Fractionally dimensioned numbers

[marraco]

No zero-divisors.

Sometimes I found myself thinking that tetration introduces zero divisors, and that some of the new numbers have a geometrical representation as lines.

In particular, a real number divided by zero should produce a line of infinite length, which is somewhat related to the Dirac delta. For example the Dirac delta is his derivative and crazy ideas like that...

So, the limit for \( \\[15pt]

^{-2}b \) is not just an asymptotic, but a number, and the square waves which are limit to \( \\[15pt]

lim_{x \to \infty}\,^{x}b \) for \( \\[15pt]

0 \leq b \leq e^{-e} \) should be continuous functions.

2) geometric

I was planning to sum along geodesics, and multiply in such a way that I get the complex numbers for dimension r=2

But for that is needed a clearly defined surface for dimensions between 1 and 2.

Tommy's spiral Numbers
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(z1,r1) * (z2,r2) = (z1 z2, r1 + r2).

(z1,r1) + (z2,r2) = (z1 + z2,[r1 + r2]\2).

Should aim to have a matrix representation with a trace equal to the dimension, because of this.

The dimension of a vector space may alternatively be characterized as the trace of the identity operator.

Note that the identity of the product is 1, and if the identity of adding angles is the period \( \\[15pt]

\Omega \pi \) then the identity operator with your definition of product is \( \\[15pt]

(1,\Omega \pi) \). That gives a definition for \( \\[15pt]

\Omega = \frac{r-1}{\pi} \), which look suspiciously like:

\( a= - \left (\frac {c_1 (1-r)}{2 \pi} \right)^2 \)

How this is gonna relate to tetration ...

Maybe this makes clear how to calculate tetration to real exponents, the same way that complex numbers facilitate calculation with polynomials, roots, and other equations.