# What is the Least Common Multiple of $(a-b)$ and $(b-a)$?

What is the Least Common Multiple of $(a-b)$ and $(b-a)$?

The question is simple. What's the answer? I'm an Engineering student, but looks like I forgot my basics.

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I assume you mean over the Integers? To start, (a-b) = -(b-a)... – Seth Jul 19 '12 at 13:48

$|a-b|$ is the LCM of $(a-b), (b-a).$ Recall LCM is the smallest positive integer that is integer multiple of both $(a-b)$ and $(b-a).$ We have $$|a-b| = +1\times(a-b) \\ |a-b| = -1\times(b-a).$$ (Of course, the signs above assuming $a > b.$ Flip the signs if $b < a.$) So $|a-b|$ is an integer multiple of both $(a-b), (b-a).$ To prove it's the smallest, assume $\ell < |a-b|$ is the LCM. Then $$\ell = k(a-b) = -k(b-a)$$ for some integer $k.$ (again with the convenient choice $a > b.$) Comparing with the equations above, we can conclude that $k < 1$ and simultaneously $-k < -1 \implies k > 1.$ So $k = 0,$ and hence $\ell$ is not an LCM. Contradiction. There exists no non-zero common multiple $< |a-b|.$
Assuming $a\geq b$, of course. – Thomas Andrews Jul 19 '12 at 13:51
A more immediate contradiction lies in the fact that $k$ is simultaneously less than and greater than $1$. – Cameron Buie Jul 19 '12 at 14:29
Hint $\$ Put $\rm\,k=-1\:$ in $\rm\ lcm(n,kn) = |kn|$