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I was thinking about how one learns about the different types of numbers. First there are just whole numbers like $1, 2, 3, ...$

Eventually you write numbers in decimal notation like this: $$6.283185307...$$

And I even found some systems that have negative digits like balanced ternary. Are there any systems where the digits can be real numbers (possibly in the interval $[0,10)$)? I call them realreals in my head.

One such number could be: $$(\pi)\;.\;(0.\overline{1})(\mathrm{e})(4.0)$$ Where I separated the digits using brackets for easier reading. I could sensibly multiply this number by $7$ to get: $$(2.0)(1.991...)\;.\;(2.\overline{7})(1.027...)(8.0)$$ multiplying exactly like you do with normal numbers - carrying wherever necessary. In a weird way this reminded me of how real numbers "fill the gaps" between rationals. These numbers with real digits seem to lie between the reals somehow.

Multiplication seems to work even if both factors are these real reals:

$$(1.0).(0.5)\; \cdot \; (2.0).(0.5) = (2.0).(1.5)(0.25)$$

Note that

  • $1\cdot 2=2$
  • $1.1\cdot 2.1 = 2.31$

and $(2.0).(1.5)(0.25)$ seems to lie between those two.

You could of course have something like a realrealreal $((2.0).(1.\overline{8})).((0.6).(\pi))...$ but that seems to open up a can of worms I'm not ready for.

NiveaNutella
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    Power series with real coefficients? – Noah Schweber Dec 22 '22 at 17:29
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    @NoahSchweber Something like $\sum_{k=-\infty}^{\infty} \frac{a_n}{10^k}$ with $a_n\in [0,10)$? Although not a number system per se, seems to describe it.. and it's also close to something that I was thinking about when that question popped into my head. – NiveaNutella Dec 22 '22 at 17:41
  • I think we should not exaggerate it. Base $b$-systems with integer $b\ge 2$ might have their merits , but number systems with non-integer bases are rather artificial. The factorial base system is perhaps still somewhat natural , but something like a base $\pi$-system or base $1$-system must be very strange indeed and is surely not worth the effort to establish it. – Peter Dec 22 '22 at 17:46
  • @Peter While I don't want to compare my "discovery" to anything of greatness here, one might've said something similar about complex numbers. "Surely it would be very strange to plug negative numbers into square roots and it would not be worth the effort to establish". This question arose naturally from a problem I am actively trying to solve as a sort of relaxation of the problem, so to me it is at least worth the effort to think about. If one doesn't think this belongs on the website, one can always use the downvote and/or report buttons. – NiveaNutella Dec 22 '22 at 17:58
  • I actually upvoted because of the effort, but I do not think that such base-system lead to anything actually useful. – Peter Dec 22 '22 at 18:00

2 Answers2

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You asked

Are there any systems where the digits can be real numbers (possibly in the interval $[0,10)$)?

An answer could be a suitably specialized sequence space which is an infinite dimensional vector space. You could modify the vector space structure and define your own addition and multiplication operations and suitable definitions of norm and convergence. This may have applications for modeling in certain problem domains, but much depends on the details. the "digits" could represent certain attributes of a model object.

Somos
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After thinking about it some more and considering the comment by Noah Schweber and Somos answer I came to the conclusion, that the "realreals" are (in a sense) just real numbers in disguise.

Let's look at some addition to make things more clear.

$\begin{align}(8.8).(0.0)\\+(6.6).(0.0)\\\overline{\quad\quad\quad\quad\quad}\\(15.4).(0.0) \end{align}$

Because we have a number greater than 10, we proceed to "carry" it over to the next digit to the left, yielding:

$(1.0)(5.4).(0.0)$

But in the same manner we could "carry" the $0.4$ over to the digit on the right, finally giving us:

$(1.0)(5.0).(4.0)$

Which looks an awful lot like $15.4$, which is the same, if we just added $8.8$ and $6.6$ in the first place. In this way, you can recover a real number from any realreal number; carrying tens to the left and tenths to the right. Looking at the example from the original question:

$$(π).(0.\overline{1})(e)(4)=$$ $$(3).(0.\overline{1} + 1)(e + 4)(4 + 1)...=$$ $$(3).(1)(e + 4 + 1)(4 + 1 + 1)...=$$ $$(3).(1)(2 + 4 + 1)(4 + 1 + 1 + 7)...=$$ $$(3).(1)(7)(13)...=$$ $$(3).(1)(8)(3)...=$$

Which is the same, if we considered the original number as:

$$\frac{\pi}{10^0} + \frac{0.\overline{1}}{10^1} + \frac{e}{10^2} + \frac{4}{10^3} \approx 3.183...$$

This looks a lot like what Noah Schweber suggested in their comment, but I wanted to make it more explicit and also recover the usual "rules" for digit-wise addition you learn in school.

So, I guess these numbers aren't really between the reals as this could (I assume) be made into a rigorous bijection between them and the $\mathbb{R}$eal numbers.

Happy Holidays!

NiveaNutella
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