A Proof of Goldbach’s Conjecture
This paper sets out to prove Goldbach’s Conjecture using elementary linear
algebra and the existence of at least one prime number in the interval (3n, 4n)
with n>1 and integral.
We can represent any even number as 2N where N is a natural number.
Clearly when N is prime, Goldbach’s Conjecture is true.
Otherwise, defining two vector spaces over the same field as follows:
U = log (to any base) of rational powers of multiples of all primes < N.
V = log (to same base) of rational powers of multiples of all primes < 2N.
Field F = the rational numbers.
One can define a linear transformation as follows:
T:U→V = log{(2N-p1)α1(2N-p2)α2 x…x (2N-pπ(N)) α π(N)}with αi ε F, 1≤i≤π(N)
and u ε U with u = log{ p1α1 p2α2 x…x pπ(N)α π(N) } and N > 1.
By considering properties of the matrices of the above linear transformations I
will eventually show that for certain values of N there has to exist log(pi) ε U
and log(pj) ε V such that log(2N- pi) = log(pj) and in doing so prove
Goldbach’s Conjecture.
I have proved below that for integral n with n>1 there always has to exist at
least one prime number in the interval (3n, 4n). We can use this fact to show
that above some value of N there always has to exist at least one prime
number in the interval ([2N/3], N), where [ ] denotes the integral part of the
enclosed number. Clearly this is true when N is a multiple of 4 and N > 1. If
we consider the “worst case” scenario in which N=4n+2 and 3n+1 is the only
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prime in the interval (3n, 4n) and 4n+3 is the next largest prime, being in the
interval (3(n+1), 4(n+1)), this is true more generally if:
3n+1 ≥ 2(4n+2)/3


