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Goldbach Conjecture: Closer To Solved?

timothy posted about 2 years ago | from the eventually-knock-it-down-to-one dept.

Math 170

mikejuk writes "The Goldbach conjecture is not the sort of thing that relates to practical applications, but they used to say the same thing about electricity. The Goldbach conjecture is reasonably well known: every integer can be expressed as the sum of two primes. Very easy to state, but it seems very difficult to prove. Terence Tao, a Fields medalist, has published a paper that proves that every odd number greater than 1 is the sum of at most five primes. This may not sound like much of an advance, but notice that there is no stipulation for the integer to be greater than some bound. This is a complete proof of a slightly lesser conjecture, and might point the way to getting the number of primes needed down from at most five to at most 2. Notice that no computers were involved in the proof — this is classical mathematical proof involving logical deductions rather than exhaustive search."

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170 comments

and here is the proof for every even number (4, Funny)

Anonymous Coward | about 2 years ago | (#39989619)

I hereby prove that every even number is a sum of no more than six primes, one of those is 1.

Re:and here is the proof for every even number (5, Informative)

santiago (42242) | about 2 years ago | (#39989681)

I hereby prove that every even number is a sum of no more than six primes, one of those is 1.

Psst, 1 isn't prime. Or composite. It's neither.

Re:and here is the proof for every even number (4, Insightful)

Smurf (7981) | about 2 years ago | (#39989739)

I hereby prove that every even number is a sum of no more than six primes, one of those is 1.

Psst, 1 isn't prime. Or composite. It's neither.

True, but you can change the GP's proof to "every even number n (where n > 4) is a sum of no more than six primes, because m = n - 3 is an odd number".

Re:and here is the proof for every even number (1)

theshibboleth (968645) | about 2 years ago | (#39991481)

I'm not sure what your point regarding m is, but if you just take out the 'one of these is 1' predicate it's still true since 5 primes is in the domain of 'no more than 6'.

Re:and here is the proof for every even number (0)

Anonymous Coward | about 2 years ago | (#39991779)

You missed "even" and "odd".

Re:and here is the proof for every even number (2)

Geoffrey.landis (926948) | about 2 years ago | (#39989745)

Every even number greater than 1 is the sum of no more than six primes, one of which is three.

Re:and here is the proof for every even number (1)

Anonymous Coward | about 2 years ago | (#39989781)

2 = 3 + ?

Re:and here is the proof for every even number (1)

Anonymous Coward | about 2 years ago | (#39989805)

an imaginary number squared, obviously

The numbers less than 3 (3, Insightful)

gringer (252588) | about 2 years ago | (#39989997)

If you're talking about integers (which this conjecture refers to), then that's easy:

2 = 5 + -3

0 is trivial:

0 = p + -p for all prime numbers p

1 is also fairly easy:

1 = 3 + -2

And just to complete this, here's 3:

3 = 5 + -2

[multiplication by a unit, in this case -1, does not change the "primeness" of a number]

Re:The numbers less than 3 (2)

gringer (252588) | about 2 years ago | (#39990151)

of course, Goldbach was a bit before Ring theory [wikipedia.org], so may not have been referring strictly to "todays" integers, or prime elements in the set of integers (i.e. including negative numbers).

Re:The numbers less than 3 (3, Informative)

mark-t (151149) | about 2 years ago | (#39990297)

fwiw, it's my understanding that negative numbers are not considered primes, since allowing primes to be negative would allow composite numbers to have non-unique prime factorization.

Re:The numbers less than 3 (1)

sexconker (1179573) | about 2 years ago | (#39990455)

Correct.
Negative numbers are not prime.
1 is not prime (though it never fucking matters since it's always the trivial/base case when you're doing anything useful).

Dis-proof of Goldbach as stated? (1)

Roger W Moore (538166) | about 2 years ago | (#39991177)

That was my understanding too - I'm a physicist not a mathematician though. However the article states that the Goldbach conjecture is:

every integer can be expressed as the sum of two primes

but this seems trivially easy to disprove. There is only one even prime, 2, so if I take an odd integer I have to construct it from the sum of an even and an odd number hence if N-2 is not a prime number Goldbach (as stated) cannot be correct. Now consider '11': since 9 is not a prime number and '2' is the only even prime this cannot hold true for all integers, only even integers which are constructed from the sum of two odd numbers.

