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The most difficult equation in mathematics

  1. Dec 11, 2014 #1

    Demystifier

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    What, in your opinion, seems to be (one of) the most difficult equation(s) in mathematics?

    Here is my choice:
    Find the solution ##k## of the equation
    $$2^{\aleph_0}=\aleph_k$$
     
  2. jcsd
  3. Dec 11, 2014 #2

    ShayanJ

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    So what about finding k for any given m in [itex] 2^{\aleph_m}=\aleph_k[/itex]?
    Seems a nice challenge! Is it possible to give a more difficult equation than a given one?
    Anyway, I think no such thing(most difficult equation) exists in mathematics. Or at least its existence isn't a trivial fact!
     
    Last edited: Dec 11, 2014
  4. Dec 11, 2014 #3

    Demystifier

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    It's a trivial generalization of the solution for ##m=0##. :D
     
  5. Dec 11, 2014 #4

    ShayanJ

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    I'm wondering which is harder! Finding k for a given m or finding m for a given k?
    Has such a problem been solved in any of its cases?(or maybe in general!!!)(I was going to write 'considered' instead of 'solved' but I thought it surely is considered by someone! or not?)
     
  6. Dec 11, 2014 #5

    Doug Huffman

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    Difficult and solved by traditional methods might be Fermat's Last solved by Andrew Wiles. Difficult and unsolved might be nonsense. How is difficulty measured, modus ponens ponendo?
     
  7. Dec 11, 2014 #6

    ShayanJ

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    Demystifier used "most difficult equation"! Those are proofs of facts. Of course proving things can become arbitrarily difficult. (EDIT: This is not trivial too!)
    And about measuring difficulty. I don't think we need to get rigorous here. If a good mathematician says that something is difficult, Its enough for me to believe that thing is difficult. (I hope you don't ask for a definition of a good mathematician!!!)
     
  8. Dec 11, 2014 #7

    Demystifier

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    The problem has been considered a lot. The most important result (by Cohen) is a proof that the problem is unsolvable by using standard axioms of set theory. Different non-standard axioms of set theory may lead to different solutions, but then the problem is how to know which axioms, if any, are the "right" ones?

    For more details see
    http://en.wikipedia.org/wiki/Continuum_hypothesis
     
  9. Dec 11, 2014 #8

    ShayanJ

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    I think you should use "desired" instead of "right"!
    Anyway, why just use 2? We can generalize to [itex] n^{\aleph_m}=\aleph_k [/itex]. Right? Or there is some reason for that 2?

    P.S.
    Mathematical logic and set theory, or anything which seems that fundamental is really interesting for me but there lots of things to learn, and things with applications to physics have priority. A day will come that I'll learn those!:D
     
    Last edited: Dec 11, 2014
  10. Dec 11, 2014 #9

    Demystifier

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    Of course there is. A set with ##n## elements has ##2^n## subsets. Since the set of integers has ##\aleph_0## elements, it can be used to show that the set of reals has ##2^{\aleph_0}## elements.
     
  11. Dec 11, 2014 #10

    Demystifier

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    For a physicist friendly explanation of some basics in mathematical logic and set theory, including the origin of this basis ##2##, see
    R. Penrose, The Road to Reality
     
  12. Dec 11, 2014 #11

    ShayanJ

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    Looks like I wasn't clear enough. Actually I meant Is it that we can only place 2 as the base, because of some features of aleph numbers, or its possible to put any other number as the base? Does that base have to be a natural number? An integer? Real?
     
  13. Dec 11, 2014 #12

    ShayanJ

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    I saw that book before. It was thick enough to make me think it contains much physics but I really didn't expect it to contain such a pure math subject! Thanks.
     
  14. Dec 11, 2014 #13

    mathman

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    The continuum hypothesis says that k = 1. The continuum hypothesis is undecidable (Gödel theorem).
     
  15. Dec 12, 2014 #14

    Demystifier

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    No, that's the Cohen's theorem. And it is only valid within the standard axioms of set theory, not within all conceivable set-theory axioms.
     
    Last edited: Dec 12, 2014
  16. Dec 12, 2014 #15

    Demystifier

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    You can use any basis you like. But if you solve the problem for basis 2 (or for any other fixed basis larger than 1), other bases should not longer be a problem.
     
  17. Dec 12, 2014 #16

    Demystifier

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    Anyway, this is not a topic about continuum hypothesis. I am sure there are also other candidates for the most difficult equation in mathematics. Any suggestions?
     
  18. Dec 12, 2014 #17

    ShayanJ

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    Solving any of the equations in this list in the most general case!(Sorry for ruining the chance for others but I wanna emphasize my last sentence in post#2!)
     
  19. Dec 12, 2014 #18
    These problems were and are also quite difficult to solve:
    http://en.wikipedia.org/wiki/Hilbert's_problems
    http://www.claymath.org/millennium-problems

    A difficult problem that can be stated in simple terms: construct an algorithm that will find, if it exists, the general analytic solution of a first order ordinary differential equation y'=f(x), with f a specified (analytic) function.

    Anyway, I think we can only sum up some unsolved problems in this thread. It is difficult to say 'this one is more difficult than that one, because it requires blabla theory' without a proper definition of what makes a problem 'difficult'.
     
  20. Dec 13, 2014 #19
  21. Dec 15, 2014 #20

    Demystifier

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    How about $$y=\int dx\,f(x) + const$$
     
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