Goldbach's Conjecture (Theorem?)

[Skainstein]

Goldbach's Conjecture (Theorem??)

Hi guys,

I just need to know if Goldbach's Conjecture has been demonstrated or not!

Thx! :D

 

[Moridin]

No mathematical proof yet. It is very upsetting.

 

[al-mahed]

Hi guys,

I just need to know if Goldbach's Conjecture has been demonstrated or not!

Thx! :D

are you trying to find a proof?

 

[al-mahed]

what about if we write the even numbers as a sum of the form 3a + 5b? N = 3a + 5b

consider gcd(3a,5b)=1, then "split" a and b obtaining a = a' + a", b = b' + b"

N = (3a' + 5b') + (3a" + 5b")

If we suceed to prove that one of these pairs (3a' + 5b') + (3a" + 5b"), when we test all the possible values for a', a", b' and b", are exactly two prime numbers, for every even number, the conjecture is proved.

Of course this is far way of a proof for the conjecture, but should give us another angle to think about the problem.

I made some observations but nothing pretty much beyond this point.

 

[HallsofIvy]

In other words you have just changed to a slight variation, which looks to me to be more complicated and harder to prove than the original. You do understand, don't you, that testing "all the possible values for a', a", b', and b" " is no more possible than checking every even number?

 

[al-mahed]

In other words you have just changed to a slight variation, which looks to me to be more complicated and harder to prove than the original. You do understand, don't you, that testing "all the possible values for a', a", b', and b" " is no more possible than checking every even number?

Let me try to explain a little bit better.

What I meant saying "all the possible values for a', a", b', and b" " is: try to find some property for a', b' such that 3a' + 5b' is a prime number when 3a" + 5b" is a prime number too. Of course, as I said, this is far way a proof, which means that perhaps this is a dead-end. But this is not the same thing as "checking every even number".

Pick a even number, like 100.

100 = 3*5 + 5*17, then a=5 and b=17, let's split a and b in a'=3, a"=2, b'=8, b"=9

100 = (3*3 + 5*8) + (3*2 + 5*9)

of course gcd(a',b') = gcd(a'',b'') = 1 when both expressions are prime numbers, but this is a necessary condition, not enough (I don't know if the correct word in english is this...)

I was thinking on doing a kind of research to find the pairs of "a's" and "b's" of each expression such that their gcd() are both = 1, make a list, and try to find some pathern or property of the pairs with gcd()=1 when these pairs are prime numbers.

I know this is a little bit vague, but could be a start... perhaps that other conjecture or mine about the numbers of the p + 2^n form, when proved, could be useful

 

[CRGreathouse]

of course gcd(a',b') = gcd(a'',b'') = 1 when both expressions are prime numbers, but this is a necessary condition, not enough (I don't know if the correct word in english is this...)

"Necessary but not sufficient" was the phrase you were looking for. But I don't think that's true -- what about (3*2 + 5*4) + (3*2 + 5*1) = 37? gcd(4, 2) is not 1, but 37 is prime. So the condition appears to be neither necessary nor sufficient.

 

[al-mahed]

sorry, my mistake, the correct is gcd(3a',5b') = gcd(3a'', 5b'') = 1, it is a necessary condition to both expression represents both prime numbers, but obviously it is not sufficient

 

[al-mahed]

another point: what I mean wans't that (3a' + 5b') + (3a'' + 5b'') = prime, instead I meant that 3a' + 5b' = prime and 3a'' + 5b'' = prime each of then

Call N = 3a + 5b an even number. When we split a and b, the question is: looking at the several pairs (3a' + 5b') + (3a'' + 5b'')*, such that gcd(3a',5b') = gcd(3a'', 5b'') = 1, obtained when we vary the "a's" and "b's", is there at least one of these pairs such that 3a' + 5b' and 3a'' + 5b'' are prime numbers?

* I call a pair (3a' + 5b') + (3a'' + 5b'')

 

[CRGreathouse]

Ah, I misunderstood you, sorry.

What are you saying then? Are you trying to use this to disprove the Goldbach conjecture by showing that there is an even number that cannot be split into coprime pairs (a, b), (a', b')? Are you trying to lessen the search space for experimental verification of a finite portion of the conjecture?

 

[al-mahed]

I don't believe that the conjecture is wrong. If it is wrong, showing that there is an even number such that it cannot be split into coprime pairs might prove that the conjecture is false, yes.

Unfortunally, I don't believe that such even number exist.

I was thinking on another aproach, a little vague yet...

 

[CRGreathouse]

I'm just saying that a proof requires something that's sufficient, not something that's necessary. Any further thoughts on the problem?