So is the article wrong? Should Goldbach actually be limited to 'every even integer' or does the mistake lie somewhere else?

Re:Dis-proof of Goldbach as stated? (5, Informative)

rpresser (610529) | about 2 years ago | (#39991305)

http://mathworld.wolfram.com/GoldbachConjecture.html [wolfram.com]

Goldbach's original conjecture (sometimes called the "ternary" Goldbach conjecture), written in a June 7, 1742 letter to Euler, states "at least it seems that every number that is greater than 2 is the sum of three primes" (Goldbach 1742; Dickson 2005, p. 421). Note that here Goldbach considered the number 1 to be a prime, a convention that is no longer followed. As re-expressed by Euler, an equivalent form of this conjecture (called the "strong" or "binary" Goldbach conjecture) asserts that all positive even integers >=4 can be expressed as the sum of two primes.

Slashdot Summaries Again (5, Informative)

TaoPhoenix (980487) | about 2 years ago | (#39992063)

Dammit, Slashdot you have some of the best commenters here but you're wasting our time making us get about 30 comments in before someone posts the correction to the flawed summaries.

From what I can see in a quick glance, the summary is at least partially wrong. The "regular" Goldbach conjecture seems to apply to every *even* integer greater than 2. So your odd number question disappears into another heading, which is apparently called variously the odd-number or three-primes version of the Goldbach.
http://primes.utm.edu/glossary/page.php?sort=goldbachconjecture [utm.edu]
http://primes.utm.edu/glossary/xpage/OddGoldbachConjecture.html [utm.edu]

(Rant)

So for a community that is expert on Forks, why can't we just Fork Slashdot? *We* are the "value". The only value they offer is the "summaries" and *every single one is wrong*. We lost our leader anyway, and we've all seen what the successors are up to, and Slashcode is sorta/mostly open source right? (Dunno if they bolted on something.)

So why can't we Fork Slashdot? Are we so exhausted and burnt out from the days when fighting IE6 and Vista mattered, that we just don't care anymore? Oh and by the way, every new user would start at the *bottom* of the thread so those new breeds of shills with names like SunriseVista and BoldBraveBalmer don't hijack the top real estate of the conversation. P.S. Sorry, AC's, the top 10 memes of 2003 Slashdot have to go to now. Basically no other forum on the entire net has the First Post thing, and while I get the low level "test against censorship thing", we need a *user option* to flip the entire first post thread and any matching titles to the *bottom* of the post set. Then the *second thread in* which tries to deal with the article can do some work.

(/Rant)

Re:The numbers less than 3 (0)

Anonymous Coward | about 2 years ago | (#39991897)

1 is not always the trivial/base case! 0 often is.

Re:The numbers less than 3 (0)

Anonymous Coward | about 2 years ago | (#39992029)

Correct.
Negative numbers are not prime.
1 is not prime (though it never fucking matters since it's always the trivial/base case when you're doing anything useful).

If you are doing something where it for some reason is important to note that 1 isn't a prime my first guess would be that the hardest part will be finding an actual real world application for it.

Re:The numbers less than 3 (1)

Pseudonym (62607) | about 2 years ago | (#39991175)

As others have noted, it depends. It's sometimes convenient to consider -1 as being prime, for example, because it allows you to extend the notion of squarefree numbers to negative integers.

Re:and here is the proof for every even number (2)

Sussurros (2457406) | about 2 years ago | (#39990157)

Given that 1 is divisible only by itself and 1 I hearby nominate it to be an honorary prime.

Re:and here is the proof for every even number (1)

JoshuaZ (1134087) | about 2 years ago | (#39990433)

Minor note: While that's true, in Goldbach's day it was actually common to include 1 as a prime. So the original conjectures allowed 1.

Re:and here is the proof for every even number (1)

Anonymous Coward | about 2 years ago | (#39991171)

1 was a prime when the Goldbach conjecture was first stated. Read some older mathematical papers. Mathematicians now tend to exclude it from the list of primes because that way is more useful, but the definition is not absolute.

Re:and here is the proof for every even number (1)

tixxit (1107127) | about 2 years ago | (#39990361)

Unfortunately (for Ramare), this was already proven before Terrence Tao's result:

We prove that every odd number N greater than 1 can be expressed as the sum of at most ve primes, improving the result of Ramare that every even natural number can be expressed as the sum of at most six primes.

