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Topic: 9th stab: proof of AP-postulate and thus proof of Goldbach #1408
Correcting Math

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plutonium.archimedes@gmail.com

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Registered: 3/31/08
9th stab: proof of AP-postulate and thus proof of Goldbach #1408
Correcting Math

Posted: Jan 11, 2014 6:35 PM
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9th stab: proof of AP-postulate and thus proof of Goldbach #1408 Correcting Math
______________________________
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.

Both Bertrand and Goldbach are statements about the density of primes, a constant steady density.

What is equal to Goldbach is the AP-postulate which says between one perfect-square and the successor perfect square always exists at least a 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? Why is the density the same? 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:

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 intervals as their even numbers.

By Bertrand, between 2 and 4 is a prime and between 3 and 6 is a prime. Now what if I did not know what those primes were in general and could not find those two primes product. Can I say anything at all about those two primes and their product? Well I could use the 4 as a lower bound and the 6 as an upper bound and thus 4x6 = 24 and 24 is in between the two perfect squares of 16 and 25.

Continuing in this pattern, between 3 and 6 is a prime and between 4 and 8 is a prime and what if I did not know what they are, then I could take 6 as the lower bound and 8 as the upper bound and have 6x8 = 48 and the previous perfect squares were from 16 to 25 and the next one after 25 is 36, so 48 jumps the mark too far. But here we have a remedy, since the prime from 2 to 4 can be used but since we do not know it we use the 4 as lower bound rather than the 6 as lower bound and we have 4x8 =32 so here we manufactured a solution of 24 intersliced between 16 and 25 and now we have 32 intersliced between 25 and 36.

Continuing, between 4 and 8 is a prime and between 5 and 10 is a prime and what if I did not know what they are, then I could take 8 as the lower bound and 10 as the upper bound and have 8x10 = 80 but 80 is far from lying in between 36 and 49 in our sequence of perfect squares. So I do the trick again of falling back and picking a smaller N to 2N interval to garner a lower bound. And the interval I pick is the prior 2 and 4 interval which gives me the lower bound of 4 and thus 4x10 is 40 and that surely lies in between the perfect squares of 36 to 49.

Now all the above can be switched over into general mathematical language without the numbers involved, so that for the General case we build up lower and upper bounds intersliced in between two successive perfect squares and since those bounds are larger than what the primes that actually are involved, that the proof is completed.

What I have proven here is that the sequence of Perfect Squares is intersliced with another sequence of two-prime-composites (not including when the pxq is say 3x3 or 5x5) :

4, pxq, 9, pxq, 16, pxq, 25, pxq, 36, pxq, . . .

QED
--     

Recently I re-opened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.      

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe      


Archimedes Plutonium




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