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Topic: professors of Stanford endorsing proof of Goldbach to arxiv
Replies: 20   Last Post: Sep 5, 2014 4:14 AM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
professors of Stanford endorsing proof of Goldbach to arxiv
Posted: Aug 7, 2014 1:56 AM

GOLDBACH

Proof of Bertrand's Postulate and AP-Postulate

Alright, in the proof of Goldbach, I managed to have to prove 4 items:
a) Bertrand's postulate
b) AP postulate
c) Goldbach
d) Generalized Goldbach

_______________________
Proving Bertrand's postulate
_______________________

Bertrand's says that between N and 2N always exists a prime.

The prime counting function says that given a number N, the amount of primes below N is N/Ln(N).

So to prove Bertrand's we are asking for a prime p to exist between each of this generalized sequence:

N to 2N to 4N to 8N to 16N . . .

And so Bertrand's becomes this for a generalized prime call it p:

N p 2N p 4N p 8N p 16N . . .

So, how does N/Ln(N) prove Bertrand's? Well, the function is a monotone increasing function and really that is all that is needed to say to prove Bertrand's. But let us go into the details.

Let us take some particular sequences and plug the function N/Ln(N) into the sequence:

2 to 4 then 8 then 16 then 32 then 64 etc etc

4/Ln(4) = 4/1.38 = approx 2.8
8/Ln(8) = 8/2.07 = approx 3.8
subtracting we have 1 prime between 4 and 8, yet actually we have two

8/Ln(8) = 8/2.07 = approx 3.8
16/Ln(16) = 16/2.77 = approx 5.7
subtracting we have 1.9 primes between 8 and 16, yet we actually have two

16/Ln(16) = 16/2.77 = approx 5.7
32/Ln(32) = 32/3.46 = approx 9.2
subtracting we have 3.5 primes between 16 and 32, yet we actually have five

So one easily sees how any sequence of doubling a number has a increasing value for the number of primes.

Let me show it for a larger number N:
Ln(100) = 4.605.. and dividing into 100 delivers 21.7.. primes approximately
Ln(200) = 5.298.. and dividing into 200 delivers 37.7.. primes approx
Now we do a subtraction which reminds me alot about integrals where we subtract to get the area of the region we wanted. Here we subtract 37.7- 21.7 is 16 so we have approx 16 primes between 100 and 200 to satisfy Bertrand's which wanted only 1 prime to satisfy.

Repeating that exercise of arithmetic for Ln(400) then for Ln(800) then for Ln(1600) etc etc we see a monotonic increase in primes for which Bertrand's demand was one single solo prime yet we supply Bertrand with a multitude of primes.

The same can be done for the generalized N doubling:
N p 2N p 4N p 8N p 16N . . .

In a sentence, Bertrand is proven true because, for subtracting the primes from 0 to N from the primes of 0 to 2N involves x/Ln(x) which is monotonic increasing function.

QED

__________
Proof of Goldbach
________________

To summarize the proof of Goldbach is simple and easy.

Notice that all even numbers starting with 4 have addition arrays such as this:

4 array
0  4
1  3
2  2

6 array
0  6
1  5
2  4
3  3

8 array
0  8
1  7
2  6
3  5
4  4

And notice that they end the array with a perfect square when we replace add by multiply. So that 4 array ends with 2x2 and 6 array ends in 3 x 3, etc

Now we proved that in the Perfect-Squares there exists a two prime composite in between every successive perfect squares, so that 9 rests between 4 and 9 and that 15 = 3x5 rests between 9 and 16. So we always get a two prime composite that exists between successive perfect squares and hence we get the proof of Goldbach. Really an easy proof once you focus on perfect-squares and two prime composites.

GOLDBACH & Generalized Goldbach

Proof of Goldbach conjecture using successive perfect squares and x/Ln(x)

_____
Proof
_____

Now, what I have done here is try to list all the two-prime-composites in order of their product. Here are all possible combinations of the odd primes listed according to their product. And I separated them according to Perfect Squares

2  2  = 4  (4)

4

3  3  = 9  (6)

9

3  5  = 15  (8)

16

3  7  = 21 (10)
5  5  = 25 (10)

25

3  11 = 33  (14)
5  7  = 35  (12)

36

3  13 = 39  (16)
7   7  = 49  (14)

49

3  17 = 51  (20)
5  11 = 55   (16)
3  19 = 57  (22)

64

5  13  = 65  (18)
3  23  = 69  (26)
7  11  = 77  (18)

