mm-80
===
I think I've figured out a way to show basically
all of you, including> people who think they don't know any
math that mathematicians have> been lying about my work. It's
so trivial you *should* wonder why> they thought they could
get away with it.Okay. (I've always *wanted* to see all those
lying mathematicians shownup, ;) but maybe there's something
to it.) Let's see this easy proofof mathematician lies.>
Here goes.>> My paper Advanced Polynomial Factorization
depends on considering a> factor of a polynomials that I call
g.ITYM factor of a polynomial, singular. These sorts of
syntacticissues can be very important in any hard science. Or
math, too. :)> (Paper linked to at
http://groups.msn.com/AmateurMath as usual.)I'm not going to
read a whole paper. This is supposed to be an *easy*proof. (I
haven't visited the link.)> And in my paper I start by showing
that I can write that as>> g = r + c>> where either r=0, or r
changes as the polynomial's value changes,> while c does
not.So, like if the original polynomial was F(x) = x^2 - 1,
then g could be (x-1) or (x+1), and r=x and c=-1 (or +1).
Makes sense, although it'skind of a funny way to write it.>
Now you can consider all factors of a given polynomial using
g's, with> something like>> g_1...g_k = P(x)>> where you have
k factors. (x-1)(x+1) = x^2 - 1. Yep. Gotcha. It raises a
slight problem if theoriginal polynomial is, say, 2x^2-2, or
something, since then r doesn'tequal x; it equals some factor
times x. But all right.> For instance, for P(x)=x^2 + 2x +
1,>> g_1 = x+1, g_2 = x+1, gives you 2 factors.>> Those are
polynomial factors, but I'm generalizing in a simple way to>
say that for the factors g, in general, you have an element I
call r,> which changes as the independent variable changes,
and you have> another element I call c, which does
not.Generalizing *beyond* polynomials? To what? Integers? :)>
For my example up above it's easy, as with g_1 = x + 1, x
varies as x> varies, while 1 does not.True dat.> Now that's
enough that the proof in the paper is straightforward, but>
posters have argued with me anyway, with some trying to argue
over the> de?ition of polynomial, amazingly enough.>>
However, consider that the g's have an important feature,
which is> that when x=0, I have>> g_1...g_k =
P(0).Obviously.> For instance, with my simple example, with
P(x) = x^2 + 2x + 1, with> x=0,>> P(0) = 1 = g_1(g_2), where
g_1 = g_2 = x+1 = 1.>> You see, P(0) gives the constant term,
so at x equal 0, the g's must> multiply to give the constant
term.Yep.> So then, maybe you still want to believe the
mathematicians and> question that I can write g = r + c.Why
would anyone dispute it?> Well consider that substituting
gives me>> g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which
gives>> r_1...r_k +...+c_1...c_k = P(x), which is>> r_1...r_k
+...+P(0) = P(x),Geez, that's confusing. Must be the lack of
whitespace. I'llstick to Perl, if you don't mind. Anyway, it
seems like some sort ofexample of why the above bit of text
is right, and I already agree withthat. So I'll skip it.>
which means that if you believe the mathematicians then
they've> convinced you to doubt algebra itself, as then you
must believe that> everything to the left of P(0) above can
*maybe* be constant, but also> *maybe* vary as x varies.>> So
why would mathematicians argue against such a simple
result?blah, blah, blah... I wanna see the proof! So I skip
the polemicizing.> Two reasons I suggest. First because they
wish to disagree with me.> Second because they probably
believe that they can get away with it.>> That is, MOST of
you will doubt algebra itself rather than consider> that
mathematicians, whom you probably don't even personally
know,> would lie.>> So where does this lead?Whew. Okay, where
*does* it lead?> Well the polynomial I show in the paper is>>
P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf>> which seems to be
just complicated enough to give mathematicians room> to
lie.You can say *that* again (but without the political
innuendos, please).Lessee... P(m) = ((mf-1)^3 + 1) * x^3 -
(3*(mf-1)*y^2) * x + (y^3)Looks reasonable, although I don't
see the point of all those extravariables. I assume they
*are* variables, right?> For instance, you may be saying,
HEY, what's with the ?m' when you> had ?x'
before??!!!Wait... you still *have* ?x'! Is ?m' supposed to
be the independentvariable here? That makes the algebra
yuckier.> Well, there's no rule that says that you have to
use the letter x as> the variable label for a polynomial.
Also, there are historical> reasons for my usage as it goes
back to my work with FLT where x, y> and z are used with x^p
+ y^p = z^p.(A little egotistical. You're not supposed to
quote *yourself* as ahistorical reason.)> Finally, the
weirder thing is that one poster in particular got a lot> of
mileage out of questioning my ?ding the constant term with
an> expression like the above by using m=0, as that gives
me>> P(0) = 3xy^2 + y^3>> and he got a lot of mileage for
YEARS (before I had discovered proof I> want to add and
after) by emphasizing that two of the ROOTS of such an>
expression considered as a polynomial with respect to x are
not> de?ed at that point.Huh? Nothing wrong with a
polynomial that only has one root. It's justlinear in x,
one might say.> Well that's easy enough to see as the
original expression is>> (v^3+1)x^3 - 3vxy^2 + y^3>> which if
you *wish* to see it as a polynomial with respect to x, is of>
degree 3, but when v=-1, it's of degree 1, so if you solve for
the> roots, you'll get funky stuff.>> Now when I was ?ding
the proof of FLT...remember the process took> some
years...(Danger, Will Robinson! Danger!)> at times I'd talk
of polynomials with respect to other> than m, but I re?ed my
discourse as my understanding improved.>> However, people
arguing with me did not.>> You may *choose* to believe that
they did not because they don't know> enough mathematics to
follow, but we're talking about actual> mathematicians
here.(Danger!)> What's more rational?>> I say it's more
rational to suppose that they *did* ?ure out that it> worked
as described, but also noticed that as long as they
disagreed,> no one seemed to call them on making false
statements, except me, and> they knew my credibility wasn't
so great.>> For most of you, there's probably the belief that
there's some funky> higher math involved that your pitiful
brain(Hoo boy. Let's just ride this one out.)> can't follow
or you> don't know about, as you may suppose that
mathematicians either> wouldn't lie, or they wouldn't lie in
such a dumb way where I could> catch them so easily.>> But
consider what's in the balance:>> 1. I discovered a proof of
Fermat's Last Theorem that's more> available to people in
general than most of what mathematicians have> been producing
lately.Available how? I've never seen any proof of FLT,
although I knowthat Wiles' proof is probably written up
somewhere. You'd think aneat FLT proof would make big waves.
(Oh, never mind.)> 2. Worse I did so having said I'd ?d it
years ago, and having spent> years looking for it posting a
lot of my ideas, and getting in insult> battles with posters,
quite a few who happened to be mathematicians.>> 3. Then to
add insult to injury, I keep questioning mathematicians in>
terms of their ethics, and maybe *extremely importantly* I
doubt that> Wiles found a proof of Fermat's Last Theorem.What
do you *think* he proved, then?> And those are just highlights
as there's more but I think that kind of> gives my point.>>
For mathematicians the situation could be considered one of
the worst> possible disasters they can imagine.>> EXCEPT, it
looks like all they have to do is either stay quiet, or>
*claim* I'm wrong.They could also claim you're right. That
just about exhausts thepossibilities, there. Unfortunately,
nobody thinks that you know anymath, which means you're
almost certainly *not* right. Hmm.> Many of you simply
believe them, and question algebra itself, which is> rather
sad. I'd think at least some of you valued your
educations.Oh, I do. My education *taught* me algebra. Did
yours?Seriously, though. Stop making these oblique general
insultsand give me that easy proof you promised.> Others of
you may ?ure it doesn't matter, maybe because Western>
civilization seems to be based on lying anyway, and maybe you
?ure I> should just grow up, accept that everybody lies and
move on. And I> don't have to talk about Enron or pedophile
Catholic priests or things> in that vein.blah blah blah...> I
mean, look at George W. Bush and Iraq. If people can be
*killed*> over lies, without consequences to the liars, then
what's with some> freaking stupid math?>> Good point.>>
Mathematicians *are* a part of society after all. Why should
they> tell the truth now? It'd be like Bush owning up. They
can just sit> tight, and be quiet, like so many American
citizens or they can out> and out lie, like so many other
patriots.blah blah blah...> After all, that's so easy, now
isn't it?>> Which is why you need to understand why I talk
about mathematicians> potentially being prosecuted. Liars
don't just stop because the gig> is up, as then, they
wouldn't necessarily be liars, then eh?blah blah blah...>>
It'd be against their *true* natures.>> James Harris>Wait!
*WHAT*!?! Where's that proof you promised? I waded through
thatwhole diatribe for *nothing*?! Dangit, Harris, you said
you had an easyproof that mathematicians lied, and all I get
is a discussion of how youchoose to express factors of
polynomials with ?g' and ?r' and ?c' insteadof normal-people
letters? Where's the beef?-ArthurIf anybody has any time,
I wouldn't mind you guys trying
this:http://www.chriscentral.com .... it's a message board
which supportsentering equations/math in standard
notation...so you can put maths in yourposts.Give it a go
please post any problems you ?d either there or
here.Chris.I've long thought that was a major opportunity
for improvement inUSENET, but unlike me, you took some action!
Good job! For your nextproject, why don't you ?d out what
impossible feat would be requiredto add something like this
to USENET itself. I used to think thiswould be
anti-democratic, as part of the beauty of USENET is that
youcould be a full and equal citizen writing in from the
African bush ona KAYPRO over a 1200 BAUD modem. Well, it's a
nice fantasy: but sinceit has nothing to do with reality, may
as well up the technicalstandard.I'm afraid your Java applet
overwhelms my old machine, but I'm gettinga more up to date
one any day ...Dear Edward Green:...> For your next>
project, why don't you ?d out what impossible feat would be
required> to add something like this to USENET itself. I used
to think this> would be anti-democratic, as part of the beauty
of USENET is that you> could be a full and equal citizen
writing in from the African bush on> a KAYPRO over a 1200
BAUD modem. Well, it's a nice fantasy: but since> it has
nothing to do with reality, may as well up the technical>
standard.A lot of web pages use imbedded graphics for the
formulae.Usenet accepts HTML posts, which can have imbedded
graphucs. It isconsidered bad manners to post this way,
however.David A. Smith>Usenet accepts HTML posts, which
can have imbedded graphucs. It is>considered bad manners to
post this way, however.In fact, both graphics and HTML are
forbidden by Usenet standards (outside of groups with
binaries in their names). This is not just theory: many
news servers enforce this rule by silently graphics and such
will be drastically limiting their audience.If you need HTML
and graphics, put up a Web page and then post the URL here.
But unless your equations or ?ures are quite complex, it's
probably better to do the best you can in plain text.-- Stan
Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com/Walrus meat as a diet is less
repulsive than seal.Actually, all considering, it
wouldn't be very hard. A standard knownas MathML allows for
the encoding of math in web pages. The images aremade by my
server to make sure everyone can see them, BUT, you
coulddownload a plugin that lets you view mathml properly,
and subsequentlyyou could put mathml into usenet without too
much dif?ulty.I welcome you to use my site where it helps
:-)Chrishttp://www.chriscentral.com Message Board with
Equation Support>Usenet accepts HTML posts, which can have
imbedded graphucs. It is>considered bad manners to post
this way, however.> In fact, both graphics and HTML are
forbidden by Usenet standards > (outside of groups with
binaries in their names). This is not > just theory: many
news servers enforce this rule by silently > graphics and
such will be drastically limiting their audience.> If you
need HTML and graphics, put up a Web page and then post the >
URL here. But unless your equations or ?ures are quite
complex, > it's probably better to do the best you can in
plain text.> If anybody has any time...Yes, you realy
have to have time. It takes minutes (760kb/s)until the java
applet WebEQ Editor is loaded.Than it takes time to type your
formula with thenot very user friendly WebEQ Editor and as a
resultyou will see a link to an image in your text, butyou
can not correct the math anymore.By the way WebEQ Editor has
a very limited font set.It only uses the symbol font.There
are several existing message boards that useLaTeX-like input
code for equations, that will than be rendered on the ?he s
erver. I think that isa much better solution than using
a fat java UI, withlots of limitations, like WebEQ
Editor.Bernhard> If anybody has any time, I wouldn't mind
you guys trying this:> http://www.chriscentral.com .... it's
a message board which supports> entering equations/math in
standard notation...so you can put maths in your> posts.Give
it a go please post any problems you ?d either there or
here.> Chris.>Dear Edward Green:>>...>> For your
next>> project, why don't you ?d out what impossible feat
would be required>> to add something like this to USENET
itself. I used to think this>> would be anti-democratic, as
part of the beauty of USENET is that you>> could be a full
and equal citizen writing in from the African bush on>> a
KAYPRO over a 1200 BAUD modem. Well, it's a nice fantasy: but
since>> it has nothing to do with reality, may as well up the
technical>> standard.>>A lot of web pages use imbedded
graphics for the formulae.>Usenet accepts HTML posts, which
can have imbedded graphucs. It is>considered bad manners to
post this way, however.Any time you deviate from de?ed
Usenet standards, be aware that some people will be using
software that can't view it, and some servers simply will not
pass the message along.-- Is that plutonium on your
gums?Shut up and kiss me! -- Marge and Homer SimpsonIn
sci.physics, Chris
Muktar<isotope@softhome.net><73d480e3.0306202246.2ac1705d@
posting.google.com>:> Actually, all considering, it wouldn't
be very hard. A standard known> as MathML allows for the
encoding of math in web pages. The images are> made by my
server to make sure everyone can see them, BUT, you could>
download a plugin that lets you view mathml properly, and
subsequently> you could put mathml into usenet without too
much dif?ulty.With about the same results as one gets when
putting HTML into Usenetnow: a general response of
eewwwwwwwwwwww, gross.:-PI use SLRN. It doesn't understand
HTML *or* base64-encoding.I suspect a number of others are in
the same boat.[rest snipped]-- #191, ewill3@earthlink.netIt's
still legal to go .sigless.> I've long thought that was
a major opportunity for improvement in> USENET, but unlike me,
you took some action! Good job! For your next> project, why
don't you ?d out what impossible feat would be required> to
add something like this to USENET itself. I used to think
thisUse latex (http://www.tex.ac.uk) in your posts, then if
someone feels the need to work out what
$frac{partialvec{B}}{partial t}&=&0$ means they can
always run it through their latex programme.Latex is free for
Unix and Windows and any other system you care to think
about.-- Frodo Morris http://users.ox.ac.uk/~wadh1342All your
bast are belong to us AKA Graham Lee, Wadham
CollegeSpectrumSofts currently on show at URL/speccy/:
Speccy@Home SETI ClientAlso the home of iloveyou.bas, the
?st PC virus ported to the ZX82!!!>Dear Edward Green:>
>>...>> For your next>> project, why don't you ?d out
what impossible feat would be required>> to add something
like this to USENET itself. I used to think this>> would be
anti-democratic, as part of the beauty of USENET is that you>
>> could be a full and equal citizen writing in from the
African bush on>> a KAYPRO over a 1200 BAUD modem. Well,
it's a nice fantasy: but since>> it has nothing to do with
reality, may as well up the technical>> standard.>>A
lot of web pages use imbedded graphics for the formulae.>
>Usenet accepts HTML posts, which can have imbedded graphucs.
