- Hi, this may be a trivial (or well known) result, but

when I first observed it I was quite surprised.

NOTE: [] indicates subscript.

Let c = q[1]q[2]...q[n], where q is an odd prime

number then c is an odd composite.

Let P = {p[1],p[2],...,p[m]}, where p is an odd prime

and gcd(p,q)=1, p[m] < 2c < p[m+1]

Let A and B be partition cells of P, such that {p:p in

A, p < c} and {p':p' in B, c < p' < 2c}.

Case 1.

If p[1]^2 > 2c, p[1] in A, then 2c - p = p', where p

in A, and p' in B.

Ex.

c = 15 = 3*5

2c = 30

P = {7,11,13,17,19,23,29}

A = {7,11,13}

B = {17,19,23,29}

30 - 7 = 23

30 - 11 = 19

30 - 13 = 17

Case 2.

If c < p[1]^2 < 2c, p[1] in A, then 2c - p' = p

(except when p = 1), where p in A, and p' in B.

Ex.

c = 21 = 3*7

2c = 42

P = {5,11,13,17,19,23,29,31,37,41}

A = {5,11,13,17,19}

B = {23,29,31,37,41}

42 - 23 = 19

42 - 29 = 13

42 - 31 = 11

42 - 37 = 5

What does this suggest? This suggests that there is a

fundamental symmetry within the primes.

Proof:

NOTE: this proof is only valid for case 1 and with

slight modification for case 2.

Using the previous definitions unless specified:

Let q' be the mirror image of q with respect to c,

meaning that q'-c=c-q or q'=c+c-q (i.e. the distance

from q to c is equal to the distance from q' to c).

Recall that c=q[1]q[2]...q[n].

Let p' be the mirror image of p with respect to c,

meaning that p'-c=c-p or p'=c+c-p (i.e. the distance

from p to c is equal to the distance from p' to c).

Recall that p is an element in A and p' is an element

in B.

Since q|c and q|(c-q), then it follows that q|(c+c-q)

or q|q'.

Since p does not divide c and p does not divide c-p,

then it follows that p does not divide c+c-p or p does

not divide p'.

Therefore p'=c+c-p or p'=2c-p and a p with equal

distance from c exists on either side of c as was

defined in case 1. Hence the symmetry.

Q.E.D.

A similar proof could be shown for case 2 but a

general case, where p[1]^ < c, would begin to be

incoherent because the elements in A would create

composite numbers that when are subtracted from 2c

yeild an element in B.

Ex.

c=25=5*5

2c=50

P={3,7,11,13,17,19,23,29,31,37,41,43,47}

A={3,7,11,13,17,19,23}

B={29,31,37,41,43,47}

But...

3*3=9 and 50-9=41, which is an element of B, therefore

50-41 does not produce a prime.

When all goes well, it is obvious that p+p'=2c, which

implies Goldbach's Conjecture. And if GC is false,

then all of the p must produce composite numbers that

equal 2c-p'. Is this possible?

-Mike

__________________________________

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http://search.yahoo.com - Hi Mike:

I think that you haven't prove that p' is prime, only that it is not divisible by p.

Jose----- Original Message -----

From: Mike Antczak

To: primenumbers@yahoogroups.com

Sent: Wednesday, July 09, 2003 5:21 PM

Subject: [PrimeNumbers] prime symmetry

Hi, this may be a trivial (or well known) result, but

when I first observed it I was quite surprised.

NOTE: [] indicates subscript.

Let c = q[1]q[2]...q[n], where q is an odd prime

number then c is an odd composite.

Let P = {p[1],p[2],...,p[m]}, where p is an odd prime

and gcd(p,q)=1, p[m] < 2c < p[m+1]

Let A and B be partition cells of P, such that {p:p in

A, p < c} and {p':p' in B, c < p' < 2c}.

Case 1.

If p[1]^2 > 2c, p[1] in A, then 2c - p = p', where p

in A, and p' in B.

