- #1

- 12,579

- 4,944

Here is my choice:

Find the solution ##k## of the equation

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Demystifier
- Start date

- #1

- 12,579

- 4,944

Here is my choice:

Find the solution ##k## of the equation

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

- #2

ShayanJ

Gold Member

- 2,809

- 604

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!

Last edited:

- #3

- 12,579

- 4,944

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

Gold Member

- 2,809

- 604

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

Gold Member

- 805

- 111

- #6

ShayanJ

Gold Member

- 2,809

- 604

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

- 12,579

- 4,944

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

Gold Member

- 2,809

- 604

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

Last edited:

- #9

- 12,579

- 4,944

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

- 12,579

- 4,944

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

Gold Member

- 2,809

- 604

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

Gold Member

- 2,809

- 604

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

Science Advisor

- 8,046

- 535

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

- 12,579

- 4,944

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).

Last edited:

- #15

- 12,579

- 4,944

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?

- #16

- 12,579

- 4,944

- #17

ShayanJ

Gold Member

- 2,809

- 604

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

Gold Member

- 654

- 177

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'.

- #19

- 815

- 70

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

- #20

- 12,579

- 4,944

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

Gold Member

- 654

- 177

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

Gold Member

- 2,809

- 604

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

- 12,579

- 4,944

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).

- #24

- 12,579

- 4,944

- #25

- 944

- 394

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

- #26

ShayanJ

Gold Member

- 2,809

- 604

If you mean varying it w.r.t. to all the dynamical variables to get all the equations of motion, I should say this is only a tedious thing to do. The algorithm for doing it is very well understood(So much that I won't wonder if there is a software able to do that), its just that its complicated and time consuming and messy calculation. So it isn't even a candidate for the hardest problem in math.

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

- #27

BiGyElLoWhAt

Gold Member

- 1,577

- 119

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!)

Haha, I don't even know what the solution is that we're looking for in most cases. Would many of these actually require multiple solutions? Some of these PDE's have multiple functions (and differentials thereof) with constraints on the equations; so I guess the first step, at least for me haha, would be to know what the solution would look like.

- #28

BiGyElLoWhAt

Gold Member

- 1,577

- 119

- #29

ShayanJ

Gold Member

- 2,809

- 604

When I saw the list the first time, I felt I know nothing!!!

Well, actually this isn't far from reality!

- #30

- 8,942

- 2,931

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?

Well, if [itex]\alpha[/itex] is infinite, then there is no difference between [itex]2^\alpha[/itex] and [itex]n^\alpha[/itex].

The number [itex]x^y[/itex] can be defined as the number of functions from a set with [itex]y[/itex] objects into a set with [itex]x[/itex] objects. If [itex]y[/itex] is infinite, then this number is independent of [itex]x[/itex], as long as [itex]0 < x < \aleph_0[/itex]

Share:

- Replies
- 6

- Views
- 2K

- Replies
- 16

- Views
- 4K