Slashdot Mirror
Grok Goldbach, Grab Gold
Caseman writes, "Are you a closet mathematician who wants to come out? British publisher Tony Faber is offering a cool million bucks to the first would-be math head to prove the infamous Goldbach conjecture. Yeah, the one about every even number being the sum of two primes. Impossible, you say? Remember Fermat's Last Theorem? It stood unproven for 350 years until a recent effort yielded a proof. Read The London Times article about the challenge. "
It says on like the 4th line that the numbers must be greater than 4
1 is a "unit" in the ring of integers. It is not prime. The following definitions/discussion should clarify why this is so: 1) The multiplicative identity of a ring K, e, is an element such that a*e=a for all elements a of K. In the case of the integers, 1 is the multiplicative identity. (if you don't know what a ring is, don't worry) 2) Element u of K is a unit if there exists some other element v in K so that u*v = e In the case of the integers, the only units are 1 and -1. 3) Assuming the object we're dealing with allows us to (uniquely) factor things into primes (like the integers, for instance), we can say a prime is a number that is not a unit and cannot be expressed as a product of other non-units. This notion of a prime useful because it's general: of course 5=1*(-1)*5*(-1), but it's still prime.
Okay, your first sentence is correct, but only because of the qualifier "an integer greater than one" (negative integers are obviously not greater than one). Your second sentence is incorrect, and does not follow from the first sentence. Formally, -2, -3, -5, ... are considered prime integers. But they correspond precisely to the positive primes, multiplied by a unit (-1), and therefore aren't very interesting. So when talking about primes, we usually only consider those which are positive.
The definition of a prime that I've seen most commonly is as follows (and notice that, unlike yours, it does not require any notion of one element being greater than another, and works equally well in rings other than the ring of integers - for example, the ring of polynomials).
Let p = a*b, where p is not a unit
Then p is prime if and only if p divides a or p divides b.
Burley, you know lots about fortran but your wrong, the operators cant be evaluated in the wrong order, they use the return value of the next operator in.
Hmm. Nifty. Thanks.
> Yep, 5 is sum of two primes. Yep, 7 is the sum of two primes."
:).
Now I know why this article was marked as "insightful"
I am thoroughly disgusted that due to the large monetary incentive hundred thousands of brains will be occupied with proving this theorem while this brainpower could be spent far more productively - e.g. reading /. Any TV quiz show has a higher benefit for mankind - at least some knowledge can be derived by someone.
Actually, I have discovered a truly remarkable proof of this theorem, which this Slashdot comment is too small to contain.
A graph of primes can't be a pattern, otherwise you could extrapolate the pattern and predict prime numbers.
He didn't claim zero was odd or prime. What he said was that the probability of an even number not being the sum of two odd numbers is zero; n being an even number: (n - 1) + 1 = n.
It's valid up to the statement:
(x-5)(x+3)=(x-5)(x+2),
but you can't divide out the (x-5) term, since (x-5) = 0. You may as well simplify the proof to:
0 = 0
->0*0 = 0*1
->0 = 1
How do you know that n even exists except by assertion? The only thing that fits your equation is 2.
> now, a+1 and b+1 are even (because primes are
> odd, excluding 2 which is why n is large
Why excluding 2? Why does 2 have to be excluded if n is large?
a=a++ is not undefined. The a++ has to be evaluated first because the a is assigned the return value of the a++. Things like a[i]=i++ are undefined because you can evaluate either i first, but you can't do that with a=a++.
Most Slashdotters are "hands on" types, facing cranky real world network problems and configuration files, not abstract thinkers except in a pragmatic "OK, how the hell do I get myself out of this one" mode. Think Soviet astronaut banging on Mir's pipes, not Albert Einstein trying to think of a way out of using the Cosmological Constant. Doesn't make us stupid, just a little less elegant in the mathematical sense. Excuse me while I wipe my hands on the tablecloth...
"...and he said it in the context of trying to differentiate himself from Bradley and other challengers."
Which, in your own immunocompromised mind, makes it true?
n|t
I don't know. I live in the UK and I've always found Regis pretty educational.
Am I the only one?
Hmm. I would hope that someone so smart would be able to choose his words a little more carefully; wouldn't you?
Al Gore took the initiative in creating the Internet in precisely the same way I took the initiative in inventing the light bulb, when I changed the bulb over my kitchen sink yesterday. No amount of spin-doctoring on your part (or Brill's) can change the fact that this was a hideously stupid thing for anyone to say.
_Why_ is the prime factorization theorem so fundamental to arithmetic? I remember being taught to break down numbers into their prime factors as early as 3rd grade, but nobody ever explained why this is useful to anyone outside of number theory. I've been a professional software engineer for over a decade now, doing some pretty diverse work, and have never needed to leverage this knowledge. Is there a simple statement of "here's why this is so insanely important" that I've managed to miss?
.sig supposed to do? I don't see any interesting or elegant relationships between the lvalue and rvalue sides of your -= assignment.
Second, what's the piece of code in your
Actually the proof of FLT used induction too.
haha. Yeah, if you want to know about 1950's movie stars or specific information focusing on who invented what when it's all really irrelevant.
Perhaps this could be solved geometrically?
goin out to my homeys. Peace out.
Actually, PGP style public key encryption usually relies on the hardness of the Dffie-Hellman Problem (at least for the non-symmetric encryption stuff). The Diffie Hellman problem involves finding discrete logarithms in prime order *multiplicative* groups, and hence has no obvious (at least not that I can see) connections with proving Goldbach. Even then, proving Goldbach would not immediately render DHP insecure immediately. You still need to find a way to reduce the computational problem of Discrete Logs. Proving Goldbach's certainly does not give this to you for free. It may, but it doesn't have to. All that being said, you may actually be referring to RSA encryption, which is based on modular exponentiation and the difficulty of prime factorization (in order to find the decryption exponent). If you manage to reduce that problem, then you have essentially solved prime factorization for free too. And the sheer amount of work that has gone on to try and find such a method (ie: to make factorization of integers efficient) is astronomical. There is a very large amount of evidence to support the conjecture that prime factorization is in fact a hard problem.
Unity is usually excluded in the definition of a prime so that each number will have a unique prime factorization -- if 1 was a prime, the factorization wouldn't be unique. Of course the fundamental theorem of arithmetic could be reformulated to say "after removing any 1's the factorization is unique" but doing it the usual way keeps the proofs in elemental number theory simpler.
6=3+3
Thank you for saying this! I almost thought that nobody would see that this is more damaging to the cooperative nature of math research than anything else. There is already a problem with finding every possible prime number, since we have still no formula to do just that. There are formulas that can give you an infinite sequence of primes, but none that give you every one of them IIRC. I think that solving that problem would go a great way towards a proof of the GC.
around 25
Clearly, fermat's last theorem has been proved...
The joke is that Fermat wrote his last theroem in the margin of a book. After describing it he writes the he has found a great proof for it, however the margin of this book is too small to write it down. The guy died and it took 350 years to come up with the proof. Is the joke still funny? guess it depends on how many times you've seen it.
Can you prove that? ;) That no pattern has been found doesn't mean that one doesn't exist. Many patterns have been found that categories of primes adhere to, and it's perfectly conceivable that a pattern can be found...
The flaw? Your inability to read the terms of the contest and consequently, the conjecture.
Thankyouverymuch, I'll be here all week. . .
AC
Yes, He certainly did write that, but people who have looked into it seem to think that he didn't really have a proof. I don't have a link, but it seems the way it goes is:
-Ok, Here's the theorem. I thought of this simple proof, I write it on the margin.
-Ooops, the proof is not correct (There was a simple non-proof which tricks people at first). I wrote something on the margin, but it's not like anybody is going to read that.
He used a simpler version of the theorem as a challenge to other people. If he really had a proof of the full theorem, he would've use that full theorem instead. Therefore, the conclusion is that he was only able to correctly prove simpler forms of it.
Still, that is not conclusive, and the challenge remained.
AC
From the description in the post, I can DISPROVE it.
1 is by definition not a prime. 2 is even. 2 is only the sum of 1 and 1 (assuming negative numbers don't count, obviously). Hence, the theory is false.
Where do I collect my money? And is it close to the 'learn to post to Slashdot' center?
--
AC
What is closet mathematics? I've never heard of that field.
At least now I have something to do while I'm awake.
**RANT MODE**
Argghhh - moderate this down to -100 or whatever but there is no such paper as "The London Times". There is The Times and The Sunday Times. Both of these papers are National papers in Britain (you know - like USA Today). Just 'cos it's the New York Times and The LA Times doesn't mean Americans have to impose their idea of how things work on the UK. Oh BTW - it's The Sun, The Mirror, The Express, The Telegraph and even The Evening Standard - none of them have the word "London" in the title.
If you want Americans not to realise please say this is an article in The Times (Britain) or something similar.
I already posted my proof in a reply to one of Jon Katz's articles. But it got moderated down to -1, flamebait, so nobody saw it.
- Take all the prime numbers that you know.
- Multiply them together.
- Add 1.
Now think about the resulting number. Is it prime? Then it's a new largest known prime. Is it composite? Well, then none of the primes you know is a divisor (because they all leave a remainder of 1). So its prime factors must all be new large unknown primes.Either way, this shows that the set of primes is not finite and that there is no "largest prime".
And that's one way to prove a statement about an infinite set of numbers. Proof due to Euclid, a couple of thousand years ago.
So my questions are: how big does d.net have to get, and how big do the nodes have to get, in order for it to process mathematics as good as a single human being? As good as a professional mathematician? As good as Gauss and Ramunajan?
One characteristic of mathematics is that the processing entity doesn't have to respond in real time. It can tolerate the high latencies of embarrassingly parallel computation.
Wow, now I want to stare at the wall and think about meta-mathematics and mathematical logic. I bet that the right software is more important than a trivial parallel speedup of 1e6 or whatever d.net has.
Unfortunately, as of a few years ago, we in the UK can get away with picking just six random numbers between 1 and 49. THere's also a bonus ball: the prize for getting the main six right is higher than getting five main + bonus right.
