# $\,m = {\rm lcm}(a,b)\iff a,b\mid m\ \, \& \ \gcd(m/a,m/b)=1$

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)

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..

• – Lord Shark the Unknown May 23 at 18:42
• It doesn't solve my problem. Most of the solutions there use the expression (a∧b)(a∨b)=ab that I'm not allowed to use... – Mrkinix May 23 at 19:51
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• Shouldn't that be $\mu = a \vee b$? – steven gregory May 24 at 5:36
• Did my alternative proof not work for you? – Bill Dubuque May 27 at 14:28

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$$.

• 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. – Mrkinix May 24 at 0:47
• @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. – Bill Dubuque May 24 at 0:50
• @Mrkinix I added an alternative proof. – Bill Dubuque May 24 at 2:18