Re:and here is the proof for every even number (1)

Arancaytar (966377) | about 2 years ago | (#39992213)

No need to make up fake primes. With 3, every even number n>4 is 3 plus an odd number n>1; 4 and 2 are, obviously, trivial.

Terry Tao (4, Interesting)

bgeezus (1252178) | about 2 years ago | (#39989653)

Terry Tao always amazes me with the scope of his knowledge. Contributions in mathematical areas as diverse as random matrix theory, harmonic analysis, and number theory. I look forward to what comes next!

Re:Terry Tao (2)

EuclideanSilence (1968630) | about 2 years ago | (#39989719)

I've also heard Tao say in lecture that he doesn't even like using computer assistance when he's working out theory. I found some of his lectures to be great for getting the scope of ideas, but unless you really know the subject of number theory he can be hard to follow.

Re:Terry Tao (0)

Anonymous Coward | about 2 years ago | (#39990509)

Yeah, nice work for a boy from Adelaide, South Australia. I went to high school with his younger brother Trevor, who is a maths/chess/music virtuoso in his own right. Congratulations Terry!

Re:Terry Tao (1)

Anonymous Coward | about 2 years ago | (#39991169)

you think those areas are diverse? All I see is analysis, analysis, and analytic number theory.

Every Integer? (0)

Anonymous Coward | about 2 years ago | (#39989669)

Really.

Re:Every Integer? (-1)

EuclideanSilence (1968630) | about 2 years ago | (#39989709)

Parent probably isn't a troll. The summary should say every even integer greater than 2. For example, 9 can't be expressed as the sum of 2 primes.

Re:Every Integer? (1)

Anonymous Coward | about 2 years ago | (#39989727)

Dude... what's 2 + 7?

Re:Every Integer? (-1)

Anonymous Coward | about 2 years ago | (#39989735)

Yes, yes it can. 7+2. 27 can't, though.

Re:Every Integer? (0)

Anonymous Coward | about 2 years ago | (#39990437)

Yes, yes it can. 7+2. 27 can't, though.

We can do it with 3 primes tho ;) 7+7+13

Re:Every Integer? (5, Insightful)

jejones (115979) | about 2 years ago | (#39989755)

7 + 2 = 9

Re:Every Integer? (5, Funny)

Old Wolf (56093) | about 2 years ago | (#39989917)

Wow, what has slashdot come to when posts are getting modded up for posting basic arithmetic :)

Re:Every Integer? (0)

konohitowa (220547) | about 2 years ago | (#39991975)

The story passed by the editor(s) (granted, it's timothy) with a summary that was quite excited to point out that computation isn't the same as proof. Perhaps it's time to relegate this site to r/slashdot and replace it with a nice stream of cats and ponies.

Re:Every Integer? (0)

Anonymous Coward | about 2 years ago | (#39990317)

2 is prime. 7 is prime. So, no, you haven't found a counter-example. Look again what the conjecture claims.

Re:Every Integer? (3, Funny)

Burpmaster (598437) | about 2 years ago | (#39990483)

7 + 2 = 9

Damn, that's the most intelligent post I've seen on Slashdot all day, and I mis-clicked and chose 'redundant' when moderating...

Re:Every Integer? (0)

Anonymous Coward | about 2 years ago | (#39989779)

For example, 9 can't be expressed as the sum of 2 primes.

7 + 2. I've done the impossible!

Re:Every Integer? (0)

Anonymous Coward | about 2 years ago | (#39989851)

Parent probably isn't a troll. The summary should say every even integer greater than 2. For example, 9 can't be expressed as the sum of 2 primes.

What about 7+2?

Re:Every Integer? (1)

EuclideanSilence (1968630) | about 2 years ago | (#39990039)

Sorry I mean 11 can't be, because one of the terms would have to be 9 which isn't prime.

Re:Every Integer? (0)

Anonymous Coward | about 2 years ago | (#39990073)

7 + 2 + 2

Re:Every Integer? (3, Funny)

sexconker (1179573) | about 2 years ago | (#39990487)

7 + 2 + 2

Ah, Mexican Math, we meet again. That's not two primes. That's three primes, two of which are 2.

Re:Every Integer? (0)

Anonymous Coward | about 2 years ago | (#39991685)

13 - 2

Re:Every Integer? (3, Informative)

jd (1658) | about 2 years ago | (#39991491)

Looked the conjecture up on Wikipedia. It's actually a little more specific still - every even number is a Goldbach Number, where a Goldbach number is a number that can be written as the sum of two odd primes.