There have been enough draws thus far that we would expect someone to have picked all seven correct numbers: anyone know if this has actually happened?
ac.uk
Hmmm ... the sum of two prime numbers and generalize proofs about them ... has anyone noticed the similarities between this and public key encryption theory? The person that proves this conjecture is probably 99% of the way towards breaking PGP style encryption. Good luck, and watch your back ... and if you do prove this, I bet you could sell it for a *lot* more than $1 million dollars. Back in the 60's when Austin Power's nemisis was first on the scene, this may have been a lot of money ...
this sounds like a job for...LISP RECURSION MAN! but thats not me, it's someone else, so they'll have to recurse, then they'll recurse the day they started to try to prove this conjecture :)
"There is no spoon"-Neo, The Matrix
"SPOOOOOOOOON!"-The Tick, The Tick
That's a most unflattering portrait of you. That is your *face*, isn't it?
I see even classic Slashdot is now pretty much unusable on dial up anymore.
Doesn't somebody (or five people, or six) make this exact same joke any time proofs are mentioned?
Maybe I've been hallucinating, but I don't think so. Really, is anyone else still amused by this? I'm not.
So you can just take your proof and suck it.
But really, what keeps n+2=q+r from being true (where q+r are prime), and n+4=s+t, and n+6=u+v, etc.?
> Lots of work has been done that could be useful
...
;-)
> to the GC, just the same was as done for FLT.
> The winnings will go not to the person who does
> the most work towards it, but whoever supplies
> the piece of the puzzle that happens to be last.
But in mathematics it can be just as difficult (if not more so) to realise that there is some connection between two hitherto unrelated pieces of mathematics as it is to think up something original. Everybody recognises Wiles(*) as the guy who proved FLT, and everybody recognises that WIles' work couldn't have been done without earlier work of Ribet, Frey, Serre,
((*) Actually, the proof of FLT is more correctly attributed to Wiles and Taylor. Wiles uses a property of Hecke algebras that is proved in a companion paper to Wiles' `Modular forms, elliptic curves and Fermat's Last Theorem' that is contained in a supplementary paper by Taylor and WIles.)
> This might encourage work on the GC, but it
> also might discourage publication of such work,
> because the mathematicians haven't quite
> finished the proof.
Unlikely. GC is sufficiently important to be independent of some monetary prize being given away as a marketing gimmick. It will give amateur mathematicians and cranks something to do for the next few months though...
As for discouraging publication, this certainly wouldn't happen in the UK. Our universities are publicly funded and every 4 years have to go through the `Research Assessment Exercise' (RAE). Essentially, each university and each department within each university is graded according to how good the research output from the academic staff is. The grade each department gets determines how much money the government is willing to give in grant income. Doing well in the RAE is far more important (in terms of prestige for the university and promotion prospects for the staff involved) than deliberately witholding papers from publication because of some prize (no matter how large!). Any substantial progress towards GC would be of value when it comes to the RAE.
Going back to the article, the time-frame seems unrealistic. A paper must be submitted to a journal with 2 years - the only way this is likely to happen is if somebody is almost all the way to proving GC already and has just not announced it yet. But to ask for publication within 4 years is pushing it. GC is sufficiently important to attract a lot of attention and interest and the journal to which it is submitted will want to make sure the proof is correct. Hence the refereeing process is likely to be lengthy. Then there's the fact that most journals have large backlogues meaning that it can sometimes take 3 years for a paper that has been accepted for publication to actually appear. Maybe GC is sufficiently important to be `fast-tracked' though...
The most interesting thing about the article is that the prize is a 6 figure sum but the insurance premium is a 5 figure sum. So, roughly speaking, the insurance company thinks there is a 1 in 10 chance of somebody actually proving GC and getting it in print within 4 years! These insurers sound like they know something the mathematical community doesn't - maybe we should hire them for the next RAE
certainly not rocket science but since we're on the topic I thought folks might enjoy it
#include <stdio.h>
#include <limits.h>
#include <math.h>
static int isPrime(int x)
{
int result = 1;
if(x > 1)
{
int middle = sqrt((double)x);
int i = 0;
for(i = 2; result && i < middle; i ++)
if(x % i == 0)
result = 0;
}
else
result = 0;
return(result);
}
int main(int argc, char **argv)
{
int i = 0;
for(i = 4; i < INT_MAX; i += 2)
{
int j = 0;
int found = 0;
for(j = 2; !found && j < i; j ++)
if(isPrime(j) && isPrime(i - j))
found = 1;
if(found)
printf("%d = %d + %d\n", i, j, i - j);
else
printf("Goldbach Conjecture FALSE for: [%d]\n", i);
}
return(0);
}
there are two kinds of people in this world - those who divide people into two groups and those who don't
For a proof by induction, you express what you're trying to prove as a statement with a parameter n - call it P(n). Then you show that it is true for P(1) (or some other n), and that if P(n) is true (for any n) then it implies that P(n+1) is true.
Then, for example, P(378) must be true because it is implied by P(377), which is implied by P(376), and so on down to P(2), which is implied by P(1), which you have shown is true separately.
If you've proved that P(n) implies P(n+1) for any n, without assuming any special cases such as that n is prime, then it must be true for all n >= 1.
Proving induction: the proof I've seen is a proof by contradiction based on the principle that the natural numbers are well-ordered. This means that any non-empty set of natural numbers must contain a lowest element.
Now suppose you have a statement, P, that P(1) is true, and that P(n) implies P(n+1). To prove that this is sufficient to show that P(n) is true for all n, first assume it isn't - ie there are some values, k, for which P(n) is not true.
Now let S be the set of all of these values, ie S = {k : P(k) is false}. As we have assumed that P(n) is not true for every n, S cannot be empty, so it must have a least element, x, by well-ordering. x cannot be 1 since we have shown P(1) is true separately.
Therefore we have that P(x) is false, but P(x-1) is true since x is the least element of S. But x-1 is a natural number because x is, so by the inductive step P(x-1) must imply P(x). Therefore P(x) must be true, but by definition it is false. This is a contradiction, and as all the logical reasoning since the assumption was sound, the assumption must be wrong - ie P(n) is true for all natural numbers n, so S is empty and x doesn't exist.
AndyThis is picky, but his proof appeared when he was 19, not 20. His proof is a thing of beauty.
--j
I'm a nature photographer.
Some people are suggesting brute-force attacks, but this is laughable to anyone with even a passing interest in mathematics. Automatic Theorem Proving has been suggested here as well, but ATP has a lot of practical problems, especially in a more complicated domain such as mathematics or more specifically, number theory.
Maybe in XEmacs 22?
See Mathworld on the issue.
Similarly, if we assume:
;-)
1. 2 = 3
we can multiply through by 0 to get:
2. 0 = 0
So (1) must be true.
But seriously, though, even numbers can be made by adding two odd numbers together. Any prime bigger than 2 is going to be an odd number, otherwise it would have at least 3 factors, now I'm starting to confuse myself, anybody know what sort of pattern a graph of primes makes?
I see even classic Slashdot is now pretty much unusable on dial up anymore.
Okay, where I was going with this is that you can generate even numbers by adding a positive odd number and a negative odd number, but I just realised that any negative odd number has at least 3 factors, +1, -1, and itself, so now I remember why I didn't major in math (other than that pesky having to get up and go to class thing).
I see even classic Slashdot is now pretty much unusable on dial up anymore.
Is 1 not a prime number (even though it is divisible only by 1 and itself) just because whoever wrote the rule said it wasn't or does making 1 a prime screw up something else (like letting division by zero be defined)?
I see even classic Slashdot is now pretty much unusable on dial up anymore.
Maybe if I'm lucky all the moderators are sleeping in this morning.
I see even classic Slashdot is now pretty much unusable on dial up anymore.
I am all too familiar with that nice feeling.
I see even classic Slashdot is now pretty much unusable on dial up anymore.
Actually, the incorrect operation is division by zero. Multiplication by zero is fine.
ebw
Remember Fermat's Last Thereom was proven by someone who work alone and had very little help if any.
This will encourage those lonely mathematician to do something come out of their shell. Anyways competion always help make things more interesting.
http://theotherside.com/dvd/
I am curious what the math department at my school are going to say about this. :)
Being one of 20 math majors at school I wonder if this is a challenge the other math majors
(And maybe some CS people ) are willing to partake in. I guess I need to read up on my Number Theory
I think that is one of my next classes to take
http://theotherside.com/dvd/
ehh... something bugs me about mathematical induction. How do you know that n+1 is going to work for all numbers without testing? Suppose there is a number somewhere along the way that behaves differently. Primes, for example, aren't evenly distributed in the integers. How do we know that "if p(k) AND p(k+1) then for every n, p(n)"? Do you have to use induction to prove induction? How does this work in math? My proof skills are a bit weak.
I figured it was probably possible - my previous message was just politely trying to find out whether the first poster knew what he was talking about.
I'd be interested to see that proof, since a formula for calculating the nth digit of pi does exist. The only catch is, the formula calculates the nth digit of pi in hexadecimal.
If there really is a proof that it's impossible, then presumably that's for base 10 numbers? Do you have a reference?
Poked around with this thing for an hour or so. Seems like if you just start enumerating all the pairs of prime numbers and sum them up you construct all the even numbers. That won't float by itself though, cause the primes logarithmically taper out as you go...
I asked the google miester whether it had any ideas, and it came up with this site's proof.
It's only 8 pages, and it seems pretty decent. I followed it till about page 4, and then I saw those capital pi product symbols and decided to run away.
-Snoot
normality.
Other sources of good, understandable infinity:
Mind Tools and Inifinity and the Mind by Rudy Rucker
The Book of Numbers by John H. Conway and Guy (don't know his first name, offhand.
Cantor's Continuum Hypothesis states that the "cardinality" (roughly count) of the real numbers is 2^(cardinality of integers). This is usually written using the Hebrew alpha, but I can't find this on my keyboard. See any of the references I gave in an earlier comment for more details.
Anyway, it has been proven that this hypothesis cannot be proven or disproven with current set theory. I won't pretend to understand either proof at all, but they are generally well accepted.