Ex.

c = 15 = 3*5

2c = 30

P = {7,11,13,17,19,23,29}

A = {7,11,13}

B = {17,19,23,29}

30 - 7 = 23

30 - 11 = 19

30 - 13 = 17

Case 2.

If c < p[1]^2 < 2c, p[1] in A, then 2c - p' = p

(except when p = 1), where p in A, and p' in B.

Ex.

c = 21 = 3*7

2c = 42

P = {5,11,13,17,19,23,29,31,37,41}

A = {5,11,13,17,19}

B = {23,29,31,37,41}

42 - 23 = 19

42 - 29 = 13

42 - 31 = 11

42 - 37 = 5

What does this suggest? This suggests that there is a

fundamental symmetry within the primes.

Proof:

NOTE: this proof is only valid for case 1 and with

slight modification for case 2.

Using the previous definitions unless specified:

Let q' be the mirror image of q with respect to c,

meaning that q'-c=c-q or q'=c+c-q (i.e. the distance

from q to c is equal to the distance from q' to c).

Recall that c=q[1]q[2]...q[n].

Let p' be the mirror image of p with respect to c,

meaning that p'-c=c-p or p'=c+c-p (i.e. the distance

from p to c is equal to the distance from p' to c).

Recall that p is an element in A and p' is an element

in B.

Since q|c and q|(c-q), then it follows that q|(c+c-q)

or q|q'.

Since p does not divide c and p does not divide c-p,

then it follows that p does not divide c+c-p or p does

not divide p'.

Therefore p'=c+c-p or p'=2c-p and a p with equal

distance from c exists on either side of c as was

defined in case 1. Hence the symmetry.

Q.E.D.

A similar proof could be shown for case 2 but a

general case, where p[1]^ < c, would begin to be

incoherent because the elements in A would create

composite numbers that when are subtracted from 2c

yeild an element in B.

Ex.

c=25=5*5

2c=50

P={3,7,11,13,17,19,23,29,31,37,41,43,47}

A={3,7,11,13,17,19,23}

B={29,31,37,41,43,47}

But...

3*3=9 and 50-9=41, which is an element of B, therefore

50-41 does not produce a prime.

When all goes well, it is obvious that p+p'=2c, which

implies Goldbach's Conjecture. And if GC is false,

then all of the p must produce composite numbers that

equal 2c-p'. Is this possible?

-Mike

__________________________________

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[Non-text portions of this message have been removed] - --- Jose_Ram�n_Brox <ambroxius@...> wrote:
> Hi Mike:

Right, first of all p' cannot be even because

>

> I think that you haven't prove that p' is prime,

> only that it is not divisible by p.

>

> Jose

> ----- Original Message -----

> From: Mike Antczak

> To: primenumbers@yahoogroups.com

> Sent: Wednesday, July 09, 2003 5:21 PM

> Subject: [PrimeNumbers] prime symmetry

>

>

> Hi, this may be a trivial (or well known) result,

> but

> when I first observed it I was quite surprised.

>

> NOTE: [] indicates subscript.

>

> Let c = q[1]q[2]...q[n], where q is an odd prime

> number then c is an odd composite.

>

> Let P = {p[1],p[2],...,p[m]}, where p is an odd

> prime

> and gcd(p,q)=1, p[m] < 2c < p[m+1]

>

> Let A and B be partition cells of P, such that

> {p:p in

> A, p < c} and {p':p' in B, c < p' < 2c}.

>

> Case 1.

> If p[1]^2 > 2c, p[1] in A, then 2c - p = p', where

> p

> in A, and p' in B.

>

> Ex.

> c = 15 = 3*5

> 2c = 30

> P = {7,11,13,17,19,23,29}

> A = {7,11,13}

> B = {17,19,23,29}

>

> 30 - 7 = 23

> 30 - 11 = 19

> 30 - 13 = 17

>

> Case 2.

> If c < p[1]^2 < 2c, p[1] in A, then 2c - p' = p

> (except when p = 1), where p in A, and p' in B.

