Most of us know that multiplication is repeated addition and that exponentiation is repeated multiplication.

You will notice if you begin solving certain problems that addition and multiplication are commutative, but suddenly exponentiation is not commutative. For example:

$2^3 = 8$

$3^2 = 9$

$8$ is not equal to $9$

My question is, are there operations beyond exponentiation that are commutative (I know tetration is the next operation after exponentiation, but I'm pretty sure it's not commutative)? If there are, what are they? If not, do we know why there aren't any?

  • 1
    $\begingroup$ Be careful with the statement that "Multiplication is repeated addition," particularly when you claim that this is something that everybody knows. It simply isn't true. More constructively, you might want to look into a branch of mathematics called "algebra" (abstract algebra, in particular---the study of groups, rings, and fields, among other mathematical doohickies). $\endgroup$
    – Xander Henderson
    Oct 3, 2017 at 2:25
  • $\begingroup$ on numbers right ? because on matrix multiplication isn't commutative in most cases. $\endgroup$
    – user451844
    Oct 3, 2017 at 2:26
  • $\begingroup$ "Are there operations beyond exponentiation" what do you mean by beyond exponentiation here? Surely, one can define a symbol to mean whatever they want, for example $x\spadesuit y:=\min(x,y)\uparrow\uparrow\max(x,y)$ which will of course be commutative. $\endgroup$
    – JMoravitz
    Oct 3, 2017 at 2:26
  • $\begingroup$ Well, tetration would be the next operation beyond exponentiation. Pentation would be the next after tetration. Hopefully this makes more sense if my question was too vague. $\endgroup$ Oct 3, 2017 at 2:28
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    $\begingroup$ But you are specifically wanting to use an operation in the sequence, I'll call it $\odot_n$, where $a\odot_n b$ is repeated application of of $a\odot_{n-1}a$ a total of $b$ number of times and $\odot_1$ is addition? $\endgroup$
    – JMoravitz
    Oct 3, 2017 at 2:30

1 Answer 1


The comments indicate that the question is about commutativity of hyperoperations, as defined in https://en.wikipedia.org/wiki/Hyperoperation.

There the notation is $H_n(a,b)$ instead of $a ⊙_n b$, so I'll use that. The quick answer is that $H_n(3,2)$ is a power of $3$ for higher $n$ than exponentiation (i.e. for $n \geq 3$). Similarly, $H_n(2,3)$ is a power of $2$ and thus $H_n(3,2) \neq H_n(2,3)$ by the fundamental theorem of arithmetic. Thus $H_n$ is not commutative for $n \geq 3$.

While this is clear from how these operations are defined, a more explicit induction argument can be made using the recursive definition given in the wiki. (I can provide one if this isn't sufficiently clear.)

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    $\begingroup$ This is not quite correct answer. Addition and multiplication are commutative because $\Gamma(1)=\Gamma(2)=1$. Hyperoperation of zeroth order (increment by one) is not dependent on the second argument at all because $\Gamma(0)=\infty$. This means that some hyperoperations of negative non-integer order $p$ such that $\Gamma(p)=1$ are commutative. Some operations of imaginary order as well. $\endgroup$
    – Anixx
    Feb 13, 2021 at 16:23
  • $\begingroup$ There was some vagueness to "operations beyond exponentiation" posed in the question. When asked in the comments, the user specifies tetration and pentation as examples. In the above answer, when specifying $n \geq 3$, I am referring to $n$ as a positive integer. The answer is correct in that very specific interpretation of the question. It sounds like it is reasonable to interpret the examples you mention as "beyond exponentiation". Adding your answer (with details) will improve this page. $\endgroup$ Feb 13, 2021 at 18:34

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