9n+3 ≥ 8n+4
n≥1
Taking n ≥ 2
; because our intervals are not valid when n=1.
This means that when N isn’t prime there always has to exist at least one
prime number in the interval ([2N/3], N) when N ≥ 8.
For a momentary diversion, when N isn’t prime and N ≥ 8 we will now
consider the properties of an algorithm that begins by taking a prime in the
interval ([2N/3], N) and subtracting this from 2N, factorising the difference
into it’s prime factors and subtracting each unique prime factor from 2N as
separate entities, factorising all these differences again, and repeating this
process with all new prime factors that are generated in turn until no new
prime factors are generated. If this doesn’t compute any instances of
Goldbach’s Conjecture before terminating then from these series of equations
we will have generated some primes q1, q2,…,qn in the interval (2, [2N/3]] for
arbitrary n < π([2N/3]), together with our initial prime qn+1 in the interval
([2N/3], N). We know this must be true because 3 is the smallest odd prime
factor we could generate and (2N-p)/3 must be less than [2N/3] for some p in
(2, N). If we were to plug the logarithm of these primes into our linear system
as defined above, then log(q1),…,log(qn),log(qn+1) would form a basis for this
subspace U’ of U and log(q1),…,log(qn) would form a basis for the
corresponding subspace V’ of V over the field F. This linear map has to be
surjective because due to the fundamental underlying properties of the
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algorithm there could not exist any other basis element of V' that hasn't been
created by it to form the range of our linear map. However the set {q1, q2,…,qn}
could not contain all the primes in the interval (2, [2N/3]]. For if it did the
domain of our linear map would be at it’s maximum possible size and fully
defined on all our elements. If this was to occur we could consider a second
linear transformation that maps U’ to a new vector space W with a change of
basis that is just a subspace of the logarithm of the integers generated by all
the subtractions of qi from 2N, with 1≤i≤(n+1), over the same field F. The
matrix of this new linear transformation would be square of dimension (n+1)
and would be zero except for a diagonal consisting entirely of the integer one
on the non-leading diagonal. Clearly all the rows and columns of this new
matrix would be linearly independent and thus it would represent a bijective
linear transformation. As a consequence of this the dimensions of U’ and W
would be equal and clearly the dimension of U’ has to be (n+1) due to the
Fundamental Theorem of Arithmetic. It follows that any (n+1) vectors that
span W must form a basis of W and so the logarithm of all these (n+1)
integers we have generated in this way would be linearly independent. From
this we can conclude that all the columns of the matrix representing our linear
transformation from U’ to V’ must be linearly independent as well for they
effectively represent these integers expressed as a product of all feasible
prime factors q1,…,qn if Goldbach’s Conjecture doesn’t hold true for 2N. Thus
we have proven that this initial matrix has to have full column rank. Now
there is a basic theorem of linear algebra which states that the column and row
rank of a matrix of a linear transformation have to be equal. Clearly this
would be impossible for the number of columns in our matrix representing our
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linear transformation from U’ to V’ would always exceed the number of rows
by one, due to the way we initially constructed the algorithm above. Thus we
have arrived at a contradiction and so our algorithm would have to compute at
least one instance of Goldbach’s Conjecture in this scenario which contradicts
what we are assuming to begin with. If we now consider the matrix
representing the entire linear map from U to V, but containing just the element
log(qn+1) from the interval ([2N/3], N) in U, this too cannot be surjective
which means that it must have at least one linearly dependent row. However,
these rows cannot be a linear combination of the others for if they were our
algorithm would have added the primes corresponding to these rows into the
set {q1, q2,…,qn}and our matrix from U’ to V’ containing these primes would
be surjective and thus have full row rank. The only way we can now have at
least one linearly dependent row in this matrix from U to V is if these rows
are entirely zero.
For a momentary diversion, if we now consider the properties of the matrices
representing the entire general linear maps from U to V for a particular value
of 2N, we know that the row and column ranks of these matrices have to be
equal. We can use this fact to say that for a particular value of 2N, # instances
of Goldbach’s Conjecture for 2N = # primes in the interval ([2N/3], N) - #
linearly dependent matrix columns + # linearly dependent matrix rows, where
# denotes “the number of”.
If we now consider the matrix of the aforementioned linear subtransformation from U to V in which the subspaces U’ and V’ are at their
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maximum size we would have one prime in the interval ([2N/3], N), one
linearly dependent column and at least one linearly dependent row so we
would have to have at least one instance of Goldbach’s Conjecture. If we now
consider the matrix representing the entire linear map from U to V, if these
zero rows happened to be spanned by another prime in the interval ([2N/3], N)
when they are transformed by our linear map, then at least one instance of
Goldbach’s Conjecture would still have to exist. For if that situation was to
occur and if we were just to consider the relevant primes from ([2N/3], N) as
and addition to our linear map on U we would not be adding any linearly
dependent rows or columns to our matrix but we would obviously have at
least one more prime contained in that interval. Our instances of Goldbach’s
Conjecture could not occur in the interval ([2N/3], N) for there are only a
finite number of primes contained within it and every time we added one to
that interval we would need to add another one to it if it wasn’t to exist in the
interval (2, [2N/3]]. However, the only way our algorithm could not have
computed an instance of Goldbach’s Conjecture in (2, [2N/3]] would be if the
primes corresponding to these zero row formed half of a prime pair summing
to 2N and thus proving Goldbach’s Conjecture for that particular scenario
involving our algorithm. When the subspaces U’ and V’ are not at their
maximum size we would still have to have at least one zero row in the matrix
representing this map from U to V. It could be the case that one or more of the
primes corresponding to these zero rows does not form part of a prime pair
summing to 2N as we’ve done above. However, if that was the case then all
zero rows corresponding to these primes would have to be spanned by another
prime in the interval ([2N/3], N) when we consider our linear map as a whole.
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This means that when we consider extending the matrix of the map from U’ to
V’ to that of this map from U to V we can compensate for each addition of a
linearly dependent column arising from the above cases by adding the
relevant linearly independent column from ([2N/3], N) to our matrix.
However, we would still have to have at least one zero row when we have
completed the extension of our matrix and so we would still have to have at
least one instance of Goldbach’s Conjecture arising in the same way as above.
If our algorithm calculates just one instance of Goldbach’s Conjecture before
terminating then we will now have created a bijective linear map between U’
and V’. If it calculates more than one instance of Goldbach’s Conjecture
before terminating then we will now have an injective map between U’ and
V’.
Using the above arguments we can conclude that Goldbach’s Conjecture
holds true for 2N≥16. Clearly Goldbach’s Conjecture holds true, by
inspection, for all even numbers between 4 and 14 as well.
Corollaries
1) If there is more than one prime number in the interval ([2N/3], N) for
any value of N ≥ 8 we can choose anyone of them to start our
algorithm and our arguments still hold. It could be the case that this
initial prime we use to start our algorithm satisfies an instance of
Goldbach’s Conjecture.
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2) In the matrix of the entire linear map from U to V there could be
instances of Goldbach’s Conjecture that are impossible to compute as
in the situation with our initial surjective map. However, as long as
there is at least one instance of Goldbach’s Conjecture that doesn’t
comply with those conditions then our algorithm will always compute
at least one instance of Goldbach’s Conjecture and produce an
injective map between U’ and V’.
3) If we are interested in being able to run our algorithm when N is prime
then we can change our inequality from above about primes in the
intervals (3n, 4n) with n≥2. Assuming N is prime, thus making
N=4n+3 instead of N being 4n+2 from above, our inequality changes
to:
3n+1 ≥ 2(4n+3)/3

