Update: Since the question was put on hold, work on this will continue on this notebook.

So long, and thanks for all the fish!


First, let me state that I am not a formally-trained mathematician, therefore please forgive my approximations and help me correct my errors. Second, I am looking for prior work related to what I would like to call Hypertype Theory. What follows is a short introduction to the topic. But before that, here are a few specific questions that will hopefully make this question less open-ended.



  1. Where can we find prior work on the subject?
  2. Does $\mathbb{H}_n$ converge toward $\mathbb{R}$?
  3. What would be a simple way of establishing such a convergence?
  4. What are examples of important numbers that appear in $\mathbb{H}_{n\ge4}$?
  5. Can $\pi$ be naturally introduced as a term of $\mathbb{H}_n$?
  6. Could the proposed notation be improved upon?
  7. Is this interesting?  



The basic idea is this: a group is roughly defined with a triplet of two operators and an identity term, while a field is made of two such triplets plus a distributivity law. But what do we get with a third triplet? And what if an unlimited number of such triplets could be added in a recursive manner? Would such an object exhibit any valuable properties?

Interestingly, there is one such object, and it can be constructed in a quite straightforward fashion by using Albert Bennett’s commutative hyperoperations. Most importantly, this object is isomorphic to a strict superset of $\mathbb{Q}$ and to a strict subset of $\mathbb{R}$, while possibly converging toward $\mathbb{R}$.


Commutative Operators

$a\overset{1}{\oplus}b = e ^ {ln(a)} + e ^ {ln(b)} = a + b$

$a\overset{2}{\oplus}b = e ^ {ln(a)} × e ^ {ln(b)} = a × b$

$a\overset{3}{\oplus}b = e ^ {ln(a) × ln(b)} = a ^ {ln(b)} = b ^ {ln(a)}$

$a\overset{4}{\oplus}b = e ^ {\displaystyle e ^ {ln(ln(a)) × ln(ln(b))}}$


$\displaystyle a\overset{n+1}{\oplus}b = e ^ {ln(a)\overset{n}{\oplus}ln(b)}$


Non-Commutative Operators

$a\overset{1}{\ominus}b = e ^ {ln(a)} - e ^ {ln(b)} = a - b$

$a\overset{2}{\ominus}b = e ^ {ln(a)} ÷ e ^ {ln(b)} = a ÷ b$

$a\overset{3}{\ominus}b = e ^ {ln(a) ÷ ln(b)} = a ^ {1 ÷ ln(b)}$

$a\overset{4}{\ominus}b = e ^ {\displaystyle e ^ {ln(ln(a)) ÷ ln(1 ÷ ln(b))}}$


$a\overset{n+1}{\ominus}b = e ^ {ln(a)\overset{n}{\ominus}ln(b)}$


Identity Terms

$i_1 = 0$

$i_2 = 1$

$i_3 = e$

$i_4 = e ^ e$


$i_{n+1} = e ^ {i_n} = e \upuparrows (n - 1) \quad (n > 1)$


Core Identities

From these definitions, we can recursively prove the following core identities:



$a\overset{n}{\ominus}i_n = i_n$



$a = b \Longleftrightarrow a\overset{n}{\ominus}b = i_n$



$a\overset{n}{\ominus}(b\overset{n}{\ominus}c) = c\overset{n}{\ominus}(b\overset{n}{\ominus}a)$



$(a\overset{n}{\oplus}b)\overset{n}{\ominus}c = a\overset{n}{\ominus}(c\overset{n}{\ominus}b)$



$(a\overset{n}{\oplus}b)\overset{n+1}{\oplus}c = (a\overset{n+1}{\oplus}c)\overset{n}{\oplus}(b\overset{n+1}{\oplus}c)$


In general, these core identities can be used as axiomatic definitions, but it just so happens that we can prove them when the operators are defined using commutative hyperoperators. This ensures that we are not dealing with purely-theoretical objects.


