2
$\begingroup$

This question already has an answer here:

The hyperoperation heirarchy in the naturals starts with addition, then multiplication, then exponentiation, then tetration, and so on. Each operation is defined as repeated application of the previous in the sequence. Here, we are defining 0^0, 0^^0, etc, x^0, x^^0, etc as 1. In a discussion on some online forum, someone was remarking that exponentiation satisfies the identity (x^y)^z=(x^z)^y but tetration does not. And of course, exponentiation fails to satisfy the commutative law. He wondered if this process keeps on happening as you ascend the hyperoperation heirarchy. So, what I would like to know is if the sequence of sets of identities of the hyperoperation heirarchy starting from multiplication, is strictly descending. By identities I mean universally quantified equations without constants. If anyone can link me to a paper on this or a similar topic, I would be immensely delighted.

$\endgroup$

marked as duplicate by Simply Beautiful Art, Namaste logic Aug 4 '18 at 21:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Are there even any non-trivial identities for tetration and above? And are there any other identities for exponentiation? $\endgroup$ – Ibrahim Tencer Feb 28 '15 at 8:20
  • $\begingroup$ It seems impossible to deduce any particular identity from the same one at the next level up. Maybe if we know more about the structure of the identities. For one thing, the leftmost variable in either side of an identity for $H_n$, $n \ge 3$ must be the same. I think a more promising strategy is to look at how the identities are proven. For example, the proof of $(x^y)^z = (x^y)^z$ which uses induction on $z$ also goes through for the same identity in $\times$. However, you can also prove it using $(x^y)^z = x^{yz}$, but clearly $(xy)z \ne x(y+z)$. $\endgroup$ – Ibrahim Tencer Feb 28 '15 at 9:56
  • $\begingroup$ Sorry, that should say $(x^y)^z = (x^z)^y$. $\endgroup$ – Ibrahim Tencer Feb 28 '15 at 17:25
  • $\begingroup$ Information about this seems scant - you should cross-post to MO. $\endgroup$ – Ibrahim Tencer Mar 18 '15 at 9:43
  • $\begingroup$ How to cross-post to MO? $\endgroup$ – user107952 Mar 22 '15 at 23:10
1
$\begingroup$

It turns out this question (or something very similar to it) has been resolved. See this paper for details. It states that "[Martin, 1972] showed that there are no nontrivial equations for $(\mathbb{N}, Ack(n,-, -))$ if $n>2$ (p2). The Ackermann function is defined a little differently from the hyper operators, though.

And "he showed that the identity (E6) [$a^{b × c} = (a^b)^c$] is complete for the standard model $(\mathbb{N}, ↑)$ of positive natural numbers with exponentiation." I think this implies that the identity mentioned in your other question ($(a^b)^c = (a^c)^b$) is also complete, which means the identities for $\wedge$ are a subset of the identities for $\times$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.