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.
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.
$|a-b|$ 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 $< |a-b|.$
Hint $\ $ Put $\rm\,k=-1\:$ in $\rm\ lcm(n,kn) = |kn|$