That means that every odd number can always be written as the sum of three primes or less. Numbers like 9 are the sum of two primes but are NOT Goldbach numbers since one of the primes is 2 and the requirement is that both primes be odd.

Errors in this post are due to Wikipedia, blame them if there are any.

It's every *even* number (5, Informative)

MrKevvy (85565) | about 2 years ago | (#39989705)

"...every integer can be expressed as the sum of two primes."

It should be every even integer. Note TFA has sums for 52, 54, 56, 58 and 60.

Re:It's every *even* number (1)

Derekloffin (741455) | about 2 years ago | (#39989717)

Ah, yes, I knew something was missing or I disproved it inside of a couple seconds :P. Yeah, it would have to be even as all primes are odd save 2, and the sum of any 2 primes is even, so you'd be forced to use 2 as one of those primes all the time for all even number and that very quickly breaks.

Re:It's every *even* number (1)

Derekloffin (741455) | about 2 years ago | (#39989733)

Bah, that should be "For all odd numbers". That will treat me to reed my messages before posting.

Re:It's every *even* number (2, Funny)

Anonymous Coward | about 2 years ago | (#39991041)

That will treat me to reed my messages before posting.

Alas, it did not treat you.

Re:It's every *even* number (2)

cupantae (1304123) | about 2 years ago | (#39989905)

Another way to say it, which just occurred to me now, is:
"Every natural number is halfway between two primes."

Re:It's every *even* number (5, Informative)

FrootLoops (1817694) | about 2 years ago | (#39990245)

Indeed, it should actually say, "every even integer greater than 2 can be expressed as the sum of two primes". 2 is degenerate. For the purposes of the conjecture calling 0 prime (this is non-standard) gets rid of that little wrinkle, though the cost of a more involved statement of the fundamental theorem of arithmetic is not worth it (which is incidentally a good reason why 1 isn't prime).

For anyone interested, an actual theorem that's similar to the Goldbach conjecture is Lagrange's four-square theorem [wikipedia.org]. It states that any non-negative whole number is the sum of the squares of four whole numbers. There are numerous proofs, though I wouldn't recommend trying to find one yourself if you don't have a background in algebra or number theory.

Re:It's every *even* number (0)

Anonymous Coward | about 2 years ago | (#39992255)

Sorta obviously, since the only even prime is 2, it means any ODD number resulting from the sum of two primes would have to be prime+2, which is defintely not true.

Should be every even integer (-1)

Anonymous Coward | about 2 years ago | (#39989747)

Hi,
      The conjecture actually states every even integer can be expressed as the sum of 2 primes. It would be hard to make odd integers work out as most primes are odd and the sum of two odd numbers is even. Anyway, 11 cannot be expressed as the sum of two primes so if the conjecture included odd numbers I suspect it would have been disproved a long time ago.

Re:Should be every even integer (1)

FrootLoops (1817694) | about 2 years ago | (#39990165)

I suspect it would have been disproved a long time ago.

Long, long ago. If the conjecture really was as written, it would require every odd integer to be prime, which is patently ridiculous (eg. 3^k would be prime). Suppose the number n is odd. n+2 is odd too and, if the misstated conjecture were true, n+2 = p+q for primes p and q. Since all primes except 2 are odd, to get the sum to be odd, one of p or q must be 2, say q=2. But then n+2 = p+2, so n = q, and n is prime since q was.

Re:Should be every even integer (1)

shentino (1139071) | about 2 years ago | (#39992201)

All primes except 2 are odd.

Otherwise 2 itself would be a factor which would make it not so prime anymore.

Exhaustive search... (2, Interesting)

Anonymous Coward | about 2 years ago | (#39989767)

Notice that no computers where involved in the proof — this is classical mathematical proof involving logical deductions rather than exhaustive search.

Exhaustive search for a result that holds for every integer? Good luck with that one.

Re:Exhaustive search... (4, Funny)

sexconker (1179573) | about 2 years ago | (#39990505)

Notice that no computers where involved in the proof — this is classical mathematical proof involving logical deductions rather than exhaustive search.

Exhaustive search for a result that holds for every integer? Good luck with that one.

Everyone knows integers only go from 0 to 4294967295!

Re:Exhaustive search... (1)

thoughtspace (1444717) | about 2 years ago | (#39991193)

You may write that in jest. You can use a theorem if it is not proven for the general case - only need it to 32 bits.