 

[al-mahed]

I'm just saying that a proof requires something that's sufficient, not something that's necessary. Any further thoughts on the problem?

Ok, let's work a little bit more on our definitions.

if a given number n =3a+5b is such that gcd(3a,5b)=1, we cannot say that n is a prime number, right? for instance,

a=1 and b=11, gcd(3,55)=1 and n = 58,
a=12 and b=11, gcd(36,55)=1 and n = 91 = 7*13

then gcd(3a,5b)=1 is a necessary condition to n be a prime number, but not sufficient

Given a even number written of the form N = 3a + 5b

lets split this number like I said before in two other numbers m = 3a'+5b' and k = 3a''+5b'' such that N = m+k

if we prove that for an any N there is no m+k pairs such that gcd(3a',5b')=1 AND gcd(3a'',5b'')=1 then this is a proof that goldbach's conjecture is FALSE, since gcd(3a',5b')=1 AND gcd(3a'',5b'')=1 is a necessary condition to m AND k be prime numbers, since there is no prime numbers represented by 3a+5b such that gcd(3a,5b)>1 (note that all numbers >7 can be expressed in the 3a+5b form)

so this particular proof, in this particular case, requires a necessary condition, not a sufficient

 

[Dragonfall]

Has it not been proven that Goldbach's conjecture can not be proven in Peano arithmetic?

 

[al-mahed]

Has it not been proven that Goldbach's conjecture can not be proven in Peano arithmetic?

I don't know about it... you mean that someone have proved that Goldbach's conjecture is undecidable? If yes, the conjecture is true.

Could you give the link?

 

[Dragonfall]

I'm only guessing. There have been theorems about the natural numbers which are not provable in PA, but are provable in ZFC.

What do you mean by "If yes, the conjecture is true"?

 

[CRGreathouse]

What do you mean by "If yes, the conjecture is true"?

I imagine the meaning was thus, as I was going to post a similar note:

Suppose the proposition "There exists no positive number n such that P(n)" for some predicate P. If the proposition is independent of Peano arithmetic (or some other framework, as long as the framework is strong enough to state the predicate P and express all natural numbers), then it cannot be disproved in that system by definition. But no natural number n exists that makes P(n) true, since if it did then the proposition could be disproved with the counterexample P(n).

 

[al-mahed]

Any further thoughts on the problem?

I did search my notes about it, but unfortunally I lost the thing... I was just working on a approach about these ideas I had put here.

I noticed that almost all numbers such that gcd(3a,5b)=1 are primes or semi-primes (meaning two prime factors), but I don't know if I am wrong or not

This is important if there is some pattern, even statistical pattern, such that prime numbers are more closer (or near?) to semiprime numbers than other numbers, but I can't explain why without my notes

This is pretty much vague... if you have some insight about it reading this please post here (or tell me if you think that this will lead us to nowhere)

 

[Hurkyl]

I imagine the meaning was thus, as I was going to post a similar note:

Suppose the proposition "There exists no positive number n such that P(n)" for some predicate P. If the proposition is independent of Peano arithmetic (or some other framework, as long as the framework is strong enough to state the predicate P and express all natural numbers), then it cannot be disproved in that system by definition. But no natural number n exists that makes P(n) true, since if it did then the proposition could be disproved with the counterexample P(n).

An attractive thought, but it doesn't work out.

The most obvious problem is that "Goldbach's conjecture is undecidable in PA" means that "PA + Goldbach's Conjecture is false" is a consistent theory!


The more subtle problem is that the following could be possible:
ZFC is able to prove a particular finite ordinal x is a counterexample to Goldbach's conjecture.
ZFC is able to prove that PA is incapable of proving that x is a counterexample.

or similarly
ZFC is able to prove a particular finite ordinal x is a counterexample to Goldbach's conjecture.
ZFC is able to construct another model of PA where the corresponding x is not a counterexample to Goldbach's conjecture.

I imagine even more pathological things could happen.

 

[CRGreathouse]

Of course the reason I didn't post my original thoughts was that I was unable to properly cover all of the pathologies, and as the complexity rose I had trouble justifying it for a one-off answer to the question.

But I will admit it's hard to imagine these pathologies. It's like nonprinciple ultrafilters -- sure, ZFC has them, but you'll never run into one.

 

[CRGreathouse]

An attractive thought, but it doesn't work out.

I guess the only rigorous thing I can pull out of my statement is that if the Goldbach conjecture is independent of PA then no consistent system can (1) construct an explicit natural number counterexample to the Goldbach conjecture, or even (2) construct an explicit upper bound for such a counterexample. In the latter case PA could take this upper bound and with finite effort check all partitions of all natural numbers up to it.