81

5  17  = 85  (22)
3  29  = 87  (32)
7  13  =  91 (20)
3  31  =  93 (34)
5  19  =  95  (24)

100

3  37  =  111 (40)
5  23  = 115  (28)
7  17  =  119 (24)
11 11 = 121 (22)

121

3   41 = 123 (44)
3   43 = 129 (46)
7  19  = 133 (26)
3   47  =  141 (50)
11 13  =  143 (24)

144

5   29  =  145 (34)
3   51  =  153  (54)
5   31  =  155  (36)
7  23  = 161  (30)
13 13 =  169 (26)

169

5    37  = 185  (42)
11  17  = 187 (28)

196

7    29  = 203 (36)
5   41   = 205 (46)
11  19  = 209 (30)
5   43  =  215  (48)
7   31  =  217  (38)
13 17 =  221 (30)

225

.
.
.

Now to prove Goldbach, I need to use the AP-postulate that between successive perfect squares always exists two distinct primes starting with the perfect squares of between 4 and 9 and we have primes 5,7 and the AP-Postulate proved to be true.

Here is the proof of the AP-postulate which is a stronger version of the Bertrand's postulate.

_____________________
Proving the AP postulate
_____________________

The AP-Postulate is making the intervals between primes smaller than in the doubling of intervals in the Bertrand's, but, also by increasing the number of primes to be two rather than a solo one prime:

The intervals are successive perfect-squares as this:

2^2   to  3^2   to 4^2    to  5^2   to  6^2 . . .

And calling a pair of generalized primes as p,q, I need to show this is true:

2^2   p,q   3^2   p,q  4^2   p,q  5^2  .  .  .

So, where Bertrand's postulate is one prime interspliced between successive N to 2N to 4N etc, the AP postulate is a pair of primes interspliced between successive perfect squares. The AP postulate is far more demanding for it involves smaller intervals and involves the existence of a pair of primes not a singular solo prime.

Now in the proof of AP-postulate the two primes are distinct.

For the generalized pair of primes p,q, I must show that we can do this:

2^2   p,q   3^2   p,q  4^2   p,q  5^2  p,q  6^2  p,q  7^2 .  .  .

Now because the Prime Counting Function (PCF) produces a lot more primes than needed in Bertrand's it is obvious PCF produces a lot more primes than needed for AP-postulate because, again, for the simple explanation that N/Ln(N) is monotonic increasing function. And even though the intervals are shortened compared to Bertrand and we require two distinct primes, the function N/Ln(N) is still capable of a lot of room to satisfy those requirements.

4/Ln(4) = 4/1.38 = approx 2.8
9/Ln(9) = 9/2.19 = approx 4.1
subtracting we have 1.3 prime between 4 and 9, yet actually we have two

9/Ln(9) = 9/2.19 = approx 4.1
16/Ln(16) = 16/2.77 = approx 5.7
subtracting we have 1.6 primes between 9 and 16, yet we actually have two

16/Ln(16) = 16/2.77 = approx 5.7
25/Ln(25) = 25/3.21 = approx 7.7
subtracting we have 2 primes between 16 and 25, yet we actually have three

25/Ln(25) = 25/3.21 = approx 7.7
36/Ln(36) = 36/3.58 = approx 10.0
subtracting we have 2.7 primes between 25 and 36, and we actually have two

But our computations stop there because clearly from 16 to 25 and 25 to 36 because N/Ln(N) is monotonic increasing the number of primes between successive perfect squares is going to be always larger than 2.

QED

And I need the AP-Postulate applied to two-prime-composites. So I have to adapt the AP-Postulate to that of the multiplication of two primes, distinct or the same two primes as seen in the table above.

Now in the proof of the AP-postulate, I used the Prime Counting Function N/Ln(N) to see how many primes exist between successive perfect squares.

To prove Goldbach, all I need is prove that 1 set of primes (distinct or the same) forms a two-prime-composite between each successive perfect square.

Now I am going to make a rather dazzling twist of logic. We know the amount of Primes follows this function of N/Ln(N). We know every Two-Prime-Composite is a composite consisting of only two primes, and since they are composed of only two primes, we can say that their Product follows the function of N/Ln(N).

So let us count up the number of Two-Prime-Composites above from 4 to 100
and we have a total of 20 such Two-Prime-Composites starting with 2x2 and ending with 5x19.

Now the function N/Ln(N) for 100 is 100/4.60 = approx 21.7 and the regular primes are 25 in that interval and the Two-Prime-Composites are 20 in that interval, so we see the Two-Prime-Composites coming closer to the function than the regular primes.

