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Doh (Score:2, Funny)
by Frijoles on Thursday May 25, @10:12AM EDT (#3)
(User Info) |
Time to go buy that 'Dummies guide to math' book.
-=- Frijoles |
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Re:Doh (Score:2, Funny)
by G-funk ([email protected]) on Thursday May 25, @10:42AM EDT (#75)
(User Info) http://g-funk.isa.net.au/
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Time to go buy that 'Dummies guide to math' book. Ooooooh, now you've done it... IDG are gonna have to close down slashdot now... I hope you're happy!
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Re:Doh (Score:5, Funny)
by SgtPepper (sgtpepper@spam^H^H^H^Hprokrams.com) on Thursday May 25, @11:46AM EDT (#159)
(User Info) http://www.prokrams.com
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Naw....
"Unsolvable Obscure Math Problems For Dummies"
THERE! NOW IDG will close down slashdot...
shit...maybe i shoulda posted that AC...
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Re:Doh (Score:1)
by dparks1 on Thursday May 25, @11:59AM EDT (#176)
(User Info) |
Dear Mr./Ms. Frijoles:
We noticed with great interest your recent on-line publication using the word "Dummies."
You may be aware that the term "Dummies" represents the brand, image, & reputation of a certain large corporation. As such, your use of this mark is not authorized. You must understand that we vigorously defend against misuse & dilution of our trustmark.
We here at Wankum, Yankum & Shankum represent the owners of the registered term "Dummies." This notice serves as your official warning that we consider your use of the term to infringe on our mark.
Please cease & desist from use of the term "Dummies," except when you intend to refer directly to us. Thank you for your cooperation.
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Re:Doh (Score:1)
by Alkivar (/dev/null) on Thursday May 25, @03:12PM EDT (#280)
(User Info) |
AHAHAHAHA
I personally couldn't think of a better name for a lawfirm to represent all those stupid lawsuits IDG files on behalf of their "For Dummies" trademark.
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Uh.... the answer is (Score:1)
by crisper ([email protected]) on Thursday May 25, @10:16AM EDT (#4)
(User Info) http://www.whatsitcalled.com
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according to my pet monkey "Sweet" Jimmy who is a genius in mathematics and currently typing up the complete works of william shakespeare. The answer is pi, blueberry pi. Sean
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I have a solution to Fermat's problem (Score:2, Funny)
by georgeha on Thursday May 25, @10:16AM EDT (#5)
(User Info) http://www.frontiernet.net/~ghaberbe/george2.htm
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But the text enter box at /.is too small to post it.
George
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Re:I have a solution to Fermat's problem (Score:1)
by vr ([email protected]) on Thursday May 25, @10:25AM EDT (#31)
(User Info) http://www.ifi.uio.no/~vetler
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But the text enter box at /.is too small to post it.
Well.. why would you need to post it?
You do know it already has been solved.. don't you?
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Re:I have a solution to Fermat's problem (Score:1)
by georgeha on Thursday May 25, @10:30AM EDT (#48)
(User Info) http://www.frontiernet.net/~ghaberbe/george2.htm
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Emperically, with a computer.
Isn't this the one where someone with a sense of humor wrote
"I have an elagant solution to Fermat's theorem, unfortunately, the margin it too small to write it."
Hence my post.
But no one got the joke.
George
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Re:I have a solution to Fermat's problem (Score:1)
by vr ([email protected]) on Thursday May 25, @10:37AM EDT (#69)
(User Info) http://www.ifi.uio.no/~vetler
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Emperically, with a computer.
Isn't this the one where someone with a sense of humor wrote
"I have an elagant solution to Fermat's theorem, unfortunately, the margin it too small to write it."
Well.. actually it was Fermat himself that wrote it in a book he had. He claimed to have solved it, but he never distributed his proof. In those days, mathematicians were very secretive.
In many years before Fermat's Theorem was solved, people actually based their work on it being true. People would start by first saying that if the theorem is true, then .. blah blah...
:-)
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Re:I have a solution to Fermat's problem (Score:1)
by susano_otter on Friday May 26, @12:13AM EDT (#386)
(User Info) |
Right. So what we really want to know is: What was Fermat's proof? Did he prove it by way of the so-called "Taniyama-Shimura conjecture", or did he have some other, more elegant (or simply more spooky) proof in mind? Well, that's what I want to know, anyway...
Americans are conditioned by junk scam mail, just like the Fremen were conditioned by the desert. |
I got it (Score:1)
by luckykaa ([email protected]) on Thursday May 25, @10:39AM EDT (#70)
(User Info) |
I just think I would have been modded down if I posted "Ha Ha"
Fermat postulated the theorem and wrote "Astounding proof" in the margin. In fact he wrote a lot of stuff in out and across the margin. It was the mathematical equivalent of Mozilla source code.
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Re:I got it (Score:1)
by georgeha on Thursday May 25, @10:47AM EDT (#86)
(User Info) http://www.frontiernet.net/~ghaberbe/george2.htm
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Thanks,
I was afraid I was losing another chunk of my memory.
George
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Re:I have a solution to Fermat's problem (Score:2, Informative)
by Mr_Dyqik on Thursday May 25, @12:42PM EDT (#204)
(User Info) |
Emperically, with a computer.
uh, no. Fermat's Last Theorem was solved analytically by Andrew Wiles, by hand, over a seven year period.
The problem cannot be solved empirically, and neither can any of the others above (except maybe by counter-example) as they involve absolute proof of general cases
And it was Fermat who wrote something like "I have a marvellous proof of this, which this margin is too small to contain"
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Re:I have a solution to Fermat's problem (Score:1)
by Old Wolf on Friday May 26, @07:59AM EDT (#412)
(User Info) |
Uh. Every man and his dog and its fleas got it. It wasn't funny cos a few dozen lamers make this joke every time a mathematical story gets posted on slashdot.
If you look further down, you will see that someone made a joke likening Fermat's proof to Mozilla source code. This, on the other hand, is funny now because it is original and has humour potential. If several people made that joke afterwards, it would cease to be funny.
Comprende ?
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Re:I have a solution to Fermat's problem (Score:1)
by Old Wolf on Friday May 26, @08:04AM EDT (#414)
(User Info) |
Every man and his dog and its mangy fleas got the joke. It was modersately funny the first time, but not now. Every time a mathematical story gets posted on slashdot, dozens of lamers reply and make this miserable joke.
There is a post just down from here that likens Fermat's proof to Mozilla source code. It is funny because it both has humour potential, and is original.
Comprende ?
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Re:I have a solution to Fermat's problem (Score:1)
by pugugly ([email protected]) on Friday May 26, @01:33AM EDT (#390)
(User Info) |
| Actually, according to the 'Urban Legends' site (whose URL I don't recall at the moment. Probably somnething simple like urbanlegends.org), Fermat's last theorem still stands. The claimant had a theory that fell into some really nasty holes in realms of high mathematics that I'm not qualified to think about, never mind explain. Hey, I aced Calculus I, I survived Calc II. Past that I wimp out. So, if anybody has a good proof of Fermats' last theorem, go for it and get yourself a Fields medal. I've got my own that I haven't managed to prove in twenty years. Of course, I have a prime number generator that I can't prove works too - life sucks - [G].
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Re:I have a solution to Fermat's problem (Score:1)
by pugugly ([email protected]) on Friday May 26, @01:47AM EDT (#391)
(User Info) |
| Eureka - I have found it . . . The URL is http://www.urbanlegends.com/ and Fermats last theorem info As to the theorem itself -
There are no positive integers such that x^n + y^n = z^n for n>2.
I've found a remarkable proof of this fact, but there is not enough space in the margin [of the book] to write it To summarize the Urban Legends site, Andrew Wiles has tried two different approaches, both of which have significant shortcomings, and both of which are into very esoteric realms of mathematics. The summary starts Both manuscripts have been published. Thousands of people have a read them. About a hundred understand it very well. For further information, apply to the Annals of Mathematics or Urbanlegends.com
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Re: the solution to Fermat's last theorum (Score:1)
by Tarquin ([email protected]) on Thursday May 25, @10:25AM EDT (#33)
(User Info) http://money4nothing.peji.com
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| Did it actually get solved, or is it just me? I can't remember if, after a couple of wrong answers, somebody actually got it right... Anybody know? There's a statistical theory that if you gave a million monkeys typewriters and set them to work, they'd eventually come up with the complete works of Shakespeare. Thanks to the Internet, we now know this isn't true.
-Ian Hart, British Actor
--
It's not the rambling I object to, so much as the mumbled incoherancies... |
Re: the solution to Fermat's last theorum (Score:1)
by Anonymous._.Coward (print reverse split(/ */,"moc.buprenroc\@demmahom") on Thursday May 25, @10:31AM EDT (#51)
(User Info) |
| Yes it got solved by Andrew Wiles of Princeton University. After 7 years of trying he announced a solution 1993 and a flaw was discovered in the proof. (heh-heh) Wiles had to work for another year to establish that he had solved the problem. When he came out his dark room after 8 years the world was a very different place...
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Re: the solution to Fermat's last theorum (Score:1)
by vr ([email protected]) on Thursday May 25, @10:33AM EDT (#59)
(User Info) http://www.ifi.uio.no/~vetler
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Did it actually get solved, or is it just me?
yes, it did get solved. it was a british guy who solved it.. or rather he solved a problem, named after a couple of Japanese mathematicians, that implied that Fermat's Theorem had to be true.
Simon Singh wrote a book about it. I can recommend it.. it is very entertaining. even for people who do not know much (or anything!) about mathemathics.
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Re: the solution to Fermat's last theorum (Score:2, Informative)
by nowonder (nowonder-at-nowonder-dot-de) on Thursday May 25, @10:57AM EDT (#105)
(User Info) http://linux-security.org
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Actually these japanese guys were Taniyama und Shimura and the conjecture is called the - guess -
Taniyama-Shimura conjecture.
It is about elliptic curves and modular forms (or whatever that is in englisch).
-- NoWonder of WonderWorks/OmegaProject |
Re: the solution to Fermat's last theorum (Score:2, Informative)
by ch2 ([email protected]) on Thursday May 25, @12:30PM EDT (#195)
(User Info) http://hodcroft.co.uk/
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Modular forms are strange four-dimensional things.
The Taniyama-Shimura conjecture was that every elliptic curve was a modular form. This implied that Fermat's Last Theorem was true in a roundabout way. Wiles proved Taniyama-Shimura and thus Fermat.
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Re: the solution to Fermat's last theorum (Score:1)
by Anonymous._.Coward (print reverse split(/ */,"moc.buprenroc\@demmahom") on Thursday May 25, @12:48PM EDT (#212)
(User Info) |
| Elliptic curves are simple functions that can be drawn as gently looping lines in the (x,y) plane. They look very simple. But you get interesting patterns at the points where the curve exactly crosses integer (x,y) coordinates. Elliptic curves got into the press because they're good for cryptography and some smart arse's cracked one landing themselves $10k.
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Wiles Nifty Proof (Score:1)
by nereidian on Thursday May 25, @01:17PM EDT (#226)
(User Info) |
This also involved the Taniyama-Shimura conjecture, which is probably more interesting than Fermat's silly riddle. I think the conjecture said something like "every elipse over Q is also a modular form" (an ellipse is a conic with an eccentricity between 0 and 1). Wiles paper was called, I think, "Modular Forms and Fermat's Last Theorem."
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Re:I have a solution to Fermat's problem (Score:1)
by omegamaid on Thursday May 25, @12:15PM EDT (#185)
(User Info) |
I'm rolling on the floor, laughing my ass off, and a bunch of other stupid acronyms. How original. How ingenious. You've managed to make the same joke that a dozen people make each and every time anything vaguely related to mathematics appears on Slashdot. Congratulations.
Yes, I'm in sarcastic bitch mode. Apologies.
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Re:I have a solution to Fermat's problem (Score:1)
by georgeha on Thursday May 25, @12:49PM EDT (#213)
(User Info) http://www.frontiernet.net/~ghaberbe/george2.htm
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Don't hold back, tell me how you really feel.
Admit it, at least it's funnier than 42.
George
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Aleph1 = C? (Score:1)
by FascDot Killed My Pr on Thursday May 25, @10:17AM EDT (#6)
(User Info) |
I have a vague recollection that someone proved that "Aleph1 = C" couldn't be proved or disproved. Or was that fiction? Or am I even more confused than I think I am?
As I recall, Aleph0 = the set of all integers. C is the set of all reals. But (at one time, anyway), the question of whether C was the same as Aleph1 existed.
As for the other problems, the only one I'm familiar with (or even heard of) is P=NP--but I guess that's what I get for being a "specialized mathematician" (i.e. computer scientist).
--
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Re:Aleph1 = C? (Score:2, Informative)
by Kinthelt on Thursday May 25, @10:21AM EDT (#17)
(User Info) http://www.undergrad.math.uwaterloo.ca/~db2nesbi
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Actually, R is the set of all Reals. C is the set of all Complex numbers (which include the Reals, Imaginary numbers, and sums of the two).
"Evil will always triumph over good, because good is dumb." - Dark Helmet (Spaceballs) |
Re:Aleph1 = C? (Score:1)
by richieb ([email protected]) on Thursday May 25, @10:22AM EDT (#21)
(User Info) http://www.netlabs.net/~richieb
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| This is the "Continum Hypothesis". It was shown to be independent of the axioms of set theory. So you can assume that Aleph1 = c or that Aleph1 /= c and get two different set theories. Sort of like Euclidean and non-Euclidean geometries. ...richie
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That's what I thought I remembered.... (Score:1)
by FascDot Killed My Pr on Thursday May 25, @10:23AM EDT (#26)
(User Info) |
...but I also remembered (right or wrong) that I read it in GEB, so I thought it might have been fictional.
--
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Re:That's what I thought I remembered.... (Score:1)
by carlos_benj on Thursday May 25, @11:33AM EDT (#146)
(User Info) |
| "...but I also remembered (right or wrong) that I read it in GEB, so I thought it might have been fictional." I'm pretty certain I saw the same thing on the cover of the National Enquirer last year. carlos
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Re:That's what I thought I remembered.... (Score:2)
by SimonK ([email protected]) on Thursday May 25, @11:53AM EDT (#169)
(User Info) |
The maths and mathematical history in GEB is all correct, as far as I am aware.
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Nope (Score:1)
by FascDot Killed My Pr on Thursday May 25, @01:06PM EDT (#223)
(User Info) |
The interspersed Dialogues have a LOT of fictional math history in them.
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Re:Aleph1 = c? (Score:1)
by richieb ([email protected]) on Thursday May 25, @05:11PM EDT (#305)
(User Info) http://www.netlabs.net/~richieb
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| Aleph_1 = the size of R = the size of C I think you misunderstood my comment. "C" is not meant to be the set of complex numbers, but the cardinality of the set of reals (it's usually written as 'c' - note lower case) Cardinality of real numbers and complex numbers is the same (as you say). But the definition of Aleph1 is the cardinality of the power set (that is a set of all subsets) of a set with Aleph0 elements. So the Continum hypothesis asks: Is the count of all subsets of integers the same as the number of real numbers. This turns out to be unprovable from the usual axioms of set theory. BTW, Cantor proved early on that the cardinality of a power set of a non-empty set is strictly greater than the initial set (in the transfinite sense). Hope this is clearer. ...richie
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Re:Aleph1 = C? (Score:5, Informative)
by smileyy on Thursday May 25, @10:32AM EDT (#55)
(User Info) http://fitterhappier.nu/
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You're pretty close there. Aleph0 is the cardinality of the set of all integers, c is the cardinality of the set of all reals. Cantor's diagonalization proof demonstrates that c > Aleph0. But the exact "value" of c, in relation to other transfinite cardinals was not determined. In particular, its relationship to Aleph1 (which I believe is the cardinality of the power set of Aleph0) is formally undecidable. That is to say, it doesn't make any difference to your system of mathematics if you choose c It's been a while since I've studied this stuff, but I've found this a decent introduction to the subject.
-- smileyy |
Re:Aleph1 = C? (Score:2)
by smileyy on Thursday May 25, @10:35AM EDT (#64)
(User Info) http://fitterhappier.nu/
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Oops...it does make a difference to your system of mathematics, but either way, the system generated by either assumption cannot be shown to be inconsistent.
-- smileyy |
Re:Aleph1 = C? (Score:1)
by whovian on Thursday May 25, @10:51AM EDT (#93)
(User Info) |
IANAMBP (I am not a mathematician by profession), but I have read Rudy Rucker's book Infinity and the Mind. His explanation of the levels of infinities is almost graspable. What do you people think?
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Re:Aleph1 = C? (Score:3, Informative)
by bifurcator on Thursday May 25, @11:57AM EDT (#172)
(User Info) |
But the exact "value" of c, in relation to other transfinite cardinals was not determined. In particular, its relationship to Aleph1 (which I believe is the cardinality of the power set of Aleph0) is formally undecidable. You're pretty close there. But not quite. The cardinality of the power set of Aleph0 is C (the cardinality of the set of all real numbers). Put simply, Aleph 1 is the cardinal that comes right after Aleph0. C looked like a pretty good candidate for some time. C is a cardinal, and it comes after Aleph0. The question was, does it come right after Aleph0, or is there another cardinal in between Aleph0 and C? If my memory serves me right, an American named Cohen proved that you cannot answer 'yes' or 'no' to this question within an axiomatic system like ZFC (Zermelo-Frenkel plus the axiom of choice.)
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Re:Aleph1 = C? (Score:2, Interesting)
by Mark Histed (histed(replacethiswithanatsign)mit.edu) on Thursday May 25, @01:27PM EDT (#231)
(User Info) http://web.mit.edu/histed/www
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Yes. Godel showed that the continuum hypothesis (CH, which says aleph_1 = |R| = pow(aleph_0)) is consistent with ZFC -- therefore, it cannot be DISproved. Cohen in 1965 or so showed that CH is not decidable within ZFC: that it cannot be proved, either.
[ZFC is set of axioms used as the basis for a (the?) standard formulation of set theory: you start with these 9 axioms and rules of logic and all of your set theory follows.]
Therefore CH is an axiom that is independent of ZFC -- you can choose to adjoin it to your axioms or not, and you get a different set theory depending on whether you use it or not. Some might say that you choose whether to 'believe' CH -- more accurately, you choose to use CH depending on what kind of model of the real world you want your theory to be.
[btw, one can also prove that the 8 axioms of ZF are independent of C (the axiom of choice). So as above, you can choose whether to adjoin it to your set of axioms and make ZFC or just stick with ZF.]
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Re:Aleph1 = C? (Score:1)
by Hitokage_Nishino (kurai[TheGPwillPunishSpammers@]galaxypolice.com) on Thursday May 25, @02:35PM EDT (#265)
(User Info) |
There is a very simple reason that the set of all real numbers is greater than the set of all integers.
Every integer is included in the set of all real numbers, and for each integer in that set, there is an infinite number of portions between any 2 integers, and each is a real number.
Integers: 1, 2, 3, ...
Reals:
1 + .0, .1, .2, .3, ...
2 + .0, .1, .2, .3, ...
3 + .0, .1, .2, .3, ...
.
.
.
Thus, while the set of integers represent infinity, the set of real numbers represet infinity times infinity( or Infinity^2) Since everyone agrees that infinity is greater than 1, n^2 > n holds true in this case. But, when will people ever figure out that pi = crust + filling?
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Wrong (Score:1)
by Nit Picker on Thursday May 25, @03:55PM EDT (#291)
(User Info) |
Math gets wierd when you deal with infinities. If your proof were correct were correct then the cardinality of ordered pairs of positive integers would exceed the cardinality of the positive integers. However, a 1 to 1 correspondence can be constructed by laying out the ordered pairs of positive integers in a 2D array and counting them along the diagonals:
1 <-> (1,1)
2 <-> (1,2)
3 <-> (2,1)
etc.
A correct proof is Cantor's diagonalization theorem, mentioned elsewhere.
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Re:Wrong (Score:1)
by Hitokage_Nishino (kurai[TheGPwillPunishSpammers@]galaxypolice.com) on Thursday May 25, @07:49PM EDT (#333)
(User Info) |
There are more permutations of a set than there are numbers in a set. Thus, as you expand the set of integers and the set of real numbers from 0 to infinity using 1 integer at a time then all real numbers which use that integer... each expansion yeilds far more real numbers than integers, and infinitly more so as well... thus, at that ratio, the function of which diverges in relation to the integers. Of course, given you accept the values of Omega and Iota, the surreal value of infinity and it's reciprocal, this explanation is valid. Maybe that's where we disagree. Or perhaps you don't realize that using that method, it would take the entire infinite set of integers just to define the possible real numbers between 1 and 2! This is why I say the set of real numbers is square of the set of integers.
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Re:Wrong (Score:1)
by Old Wolf on Friday May 26, @08:10AM EDT (#416)
(User Info) |
To speak your language (*):
infinity^2 = infinity
(similar to how infinity + 1 = infinity)
so you have shown that there are only as many natural numbers as real numbers.
(*) As a mathematician, I disclaim all rights to and knowledge of this post.
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Re:Wrong (Score:1)
by krimson on Monday May 29, @07:37AM EDT (#457)
(User Info) |
It is not correct to say that c = infinity^2. However, it could be said that c = 2^infinity. This is in anology with sets of finite cardinality, since the power set of a set with the finite cardinality n has 2^n members. Still, saying that c = infinity^2 is obviously false.
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Re:Wrong (Score:1)
by krimson on Friday June 02, @07:41AM EDT (#466)
(User Info) |
One more thing: The omega notation is only used for ordinal numbers. When discussing the size of sets we are dealing with cardinal numbers. Thus the reasoning is invalid.
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Re:Aleph1 = C? (Score:1)
by darkside13 ([email protected]) on Thursday May 25, @02:54PM EDT (#274)
(User Info) |
I think you are very confused. Aleph0 is not a set, it is the size of the real numbers. The size of the Integers is considered to be infinity, at least a countable level of infinity. The interesting and surprising thing is that there are the same amout of Rationals as Even Integers and the same amount of Even Integers as Odd Integers and the same amount of Odd Integers as Integers. However, there are more Reals in the open interval from 0 to 1 than there are Integers, even or odd, or Rationals. It's all very interesting and counter-intuitive.
This is all a result of the idea of countability. That is, you could conceivable count the integers by putting them in a 1-1 correspondence with themselves. Therefore, you could, for any integer, determine which member it was of the ordered set of integers, i.e. 3 is the third member of the set of integers. The same can be done with the even integers, i.e. 4 is the second member, because it maps to 2. The rationals are a bit more tricky, it helps to look at a graph of ratios of integers to see how the rationals can be done.
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Re:Aleph1 = C? (Score:1)
by jejones on Thursday May 25, @05:58PM EDT (#314)
(User Info) |
| Nope, you're not confused--OK, a little bit, because aleph-0 is the cardinality of the set of natural numbers, C the cardinality of the set of reals, etc. Paul Cohen in the 1960s proved that you can either assume the continuum hypothesis or its negation, and either way you don't get any contradictions that weren't in set theory without either of those assumptions. (Ditto for the axiom of choice, if memory serves.) Aleph-null bottles of beer on the wall, aleph-null bottles of beer...
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Re: |power set of integers| = |reals|? (Score:2)
by kevin805 ([email protected]) on Friday May 26, @12:56AM EDT (#389)
(User Info) |
I thought this through a while back, and here's what I came up with. I'd appreciate if someone could tell me where it breaks down.
We know that the reals can be mapped onto the reals from 0 to 1. So, consider the following sequence of numbers:
000000000000000...
100000000000000...
010000000000000...
110000000000000...
001000000000000...
...
Just a progression of binary integers. Okay. Now, consider these as binary expansions of a real number less than one:
0.10000000000000...
So we have a mapping from the above to all reals less than one. We can also consider this a mapping onto the power set of the integers:
010010010000000... {1, 4, 7}
So we have the same progression mapping onto both sets of interest. But, aleph0 of the reals (terminating fractions) will be encoded twice. It seems like we could in some way "subtract these out", since subtracting a lower order of infinity from a higher order should still give the higher order. Also, the diagonalization argument that Cantor used doesn't work, because the number generated is:
0.1111111111111111111111... = 1
But we said that we are limiting it to numbers less than one.
But, each of the bit sequences we are using to encode a subset or a real also encodes an integer. So the above shows that:
|reals| = |power set of integers| = |integers|
Which we know is false because everyone says so. So, where is the flaw?
--Kevin
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Re: |power set of integers| = |reals|? (Score:1)
by tbarrie on Friday May 26, @03:45AM EDT (#401)
(User Info) |
Well, intuitively, you need infinite bit strings to represent the real numbers, whereas the natural numbers correspond to finite bit strings. The same set of strings can't represent both.