 

[husseinshimal]

some ideas may help.

ok,here is an idea may help,
even number ends with alast digit from the left,(0,2,4,6,8).prime number ends with(1,3,7,9) except,5.now the sum of any 2 prime numbers must end with(0,2,4,6,8).the two statements are equivalent to each others.
another idea.the even numbers can be expressed as 2(p+q+1),p&q.are positive integers.now,prime numbers are odd and odd numbers can be expressed as,2p+1 or 2q+1,we get even numbers can be expressed as asum of two prime numbers.you may say that prime number cannot be expressed as,2p+1,because it includes all of the odd numbers.but the real question is,if the set,A,belongs to set,B,then not all the judgements that can be applied on,B,are applicable on,A,like countablity.if you can prove that the statement"ALL ODD NUMBERS CAN BE EXPRESSED AS,2p+1"can be applied on the prime numbers set.then the conjecture would be clear.i think you will have aproblen because our unwell defined statements seem to be tricky,for example,when i say,all even numbers can be expressed as 2n,i can also say,not all even numbers can be expressed as 2n.some of even numbers can be expressed as,3n+1.end of talk.

 

[CRGreathouse]

even number ends with alast digit from the left,(0,2,4,6,8).prime number ends with(1,3,7,9) except,5.

The rightmost digit of an even in base 10 is one of {0, 2, 4, 6, 8}, yes. For the primes you forgot to exclude 2: prime numbers other than 5 and 2 have a rightmost digit in base 10 of one of {1, 3, 7, 9}.

the two statements are equivalent to each others.

The statement that the sum of any two primes greater than 2 is even follows, but that doesn't show Goldbach's (strong) conjecture, which is that any even number can be additively split into two primes.

For example, consider n = 10^1000. n - 2 is composite, n - 3 is composite, n - 5 is composite, ..., n - (10^1000 - 1769) is composite. If all the omitted numbers in the ... are composite as well, then Goldbach's conjecture fails. Sure, all the numbers in the middle (all numbers except n - 2) are odd, but that doesn't help.

the even numbers can be expressed as 2(p+q+1),p&q.are positive integers.

Okay, all even numbers greater than 4 can be expressed in this form, sure.

now,prime numbers are odd and odd numbers can be expressed as,2p+1 or 2q+1,we get even numbers can be expressed as asum of two prime numbers.

Backward again. Yes, each odd prime can be expressed in the form 2n+1 for some positive integer n, but that doesn't mean that n = p for the p you've already used. It could be a different positive integer.

if you can prove that the statement"ALL ODD NUMBERS CAN BE EXPRESSED AS,2p+1"can be applied on the prime numbers set.then the conjecture would be clear.

No. All primes greater than 2 can be expressed in the form 2n+1 for some positive integer n, but that doesn't imply the Goldbach conjecture -- it's not nearly a proof.

i think you will have aproblen because our unwell defined statements seem to be tricky,for example,when i say,all even numbers can be expressed as 2n,i can also say,not all even numbers can be expressed as 2n.some of even numbers can be expressed as,3n+1.

All even numbers can be expressed as 2n for some integer n. Some even numbers can also be expressed as 3n+1 for some integer n -- these are precisely the integers congruent to 4 mod 6.

 

[Dragonfall]

ZFC is able to prove a particular finite ordinal x is a counterexample to Goldbach's conjecture.

ZFC could prove that such a counterexample exists without constructing it. In terms of finite sets/numbers, how strong is ZFC compared to PA? I'm trying to remember the theorem that was proved in ZFC but now PA; it involved constructing a sequence by taking the previous number, using it to get an expression involving towering powers of 2, then adding 1 -- I think -- and showing that the sequence eventually ends at 1.

 

[Hurkyl]

http://en.wikipedia.org/wiki/Goodstein's_theorem

 

[Dragonfall]

Yes that's it!

 

[JasonRox]

Wow, that's the first I ever heard of Goodstein's Theorem. Really interesting.

I derived a sequence out of Goldbach's Conjecture awhile back, and it had a clear pattern. A simple one too. I tried running the program the other day that created the sequence and it wasn't working (been 4 years and newer computers). I'll pull out the code and put it in C++. Send it in for publication and let you guys know about it.

(I'm really slow for publishing stuff. I'm just typing out an alternative proof of Tychonoff's Theorem that I wrote last year! I'm excited to finish it though.)

 

[asar]

Goldbach's Conjecture is "true"!