It is>considered bad manners to post this way, however.>
Any time you deviate from de?ed Usenet standards, be aware
that some > people will be using software that can't view it,
and some servers simply > will not pass the message
along.Noted. Maybe that's why I added impossible. I suspect
or know itwouldn't be terribly technically dif?ult for
others, if not for me,to implement something like this, but
standard is the bugaboo: youwould need a critical mass of
people on the same page before trying tochange things.It
seems to me what's really wanted (for some
hyptotheticalNeoNet(SM)) is a standard of embedding
(compressed) binary images intext, plus, say, index card size
touch screen peripheral writingtablets -- you are writing your
post, you de?e a box with a few keystrokes or mouse clicks,
and then put your equations or sketchesinside it, by hand,
with the writing tablet. Transparent, and no needto
standardize on some math typsetting language. This would
mimic theway people communicate using a blackboard or a pad
of paper.Given the existence of so-called binary groups,
bandwidth reallywouldn't really be an issue: such a group
would use data/postcomparable to an ordinary binary. The
issues are inventing,promulgating and propagating a technical
standard. Interesting, thetouch sensitive writing tablet seems
like a natural peripheral, butdoesn't seem to be widely known
or used.Of course it might be problematic competing for disk
space with imagesof moose genetalia and all the other
material whose promulgation mustbe maintained to preserve the
right for free speech.impossible after all: given a standard,
a NeoGroup(SM) could ?neatly inside existing networks and
hardware, blah, blah, blah, andhave about the same overhead
as a binary group. In fact, maybe itcould be distributed as a
binary with some embedded coding which wouldsimply tell end
machines with the right software how to read theposts. So one
wouldn't have to mess with the Sacred Structure ofUsenet much
after all ... a new standard could sort of insinuateitself
inside the old, waiting for converts. I think such a
thingwould be a big hit on all the science groups: text +
handwrittenequations and sketches would be an order of
magnitude more powerfuland ?not to mention luminous ;-)
as a medium than text + asciimath or text + web page
references.Any Usenet savvy programmers out there want to
help me develop thisfor love?>> ...By the way WebEQ
Editor has a very limited font set.> It only uses the symbol
font.That has never been true, but it is certainly not true
for the currentversion of WebEQ, which supports all the
characters in the MathML 2.0Recommendation -- and then
some.Bob Mathews bobm@dessci.comDirector of Training
830-990-9699http://www.dessci.com/free.asp?free=newsDesign
Science, Inc. -- How Science CommunicatesMathType, WebEQ,
MathPlayer, MathFlow, Equation Editor, TeXaide>> ...By
the way WebEQ Editor has a very limited font set.> It only
uses the symbol font.And I replied:> That has never been
true, but it is certainly not true for the> current version
of WebEQ, which supports all the characters in> the MathML
2.0 Recommendation -- and then some.Actually that's not
correct either. Currently WebEQ supports around400
characters, with more planned. That's a great deal more than
usingonly the symbol font, but less than the full MathML
2.0 spec.Sorry for the confusion.Bob Mathews
bobm@dessci.comDirector of Training
830-990-9699http://www.dessci.com/free.asp?free=newsDesign
Science, Inc. -- How Science CommunicatesMathType, WebEQ,
MathPlayer, MathFlow, Equation Editor, TeXaideHere's a
look at the methods highlighted in my paper
AdvancedPolynomial Factorization where I bring down the
number of symbols. Notice if you can follow to the conclusion
you're looking at an*error* in taught mathematics.If you can't
follow when the only symbol left is m, then I don't knowhow
you could ever follow.The ring is algebraic integers, though
at one point you're pushed outof the ring by the ?
taught mathematics. I'm curious to know ifany of you can ?d
that point.Ok I have P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 +
3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)where f is a prime integer
other than 3, and u is coprime to f, andlooking at that x, I
see the possibility for the factorization P(m) = (a_1 x +
uf)(a_2 x + uf)(a_3 x + uf).However, now I'll let x=2, f=5,
u=1, so that I have P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)-
3(2)(-1 + 25m ) + 5)and P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x
+ 5).Now for P(0), I have P(0) = 25 (3(2) + 5), which is that
constant termI talk about a lot. And now notice that P(0)/25
= 3(2) + 5, which iscoprime to 5(yes, f is coprime to 3, x,
and u which is given in thepaper).Here's where it gets
interesting as while the key variable is m, Inotice that
focusing on those x's it looks like another cubic, so
Iconsider the factorization P(m) = (2 a_1 + 5)(2 a_2 + 5)(2
a_3 x + 5)and speci?ally focus on the factor 2 a_1 x + 5,
which I'll call g.I notice that if a_1 is 0, when m=0, as at
least one of the a's mustbe, then at that point g=5, and has
a factor that is 5.However, P(m)/25 has a factorization as
well, but its constant termis P(0)/25 = 3(2) + 5which is
coprime to 5, and considering g I see that factor of 5 thatis
visible with g = a_1 x + 5has to go away, and in going away it
must go through a_1.That is, I'm considering P(m)/25 = 8(625
m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5and I have to do it
somewhat obliquely because I don't know what thefactors of
the a's are right off just from looking at P(m) = (2 a_1 +
5)(2 a_2 + 5)(2 a_3 x + 5)though because P(m) = 25 (8(625 m^3
- 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)I know that the a's have
some factors in common with 5, which *should*be trivial.Now I
hope that putting in some numbers helped.James Harris>
Here's a look at the methods highlighted in my paper
Advanced> Polynomial Factorization where I bring down the
number of symbols. > Notice if you can follow to the
conclusion you're looking at an> *error* in taught
mathematics.> No. Your conclusion is incorrect. You continue
to insist that in thepolynomial factorization 65 x^3 - 12 x +
1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)with a1,a2,a3 algebraic
integers, at least one of the a's is coprime to5. More
recently, you have insisted that *all* of the a's are coprime
to5. I show that none of them is coprime to 5, that there is,
for each ai,a factor r(-ai) which is (1) an algebraic
integer, and (2) a factor of5. If you insist that all three
are units, then you are compelled toclaim that 5 is a unit in
the ring of algebraic integers.What is the error in the
following computation? It is a *DIRECT*calculation that shows
that your claim regarding the a's is incorrect.Do you retract
that claim? If not, why not?If you do not retract the claim,
you must disagree with my assertionthat is easily veri?d.You
must disavow the truth. Are you doing that?Why are you
disavowing the truth? De truth been berry berry good to
me.I'll repeat, for the record, that I have produced, for
EACH of the a's,two factorizations: 5 = q(-ai)*r(-ai) ai =
r(-ai)*s(-ai)That is, there are three a's. There are three
factorizations, and thereare three common factors (one for
each ai, a factor in common between aiand 5).For the sake of
completeness, here are the polynomials q, r, and s: q(x) = 8
x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x
- 104A quick bit of arithmetic will show the following:
q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x +
72)*p(x) + x,where p(x) = x^3 - 12 x^2 + 65. Note that this
immediately showsthat for z = any root of p(x), q(z)*r(z) =
5, r(z)*s(z) = z.Despite your ranting, you do not appear to
be up to the task ofcomputing the values of q,r,and s for the
roots, so I'll do thathere (courtesy of DOE Macsyma [version:
Maxima 5.9.0]), for oneof them.Recall that I have given these
results in terms of roots of thepolynomial: p(x) = x^3 - 12
x^2 + 65Those roots are all (-1) times the coef?ients ai
that you havemade these foolish claims about. Here are the
roots:Let u = (63 + i*sqrt(12415))/2, ubar = (63 -
i*sqrt(12415))/2, zeta = (-1 + i*sqrt(3))/2, zetabar = (-1 -
i*sqrt(3))/2.Then the roots of p(x) can be given as follows:
x1 = u^(1/3) + ubar^(1/3) + 4 x2 = zeta*u^(1/3) +
zetabar*ubar^(1/3) + 4 x3 = zetabar*u^(1/3) + zeta*ubar^(1/3)
+ 4where the (...)^(1/3) above are the values with argument =
1/3 timesthe argument of (...). Note that this convention
forces the threeroots to be real (as they must be, for this
polynomial). The roots areapproximately: x1 ~ 11.50930 x2 ~
-2.14375 x3 ~ 2.63445The factorizations claimed above are as
follows, in the case of the rootx1 = -a1: q1 = q(x1) = 8(63 +
i sqrt(12415))^(2/3)/2^(2/3) + 8 (63 - i
sqrt(12415))^(2/3)/2^(2/3) + 256/2^(2/3) - 12 (63 + i
sqrt(12415))^(1/3)/2^(1/3) - 12 (63 - i
sqrt(12415))^(1/3)/2^(1/3) - 361 r1 = q(x1) = 8(63 + i
sqrt(12415)^(2/3)/2^(2/3) + 8(63 - i
sqrt(12415)^(2/3)/2^(2/3) + 256/2^(2/3) + 60 (63 + i
sqrt(12415))^(1/3)/2^(1/3) + 60 (63 - i
sqrt(12415))^(1/3)/2^(1/3) + 67 s1 = s(x1) = 4(63 + i
sqrt(12415)^(2/3)/2^(2/3) + 4(63 - i
sqrt(12415)^(2/3)/2^(2/3) + 128/2^(2/3) + 5 (63 + i
sqrt(12415))^(1/3)/2^(1/3) + 5 (63 - i
sqrt(12415))^(1/3)/2^(1/3) - 188 While veri?ation that q1*r1
= 5 and r1*s1 = x1 is tedious, the aboveidentities: q(x)*r(x)
= (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) +
xsuf?e to produce the desired result (given that p(x1) =
0).> If you can't follow when the only symbol left is m, then
I don't know> how you could ever follow.It doesn't matter what
you're saying, because your methodology is? since it
gives incorrect results (namely: all a's coprimeto 5). ...
irrelevant nonsense deleted ...> I know that the a's have
some factors in common with 5, which *should*> be trivial.>
Please address the question, James Harris.> Now I hope that
putting in some numbers helped.> Ditto, James Harris. Answer
the question.What is the ? my DIRECT calculation (no
complicated logicinvolved, only arithmetic) that shows you're
wrong?> James HarrisDale.> Here's a look at the methods
highlighted in my paper Advanced> Polynomial Factorization
where I bring down the number of symbols. > Notice if you can
follow to the conclusion you're looking at an> *error* in
taught mathematics.> If you can't follow when the only
symbol left is m, then I don't know> how you could ever
follow.I was in a bit of a rush earlier. That should be
leaving only m andthe a's, as you have a_1, a_2 and a_3
below. > The ring is algebraic integers, though at one point
you're pushed out> of the ring by the ? taught
mathematics. I'm curious to know if> any of you can ?d that
point.> Ok I have> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 +
3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)> where f is a prime
integer other than 3, and u is coprime to f, and> looking at
that x, I see the possibility for the factorization> P(m) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf).> However, now I'll let
x=2, f=5, u=1, so that I have> P(m) = 25 (8(625 m^3 - 75 m^2
+ 3m)- 3(2)(-1 + 25m ) + 5)> and> P(m) = (2 a_1 + 5)(2 a_2
+ 5)(2 a_3 x + 5).Should be P(m) = (2 a_1 + 5)(2 a_2 + 5)(2
a_3 + 5).> Now for P(0), I have P(0) = 25 (3(2) + 5), which
is that constant term> I talk about a lot. And now notice that
P(0)/25 = 3(2) + 5, which is> coprime to 5(yes, f is coprime
to 3, x, and u which is given in the> paper).> Here's where
it gets interesting as while the key variable is m, I> notice
that focusing on those x's it looks like another cubic, so I>
consider the factorization> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2
a_3 x + 5)Again should be P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3
+ 5).> and speci?ally focus on the factor 2 a_1 x + 5,
which I'll call g.That should be 2 a_1 + 5.> I notice that if
a_1 is 0, when m=0, as at least one of the a's must> be, then
at that point g=5, and has a factor that is 5.> However,
P(m)/25 has a factorization as well, but its constant term>
is> P(0)/25 = 3(2) + 5> which is coprime to 5, and
considering g I see that factor of 5 that> is visible with>
g = a_1 x + 5Should be g = 2 a_1 + 5.> has to go away, and in
going away it must go through a_1.> That is, I'm
considering> P(m)/25 = 8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 +
25m ) + 5> and I have to do it somewhat obliquely because I
don't know what the> factors of the a's are right off just
from looking at> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x +
5)Should be P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). >
though because> P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)-
3(2)(-1 + 25m ) + 5)> I know that the a's have some factors
in common with 5, which *should*> be trivial.> Now I hope
that putting in some numbers helped.But I'm wary. It
shouldn't have been necessary for me to put innumbers, and
I'm wary that some poster is going to come along, throwup a
bunch of technical b.s and people will just believe yet
again.Mathematicians CAN lie to you. There's no rule written
into the lawsof physics that prevents them from LYING TO
YOU.James Harris> Here's a look at the methods
highlighted in my paper Advanced> Polynomial Factorization
where I bring down the number of symbols. > Notice if you can
follow to the conclusion you're looking at an> *error* in
taught mathematics.> If you can't follow when the only
symbol left is m, then I don't know> how you could ever
follow.> The ring is algebraic integers, though at one
point you're pushed out> of the ring by the ? taught
mathematics. I'm curious to know if> any of you can ?d that
point.> Ok I have> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 +
3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)> where f is a prime
integer other than 3, and u is coprime to f, and> looking at
that x, I see the possibility for the factorization> P(m) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf).> However, now I'll let
x=2, f=5, u=1, so that I have> P(m) = 25 (8(625 m^3 - 75 m^2
+ 3m)- 3(2)(-1 + 25m ) + 5)> and> P(m) = (2 a_1 + 5)(2 a_2
+ 5)(2 a_3 x + 5).> Now for P(0), I have P(0) = 25 (3(2) +
5), which is that constant term> I talk about a lot. And now
notice that P(0)/25 = 3(2) + 5, which is> coprime to 5(yes, f
is coprime to 3, x, and u which is given in the> paper).>
Here's where it gets interesting as while the key variable is
m, I> notice that focusing on those x's it looks like another
cubic, so I> consider the factorization> P(m) = (2 a_1 +
5)(2 a_2 + 5)(2 a_3 x + 5)> and speci?ally focus on the
factor 2 a_1 x + 5, which I'll call g.> I notice that if
a_1 is 0, when m=0, as at least one of the a's must> be, then
at that point g=5, and has a factor that is 5.> Note for the
discussion below: you are saying explicitly herethat when m =
0, a_1 = 0; and that you are saying that therefore* 5 is a
factor of a_1 *.> However, P(m)/25 has a factorization as
well, but its constant term> is> P(0)/25 = 3(2) + 5>
which is coprime to 5, and considering g I see that factor of
5 that> is visible with> g = a_1 x + 5> has to go away,
and in going away it must go through a_1.> As you have said
above, when m = 0, a_1 = 0. Of course0 is a multiple of 5,
albeit in a rather trivial sense. And right here is the key
to the error in your reasoning.You implicitly argue that
since a_1 is a multiple of 5 when m = 0, it must also be such
when m <> 0. The key thing here is, *** a_1 is not a constant!
***. It is an algebraic integer *variable* whose values depend
on the integer m. The fact that a_1 is 0 when m = 0 tells you
absolutely nothing about what its values may be, or how they
are related to the number 5, when m is not equal to zero. You
have made a giant leap here, based on an intuitive hunch -
no proof, no underlying mathematical principle. And as
demonstrated by both me and W. Dale Hall, your hunch here is
dead wrong. See below.> That is, I'm considering> P(m)/25 =
8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5> and I have to
do it somewhat obliquely because I don't know what the>
factors of the a's are right off just from looking at> P(m)
= (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5)> though because>
P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)>
I know that the a's have some factors in common with 5, which
*should*> be trivial.> Correct. In fact all of the a's share
factors with 5.> Now I hope that putting in some numbers
helped.> James Harris Putting in numbers just makes it
harder in this case foryou to see the forest for the trees.
