prove that $\operatorname{lcm}(n,m) = nm/\gcd(n,m)$ I'm trying to prove that $\operatorname{lcm}(n,m) = nm/\gcd(n,m)$
I showed that both $n,m$ divides $nm/\gcd(n,m)$ 
but I can't prove that it is the smallest number. 
Any help will be appreciated.
 A: Hint: For any $a,b$ real numbers: $\min(a,b)+\max(a,b)=a+b$.
Now, if we have $a=a_1^{p_1} a_2^{p_2}\ldots$ and similarly with $b$, if you use the equation I just mentioned for all $p_i$, you will get, that $\gcd(a,b)\cdot\operatorname{lcm}(a,b)=ab$.
A: Hint $\,\ n,m\mid j \!\iff\! nm\mid nj,mj\!$ $\overset{\ \rm\color{darkorange}U}\iff\! nm\mid (nj,mj) \overset{\ \rm \color{#0a0}D_{\phantom |}}= (n,m)j\!$ $\iff\! nm/(n,m)\mid j$
where above we have applied $\,\rm \color{darkorange}U = $ GCD Universal Property and $\,\rm\color{#0a0} D =$ GCD Distributive Law.
Remark $ $ If we bring to the fore implicit $\rm\color{#0a0}{cofactor\  reflection}$ symmetry we obtain a simpler proof: $ $ it is easy to show  $\,d\,\color{#0a0}\mapsto\, mn/d\,$ bijects common divisors of $\,m,n\,$ with common multiples $\le mn.$ Being order-$\rm\color{#c00}{reversing}$, it maps the $\rm\color{#c00}{Greatest}$ common divisor to the $\rm\color{#c00}{Least}$ common multiple, i.e. $\,{\rm\color{#c00}{G}CD}(m,n)\,\color{#0a0}\mapsto\, mn/{\rm GCD}(m,n) = {\rm \color{#c00}{L }CM}(m,n).\,$
See here and here more on this $\:\!\rm\color{#0a0}{involution\ (reflection)}$ symmetry at the heart of gcd, lcm duality.
A: Here is one way without using the Fundmental theorem of arithmetic just using the definitions 
The definition of lcm(a,b) is as follows:
t is the lowest common multiple of a and b if it satisfies the following:
i)a | t and b | t 
ii)If a | c and b | c, then t | c.
Similiarly for the gcd(a,b).
Here is my proof:
Case I: gcd(a,b) $\neq$ 1
Suppose gcd(a,b) = d.
Then $ab = dq_1b = dbq_1  = d*(dq_1q_2)$
Claim: $lcm(a,b) = dq_1q_2$
$a = dq_1$ | $dq_1q_2$ 
$b = dq_2$ | $dq_2q_1$.
Supppose lcm(a,b) = c.
Hence c $\leq$ $dq_1q_2$ .
To get the other inequality we have $dq_1$ | a and $dq_2$ | b. Hence $dq_1$ $\leq$ a $\leq$ c $\leq$ $dq_1q_2$  similiarly for $dq_2$.
Suppose that c is strictly less than $dq_1q_2$, so we have $dq_1q_2$ < $cq_2$ and $dq_1q_2$ < $cq_1$.
So $dq_1q_2$ < c < $cq_2$ < $dq_2^2q_1$ and $dq_1q_2$ < c < $cq_2$ < $dq_1^2q_2$, but $dq_1^2q_2$ > $dq_1q_2$ so c < $dq_1q_2$ and 
c > $dq_1q_2$ contradiction. Hence c = d$q_1q_2$ 
Notice that the case where gcd(a,b) = 1 we can just set $q_1 = a$ and $q_2$ = b, and the proof will be the same.
A: Let's just do this directly. Let $g = \gcd(m,n)$. We need to prove that
$\operatorname{lcm}(m,n) = \dfrac{mn}{g}$.

STEP $0$. (Preliminary stuff.)
DEFINITION $1$. $L = \operatorname{lcm}(m,n)$ if and only if
 1. L is a multiple of m and of n.
 2. If C is a multiple of m and of n, then C is a multiple of L.

LEMMA $2$. If $\gcd(a,b) = 1$ and $a \mid bc$, then $a \mid c$.
PROOF. If $\gcd(a,b) = 1$, then there exists integers $A$ and $B$ such that
$aA + bB = 1$. It follows that $acA + bcB = c$. Since $a | acA$ and $a \mid bcB$, then $a \mid c$.

STEP $1$. $\dfrac{mn}{g}$ is a common multiple of $m$ and of $n$.
This is true because $\dfrac m g$ and $\dfrac n g$ are integers and
$\dfrac{mn}{g} = \dfrac{m}{g}n = m \dfrac{n}{g}$.

STEP $2$. If $G$ is a common multiple of $m$ and of $n$, then $G$
is a multiple of  $\dfrac{mn}{g}$. 
Suppose $G = mM = nN$ for some integers $M$ and $N$. Then 
$\dfrac G g = \dfrac m g M = \dfrac n g N$.
Since $\gcd\left( \dfrac m g, \dfrac n g \right) = 1$, and 
$\dfrac m g M = \dfrac n g N$, then, by LEMMA $2$, $\dfrac m g \mid N$, say 
$N = \dfrac m g N'$ for some integer $N'$.
So $\dfrac{G}{g} = \dfrac{n}{g} N = \dfrac{m}{g} \dfrac{n}{g} N'$. It follows that $G = \dfrac{mn}{g} N'$ and so $G$ is a multiple of $\dfrac{mn}{g}$.

From STEP $1$, STEP $2$, and DEFINITION $0$, we can conclude that
$\operatorname{lcm}(m,n) = \dfrac{mn}{\gcd(m,n)}$.