Now look at how many two-prime-composites from 2x2 to 225 perfect-square and we have 42 of them. Now let us see what N/Ln(N) gives and we have 225/5.41 = approx 41.5. So that our replacement of regular primes by that of two-prime-composites product is a good choice of replacement for both are monotonic increase as a function.

So, now, what is needed to prove Goldbach is simply show that at least one pair of two-prime-composites exists in between two successive perfect squares.

Between 4 and 9 is 3x3 or 3+3 and between 9 and 16 is 3x5 or 3+5.

4

3  3  = 9  (6)

9

3  5  = 15  (8)

16

Now let me compute with the N/Ln(N) for successive perfect squares:

25/Ln(25) = 25/3.21 = 7.7 approx
16/Ln(16) = 16/2.77 = 5.7 approx
subtracting leaves 2

36/Ln(36) = 36/3.58 = 10.0 approx
25/Ln(25) = 25/3.21 = 7.7 approx
subtracting leaves 2.3

Successive counts delivers 2 or greater than 2 primes, either distinct or the same primes for a two-prime-composite residing between those two successive perfect squares. And since the function N/Ln(N) is monotonic increasing we have our adaption of the AP-Postulate, that between every successive perfect-square starting with 4, there exists at least one of two-prime-composites.

Now we are not quite finished with the proof for we need one last thing to do. To relate the two-prime-composites with the even numbers. And that relationship is very easy to see in the Addition Columns below of all the even numbers. For every even number ends its column as a perfect square. So the proof of Goldbach is that there exists, always at least one two-prime-composite inside a even number column, matched and paired up, hence Goldbach.

0  4
1  3
2  2

0  6
1  5
2  4
3  3

0  8
1  7
2  6
3  5
4  4

0  10
1  9
2  8
3  7
4  6
5  5

0  12
1  11
2  10
3  9
4  8
5  7
6  6

0  14
1  13
2  12
3  11
4  10
5  9
6  8
7  7

0  16
1  15
2  14
3  13
4  12
5  11
6  10
7  9
8  8
.
.
.

QED

Proof of Generalized Goldbach conjecture-- beyond 12, at least a two set solutions

_______
PROOF
_______

Now, what I have done here is try to list all the two-prime-composites in order of their product. Here are all possible combinations of the odd primes listed according to their product. And I separated them according to Perfect Squares

2 2 = 4 (4)

4

3 3 = 9 (6)

9

3 5 = 15 (8)

16

3 7 = 21 (10)
5 5 = 25 (10)

25

3 11 = 33 (14)
5 7 = 35 (12)

36

3 13 = 39 (16)
7 7 = 49 (14)

49

3 17 = 51 (20)
5 11 = 55 (16)
3 19 = 57 (22)

64

5 13 = 65 (18)
3 23 = 69 (26)
7 11 = 77 (18)

81

5 17 = 85 (22)
3 29 = 87 (32)
7 13 = 91 (20)
3 31 = 93 (34)
5 19 = 95 (24)

100

3 37 = 111 (40)
5 23 = 115 (28)
7 17 = 119 (24)
11 11 = 121 (22)

121

3 41 = 123 (44)
3 43 = 129 (46)
7 19 = 133 (26)
3 47 = 141 (50)
11 13 = 143 (24)

144

5 29 = 145 (34)
3 51 = 153 (54)
5 31 = 155 (36)
7 23 = 161 (30)
13 13 = 169 (26)

169

5 37 = 185 (42)
11 17 = 187 (28)

196

7 29 = 203 (36)
5 41 = 205 (46)
11 19 = 209 (30)
5 43 = 215 (48)
7 31 = 217 (38)
13 17 = 221 (30)

225

.
.
.

Now to prove both the Regular Goldbach and the Generalized Goldbach
I need to use almost the very same method in the proof of both. And that method involves the AP-Postulate that between successive perfect squares, exists at least two primes for the Regular Goldbach and at least 2 sets of two-prime-composites between successive perfect squares starting with 14 for the Generalized Goldbach.

Now to prove the AP-postulate, I use the Prime Counting Function x/Ln(x) to see how many primes exist between successive perfect squares.

Now in this Generalized Proof, I am going to make a rather dazzling twist of logic. We know the Primes follow this function of x/Ln(x). We know every Two-Prime-Composite is a composite consisting of only two primes, and since they are composed of only two primes, we can say that their Product follows the function of x/Ln(x).