Group Properties

From the core identities stated above, we can prove the following properties:


$\overset{n}{\ominus}$ Anticommutativity

$a\overset{n}{\ominus}b = i_n\overset{n}{\ominus}(b\overset{n}{\ominus}a)$


$\overset{n}{\oplus}$ Left and Right Identity

$i_n\overset{n}{\oplus}a = a\overset{n}{\oplus}i_n = a$


$\overset{n}{\oplus}$ Associativity

$(a\overset{n}{\oplus}b)\overset{n}{\oplus}c = a\overset{n}{\oplus}(b\overset{n}{\oplus}c)$


$\overset{n}{\oplus}$ Commutativity

$a\overset{n}{\oplus}b = b\overset{n}{\oplus}a$


Additional Properties

Along the way, the following properties can be established as well:


Double Reversibility

$i_n\overset{n}{\ominus}(i_n\overset{n}{\ominus}a) = a$


Associative Commutativity

$(a\overset{n}{\ominus}b)\overset{n}{\ominus}c = (a\overset{n}{\ominus}c)\overset{n}{\ominus}b$


Affine Equivalence

$a\overset{n}{\ominus}b = c \Longleftrightarrow a\overset{n}{\ominus}c = b$


Reverse Substitutivity

$a\overset{n}{\oplus}b = a\overset{n}{\ominus}(i_n\overset{n}{\ominus}b)$



$(a\overset{n}{\ominus}b)\overset{n}{\oplus}b = a$

$(a\overset{n}{\oplus}b)\overset{n}{\ominus}b = a$



We start by defining $\mathbb{H}_1$ in a coinductive fashion, as described in this article. We then define a sequence of hypertypes by granting the hypertype $\mathbb{H}_n$ the triplet $(\overset{n}{\oplus}, \overset{n}{\ominus}, i_n)$ and all the triplets granted to its predecessors. This makes $\mathbb{H}_1$ isomorphic to $\mathbb{Z}$ and $\mathbb{H}_2$ isomorphic to $\mathbb{Q}$, while $\mathbb{H}_3$ is isomorphic to a strict subset of $\mathbb{R}$, which we call exponential numbers $\mathbb{E}$. Most importantly, $\mathbb{H}_n$ defines $n$ groups and $n-1$ fields.

We call our structure Hypertypes instead of Hypergroups for three main reasons: first, Hypergroups refer to an already-existing collection of objects that have nothing to do with Hypertypes; second, we are interested in developing the Hypertype Theory in the context of Type Theory, without relying on any of the axioms of Set Theory; third, we are interested in treating commutative operators separately from their non-commutative duals.

The third point is of critical importance: in order to properly handle real-world measures like temperatures that are non-additive (physicists and statisticians call them intensive), we must be able to distinguish them from measures that are additive (extensive). Therefore, our hierarchy of types should add one operator at a time, always starting with the non-commutative operator. Therefore, for every hypertype $\mathbb{H}_n$, we will have an intensive sub-hypertype $\mathbb{H}_{\bar{n}}$ defined with the non-commutative operator but without its commutative dual.

This approach is nicely supported by the fact that the Core Properties defined above are written with affine-style identites focused on the non-commutative operator. Therefore, we can properly define the commutative operator from its non-commutative dual. This approach also reduces the number of axiomatic definitions that need to be stated when defining the two operators.



The notation introduced above is especially useful for $n = 3$. Therefore, we suggest that the index $n$ can be ommitted whenever it is equal to $3$. Doing so, we would benefit from a single-symbol notation for the commutative power law and its inverse.

$a \oplus b = a\overset{3}{\oplus}b = e ^ {ln(a) × ln(b)} = a ^ {ln(b)} = b ^ {ln(a)}$

$a \ominus b = a\overset{3}{\ominus}b = e ^ {\frac{ln(a)}{ln(b)}} = a ^ {\frac{1}{ln(b)}}$

Alternatively, if we were to limit ourselves to symbols found on most keyboards, we would suggest:

$a \# b = a\overset{3}{\oplus}b = e ^ {ln(a) × ln(b)} = a ^ {ln(b)} = b ^ {ln(a)}$

$a \backslash b = a\overset{3}{\ominus}b = e ^ {\frac{ln(a)}{ln(b)}} = a ^ {\frac{1}{ln(b)}}$



Many thanks to the following people for their contributions:

  • @Henry for having found the triplet for $\mathbb{H}_3$.
  • Prof N J Wildberger for being such an inspiration to young contrarian mathematicians.

