Could we define negative hyper operators?

Question

So, I've researched a little bit about this topic, and hyper operators are written $H_n(a, b)$ with :

1. $H_0(a, b) = b+1$ (succession)
2. $H_1(a, b) = a+b$ (addition)
3. $H_2(a, b) = ab$ (multiplication)
4. $H_3(a, b) = a^b$ (exponentiation)
5. $H_4(a, b) = {^b}a$ (tetration)
6. ...

With this logic, is it possible to extend this list into the negative numbers? By taking the inverses of the function. There's only one problem with this idea (actually 2). The opposite of counting up is...counting down. But the opposite of 0 is 0. So that would mean that $H_0(a,b)$ would be both $a+1$ and $a-1$. Same for inverse functions

1. Addition $\rightarrow$ subtraction
2. Multiplication $\rightarrow$ division
3. Exponentiation $\rightarrow$ both roots and logs
4. Tetration $\rightarrow$ both super-roots and super-logs.
   ...

It seems like we have 2 functions for $n≤-2$,and for $n=0$. That can be considered a problem, unless we just allow it, like in the $W$ Lambert function which can give 2 results for certain inputs.
So for $H_0$, to differentiate succession $(b+1)$ and counting down $(b-1)$, I suggest writing $H_{0^+}$ for succession and $H_{0^-}$ for counting down. And to differentiate between roots and logs, I suggest putting a little "$R$" for roots and "$L$" for logs like : $H_{-3_L}$ for logarithm and $H_{-3_R}$ for roots.

So we could logically say that :

1. $H_{0^+}(a, b) = a+1,~H_{0^-}(a, b) = a-1$
2. $H_{-1}(a, b) = a-b$
3. $H_{-2}(a, b) = a/b$
4. $H_{-3_R}(a, b) = \sqrt[b]{a}$
5. $H_{-3_L}(a, b) = \log_b({a}) $
6. $H_{-4_R}(a, b) = \sqrt[b]{a_s}$ (super-root)
7. $H_{-4_L}(a, b) = \text{slog} _b({a})$ (super-log)

Is that reasoning logic and coherent?

Answer 1

Let's see what happens to $H_n(a, b)$ when you increment $b$.

$$H_0(a, b + 1) = a + 1$$ $$H_1(a, b + 1) = a + b + 1 = H_0(H_1(a, b), ?)$$ $$H_2(a, b + 1) = a(b+1) = ab + a = H_1(a, H_2(a, b))$$ $$H_3(a, b + 1) = a^{b+1} = a(a^b) = H_2(a, H_3(a, b)) $$ $$H_4(a, b + 1) = a^{H_4(a, b)} = H_3(a, H_4(a, b))$$

Note the pattern: For $n \ge 2$, we have

$$\boxed{H_n(a, b + 1) = H_{n-1}(a, H_n(a, b))}$$

We can make this recursive formula work for $n = 1$ as well with a slight redefinition of $H_0$:

$$H_0(a, b) := b + 1$$ $$H_1(a, b + 1) = a + b + 1 = H_1(a, b) + 1 = H_0(a, H_1(a, b))$$

Anyhow, let's consider tetration. By repeatedly solving the equation $H_4(a, b + 1) = a^{H_4(a, b)}$ for $H_4(a, b)$ in terms of $H(a, b + 1)$, we get:

$$^0a = 1$$ $$^{-1}a = 0$$

But we run into a snag for $b = -2$, when our recurrence relation gives:

$$0 = a^{H_4(a, -2)}$$

How do you raise a nonzero number (whether real or complex) to a power and get zero? Maybe $^{-2}a = -\infty$? But then what's $^{-3}a$?

Maybe we could introduce a new type of number with different "levels" of infinity, so that tetration can have a proper inverse. Just like we have negative numbers so that addition has an inverse, and rational numbers so that multiplication has an inverse.

Answer 2

I attempted this with the join operation (equivalent to a maxplus operation) such that its repetition becomes addition. See my paper.

Albert A. Bennett also explored this topic in 1915: Bennett, Albert A. "Note on an Operation of the Third Grade." The Annals of Mathematics 17.2 (1915): 74-75.