| View previous topic :: View next topic |
| Author |
Message |
James Harris
Guest
|
 Posted: Thu Nov 06, 2003 9:16 am Post subject: Radical math ideas, difficult acceptance |
 |
|
Assume that there>s some embarrassing error within the highly
structured discipline of mathematics found by a non-mathematician,
what are the chances of ever getting it known if mathematicians refuse
to aknowledge it?
What if the error is easy enough to show as rather basic algebra does
the trick, and more importantly, the techniques used to show it,
represent an advancement of human knowledge?
My guess is that the scenario sounds impossible, so you guessed it, I
have the key proof that forms the linchpin of an argument showing a
fascinating error that crept into what is usually called mathematics
over a hundred years ago.
If you remember your algebra from school, it shouldn>t be too hard to
follow.
Note that ultimately the proof relies on 22 NOT having 7 as a factor,
and constant terms like 7 and 22, being constant, and not variables
dependent on x, which may seem like odd things to emphasize, but I>ve
faced posters who>ve gotten away with challenging those truths because
people seem unaware that>s what they>re doing.
1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is
in the ring of algebraic integers, notice that P(x) has a constant
term that is 1078.
2. It can be shown that
P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3
where you should note that using v = -1 + 49x, gives
P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3
where the *same* polynomial has been put in a form which allows a
factorization into non-polynomial factors so that I have
P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
where the a>s are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
3. Now let x=0, so
P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)
as the cubic defining the a>s at x=0 is
a^3 - 3a^2, which has roots, 0, 0 and 3, and I>ve picked a_1(0) and
a_2(0) to equal 0, which leaves a_3(0) with a value of 3.
4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I
have
P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
5. Now P(x) has a factor of 49 as
P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22
which means that
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)
has a factor of 49.
6. However, the constant term of P(x)/49 is 22, which is verified by
again setting x=0, which gives P(0)/49 = 22.
But for two of the factors of P(x), the constant terms is 7, which is
NOT a factor of 22. Therefore, *none* of the constant terms of
P(x)/49 as they multiply to give 22 can have 7 as a factor.
(By saying that 7 is NOT a factor of 22, I>m making a choice as to
where the proof is going. Since I>ve been talking about algebraic
integers, where 7 is NOT a factor of 22, it>s natural to go with a
choice where 7 is NOT a factor of 22.)
Given that the constant terms are independent of x>s value, it must be
the case that dividing P(x) by 49 divides the two constant terms equal
to 7, by 7.
7. But to divide 7 from those constant terms requires dividing
through two of the factors, so
(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
300125 x^3 - 18375 x^2 - 360 x + 22
from reverse use of the distributive property, which gives constant
terms that don>t have 7 as a factor, as required.
Notice that it>s a rather short and direct argument, where if you
accept that 22 does not have 7 as a factor, it>s obvious enough what
the constant terms of the factors must be as you go from 7, 7 and 22,
necessarily to 1, 1, and 22, when you divide P(x) by 49.
James Harris
http://mathforprofit.blogspot.com/ |
|
| |
|
 Back to top |
Joe Legris
Guest
|
| Posted: Thu Nov 06, 2003 5:24 pm Post subject: Re: Radical math ideas, difficult acceptance |
|
|
James Harris wrote:
[quote]Assume that there>s some embarrassing error within the highly
structured discipline of mathematics found by a non-mathematician,
what are the chances of ever getting it known if mathematicians refuse
to aknowledge it?
What if the error is easy enough to show as rather basic algebra does
the trick, and more importantly, the techniques used to show it,
represent an advancement of human knowledge?
My guess is that the scenario sounds impossible, so you guessed it, I
have the key proof that forms the linchpin of an argument showing a
fascinating error that crept into what is usually called mathematics
over a hundred years ago.
If you remember your algebra from school, it shouldn>t be too hard to
follow.
Note that ultimately the proof relies on 22 NOT having 7 as a factor,
and constant terms like 7 and 22, being constant, and not variables
dependent on x, which may seem like odd things to emphasize, but I>ve
faced posters who>ve gotten away with challenging those truths because
people seem unaware that>s what they>re doing.