Goldbach's Conjecture is "True". The problem is that "what is true is contestable. There are problems like what is a prime number,etc. A simple case is mathematical induction proof for the first even number 2, then n=6 another even number and its successor also true: 2, 6, 8.

2(1,1), 6(1,5;3,3), 8(1,7;3,5).

My approach to "proving" Goldbach is the quadratic equation. The non complex roots of the quadratic equation are the sum of two real conjugates. These roots cover all the real numbers and includes the primes. The quadratic equation includes all the real numbers.

The problem with proving Goldbach is that most people consider the issue of primes as the key. Goldbach's Conjecture is not about proving a number is prime, it is about finding primes that sum to a even number. Proving a number is prime is another issue. The Goldbach problem given an even number can two primes that sum to the number always be found. The answer is yes using the quadratic equation.

x^2 + (p + p')x + pp' = x^2 + (Even-number)x + pp'=0

Proof: p=s +a, p'=s -a then x^2 + (p + p') +pp'= x^2 (s+a +s-a)x + s^2 -a^2=0
x^2 + 2sx + (s^2 -a^2)= 0 GIVEN 2s, find 'a' such that p and p' prime. There is always a solution in the quadratic equation.

If the qudratice equation is true, the Goldbach Conjecture is true!

 

[CRGreathouse]

Proof: p=s +a, p'=s -a then x^2 + (p + p') +pp'= x^2 (s+a +s-a)x + s^2 -a^2=0
x^2 + 2sx + (s^2 -a^2)= 0 GIVEN 2s, find 'a' such that p and p' prime. There is always a solution in the quadratic equation.

If the qudratice equation is true, the Goldbach Conjecture is true!

If you choose your p and p' to be arbitrary primes, there's no reason to think that you will cover all possible values of s. If you choose s arbitrarily, there's no reason to think that there is a prime p for which p' is prime. Further, your proof does not use the fact that s > 1, which requires the absurdity of two primes adding to 2.

 

[asar]

"Further, your proof does not use the fact that s > 1, which requires the absurdity of two primes adding to 2."

The absurdity is that the first prime is not considered a prime by some. 1 + 1 =2 two primes 1+1 =2.

Thanks for your comments. Why choose s arbitraily. S is half the even number, the object for the primes are sought.

 

[CRGreathouse]

Thanks for your comments. Why choose s arbitraily. S is half the even number, the object for the primes are sought.

Choosing s arbitrarily is the right way, no problem there. But your proof does not show that either p or p' is prime. Now it's easy to show that one of p or p' can be chosen to be prime for s >= 1, by the infinitude of the primes and the fact that 2 is prime, respectively. But there's no reason to think that they can be chosen so that both are prime, and your proof does not even attempt to show that they are prime.

 

[namus]

Hi Dragonfall,

What al-mahed means is that for a question like goldbach conjecture proving undecidability is as hard as proving the theorem itself. Well, that is a bit erroneous, in fact proving it is undecidable is in fact fallacious. For if the conjecture is false there exists a proof (the very number that is not sum of two primes, together with the list of primes upto it and their pairwise sums, and pair-by-pair verification that neither of the sums is the number, establishes a (possibly very long) proof of its falsity. So when you prove that it is undecidable, you consequently prove that is it is true, and thereby contradict your own theorem.

And the above is a proof that it cannot be proven undecidable. Is this too confusing?

However, there exists a faint possibility of the following (let us go into terms more concrete that "undecidable" which can mean "neither can it be proven true, nor untrue", and just consider statements like "xyz can be proven true"):

Consider the theorems P_0, P_1, P_2, ...

where P_0 = the goldbach conjecture.

and P_{i+1} = "The statement P_{i} cannot be proven true".

Now there is a possibility that all the statements P_0, P_1, ... are true, but therefore, none of them can be established.

The morale of the story is that if you want to enter into any investigation around goldbach conjecture, you better invest your time for the conjecture itself, no questions on its decidability, and decidability thereof in higher order, can be attained-- if true. :)

 

[namus]

I am terribly sorry. I seem to have posted a comment into the wrong thread... multitab parallel processing... :) sorry guys.

 

[namus]

Hi Dragonfall,

What al-mahed means is that for a question like goldbach conjecture proving undecidability is as hard as proving the theorem itself. Well, that is a bit erroneous, in fact proving it is undecidable is in fact fallacious. For if the conjecture is false there exists a proof (the very number that is not sum of two primes, together with the list of primes upto it and their pairwise sums, and pair-by-pair verification that neither of the sums is the number, establishes a (possibly very long) proof of its falsity. So when you prove that it is undecidable, you consequently prove that is it is true, and thereby contradict your own theorem.

