M Theory Lesson 55

As Matti Pitkanen has commented, a major motivation for such an approach to elementary axioms for n-logoses is the possibility of easy number construction. In the 3-logos case, 3-adic numbers arise, much as the dyadic numbers appear on finite segments of the surreal tree for binary logic.

It follows that ordinary real numbers are difficult to understand, appearing both as infinite sequences in the dyadic case, and in the $\omega$-logos setting at the infinite prime. Models of the axioms that use real and complex geometry must therefore respect the p-adic heirarchy. In the case of the Jordan algebra M Theory, for instance, this is at least partly achieved via triality for the prime 3.

3 Responses so far »

  1. 1

    Anonymous said,

    A “(sur)real tree” of Finnish homophonic logic, representing both aspects (aka “too times tree”) of the string “kuusi”:

    A
    A A
    A A A
    6

    And when Yule Goat leaves under the t(h)ree his 2^2 presents (Timeo danaos et dona ferentes!) Pythagoreans can feast their Tetractys. Morfeus, God of Dreams, with his magic wand creates a Tetrahedron, the element of Fire, which Heracleitus then doubles and combines into Logos of a Tetractys and 2*Tetrahedrons (pointing to opposite directions), creating a non-platonic hexaedron, which can only be called the Cube of Heracleitus, with sum of 030. Division of the surreal cube by the (sur)real tree “kuusi” reduces the Heraclitian cube into 5 or “…1,3,1…” points, the next prime.

  2. 2

    Kea said,

    Thanks, anonymous! Another really cool person visits the blog. Hmmm … I’m going to have to think about this … I’m really very slow, you know.

  3. 3

    Anonymous said,

    The entropic software left perhaps too much for i(matinative) cognition to i(interprete), here is “kuusi” again:

    __A__
    _A A_
    A A A
    __6__

    Hermeneutic phenomenology can also immediately see that this homophonic A(rcadian functor) “kuusi-tree” plus “kuusi-6” makes 7.

    So both Tetraclys minus 7, and Platonic cube of 2^3 minus Heraclitian cube of 5 produce t(h)ree. 🙂


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