1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is
in the ring of algebraic integers, notice that P(x) has a constant
term that is 1078.
2. It can be shown that
P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3
where you should note that using v = -1 + 49x, gives
P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3
where the *same* polynomial has been put in a form which allows a
factorization into non-polynomial factors so that I have
P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
where the a>s are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
3. Now let x=0, so
P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)
as the cubic defining the a>s at x=0 is
a^3 - 3a^2, which has roots, 0, 0 and 3, and I>ve picked a_1(0) and
a_2(0) to equal 0, which leaves a_3(0) with a value of 3.
4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I
have
P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
5. Now P(x) has a factor of 49 as
P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22
which means that
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)
has a factor of 49.
6. However, the constant term of P(x)/49 is 22, which is verified by
again setting x=0, which gives P(0)/49 = 22.
But for two of the factors of P(x), the constant terms is 7, which is
NOT a factor of 22. Therefore, *none* of the constant terms of
P(x)/49 as they multiply to give 22 can have 7 as a factor.
(By saying that 7 is NOT a factor of 22, I>m making a choice as to
where the proof is going. Since I>ve been talking about algebraic
integers, where 7 is NOT a factor of 22, it>s natural to go with a
choice where 7 is NOT a factor of 22.)
Given that the constant terms are independent of x>s value, it must be
the case that dividing P(x) by 49 divides the two constant terms equal
to 7, by 7.
7. But to divide 7 from those constant terms requires dividing
through two of the factors, so
(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
300125 x^3 - 18375 x^2 - 360 x + 22
from reverse use of the distributive property, which gives constant
terms that don>t have 7 as a factor, as required.
Notice that it>s a rather short and direct argument, where if you
accept that 22 does not have 7 as a factor, it>s obvious enough what
the constant terms of the factors must be as you go from 7, 7 and 22,
necessarily to 1, 1, and 22, when you divide P(x) by 49.
James Harris
http://mathforprofit.blogspot.com/
[/quote]
Hey Lester, look! A soul mate! |
|
| |
|
| Back to top |
Lester Zick
Guest
|
| Posted: Thu Nov 06, 2003 10:37 pm Post subject: Re: Radical math ideas, difficult acceptance |
|
|
On Thu, 06 Nov 2003 06:24:32 -0500, Joe Legris <jalegris@xympatico.ca>
in sci.cognitive wrote:
[quote]James Harris wrote:
Assume that there>s some embarrassing error within the highly
structured discipline of mathematics found by a non-mathematician,
what are the chances of ever getting it known if mathematicians refuse
to aknowledge it?
What if the error is easy enough to show as rather basic algebra does
the trick, and more importantly, the techniques used to show it,
represent an advancement of human knowledge?
My guess is that the scenario sounds impossible, so you guessed it, I
have the key proof that forms the linchpin of an argument showing a
fascinating error that crept into what is usually called mathematics
over a hundred years ago.
If you remember your algebra from school, it shouldn>t be too hard to
follow.
Note that ultimately the proof relies on 22 NOT having 7 as a factor,
and constant terms like 7 and 22, being constant, and not variables
dependent on x, which may seem like odd things to emphasize, but I>ve
faced posters who>ve gotten away with challenging those truths because
people seem unaware that>s what they>re doing.
1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is
in the ring of algebraic integers, notice that P(x) has a constant
term that is 1078.
2. It can be shown that
P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3
where you should note that using v = -1 + 49x, gives
P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3
where the *same* polynomial has been put in a form which allows a
factorization into non-polynomial factors so that I have
P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
where the a>s are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
3. Now let x=0, so
P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)
as the cubic defining the a>s at x=0 is
a^3 - 3a^2, which has roots, 0, 0 and 3, and I>ve picked a_1(0) and
a_2(0) to equal 0, which leaves a_3(0) with a value of 3.
4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I
have
P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
5. Now P(x) has a factor of 49 as
P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22
which means that
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)
has a factor of 49.