- 2*N-prime = prime
for your equation.A language that might be well suited to genetic programming and proving math theorems is Erlang (http://www.erlang.org). It supports all sorts of neat stuff, like hot swap code changes, and the ability to run both the new and old code at the same time. Your program could evolve.
There's a lot of info on genetic programming available on http://www.geneticprogramming.com.
Need Free Juniper/NetScreen Support? JuniperForum
- Show empirically that the conjecture is true for some specific value o
- Find a logical argument that shows that anytime the conjecture is true for any value n, it must also be true for n+1
- If 1 & 2 are true, then the conjecture must be true every value greater than o.
Of course for this particular conjecture you would use n+2 in step two, since you are only interested in every second number.2.5) We all know that 1 != 0
2.75) Combining steps 2 and 2.5, it must be the case that For all N, N=4
--
Linux MAPI Server!
http://www.openone.com/software/MailOne/
(Exchange Migration HOWTO coming soon)
this may futher your point of "anyone interested in Slashdot should still be for [gore]."
this is made by one of those "people with relatively weak intellectual immune systems."
The following quote comes at the end of a recent email from Bush to Gore.
"Thank you for your email. This Internet of yours is a wonderful invention"
take it as you will, i wanted McCain.
I can just see some old grandma coming out of the wood work announcing that she has a proof. she then goes on to show it using lemon drops and apple pie, therebye stunning the mathematics community.
For some reason this post brings up an interesting question. suppose within two years i create a sufficiently(sp?) capable ai, that then goes on to create a proof for GC.
do i still get the prize money?
And more than that, Wiles' proof, IIRC, uses plenty of number theory and other unrelated disciplines which didn't even exist, and where far from suspected in Fermatt's time... so, if Fermatt had a proof using the number theory in his time, it has to be very inteligent and/or full of ingenuity.
:)
Fermatt was a genius. And he was a joker. So, in a way, Fermatt's Last Theorem is *still* unresolved. Wiles' proof is NOT the proof Fermatt had, if he had one.
Taken from a lecture this winter by Carl Pomerance of Bell Labs/Lucent:
- The conjecture is true for numbers less than 10^14
- All sufficiently large even numbers are the sum of a prime and (the product of 2 primes). Note that this product is thus composite, but not very!! So this fact is kinda close to the GC.
- All sufficiently large odd numbers are the sum of 3 primes.
- If the generalized Riemann Hypothesis is true, all odd numbers bigger than 5 are the sum of 3 primes.
- There will generally be multiple representations of a given number as the sum of two primes. The count of these representations is large, and is suspected to go as n/( log(n)*log(n) )
This is probably too late to reach many people, but oh, well.
- Brian
My friend Jon wrote a C program the other day that made a list of prime numbers. He's currently past 1,000,000. I'm sure that his program could be modified in some way as to test these numbers. However, I don't think he'd enjoy using his CPU for that.
Hah, your frend must be a pretty crappy programmer then, I've written a program that searches all the primes in prettymuch linear time, 10,000,000 takes about 3 or so seconds (on a p200, when I last ran it, btw)
ReadThe ReflectionEngine, a cyberpunk style n
Actualy, I belive that you only have to prove for n > 4...
ReadThe ReflectionEngine, a cyberpunk style n
Programming may not prove the theorem true, but it can certainly prove it false by finding one counter example.
Also it's worth mentioning some non-obvious math theorems have been "proven" by symbolic math systems without human intervention. Of course it takes a human to recognize weither all the proofs that were calculated had any significance or not.
-- Virtual Windows Project
In some cases its quite easy. Here's a proof that there are an infinite number of prime numbers.
Q = (1*2*3*4*....*(P-1)*(P))+1
No computing, no billions of cacluations or systems with 64-bit integers needed :)
--- Hot Shot City is particularly good.
IIRC the three schools of thought in the study of the history of mathematics are:
a) he was joking
b) he really thought he had found a proof
c) he really thought he had found a proof and was teasing his readers by not printing it.
There is absolutely no way his proof (if it existed) was anything remotely similar to the one Wiles found. Wiles' proof is a thing of incredible beauty. Far from being just a pure exercise in theory, it links hitherto unrelated disciplines in mathematics together.
--- Hot Shot City is particularly good.
Just for kicks... attack the problem from the opposite direction, and this one is easy to prove.
The sum of 2 prime numbers > 2 is an even number greater than 4. For obvious reasons, neither prime can be 2 because the sum will then be odd.
I forget what its called, but there are gaps in between prime numbers and these gaps grow as the numbers get higher. Up to the number 9, primes are spaced 2 apart: 3, 5, 7. Then there is a 4 number gap from 7 til 11. A 6 number gap appears between 23 and 29.
In order for a number to be the sum of 2 primes, either the sum of that number divided by 2 must be a prime, otherwise one number must be greater than half and the other must be less than half. Since gaps become more prevalant as the numbers get higher, there will likely be fewer prime numbers in between the halfway mark and the even number in question, than there are before the halfway mark.
You have to be able to prove that there MUST be a prime number P between any 2X > 4 and X.
Then there must also be a prime number p such that P+p=2X. Another quality is that P and p are both equidistant from the halfway mark.
This means, it must be proven that for any number Z >= 3, there must be a number Y >= 0 for which Z-Y is prime and Z+Y is prime. For example, in the case of Z=3, Y=0. In the case of Z=4, Y=1 (3 and 5 are both primes).
Now we get back to the gaps I was talking about earlier. As numbers get larger, the primes will occur with less frequency because there are more primes below it to help fill in the gaps. However, primes will still occur with a certain pattern. Primes are likely to occur when the numbers before and after it are rich in different prime factors. No 2 consecutive numbers share any prime factors. Therefore, the more unique factors both the adjacent numbers have, the less possible factors the middle number can have. If both adjacent numbers include all prime factors up to the square root of the largest number, then the middle number MUST be prime. Prove that first, if it isn't already.
In order to get a prime number, you need to get 2 factor rich even numbers on either side of it. This will occur with a predictable pattern. Prime numbers leave a trail, so to speak. You get a number that is prime and every multiple of that number from this point on will be composite. You are looking for points where these nth multiples all converge on 2 consecutive even numbers. That will get you ONE prime, but for this proof you ALSO need those multiples to converge an equidistant distance greater on the number scale from another number. Develop a system to determine what those numbers are, and if there is a set pattern to those numbers. If there IS a set pattern, and that pattern can be proven, then the numbers can be sieved out, and that seive can be used as a base proof for the other questions I asked earlier.
Ok. I haven't had any sleep in a while. Time for bed. Hope I didnt' screw anything up TOO majorly.
-Restil
Play with my webcams and lights here
An interesting question is: assuming prime numbers are "randomly" distributed, what is the probability that an even number won't be the sum of two odd numbers?
Oops ... a typo. I ment "two prime numbers" - not "two odd numbers" ...
Sela
I find the smartest people on the net on slashdot, but sometimes, I likewise find the dumbest folks. Dumb folks! You cannot freaking brute force this problem! you are not trying to prove for a certain range of numbers, but for infinite numbers, this is a case where computers can never beat mathematics, with one proff you can solve an for an infinite numbers, whereas with your computer, you will run it forever and yet can only be certain for those numbers you tested... blah blah blah.
------ Curiosity killed the cat. {satisfaction brought it back | it didn't die ignorant | lack of it is killing mankind
On the contrary, I'm a recovering mathematician. I'm in an n-step program.
I don't know exactly what Schnirelman proved, but it seems that he meant 300000 distinct primes (The number 3 * 300002 is even and the sum of more than 300000 primes). So you're proof is wrong.
"I suspect that a proof of Goldbach's conjecture might entail a more fundamental formula or series that could generate the set of all primes. That would be a much more important finding since it would shatter the mathematical security of many cryptographic algorithms --- particularly the RSA PK (public key) system. (There's billions of $ at stake in that case)."
It has already been proven that no formula can generate the Nth digit of PI. I believe (but I'm not sure) there such a proof for the non-existence of any formula for finding the Nth prime (without the preceding primes) as well. Can anyone verify that?
Understood: gcc's job is to compile, not to validate. But is there any foolproof way to validate your code (manually or automatically) without having a copy of the ISO standard or any other non-free documentation? (I'm not saying others should consider non-free documentation bad, I just don't want to use it myself). One great thing about HTML is being able to automatically validate it against the DTD using, say, validator.w3.org.
perl -e 'fork||print for split//,"hahahaha"'
Checking the whole list of numbers would be a proof in the mathematical sense. But since theres an infinite number of even numbers its somewhat difficult checking every single number.
This is too trivial.
n is even and large
if the conjecture holds then
n = a + b
where a and b are prime and
n + 2 = a + b + 2
so,
n + 2 = a+1 + b+1
now, a+1 and b+1 are even (because primes are odd, excluding 2 which is why n is large)
so,
n + 2 = c + d + e + f
where c,d,e and f are prime, so
n + 6 = e + f + g + h + 4
or
n + 6 = e+1 + f+1 + g+1 + h+1
and we can split again,
n + 6 = i + j + k + l + m + n + o + p
and it is obvious that we can keep doing this forever. However! Schnirelmann (1931) proved that every Even number can be written as the sum of not more than 300,000 Primes (Dunham 1990) which contradicts our finding! Thus, the Goldbach Conjecture is disproved by contradiction.
Now is someone would like to slap some mathematical grammer around that and get it published, I'll split the prize money with ya
How we know is more important than what we know.
you have to be a total whore.. best way to do this is to reply to every comment on a single post, then reply to everyone who replies to your comments, especially if it is a few days later because they may reply to your reply and then you can reply again. long/annoying posts dont increase your karma but are damn fun.
How we know is more important than what we know.
oooooowww.. he's using a seive to find prime numbers.. knuth invented the probablistic prime test baby.. use it!
How we know is more important than what we know.
There was actually a case where something along those lines happened. A man stood up, wrote on the chalkboard for about 15 minutes, turned around and got a standing ovation.
If I remember right, he found the factors to a really large number thought to be prime. When they asked him how he did it they expected some complicated answer. Instead, it turned out he had spent the last 20 years or so multiplying out prime numbers - by hand - in his spare time. No real technical math background.
