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hyperoperation below addition

Anonymous Sun 07 Mar 03:31:21 2021 No.12800466 View Reply Original Report

Quoted By: >>12800468 >>12801124 >>12801133 >>12802929 >>12803594

Is there any mathematical operation that would satisfy the bottom equation if placed in the red square?

Anonymous

Anonymous Sun 07 Mar 2021 03:32:51 No.12800468 Report

Quoted By:

>>12800466
you can make one up if you want. You can even define axioms for it.

El Arcón

El Arcón Sun 07 Mar 2021 03:32:54 No.12800469 Report

Quoted By: >>12802949

+(2/3)*

Anonymous

Anonymous Sun 07 Mar 2021 07:43:01 No.12801124 Report

Quoted By:

>>12800466
3+2 = 3+3 (mod 1)

Anonymous

Anonymous Sun 07 Mar 2021 07:47:52 No.12801133 Report

Quoted By: >>12801426 >>12801479 >>12802763

>>12800466
successor
S(S(2))=2+2

Anonymous

Anonymous Sun 07 Mar 2021 10:06:37 No.12801426 Report

Quoted By: >>12802763

>>12801133
This, but
$S \circ^2 (3)$
so the 2 can be abstracted away. Succession is fundamental in ZF.

Anonymous

Anonymous Sun 07 Mar 2021 10:24:21 No.12801479 Report

Quoted By: >>12802763

>>12801133
3 successor 3 = 3+2 right

Anonymous

Anonymous Sun 07 Mar 2021 18:30:38 No.12802763 Report

Quoted By:

>>12801133
>>12801426
>>12801479
++ doesn't fit n@n, it's not a binary operation

Anonymous

Anonymous Sun 07 Mar 2021 19:01:50 No.12802929 Report

Quoted By:

>>12800466
By looking at strings of powers (addition) you can almost fully derive the operation except when the operands differ by 1: 0 and 1, 1 and 2, 2 and 3,...
If you start indexing at 0 instead, so $3+2= 3 \boxdot 3 \boxdot 3$ you get $a \boxdot b = max(a,b)+1$ .

$3 \boxdot 3 = max(3,3)+1 = 4 = 3+1$
$3 \boxdot 3 \boxdot 3 = 4 \boxdot 3 = max(4,3)+1=5 = 3 + 2$

Not sure but I think filling in the undefined bits of your operation with this one makes everything work.

Anonymous

Anonymous Sun 07 Mar 2021 19:06:01 No.12802949 Report

Quoted By:

>>12800469
Schizo bro isn't this a little too simple for your taste?

Anonymous

Anonymous Sun 07 Mar 2021 19:18:09 No.12803020 Report

Quoted By: >>12803503

The general relation you're seeking is
$\begin{align} n^{n-1} &= \underbrace{n\times n\times ...\times n}_{n-1 \text{ times}} \\ n\times (n-1) &= \underbrace{n + n + ... + n}_{n-1 \text{ times}} \\ n + (n - 1) &\stackrel{?}{=} \underbrace{n \square n \square ... \square n}_{n-1 \text{ times}} \end{align} $
So the goal here is to construct an operator $\square$ such that, given a number $n$ it returns the number $2n - 1$ via $n -1$ repeated iterations.
It can certainly be done.

Anonymous

Anonymous Sun 07 Mar 2021 21:14:24 No.12803503 Report

Quoted By: >>12803515

>>12803020
I don't think you can generalize a binary operation over the natural numbers to which addition is a hyperoperation. All you have left after you "take the derivative" of addition is a big bag of ones. a@b | a>1 V b>1 isn't even in the domain.

Anonymous

Anonymous Sun 07 Mar 2021 21:18:11 No.12803515 Report

Quoted By: >>12803586

>>12803503
>>12803503
I'm not sure. It seems to have some structure. Consider for instance
$\begin{align} 3\square 3 &= 5 \\ 4\square 4\square 4 &= 7 := 4\square^2 4\\ 5\square^3 &=9 \\ 6\square^4 &= 11 \end{align} $

Anonymous

Anonymous Sun 07 Mar 2021 21:34:57 No.12803586 Report

Quoted By:

>>12803515
I think it's a local solution to the 2n-1 case. I agree that you can solve the OP image. But I don't think you can define a binary operator that works as a generalized "hypo-operation" to addition.

Anonymous

Anonymous Sun 07 Mar 2021 21:36:31 No.12803594 Report

Quoted By: >>12803601 >>12803624

>>12800466
what is above exponent?

Anonymous

Anonymous Sun 07 Mar 2021 21:39:10 No.12803601 Report

Quoted By:

>>12803594
A number of repeated exponents.
3@5 = 3^3^3^3^3, 5@3 = 5^5^5, etc.

Anonymous

Anonymous Sun 07 Mar 2021 21:45:41 No.12803624 Report

Quoted By:

>>12803594
https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

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