6. However, the constant term of P(x)/49 is 22, which is verified by
again setting x=0, which gives P(0)/49 = 22.
But for two of the factors of P(x), the constant terms is 7, which is
NOT a factor of 22. Therefore, *none* of the constant terms of
P(x)/49 as they multiply to give 22 can have 7 as a factor.
(By saying that 7 is NOT a factor of 22, I>m making a choice as to
where the proof is going. Since I>ve been talking about algebraic
integers, where 7 is NOT a factor of 22, it>s natural to go with a
choice where 7 is NOT a factor of 22.)
Given that the constant terms are independent of x>s value, it must be
the case that dividing P(x) by 49 divides the two constant terms equal
to 7, by 7.
7. But to divide 7 from those constant terms requires dividing
through two of the factors, so
(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
300125 x^3 - 18375 x^2 - 360 x + 22
from reverse use of the distributive property, which gives constant
terms that don>t have 7 as a factor, as required.
Notice that it>s a rather short and direct argument, where if you
accept that 22 does not have 7 as a factor, it>s obvious enough what
the constant terms of the factors must be as you go from 7, 7 and 22,
necessarily to 1, 1, and 22, when you divide P(x) by 49.
James Harris
http://mathforprofit.blogspot.com/
Hey Lester, look! A soul mate!
[/quote]
For you or me, Joe? See, now I>m beginning to feel the love. I knew
you couldn>t stay away. I don>t know whether it>s masochism or envy or
a combination on your part but I just knew you>d miss the abuse. This
is why organized boycotts and blacklists are so childish. People who
have nothing to say to one another usually just don>t say it. You on
the other hand seem to have lots to say to me but just don>t have the
cujones to make any kind of sensible case for what you would like to
say. But then instead of not saying it you run around advertising the
fact that you>re not saying it. Very childish indeed.
As far as James>s proof is concerned, sci.math seems a more
appropriate venue and if he wants to cross post it for some reason
that>s his business. And despite his lamentations at the beginning I
can>t really tell what he>s proving or trying to prove since he
doesn>t say. Poor technique to say the least especially if he>s
suggesting he>s uncovered some basic flaw in mathematical thinking. I
think you>d agree that I>ve never been shy about spelling out what it
was I was after.
I>d definitely be interested in an analytical proof that 7 is not a
factor of say 29 in addition to 22 but this proof just seems to rely
on the ostensible contention that 7 is not a factor of 22. However
since you>ve made such a point of asking my opinion I suppose I>ll
have to take a look to see if I can make anything of it.
[quote]
[/quote]
Regards - Lester |
|
| |
|
| Back to top |
James Harris
Guest
|
| Posted: Thu Nov 06, 2003 10:38 pm Post subject: Re: Radical math ideas, difficult acceptance |
|
|
Joe Legris <jalegris@xympatico.ca> wrote in message news:<3FAA2F70.4080302@xympatico.ca>...
[quote]
Hey Lester, look! A soul mate!
[/quote]
Who is Lester? And what relevance those that person have to a
troubling math discovery?
I>m sure some of you are thinkers about thinking based on the
newsgroup having "cognitive" as part of its name.
So I>d think that you>d be intrigued by something so easily testable,
as an assertion of a problem in the math world when a linchpin proof
uses basic algebra.
Or do you need more than mathematical logic?
Do you have *social* needs as well? Need notice that a high I.Q.
group is set to publish a paper on the subject? Need to know that top
mathematicians have looked it over and basically ran away in terror?
What are you looking for, an easy answer?
James Harris
http://mathforprofit.blogspot.com/ |
|
| |
|
| Back to top |
Lester Zick
Guest
|
| Posted: Fri Nov 07, 2003 3:03 am Post subject: Re: Radical math ideas, difficult acceptance |
|
|
On 6 Nov 2003 08:38:05 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
[quote]Joe Legris <jalegris@xympatico.ca> wrote in message news:<3FAA2F70.4080302@xympatico.ca>...
Hey Lester, look! A soul mate!