Granted, much easier than this problem. Interestingly enough, plotting the number of matches for each even number seems to increase almost linearly. Proving no dips to zero on the hand is harder =)
I'm not a journalist, but I play one on slashdot
Are you sure you understand it?:)
:)
The tricky step in Cantor's proof is this one:
"the number on the diagonal is both in the list (since it is a real number), and not in the list (since it differs from everything in the list). this is a contradiction, therefore the list cannot exist."
Studying this led Bertrand Russell to state the following very serious paradox that uses the same logic:
[a definition first] to any intelligible condition there corresponds a class whose members are all and only those things that meet that condition; for example Torvalds is a member of the class of "men", and also of the class "non-goats". Also, the class "men" is itself a member of the class "non-goats" (a class is hardly a goat). Note that the class "non-goats" is a member of itself.
Now for the paradox:
Let R be the class of all classes which are not members of themselves.
Is R a member of itself?
In view of Cantor's argument, one is forced to reason:
"R is a member of itself (because if it isn't, then it is by definition) and R is not a member of itself (because if it is then it isn't), therefore R cannot exist."
However, R is a well-defined class ! Whoops.
The upshot of all this is that people have tried to find a way to fix Russell's paradox without disturbing Cantor's proof (since Cantor's proof is quite beautiful), and have come up with all sorts of messy modifications to Russell's but nothing that actually resolves it.
A good place to get your thinking started
Infinity is often confusing to non-mathematicians, and downright incomprehensible to computers. Hence it is no surprise that trying to write a computer program to "carry out" a proof will fail miserably.
A human can make judgments such as "The diagonal number is not a member of the list" (which I hope you agree is rather obvious), whereas a computer cannot judge this without completing a search of the whole list (which is where your loop problem comes in).
And now for something different.. it is situations like this that have convinced me that traditional computers will never be able to think like a human, they are incapable of making "insights" such as that fact about the diagonal number. Try explaining Russell's paradox in C.
You mean, he used unrelated disciplines that existed but hadn't been discovered in Fermat's time.
Duh. It's not a non-ANSI extension. It's legal code. "Undefined" merely means that it could give an unexpected result (depending on the compiler).
Similar to things like: a = a++;
The article said that all numbers up to 400,000,000,000,000 had been checked.
However this is less than 1% of all numbers.
There's still quite a ripe picking field if you want to look for a counterexample.
You aren't a VB programmer by any chance, are you?
prime numbers are a subset of naturals, which do not contain negative integers
In Capitalist America, bank robs you!
I can't remember the details, but I know back in the 1800s someone tripped across a what looked to be a short and elegant proof of FLT, but which turned out to be subtly wrong. Most likely this almost-proof is what Fermat found.
--
"HORSE."
"HORSE."
-Flaming Carrot
No, because you only pick six numbers!
i suppose "unattackable" means there's no good way to prove the theroem, its converse, or anything about existence of such proofs. after all, those are all nice ways of attacking the problem.
just imagine, nobody even knows how to construct any infinite family of prime numbers, though we know that they're infinite! (the sieve of eratosthenes is of course not a construction). mersenne thought he had one, but we now know he didn't.
and this is a problem with some history. all of the simple approaches are likely to have been tried. i remember doing a project on prime numbers in high school, and it was just frustrating how few real results there were out there.
so, all i can do is appeal to history, authority, and personal experience. i can't prove a damn thing. but then again, that was my original point wasn't it
cheers,
sh_
Interested in learning Chinese or Japanese? check out Chinese/Japanese-English Dictiona
whereas fermat may or more likely may not have had a proof to his conjecture, i am fairly certain that you don't have such a proof.
after all, this is one of the four problems that Landau called unattackable. in fact, i almost think it's disingenuous of them to offer the prize for this particular problem. they know they're never gonna part with that cash.
cheers,
sh_
Interested in learning Chinese or Japanese? check out Chinese/Japanese-English Dictiona
There's nothing stating the two primes must be different :-)
As an aside the Goldbach Conjecture explicitly states that it holds only for even numbers greater than 2, so all the people shouting about 2 not being the sum of two primes, and claiming the conjecture is false are pointless. But at least some people manage to realise that a trivial disproof is most likely flawed...
NP
Can you sum it up in a word? *No.* In a noise? *Whuuuurghhhhh!*
You probably realize that your conclusion is too weak. The probability of finding a counterexample beyond 10^15 decreases so rapidly for every number searched that given a "random" distribution you will not, with almost complete certainty, _ever_ find a counterexample beyond 10^15.
Asymptotically, your probability is e^(-2n/(log n)/(log n)). The chance of finding a counterexample beyond 10^15 is less than the limit as (N -> infinity) of the integral from (10^15 to N) of e^(-2x/(log x)/(log x)) dx. Since this is hard to integrate, take a simple function that dominates it beyond 10^15, such as:
10^(-4 * 10^11) * (e^(-sqrt(x))) * 2 * 10^7.5) / e^(-10^7.5) / ( 2 * sqrt(x)).
The integral of this for x between 10^15 and infinity, if my scribblings are correct, is 10^(-4*10^11) * 2 * 10^7.5 = 10^(-3.99999999992 * 10^11). In other words, the odds are worse than 10^(100000000000) to 1 against finding a counterexample beyond 10^15 (or beyond 10^11, for that matter).
Heh, sounds a lot like a certain EECS 303 course I am taking! :P
The higher, the fewer.
Nope.. the probability is zero. Sum of two odd numbers is always even! Consider (2n+1) + (2n+1) = 2(2n+1). If you can take a factor two in front it means you have an even number.. But then again the original poster probably meant two primes(didn't write so though..) and again according to this conjecture the probability is zero..
Yup.. now just show that there are no two primes that add up to this number..
What are you trying to get to? Generating even numbers? even number = 2 + prime + prime(prime > 2).
Interesting! While this works, I'm not sure how much better this is than exhaustive search, though...good idea!
Check out Greg's Bridge Page!
I've been thinking, the (strong Goldbach) conjecture states, "all Positive Even Integers greater and/or equal to 4 can be expressed as the Sum of two Primes.".
A Prime Number is defined as, "a Positive Integer p greater than 1 which has no positive integer Divisors other than 1 and p, itself. (More concisely, a prime number is a positive integer having exactly one positive divisor other than 1.)". Consequently, 1 is not prime number.
The Goldbach conjecture as stated implies that the sum is comprised of only two numbers. Then what about 6? It is a positive, even integer greater than 4, yet:
6 = 5+1 (not valid since one is not a prime)
6 = 2+3+1 (not valid since more than two numbers are used)
Since there is no sum comprised of only two primes, therefore, the conjecture is disproven by counter-example!
This cannot be right. I believe I've misinterpreted the conjecture. Anyone care to flush out my error?.
-----
"I will be as a fly on the wall... I shall slip amongst them like a great
Higher Logics: where programming meets science.
The conjecture states that the integer must greater than and/or equal to 4. Sorry! :)
-----
"I will be as a fly on the wall... I shall slip amongst them like a great
Higher Logics: where programming meets science.
I've always had a problem with "proof"s like this. They pretend, like you do, that there is an end/bottom of the list whereas surely the point of infinity is that there isn't. It's easier to see expressed as C loops. It's a bit like saying: for (int i = 0; ; i++) { for (int j = 0; ; j++) { add_to_list[i + j]; } } and then pretending that the inner loop actually completes, you get a different (bigger) type of infinity from: for (int i = 0; ; i++) { add_to_list[i]; } In your mind you effect a closure of the inner loop, wrap it up in a box. But seen as a C program it's clearly absurd. I guess people imagine it more as: for (int i = 0; i infinity; i++) { for (int j = 0; j infinity; j++) { add_to_list[i + j]; } }
I wasn't trying to prove it by writing a program, the program was meant more as an analogy to demonstrate what I think is flawed visualization with regard to infinity. Setting it down as a program brings home the point. I've read a bit about infinity - it appears that even amongst mathematicians some of the uses of it are not universally accepted.
The article, and specifically your quote from it (thanks, seriously) makes the very mistake I was pointing out. Namely, it actually says that Gore claimed to be the "inventor of the Internet." However, if you read Gore's quote, he says no such thing.
Your article notes that Gore has a 133-134 IQ and went to college with a 1355 SAT. Please don't say you think that someone this smart says or thinks he "invented the Internet." I know you're too smart for that.
Reread Wired magazine's original post and, if you feel like it, CmdrTaco's fateful and uncharacteristically shallow propagation of the article to Slashdot.
If you read it, you will begin to see why Brill's Content is such a great magazine. It's sad to see someone like Gore, who was legitimately among the first to even use the term "information superhighway" in politics, suffer so much from the media's techno-illiteracy and incompetence.
Quite simply, after Gore said he "took the initiative in creating the Internet," Wolf Blitzer could have sought clarification. Any reader of Slashdot would have, but Blitzer fumbled. Blitzer should have said, "You're not suggesting you created the Internet yourself, are you?"
Had he done that (or had Gore chosen his words more carefully), the issue would be where it belongs, squarely in Gore's asset column.
Just because you can write a big number as a sum of lots and lots of primes, doesn't mean you can't do so with less. For example, if a+6=i+j+k+l+m+n, and i+j+k is also prime, then a+6 can be written as the sum of 4 primes.
The "graph" of primes does NOT make a pattern! If it even remotely resembled a pattern, we wouldn't have all of this hub-bub about finding primes.... Here is a graph of the first 15
Now get to work
Anybody want to start one with me? :)
"Evil will always triumph over good, because good is dumb." - Dark Helmet (Spaceballs)
Somehow I doubt it...
The self-proclaimed inventor of the Internet avoided all courses in mathematics and logic throughout college...
"The best way to do mathematics is to be creatively lazy." -I. M. Isaacs
In a word: two.
Two is an even number, two is not the sum of two primes, Q.E.D. It isn't a proof, but it does disprove it, which make the challenge meaningless.
