# Re: MATH/COMP/PHIL: "Omega Man"

From: scerir (scerir@libero.it)
Date: Fri Apr 06 2001 - 15:19:24 MDT

[FWD]

A proof for Goldbach's conjecture

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

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