"PROOF" OF GOLDBACH'S CONJECTURE

Awojobi · 2019-06-20, 11:23

PROOF OF GOLDBACH’S CONJECTURE

Goldbach’s conjecture states that every even integer > 2 is the sum of 2 primes. An example is 100 = 3 + 97 = 11 + 89 = 17 + 83 = 29 + 71 = 41 + 59 = 47 + 53. An alternative statement of the conjecture is as follows. Every integer > 3 is the arithmetic mean of 2 primes. This means that every integer > 3 is midway between 2 primes. For instance, 9 is midway between 5 and 13. 9 is also midway between 7 and 11.

Imagine a finite part of the number line up to some integer, N. Let this number line be divided up into roughly 4 equal parts. If it can be shown that every integer in the 2nd quarter of this number line is the arithmetic mean of 2 primes, then Goldbach’s conjecture is proved. So prime number 3 (in the 1st quarter of the number line) and primes mainly in the 3rd and 4th quarters of the number line will produce arithmetic means in the 2nd quarter of the number line. The next prime number, 5 (in the 1st quarter of the number line) and primes mainly in the 3rd and 4th quarters of the number line will produce arithmetic means in the 2nd quarter of the number line. Some of these arithmetic means will be repetitions of those produced using prime number 3. The number of new arithmetic means produced by prime number 5 in the 2nd quarter of the number line is equal to the number of arithmetic means produced by prime number 5 minus the number of twin primes (5 – 3 = 2) mainly in the 3rd and 4th quarters of the number line. The next prime number, 7 (in the 1st quarter of the number line) and primes mainly in the 3rd and 4th quarters of the number line will produce arithmetic means in the 2nd quarter of the number line. Some of these arithmetic means will be repetitions of those produced using prime number 3 and prime number 5. The number of new arithmetic means produced by prime number 7 in the 2nd quarter of the number line is equal to the number of arithmetic means produced by prime number 7 minus the number of twin primes (5 – 3 = 2), mainly in the 3rd and 4th quarters of the number line, minus the number of primes that differ by 4 (7 – 3 = 4), mainly in the 3rd and 4th quarters of the number line. This process carries on until all arithmetic means in the 2nd quarter are produced and any further process will produce only repetitions of arithmetic means.

The phrase, ‘mainly in the 3rd and 4th quarters of the number line’, is used because primes in the second quarter will also be involved in producing arithmetic means in the second quarter. The mathematical justification of the last but one sentence will now be discussed.
The number of primes less than N is approximately equal to N / log e N. This is a lower bound for N ≥ 17. An upper bound for the number of prime difference, h (not necessarily consecutive primes) < N is 8N / (log e N)^2. For instance, for the twin primes, h = 2 and therefore the number of twin primes < N cannot exceed 8N / (log e N)^2. This can be shown using sieve theory. It should be clear from the discussion so far that for each cycle of production of arithmetic means, the primes involved in the 2nd, 3rd and 4th quarters are such that the difference between the largest and smallest prime is approximately N/2. So, prime number 3 uses approximately N / log e N – [N/2]/[ log e (N/2)] primes to produce the same number of arithmetic means in the 2nd quarter of the number line. For large enough N, N / log e N – [N/2]/[ log e (N/2)] can be approximated to [N/2]/[ log e (N/2)] since log e 2 = 0.693, which is small compared to log e N. For large enough N, approximately [N/2]/[ log e (N/2)] primes in the 2nd, 3rd and 4th quarters will be used for each cycle. Let c be the number of cycles required to produce all the N/4 arithmetic means in the 2nd quarter. The following equation can now be formed using the upper bound 8N / (log e N)^2.
c[N/2]/[ log e (N/2)] – 4cN / (log e N)^2 = N/4
The approximate number of cycles, c, required using an upper bound of 8N / (log e N)^2 is
c ≈ [ log e (N/2)]/2

So the number of cycles required (which is the same as the number of primes required in the first quarter) to produce all N/4 arithmetic means is much less than the number of primes in the first quarter i.e. c ≈ [ log e (N/2)]/2 < [N/4]/[ log e (N/4)]. This proves that Goldbach’s conjecture is true.

2M215856352p1 · 2019-06-20, 14:55

This 'proof' looks more like a heuristic argument at first glance.

2M215856352p1 · 2019-06-20, 15:33

Quote:

An upper bound for the number of prime difference, h (not necessarily consecutive primes) < N is 8N / (log e N)^2. This can be shown using sieve theory.

This part is not clear. Can you please explain how it can be shown using sieve theory? Frankly, I cannot understand sieve theory but this seems to be a major issue in the attempted proof.

