03/23/2015, 02:31 PM

(03/23/2015, 01:39 PM)marraco Wrote:Ure misunderstanding the expression(03/21/2015, 11:11 PM)tommy1729 Wrote: a*b = b*aThat would mean

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

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

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

Quote:General case

- Case q=1 with , and

- Case q=2 with , , and

- Case q=0 with and , but in this case how we get the following

Quote:-∞ is used on all the basic operations:Agree with this last point...

the neutral of addition is ÷∞=1/∞=0

the neutral of product is ∞√=∞√n=n^÷∞=1

the neutral of exponentiation is

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 are transfinite.

It really deserves some attention imho.

MSE MphLee

Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)