# Can a number be its own least common multiple?

I almost feel foolish asking this, but I've stared at this problem for so long and can't make sense of it...

Given the problem:

Consider $\mathbb N$ with the partial ordering of ”divides”, i.e. define a ≤ b if and only if a|b. For $a, b \in \mathbb N$, let gcd(a, b) denote the greatest common divisor of a and b, and let lcm(a, b) denote the least common multiple of a and b.

prove that if $a,b \in \mathbb N$ then the supremum (i.e. the least upper bound) of {$a,b$} is the lcm(a,b).

Please correct me if I'm wrong but I am reading this as that $3|6$ so $3 ≤ 6$ but the while the supremum of the set {$3,6$} is 6, the lcm(3,6) is 18? Am I missing some key detail? for instance does the set denoted {$a,b$} include more numbers than just a and b?

• No, $\{a,b\}$ includes only $a$ and $b$. By the way, $lcm(3,6) = 6$. And yes, it is true that $lcm(a,a) = a$. – BrianO Nov 7 '15 at 18:03
• Perhaps, the formula $lcm(a,b)=\frac{a\times b}{gcd(a,b)}$ helps. – Peter Nov 7 '15 at 18:05
• @Peter @ Unit I know I saw it myself, too late to avoid public display of stupidity :) – BrianO Nov 7 '15 at 18:06
• So what you're asked to prove is: (1) $a \mid lcm(a,b)$ and $b \mid lcm(a,b)$, and (2) if $a \mid c$ and $b \mid c$, then $lcm(a,b) \mid c$. – BrianO Nov 7 '15 at 18:07
• I see now, I just had a naive understanding of lcm. I new I'd end up feeling foolish about asking this, but thank you very much of clarifying. – user3256725 Nov 7 '15 at 18:17