Now, let's suppose that somebody only knows a number n, and the product operation.

Then he can operate n.n and get n²

he would get a "natural" sequence of numbers n, n², n³.... nᵐ

then, if he wants to solve any equation, he will discover the neutral 1 and the inverses of product, which he may name the "integer" sequence: nˉᵐ, nˉ³, nˉ², nˉ¹, 1, n, n², n³.... nᵐ

Then, he may ask what operation lies before the product, what is addition.

He will have a hard time, because he does not know what we call integer numbers. He only knows powers of n, so he will try to assign

a=1.a

a+a=n.a

a+a+a=n².a

a+a+a+a=n³.a

and of course, he will go nowhere, because we know that this is the answer:

0=nᶫᶰ°.a

a=nᶫᶰ¹.a

a+a=nᶫᶰ².a

a+a+a=nᶫᶰ³.a

^ LNx means ln(x)/ln(n), but I'm using "our" representation for x, not "his" x

We have a similar problem. If we define zeration to satisfy ln(a+b)= ln(a)°ln(b) we find that a₁°a₂°a₃°...°aₓ=a+ln(x)

but because I'm using "our" x instead of the x element of zeration, it should be

-∞=a+ln(-∞)

a=a+ln(f)

a°a=a+ln(f°f)

a°a°a=a+ln(f°f°f)

a°a°a°a=a+ln(f°f°f°f)

where f is the first number known by somebody who his first operation is zeration, and maybe "that" logarithm is not what we know as "our" logarithm.

Then he can operate n.n and get n²

he would get a "natural" sequence of numbers n, n², n³.... nᵐ

then, if he wants to solve any equation, he will discover the neutral 1 and the inverses of product, which he may name the "integer" sequence: nˉᵐ, nˉ³, nˉ², nˉ¹, 1, n, n², n³.... nᵐ

Then, he may ask what operation lies before the product, what is addition.

He will have a hard time, because he does not know what we call integer numbers. He only knows powers of n, so he will try to assign

a=1.a

a+a=n.a

a+a+a=n².a

a+a+a+a=n³.a

and of course, he will go nowhere, because we know that this is the answer:

0=nᶫᶰ°.a

a=nᶫᶰ¹.a

a+a=nᶫᶰ².a

a+a+a=nᶫᶰ³.a

^ LNx means ln(x)/ln(n), but I'm using "our" representation for x, not "his" x

We have a similar problem. If we define zeration to satisfy ln(a+b)= ln(a)°ln(b) we find that a₁°a₂°a₃°...°aₓ=a+ln(x)

but because I'm using "our" x instead of the x element of zeration, it should be

-∞=a+ln(-∞)

a=a+ln(f)

a°a=a+ln(f°f)

a°a°a=a+ln(f°f°f)

a°a°a°a=a+ln(f°f°f°f)

where f is the first number known by somebody who his first operation is zeration, and maybe "that" logarithm is not what we know as "our" logarithm.