Re:Exhaustive search... (0)

Anonymous Coward | about 2 years ago | (#39992145)

Damn, you are right. Are there any practically useful theorems that would be valid for 32-bit integers only?

Re:Exhaustive search... (2, Funny)

Anonymous Coward | about 2 years ago | (#39991469)

They recently discovered a few more: 4294967296 through 18446744073709551615. Just in time too--we were starting to run out in some computations. Unfortunately it'll take a bit longer to verify the conjecture for these newly discovered specimens. At least there's only a finite number of primes...

Re:Exhaustive search... (1)

Kjella (173770) | about 2 years ago | (#39991911)

Everyone knows integers only go from 0 to 4294967295!

Hey, on my computer INT_MAX is only 2137483647. Damn store must have cheated me, giving me crippled integers. I'm going to down there right now and demand one where integers go all the way up to 0x11..11..11..11.

This way more useful than electricity (0)

Anonymous Coward | about 2 years ago | (#39989769)

Think of all of the applications this can be used for...

Ok ... while I'm having difficulty at the moment coming up with one that's probably because of the electrodes they attached to my head to make me happy.

Exhaustive (2)

lurker1997 (2005954) | about 2 years ago | (#39989775)

every odd number greater than 1 is the sum of at most five primes

Notice that no computers where involved in the proof — this is classical mathematical proof involving logical deductions rather than exhaustive search."

That would have been a pretty long "exhaustive search".

Re:Exhaustive (0)

Beetle B. (516615) | about 2 years ago | (#39991239)

I'm not sure if you meant it as a joke, but exhaustive search can still be used for problems like these. An example would be to categorize all the integers into, say, 10000 groups and then write a program to prove each of those groups...

and yet another plagiarized slashdot summary (0)

bcrowell (177657) | about 2 years ago | (#39989777)

Why can't submitters get it through their head that when you quote someone, you need to put the quote in quotation marks and give credit to the source?

Re:and yet another plagiarized slashdot summary (1)

cupantae (1304123) | about 2 years ago | (#39989937)

What?
The entire summary is quoted (error and all) from the only linked article. How is that not giving credit?

Re:and yet another plagiarized slashdot summary (2)

bcrowell (177657) | about 2 years ago | (#39991493)

What?
The entire summary is quoted (error and all) from the only linked article. How is that not giving credit?

It's not giving credit because it says: 'mikejuk writes "...",' where ... is a collection of sentences grabbed from various places in the article, and none of those sentences were written by mikejuk.

Similarly --

bcrowell writes:

O Romeo, Romeo! wherefore art thou Romeo? Deny thy father and refuse thy name. Parting is such sweet sorrow, that I shall say good night till it be morrow. Courage, man; the hurt cannot be much.

Re:and yet another plagiarized slashdot summary (0, Flamebait)

Anonymous Coward | about 2 years ago | (#39991113)

"The parent is a faggot." --AC

Could be used for many practical things. (-1)

Anonymous Coward | about 2 years ago | (#39989801)

Every equation has a purpose to something.
This is just another source of numbers, which could be expressed in various different bases, different decimal positions for various uses or a whole host of things.
And that is just alone. They could likely be combined as part of a larger algorithm for procedural generation purposes. (note, must try that)

There never used to be a practical use for a series of powers just as the numbers themselves.
But then people figured out these could be added together in to one value and never overlap. (at least for 2, I haven't actually checked for other sets)
Found use all over from input / output codes to very basic (and limited) compression.

Someone will find a use for it. Whether it is now or 100 years from now.

This is part of a very long trend (5, Informative)

JoshuaZ (1134087) | about 2 years ago | (#39989803)

Work on this problem has been ongoing for about a hundred years now. First, Schnirelmann proved that there was some k such that every even integer could be expressed as a sum of at most k primes. The value for k had then been reduced over time. Vinogradov's proved that the Odd Golbach Conjecture (that every odd integer greater than 7 is the sum of three primes) was true for sufficiently large n. How large sufficiently large is has been slowly reduced. Later in the 1970s, Chen proved that every sufficiently large even integer is the sum of a number that is prime and another number that is either prime or a product of two primes. At this point, Chen's result is the strongest result known.

