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...

Update After 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.

  • $\begingroup$ I changed $gcd$ to $\gcd$, $lcm$ to $\operatorname{lcm}$, and $a\ | \ b$ to $a\mid b$. That is standard $\TeX$ usage. $\endgroup$ Nov 11 '12 at 20:32

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) $\lcm(dm,dn)=d \lcm(m,n)=dmn = \frac{(dm)(dn)}{d} = \frac{(dm)(dn)}{\gcd(dm,dn)}$.


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.


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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