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[FWD]

A proof for Goldbach's conjecture

The purpose of this article is proving that each even number is sum of

two prime numbers. For this, we first should prove that for an infinitely

large even number there exist infinite pairs of odd numbers that the sum of

the members of each pair is equal to the even number.

Let's write the series 3, 5, 7, ... in a row named "A"

and just under it write the series of the even numbers 6, 8, 10, ... in a

row named "B":

_A: 03 05 07 09 11 13 15 17 ...

_B: 06 08 10 12 14 16 18 20 ...

C3: 01 01 01 01 01 01 01 01 ...

C5: 00 00 01 01 01 01 01 01 ...

C7: 00 00 00 00 01 01 01 01 ...

C9: 00 00 00 00 00 00 01 01 ...

_D: 01 01 02 02 03 03 04 04 ...

There is a relation between the numbers in the row A and in the row B:

3+3 from A is 6 from B, 3+5 from A is 8 from B, 3+7 from A is 10 from B,

...

We show this relation by the numbers of 1 in the row C3. Also 5+5 from A is

10 from B, 5+7 from A is 12 from B, 5+9 from A is 14 from B, .... We show

this relation by the numbers of 1 in the row C5. Also 7+7 from A is 14 from

B, 7+9 from A is 16 from B, 7+11 from A is 18 from B, .... We show this

relation by the numbers of 1 in the row C7.

We can continue this procedure.

Each item in a column of the row D is the sum of all the items in the same

column of the rows Cn. In this manner each item in a column of the row D

is the number of all the pairs of odd numbers that the sum of the members

of each pair is the even number of this column (of the row B).

As we can easily see, in the row A the horizontal distance between each odd

number and the first next odd number in the column of which there is an even

number which is sum of this odd number and itself, is as large as the

horizontal distance between this odd number and the first number in the row

A. (For example the horizontal distance between 9 and the column containing

18(=9+9) is three units, and the horizontal distance between 9 and the first

number in the row A, ie 3, is also three units.) Let's call such a distance

as a "triangular distance" related to the odd number (eg the triangular

distance related to 9 in the row A is (9 8 11 13 15) the length of

which is three units).

It is clear that the series 1, 1, 2, 2, 3, 3, ... approaches infinity.

Let's choose an odd number and make the triangular distance of it and as

soon as such making subtract one unit from each number in the row D which is

in the same column in which a member of this triangular distance is, and

let's substitute the obtained series for the old series (here the series D).

By this work we are subtracting the number of all the pairs of odd numbers,

the bigger member of each being our chosen odd number, from the numbers of

pairs (of odd numbers) written in the row D. Now let's choose a bigger odd

number and perform the same above procedure to obtain a new series. For

instance for the odd numbers 9, 11, 15, 19 and 23 we get the following sets

of series Dn from the initial series D:

__A: 03 05 07 09 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 ...

__B: 06 08 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 ...

__D: 01 01 02 02 03 03 04 04 05 05 06 06 07 07 08 08 09 09 10 10 11 11 ...

_D9: 00 00 00(01 02 02 03)04 05 05 06 06 07 07 08 08 09 09 10 10 11 11 ...

D11: 00 00 00 00(01 01 02 03 04)05 06 06 07 07 08 08 09 09 10 10 11 11 ...

D15: 00 00 00 00 00 00(00 02 03 04 05 05 06)07 08 08 09 09 10 10 11 11 ...

D19: 00 00 00 00 00 00 00 00(02 03 04 04 05 06 07 07 08)09 10 10 11 11 ...

D23: 00 00 00 00 00 00 00 00 00 00(03 03 04 05 06 06 07 08 09 09 10)11 ...

As it is seen, the first number in each series which is right-adjacent

to the triangular distance is equal to the number of the members of the

triangular distance. It's evident that for an infinite odd number this

first number, right-adjacent to the triangular distance of this odd

number, is an infinitely large number. If the triangular distances of the

successive odd numbers bigger than our odd number are to subtract one by

one from this infinite number through one-by-one rightward advance of the

triangular distances (in the successive stages ie in the successive rows

caused by the next triangular distances, just as eg the second 4 in the

row D has been changed into 2 in the row D15 after three times rightward

advancing of the triangular distances), we can say that the advance speed of

the left border of the triangular distance (ie eg the mark "(" in the

above rows) toward this infinite number is equal to the rate of subtraction

from infinity, and finally, when this border reaches this infinite number,

there will be no infinite number but a finite number (because as we saw, the

number of the members of the triangular distance is equal to that number

which is immediately right-adjacent to this distance).