It obscures thefact that a_1, a_2, and a_3 are FUNCTIONS of
m. If a_1were a constant with respect to m, everything would
workout. But obviously it cannot be. When m = 0, a_1 = 0.When
m <> 0, a_1 also cannot be equal to 0. The key thing here is,
you CANNOT use the fact that a_1 = 0when m = 0 to deduce
properties of a_1 when m <> 0. Think of it this way. When m =
0, a_1 = 0, so notonly is it divisible by 5, it is also
divisible by 7, 11, 137, 10^9 + 1, etc.. Does that imply that
when m <> 0, a_1 is also divisible by 7, 11, etc. ??? Of
course not! But that is exactly where your logic leads. You
have started a new thread here rather than respondto my proof
and comments, and Dale Hall's proof, in other threads. Dale
has forcefully reminded you of his proof that themain
conclusion of APF is wrong, and that you can verify this for
yourself with simple computations. Dalespeci?ally challenges
you to ?d an error in hisargument. Similarly I remind you of
my independent proof, reproduced below, to which you have not
provided a valid
objection:
===

The claim is made in Advanced
Polynomial Factorization that if> P(x) = 65*x^3 - 12*x +
1> is factored in the form> P(x) = (a1*x + 1)*(a2*x +
1)*(a3*x + 1),> where a1, a2, and a3 are algebraic
integers, then at least one> of a1, a2, and a3 must be
coprime to 5 in the algebraic integers.> Thus in the
following, assume that one of the ai, > say a1, is, as
claimed, coprime to 5 in the algebraic integers. > That is,
there exist algebraic integers s and t such that> s*a1 +
t*5 = 1. [1]> Let x = 1/u. Then P(x) = 0 implies that>
P(x) = 65/u^3 -12/u + 1 = 0, or> Q(u) = u^3 - 12*u^2 + 65 =
0.> Let r1, r2, and r3 be the roots of this> equation. All
of r1, r2, and r3 are algebraic> integers. Note that r1, r2,
and r3 happen to> be the negatives a1, a2, and a3 in some
order -> say, a1 = -r1, a2 = -r2, a3 = -r3.> Note that Q(u)
is irreducible over the rationals.> Let H be the ?ld of
algebraic numbers; clearly> r1, r2, and r3 are in H. Let F12
be an automorphism> of H such that F12(r1) = r2, and F12
leaves ?ed> the sub?ld of rational numbers. Note also that>
F12(a1) = a2, since a1 = -r1 and a2 = -r2.> {That such an
automorphism F12 exists is well-known,> and is described in:>
> http://www.math.niu.edu/~beachy/aaol/galois.html,> see
especially Proposition 8.6.2 on that page. Or> see the
excellent textbook, Abstract Algebra, by> John Beachy and
William D. Blair.}> Now apply the automorphism F12 to
both sides of> equation [1]:> F12(s)*F12(a1) +
F12(t)*F12(5) = F12(1).> But F12(1) = 1, F12(5) = 5, and
F12(a1) = a2. Thus> we obtain> s' * a2 + t' * 5 = 1, [2]>
where s' = F12(s) and F12(t) = t'. Note that the> automorphism
F12 preserves the property of being> an algebraic integer;
that is, s' and t', like> s and t, are algebraic integers.
Therefore> equation [2] implies that a1 is coprime to 5>
also.> Similarly one shows that a3 is coprime to 5.>
Thus, from the assumption that one of a1, a2, > or a3 is
coprime to 5, we deduce that all three> of them must be
coprime to 5.> But a1 * a2 * a3 = 65. Therefore at least
one> of a1, a2, or a3 is NOT coprime to 5. Therefore > a
contradiction. The source of the contradiction> was the
initial assumption that one of a1, a2, or> a3 is coprime to
5. Therefore NONE of a1, a2, and > a3 are coprime to 5,
contradicting the main result > of Advanced Polynomial
Factorization.>
=
== Thus the main claim in APF is absolutely false.
There are two different, independent proofs of its falsity.I
think it may be the case that essentially your same APF
argument is employed in your claimed proof of Fermat's Last
Theorem. This clearly should be the end of this discussion.
There is nohope now of retrieving what you wanted from APF.
In starting this thread, you have only driven more nails into
the cof?: you havemade it even more clear how your thinking
went astray. If youventure to add anything to this you will
simply be beating a horse that died on you some time ago.
Finally, I believe you submitted APF to one of the journals
of theAmerican Mathematical Society, and that it was
rejected. I wonderif you might be willing to share with us
what the editors orreviewers gave as their reasons for
rejection? Nora Baron[snip analysis and refutation of
James' latest failed attempt to salvage his proof]> Finally,
I believe you submitted APF to one of the journals of the>
American Mathematical Society, and that it was rejected. I
wonder> if you might be willing to share with us what the
editors or> reviewers gave as their reasons for rejection?>>
Nora BaronIf he does answer your above question, which I
doubt, he is most likely to respond that his manuscript
wasrejected because the reviewers were biased members of a
vast conspiracy dedicated to repelling invasions
by'outsiders'. James Harris believes he is infallible and his
proofs irrefutable. I wonder if anyone has everseen a more
spectacular example of self-imposed blindness before?--There
are two things you must never attempt to prove: the
unprovable -- and the obvious.--Democracy: The triumph of
popularity over principle.--http://www.crbond.commust ...
start ... another ... item!(I have no hands, but I must type
.-)> Here's a look at the methods highlighted in my paper
Advanced> Polynomial Factorization where I bring down the
number of symbols. > Notice if you can follow to the
conclusion you're looking at an> *error* in taught
mathematics.> No. Your conclusion is incorrect. You
continue to insist that in the> polynomial factorization>
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)> with
a1,a2,a3 algebraic integers, at least one of the a's is
coprime to> 5. More recently, you have insisted that *all* of
the a's are coprime to> 5. I show that none of them is coprime
to 5, that there is, for each ai,> a factor r(-ai) which is
(1) an algebraic integer, and (2) a factor of> 5. If you
insist that all three are units, then you are compelled to>
claim that 5 is a unit in the ring of algebraic integers.--A
church-school McCrusade (Blair's
ideals?):Harry-the-Mad-Potter want's US to kill Iraqis?...For
a 1000-year anglo-american hegemony?HEY, JIMMY; LET'S US
and SU FIGHT -then-PM of England & Zbiggy
http://www.tarpley.net/bush25.htm (Thyroid Storm ch.)
http://www.rwgrayprojects.com/synergetics/plates/plates.html
http://quincy4board.homestead.com/?es/curriculum/Cosmo.PCX==
==Well I'm going to try and break it down even more to try
and see ify'all will accept the mathematics:Ok I have P(m) =
f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3
f)where f is a prime integer other than 3, and u is coprime
to f, andlooking at that x, I see the possibility for the
factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf).That's what I've been putting up a lot where you see a
LOT of symbols,which seem to confuse people. The ring is
algebraic integers, and letme get rid of as many of those
symbols as I can:Let x=2, f=5, u=1, so that I have P(m) = 25
(8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)which is what
I put up earlier, but I'm wary about some of you still?ding
that confusing, so I'll work it out more to get P(m) =
25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is P(m) =
25(5000m^3 - 600 m^2 - 126m + 11), and now you can see what
the polynomial P(m) looks like without somany symbols.Now
from before where I had x, I *still* have that P(m) = (2 a_1
+ 5)(2 a_2 + 5)(2 a_3 + 5).So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5) = 25(5000m^3 - 600 m^2 - 126m + 11).Now setting m=0 gives
me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).Now let's say
you accept that any factor of a polynomial can bewritten like
r+c, where r=0, or r varies as the polynomial variablevaries,
while c remains constant and is a factor of the constant
term.Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),
which equals 0, when m=0, so at least one of the a's must
equal 0,when m=0.And to get that factor that is 25, you must
have two a's that go to 0,when m=0.(Note: Some posters have
gotten a lot of mileage out of calling that adegenerate
case, but they were just fooling you into forgetting
yourbasic algebra and what you know about polynomials. I
think they didso deliberately as the math isn't
complicated.)Now comes the question of what happens when m is
NOT 0, and answeringthat question requires looking again at (2
a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m +
11)and noticing that the constant term is 25(11), but you can
divide offthat 25 to get P(m)/25 which gives you a constant
term that's 11. And11 and 5 are coprime. That's very
important. In fact, that's the*key* fact which should stick
in your mind.So now looking at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5)/25 = (5000m^3 - 600 m^2 - 126m + 11)I know that if 2 a_1 +
5 has a factor of 5, when m=0, then that factoris a factor of
the constant term, and in fact, it'd have a factor ofthe
constant term that is 5. So why would you think that the
factorof the constant term would move or change when m
changes?Well it can't.Given that with 2 a_1 + 5, that 5 in
there is a factor of the constantterm of 25(5000m^3 - 600 m^2
- 126m + 11)you *still* have a factor of the constant term
when 25 is separatedoff.But now your constant term is 11.That
forces all the factors of 5 to go away from 2 a_1 + 5.Now you
may wish for there to be someway for some factors to
remain,but what actuall happens is you get (2 a_1/5 + 1)(2
a_2/5 + 1)(2 a_3 + 5) = (5000m^3 - 600 m^2 - 126m + 11)while
posters have argued that *all* the a's would have some factor
of5.But let's let m=0 again and see what happens now, as that
gives (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11and you'll
notice I have 1 where I had 5 before, so it can all work.That
is, the a's that went to 0 before will go to 0 again.Some
posters have claimed otherwise which I've called voodoo
math.Of course, if you think about it, you'll understand
that they probablyknew the truth and just lied to you as they
didn't think you could?ure it out, and probably didn't expect
me to simplify in this way.Those people aren't your friends.
You may trust them. You may admirethem, but they clearly
don't think much of you.So go back now and read posts from
Arturo Magidin, Keith Ramsay, andNora Baron among others, and
ask yourself: What do they think of me,really? James
Harris[snip]> So go back now and read posts from Arturo
Magidin, Keith Ramsay, and> Nora Baron among others, and ask
yourself: What do they think of me,> really?Harrigance - the
guy on the balcony:
http://users.pandora.be/vdmoortel/dirk/Stuff/Arrogance1.jpg
http://users.pandora.be/vdmoortel/dirk/Stuff/Arrogance2.
jpgDirk Vdm[snip techno-babble]> So go back now and read
posts from Arturo Magidin, Keith Ramsay, and> Nora Baron
among others, and ask yourself: What do they think of me,>
really?I can't speak for them, but I think you are an
arrogant imbecile with anattention span of about a
nanosecond, and a serious disorder of thecognitive
faculties.--There are two things you must never attempt to
prove: the unprovable --and the obvious.--Democracy: The
triumph of popularity over
principle.--http://www.crbond.com [.snip.]>So go back now
and read posts from Arturo Magidin, Keith Ramsay, and>Nora
Baron among others, and ask yourself: What do they think of
me,>really?What does it matter? Your arguments fall on their
own.Now, go back and read posts from James Harris, and ask
yourself: whatdoes he think of me, really? And
why?=
=Why do you take so much trouble to expose such
a reasoner as Mr. Smith? I answer as a deceased friend of
mine used to answer on like occasions - A man's capacity is
no measure of his power to do mischief. Mr. Smith has
untiring energy, which does something; self-evident honesty
of conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
?ures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
inde?ite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by
Augustus de
Morgan
===


===
Arturo Magidinmagidin@math.berkeley.edu>
Well I'm going to try and break it down even more to try and
see if> y'all will accept the mathematics:> Ok I have>
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 +
u^3 f)> where f is a prime integer other than 3, and u is
coprime to f, and> looking at that x, I see the possibility
for the factorization> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3
x + uf).> That's what I've been putting up a lot where you
see a LOT of symbols,> which seem to confuse people. The ring
is algebraic integers, and let> me get rid of as many of those
symbols as I can:> Let x=2, f=5, u=1, so that I have> P(m)
= 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)> which
is what I put up earlier, but I'm wary about some of you
still> ?ding that confusing, so I'll work it out more to
get> P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5),
which is> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > and
now you can see what the polynomial P(m) looks like without
so> many symbols.> Now from before where I had x, I *still*
have that> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).> So>
> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > 25(5000m^3 - 600
m^2 - 126m + 11).> Now setting m=0 gives me> (2 a_1 +
5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).> Now let's say you
accept that any factor of a polynomial can be> written like
r+c, where r=0, or r varies as the polynomial variable>
varies, while c remains constant and is a factor of the
constant term.> Notice that > a_1 a_2 a_3 = 25 (8(625 m^3
- 75 m^2 + 3m)), I think this should be a_1(m) a_2(m) a_3(m) =
25(625 m^3 - 75m^2 + 3m)> which equals 0, when m=0, so at
least one of the a's must equal 0,> when m=0.> And to get
that factor that is 25, you must have two a's that go to 0,>
when m=0.Why can't a_2(0) = 5, a_3(0) = -2/3? You have made
claims to the effect that you think these functions are
continuous, so they certainly should take on non-algebraic
integer values.Also, why do you INSIST on not using the
notation a_i(m) or a_i(0)? It would make your work easier to
read and much clearer.-- Will Twentyman[snip]> Also, why
do you INSIST on not using the notation a_i(m) or a_i(0)? It>
would make your work easier to read and much clearer.... and
much easier to debunk, so he won't do it.Dirk Vdm>
[snip]>>Also, why do you INSIST on not using the notation
a_i(m) or a_i(0)? It>>would make your work easier to read and
much clearer.> ... and much easier to debunk, so he won't
do it.> Maybe if he did it he could ?d his own mistakes and
not bother us with them. Or maybe it has something to do with
his Object Math. I looked at his website but couldn't make
much sense of it. Lack of clear de?itions with examples,
perhaps.-- Will Twentyman> [snip]>Also, why
do you INSIST on not using the notation a_i(m) or a_i(0)?
It>would make your work easier to read and much clearer.>>
> ... and much easier to debunk, so he won't do it.>>
>>Maybe if he did it he could ?d his own mistakes and not
bother us with >them. Or maybe it has something to do with
his Object Math. I looked >at his website but couldn't make
much sense of it. Lack of clear >de?itions with examples,
perhaps.The shortcomings of his de?ition of object have
been pointed outnumerous times. While denying any problems
repeatedly, he hasnonetheless made subtle changes to it; yet,
it still remains aproblem. Since it is supposed to be the
basis of his whole enterprise,that is enough to establish
that it tumbles down.At one point, I tried to explain exactly
why his de?ition of objectwas faulty, and why even if one
were to give it its most liberalreading, the de?ition is
such that it only includes integers. Henever
replied.http://groups.google.com/groups?selm=b8p0q2%241ahu%
241%40agate.berkeley.eduJames is not interested in ?ding his
own mistakes. Never hasbeen. In fact, he will avoid
confronting them as long as humanlypossible, if not
longer.==
Why do you take so much trouble to expose
such a reasoner as Mr. Smith? I answer as a deceased friend
of mine used to answer on like occasions - A man's capacity
is no measure of his power to do mischief. Mr. Smith has
untiring energy, which does something; self-evident honesty
of conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
?ures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
inde?ite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by
Augustus de
Morgan
===


===
Arturo Magidinmagidin@math.berkeley.edu>
>> [snip]>>Also, why do you INSIST on not using the
notation a_i(m) or a_i(0)? It>>would make your work easier
to read and much clearer.> ... and much easier to
debunk, so he won't do it.> Maybe if he did it he could
?d his own mistakes and not bother us with> them. Or maybe
it has something to do with his Object Math. I looked> at his
website but couldn't make much sense of it. Lack of clear>
de?itions with examples, perhaps.IMO the only thing someone
who suffers from this kind o?lness is interested in, is
attention - it doesn't matter if it ispositive or negative.
He's getting plenty of it, so I guess weare all very kind to
James :-)Dirk Vdm>>[snip]>Also, why
do you INSIST on not using the notation a_i(m) or a_i(0)?
It>>would make your work easier to read and much
clearer.>... and much easier to debunk, so he won't
do it.>Maybe if he did it he could ?d his own mistakes
and not bother us with >>them. Or maybe it has something to do
with his Object Math. I looked >>at his website but couldn't
make much sense of it. Lack of clear >>de?itions with
examples, perhaps.> The shortcomings of his de?ition of
object have been pointed out> numerous times. While denying
any problems repeatedly, he has> nonetheless made subtle
changes to it; yet, it still remains a> problem. Since it is
supposed to be the basis of his whole enterprise,> that is
enough to establish that it tumbles down.> At one point, I
tried to explain exactly why his de?ition of object> was
faulty, and why even if one were to give it its most liberal>
reading, the de?ition is such that it only includes integers.
He> never replied.>
http://groups.google.com/groups?selm=b8p0q2%241ahu%241%
40agate.berkeley.edu> James is not interested in ?ding his
own mistakes. Never has> been. In fact, he will avoid
confronting them as long as humanly> possible, if not
longer.tendency to start new threads and simply reassert his
claims is something I've become all too familiar with in the
past month. I still haven't ?ured out why he wants all of us
to join MSN's Amateur Math to just post direct links
there.Right now I wonder how long it will be before I tire of
reading his and the practice serves no purpose.Do you think
there's a chance of getting him to play around with Set
Theory or Computability Theory?-- Will Twentyman>
>>[snip]>Also, why do you INSIST on not
using the notation a_i(m) or a_i(0)? It>>would make your
work easier to read and much clearer.>... and much
easier to debunk, so he won't do it.>Maybe if he did it
he could ?d his own mistakes and not bother us with>>them. Or
maybe it has something to do with his Object Math. I
looked>>at his website but couldn't make much sense of it.