As it happens, your list is only the natural numbers. Note that for the nth string, all the digits from position n on are 0 (as your comment re: the diagonal slash requires). Thus, the nth string in your list contains fewer than n 1's; in other words, every string in your list has only finite 1's. Some real numbers in [0,1) have binary expansions with infinite 1's. Ergo, your list does not include the binary expansions for all such real numbers.
Similarly, you haven't encoded all of the subsets of the natural numbers, just the FINITE subsets. Those are in fact equinumerous to the natural numbers themselves.
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Re: |power set of integers| = |reals|? (Score:1)
by Old Wolf on Friday May 26, @07:48AM EDT (#411)
(User Info) |
What is:
0100000000...
pray tell?
It certainly doesn't look like an integer from here
Now, supposing you have replied with a definition:
How is it different to 00100000000... ?
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Re: |power set of integers| = |reals|? (Score:1)
by jejones on Friday May 26, @11:04AM EDT (#428)
(User Info) |
The mapping is not onto [0,1); you'll never hit any of the irrationals, which have nonterminating expansions (so that there's always at least one 1 bit any given finite integer will have missed), with your sequence.
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Re: |power set of integers| = |reals|? (Score:1)
by krimson on Monday May 29, @07:42AM EDT (#458)
(User Info) |
The flaw is that you assume that Cantor's diagonalisation theorem can only be applied to one ordering of the image of the integers. By reordering them and appllying the theorem you can show that there are many different real numbers that are not in the image of the mapping. Most of these will in fact be members of ]0,1[.
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*NEW* problems? (Score:5, Interesting)
by Kinthelt on Thursday May 25, @10:17AM EDT (#7)
(User Info) http://www.undergrad.math.uwaterloo.ca/~db2nesbi
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| We haven't even finished solving the all of the original 23 problems! One great example is Goldbach's Conjecture: All even numbers greater than 2 are the sum of two prime numbers. Mathematicians have tried for years to prove or disprove this one (or to prove it is unprovable), but still haven't come up with an answer. I think the thing that irks most mathematicians is the fact that it is so short and elementary.
"Evil will always triumph over good, because good is dumb." - Dark Helmet (Spaceballs) |
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Re:*NEW* problems? (Score:2, Funny)
by DGregory ([email protected]) on Thursday May 25, @10:23AM EDT (#25)
(User Info) http://www.feick.net
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Isn't there a "largest" known prime number? Take double that + 2 and isn't it disproven?
Of course I'm the furthest thing from a mathematician and I was really happy when I knew the answer to "the longest side of a right triangle" on Who Wants to Be a Millionaire, so I'm probably wrong but that's what comes to mind if I recall some of the math I had way back when.
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Re:*NEW* problems? (Score:1)
by groen ([email protected]) on Thursday May 25, @10:30AM EDT (#44)
(User Info) |
| Isn't there a "largest" known prime number? Take double that + 2 and isn't it disproven? Wow, you've solved goldbach! Just kidding. No, there are infinitly many primes due to a trick of Euclid: suppose there are finitely many: P1,...pn. Multiply and add one: p1...pn+1. This number is divisible by a prime number (as are all numbers) and this number cannot be p1 or p2 or ... or pn. So, this is a new prime number. Contradiction. Richard
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The Infinitude of Primes (Score:2, Interesting)
by FascDot Killed My Pr on Thursday May 25, @10:31AM EDT (#50)
(User Info) |
"Isn't there a "largest" known prime number?
"Known", yes. Just plain largest, no. Proof:
Let n1, n2, n3...nx be all known prime numbers. Construct M = n1 * n2 * n3 * ... * nx + 1. Is M a compound number (i.e. non-prime?)? Well, it isn't factorable by any known primes (it would always leave a remainder of 1). So we have two cases:
Case 1: M is prime. But it must be a previously unknown prime and it is clearly larger than any previously unknown prime. Therefore there exists another largest prime.
Case 2: M is not prime, but isn't factorable by any known prime. Therefore it is factorable by some unknown prime larger than all known primes.
Since this process can be repeated indefinitely, there must be an infinitude of primes.
--
Have Exchange users? Want to run Linux? Can't afford OpenMail?
Try MailOne for Linux! |
Re:The Infinitude of Primes (Score:1)
by Wiktor Kochanowski (alterego$cheerful,com (you know what to alter)) on Thursday May 25, @08:44PM EDT (#357)
(User Info) |
Case 2: M is not prime, but isn't factorable by any known prime. Therefore it is factorable by some unknown prime larger than all known primes. Untrue. It might be factorable by an unknown prime smaller than the larger known prime; there may exist such primes (and do now). So your proof is erroneus and anyway doesn't prove anything useful (that there is an infinite number of primes has been known for quite a while, heh). Let's toy with the idea that there may actually be a largest possibly known prime. Of course, if we knew such a prime, it would be theoretically possible to examine all the numbers smaller than it for being a prime, and according to the Euclidean disproof construct the next prime, but such operation might involve checking so many numbers that it would not be computable with the resources available in the universe. Of course, there are smarter ways than that to do it and this is why we are still finding new primes, even though the currently known largest primes are far into the realm of incomputable using e.g. the Eratosthenes' sieve.
-- Wiktor |
Re:The Infinitude of Primes (Score:1)
by Old Wolf on Friday May 26, @08:18AM EDT (#417)
(User Info) |
| Uh, grow a brain. It might be factorable by an unknown prime smaller than the larger known prime; Go back and read his proof. Just near the start, he places the assumption that all primes up to the largest known prime.
What's more, this is a safe assumption because if a prime is known, all numbers smaller can then be checked for primeness. So your proof is erroneus and anyway doesn't prove anything useful (that there is an infinite number of primes has been known for quite a while, heh). 1) It's exceedingly useful in pure mathematics to know that an infinitude of primes exists,
2) the point of this proof was to prove that, so I think he has succeeded admirably Let's toy with the idea that there may actually be a largest possibly known prime. There is a largest known prime. And it is definitely known, not just possibly.
At some point in the future, there will be a new largest known prime. The rest of your post seems to be degenerate drivel which talks about computing times (what have they got to do with anything?)
I don't think anyone here would be stupid enough to say that there are not a finite, fixed number of primes in a given finite range. (for example, between 1 and largest-known-prime), even if we don't know what they are yet.
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Re:The Infinitude of Primes (Score:1)
by Wiktor Kochanowski (alterego$cheerful,com (you know what to alter)) on Saturday May 27, @05:02PM EDT (#449)
(User Info) |
>>It might be factorable by an unknown prime
>>smaller than the larger known prime;
>Go back and read his proof. Just near the start,
>he places the assumption that all primes up to
>the largest known prime.
That's right - I missed it. I thought his post declared that we only know the largest prime, not necessarily all the primes up to it.
Anyway,
>What's more, this is a safe assumption because if
>a prime is known, all numbers smaller can then be
>checked for primeness.
But this is your own completely unwarranted assumption. Try checking all the numbers up to 2^234567-1, or whatever the current largest is. Which is precisely why I am talking about computing times.
>1) It's exceedingly useful in pure mathematics to
>know that an infinitude of primes exists,
>2) the point of this proof was to prove that, so
>I think he has succeeded admirably
Actually, it was one of the ancient Greeks, can't remember which one, who succeeded admirably on that. The point being that there's nothing useful in quoting a proof that a) has been known for 2500 years b) has been quoted a few times before in this discussion c) everyone has been learning about in Number Theory 101.
>>Let's toy with the idea that there may actually
>>be a largest possibly known prime.
>There is a largest known prime. And it is
>definitely known, not just possibly.
Now, sir, I'd recommend the brain-growing exercise for you (I'm told yeast does it). I'm speaking here of a largest possibly known prime in the sense that:
a) it may be possible to prove that it's impossible to find any prime beyond a certain number by any means other than the brute force of e.g. Eratosthenes' Sieve. Admittedly, it's not very likely, but that's why I'm saying "Let's toy with the idea".
b) in which case, computation times have a lot to do about whether the prime is actually the largest possible or not.
>The rest of your post seems to be degenerate
>drivel which talks about computing times (what
>have they got to do with anything?)
Well, read up about how the primes are being found nowadays? If you think that computation has nothing to do with proofs in contemporary mathematics, I suggest that you read up on the theorem about coloring the map with only four colors.
>I don't think anyone here would be stupid enough
>to say that there are not a finite, fixed number
>of primes in a given finite range. (for example,
>between 1 and largest-known-prime), even if we
>don't know what they are yet.
But it may, just may still be impossible to find them all. See above.
PS. How about getting a life instead of spewing insults around?
-- Wiktor |
Re:The Infinitude of Primes (Score:1)
by Old Wolf on Monday May 29, @10:51AM EDT (#460)
(User Info) |
This is pure mathematics, not computer science.
Computation questions and times are of no relevance whatsoever.
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Re:*NEW* problems? (Score:1)
by pe1rxq ([email protected]) on Thursday May 25, @10:32AM EDT (#54)
(User Info) http://www.pe1rxq.ods.org/
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| Isn't there a "largest" known prime number? No, there will always be bigger numbers, and some of them will be primes. Just like you can't define the value of infinity. Jeroen
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Re:*NEW* problems? (Score:1)
by Kinthelt on Thursday May 25, @10:34AM EDT (#62)
(User Info) http://www.undergrad.math.uwaterloo.ca/~db2nesbi
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| Isn't there a "largest" known prime number? Take double that + 2 and isn't it disproven? Actually, there is a proof that there is no largest prime number. To use it, you need to know the Unique Factorization Theorem (which I'm not ready to explain on /.) I just realized that you had a "known" in there... Yeah. The largest known prime number is of the form (2^p)-1, also known as a Mersenne prime.
"Evil will always triumph over good, because good is dumb." - Dark Helmet (Spaceballs) |
Re:*NEW* problems? (Score:1)
by krimson on Monday May 29, @07:46AM EDT (#459)
(User Info) |
I do not think you have to know the unique prime factorisation theorem. It is enough to know that every non-prime has a smallest divisor which is a prime.
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Score -1, stupid (Score:1)
by st.n. (st.n at gmx net) on Thursday May 25, @10:46AM EDT (#83)
(User Info) http://2130706433/
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Isn't there a "largest" known prime number? Take double that + 2 and isn't it disproven? Sometimes I wish there would be a moderation option -1 stupid. It's an enormous difference between largest known and largest! And it's a very basic proof to show that there is no largest prime, you will always find a bigger one if you search long enough. That was already proven by some ancient greek more than 2000 years ago. - Stephan. --
Carpe diem!
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No (Score:2)
by SimonK ([email protected]) on Thursday May 25, @10:59AM EDT (#109)
(User Info) |
There is no largest prime number. This is one of the very few things about prime numbers that can be proven.
If there were a largest prime number, there would be finite list of primes you could write down. If you took all of these numbers and multiplied them together, then added one, you would get a new prime number.
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Re:No (Score:1)
by Tony-A on Thursday May 25, @01:40PM EDT (#240)
(User Info) |
or a composite number with prime factors not in the list of primes.
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Nitpicking Re:No (Score:1)
by Pyotri on Thursday May 25, @07:58PM EDT (#335)
(User Info) |
You missed the point. - Assume that there are finite primes.
- It is possible to multiply all of these primes, and add one to it. This new number exists.
- This number does not have any of the other primes as a factor, so it doesn't exist. (Prime factor theorum.)
- Contradiction.
QED
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Nit well picked. (Score:1)
by Tony-A on Saturday May 27, @08:11PM EDT (#450)
(User Info) |
I like yours better. If there were a finite number of primes, then p1*p2*...*pn+1 could not exist.
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Re:*NEW* problems? (Score:2, Interesting)
by (void*) (voice@void.) on Thursday May 25, @11:14AM EDT (#131)
(User Info) |
| Seeing how that nobody answered your question, let me try. Let n be the largest known prime number. Now is n+2 itself prime? (You have to prove/disprove this one)If it is, (just like 5 and 7) then you've just increased the largest known prime number by 2. Then it is just another instance of Goldbach's conjuecture. If it is not prime, things become more interesting. Is there some primes less then n, such that when you add them together, you get n+2? If you can prove that there isn't, then you have disproven Goldbach's conjecture. But chances are, there is, since we can do (for example) 3 + 3 + 3 + 3 + ... and use that hit a lot of numbers. Good luck finding that number missed by ALL such sequences ....
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Re:*NEW* problems? (Score:2)
by Dr. Evil (http://) on Thursday May 25, @01:10PM EDT (#224)
(User Info) |
He said double the largest known prime and add two. That way the result is gauranteed not to be a prime (it's even!) and it is not the sum of two primes. This and the original post said the sum of two primes. Otherwise we could just keep adding one to our heart's content. What may also apply is that the primes may be positive and negative. I don't know. Now it's just a matter of finding two prime factors which sum to a number twice the largest known prime plus two :-)
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Re:*NEW* problems? (Score:1)
by tbarrie on Friday May 26, @02:47AM EDT (#394)
(User Info) |
| This and the original post said the sum of two primes. Otherwise we could just keep adding one to our heart's content. 1 is not prime. OTOH, you're right about being limited to only 2 primes... you can trivially construct any odd number greater than 1 by adding 3 and some number of 2's, and construct any even number by adding 2's.
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Re:wrong (Score:2)
by Dr. Evil (http://) on Thursday May 25, @05:40PM EDT (#311)
(User Info) |
He said double the largest known prime and add two. That way the result is garaunteed not to be a prime (it's even!) and it is not the sum of two primes. wrong, its not going to be the sum of two known primes. Oh that's just splitting hairs... I actually meant "the" two primes. You're absolutely correct, but I think it was quite clear that I didn't think it a solution for the problem when I said that all you had to do was calculate which two primes sum to that total. What may also apply is that the primes may be positive and negative. I don't know. All primes are positive by definition. When my statement is qualified with the words "I don't know", you certainly can't fault me for claiming to be an authority on the matter :-) But thanks for the correction.
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Re:wrong (Score:1)
by Old Wolf on Friday May 26, @08:27AM EDT (#418)
(User Info) |
It isn't splitting hairs, it's the crux of the problem.
Goldbach's conjecture doesn't require that you know what the primes are, you only have to prove that they exist.
It's conceivable that there is a proof that Goldbach's conjecture is false, but it won't be able to tell you what the numbers involved are. (This sort of thing happens in number theory all the time.)
BTW, I can't believe the exceptional lameness that is going on in this thread. Whichever is the largest "known" prime is completely irrelevant. We are not trying to answer questions about a certain subset of numbers that we happen to know at the moment - these problems apply to *all* natural numbers!
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Re:*NEW* problems? (Score:1)
by throbbin on Thursday May 25, @07:05PM EDT (#328)
(User Info) |
I seem to be confused with the proof that if there are only n prime numbers then M=p1*...*pn + 1 is prime. Just because M isn't divisible by prime numbers doesn't mean that it is prime. It also has to not be divisible by any non-prime numbers as well. This rule doesn't work for the primes p1=1, p2=3, p3=5 where M=16 which is obviously not prime.
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Re:*NEW* problems? (Score:2)
by MalcolmT ([email protected]) on Thursday May 25, @10:41PM EDT (#372)
(User Info) |
It is not the case that M (in the previous) post must be prime. It is possible for it to be divisible by some prime greater than pn. The smallest such example is 2*3*5*7*11*13 + 1 which is 59*509 (both of which are primes). Secondly, checking divisibility by primes is enough, because if some non prime N was a factor of M, then any prime that divided into N would also divide into M. So the only factors of N (and it must have non-unit factors, since it is not a prime) can be composites (non primes). Then apply the same process to them (i.e. they can only have composite factors) and so on. The problem is that your factors are always getting smaller and are always positive. Eventually you run out of positive integers, so the original assumption that N has only composite factors must be wrong. (OK, that last paragraph was a little more complicated that I was intending it to be.)
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Re:*NEW* problems? (Score:1)
by morat on Friday May 26, @05:12AM EDT (#407)
(User Info) |
There is indeed a largest prime number obtainable by humans, it just cannot be bounded. Consider this. The number of integers ever considered by humans, including computers, are finite. Therefore, the number of primes ever known to man is finite, and one of those are the largest one. Of course, once you have a large number, you can construct an even larger prime. So if you have a bound, you can construct a larger prime, thus the largest prime cannot be bounded. 8-) - K
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Re:*NEW* problems? (Score:1)
by Jester99 ([email protected]) on Thursday May 25, @10:27AM EDT (#37)
(User Info) http://www.concentric.net/~kimballa/
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One great example is Goldbach's Conjecture: All even numbers greater than 2 are the sum of two prime numbers. Greater than 2? 1 is a prime number; 1+1=2... #include
gcc: something/funny.h not found
-- Aaron Kimball
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Re:*NEW* problems? (Score:1)
by Kinthelt on Thursday May 25, @10:29AM EDT (#40)
(User Info) http://www.undergrad.math.uwaterloo.ca/~db2nesbi
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| Greater than 2? 1 is a prime number; 1+1=2... I said "Greater Than", not "Greater Than or Equal To". So two is not inclusive. Start the even numbers at 4. And 1 is not a prime number.
"Evil will always triumph over good, because good is dumb." - Dark Helmet (Spaceballs) |
Re:*NEW* problems? (Score:1)
by norton_I ([email protected]) on Thursday May 25, @10:52AM EDT (#96)
(User Info) |
Acording to number theory ( I repeat, number theory. Not necissarily your idea of common sense) 1 is not a prime.
All elements of a ring (type of set, a notable element being the integers) are either units, prime, or composite. For the integers, the "units" are {1, -1}, and all numbers can be written as the product of a unit and a unique set of positive primes.
The important thing is the unique representation, which you can't get unless you use the "unit", otherwise 6 = 2*3 = 1*2*3 = 1*-1*-1*2*3. While it is sort of a hand waving excercise for the integers, it is more important for other rings, such as polynomials.
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Re:*NEW* problems? (Score:1)
by pal on Thursday May 25, @12:34PM EDT (#197)
(User Info) |
1 is _not_ a prime number. there are many different, but equivalent, definitions of a prime number. none of them include 1.
basically, a prime is a non-unit that cannot be written as the product of two non-units. a non-unit is something that doesn't have a multiplicative inverse in the system (ring) we are looking at. for instance, 2 has no inverse (which we usually would call 1/2) in the integers, and any product that includes two non-units cannot be 2. so 2 is prime.
however, 1 has an inverse -- itself.
if we permitted units to be considered prime, we would totally mess up the idea of "unique factorization," which the primes are pretty much designed for.
- pal
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Proof for Goldbach's Conjecture (Score:2, Interesting)
by Cynic (cynic@(the ever-so-wonderful)execpc.com) on Thursday May 25, @10:58AM EDT (#108)
(User Info) |
A quick Google search:
http://www.mathematical.com/mathgold1.html
Now I'm no mathematician, so I obviously didn't check to make sure they carried all of their ones, but at a glance, it looks ok. (How's that for an endorsement?)
Cynic
http://napalm.firest0rm.org/
"You don't burn out by going too fast. You burn out by going too slow and getting bored." -Cliff Burton |
Goldbach *not* undecideable. (Score:5, Informative)
by divec on Thursday May 25, @11:07AM EDT (#125)
(User Info) http://3334130452/
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Goldbach's conjecture is a first-order statement, so by G�del's Completeness Theorem (which is *not* the same as his Incompleteness Theorem) there is either a proof or a disproof.
A "first-order statement" is one which can be formulated without saying "For all X" or "Exists X" for any infinite set X. (It's ok to say "for all/exists x" where x is just a number, or a set element. Also I'm not stating this mathematically precisely).
Goldbach is stated as "for all x there exists y,z such that y prime and z prime and y+z=x". "y prime" is a first order statement: "for all j, j > y or exists k such that jk y".
So there *is* a (dis)?proof of Goldbach, just waiting for someone to find it.
[Sig obliterated ;-) ] |
Re:Goldbach *not* undecideable. (Score:2)
by divec on Thursday May 25, @11:10AM EDT (#128)
(User Info) http://3334130452/
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Grr I forgot about > being munched. That last line should have read:
"y prime" is "for all j, j > y or exists k such that jk < y and j(k+1) > y".
[Sig obliterated ;-) ] |
Re:Goldbach *not* undecideable. (Score:4, Interesting)
by Chalst (cas-at-achilles.bu.edu) on Thursday May 25, @11:39AM EDT (#152)
(User Info) http://achilles.bu.edu/cas
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This is not true. G�del's completeness theorem only applies to
primitively recursive axiomatisations in first-order logic.
Arithmetic, as G�del's incompletensss theorem's show, cannot be
captured by by such an axiomatisation. Goldbach's conjecture is a
Pi^0_1 statement of arithmetic (a measure of the intrinsic logical
complexity of a statement that includes the G�del sentence), so as far
as we know, we might not know of a formalisation capable of proving it
to be true.
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Goldbach *not* first order. (Score:4, Interesting)
by sh_mmer on Thursday May 25, @11:49AM EDT (#164)
(User Info) http://www.ee.ucr.edu/~jthorpe
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Goldbach's conjecture is a first-order statement...
come on: Goldbach obviously is *not* first order. let me slightly rephrase Goldbach, and you will see:
"for every integer n > 2, there exist positive primes p and q such that p+q = 2*n."
that's the proper formulation, and hence the conjecture is apparently *not* first order!
d'accord?
sh_
GPL is not free. public domain is free. |
Re:Goldbach *not* first order. (Score:2)
by Chalst (cas-at-achilles.bu.edu) on Thursday May 25, @02:23PM EDT (#259)
(User Info) http://achilles.bu.edu/cas
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Goldbach's conjecture is first-order. There is no quantification over proerties or sets of numbers in it.
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Re:Goldbach *not* first order. (Score:2)
by orabidoo ([email protected]) on Thursday May 25, @03:02PM EDT (#277)
(User Info) http://www.iagora.com/~espel/index.html
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sure, it's first order. but there are plenty of undecidable first-order statements, so this proves nothing about this problem being decidable or not.
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Re:Goldbach *not* first order. (Score:2)
by Chalst (cas-at-achilles.bu.edu) on Thursday May 25, @03:37PM EDT (#283)
(User Info) http://achilles.bu.edu/cas
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divec's original post is simply ignorant, as I pointed out in this
post. Goldbach's conjecture is first-order, but it is a first
order sentence of a theory which cannot be given a primitive recursive
axiomatisation. A lot of people here are confused about language
vs. theory and decidability vs. completeness: back to the textbooks, I
think.
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Goldbach's Conjecture not important (Score:1)
by Habanero on Thursday May 25, @11:44AM EDT (#157)
(User Info) |
The difference, I think, is that while Goldbach's conjecture is interesting and hard; it's not important in the sense that it doesn't really have many applications. On the other hand, determining whether P=NP, or giving a mathematical foundation for quantum YM, is huge. Compare to Fermat's last theorem. Interesting, but who cares, really? Wiles' proof is celebrated because of his proof of Shimure-Taniyama-Weil which is a very important result with applications that people care about. I can almost hear the cogs grinding as you think by the same argument "The Riemann hypothesis" shouldn't have made the list since everyone knows it's true, so if it has applications they're already out there." Conceeded, but the proof of RH will most likely be very important, probably using breakthrough techniques in quantum algebra and geometry.
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In a related problem (Score:1)
by haystor ([email protected]) on Thursday May 25, @12:12PM EDT (#180)
(User Info) |
All Ip addresses are the sum of /. account id's.
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Re:*NEW* problems? (Score:1)
by randombit ([email protected]) on Thursday May 25, @12:53PM EDT (#216)
(User Info) http://www.randombit.net
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One great example is Goldbach's Conjecture: All even numbers greater than 2 are the sum of two prime numbers.
One of my favorite math quotes: "If you can state a problem in mathematics that's unsolved and over 100 years old, it is probably a problem in number theory" - Erdos
That's not exact, but as best as I can find at the moment. Very true, too.
-- (Remove the leters S,P,A,M and then rot13 to email) |
How many left? (Score:2)
by nano-second ([email protected]) on Thursday May 25, @01:27PM EDT (#230)
(User Info) |
How many are left?
... the only one I know about being solved is FLT.
---
I hope you're not pretending to be evil while secretly being good. That would be dishonest. |
Re:*NEW* problems? (Score:1)
by cvillopillil on Thursday May 25, @03:48PM EDT (#289)
(User Info) |
One great example is Goldbach's Conjecture: All even numbers greater than 2 are the sum of two prime numbers. I'm no math wiz. But The Real Steve Woston is. I'll take a shot at the prime numbers/even number sum problem, though.Now, we've got 4, right? that's the next even number after 2. So, 3=prime, 1=prime, and 3+1 = 4. So, it looks like the theory is actually true. :) :) :) Ok, on a more serious note, The Real Steve Woston is severely upset at being impersonated and trolled on Slashdot. I encourage you to vist his site and see what the man himself has to say about it. To the Steve Woston Troll: Please stop impersonating Steve Woston on Slashdot.