Who is Lester? And what relevance those that person have to a
troubling math discovery?
I>m sure some of you are thinkers about thinking based on the
newsgroup having "cognitive" as part of its name.
So I>d think that you>d be intrigued by something so easily testable,
as an assertion of a problem in the math world when a linchpin proof
uses basic algebra.
Or do you need more than mathematical logic?
Do you have *social* needs as well? Need notice that a high I.Q.
group is set to publish a paper on the subject? Need to know that top
mathematicians have looked it over and basically ran away in terror?
What are you looking for, an easy answer?
No, but it would help if you could explain what you are proving and[/quote]
why you post this exclusively to sci.cognitive and not sci.math as
well.
Regards - Lester |
|
| |
|
| Back to top |
Randolph M. Jones
Guest
|
| Posted: Fri Nov 07, 2003 3:51 am Post subject: Re: Radical math ideas, difficult acceptance |
|
|
James Harris wrote:
[quote]Assume that there>s some embarrassing error within the highly
structured discipline of mathematics found by a non-mathematician,
what are the chances of ever getting it known if mathematicians refuse
to aknowledge it?
What if the error is easy enough to show as rather basic algebra does
the trick, and more importantly, the techniques used to show it,
represent an advancement of human knowledge?
My guess is that the scenario sounds impossible, so you guessed it, I
have the key proof that forms the linchpin of an argument showing a
fascinating error that crept into what is usually called mathematics
over a hundred years ago.
If you remember your algebra from school, it shouldn>t be too hard to
follow.
Note that ultimately the proof relies on 22 NOT having 7 as a factor,
and constant terms like 7 and 22, being constant, and not variables
dependent on x, which may seem like odd things to emphasize, but I>ve
faced posters who>ve gotten away with challenging those truths because
people seem unaware that>s what they>re doing.
1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is
in the ring of algebraic integers, notice that P(x) has a constant
term that is 1078.
2. It can be shown that
P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3
where you should note that using v = -1 + 49x, gives
P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3
where the *same* polynomial has been put in a form which allows a
factorization into non-polynomial factors so that I have
P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
where the a>s are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
3. Now let x=0, so
P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)
as the cubic defining the a>s at x=0 is
a^3 - 3a^2, which has roots, 0, 0 and 3, and I>ve picked a_1(0) and
a_2(0) to equal 0, which leaves a_3(0) with a value of 3.
4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I
have
P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
5. Now P(x) has a factor of 49 as
P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22
which means that
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)
has a factor of 49.
6. However, the constant term of P(x)/49 is 22, which is verified by
again setting x=0, which gives P(0)/49 = 22.
But for two of the factors of P(x), the constant terms is 7, which is
NOT a factor of 22. Therefore, *none* of the constant terms of
P(x)/49 as they multiply to give 22 can have 7 as a factor.
(By saying that 7 is NOT a factor of 22, I>m making a choice as to
where the proof is going. Since I>ve been talking about algebraic
integers, where 7 is NOT a factor of 22, it>s natural to go with a
choice where 7 is NOT a factor of 22.)
Given that the constant terms are independent of x>s value, it must be
the case that dividing P(x) by 49 divides the two constant terms equal
to 7, by 7.
7. But to divide 7 from those constant terms requires dividing
through two of the factors, so
(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
300125 x^3 - 18375 x^2 - 360 x + 22
from reverse use of the distributive property, which gives constant
terms that don>t have 7 as a factor, as required.
Notice that it>s a rather short and direct argument, where if you
accept that 22 does not have 7 as a factor, it>s obvious enough what
the constant terms of the factors must be as you go from 7, 7 and 22,
necessarily to 1, 1, and 22, when you divide P(x) by 49.
James Harris
http://mathforprofit.blogspot.com/
[/quote]
So what>s the problem? |
|
| |
|
| Back to top |
Wolf Kirchmeir
Guest
|
| Posted: Fri Nov 07, 2003 2:53 pm Post subject: Re: Radical math ideas, difficult acceptance |
|
|
On 7 Nov 2003 05:46:13 -0800, James Harris wrote:
[quote]The last main point, point 7. is what>s being proven by what precedes
it.
