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If this is the wrong sort of question to ask here, apologies!

Not much background, I have no experience (except idle curiosity) with cardinality, cartesian products, and the like, but I was curious about the idea of some set $\mathbb{R}^{1/2}$ such that $\mathbb{R}^{1/2}\times \mathbb{R}^{1/2}=\mathbb{R}.$ I do know already that cardinality is not very helpful here because $|\mathbb{R}\times\mathbb{R}|=|\mathbb{R}|$. I was wondering if there would be any possible way to define this, and maybe generally define the notion of non-integer "$x$-tuples". Any ideas would be greatly appreciated.

  • I see no reason to talk about $\mathbf{R}$ here. You may as well try to define first ${a, b}^{1/2}$ for $a\ne b$. – Alexey Jun 01 '25 at 12:33
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    I have never seen the notation $\mathbb{R}^{1/2},$ so this is likely somewhat an original thought. – Adam Rubinson Jun 01 '25 at 12:34
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    $\mathbb R^n$ is defined inductively, and it makes it very hard to convert to fractions. There are some ways - for example, as $\mathbb Q$-vector spaces - that $\mathbb R\times \mathbb R\cong \mathbb R,$ but that way will not not let us define $\mathbb R^{1/2}$ un;ess we define all $\mathbb R^n\cong \mathbb R.$ – Thomas Andrews Jun 01 '25 at 12:35
  • @Alexey Ah, thanks! It is my end goal to see if there is any way I could have "1/2 dimensional" vectors or the like, so now I know where to start – Helpsilon Delta Jun 01 '25 at 12:37
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    It's a pretty interesting question. I haven't seen this before. Maybe you could look into the notion of a Hausdorff-Dimension. – Samuel Jun 01 '25 at 12:38
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    This question captures the spirit: https://mathoverflow.net/questions/60375/is-mathbb-r3-the-square-of-some-topological-space though it's tackled highly nontrivially with topology (the question contains an argument for $\mathbb R$ also) – George Coote Jun 01 '25 at 12:39
  • @AdamRubinson Ah darn, I was hoping someone would have seen it somewhere. – Helpsilon Delta Jun 01 '25 at 12:39
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    @Samuel Interesting! I'll have a look – Helpsilon Delta Jun 01 '25 at 12:40
  • From a pure cardinality point of view, it doesn't make sense, as Alexey says, because you'd have to have square roots of all natural numbers. Even if $X$ has $4$ elements, it wouldn't be clear what to make the set $X^{1/2}$ be. $X^2$ is well-defined as a set for any $X,$ but saying $X^{1/2}$ is "any" two-element set dosn't define a set, it defines a cardinality. – Thomas Andrews Jun 01 '25 at 12:41
  • @HelpsilonDelta, it was just my first thought. However, I think, usually this kind of generalization is done by first providing a more flexible interpretation to the standard cases, like with generalized functions. So, I am not sure, maybe such a flexible interpretation will be easier to find for the powers of some field, like $\mathbf{R}$. – Alexey Jun 01 '25 at 12:43
  • @ThomasAndrews It isn't looking too good then, thank you for taking the time to think with me! – Helpsilon Delta Jun 01 '25 at 12:44
  • @Alexey Awesome, thank you so much for pointing me in the right direction. – Helpsilon Delta Jun 01 '25 at 12:48
  • @GeorgeCoote It is exactly as I feared; some strange algebraic topology is involved. Thank you for showing me this, I'll work on gaining an understanding! – Helpsilon Delta Jun 01 '25 at 12:53
  • @Hlpsilon Delta for that $\mathbb R$ argument you just need point-set topology, however when you said you had no experience with cardinality or Cartesian products I assumed you hadn't got to this yet. Good motivation to do some more reading though! – George Coote Jun 01 '25 at 12:56
  • @GeorgeCoote Oh okay then! It's beginning to smell like a textbook purchase for me... – Helpsilon Delta Jun 01 '25 at 13:04
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    The elements of $A\times B$ are ordered pairs (no matter what $A$ and $B$ are). The elements of $\bf R$ are not ordered pairs. End of discussion. – Gerry Myerson Jun 01 '25 at 13:07
  • The question that you are asking in essence is whether it is possible for a set to have a number of dimensions that is not an integer. Under some circumstances, the answer is yes – K.defaoite Jun 01 '25 at 15:10
  • If you know what a topological space is you might find this take on the matter interesting. In the question it is explained why there is no topological space $X$ with $X^2$ homeomorphic to $\mathbb R$. – Jan Bohr Jun 03 '25 at 07:46

2 Answers2

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I will answer the easier half of your question.

