The standard Peano axioms defining the natural numbers stipulate a single operation called the "successor". This is commonly written S(n), which indicates the next natural number after n. Later on, addition is defined in terms of repeated successor operations, and so forth.

The traditional definition of zeration, per Goodstein, is: \(H_0(a, b) = b + 1\). Now when I first saw this, I was surprised and taken aback. All the other operations start with \(a\) as a "base", and then effectively apply some simpler operation \(b\) times, so it seems odd to start with the \(b\) and just add one to it. (If anything my expectation would have been to take \(a+1\), but that doesn't satisfy the regular recursive definition of \(H_n\) when you try to construct addition.)

As it turns out, when you get to this basic level, you're doomed to lose many of the regular properties of the operations hierarchy. So there's nothing to do but start arguing about which properties to prioritize as "most fundamental" when constructing the definition.

Here are some points in

**favor**of the standard definition \(b+1\): (1) It does satisfy the recursive formula that repeated applications are equivalent to addition (\(H_1\)). (2) It does looking passingly like counting by 1, i.e., the Peano "successor" operation. (3) It shares the key identity that \(H_n(a, 0) = 1\), for all \(n \ge 3\). (4) Since it is an elementary operation (addition, really), it can be extended from natural numbers to all real and complex numbers in a fashion which is analytic (infinitely differentiable).

But here are some points

**against**the standard definition (1) It is not "really" a binary operator like the rest of the hierarchy, in that it totally ignores the first parameter \(a\). (2) Because of its ignoring \(a\), it's not commutative like the other low-level operations n = 1 or 2 (yet like them it is still associative and distributive, or as I sometimes say, collective of the next higher operation). (3) For the same reason, it has no identity element (no way to recover the value \(a\), unique among the entire hyperoperations hierarchy). (4) It's the only hyperoperation which doesn't need a special base case for when \(b = 0\). (5) I might turn around favorable point #3 above and call it weird and unfavorable, in that it is misaligned in this way with operations n = 1 and 2, and it's the only case of one of the key identities being

*added*at a lower level instead of being lost. See how weird that looks below?

So as a result, a variety of alternative definitions have been put forward. I think my favorite is \(H_0(a, b) = max(a, b) + 1\). Again, this looks a lot like counting; I might possibly explain it to a young student as "count one more than the largest number you've seen before". Points in

**favor**: (1) Repeated applications are again the same as addition. (2) It is truly a binary operation. (3) It is commutative, and thus completes the trifecta of commutativity, association, and distribution/collection being true for all operations \(n < 3\). (4) It does have an identity element, in \(b = 0\). (5) It maintains the pattern of

*losing*more of the high-level identities, and in fact perfects the situation in that

*none*of the five identities hold for this zeration (all "no's" in the modified table above for \(n = 0\)). Points

**against**: (1) It isn't exactly the same as the unary Peano successor function. (2) It's non-differentiable, and therefore cannot be extended to an analytic function over the fields of real or complex numbers.

There are vocal proponents of related possible re-definition: \(H_0(a, b) = max(a, b) + 1\) if a ≠ b, \(a + 2\) if a = b. Advantage here is that it matches some identities in other operations, like \(H_n(a, a) = H_{n+1}(a, 2)\) and \(H_n(2, 2) = 4\), but I'm less impressed by specific magic numbers like that (as compared to having commutativity and the pattern of actually

*losing more identities*). Disadvantage is obviously that the possibility of adding 2 in the \(a+2\) case gets us even further away from the simple Peano successor function.

And then some people want to establish commutativity so badly that they assert this: \(H_0(a, b) = ln(e^a + e^b)\). That does get you commutativity, but at that point we're so far away from simple counting in natural numbers that I don't even want to think about it.

Final thought: While most people interpret the standard definition of zeration, \(H_0(a, b) = b + 1\) as "counting 1 more place from b", it makes more sense to my brain to turn that around and say that we are

*"counting b places from 1"*. That is, ignoring the \(a\) parameter, start at the number 1 and apply the successor function repeatedly b times: \(S(S(S(...S(1))))\), with the \(S\) function appearing \(b\) times. This feels more like "basic" Peano counting, it maintains the sense of \(b\) being the number of times some simpler operation is applied, and it avoids defining zeration in terms of the higher operation of addition. And then you also need to stipulate a special base case for \(b = 0\), like all the other hyperoperations, namely \(H_0(a, 0) = 1\).

So maybe the standard definition is the best we can do, and the closest expression of what Peano successor'ing in natural numbers (counting) really indicates. Perhaps we can't really have a "true" binary operator at level \(H_0\), at a point when we haven't even discovered what the number "2" is yet.

P.S. Can we consider defining an operation one level even lower, perhaps \(H_{-1}(a, b) = 1\) which ignores

*both*parameters, just returns the natural number 1, and loses every single one of the regular properties of hyperoperations (including recursivity in the next one up)?

If you change the base in the last version, to log₂(2ᵃ+2ᵇ), then it looks pretty interesting, actually. The result is between max(a,b) and max(a,b)+1, reaching the maximum when a=b.

ReplyDeleteIndeed, interesting!

DeleteThe floor of that plus one is the second to last definition.

ReplyDeleteIs safe to define H_(-1)=H_(0)?

ReplyDeleteAbout your PS, if you define H_(-1)(a, b) = 1, you break the recursive rule because you have to had H_0(a,b) = H_(-1)(a, H_0(a, n-1)), if H_(-1) gives 1 every time you have to conclude that H_0(a, b) = 1, so you can prove that H_n(a, b) = 1 for every n > 0 in the integers

ReplyDeleteAgreed, that's what I meant by "loses every single one of the regular properties of hyperoperations (including recursivity in the next one up)".

Delete