Awojobi · 2019-06-20, 15:40

I don't understand sieve theory myself. However, I got this upper bound from Terence Tao's blog. This proof is not a heuristic proof. It is its simplicity that probably makes it seem that way.

a1call · 2019-06-20, 16:15

I do not know if the proof is valid or not. However, personally I like your approach in redefining the conjecture.
I suggest that you define the zones more precisely and in terms of arbitrary values such as N.
I think a more precise parametric/formulaic argument will either show if there are holes to be patched or not.

a1call · 2019-06-20, 21:01

Use of the word "roughly" is problematic in a mathematical proof.
Why not say that for any given positive integer N,
Set 1 is the set of all integers 1 to N
Set 2 is the set of all integers N+1 to 2N
Set 3 is the set of all integers 2N+1 to 4N

Now if you can prove that for any given N, all the integers in set 2 are arithmetic means of 2 prime numbers which are members of sets 1 and 3, then you have indeed proven the conjecture.
There might be such a proof in OP, but it's too long of a post for my attention span to look for it.

2M215856352p1 · 2019-06-21, 00:17

Quote:

Originally Posted by a1call

Use of the word "roughly" is problematic in a mathematical proof.
Why not say that for any given positive integer N,
Set 1 is the set of all integers 1 to N
Set 2 is the set of all integers N+1 to 2N
Set 3 is the set of all integers 2N+1 to 4N

Now if you can prove that for any given N, all the integers in set 2 are arithmetic means of 2 prime numbers which are members of sets 1 and 3, then you have indeed proven the conjecture.
There might be such a proof in OP, but it's too long of a post for my attention span to look for it.

Not true, when N=10, 11 is in Set 2 and there is no valid pair which satisfy the condition.
The prime numbers in Set 1 would be 2,3,5 and 7 and the corresponding integers 20,19,17 and 15 are all in Set 2.

a1call · 2019-06-21, 00:49

Quote:

Originally Posted by 2M215856352p1

Not true, when N=10, 11 is in Set 2 and there is no valid pair which satisfy the condition.
The prime numbers in Set 1 would be 2,3,5 and 7 and the corresponding integers 20,19,17 and 15 are all in Set 2.

Yes, that is true. For the record the error is of my introduction and not the OP, since it does not state that the primes can not belong to the 2nd quadrant.

Quote:

Originally Posted by Awojobi

Imagine a finite part of the number line up to some integer, N. Let this number line be divided up into roughly 4 equal parts. If it can be shown that every integer in the 2nd quarter of this number line is the arithmetic mean of 2 primes, then Goldbach’s conjecture is proved.

Regardless, I can't see the point of defining quadrants without using them as ranges of some sort. But then again my comprehension is of no significance here.

2M215856352p1 · 2019-06-21, 01:25

Quote:

Originally Posted by Awojobi

I don't understand sieve theory myself. However, I got this upper bound from Terence Tao's blog. This proof is not a heuristic proof. It is its simplicity that probably makes it seem that way.

Up till this point, the proof seems ok. After that, the OP seems to have made assumptions (e.g. the primality of an odd number is independent of another odd number) and made approximations which he/she has yet to prove rigorously. Hence, the claimed proof appears to be a heuristic argument at best.

Awojobi · 2019-06-21, 09:19

I have not made any assumptions. All I have done is make approximations that are valid. Valid, in the sense that upper bounds of repetitions and lower bound of the number of primes less than N have been used in modelling the scenario. So, basically, in my modelling, for each cycle, less arithmetic means are being produced than the actual number of arithmetic means produced and yet all arithmetic means will be produced, if that makes any sense.

2M215856352p1 · 2019-06-21, 11:49

Quote:

Originally Posted by Awojobi

I have not made any assumptions. All I have done is make approximations that are valid. Valid, in the sense that upper bounds of repetitions and lower bound of the number of primes less than N have been used in modelling the scenario. So, basically, in my modelling, for each cycle, less arithmetic means are being produced than the actual number of arithmetic means produced and yet all arithmetic means will be produced, if that makes any sense.

May I clarify about what you mean by cycles? This is unclear to me.

