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Topic: 8th stab: proof of AP-postulate and thus proof of Goldbach #1407
Correcting Math

Replies: 0

 plutonium.archimedes@gmail.com Posts: 8,765 Registered: 3/31/08
8th stab: proof of AP-postulate and thus proof of Goldbach #1407
Correcting Math

Posted: Jan 11, 2014 3:03 PM

8th stab: proof of AP-postulate and thus proof of Goldbach #1407 Correcting Math
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Proof of AP-postulate and thus proof of Goldbach
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Here are the first Addition Columns of the numbers from 0 to 12:

0  0

0  1
1  0

0  2
1  1

0  3
1  2
2  1

0  4
1  3
2  2

0  5
1  4
2  3
3  2
4  1

0  6
1  5
2  4
3  3

0  7
1  6
2  5
3  4
4  3

0  8
1  7
2  6
3  5
4  4

0  9
1  8
2  7
3  6
4  5
5  4

0  10
1  9
2  8
3  7
4  6
5  5

0  11
1  10
2  9
3  8
4  7
5  6
6 5

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

Now it is simple to prove that if Goldbach is true then Bertrand's Postulate is true from those Addition columns because we have a k and k/2 sides of the column guaranteeing a prime existing between N and 2N.

So Goldbach implies Bertrand, but does Bertrand imply Goldbach? No, for that would mean Goldbach is equal to Bertrand and that is obviously not true.

What is equal to Goldbach is the AP-postulate which says between one perfect-square and the successor perfect square always exists at least one two-prime-composite (a composite that has just two prime factors in decomposition such as 3x5=15).

So we have Goldbach =/ Bertrand but we do have Goldbach = AP-postulate.

Why is the AP-postulate the same as Goldbach conjecture? Because if you have a two-prime-composite between every perfect square, means you have in the Addition Column above, you have two primes whose sum is that even number.

So, to prove Goldbach, we need only prove the AP-postulate.

Now here is the density of two-prime-composites omitting the prime 2

3 3 = 9 (6)

3 5 = 15 (8)

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

5 7 = 35 (12)

3 11 = 33 (14)
7 7 = 49 (14)

3 13 = 39 (16)
5 11 = 55 (16)

5 13 = 65 (18)
7 11 = 77 (18)

3 17 = 51 (20)
7 13 = 91 (20)

3 19 = 57 (22)
5 17 = 85 (22)
11 11 = 121 (22)

5 19 = 95 (24)
7 17 = 119 (24)
11 13 = 143 (24)

3 23 = 69 (26)
7 19 = 133 (26)
13 13 = 169 (26)

5 23 = 115 (28)
11 17 = 187 (28)

7 23 = 161 (30)
11 19 = 209 (30)
13 17 = 221 (30)
.
.
.

Now here is the density of the AP-postulate of how many two-prime-composites rest between successive perfect squares.

2 2 = 4 (4)

2^2

3 3 = 9 (6)

3^2

3 5 = 15 (8)

4^2

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

5^2

5 7 = 35 (12)

6^2

7 7 = 49 (14)

7^2

5 11 = 55 (16)

8^2

5 13 = 65 (18)
7 11 = 77 (18)

9^2

7 13 = 91 (20)

10^2

11 11 = 121 (22)

11^2

11 13 = 143 (24)

12^2

13 13 = 169 (26)

13^2

11 17 = 187 (28)

14^2

11 19 = 209 (30)
13 17 = 221 (30)

15^2

So, essentially, all of the Goldbach conjecture boils down to proving the AP-postulate, is there always a two-prime-composite between successive perfect-squares.

Proof of AP-postulate:
Now I cannot do the Pascal triangle in this format, so I leave it to the reader to do the Pascal triangle which was used in the proof of the Bertrand's postulate of between N and 2N always exists a prime. Here I am proving between perfect square s^2 and its successor t^2 exists a two-prime-composite (such as 3x5 = 15 exists between 3^2 and 4^2)

The Pascal triangle is used in Bertrand and in AP postulates because it is the sequence 2, 4, 8, 16, 32, . . a binomial expansion. In the proof, we use Bertrand's fact of a prime exists between N and 2N and we chose the smallest prime if more than one prime exists between N and 2N. So between 2 and 4 is the prime 3, and between 4 and 8 is the prime 5 chosen, not 7. And between 8 and 16 the prime 11 is chosen. Now we interlace the perfect-squares with the sequence 2, 4, 8, 16, . . . Such that the perfect square is aligned to the numbers of the binomial sequence, as this shows:

2
3
4 2^2
5
8
3^2 3x3=9 and 3+3 = 6
11
16 4^2 3x5=15 and 3+5 = 8
17
5^2 5x5 =25 and 5+5=10

32
6^2 5x7 =35 and 5+7= 12
37
7^2

64 8^2
.

.

.

Here I have aligned the binomial sequence with the perfect squares.

The proof is constructive for what I want to prove is merely that I can reach back in earlier successors and multiply two smaller primes to reside in a future successor. Only I am clever here, because I will not pick the primes since their identities are not known to me, but rather I pick the endpoints of the Natural Number Sequence instead of the binomial sequence.

2 to 4 is 3
3 to 6 is 5
4 to 8 is 5 since we pick the smallest prime
5 to 10 is 7
6 to 12 is 7
7 to 14 is 11
.
.
.

Now, here is where I am clever, for the identities of the smallest primes are unknown in general, so to alleviate that problem, I pick the even number as the lower bound and the even number as the upper bound so that for example in the 5 to 10, I pick the 10 and from 6 to 12 I pick the 12 since I do not know what the primes are but I surely know the endpoints of the number expansion.

So I have this sequence developed

4x6 where 4 is the lower bound and 6 the upper bound instead of the 3x5
6x8
8x10
10x12
12x14

Now, one more modification, in that the lower bound and upper bound can be mixed from earlier ones such as this:

4x4
4x6
4x8
6x6
4x10
6x8
6x10
8x8
.
.
.

So, now, almost finished. For if I can prove that this sequence above lies intersliced between this sequence 4^2, 5^2, 6^2, 7^2. . . then I will have proven the AP-postulate.

Explanation: In the general proof, we cannot know the smallest primes in intervals, but we can know the even number lower and upper bounds, and thus, instead of actually showing the primes intersliced with the perfect squares, I show the upper and lower bounds of the primes intersliced with the perfect squares.

In my proof, I dispense with the arcane and obfuscation of the proof of AP-postulate or the Bertrand's postulate by replacing the upper and lower bounds with a known even number. My proof is identical to Chebyshev's or Ramanujan's proof of Bertrand's postulate, only I substitute even numbers instead of primes as the bounds.

QED
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