In general, there are two general methods of attack on this problem, one which uses Schinerlmann's method and variants thereof, and the other which uses sieve theoretic approaches with the Hardy-Littlewood circle method http://en.wikipedia.org/wiki/Hardy-Littlewood_circle_method [wikipedia.org] (Chen used a version of this for his result and Tao's work uses a similar approach). Unfortunately, there's not much work on actually connecting the two methods. There's an excellent piece of Tao at his blog where he discusses his work on the problem and is understandable without much background. http://terrytao.wordpress.com/2012/02/01/every-odd-integer-larger-than-1-is-the-sum-of-at-most-five-primes/ [wordpress.com]. Note that TFA is a little out of date since he announced this result with a preprint a few months ago, and it is only that now the result is being published.

It does not seem that this result really does put us much closer to proving the full Golbach Conjecture. At most this could be used to prove some version of the odd Goldbach Conjecture. The methods used would have a large amount of trouble dropping from 5 to 3. There's some bit of leeway, and if anyone is going to do it, it is going to to be Tao, but right now, I'm not optimistic.

Re:This is part of a very long trend (2)

gnasher719 (869701) | about 2 years ago | (#39990021)

It is astonishing how weak the result is, and how hard it is to prove.

The "ordinary" Goldbach conjecture is: Every even number N >= 4 is the sum of two primes. For example, 100 = 3 + 97 = 11 + 89 = 17 + 83 = 29 + 71 = 41 + 59 = 47 + 53, so we see that sum numbers can be written as the sum of two primes in many different ways. We call a number that is the sum of two primes a "Goldbach number", then the conjecture says that every even integer N >= 4 is a Goldbach number.

The "weak" Goldbach conjecture is: Every odd number N >= 7 is the sum of three primes. At least one of these primes nust be odd, so we can express the weak conjecture as: For every odd N >= 7, there is an odd prime p = N - 4 such that N - p is a Goldbach number. Since there are many odd primes p = N - 4, only one out of a huge list of numbers of the form N - p need to be a Goldbach number.

Re:This is part of a very long trend (4, Informative)

FrootLoops (1817694) | about 2 years ago | (#39990025)

and if anyone is going to do it, it is going to to be Tao, but right now, I'm not optimistic.

Agreed. I imagine Terry Tao isn't well-known outside of mathematics, but for those who don't know, he's certainly one of the most famous and skilled living mathematicians. He's originally Australian and is currently at UCLA. His list of high profile awards is ridiculously long, but aside from top-notch research, he's also an excellent teacher. His blog is mainly pitched at math grad students and higher, but some of it is very accessible. There's of course more biographical details at his Wikipedia page [wikipedia.org]. The statement of the Green-Tao theorem [wikipedia.org] is also accessible and interesting.

I totally have a researcher-crush on him, or more specifically his math skills.

Re:This is part of a very long trend (0)

Anonymous Coward | about 2 years ago | (#39991337)

I totally have a researcher-crush on him, or more specifically his math skills.

It's OK of if you have real crush on him. Like a man-to-man crush. We understand.

And the two primes are? (1)

Anonymous Coward | about 2 years ago | (#39989807)

Optimus and Giedi?

Re:And the two primes are? (1)

sexconker (1179573) | about 2 years ago | (#39990515)

Optimus and Giedi?

Optimus and Rodimus. Deal with it.

Rodimus? I can dig it. (0)

Anonymous Coward | about 2 years ago | (#39991025)

Still better than Sentinel Prime.

Re:And the two primes are? (0)

Anonymous Coward | about 2 years ago | (#39992245)

Never! The Prime Must Flow!

Corrections (-1)

Anonymous Coward | about 2 years ago | (#39989911)

First, the title is in Engrish, not English. Shouldn't it be "Closer to solution?" rather than "Closer to Solved?"

Second, the Goldbach's conjecture states that every even integer can be expressed as a sum of two primes.

ZERO? (0)

Anonymous Coward | about 2 years ago | (#39989921)

What about the integer of ZERO? What two primes represent that, noting that primes are >1 (ie. not ZERO and not negative). I'm no mathematician and this occurred to me in about 5 seconds, so I'm sure it's flawed thought from me.

AC

Re:ZERO? (3, Informative)

Spodi (2259976) | about 2 years ago | (#39989987)

Goldbach's conjecture: "Every even integer greater than 2 can be expressed as the sum of two primes." (source [wikipedia.org])

Simple disproof (-1)

Anonymous Coward | about 2 years ago | (#39989929)

If all you have to do is find a number that isn't the sum of 2 primes, here it is: eleventy bajillion. Thanks, I'll take my prize money now.

