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Exponentiation of course satisfies a number of nontrivial identities:

  • $x^{y+z}=x^yx^z$

  • $(x^y)^z=x^{yz}$

  • $x^0=1$, $x^1=x$

However, these identities all involve functions other than exponentiation (I'm thinking of $0$ and $1$ as nullary functions, here). My question is what identities hold of exponentiation alone. That is:

What is the equational theory of $(\mathbb{N}, exp)$?

To be clear, I mean "identity" in the strict, universal-algebraic sense: one term equals another term, where each term is built from variables and exponentiation alone. Also, an identity has to hold on all of $\mathbb{N}$: identities which hold only on, say, numbers divisible by $17$ don't count.

A related question:

Is that theory axiomatized by finitely many equations?

Note: A previous version of this question asked whether there were any nontrivial identities at all. This was extremely silly of me, as pointed out almost immediately by Stefan Perko below: $(x^y)^z=(x^z)^y$.

Noah Schweber
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  • Hmm... how do you feel about $x^p \equiv x \bmod p$, for primes $p$? – hardmath Dec 31 '16 at 23:14
  • @hardmath Definitely doesn't count. I'm talking about identities in the strict, universal-algebraic (or logical) sense: a term on the left, a term on the right, and an "$=$" in the middle. That expression would require a function for "$a$ mod $b$", a predicate for "primeness", and the Boolean operation "$\implies$". I've edited to clarify this. – Noah Schweber Dec 31 '16 at 23:15
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    What about $(x^y)^z = (x^z)^y$? – Stefan Perko Dec 31 '16 at 23:18
  • @StefanPerko . . . Wow. Boy is my face red. I've edited to indicate that the original question was trivial, and to focus on the second question (normally I wouldn't do this, but since nobody's wasted effort on an answer, I feel it's ok in this instance). – Noah Schweber Dec 31 '16 at 23:19
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    @NoahSchweber No problem. I was in a similar situation before, multiple times actually. – Stefan Perko Dec 31 '16 at 23:20
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    @StefanPerko Interestingly, that identity "comes from" a lower-level identity, namely the commutativity of multiplication; but going up, I don't see any way to lift this to a nontrivial identity about tetration. I wonder if tetration satisfies any nontrivial equations . . . – Noah Schweber Dec 31 '16 at 23:24
  • @NoahSchweber Of course this question can be generalized. - I don't know whether there is such an identity. But what comes to mind is that $+,\cdot$ and taking powers all fulfill a nice universal property (in the last case it is w.r.t to $\cdot$, there is the connection), though afaik (correct me if im wrong) there is nothing nice like this known for tetration, pentation, etc.. – Stefan Perko Dec 31 '16 at 23:30
  • I guess I was thinking of this question and answer. – Stefan Perko Dec 31 '16 at 23:38
  • @StefanPerko Neat - I was definitely unaware of that. Thanks! – Noah Schweber Dec 31 '16 at 23:38
  • I am wondering if the $\mathbb N$ restriction will stop Euler's formula from taking a role. – Simply Beautiful Art Jan 01 '17 at 00:10
  • @SimpleArt Also, note that any identity which holds in $(\mathbb{C}, exp)$ will hold in $(\mathbb{N}, exp)$ since the latter is a substructure of the former; so if anything, the structure of $(\mathbb{N}, exp)$ is more intricate. – Noah Schweber Jan 01 '17 at 00:27
  • $a^0=1$, $1^a=1$, $0^a=0$ unless $a=0$ come to mind. – Akiva Weinberger Jan 01 '17 at 00:29
  • @AkivaWeinberger None of those are identities in exponentiation alone, since they each use nullary function symbols (=constant symbols) in addition to exponentiation; also, the last one isn't an equation in the strict sense of my question. – Noah Schweber Jan 01 '17 at 00:32
  • @NoahSchweber I most certainly wasn't thinking just about Euler's formula. I was thinking about it's implications about the nature of exponentiation. – Simply Beautiful Art Jan 01 '17 at 00:58
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    @SimpleArt Ah (I've deleted that comment). Then the point about $\mathbb{C}$ vs. $\mathbb{N}$ stands: any equation about exponentiation which is implied by Euler's formula will hold in $\mathbb{N}$. – Noah Schweber Jan 01 '17 at 01:05
  • I'm not sure if this counts, but$$2^n=\sum_{k=0}^n\binom nk$$Similar such exist for arbitrary $a^b$ with $a,b\in\mathbb N$ and generalize nicely for arbitrary complex $b$ with the generalized binomial expansion theorem. – Simply Beautiful Art Jan 01 '17 at 01:26
  • Note about tetration, write exponentiation as $a \diamond b$ then $\diamond$ it is NOT associative and NOT commutative, as $a \diamond (b \diamond c) \ne (a \diamond b) \diamond c$ and $a \diamond b \ne b \diamond a$, while $+$ and $.$ ARE associative and commutative.

