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#1 2014-08-16 15:45:10

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

I have proved Goldbach's theorem can someone check it for me?

For Goldbach's Theorem it states that any even no. can be made up of two primes.
e =the even no. then if there are more than 1 prime occurring in every square root(e) then the probability of this being true is 1/square root(e) timesed by 1/square root(e)
so if the gap between each prime is always smaller than sqrt(e) minus 1, then goldbach's conjecture will be true.
This is the case because;

2+/-1..
6+/-1 5..
30+/- 1 7..
210 +/- 1 11..
2310 +/- 1 13..

equals primes......to a certain point above highest prime squared. e.g.>9,25,49,121 or 169 accordingly. And so on but the numbers get very big but you can do this up to infinity.

As you can see the biggest gap occurs around the beginning which could possibly be bigger than e but this is not the case because those gaps are filled with primes normally. The second time this gap will occur is around the previous prime squared. This is because the sets are interlinked where if I wanted the next set after 2+1 all I have to do is repeat it 3 times and remove any no.'s in the previous set x3 i.e. 1 x 3 = 3.... to get 1 and 5. So as there could be a gap bigger than e - 1 that is not a problem if you ignore no.'s around that gap and think that the rest of the primes will have a density needed to create e then goldbach's conjecture is true. Especially if you think there is only one gap and not two so they can't affect two primes.

Last edited by Primenumbers (2014-08-16 16:57:48)


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#2 2014-08-16 20:05:38

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: I have proved Goldbach's theorem can someone check it for me?

hi Primenumbers

Well it would be marvellous if you've done it, and thanks for posting it here first.  But, to stand up to scrutiny from mathematicians far cleverer than me, you'll have to fill in all the details.

Here's a few things that seem vague to me at the moment (apologies if it's just my limited brain size):

occurring in every square root(e)

If this an interval ? 

probability of this being true is 1/square root(e) timesed by 1/square root(e)

Can you justify this ?

2+/-1..
6+/-1 5..
30+/- 1 7..
210 +/- 1 11..
2310 +/- 1 13..

I've no idea what you are saying here.  Are several lines of explanation missing ?

To show this proof fails, one would only have to find an 'e' for which this procedure falls apart.  So you have to show in a 'watertight' way that no one can do this.

Good luck with it.  smile

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2014-08-16 21:07:22

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

Re: I have proved Goldbach's theorem can someone check it for me?

(1) All I mean is; say e = a + b and there was only 1 prime that could = a and one prime that could = b. Then there would be only one occurrence of this happening in e and that's all we are looking for...at least one.


(2) Secondly (1/sqrt(e))*(1/sqrt(e)) would equal this.
If you round down......then you will equal a No. greater than sqrt(e)...........which means you can then say that the percentage must equal this. I can explain it as follows; if there were one prime occurring every 5 no.'s in the whole of 100. Then when I half 100..I get 50 and if there's one occurring every 5 No.'s on one side and one prime occurring every 5 No.'s on the other. Then +/- c = a or b, and a and b must occur because the probability of c causing a and b to equal primes is the above probability. There's a physics formula where if the No. of trees in a forest reaches a certain percentage, then if one of the trees catches fire they all do.


(3) Thirdly the last question is really complicated to explain but is all true:

To get a No. not factorable by a bunch of primes, you have to get those numbers multiplied together. Then minus....or add, a no. NOT factorable by them. Once you've grasped that, it's simple, to get these formula's such as 2m +/- 1 or 6m+/- 1 or 5 you just repeat the formula (x) no. of times and then remove all no.'s factorable by x....which will be exactly the PREVIOUS set multiplied by x. All primes will be found in < ((the next prime above (x))squared) otherwise they are at risk of being factorable by primes>x.

2m+/-1
6m+/-1 or 5
30m+/- 1 or 7 or 11 or 13 or 17 or 19 or 23 or 29
210+/- 1 or 11......all no.'s not factorable by 2,3,5 or 7....(therefore prime if<11 squared)


(4) Also the gap that I talked about alters the percentage possibly, but everywhere around that gap will have the correct percentage. e +/- c = a or b, how big does c have to be? sqrt(e) rd. dwn. I think because then you will have enough occurences for probability to take effect. This is just right because the big gap will occur exactly at (x) squared. So maybe round e down to the nearest PRIME and it's true.

Thanks.


"Time not important. Only life important." - The Fifth Element 1997

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#4 2014-08-16 21:34:19

anonimnystefy
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From: Harlan's World
Registered: 2011-05-23
Posts: 16,049

Re: I have proved Goldbach's theorem can someone check it for me?