Why couldn't Microsoft have made NT a multiuser/multitasking version of MS DOS ? |
Re:*NEW* problems? (Score:2, Informative)
by pe1rxq ([email protected]) on Thursday May 25, @10:29AM EDT (#42)
(User Info) http://www.pe1rxq.ods.org/
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| All even numbers greater than 2 are the sum of two prime numbers. Well, there is the obvious number theory rule that odd+odd=even. Since all primes >2 are odd... Try to read before replying.... Jeroen
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Re:*NEW* problems? (Score:1)
by pe1rxq ([email protected]) on Thursday May 25, @11:44AM EDT (#158)
(User Info) http://www.pe1rxq.ods.org/
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This would no prove anything (you could however disprove it if you are lucky).
You can only prove that it is correct for all numbers up to the highest that you checked.
Jeroen
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Re:*NEW* problems? (Score:1)
by DEFCONry on Thursday May 25, @12:44PM EDT (#206)
(User Info) http://www-math.cudenver.edu/~fconry
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... because it would probably then indicate some sort of pattern in the generation of the list of primes ...
I thought about this problem a bit cause I think some one is also offering a million to solve this and I think I would prove it as follows (avioding the pattern in primes):
1)(this is the catch) get some kind of lower bound on the number of primes lower than a given even number.
2)Now, what we want is say our even number is m we want a prime p such that m-p is also prime. So first throw out all the even numbers and pair up the rest of the numbers less than m with thier 'm-p partners'.
3) Now we have say n primes and we can put one and only one into each of the [m/4] pairs so we if we have more than [m/4] primes two must go in the same slot (pigeon hole priciple) and blamo, we're done.
Well I guess we've reduced it to showing that the number of primes less than a given number, , must be greater than [m/4].
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Re:*NEW* problems? (Score:1)
by FascDot Killed My Pr on Thursday May 25, @10:35AM EDT (#63)
(User Info) |
Goldbach's Conjecture: All even numbers greater than 2 can be expressed as the sum of two primes.
What you've proved: All primes sum to an even number.
What you've left out: There may be some even numbers that don't "get hit" by summing any two primes.
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Missed one. (Score:3, Funny)
by Black Parrot on Thursday May 25, @10:17AM EDT (#8)
(User Info) |
I don't see anything about the famous and all-important karma maximization problem.
--
You can tell how desperate they are by counting the number of times they say "innovate" in their press releases. |
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It's been solved. (Score:3, Funny)
by FascDot Killed My Pr on Thursday May 25, @10:21AM EDT (#15)
(User Info) |
More than once, in fact.
Method 1: Follow story link. Cut random text and past to /..
Method 2: Create post such that title = "Why this is good for Linux" and body contains random text.
Method 3: Create post such that body contains "I am post anonymously because (insert random reason here)."
Method 4: Post something critical of Jon Katz (only works temporarily, karma rapidly falls off after 5 minutes).
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Re:It's been solved. (Score:1)
by saridder ([email protected]) on Thursday May 25, @10:23AM EDT (#24)
(User Info) |
You are so right.
"The way that all the bootleggers got shut down was by the end of Prohibition." --Rob Glaser on how to end digital pirating |
Re:It's been solved. (Score:1)
by Kinthelt on Thursday May 25, @10:25AM EDT (#30)
(User Info) http://www.undergrad.math.uwaterloo.ca/~db2nesbi
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Or you could always make a post claiming to have a solution to the Karma Maximization problem. That'd work, too. :)
"Evil will always triumph over good, because good is dumb." - Dark Helmet (Spaceballs) |
Re:It's been solved. (Score:1)
by theonetruekeebler ([email protected]) on Thursday May 25, @11:24AM EDT (#139)
(User Info) http://keebler.home.mindspring.com/
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Or you could always make a post claiming to have a solution to the Karma Maximization problem. Really? Let me try: I've solved the Karma Maximization problem! Unfortunately my proof is too large to fit in this Internet...
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It's never too late to panic. |
Re:It's been solved. (Score:1)
by theonetruekeebler ([email protected]) on Thursday May 25, @06:03PM EDT (#315)
(User Info) http://keebler.home.mindspring.com/
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| Well, it's been eight whole hours and no positive (or negative) moderation, so it looks like the theorem has been disproved. Just doing my part for science.
--
It's never too late to panic. |
None of you could solve these things, including me (Score:2, Interesting)
by saridder ([email protected]) on Thursday May 25, @10:17AM EDT (#9)
(User Info) |
Although it's interesting to see an academic body offer a lot of cash to solve these problems, they're asking the wrong crowd. The people who have a shot at solving these problems are probably the same people who are offering the cash. If they can't do it, how could the non-mathematicians do it? There are no Will Hunting's out here.
"The way that all the bootleggers got shut down was by the end of Prohibition." --Rob Glaser on how to end digital pirating |
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Re:None of you could solve these things, including (Score:1)
by ZarKov (zarkov{\at}netnitco{\dot}net) on Thursday May 25, @10:24AM EDT (#27)
(User Info) |
They didn't exactly hold this conference with the primary intention of showing these problems to slashdot. These are problems that (nearly) all mathematicians agree are important to solve. And there are plenty of good mathematicians who aren't the ones offering up the cash.
Anyway, what makes you so sure that "None of you could solve these things..."? Plenty of mathematicians read slashdot.
C:\DOS\CRASH>
Zarkov |
Could really none of you solve these things? (Score:1)
by st.n. (st.n at gmx net) on Thursday May 25, @10:32AM EDT (#53)
(User Info) http://2130706433/
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Although it's interesting to see an academic body offer a lot of cash to solve these problems, they're asking the wrong crowd. The world contains more Mathematicians that just those at the Clay Mathematics Institute. Considering the number of readers /. has, there are certainly quite a few good Mathematicians among them. I'm certainly not qualified enough even to try to solve any of these, but I always found those unresolved problems quite interesting, and it's refreshing to see this article on /., among the huge amount of political one! - Stephan. --
Carpe diem!
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Re:None of you could solve these things, including (Score:1)
by kill -9 $$ on Thursday May 25, @02:26PM EDT (#261)
(User Info) |
Boy are you a little negative and that is a very naive statement. What's to say that somebody (possibly even a non-mathematician) couldn't approach these problems with a different mindset. Consider a CS student who is really good at writing algorithms. Perhaps this student with a little mathematics background could come up with one hell of a program for performing numerical analysis/crunching on a given problem. (Possibly an angle not considered by an "expert") It may not prove the theorem, but it may provide some inroads into solving the larger problem. I'll agree that chances are a non-mathematician or a mediocre mathematician solving these is pretty slim, but to outright say it can't be done, to me is just dumb... Open your mind a little. One of the most influential statements I've seen was on my high school Civics teacher classroom wall: 'You never fail until you stop trying.' As corny as it may sound, I've lived by this philosophy and you be surprised how far a little persistance can carry you. One more for you, '...(we) do the other things, not because they are easy, but because they are hard.' (If you don't know who said it, do a web search on the saying and possibly include 1962 and moon) Think about em, they just might change your outlook on life...
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Re:None of you could solve these things, including (Score:1)
by Pyotri on Thursday May 25, @08:21PM EDT (#343)
(User Info) |
I'm afraid that I mostly disagree...
I'm a refugee from the Mathematics course at Cambridge, England. It was too tough for me, but I stuck around long enough to appreciate the sort of insight and analytical tools that a "real" Mathematician wields. IMHO, the only discipline that comes close is Physics.
Having said that, I dimly recall one exception. I wish I could remember the details, but it involved somebody who was asked to write a computer program to help analyse a particularly knotty conjecture. The guy who wrote the program came up with some pretty nifty shortcuts, and eventually used the insight that he'd obtained to prove the conjecture.
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Re:None of you could solve these things, including (Score:1)
by pugugly ([email protected]) on Friday May 26, @02:18AM EDT (#392)
(User Info) |
| Will Hunting don't exist. The occasional Ramnujen does though. The solutions of these kinds of problems seem (to me) to either rise from somebody that extrapolates from the most esoteric realms (My word for the day - esoteric. But in Mathematics, it fits) or somebody who takes relatively basic mathematics and goes into an entirely new direction with it. That second category is still possible out here in the real world. Or so I believe - [g]
This has been a test of the Slashdot Broadcast Network . . .
Had this been a real message, there would be real content . . . |
I'll do it (Score:1)
by MSisNOT4Sale on Thursday May 25, @10:18AM EDT (#10)
(User Info) |
First I need $500,000 up front before I would be able to finish the problems. I need the money to transplant einstein's brain.
When death looks you in the eye, smile. Someone needs to cheer him up. |
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Re:Einstein couldn't do simple math... (Score:1)
by t0m f00l on Thursday May 25, @11:15AM EDT (#132)
(User Info) |
Yeah, just like he failed primary school, because he was misunderstood, and the theory of relativity was just a good guess that anyone could have had. Wow. Now that we've established that, I could be a genius in disguise! *sigh of relief* And I thought I was just a mediocre mind. I feel so good now.
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Good thing this math isn't simple, eh? ;) (Score:1)
by Noel ([email protected]) on Thursday May 25, @01:11PM EDT (#225)
(User Info) |
Einstein's genius was in finding new (and better) ways of looking at things. Seems to me that's what's needed here, too...
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Re:Einstein couldn't do simple math... (Score:1)
by Gerund on Thursday May 25, @04:10PM EDT (#292)
(User Info) |
That story about Einstein doing terribly in primary school was never true. It was a joke he made at a press conference, because somebody asked him how he did in school. Most of the reporters there failed to realise he was pulling their leg, and reported it as true. People who like to repeat surprising "facts" have been spreading the story ever since. If Einstein had trouble with things like simple math, he would never have come close to his theory of relativity.
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Re:Einstein couldn't do simple math... (Score:2)
by mvw ([email protected]) on Thursday May 25, @05:06PM EDT (#303)
(User Info) http://www.freebsd.org/~3d
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| Einstein couldn't do simple math. He had to get someone to help him with geometry for his own theories... Nope. Replace 'simple' by 'esoteric non-trivial' and you come closer to real thing. Einstein was pondering about how to extend his theory of special relativity, which is suited for uniform motion to be extended to the case where acceleration happens. He was a genius in the art of thought experiments, and realized that if you are closed into an elevator without the means to look outside, you can't distinguish if you experience acceleration to the bottom of the elevator by a source of gravity or by someone accelerating/pulling the elvevator. More abstract, he concluded an eminent physical priciple: Dynamical and gravitational mass are the same! That the 'm' for mass that shows up in Newton's equation of motion is the same as the 'm' that shows up in the law of gravitation, the gravitational charge of materia so to say. He went one and found that this narrows the set of all possible physical laws down considerably. (Physical laws must be covariant). So he talked to his friend Marcel Grossman, that he should go to the library and look for a mathematical framwork that had this and that features. Like other people order pizza, Einstein orderd non-Eucledean geometry this way. Grossman could report to him that Bernhard Riemann, based on work of Carl Friedrich Gauss, had already developed a suitable theory a hundered years before. Differential Geometry on Manifolds in a modern speaking, Tensor calculus in more old fashioned formulation. That stuff is not easy. It is not standard curriculum in Physics and except for mathematicians, only some mechanical engineers in fluid and solid dynamics use that framework.
- Regards, Marc - |
whew! (Score:1)
by TheTick21 ([email protected]) on Thursday May 25, @10:19AM EDT (#11)
(User Info) http://www.apartment6.org
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thank goodness I though they were going to be hard problems....gimme 20 minutes gotta find my ti-89 and get out my big chief tablet and crayons
My Home: Apartment6
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Most of these are much harder than they seem. (Score:5, Insightful)
by Uruk ([email protected]) on Thursday May 25, @10:20AM EDT (#13)
(User Info) http://opop.nols.com/index.shtml
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Taking number theory in college and persuing math in my spare time, I've found out that sometimes the ones that look the easiest are actually the most nasty, bastardly, evil problems you've ever seen. Some may look easy, but you have to figure there's a reason why there's a $1 million prize on each question.
Some of the toughest ones are the "prove or disprove" type problems - for example, before fermat's last theorem was proven, (some people are sticklers for long periods of peer review and say that it hasn't been officially proven yet) it was pretty easy with the use of computers to prove that fermat's last theorem held for all integers up through some ungodly number. But proving through "some ungodly number" != proving through all numbers. Induction is the name of the game.
Math is always news for nerds, because programming is very much like the activity proving theorems - you go for simplicity, generality, and ABOVE ALL, conceptual elegance. The really cool part is that when you find a truly elegant, awesome proof, I almost always have that feeling of "damn! It was so OBVIOUS!" -- even though it wasn't, the elegance of the proof makes the concepts surrounding the program just gel perfectly.
I hope to see solutions like that for some of these problems. I'm not really all that great at number theory, but I really enjoy doing it and watching the big boys do it.
There are three types of people in the world; those who can count, and those who can't. http://opop.nols.com/index.shtml |
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Re:Most of these are much harder than they seem. (Score:3, Interesting)
by El Cabri on Thursday May 25, @10:51AM EDT (#94)
(User Info) |
>> Math is always news for nerds, because programming is very much like the activity proving theorems
Actually proving theorem and programming _are_ the same things under certain constraints (basically that you don't prove that the [the property being false] is false in order to prove that the property is true). This is called the Curry-Howard isomorphism and is the general idea behind many formalizations of programming.
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Re:Most of these are much harder than they seem. (Score:2, Interesting)
by Nyarly ([email protected]) on Thursday May 25, @12:24PM EDT (#190)
(User Info) http://slashdot.org/comments.pl?sid=azathoth
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| Math is always news for nerds, because programming is very much like the activity proving theorems - you go for simplicity, generality, and ABOVE ALL, conceptual elegance. And, actually, the Is P=NP question is vitally important to any sort of reasonable algorithms work. Travelling salesman is one problem that's easy to explain, and even easier to tie up huge amounts of resources with. But it's equivalent to natural language processing, for instance, and a lot of useful AI applications. Reducing any of those problems to polynomial time would be the same as reducing all of them, and the fact of the matter is that that is almost certainly worth far more than a measely $1e6. More like $1e(8 or 9). So, yeah, these are incredibly hard problems. And like all of the real stumpers in math, they begin with one question: where to start?
Ushers will eat latecomers.
Nyarly is annoyed that moderation is costing him karma. |
Re:Most of these are much harder than they seem. (Score:2, Funny)
by toofani on Thursday May 25, @02:22PM EDT (#258)
(User Info) |
P = NP if N = 1, or P is 0.
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Re:Most of these are much harder than they seem. (Score:1)
by snarkh ([email protected]) on Thursday May 25, @11:19PM EDT (#379)
(User Info) |
Most people have a misconception as to the practical importance of P=NP. It seems almost certain that even if P=NP, the polynomials you obtain will be of very high degree. From the point of view of practical computations there is no difference between time that is proportional to (say) x^20 and exponential time. Both make computations unfeasible.
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Re:Most of these are much harder than they seem. (Score:1)
by Nyarly ([email protected]) on Friday May 26, @04:28PM EDT (#437)
(User Info) http://slashdot.org/comments.pl?sid=azathoth
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| Actually, the real difference between polynomical and exponential time is in large problems, where it's going to take a while anyway. Functionally, though, if N is Large Enough, a polynomial begins to look linear, whereas an exponential never slows down. And large problems are exactly what these apply to. Say, for instance, natural language processing. N is the size of the recognized vocabulary and the complexity of the grammar, roughly. Imagine being able to conceivably using the OED as your vocabulary/grammar set. P=NP relates directly not necessarily to the speed of solutions (which will be slow, usually) but to the possibilty of reasonable solutions. That's a big difference.
Ushers will eat latecomers.
Nyarly is annoyed that moderation is costing him karma. |
Re:Most of these are much harder than they seem. (Score:1)
by snarkh ([email protected]) on Thursday May 25, @11:05PM EDT (#377)
(User Info) |
Gee, you don't say. People have only been trying to prove most of them for less than 200 years, they cannot be all that hard.
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Re:Most of these are much harder than they seem. (Score:2)
by Uruk ([email protected]) on Thursday May 25, @10:34AM EDT (#61)
(User Info) http://opop.nols.com/index.shtml
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Nonononon...the solution to fermat's theorem was most definately not obvious - I didn't mean that one, I just meant some other proofs that I've seen. You're right - the way the guy tied everything together to prove fermat's last theorem was the genius in that proof, but any proof so long as to require several PhD's years to read the hundreds of pages it contains is not what I would call obvious. :)
Ah, and the old 0.999999999... thing. Most people don't realize that it is 1 until you tell them this:
What does it mean for one number to be equal to another number? Well, if x = y, then x - y = 0. In this case, what's 1 - 0.9999999999999...? The answer is 0.000000000000....(infinite number of 0's). Since 0.000000000.... = 0, they are indeed the same number.
There are three types of people in the world; those who can count, and those who can't. http://opop.nols.com/index.shtml |
Re:Most of these are much harder than they seem. (Score:1)
by Xofer D on Thursday May 25, @10:47AM EDT (#84)
(User Info) |
Hmm...
What does it mean for one number to be equal to another number? Well, if x = y, then x - y = 0. In this case, what's 1 - 0.9999999999999...? The answer is 0.000000000000....(infinite number of 0's). Since 0.000000000.... = 0, they are indeed the same number.
I beg to differ. After that infinite number of 0's is the digit "1" at position infinity+1 (or -infinity-1 depending on how you index) - that is, perhaps a more helpful notation would be 0.00000...001, which != 0. More true would be to say that if x=1 and as y approaches 1, x-y approaches 0 so we can select a y to get x-y arbitrarily close to 0. Hence, as we all know, we can get close enough so that nobody cares any more, but mathematically speaking I'm afraid what you've been telling people has been subject to the underflow error! Ever work for Intel? :)
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Re:Most of these are much harder than they seem. (Score:1)
by Anomalous Canard (murphy(at)panix(dot)com) on Thursday May 25, @11:06AM EDT (#124)
(User Info) |
He's right, but in an intuitive and not a rigorous way. You're right in a more rigorous way (Where are the epsilons and deltas, man?!? That's how to "do" limits! Or Cauchy sequences, even!! ;^)
Recognize the same fundamental truth underlying your two different expressions of it.
Anomalous: inconsistent with or deviating from what is usual, normal, or expected
Canard: a false or unfounded report or story |
Re:Most of these are much harder than they seem. (Score:1)
by radish on Thursday May 25, @11:11AM EDT (#129)
(User Info) |
First of all infinity+1 = infinity. So we have your digit "1" at position infinity.
Now we can say that 1 - 0.99999... = 0.000...01
So how big is 0.000...01?? Well, 0.1 = 1/10, and 0.01 = 1/100, so it stands to reason that 0.000...01 = 1/infinity. Which is as we all know ZERO.
So, 1 - 0.9999..... = 0.0000...001 = 0
therefore 0.9999... = 1
QED ;-)
---- No I probably don't know what I'm talking about... |
Re:Most of these are much harder than they seem. (Score:1)
by RedGuard on Thursday May 25, @11:51AM EDT (#167)
(User Info) |
s/infinity/arbitrary large number/g
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Re:Most of these are much harder than they seem. (Score:1)
by spiro_killglance on Thursday May 25, @12:37PM EDT (#200)
(User Info) |
Re. .99999 = 1 Proof 1/3 = 0.3333333..... 3* LHS = 1, 3* RHS = 0.99999.... Therefor 1 = 0.9999999.... In ordinary Arthimatic anyway. But i think this may not be true if your using sureal or hyperreal numbers, where 1+infinity != infinity and infinitemissals are considered as valid numbers just like integers and real
If looks could kill, i'd be alright. |
Re:Most of these are much harder than they seem. (Score:1)
by Tony-A on Thursday May 25, @01:59PM EDT (#249)
(User Info) |
errrr,
1 - 0.9999... = 0.0000...
There is no ...001 at the end.
There is no end to put the 1 in.
It is just 0.0000... 000... 000... with no end.
The set of integers is infinite, but there is no infinity in the set. If you add infinity to the set of numbers, you break things like a+b=a+c implying b=c.
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Re:Most of these are much harder than they seem. (Score:1)
by kel-tor ([email protected]) on Thursday May 25, @10:09PM EDT (#367)
(User Info) |
| .333... = 1/3 = ��.3333
1/ 3
��� 10 ............
=.33333333 to infinity always ending in 1/3 a non decimal value that can only be expressed as .3333 to infinity. and therefore .3333... +1/3 x 3 == .9999... +3/3 or== .9999...+1 turning the final 9 in a non final series to 0 and adding 1 to the previous nine all the way back from infinity to arrive at ==1 division is a translation from fractions to decimal and limits reverse translation isnt something always lost in translation?
an image had better be worth a 1000 words-- it takes longer to download. (this message posted from my Debian X-box) |
Re:Most of these are much harder than they seem. (Score:1)
by digitalhermit on Thursday May 25, @11:17AM EDT (#133)
(User Info) |
Eh!? Infinity + 1? Index on infinity? There is no 1 at that position. The zeros stretch on interminably to the horizon and beyond. They don't end. They stretch beyond Aunt Harriet's farm, beyond the State Fair, beyond the Capitol, beyond the ocean, beyond the exotic, whispered lands of Katmandu or Nepal, across mountains, and still no sign of the end. Zeroes, zeroes everywhere, and not a nth in sight. Bearded gurus have conjectured that the One must exist, and all passed into Eternity, ever peering across the landscape, till Time took them. And still no One!
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Re:Most of these are much harder than they seem. (Score:1)
by Xofer D on Thursday May 25, @12:23PM EDT (#189)
(User Info) |
Eh!? Infinity + 1? Index on infinity? There is no 1 at that position. The zeros stretch on interminably to the horizon and beyond.
Ah yes, the road goes ever, ever on. I was trying to make a point, to a crowd who's used to dealing with arrays. Since I've seen a few of these comments popping up I'll handle it here:
Infinity + 1 is easy. Let's say x is an integer and y=x+1 and just for illustration let's say z=x. Now imagine the graph of y vs x imposed on z vs x - you'll have two lines (one for y, one for z) and everywhere on the graph, y will be higher than x, for any x. So if we send x out to infinity, y will still be bigger by exactly 1. y=infinity + 1. Although, just *try* getting a address space that size in any language but Haskell!
If you're still having trouble, try this thought exercise: Let's construct our infinitely repeating number like this: write 0.01, then insert 0's just to the right of the decimal forever. You send the 1 off "infinitely" to the right, but it is still there... see, I'm really describing a limit - you know that for all intents and purposes, it's close enough but the hair-splitting truth is that unless you *explicitly* neglect that 1 you cannot say that 0.0000...001 = 0. It is never true, no matter how many (even infinitely many) 0's you put in front of it. All you can say is that it's so similar nobody but Zeno cares.
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Re:Most of these are much harder than they seem. (Score:1)
by digitalhermit on Thursday May 25, @12:47PM EDT (#211)
(User Info) |
I think you're confusing a couple aspects of infinity. You cannot use standard arithmetic when dealing with infinity. I.e., infinity + 1 is still infinity. Here's a link to an explanation: Why Infinity...
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Re:Most of these are much harder than they seem. (Score:1)
by Xofer D on Thursday May 25, @02:42PM EDT (#268)
(User Info) |
infinity + 1 is still infinity.
I know that, which is why I said, "send x out to infinity", by which I meant take the limit of these two equations as x->infinity. They "are" both infinity, but y is still larger than z because we have y > z for all x hence since z=x then y > x for all x. I'm not using standard arithmetic, which of course doesn't make sense.
To give an example from that link you provided: So a polynomial like "x+2" would be considered larger than any constant (no matter how big that constant was). And I'm saying, that (replace "contstant" with "y") if you take the limit as x->infinity *and* as y->infinity then both will be "infinity" but that for the whole journey there then x+2 will be greater than y. Hence:
lim 1-(1-r) = 0
r->0
BUT
1-0.999... != 0, see? No limit, no go. The limit allows one to "cut to the chase" as it were, but we can't do that implicitly. We could say, "When 0.999... becomes 1, then it is equal to 1, but for the entire way there, it is less than 1".
Look, I'm starting to get the feeling that we're not communicating properly here. I'm talking limits of functions, and I think you think I'm talking arithmetic. Is what I've said here strictly wrong? Explain how - maybe I'm actually making a mistake in my thinking (although it doesn't seem to manifest in my calculus! :) ).
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Re:Most of these are much harder than they seem. (Score:1)
by digitalhermit on Thursday May 25, @04:17PM EDT (#295)
(User Info) |
Fire your calculus teacher. :)
We are most definitely not communicating -- different languages apparently. The error is in thinking that 1-0.999... 0. As we're not speaking the same language I can only offer an example:
Do you remember Philosophy 101 and Zeno's little problem of moving across a room? Say that you want to cover a distance and travel at x units per time unit. According to Zeno, you'd first need to cover half the distance. Then you cover half the remaining distance. Then half the remaining again, ad infinitum. Zeno believed that you would never reach your destination because there would always be a tiny half that would need to be crossed.
