We define $\epsilon\text{-LCM}(a,b)$ for $a,b \in \mathbb R$ as the least non-negative real number such that there exists $n,m \in \mathbb Z$ (not both $0$) with $|\epsilon\text{-LCM}(a,b) - na| \le \epsilon$ and $|\epsilon\text{-LCM}(a,b) - mb| \le \epsilon$. This is always defined if $\epsilon \gt 0$. Additionally, $0\text{-LCM}(a,b)$ is the regular least common multiple, if it exists. Otherwise $\lim_{\epsilon \to 0^+}\epsilon\text{-LCM}(a,b)=\infty$.

My question is, given $a$ and $b$, what is the asymptotic growth rate of $\epsilon\text{-LCM}(a,b)$ as $\epsilon$ approaches $0^+$ (particularly when $\frac ab$ is irrational).


  • $1\text{-LCM}(e,\pi)=\pi-1$, since $|(\pi-1) - e| \le 1$, and $|(\pi-1)-\pi|=1\le1$, and there is no number smaller than $\pi-1$ that satisfies this property.
  • $(0.1)\text{-LCM}(2,3.1)=6.1$
  • $\epsilon\text{-LCM}(a,b)=0$ if $|a|\le\epsilon$
  • $\begingroup$ It seems to me the algorithm for computing this would be to find $na$ close to $mb$, starting with the smaller one and multiplying up to the larger one, then multiplying the larger one if $na$ 'misses' $b$, etc. repeat if $na$ misses $2b$, etc. $\endgroup$ Commented Dec 6, 2017 at 2:46
  • $\begingroup$ Multiply both sides of the inequalities by $x>0$ to get$$|x(\epsilon-{\rm LCM}(a,b))-n(xa)|\le x\epsilon$$and likewise for $b$. This should in theory give us $$\epsilon-{\rm LCM}(a,b)=\frac1x(x\epsilon)-{\rm LCM}(xa,xb)$$ $\endgroup$ Commented Dec 6, 2017 at 2:57
  • $\begingroup$ @SimplyBeautifulArt Hmm, that suggests we only need to worry about $\epsilon\text{-LCM}(a,1)$ (by setting $x$ to $\frac 1b$). $\endgroup$ Commented Dec 6, 2017 at 3:56


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