Unless you feel like calling 1 a prime, which I don't. Is this so glaringly wrong that no one else mentioned it? If so, explain the flaw, I beseech you. =)
Then again, I could be wrong.
x=5
-> 2x+15=3x+10
-> x^2-2x-15=x^2-3x-10
->(x-5)(x+3)=(x-5)(x+2)
-> x+3=x+2
-> 3=2
-> 1=0
This works for any specific value of x, and is easily extensible into the general case.
(Spot the mistake in the proof...)
--
I don't suffer from insanity- I enjoy it immensly!
This number is the proof -- Goldbach was wrong! VISA Card: 3472569876-97846957824-356346-53747-5666-66
5 698769784695782435634653747566247 5 698769784695782435634653747566013 2 5698769784695782435634653747565479 2 5698769784695782435634653747565047 2 5698769784695782435634653747564759 2 5698769784695782435634653747564663 2 5698769784695782435634653747563979 2 5698769784695782435634653747563127 2 5698769784695782435634653747562527 2 5698769784695782435634653747562119 2 5698769784695782435634653747561519 2 5698769784695782435634653747560637 2 5698769784695782435634653747560619 2 5698769784695782435634653747559803 2 5698769784695782435634653747559623 2 5698769784695782435634653747558759 2 5698769784695782435634653747558393 2 5698769784695782435634653747558069 2 5698769784695782435634653747557247 7 25698769784695782435634653747556599 7 25698769784695782435634653747555693 7 25698769784695782435634653747555177 7 25698769784695782435634653747555009 7 25698769784695782435634653747554319 7 25698769784695782435634653747553479 7 25698769784695782435634653747552867 7 25698769784695782435634653747552219 7 25698769784695782435634653747552177 7 25698769784695782435634653747552039 7 25698769784695782435634653747551529 7 25698769784695782435634653747551397 7 25698769784695782435634653747551223 7 25698769784695782435634653747550857 7 25698769784695782435634653747550437 7 25698769784695782435634653747550173 7 25698769784695782435634653747549879 7 25698769784695782435634653747549729 7 25698769784695782435634653747548163
Sorry, but you missed a few primes. My calculator gives the following as the first few examples:
34725698769784695782435634653747566666=419+3472
34725698769784695782435634653747566666=653+3472
34725698769784695782435634653747566666=1187+347
34725698769784695782435634653747566666=1619+347
34725698769784695782435634653747566666=1907+347
34725698769784695782435634653747566666=2003+347
34725698769784695782435634653747566666=2687+347
34725698769784695782435634653747566666=3539+347
34725698769784695782435634653747566666=4139+347
34725698769784695782435634653747566666=4547+347
34725698769784695782435634653747566666=5147+347
34725698769784695782435634653747566666=6029+347
34725698769784695782435634653747566666=6047+347
34725698769784695782435634653747566666=6863+347
34725698769784695782435634653747566666=7043+347
34725698769784695782435634653747566666=7907+347
34725698769784695782435634653747566666=8273+347
34725698769784695782435634653747566666=8597+347
34725698769784695782435634653747566666=9419+347
34725698769784695782435634653747566666=10067+34
34725698769784695782435634653747566666=10973+34
34725698769784695782435634653747566666=11489+34
34725698769784695782435634653747566666=11657+34
34725698769784695782435634653747566666=12347+34
34725698769784695782435634653747566666=13187+34
34725698769784695782435634653747566666=13799+34
34725698769784695782435634653747566666=14447+34
34725698769784695782435634653747566666=14489+34
34725698769784695782435634653747566666=14627+34
34725698769784695782435634653747566666=15137+34
34725698769784695782435634653747566666=15269+34
34725698769784695782435634653747566666=15443+34
34725698769784695782435634653747566666=15809+34
34725698769784695782435634653747566666=16229+34
34725698769784695782435634653747566666=16493+34
34725698769784695782435634653747566666=16787+34
34725698769784695782435634653747566666=16937+34
34725698769784695782435634653747566666=18503+34
Tarsnap: Online backups for the truly paranoid
First of all, the empty product equals 1 by _convention_. But of course you do not make convention out of the blue. Here is the reason that the empty product equals one:
/. eats all my white space. (I can not write n, m and m+n the proper places.)]
The following is true for positive n and m:
(a*...*a) * (a*...*a) = a*...*a (+)
The parenthesises contains n, m and m+n a-s respectively.
[Damn,
or written using powers:
a^n a^m = a^(n+m) (++)
Now in math we to extend the symbol a^n to the case n=0. Thus we need to give a^0 a value (compare this to (+), the value a^0 is the value of the empty product).
We would like equation (++) to stay true, so we have to require:
a^0 a^m = a^(0+m) = a^m
Thus a^0 _have to_ be 1.
If we formulate it this way:
The empty product equals the neutral element of multiplication.
Then it is clear that the empty sum must equal 0, as 0 is the neutral element of addition.
-- A Mathematician is a machine for turning coffee into theorems. - Paul Erdös
So what happens when you plug 10^-(4e11) into an improbability drive?
Mangos?
no
Mail several coppies of it to yourself. The postmark will serve as a date, so you have a sealed dated document. A public notarty might also help, but INAL.
I have heard somewhere that the predicate has been tested for all numbers up to some big number (in the range of 10^12 i think).
Trusting GCC to reliably catch all violations of the standard is unwise. -pedantic is not really about catching coding errors, anyway. You really want some kind of -W option to catch this and, yes, it should ideally be on by default.
It means GCC doesn't catch this sort of bug for you by default. Maybe it would if you specified -Wall -W -O2 or something.
No, because I don't track the standardization efforts (not even of C) closely. We're not talking about a "feature" so much as a detailed rule of behavior, designed to allow certain optimizations to take place. C is not Java; it does not mandate many sorts of ordering of things like side effects, the result being various things are left undefined. "Undefined" means not valid C code.
I believe I gave the basics of a good reference -- section 3.3 "Expressions", which might be numbered or even titled differently since 1988-12-07.
Nothing, if any of a, b, c, d refer to the same storage. Assuming they don't, though, which is likely what you mean to do, then what you're missing is that the sample statement does not violate this wording: "Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression."
And, from an anoncwrd:
First, I really know only FORTRAN 77 well, and that as an implementor, not a user. I know far more about ISO C than I do the present Fortran standard (95). (Which doesn't say much, I'll admit.)
Second, I wasn't objecting to evaluation order, but to the assumption of modification order in the original code.
A statement like "b += b += 2;" has only two pertinent sequence points, the one prior to the beginning of the execution of the statement, and the one at the end. There are no sequence points within the expression itself.
Now, y'all are right that "b += 2" must be evaluated before the rest of the expression.
But you're wrong in assuming the result of this evaluation is anything more than as if "b + 2" had been written instead.
Since there's another (outer) b += ... in the same expression, the C standard (draft copy I have anyway) says the expression is undefined. (I.e. it specifies conditions for the expression such that it is defined, and the statement in question violates those conditions, therefore it is not defined according to the standard.)
From a technical, practical-implementation point of view, the expression could be implemented (compiled as) "tmp2 = b + (tmp1 = b + 2);" with the following side-effects specified:
Specifically, the order in which those two assignments, vis-a-vis the evaluation of the expression, happen is unspecified, and an implementation could choose any order at any time, assuming it meets the obvious def-use ordering (i.e. tmp N must be defined before it can be referenced). You can't assume the b = tmp1; modification happens before the tmp2 = b + ... evaluation, for example.
So, you can't rely on the inner assignment to b being completed (i.e. the modification of b itself as a side effect) before the subsequent reference to it as part of the outer "b += ..." operation. So it might have the old value or, strictly speaking, any other value, or start World War III, or whatever.
It is undefined, according to my draft copy, because a gets modified twice, not once, between the two sequence points. a=(a++,a) fixes that, I think, since there's now a sequence point between the two modifications.
You might say "but the same value gets written to a regardless of whether the a= or a++ modification happens first", and I'd agree with that, but the wording in the standard does not make a distinction between multiple modifications of an object with the same value versus with different values.
So, you might get away with it on more implementations (compilers, machines, etc.), and that's all nice and well, but it isn't any more "defined" than the clever XOR-hack .sig.
BTW, read also the wording under "Assignment operators". In my draft copy it says "The side effect of updating the stored value of the left operand shall occur between the previous and the next sequence point." Y'all are assuming it occurs between the evaluation of the assignment operator and the evaluation of any lower-precedence (outer) operator -- but it doesn't, it might occur later than that.
Finally, hey, couldn't y'all just read the comp.lang.c FAQ? I just pulled it up in no time, and questions around 3.3 seem to pretty succinctly say what I've been saying more long-windedly.
P.S. I'm clicking "No Score +1 Bonus" again on my post. It seems like canonical /. practice to do so in cases like this (off-topic posts).
Practice random senselessness and act kind of beautiful.
Not that I'm aware of. HTML and C code are two different things, in the sense that both have static semantics, but C has dynamic semantics as well. (HTML does, presumably, when you expand it to include Java, Perl, other Turing-complete CGI stuff...but I'm getting out of my element here, maybe HTML these days is dynamic too. I use it only statically.)
So, a static analysis tool can catch "b += b += 2;", and tools like lint and GCC (with its various warning options) try to be good static analysis tools.
But to really be foolproof, one needs static, dynamic, and some other sort of analysis.
Dynamic analysis would catch "b->foo += c->foo += 2;" when (b == c), but only at run time, unless of course static analysis can prove that it'll always be true. (Remember, run-time analysis is better than nothing, but isn't "foolproof" -- you want to find all bugs in your code before it's deployed, not when it goes to drop the oxygen masks during an emergency decompression or something. ;-)
The other sort of analysis boils down to the halting problem, but can in practice be quite useful, just not entirely foolproof. Here the statement is analyzed statically, but in a complete context so logical relations can be asserted in boiled-down form for the program as a whole. (E.g. "If the input file contains a zero byte, the program is undefined because it'll perform a divide by zero in line 1234.")
Anyway, yeah, a web site that validates C (etc.) code against prevailing standards would be pretty cool! Maybe few would use it, but they'd be pretty grateful, and it could settle all sorts of tedious arguments more quickly. Feel free to send me URLs to such things.... ;-)
Practice random senselessness and act kind of beautiful.
