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Here is my choice:

Find the solution ##k## of the equation

$$2^{\aleph_0}=\aleph_k$$

- Thread starter Demystifier
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- #1

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Here is my choice:

Find the solution ##k## of the equation

$$2^{\aleph_0}=\aleph_k$$

- #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!

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!

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- #3

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It's a trivial generalization of the solution for ##m=0##. :DSo what about finding k for any given m in 2ℵm=ℵk2^{\aleph_m}=\aleph_k?

- #4

ShayanJ

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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?)

- #5

Doug Huffman

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- #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!)modus ponens ponendo?

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!!!)

- #7

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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?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?)

For more details see

http://en.wikipedia.org/wiki/Continuum_hypothesis

- #8

ShayanJ

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I think you should use "desired" instead of "right"!but then the problem is how to know which axioms, if any, are the "right" ones?

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

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- #9

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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.Or there is some reason for that 2?

- #10

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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##, seeMathematical 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

R. Penrose, The Road to Reality

- #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?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.

- #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.For a physicist friendly explanation of some basics in mathematical logic and set theory, including the origin of this basis 22, see

R. Penrose, The Road to Reality

- #13

mathman

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The continuum hypothesis says that k = 1. The continuum hypothesis is undecidable (Gödel theorem).

Here is my choice:

Find the solution ##k## of the equation

$$2^{\aleph_0}=\aleph_k$$

- #14

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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.The continuum hypothesis is undecidable (Gödel theorem).

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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.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?

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- #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!)Demystifier said:

- #18

bigfooted

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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'.

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Edit: Not sure if this really counts as an equation...

- #20

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

- #21

bigfooted

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You're a genius!How about $$y=\int dx\,f(x) + const$$

I should have used a slightly more general first order ODE of course: y'=f(x,y).

- #22

ShayanJ

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You should still note that no single algorithm exists for finding the integral of a function, even if we exclude integrals which can't be expressed in terms of elementary functions(Oops...its not applicable to such integrals, right?). So finding such an algorithm is an open problem and it seems to be very hard.How about $$y=\int dx\,f(x) + const$$

- #23

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You are sarcastic. :)You're a genius!

That's much harder, I admit.I should have used a slightly more general first order ODE of course: y'=f(x,y).

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Though someone more well-versed than me may be able to elaborate on it.

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