There is $\gcd(a,b)$, written as $\gcd(a,b)$. How I can write lcm(a,b),
the command $\lcm(a,b)$ gives an error.
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ It's strange that such common operator has no built in command already, but I guess there are too many, related: tex.stackexchange.com/questions/112037/…. Anyway for these it is point 11. in math.meta.stackexchange.com/questions/5020/…. $\endgroup$– SilAug 18, 2020 at 8:47
-
$\begingroup$ @Sil I just missed this link because I typed $5040$ instead $5020$. Thank you! $\endgroup$– Michael RozenbergAug 18, 2020 at 8:50
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
\operatorname{lcm} or if you use it a lot in a given answer, \DeclareMathOperator{\lcm}{lcm} will make the \lcm command available, but only for that answer.
answered Aug 18, 2020 at 8:12
-
1$\begingroup$ I've seen also
$\newcommand{\lcm}{\operatorname{lcm}}$for the same functionality. $\endgroup$– SilAug 18, 2020 at 8:51 -
$\begingroup$ We can also talk about
\newcommand{\lcm}{\mathop{\mathrm{lcm}}}if you want to start listing all the equivalent ways of doing that... :-) $\endgroup$– Asaf Karagila ModAug 18, 2020 at 9:10 -
1$\begingroup$ Not really, I am sure someone has already did that somewhere (I wonder if they included
\text{lcm}there? I wouldn't recommend that one) $\endgroup$– SilAug 18, 2020 at 9:15 -
$\begingroup$ @Sil I'm agree with you. But if you use
\textin many cases it is necessary a bit of the space. $\endgroup$ Aug 20, 2020 at 22:01