Not in ISO (Standard) C it doesn't; it's undefined. In "pidgin C" it does, if the reader assumes sequence points within the expression as if you'd written:
Specifically, operators like ^= do not define sequence points, so there's no assurance that the inner assignment to b will happen before the outer one. In my oldish draft copy of ANSI C, the second paragraph of 3.3 "Expressions" makes it clear the two modifications of b in your no-sequence-point-containing expression renders it undefined.
(Of course, my version hasn't been tested or carefully analyzed, but I wouldn't make it my .sig. ;-)
Practice random senselessness and act kind of beautiful.
onwards!
"The new wave is not value-added; it's garbage-subtracted" - Esther Dyson, Dec 1994
$ ps aux ./goldbach
./goldbach
...
timmy 16418 88.1 37.3 132764 95888 pts/7 R 21:47 3:30
Don't bother to find a counter-example
You ruined my plans!
timmy:~/prgm/goldbach/src$
Enter the number of prime numbers you need: 7000
Creating a vector with 7000 primes...
Getting all possible even numbers...
Sorting...
Removing duplicate even numbers...
Printing first missing number. Note: not correct if not enough primes found.
140792
that means i know that two primes can be added to get all the even numbers below 140792! almost there! but when my stupid program takes up ~40% of my ram using only 7000 primes, i got a real problem.
i was just writing the program for fun. It would do about 5000 primes in less than 10 seconds but when i decided to jump to 7000, I got some error, so i used a vector of a biginteger class rather than unsigned ints. if i thought i had a change at finding an exception, then i would have done it differently. however, i don't think i would have been able to use the bitshifting method because i was using a std::vector. thanks for the idea.
If the conjecture holds for k then it holds for k+1.
Now, we can obviously test this for small numbers. For example, it holds for 5 (=3+2). However, we can prove a conjecture by proving the inductive step - which means you don't have to try every number up to infinity. Since that might take some time... :-)
Of course, the difficult part is proving the inductive step :-) Indeed, _finding_ the right inductive step is often hard enough...
If that is true, it could be used for increasingly finding largest known prime numbers. Just take the largest known prime, multiply by two, add two, the resultant even number, being the sum of two primes, one of which must be greater than the previously largest known. Start subtracting previously known primes from that, then test the other of the pair for being prime. I wonder if that is how they do it. Of course crunching those numbers to test would take a hefty computer (distributed.net?).
With a conjecture such as this, isn't it impossible to find a proof without trying every even number bigger than 4, up to infinity?
This post is brought to you by the letters T and A, and the number 69
An interesting question is: assuming prime numbers are "randomly" distributed, what is the probability that an even number won't be the sum of two odd numbers?
Zero.
For all intensive purposes, "whom" is no longer a word. That begs the question, "who cares"?
They're looking for an abstract proof, not a massive crunch. Now if somebody programs d.net to handle random logical arguments, and test those logical arguments for validity, that would be interesting. OTOH, I have a feeling that the number of possible symbolic logical arguments that conclude with an assertion of the Goldbach conjecture is *considerably* larger than a keyspace :) . Also, the program would have to include a knowledge bank containing all the mathematical proofs (which these professors will have at their disposal). This is a clear cut case where the human brain still wins, because the brain can intelligently determine what theorems need to be cited, what papers to look at, what aspects of prior research are important, etc.
For all intensive purposes, "whom" is no longer a word. That begs the question, "who cares"?
That means that -1 is a prime number. It is divisible by 1 and itself. 1 * -1 = -1. I've never seen it listed as such, but by all technicality it is.
Aah. Thank you. I had this nice feeling that I was incorrect. :)
Plus, it's been a while for me anyways. Maybe I'll learn some more of this stuff this summer. I'm taking Number Theory *grin*.
All you need to do is write a program that will produce every possible combination of characters (starting at 5 characters, then 6, then 7, and so on). Then you put it through a spell checker to eliminate the completely random characters, filter out everything that doesn't have the word "prime", and read all the combinations generated to see what works.
They that quote Benjamin Franklin on liberty and safety deserve neither.
I tried doing a "Proof by Asumption" in Grade 12.
I wrote in on a board right before a test on "proofs" She told me to take it off quickly before The other students got confused.
Prove X = Y
1. Assume X = y
2. therefore X = Y
-- I Beowulfed my left and right brains!
I do remember that negative numbers are only numbers with properties therefore they cannot be primes. Example -2 is the value 2 with a math equation of subtraction attached to it. So in theory positive 2 and -2 have the same value. however -2 is not a true number it is part of an equation. Man I remember failing that class and it turns out i actually retained some of it. Only took 9 years to remember it. :(.
This number is the proof -- Goldbach was wrong! VISA Card: 3472569876-97846957824-356346-53747-5666-66
Unfortunately, the margin of the page was too small to show a diagram of his time machine.
Well, it's already been proven true for all even numbers up to 400,000,000,000,000. Good luck with finding a counter-example :-)
> If on the other hand the above proof does not > exist, then there must exist a counterexample. Not quite: there's a third alternative; that the conjecture is true, but no proof exists. Godel showed us that truth and provability are not the same thing.
--Lee Daniel Crocker : http://www.etceterology.com My life is in the public domain.
Why isn't 1 considered prime?
My teacher explained that when you factorised a random number into its prime composants you are obtening something like:
number_to_factorize = prime_componant_1**n1 * prime_componant_2**n2 ...
Well, but wait a minute, if 1 is prime then I can say that 4 = 2**2 * 1**n, n being (at least) any natural number, we could even say that 4 - 2**2 * 1**INFINI.
Of course this is quite stupid, so 1 isn't considered as prime (I think there was another reason but I can't remember).
So if 1 isn't considered to be prime by convention, then you can't say that 3 = 2 + 1 to obey the theorem, due to 1 non-primeness.
Given that I suppose that the guy that made the theorem was WAY much better than me at math I suppose that either:
1. When he specified that, 1 was considered a prime number.
2. Maybe not all countries recognize 1 as not being prime.
3. Maybe the theorem states that it begins at 5 = 3 + 2?
Can someone tell me which is it or wether I smoked to much pot and am completely wrong on 1 non-primeness (web references would be appreciated), thanks.
"The obvious mathematical breakthrough would be development of an easy way to factor large prime numbers." Bill Gates,
One guess is that his proof assumed that the algebraic integers had unique
factorization into primes. (Algebraic ntegers=roots of polynomials
with integer coefficients where the highest power term has coefficient 1.)
They don't. Kummer &(actually vs.) Dedekind worked out their properties
& then Emma Noether and others generalized their work and created a big
part of modern algebra.
Probably this is why Fermat's conjecture was not on the `unapproachable' list.
No good. If you read the article, you'd know that evens up to 400,000,000,000,000 have already been tested. I think that's reason enough to conclude that brute force will not produce a counterexample. No matter your computing power, it would take an infinite amount of time to check for aleph null numbers. Seems like a formal proof is the correct method instead.
A close-to-improvable conjecture
A bunch of mathematicians not willing to waste their time on a close-to-improvable conjecture
Some loser issuing a contest
It then clearly follows that NOTHING HAPPENS
I can assure you all that this is as close to a proof of anything that will follow as a consequence of this pathetic contest
Things are more like they are now than they ever were before. - Dwight D. Eisenhower
If you've never heard of them, they run what can be quickly described as the fastest computer on earth. With the processing power of 111.65 gigakeys/second (RC5-64 rate), I think they could do it.
After reading this article and all of the comments attached to it, I have come to the conclusion that I'm a complete idiot . I can't think of a single way at all =) Damn math gives me a headache.
(No i'm not trolling- in maths it really is sometimes possible to write a proof that something else can't be proven... for example there's no general equation to find the solutions to a 5th order equation, that was proven.)
In that case presumably they wouldn't pay up... would that suck or what? ;-)
-WolfWithoutAClause
"Gravity is only a theory, not a fact!"I think it would be a good chance for a programmer (and anyone who supports them) to begin a SETI@home-like client to figure this out. But this begs the question: if the solution was figured out in the prequisite 2 years, how would the money be distributed?
I happen to have a copy of the Encyclopedic Dictionary of Mathematics (Mathematical Society of Japan, translated, 1980 by MIT Press) here in my den.
Looking up Goldbach I find:
The article then goes on to summarize the gist of this proof which involves a summation of (exp( - 2 * pi * i * ( a/q ) * N)) and a convergent series S(N)= sum(q=1, inf, A(q,N)) (which is "absolutely convergent and is equal to" [a copy of product expressions than can't be typeset here]).It doesn't mention the magnitude of "sufficiently large" (read articles at Wolfram Research as referenced by the /. "related links" box near this article for more on that). Anyway --- Vinogradov solved Goldbach's conjecture for "sufficiently large odd integers."
Apparently "a finite or infinite sum of exponential functions such as this is called a trigonometric sum."
The article then discusses the case of even integers: "I.G. van der Corput, T. Estermann, and N. G. Cudakov proved simultaneously (1938) that almost all even integers (i.e., except a set of density 0) can be expressed as the sum of two primes." [emphasis mine].
All of this is in section 5.C of EDM.
The article goes on to comment on further work by Cudakov and Linnick, their introduction of "function-theoretic methods" and "the density theorem concerning the zeros of L-series" (which is greek to me).
They conclude with reference to the work of Zulauf (1952) which "continued along the same lines" as suggested by G.H. Hardy and J.E. Littlewood (1919).
I'm sure that some of the more academic participants here at /. might have access to a more comprehensive library than the bookshelf in my den. I'm not a mathematician, my education of the subject pretty much ended with high school Calculus, one audited class in combinatorics at Reed College and a few forays into Mathematica (tech support for my father, who is the arm chair mathematician and physicist of the family).
Personally I'd recommend starting with a thorough review of the literature to see what approaches have already failed. When you understand why those approaches failed then you can apply those tests to any methods you propose as a solution.
For example, if one could show that all even numbers less than some number N are the sum of pairs of primes less then N' and you could show that N and N' are related --- that N+x results in N'+y --- you might have a inductive proof in there somewhere.
If you win the $1M (or was that in pounds?) then feel free to donate a few grand to the FSF in my name.