Lack of clear>>de?itions with examples, perhaps.> IMO
the only thing someone who suffers from this kind of> illness
is interested in, is attention - it doesn't matter if it is>
positive or negative. He's getting plenty of it, so I guess
we> are all very kind to James :-)> Dirk VdmIf only he
would reciprocate and give each of us as much attention as we
give him. Answers to objections might be nice. Minus the
generalized attacks on character would be nicer.-- Will
Twentyman>[snip]>>
>>Also, why do you INSIST on not using the notation a_i(m)
or a_i(0)? It>>would make your work easier to read and
much clearer.>... and much easier to debunk, so
he won't do it.>Maybe if he did it he could ?d
his own mistakes and not bother us with>>them. Or maybe it
has something to do with his Object Math. I looked>>at his
website but couldn't make much sense of it. Lack of clear>
>>de?itions with examples, perhaps.> IMO the only
thing someone who suffers from this kind of> illness is
interested in, is attention - it doesn't matter if it is>
positive or negative. He's getting plenty of it, so I guess
we> are all very kind to James :-)>> Dirk Vdm>> If only
he would reciprocate and give each of us as much attention as
we> give him. Answers to objections might be nice. Minus the
generalized> attacks on character would be nicer.That would
be nice indeed. It would also be nice if mentaldisease would
be effectively curable with proper medication.But don't let
me put you off - enjoy it while you can :-)Dirk Vdm>Well
I'm going to try and break it down even more to try and see
if>y'all will accept the mathematics:>>[...]>>So go back now
and read posts from Arturo Magidin, Keith Ramsay, and>Nora
Baron among others, and ask yourself: What do they think of
me,>really?Fascinating argument: I know, I'll behave like a
complete and totalass. Then people will hate me, and then
when they say I'm wrongabout the math I can explain that it's
just because they hate me.Too bad nobody's stupid enough to
buy it. >James Harris************************David C.
Ullrich> Well I'm going to try and break it down even
more to try and see if> y'all will accept the mathematics:>
> Ok I have> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 +
3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)> where f is a prime
integer other than 3, and u is coprime to f, and> looking
at that x, I see the possibility for the factorization>
P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf).> That's
what I've been putting up a lot where you see a LOT of
symbols,> which seem to confuse people. The ring is
algebraic integers, and let> me get rid of as many of those
symbols as I can:> Let x=2, f=5, u=1, so that I have>
> P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) +
5)> which is what I put up earlier, but I'm wary about
some of you still> ?ding that confusing, so I'll work it
out more to get> P(m) = 25(5000m^3 - 600 m^2 + 24m + 6
- 150m + 5), which is> P(m) = 25(5000m^3 - 600 m^2 -
126m + 11), > and now you can see what the polynomial
P(m) looks like without so> many symbols.> Now from
before where I had x, I *still* have that> P(m) = (2
a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).> So> (2 a_1 +
5)(2 a_2 + 5)(2 a_3 + 5) = > 25(5000m^3 - 600 m^2 -
126m + 11).> Now setting m=0 gives me> (2 a_1 +
5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).> Now let's say you
accept that any factor of a polynomial can be> written like
r+c, where r=0, or r varies as the polynomial variable>
varies, while c remains constant and is a factor of the
constant term.> Notice that > a_1 a_2 a_3 = 25
(8(625 m^3 - 75 m^2 + 3m)), > I think this should be a_1(m)
a_2(m) a_3(m) = 25(625 m^3 - 75m^2 + 3m)That's like saying, if
I write y=mx+b, that you think it should bey(x)=mx+b.> which
equals 0, when m=0, so at least one of the a's must equal 0,>
> when m=0.> And to get that factor that is 25, you
must have two a's that go to 0,> when m=0.> Why can't
a_2(0) = 5, a_3(0) = -2/3? You have made claims to the effect
> that you think these functions are continuous, so they
certainly should > take on non-algebraic integer values.The
a's are roots of a monic polynomial with integer
coef?ients,given that m is an integer.Now if you wish to
make m rational then you can explore continuity. > Also, why
do you INSIST on not using the notation a_i(m) or a_i(0)? It
> would make your work easier to read and much clearer.Why oh
why do people ever insist on writing y=mx+b instead
ofy(x)=mx+b?It seems to me that you need to begin a crusade
to make certain thatsuch a horror is no longer committed.You
can be a paragon of mathematical style on a quest to save
theignorant masses from this travesty of ever writing such a
thing asy=mx+b as instead you make certain that people write
y(x) = mx + b!!!James Harris> Well I'm going to try and
break it down even more to try and see if> y'all will accept
the mathematics:>> Ok I have>> P(m) = f^2 ((m^3 f^4 - 3m^2
f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)>> where f is a prime
integer other than 3, and u is coprime to f, and> looking at
that x, I see the possibility for the factorization>> P(m) =
(a_1 x + uf)(a_2 x + uf)(a_3 x + uf).>> That's what I've been
putting up a lot where you see a LOT of symbols,> which seem
to confuse people. The ring is algebraic integers, and let>
me get rid of as many of those symbols as I can:>> Let x=2,
f=5, u=1, so that I have>> P(m) = 25 (8(625 m^3 - 75 m^2 +
3m)- 3(2)(-1 + 25m ) + 5)>> which is what I put up earlier,
but I'm wary about some of you still> ?ding that confusing,
so I'll work it out more to get>> P(m) = 25(5000m^3 - 600 m^2
+ 24m + 6 - 150m + 5), which is>> P(m) = 25(5000m^3 - 600 m^2
- 126m + 11),>> and now you can see what the polynomial P(m)
looks like without so> many symbols.>> Now from before where
I had x, I *still* have that>> P(m) = (2 a_1 + 5)(2 a_2 +
5)(2 a_3 + 5).>> So>> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =>>
25(5000m^3 - 600 m^2 - 126m + 11).This is greatly simpli?d.
I'm with you so far.>> Now setting m=0 gives me>> (2 a_1 +
5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).>> Now let's say you accept
that any factor of a polynomial can be> written like r+c,
where r=0, or r varies as the polynomial variable> varies,
while c remains constant and is a factor of the constant
term.>> Notice that>> a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 +
3m)),Well, no, I don't notice that. But I don't think it
matters.>> which equals 0, when m=0, so at least one of the
a's must equal 0,> when m=0.>> And to get that factor that is
25, you must have two a's that go to 0,> when m=0.>OK. So
you've declared that two of the a's are 0 at m=0.Therefore,
the third a=3 at m=0. I'll simplify if you don't mind; makes
iteasier for me:a_3 = b_3 + 3(2 a_1 + 5)(2 a_2 + 5)(2 b_3 +
11) = 25(11).OK. Still with you. And all of a_1, a_2, b_3 are
0 at m=0.> (Note: Some posters have gotten a lot of mileage
out of calling that a> degenerate case, but they were just
fooling you into forgetting your> basic algebra and what you
know about polynomials. I think they did> so deliberately as
the math isn't complicated.)>> Now comes the question of what
happens when m is NOT 0, and answering> that question requires
looking again at>> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =>>
25(5000m^3 - 600 m^2 - 126m + 11)>> and noticing that the
constant term is 25(11), but you can divide off> that 25 to
get P(m)/25 which gives you a constant term that's 11. And>
11 and 5 are coprime. That's very important. In fact, that's
the> *key* fact which should stick in your mind.>> So now
looking at>> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 =>>
(5000m^3 - 600 m^2 - 126m + 11)> I know that if 2 a_1 + 5
has a factor of 5, when m=0, then that factor> is a factor of
the constant term, and in fact, it'd have a factor of> the
constant term that is 5. So why would you think that the
factor> of the constant term would move or change when m
changes?>> Well it can't.I agree with you. I think I'm
following. 5 is a factor of 5 for any value ofm.>> Given
that with 2 a_1 + 5, that 5 in there is a factor of the
constant> term of>> 25(5000m^3 - 600 m^2 - 126m + 11)>> you
*still* have a factor of the constant term when 25 is
separated> off.>> But now your constant term is 11.>> That
forces all the factors of 5 to go away from 2 a_1 + 5.Oops.
Think you just lost me. You've shown that 5 is a factor of
2a_1 + 5when a_1 is zero; not for a_1non-zero. Could you
clarify what you mean here?>> Now you may wish for there to
be someway for some factors to remain,> but what actuall
happens is you get>> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)
=>> (5000m^3 - 600 m^2 - 126m + 11)>> while posters have
argued that *all* the a's would have some factor of> 5.>Well,
not quite. They independently proved that *none* of the a's
werecoprime to 5 for acompletely different polynomial. I
can't see the relevance.> But let's let m=0 again and see
what happens now, as that gives>> (2 a_1/5 + 1)(2 a_2/5 +
1)(2 a_3 + 5) = 11>> and you'll notice I have 1 where I had 5
before, so it can all work.>> That is, the a's that went to 0
before will go to 0 again.>> Some posters have claimed
otherwise which I've called voodoo math.You're being far
too polite. All your doing is dividing by 5, and
assertingthat if a is zero then a/5 iszero. If people are
claiming otherwise they're complete liars.>> Of course, if
you think about it, you'll understand that they probably>
knew the truth and just lied to you as they didn't think you
could> ?ure it out, and probably didn't expect me to
simplify in this way.>Yep, they're liars.> Those people
aren't your friends. You may trust them. You may admire>
them, but they clearly don't think much of you.>Yep, they're
not my friends.> So go back now and read posts from Arturo
Magidin, Keith Ramsay, and> Nora Baron among others, and ask
yourself: What do they think of me,> really?>It's not my
place to opine what they think of you. I haven't noticed any
ofthem claiming that a/5 is not zeroif a=0 however.But.
There's no conclusion here.What are you saying?Given: (2 a_1
+ 5)(2 a_2 + 5)(2 b_3 + 11) = 25(5000m^3 - 600 m^2 - 126m +
11)That some of the a's are coprime to 5?How about (2 c_1 +
7)(2 c_2 + 7)(2 c_3 + 7) = 25(5000m^3 - 600 m^2 - 126m +
11)By exactly the same reasoning, some of the c's are coprime
to 7?I think there's something missing in all this.>> James
HarrisPhil Nicholson> Maybe if he did it he could ?d his
own mistakes and not bother us with them.What you call
bothering, he proudly calls modern problem solving
techniques. Take a look
-http://groups.google.com/groups?selm=
3c65f87.0304261407.7bb46534@posting.google.com> That
would be nice indeed. It would also be nice if mental>
disease would be effectively curable with proper medication.>
But don't let me put you off - enjoy it while you can :-)I
enjoy the entertainment JSH provides as much as anyone, but
... I have a nagging feeling that this is a modern
intellectual version of the medieval court enjoying the
antics of the fool, who was a cripple or a hunchback or
otherwise handicapped.Gib> That would be nice indeed.
It would also be nice if mental> disease would be
effectively curable with proper medication.> But don't let
me put you off - enjoy it while you can :-)>> I enjoy the
entertainment JSH provides as much as anyone, but ... I have>
a nagging feeling that this is a modern intellectual version
of the> medieval court enjoying the antics of the fool, who
was a cripple or a> hunchback or otherwise handicapped.... or
the village idiot providing the necessaryentertainment on a
hot Sunday afternoon.Of course, that's very obvious.Dirk
Vdm>> That would be nice indeed. It would also be
nice if mental>> disease would be effectively curable with
proper medication.>> But don't let me put you off - enjoy it
while you can :-)> I enjoy the entertainment JSH provides
as much as anyone, but ... I have > a nagging feeling that
this is a modern intellectual version of the > medieval court
enjoying the antics of the fool, who was a cripple or a >
hunchback or otherwise handicapped.That does seem to be the
case. The difference is: in medieval courts, the fool was
aware of his situation, as well as the true purpose of his
being there.I think similar analogies could be made to people
throwing rocks in the school yard. There are times when I have
to wonder if we wouldn't perform a better service by
responding only to his math and ignoring/deleting
everything else that comes from his mouth. Maybe then he
would learn to focus on the math as well.-- Will
Twentyman>>Maybe if he did it he could ?d his own
mistakes and not bother us with them.> What you call
bothering, he proudly calls modern problem solving
techniques. Take a look ->
http://groups.google.com/groups?selm=
3c65f87.0304261407.7bb46534@posting.google.com>
disturbing.-- Will Twentyman>Well I'm going to try
and break it down even more to try and see if>y'all will
accept the mathematics:>>Ok I have>> P(m) = f^2 ((m^3
f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)>>where
f is a prime integer other than 3, and u is coprime to f,
and>looking at that x, I see the possibility for the
factorization>> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf).>>That's what I've been putting up a lot where you
see a LOT of symbols,>which seem to confuse people. The
ring is algebraic integers, and let>me get rid of as many
of those symbols as I can:>>Let x=2, f=5, u=1, so that I
have>> P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m
) + 5)>>which is what I put up earlier, but I'm wary about
some of you still>?ding that confusing, so I'll work it out
more to get>> P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m
+ 5), which is>> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),
>>and now you can see what the polynomial P(m) looks like
without so>many symbols.>>Now from before where I had
x, I *still* have that>> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2
a_3 + 5).>>So>> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
>> 25(5000m^3 - 600 m^2 - 126m + 11).>>Now setting
m=0 gives me>> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) =
25(11).>>Now let's say you accept that any factor of a
polynomial can be>written like r+c, where r=0, or r varies
as the polynomial variable>varies, while c remains constant
and is a factor of the constant term.>>Notice that >>
a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), >>I think this
should be a_1(m) a_2(m) a_3(m) = 25(625 m^3 - 75m^2 + 3m)>
That's like saying, if I write y=mx+b, that you think it
should be> y(x)=mx+b.You say this as if y(x)=mx+b is
uncommon. When teaching about linear functions in algebra,
it's usually written as y=mx+b. In calculus, I've seen both.
This is especially true when switching rapidly back and forth
between y as a function of x and y evaluated at a particular
value of x.>which equals 0, when m=0, so at least one of
the a's must equal 0,>when m=0.>>And to get that factor
that is 25, you must have two a's that go to 0,>when
m=0.>>Why can't a_2(0) = 5, a_3(0) = -2/3? You have made
claims to the effect >>that you think these functions are
continuous, so they certainly should >>take on non-algebraic
integer values.> The a's are roots of a monic polynomial
with integer coef?ients,> given that m is an integer.I
missed that m is an integer. What does that have to do with
the possible values of a_i(0) though? You've said the a's are
roots of a monic polynomial, but also that they don't need to
be polynomials themselves. As a result, I see no reason for
assuming anything about the codomain of those
functions.>>Also, why do you INSIST on not using the notation
a_i(m) or a_i(0)? It >>would make your work easier to read and
much clearer.> Why oh why do people ever insist on writing
y=mx+b instead of> y(x)=mx+b?Because it's a shortcut AND then
don't use the symbol y to refer to both mx+b and m*0+b with
little warning as to which it will be in the next line.> It
seems to me that you need to begin a crusade to make certain
that> such a horror is no longer committed.I'll take that
under advisement. Would you care to assist me?> You can be a
paragon of mathematical style on a quest to save the>
ignorant masses from this travesty of ever writing such a
thing as> y=mx+b as instead you make certain that people
write y(x) = mx + b!!!Can I use you as a model of the
increased clarity that happens? Maybe a before and after
snapshot.-- Will Twentyman > Well I'm going to try and
break it down even more to try and see if> y'all will
accept the mathematics:>> Ok I have>> P(m) = f^2
((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)>>
> where f is a prime integer other than 3, and u is coprime to
f, and> looking at that x, I see the possibility for the
factorization>> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf).>> That's what I've been putting up a lot where you
see a LOT of symbols,> which seem to confuse people. The
ring is algebraic integers, and let> me get rid of as many
of those symbols as I can:>> Let x=2, f=5, u=1, so that I
have>> P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m
) + 5)>> which is what I put up earlier, but I'm wary
about some of you still> ?ding that confusing, so I'll
work it out more to get>> P(m) = 25(5000m^3 - 600 m^2 +
24m + 6 - 150m + 5), which is>> P(m) = 25(5000m^3 - 600
m^2 - 126m + 11),>> and now you can see what the
polynomial P(m) looks like without so> many symbols.>>
Now from before where I had x, I *still* have that>> P(m)
= (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).>> So>> (2 a_1 +
5)(2 a_2 + 5)(2 a_3 + 5) =>> 25(5000m^3 - 600 m^2 - 126m
+ 11).> This is greatly simpli?d. I'm with you so
far.Good. >> Now setting m=0 gives me>> (2 a_1 + 5)(2
a_2 + 5)(2 a_3 + 5) = 25(11).>> Now let's say you accept
that any factor of a polynomial can be> written like r+c,
where r=0, or r varies as the polynomial variable> varies,
while c remains constant and is a factor of the constant
term.>> Notice that>> a_1 a_2 a_3 = 25 (8(625 m^3 -
75 m^2 + 3m)),> Well, no, I don't notice that. But I don't
think it matters.It's very important as it shows part of the
relationship between thea's and m, which is fully de?ed by
three expressions, but I'mfocusing on one.Here the point is
that at *least* one of the a's is equal to 0,
whenm=0.Remember the a's come from the factorization of P(m),
where P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2
)xu^2 + u^3 f),and I've put in some numbers for f, x, and u,
as f=5, x=2, and u=1.The factorization isP(m) = (a_1 x +
uf)(a_2 x + uf)(a_3 x + uf).Note to readers: I'm leaving
everything in from the previous post,which makes things kind
of bulky, but I'm hoping that in this caseyou'll be better
served by having all the information readilyavailable. Though
I do fear there may be a bit of overload for someof you.>>
which equals 0, when m=0, so at least one of the a's must
equal 0,> when m=0.>> And to get that factor that is
25, you must have two a's that go to 0,> when m=0.>> OK.
