Prove that
If $m'$ is a common multiple of $s$ and $t$, then $m | m'$. Here $m$ is the LCM of $s$ and $t$.
Although the statement is intuitively clear to me I don't know how to prove.
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mathemather
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Hint: since $m$ is the least common multiple of $s$ and $t$, we know $m \leqslant m'$. Perform Euclidean division to write $m' = qm + r$ for some $q \in \mathbb{Z}, 0 \leqslant r < m$. Since $s, t$ divide $m$ and $m'$, they divide $r = m' - qm$. What can you conclude using the minimality of $m$?
Alex Wertheim
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It is tempting that r must be zero. But why? – mathemather Jul 09 '16 at 05:43
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1@arutoregni: Indeed, $r$ must be zero. Otherwise, as argued above, $r$ is a positive number which is divisible by both $s$ and $t$ but is less than $m$, contradicting the minimality of $m$! – Alex Wertheim Jul 09 '16 at 05:47
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That $\ a,b\mid m\,\Rightarrow\,{\rm lcm}(a,b)\mid m\ $ may be conceptually proved by Euclidean descent as below.
The set $M$ of all positive common multiples of $\,a,b\,$ is closed under positive subtraction, i.e. $\,m> n\in M$ $\Rightarrow$ $\,a,b\mid m,n\,\Rightarrow\, a,b\mid m\!-\!n\,\Rightarrow\,m\!-\!n\in M.\,$ By induction, we deduce that $\,M\,$ is closed under mod, i.e. remainder, since it arises by repeated subtraction, i.e. $\ m\ {\rm mod}\ n\, =\, m-qn = ((m-n)-n)-\cdots-n.\,$ Thus the least $\,\ell\in M\,$ divides every $\,m\in M,\,$ else $\ 0\ne m\ {\rm mod}\ \ell\ $ is in $\,M\,$ and smaller than $\,\ell,\,$ contra minimality of $\,\ell.$
Remark $ $ The innate structure exploited above, that sets of integers closed under subtraction are multiples of their least positive element plays a fundamental role in number theory and algebra. Such (principal ideal) structure is highlighted further further in other posts here.
Bill Dubuque
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