Zeration

[MphLee]

a*b = b*a

"add+b"^[a-1](1*b) = "add+a"^[b-1](1*a)

That would mean
b*(a-1)=a*(b-1)

but if "-" is an inverse operator of [q], such that x-y is defined as x-y=x[q]-y, and -y is defined as y[q]-y=N(q,y) then

for [q]=product, "-" would be division, so

Ure misunderstanding the expression \( f^{n}(x) \)

In fact Tommy tryes to replace g and f with "add_b" and "add_1" and the interpretation becomes the following

General case \( f^{a - N_q}(N_q[q]b) = g^{b - N_q}(N_q[q]a) \)

- Case q=1 with \( [1]=+ \), \( f(x)=g(x)={\rm add}_1(x)=x+1 \) and \( N_1=0 \)

\( {\rm add}_1^{a - N_1}(N_1[1]b) = {\rm add}_1^{b - N_1}(N_1[1]a) \)
\( {\rm add}_1^{a - 0}(0+b) = {\rm add}_1^{b -0}(0+a) \)
\( {\rm add}_1^{a }(b)=a+b =b+a= {\rm add}_1^{b }(a) \)


- Case q=2 with \( [2]=\cdot \), \( f(x)={\rm add}_b(x)=b+x \), \( g(x)={\rm add}_a(x)=a+x \) and \( N_2=1 \)

\( {\rm add}_b^{a - N_2}(N_1[2]b) = {\rm add}_a^{b - N_2}(N_2[2]a) \)
\( {\rm add}_b^{a - 1}(1 \cdot b) = {\rm add}_1^{b -1}(1\cdot a) \)
\( (b+...+b)_{a-1 times}+(b)=(a+...+a)_{b-1 times}+(a) \)
\( b(a-1)+(b)=a(b-1)+(a) \)
\( (ba-b)+(b)=ba=ab=(ab-a)+(a) \)

- Case q=0 with \( [0]=\circ \) and \( N_0=-\infty \), but in this case how we get the following

\( f^{a -(-\infty)}((-\infty)[0]b) = g^{b - (-\infty)}((-\infty)[0]a) \)

\( f^{+\infty}(b) = g^{+\infty}(a) \)

-∞ is used on all the basic operations:

the neutral of addition is ÷∞=1/∞=0
the neutral of product is ∞√=∞√n=n^÷∞=1
the neutral of exponentiation is \( ^{-\infty}n=n^{\frac{1}{n}} \)

from product viewpoint, all numbers smaller than 0 are transfinite.
from exponentiation of n viewpoint, all numbers smaller than 1 are transfinite.
from tetration of n viewpoint, all numbers smaller than \( n^{\frac{1}{n}} \) are transfinite.

Agree with this last point...
It really deserves some attention imho.