18
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$\begingroup$ These properties say that $\mathbb N$ with the divisor relation is a distributive lattice. $\endgroup$ – lhf Sep 14 '11 at 14:48
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$\begingroup$ 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 domain? $\endgroup$ – wxu Sep 14 '11 at 15:05
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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.
answered Sep 14 '11 at 17:24
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$\begingroup$ Ah of course, $(a,b)(c,d) = (a(c,d),b(c,d)) = ((ac,ad),(bc,bd)) = ((ac,ad),bc,bd) = (ac,ad,bc,bd)$ so you can just multiply them out. I should give up maths for today :( $\endgroup$ – TMM Sep 14 '11 at 18:32
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$\begingroup$ @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 $\rm\ (A+B)^n\:=\:A^n + B^n$ $\endgroup$ – Bill Dubuque Sep 14 '11 at 18:46
