## prime symmetry

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• 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],
Message 1 of 4 , Jul 9, 2003
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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• Hi Mike: I think that you haven t prove that p is prime, only that it is not divisible by p. Jose ... From: Mike Antczak To: primenumbers@yahoogroups.com
Message 2 of 4 , Jul 9, 2003
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
Sent: Wednesday, July 09, 2003 5:21 PM

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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• ... 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
Message 3 of 4 , Jul 9, 2003
--- Jose_Ram�n_Brox <ambroxius@...> wrote:
> 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
> Sent: Wednesday, July 09, 2003 5:21 PM
>
>
> 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

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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• ... To show you what I mean... The best possibility for case 2 is c=105=3*5*7, 2c=210. Note that 11^2=121 which 105
Message 4 of 4 , Jul 9, 2003
> --- Jose_Ram�n_Brox <ambroxius@...> wrote:
> > 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
> > 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

To show you what I mean...

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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