Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Are there unsolved problems in math that are large believed to be true, but for reasons other then statistical justification?

It seems that Goldbach should be true, but this is based on heuristic justification.

I am looking for conjectures that seem to be true, but where the 'why' is something other then a statistical justification, and I want know what exactly that 'why' is.

Edit: Can you please include the reason it is widely believed to be true in your posting? That is the interesting part

share|improve this question
4  
Can you give a good example of an obviously true but unproved statement? I really don't think there is anything obvious about Golbach, P=NP or the Riemann hypothesis---at all. "Largely believed to be true" and "obvious" seem to me very, very different things! How can something be obvious if no one can come up with a convincing proof? –  Mariano Suárez-Alvarez Feb 1 '11 at 14:08
2  
@Mari Ofcourse it does, entire physics and most of science is based on it –  user1708 Feb 1 '11 at 16:38
3  
LOL that's why I do math and not science –  Aaron Mazel-Gee Feb 1 '11 at 16:49
2  
The argument you are using, kakemonsteret, to argue that the truth of the Goldbach conjecture is true is the same one one can use to show that all natural numbers are less than ${{{{{{{{{{{{{{{{{{{{{2^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}‌​^2}^2}^2$ –  Mariano Suárez-Alvarez Feb 1 '11 at 18:38
2  
@Mari I never used the word proof or true. I'm not looking for theorems. Conjectures. Heuristics dont show that statements must be true. It's called a heurisitc justification. –  user1708 Feb 1 '11 at 18:42

9 Answers 9

up vote 4 down vote accepted

Most of us believe that $e+\pi\notin\mathbb{Q}$, but this is not proved yet. Otherwise, this will lead to very interesting consequences.

share|improve this answer
3  
can anyone list some of the consequences? @norbert ? –  Bhargav Jan 2 '12 at 14:24
1  
For example one of this number would become of no use, because it can be expressed in terms of another. But I think such an expression would be bulky. –  no identity Jan 2 '12 at 14:39
    
It's worthwhile noting that, while we don't know if $e+\pi$ nor $e-\pi$ are irrational, at least one of them is. –  F M Jan 2 '12 at 17:28

The Riemann hypothesis is largely believed to be true, and further conjectures have been made based on its truth (e.g. statements about the distribution of prime numbers) but no one has ever proved it.

share|improve this answer
2  
Isn't that just because it has been checked for large numbers? Or is there some deep reasons it should be true ? –  user1708 Feb 1 '11 at 18:46

It is widely believed that the fundamental axioms of set theory ZFC are consistent, but this has not been proved in ZFC and in fact provably cannot be proved in ZFC itself unless ZFC is inconsistent, by the second Gödel Incompleteness theorem.

Indeed, whatever fundamental axioms you favor, whether PA or KP or Z or ZF or ZFC or ZFC+large cardinals, then it is natural to suppose also that since you believe that those axioms are true that you also believe that those axioms are consistent, but this is provably not provable from your axioms, unless they are inconsistent.

share|improve this answer

What about $P \neq NP$? Scott Aaronson has made some excellent points at here

share|improve this answer
3  
Actually this is not 'obviously' true, but a heavily discussed problem. Noone yet knows if there isn't some polynomial algorithm that just has a very high constant/factor to solve some NP-complete problem. –  Listing Feb 1 '11 at 10:56
1  
The impression I've had from discussing with people in TCS, most people tend to think it's obviously true, but of course, one shouldn't be all to sure. I'd say that it's "obviously true", though. See Scott's post. –  Dedalus Feb 1 '11 at 11:02
1  
$P \neq NP$ is consensus as sensible assumption because it allows whole theories to be constructed that try to deal with (by assumption) hard problems. All that would be (in theory) meaningless would we assume $P = NP$, but we want to do these things in order to solve problems as well as possible now. Wether the assumption is true or not is still discussed, though. Of course, many people would rather have no polynomial time algorithms for important problems than admit they were just not able to to find them. –  Raphael Feb 1 '11 at 12:27
    
@user3123 and Raphael: Here's the short reason - once you get past definitions (which hinder 'obviousness') it is 'obviously' true that $P \neq NP$ because there is an obvious exponential algorithm for SAT, and it's not obvious how to remove the 'trying every possibility' approach. There is doubt (we don't -know it's true) only because there is no proof (and proving a negative is hard), and there are other examples of where nondeterminism (-very- surprisingly) doesn't get you anything (language definiability in regular expressions and in TMs). –  Mitch Feb 1 '11 at 17:11

I believe that the Jacobian Conjecture is widely considered true (is even given as a double-starred problem in Hartshorne's Algebraic Geometry book) but it has resisted attempts.

