Why is there no symbol for rounding to different precisions?

Question

I was taught in a school that one uses rounding for different precision with just the symbol $\approx$, like $13.46\approx 13.5$ or $13.46\approx 13$ depending on the situation. Why are there no different notations to round to different precision, like $13.46\approx_{0.1} 13.5$, $13.46\approx_{1} 13$, $13.46\approx_{10} 10$, at least not in use in schools? And how can one define the notation $\approx_\epsilon$ for fixed $\epsilon>0$ rigorously if one wants that notation to mean rounding to the nearest number from the set $\{n\epsilon|n\in\mathbb Z\}$?

Answer 1

If your question is really "why doesn't so-and-so notation exist," then you're asking an unanswerable question. How should I know? More to the point, how should anyone know? And even if somebody knew, what good would that do?

On the other hand, perhaps what you're really asking is:

How can I define $\approx_\varepsilon$ rigorously, and how should I think about it?

These are perfectly reasonable questions, in fact they are very good questions, and these are the ones I will attempt to answer.

In any metric space $X$, we can define $x \approx_\varepsilon y$ to mean that $d(x,y) \leq \varepsilon$. Then $\{\approx_\varepsilon \mid \varepsilon \in \mathbb{R}_{>0}\}$ induces a uniform structure on $X$ as follows: we say that $R \subseteq X \times X$ is an entourage of $X$ iff $R \supseteq \;\approx_\varepsilon$ for some $\varepsilon \in \mathbb{R}_{>0}$. It is then routine to check that this resulting collection of entourages satisfy the axioms of a uniform structure on $X$.

Note that if $f : X \rightarrow Y$ is a metric map (i.e. a non-expansive map), then from $x \approx_\varepsilon x'$ we can deduce $f(x) \approx_\varepsilon f(x')$. This allows us to show that every metric map is uniformly continuous. Hence there is a forgetful functor from the category of metric spaces and metric maps to the category of uniform spaces and uniformly continuous maps.

For the special case where $X=\mathbb{R}$, there's also stuff about normal forms worth thinking about, which appears to motivate your question. In particular, for all $\varepsilon \in \mathbb{R}_{>0}$, it holds that for almost all $x \in \mathbb{R}$, there is a unique $y \in \varepsilon \mathbb{Z}$ such that $x \approx_\varepsilon y$. With a bit of care for the real numbers dead-on between two elements of $\varepsilon \mathbb{Z},$ this can be used to define a function $f_\varepsilon : \mathbb{R} \rightarrow \varepsilon \mathbb{Z}$ that accepts a real number and returns its best $\varepsilon$-approximation.

Another interesting point is that that for any $\varepsilon \in \mathbb{R}_{>0}$, we get a corresponding "floor" function $\mathrm{floor}_\varepsilon : \mathbb{R} \rightarrow \varepsilon\mathbb{Z}$ and a corresponding "roof" function $\mathrm{roof}_\varepsilon : \mathbb{R} \rightarrow \varepsilon\mathbb{Z}$. Using the language of order theory, these can be defined as adjoints. In particular: there is an inclusion $\mathrm{incl}_\varepsilon : \varepsilon\mathbb{Z} \rightarrow \mathbb{R}.$ The left adjoint is $\mathrm{roof}_\varepsilon$ and the right-adjoint it $\mathrm{floor}_\varepsilon.$

If anything in my answer sounds interesting but doesn't make complete sense, feel free to ask for clarification.

Answer 2

Well, as far as I know, physicists use a couple of ways to make the uncertainity explicit, when strictly needed. One may write your "$13.5015\approx_{0.1}13.5$" as "$13.5\pm0.1$", or assume the number of decimal cyphers to be the precision (writing "$13.50$" instead of your "$13.5015\approx_{0.01}13.5$").

I don't know of a mathematical interest in this kind of notation, though.