[/quote]
Yes, and?????????????????
IOW, what do you think it _should_ be, and why?
--
Wolf Kirchmeir, Blind River ON Canada
"Nature does not deal in rewards or punishments, but only in consequences."
(Robert Ingersoll) |
|
| |
|
| Back to top |
James Harris
Guest
|
| Posted: Fri Nov 07, 2003 7:46 pm Post subject: Re: Radical math ideas, difficult acceptance |
|
|
lesterDELzick@worldnet.att.net (Lester Zick) wrote in message news:<3faab77b.81687303@netnews.att.net>...
[quote]On 6 Nov 2003 08:38:05 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
Joe Legris <jalegris@xympatico.ca> wrote in message news:<3FAA2F70.4080302@xympatico.ca>...
Hey Lester, look! A soul mate!
Who is Lester? And what relevance those that person have to a
troubling math discovery?
I>m sure some of you are thinkers about thinking based on the
newsgroup having "cognitive" as part of its name.
So I>d think that you>d be intrigued by something so easily testable,
as an assertion of a problem in the math world when a linchpin proof
uses basic algebra.
Or do you need more than mathematical logic?
Do you have *social* needs as well? Need notice that a high I.Q.
group is set to publish a paper on the subject? Need to know that top
mathematicians have looked it over and basically ran away in terror?
What are you looking for, an easy answer?
No, but it would help if you could explain what you are proving and
why you post this exclusively to sci.cognitive and not sci.math as
well.
[/quote]
The last main point, point 7. is what>s being proven by what precedes
it.
I>m posting here and not to sci.math because mathematicians are
running from the result because of its implications.
I have also posted the argument on sci.math and you can go there to
see the discussions to see what I mean.
James Harris
http://mathforprofit.blogspot.com/ |
|
| |
|
| Back to top |
Lester Zick
Guest
|
| Posted: Fri Nov 07, 2003 10:13 pm Post subject: Re: Radical math ideas, difficult acceptance |
|
|
On 7 Nov 2003 05:46:13 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
[quote]lesterDELzick@worldnet.att.net (Lester Zick) wrote in message news:<3faab77b.81687303@netnews.att.net>...
On 6 Nov 2003 08:38:05 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
Joe Legris <jalegris@xympatico.ca> wrote in message news:<3FAA2F70.4080302@xympatico.ca>...
Hey Lester, look! A soul mate!
Who is Lester? And what relevance those that person have to a
troubling math discovery?
I>m sure some of you are thinkers about thinking based on the
newsgroup having "cognitive" as part of its name.
So I>d think that you>d be intrigued by something so easily testable,
as an assertion of a problem in the math world when a linchpin proof
uses basic algebra.
Or do you need more than mathematical logic?
Do you have *social* needs as well? Need notice that a high I.Q.
group is set to publish a paper on the subject? Need to know that top
mathematicians have looked it over and basically ran away in terror?
What are you looking for, an easy answer?
No, but it would help if you could explain what you are proving and
why you post this exclusively to sci.cognitive and not sci.math as
well.
The last main point, point 7. is what>s being proven by what precedes
it.
[/quote]
I can tell that. What I don>t see is any explanation for what point 7
means in general mathematical terms. If you>re going to prove
something by means of algebraic manipulation, you really need an
abstract in the beginning that states what it is you are proving in
general terms and its implications with respect to mathematics.
[quote]
I>m posting here and not to sci.math because mathematicians are
running from the result because of its implications.
[/quote]
Yes, but what are its implications?
[quote]
I have also posted the argument on sci.math and you can go there to
see the discussions to see what I mean.
James Harris
http://mathforprofit.blogspot.com/
[/quote]
Regards - Lester |
|
| |
|
| Back to top |
James Harris
Guest
|
| Posted: Sat Nov 08, 2003 12:30 am Post subject: Re: Radical math ideas, difficult acceptance |
|
|
"Wolf Kirchmeir" <wwolfkir@sympatico.can> wrote in message news:<jbysxveflzcngvpbpna.hnzldd0.pminews@news1.sympatico.ca>...