So you've declared that two of the a's are 0 at m=0.No that's
required by the given expressions.> Therefore, the third a=3
at m=0. I'll simplify if you don't mind; makes it> easier
for me:> a_3 = b_3 + 3Whatever works for you is ?e with me,
as long as it works, and ismathematically correct.That looks
ok as you're identifying part of the constant term for
thefactor 2a_3 + 5as the full value is 11 for its constant
term is 11.Here you have b_3 as part of the varying portion.
> (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) = 25(11).> OK. Still
with you. And all of a_1, a_2, b_3 are 0 at m=0.Yup.>
(Note: Some posters have gotten a lot of mileage out of
calling that a> degenerate case, but they were just
fooling you into forgetting your> basic algebra and what
you know about polynomials. I think they did> so
deliberately as the math isn't complicated.)>> Now comes
the question of what happens when m is NOT 0, and answering>
> that question requires looking again at>> (2 a_1 + 5)(2
a_2 + 5)(2 a_3 + 5) =>> 25(5000m^3 - 600 m^2 - 126m + 11)>
>> and noticing that the constant term is 25(11), but you
can divide off> that 25 to get P(m)/25 which gives you a
constant term that's 11. And> 11 and 5 are coprime. That's
very important. In fact, that's the> *key* fact which
should stick in your mind.>> So now looking at>> (2
a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 =>> (5000m^3 - 600 m^2
- 126m + 11)> I know that if 2 a_1 + 5 has a factor
of 5, when m=0, then that factor> is a factor of the
constant term, and in fact, it'd have a factor of> the
constant term that is 5. So why would you think that the
factor> of the constant term would move or change when m
changes?>> Well it can't.> I agree with you. I think
I'm following. 5 is a factor of 5 for any value of> m.Ok,
let see what's next.>> Given that with 2 a_1 + 5, that 5
in there is a factor of the constant> term of>>
25(5000m^3 - 600 m^2 - 126m + 11)>> you *still* have a
factor of the constant term when 25 is separated> off.>>
> But now your constant term is 11.>> That forces all the
factors of 5 to go away from 2 a_1 + 5.> Oops. Think you
just lost me. You've shown that 5 is a factor of 2a_1 + 5>
when a_1 is zero; not for a_1> non-zero. Could you clarify
what you mean here?Hmmm...how about this explanation?Remember
that the polynomial is P(m) = 25(5000m^3 - 600 m^2 - 126m +
11),and you see that factor 25. Now imagine that I have the
polynomialQ(m), where P(m) = 25 Q(m)so you have as a
factorization Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25and
the question is, how does that factor of 25 divide out? Well,
checking at Q(0), gives 11, so how can those factors of
25divide through Q(m) in such a way as to give 11, at
m=0?There's only one way, which gives you Q(m) = (2 a_1/5 +
1)(2 a_2/5 + 1)(2 a_3 + 5).If that bothers you, remember that
at m=0, two of the a's equal 0.If it still bothers you, try
and get everything to work some otherway.And yes, it can be
shown rigorously, but right now there's the problemof getting
that feel for the math.Then you can go to the paper Advanced
Polynomial Factorization.>> Now you may wish for there to
be someway for some factors to remain,> but what actuall
happens is you get>> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 +
5) =>> (5000m^3 - 600 m^2 - 126m + 11)>> while
posters have argued that *all* the a's would have some factor
of> 5.>> Well, not quite. They independently proved that
*none* of the a's were> coprime to 5 for a> completely
different polynomial. I can't see the relevance.Nope. It
turns out that they're making their claim for the
polynomialI've given you, but in the past there have been
*symbols* where I'venow given numbers.> But let's let m=0
again and see what happens now, as that gives>> (2 a_1/5
+ 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11>> and you'll notice I
have 1 where I had 5 before, so it can all work.>> That
is, the a's that went to 0 before will go to 0 again.>>
Some posters have claimed otherwise which I've called voodoo
math.> You're being far too polite. All your doing is
dividing by 5, and asserting> that if a is zero then a/5 is>
zero. If people are claiming otherwise they're complete
liars.Hmmm...looks like I screwed up in my statement, as what
I should havesaid is that they've made that claim when m does
NOT equal 0. Myapologies.The voodoo math has come in when m
doesn't equal 0, as if suddenlyconstant factors shift around.>
>> Of course, if you think about it, you'll understand that
they probably> knew the truth and just lied to you as they
didn't think you could> ?ure it out, and probably didn't
expect me to simplify in this way.>> Yep, they're
liars.Well that is true as can be seen by the replies from
some of them inseveral threads I created yesterday, like this
one.My simpli?ation removes areas for doubts, and if they
weren't liars,they'd simply acknowledge the truth at this
point.> Those people aren't your friends. You may trust
them. You may admire> them, but they clearly don't think
much of you.>> Yep, they're not my friends.Hmmm...so much
agreement puts me somewhat at a loss for words.> So go back
now and read posts from Arturo Magidin, Keith Ramsay, and>
Nora Baron among others, and ask yourself: What do they think
of me,> really?>> It's not my place to opine what they
think of you. I haven't noticed any of> them claiming that
a/5 is not zero> if a=0 however.That is true.> But. There's
no conclusion here.> What are you saying?> Given:> (2 a_1 +
5)(2 a_2 + 5)(2 b_3 + 11) 25(5000m^3 - 600 m^2 - 126m +
11)> That some of the a's are coprime to 5?It turns out
that in the ring of algebraic integers they all are,which is
why the ring has problems.> How about> (2 c_1 + 7)(2 c_2 +
7)(2 c_3 + 7) 25(5000m^3 - 600 m^2 - 126m + 11)> By
exactly the same reasoning, some of the c's are coprime to
7?No. > I think there's something missing in all this.That's
ok. My suggestion is to consider my reply and see if you
havethat same feeling.Remember what I'm doing is considering
a polynomial factorization,with speci? reference to its
constant term *because* I'm using thatas my anchor given that
I'm using non-polynomial factors of thatpolynomial.That is,
see the constant term as a lighthouse in the darkness.Or
better yet, see the math here as requiring that you ?ouri
nstruments, as otherwise you may spiral into the ground,
?urativelyspeaking.James HarrisFair bit of snippage. I've
got some work to do now to follow this upproperly.<Snip>>
Notice that>> a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 +
3m)),>> Well, no, I don't notice that. But I don't think
it matters.Sorry, poorly worded on my part. I *DID* notice
that a_1 a_2 a_3 *DOESN'T*in fact equalyour main point.>>
It's very important as it shows part of the relationship
between the> a's and m, which is fully de?ed by three
expressions, but I'm> focusing on one.>> Here the point is
that at *least* one of the a's is equal to 0, when> m=0.Which
was this. And agreed.<Snip>> And to get that factor that
is 25, you must have two a's that go to 0,> when m=0.>
>> OK. So you've declared that two of the a's are 0 at
m=0.>> No that's required by the given expressions.>OK. I'm
guessing that you're right. But I'll check this and get back
ifnecessary.> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) =>
>> (5000m^3 - 600 m^2 - 126m + 11)>> while
posters have argued that *all* the a's would have some factor
of> 5.> Well, not quite. They independently
proved that *none* of the a's were> coprime to 5 for a>
completely different polynomial. I can't see the relevance.>>
Nope. It turns out that they're making their claim for the
polynomial> I've given you, but in the past there have been
*symbols* where I've> now given numbers.I don't recall any
claims being made by others. I did see proofs made byothers
thatfor the polynomial 65x^3 -12x +1, none of the a's are
coprime to 5.The polynomial under discussion here is 5000m^3
-600m^2 -126m +11.<Snip>> (2 a_1 + 5)(2 a_2 + 5)(2 b_3 +
11) 25(5000m^3 - 600 m^2 - 126m + 11)>> That some of
the a's are coprime to 5?>> It turns out that in the ring of
algebraic integers they all are,> which is why the ring has
problems.Surely not. What about when m=5? What are the a's in
that case?>> How about> (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7)
25(5000m^3 - 600 m^2 - 126m + 11)>> By exactly the
same reasoning, some of the c's are coprime to 7?>> No.I
believe you'd be right.>> James Harris> Fair bit of
snippage. I've got some work to do now to follow this up>
properly.> <Snip>> Notice that>> a_1
a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)),>> Well, no,
I don't notice that. But I don't think it matters.> Sorry,
poorly worded on my part. I *DID* notice that a_1 a_2 a_3
*DOESN'T*> in fact equal> your main point.Yeah, you're
correct. That should be a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 +
3m). >> It's very important as it shows part of the
relationship between the> a's and m, which is fully de?ed
by three expressions, but I'm> focusing on one.>> Here
the point is that at *least* one of the a's is equal to 0,
when> m=0.> Which was this. And agreed.> <Snip>>
> And to get that factor that is 25, you must have two a's
that go to 0,> when m=0.>> OK. So you've
declared that two of the a's are 0 at m=0.>> No that's
required by the given expressions.>> OK. I'm guessing that
you're right. But I'll check this and get back if>
necessary.Well I guess you're thinking that I just picked the
possibility thattwo of the a's are 0, and maybe it's possible
for, say, only one ofthe a's to equal 0.But if you do that
then you can only get a factor of 5 that is 5, forthe
result.> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) =>
>> (5000m^3 - 600 m^2 - 126m + 11)>> while
posters have argued that *all* the a's would have some factor
of> 5.> Well, not quite. They
independently proved that *none* of the a's were>
coprime to 5 for a> completely different polynomial. I
can't see the relevance.>> Nope. It turns out that
they're making their claim for the polynomial> I've given
you, but in the past there have been *symbols* where I've>
now given numbers.> I don't recall any claims being made by
others. I did see proofs made by> others that> for the
polynomial 65x^3 -12x +1, none of the a's are coprime to 5.>
The polynomial under discussion here is 5000m^3 -600m^2 -126m
+11.Then you haven't looked closely. The claim is against the
polynomialfactorization (v^3+1)x^3 - 3v xy^2 + y^3 = (a_1 x +
y)(a_2 x + y)(a_3 x + y)where v=-1+mf^2, and readers can get
from there to 25(5000m^3 -600m^2 -126m +11)easily enough by
plugging in f=5, x=2, y=5. > <Snip>> (2 a_1 + 5)(2 a_2 +
5)(2 b_3 + 11) 25(5000m^3 - 600 m^2 - 126m + 11)>>
> That some of the a's are coprime to 5?>> It turns out
that in the ring of algebraic integers they all are,> which
is why the ring has problems.> Surely not. What about when
m=5? What are the a's in that case?The problem is with the
ring of algebraic integers as I've explained.Consider the
following which you deleted out from my previous
post:Hmmm...how about this explanation?Remember that the
polynomial is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),and
you see that factor 25. Now imagine that I have the
polynomialQ(m), where P(m) = 25 Q(m)so you have as a
factorization Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25and
the question is, how does that factor of 25 divide out? Well,
checking at Q(0), gives 11, so how can those factors of
25divide through Q(m) in such a way as to give 11, at
m=0?There's only one way, which gives you Q(m) = (2 a_1/5 +
1)(2 a_2/5 + 1)(2 a_3 + 5).If that bothers you, remember that
at m=0, two of the a's equal 0.If it still bothers you, try
and get everything to work some otherway.>> How about>
> (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) 25(5000m^3 - 600
m^2 - 126m + 11)>> By exactly the same reasoning,
some of the c's are coprime to 7?>> No.> I believe you'd
be right.The math should be simple now that so many symbols
have been replacedwith numbers.The issue is an error in
taught mathematics revealed by what I'veshown here.My
position is that mathematicians should be diligent in
teaching thetruth, and should acknowledge an error, so that
it is no longertaught.James Harris> Well I'm going to try
and break it down even more to try and see if> y'all will
accept the mathematics:> Ok I have> P(m) = f^2 ((m^3 f^4
- 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)> where f is
a prime integer other than 3, and u is coprime to f, and>
looking at that x, I see the possibility for the
factorization> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf).> That's what I've been putting up a lot where you see
a LOT of symbols,> which seem to confuse people. The ring is
algebraic integers, and let> me get rid of as many of those
symbols as I can:> Let x=2, f=5, u=1, so that I have>
P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5)>
which is what I put up earlier, but I'm wary about some of
you still> ?ding that confusing, so I'll work it out more to
get> P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5),
which is> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > and
now you can see what the polynomial P(m) looks like without
so> many symbols.> Now from before where I had x, I *still*
have that> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).>
Interesting. When the x's were left unspeci?d,the form of
the factorization was P(m) = (a_1*x + 5)*(a_2*x + 5)*(a_3*x +
5).Now that you have substituted in x = 2, it is nolonger a
factorization of a polynomial in x. It isjust a factorization
as a product of three numbers.So 2*a_1 + 5, for example, is
just an algebraic integer. The factorization is [1] P(m) =
(2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5).So then what you are
asserting below is that if you factorthe number P(m) as in
[1], then one of a1, a2, ora3 must be coprime to 5.
Right?Let's take m = 1. Then P(m) = 25 * 4285.This can be
factored as 5 * 5 * 4285,which yields a1 = 0, a2 = 0, and a3
= 2140.None of these is coprime to 5. End of story.Don't like
a1 = a2 = 0 ? Other things work too - e.g., a1 = -5, a2 = -5,
a3 = 2140. Nonecoprime to 5, as before.Read on, however ->
So> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > 25(5000m^3 -
600 m^2 - 126m + 11).> Now setting m=0 gives me> (2 a_1 +
5)(2 a_2 + 5)(2 a_3 + 5) = 25(11).> Now let's say you accept
that any factor of a polynomial can be> written like r+c,
where r=0, or r varies as the polynomial variable> varies,
while c remains constant and is a factor of the constant
term.> Notice that > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2
+ 3m)), > which equals 0, when m=0, so at least one of the
a's must equal 0,> when m=0.> And to get that factor that
is 25, you must have two a's that go to 0,> when m=0.>
(Note: Some posters have gotten a lot of mileage out of
calling that a> degenerate case, but they were just fooling
you into forgetting your> basic algebra and what you know
about polynomials. I think they did> so deliberately as the
math isn't complicated.)> Now comes the question of what
happens when m is NOT 0, and answering> that question
requires looking again at> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5) = > 25(5000m^3 - 600 m^2 - 126m + 11)> and noticing
that the constant term is 25(11), but you can divide off>
that 25 to get P(m)/25 which gives you a constant term that's
11. And> 11 and 5 are coprime. That's very important. In fact,
that's the> *key* fact which should stick in your mind.> So
now looking at> (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = >
(5000m^3 - 600 m^2 - 126m + 11)> I know that if 2 a_1 + 5
has a factor of 5, when m=0, then that factor> is a factor of
the constant term, and in fact, it'd have a factor of> the
constant term that is 5. So why would you think that the
factor> of the constant term would move or change when m
changes?> Well it can't.> Given that with 2 a_1 + 5, that
5 in there is a factor of the constant> term of> 25(5000m^3
- 600 m^2 - 126m + 11)> you *still* have a factor of the
constant term when 25 is separated> off.> But now your
constant term is 11.> That forces all the factors of 5 to
go away from 2 a_1 + 5.> Now you may wish for there to be
someway for some factors to remain,> but what actuall happens
is you get> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = >
(5000m^3 - 600 m^2 - 126m + 11)> while posters have argued
that *all* the a's would have some factor of> 5.> Yep, sure.
