# Are there commutative operations beyond exponentiation?

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?

• 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). – Xander Henderson Oct 3 '17 at 2:25
• on numbers right ? because on matrix multiplication isn't commutative in most cases. – user451844 Oct 3 '17 at 2:26
• "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. – JMoravitz Oct 3 '17 at 2:26
• 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. – Risonow Quelbrij Oct 3 '17 at 2:28
• 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? – JMoravitz Oct 3 '17 at 2:30

## 1 Answer

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.)