Joel David Hamkins on Gödel's Incompleteness, Set-Theoretic Multiverse & Foundations of Mathematics

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  • เผยแพร่เมื่อ 22 พ.ย. 2024

ความคิดเห็น • 11

  • @GuillermoPSKrebs
    @GuillermoPSKrebs 5 วันที่ผ่านมา

    Awesome!

    • @RahulSam
      @RahulSam  4 วันที่ผ่านมา

      Cheers!

  • @StephenPaulKing
    @StephenPaulKing 10 หลายเดือนก่อน

    I am very interested in opinions on the Axiom of anti-foundation and the work of Jon Barwise. Can we imbed ZFC in some version of non-well founded sets?

    • @RahulSam
      @RahulSam  10 หลายเดือนก่อน

      Interesting... I'm not familiar with the work of Jon Barwise. Any good reading material you recommend?

  • @BreezeTalk
    @BreezeTalk 10 หลายเดือนก่อน

    Mhm okay I’ll give it a chance.

    • @RahulSam
      @RahulSam  10 หลายเดือนก่อน +1

      Please do 😉

  • @BuleriaChk
    @BuleriaChk 7 หลายเดือนก่อน

    Godel expresses wff's in odd numbers
    every number is prime relative to its own base n = n(n/n)=n(1_n) (primes do not include division by other numbers)
    Goldbach's Conjecture "every even number is the sum of two primes" n + n = 2n
    Godel's expression does not include even numbers in his defintion of wff's - they are therefore "undecidable"
    (o + e) = o is always odd so is undecidable because of the existence of even numbers (e+e) = e
    (o and e are sets of numbers).
    Note that the product of differing powers of prime numbers is zero, since the graphs of x^m and x^n only intersect at x = 0 so (x^n)(x^m) = 0. (the reason powers form a basis in polynomial space).
    Proof of Fermat"s Theorem for Village Idiots
    c = a + b
    c^n = [a^n + b^n] + f(a,b,n) (Binomial Expansion)
    c^n = a^n + b^n iff f(a,b,n) = 0
    f(a,b,n) 0
    c^n a^n + b^n QED
    Pythgoras is wrong, Fermat is correct even for n = 2. Someone go tell the physicists (Especially Einstein and Pauli)
    and also for multinomials (tell the cosmetologists..)
    (Hint: Wiles had to use modular functions, which are only defined on the positive half of the complex plane.)
    there are no negative numbers: -c= a-b, b>a iff b-c=a, a >0, a-a = 0, a=a
    if there are no negative numbers, there are no square roots of negative numbers. The ""complex" plane is affine to the real plane (1^2 1, sqr(1^2) = 1 2qr(1) (Russsell's Paradox; a number can't both multiply and not multiply itself).
    more on this on the physicsdiscussionforum (dot org)

    • @RahulSam
      @RahulSam  7 หลายเดือนก่อน

      I will have to read this comment more carefully. Thanks. Bookmarked!

    • @BuleriaChk
      @BuleriaChk 7 หลายเดือนก่อน

      @@RahulSam If you want to discuss it further, let me know. I'll buy the beer and pizza!

    • @98danielray
      @98danielray 6 หลายเดือนก่อน +2

      ​@@RahulSamyou cant even distinguish this as crankyness?