Let $\{a_1,a_2,\ldots,a_k\}$ be a set of integers and let $m=\text{lcm}(a_1,\ldots ,a_k)$
What I am trying to prove is that if $a_1|n, a_2|n,\ldots, a_k|n$ then $m|n$.
I understand that $m$ is the lowest common multiple, and that $n$ dividies into all $a_1, a_2, \ldots a_k$.
What I am trying to prove is that $m$ divides into $n$, or in other words, that the lowest common denominator divides into $n$.
However, I have no idea how to prove this algebraically. What steps should I do?
On
The intersection of the ideal $(a_1),...,(a_k)$ is $(m)$, $a_i|n, i=1,...,k$ implies $n\in (a_i)$ henceforth $n\in (m)$.
On
That $\ a_1,\ldots,a_k\mid m\,\Rightarrow\,{\rm lcm}(a_1,\ldots,a_k)\mid m\ $ is a prototypical Euclidean descent.
The set $M$ of all positive common multiples of all $\,a_i$ is closed under positive subtraction, i.e. $\,m> n\in M$ $\Rightarrow$ $\,a_i\mid m,n\,\Rightarrow\, a_i\mid m\!-\!n\,\Rightarrow\,m\!-\!n\in M.\,$ So $\,M\,$ is closed under mod, i.e. remainder, by mod = repeated subtraction: $\,m\bmod n\, =\, m\!-\!qn = ((m\!-\!n)\!-\!n)-\cdots-n.\,$ 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.
Suppose not. Then we use the division algorithm, with remainder, to write $n=Qm+R$ (where $Q$ is the quotient, and $R$ the remainder.) Of course we have $0<R<m$. But it is easy to see that $a_i$ divides $R$ for each $i$, hence $R$ is a common multiple of the $a_i$, contradicting the assumption that $m$ was the least such.