And the above is a proof that it cannot be proven undecidable. Is this too confusing?

However, there exists a faint possibility of the following (let us go into terms more concrete that "undecidable" which can mean "neither can it be proven true, nor untrue", and just consider statements like "xyz can be proven true"):

Consider the theorems P_0, P_1, P_2, ...

where P_0 = the goldbach conjecture.

and P_{i+1} = "The statement P_{i} cannot be proven true".

Now there is a possibility that all the statements P_0, P_1, ... are true, but therefore, none of them can be established.

The morale of the story is that if you want to enter into any investigation around goldbach conjecture, you better invest your time for the conjecture itself, no questions on its decidability, and decidability thereof in higher order, can be attained-- if true. :)

 

[ramsey2879]

sorry, my mistake, the correct is gcd(3a',5b') = gcd(3a'', 5b'') = 1, it is a necessary condition to both expression represents both prime numbers, but obviously it is not sufficient

Sorry but not all primes can be expressed in the form 3a+5b, in particular 7 can not be put into that form and (5+7 = 12) is not the sum of primes according to this form.

 

[johnny_boy]

An attractive thought, but it doesn't work out.

The most obvious problem is that "Goldbach's conjecture is undecidable in PA" means that "PA + Goldbach's Conjecture is false" is a consistent theory!

I know this is super old school, but in case someone else finds this thread, I thought it might be worth clarifying this highly interesting discussion.

It's true that if GC is independent of PA then it's true. It's also true that PA + ~GC is consistent, but that's irrelevant to the previous fact. Consider a usual Godel sentence G, saying of itself that it's not provable. Well clearly G is both independent of PA and true (!) even though PA + ~G is consistent. Of course, any model for PA + ~G must be non-standard since there can be no standard witness to ~G (which is equivalent to a Sigma_1 formula).

EDIT: I thought it might be helpful to give an informal proof of why GC is true if independent. Suppose GC is independent and, for reductio, false. Then there's an n that is a counterexample to GC. Note that GC is Sigma_2, having the form ExAyAzF(n), so AyAzF(n/x) is true. But given F, obviously the universal quantifiers Ay and Az can be bound by n, and clearly Ay<nAz<nF(n/x) is decidable, whence PA-provable.

 

[Hurkyl]

It's true that if GC is independent of PA then it's true.

... for the finite ordinals, or some other isomorphic model. (in the set theory we've used to prove it)

It always bothers me when someone says "it's true" in a situation that could be interpreted as "it's true for the formal theory under consideration".

But given F, obviously the universal quantifiers Ay and Az can be bound by n, and clearly Ay<nAz<nF(n/x) is decidable, whence PA-provable.

I think your plan is basically to, by external induction on n, internally prove that

Ay<n: P(y)​

is equivalent to

\bigwedge_{y = 0}^{n-1} P(y)​

(both of which make sense because n is a finite ordinal and our formal logic allows up-to-some-finite-ordinal-indexed operations)

And so the claim "n is a counterexample to Goldbach's conjecture" can be proven equivalent to an (externally) finite conjunction of arithmetic equations involving only finite ordinals, which can then be straightforwardly proven true.

And so we have the (external) theorem "If a finite ordinal n is a counterexample to Goldbach's conjecture, then there is a proof in PA of that fact".

I think I'm happy with the argument...


(P.S. I've had no luck finding an internet reference on the properties of Sigma_n sentences. Do you know of one?)

 

[TylerH]

[English] Re: Goldbach's Conjecture (Theorem??)

not enough (I don't know if the correct word in english is this...)

"enough" makes sense, but "sufficient" is more often used. For example, if a \Rightarrow b, then you could say "a is necessary, but not sufficient for b."

I'm not picky. What you used was fine. I just like to help non-native speakers (and native speakers who didn't pay attention in language class). :)

 

[johnny_boy]

... for the finite ordinals, or some other isomorphic model. (in the set theory we've used to prove it)

It always bothers me when someone says "it's true" in a situation that could be interpreted as "it's true for the formal theory under consideration".

In the case of arithmetic "it's true" means "it's true in the standard model". I don't know anyone else who uses it differently. Likewise, for any theory that has a standard model, whenever someone says it's true relative to that theory, they almost always mean true in the standard model of the theory.

(P.S. I've had no luck finding an internet reference on the properties of Sigma_n sentences. Do you know of one?)

Look http://en.wikipedia.org/wiki/Arithmetical_hierarchy" .