As I understand it, addition and multiplication are defined on the reals as having identity elements 0 and 1 and being commutative and associative. Multiplication is also distributive over addition.

With exponentiation we lose the properties of commutativity, associativity, and identity but maintain closure over the positive integers.

Fixing the identity issue seemed difficult since there is no clear extension to the sequence $0,1\ldots$, so I just decided to use $2$.

I decided to try to define a new "exponentiation" ($\star_2$) over the positive reals with the following properties:

$\newcommand{\ex}{\star_2} \begin{align*} a\ex b &= b\ex a\\ a\ex (b\ex c) &= (a\ex b)\ex c\\ a\ex(b \star_1 c)&=(a\ex b)\star_1(a\ex c)\\ a\ex2&=a\\ \lim_{x\to x_0}(a\ex x)&=a\ex x_0 \end{align*}$

where $\star_1$ is multiplication.

After some manipulation, I concluded that the only operation satisfying these axioms (commutativity, associativity, distributivity over multiplication, identity element 2, and continuity) was $a\ex b = a^{\log_2 b}$.

I was wondering if there is there some series $\star_n$ that satisfies the following axioms and if it has been studied what it is called.

$ \begin{align*} a \star_0 b &= a + b\\ a\star_n b &= b\star_n a\\ a\star_n (b\star_n c) &= (a\star_n b)\star_n c\\ a\star_n (b \star_{n-1} c)&=(a\star_n b)\star_{n-1}(a\star_n c)\\ a\star_n n &=a\\ \lim_{x\to x_0}(a\star_n x)&=a\star_n x_0 \end{align*}$

  • $\begingroup$ Your fourth axiom is not how an identity works. It would have to read $a\star_2 2=2\star_2 a=a$. $\endgroup$ Commented Jan 22, 2015 at 4:00
  • $\begingroup$ Whoops. I meant $a\star_2 2 = a$. I think that commutativity implies left inverse = right inverse, right? $\endgroup$
    – k_g
    Commented Jan 22, 2015 at 4:26
  • $\begingroup$ Yep, you're right. $\endgroup$ Commented Jan 22, 2015 at 4:47
  • $\begingroup$ I suggest you this "almost-duplicate" question: math.stackexchange.com/questions/383168/… Here you can find some references to Bennet and other who studied this. $\endgroup$
    – MphLee
    Commented May 23, 2015 at 15:14

1 Answer 1


If we define $\exp^{\circ n}(x) = \exp(\exp(...(n\,times)...(\exp(x))$ and $\log^{\circ n}(x) = \log(\log(...(n\,times)...(\log(x))$

then $a [n] b = \exp^{\circ n}(\log^{\circ n}(a) + \log^{\circ n}(b))$

these are called the commutative hyper operations. They interpolate addition and multiplication and satisfy:

$a [n] b = b[n]a$

$a[n+1] (b [n] c) = (a [n+1] b) [n] (a[n+1] c)$

They can be defined recursively through the following algorithm:

$a [0] b = a + b$

$a [n+1] b = \exp(\log(a) [n] \log(b))$

and they do have an identity element, namely $e = \exp^{\circ n}(0)$

There is some literature on these, they're analytic on some whacky domains because $\exp$ and $\log$ have some whacky behaviour when iterated.

  • $\begingroup$ The "whacky-domain-problem" can be smoothed using a different base for logarithm and exponentiation; say $b=\sqrt 2$ Then for a certain range of $x$ one can iterate infinitely often, and also apply the interpolation to fractional iteration heights using the Schröder-mechanism. Another question where I detailed about this is at math.stackexchange.com/a/1272791/1714 $\endgroup$ Commented Apr 28, 2019 at 9:36

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