Open Science Repository Mathematics
doi: dx.doi.org/10.7392/Mathematics.70081940
Goldbach Conjecture, Uncorrupted
Jarmo Kiukkonen
University of Eastern Finland
Abstract
I used Cantor-like rational number matrix to positive numbers, so that on the diagonal goes number pairs, which sum is even. To each even number pair I put perpendicular line, and noticed that on this line the sum of number pairs the same as in the sum of middle number pairs. And at last I proved by pigeonhole axiom theorem.
Keywords: Goldbach conjecture, Cantor-like rational number matrix, mathematics.
Citation: Kiukkonen, J. (2013). Goldbach Conjecture, Uncorrupted. Open Science Repository Mathematics, Online(open-access), e70081940. doi:10.7392/Mathematics.70081940
Received: February 12, 2013
Published: March 11, 2013
Copyright: © 2013 Kiukkonen, J. Creative Commons Attribution 3.0 Unported License.
Contact: [email protected]
Proposition
Let p > 3 and even. (I apoIogise, dear readers, that I cannot use beIow index, so that I sign a(I) vaIue a with beIow index I).
Every even natural number p, p>3 is sum of two primes.
4 = 2
+ 2,
6 = 3
+ 3,
8 = 5
+ 3,
10 =
5 + 5.
I study intervaIs
[(2m-1)^2, (2m+1)^2[, m = 2, 3, 4, ...
= [9,
25[, [25, 49[ … = [3*3, 5*5,[ , [5*5, 7*7[ ,
[7*7, 9*9[
and so on.
I applicate Cantor’s rational number matrix A:
1,1
1,2 1,3
1,4 1,5 …
2,1
2,2
2,3
2,4 2,5 …
3,1
3,2
3,3
3,4 3,5 …
4,1
4,2
4,3
4,4 4,5
In this matrix are all natural number pairs and on diagonal are two same
numbers 1,1; 2,2; 3,3; and so on. The sum
of these give all positive even numbers beginning from
1+1=2, 2+2=4, 3+3=6, and so on.
And I have done observation, that on perpendicular lines to diagonal
behave the numbers very neatly. On one line are all the number pairs which sum
is the same as the sum on cutting point of diagonal:
1 + 5
= 2 + 4 = 3 + 3 = 4 + 2 = 1+ 5.
This means for example 3+3=6 6/2 odd pairs:
{ 1, 3, 5 }. Odd numbers on half of this are
6/4 +
½,
Because 6/2 is odd and odd number 3 must I count with.
For example when 8/2 is even, odd numbers are {
1, 3, 5, 7 }. Odd numbers on half of this set are
8/4 =
2.
If now primes are more than odd numbers on the other left-hand side of
cutting point of diagonal line and perpendicular line, there must be two primes
on both side of the cutting point on perpendicular line.
And because of the structure of perpendicular the sum is same as sum in
cutting point’s even sum.
Another axiom which I use is pigeon hole axiom:
If there are amount n of pigeon holes
and at least n+1 pigeons, then two of pigeons must take the samehole. This I use in above clarifying.
I am intrusted onIy in odd numbers between
intervals
[(2m-1)^2, (2m+1)^2[, m = 2,3,4, ...,
because even numbers
except 2 are non-prime.
First I caIcuIate amount of connected numbers
of odd numbers between my intervaIs
[(2m-1)^2 , (2m+11)^2[ , m = 2, 3, 4, ...
and then amount of aII odd numbers between my intervaIs
[(2m-1)^2, (2m+1)^2[, m = 2, 3, 4, ...
Then I subtract from amount of aII odd numbers amount of aII connected odd numbers between the same intervaI. So that triviaIIy aII odd numbers - aII odd
connected numbers gives to me amount of aII odd prime
numbers between my intervaI
[(2m-1)^2, (2m+1)^2[, m= 2, 3, 4, and so on.
I write by smaIIer-odd* bigger-odd-product of
two odd numbers, which are equaI or first odd number is smaIIer than second odd
number.
By p(j) I mean greatest even number between my intervaI
[(2m-1)^2, (2m+1)^2[, m=2,3,4,... and by p(k)
what so ever even number between my intervaI
[(2m-1)^2, (2m+1)^2[, m= 2,3,4, and so on.
For exampIe:
p(j) = 24, p(k) = 10, 12, 14, 16, 18,
20,
22 or 24 for intervaI [(2m-1)^2,
(2*2+1)^2[ =[9, 5^2[
= [9, 25[.