I suspect that a proof of Goldbach's conjecture might entail a more fundamental formula or series that could generate the set of all primes. That would be a much more important finding since it would shatter the mathematical security of many cryptographic algorithms --- particularly the RSA PK (public key) system. (There's billions of $ at stake in that case).
Does that mean that any even number >=6 can also be the difference of two positive integers?
example: 11 - 5 = 6
What if you disprove the theorem? Then what, huh? No money?
1 + 3 + ... + n = (n + 1)^2 / 4
We know that, for n = 5, 1 + 3 + 5 = (5 + 1)^2 / 4 = 9, and so the formula is true for n = 5. Now assume that the formula is correct for all n. What happens if we work it for n + 2?
1 + 3 + ... + n + (n + 2) = ((n + 2) + 1)^2 / 4 - substitute n+2 for n in the formulas ... + n
(n + 1)^2 / 4 + (n + 2) = ((n + 2) + 1)^2 / 4 - substitute the assumed true value in for 1 + 3
(n + 1)^2 / 4 + (n + 2) = (n + 3)^2 / 4 - add n + 2 and 1
(n + 1)^2 + 4(n + 2) = (n + 3)^2 - Multiply through by 4
(n^2 + 2n + 1) + (4n + 8) = n^2 + 6n + 9 - multiply everything out
n^2 + 6n + 9 = n^2 + 6n + 9 - add everything up. This is true for all n.
What all that means is that if the formula holds for n, then it must hold for n + 2. And since it holds for n + 2, it must hold for n + 2 + 2, and so on into infinity. I came up with one n (5) for which the formula works, so it must work for all odd numbers larger than five. I don't have to test all the odd numbers into infinity - the infinitely recursive nature of induction does that for me in something that can be expressed in a noninfinite amount of data (and a good thing, too - I don't have enough bandwidth to post an infinite-length comment to /.)
This, in turn, means that for every integer I, there is a circle, such that the centre is at I and the intersection of the circle with the positive natural numbers will occur at two prime numbers.
Now, how does this make things any easier? Simple. If the distribution of primes is =totally= random, this statement HAS to be false. Randomness means that by knowing any number of points, you cannot deduce an unknown point. Here, we have shown that by knowing one point and a centre, we can deduce another point. This implies an inherent relationship between the prime numbers.
Thus, IF AND ONLY IF the prime numbers are non-random, Goldbach's Theorum is correct. If, however, they ARE random, it would be impossible to define a geometric relationship, which (as shown) does exist.
It's a small world and it smells funny; I'd buy another if it wasn't for the money; Take back what I paid (SoM)
You are holding waaay too much in memory. If you have 7000 primes, and you consider all pair of them, that is 49,000,000 numbers. If you were slightly smart about it you could cut that in half.
As you noticed, this is pretty inefficient.
A much better way to slice this problem is to look at the numbers as an array of bits. For instance the first 800,000 numbers is a 100 K string. Start off with a block of 0's. Generate a vector of primes. Shift the vector by 2. OR it with your first vector. Shift by 1 more. OR it with your first vector. Shift it by 2 more. OR it again. Shift it by 2 again, then 4, then 2... (We are walking through cumulative shifts equivalent to the number of primes.)
You can make it faster still by having your bitmap being only the even numbers. You will have to do a bit of thinking. But you should find that this approach is significantly faster and cuts down tremendously on the memory. You will also find that after a few shifts you will have covered most of the territory. At some point it becomes more efficient to track down each one you have not tracked down and search for primes that add up to that one...
Cheers,
Ben
My usual seat in the cluetrain is at A HREF="http://pub4.ezboard.com/biwethey.ht
With the probability reasoning it is trivial that there are (100% probability) only a finite number of exceptions. The odds of there being any after you have verified the first 100,000 cases is pretty small. But it gives you no idea how to prove it.
:-/
Likewise the probability result strongly suggests that the Riemann conjecture is true. But again, it gives no insight into the real problem.
However when you compute statistics, perfect match with the probability prediction.
Cheers,
Ben
My usual seat in the cluetrain is at A HREF="http://pub4.ezboard.com/biwethey.ht
An interesting counter-proposal to address this dilemma is proposed by a New Zealand economist named Ronnie Horesh, at http://www.geocities.com/WallStreet/9856/, albeit in a different context. Horesh's idea is a social policy bond issued by a government or somebody else with deep pockets and a social agenda. The bond is redeemable for a large fixed sum when the goal is achieved. Until then, the bond is a good that can be bought and sold like a share of stock. Like a share of stock, its price will rise as the goal moves closer to being accomplished. The purpose of inventing these bonds is to delegate the implementation of social goals to free market forces.
If a mathematician felt confident that he had made an important contribution, but didn't feel confident that he'd complete the proof himself, he could buy bonds cheaply as soon as they were issued, and hold onto them until the proof was complete. When the proof was made public, each bond would become redeemable for a large fixed sum, and the guy who bought early would make big bucks.
WWJD for a Klondike Bar?
Well, you can just e-mail that to me and I'll be sure to pass it along to Rob Malda so he can post it elsewhere on the page :) ...just take out the TINLC part, there is no lumber cartel.... :)
My journal has hot
That is - an extremely small probability!!!
:)
Yeah, but close only counts in horseshoes, hand grenades, and Inter-Continental Ballistic Missiles.....
My journal has hot
An odd number is some integer (n*2)+1. There is no integer that will satisify 0 = (n*2)+1. Therefore, zero is not odd (it's even).
A prime number is some integer that is divisble only by 1 and itself. Since a denominator of zero is undefined, zero cannot divided itself. Therefore, zero is not prime.
cpeterso
Andrew Wiles' proof of FLT runs into the hundreds of pages and makes use of vast tracts of number theory, some dating back hundreds of years. It can probably only be understood in full by a handful of top mathematicians.
;-)
Now that FLT has been proven, what about Fermat's comment about his "great proof"? Do modern mathematicians believe he was joking? Or did he have some different, simpler proof? That he would have the same lengthy page proof seems unlikely, but it would definitely fit the description of "too large for the margin of this book"!
cpeterso
What do you mean "such as this"? The whole point of a proof is that it is deeper than a whole list of numbers with checkmarks beside them. "Yep, 5 is sum of two primes. Yep, 7 is the sum of two primes." is just a demonstration, not a proof.
--
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"Who Wants To Be A Millionaire" is the American version of a British show. Only... nobody there ever won the big prize (the insurance company that pays the prizes is rather upset).
If we find a counterexample, thereby proving the conjecture false, do we also collect the million?
Time to get a cluster searching, eh?
Well I don't have a copy of the standard, but my fragment compiles without warnings on gcc 2.95.2 using the -pedantic option. Since that's meant to make gcc complain about non-ANSI (i.e. ISO) extensions, does that mean this is a bug in gcc? Are you sure it hasn't been added to the standard since the version you've seen?
Hmm the standard says that ^= and -= are right-associative and have the same precedence. The associativity should force the expression to be evaluated as I claim.
Would you care to provide a reference saying where in the ISO standard I should look to see what you're saying?
perl -e 'fork||print for split//,"hahahaha"'
But = also does not define a sequence point, so by that argument "a=b=c=d=2" is also undefined. What am I missing here?
perl -e 'fork||print for split//,"hahahaha"'
Who says mathematicians can't write poetry? Hmmm.
perl -e 'fork||print for split//,"hahahaha"'
they count as integers :)
but I dont think so
How we know is more important than what we know.
what I did first was to figure out the number of primes that sum to the first 20 or so evens.. obviously it's been done before (hey, I was hoping for a fibonachi sequence or something groovy like that).. anyways
.att.com/~njas/sequences/Sindx_G.html#Goldbach
http://www.research
How we know is more important than what we know.
The k p (calc-prime-test) command checks if the integer on the top of the stack is prime. For integers less than eight million, the answer is always exact and reasonably fast. For larger integers, a probabilistic method is used (see Knuth vol. II, section 4.5.4, algorithm P). The number is first checked against small prime factors (up to 13). Then, any number of iterations of the algorithm are performed. Each step either discovers that the number is non-prime, or substantially increases the certainty that the number is prime. After a few steps, the chance that a number was mistakenly described as prime will be less than one percent. (Indeed, this is a worst-case estimate of the probability; in practice even a single iteration is quite reliable.) After the k p command, the number will be reported as definitely prime or non-prime if possible, or otherwise "probably" prime with a certain probability of error.
How we know is more important than what we know.
Faber has stipulated that the proof must be submitted to a respectable mathematical journal within two years of the book's publication next week, and published within four years. A panel of world-renowned mathematicians will be appointed to decide whether the proof is valid (Faber is refusing to disclose the panel's names, for fear that they will be flooded with letters from amateurs).
I have a serious question here, and I hope that someone might be able to answer it.
From the tone of the article, it sounds like they are expecting somebody in the mathematical community to submit it, if anybody does. How would say, me, your average slashdot reader, go about claiming such a prize? I have no idea how one would go about publishing something in a mathematical journal, and with such a prize at stake how would one prevent one's proof from being stolen by someone else (like a person you find to "help" you get it published)?
I'm not a journalist, but I play one on slashdot
It does make sense to have lots of people offering lots of little prizes all over the place.
As I've pointed out in The Bowery Prize for Amateur Rocketry:
The Internet provides a unique opportunity to promote space technology via prize awards. Individuals can post a public pledge of money as a prize award for any technical objective they see fit, to be disbursed at their sole discretion. With enough diversity of people and technical objectives, there would be a "fuzzy" gradient of incentive created for ever higher performance amateur rockets, not dependent on the credibility of any one organization's political structure for "fairness" or good technical judgement. A periodic posting of all such prize award offers in this, and related, newsgroups, would serve as an ongoing challenge to technical excellence and as an inspiration for young people.
This sort of approach to what should be called "Contest Philanthropy" will address the concern that people will hold off publishing important but obscure work.
Here's how:
If a bunch of mathematicians believe a particular popular conjecture is critically dependent on some obscure, but hard, math getting done, then all they need to do is post a prize award that promises they will give half their winnings for the popular conjecture to the mathematician who does the obscure work. If enough credible mathematicians target a particular obscure piece of work, then the philanthropists can start providing direct prize awards for that support work.