>

> Ex.

> c = 21 = 3*7

> 2c = 42

> P = {5,11,13,17,19,23,29,31,37,41}

> A = {5,11,13,17,19}

> B = {23,29,31,37,41}

>

> 42 - 23 = 19

> 42 - 29 = 13

> 42 - 31 = 11

> 42 - 37 = 5

>

> What does this suggest? This suggests that there

> is a

> fundamental symmetry within the primes.

>

> Proof:

>

> NOTE: this proof is only valid for case 1 and with

> slight modification for case 2.

>

> Using the previous definitions unless specified:

>

> Let q' be the mirror image of q with respect to c,

> meaning that q'-c=c-q or q'=c+c-q (i.e. the

> distance

> from q to c is equal to the distance from q' to

> c).

> Recall that c=q[1]q[2]...q[n].

>

> Let p' be the mirror image of p with respect to c,

> meaning that p'-c=c-p or p'=c+c-p (i.e. the

> distance

> from p to c is equal to the distance from p' to

> c).

> Recall that p is an element in A and p' is an

> element

> in B.

>

> Since q|c and q|(c-q), then it follows that

> q|(c+c-q)

> or q|q'.

>

> Since p does not divide c and p does not divide

> c-p,

> then it follows that p does not divide c+c-p or p

> does

> not divide p'.

>

> Therefore p'=c+c-p or p'=2c-p and a p with equal

> distance from c exists on either side of c as was

> defined in case 1. Hence the symmetry.

>

> Q.E.D.

>

> A similar proof could be shown for case 2 but a

> general case, where p[1]^ < c, would begin to be

> incoherent because the elements in A would create

> composite numbers that when are subtracted from 2c

> yeild an element in B.

>

> Ex.

> c=25=5*5

> 2c=50

> P={3,7,11,13,17,19,23,29,31,37,41,43,47}

> A={3,7,11,13,17,19,23}

> B={29,31,37,41,43,47}

>

> But...

> 3*3=9 and 50-9=41, which is an element of B,

> therefore

> 50-41 does not produce a prime.

>

>

> When all goes well, it is obvious that p+p'=2c,

> which

> implies Goldbach's Conjecture. And if GC is

> false,

> then all of the p must produce composite numbers

> that

> equal 2c-p'. Is this possible?

>

> -Mike

p'=c+c-p. Since c is odd, odd+odd is even-odd is odd.

Also, the other possiblity is that q|p'. Since

p'=c+c-p then it follows that q|(c+c-p). Although

q|c, q cannot divide c-p because p is a prime such

that gcd(p,q)=1 (i.e. p and q are relatively prime).

Therefore, since p' is odd and not divisible by p or

q, p' is prime.

And remember, I'm only proving case 1.

__________________________________

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http://search.yahoo.com > --- Jose_Ram�n_Brox <ambroxius@...> wrote:

To show you what I mean...

> > Hi Mike:

> >

> > I think that you haven't prove that p' is prime,

> > only that it is not divisible by p.

> >

> > Jose

> > ----- Original Message -----

> > From: Mike Antczak

> > To: primenumbers@yahoogroups.com

> > Sent: Wednesday, July 09, 2003 5:21 PM

> > Subject: [PrimeNumbers] prime symmetry

> >

> >

> > Hi, this may be a trivial (or well known)

> result,

> > but

> > when I first observed it I was quite surprised.

> >

> > NOTE: [] indicates subscript.

> >

> > Let c = q[1]q[2]...q[n], where q is an odd prime

> > number then c is an odd composite.

> >

> > Let P = {p[1],p[2],...,p[m]}, where p is an odd

> > prime

> > and gcd(p,q)=1, p[m] < 2c < p[m+1]

> >

> > Let A and B be partition cells of P, such that

> > {p:p in

> > A, p < c} and {p':p' in B, c < p' < 2c}.

> >

> > Case 1.