Now suppose that instead of the above successive triangular distances of

successive odd numbers we have only triangular distances of those odd

numbers that are not prime. In such a case since we know the number

of primes is infinite, the row by row triangular subtractions are not one

by one, ie this is not the case that the left border of the triangular distance

(ie "(") in a row is always only one unit farther than this border in the previous

row (towards right), but the advance of this left border towards right is

sometimes impulsive (not one by one), because it is to consist only of the

(eliminating) triangular distances of the odd numbers which are not prime

and we know we have infinite prime (odd) numbers up to infinity. Thus, when

this left border reaches our infinite number, it will have subtracted an

amount from our infinite number smaller than the largeness of this infinite

number. It's clear that the amount remained of this infinite number (after

these subtractions) is equal to the total number of the primes existent in

an infinite continuous distance of natural numbers. And we know this number

is itself an infinite number. Thus in fact when all the possible triangular

distances have done their subtractions from the infinite number during their

advance towards this number the remaining number is still infinitely large.

(As the number of primes in the distance (1,infinity) is infinite, the

number of primes in the distance (infinity/2,infinity) is also infinite,

because otherwise we must accept that there is an upper bound for

the primes in this distance which is not smaller than the infinity existent in

(infinity/2,infinity), and consequently it must be also an upper bound (for

the primes) in (1,infinity) (since the infinity existent in (1,infinity) is

the same infinity existent in (infinity/2,infinity)), but we know existence

of any upper bound for the primes in (1,infinity) is not possible.) This

means that for an infinite even number the numnber of pairs of odd numbers,

the sum of the numbers of each being the even number and the bigger number

of each pair being a prime, is infinite. Let's call each pair of odd numbers the

sum of the numbers of which is an even number as a constructive pair of the

mentioned even number. Therefore, if an even number approaches infinity the

number of its constructive pairs, the bigger number of each of which is a

prime, will also approaches infinity. (Notice that in the eliminating

mechanism presented above, pairs like (9,11) or (15,19) won't be eliminated,

because in each pair while the smaller number is not a prime the bigger

number is a prime.)

In addition, as it is evident in the process of the proof, for an even

number there are no prime numbers other than those constituting the set of

the bigger numbers of its constructive pairs between the smallest and

biggest numbers of these bigger numbers of these pairs. This means that

the mentioned set is a continuous set of primes. Is it possible that for

an infinitely large even number none of the smaller numbers of its

constructive pairs is prime? No, it is not possible, because infinity/2

is not a finite number. Explanation: Suppose that the statement P is

true for each member of the infinite distance (1,infinity). In such a case

we can consider infinity as a member of this distance and say that P is true

for it, and of course this means that P is true for every number of this

distance which is greater than a chosen number. It is possible that several

statements are true separately for different members of this distance,

for instance P1 for the distance (1,100) and P2 for (100,1000000) and P3

for (1000000,infinity) in which again we can consider infinity as a member

for which the statement P3 is true. What is important in dealing with

infinity as a member is that the statement for infinity must be clear and

determined. Now suppose that for the finite distance (1,n) the statement is

that firstly n/2 is determined and secondly (after performing the above

eliminating mechanism) all the bigger numbers of the constructive pairs

of the number which is immediately right-adjacent to this distance

construct all the members of the distance (n/2,n) and none of the smaller

numbers of these pairs belongs to the distance (1,n/2). In which case can

this statement be true for the infinite distance (1,infinity)? When at least

infinity/2 is determined, but we know this is not the case. In a similar

manner we must conclude that in principle the set of the prime numbers of

these smaller numbers of the pairs is not a finite set since otherwise we

can eliminate the pairs related to them and again have an infinite set the

center of which must be determined.

In this manner we showed that when the even number approaches infinity

the number of its constructive pairs, both numbers of which being primes,

approaches infinity. Now consider the function of the number of those

constructive pairs of even numbers both members of each of which are

primes. We proved that this function has an (infinite) limit in infinity;

then it is a well-behaved function (ie its general tendency on sufficiently

large continuous distances of its domain is not unpredictable). On the

other hand, as initial values, we know that on a large continuous

distance of the first portion of the domain of this function

this function takes positive values which are ascending on average.

Since this function is well-behaved and is positive and averagely

ascending at the first and has a tendency towards infinity at the end,

we conclude that it cannot have an abrupt fall toward zero at any other

point of its domain.

(If through some way other than that the function has an (infinite) limit

in infinity we accepted that the function was well-behaved, could we

conclude only from its general ascending tendency at the first that,

up to infinity, it would have an ascending tendency? Obviously not,

because in such a case we couldn't be certain that our well-behaved

function not to start falling towards zero at a point.) In this

manner we have shown that every even number is sum of (the members of

at least one pair consisting of) two primes.

Hamid V. Ansari

http://www.netnormal.com/users/elmiran/03.doc

**Next message:**Bill Douglass: "Re: Anyone read Rumanian?"**Previous message:**Spike Jones: "Re: HELP: attention aero nuts..."**In reply to:**John Clark: "Re: MATH/COMP/PHIL: "Omega Man""**Next in thread:**Mitchell J Porter: "Re: MATH/COMP/PHIL: "Omega Man""**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]

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