Prove that $\mathrm{lcm}(m,n)$ exists, that $\mathrm{lcm}(n,m)=\mathrm{lcm}(|n|,|m|)$, and that $|n|*|m|=\gcd(m,n)*\mathrm{lcm}(m,n)$.
I have been able to proof the first and third part of the statement above however I am stuck proving the $\mathrm{lcm}(n,m) = \mathrm{lcm}(|n|,|m|)$. Can some one guide me to the proof of this question?
$\newcommand{\lcm}{\operatorname{lcm}}$Say $\alpha=\lcm(m,n)$ and $\beta=\lcm(|m|,|n|)$. We may assume that both $m,n \neq 0$ (otherwise the LCM will undefined or we may take it to be $0$). Furthermore we may also assume that at least one of $m$ or $n$ is negative because otherwise the result is trivially true. Say $m <0$.
By the definition of LCM $\alpha >0$ and $\alpha=mk=nj$ for some $k,j \in \mathbb{Z}$ (in fact $k<0)$, also $\beta > 0$ and $\beta=|m|s=|n|t$ for some $s,t \in \mathbb{Z^+}$. Suppose $\alpha > \beta$. Then $mk>|m|s$. But this implies that $k>-s$. Likewise we will have $j>-t$. In which case $\alpha$ cannot be the $\lcm(m,n)$ because we can have $m(-s)=n(-t)$ as a smaller common multiple of $m$ and $n$.
Now you can check the other case when $\alpha < \beta$.
If you think of $\lcm(m,n)$ as the generator of the subgroup $m\mathbb{Z} \cap n\mathbb{Z}$ then this follows almost trivially.
$\ n,m\mid k\iff |n|,|m|\mid k,\ $ so $\,n,m\,$ and $\,|n|,|m|\,$ have the same set $\,K\,$ of common multiples $k>0,\,$ hence they have the same least common multiple $(= \min\,K).$
