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I'm an extreme mathematical layman, so please excuse the probable ignorance and awkward phrasing of this question.

Is there such thing as a kind of number which has an uncountably infinite amount of digits after the decimal point?

To give an example of what I mean, say such numbers exist and we'll call one of them S. The digits of S are such that every rnth place (and rn can be any positive real number) corresponds to the first decimal place of sin(rn).

The 0th place of of S is 0. What follows are infinitely many zeros until the digit place that corresponds to sin^-1(.1) is reached, and the digit of that place will obviously be 1. Long after that, the .5th place is 4, the .61999th place is 5, the 1st place is 8, the πth place is 0, the 6000π/e-th place is 7.

So, S will have a rather dull repeating pattern of 0.000...111...222...333..., eventually returning to 0s at the πth place and then cycling on again.

Another such number could be C, which has a similar rule but the rnth place corresponds to the first decimal of cos(rn) instead. C would look like 1.999...888...777, and so on.

I know S and C can't be real numbers (if these can even be considered numbers at all), because the cardinality of a real number's digits is equal to the set of the natural numbers. The cardinality of the digits of S and C is equal to the set of real numbers.

Now, is all this just useless mumbo-jumbo hiding behind ellipses? Supposing it's not, what would the cardinality of the set of all these weird numbers like S and C be? Would it be larger than the set of all real numbers?

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    $\begingroup$ Well, you can define anything you want, but it's not clear how you would add or multiply such numbers which makes it... less useful. (Also are you aware that the powerset of a set is always larger than the original set? This would answer your cardinality question) $\endgroup$
    – Milo Brandt
    Commented Jul 28, 2015 at 22:53
  • $\begingroup$ objects like the ones you discuss might exist, but I see no reason to call these entries digits. $\endgroup$
    – Mirko
    Commented Jul 28, 2015 at 22:57
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    $\begingroup$ To Milo's comment on addition, since the digits of S correspond to the first decimal of sin(rn) and the digits of C correspond to the first decimal of cos(rn), wouldn't the digits of S+C correspond to the digits of sin(rn) + cos(rn)? $\endgroup$
    – Michael
    Commented Jul 28, 2015 at 23:00

4 Answers 4

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Decimal representations of a number is just a representation - ie, it is a way of specifying a number by giving a certain amount of information about it.

In the case of real or complex numbers, since there are strictly more than $\aleph_0$ many of them, you need $\aleph_0$ (countably-infinite)-many digits to describe an arbitrary real/complex number. Similarly, for rational numbers, since there are $\aleph_0$ many of them, you only need finitely many digits to specify any rational number (note that non-terminating rational numbers must repeat their digits, so you only need to know the digits up to the first repeat).

There do exist mathematical structures (fields, rings, vector spaces, groups) that have cardinality greater than $|\mathbb{R}|$, in which case you will need at least $|\mathbb{R}|$-many digits to specify an element of that structure.

See for example:

https://mathoverflow.net/questions/44705/cardinalities-larger-than-the-continuum-in-areas-besides-set-theory

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    $\begingroup$ The assertion that the number of real numbers is $\aleph_1$ is rather bold. In fact the number is $2^{\aleph_0}$. Whether that's the same as $\aleph_1$ is in fact independent of the usual axioims of set theory. $2^{\aleph_0}$ is the number of subsets of a set of size $\aleph_0$. On the other hand $\aleph_1$ is the number of countable ordinals. These conventions date back to Georg Cantor himself. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented Jul 29, 2015 at 0:38
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    $\begingroup$ I think you're confusing $\aleph_1$ with $2^{\aleph_0}$ and $\aleph_2$ with $2^{\aleph_1}$. It is actually customary to use the letter $\beth$ for that: $\beth_{\alpha+1}$ is defined as $2^{\beth_\alpha}$ and $\beth_0$ is the same as $\aleph_0$, and if $\alpha$ is a limit ordinal then $\beth_\alpha$ is $\sup\left\{\beth_\beta : \beta<\alpha\right\}$. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented Jul 29, 2015 at 0:45
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    $\begingroup$ Thanks for pointing out my mistake. I've edited my answer correspondingly. $\endgroup$
    – oxeimon
    Commented Jul 29, 2015 at 4:36
  • $\begingroup$ I believe there are only 10 digits in common use: 0,1,2,3,4,5,6,7,8,9. $\endgroup$
    – Mirko
    Commented Jul 29, 2015 at 4:57
  • $\begingroup$ The examples at that mathoverflow link aren't really sets where someone does something like use more than countably many digits, because nothing like digital expansions really arise. That said, sign expansions of the Surreal numbers (or a field naturally sitting inside) are pretty close to binary expansions which can involve a an expansion of uncountable length. $\endgroup$
    – Mark S.
    Commented Aug 2, 2015 at 17:21
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A number can be seen as an infinite sequence of digits (or bits, if you write in binary). In other words, it is an infinite word on the alphabet $\{0, \dotsm, 9\}$ (or $\{0, 1\}$ if you write in binary).

Now, a theory of words over ordinals and even words over linear orders has been proposed in [1]. Given a finite alphabet $A$ and a totally ordered set $I$, a word $(a_i)_{i \in I}$ is simply a function from $I$ to $A$. Usual finite words are the words indexed by finite orderings $I = \{1,2,...,n\}$. An infinite word is a word over $I = \mathbb{N}$, but you can define words over $\mathbb{R}$ if you wish.

This definition and the main results about these words belong to automata theory, which is probably not what you are looking for, but you can have an "uncountably infinite amount of digits" in this way.

[1] A. Bès and O. Carton, A Kleene theorem for languages of words indexed by linear orderings, Int. J. Found. Comput. Sci., vol. 17, no. 3, pp. 519-542, 2006.

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If you want each element of your field $F$ to be represented by a unique sequence $\alpha \rightarrow \mathbb{N}$ (where $\alpha$ is the number of digits you require), its cardinal must be $2^{\alpha}$.

It seems to me that the field of surreal numbers with birth date $< \omega_1$ could satisfy your condition. Its cardinal is $(2^{\aleph_0})^{\aleph_1} = 2^{\aleph_1}$. I don't know if it's stricly greater than $2^{\aleph_0}$.

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  • $\begingroup$ Whether or not $2^{\aleph_0} < 2^{\aleph_1}$ is independent from $ZFC$. $\endgroup$
    – Stefan Mesken
    Commented Jul 29, 2015 at 4:50
  • $\begingroup$ Ok; so the cardinal of the "set of numbers" Michael is looking for may or may not be stricly greater than the continuum. $\endgroup$
    – nombre
    Commented Jul 29, 2015 at 21:51
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An uncountable sum cannot converge in a standard sense of the word. Let $\{a_j: j \in J\}$ where $J$ I an uncountable index: Define $K_n:=\{ a_k \in a_j : a_k >1/n\}$. Then, by a cardinality argument, one of the sets $K_n$ will have infinitely -many terms, and the infinite sum $\sum K_n > 1/n+ 1/n +\cdots$ will diverge. So there must be a countable index beyond which all terms must be zero, unless you have a different notion of convergence.

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