Posers and Puzzles

Posers and Puzzles

  1. DonationAcolyte
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    02 Feb '04 11:06
    Prove the existence of transcendental numbers (ie, numbers which are not roots of polynomials with integer coefficients).
  2. Melbourne
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    02 Feb '04 14:59
    42
  3. Standard memberroyalchicken
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    02 Feb '04 19:35
    Fine then 😛. I'll try to post an erroneous proof later and make you all find the problem with it...
  4. Standard memberTheMaster37
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    02 Feb '04 20:581 edit
    Pi is transcendental, but the proof for it is hard 🙂

    EDIT:

    A non-transcedental number, eg algebraic number, is the root of a polynomial of finite order with integer (positive and negative whole numbers) coefficients, right?

    If that is so then;

    given a certain order O, there is a countable number of polynomials of that order with integer coefficients (integers are countable, and for a polynomial of order O you require O+1 integers).

    The order is an element of the natural numbers (positive integers) so there is only a countable set of polynomials of finite order with integer coefficients.

    This means there is only a countable set of algebraic numbers.

    The reals form an over-countable set of numbers, and therefore some reals must be transcedental.
  5. Donation!~TONY~!
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    02 Feb '04 21:55
    e is also included! WHOOO! No way in hell I am provin' it though! 😀
  6. DonationAcolyte
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    02 Feb '04 23:33
    Originally posted by TheMaster37
    Pi is transcendental, but the proof for it is hard 🙂

    EDIT:

    A non-transcedental number, eg algebraic number, is the root of a polynomial of finite order with integer (positive and negative whole numbers) coefficients, right?

    If that is so then;

    given a certain order O, there is a countable number of polynomials of that order with integer co ...[text shortened]... The reals form an over-countable set of numbers, and therefore some reals must be transcedental.
    That works. However, can you prove it without showing or assuming that R is uncountable?
  7. Standard memberroyalchicken
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    02 Feb '04 23:41
    Originally posted by Acolyte
    That works. However, can you prove it without showing or assuming that R is uncountable?
    Why don't you issue a formal challenge, from the Cult of Maths to those mathematical usurpers, the RUNning Pawn clan, asking them to show e is transcendental? We could taunt them mercilessly 😉.
  8. Standard memberTheMaster37
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    03 Feb '04 08:37
    Originally posted by royalchicken
    Why don't you issue a formal challenge, from the Cult of Maths to those mathematical usurpers, the RUNning Pawn clan, asking them to show e is transcendental? We could taunt them mercilessly 😉.
    You're on
  9. Zeist, Holland
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    03 Feb '04 09:31
    RUNning Pawn wins. I can't show it myself, but look for the transcendence of e and pi here:

    http://www.math.sc.edu/~filaseta/gradcourses/Math785/Math785Notes6.pdf

    Signed: Pi-derman
  10. Standard memberroyalchicken
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    03 Feb '04 23:42
    Originally posted by piderman
    RUNning Pawn wins. I can't show it myself, but look for the transcendence of e and pi here:

    http://www.math.sc.edu/~filaseta/gradcourses/Math785/Math785Notes6.pdf

    Signed: Pi-derman
    Hah! The ''e'' bit of that proof is almost the same as mine.

    However, since the RUNning Pawn clan is very skilled, I challenge them to solve the following:

    Let h(n) be defined as 1+ 1/2 + 1/3 + 1/4 +...+1/n

    Now prove that lim(n->oo) [h(n) - log n] is transcendental.
  11. Standard memberFiathahel
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    04 Feb '04 15:30
    Originally posted by royalchicken
    Hah! The ''e'' bit of that proof is almost the same as mine.

    However, since the RUNning Pawn clan is very skilled, I challenge them to solve the following:

    Let h(n) be defined as 1+ 1/2 + 1/3 + 1/4 +...+1/n

    Now prove that lim(n->oo) [h(n) - log n] is transcendental.
    Hmm, I've never learned anything about trancendentals, only their definition, but I'll work on it.

    Have you allready thought about Q(sqrt(6))?
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