[quote]On 7 Nov 2003 05:46:13 -0800, James Harris wrote:
The last main point, point 7. is what>s being proven by what precedes
it.
Yes, and?????????????????
IOW, what do you think it _should_ be, and why?
[/quote]
Think of my original post as a logical construct. It has a beginning,
a middle and an end. I>m saying that the ending follows from what
came before, so that the work can be considered by itself, and checked
for consistency.
I want you to do that first because the conclusion reached is HUGE,
but you may not quite understand yet why it is.
But if you understand the argument first, then when some mathematician
tries to cast doubt on your understanding, you can hopefully hold on
to it.
In case you>re wondering, those arguing with me are challenging main
points 6. and 7. claiming that 49 can divide off as a function of the
value of x, because they don>t want to accept that the factorization
given is the only one possible if you have that 22 does not have 7 as
a factor, despite the proof of it, focusing on constants.
The situation is bizarre because the logical position is being
challenged by mathematicians for social reasons. If they admit that
what I posted is correct then it changes their world, and how they
appear to the rest of the world.
I like the example because it>s within range of those of you who have
taken algebra and remembered the basics, which gives me hope that the
mathematicians can be forced to face the truth.
James Harris
http://mathforprofit.blogspot.com/ |
|
| |
|
| Back to top |
Lester Zick
Guest
|
| Posted: Sat Nov 08, 2003 3:30 am Post subject: Re: Radical math ideas, difficult acceptance |
|
|
On 7 Nov 2003 10:30:18 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
[quote]"Wolf Kirchmeir" <wwolfkir@sympatico.can> wrote in message news:<jbysxveflzcngvpbpna.hnzldd0.pminews@news1.sympatico.ca>...
On 7 Nov 2003 05:46:13 -0800, James Harris wrote:
The last main point, point 7. is what>s being proven by what precedes
it.
Yes, and?????????????????
IOW, what do you think it _should_ be, and why?
Think of my original post as a logical construct. It has a beginning,
a middle and an end. I>m saying that the ending follows from what
came before, so that the work can be considered by itself, and checked
for consistency.
I want you to do that first because the conclusion reached is HUGE,
but you may not quite understand yet why it is.
But if you understand the argument first, then when some mathematician
tries to cast doubt on your understanding, you can hopefully hold on
to it.
In case you>re wondering, those arguing with me are challenging main
points 6. and 7. claiming that 49 can divide off as a function of the
value of x, because they don>t want to accept that the factorization
given is the only one possible if you have that 22 does not have 7 as
a factor, despite the proof of it, focusing on constants.
The situation is bizarre because the logical position is being
challenged by mathematicians for social reasons. If they admit that
what I posted is correct then it changes their world, and how they
appear to the rest of the world.
I like the example because it>s within range of those of you who have
taken algebra and remembered the basics, which gives me hope that the
mathematicians can be forced to face the truth.
This begins to remind me of certain arguments advanced recently by[/quote]
another author wherein the precondition for explaining his theory was
that you first understand his claims. There was also a beginning and
middle but unfortunately no end. And I>m inclined to suspect the same
in this case as well.
Regards - Lester |
|
| |
|
| Back to top |
James Harris
Guest
|
| Posted: Sat Nov 08, 2003 4:29 am Post subject: Re: Radical math ideas, difficult acceptance |
|
|
lesterDELzick@worldnet.att.net (Lester Zick) wrote in message news:<3fabc548.92112709@netnews.att.net>...
[quote]On 7 Nov 2003 05:46:13 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
lesterDELzick@worldnet.att.net (Lester Zick) wrote in message news:<3faab77b.81687303@netnews.att.net>...
On 6 Nov 2003 08:38:05 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
Joe Legris <jalegris@xympatico.ca> wrote in message news:<3FAA2F70.4080302@xympatico.ca>...
Hey Lester, look! A soul mate!
