unforum

[pheebs]

has anyone tried FIVECOWS??

[Boomanda]

Gödel's Incompleteness Theorem

[jaikaiman]

 

[Boomanda]

Okay. I just read a book on Fermat's Last Theorem, which stated more or less the same thing, (The Theorem was that x^3+y^3=z^3 had no solutions, by the way.) that should the theorem be unprovable, it would be true, as finding a counterexample would be impossible. That should apply to the Riemann Hypothesis as well, I'd imagine. So... knock yourself out solving it. ^^ As you might know, it's soon been 150 years since the hypothesis was created.

( In case you want to read about the incompleteness theorem... http://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorem)

[GasparLewis]

Fermat's was not just for cubes, but for any power greater than 2.

x^n + y^n = z^n has no real answers for any n > 2

Andrew J. Wiles proved it in the nineties, formally; he's head of the Princeton University Mathematics Department as we speak.

-------------------

But, going back to the point; yes. There are four possibilities of the proavbility of the Riemann hypothesis: true, false, unprovably true, unprovably false. BUT, since being false would indicate a zero off the critical line, one must be able to find that zero, or else it's not there. Thus, it cannot be unprovably false.

True, false, or unprovably true: that is the question.

[jaikaiman]

Surely if it was unprovably false then it cannot be false either...oooh my head hurts!

[GuyIncognito]

For completeness sake, I would like to quote GuyP of Mind Candy-fame (from the PerplexCity.com forums, 29.09.06, emphasis by me):

[si]

This may have been suggested already, apologies if it has, but as the card is a statement as pointed out in the earlier pages and there isn't actually a direct question or puzzle to solve (apart from the $1 million prize) has anyone tried on the solve page:

'What is the question'


Maybe it is as simple as that, realistically the answer to the problem on the card will never be proven in our lifetime and I cannot see the point of setting a card with no answer when the other 255 cards are all possible?

Worth a try !

[friscodude]

Could always try and hack the MindCandy answer database if worst comes to worse.

Of course, that's probably nearly impossible, but I agree with the idea that you can't have to solve the hypothesis. I of course, don't physically have the card, so I can't try anything with it (burning, 3D glasses, anything stupid) and I'm stumped as to what the answer could be.

[gunny]

The problem is that, assuming that Mind Candy hasn't actually solved Riemann themselves, (which seems very unlikely) is what the actual question is, because if it is Riemann, how would MC know the answer? I think we can safely assume that the answer to the Riemann problem is not the answer to the card.

But at the moment, cracking the MC Dbases seems the only way to find the answer.

[Mindez]

[Perplexed_Rug]

Riemann Schmieman

[themandotcom]

If anyone wants to *attempt* to do this card, i suggest reading Prime Obsession by John Derbyshire. Its a great book and its very informative, and you dont need a big math background to understand it

[Si]

How can anyone solve the card when we don't know what the question is?

If the answer to the card is to solve the Reimann hypothesis then we all know that's not going to happen so card in the bin time, but I can't understand why Mind Candy would put a card in that is (as far as we are concerned) is unsolveable.

Surely, the 256 cards are all solveable otherwise A - We are wasting money buying them, B - Mind Candy are very cruel or C - There is an answer but we are missing something.

Also, why choose this puzzle out of the Millenium puzzles - why not choose another one that could at least be attempted or possibly solved by someone in the near future?

I am probably wrong but something is not right with this card, I still believe that there is a solution (otherwise whats the point of having a solve page for the answer when Mind Candy wouldn't know the solve for Reimann as pointed out lots of times) and I hope someone solves it (if it can be lol) soon.

[Si]

Guys, has anyone tried this as a solution :

The Bombieri-Vinogradov theorem is one of the major applications of the large sieve method. It improves Dirichlet's theorem on prime numbers in arithmetic progressions, by showing that by averaging over the modulus over a range, the mean error is much less than can be proved in a given case.

This result can sometimes substitute for the still-unproved generalized Riemann hypothesis.

Worth trying '(The) Bombieri-Vinogradov Theorem' as a solution?