So let us count up the number of Two-Prime-Composites above from 4 to 100
and we have a total of 20 such Two-Prime-Composites starting with 2x2 and ending with 5x19.

Now the function x/Ln(x) for 100 is 100/4.60 = approx 21.7 and the regular primes are 25 in that interval and the Two-Prime-Composites are 20 in that interval, so we see the Two-Prime-Composites coming closer to the function than the regular primes.

Now look at how many two-prime-composites from 2x2 to 225 perfect-square and we have 42 of them. Now let us see what x/Ln(x) gives and we have 225/5.41 = approx 41.5. So that our replacement of regular primes by that of two-prime-composites product is a good choice of replacement for both are monotonic increase as a function.

So, now, what is needed to prove the Generalized Goldbach that there are at least a two solution set of primes for every even number greater than 12? All the proof needs is at least two solution sets between every perfect square and the function x/Ln(x) forces that to be true because as we take that function to 16, 25, 36 the answer is greater than 2 in each successive interval. And then from 25, 36, 49 the answer is greater than 2 in each successive interval. As seen here:

25/Ln(25) = 25/3.21 = 7.7 approx
16/Ln(16) = 16/2.77 = 5.7 approx
subtracting leaves 2

36/Ln(36) = 36/3.58 = 10.0 approx
25/Ln(25) = 25/3.21 = 7.7 approx
subtracting leaves 2.3

Successive counts delivers 2 or greater than 2

Now, I am not done yet, for I have to relate back to the Even Numbers, and the relationship of the Perfect Squares involved above. Well if you look at the even numbers, you see they end each column with a perfect square. So that if we are guaranteed two prime set solutions, means we have two solutions in each even number starting with 14 as shown below:

0  0

0  2
1  1

0  4
1  3
2  2

0  6
1  5
2  4
3  3

0  8
1  7
2  6
3  5
4  4

0  10
1  9
2  8
3  7
4  6
5  5

0  12
1  11
2  10
3  9
4  8
5  7
6 6

0  14
1  13
2  12
3  11
4  10
5  9
6 8
7 7

0  16
1  15
2  14
3  13
4  12
5  11
6 10
7 9
8 8
.
.
.

Proof of Generalized Goldbach conjecture-- beyond 1090, every even number has at least a 3 set solutions

_______
PROOF
_______

I am not going to repeat what was done of Goldbach and Goldbach beyond 12
because the proof for beyond 1090 with 3 set solutions follows in the same manner as beyond 12 with 2 set solutions.

Now to prove both the Regular Goldbach and the Generalized Goldbach
I need to use almost the very same method in the proof of both. And that method involves the AP-Postulate that between successive perfect squares, exists at least two primes for the Regular Goldbach and at least 2 sets of two-prime-composites between successive perfect squares starting with 14 for the Generalized Goldbach, and at least 3 sets of solutions starting with 1090, etc etc. The reader should be aware that as the even numbers get larger and larger there is at least a 4 set solution then a 5 set solution of Goldbach primes and when we get near to the infinity borderline of 1*10^603 we find there are approx 10^599 Goldbach solutions to this even number of 1*10^603. Reasoning? Because the Prime Counting Function x/Ln(x) is monotonic increasing.

So, now, what is needed to prove the Generalized Goldbach of 1090 and beyond has at least 3 sets of Goldbach solutions? All the proof needs is at least 3 solution sets between every successive perfect-square starting at 32^2 and the function x/Ln(x) forces that to be true because as we take that function to 32^2, then 33^2, then 34^2, then 35^2 the answer is greater than 8 primes or 4 sets in each successive interval. As seen here:

1024/Ln(1024) = 1024/6.93 = 147.7 approx
1089/Ln(1089) = 1089/6.99 = 155.7 approx
subtracting leaves 8 two-prime-composites for 4 sets of Goldbach solutions

1089/Ln(1089) = 1089/6.99 = 155.7 approx
1156/Ln(1156) = 1156/7.05 = 163.9 approx
subtracting leaves 8.2 two-prime-composites for 4 sets or more

Successive counts delivers 4 or greater than 4 because x/Ln(x) is monotonic increasing.

Now, I am not done yet, for I have to relate back to the Even Numbers, and the relationship of the Perfect Squares involved above. Well if you look at the even numbers starting at 1090 they end in a perfect square and they have 3 set solutions for that even number column.

Now using the very same methods, the function N/Ln(N) proves that there is a even number where the solutions are all 4 sets at minimum, then a larger even number where the Goldbach solutions are 5 sets at minimum. Hence, the Generalized Goldbach is proven true.

--

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