See above. With actual numbers!> But let's let m=0 again and
see what happens now, as that gives> (2 a_1/5 + 1)(2 a_2/5
+ 1)(2 a_3 + 5) = 11> and you'll notice I have 1 where I
had 5 before, so it can all work.> That is, the a's that
went to 0 before will go to 0 again.> Some posters have
claimed otherwise which I've called voodoo math.> Of
course, if you think about it, you'll understand that they
probably> knew the truth and just lied to you as they didn't
think you could> ?ure it out, and probably didn't expect me
to simplify in this way.> Those people aren't your friends.
You may trust them. You may admire> them, but they clearly
don't think much of you.> So go back now and read posts
from Arturo Magidin, Keith Ramsay, and> Nora Baron among
others, and ask yourself: What do they think of me,> really?>
> James Harris> It's almost funny until you realize why
there was so much frenetic> energy, and why these people are
so desperate to hide the math that> they continually delete
it out:> The mathematics is rather simple, the algebra is
basic, but the> conclusion is dramatic.> So let's say you
were facing people who needed to keep people from> looking
closely, and every time you tried to simplify and show>
details, they'd jump in and hide details and make things
more> convoluted?> Well you might do what I've done today
and stretch them out.> What this means is: start a lot of new
threads in the hope that people won't notice that the problems
in theold ones were not answered. Mr. Harris imagines that
heis playing to a huge audience of silent lurkers, perhaps
even math undergrads. I am sure they will be impressed bythis
clever evasive maneuver! See below for what happens when
people look closely andshow details ... be sure to note where
I have hidden detailsand made things more convoluted ... where
I was so desperateto hide the math that I continually deleted
it out ...> What I'm doing is not very complicated as I'm
using the extra symbols> to factor a polynomial into
non-polynomial factors.> Now that's a fascinating idea for
factoring polynomials that I guess> is new to the math
world.> And yes, bad apples can take advantage when there's
something new.> What is a surprise though is that
mathematical society can be taken in> by people like Arturo
Magidin or Nora Baron, when they're posting so> desperately,
so quickly when stretched out, that they can barely get> to
the math, or say things that are clearly false.> But what
if they never really believed in mathematics?> What if for
them it is a fashion show?> Remember mathematics can be
about appearance for some people. They> might have gotten a
lot of mileage out of being able to put down math> that
looked good.> That is, you may have people who are cons in
your midst who are used> to playing a game upon other
mathematicians, and now they're> trapped--stretched out--by a
lot of threads and math where I did the> sudden move of
putting in numbers where once there were symbols.> It seems
to me that in mathematics there is a certain respect that is>
to be given to ideas, logic, and mathematical truth. Sure
being a> rather large society it's not surprising that
corrupt people can slip> in, and maybe get some stature, or
get through a Ph.d program.> But in society when the
corrupt people reveal themselves clearly by> ?ally pushing
beyond obvious limits--like denying basic algebra--it>
becomes time to clean house. Look over those threads, look at
their> desperation, and their contempt for algebra and your
algebra> knowledge, then please ask yourselves how it cannot
now be that time.> James Harris I looked a little more
closely at Advanced Polynomial Factorization. It turns out
that, in addition to basic conceptual problems, it alsohas
careless mistakes in algebra. Start with P(m) = f^2
*((m^3*f^4 - 3*m^2*f^2 + 3*m)*x^3 - 3*(-1 + m*f^2)*x*u^2 +
u^3*f), as you give early in Section 2 of APF. You note that
P(m) hasa factor of f^2. Nothing wrong with this particular
formula. However, on the lastpage, when you substitute in m =
1, f = sqrt(5), and u = 1, you do NOT obtain 65*x^3 - 12*x +
1. Instead you get 65*x^3 - 60*x + 5*sqrt(5). Now if this is
factored in the form (a1*x + 1)*(a2*x + 1)*(a3*x + 1),it
obviously doesn't work: 1*1*1 is not equal to 5*sqrt(5). Back
on the second page, in a parenthetical note atthe bottom, you
say: Note: the a's are roots of a monicpolynomial with
algebraic integer coef?ients so theyare algebraic integers.
Not true at all. Note that the polynomial P(m), as given
above, if considered as a polynomial in x, does not have
constant term equal to 1 or -1. The constant term with
respect to x is u^3*f^3. Thus the polynomial of which the a's
are roots is not monic.I am not sure what you actually
intended. This is carelessly written, even independentof the
unsoundness of the ideas. Parts of your postsfrom yesterday
on this topic also contained similaralgebraic errors. I think
it was still your intention in APF however to show that if
65*x^3 - 12*x + 1were factored in the form (a1*x + 1)*(a2*x +
1)*(a3*x + 1),where a1, a2, and a3 are algebraic integers,
then oneof a1, a2, or a3 is coprime to 5. No matterhow you
cut it, that is still false, as shown by the proofs given in
other threads by me and W. DaleHall. Yes, you have started
other several new threadstoday, but you have explicitly
avoided a direct response toeither me or Dale Hall on this.
Even if you are unable to perform the computations thatDale
proposes and are incapable of understanding the proof thatI
presented, you cannot ignore the simple algebraic errorsin
APF. Something there has to change. Nora B.> Fair bit
of snippage. I've got some work to do now to follow this up>
> properly.>> And again.>> <Snip>> (2
a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) 25(5000m^3 - 600
m^2 - 126m + 11)>> That some of the a's are
coprime to 5?>> It turns out that in the ring of
algebraic integers they all are,> which is why the ring
has problems.>> a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 +
3m)> For any integer m, a_1 a_2 a_3 is a multiple of 25.
i.e. not coprime to 5.> Perhaps you mean that none of the
a's are coprime to 5?> Which is trivially obvious and
uninteresting after all.> Sorry to have bothered you.It's
actually fascinating as it's a bizarre result, and I think
yourealize that I'm correct as you *again* deleted out the
followingwonderful simpli?ation which shows clearly that I'm
right.The problem is with the ring of algebraic integers as
I've explained.Consider the following which you deleted yet
again from my previouspost:Hmmm...how about this
explanation?Remember that the polynomial is P(m) = 25(5000m^3
- 600 m^2 - 126m + 11),and you see that factor 25. Now imagine
that I have the polynomialQ(m), where P(m) = 25 Q(m)so you
have as a factorization Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5)/25and the question is, how does that factor of 25 divide
out? Well, checking at Q(0), gives 11, so how can those
factors of 25divide through Q(m) in such a way as to give 11,
at m=0?There's only one way, which gives you Q(m) = (2 a_1/5 +
1)(2 a_2/5 + 1)(2 a_3 + 5).If that bothers you, remember that
at m=0, two of the a's equal 0.If it still bothers you, try
and get everything to work some otherway.Readers who want
some sense of what I'm up against should read thevarious
posts in this thread from people trying to escape that
obviousconclusion.Above the *only* thing that works at m=0,
is a_1=a_2=0, while a_3 = 3.That's trivial enough and has
been admitted, but posters seem to wantthe constants to move
around if m doesn't equal 0, which is what Icall voodoo
math.James Harris<deleted>> a_1 a_2 a_3 = 25 (625
m^3 - 75 m^2 + 3m)> For any integer m, a_1 a_2 a_3 is a
multiple of 25. i.e. not coprime to 5.> Perhaps you
mean that none of the a's are coprime to 5?> Which is
trivially obvious and uninteresting after all.> Sorry to
have bothered you.> It's actually fascinating as it's a
bizarre result, and I think you> realize that I'm correct as
you *again* deleted out the following> wonderful
simpli?ation which shows clearly that I'm right.Yuck.
Unfortunately in my haste I neglected to refute the claim,
asit is in fact NOT true that none of the a's are coprime to
5, as infact the fascinating as it's a bizarre result is
that they ALL arecoprime to 5 in the ring of algebraic
integers.> The problem is with the ring of algebraic integers
as I've explained.> Consider the following which you deleted
yet again from my previous> post:> Hmmm...how about this
explanation?> Remember that the polynomial is> P(m) =
25(5000m^3 - 600 m^2 - 126m + 11),which gives the reader a
chance to see P(m) with only the m left asa symbol> and you
see that factor 25. Now imagine that I have the polynomial>
Q(m), where> P(m) = 25 Q(m)> so you have as a
factorization> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25>
> and the question is, how does that factor of 25 divide out?
I want readers to consider how many posters have apparently
attemptedto make them believe that the factor of f^2, here
25, was in somesense welded into the expression, when in fact
it's a factor of P(m)such that I can write P(m) = 25 Q(m)as
I've done here, or P(m) = f^2 Q(m), in general.> Well,
checking at Q(0), gives 11, so how can those factors of 25>
divide through Q(m) in such a way as to give 11, at m=0?>
There's only one way, which gives you> Q(m) = (2 a_1/5 +
1)(2 a_2/5 + 1)(2 a_3 + 5).> If that bothers you, remember
that at m=0, two of the a's equal 0.And remember that posters
in trying to refute have continuously calledm=0 a degenerate
case when in fact they need you to ignore theobvious.That is,
given that f^2 is this factor that's multipled times P(m)
asit is, then of course, it can be separated off, and it's
not socomplicated and extraordinary that you need Galois
Theory or any ofall that extra technicality.> If it still
bothers you, try and get everything to work some other>
way.Indeed.> Readers who want some sense of what I'm up
against should read the> various posts in this thread from
people trying to escape that obvious> conclusion.> Above
the *only* thing that works at m=0, is a_1=a_2=0, while a_3 =
3.> That's trivial enough and has been admitted, but posters
seem to want> the constants to move around if m doesn't equal
0, which is what I> call voodoo math.But what's at stake is
the *belief* that mathematicians could not havetaught
erroneous mathematics for over a hundred years.And in fact
the ?mathematics is STILL in the textbooks.It is rather
weak-minded that mathematicians would continue to ?htover
this, but I'm unfortunately familiar with how weak people
oftenare.The truth can be a bitter pill, expecially for fakes
?hting topreserve the lie.James Harris[snip]> The truth
can be a bitter pill, expecially for fakes ?hting to>
preserve the lie.>> James HarrisWell, enjoy the pill! You've
earned it. The argument you have given is erroneous. It
leads to a falseconclusion. The problem is not any
contradiction in math, or incompleteness of any ring, but in
the errorsyou made in your argument. You are now ?g a
dead horse. Please mount the beast and ride off into
thewilderness where you belong.--There are two things you
must never attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com> <deleted>>
>a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)>>For any
integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime
to 5.>>Perhaps you mean that none of the a's are coprime
to 5?>Which is trivially obvious and uninteresting after
all.>Sorry to have bothered you.>>It's actually
fascinating as it's a bizarre result, and I think
you>>realize that I'm correct as you *again* deleted out the
following>>wonderful simpli?ation which shows clearly that
I'm right.> Yuck. Unfortunately in my haste I neglected
to refute the claim, as> it is in fact NOT true that none of
the a's are coprime to 5, as in> fact the fascinating as
it's a bizarre result is that they ALL are> coprime to 5 in
the ring of algebraic integers.> Pretty much like the a's in
the factorization 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3
x + 1)are all coprime to 5?Why should your argument work in
the one case, and not in the other?Oh, in case you forgot: I
have shown explicit factors that are sharedby 5, and *EACH*
of the a's in this example. As a result, 5 CANNOT becoprime
to any of these a's.Why do you ?om the case you should
be ?ing?>>The problem is with the ring of algebraic
integers as I've explained.>>You have not proven a
thing.>>Consider the following which you deleted yet again
from my previous>>post:>>Hmmm...how about this
explanation?>>Remember that the polynomial is>> P(m) =
25(5000m^3 - 600 m^2 - 126m + 11),> which gives the
reader a chance to see P(m) with only the m left as> a
symbol>>and you see that factor 25. Now imagine that I
have the polynomial>>Q(m), where>> P(m) = 25 Q(m)So Q(m) =
5000 m^3 - 600 m^2 - 126 m + 11?>>so you have as a
factorization>> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5)/25Rather, a factorization: P(m) = (2 a_1 + 5)(2 a_2 + 5)(2
a_3 + 5)?What result guarantees that such a factorization is
possible?>>and the question is, how does that factor of 25
divide out? > What happened to the extra dollar?> I want
readers to consider how many posters have apparently
attempted> to make them believe that the factor of f^2, here
25, was in some> sense welded into the expression, when in
fact it's a factor of P(m)> such that I can write> P(m) =
25 Q(m)> as I've done here, or P(m) = f^2 Q(m), in
general.> I'm sorry, I was thinking about your other case (I
mentioned it above)where you said all the a's were coprime to
5, yet I *showed* you thecommon factors.I was wondering why
you wouldn't even address that failure of yourmethod, rather
than moving to another (most likely incorrect) case?Why
wouldn't a person address what can be computed directly,
ratherthan running to another case where the computations are
almost surelymore complicated? Is it the very intractibility
of the new problemthat makes you safe? If no one can tell
you're wrong, does that makeyou right?What if you could be
shown wrong with Galois theory? I see in your paperthat you
call such applications *overinterpretations* of Galois
theory,although you have admitted that you know less about
Galois theory thanvirtually any human, & you prove that point
by pointing out (as thoughitwere some ?hat Galois theory
typically deals with ?lds, whereyou're working with rings.>
>>Well, checking at Q(0), gives 11, so how can those factors
of 25>>divide through Q(m) in such a way as to give 11, at
m=0?>>There's only one way, which gives you>> Q(m) = (2
a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).>>If that bothers you,
remember that at m=0, two of the a's equal 0.> And
remember that posters in trying to refute have continuously
called> m=0 a degenerate case when in fact they need you to
ignore the> obvious.> The case is degenerate in that the
degree of the polynomial drops from 3to 1 when m=0.
Frequently, calculations change in degenerate cases.> That
is, given that f^2 is this factor that's multipled times P(m)
as> it is, then of course, it can be separated off, and it's
not so> complicated and extraordinary that you need Galois
Theory or any of> all that extra technicality.> Do you mean
divided? Why say separated? Is it necessary to use your
ownprivate language?>>If it still bothers you, try and get
everything to work some other>>way.> Indeed.>
>>Readers who want some sense of what I'm up against should
read the>>various posts in this thread from people trying to
escape that obvious>>conclusion.>>Above the *only* thing
that works at m=0, is a_1=a_2=0, while a_3 = 3.>>What do you
mean? My PC works when m=0, and it has nothing whatsoeverto
do with those a's. My car works, and it is totally unaware of
thevalues of m and the a's. I work, for that matter, and I
haven't worriedabout your m and a's, not ever.>>That's
trivial enough and has been admitted, but posters seem to
want>>the constants to move around if m doesn't equal 0,
which is what I>>call voodoo math.> What has been
admitted?> But what's at stake is the *belief* that
mathematicians could not have> taught erroneous mathematics
for over a hundred years.> Could you please address your
failure to ? the claim you had about theabove polynomial
factorization? Please make correct assertions beforeyou claim
that mathematicians have taught erroneous mathematics.> And
in fact the ?mathematics is STILL in the textbooks.>
It is rather weak-minded that mathematicians would continue
to ?ht> over this, but I'm unfortunately familiar with how
weak people often> are.You *are* familiar with how weak a
*certain person* often is. You arethe source of dishonesty,
error, and cynicism in these threads, yet youmake claims of
being entirely honest and forthright. Doesn't yourpractice of
hypocrisy cause you any discomfort? What do your parentsthink
of your hypocrisy? Are they hypocrites as well?> The truth
can be a bitter pill, expecially for fakes ?hting to>
preserve the lie.> Fighting to preserve the lie. That's how
JSH refers to a personpointing out that he [JSH] is wrong;
pointing this fact out for dayson end, and with algebra that
is as simple and direct as anyone couldimagine.Fakes.