While there won’t exist a canonical $\mathbb{R}^{\frac{1}{2}}$ there is a notion that generalizes “non integer” tuples.

I’ll assume some familiarity with fractals here (if not just make a comment and I can add an introduction).

We consider 3 fractals: the regular real line $\mathbb{R}$, the $\log_{3}(5)$ dimensional vicsek fractal and then the plane $\mathbb{R}^2$ although for convenience we will consider this to be the same as $\mathbb{C}$.

Elements on the real line have a trinary representation where the elements can be written as $\sum a_n 3^n$ where the $a_n \in \lbrace -1, 0, 1 \rbrace$. Notice the base of the powers is 3, and there are 3 symbols (-1,0,1) to pick from at each digit and $\log_3(3) = 1$. This will be important later.

We might desire a similar representation on the complex plane but it’s not so obvious since we have a pair of real numbers and we want a single base-3 expansion. With a little cleverness we might observe that if we let our symbols range from $S = \lbrace -1-i,-1,-1+i,-i,0,i,1-i, 1, 1+i \rbrace$ suddenly it becomes possible to represent every point on the plane as $\sum a_n 3^n$ where $a_n \in S$ and here we have 9 symbols to pick from at each digit and $\log_3(9)=2$

With this established it becomes clear that “n-tuples of real numbers” become “3^n characters in a single base 3 expansion” and so we can finally try to ask “what does a $\log_3(5)$ tuple look like?” turning our attention to the vicsek fractal.

Here our set of characters are $S = {-1,-i,0,i,1}$ and our base-3 representation of each element is the usual $\sum a_n 3^n$.

So what have we accomplished? We turned integer sized tuple into a base $3$ representation and then found a new base $3$ representation that doesn’t have an integer sized tuple analogue.

This suggests that instead of thinking in terms of tuples we should instead think in terms of base-N representations and suddenly fractional dimensions and integer dimensions are all treated equally nicely on the same footing.

We also get a hint about what $\mathbb{R}^n$ really is. It is those base-$B$ representations for which there exists a natural addition, multiplication, and “carry” rule.

If you relax the notion of “natural” enough you might then find many candidates for a $\mathbb{R}^{\frac{1}{2}}$ though I suspect none will have ALL the properties you want

  • Thank you so much, this is some really cool maths. While it didn't reach the conclusion I was hoping for, maybe the journey was more important than the destination – Helpsilon Delta Jun 02 '25 at 01:14
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You write:

"I do know already that cardinality is not very helpful here because |R×R|=|R| ."

There is a kind of category error involved here, in the following sense. One usually speaks of two types of "infinite number": the Cantorian (infinite) cardinals and ordinals. But there is a third type of "infinite number" where extracting roots does not pose a problem. This type has been called ringinal in some recent publications. Thus, instead of using ordinals or cardinals, one should use ringinals.

Ringinals are the "infinite numbers" of nonstandard analysis, where root extraction works just as any operation of classical analysis. In particular, given a ringinal $H$, its square root $\sqrt{H}$ is entirely unproblematic. One only runs into problems if one looks for radicals of the wrong type of infinite number.

Mikhail Katz
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  • Thanks for this answer, I haven't encountered the concept of a ringinal before. If I give some kind of ring (say a subset of R^2 and some operations for +, X, 1) is there an algorithm to compute what the natural "ringinal" of that ring is? It feels like I am asking for a "continuous" version of @anixx's numerosities. – Sidharth Ghoshal Jun 03 '25 at 21:13
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    The terminology of ringinals is mostly intended for ordered rings. Leibniz referred to these as linea infinita terminata. These can occur both in a discrete and a continuous context. From the viewpoint of axiomatic nonstandard analysis, a nonstandard integer in $\mathbb N$ is greater than every standard integer. Similarly, a real number greater than every positive standard real in $\mathbb R$ would be such a ringinal. @SidharthGhoshal – Mikhail Katz Jun 05 '25 at 13:02
  • What sort of courses should I be aiming for to understand these sorts of things? I'm only a first year but these sorts of ideas really fire me up – Helpsilon Delta Jun 07 '25 at 01:05
  • @HelpsilonDelta, you will find some suggestions here. – Mikhail Katz Jun 08 '25 at 10:51