This all goes out the window when limits come into play. You can write the above as the limit of n,n(1)-->n(infinity) of 1/2^n. I.e., 1 + 1/2 + 1/4 + 1/8. If you perform the mathematics you get 2. Not a number arbitrarily close to 2, but 2.
This is borne out if you consider the time involved. For a given (x distance unit)/(time unit) you can cover x units in t units. E.g., if your rate is 2 m/s, you'll travel 2 metres per second. So your time to travel across Zeno's distance is the sum of the t's, just as the distance is the sum of the x's. But you don't end up in a state of suspended motion verging on that last 1/2^infinity. 2 seconds comes and goes.
As someone mentioned, this limit mathematics is what made the concept of infintessimals and paradoxes like Zeno's irrelevant.
To bring it back to .9999..., we only need to change the formula from 1/2^n to an expression yielding: 9(.1)^1 + 9(.1)^2 + 9(.1)^3.... 9(.1)^infinity. Do the summation and you get... 1.
Hope this clears it up.. KL
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Re:Most of these are much harder than they seem. (Score:2)
by dillon_rinker (dillonunderscorerinkerathotmaildotcom) on Thursday May 25, @12:53PM EDT (#215)
(User Info) http://slashdot.org/comments.pl?sid=dillon_rinker
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So if we send x out to infinity, y will still be bigger by exactly 1.
You seem to be laboring under the rather common misconception that infinity is a number (at least, not in the ordinare "real" number system). It is not. You cannot set x equal to infinity, so you cannot have y=infinity + 1.
If you're still having trouble, try this thought exercise: Let's construct our infinitely repeating number like this: write 0.01, then insert 0's just to the right of the decimal forever.
"Zero followed by a decimal point followed by an infinite number of zeroes followed by a one" does not represent a real number.
If you're interested in this sort of thing, I REALLY think you ought to study calculus (and I don't just mean take the class). The sorts of mathematical and logic errors that you are making peaked in the mathematical world in the 19th century and were put to rest when calculus was finally put on a mathematically rigorous foundation. I think if you understood the foundations of calculus you'd have a better idea of why you are wrong. If you don't understand those foundations, there is no way I can really describe to you why you are wrong. It'd be like trying to explain pointers to a non-programmer who's doesn't speak your language.
My personal /. discussion page |
Re:Most of these are much harder than they seem. (Score:1)
by Tungz10 ([email protected]) on Thursday May 25, @11:11PM EDT (#378)
(User Info) http://www.buffalo.edu/~jrross
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Real numbers are continuous. So, for any two numbers to be distinct, there always exists some number in between them.
You cannot describe a number between 0 and 0.00...01 so they must be the same number.
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Re:Most of these are much harder than they seem. (Score:1)
by Old Wolf on Friday May 26, @08:41AM EDT (#420)
(User Info) |
That's the most sensible thing I've seen in a long time. Have on one me, bud.
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Re:Most of these are much harder than they seem. (Score:1)
by pugugly ([email protected]) on Friday May 26, @02:34AM EDT (#393)
(User Info) |
| Infinity + 1 is easy. It's easier than that - Infinity +1 = 0, after you account for the infinite number of ones that roll over into the bit bucket. Unless your dealing in 2's complement notation, where where 0111 ... 111 +1 would roll over to 10000 ... 0000, or negative infinity. so yeah, that should work . . . as long as you remember that the two's complement inifinity is only half the size of the standard inifnity.
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Re:Most of these are much harder than they seem. (Score:2)
by dillon_rinker (dillonunderscorerinkerathotmaildotcom) on Thursday May 25, @12:22PM EDT (#188)
(User Info) http://slashdot.org/comments.pl?sid=dillon_rinker
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I beg to differ. After that infinite number of 0's is the digit "1" at position infinity+1 (or -infinity-1 depending on how you index) - that is, perhaps a more helpful notation would be 0.00000...001, which != 0.
Ummm - no, that'd be the case only if 0.999... = 0.9999...990. The sequence of nines does not terminate. The sequence of zeroes does not terminate either. Granted, the poster you are replying to is giving an intuitive demonstration rather than a rigorous proof, but you are proposing mathematical nonsense.
More true would be to say that if x=1 and as y approaches 1, x-y approaches 0 so we can select a y to get x-y arbitrarily close to 0.
This is true for a finite number of nines following the decimal point. The logic does not hold up if you make the leap to an infinite number of nines. In other words, 0.9999.... != 0.999....9990.
If I remember my infinite series correctly (it's been 10 years since calc 2 for me), 0.999 represents
oo -----> (think 8 on its side)
---
\ 9
� \ -------
� / 10^(-n)
/
---
n=1
I've long since forgotten how to sum the series rigorously, but I haven't forgotten the result; 0.99... = 1.
In summary, the arithmetic of infinite series is intuitively nonobvious, and slightly different from the arithmetic of
My personal /. discussion page |
Re:Most of these are much harder than they seem. (Score:1)
by Xofer D on Thursday May 25, @02:16PM EDT (#255)
(User Info) |
Interesting....
Granted, the poster you are replying to is giving an intuitive demonstration rather than a rigorous proof, but you are proposing mathematical nonsense.
Well, I sure as heck am not trying to prove anything, so I suppose that throws us both in that camp. As for nonsense, I think you're dismissing me a little out of hand.
I'm not going to beat around the bush with you here - I think you're misunderstanding because of a notational misrepresentation. What I think we both agree on is that
0.99 ~ 1 (that is, "is approximately equal to")
and that
lim (1-x) = 0
x->1
and even that, in pseudocode,
makeone()
{
real x = 0;
real i;
for(i=0.9;0;i/10)
x += i;
return x;
}
and makeone() ~ 1
But you won't convince me short of a real proof that 1 = 0.999... - I would be delighted for you to show me wrong, but please keep nonsense out of this until you can, Mr. Rinker. I mean, come on - if 0.999... was 1, we could just call it 1, couldn't we? Oh, WAIT, that's true by definition. Odd, that.
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The proof I remember... (Score:1)
by Snoochie Bootchie on Thursday May 25, @03:38PM EDT (#284)
(User Info) |
The "proof" I remember seein for this one uses a method for converting decimals (usually infinitely rrepeating) into fractions:
x = 0.9999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1
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Re:The proof I remember... (Score:1)
by kel-tor ([email protected]) on Thursday May 25, @09:52PM EDT (#361)
(User Info) |
actually didnt you just prove something new... that fractions != dot decimals, that they are very close, infinitely close, but that the actual decimal representation of .9999... is understood to equate to 3/3 ... but only becuase the answer to 1/3 times 3 is .99999 and actually depends on where you do the alegebretic (i cant spell) solution at (3x1)/3 or (1/3)x3... that alegebra says that the two are equal but that you just proved that they are only extremely close? Just wondering.
an image had better be worth a 1000 words-- it takes longer to download. (this message posted from my Debian X-box) |
Re:Proof (empirical) that .9999 ... 999 = 1 (Score:1)
by pugugly ([email protected]) on Friday May 26, @02:51AM EDT (#395)
(User Info) |
| Lets see if I remember my grammar school correctly - [g] 1) 0.99999999... * 10 = 9.99999999... 2) 9.99999999... - .99999999 = 9.00000000...
2b) Therefore 9.00000... = 9 * .999999... 3a) 9.00000.../9 = .999999 ....
- and
3b) 9.00000.../9 = 1 4) Therefore .999999.... = 1 QED It's an old trick for converting repeating decimals to fractions - Frankly I'm surprised I remembered it after this many years. I believe I'll be insufferable with no good cause now - [g] Pug
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Re:Most of these are much harder than they seem. (Score:1)
by mathfreak ([email protected]) on Thursday May 25, @12:47PM EDT (#210)
(User Info) |
One way a teacher told me to visualize infinity was to picture a hotel with an infinite number of rooms and all of them are booked. A man comes in and asks to get a room. Can he fit?
Yes, he can. You just take the man in the first room and move him to the second, take the man from the second and move him to the third, etc. Then, you have a free room.
It's not very intuitive, but then again, neither is infinity. For instance, in the hotel example, it seems weird because with a finite number of rooms - even if you keep increasing the number to some pornographic amount - you will never be able to fit that one extra man. But if there are an infinite number, you can alway keep switching rooms.
Now, what if an inifinite number of people showed up and wanted a room?
Chris
the MathFreak (not to be confused with a mathematician).
int a=10000,b,c=2800,d,e,f[2801],g;main(){for(;b-c;)f[b++]=a/5;for(;d=0,g=c*2;c-=14,printf("%. 4d",e+d/a),e=d%a)for(b=c;d+=f[b]*a,f[b]=d%--g,d/=g--,--b;d*=b);} |
Re:Most of these are much harder than they seem. (Score:1)
by kallisti ([email protected]) on Thursday May 25, @01:22PM EDT (#227)
(User Info) |
| Now, what if an inifinite number of people showed up and wanted a room? Move everyone to a room twice their current room, and interleave the new guests with the old. This works for "omega" guests where omega is "countable", there are larger cardinals which cannot fit into the hotel, but omega is easy. For much more, check out Rudy Rucker's Infinity and the Mind or Raymond Smullyan's Satan, Cantor and Infinity.
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Re:Most of these are much harder than they seem. (Score:1)
by Tony-A on Thursday May 25, @02:08PM EDT (#252)
(User Info) |
Simple. Just move the occupant of room 1 into 2, the former occupant of room 2 into room 4, the former occupant of room 3 into room 6, and so forth. You can now fit the new arrivals into the odd numbered rooms.
It's worse, actually. There are exactly as many fractions as there are prime integers. Count off fractions as 1/1, 1/2, 2/1, 1/3, 2/3, 3/2, 3, ... (or some such).
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Here's another way to look at .9999...=1 (Score:1)
by Metuchen ([email protected]) on Friday May 26, @12:51AM EDT (#388)
(User Info) |
Think about this: 1/3 + 2/3 = 1 ==> .33333... + .66666... = .99999... Therefore, .9 repeating must equal 1.
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Re:Most of these are much harder than they seem. (Score:2)
by LordNimon on Thursday May 25, @11:06AM EDT (#122)
(User Info) |
| What does it mean for one number to be equal to another number? Well, if x = y, then x - y = 0. In this case, what's 1 - 0.9999999999999...? The answer is 0.000000000000....(infinite number of 0's). Since 0.000000000.... = 0, they are indeed the same number. Actually, there's a much easier to understand proof that 0.999... is equal to 1. What's 1/9? 0.111111.... What's 2/9? 0.22222.... So what's 9/9? 0.999999..... Of course, 9/9 is 1.
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0.9 repeating [OT] (Score:1)
by hawkestein on Thursday May 25, @11:09AM EDT (#126)
(User Info) |
Ah, and the old 0.999999999... thing. Most people don't realize that it is 1 until you tell them this:
What does it mean for one number to be equal to another number? Well, if x = y, then x - y = 0. In this case, what's 1 - 0.9999999999999...? The answer is 0.000000000000....(infinite number of 0's). Since 0.000000000.... = 0, they are indeed the same number.
The real question is not "what does it mean for one number to be equal to another", but "What does a repeating decimal mean?". Mathematicians like well-defined things. Saying 0.9999999999... isn't too well-defined. However, if you instead say
Lim n->infinity Sum from 1 to n 0.9/10^n
then you have a more mathematically well-defined beast, which we can evaluate properly (to 1).
However, strictly, all you can say is, "You can get as close to 1 as you like without going over by adding enough 9's". This fact alone is fairly obvious to most people.
I think anything else is hand-waving, which can often get you into mathematical trouble, like when you try to reorder with alternating series.
Of course, I'm not a mathematician...
Will quantum computers run imaginary-time operating systems? |
Re:0.9 repeating [OT] (Score:2)
by dillon_rinker (dillonunderscorerinkerathotmaildotcom) on Thursday May 25, @12:57PM EDT (#218)
(User Info) http://slashdot.org/comments.pl?sid=dillon_rinker
|
I think anything else is hand-waving, which can often get you into mathematical trouble, like when you try to reorder with alternating series.
I'm not a mathematician either - never finishe dup the graduate degrees. But it looks to me like his proof could easily be made rigorous in the form which he describe. Define the notation so that 0.999... and 1.000... represent the appropriate infinite series, define the procedure for subtracting and borrowing, and presto! The outline he's given becomes fairly rigorous.
My personal /. discussion page |
Re:Most of these are much harder than they seem. (Score:1)
by perky ([email protected]) on Thursday May 25, @11:12AM EDT (#130)
(User Info) |
| In fact Peano defined equality of natural number slightly differently. 1 is a natural number (axiom 1). for any number x there exists exactly 1 number x' that is its successor (axiom 2). x' != 1 (axiom 3). If y' = x' , then x=y (axiom 4). that's how he defined number equality anyway, and went on to derive the whole of number theory of his time on the basis of these and the axiom of induction alone. Poincar� argued, however, that he used a cyclical arguament to show the principle of induction.
"Freedom is the by-product of economic surplus" - Aneurin Bevan
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Re:Most of these are much harder than they seem. (Score:2)
by dillon_rinker (dillonunderscorerinkerathotmaildotcom) on Thursday May 25, @01:01PM EDT (#219)
(User Info) http://slashdot.org/comments.pl?sid=dillon_rinker
|
Poincar� argued, however, that he used a cyclical arguament to show the principle of induction.
Which is why the principle of induction is generally introduced as a postulate in most courses that start with Peano's postulates.
My personal /. discussion page |
Re:Most of these are much harder than they seem. (Score:1)
by sconeu on Thursday May 25, @11:27AM EDT (#142)
(User Info) http://alumni.cse.ucsc.edu/~scottn
|
Simple proof.
x = 0.99999999...
10x = 9.99999999...
9x = 10x - x = 9.99999... - 0.9999999... = 9
9x = 9 => x = 1
"They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety." - Ben Franklin |
Re:Most of these are much harder than they seem. (Score:1)
by Alphons Clenin on Thursday May 25, @12:09PM EDT (#179)
(User Info) |
Use a geometric series to represent 0.99999999999999999999...
we get:
0.9999999...=9*1/10 + 9*1/100 + 9*1/1000 +...=
9*(1/10 + 1/100 + 1/1000 + ...)=
9*(1+1/10+1/100+...)-9 =
9*(Sum_{i=1}^infty (1/10)^i)-9
It is a well known fact (see http://mathworld.wolfram.com/GeometricSeries.html) that a geometric series converges to 1/(1-r) where r is the ratio of the nth to the (n-1)th terms and r is between zero and 1.
(Sum_{i=1}^infty (1/10)^i) is such a series, and so it converges to 1/(1-1/10). This gives us the following
.99999...=9*1/10 + 9*1/100 + 9*1/1000 +...=
9*(1/10+1/100+1/1000...)=
9*(1+1/10+1/100+...)-9=
9*(Sum_{i=1}^infty (1/10)^i)-9=
9*(1/(1-1/10))-9=
9*10/9 - 9 =
9*10 - 9 = 1 qed beotch
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Re:Most of these are much harder than they seem. (Score:1)
by mw ([email protected]) on Friday May 26, @12:01PM EDT (#430)
(User Info) |
> 9*(1/(1-1/10))-9= > 9*10/9 - 9 = > 9*10 - 9 = 1 qed beotch ^^^^^^^^^^^^ Hmm..., I thought 9*10-9=81, but I'm no mathematician...
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Re:Most of these are much harder than they seem. (Score:1)
by kzinti ([email protected]) on Thursday May 25, @12:44PM EDT (#208)
(User Info) http://jimthompson.org/
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Ah, and the old 0.999999999... thing
Wow, so many posts over a simple summation of an infinite series. I wonder what this crowd would do with the old "Monty Hall" problem?
--Jim
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What would we do with Monty Hall? (Score:1)
by phossie on Thursday May 25, @06:35PM EDT (#324)
(User Info) |
Well, I'll show you. First, trot out the problem, a problem which most people who don't already understand it will think is nonsensical and obvious. Like this:
You are allowed to choose between three doors. Behind one door is a prize - behind the other two, goats. After you choose a door, Monty Hall (the gameshow host) will reveal what's behind one of the other doors (a goat), and then ask you whether you'd like to stay with your original choice or choose the other door. The problem: should you choose the other door? Most people will say that it doesn't matter, as it's a "50/50 chance" that you'll win or lose. Most people will be wrong, since they haven't thought about the problem in a set-oriented way. (Excuse my forthcoming explanation, I'm into this by way of philosophy and formal logic.) The problem seems simple on a small scale. Consider what happens when you have 100 doors, and the process keeps iterating until you're down to (again) two doors. At first, the likelyhood you'll choose the correct door is 1/100. If you hold on to that choice through all the subsequent rounds, including the last round, your likelyhood of correctness is still 1/100. On the other hand, if you choose a different door in the last round, you are effectively using Monty's knowledge against him - he will have eliminated many many bad choices, and the likelyhood that you initially picked the right door is much lower than the likelyhood that he had to consciously avoid a (1 of 99) different door. If you picture this with actual doors, it's a bit more intuitive (if that means anything to you). For the mathematical basis and the actual relationships, which I'm not qualified to give, try this site, or search google. There's quite a large community trying to educate the world about this one. :)
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Re:Most of these are much harder than they seem. (Score:1)
by Fyndo ([email protected]) on Thursday May 25, @06:29PM EDT (#322)
(User Info) http://fyndo.myip.org/
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| Here's a alightly more rigorous proof: x = 0.9999.....
10x = 9.9999....
10x - x = 9
9x = 9
x = 1
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Re:Most of these are much harder than they seem. (Score:1)
by Old Wolf on Friday May 26, @08:50AM EDT (#421)
(User Info) |
I don't really see how you can say "slightly more rigorous", since these two both have no rigour whatsoever. Your argument can have the same criticisms levelled against it as the original argument (and in fact I see that someone has).
Anything rigorous will, for starters, not include "...".
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Re:Most of these are much harder than they seem. (Score:1)
by Old Wolf on Friday May 26, @08:35AM EDT (#419)
(User Info) |
Actually, that won't even convince most people (I know this from experience), there are some extremely thick heads out there.
In fact, scrolling down, I see there are all these posts on the matter. *laughs heartily* This is going to be a fun read. One to save for posterity.
Curiously, some people are more convinced by the argument that 1/3 = 0.333333333333....
so that 3/3 = 0.9999999999999999....
and 1 = 0.999999999999.....
One can only speculate about how this is more convincing... perhaps because people are more familiar with "1/3" than they are with 0.99999..
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That's easy enough... (Score:2, Funny)
by Dr Caleb (thedarkknight at(spam&eggs)hushmail dot com) on Thursday May 25, @10:21AM EDT (#14)
(User Info) |
| I just want to be able to prove (or disprove) that the amount I think is in my chequing account has a direct relationship to the amount that is actually in the account. I'd pay good money for that!
bash: (e_non-seqitur) Program not derived from: printf("hello world!\n");
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Re:That's easy enough... (Score:1)
by ocelotbob ([email protected]) on Thursday May 25, @10:31AM EDT (#52)
(User Info) |
Hey, it really is easy...
c = money in account
t = money thought to be in account
p = percentage of the moon that is full
the equation is:
c = [t^2 * (.80 ^ p)] / [t * (2p)^2 + 1]
This explains why the amount you think you have stays steady while the amount that's really in the bank cycles seemingly randomly
Reality is all in your head |
Re:That's easy enough... (Score:1)
by Hooptie (ten.etg@2reprah) on Thursday May 25, @10:50AM EDT (#91)
(User Info) |
| I always thought that the ammount of money in my checking account was related directly to the number of checks I had remaining. Is this not true? Hooptie
A.D. 1517: Martin Luther nails his 95 Theses to the church door and is promptly moderated down to (-1, Flamebait). -- Yu Suzuki |
Uber-Math (Score:1)
by jayhawk88 (rockchalk88[AT]yahoo.com) on Thursday May 25, @10:23AM EDT (#22)
(User Info) |
Ouch, my brain hurts just looking at the descriptions of those problems. Funny thing is, though, that despite my total lack of of theoreticaly math knowledge, I probably have a better chance at solving one of these than hitting the lottery.
Seriously, though, how tough are these problems? Obviously hard, if their offering a million dollars, but anyone out there in-the-know care to comment on whether these are feasible? How many people in the world would have a legitimate shot at solving one?
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Re:Uber-Math (Score:1)
by TheSage on Thursday May 25, @10:43AM EDT (#78)
(User Info) |
I can not comment on all of the problems, but I give you my 0.02 euro on two of them. The Riemann Hypothesis: There is much progress beeing made torwards proving that. However, we still seem to be quite a long way from solving it. This is a very hard problem. I doubt, that this will be solved in the next 20 years. Navier-Stokes: J.L. Lions has already gotten us halfway to this one. From a cursory glance, this seems to be the "easiest" one of the bunch. My guess is, that this will be solved in this dekade (hopefully by me :).
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Re:Uber-Math (Score:1)
by Dr Caleb (thedarkknight at(spam&eggs)hushmail dot com) on Thursday May 25, @10:49AM EDT (#87)
(User Info) |
| I studied some higher math in University. These are very tough. They are generally accepted as true, but no one has been able to prove/disprove them yet using a general all-encompassing proof. Like we take Einstien's General relativity as true, yet no one has travelled at the speed of light to see if he was right. Anyone has a shot at solving them. The problem is creativity. Feremat's theory [X^2+Y^2=Z^2] was proven by a mathmetician by relating two previously unrelated theories and blending them to prove Feremat's Theory. His effort was most creative, and a little mind wobbling. The only problem for proving them is coming up with the right blend of other proofs and combining them in such a way that has never been done before. Or in being able to come up with your own, right out of the blue, such as Sir Isacc Newton did when he tried to study gravity, and had to invent Calculus! I'll stick to trying to balance my chequebook...
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Re:Uber-Math (Score:1)
by Tower (/dev/whoop-ass) on Thursday May 25, @11:21AM EDT (#137)
(User Info) |
>Feremat's theory [X^2+Y^2=Z^2]
Isn't that [X^3+Y^3=Z^3] (being impossible for integers?
-- "Funk the Dumb Stuff!" - ToP |
Re:Uber-Math (Score:1)
by sconeu on Thursday May 25, @11:50AM EDT (#166)
(User Info) http://alumni.cse.ucsc.edu/~scottn
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| Like we take Einstien's General relativity as true, yet no one has travelled at the speed of light to see if he was right. BZZZT! And thank you for playing! Subatomic particles accelerated to a large fraction of the speed of light do, in fact, behave as Special Relativity predicts. Clocks in free fall do in fact behave differently from those in a stable gravitational field (experiments with clocks in space). Different subject: Fermat. Fermat's last theorem: The equation a^n + b^n = c^n has no solutions a,b,c,n in the positive integers for n > 2. But you do get some lovely parting gifts.
"They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety." - Ben Franklin |
Re:Uber-Math (Score:1)
by Dr Caleb (thedarkknight at(spam&eggs)hushmail dot com) on Thursday May 25, @12:05PM EDT (#178)
(User Info) |
| It's been a while since my University days - please forgive the equation errors :-( But I was right in part! No one - being a person - has travelled at any velocity reasonabally approaching the speed of light, to experience the time-dialation effect! Do I get a copy of the board game? ;-)
bash: (e_non-seqitur) Program not derived from: printf("hello world!\n");
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Re:Uber-Math (Score:2, Funny)
by alkali ([email protected]) on Thursday May 25, @12:30PM EDT (#196)
(User Info) |
Hey, just got back. Why are you guys all so old?
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Re:Uber-Math (Score:1)
by Nate Eldredge on Thursday May 25, @01:33PM EDT (#236)
(User Info) http://www3.hmc.edu/~neldredge/
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| But I was right in part! No one - being a person - has travelled at any velocity reasonabally approaching the speed of light, to experience the time-dialation effect! True, but why should it have to be a person? AFAIK what's been done is this: You take a particle that has a known lifespan; it lasts for time T and then decays. You then create one moving at relativistic speeds. It has been observed, I believe, that the time it takes to decay in the observer's frame is greater than T.
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Re:Uber-Math (Score:1)
by drinkypoo on Thursday May 25, @02:43PM EDT (#269)
(User Info) http://down.on.fascination.st/~drink/
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The problem with this is that it is very difficult (some might say impossible, but nothing is impossible) to prove that something does not exist, or is not possible, whereas it is very easy (in the broadest variety of cases) to show that something IS possible under the proper set of circumstances, or at least to postulate about it. However, even in modern physics we have found ways around a number of things which we previously thought to be false (or true), and since we have no practical experience with anything other than light traveling at C, we really can't say whether or not Einstein was right. I lack the math to describe this elegantly, but what if your mass DOES increase as you approach C, but at (near) the final instant, it tapers off to almost nothing, or nothing? (Or even less than nothing, which might be interesting; Though probably not so interesting if you're experiencing it firsthand.) We can't say that it doesn't curve back the other way, because we haven't done it. Incidentally, this is why it's called the theory of relativity. It hasn't really been scientifically proven, though it may have been mathematically proven. One is not necessarily the other (And while I'm sure that will generate flames if it's read, it's NOT flamebait. I'm serious.) Now, I don't know if this has been debunked or not, but supposedly precision time devices flown around the planet a few times will show a discrepancy in time, so small that you couldn't measure it if it were just a Teiko Sports Watch(tm). Nonetheless, however, the difference is there, and can be measured, so we HAVE supposedly (I haven't seen it happen) measured relativistic time effects in the past. Again, I assert that that doesn't necessarily mean that the theory covers you all the way up to C, and I defy you to prove it to me without actually making something with significant mass (I'll settle for a gram!) move at C.