That's who can show what JSH says doesn't exist, and who
canpoint those non-existent things out in a simple fashion.>
> James HarrisPerhaps our esteemed Mister Harris is warning
everyone about himself.Dale.> <deleted>>
a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)> For any
integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime
to 5.> Perhaps you mean that none of the a's are
coprime to 5?> Which is trivially obvious and
uninteresting after all.> Sorry to have bothered you.>
> It's actually fascinating as it's a bizarre result, and I
think you> realize that I'm correct as you *again* deleted
out the following> wonderful simpli?ation which shows
clearly that I'm right.> Yuck. Unfortunately in my haste I
neglected to refute the claim, as> it is in fact NOT true
that none of the a's are coprime to 5, as in> fact the
fascinating as it's a bizarre result is that they ALL are>
coprime to 5 in the ring of algebraic integers.> The
problem is with the ring of algebraic integers as I've
explained.> Consider the following which you deleted
yet again from my previous> post:> Hmmm...how about
this explanation?> Remember that the polynomial is>
> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),> which gives the
reader a chance to see P(m) with only the m left as> a
symbol> Yes! I like this example. See below.> and you see
that factor 25. Now imagine that I have the polynomial>
Q(m), where> P(m) = 25 Q(m)> so you have as a
factorization> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5)/25> and the question is, how does that factor of 25
divide out? > I want readers to consider how many posters
have apparently attempted> to make them believe that the
factor of f^2, here 25, was in some> sense welded into the
expression, when in fact it's a factor of P(m)> such that I
can write> P(m) = 25 Q(m)> as I've done here, or P(m) =
f^2 Q(m), in general.> Well, checking at Q(0), gives 11,
so how can those factors of 25> divide through Q(m) in such
a way as to give 11, at m=0?> There's only one way,
which gives you> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2
a_3 + 5).> If that bothers you, remember that at m=0,
two of the a's equal 0.> Proving what, exactly? Clearly a_1,
a_2 and a_3 are functionsof m. When m = 0, two of them are 0.
However when m is not zero,none of them are zero. So how do
their values when m = 0 determine what they will be when m <>
0 ?> And remember that posters in trying to refute have
continuously called> m=0 a degenerate case when in fact
they need you to ignore the> obvious.> That is, given that
f^2 is this factor that's multipled times P(m) as> it is,
then of course, it can be separated off, and it's not so>
complicated and extraordinary that you need Galois Theory or
any of> all that extra technicality.> If it still bothers
you, try and get everything to work some other> way.>
Indeed.> Readers who want some sense of what I'm up
against should read the> various posts in this thread from
people trying to escape that obvious> conclusion.>
Above the *only* thing that works at m=0, is a_1=a_2=0, while
a_3 = 3.> That's trivial enough and has been admitted,
but posters seem to want> the constants to move around if m
doesn't equal 0, which is what I> call voodoo math.> But
what's at stake is the *belief* that mathematicians could not
have> taught erroneous mathematics for over a hundred years.>
Let's go back to your original polynomial, P(m) = 25(5000m^3 -
600 m^2 - 126m + 11),and assume it is factored *as you
propose* in the form[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5). Now instead of m = 0, let's try m = 1: P(m) = 25 * (5000 -
600 - 126 + 11) = 25 * 4285. This can be factored in the form
[*] by letting a_1 = -5, a_2 = -5, and a_3 = 2140. Note that
EACH ONE of these is divisible by 5: NONE ARE COPRIME TO 5.
Clearly m = 0 is a special case. It is differentin an
essential way from m = 1 and other nonzerovalues of m. Now,
what's your explanation? Nora B.> And in fact the ?athe
matics is STILL in the textbooks.> It is rather
weak-minded that mathematicians would continue to ?ht> over
this, but I'm unfortunately familiar with how weak people
often> are.> The truth can be a bitter pill, expecially for
fakes ?hting to> preserve the lie.> James Harris> But
what's at stake is the *belief* that mathematicians could not
have> taught erroneous mathematics for over a hundred
years.No, it's not at stake. It's not even a serious
consideration foranyone but you.-- Wayne Brown | When your
tail's in a crack, you improvisefwbrown@bellsouth.net | if
you're good enough. Otherwise you give | your pelt to the
trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock> <deleted>> a_1 a_2
a_3 = 25 (625 m^3 - 75 m^2 + 3m)> For any
integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime
to 5.> Perhaps you mean that none of the a's
are coprime to 5?> Which is trivially obvious and
uninteresting after all.> Sorry to have bothered you.>
> It's actually fascinating as it's a bizarre result,
and I think you> realize that I'm correct as you *again*
deleted out the following> wonderful simpli?ation which
shows clearly that I'm right.> Yuck. Unfortunately in
my haste I neglected to refute the claim, as> it is in fact
NOT true that none of the a's are coprime to 5, as in> fact
the fascinating as it's a bizarre result is that they ALL
are> coprime to 5 in the ring of algebraic integers.>
> The problem is with the ring of algebraic integers as I've
explained.> Consider the following which you
deleted yet again from my previous> post:>
Hmmm...how about this explanation?> Remember that
the polynomial is> P(m) = 25(5000m^3 - 600 m^2 -
126m + 11),> which gives the reader a chance to see
P(m) with only the m left as> a symbol> Yes! I like
this example. See below.Fascinating.> and you see that
factor 25. Now imagine that I have the polynomial> Q(m),
where> P(m) = 25 Q(m)> so you have as a
factorization> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3
+ 5)/25> and the question is, how does that factor
of 25 divide out? > I want readers to consider how many
posters have apparently attempted> to make them believe that
the factor of f^2, here 25, was in some> sense welded into
the expression, when in fact it's a factor of P(m)> such
that I can write> P(m) = 25 Q(m)> as I've done
here, or P(m) = f^2 Q(m), in general.> Well, checking
at Q(0), gives 11, so how can those factors of 25> divide
through Q(m) in such a way as to give 11, at m=0?>
There's only one way, which gives you> Q(m) = (2
a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).> If that
bothers you, remember that at m=0, two of the a's equal 0.>
> Proving what, exactly? Clearly a_1, a_2 and a_3 are
functions> of m. When m = 0, two of them are 0. However when
m is not zero,> none of them are zero. So how do their values
when m = 0 > determine what they will be when m <> 0 ?Well you
have P(m) = 25 Q(m) and that factor 25, which has been
thefocus of all the arguing. Now considering Q(m) it turns
out thatlooking at the factorization Q(m) = (2 a_1 + 5)(2 a_2
+ 5)(2 a_3 + 5)/25the a's are in the way, and it's not clear
how you divide out.Now you *know* that the 25 divides through
some way, and it's clearthat whatever way that is would be
true for any m, just like with 2(x^2 + 2x + 1) =
(x+1)(2x+2)when you divide off 2, you do it for all x.For
readers, people like Nora Baron have basically been arguing
thatyou can have something like Q(m)=(2 a_1/h_1 + h_2 h_3)(2
a_2/h_2 + h_1 h_3)(2 a_3/h_3 + h_1 h_2)where h_1 h_2 h_3 = 5,
and each is not a unit.But notice then that setting m=0, gives
(h_1 h_2 h_3)^2 = 25, when infact Q(0)=11.Given that rigorous
fact they claim that m=0 is a some kind of specialcase and
try to cast doubt on the result, which is an attack onalgebra
that I call voodoo math.Surprisingly they've been quite
successful from what I've gathered inconvincing people, which
makes you wonder about people's understandingof
algebra.Because from algebra, it turns out that you can just
set m=0, to?ure out how that 25 divides out, as I've done.>
> And remember that posters in trying to refute have
continuously called> m=0 a degenerate case when in fact
they need you to ignore the> obvious.> That is, given
that f^2 is this factor that's multipled times P(m) as> it
is, then of course, it can be separated off, and it's not so>
> complicated and extraordinary that you need Galois Theory or
any of> all that extra technicality.> If it still
bothers you, try and get everything to work some other>
way.> Indeed.> Readers who want some sense of
what I'm up against should read the> various posts in
this thread from people trying to escape that obvious>
conclusion.> Above the *only* thing that works at
m=0, is a_1=a_2=0, while a_3 = 3.> That's trivial
enough and has been admitted, but posters seem to want>
the constants to move around if m doesn't equal 0, which is
what I> call voodoo math.> But what's at stake is
the *belief* that mathematicians could not have> taught
erroneous mathematics for over a hundred years.> Let's
go back to your original polynomial,> P(m) = 25(5000m^3 -
600 m^2 - 126m + 11),> and assume it is factored *as you
propose* in the form> [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2
a_3 + 5).> Now instead of m = 0, let's try m = 1:Ok.> P(m)
= 25 * (5000 - 600 - 126 + 11) = 25 * 4285.> This can be
factored in the form [*] by letting> a_1 = -5, a_2 = -5,
and a_3 = 2140.Um, Nora Baron, now you willy-nilly give the
a's values. Doesn't thatseem to be just a tad bit strange to
you?> Note that EACH ONE of these is divisible by 5:Ok, you
*pick* values from the a's as if they're not de?ed by
acubic, which they are, and your picked values are supposed
to provesomething? > NONE ARE COPRIME TO 5.So are your
claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 =
2140?Remember you simply tossed those numbers out there, but
there *is* away to calculate the a's so your assertion can be
tested, as they areroots to a cubic.Now if you can do that
Nora Baron then clearly I made some kind ofmistake and
there's no more room for me to argue.So why don't you go back
to the cubic which de?es the a's, stick inall the values and
see if the a's come out as you have above, and thenit's
over.> Clearly m = 0 is a special case. It is different> in
an essential way from m = 1 and other nonzero> values of
m.Clearly you haven't proven your assertion.> Now, what's
your explanation?> Nora B.Well I've explained above, and
again, of course, you can't just *pick*values for the a's for
a particular m, as their values are setrigorously. What I'm
curious about are the people who believe NoraBaron had a
valid point.Speak up, and don't be shy as I suspect you may
believe she still hasa valid point and I need to understand
why she's so effective inconvincing people.Come on, speak up,
do you think I answered her objections? Do youbelieve they
were valid in the ?st place?James Harris>and
you see that factor 25. Now imagine that I have the
polynomial>>Q(m), where>> P(m) = 25 Q(m)>>so
you have as a factorization>> Q(m) = (2 a_1 + 5)(2 a_2
+ 5)(2 a_3 + 5)/25>>and the question is, how does that
factor of 25 divide out? >>I want readers to consider how
many posters have apparently attempted>to make them believe
that the factor of f^2, here 25, was in some>sense welded
into the expression, when in fact it's a factor of
P(m)>such that I can write>> P(m) = 25 Q(m)>>as
I've done here, or P(m) = f^2 Q(m), in
general.>>Well, checking at Q(0), gives 11, so how
can those factors of 25>>divide through Q(m) in such a way
as to give 11, at m=0?>>There's only one way, which
gives you>> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 +
5).>>If that bothers you, remember that at m=0, two of
the a's equal 0.> Proving what, exactly? Clearly a_1, a_2
and a_3 are functions>>of m. When m = 0, two of them are 0.
However when m is not zero,>>none of them are zero. So how do
their values when m = 0 >>determine what they will be when m
<> 0 ?> Well you have P(m) = 25 Q(m) and that factor 25,
which has been the> focus of all the arguing. Now considering
Q(m) it turns out that> looking at the factorization> Q(m) =
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25> the a's are in the
way, and it's not clear how you divide out.> Now you *know*
that the 25 divides through some way, and it's clear> that
whatever way that is would be true for any m, just like with>
> 2(x^2 + 2x + 1) = (x+1)(2x+2)> when you divide off 2, you
do it for all x.> For readers, people like Nora Baron have
basically been arguing that> you can have something like>
Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 +
h_1 h_2)> where h_1 h_2 h_3 = 5, and each is not a unit.>
But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25,
when in> fact Q(0)=11.> Given that rigorous fact they claim
that m=0 is a some kind of special> case and try to cast doubt
on the result, which is an attack on> algebra that I call
voodoo math.> Surprisingly they've been quite successful
from what I've gathered in> convincing people, which makes
you wonder about people's understanding> of algebra.>
Because from algebra, it turns out that you can just set m=0,
to> ?ure out how that 25 divides out, as I've done.>And
remember that posters in trying to refute have continuously
called>m=0 a degenerate case when in fact they need you
to ignore the>obvious.>>That is, given that f^2 is this
factor that's multipled times P(m) as>it is, then of course,
it can be separated off, and it's not so>complicated and
extraordinary that you need Galois Theory or any of>all
that extra technicality.>>If it still bothers you,
try and get everything to work some
other>>way.>>Indeed.>>Readers who want some
sense of what I'm up against should read the>>various posts
in this thread from people trying to escape that
obvious>>conclusion.>>Above the *only* thing that
works at m=0, is a_1=a_2=0, while a_3 = 3.>>That's
trivial enough and has been admitted, but posters seem to
want>>the constants to move around if m doesn't equal 0,
which is what I>>call voodoo math.>>But what's at
stake is the *belief* that mathematicians could not
have>taught erroneous mathematics for over a hundred
years.> Let's go back to your original polynomial,>>
P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),>>and assume it is
factored *as you propose* in the form>>[*] P(m) = (2 a_1 +
5)(2 a_2 + 5)(2 a_3 + 5).>> Now instead of m = 0, let's try
m = 1:> Ok.>> P(m) = 25 * (5000 - 600 - 126 + 11) = 25
* 4285.>> This can be factored in the form [*] by
letting>> a_1 = -5, a_2 = -5, and a_3 = 2140.> Um, Nora
Baron, now you willy-nilly give the a's values. Doesn't that>
seem to be just a tad bit strange to you?You haven't provided
a way to de?e what the a_i(m) are, so of course she can
simply select values that work.>> Note that EACH ONE of these
is divisible by 5:> Ok, you *pick* values from the a's as if
they're not de?ed by a> cubic, which they are, and your
picked values are supposed to prove> something?They AREN'T
de?ed by a cubic. They are constrained by a cubic. If you
want them to be de?ed by a cubic, you'll have to say more
about how they are supposed to be constructed.>> NONE ARE
COPRIME TO 5.> So are your claiming that at m=1, a_1 =
-5, a_2 = -5, and a_3 = 2140?> Remember you simply tossed
those numbers out there, but there *is* a> way to calculate
the a's so your assertion can be tested, as they are> roots
to a cubic.Not that you've provided. If you have a way in
mind, please provide it to us and include it in your paper.
You've only claimed this things exist and have certain
properties. If you want to change them, so be it. But that
may cause more problems for you.> Now if you can do that Nora
Baron then clearly I made some kind of> mistake and there's no
more room for me to argue.room for you to argue.> So why don't
you go back to the cubic which de?es the a's, stick in> all
the values and see if the a's come out as you have above, and
then> it's over.For each value of m, she can de?e a_i(m) to
be whatever three values makes both sides equal. You've given
nothing more speci? about the values to require more.>> Now,
what's your explanation?>> Nora B.> Well I've explained
above, and again, of course, you can't just *pick*> values for
the a's for a particular m, as their values are set>
rigorously. What I'm curious about are the people who believe
Nora> Baron had a valid point.Where have you established how
to rigorously compute them. What are they?> Speak up, and
don't be shy as I suspect you may believe she still has> a
valid point and I need to understand why she's so effective
in> convincing people.Because you have not stated all of the
conditions you want to exist. Once you do that, you will lose
the results you are trying to achieve.> Come on, speak up, do
you think I answered her objections? Do you> believe they
were valid in the ?st place?I think you didn't understand
her objections.-- Will Twentymanthe only answer is,JSH is
always-and-only every where dense. > P(m) = 25(5000m^3 - 600
m^2 - 126m + 11),> and assume it is factored *as you
propose* in the form> [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2
a_3 + 5).> Now instead of m = 0, let's try m = 1:> P(m) =
25 * (5000 - 600 - 126 + 11) = 25 * 4285.> This can be
factored in the form [*] by letting> a_1 = -5, a_2 = -5,
and a_3 = 2140.> Note that EACH ONE of these is divisible
by 5:> NONE ARE COPRIME TO 5.> Clearly m = 0 is a special
case. It is different> in an essential way from m = 1 and
other nonzero> values of m.> Now, what's your
explanation?--UN HYDROGEN (sic; Methanex (TM) reformanteurs)
ECONOMIE?...La Troi Phases d'Exploitation de la Protocols des
Grises de Kyoto:(FOSSILISATION [McCainanites?]
