It's a trivial generalization of the solution for ##m=0##. :DSo what about finding k for any given m in 2ℵm=ℵk 2^{\aleph_m}=\aleph_k?
Demystifier used "most difficult equation"! Those are proofs of facts. Of course proving things can become arbitrarily difficult. (EDIT: This is not trivial too!)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?
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?)
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?
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?
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
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.
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
The continuum hypothesis says that k = 1. The continuum hypothesis is undecidable (Gödel theorem).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
$$2^{\aleph_0}=\aleph_k$$
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).
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?
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: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?
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.
You're a genius!How about $$y=\int dx\,f(x) + const$$
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 $$y=\int dx\,f(x) + const$$
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).