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Is it possible to proof: when $gcd\left(m_1,m_2,m_3\right)=1$, then $\frac{lcm\left(m_1,m_2\right)m_3}{gcd\left(lcm\left(m_1,m_2\right),m_3\right)}=\frac{lcm\left(m_1,m_2,m_3\right)}{gcd\left(m_1,m_2,m_3\right)}$ is true for any positive integers $ m_1,m_2, m_3 $?

I have test many numbers looks it is actually true.

Bill Dubuque
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xMath
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    Consider $a=8$, $b=4$, and $c=2$. – Sahaj Sep 14 '23 at 09:13
  • You are right, thanks so much. I have missed a condtion, $gcd(m_1,m_2,m_3)=1$, they are setwise prime. – xMath Sep 14 '23 at 09:34
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    This is a nice way to reduce it to checking a specific truth function is a tautology. It also makes obvious why you need setwise primality. – J.G. Sep 14 '23 at 10:40
  • This is a special case of the linked gcd $\times$ lcm law (put $,a = {\rm lcm}(m_1,m_2),\ b = m_3),$ and use lcm associativity. $\ \ $ – Bill Dubuque Sep 15 '23 at 19:32

1 Answers1

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Your conjecture is indeed correct. I suppose you can prove the following:

$$1) \ \ ab = gcd(a,b)\cdot lcm(a,b)$$ $$2) \ \ lcm(a,b,c) = lcm(lcm(a,b), c)$$ These are standard results, you can look them up.

From (1) you get:

$$gcd(lcm(a,b),c) \cdot lcm(lcm(a,b),c) = lcm(a,b)\cdot c$$ or that $$gcd(lcm(a,b),c) = \frac{lcm(a,b)\cdot c}{lcm(a,b,c)}$$ Now consider: $$\frac{lcm(a,b)\cdot c}{gcd(lcm(a,b), c)}$$ Put the value you obtained already and the it becomes: $$lcm(a,b,c)$$

Hence proving the proposition.

More generally, $$\frac{lcm(a,b)\cdot c}{gcd(lcm(a,b),c)}=lcm(a,b,c)\ \ \ \ \ \forall a,b,c \in Z^+$$

Sahaj
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