> > If p[1]^2 > 2c, p[1] in A, then 2c - p = p',

> where

> > p

> > in A, and p' in B.

> >

> > Ex.

> > c = 15 = 3*5

> > 2c = 30

> > P = {7,11,13,17,19,23,29}

> > A = {7,11,13}

> > B = {17,19,23,29}

> >

> > 30 - 7 = 23

> > 30 - 11 = 19

> > 30 - 13 = 17

> >

> > Case 2.

> > If c < p[1]^2 < 2c, p[1] in A, then 2c - p' = p

> > (except when p = 1), where p in A, and p' in B.

> >

> > Ex.

> > c = 21 = 3*7

> > 2c = 42

> > P = {5,11,13,17,19,23,29,31,37,41}

> > A = {5,11,13,17,19}

> > B = {23,29,31,37,41}

> >

> > 42 - 23 = 19

> > 42 - 29 = 13

> > 42 - 31 = 11

> > 42 - 37 = 5

The best possibility for case 2 is c=105=3*5*7,

2c=210. Note that 11^2=121 which 105 < 121 < 210,

this means that 2c-p'=p which corresponds to the

definition of case 2.

All the primes greater than 105 and less than 210

subtracted from 210 will yield a prime less than 105.

Check it out.

> > What does this suggest? This suggests that

__________________________________

> there

> > is a

> > fundamental symmetry within the primes.

> >

> > Proof:

> >

> > NOTE: this proof is only valid for case 1 and

> with

> > slight modification for case 2.

> >

> > Using the previous definitions unless specified:

> >

> > Let q' be the mirror image of q with respect to

> c,

> > meaning that q'-c=c-q or q'=c+c-q (i.e. the

> > distance

> > from q to c is equal to the distance from q' to

> > c).

> > Recall that c=q[1]q[2]...q[n].

> >

> > Let p' be the mirror image of p with respect to

> c,

> > meaning that p'-c=c-p or p'=c+c-p (i.e. the

> > distance

> > from p to c is equal to the distance from p' to

> > c).

> > Recall that p is an element in A and p' is an

> > element

> > in B.

> >

> > Since q|c and q|(c-q), then it follows that

> > q|(c+c-q)

> > or q|q'.

> >

> > Since p does not divide c and p does not divide

> > c-p,

> > then it follows that p does not divide c+c-p or

> p

> > does

> > not divide p'.

> >

> > Therefore p'=c+c-p or p'=2c-p and a p with equal

> > distance from c exists on either side of c as

> was

> > defined in case 1. Hence the symmetry.

> >

> > Q.E.D.

> >

> > A similar proof could be shown for case 2 but a

> > general case, where p[1]^ < c, would begin to be

> > incoherent because the elements in A would

> create

> > composite numbers that when are subtracted from

> 2c

> > yeild an element in B.

> >

> > Ex.

> > c=25=5*5

> > 2c=50

> > P={3,7,11,13,17,19,23,29,31,37,41,43,47}

> > A={3,7,11,13,17,19,23}

> > B={29,31,37,41,43,47}

> >

> > But...

> > 3*3=9 and 50-9=41, which is an element of B,

> > therefore

> > 50-41 does not produce a prime.

> >

> >

> > When all goes well, it is obvious that p+p'=2c,

> > which

> > implies Goldbach's Conjecture. And if GC is

> > false,

> > then all of the p must produce composite numbers

> > that

> > equal 2c-p'. Is this possible?

> >

> > -Mike

>

> Right, first of all p' cannot be even because

> p'=c+c-p. Since c is odd, odd+odd is even-odd is

> odd.

> Also, the other possiblity is that q|p'. Since

> p'=c+c-p then it follows that q|(c+c-p). Although

> q|c, q cannot divide c-p because p is a prime such

> that gcd(p,q)=1 (i.e. p and q are relatively prime).

>

> Therefore, since p' is odd and not divisible by p or

> q, p' is prime.

>

> And remember, I'm only proving case 1.

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