You are what you do when you count --Steakley |
Re:Uber-Math (Score:1)
by QuantumET ([email protected]) on Thursday May 25, @04:20PM EDT (#296)
(User Info) |
We do not have practical experience with anything traveling at C, because nothing with mass _can_ travel at C. Showing you the proof you want would invalidate the theory, since nothing with mass can reach C.
We have, however, accelerated protons, electrons, etc, to 0.999999% C, certainly close enough to detect any irregularities. And while there's always a gap, because there has to be, something like what you describe would probably violate enough laws of energy conservation to send physicists into the mental ward.
Not that energy conservation isn't violated in the quantum scale, but in the macroscopic level, it's not something you dismiss lightly.
And the theory of relativity has been proven just as much as any other scientific theory. The whole point of a scientific theory is that you can't prove it correct, you can simply show that nothing disproves it yet. A theory is as good as scientific thought gets. Every once in a while, something is called a law, like Newton's Laws of Gravity. But they're still theories, suspect to reinspection and study. (see what happened to Newton's laws?)
There are large amounts of experimental evidence for both theories of relativity. Special relativity predicts bending of light rays, and the slowing of clocks on moving platforms from your point of view, the relativistic increase in mass at high velocities, and some other things, as well. All have been observed in nature and in the laboratory. Particle physics wouldn't work at all if these equations weren't right.
General relativity is harder to test, but its predictions of gravitational redshift, black holes, and so on, have evidence. Some of the more estoteric predictions (like the twisting of spacetime around a massive spinning body) will most likely soon be tested. (The twisting bit will be tested by Gravity Probe B)
There is nothing unproven about the theories of relativity, at least when compared to any other scientific theory.
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Re:Uber-Math (Score:2)
by mvw ([email protected]) on Thursday May 25, @05:37PM EDT (#308)
(User Info) http://www.freebsd.org/~3d
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| One of the truly remarkable things about Einsteins theory of general relativity is that is not ivory tower stuff, pleasing some physicists, but actually this is put to use in ordinary every day devices: The GPS system (basically a bunch of high precision atomic clocks that send down their time) uses general relativistic corrections to achieve its high position resolution! On the other hand I know no application of the theory of strong interaction (Quantumn chromodynamics - think quarks) of any kind.
- Regards, Marc - |
Re:Uber-Math (Score:1)
by Fyndo ([email protected]) on Thursday May 25, @07:33PM EDT (#331)
(User Info) http://fyndo.myip.org/
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Well, general relativity is also a lot older than QCD, having been formulated in 1915,a full 57 years before QCD. It does take time for theoretical advances to be useful in ordinary every day devices. General relativity didn't have many practical applications in ordinary every day devices in 1943 either.
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Re:Uber-Math (Score:1)
by drinkypoo on Thursday June 08, @08:17PM EDT (#468)
(User Info) http://down.on.fascination.st/~drink/
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Actually, it says that you can't accelerate something with mass to C, right? Not that you can't make something move at C? ObNote: I am not a physicist, outside of hobbyism
You are what you do when you count --Steakley |
Re:Uber-Math (Score:1)
by Pyotri on Thursday May 25, @09:21PM EDT (#360)
(User Info) |
Moderate that loser as flamebait. The only reason I'm responding to this is in case somebody actually believes him.
Tachyons are a tentative hypothesis, and I don't think anybody believes that they exist. They are particles with imaginary mass (i.e. their mass is equal to the square root of a negative number.) A tachyon with no kinetic energy would be moving at infinite speed. As their energy increases, they slow down, and the lower limit on their speed is the speed of light.
Ordinary everyday particles with real mass are called tardons, and particles that move at the speed of light are called luxons. It all has a nice symmetry to it...
I've been trying to remember what little I know of General Relativity to show how this works, but have failed. If anybody else is interested, perhaps they could educate us all on this subject? :-)
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Re:Uber-Math (Score:1)
by Pyotri on Friday May 26, @06:37PM EDT (#441)
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| Fair comment. :-) You're right to pull me up over my earlier comments; I was quite arrogant, although IMHO the little twerp deserved it. Nevertheless, since I have stated a position, I feel honour bound defend it: Firstly, although C makes solving Physics & Engineering problems a lot easier, AFAIK the question is always in R, the answer is always in R, and it could, in principle, have been solved in R. (Excepting, of course, those who are exploring the frontiers of Theoretical Physics.) Secondly, just because the Mathematics works, doesn't mean it's true. Alright, that's a little facetious, but turn that on it's head: Why should Mathematics be so good at describing the Physical universe? Although your Tip of the Day is supported by the likes of Einstein's blunder and Maxwell's superlatively brilliant insight, I believe that, unless we are going to just cite anecdotes at each other, we need a philosophical basis. ("Science, separated from Philosophy, is the opium of the suburbs.") The problem is, unless we can make some headway with the question of why Mathematics is such a powerful tool, I don't really know where we can start. Of course, if you really want to go that way, don't forget that if we modify the number axioms to include i, we must let all the rest of C in too. Particles with complex mass...? Scary. -- I'm not one to name drop, but don't hang around while Dr Hawking has his elevenses.
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Re:Uber-Math (Score:2, Insightful)
by Trojan on Thursday May 25, @08:03PM EDT (#336)
(User Info) |
Most likely you have a much better chance at winning the lottery, unless you don't compete in the lottery in which case both chances equal nil :)
These problems are _hard_.
IMHO the most notorious one is the Riemann Hypothesis. Some of the best mathematicians have been working for years on that one. It has very broad implications (but so have most of the others).
For example, one can show that there's a strip around this critical axis in which all non-trivial zeroes are located. The smaller this strip, the more accurate are our estimates on things like 'the number of primes not larger than x'.
If the Riemann hypothesis were disproved, a lot of mathematics would immediately become invalid (many papers assume the validity of the hypothesis to prove deep results). In this respect it is quite similar to the Taniyama conjecture (from which the truth of Fermat's last theorem followed, together with lots of other theorem based on this conjecture).
Should the two number-theoretic conjectures (Riemann and Birch Swinnerton-Dyer) be proven in the first half of this century, I suggest that number theorists concentrate on the abc conjecture in the 2nd half of this century.
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Re:Uber-Math (Score:2)
by bluGill ([email protected]) on Friday May 26, @02:14PM EDT (#433)
(User Info) http://www.black-hole.com/users/henrymiller/
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Most likely you have a much better chance at winning the lottery, unless you don't compete in the lottery in which case both chances equal nil :) Accually your odds of winning the lottery (By which we mean the biggest jackpot in the largest drawing, since smaller lotteries are winnable) is almost exactly the same wither you buy a ticker or not. Remember you can find the winning ticket on the sidewalk. For normal statistics you can ignore those events as insignificant, but winning the lottery is so statisticly unlikely at you have to factor finding the winning ticket it, which is only slightly smaller. (Order of mangnitude yes, but not significant) The odds that I will come up with any solution in my lifetime are very low as I currently stand - I don't think about math in general much. If I decided to solve one problem my odds of doing so in my lifetime increase dramaticly. Of course deciding that I want to solve one of those problems means that spend all my freetime on it. I'd sell my boat since I wouldn't use it, move into a tiny house (preferabbly near a major university so I can access their library and professors... perhaps even become a professor though as others have said I'd be unlikely to make tenure) I'm smart enough to make reasonable progress in a solution if I dedicated my life to it. I may or may not find a solution, but I have a chance. Dedicate your life to being the big winnder in the lottery doesn't depend much on your brains. While it is true that I can increase my odds of winning the lottery, to do so significantly would mean leaving cheaply (again, no boat...) and making enough money that in the end I'd have more money saving it all then I'd get when I got the winning ticket.
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Hilbert's problems and undecidability (Score:5, Insightful)
by cpeikert (cpeikert @ mit dot edu) on Thursday May 25, @10:23AM EDT (#23)
(User Info) |
Several of Hilbert's problems in 1900 were of the form "find an algorithm for such-and-such", where such-and-such was something like "determining whether a polynomial of arbitrary degree and arbitrary number of variables has integer roots." These problems were closed, but not strictly solved, since it was proved that there were no such algorithms to do what he wanted. I would find it extremely interesting if, in the 21st century, it were shown that many of these conjectures were also proved to be undecidable. It would certainly explain why some of them have taken so long to figure out. If you show that the conjecture is undecidable, you haven't proven or disproven it - so do you still get the million $$? :)
This is my sig. There are many like it, but this one is mine. |
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Re:Hilbert's problems and undecidability (Score:1)
by Salsaman ([email protected]) on Thursday May 25, @10:33AM EDT (#58)
(User Info) |
You mean like a meta-proof :-) ?
eXistenZ is *paused* |
Re:Hilbert's problems and undecidability (Score:1)
by platypus on Thursday May 25, @10:46AM EDT (#81)
(User Info) |
I would find it extremely interesting if, in the 21st century, it were shown that many of these conjectures were also proved to be undecidable. It would certainly explain why some of them have taken so long to figure out.
Not really. The question would still remain why it had taken so long to figure out their undecidebality.
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Re:Hilbert's problems and undecidability (Score:1)
by Totto on Thursday May 25, @11:09AM EDT (#127)
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| >If you show that the conjecture is undecidable, you haven't proven or disproven it - so do you >still get the million $$? :) Uhm, if you show that it is undecidable, you have at least shown which way it goes, since there cannot be any counter-examples.
Example; assuming Fermat's Last Theorem had been undecidable, it would have been true, but there would have been no way to prove it.... ... wouldn't that indirectly mean that you have proved it to be true? - Totto
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Re:Hilbert's problems and undecidability (Score:1)
by cpeikert (cpeikert @ mit dot edu) on Thursday May 25, @11:38AM EDT (#150)
(User Info) |
If you show that a statement T is undecidable, then it is "true" in the sense that there are no counterexamples, but the "way that it goes" is up to you. You can extend your system into two logically consistent systems: one in which T is provable (by making T an axiom, for example), and one in which T is false (by making "not T" an axiom). So, in conclusion, proving the undecidability of T does not constitute a proof of T.
This is my sig. There are many like it, but this one is mine. |
Re:Hilbert's problems and undecidability (Score:2)
by dillon_rinker (dillonunderscorerinkerathotmaildotcom) on Thursday May 25, @01:44PM EDT (#243)
(User Info) http://slashdot.org/comments.pl?sid=dillon_rinker
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So Q is a number theoretical conjecture of the form "All A have property B." A is a member of a countably infinite set - positive integers greater than three, for example. This is an important point, as it means that a suitable programmed computer would find a counterexample, if it existed, in a finite number of steps.
1. If Q is provably true, then there exists a series of statements that constitutes a proof of Q.
2. If Q is false, then there exists a counterexample.
2a (contrapositive of 2) If a counter example does NOT exist, then Q is NOT false. (ie no counterexample == true)
If Q undecidable, then there does not exist a series of statements that constitute a proof, and there does not exist a counterexample. But wait! If there's no counterexample, the conjecture must be true. Proving the conjecture's undecidability thus produces a contradiction - you have proved it true AND you have proved it undecidable. So the conjecture is not undecideable.
Either there's a flaw in my logic, or yours. Which, and where? Or were you merely being more general than I?
My personal /. discussion page |
Re:Hilbert's problems and undecidability (Score:1)
by cpeikert (cpeikert @ mit dot edu) on Thursday May 25, @02:06PM EDT (#250)
(User Info) |
I think the problem derives from you think "true" means. When you prove that T is undecidable, you're formally proving that there is no finite sequence of statements that prove T, and no finite sequence of statements that prove "not T". In other words, you can't prove that every system containing the axioms has a counterexample. There is, in fact, a counterexample in a suitably extended system! The example of (non-)Euclidean plane geometry and the parallel postulate is a good one. That axiom is undecidable wrt the other axioms. In other words, it's not provably true, but you can't prove that there is a counterexample in every logically consistent geometry that includes the other axioms. Therefore good ol' high school geometry tacks on the parallel postulate, but non-Euclidean geometries (which contain all the other axioms) define lines and points such that there is greater/less than one parallel line through the point. Conversly, the statement "for all x, x>=2" is provably false in every system which obeys Peano's axioms for the natural numbers, because the element known as "1" is a counterexample.
This is my sig. There are many like it, but this one is mine. |
Re:Hilbert's problems and undecidability (Score:2)
by YU Nicks NE Way ([email protected]) on Thursday May 25, @02:25PM EDT (#260)
(User Info) |
The flaw's in your logic.
You're conflating what's usually referred to as "Proof theory" with what's usually referred to as "Model theory". By Godel's Completeness Theorem (not Incompleteness, Completeness -- they're different), if a statement is provable in first order logic from a theory, then it is true in all models of that theory -- and vice versa. (If the statement is provable, then it's true everywhere.) 2a doesn't actually follow from 2. 2a follows (after a highly non-trivial proof!) from the stronger statement "If Q is false, then there is a counterexample in all models."
Think about the axioms for the Theory of Groups. There are perfectly good groups in which multiplication isn't commutative. That means that the Abelian axiom (for all x and y, xy == yx) isn't provable from the standard axioms of group theory. It's not a proof of a contradiction, it's just a proof that a sentence isn't decided.
A theory in which all sentences are decided is called "complete". Very few theories are complete, and, in fact, what Godel's incompleteness theorem shows is that any theory which codes up the notion of proof in a non-trivial way relative to itself -- pretty much, any interesting theory -- is, ipso facto, not complete.
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Irrelevant for several of these problems (Score:2)
by tilly on Thursday May 25, @08:43PM EDT (#356)
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Many number theoretic results, if they are undecidable, cannot be proved to be undecidable.
Others can.
The Goldbach problem is an example of something which if it is undecidable, cannot be proved to be undecidable.
P=NP is something that could theoretically be undecidable and be proven undecidable.
In both cases the potential counter-examples can be enumerated (ie placed into a list). The difference is as follows. There might be an algorithm that actually works for P=NP which you cannot actually prove works. However you can always prove one way or another whether a specific integer is the sum of two primes.
Cheers,
Ben
My usual seat in the cluetrain is at IWETHEY |
Re:Hilbert's problems and undecidability (Score:1)
by The Pim on Thursday May 25, @12:36PM EDT (#199)
(User Info) |
You are exactly right. But you have proved it to be true using meta-mathematical reasoning! So, it is not a mathematical proof.
More precisely, a mathematical proof is a set of mathematical statements, each implying the next by the rules of logic and arithmetic. If FLT is false, there exists a, b, c, n, such that yada yada. It is obvious that in this case, a trivial mathematical proof could be constructed. The contrapositive of this is that if a mathematical proof cannot be contructed, FLT is true.
If this intrigues you at all, go read Godel, Escher, Bach at your earliest convienience.
The evaluation of an action as 'practical' . . . depends on what it is that one wishes to practice. |
P=NP may very well be one (Score:1)
by Habanero on Thursday May 25, @11:56AM EDT (#171)
(User Info) |
It may very well turn out that whether P=NP is an undecidable. And yes. I would most certainly say that proving a conjecture is undecidable gets you the mil. On a side note, I'm sure that Hilbert was turning over in his grave when it was finally proved (in 1963, I think) that the Continuum Hypothesis is independent of ZFC. I mean, Hilbert was one of the greatest ever, and certainly the greatest of the gay, British, atheists, but the guy thought everything was decidable. Today, it wouldn't be as much of a shock, but I'm sure a lot of people thought that they left Cohen forcing back in the sixties. I can hear them saying "Bringing independence proofs back in fashion?" with about the same distaste as "Bringing Boy George back in fashion?"
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Re:P=NP may very well be one (Score:2)
by Chalst (cas-at-achilles.bu.edu) on Thursday May 25, @12:01PM EDT (#177)
(User Info) http://achilles.bu.edu/cas
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Hilbert a gay British atheist? I think you have him confused with someone else...
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You are thinking Alan Turing (Score:2)
by tilly on Thursday May 25, @08:36PM EDT (#351)
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Alan Turing was a gay, British, atheist.
David Hilbert was a heterosexual (even went after his grad student's wives) German. I don't know whether he was a Christian or not though.
Cheers,
Ben
My usual seat in the cluetrain is at IWETHEY |
ooops. my error. (Score:1)
by Habanero on Thursday May 25, @10:52PM EDT (#374)
(User Info) |
Right. I don't know where that idea that Hilbert was a gay, British atheist came from. My apologies.
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Re:Hilbert's problems and undecidability (Score:2)
by MalcolmT ([email protected]) on Thursday May 25, @05:50PM EDT (#313)
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Since I haven't seen any links posted to Hilbert's original list yet: a copy of his original paper is avaiable for those interested.
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Re:Hilbert's problems and undecidability (Score:1)
by Alpha State ([email protected]) on Thursday May 25, @07:41PM EDT (#332)
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What if you can prove that there's no proof for whether it is decideable or not?
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Re:Hilbert's problems and undecidability (Score:2)
by jesser on Friday May 26, @12:05AM EDT (#385)
(User Info) http://www.palosverdes.com/jesse/
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What if you can prove that there's no proof for whether it is decideable or not?
I claim to have a proof that this situation is impossible. There could be a flaw in my logic.
suppose that the original statement is decidable. then there's a proof that it's true, or there's a counterexample. therefore its decidability is definately true (and therefore decidable).
now suppose the decidability isn't decidable. then the original statement can't be decidable (see above). therefore, the original statement is not decidable.
--
Help the lizard fight annoying security problems. |
Re:Hilbert's problems and undecidability (Score:1)
by S7urm ([email protected]) on Thursday May 25, @10:10PM EDT (#368)
(User Info) |
well think of the concept of a non-lineaur equation and the concept of chaos theorum and period oscillations in comparison to non-linear equations. Read any work put out by Mandlebrot or Lorenz. When you deal with a non-lineaur equation you deal with an equation that has no set solution in a "logical" sense, you can incorporate this line of thought when thinking on a chaotic part of nature, or what may SEEM to be chaotic, Perception is the key to solving problems of this caliber, to be able to perceive the problem from not one or two or X angles but to perceive from ALL possible angles is to truly formulate a working equation to solve this caliber problem. Another thing to think on along those lines is "Is anything random" My answer would be NO
"If man, in his greatness, could comprehend infinity we could become god, 46 & 2 just ahead of me" Securify, modify, pertfiy, and die |
Nice for a change (Score:2)
by vvulfe on Thursday May 25, @10:25AM EDT (#28)
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I find it reassuring that in a time so obsessed with giving money away for just about any non-intellectual endevor, that a group has taken it upon themselves to reward those who accomplish with their minds, a trait which seems to be rather overlooked and undervalued these days.
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Math 101 (Score:3, Funny)
by Anonymous._.Coward (print reverse split(/ */,"moc.buprenroc\@demmahom") on Thursday May 25, @10:25AM EDT (#29)
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| Excellent, now I don't have to spend any time making up the finals exam questions for my math students. Prof. AC
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Re:Math 101 (Score:1, Funny)
by Anonymous Coward on Thursday May 25, @11:18AM EDT (#135)
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Reminds me of an extra credit problem I heard about on a test.
This is extra credit. Answer all parts. No partial credit will be given.
a) Prove or disprove Goldbach's Conjecture.
b) Prove or disprove the RSA Assumption
� � � � (i.e. breaking RSA is hard)
c) Prove or disprove : P = NP.
The way I heard it, some people took it seriously
and complained to the prof that it was "too hard..."
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Re:Math 101 (Score:1)
by jbarnett ([email protected]) on Thursday May 25, @11:47AM EDT (#161)
(User Info) http://axil.netmate.com/~jbarnett
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One of my prof (in Intro to CS) made ALL correct answers on a multichoice test 'A'. If you wanted to check a prefect grade all you had to do was mark 'A' on every question.
Over %80 of the class failed. In that class on that test before, it was around %5-10 that failed, here it was over %80 that failed, kinda funny.
It was extremely cruel though. Make you math test all come out to the answer '42' and see what happens.
Or have the final test that is %100 of the students grade
True or False:
___ P = NP
"Science is like sex: sometimes something useful comes out, but that is not the reason we are doing it" -- Richard Feynman |
Well, the first thing... (Score:4, Funny)
by Tebriel on Thursday May 25, @10:26AM EDT (#35)
(User Info) |
NP = P
NP * 0 = P * 0
0 = 0
q.e.d.
Gimmie my million!
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Re:Well, the first thing... (Score:1, Funny)
by Anonymous Coward on Thursday May 25, @10:54AM EDT (#99)
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Here is another famous mathematical formula which combines the contants e, i, pi, 1, and 0. e*i*pi*1*0=0
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Re:Well, the first thing... (Score:1)
by omegamaid on Thursday May 25, @12:13PM EDT (#181)
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It's funny you should mention those three constants... they really are related by Euler's formula, e^(pi i) = -1.
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My proof is much better (Score:1)
by Habanero on Thursday May 25, @11:58AM EDT (#174)
(User Info) |
Unfortunately, this form is too small to contain it.
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Re:Well, the first thing... (Score:4, Funny)
by dillon_rinker (dillonunderscorerinkerathotmaildotcom) on Thursday May 25, @02:09PM EDT (#253)
(User Info) http://slashdot.org/comments.pl?sid=dillon_rinker
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Dear ___Tebriel___,
The first error in your proof occurs on page ____1____, in line _____2____.
Sincerely,
Dillon
When the math department where I went to school received a crank proof, this form letter would be stapled to the front, a grad student would fill it out, and the proof would be returned to its sender.+
My personal /. discussion page |
damn (Score:1)
by DGregory ([email protected]) on Thursday May 25, @10:26AM EDT (#36)
(User Info) http://www.feick.net
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If I had any hope of solving any of these, I sincerely doubt I would have any money problems in the first place. I'd probably be a multibillionaire genius working for a hotshot company who will do whatever I want since I'm so irreplaceable. And then I'd be doing problems like that in my spare time because I'm working on some REAL problems for work.
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another million (Score:3, Informative)
by jabkie on Thursday May 25, @10:30AM EDT (#45)
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| don't forget that proving Goldbach's conjecture is also worth a million bucks the /. story is here. the conjecture itself (for the link-following-impaired) is:
Every even number greater than two is the sum of two primes. Enjoy!
--
.signature fault. joke dumped. |
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Re:another million (Score:1)
by Old Wolf on Friday May 26, @09:27AM EDT (#423)
(User Info) |
The problem doesn't mention infinity, you dumb fuck
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Re:another million (Score:1)
by S7urm ([email protected]) on Friday May 26, @06:05PM EDT (#439)
(User Info) |
your right, you dumb fuck, BUT in order to prove a problem of that caliber requires going to infinty to prove.
"If man, in his greatness, could comprehend infinity we could become god, 46 & 2 just ahead of me" Securify, modify, pertfiy, and die |
Re:another million (Score:1)
by Old Wolf on Saturday May 27, @04:14AM EDT (#445)
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No it doesn't. The problem only refers to natural numbers, which doesn't include infinity.
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Re:another million (Score:1)
by S7urm ([email protected]) on Saturday May 27, @10:19AM EDT (#446)
(User Info) |
The problem stated "That every" when you use the word EVERY then you use ALL, what is the connotation of the word every in relation to numbers?
"If man, in his greatness, could comprehend infinity we could become god, 46 & 2 just ahead of me" Securify, modify, pertfiy, and die |
Re:another million (Score:1)
by Old Wolf on Saturday May 27, @11:38AM EDT (#447)
(User Info) |
If you are going to get pedantic, then "For every" refers to the universal logical operator, whose symbol looks like an upside down 'A'.
Infinity does not come into this. Any attempt to introduce infinity will just complicate the problem, and I will bet more than a million dollars (Well, I would if i had it) that a proof or disproof would also not involve infinity.
Here is an example of another problem that involves the universal operator, but does not require infinity:
"Every multiple of 9 has its digits sum to a multiple of 9."
It is easy to prove this without using infinity.
(I can do so if you like, but it might confuse you, but by all means reply to this and request it if you need to.)
