The set $\mathbb{N}$ can be viewed as a mathematical structure with operations off addition, multiplication and exponentiation. Observe that:

  1. It forms an Abelian monoid under both addition and multiplication.
  2. Multiplication distributes over addition.
  3. $1^x = x^{0} = 1$
  4. $x^{a+b}=x^{a}x^{b}$
  5. $x^{ab} = (x^a)^b$

Furthermore, the set $[0,\infty)$ can also be viewed in this way. Are there other interesting examples of this sort of thing? I am especially looking for examples lacking a natural total order.

Remark. A few more examples occur to me. However, they're both naturally ordered.

Firstly, for every strong limit cardinal $\kappa$, I think that the set $\{\nu < \kappa\}$ is an example of such a structure.

Secondly, if we drop the requirements that addition and multiplication be commutative, and require distributivity only on the left, as in $$x(a+b)=xa+xb,$$ then every non-trivial ordinal that is closed under exponentiation is an example of such a structure.


The integers modulo $n, \Bbb Z/n\Bbb Z$ are an example. One could argue whether they have a natural total order because $-1\equiv n-1 \pmod n$, but "natural" is a matter of taste.

  • $\begingroup$ Sorry to revive this so late, but I don't think the integers modulo $n$ constitute a valid example for all $n$. For example, $2^2$ is not congruent to $2^5$ modulo $3$ despite $2$ being congruent to $5$, so one can't consistently define exponentiation in $\mathbb{Z}/3\mathbb{Z}$ without breaking rule 4 above. $\endgroup$ – pregunton Dec 22 '17 at 10:09

There is such a thing called an Exponential Field, though the exponential here is more indicative of the unary function rather than a binary one, however. This is essentially precisely what you describe if we replace monoid with group. Since $\mathbb{N}$ and $[0,\infty)$ are semirings, I would guess the analogous term would be Exponential Semiring.

The complex numbers form such an exponential field, and they lack a total order.

Going with the Exponential semiring idea, I believe that you could take the semiring of $n\times n$ matrices with positive coefficients and use the exponential operator. This doesn't have the property that multiplication commutes or that $e^{A+B}=e^Ae^B$, however, since this requires the the matrices commute. This does not satisfy (5) either.

To correct this, we can take the semiring of all $n\times n$ diagonal matrices. In this case, we have commutative multiplication and addition, and $e^{A}$ is another diagonal matrix. This satisfies (5) when we examine the exponentiation as a power of a matrix rather than exponential, meaning that in your $x^{ab}=(x^a)^b$, we regard $b$ as a scalar.

To finally finish constructing an example, we can make the semiring of all $n\times n$ diagonal matrices with non-negative diagonal entries. We define an exponential operator that is binary by defining $\log(I+M)=\sum_{n\geq 1}{\frac{(-1)^{n-1}M^n}{n}}$ a la the Mercator series and defining $M^N=\exp(N\log M)$. This satisfies the properties $I^M=I=M^0$ and $M^{N+N'}=M^NM^{N'}$. To see that this satisfies (5), we notice that both the exponential and logarithm operation applied to a diagonal matrix are simply applications of the $\exp$ and $\log$ functions on the reals applied component-wise to the diagonal entries of the diagonal matrices. This is why we must take non-negative diagonal matrices (and positive diagonal matrices for the logarithm). When we do this, we find that it naturally fulfills (5). There is a slight-issue here though: this has a total ordering on it by giving it the dictionary ordering.

Perhaps taking complex diagonal matrices and taking principal values might give you what you're looking for, but there is a total order given on this as well by using the dictionary ordering as a total order on the complex numbers (with the natural ordering on $\mathbb{R}$) and then taking the dictionary ordering on the diagonal matrices.

Theoretically, since every set can be totally ordered, without making additional stipulations regarding what properties the order needs, not much can be said about exponential semirings with or without a total order. If we want that order to be compatible with the monoid operations, then you have a chance to make the object you want.

I'll update with examples as I think of them.

  • $\begingroup$ I added to the answer to address that. $\endgroup$ – Hayden Dec 9 '13 at 3:13
  • $\begingroup$ Why doesn't it satisfy (5)? The power of a diagonal matrix is just the diagonal of the powers of the elements. $\endgroup$ – Ross Millikan Dec 9 '13 at 3:29
  • $\begingroup$ Good point, you're quite right. It's a different kind of power, but it is certainly true. Nice observation. $\endgroup$ – Hayden Dec 9 '13 at 3:33
  • $\begingroup$ Cool example with the matrix log / matrix exponential. Any ideas how to understand these operations intuitively, or geometrically? I don't really understand what they "are" at the moment, if that makes sense. Edit. Perhaps I will ask another question to that end. $\endgroup$ – goblin Dec 9 '13 at 3:43
  • $\begingroup$ I can't really think of the geometric intuition either. When dealing only with diagonal matrices, the operations are almost trivial, so they gain all the geometric intuition that would be expected from their single variable variants, since we essentially just apply those functions component-wise on the diagonal entries. $\endgroup$ – Hayden Dec 9 '13 at 3:47

There's study of the system you describe under the name of Tarski's high school algebra problem. Tarski wrote down a system of 11 identities that correspond to your axioms, and asked if there were any identities that were true over the non-negative integers that couldn't be derived from those identities. The answer turns out to be yes -- there are true identities that cannot be proven. The link gives an explicit example.

  • $\begingroup$ Thanks for the link. Technically the system I listed does not include $(xy)^a = x^a y^a$, although mostly just because I forgot to include it... $\endgroup$ – goblin Dec 9 '13 at 23:44
  • $\begingroup$ I could tell by your body language that you had it in mind. $\endgroup$ – arsmath Dec 10 '13 at 0:35

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