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This question is inspired by xkcd #2585 (Rounding):

Let $u_0,\ldots,u_n$ be positive real numbers (we can assume w.l.o.g. that $u_0=1$) or “units”.

Consider the following directed graph: its vertices are pairs $(i,k)$ where $0\leq i\leq n$ designates a unit, and $k\in\mathbb{Z}$ is a “measurement” performed in that unit. We place a directed edge from $(i,k)$ to $(j,\ell)$ whenever $$ \ell = \lfloor k \cdot u_i/u_j \rceil $$ where $\lfloor—\rceil$ means “closest integer to” (i.e., $\lfloor x\rceil = \lfloor x+\frac{1}{2}\rfloor$). Maybe assume that all $u_i/u_j$ are irrational so there is never any ambiguity as to what the closest integer means. In other words, we can get from $(i,k)$ to $(j,\ell)$ by converting unit $u_i$ into unit $u_j$ and rounding the result to the nearest integer.

Question: does there exist $u_0,\ldots,u_n$ such that the graph just defined has an infinite strongly connected component?

(In other words, can we find units such that infinitely many different values are reachable from one another by converting between these units and rounding to the nearest integer?)

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1 Answer 1

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For a vertex $(i,k)$, it is reasonable to define its value as $ku_i$. We show that the value cannot grow too large, which establishes a negative answer to the question.

Let $v$ be the current value, and let $V>v$ be a real number such that $\{V/u_i\}<1/2$ for all $i$. Such a $V$ exists: we can start by choosing a very small $w>0$ such that $0<\{w/u_i\}<\frac12$, and then approximate those values by $\{V/u_i\}$ due to Kronecker.

We claim that the value cannot exceed $V$. Indeed, if the current value $ku_i$ does not exceed $V$, then $ku_i/u_j$ rounds to at most $\lfloor V/u_j\rfloor$ (check the brackets’ corners!), so the new value does not exceed $V$.

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    $\begingroup$ It seems this estimate is sharp. if we let $V_1$ be the first such $v$, it seems we can always reach $V_1$, since for any $v'$ with $v \leq v' < v_1$, some $\{ v'/u_i\}$ is at least $1/2$, and thus by the irrationality $>1/2$, and then we can round up to increase. Because there are only finitely many possible values, we reach $V_1$ after finitely many increases. $\endgroup$
    – Will Sawin
    Commented Feb 26, 2022 at 13:49
  • $\begingroup$ @WillSawin Right! I had in mind this equivalence but forgot to say anything on it —- thanks! $\endgroup$ Commented Feb 26, 2022 at 14:14

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