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Re:another million (Score:1)
by S7urm ([email protected]) on Saturday May 27, @10:14PM EDT (#451)
(User Info) |
I think we are just thinking along different lines man. All I am saying is that you can never prove that those theories are correct. Think of it like this, if those theories worked to the largest number computable by man/machine (say to the 100 billionth digit) and for some whacked out reason it changed and the theory become untrue. Then it would be false. I'm not saying it would happen, but you can't PROVE it because every integer can have a number added to it or subtracted from it. It may seem illogical that a conjecture had truth all the way to a billionth digit, then changed BUT it could happen
"If man, in his greatness, could comprehend infinity we could become god, 46 & 2 just ahead of me" Securify, modify, pertfiy, and die |
Re:another million (Score:1)
by Old Wolf on Saturday May 27, @10:53PM EDT (#452)
(User Info) |
That's why you PROVE it for all possible natural numbers, so that you don't have to go and check each one!
The statement I gave in my previous post on this thread is such an example: it is true no matter how high you go or how long you go on for, there is no counterexample, it does not change after a billion digits, and it is 100% totally rock solid undeniable truth.
In case my example was too complicated for you, here is a simple one:
"Every number greater than 5, is greater than 2."
Are you going to tell me that this might suddenly become untrue after a billion digits?
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Re:another million (Score:1)
by S7urm ([email protected]) on Sunday May 28, @02:53PM EDT (#454)
(User Info) |
I do agree, and you can stop making feeble attempts to mock my intelligence, I can grasp your conjectures, I just see things in a different way . I know that those statements have an "undeniable truth", I was simply trying to state that when you have a conjecture that states a theory that is true to the end of numbers then you simply can't give an example to the END of numbers because there isn't one.
"If man, in his greatness, could comprehend infinity we could become god, 46 & 2 just ahead of me" Securify, modify, pertfiy, and die |
Re:another million (Score:1)
by S7urm ([email protected]) on Sunday May 28, @02:57PM EDT (#455)
(User Info) |
As a related example, take euclidian geometery (in case you didn't know, what they teach high school kids). When you solve for the radius of a circle or the diameter of a circle you use Pi. Well sure you can get an answer, but that answer is an aproximation because Pi (as im sure you know) is an infinite number, so when sitting in your 10th grade geometery class in high school kids and Mr./Mrs. Smith says "Alright kiddies, solve for the radius of this circle" you can state that you would, but you don't like giving only an approximate answer :P
"If man, in his greatness, could comprehend infinity we could become god, 46 & 2 just ahead of me" Securify, modify, pertfiy, and die |
Re:another million (Score:1)
by Old Wolf on Monday May 29, @11:05AM EDT (#461)
(User Info) |
I think you mean "Euclidean geometry".
Pi isn't infinite; it's in fact between 3 and 4 - two finite numbers.
You can get an exact answer - if a circle has radius 1 then its circumference is (2*pi). It's true that I can't express this exactly by using decimal expansions, but I have no wish to use decimal expansions (which didn't even exist in Euclid's time).
You aren't going to argue that pi is somehow a "fuzzy" number and it doesn't have an exact value, are you?
Re. the previous post.. I agree (sort of ) that you can't "give an example to the end of numbers", but I have no desire or need to give such an example, since I can prove the theorem without the need for examples.
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Re:another million (Score:1)
by S7urm ([email protected]) on Tuesday May 30, @09:02AM EDT (#463)
(User Info) |
Of course Pi isn't an infinite number but how can you say that it isn't a "fuzzy" number since with decimal expansion you CAN"T get the exact number to the exact decimal for Pi. So when you do you 2*Pi equation you aren't gettin an EXACT number, your making an approximation which is what I was trying to state
"If man, in his greatness, could comprehend infinity we could become god, 46 & 2 just ahead of me" Securify, modify, pertfiy, and die |
Re:another million (Score:1)
by Old Wolf on Tuesday May 30, @06:48PM EDT (#465)
(User Info) |
*LOL*
Go back to third grade, boy.
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The best way to solve on of these problems... (Score:1)
by CokeBear (http://publish.uwo.ca/~djfox/) on Thursday May 25, @10:30AM EDT (#46)
(User Info) http://publish.uwo.ca/~djfox/
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...is to assign one to a class of college math students, without telling them that its never been solved before. Tell them it is required that they solve this problem for some percentage of their final grade. Some will stay up all night, but in a class of 400 students, 1 might get it.
Its amazing what a human mind can do when its not aware of what is impossible.
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Re:The best way to solve on of these problems... (Score:1)
by b_pretender ([email protected]) on Thursday May 25, @10:59AM EDT (#111)
(User Info) |
...better yet, give them all cell phones and tell the whole class that they only have to turn in 1 problem. That way you would have a Beowulf cluster of students working on the problem.
Of course, students would make _very_ unreliable nodes. That's where monkeys come in to the picture.
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Re:The best way to solve on of these problems... (Score:1)
by norton_I ([email protected]) on Thursday May 25, @11:06AM EDT (#123)
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The problem is, realistically speaking by the time students have enough background to understand the questions, let alone know enough about higher math to solve these problems without many years of redeveloping advanced mathematical concepts, they already know about these problems.
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Re:The best way to solve on of these problems... (Score:1)
by sconeu on Thursday May 25, @11:57AM EDT (#173)
(User Info) http://alumni.cse.ucsc.edu/~scottn
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| Huffman coding (minimal redundancy coding) was developed in this manner. I went to UC Santa Cruz and had the (dubious) privilige of having the late Dr. David Huffman as a professor. He always claimed that when he was at MIT, he was flunking an information theory class, so the professor told him to solve the minimal redundancy coding problem and he would pass... So he solved it... [for the humor impaired... BEGIN JOKE]
Of course, at UCSC, we were always of the opinion that it was actually Huffman's roommate Frank who solved it, and that Frank disappeared that same night under "mysterious circumstances"... :-)
[END JOKE]
"They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety." - Ben Franklin
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Re:The best way to solve on of these problems... (Score:1)
by pugugly ([email protected]) on Friday May 26, @03:25AM EDT (#398)
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| I seem to recall from Stephen Hawkings Biography that a classmate (Penrose? I think) was discussing how good Hawking actually was at mathematics. For an assignment, the students were gvien 10 incredibly difficult proofs to do. Penrose and three other students worked together over the week, and managed three of them, with most of the rest of the class being in the same general boat. After Hawking got back from sports over the week and partying over the weekend, he hit the problems Sunday night. Needless to say, monday his classmate ragged him for not getting the problems done, and Hawking blushed and admitted he hadn't gotten the last one done. Penrose - 'Suddenly, I realized . . . he wasn't joking. That was when I realized, not only were we not in the same league Stephen was in, we weren't on the same planet Stephen was on' This is heavily paraphrased of course - Pug
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Public Interest (Score:1)
by Kexel on Thursday May 25, @10:32AM EDT (#57)
(User Info) |
| Really a great idea. "Good Will Hunting", but for real. You know this is a great way to try to get people interested in mathematics, that never would have bothered to find out what a Zeta function is. Best of luck to those larger brains that are reading these pages.
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Re:Public Interest (Score:1)
by Tower (/dev/whoop-ass) on Thursday May 25, @11:24AM EDT (#140)
(User Info) |
Heh--- almost all of the problems shown in "Good Will Hunting" were rather basic discrete math. Anyone with a month or two of class could do them.
Some were easy enough to solve in real-time while watching the movie...
-- "Funk the Dumb Stuff!" - ToP |
Re:Public Interest (Score:1)
by maugt ([email protected]) on Thursday May 25, @12:28PM EDT (#192)
(User Info) |
But you try telling anyone in the general public that and they don't believe you. I hate that movie precisely because the mathematics in it was overblown. I mean, getting all excited over cancelling variables in an equation? Is that what people think is difficult? Duh.
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P=NP? (Score:1)
by levendis on Thursday May 25, @10:36AM EDT (#67)
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I've been fascinated with P=NP for a long time now. If, in fact, P=NP, then a whole lotta computer-oriented things are gonna change, probably for the worse. For example, standard RSA encryption will pretty much be useless, since at its core it depends on P!=NP. So, I'm just curious - what's the /. community's take on P=NP? Is it true or false (in your humble opinions)? If its true, what are the consequences?
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Re:P=NP? (Score:1)
by FascDot Killed My Pr on Thursday May 25, @10:43AM EDT (#77)
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False, IMHO.
But I'm talking about the mathematical problem, not the real world issues. Encryption doesn't just depend on P!=NP. They also rely on NP being hard. If this stops being the case, all hell breaks loose.
By the Church-Turing thesis NP for all Turing Machines equivalents and nothing's more powerful than a TM. But CT is not a rigorous proof, just a semi-soft conjecture. So if we ever found anything more powerful than a Turing Machine, problems that are non-polynomial on TMs might turn out to be polynomial somewhere else.
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Re:P=NP? (Score:1)
by levendis on Thursday May 25, @10:55AM EDT (#100)
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Not necessarily... Turing Machines aren't necessarily powerful, they are just the bare minumum you need to solve any computable problem. Even your brand-spanking-new Athlon 1Ghz is still, theoretically at least, a (deterministic) Turing Machine. Anything one can do, the other can do, in the same amount of time (poly time versus non-det poly time). NP problems run in poly time, but on a non-deterministic Turing Machine. So stating that "problems that are non-polynomial on TMs might turn out to be polynomial somewhere else" is pretty meaningless; the only machine any such problem will run in poly time on is a non-deterministic machine, which to the best of my knowledge we don't have any of.
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Re:P=NP? (Score:1)
by Tom7 (spam-sucks) on Thursday May 25, @10:59AM EDT (#110)
(User Info) http://www.andrew.cmu.edu/~twm/
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Except perhaps on a quantum computer.
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Re:P=NP? (Score:1)
by levendis on Thursday May 25, @11:02AM EDT (#116)
(User Info) file://localhost/dev/random
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Not even. The main advantage of quantum computing is massive parallelism. If the algorithm its running is still NP-hard, its still going to be slower than a polynomial time algorithm. Fundamentally, a quantum computer is still just a Turing Machine.
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Re:P=NP? (Score:1)
by Tom7 (spam-sucks) on Monday May 29, @11:36AM EDT (#462)
(User Info) http://www.andrew.cmu.edu/~twm/
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I don't know that much about quantum machines, really, but what I've heard is that they are fundamentally different because bits can be in multiple states at once. A turing machine is deterministic while a quantum machine is not.
A problem in NP has P-verifiable solutions. If a quantum computer can check all P-size solutions simultaneously (I think this is the point), then the problem can be solved in (quantum) P time.
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Re:P=NP? (Score:1)
by FascDot Killed My Pr on Thursday May 25, @11:03AM EDT (#117)
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"...the only machine any such problem will run in poly time on is a non-deterministic machine, which to the best of my knowledge we don't have any of."
Hey, what do you know! You've recreated MY POINT.
No, we don't have any of. But we might have any of in the future. Quantum computer (or something stranger still) may come along and blow all us TMers away.
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Re:P=NP? (Score:2)
by (void*) (voice@void.) on Thursday May 25, @10:51AM EDT (#95)
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Not true. It is widely believed, but not proven, that the problem of factoring a large number is NP. This difficulty (even if it is not NP) is what RSA depends upon. If P = NP, then obviously, there is a polynomial time algorithm to decrypt RSA encrypted messages. But if P != NP, we still have to prove that decrypting RSA is really NP.
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Re:P=NP? (Score:2)
by (void*) (voice@void.) on Thursday May 25, @10:55AM EDT (#103)
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Sorry - I mean to say "It is widely believed, but not proven, that the problem of factoring a large number is NP-hard". That is, it believed to be in NP, but outside of P.
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Re:P=NP? (Score:1)
by pal on Thursday May 25, @12:49PM EDT (#214)
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NP-hard is a larger class than NP. the intersection of NP-hard and NP is NPC. one would call "in NP but outside of P" simply "NP-P".
- pal
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Re:P=NP? (Score:1)
by levendis on Thursday May 25, @10:59AM EDT (#112)
(User Info) file://localhost/dev/random
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That's my point. If P=NP, we are suddenly in some very deep trouble. However, if we can prove P!=NP (which is basically what all of computer sceince has been assuming for 30+ years), we still have lots of other questions to tackle.
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Re:P=NP? (Score:1)
by CMUMikey on Thursday May 25, @05:45PM EDT (#312)
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With the advent of quantum search algorithms that can turn O(n) problems into O(1), how would that effect problems in NP? If one comes up with a quantum algorithm that solves traveling salesman or some other NP-complete problems what does that mean for P=NP?
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Re:P=NP? (Score:1)
by SomeOne2 on Thursday May 25, @10:57AM EDT (#104)
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I would think P!=NP because it would be nice if P==NP. If P==NP you could solve all NP-hard problems (sadly nearly all interesting problems are NP-hard) in P, which would be a good thing, and experiences teaches us that nothing is ever easy.
Of couse you could earn a lot of money of you find a method to solve a NP problem in P... :) (RSA aside, a _lot_ of real-life problems are NP, TSP for example)
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Which one to try for (Score:5, Informative)
by YoJ on Thursday May 25, @10:37AM EDT (#68)
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| Given the background of most readers of Slashdot, if you are thinking of trying to solve one of these problems the best one might be the P=NP problem. Most of the others require some high-level knowledge of mathematics or physics. The high powered theoretic scientists have all taken a look at P=NP. I think real progress will come from a non-specialist who has really original ideas about new algorithms. What you have to do: find an algorithm to solve any of the following problems in polynomial time. - Given a graph, find a cycle (loop) that visits every vertex exactly once (Hamiltonian cycle problem)
- Given a weighted graph, find the least-weight path that visits every vertex (Travelling salesman)
- Given a phrase of boolean variables, determine if any combination of values for variables makes the phrase true (Satisfiability)
- Given a set of numbers, and a target sum, determine if any combination of the numbers add up to the target sum (Subset-sum problem)
Most CS people think that P is not equal to NP. They might be right, but I think we have vastly underestimated the power of polynomial time algorithms. Problems that are in P can be solved for sure in polynomial time. Problems in NP have solutions that can be verified in polynomial time. For example, in the subset-sum problem it is easy to verify that the sum of a certain combination of numbers adds up to the target. It is harder to come up with the right combination. But is it so much harder that it cannot be done in polynomial time? Maybe not. -Nathan Whitehead
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Re:Which one to try for (Score:1)
by jesser on Thursday May 25, @10:55AM EDT (#102)
(User Info) http://www.palosverdes.com/jesse/
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Most CS people think that P is not equal to NP. They might be right, but I think we have vastly underestimated the power of polynomial time algorithms.
According to a recent Science News article, you're right. The article described "experiments" with the satisfiability problem that demonstrate that except in rare circumstances, smart algorithms can solve satisfiability problems fairly quickly. Furthermore, various different "smart" algorithms had trouble (went exponential) under the same circumstances, and a simple heuristic could be used fairly effectively guess whether a given satisfiability problem will be hard or not.
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Help the lizard fight annoying security problems. |
Re:Which one to try for (Score:2)
by jbarnett ([email protected]) on Thursday May 25, @11:24AM EDT (#141)
(User Info) http://axil.netmate.com/~jbarnett
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Given a phrase of boolean variables, determine if any combination of values for variables makes the phrase true (Satisfiability)
Given a set of numbers, and a target sum, determine if any combination of the numbers add up to the target sum (Subset-sum problem)
Please excuse my ignorance on this, but this both seems easy if you can use really fsck bloated and slow moving code. I assume this won't be consider polynomial time? Could anyone explain what exact is 'polynomial time' is? How it is defined?
"Science is like sex: sometimes something useful comes out, but that is not the reason we are doing it" -- Richard Feynman |
Re:Which one to try for (Score:1)
by sconeu on Thursday May 25, @11:39AM EDT (#151)
(User Info) http://alumni.cse.ucsc.edu/~scottn
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Yes they are easy to solve, but they currently require exponential time.
Polynomial time: The number of operations required to solve the problem is a polynomial function of N, where N is the problem size. (i.e. aN^x+bN^(x-1)+...)
Exponential time. The number of operations required to solve the problem is an exponential function of N, where N is the problem size. (i.e. ae^N).
As the size of the problem grows, the time for an exponential solution grows much more rapidly than a polynomial solution.
For example, the Satisfiability problem: Exponential solution might be the brute force -- try every possible combination of boolean values. Because for problem size N, there are 2^N operations required to check everything, it's exponential.
"They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety." - Ben Franklin |
Re:Which one to try for (Score:5, Informative)
by Azog (thoffman1 at hotmail) on Thursday May 25, @12:36PM EDT (#198)
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P is the set of problems that can be solved in polynomial time; i.e. the set of problems for which polynomial-time algorithms are known.
So what's a polynomial time algorithm?
For some deterministic algorithm 'A', (like quicksort) which takes an input of size 'n' (the objects to be sorted), there is some formula that says how many computational steps must be carried out to complete the algorithm.
A simple example. Suppose you write an algorithm to find the maximum number in an array by scanning through the whole array once. This will take time proportional to the size of the array. In mathematical jargon, this is O(n) "big-Oh of n" where n is the size of the array.
You can have algorithms that are O(n^2) "n squared", O(n^5 + 3n^2 + 50n), or whatever. If the function of n is a polynomial in n, then the algorithm runs in polynomial time.
But if your function is like O(2^n) "two to the n" that is not a polynomial in n, and you don't have a polynomial time algorithm.
As a simple example, consider code breaking. In this case, n is usually the number of bits in the key. If you have a 128-bit key, there are 2^128 possible keys. If your "algorithm" is the obvious one of trying every single key one after another until one works, you will have to check O(2^n) keys. This is exponential time, and it's a very slow way to break codes.
Anyway... A definition of NP is a little harder to understand. NP is not "non-polynomial" although people tend to think of it that way. What it really means is "non-deterministic polynomial". Non-determinstic algorithms cannot be directly implemented in computers. However, they can be converted into deterministic algorithms which run in exponential time.
So: the P=NP question is: For all the problems where we know non-deterministic polynomial time algorithms, are there actually determinstic polynomial time algorithms that we just haven't been smart enough to figure out yet?
Probably not. Why? Well, there's this "NP-completeness" thing. If you find some interesting problem that you can't get a (determinstic) polynomial time algorithm for, you can find a non-determistic algorithm, and then then prove your problem is "equivalent" to a known NP-complete problem. That means that if someone was able to find a deterministic polynomail time to solve YOUR problem, that same solution could be used to solve ALL OTHER NP-complete problems in polynomial time. And since hundreds of people haven't been able to solve them in years of trying, there probably isn't much point in trying to solve yours.
At this point you talk to your thesis advisor and start working on fixed-parameter tractability, or polynomial time approximations, or other more esoteric computer science approaches to the problem...
Now you know more than you wanted to.
Torrey Hoffman (Azog)
"HTML needs a rant tag" - Alan Cox |
Re:Which one to try for (Score:2)
by Just Some Guy (kirk@stra%deletethis%user.com) on Thursday May 25, @12:28PM EDT (#193)
(User Info) http://www.honeypot.net/~kirk/
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In one of my CS classes, we had to write an algorithm to solve the "subset sum" problem. The naive solution is to, basically, count from 1 to 2^n (where n is the number of elements in the list). Then, use i'th bit in the counter as boolean values to determine whether or not to include the i'th value in the list to add into a guess. I.e., for count == 1, see value[0] equals the target. For count == 2, see if value[1] equals the target. Count == 3 -> value[0] + value[1]. The algorithm I came up with used recursive subsets of the (sorted) listed, as well as pre-calculated knowledge of the sum of the first i elements. Try as I might, I could *not* get an exponential-time result from sample input. The teacher "proved" that my solution was still NP, but it was kind of brushed over. I'm not saying that I wrote a P algorithm, but I certainly never saw proof to the contrary. Is there a (friendly) group of people who would be willing to look at such things from a lowly student? I'd be perfectly willing to be proved incorrect, but neither do I want to just give up simply because I'm not a world-famous scientist.
-- Where did you want to go yesterday? |
Re:Which one to try for (Score:2)
by YoJ on Thursday May 25, @05:09PM EDT (#304)
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In your algorithm you say to "count from 1 to 2^n" where n is the size of the list. That is the problem right there, looping 2^n times is exponential (2^n is an exponential, not a polynomial). Polynomial time algorithms are allowed to loop n times, or n^2 times, even n^100 times. But you're not allowed to loop 2^n times in a polynomial time algorithm.
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Re:Which one to try for (Score:2)
by Just Some Guy (kirk@stra%deletethis%user.com) on Thursday May 25, @05:40PM EDT (#310)
(User Info) http://www.honeypot.net/~kirk/
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Correct. I was referring to the "naive algorithm", with is O(2^n), and therefore NP.
-- Where did you want to go yesterday? |
Re:Which one to try for (Score:1)
by tbarrie on Friday May 26, @03:12AM EDT (#396)
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First, I'd like to note that you're a tad sloppy in your terminology. You don't talk about specific algorithms being P or NP, you talk about problems being in P or NP. Also, P is a subset of NP, so saying that the teacher proved the problem was still in NP isn't terribly helpful as far as deciding whether a polynomial-time algorithm exists goes.
And I'm just a graduate student myself, but I'm not bad at this stuff and if you want to e-mail me your algorithm I can look at it. I can't guarantee an answer one way or the other, though.
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Oh yeah (was Re:Which one to try for) (Score:1)
by tbarrie on Friday May 26, @03:13AM EDT (#397)
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Forgot that my e-mail isn't displayed on my messages... it's [email protected]
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Re:Which one to try for (Score:2)
by Just Some Guy (kirk@stra%deletethis%user.com) on Friday May 26, @11:02AM EDT (#427)
(User Info) http://www.honeypot.net/~kirk/
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You're correct, of course. What I meant to say was that if an algorithm with a polynomial runtime exists for a problem, then the problem can't be NP. Can it? We didn't go nearly as in-depth as I would have liked, so please understand that I'm approaching the discussion with an "interested amateur" point of view. On a side note, my exposure to this kind of theory prompted me to join the ACM's SIGACT. That is, the Special Interest Group on Algorithms and Computation Theory. It's a great place to be if you like this stuff.
-- Where did you want to go yesterday? |
Re:Which one to try for (Score:1)
by tbarrie on Friday May 26, @01:52PM EDT (#432)
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Actually, if a polynomial time algorithm exists for a problem, then the problem HAS to be NP. P is a subset of NP. To recap some definitions:
P is the set of problems for which a polynomial time algorithm exists. NP is the set of problems where, if you're given a purported solution, you can check whether it actually is a solution in polynomial time. (That's kind of vague, but I think it'll do. Note that NP is more commonly defined in terms of what a nondeterministic Turing Machine can do, but I find this definition easier to understand, and they're equivalent.) If you can do the latter, you can certainly do the former, so all of P is in NP.
So a polynomial time algorithm for subset sum would show that it's in P, not that it's not in NP.
Of course, subset sum is known to be NP-complete, implying that any NP problem can be reduced to it in polynomial time. So an algorithm to solve subset sum in polynomial time could be used to construct a polytime algorithm for any NP problem; in other words, the existence of such an algorithm would mean that P=NP. This is why many researchers would be skeptical of such an algorithm. There are a LOT of NP problems for which a LOT of people have spent a LOT of time trying to find polynomial time algorithms without success; many believe that if such algorithms actually exist for all of these problems, somebody would have found one by now.
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Re:Which one to try for (Score:1)
by StormyMonday on Thursday May 25, @12:43PM EDT (#205)
(User Info) http://www.aginc.net
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Given a set of numbers, and a target sum, determine if any combination of the numbers add up to the target sum (Subset-sum problem)
??
Isn't this the same as the "knapsack problem", used in some early public-key crypto? I thought that this had been solved. At least the cryptosystems were broken. Too bad -- they're a lot simpler than the DH-RSA stuff.
Note also that a problem in P is not necessarily an easy/fast problem. It just determines how the problem scales as it gets bigger. Even there, if you have a solution that scales as n^1000, it won't buy you much.
Main thing, if a problem is in P, it's worth looking for a simpler solution. Doesn't mean you'll find a usable one.
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Re:Which one to try for (Score:1)
by kallisti ([email protected]) on Thursday May 25, @01:36PM EDT (#237)
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Schneier's (sp?) Applied Cryptography has an explanation of what went on here. In order to use as a cryto-system there needs to be a solvable version of the knapsack, which is converted to the complete version. It is the weaker knapsack which was broken, the general problem remains NP-complete.
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Re:Which one to try for (Score:1)
by Furry Ice on Thursday May 25, @02:16PM EDT (#256)
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Maybe someone will find a polynomial algorithm for an NP-complete problem. A lot of effort has already gone into the search for such solutions, however, with nothing to show. What we really need is some technique (as yet undiscovered) for proving that no algorithm exists in a specific complexity class which solves a particular problem, much like the techniques for proving undecidability had not been discovered when Hilbert proposed his problems for the 20th century.
