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The set of real numbers is closed under multiplication, but not under exponentiation (Eg. square root of negative numbers). That is, $\exists a, b \in R \mid {a^b} \notin R$. Then we introduced complex numbers and it is closed under exponentiation.

Now that I came across tetration in Wikipedia, is the set of complex numbers closed under tetration? or can tetration between any two complex numbers does not exist in C? That is, $\exists a, b \in C \mid {^ba} \notin C$ is true for any $a,b$?

Sourav Kannantha B
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    Can you elaborate on why you are not satisfied with the information provided by the Wikipedia page you linked? In my view, Wikipedia seems to answer your question, so explaining why it doesn't would help us clear things up for you. – Mark S. Mar 28 '22 at 11:36
  • @MarkS. see the changes. Also, I wasn't able to find closure property of tetration in Wikipedia. – Sourav Kannantha B Mar 28 '22 at 13:12
  • You are mistaken. The complex numbers are not closed under exponentiation. Exponentiation is not even a binary operation on the complex numbers. There are no axiom that even define what such an operation is supposed to be. Sure, $z^n$ is well defined for all $z\in\mathbb{C}$ and $n\in\mathbb{N},$ but that only gives you a function $\mathbb{N}\times\mathbb{C}\to\mathbb{C},$ not a function $\mathbb{C}^2\to\mathbb{C}.$ There is no meaningful way to the extend the former to the latter. Tetration is a function $\mathbb{N}^2\to\mathbb{N}.$ – Angel Mar 28 '22 at 14:21
  • Perhaps a better question is to ask whether it is possible to define a binary function $u:\mathbb{C}^2\to\mathbb{C}$ such that $u(\alpha,\beta+1)=\exp[\alpha\cdot{u}(\alpha,\beta)]$ for $\alpha,\beta\in\mathbb{C},$ and this is, in a very loose sense, close to the notion of tetration, but it is by no means a natural extension of the concept. I would call it a related concept, rather than an extension. – Angel Mar 28 '22 at 14:25
  • @Angel There is extension of exponentiation in C2→C. It is multivalued, but a principal value is defined. And so is logarithm. See this Wikipedia section. – Sourav Kannantha B Mar 28 '22 at 19:31
  • @SouravKannanthaB No. You are still mistaken. A multivalued relation cannot be part of an algebraic structure. A principal branch can be chosen, but that does not address anything about structure. Besides, you are mistaken that such an extension exists for all of $\mathbb{C}^2.$ Take a look at $0^z,$ which is undefined for an entire plane. $\log(0)$ is undefined. The Wikipedia article oversimplifies the topic to the point of error. – Angel Mar 29 '22 at 11:56
  • How do I move this to chat? There is something I want to add, but it could get too lengthy and it will only be relevant for what I have already said on exponentiation and not tetration. – Angel Mar 30 '22 at 01:28
  • @Angel isn't $0^z, z\neq0$ is $0$? Only $0^0$ is undefined isn't it? And probably I need to see the definition of an algebraic structure. But with principal branch, exponention seems to be well defined. If being multivalued is a problem, then exponentiation isn't even defined on complete real numbers. – Sourav Kannantha B Mar 30 '22 at 04:32
  • @Angel When there is a lot of to and fro comments, site would ask us to move the chat. I'm trying.. – Sourav Kannantha B Mar 30 '22 at 04:34
  • @SouravKannanthaB $0^{-1}$ is not defined, since $0$ has no multiplicative inverse. $0^i$ is regarded as not defined either. $0^0$ is not regarded as undefined by most mathematicians in abstract algebra. But either way, if you define $x^y=\exp[y\log(x)]$ for complex $x$ and $y,$ as you seem to want, then $0^y$ is undefined for all complex $y.$ As for the branching issues, that is precisely what I want to address in the chat. And I agree with you: exponentiation is not defined on the real numbers either. – Angel Mar 30 '22 at 14:14
  • Well, I guess I will just try to address it here, and if it gets too long, someone will move this to the chat for us. Exponentiation, as a general abstract algebraic concept, would be defined by a function $\uparrow:A^2\to{A}$ satsifying the property $a\uparrow{(b+c)}=(a\uparrow{b})\cdot(a\uparrow{c}).$ This follow very straightforwardly when you think about exponentiation on the natural numbers. Of course, this would require $A$ to already have $+$ and $\cdot$ as operations. When you work with the principal branch of multifunctions, you do not have this. – Angel Mar 30 '22 at 14:25
  • And besides, the specific concept of branch cuts is only applicable to the complex numbers, it does not provide a framework for an algebraic structure. For real numbers, the problem is that there is no way of doing it for $a\lt0.$ But you run into problems even just when you try to account for $a=0,$ so even for a ring, you have issues. And so, when you try to explore in depth, you will find that there really are not many algebraic structures where you can even define exponentiation. The natural numbers are one of those few. – Angel Mar 30 '22 at 14:29
  • For one, it must be limited to structures where there are no additive inverses (no subtraction). Why? Because if there is subtraction, then you run into the $0^{-1}$ problem, and this is undefined. And so $0^{-2}$ is undefined, and a bunch of other powers of $0$ are undefined. But that is not enough to solve the problem. Suppose you allowed elements $u$ that have multiplicative inverse (other than $1$.) Then you have to ask, what is $a^{\frac1{u}}$? And this would require some weak notion of algebraic closure. But that still would not be enough. – Angel Mar 30 '22 at 14:32
  • It would not be enough because then you have to have $a^{a^{1/u}},$ and algebraic closure does not cover this. So you must want some kind of "exponent closure," but this really is not a straightforward concept: it is not even clear that such a thing is well-defined. – Angel Mar 30 '22 at 14:34
  • Let us continue this discussion in chat. – Angel Mar 30 '22 at 14:41

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