Who is Lester? And what relevance those that person have to a
troubling math discovery?
I>m sure some of you are thinkers about thinking based on the
newsgroup having "cognitive" as part of its name.
So I>d think that you>d be intrigued by something so easily testable,
as an assertion of a problem in the math world when a linchpin proof
uses basic algebra.
Or do you need more than mathematical logic?
Do you have *social* needs as well? Need notice that a high I.Q.
group is set to publish a paper on the subject? Need to know that top
mathematicians have looked it over and basically ran away in terror?
What are you looking for, an easy answer?
No, but it would help if you could explain what you are proving and
why you post this exclusively to sci.cognitive and not sci.math as
well.
The last main point, point 7. is what>s being proven by what precedes
it.
I can tell that. What I don>t see is any explanation for what point 7
means in general mathematical terms. If you>re going to prove
something by means of algebraic manipulation, you really need an
abstract in the beginning that states what it is you are proving in
general terms and its implications with respect to mathematics.
[/quote]
The proof shows that the ring of algebraic integers is too small.
That ring is defined as having the roots of monic polynomials with
integer coefficients.
However, for certain values of x, you can show that a_1(x)/7 and
a_2(x)/7 are not algebraic integers though the proof shows that they
have 7 as a factor for any algebraic integer x, only not in the ring
of algebraic integers.
[quote]
I>m posting here and not to sci.math because mathematicians are
running from the result because of its implications.
Yes, but what are its implications?
[/quote]
Mathematicians assumed that the ring of algebraic integers was
complete in that it covered all numbers with certain properties of
integrality, like having factors in a meaningful sense like integers,
where you can say that 2 is a factor of 3 in a meaningful way.
Notice I say "meaningful" because in the field of complex numbers, 2
is trivially a factor of 3 as 2(3/2) = 3.
The proof I>ve posted isolates one factorization out of an infinity by
focusing on a key property of certain rings: the ability to have
meaningful factors.
That is, it focuses on rings where 7 is NOT a factor of 22.
My blog goes into some historical detail:
http://mathforprofit.blogspot.com/
The problem for math society is that I>ve found a ding, a blemish that
proves that their foundations are not perfect.
It may be a tiny ding, but apparently they>d rather run away from it
than face it, so I need--you know, like an intervention.
James Harris |
|
| |
|
| Back to top |
James Harris
Guest
|
| Posted: Sat Nov 08, 2003 9:14 am Post subject: Re: Radical math ideas, difficult acceptance |
|
|
lesterDELzick@worldnet.att.net (Lester Zick) wrote in message news:<3fac0e19.1430132@netnews.att.net>...
[quote]On 7 Nov 2003 10:30:18 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
"Wolf Kirchmeir" <wwolfkir@sympatico.can> wrote in message news:<jbysxveflzcngvpbpna.hnzldd0.pminews@news1.sympatico.ca>...
On 7 Nov 2003 05:46:13 -0800, James Harris wrote:
The last main point, point 7. is what>s being proven by what precedes
it.
Yes, and?????????????????
IOW, what do you think it _should_ be, and why?
Think of my original post as a logical construct. It has a beginning,
a middle and an end. I>m saying that the ending follows from what
came before, so that the work can be considered by itself, and checked
for consistency.
I want you to do that first because the conclusion reached is HUGE,
but you may not quite understand yet why it is.
But if you understand the argument first, then when some mathematician
tries to cast doubt on your understanding, you can hopefully hold on
to it.
In case you>re wondering, those arguing with me are challenging main
points 6. and 7. claiming that 49 can divide off as a function of the
value of x, because they don>t want to accept that the factorization
given is the only one possible if you have that 22 does not have 7 as
a factor, despite the proof of it, focusing on constants.
The situation is bizarre because the logical position is being
challenged by mathematicians for social reasons. If they admit that
what I posted is correct then it changes their world, and how they
appear to the rest of the world.
I like the example because it>s within range of those of you who have
taken algebra and remembered the basics, which gives me hope that the
mathematicians can be forced to face the truth.