I caIcuIate on intervaIs
[(2m-1)^2, (2m+1)^2[, m= 2, 3, 4 ,....,
aII odd*odd-products, because all even numbers except 2 are
connected.
Because of for example 5*3=3*5, I must
take smaller-odd*bigger-odd-products. In Iater rows I
will prove that products which have more than two odd factors I can aIso reduce to form smaller-odd*bigger-odd-product.
And all products which have another factor even are even.
I take with to smaIIer-odd*bigger-odd-products
between intervaIs
[(2m-1)^2, (2m+1)^2[
3*bigger-odd,
5*bigger-odd, ...(2m-1)*(2m+3) , m=2,3,4,... .
because products
(2m+1)* bigger-odd
is over edge (2m+1)^2, m=2,3,4, ...
.
I count some examples of some odd products:
5^2 =
25, 3*7 = 21, 25-21=4;
7^2 =
49, 5*9 = 45, 49-45=4;
9^2 =
81, 7*11=77, 81-77 = 4,… .
The smallest difference of odd*odd-products is
3*2 =
6.
From this acknowledgment and above I can count greatest
smaller-odd*bigger-odd-product
= 21,
45, 77, …, (2m-1)*(2m+3).
Odd*odd-products leaves the even products of odd*even values between.
For smallest interval there are m = 2 and
only products of 3 which makes on 1 = m-1 kind of odd*odd-products,
named products of kind 3*x. This is valid
for arbitrary interval as I will prove in later rows.
Because on arbitrary intervaI
((2m+1)^2)/2 -((2m+1)^2)*(m-1)/6 >
((2m+1)^2)/4 (+½) (inequation 1)
and for smaller even
p(k)/2-p(k)/4 (-½) > (p(k)*(m-1))/6,
6/4 > (m-1).
The value ½ comes with if p(k)/2 is odd.
On the left side of inequation (1) is amount
of primes at least = pigeons. And on the right side is amount of the
pigeonholes.
Now I get
6/4-
1 / ((p(k)*2) > 2-1 p(k) > 0
for m=2,
even p(k) belongs to [9,25[.
By little algebra I can prove, that this is true:
6/4 -
1/(2*10) > 1.
6/4 +
10/4- 1/((p(k)*2) > 3-1 for m=3,
when even p(k) belongs to
interval [26, 48[.
6/4 +
10/4 + 14/4 -1/((p(k)*2). > 4-1,
when even p(k) belongs to
interval [48, 80[
and so on between intervals
[(2m-1)^2, (2m+1)^2[,
2*3 =
6,
2*5 =
10.
2*7 =
14.
General proof
If p(i)
is between arbitrary of my interval, and if
6/4+10/4+14/4+… -1/((p(i)*2)
> m-1
then here p(i)
is smallest even between intervals
[(2m-1)^2, (2m+1)^2[ , m=2, 3, 4, and so on,
is then
6/4+10/4+14/4+…2*(2m-1) /4 > m-1.
Depending which is value m comes on
left-hand side m-1 addition terms, which are all >1.
Little negative disturbing can I count out:
6/4 -
1/(2*10).
This can I by little algebra count to be > 1.
Above shows that there are more pigeons = primes than pigeonholes on
half of perpendicular line on diagonal line of matrix A.
Amount of smaller-odd’s is
3, 5,
7, ... , 2m-1 is
m-1,
when vaIue
m becomes from 2m-1.
I prove by mathematical induction, that aII odd(1)*odd(2)*odd(3)*...*odd(n)-products can I reduce into
form smaIIer-odd*bigger-odd.
Proof
Step 1
I take three odd numbers. TriviaIIy I
can then multipIy two of these together and get again
odd number. Now I have two odd numbers, which I order into greatnessorder
and put into form smaIIer-odd*bigger-odd.
Step 2
Induction hypothesis: AII products of odd numbers of Iongitude
n-1 I can reduce to form smaIIer-odd*bigger-odd.
Induction insists: AII products of odd numbers of Iongitude
n I can reduce to form smaIIer-odd*bigger-odd.
I put into induction hypothesis one odd number more and muItipIy this extra number by bigger-odd in induction
hypothesis. I get odd number, because odd-number*odd-number is odd number.
Then I have odd numbers of Iongitude n-1 and triviaIIy, when I have ordered these n-1 odd numbers into
greatness order, I get smaIIer-odd*bigger-odd-product
by induction hypothesis.
Now I have proved my mathematical induction.
And differences between smaIIest smaIIer-odd*bigger-odd-products are
2*3 =
6
and differences between biggest smaIler-odd*bigger-odd-products are
2*(2m-1). m=2,3,4,5, ... .