9n+3 ≥ 8n+6

n≥3

N ≥ 15
Thus we can conclude with certainty that when 2N≥30 and N is prime,
there always has to exist at least one non-trivial instance of Goldbach’s
Conjecture.
Lemma
Proof of existence a prime in the interval (3n, 4n) with n ≥ 2
and integral
This is an adaptation of Erdos’s proof of Bertrand’s postulate.
If 4n/5 < p ≤ n then p ┼  4n  for p occurs once in the prime factorisation of n!
 3n 
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and 4 times in (4n)! as 5p > 4n. Therefore, if we assume the converse of what
we’re trying to prove then all prime factors p of  4n  satisfy p ≤ 4n/5.
 3n 
 n / p  where p
s( p)
The exponent of p in n! is
j
s( p)
≤ 4n < p s ( p )1 and so the
j 1
exponent of p in  4n  is
 3n 
s( p)

 
 
 1 = s(p) using
{ 4n / p j  3n / p j  n / p j }≤
j 1
s( p)
j 1
[x]+ [y] ≤ [x+y].
Hence, if s(p) > 1 then p ≤ 4n and it follows that no more than [2 n ]
primes
occur in  4n  with exponent larger than 1. Using this and p s ( p ) ≤ 4n we have
 3n 
4
n
  ≤ ( 4n) [ 2
 
 3n 
n]
 p.
p4n / 5
 4n  > p 2 n /(4n+1) because
 
 2n 
 4n  is the largest term in (1  1) 4 n with 4n+1
 
 2n 
summands.
 4n 
 
3n(3n  1)...( 2n  1)
 2n  =
≤ 2 n thus  4n  ≥ 2  n
2n(2n  1)...( n  1)
 3n 
 4n 
 
 3n 
Thus 8 n /(4n+1) < ( 4n) [ 2
n]
 p< 4
4n / 5
(4n)[ 2
p4n / 5
n]
n
 4n  > 8
 
 2n  4n  1
as
p< 4
n
using
p n
induction.
4n+1 < (4n) 2 and so 4 6 n / 5 < (4n) 2 2
n

6n
log 4 < (2+2 n ) log(4n).
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When n=55: 39.74 < 39.43 and this is contradicted when n>55 too.
One can test each value of 1<n<55 or for our purposes the following sequence
of primes will satisfy our conditions:
7,11,13,19,23,31,43,59,79,107,139,181,241
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References
1) “Selected topics in Number Theory” by H Gupta, Abacus Press 1980
2) “A course in Number Theory” by H. E Rose, Oxford Science
Publications 1988
3) www.geocities.com/capecanaveral/lab/8266/par_e.html
I learnt about the existence of a prime in the interval (3n, 4n) with n>1 from
1) but had to prove it myself by adapting a proof of Bertrand’s Postulate
(which states that there has to exist a prime in the interval (n, 2n) with n>1)
from 2). 3) was a source of inspiration for my ideas but was, unfortunately, an
incorrect proof of Goldbach’s Conjecture.
Mr David Cowham,
Eccles,
Greater Manchester,
England.
1st March, 2012.
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