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I asked a question about operations and one comment puzzled me.

Given a binary operation $\ast$ on integers at least $2$, define $\ast'$ by $$m\ast' n = \overbrace{m\ast m\ast \cdots \ast m}^{n\text{ times}}.$$

Example :

  • if $*$ is $+$ , $*'$ is $×$. Multiplications are a lot of additions.

  • if $*$ is $×$ , $*'$ is $^$. Exponential are a lot of additions.

The question is what's $*$ when $*'$ is $+$ ?

Additions are a lot of... what?

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Successor operations. But the successor operation is a unary operation. – Daniel Fischer Sep 14 '13 at 19:29
So in a sense "successor operatioans" is right, but it doesn't answer the question since it's a unary operation. – Michael Hardy Sep 14 '13 at 20:09
@MichaelHardy Can't the successor function be made into a binary function very easily? $$f(a,b)=a+1$$ – Zubin Mukerjee Jan 7 '15 at 0:24
@ZubinMukerjee : I think Daniel Fischer was saying that multiplication is in a sense the "successor" of addition; he wasn't talking about the successor function that adds $1$ to a number. – Michael Hardy Jan 7 '15 at 19:05
Okay, thanks! $\,$ – Zubin Mukerjee Jan 8 '15 at 5:38
up vote 2 down vote accepted

You can define operation $*_1 = +$, $$x *_{n+1} y = e^{\log x *_n \log y},$$ and $$x *_{n-1} y = \log\left(e^x *_n e^y\right).$$ Then we have $*_1 = +$, $*_2 = \times$, and $x *_3 y = x^{\log y}$ which isn't quite what you're looking for but is close. We can continue forever in both directions. (Be careful about the domains of these operations.)

In particular, the operation you're interested in, the predecessor of $+$, is $*_0$, defined as: $$x *_0 y = \log\left(e^x + e^y\right).$$ I'm not aware of a name for this operation but you can feel free to explore it.

As an aside, note that all of these operations are commutative. Specifically, $*_3$ is like a commutative version of exponentiation.

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Great answer, +1. I did some searching and found that these are called "commutative hyperoperations". – Zubin Mukerjee Jan 6 '15 at 17:38
@ZubinMukerjee Very interesting - I knew these operations must have been studied before but didn't know where to begin looking for a reference. – Solomonoff's Secret Jan 6 '15 at 17:40
Is this correct? $$x *_{-1} y=\log\log\left(e^{e^x}+e^{e^y}\right)$$ In any case, I feel as though there are some ways your $*_3$ makes more sense than exponentiation ... – Zubin Mukerjee Jan 6 '15 at 17:48
That equation looks right to me. $*_3$ is in a sense more natural than exponentiation, although exponentiation has an extremely rich set of properties. I wonder what properties $*_3$ has - perhaps there is an opportunity for a good question for this site about it. – Solomonoff's Secret Jan 6 '15 at 17:54
@Solomonoff'sSecret If you want to do it, do it ! It's your idea. Thank you for this very interesting answer, it gives me more hint about the possible answers. – Pyrofoux Jan 9 '15 at 16:03

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