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the symbols $(a,b,c,...,g)$ and $[a,b,c,...,g]$ are denote the greatest common divisor and the least common multiple, respectively for the positive integers $a,b,c,...g$.

Example :

$(3,6,18)=3$ and $[6,15]=30$ Prove that:

$${{[a,b,c]}^2\over{[a,b][b,c][c,a]}}=\frac{{(a,b,c)}^2}{(a,b)(b,c)(c,a)}$$ I don't know how to start can you help me please ?

Thanks in advance

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  • $\begingroup$ I edited your tags a bit, I think they are more accurate now. $\endgroup$ – Ben Millwood May 15 '16 at 16:19
  • $\begingroup$ ok thanks Ben Millwood $\endgroup$ – xAminex May 15 '16 at 16:43
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To my mind, this sort of problem becomes easier to think about when you think of $(\ldots)$ and $[\ldots]$ as $\min$ and $\max$, respectively, applied separately to the exponents of all primes. Then your equation becomes

\begin{align} &2\max(a,b,c)-(\max(a,b)+\max(b,c)+\max(c,a))\\={}&2\min(a,b,c)-(\min(a,b)+\min(b,c)+\min(c,a)) \end{align}

This you can easily check by assuming without loss of generality that $a\le b\le c$; the equation then becomes

$$ 2c-(b+c+c)=2a-(a+b+a)\;. $$

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  • $\begingroup$ Hi Joriki, thanks for you answer, but I don't understand what is the relation or the similarity between ${{[a,b,c]}^2\over{[a,b][b,c][c,a]}}$ and $2\max(a,b,c)-(\max(a,b)+\max(b,c)+\max(c,a))\\$ can you explain it to me ? $\endgroup$ – xAminex May 15 '16 at 17:52
  • $\begingroup$ @xAminex: Your equation holds if and only if it holds for each prime $p$ separately. Write $p^a$ instead of $a$ and so on, use $(p^a,p^b,p^c)=p^{\min(a,b,c)}$ and $[p^a,p^b,p^c]=p^{\max(a,b,c)}$ and equate the exponents on the two sides. You can think of natural numbers as infinite vectors of prime exponents, and $(\ldots)$ and $[\ldots]$ as $\min$ and $\max$, respectively, applied to these vectors componentwise. $\endgroup$ – joriki May 15 '16 at 17:54
  • $\begingroup$ Hi @Jorki, Thanks a lot , I got it $\endgroup$ – xAminex May 15 '16 at 17:58
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If you want a hint how to start, then you could replace all lcm's on the LHS by gcd's. For two elements this is clear, and for three you can use the following result:

Theorem $$\operatorname{lcm}(a,b,c)=\frac{a\cdot b\cdot c\cdot \operatorname{gcd}(a,b,c)}{\operatorname{gcd}(a,b)\cdot \operatorname{gcd}(a,c)\cdot \operatorname{gcd}(b,c)}$$

Reference: This question.

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