    And because exponentiation is NOT associative - tetration is NOT unique defined...

     – johannesvalks Jan 01 '17 at 02:01
  • Interesting question! The first-order theory of exponentiation is not even recursively axiomatizable (we can express $x\cdot y=z$ as $\forall a:(a^x)^y = a^z$ and then $x+y=z$ as $\forall a:a^x\cdot a^y = a^z$, and suddenly Gödel), but that doesn't necessarily tell us anything about its universal fragment. – hmakholm left over Monica Jan 01 '17 at 02:06
  • @SimpleArt Those identities use addition and multiplication, so they don't count. – Noah Schweber Jan 01 '17 at 03:21
  • @johannesvalks Tetration is uniquely defined - it associates to the right. So $2\uparrow 4$ is $2^{(2^{(2^2)})}$. (Note that usually the parentheses are left off - exponentiation without parentheses is understood as associating to the right.) You could define different versions of tetration which associate in different ways, but they would not be tetration. – Noah Schweber Jan 01 '17 at 03:22
  • @HenningMakholm Indeed, the equational theory is in general much simpler - in fact, the equational theory of $(\mathbb{N}, +, \cdot, exp)$ is decidable! – Noah Schweber Jan 01 '17 at 03:25
  • @NoahSchweber I think he means that it does not uniquely define $2\uparrow3.5$. On the other hand, $2^{3.5}$ is uniquely defined since $a^x\cdot a^x=a^{2x}$ – Simply Beautiful Art Jan 01 '17 at 12:42
  • @SimpleArt Well, that's fine - but I never mention tetrating non-natural values in my quesiton. – Noah Schweber Jan 01 '17 at 23:18

1 Answers1

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I learned indirectly that Martin [1] showed that the identity $(x^z)^y = (x^y)^z$ is complete for the standard model ⟨N, ↑⟩ of positive natural numbers with exponentiation. Unfortunately, I don't have access to this article. Could someone confirm this information?

[1] Charles F. Martin. Axiomatic bases for equational theories of natural numbers. Notices of the Am. Math. Soc., 19(7), 778 (1972).

J.-E. Pin
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  • I can't find a copy of the paper by Martin online, but this appears correct - see page 2 of this paper. And, this is totally awesome - +1!. I'm going to hold off on accepting for a bit, to see if I can get a copy of the paper and verify things, but if I can't do that in a reasonable amount of time I'll accept. – Noah Schweber Jan 02 '17 at 19:21
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    Well, I actually found this paper myself online and wrote my colleague Roberto Di Cosmo to ask him about the identity (E6). Roberto confirmed that the identity used by Martin was not (E6) but $(x^z)^y = (x^y)^z$. – J.-E. Pin Jan 02 '17 at 21:43
  • Well, yes - that's the exponentiation-only version of (E6) (I assumed that was what was meant in the paper). – Noah Schweber Jan 02 '17 at 22:03