2019-06-20, 11:23	#1
Awojobi Feb 2019 94₁₀ Posts	"PROOF" OF GOLDBACH'S CONJECTURE PROOF OF GOLDBACH’S CONJECTURE Goldbach’s conjecture states that every even integer > 2 is the sum of 2 primes. An example is 100 = 3 + 97 = 11 + 89 = 17 + 83 = 29 + 71 = 41 + 59 = 47 + 53. An alternative statement of the conjecture is as follows. Every integer > 3 is the arithmetic mean of 2 primes. This means that every integer > 3 is midway between 2 primes. For instance, 9 is midway between 5 and 13. 9 is also midway between 7 and 11. Imagine a finite part of the number line up to some integer, N. Let this number line be divided up into roughly 4 equal parts. If it can be shown that every integer in the 2nd quarter of this number line is the arithmetic mean of 2 primes, then Goldbach’s conjecture is proved. So prime number 3 (in the 1st quarter of the number line) and primes mainly in the 3rd and 4th quarters of the number line will produce arithmetic means in the 2nd quarter of the number line. The next prime number, 5 (in the 1st quarter of the number line) and primes mainly in the 3rd and 4th quarters of the number line will produce arithmetic means in the 2nd quarter of the number line. Some of these arithmetic means will be repetitions of those produced using prime number 3. The number of new arithmetic means produced by prime number 5 in the 2nd quarter of the number line is equal to the number of arithmetic means produced by prime number 5 minus the number of twin primes (5 – 3 = 2) mainly in the 3rd and 4th quarters of the number line. The next prime number, 7 (in the 1st quarter of the number line) and primes mainly in the 3rd and 4th quarters of the number line will produce arithmetic means in the 2nd quarter of the number line. Some of these arithmetic means will be repetitions of those produced using prime number 3 and prime number 5. The number of new arithmetic means produced by prime number 7 in the 2nd quarter of the number line is equal to the number of arithmetic means produced by prime number 7 minus the number of twin primes (5 – 3 = 2), mainly in the 3rd and 4th quarters of the number line, minus the number of primes that differ by 4 (7 – 3 = 4), mainly in the 3rd and 4th quarters of the number line. This process carries on until all arithmetic means in the 2nd quarter are produced and any further process will produce only repetitions of arithmetic means. The phrase, ‘mainly in the 3rd and 4th quarters of the number line’, is used because primes in the second quarter will also be involved in producing arithmetic means in the second quarter. The mathematical justification of the last but one sentence will now be discussed. The number of primes less than N is approximately equal to N / log e N. This is a lower bound for N ≥ 17. An upper bound for the number of prime difference, h (not necessarily consecutive primes) < N is 8N / (log e N)^2. For instance, for the twin primes, h = 2 and therefore the number of twin primes < N cannot exceed 8N / (log e N)^2. This can be shown using sieve theory. It should be clear from the discussion so far that for each cycle of production of arithmetic means, the primes involved in the 2nd, 3rd and 4th quarters are such that the difference between the largest and smallest prime is approximately N/2. So, prime number 3 uses approximately N / log e N – [N/2]/[ log e (N/2)] primes to produce the same number of arithmetic means in the 2nd quarter of the number line. For large enough N, N / log e N – [N/2]/[ log e (N/2)] can be approximated to [N/2]/[ log e (N/2)] since log e 2 = 0.693, which is small compared to log e N. For large enough N, approximately [N/2]/[ log e (N/2)] primes in the 2nd, 3rd and 4th quarters will be used for each cycle. Let c be the number of cycles required to produce all the N/4 arithmetic means in the 2nd quarter. The following equation can now be formed using the upper bound 8N / (log e N)^2. c[N/2]/[ log e (N/2)] – 4cN / (log e N)^2 = N/4 The approximate number of cycles, c, required using an upper bound of 8N / (log e N)^2 is c ≈ [ log e (N/2)]/2 So the number of cycles required (which is the same as the number of primes required in the first quarter) to produce all N/4 arithmetic means is much less than the number of primes in the first quarter i.e. c ≈ [ log e (N/2)]/2 < [N/4]/[ log e (N/4)]. This proves that Goldbach’s conjecture is true.

2019-06-20, 14:55	#2
2M215856352p1 May 2019 3·37 Posts	This 'proof' looks more like a heuristic argument at first glance. Last fiddled with by 2M215856352p1 on 2019-06-20 at 15:02

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2019-06-20, 15:40	#4
Awojobi Feb 2019 1011110₂ Posts	I don't understand sieve theory myself. However, I got this upper bound from Terence Tao's blog. This proof is not a heuristic proof. It is its simplicity that probably makes it seem that way.

2019-06-20, 16:15	#5
a1call "Rashid Naimi" Oct 2015 Remote to Here/There 2·11·109 Posts	I do not know if the proof is valid or not. However, personally I like your approach in redefining the conjecture. I suggest that you define the zones more precisely and in terms of arbitrary values such as N. I think a more precise parametric/formulaic argument will either show if there are holes to be patched or not.

2019-06-20, 21:01	#6
a1call "Rashid Naimi" Oct 2015 Remote to Here/There 100101011110₂ Posts	Use of the word "roughly" is problematic in a mathematical proof. Why not say that for any given positive integer N, Set 1 is the set of all integers 1 to N Set 2 is the set of all integers N+1 to 2N Set 3 is the set of all integers 2N+1 to 4N Now if you can prove that for any given N, all the integers in set 2 are arithmetic means of 2 prime numbers which are members of sets 1 and 3, then you have indeed proven the conjecture. There might be such a proof in OP, but it's too long of a post for my attention span to look for it.

2019-06-21, 09:19	#10
Awojobi Feb 2019 1011110₂ Posts	I have not made any assumptions. All I have done is make approximations that are valid. Valid, in the sense that upper bounds of repetitions and lower bound of the number of primes less than N have been used in modelling the scenario. So, basically, in my modelling, for each cycle, less arithmetic means are being produced than the actual number of arithmetic means produced and yet all arithmetic means will be produced, if that makes any sense.