By the way, I haven't checked in a while, so what is the current status of the Jacobian Conjecture?

share|improve this answer
3  
Still as open as ever :) –  Mariano Suárez-Alvarez Feb 1 '11 at 18:06
1  
@Mariano: I guess I knew that :-) What I meant is that if there has been some kind of progress lately. –  Andrea Mori Feb 2 '11 at 9:58

The rank conjecture for rational elliptic curves (for every $n\gt 0$ there is an elliptic cuve over $\mathbb{Q}$ whose group of rational points has rank at least $n$).

share|improve this answer
    
Do you know the reason why it is believed that this is so? Is there some sort of heuristic argument that makes it plausible? I was able to find this page about ranks where it actually says that this is the "folklore" conjecture but does not provide a specific reference. –  Adrián Barquero Feb 1 '11 at 17:32
    
@Adrián: I don't know why it is believed to be true, but my impression is that the general feeling is "there is no good reason why it would be false." –  Arturo Magidin Feb 1 '11 at 18:24
    
Hmmm, maybe I'll try asking this at MO. Thanks. –  Adrián Barquero Feb 1 '11 at 18:35
    
I guess there's no need for that. It has already been asked here. –  Adrián Barquero Feb 1 '11 at 18:38
    
@Adrián: Ah, well, clearly there are good reasons I did not know about. Thanks! –  Arturo Magidin Feb 1 '11 at 19:05

The Kakeya conjecture. It states that a set in $\mathbb{R}^d$ containing a line in every direction has Hausdorff dimension $d$. It is solved for $d = 2$ and open for $d \geq 3$.

The finite field analogue was recently solved by Dvir (Tao made a very nice post about it, here). Dvir's contributions have also led to a deeper understanding and an ultimate resolution by Guth and Katz of the Erdos distance problem in $\mathbb{R}^2$ (of which Tao also has a nice post, here).

When one constructs a Kakeya set of measure zero, it is typically visualized by Bourgain's "Bush" construction or Wolff's "Hairbrush" argument. The iterations of these constructions has full Hausdorff dimension.

I believe a great number of people would be surprised if it were false.

share|improve this answer

The inverse Galois problem is to determine whether every finite group is the Galois group of some Galois extension of $\mathbb{Q}$. It has been proved that all finite soluble groups and all the sporadic groups except for $M_{23}$ (whose status is unknown) appear as the Galois group of some Galois extension of $\mathbb{Q}$; likewise it has been proved that all finite groups are the Galois group of some Galois extension of other fields, $\mathbb{C}(t)$, for example.

I'd be interested to know if there is a consensus about the truth or falsity of this problem, and if so what the consensus is.

share|improve this answer
    
@EricNaslund: depends where you grew up, I think. "Able to be solved" is usually the second definition in most dictionaries, and I've myself heard people in the UK and Canada use "soluble group". –  DSM Jan 2 '12 at 17:01
    
@Eric: As DSM says, I've come across both variants studying in the UK, but primarily soluble. –  Clive Newstead Jan 2 '12 at 17:26
    
@DSM, Clive: I apologize then, I was unaware of the terminology. –  Eric Naslund Jan 2 '12 at 18:05

If for some real x > 0 both $2^x$ and $3^x$ are rational integers then so is $x$.

It is obviously true (or seems so to me at any rate); but I think it's fair to say that a proof is nowhere in sight, unless there has been some recent progress.

I seem to recall that Ramanujan proved the similar but weaker result with $2^x$, $3^x$, and $5^x$ all rational integers.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.