Note: Some credits might detract some readers, but credit should be given when it is due.


closed as unclear what you're asking by abx, YCor, Chris Godsil, Andy Putman, Harry Gindi Jan 7 at 15:49

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Somewhat related are hyperidentities. You are looking for similar identities holding for certain terms, but interesting things happen when you ask them to hold for all terms, or for certain closed subsets of terms. Start with work of Denecke, Schweigert, and Wismath. Gerhard "Can Provide Some More Names" Paseman, 2019.01.05. $\endgroup$ – Gerhard Paseman Jan 6 at 1:57
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    $\begingroup$ I don't think this is a good question for MO. This posts basically asks for open-ended discussion without any specific problem, while MO (and StackExchange in general) are suited for well-defined problems with verifiable solutions. This post is more of a research article posted on a forum. Regarding your question, I find missing in your observations any theorems that would connect your hyperoperations at different level. It's not hard to make up a new group, a family of groups with some strong connections is a much more interesting (a recursive definition on its own is too weak a connection). $\endgroup$ – Anton Fetisov Jan 6 at 13:06
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    $\begingroup$ The precise definition of $\mathbb{H}_n$ is not clear to me. Is it an inductive definition, where you set $\mathbb{H}_1 = \mathbb{Z}$, and then $\mathbb{H}_{n+1}$ equal to the closure of $\mathbb{H}_n\cup \{i_n\}$ (inside $\mathbb{R}$) under the operations $\oplus^{n+1}$ and $\ominus^{n+1}$? $\endgroup$ – Alex Kruckman Jan 6 at 18:06
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    $\begingroup$ Regarding Questions 2 and 3: It's also not totally clear what you mean by "$\mathbb{H}_n$ converges to $\mathbb{R}$". The most natural meaning is $\bigcup_{n} \mathbb{H}_n = \mathbb{R}$. If this interpretation (and the definition of $\mathbb{H}_n$ in my previous comment) is correct, then the answer is no: each $\mathbb{H}_n$ is countably infinite, so $\bigcup_{n} \mathbb{H}_n$ is countably infinite, while $\mathbb{R}$ is uncountable. $\endgroup$ – Alex Kruckman Jan 6 at 18:09
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    $\begingroup$ Based on the reference to Wildberger at the end of your question, I'm worried there's a real danger you're going to come back and tell me that this argument is invalid because infinite sets "don't make sense". (Just a tip: linking to Wildberger is not the best idea if you want to be taken seriously by mathematicians.) $\endgroup$ – Alex Kruckman Jan 6 at 18:14

1. Where can we find prior work on the subject?

The model of exponential fields seems to be related, but these are defined with an exponential function instead of a commutative operator, and no inverse function. Furthermore, an exponential field defines a single field structure, while $\mathbb{H}_n$ defines $n-1$ field structures on the same object, alongside $n$ group structures. Therefore, exponential fields and hypertypes are quite different.

Many of the ideas outlined above where first proposed by @goblin back in 2013, but only in the context of $\mathbb{R}$, while distributivity was questioned but not established. One of the main benefits of the Hypertype Theory prposed here is to provide a recursive sequence of larger and larger objects that are isomorphic to subsets of $\mathbb{R}$ while providing recursive group and field structures.


5. Can $\pi$ be naturally introduced as a term of $\mathbb{H}_n$?

If my understanding of Schanuel’s conjecture is correct and if the conjecture is true, then $\pi$ is not a term of $\mathbb{H}_n$.


7. Is this interesting?

This is interesting for several reasons:

First, it provides a recursive set of objects $\mathbb{H_n}$ that have very simple definitions yet unique properties. Among them, the existence of $n$ group structures and $n-1$ field structures. And because of the recursive nature of the construction, many properties exhibited at level $n$ become true at any level. This provides a very powerful model to work with.

Second, the construction centered around $e$ introduces this transcendental number in a very natural manner. This is important, because $e$ is one of the few common numbers that is believed not to be a period. Therefore, the combination of Hypertype Theory and Period Theory provides a solid framework for the classification of transcendental numbers.

Third, the adoption of a commutative power law sheds a new light on the notion of power laws and exponential operators, which are traditionally non-commutative. The introduction of a different viewpoint on a fundamental operation might foster innovation or lead to some interesting discoveries.


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