Seastead this.
Unfortunately, that does nothing to make finding prime numbers any easier. IE, knowing the approximate shape of the graph (and the graph isn't perfect) still doesn't give you any information for calculating the next prime number. You have to pick a number (ending in 1,3,7, or 9), brute force it to test its primitude (I just made up a word!) by dividing it by all previous primes = the square root of the number you're testing.
-CausticPuppy "Of all the people I know, you're certainly one of them." -Somebody I don't know
For those banging their heads against the monitor. He multiplies by zero at one point. Then, all bets are off.
But your assumption is wrong. How about:
1) Assume 1 == 1
2) Either 1 == 1 or N = 4 for all N
3) 4 is the sum of two primes
4) by 2 and 3, for all N, N is the sum of two primes
5) Since all numbers are the sum of two primes, all odd numbers are the sum of two primes (QED)
Finding that number is left as an exercise for the reader. Heh.
Best regards,
SEAL
An integer greater than one is called a prime number if its only positive divisors (factors) are one and itself.
Thus, negative numbers are not prime.
Best regards,
SEAL
Slashdotter's should consider reading "Brill's Content" magazine for its recent story on the media's incompetent handling of the "Gore invented Love Canal" story. Then they should ask themselves if a guy as smart as Gore, who actually does have well-reported techno-wonk credentials, would really have claimed to have "invented the Internet."
Gore was one of the leading proponents of the "information superhighway." Now, if he wants to talk more about it, he runs the risk of a bunch of know-it-nothings laughing at him and hurling once-funny cheap shots.
This "Gore says he invented the Internet (har-har)" stuff is a silly mental virus propagated by people with relatively weak intellectual immune systems, IMHO. That's not to say they are stupid, just susceptible. Gore fouled up his line, true. But anyone interested in Slashdot should still be for him.
at least in the UK you seem to value education and things such as this. Here in the states we prefer to have people win a million dollars by either picking 7 random numbers and watching ping pong balls or answering 10 questions or so from Regis.
Goldbach's conjecture is very probably true; mathematicians and number theorists don't need to be reminded of that and likewise don't need to be bothered by "amateurs" (really, lay-people without formal education in number theory) who suggest that the conjecture is false. In all matters of fact, while I am sure that Faber and his erudite associates would be happy to be presented with a valid counter-example, said mathematicians probably aren't losing any sleep over the possibility. What Tony Faber is looking for and what's worth $1m to him is not the labors of the unitiated towards a hopeless cause but is instead the establishment of the conjecture's logical truth by means of established principles of number theory -- he wants to know how this elegantly simple and ostensibly true statement may be derived from postulates and previous proofs, not, regrettably, whether or not it is true.
There has been some grumbling over the apparent fact that Faber is trying to keep the proof of Goldbach's conjecture within the mathematical community. If the conjecture is proved, it will be through the collaborative efforts of the professional mathematic community -- I simply can't see someone without extensive background in number theory really having a chance of winning the contest. Remember that the proof of Fermat's theorem was by no stretch of the imagination simple and elegant. If there were an elegant proof to Goldbach's conjecture, it would almost certainly have already been realized by mathematicians who are just as intelligent and, if anything, better-versed in matters of number theory than folks like you and me, and if there is a proof, I suggest that its development will not be a feat accomplishable by anyone who doesn't make their living in mathematics.
And before anyone tries to make this proof a black&white, true/false issue (too late?), let us remember the work of mathematician Kurt Gödel:
Do I underrate the ingenuity of the human mind? I hope so; however, I would like to think I am merely being realistic. It seems to me that blind beginner's luck has been drastically overrated in response to the proposed contest.
To those of you interested by the prospect of testing every reasonably-large even number to find a counter-example and thus disprove the conjecture, and especially to those of you who have suggested that the resources of distributed.net be used in so doing, please refer back to sela, 3/19. I'm afraid you'll be wasting your time.
_subterfugitive
Do negative numbers count as primes?
I see even classic Slashdot is now pretty much unusable on dial up anymore.
Honest,
Al Gore told me he's got the proof to this theory - he did it while he was inventing the Internet, and Open Source and all that.
1) Assume 1 == 0.
2) Then clearly this follows Either 1 == 0 OR N = 4 for all N
3) Obviously 4 is the sum of two primes (2+2)
4) Combining steps 2 and 3 we get For all N, N is the sum of two primes
5) Step 4 states that all numbers are the sum of two primes, therefore all odd numbers are the sum of two primes.
QED
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The first step any Goldbach-conjecture-prover-wannabee should take is to read the book ``Additive Number Theory: the Classical Bases'' by Melvyn B. Nathanson, in Springer's Graduate Texts in Mathematics (number 164). It contains a description of the sieve methods and the Hardy-Littlewood-Ramanujan circle methods used to prove that sort of results in additive number theory. It gives a proof of the closes results known to Goldbach's conjecture: Vinogradov's theorem that every sufficiently large odd integer is the sum of three primes, and Chen's theorem that every even integer is the sum of a prime and a number which is at most product of two primes.
If you think it must be interesting reading, you're dead wrong. I have read many boring proofs in my life, but none of them came even close to the total lack of interest, the absolute and utter dullness of the analytic estimations of additive number theory, as found in the book in question. Even the (much easier) proof of Waring's problem (every integer is the sum of a bounded number of k-th powers, where the bound depends only on k) is unconceivably dull, and it is nothing compared to the more subtle results in the second part of the book.
Now why couldn't they offer a million dollars for proving the Riemann hypothesis. At least that's a worthwile result (nobody cares in the least about Goldbach's conjecture, but the Riemann hypothesis has many fascinating consequences).
// usage: goldbach
// 2 is prime
// Lists all pairs of primes that sum to
// Copyright Chaz Randles 2000
// Licence: GPL
#include
#include
using std::cout;
using std::endl;
using std::vector;
vector * get_primes(int maxPrime) {
vector* result = new vector();
vector::iterator it;
bool isPrime;
result->push_back(2);
for (int candidate = 3; candidate begin(); it!=result->end(); ++it) {
if (!(candidate % *it)) {
isPrime=false;
}
}
if (isPrime) result->push_back(candidate);
}
return result;
}
bool find(vector* vec, int data) {
vector::iterator it;
bool found=false;
for (it=vec->begin(); it!=vec->end(); ++it) {
if (*it==data) {
found=true;
break;
}
}
return found;
}
int main(int argc, char** argv) {
if (argc!=2) {
cout " * primes=get_primes(num);
vector::iterator it;
for (it=primes->begin(); it!=primes->end(); ++it) {
if(find(primes, num - (*it))) {
cout num '\t'*it'\t'num - (*it)endl;
}
}
delete primes;
return 0;
}
// Apologies for poor quality code - done in 10 mins
No, you can do tricks like "proof by contradiction". That works as follows:
perl -e 'fork||print for split//,"hahahaha"'
That means that "there is some way to get to any even number by adding together less than 300,000 primes". It doesn't mean "no number can be obtained by adding more than 300,000 primes". The second statement is clearly false. e.g. if
then a is the sum of 300,001 primes.
perl -e 'fork||print for split//,"hahahaha"'
Lots of work has been done that could be useful to the GC, just the same was as done for FLT. The winnings will go not to the person who does the most work towards it, but whoever supplies the piece of the puzzle that happens to be last.
Mathematics is a great cooperative venture. It wouldn't be easy to identify one mathematician who did the largest share of the work even if they tried to do it that way, and if they did it would probably be Gauss or Poincare or some other dead heavyweight genius.
This might encourage work on the GC, but it also might discourage publication of such work, because the mathematicians haven't quite finished the proof.
--
Xenu loves you!
Unfortunately, the slashdot comment submission box is too small to post the whole proof. I shall leave the proof to the reader.
"Pinky, you've left the lens cap of your mind on again." - P&TB
"I can see my house from here!" - ST:
Yes, it is possible that the Goldbach conjecture may be proved or disproved by a lay person but it's far more likely to be solved by someone with a sound grasp of lots of number theory who is already studying the properties of numbers for a living.
BTW I have a truly marvellous proof of the Goldbach conjecture but there is no margin on the submit form to write it in :)
--- Hot Shot City is particularly good.
Disclaimer: long boring math stuff with a conclusion that may be interesting for some at the end ...
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An interesting question is: assuming prime numbers are "randomly" distributed, what is the probability that an even number won't be the sum of two odd numbers?
If the probability is rather high, yet we did not find any such even number, it can be an indication there may be some "meat" in goldbach theorem. On the other hand, if such probability is exremely low, than the verification of goldbach's theorem up to 4e+15 means nothing
Well, according to Gauss' estimate for prime number density around sufficiently large n, the density is 1/log(n). To find pi(n), which is the number of prime numbers up to n, we need to integrate this, but pi(n)=n/log(n), so we can make life simpler and over-estimate the probability by taking the smaller number.
Now, given the number of prime numbers up to n, pi(n), if we choose x to be a prime number smaller than n, than the probability its pair: (n-x) is a prime number is: 2*pi(n)/n. (I multiplied by 2 since itcannot be possibly an even number, so no need to consider them).
The probability that it is not a prime number is therefore (1-2*pi(n)/n).
Since there are pi(n) different prime numbers to choose from, the probability that an even number is not the sum of two prime numbers is:
(1-2*pi(n)/n)^pi(n).
So what is the probability for numbers around 1e+15? If we use the estimate for pi(n), we get:
pi(n)=2e+13
(1-2*pi(n)/n) = 0.959
Taking the last number by the power of pi(n), the result is that the probability is around:
10^-(4e11)
That is - an extremely small probability!!!
Conclusion:
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If Goldbach's theorem is false, the probability that an even number won't be the sum of two prime numbers is so small that we may need to check around 10^(10^11) numbers to find a counter example. This number is so big no computer could possibly handle it now or any time soon
In other words: Don't bother to find a counter-example.
Sela
It's stunning how many people want to approach this as a programming problem; i.e. a fast way of checking lots of numbers. We're talking about a *proof*. Guessing and checking is missing the point.