$\newcommand{\lcm}{\operatorname{lcm}}$
Let $m,n$ $\in$ $\Bbb N$. The least common multiple ($\lcm$) of $m,n$ is the smallest natural number $x$, such that $m \mid x$ and $n \mid x$. Prove that the $\lcm$ of $m,n$ is equal to ${m,n\over \gcd(m,n)}$.
Here is my proof so far:
${m,n\over \gcd(m,n)}$ is a common multiple of m and n.
I am assuming that $q_m$ and $q_n$ are coprime.
If $m = q_m \gcd(m,n)$ & $n = q_n \gcd(m,n)$, then ${m,n\over \gcd(m,n)}$ becomes ${q_m \gcd(m,n)q_n \gcd(m,n)\over \gcd(m,n)}$ which equals $q_m q_n \gcd(m,n)$.
So, $m \mid q_m \gcd(m,n)q_n$ & $n \mid q_n \gcd(m,n)q_m$.
Taking $x$ into consideration, $m \mid x$ and $n \mid x$, $x = k_1 m$ and $x = k_2 n$ which is $x = k_1 q_m \gcd(m,n)$ and $x = k_2 q_n \gcd(m,n)$. Since $x$ is equal to itself, we set these equal to each other:
$k_1 q_m \gcd(m,n) = k_2 q_n \gcd(m,n)$
$k_1 q_m = k_2 q_n$
${k_1 \over q_n} = {k_2 \over q_m}$.
So now I'm not really sure where to go from this point. It doesn't really feel like I've proven anything at all...
UpdateAfter using the suggestions below and amending my homework, I turned it in to my professor and he said absolutely none of what I said made any sense and refused to give me any homework points. Without a doubt in my mind I believe that my answer was correct, or at least made sense, so I am leaving this question up here to assist other users. I'm just forewarning you that my professor made it fairly clear to me that none of what I had written made any sense so please use this problem with caution. Thank you to everyone that helped me better understand this proof, I really appreciated it.
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$\begingroup$When $m,n$ are coprime $\newcommand{\lcm}{\operatorname{lcm}}\lcm(m,n)=mn$, now consider $dm,dn$ (every pair of numbers may be written this way) $\lcm(dm,dn)=d \lcm(m,n)=dmn = \frac{(dm)(dn)}{d} = \frac{(dm)(dn)}{\gcd(dm,dn)}$.
$\endgroup$ $\begingroup$Hint $\rm\,\ m,n\mid x\!\iff\! mn\mid mx,nx\!\iff\! mn\mid\overbrace{(mx,nx)}^{\textstyle (m,n)x}\!$ $\rm\iff\! mn/(m,n)\mid x\!\iff\!\ell\mid x$
for $\rm\:\ell = mn/(m,n).\ $ $\rm\:x = \ell\:$ in $\,(\Leftarrow)\,$ shows $\rm\:m,n\mid \ell,\:$ i.e. $\rm\:\ell\:$ is a common multiple of $\rm\:m,n,\:$ necessarily the least common multiple, since $(\Rightarrow)$ shows $\rm\:m,n\mid x\:\Rightarrow\:\ell\mid x\:\Rightarrow\:\ell\le x.$
Remark $\ $ From above we deduce $\rm\: m,n\mid x\iff lcm(m,n)\mid x.\:$ This is the definition of the lcm in more general rings. See this answer for this efficient universal approach to LCMs and GCDs.
$\endgroup$ $\begingroup$You are almost at the end! We have $k_1q_m=k_2q_n$. But $q_m$ and $q_n$ are relatively prime. That will tell us something about $k_1$, and therefore about the size of $x$.
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