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Re:Which one to try for (Score:1)
by bobv-pillars-net ([email protected]) on Thursday May 25, @08:20PM EDT (#342)
(User Info) http://www.pillars.net/
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Yoj wrote: - Given a set of numbers, and a target sum, determine if any combination of the numbers add up to the target sum (Subset-sum problem)
This sounds like it may be equivalent to the problem I've carried around for fifteen years, namely: Given "n" number of "d"-sided dice, compute the odds of rolling them and coming up with a total of "t" number of pips on top. Partial credit for finding a reasonable approximation (say, three significant digits) for f(n,d,t) where n=1..1000 and d=1..100 Oh yeah, and the function should be computable in polynomial time.
[email protected], geek@large |
Re:Which one to try for (Score:1)
by YoJ on Saturday May 27, @01:00AM EDT (#443)
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Summing multiple random sources usually gives you a normal distribution, i.e. bell-shaped probability curve. The bell curve is generated by e^(-x^2), so scaling this curve and shifting it, then rounding to the nearest integer should solve this problem. As you add more and more dice, the probability of the sums looks more and more like a bell curve. (I remember doing some stuff with this when I was into D&D).
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Re:Which one to try for (Score:1)
by S7urm ([email protected]) on Thursday May 25, @09:59PM EDT (#365)
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Well think of it like this I know I'm wrong, but what the hell P=NP, looks like a basic algebraic eqaution So break it down The variable P is always equal to the variable N multiplied by the variable P.....so why not make the variable N always "1"
"If man, in his greatness, could comprehend infinity we could become god, 46 & 2 just ahead of me" Securify, modify, pertfiy, and die |
Rieman or Artin? (Score:1)
by unique123 on Thursday May 25, @10:40AM EDT (#72)
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Often the Riemann Hypothesis is called Artin's conjecture. Personally I think Riemann's name is overused in mathematical circles. Emil Artin was a much better clavichord player after all...
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P vs. NP (Score:1)
by a.out ([email protected]) on Thursday May 25, @10:46AM EDT (#82)
(User Info) http://xyu.dhs.org/brad/
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I had a hard time following that postscript file on the side and since I've studied P and NP stuff I just dug out my notes and hence will spew forth a little background:
P is the set of problems that can be solved in polynomial time.
If one problem is in P then the whole class is in P.
Nondeterminism: after we do one step we then can decide the next. (bad for parallel algorithms for MPI, MPICH etc.)
NP: The set of all problems that can be solved by a nondeterministic algorithm in polynomial time.
Well we know that it's infact trivial to prove that P (problems that can be solved in polynomial time) is contained in NP.
P = NP ?
Well how can we prove this ..
NP-Hard: A problem X is NP-hard if every problem in NP is polynomially reducible to X.
Then by definition, if any NP-hard problem is ever proved to belong to P, then P == NP!
Sync Linux and CE: SyncLICE |
Where's my cheque? (Score:1)
by Andy_R on Thursday May 25, @10:49AM EDT (#89)
(User Info) |
| Is P=NP? Nah, there's an extra N on the right hand side Seriously, though, I can't work out if I should be worried that I can't understand any of the questions, or pleased that mathematicians are now only left with things I can't understand still to do? - Andy R.
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P = NP? (Score:1)
by sid_vicious on Thursday May 25, @10:58AM EDT (#107)
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| Who cares? Quantum computing will eventually collapse out these polynomially "hard" problems so they can be solved almost instantly, and it won't matter anymore.
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Re:P = NP? (Score:1)
by VaporX on Thursday May 25, @12:56PM EDT (#217)
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Stop saying those two words! My dick is getting tired...
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Does P=NP? (Score:2, Funny)
by MasteroftheVoxel on Thursday May 25, @11:02AM EDT (#115)
(User Info) |
I remember way back in freshman year when I took CS121 (Intro to Formal Systems and Computation)...
The night before the final I posted a message on comp.theory asking for a proof that P=NP because it would greatly help on the problem set I was working on (namely invalidating the proofs I had to do to show that several different problems were NP-Complete).
Well, Unfortunately everyone on the newsgroup took me seriously and there was some angry responses along the lines of: "Are you stupid? Your going ot fail your final tommorrow if you don't already know that this is the biggest unsolved porblem in computer science..."
The best was that a few weeks later (after the grades were turned in, thank god), my TA noticed the post and E-mailed me in disbelieve asking if I had really posted that question...
I don't kow how she knew it was me since I only signed with my initials...
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Re:Does P=NP? (OT) (Score:1)
by kjeldar (jhobbs#mail/orion/org) on Thursday May 25, @03:09PM EDT (#279)
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| I remember way back in freshman year when I took CS121 (Intro to Formal Systems and Computation)... Damn, brother, where'd you go to school? All my school did in CSC131 was dick around with some basic Java methods.
J |
Hey wake up. We're in year 2000 (Score:5, Funny)
by El Cabri on Thursday May 25, @11:04AM EDT (#118)
(User Info) |
Never heard of the New Economy ?
Giving $1M when the problem _is_ solved is an old-fashioned, actual-results-based, old-economy way of doing things.
CNNfn taught me that the New Economy way of doing it is :
-- give me $10M now. I promise to hire the brightest people on earth.
-- If I don't get anything done, I'll fill an IPO and screw some shareholders.
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P=NP (Score:2, Troll)
by roman_mir on Thursday May 25, @11:04AM EDT (#119)
(User Info) |
I don't know about other problems BUT if anyone of you, guys know how to solve this one, I will personally get 10,000,000 into your pocket. If this problem is solved I will buy the solution and the rights to the solution for 10 million dollars US from ANYBODY as long as it is shown completely and with code example that this is doable.
Why such extreme numbers? Well, basically, if the solution to this problem is found, then I can easily reduce the solution into ATM or into Turint Halting problem and by doing that I can basically solve every single protection algorithm on this planet in polynomial time (in no time - rather.)
So, if any of you out there, who wants to make 10million USD, you should solve this problem and contact me.
(I would like to insert a picture with thousands of hands holding burning candlesticks and lighters right here...)(tm, pending patent, cr) |
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Re:P=NP (Score:3, Insightful)
by anatoli ([email protected]) on Thursday May 25, @11:37AM EDT (#149)
(User Info) file:///unix
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If I prove to you that Halting Problem is not in NP, will I get my $10M?
--
Industrial space for lease in Flatlandia. |
Re:P=NP (Score:1)
by roman_mir on Thursday May 25, @11:48AM EDT (#163)
(User Info) |
Well, then you'll only get 1,000,000
(I would like to insert a picture with thousands of hands holding burning candlesticks and lighters right here...)(tm, pending patent, cr) |
Re:P=NP (Score:2)
by anatoli ([email protected]) on Thursday May 25, @11:56AM EDT (#170)
(User Info) file:///unix
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| Send me a binding offer, signed and all. My email is posted on ./; send me yours; I'll send you a fax #; you fax me your offer. I'm dead serious. Not that I'm hoping to ever collect that $1M from you. I just want to frame the offer and hang it from my wall. You'll get the proof anyway.
--
Industrial space for lease in Flatlandia. |
Re:P=NP (Score:1)
by seeken ([email protected]) on Thursday May 25, @04:45PM EDT (#300)
(User Info) |
I'll take 10% to keep quiet, please.
Surfing the net and other cliches...
(Who Meta-Meta-Moderates the Meta-Moderators?) |
Re:P=NP (Score:1)
by jekk ([email protected]) on Thursday May 25, @01:23PM EDT (#228)
(User Info) |
| Be careful what you offer. Suppose I were to develop a proof that P=NP, but it was an existance proof... proved that they were equivalent without showing how to achieve that equality. Or suppose that I discovered an algorithm to actually perform NP problems in polynomial time... specifically in time proportional to 1*N^(100000000) + 100000000*N^2. This is polynomial time, but it is SO BAD, that the size of N at which it becomes better then exponential is SO LARGE, that it is still not useful for real-world problems. I guess my point is that you shouldn't make the assumption that a mathematical proof automatically has immediately applicable consequences. And you shouldn't run around offering $10M USD unless you mean it. -- Michael Chermside
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Re:P=NP (Score:1)
by Bassthang on Thursday May 25, @01:32PM EDT (#233)
(User Info) http://www.bath.ac.uk/~mapjpw
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| I will buy the solution and the rights to the solution ... Maths doesn't work like that (yet!). You might be able to patent a particular algorithm (gggrrrrr!). Say you have solved the traveling salesman problem. No one will believe you until you publish; when you do the cat is out of the bag with regard to all the other NP complete problems.
"What I look forward to is continued immaturity followed by death."
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Re:P=NP (Score:1)
by Furry Ice on Thursday May 25, @02:27PM EDT (#262)
(User Info) |
You won't be able to reduce a solution proving P=NP to ATM or the Halting problem because those problems are neither in P nor NP. They are undecidable. *No* algorithm solves them. There's no need to worry about how efficiently an algorithm *doesn't* solve them. :-)
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Re:P=NP: It's not that easy (Score:1)
by TrixX on Thursday May 25, @05:04PM EDT (#302)
(User Info) |
First point: By "solving" it, it means that proving that P!=NP IS A SOLUTION. And that won't help you to solve "every protection algorithm in polynomial time".
Even if you proved that P=NP, and, say, RSA can be cracked in O(P(n)), it could still take a very long time to crack it, even with massive parallel efforts.
O(P(n)), means that
There are some m,r, that, for all n>m, r.P(n)>(the time of your algorithm for n bits).
That indicates several things:
* if m>(bits in RSA), the upper bound (r.P(n))in the solving time of that algorithm is not useful
* if r=bignum, and P(n)=x^bignum2, that only says that the cracking time is polynomial, but it could take bignum.x^bignum2 time units... and that could be a looong time (even with all the computer power of today acting together).
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I thought Navier-Stokes were unsolvable? (Score:1)
by linuxdevil (linuxdevil_at_fastq_dot_com) on Thursday May 25, @11:06AM EDT (#121)
(User Info) |
Being a formal aerospace engineering grad, I thought that the Navier-Stokes equations were generally unsolvable unless gross approximations were made and powerful computers were used for to calculate the results to any given degree of accuracy. 3rd order, partial, non-linear differential equations are fairly tough nuts to crack!
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Re:I thought Navier-Stokes were unsolvable? (Score:1)
by Shillo ([email protected]) on Thursday May 25, @02:39PM EDT (#267)
(User Info) |
Actually they are solvable in the sense that 'solution exists', not in the sense 'we can find the solution'. This is a distinction that was a topic of a rather sharp flamewar in the math community around 100 years ago. IIRC it was Hilbert who gave the first proof that was purely existential (i.e. it proved that something existed without a slightest hint as to how it might be found - the problem was one in algebraic geometry).
I refuse to use .sig |
Mistake 1 (Score:1)
by curril on Friday May 26, @11:27AM EDT (#429)
(User Info) |
Now try to prove or disprove that this has a solution:
find a,b,c,n in N:
a^n+b^n=c^n
Ah, but it does have solutions for n=2
;-)
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More easy $$$ :-) (Score:1)
by dped on Thursday May 25, @11:22AM EDT (#138)
(User Info) |
There is also a $1,000,000 prize for anyone solving Goldbach's Conjecture which is another of the 23 problems Hilbert stated at the ICM. See this article , it's an unrelated contest to bring attention to a book, Uncle Petros and Goldbach's Conjecture which I definetely recommend. It's more of a story/biography than a math book though.
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My final answer... (Score:1)
by PDHoss ([email protected]) on Thursday May 25, @11:32AM EDT (#144)
(User Info) |
Do I get any lifelines?
======================================
Writers get in shape by pumping irony. |
Navier Stokes Equation (Score:2, Interesting)
by mathematician ([email protected]) on Thursday May 25, @11:39AM EDT (#153)
(User Info) http://www.math.missouri.edu/~stephen
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Having worked on the Navier Stokes equation for about three years, I can say that this is a very interesting problem. Actually, the Navier Stokes equation is the equation governing an incompressible flow - that is, it is an equation that very closely approximates the flow of water. Thus to gain any understanding of this equation is to make a lot of progress in understanding turbulance. This will have great practical applications.
I know a little about some of the other problems, the Poincare conjecture, P=NP, and the Riemann hypothesis. Although these problems are hard, not too many people work on them. Basically, for an untenured professor, these are career destroyers. If one works on these problems as obsessively as a solution will require, you can kiss goodbye to tenure (unless you actually solve it).
In contrast, there are many many researchers the world over working on the Navier-Stokes problem, from graduate students to top mathematicians (like Charles Fefferman who wrote on the claymath website). One may not solve the real problem, but there is progress to be made, and many other easier problems to work on. One can make a reasonable career, while still having a dream of fame.
----
Another poster said that whoever solves this problem will probably already be very rich, working for some rich institute. This really is not the case - probably the solution will come from a moderately paid university professor. The money will be very handy, although then his/her career will be so certain that he/she will be comfortable without it. In any case, the motivation will probably be prestige rather than the money, but the amount of money raises the prestige.
Another poster commented that there are no Good Will's out there. I personally did not like the movie. I think that there are unfound genius's out there - Ramanujan is an example. But noone can solve these kinds of problems without being to some extent obsessive. Good Will did math as an afterthought, and no person can do great work like that. All the great genius's of history, in all creative endeviours, were not just brilliant, but also worked extremely hard.
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Kary Mullin (Score:1)
by lazzaro on Thursday May 25, @06:12PM EDT (#317)
(User Info) |
Kary Mullin may be an exception -- he did something (invent PCR) that seems to put him on a pretty high achievement plane (quick Nobel Prize, key technology for all of biotech, etc), and from his own accounts, he didn't really work obsessively long or hard on it. He was just driving home after a long trip, stopped by the side of the road, jotted down the basic idea for PCR, then spent a year of 40-hour weeks in the lab getting the bugs out of it.
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Re:Kary Mullin (Score:1)
by Pyotri on Thursday May 25, @09:55PM EDT (#362)
(User Info) |
That was a special case. A lot of biochemists were asking themselves "why didn't I think of that first"?
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Re:Navier Stokes Equation (Score:1)
by pittance ([email protected]) on Sunday May 28, @04:25AM EDT (#453)
(User Info) http://www.excession.swinternet.co.uk
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"Having worked on the Navier Stokes equation for about three years, I can say that this is a very interesting problem. Actually, the Navier Stokes equation is the equation governing an incompressible flow - that is, it is an equation that very closely approximates the flow of water. Thus to gain any understanding of this equation is to make a lot of progress in understanding turbulance. This will have great practical applications..."
Not sure where you got the incompresible bit from... The Navier-Stokes equations actually govern _any_ type of flow (unsteady, compressible, viscous and any other type you want) in their unreduced form.
If you want to assume incompressibility they will still be valid (for water and air flows up to about Mach 0.3 ish), but they won't be the full Navier-Stokes equations any more.
This problems may have a larger impact than you suggest... or less.
Computational aerdynamics can't solve the full N-S equations at all yet - and even if they could the computer power needed for the fineness of mesh needed would be beyond what's available now, for any reasonable run time at least.
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what's the name of this math trick? (Score:2, Insightful)
by Fr�d�ric ([email protected]) on Thursday May 25, @11:43AM EDT (#156)
(User Info) http://fly.to/tiange
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take the number 12345679 (there's no 8)
multiply by a number that the sum is 9 and see:
12345679 x 09 = 111111111
12345679 x 18 = 222222222
12345679 x 27 = 333333333
12345679 x 36 = 444444444
12345679 x 45 = 555555555
12345679 x 54 = 666666666
12345679 x 63 = 777777777
12345679 x 72 = 888888888
12345679 x 81 = 999999999
who discovered that? is there a name to this thing? do you know any others funny values that make the same thing? is there anything to proved in that?
--
BeDevId 15453 - Download BeOS R5 Lite free!
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Re:what's the name of this math trick? (Score:1, Funny)
by Anonymous Coward on Thursday May 25, @12:13PM EDT (#182)
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| Nice one, I could never remember my 9 times table. Now all I have to do is think 9*x is the same as xxxxxxxxx / 12345679 Genius!
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Re:what's the name of this math trick? (Score:1, Interesting)
by hubertf on Thursday May 25, @12:26PM EDT (#191)
(User Info) |
funny... this also works with hex numbers:
% bc -l
ib=16
ob=10
x=123456789ABCDF ���<- no "E"
x * F * 1
111111111111111
x * F * 2
222222222222222
...
x * F * 9
999999999999999
x * F * A
AAAAAAAAAAAAAAA
...
x * F * E
EEEEEEEEEEEEEEE
x * F * F
FFFFFFFFFFFFFFF
� - Hubert
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Re:what's the name of this math trick? (Score:1)
by mjc_w on Thursday May 25, @07:22PM EDT (#330)
(User Info) |
This works because:
in base 10, 1/(9*9) = .012345679...
in base 16, 1/(15*15) = .012345789ABCDF...
It works in any base.
Here's one I proved a while ago:
987654312
--------- = 8
123456789
This also works in any base.
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Re:what's the name of this math trick? (Score:1)
by DrTomorrow on Thursday May 25, @01:03PM EDT (#221)
(User Info) |
| When you rewrite it like this, it doesn't look so incredible 12345679 x 09 x 1 = 111111111
12345679 x 09 x 2 = 222222222
12345679 x 09 x 3 = 333333333
12345679 x 09 x 4 = 444444444
12345679 x 09 x 5 = 555555555
12345679 x 09 x 6 = 666666666
12345679 x 09 x 7 = 777777777
12345679 x 09 x 8 = 888888888
12345679 x 09 x 9 = 999999999
print sort split('a','roaDrawaraoamaTo'); |
I like this one better... (Score:1)
by TrixX on Thursday May 25, @05:14PM EDT (#306)
(User Info) |
x=142857
1*x=142857
3*x=428571
2*x=285714
6*x=857142
4*x=571428
5*x=714285
(If you don't get it, try to look at the diagonals...)
oh, and
7*x=999999
and obviously, 1/7=0.142857 142857 142857...
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They Forgot Goldbach. (Score:2, Funny)
by istartedi ([email protected]) on Thursday May 25, @11:47AM EDT (#160)
(User Info) http://www.vrml3d.com/
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There was a /. story a month or so back about the Goldbach conjecture being worth a million. At least I could understand that problem. These look hopeless. What is a topological manifold anyway? Whenever I hear that, I always picture mathematicians going down to the junkyard and picking apart exhaust systems... Q: Why is that math professor making all that noise and stinking up the place? A: Sounds like he has a hole in his topological manifold. :)
"Your theory of a donut shaped universe is intriguing. I may have to steal it." --Stephen Hawking to Homer Simpson. |
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Re:They Forgot Goldbach. (Score:1)
by VaporX on Thursday May 25, @01:01PM EDT (#220)
(User Info) |
I think MOST people have a hole in their manifold, they just don't make noise and emit stink from it publicly ;)
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Relevance (Score:2, Interesting)
by bsletten ([email protected]) on Thursday May 25, @11:47AM EDT (#162)
(User Info) |
This whole "contest" is fascinating. On the one hand, fewer and fewer people are able to make a living as mathematicians. On the other hand, to solve these kinds of problems, one has to throw oneself at them whole-heartedly. It does seem like the money should come up front to support those capable of such solutions; somewhat of a mathematical patronage. But, the question then becomes one of purpose. I love the notion of solving these kinds of problems, but to whom is it worth a million dollars? As the problems become harder and more abstract, solvable by fewer people, the tangible benefit seems to tend toward naught. It is a little akin to mountain climbing expeditions. It serves a corporation no direct benefit to sponsor these trips. Are we going to make the transition from contest to corporate-sponsored mathematicians?
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Re:Relevance (Score:1, Interesting)
by Anonymous Coward on Thursday May 25, @12:15PM EDT (#186)
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| Actually the job market for phd mathematicians has improved in the last couple of years. From about 1989 to 1998 it was very bad. In a typical year about 1200 phds in math are awarded in the US; a large fraction (2/3rds?) are able to get academic posts, although many of these are temporary. For a reasonably talented mathematician (i.e., can complete a phd at a good school), an academic career is a reasonable expectation. You can then obsessively spend your time working on strange problems (although the advice to stay away from famous problems is wise, and you will have to earn your keep by teaching). Speaking of strange problems: My favorite is this one. Pick a number $x>1$ not an integer. Look at its powers $x^n$, $n>1$. Look at the fractional part of its powers. Any pattern? It is known that for "almost all" numbers (in the sense of Lebesgue), the fractional parts are uniformly distributed. But no-one knows for specific numbers like $e$ or $\pi$. No one knows what the distribution is for $x=3/2$. But check out $x=\frac{1+\sqrt{5}}2$, the golden ratio. The fractional parts all are either 0.000... or 0.9999 (alternating) after awhile. Algebraic numbers with this property are called Pisot numbers. No one knows if there are transcendental numbers that do this.
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Re:Relevance (Score:2)
by J Stephen on Thursday May 25, @12:40PM EDT (#202)
(User Info) |
Come on people!!!! The minute we start questioning whether we really need to know something is the minute we cast off our Linux boxes and head back for the caves!!! Talk about asinine comments. I thought we figured out a long time ago that scientific (including mathematical) research is justified in and of itself. Sheeesh!!!
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Re:Relevance (Score:1)
by bsletten ([email protected]) on Thursday May 25, @02:47PM EDT (#271)
(User Info) |
Reading comprehension is a lost art.
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Re:Relevance (Score:1)
by J Stephen on Friday May 26, @06:11PM EDT (#440)
(User Info) |
I'm sorry. Maybe I had a little too much coffee that morning. But your comment did come across as somewhat anti-progressive. What difference does it make who gets paid and who doesn't for solving some of the most fascinating (to some) problems mathematical endeavor has yet uncovered. Your comment about ". . . to whom is it worth a million dollars?" should be answered with a resounding HUMANITY!!! Who cares why these "corporations" or "contests" want to promote mathematical research (probably to promote themselves to some sort of constituency). The important thing is that the rolling stone that is science keeps rocking!! (Don't insult my education, that pisses me off =])
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Re:Relevance (Score:1)
by bsletten ([email protected]) on Tuesday May 30, @10:26AM EDT (#464)
(User Info) |
You completely missed the point of my message. You read into it things that weren't there. Get the chip off your shoulder and re-read it. Never did I question the validity of pure research, pure solutions, etc. I was considering the economic realities of doing this stuff. So that this conversation doesn't continue to be irrelevant and unrelated to my original post, let's enumerate my points again.
There aren't many pure math jobs
There aren't many people who can solve these problems
People need to throw themselves at these problems to solve them
This requires external funding
Whence comes that funding?
Why?
Do you see that nowhere in there did I even insinuate these weren't problems worth solving? I was asking a question of who would pay people to solve them. There is a definitely a celebrity status associated with it. Celebrities get endorsements. Perhaps we will get into a model where we have corporate-sponsored mathematical celebrities like we do athletic and entertainment celebrities. That was the purpose of my post. Apparently you missed that purpose entirely. *THUS* I lamented the decline of reading comprehension out there. I'm sorry if you feel like I've insulted your education. Try reading without reacting; that's a good start.
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P != NP (Score:2, Interesting)
by Anonymous Coward on Thursday May 25, @12:14PM EDT (#184)
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Take a typical NP search space, and start numbering the states (breadth first search works well, as it will number infinite search spaces) Now the proof that finding the minimal (or maximal, ordinal) state given N of them takes O(N) time if the states contain no information about each other(they are uncorrelated). So the problem reduces to finding one such search space so that there are 2^N states, and you can prove that the states are statistically uncorrelated. That problem cannot be solved in polynomial time. Finding such a search space is an exercise left to the reader.
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Form Letter from the Math Department (Score:2)
by JoeBuck (jbuck at welsh-buck dot org) on Thursday May 25, @08:17PM EDT (#339)
(User Info) http://www.welsh-buck.org/jbuck/
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| Dear ___Anonymous Coward___, The first error in your proof occurs on page ____1____, in line _____3____. Sincerely, Joe That is, you make a fundamental error about the implications of reducing a problem to another problem. If I reduce a problem to a polynomial problem using a polynomial transformation, I know I have a polynomial algorithm, but you propose doing the reverse: "reducing" the problem to an exponential problem. That proves nothing; after all, I can provide you with an exponential sort algorithm (try every permutation until you find one that puts the values in order). The problem you transform to may not be the least expensive problem. Hence you do not prove that P != NP.
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Hilbert's recording (Score:1)
by JimH on Thursday May 25, @12:29PM EDT (#194)
(User Info) |
Does anyone know if that recording is online?
I'd very much love to hear it, and have it.
Jim
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Re:Hilbert's recording (Score:3, Informative)
by David A. Madore ([email protected]) on Thursday May 25, @08:36PM EDT (#352)
(User Info) http://www.eleves.ens.fr:8080/home/madore/
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No, it's not available on-line. (Even after hearing it, I would really like to get it a copy.) Maybe by asking Professor Alain Connes (he seems to be the one who found the tape) very politely...
