Doubt in the proof of $\operatorname{lcm}(m,n) = mn / \operatorname{gcd}(m,n)$

Question

I don't understand why $l'd = xm\beta n + yn\alpha m$ when $l' = \alpha m = \beta n$ and $d = xm +yn$.

Answer 1

It appears that your definition of the least common multiple $l$ of $m$ and $n$ is

1. $m\mid l$ and $n\mid l$
2. for all nonnegative integers $l'$, if $m\mid l'$ and $n\mid l'$, then $l\mid l'$

Assuming $m,n>0$, otherwise $ld=mn$ is obvious, your first step is to define $l=m_1n_1d=mn/d$ (with your notation) and then you prove that $l$ satisfies the two properties above.

The first property is clear, because $l=mn_1=nm_1$.

Now let's prove the second propery. If $m\mid l'$ and $n\mid l'$, we have $l'=\alpha m=\beta n$. So far, so good. By Bézout's identity, $d=xm+yn$ and therefore \begin{align} l'd &=l'(xm+yn)\\ &=l'xm+l'yn\\ &=\beta nxm+\alpha myn\\ &=mn(\beta x+\alpha y)\\ &=d^2m_1n_1(\beta x+\alpha y)\\ &=ld(\beta x+\alpha y) \end{align} Cancelling $d$ we obtain $$ l'=l(\beta x+\alpha y) $$ and therefore $l\mid l'$ as wanted.

Answer 2

$l'd=l'(xm+yn)=l'xm+l'xn=(\beta n)xm+(\alpha m)yn$

Answer 3

Said simpler: $\smash[b]{\,n,m\mid l'\,\underset{\times\ n}{\overset{\times\ m}\Longrightarrow}\, mn\mid ml',nl'}$ ${\Rightarrow \overbrace{mn\mid x(ml')\!+\!y(nl')}^{\textstyle \overbrace{mn\mid \color{#0a0}d\:\!l'}^{\textstyle\color{#c00}{mn/d\mid l'}\!\!}\! =\!(\color{#0a0}{xm\!+\!yn})\:\!l'\!\!\!\!\!}},\, $ thus

$m(n/d)\! =\! n(m/d)\,$ is a common multiple of $\,m,n\,$ $\rm\color{#c00}{dividing}$ (so $\color{#c00}{\le}$) all common multiples $\,\color{#c00}{l'},\,$ therefore $\, mn/d\,$ is the $\rm\color{#c00}{least}$ common multiple of $\,m,n$.

Remark $ $ Worth mention: $\ m,n\mid l \iff {\rm lcm}(m,n)\mid l\,$ is the definition of an $\,\rm lcm\,$ in more general domains (which may lack ambient size measures), i.e. the lcm is a common multiple that is least in the divisibility order, i.e. the supremum (least upper bound). Said equivalently in ideal language $\,m\Bbb Z, n\Bbb Z\supseteq l\Bbb Z\iff m\Bbb Z \cap n\Bbb Z\supseteq l\Bbb Z\,$ so $\,{\rm lcm}(m,n)\Bbb Z = m\Bbb Z \cap n\Bbb Z$.