I don't think "proving it" probabilistically does not actually prove it, just shows it's true in many cases, unless I'm missing something.


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#5 2014-08-16 22:52:55

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

Re: I have proved Goldbach's theorem can someone check it for me?

Let e = a + b where e is any even no.
                    where a is any prime no.
                    where b is any prime no.

((sqrt(e)) rd. dwn. to nearest prime) = z

In (z)squared there must be 1 occurrence of two primes adding up to e.
In (z) squared the biggest gap will be (z - 1) apart from around (z)squared where the gap will be bigger. Because for all gaps to contain factors of primes up to but not including z must be (z-1) at the biggest. (See earlier post).

With primes occurring every z numbers, if a is prime there is a 1/z chance and for b to be prime as well there is a (1/z)*(1/z) chance which = 1/((z)squared). So this will occur once in (z)squared. So e=a+b is true.

We can use probability in this case because it is definite that primes occur every z no.'s not that they probably will, we know they do.


"Time not important. Only life important." - The Fifth Element 1997

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#6 2014-08-16 23:03:58

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: I have proved Goldbach's theorem can someone check it for me?

Hi;

But can you say

Let e = a + b where e is any even no.

? That is what you are trying to prove.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#7 2014-08-16 23:15:33

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: I have proved Goldbach's theorem can someone check it for me?

hi Primenumbers

I am trying to follow this, but it still isn't making sense.

(1) All I mean is; say e = a + b and there was only 1 prime that could = a and one prime that could = b. Then there would be only one occurrence of this happening in e and that's all we are looking for...at least one.

That's just a statement of the conjecture.

Secondly (1/sqrt(e))*(1/sqrt(e)) would equal this.

Whatever is "this" ?  That's a fraction, and the previous things referred to was e, a, and b, all of which are integers.

To get a No. not factorable by a bunch of primes

Don't try to do that.  It has been proved that all positive integers > 1 can be factored onto prime factors.

With primes occurring every z numbers,

Really!  Have you read this:

http://en.wikipedia.org/wiki/Prime_number_theorem

Wiki wrote:

primes become less common as they become larger.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#8 2014-08-16 23:58:41

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

Re: I have proved Goldbach's theorem can someone check it for me?

I think I might have made a mistake which makes things simpler:
For No.'s<(z)squared the greatest gap between two primes will be (z - 1).
This is because the greatest prime we are concerned with is not z, it is the prime below z. Therefore you do not need to be concerned with remainders that are as big as z.

e.g. for primes <49 need only have remainders; when you try to factor primes 2,3 and 5.
You need not be concerned with a remainder >6 because; that would incorporate 7. We are not concerned with the prime 7 having a remainder at all.

Two numbers that add up to e can be prime because there is 1/z occurences of primes in (z)squared therefore the occurences of two random numbers being prime is 1 in (z)squared.


"Time not important. Only life important." - The Fifth Element 1997

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#9 2014-08-17 00:06:11

Bob
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Registered: 2010-06-20
Posts: 10,053

Re: I have proved Goldbach's theorem can someone check it for me?

hi Primenumbers

I have a suggestion which may help to take us forward.

Whenever I encounter some difficult, abstract maths, that I cannot grasp, I make up some examples to see how it goes.  I always find it easier with numbers than algebra. 

So, I'll choose an even integer (e) .

You describe carefully with a step by step algorithm, how you can find a and b.

If I've understood what you did, I'll try it for myself with some more numbers.  If it works every time, then it adds strength to your argument.  If it doesn't, then it's 'back to the drawing board' I afraid.  sad

I expect other forum members will be willing to put the algorithm to the test too.  smile

OK ?

If so, then my first choice is e = 268

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#10 2014-08-17 01:46:49

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

Re: I have proved Goldbach's theorem can someone check it for me?

INPUT e
(STEP 1) Take e, sqrt it AND THEN Rd. dwn. to nearest prime:
(STEP 2) Select a prime <e at random and minus it from e.
            IF NON-PRIME THEN REPEAT UP TO A MAXIMUM OF, (sqrt(e) rd. dwn. to nearest prime) TIMES AND IF PRIME THEN STOP AND OUTPUT.

INPUT 268
(STEP 1)16.37 (2d.p.) AND 13
(STEP 2)1st go: 268 - 23 = 245 (a non-prime) REPEAT
             2nd go: 268 -167 = 101 (prime) STOP OUTPUT (268 -167=101)

Note: The smallest probability will occur when two primes selected from <289 and >169 (Probability =1/169)
Probability from lower ranges will give a higher probability than 1/169 because primes occur more often in lower ranges.


