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For $a \in\Bbb N$, $b\in\Bbb N$, $μ \in\Bbb N^*$, we have $μ = \operatorname{lcm}(a,b) \iff μ = αa\text{ and }μ= βb$ and $\gcd(α,β)$ is $1$

Till now I succeeded to prove the left $\Rightarrow$ right implication, now I need to prove the reciprocal ($\Leftarrow$ way)

Here is my work: http://imgur.com/a/oPy6XDN

Can someone help me, or give me a hint? Rules : you can't use the expression ($a \wedge b)(a\vee b)=ab$, $(ka)\vee(vkb) = k(a\vee b)$ + You can also help me improve my redaction if possible..

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This answer proves the theorem below using basic divisibility laws (with a duality viewpoint).

Theorem $\ $ If $\,\ a,b\mid m\,\ $ then $\ \displaystyle \frac{m}{{\rm lcm}(a,b)} \,=\,\gcd\left(\frac{m}a,\frac{m}b\right),\ $ i.e. $\ {\rm lcm}(a,b)' = \gcd(a',b')$

$\begin{align}{\rm We\ seek}\ \ m = {\rm lcm}(a,b) &\iff a,b\mid m\ \ \&\ \gcd\left(\dfrac{m}a,\dfrac{m}b\right) = 1,\ \ \text{which by the above is}\\[.3em] &\iff a,b\mid m\ \ \&\ \ \dfrac{m}{{\rm lcm}(a,b)}\, =\, 1,\quad\ \ \ \text{which is clearly true.} \end{align}$

Remark $ $ The linked proof using cofactor duality reveals your equivalence boils down to

${\rm lcm}(a,b) = m \iff {\rm lcm}(a,b)' = m',\ $ via $\ {\rm lcm}(a,b)' = \gcd(a',b') = \gcd(m/a,m/b)$


Alternatively $ $ we show: $ $ a common multiple $m$ of $\,a,b\,$ is least $\iff \gcd(m/a,m/b)=1$

Proof $\ $ If the gcd $= c> 1\,$ then $\,c\mid m/a,m/b\,$ so $\,a,b\mid m/c,\,$ therefore $m$ is not least. $ $ Conversely, if $m$ is not least then $\,m = c\ell\,$ for $\,\ell\,$ least, so the gcd $= (c\ell/a,c\ell/b) = c(\ell/a,\ell/b) > 1\,$ by $\,c>1$.

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  • $\begingroup$ The problem is that my teacher won't accept this as a proof since he told us to prove it only by using divisibility laws and gcf laws, and I'm not allowed to use that expression even if it's derived as a corollary... + One more thing I'm not allowed to use fractions : notice how I didn't use any in my paper. I hope you understand, thanks. $\endgroup$ – Mrkinix May 24 at 0:47
  • $\begingroup$ @Mrkinix Again, the linked proof uses only simple divisibility laws and the definition of gcd and lcm. You can ignore the Corollary since I don't use it above. $\endgroup$ – Bill Dubuque May 24 at 0:50
  • $\begingroup$ @Mrkinix I added an alternative proof. $\endgroup$ – Bill Dubuque May 24 at 2:18

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