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Choosing a good set of problems is hard (Score:1)
by SIGFPE (0x7ff00000 plus four at sigfpe.com) on Thursday May 25, @12:40PM EDT (#201)
(User Info) http://www.sigfpe.com
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Hilbert did a fantastic job of picking his original set of problems. For example, when he asked for an algorithm to determine if there exist integer solutions to any polynomial equation it just seemed like a pretty ordinary outstanding problem to most people - an obvious thing to ask. But it turned out to be undecidable. This was a whole new category of answer that mathematicians simply hadn't anticipated - the idea that not only was the problem unsolvable - but that it was provably so. The solution to this problem had much much wider implications than the problem originally suggested. This was true of many of Hilbert's other problems. He was a genius for picking such rich and complex problems. It remains to be seen whether this new set will inspire as much great mathematics.
--
http://www.sigfpe.com |
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Re:Choosing a good set of problems is hard (Score:1)
by mathematician ([email protected]) on Thursday May 25, @10:38PM EDT (#371)
(User Info) http://www.math.missouri.edu/~stephen
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What was so good about Hilbert's problems is that they were just the right degree of hardness - not so easy as to be solved the next day, but not so hard as to be totally unsolvable. The fact that so many of his problems were solved in the twentieth century shows that they really were the problems for the twentieth century.
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Why are these important? (Score:1)
by VaporX on Thursday May 25, @01:05PM EDT (#222)
(User Info) |
Having a CS background, I understand the importance of solving P=NP, but what is the importance of the other problems? Obviously people think they are very important problems to solve. What great mysteries unravel upon solving them, though? In the case of Fermat's last theorem, proving it didn't really affect math or physics much. (or did it?) Are these any different, or just "curiosity" problems?
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Re:Why are these important? (Score:1)
by The Linguist on Thursday May 25, @03:41PM EDT (#288)
(User Info) |
Adam and Eve ate from the Tree of Knowledge of Good and Evil even though they weren't supposed because they couldn't stand the feeling of an available experience that they had not experienced themselves. That curiosity has survived in humanity to this day. So the importance of solving these math problems does not lie in "unravelling great mysteries" or breaking new ground in mathematical theory, or in finding new engineering applications. People solve these problems because they like the taste of apples, and no other reason. If engineers pick up the results and make something useful, that's good, but even if they don't we are left with something beautiful unto itself.
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Re:Why are these important? (Score:2, Interesting)
by Trojan on Thursday May 25, @08:18PM EDT (#340)
(User Info) |
The single fact that Fermat's last theorem is of mostly historical value. However, it was proved by showing that a large parts of the Taniyama conjecture hold (in fact, by now the full Taniyama conjecture has been proven). The Taniyama conjecture is important in the theory of elliptic curves, and a lot of mathematics has been built on top of it (even though for a long time it was not clear that the conjecture was actually true). The theory of elliptic curves nowadays has applications in areas from cryptography (factorization of integers) to theoretical physics.
The 7 problems at hand have similar implications. For example if we know that the Riemann Hypothesis is true, we suddenly have very precise estimates on for example the growth of the number of primes. Also, the Riemann Hypothesis has generalizations in other areas of mathematics (some of these generalizations have been proven by now).
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Didn't we see this before? (Score:2)
by grappler (thegrappler@DIE_SPAMMERS.usa.net) on Thursday May 25, @01:26PM EDT (#229)
(User Info) http://www.mines.edu/Stu_life/organ/ufo/
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Wasn't this posted months ago? Or was that a different problem? I seem to recall another math problem that a million dollar reward was offered for recently...
--
grappler
For-pay Internet distributed processing - check it out! |
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Nevermind, found it (Score:2)
by grappler (thegrappler@DIE_SPAMMERS.usa.net) on Thursday May 25, @01:32PM EDT (#235)
(User Info) http://www.mines.edu/Stu_life/organ/ufo/
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Just found the article (funny, I looked for it BEFORE I posted and didn't see it).
It's the Goldbach Conjecture, from two months ago.
--
grappler
For-pay Internet distributed processing - check it out! |
Andrew Wiles must be turning red.... (Score:1)
by efuseekay on Thursday May 25, @01:36PM EDT (#238)
(User Info) |
IF only he has held on to his proof of Fermat until they announce 1M bux for it... (Btw, all those problems also come attached a Fields Medal.) So now those Field Medalists can also say to the Nobelists "Ha! _I_ also got a Million Bucks!"
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Re:Andrew Wiles must be turning red.... (Score:1)
by Trojan on Thursday May 25, @08:19PM EDT (#341)
(User Info) |
An important difference with the Nobel prize is that one has to be under 40 to be eligible for a Fields Medal.
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Re:Andrew Wiles must be turning red.... (Score:2)
by David A. Madore ([email protected]) on Thursday May 25, @08:34PM EDT (#350)
(User Info) http://www.eleves.ens.fr:8080/home/madore/
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Professor Andrew Wiles was present at the meeting (he made a little speech to introduce Laurent Lafforgue and later John Tate - although the latter really needs no introduction). I can assure you he did not turn red.
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Re:Andrew Wiles must be turning red.... (Score:1)
by bluGill ([email protected]) on Friday May 26, @02:45PM EDT (#434)
(User Info) http://www.black-hole.com/users/henrymiller/
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There were several prizes for solving fremat's last theorom. One of the larger was set up my german mathamatitions before WWI. Unfortunatly inflation in germany in the '20s turned several million into enough for a gumball.
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Here are the answers: (Score:2)
by cwhicks ([email protected]) on Thursday May 25, @01:40PM EDT (#239)
(User Info) |
1. Elevendy-four :QED
2. 42
3. 69 (Ooh, sexy.)
4. Both 7 and 11 (it's a quantum problem)
5. 14 cats + 1 hedgehog * Larry Storch
6. No
7. E=mc^3
Please have my check ready, I will be over to get it after "Full House".
- I like pudding. |
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Re:Here are the answers: (Score:1)
by aphr0 ([email protected]) on Thursday May 25, @03:39PM EDT (#287)
(User Info) |
God damn. You are incredibly not funny.
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Re:Here are the answers: (Score:1)
by cwhicks ([email protected]) on Thursday May 25, @04:12PM EDT (#293)
(User Info) |
I love you too dillhole.
- I like pudding. |
I'm sorry, you didn't use QED (Score:1)
by PepperDude ([email protected]) on Thursday May 25, @01:43PM EDT (#242)
(User Info) |
If solution to the problem doesn't end in QED, are they disqualified, similar to not using "who is" in Jeopardy?
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Re:I'm sorry, you didn't use QED (Score:1)
by CrazyMadPsychoBandit on Thursday May 25, @03:51PM EDT (#290)
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| I had a prof who would always end his proofs in W^5 (which was what we wanted). Then I had another guy who drew a little box at the end of the proof (and he didn't like it if we used q.e.d. or anything else!)
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Mathematicians are alien infiltrators ... (Score:1)
by SatansNemesis (jamesmcl(at)mweb.co.za) on Thursday May 25, @01:44PM EDT (#244)
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Having had math to tertiary level I have come to realise that it takes a special animal to be a mathematician :D Really weird ass people ... Clever and usefull though ;)
One ring to rule them all, One ring to find them, One Ring to bring them all and in the darkness bind them. |
This show is brought to you by the number 9... (Score:1)
by akiaki007 on Thursday May 25, @01:59PM EDT (#248)
(User Info) |
Fr�d�ric posted:
take the number 12345679 (there's no 8)
multiply by a number that the sum is 9 and see:
12345679 x 09 = 111111111
12345679 x 18 = 222222222
12345679 x 27 = 333333333
12345679 x 36 = 444444444
12345679 x 45 = 555555555
12345679 x 54 = 666666666
12345679 x 63 = 777777777
12345679 x 72 = 888888888
12345679 x 81 = 999999999
Here is the answer:
this has nothing to do with 12345679...use any number. The number 9 is a special number... If you multiply any number by a numbers who's digits add to make 9 (i.e. 9, 18, 27, 63, 36, 1233, 0.003211000101) and you will get a number who's digits also add to make 9...Try it
1233 * 5 = 6165...6+1+6+5 = 18 ... 1+8=9
You can use any number who's digits add to make 9 and do this...it's a nifty trick...works great to change any number to any number you want...as long as somewhere in the line you multiply by something like 9, or any of the other numbers.
-akshay
"Time is long and life is short, so begin to live while you still can." -EV |
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I have a math problem of my own... (Score:2)
by extrasolar (klh@sedonaSPAM_TRAP.net) on Thursday May 25, @03:26PM EDT (#281)
(User Info) http://users.sedona.net/~klh/kevin
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When, again, does the 20th century start?
:)
--- Only YOU can stop the spread of non-recursive acronyms! |
Proof? (Score:1)
by PhantomCowboyGorilla on Thursday May 25, @04:23PM EDT (#297)
(User Info) |
How do we know these things can be proven at all? If they know for a fact they can be proven shouldn't they already have the answer? Prove to me why you can't tickle yourself and I'll be impressed.
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Re:Proof? (Score:1)
by x-empt ([email protected]) on Thursday May 25, @10:22PM EDT (#370)
(User Info) http://www.ispep.cx/
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Yeah, I tickle myself all the time ... All I do is stimulate the right nerves. Our methods of thinking, as human beings, are flawed. We have developed a system of math and science, that for the most part is correct but then cannot be "proven" in any existance. In reality you can NEVER have '1'in addition to '1' equal '2' (basically 1+1=2) because the process of adding two 1's together actually requires something to bond them, hence we have the plus sign (which means basically a bond... in simple mathematics). We added a new element to the equation to make 1+1 which is a reference to '2'. Anyone follow?
"School officials do not possess absolute authority over their students. Students in school as well as out of school are ?persons? under our Constitution.? -- J |
Re:Proof? (Score:1)
by Old Wolf on Friday May 26, @09:26AM EDT (#422)
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I think you're on drugs.
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Applicability (Score:1)
by LionMan ([email protected]) on Thursday May 25, @05:34PM EDT (#307)
(User Info) http://127.0.0.1/
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These problems are some of the most important problems today, as I see it. Flow of incompressible fluids for fluid mechanics, missing mass from yang-mills fields, and expecially the Reimann-Zeta hypothesis. If I am correct, Reimann-Zeta is extremely important to find the distribution of primes (will there always be more twin primes? etc). If this is solved, we have some problems: if we can easily determine the distribution of primes then we can predict the next one, and RSA falls apart because it relies upon people having a hard time finding those really-really-big primes that people use for high encryption. If it is proved unsolvable, we are happy in encryption-land, but then we can't find stuff like:
oo
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\ 1
� | ---
/ n^3
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n=1
in closed form (yes, it can be expressed as an infinite product of (1-x/root_n) or (x-root_n)).
It would be sad to see it proven unprovable, for such a cool problem.
-LLM |
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Re:Applicability (Score:1)
by LionMan ([email protected]) on Thursday May 25, @05:37PM EDT (#309)
(User Info) http://127.0.0.1/
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Sorry, my bad. Since this one needs to have roots along axes which are non-real, that should have been (1-z/root_n) or (z-root_n) in those parenthases up there.
-LLM |
Re:Applicability (Score:1)
by Trojan on Thursday May 25, @08:26PM EDT (#344)
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Not really true: in fact it is very easy to find big primes. The problem lies in factoring the product of two particular big primes. The fact that you can find lots of primes of about the right size doesn't help here at all.
(Primes are pretty common, even when you get to very large numbers you find an awful lot of primes and in fact way way way too much to check them all.)
You are correct in saying that the Riemann hypothesis is very important to find the distribution of primes. If all non-trivial zeroes are on the trivial strip, we have very good estimates for, say, the number of primes smaller than x. However, this will never be precise enough to 'predict' the next one since 'locally' the distribution of primes is inherently irregular (it's only on a large scale where the pattern becomes clear).
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Re:Applicability (Score:2)
by Signail11 on Thursday May 25, @08:38PM EDT (#353)
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Flow of incompressible fluids?!?! That's what the Navier-Stokes differential equations are for. Most mathematicians believe that the GRH is almost certainly true. There will be no new applied results that suddenly arise when GRH is finally proven. We already have a very good sense of the distribution of primes, but it is not possible to predict the next prime in any real sense (that is, a priori based on being given an arbitrary prime), except for actually finding it. The GRH offers insight into how the primes are distributed, but not where a specific prime would be; RSA depends not on the difficulty of finding large primes (this is already easy: Miller-Rabin SPPTs are entirely adaquate, use eliptic curve techniques if you wish to prove primality), but rather that of decomposing large composites into primes. GRH offers no insight into how to do this; even if it did, we can merely assume that it is true, without any proof, and use the corresponding results. Proving GRH!=breaking RSA.
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Re:Applicability (Score:1)
by LionMan ([email protected]) on Thursday May 25, @10:57PM EDT (#375)
(User Info) http://127.0.0.1/
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DUH, that's what I'm talking about!! Do you think I'm so stupid that I think Reimann-Zeta applies to fluid mechanics? Did you think i thought that of yang-mills fields too? I was referring to more than ONE of the problems! Don't think yourself higher than others just because you are higher than many (or think you are).
-LLM |
give me the 7000000$ (Score:1, Redundant)
by johoho on Thursday May 25, @06:22PM EDT (#321)
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why didn't anybody else think about it? 42!
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My own comments (Score:4, Informative)
by David A. Madore ([email protected]) on Thursday May 25, @07:53PM EDT (#334)
(User Info) http://www.eleves.ens.fr:8080/home/madore/
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All right. I posted the story, now here are a few comments of my own. The problems are all fairly old. The Riemann hypothesis is the oldest, as it dates from around 1860 and it was already part of Hilbert's 1900 list. Surprisingly, almost no progress has been made toward its resolution (except for Deligne's proof in the generalized local case, aka the Weil conjectures, but that isn't really the same problem). The Poincar� conjecture is also rather old. The study of the Navier-Stokes equation is probably as old as the equations themselves, but the mathematical foundations for a rigorous study are more recent. P=NP is the newest problem, coming as it does from computer science, and also the most likely to be of interest to Slashdot geeks. John Tate rather messed up (IMHO) in his presentation of it (he obviously didn't think it was very interesting), and I gather many mathematicians believe it isn't very rigorously refined (certainly, many French mathematicians, with their usual contempt for logic and computers science). In fact, it is perfectly rigorous and mathematicial. It is probably the easiest problem on the list, as the theory behind it is still not fully developped. For a start, I'd recommend Papadimitriou's book "Computational Complexity" for the present state of knowledge: it is very thorough, well-written and interesting. Personally, I'm convinced P=NP is undecidable (in a nutshell, my point is that for Turing machines with oracles, by judiciously choosing the oracle we can arrange so that P=NP or P!=NP, see Papadimitriou's above-mentioned book for a proof, and I don't despair of the existence of a notion of forcing on oracles of some kind). It would be a very weird situation (for Pi^1 propositions, as a previous poster pointed out, if it is undecidable then it is true - proviso the consistency of arithmetic which is the hidden assumption in the meta-proof that "unrefutable=>true"; but for P=NP it just looks... weird). I don't think any of the other statements are undecidable (if any of them is, we are light-years away from being able to prove it). Naturally, it is possible for the undecidability of a statement to be undecidable, and so on up to (ordinal) heights of staggering difficulty (though by a theorem of Tarksi, at some point the proposition must be decidable - little comfort in practice). Some have asked why Goldbach's conjecture (for which someone else is offering $1M) is not in the list. The fact is that it's not interesting. Every one of these seven problems, if solved, would have far-fetching consequences, and its proof would open vast new areas of research. Not so with Goldbach's conjecture. It is amusing because it is very simple, simple enough to be explained to laymen. But apart from a few additive number theorists, nobody is interested in it. Proving it would require ingenuity, certainly, and much technique in sieving methods and circle integrals; but nothing like opening new areas of mathematics. Don't waste your time on Goldbach's conjecture: it is highly technical and intellectually almost worthless. Anyway, getting back to the seven problems, some people I know have been rather angry that such old problems have been selected, not more recent stuff. In a way the list is rather "conservative". For example, the global case of the Langlands programme, which is probably the most ambitious programme in mathematics (at any rate, in number theory and algebra), the global analogue of what Lafforgue proved in the local ("functions field") case, was not selected. Perhaps also because it is thought to be too difficult...
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Re:My own comments (Score:1)
by Signail11 on Thursday May 25, @08:33PM EDT (#349)
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Minor technical quibble: when/if Goldbach's conjecture is finally decided, sieves will be probably not be used. The techniques of sieving tend not to be useful when the underlying assumption that the spliting residues of integers are distributed randomly (in a certain exact sense). This is not the case when the integers are close to each other or related to each other in some way. I'm sorry that I can't explain this any better, but Halberstam's book on sieve methods does a very admirable job if you are interested in the details.
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Re:My own comments (Score:1)
by streetlawyer (johnsaulmontoya@MAJORPORTALENDINGINEXCLAMATIONPOIN) on Friday May 26, @04:44AM EDT (#404)
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| certainly, many French mathematicians, with their usual contempt for logic Ever hear of Bourbaki, fuckhead?
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Re:My own comments (Score:2)
by David A. Madore ([email protected]) on Friday May 26, @05:06AM EDT (#406)
(User Info) http://www.eleves.ens.fr:8080/home/madore/
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Yes, thank you very much, my school (translate: college) happens to be where Bourbaki's office is (and, essentially, his "birth place"), and I've read a significant portion of his "�l�ments de Math�matiques". I fail to see how that is relevant. Bourbaki was (were?) not interested in logic. His (their?)treatment of logic and set theory in his first book ("Th�orie des Ensembles" - whose publication was long delayed as it remained in the state of a "fascicule de r�sultats") is sketchy at best, unelegant, and, in essence, as short as he could arrange it, just enough to provide the foundations for the other books. Anyway, that is old. Currently, there are only about two groups doing logic in France: around Krivine in Paris and around Girard in Marseille (I think). Girard is the inventor of linear logic, so that's a clame to fame, but he's more the exception than the rule: logic is held in disrepute, if not contempt, in France (and I much regret it).
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Re:My own comments (Score:1)
by Dr Dick (pfaf@[email protected]) on Friday May 26, @06:01AM EDT (#409)
(User Info) http://www.scs.leeds.ac.uk/pfaf/index.html
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| Don't waste your time on Goldbach's conjecture: it is highly technical and intellectually almost worthless. Would the same have been said of Fermat's last theorem. Again it seems to be a rather special result in number theory. Yet the proof of this turned out to be very deep indeed, and perhaps one of the outstanding mathematical results in the last centuary.
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Add this one to the list... (Score:1)
by bobv-pillars-net ([email protected]) on Thursday May 25, @08:06PM EDT (#337)
(User Info) http://www.pillars.net/
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| I've got one to add to the list. I first formulated it about fifteen years ago, when I was very into dice-throwing role-playing fantasy games and wanted to find an algorithm to generate well-behaved random numbers. One way to state the problem is this: Write a function "f(t,n,d)" that calculates the odds of arriving at a total "t" when rolling "n" number of "d"-sided dice and adding the total. For instance, f(15,3,6) would equal 10/216, or approximately 0.046. About ten years ago I figured out that the numerator in the fraction would be equal to the integral coefficient of X^t in the expansion of (x^d + x^(d-1) + x^(d-2) + ... + 1)^n. Of course, the denominator would be equal to d^n and also equal to the sum of the coefficients. Partial credit for finding a function that generates a reasonable approximation (say to three significant digits) over a reasonable range of "d" and "n" (say 1-100 and 1-1000, respectively). Oh yeah, and the function has to be computable in polynomial time. Someday when I learn how to use a symbolic math package, I'm gonna generate a huge dataset and tell it to find the nearest nth-order approximation, for n=3..20, and look for a pattern.
[email protected], geek@large
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Re:Add this one to the list... (Score:1)
by bobv-pillars-net ([email protected]) on Thursday May 25, @08:12PM EDT (#338)
(User Info) http://www.pillars.net/
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Oops -- that should have read: - ...the "t"th integral coefficient of the expansion of...
not - ...the integral coefficient of X^t in the expansion of...
Sorry.
[email protected], geek@large |
Re:Add this one to the list... (Score:2)
by Signail11 on Thursday May 25, @08:26PM EDT (#345)
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This is a trivial, solved problem. The method of generating functions solves this and a much more general set of enumeration and probability problems. Informally, a generating function is a function whose formal power series (to be distinguished from the more familiar Taylor power series) has coeffecients equal to the corresponding value of the series that the function generates. Wilf's _Generationfunctionology_ (yes, the title is real) explains this very well, much better than I could do, since I only have a cursory understanding of these methods. By the way, even the naive approach of enumerating all the possibilities is easily computing, in a certain abstract sense. The function is primitive recursive and can be computed in polynomial time using the general methods in Knuth's _TAOCP_. Volume 4, when it comes out, will go into more detail about these techniques.
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Re:Add this one to the list... (Score:1)
by bobv-pillars-net ([email protected]) on Friday May 26, @06:07AM EDT (#410)
(User Info) http://www.pillars.net/
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| Lessee... Amazon doesn't have it. Barnes and Noble has... a more intelligent search engine. Perhaps you mean - Generating Functionology
by Herbert S. Wilf Yup, that's gotta be it. Thanks -- I'll buy and read the book.
[email protected], geek@large
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Re:Add this one to the list... (Score:1)
by bertilak on Friday May 26, @03:04PM EDT (#435)
(User Info) http://whutch.home.netcom.com/
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To be pedantic, the name of the book is Generatingfunctionology. See http://www.bookfinder.com/search/?author=wilf&title=Generatingfunctionology&submit=Begin+S earch&new_used=*&binding=*&isbn=&keywords=&min_price=&max_price=&order_help=on¤cy=US D&mode=advanced&st=sr&ac=qr
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I know the awnser (Score:1, Offtopic)
by Duke of Org ([email protected]) on Thursday May 25, @08:59PM EDT (#359)
(User Info) http://www.lstrunk.homestead.com
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The awnser is 3, I do not care what the question is, the awnser is 3!
What about a beowulf cluster of these...... |
Where the real solutions lay ... (Score:1)
by x-empt ([email protected]) on Thursday May 25, @09:55PM EDT (#363)
(User Info) http://www.ispep.cx/
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In today's world we are told to think in simplified terms of +1+1=+2 and that +2-1=+1, "+" being the inverse of "-" all relative to zero. Relative to one object, zero, but then we move into Algebra class and learn of variables and that 'x' can be relative to 'z' and that 'y' is relative to 'x'... hence 'y' we know is related to 'z', although we are not told that directly.
"School officials do not possess absolute authority over their students. Students in school as well as out of school are ?persons? under our Constitution.? -- J |
evens or odds. which are there more of? (Score:1)
by classique on Thursday May 25, @11:01PM EDT (#376)
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Back in highschool, I had a thought. I thought of an axiom that 2/3 of all hole numbers are even. Here's my proof:
even * even = even
odd * even = even
odd * odd = odd
Also, I continued to say:
even + even = even
odd + odd = even
even + odd = odd
Of course, I know the fallacy in this. Do you?
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Re:evens or odds. which are there more of? (Score:1)
by Old Wolf on Friday May 26, @09:29AM EDT (#424)
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I really, really, REALLY hope this is a troll.
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I have a proof for all of them...! (Score:1)
by diamond on Friday May 26, @04:12AM EDT (#403)
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...but it will not fit in the margins of this comment box...
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Maths is pretty stupid (Score:1)
by ChristianSuburbanite on Saturday May 27, @02:21AM EDT (#444)
(User Info) http://www.ghostoutpost.8m.com
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Ok, I'm just a high school student, but I've always thought maths had serious problems. I've been told if I knew calculas it would help but anyway. Can't P=NP only when N=1? I seriously don't understand it. But my proof that maths has flaws. When you graph y=(x+2)/(x^2-4). when y=(x+2)/(x^2-4) x can't = + or - 2 right. But then you factorize (x+2)/(x^2-4) and you get 1/(x-2). so then when y=1/(x-2) can't = 2, but it can = -2 now. If you understood all that it's pretty screwed :P
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Mathematical Problems for the New Age (Score:1)
by Kywong on Monday May 29, @05:47AM EDT (#456)
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I imagine this is a really dumb question but ... How was it that the third problem, one involving deterministic and non-deterministic Turing machines, could be posed in 1900 when Alan Turing wasn't born until 1912 ?
Kywong, Elanora Heights, NSW, Australia |
Re:I think I got it! (Score:1)
by br_2k on Thursday May 25, @08:26PM EDT (#346)
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Hmm. I feel with you! I thought it was funny. Looks like nobody around here knows Ford Prefect... Let's have a Pan Galactic Gargle Blaster and forget about it.
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Re:I think I got it! (Score:1)
by Old Wolf on Friday May 26, @08:02AM EDT (#413)
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It was funny(ish) the first time anyone heard it, about a million years ago. Now it is boring.
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Posted by Hemos on Thursday May 25, @10:09AM
Mathematical Problems For The New Age