"Time not important. Only life important." - The Fifth Element 1997

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#11 2014-08-17 01:54:49

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: I have proved Goldbach's theorem can someone check it for me?

Hi;

I think it is important to remember that no one has ever found any even number greater than 2 that was not representable as the sum of two primes. Even though searches have gone to very large numbers. This does not prove that there is not some even number yet undiscovered that fails.

It might be unfeasible to try to find a number where your algorithm fails to get a and b but that does not prove that it will always work.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#12 2014-08-17 02:19:07

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

Re: I have proved Goldbach's theorem can someone check it for me?

The Algorithm will always work because 13 primes with the biggest gap between them being 13-1=12 means that 13 squared (169) is the biggest range you would need to work with to find one occurrence of a and b.


"Time not important. Only life important." - The Fifth Element 1997

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#13 2014-08-17 02:39:50

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: I have proved Goldbach's theorem can someone check it for me?

That may not be true. Like I said it has been verified computationally to very large numbers but no one knows why it will never fail. Just because there are some primes in some range does not mean that they will add up to your even number.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#14 2014-08-17 03:30:15

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: I have proved Goldbach's theorem can someone check it for me?

hi

The algorithm:

INPUT 268
(STEP 1)16.37 (2d.p.) AND 13 


(STEP 2)1st go: 268 - 23 = 245 (a non-prime) REPEAT
             2nd go: 268 -167 = 101 (prime) STOP OUTPUT (268 -167=101)

So step 2 consists of "Keep trying random primes for 'a' until we get a 'b' that is also a prime."

That's disappointing. I had hoped for an algorithm that went straight for an 'a' and 'b' that works.  Not just keep trying until you find one.  What happens if I don't?

And what is step 1 for, since it doesn't feature again.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#15 2014-08-17 03:35:59

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: I have proved Goldbach's theorem can someone check it for me?

For large numbers like 901234172653788488484888999912 it will be horrendously slow. This means finding a counterexample is almost impossible. So we have a conjecture that is extremely difficult to prove and even harder to find a counterexample to. Not too promising.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#16 2014-08-17 04:31:13

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

Re: I have proved Goldbach's theorem can someone check it for me?

But if there is a certain no. of primes in a certain range doesn't that mean that the proportion of primes is high enough to create a and b?
PROOF:
(step 1) Take 169 write it down on a piece of paper (1,2,3,4,5,6,7.........169) not literally....
(step 2) Make at least 13 holes with the biggest gap being 12 between them. These are the occurences of (a) which are unknown.
(step 3) Fold the piece of paper 13 times these are the occurences of (b) which are unknown. Then un-ravel it and look at the creases...will any of the columns not have at least one hole in it.........no. Neither will the creases.


"Time not important. Only life important." - The Fifth Element 1997

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#17 2014-08-17 04:43:56

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: I have proved Goldbach's theorem can someone check it for me?

I am not following you. How does that imply that there will be two primes that sum to e?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#18 2014-08-17 07:26:02

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

Re: I have proved Goldbach's theorem can someone check it for me?

If when (a) is prime, (b) always isn't, that would mean that (b) would have a prime within 12 numbers of b and not more. If it did have a bigger gap than 12 then there would not be a prime occurring every 13 numbers.
If primes are occurring within 12 numbers of b and a is occurring 13 times in 169, then non - primes (b) = (h,i,j,k,l,m,n,p,r,t,u,v,w,) fill in each gap with a prime and you get 14 primes i.e.don't forget h minus up to 12. Therefore one gap must =0 for there to be 13 primes and not 14. I.e. there must be at least one occurrence of (b) being prime when (a) is.

Last edited by Primenumbers (2014-08-17 07:26:50)


"Time not important. Only life important." - The Fifth Element 1997

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#19 2014-08-17 13:21:47

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: I have proved Goldbach's theorem can someone check it for me?

(b) would have a prime within 12 numbers of b and not more.

That is not true, the primes become fewer the higher you go.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#20 2014-08-17 16:25:30

Primenumbers
Member
Registered: 2013-01-22
Posts: 149

Re: I have proved Goldbach's theorem can someone check it for me?

If Goldbach's conjecture is false and in the occurrence of e= a+b, (e) being any even no. (a) is always prime and (b) is always non-prime. The below statement is always true:

"For any area of (y)squared or greater there are y no. of primes separated by <y sized gaps."

This is not true.

For the high No.'s >(y) squared these bigger gaps are eradicated by the smaller one's at the beginning. (e)/2 will never = <(y)squared where y is the biggest prime square <e.


"Time not important. Only life important." - The Fifth Element 1997

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