This begins to remind me of certain arguments advanced recently by
another author wherein the precondition for explaining his theory was
that you first understand his claims. There was also a beginning and
middle but unfortunately no end. And I>m inclined to suspect the same
in this case as well.
[/quote]
No need to get excited as I explained in another reply but will
happily explain again.
Mathematicians have relied on the assumption that the ring of
algebraic integers is a full and complete repository of certain
numbers that maintain integrality, such that you can say something
like 7 is not a factor of 22.
It turns out that the short and basic algebra argument that I give
blows that notion out of the water.
Besides that it>s also just a cool proof as it links ring properties
to a particular factorization, selecting one factorization out of
infinity.
What I do is use one polynomial to analyze the roots of another in a
rather basic and nifty way. Mathematicians should celebrate the
advancement, but are clearly shying away from the social implications.
James Harris
http://mathforprofit.blogspot.com/ |
|
| |
|
| Back to top |
Lester Zick
Guest
|
| Posted: Sat Nov 08, 2003 11:59 pm Post subject: Re: Radical math ideas, difficult acceptance |
|
|
On 7 Nov 2003 14:29:04 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
[quote]lesterDELzick@worldnet.att.net (Lester Zick) wrote in message news:<3fabc548.92112709@netnews.att.net>...
On 7 Nov 2003 05:46:13 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
lesterDELzick@worldnet.att.net (Lester Zick) wrote in message news:<3faab77b.81687303@netnews.att.net>...
On 6 Nov 2003 08:38:05 -0800, jstevh@msn.com (James Harris) in
sci.cognitive wrote:
Joe Legris <jalegris@xympatico.ca> wrote in message news:<3FAA2F70.4080302@xympatico.ca>...
Hey Lester, look! A soul mate!
Who is Lester? And what relevance those that person have to a
troubling math discovery?
I>m sure some of you are thinkers about thinking based on the
newsgroup having "cognitive" as part of its name.
So I>d think that you>d be intrigued by something so easily testable,
as an assertion of a problem in the math world when a linchpin proof
uses basic algebra.
Or do you need more than mathematical logic?
Do you have *social* needs as well? Need notice that a high I.Q.
group is set to publish a paper on the subject? Need to know that top
mathematicians have looked it over and basically ran away in terror?
What are you looking for, an easy answer?
No, but it would help if you could explain what you are proving and
why you post this exclusively to sci.cognitive and not sci.math as
well.
The last main point, point 7. is what>s being proven by what precedes
it.
I can tell that. What I don>t see is any explanation for what point 7
means in general mathematical terms. If you>re going to prove
something by means of algebraic manipulation, you really need an
abstract in the beginning that states what it is you are proving in
general terms and its implications with respect to mathematics.
The proof shows that the ring of algebraic integers is too small.
That ring is defined as having the roots of monic polynomials with
integer coefficients.
However, for certain values of x, you can show that a_1(x)/7 and
a_2(x)/7 are not algebraic integers though the proof shows that they
have 7 as a factor for any algebraic integer x, only not in the ring
of algebraic integers.
I>m posting here and not to sci.math because mathematicians are
running from the result because of its implications.
Yes, but what are its implications?
Mathematicians assumed that the ring of algebraic integers was
complete in that it covered all numbers with certain properties of
integrality, like having factors in a meaningful sense like integers,
where you can say that 2 is a factor of 3 in a meaningful way.
Notice I say "meaningful" because in the field of complex numbers, 2
is trivially a factor of 3 as 2(3/2) = 3.
The proof I>ve posted isolates one factorization out of an infinity by
focusing on a key property of certain rings: the ability to have
meaningful factors.
That is, it focuses on rings where 7 is NOT a factor of 22.
[/quote]
Based on these and the other comments and taking the proof at face
value, would it be fair to say that what your demonstration proves is
that there are relative prime integers greater than 1? And if so why
does it follow that the field of algebraic integers is too small?
[quote]
[/quote]
Regards - Lester |
|
| |
|
| Back to top |
|