For exampIe:
[9,25[
9,
9+6 = 15, 15+6=21,
[25,49[
9+6+6+6=27,
27+6=33, ... ,
25, 25 +10 =35, 35 + 10 = 45.
Even numbers on my intervaI ends at
p(j) = (2m+1)^2-1
and begins at
p(i) = (2m-1)^2+1.
At the end I caIcuIate:
10 =
5 + 5,
12 =
7 + 5,
14 =
7 + 7,
16 =
5 + 11,
18 =
7 + 11,
20 =
7+ 13,
22 =
5 + 17,
24 =
11+13,
26 =
13+13,
28 = 23 + 5,
30 = 17 + 13,
32 = 29 + 3,
34 = 17 + 17,
36 = 17 + 19,
38 = 19 + 19,
40 = 23 + 17,
42 = 23 + 19,
44 = 31 + 13
46 = 41 + 5,
48 = 41 + 7,
50 = 43 + 7,
52 = 23 + 29,
54 = 23+ 31,
56 = 37 + 19,
58 = 29 + 29,
60 = 29 + 31,
62 = 43+ 19,
64 = 53 + 11,
66 = 43 + 23,
68 = 49 + 19,
70 = 53 + 17,
72 = 53 + 19,
74 = 43 + 31,
76 = 53 + 23,
78 = 61 + 17,
80 = 61 + 19.
Connections
1. Deshouillers, J. M., te Riele, H., & Saouter, Y. (1998). New experimental results concerning the Goldbach conjecture. Algorithmic number theory, 204-215. FIND ONLINE
2. Coykendall, J., & Spicer, C. (2012, July). Cohen-Kaplansky domains and the Goldbach conjecture. In Proc. Amer. Math. Soc (Vol. 140, pp. 2227-2233). FIND ONLINE
3. Stein, M. L., & Stein, P. R. (1965). New experimental results on the Goldbach conjecture. Mathematics Magazine, 38(2), 72-80. FIND ONLINE
4. Sun, Z. W. (2010). Mixed sums of primes and other terms. Additive Number Theory, 341-353. FIND ONLINE
5. Zhi-jiu, Y. E. (2011). The Proof of Goldbach Conjecture. Science & Technology Information, 25, 495. FIND ONLINE
6. Saouter, Y. (1998). Checking the odd Goldbach conjecture up to 1020. Mathematics of computation, 67(222), 863-866. FIND ONLINE
7. Eckhoff, J., & Nischke, K. P. (2009). Morris's pigeonhole principle and the Helly theorem for unions of convex sets. Bulletin of the London Mathematical Society,41(4), 577-588. FIND ONLINE
8. Gaspar, J., & Kohlenbach, U. (2010). On Tao's “finitary” infinite pigeonhole principle. Journal of Symbolic Logic, 75(1), 355-371. FIND ONLINE
9. Gandhi, K. R. (2012). An instinct to conclude the famous twin prime and Goldbach conjectures. South Asian J Math, 2(1), 68-72. FIND ONLINE
10. Arana, A. (2010). Proof Theory in Philosophy of Mathematics. Philosophy Compass, 5(4), 336-347. FIND ONLINE
Cite this paper
APA
Kiukkonen, J. (2013). Goldbach Conjecture, Uncorrupted. Open Science Repository Mathematics, Online(open-access), e70081940. doi:10.7392/Mathematics.70081940
MLA
Kiukkonen, Jarmo. “Goldbach Conjecture, Uncorrupted.” Open Science Repository Mathematics Online.open-access (2013): e70081940.
Chicago
Kiukkonen, Jarmo. “Goldbach Conjecture, Uncorrupted.” Open Science Repository Mathematics Online, no. open-access (2013): e70081940. http://www.open-science-repository.com/goldbach-conjecture-uncorrupted.html.
Harvard
Kiukkonen, J., 2013. Goldbach Conjecture, Uncorrupted. Open Science Repository Mathematics, Online(open-access), p.e70081940. Available at: http://www.open-science-repository.com/goldbach-conjecture-uncorrupted.html.
Science
1. J. Kiukkonen, Goldbach Conjecture, Uncorrupted, Open Science Repository Mathematics Online, e70081940 (2013).
Nature
1. Kiukkonen, J. Goldbach Conjecture, Uncorrupted. Open Science Repository Mathematics Online, e70081940 (2013).
doi
Research registered in the DOI resolution system as: 10.7392/Mathematics.70081940.

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