What about negative numbers (1)

rossdee (243626) | about 2 years ago | (#39990103)

When I went to school, integers included negative numbers. Of course that may have changed.

Re:What about negative numbers (2)

FrootLoops (1817694) | about 2 years ago | (#39990319)

Integers do still include negatives. Actually, the "prime numbers" used in abstract algebra also include negatives, so for instance -5 is prime. This genuinely useful convention results in the following statement of the fundamental theorem of algebra: "Every non-zero integer can be factored as the product of prime numbers. The factorization is unique up to order and signs." (Example: -12 = 2*(-2)*3 = (-2)*(-3)*(-2).) This directly generalizes to a corresponding statement in so-called unique factorization domains [wikipedia.org], of which the integers are a particular case. Still, fiddling with negatives might confuse students, and people don't often factor negative numbers anyway, so the definition is often restricted to positive numbers, even though it's artificial.

Re:What about negative numbers (0)

Anonymous Coward | about 2 years ago | (#39990519)

Prime number [wikipedia.org] are positive [mathsisfun.com]. There is no way to get a negative number from the sum of positive numbers.

Is this progress? (4, Insightful)

spaceyhackerlady (462530) | about 2 years ago | (#39990203)

Sorry, but I can't accept this being progress toward a proof.

Consider Fermat's Last Theorem. Proving it for any particular exponent is doable. Mathematicians had proved it for various sets of exponents (Sophie Germain, Wieferich, etc.). But the proof for all exponents was based on completely different mathematics (Elliptic curves/modular forms, Taniyama-Shimura, Wiles) and didn't look like anything that had come before.

...laura

Re:Is this progress? (4, Interesting)

JoshuaZ (1134087) | about 2 years ago | (#39990309)

That's not completely true. Wiles's proof only proves it for an exponent that is a prime p>=7. So one needs the classical results of n=3,4,5,7 also. This is to some extent a minor criticism. Your essential point is correct that sometimes a proof of a theorem comes out of a completely different direction. But, very often, it does come from a straightforward way of refining the same techniques more and more. For example, Catalan's Conjecture http://en.wikipedia.org/wiki/Catalan's_conjecture [wikipedia.org] (the claim that that the only consecutive positive perfect powers are 8 and 9) was proven by what in many ways amounted to slow and steady progress.

Re:Is this progress? (2)

phantomfive (622387) | about 2 years ago | (#39990475)

Even in the worst case, and the ultimate proof doesn't look anything like this, he's still eliminated one path that someone else might try (or forged along that path to show what it could do).

Exhaustive search? (0)

Anonymous Coward | about 2 years ago | (#39990415)

"this is classical mathematical proof involving logical deductions rather than exhaustive search"

I would be very impressed with an exhaustive search covering every even integer.

Not Quite! (1)

Anonymous Coward | about 2 years ago | (#39990733)

The proof is missing a crucial part at the end...

  Q . E . D .

There!

Slashdot editors (0)

Anonymous Coward | about 2 years ago | (#39990855)

"no computers where involved in the proof"
should be
"no computers were involved in the proof"

There WERE computers involved, indirectly. (1)

kevinatilusa (620125) | about 2 years ago | (#39990895)

From the abstract of Tao's paper: Our argument relies on some previous numerical work, namely the verification of Richstein of the even Goldbach conjecture up to $4 \times 10^{14}$, and the verification of van de Lune and (independently) of Wedeniwski of the Riemann hypothesis up to height $3.29 \times 10^9$.

Richstein's work (available at http://www.ams.org/journals/mcom/2001-70-236/S0025-5718-00-01290-4/S0025-5718-00-01290-4.pdf [ams.org] ) definitely involves a computer, and I assume the Riemann hypothesis verification does as well.

Computers were used! (but/and this is still great) (0)

Anonymous Coward | about 2 years ago | (#39991145)

Computers were used in the proof! An exhaustive search was done over some small cases; and there have been previous proofs for the integers that are "large enough". Tao managed to make large enough small enough so that it falls into the rang covered by the computerized exhaustive search,

Goldbach's Conjecture was already solved (0)

Anonymous Coward | about 2 years ago | (#39991185)

Watch the movie "Fermat's Room", which is a great movie.

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