5

I mean...

  • multiplication is the repetition of addition:
    • $2*2 = 2+2$
    • $3*3 = 3+3+3$
  • exponential is the repetition of multiplication:
    • $2^2 = 2 * 2$
    • $3^3 = 3*3*3$

.. it is an obvious pattern.

I propose:

  • Addition is rank 1
  • Multiplication is rank 2
  • Exponential is rank 3
  • etc ...

This would mean

$H(r, x)$ is the $r$th rank operation on $x$.

For example: $H(3, 2) = 2^2$

and $H(4, 3) = {3^{3^3}}$

It is already kind of wired to get this value since it is quite big. But the function of the rank is certainly one that rises fast.

Does anyone know more about this kind of stuff?

Christian
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3 Answers3

10

you might check out : https://en.wikipedia.org/wiki/Hyperoperation it talks about different operations to extend this chain of operations.

5

Note also that you can go even in the opposite direction lowering what you called "rank", in a log-like way. For instance, just like addition is one rank lower than multiplication, you can devise an operation that is one rank lower than addition: the tropical addition, as Vladimir I. Arnold called it.

trying
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  • @Simply Beautiful Art I'm sorry to hear that from you. I hope you can change idea if you will have any chance to consult "Dynamics, Statistics and Projective Geometry of Galois Fields" by Arnold. – trying Jul 27 '17 at 16:25
  • You know, I'm not even sure on how far your negative ranks go. – Simply Beautiful Art Jul 27 '17 at 16:26
  • @Simply Beautiful Art Now I can fully appreciate your reasons. But those come from the way hyperoperations are defined, whereby what precedes the addition "degenerates" to a unary operation. But I was talking about tropical operations: with them the tropical operation (immediately) preceding the addition is still binary. Over this new lower operation the addition is distributive, as the multiplication is distributive over the addition. This is the abstract definition, which more than one concrete operation may satisfy. One of this is $xy=\max{(x, y)}$. You can see that $(xy)+z=(x+z)*(y+z)$ – trying Jul 27 '17 at 17:25
  • Hm, nifty $~{}~$ – Simply Beautiful Art Jul 27 '17 at 17:26
  • Do either of you have links to further references on this? I can't find anything good, and I just finished writing this and figured I was way off...! – Trevor Dec 08 '19 at 03:51
4

There is also the notation $a^{n*} $ for $\underbrace {a*\cdots*a}_{\text {$n $ factors} }$. Using this notation we have $$na=a^{n+} $$ $$a^n=a^{n\cdot} $$

md2perpe
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