(TM/sic))/BORE/GUSH/NADIR @
http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/
bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de
COUP D'ETAT, 3/30/81> <deleted>>
> a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)>
> For any integer m, a_1 a_2 a_3 is a multiple of 25.
i.e. not coprime to 5.> Perhaps you mean
that none of the a's are coprime to 5?> Which is
trivially obvious and uninteresting after all.> Sorry
to have bothered you.> It's actually fascinating
as it's a bizarre result, and I think you> realize that
I'm correct as you *again* deleted out the following>
wonderful simpli?ation which shows clearly that I'm right.>
> Yuck. Unfortunately in my haste I neglected to
refute the claim, as> it is in fact NOT true that none of
the a's are coprime to 5, as in> fact the fascinating as
it's a bizarre result is that they ALL are> coprime to
5 in the ring of algebraic integers.> The problem
is with the ring of algebraic integers as I've explained.>
> Consider the following which you deleted yet
again from my previous> post:>
Hmmm...how about this explanation?> Remember
that the polynomial is> P(m) = 25(5000m^3 - 600
m^2 - 126m + 11),> which gives the reader a chance
to see P(m) with only the m left as> a symbol>
> Yes! I like this example. See below.> Fascinating.>
> and you see that factor 25. Now imagine that I have the
polynomial> Q(m), where> P(m) = 25 Q(m)>
> so you have as a factorization>
Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25>
and the question is, how does that factor of 25 divide out? >
> I want readers to consider how many posters have
apparently attempted> to make them believe that the
factor of f^2, here 25, was in some> sense welded into
the expression, when in fact it's a factor of P(m)> such
that I can write> P(m) = 25 Q(m)> as I've
done here, or P(m) = f^2 Q(m), in general.> Well,
checking at Q(0), gives 11, so how can those factors of 25>
> divide through Q(m) in such a way as to give 11, at m=0?>
> There's only one way, which gives you>
> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).>
> If that bothers you, remember that at m=0, two of the
a's equal 0.> Proving what, exactly? Clearly a_1,
a_2 and a_3 are functions> of m. When m = 0, two of them are
0. However when m is not zero,> none of them are zero. So
how do their values when m = 0 > determine what they will
be when m <> 0 ?> Well you have P(m) = 25 Q(m) and that
factor 25, which has been the> focus of all the arguing. Now
considering Q(m) it turns out that> looking at the
factorization> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25>
> the a's are in the way, and it's not clear how you divide
out.> Now you *know* that the 25 divides through some way,
and it's clear> that whatever way that is would be true for
any m, just like with> 2(x^2 + 2x + 1) = (x+1)(2x+2)>
when you divide off 2, you do it for all x.> For readers,
people like Nora Baron have basically been arguing that> you
can have something like> Q(m)=(2 a_1/h_1 + h_2 h_3)(2
a_2/h_2 + h_1 h_3)(2 a_3/h_3 + h_1 h_2)> where h_1 h_2 h_3
= 5, and each is not a unit.> But notice then that setting
m=0, gives (h_1 h_2 h_3)^2 = 25, when in> fact Q(0)=11.>
Given that rigorous fact they claim that m=0 is a some kind
of special> case and try to cast doubt on the result, which
is an attack on> algebra that I call voodoo math.>
Surprisingly they've been quite successful from what I've
gathered in> convincing people, which makes you wonder about
people's understanding> of algebra.> Because from algebra,
it turns out that you can just set m=0, to> ?ure out how
that 25 divides out, as I've done.> No voodoo at all. What
you would like to have here is that the same a_1, a_2, and
a_3 work when m = 0 as when m <> 0. But you know, and
everyone else knows, that that is totally impossible. Why?
Because when m = 0, two of the a's must be zero. When m <> 0,
NONE of the a's can be zero. It's real simple. The a's are
functions of m. And I don't mean CONSTANT functions. Got
that? Different m's give different a's. You believe that
because a_1 is divisible by 5 when m = 0,it must be divisible
by 5 for other values of m also. It issome kind of hunch that
you have, based on intuition. It seemslike a nice parallel
construction. You think it's obvious. It isn't. In fact it is
false. Your intuition this time is wrong. Your hunch has been
disproved. What the a's are when m = 0 is a special case. It
does NOTdetermine what they are when m <> 0. Sure, when m =
0, you can assume that a_1 and a_2 are zero. Can you can
saythat 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of
courseyou can - they are both 0! After all, ANY number
divides 0. For example, 137 divides 0. Both 2*a_1/137 and
2*a_2/137 arealgebraic integers also, since both are 0. Does
that mean that 137 must divide two of the a's, when m <> 0
???? By your logic, YES! Again: ***** THE A's ARE NOT
CONSTANTS. THEY ARE FUNCTIONS OF M.WHATEVER WEIRD PROPERTIES
THEY MAY HAVE WHEN M = 0, CANNOT BE ASSUMEDTO BE TRUE WHEN M
<> 0. ***** Why is this so hard for you to understand? Why do
you keep trying toact like I am pulling some kind of trick
when I say this, or pretending that I am lying?> And
remember that posters in trying to refute have continuously
called> m=0 a degenerate case when in fact they need
you to ignore the> obvious.> That is, given
that f^2 is this factor that's multipled times P(m) as>
it is, then of course, it can be separated off, and it's not
so> complicated and extraordinary that you need Galois
Theory or any of> all that extra technicality.>
> If it still bothers you, try and get everything to work some
other> way.> Indeed.> Readers who
want some sense of what I'm up against should read the>
various posts in this thread from people trying to escape
that obvious> conclusion.> Above the
*only* thing that works at m=0, is a_1=a_2=0, while a_3 =
3.> That's trivial enough and has been
admitted, but posters seem to want> the constants to
move around if m doesn't equal 0, which is what I> call
voodoo math.> But what's at stake is the *belief*
that mathematicians could not have> taught erroneous
mathematics for over a hundred years.> Let's go
back to your original polynomial,> P(m) = 25(5000m^3 -
600 m^2 - 126m + 11),> and assume it is factored *as
you propose* in the form> [*] P(m) = (2 a_1 + 5)(2 a_2
+ 5)(2 a_3 + 5).> Now instead of m = 0, let's try m =
1:> Ok.> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 *
4285.> This can be factored in the form [*] by letting>
> a_1 = -5, a_2 = -5, and a_3 = 2140.> Um, Nora Baron,
now you willy-nilly give the a's values. Doesn't that> seem
to be just a tad bit strange to you?> Nope. It was your idea,
not mine, to replace the x'swith a constant value of 2. When
you did that you were no longer factoring a polynomial. You
are just factoringan ordinary number. There are no hidden
rules that say itmust be factored as a polynomial. You can't
have it both ways. You want it factored as a polynomial,
leave the x's in there (and see below). You want it factored
as an ordinary number (remember? VERY much simpli?d) then
one must assume that any factorization is legitimate, and you
get my example above. So let's go back to assuming you want it
factored as a polynomial. Of course you know that both I and
W. DaleHall have proofs that that is not going to work. You
havenot provided a valid objection to my proof, and you
havenot provided any response whatsoever to what Dale did.
Idon't see how you can. It's a simple computation. You do the
arithmetic, and Dale is either right or he is wrong.You don't
have to understand ANYTHING. There is no wool to be pulled
over anyone's eyes. Try it and see. See also below.> Note
that EACH ONE of these is divisible by 5:> Ok, you *pick*
values from the a's as if they're not de?ed by a> cubic,
which they are, and your picked values are supposed to prove>
something?> Yes. It proves what you said was wrong. You said
that oneof the a's had to be coprime to 5. Clearly, totally,
unambiguously false. > NONE ARE COPRIME TO 5.> So are
your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 =
2140?> Remember you simply tossed those numbers out there,
but there *is* a> way to calculate the a's so your assertion
can be tested, as they are> roots to a cubic.> They *would*
be roots of a cubic in the variable x, but you replaced x by
2. At that point there is no longer a polynomial.The
restriction that they be roots of a cubic is gone. It wasyour
idea, not mine, to oversimplify things and make the
substitution. Remember: if you want to go back to assuming
that a_1, a_2, anda_3 are roots of a cubic in the variable x,
to continue to maintain what you think is true, you are going
to have to ?d an error in the proof I have presented (and is
appended below).> Now if you can do that Nora Baron then
clearly I made some kind of> mistake and there's no more room
for me to argue.> Yep. You should not have substituted x = 2.
You want it factored as a polynomial, you had better leave it
a polynomialand not try to oversimplify. You have shown
confusion previouslyover what the difference between a
polynomial and the evaluationof that polynomial when you
choose a particular value for x. The moral here is, this
little example should show you pretty convincingly that when
people tried to get you to understand that difference, they
were not just playing with semantics. There is another more
general moral here, and you have run afoul of this one many a
time in the past also: Examples arenot a substitute for
proofs. And do NOT oversimplify when youare trying to do
mathematics. > So why don't you go back to the cubic which
de?es the a's, stick in> all the values and see if the a's
come out as you have above, and then> it's over.> I will do
better than that. I will reproduce my proof that yourclaim
that one of the a's is coprime to 5 is wrong, but I will do
it inmore generality. That way I will be heading off any
examplesyou might try to come up with in the future. What you
want isa consequence of the following claim, which you have
statedelsewhere: CLAIM: It is possible to ?d a monic
polynomial of degree 3, with integer coef?ients, irreducible
over the rationals. and with roots a1, a2, and a3, such that
at least one of a1, a2, or a3 is coprime, in the ring of
algebraic integers, to a prime factor of the constant term of
the polynomial. Right? You have claimed as recently as
yesterday. DISPROOF OF CLAIM: Let Q(x) = x^3 + a*x^2 + b*x +
c, where a, b, and c are integers and c = p*v, where p is a
prime and v is another integer. Q(x) is clearly monic. Assume
Q(x) is irreducible. Let a1, a2, and a3 be roots of Q(x). Note
that by de?ition, a1, a2, and a3 are algebraic integers. The
claim is that at least one of a1, a2, or a3 is coprime to p.
Assume a1 is coprime to p. By standard theory***, there
exists an automorphism F12 of the ?ld of algebraic numbers
such that: 1. F12 leaves the sub?ld of rational numbers
?ed, i.e., if q is rational, F12(q) = q. 2. F12(a1) = a2. 3.
If t is an algebraic integer, F12(t) is also an algebraic
integer. Now since a1 is relatively prime to p, there exist
algebraic integers r and s such that [1] r*a1 + s*p = 1. Now
apply the automorphism F12 to both sides of [1]:
F12(r)*F12(a1) + F12(s)*F12(p) = F12(1). By the properties
above, F12(p) = p and F12(1) = 1. Moreover, r' = F12(r) is an
algebraic integer, and s' = F12(s) is an algebraic integer.
Finally, F12(a1) = a2. Thus one obtains: r'*u2 + s'*p = 1,
which says: a2 and p are coprime in the algebraic integers.
Similarly one shows that a3 and p are coprime. Therefore if
one of a1, a2, or a3 is coprime to p, then they all are. But
a1 * a2 * a3 = p * v. That is, p divides the product of a1,
a2, and a3. Therefore p cannot be coprime to each of a1, a2,
and a3. Putting all this together, one concludes that NONE of
a1, a2, or a3 can be coprime to p. This directly contradicts
the CLAIM made above. Please feel free to point out any
errors in the proof I just gave.***: Reference on
automorphisms:
http://www.math.niu.edu/~beachy/aaol/galois.html or the
excellent textbook Abstract Algebra, by John Beachy and
William D. Blair > Clearly m = 0 is a special case. It is
different> in an essential way from m = 1 and other
nonzero> values of m.> Clearly you haven't proven your
assertion.> Now, what's your explanation?> Nora B.>
> Well I've explained above, and again, of course, you can't
just *pick*> values for the a's for a particular m, as their
values are set> rigorously. If you can just *pick* x = 2,
changing the problem from factoringa polynomial to factoring
an ordinary number, certainly I can*pick* any factorization
that works. Lesson here: be careful whatyou *pick*.> What I'm
curious about are the people who believe Nora> Baron had a
valid point.> Speak up, and don't be shy as I suspect you
may believe she still has> a valid point and I need to
understand why she's so effective in> convincing people.>
Come on, speak up, do you think I answered her objections? Do
you> believe they were valid in the ?st place?> On this point
I agree. Speak up, lurkers! Nora B.> James
Harris<deleted> Let's go back to your original
polynomial,>> P(m) = 25(5000m^3 - 600 m^2 - 126m +
11),>>and assume it is factored *as you propose* in the
form>>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).>
>> Now instead of m = 0, let's try m = 1:> Ok.>
>> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285.>
>> This can be factored in the form [*] by letting>>
a_1 = -5, a_2 = -5, and a_3 = 2140.> Um, Nora
Baron, now you willy-nilly give the a's values. Doesn't that>
> seem to be just a tad bit strange to you?> You haven't
provided a way to de?e what the a_i(m) are, so of course >
she can simply select values that work.That's false as the
a's are given by the factorization (v^3+1)x^3 -3vxy^2 + y^3 =
(a_1 x + y)(a_2 x + y)(a_3 x + y)where v=-1+mf^2, and y=uf.For
my example I used x=2, f=5, u=1. And now letting m=1, to
testyour claim I have 13825x^3 - 72xy^2 + y^3 = (a_1 x +
y)(a_2 x + y)(a_3 x + y)so the a's are the roots of the cubic
a^3 + 72a^2 - 13825 = 0.And checking with a=-5, gives -12150,
and not 0.So Nora Baron's claim and yours is refuted.>>
Note that EACH ONE of these is divisible by 5:That was Nora
Baron's comment for her picked values for the a's. > Ok,
you *pick* values from the a's as if they're not de?ed by a>
> cubic, which they are, and your picked values are supposed
to prove> something?> They AREN'T de?ed by a cubic. They
are constrained by a cubic. If > you want them to be de?ed by
a cubic, you'll have to say more about > how they are supposed
to be constructed.That statement is false as I've shown by
giving the cubic. >> NONE ARE COPRIME TO 5.> So
are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 =
2140?That was my question to Nora Baron.> Remember you
simply tossed those numbers out there, but there *is* a>
way to calculate the a's so your assertion can be tested, as
they are> roots to a cubic.> Not that you've provided. If
you have a way in mind, please provide it > to us and include
it in your paper. You've only claimed this things > exist and
have certain properties. If you want to change them, so be >
it. But that may cause more problems for you.I simplify and
people like you come in and try to confuse, and sadly,I think
that the readership of sci.math wants to be confused by you. >
> Now if you can do that Nora Baron then clearly I made some
kind of> mistake and there's no more room for me to argue.>
> room for you to argue.Well now that I've given the cubic, I
challenge you to admit *you*made a mistake.> So why don't
you go back to the cubic which de?es the a's, stick in>
all the values and see if the a's come out as you have above,
and then> it's over.> For each value of m, she can de?e
a_i(m) to be whatever three values > makes both sides equal.
You've given nothing more speci? about the > values to
require more.That is false as I've repeatedly given the
expression de?ing thea's.>> Now, what's your explanation?>
>> Nora B.> Well I've explained above, and
again, of course, you can't just *pick*> values for the a's
for a particular m, as their values are set> rigorously.
What I'm curious about are the people who believe Nora>
Baron had a valid point.> Where have you established how to
rigorously compute them. What are they?For m=1, f=5, the a's
are the roots of the cubic a^3 + 72a^2 - 13825 = 0.> Speak
up, and don't be shy as I suspect you may believe she still
has> a valid point and I need to understand why she's so
effective in> convincing people.> Because you have not
stated all of the conditions you want to exist. > Once you do
that, you will lose the results you are trying to achieve.I've
stated those conditions repeatedly. Are you now claiming
thatyou have never seen the factorization (v^3+1)x^3 -3vxy^2
+ y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)? > Come on, speak
up, do you think I answered her objections? Do you> believe
they were valid in the ?st place?> I think you didn't
understand her objections.I think you're lying.James
Harris [.snip.]>> What I'm curious about are the people
who believe Nora>> Baron had a valid point.> Speak up,
and don't be shy as I suspect you may believe she still has>>
a valid point and I need to understand why she's so effective
in>> convincing people.> Come on, speak up, do you think
I answered her objections? Do you>> believe they were valid
in the ?st place?> On this point I agree. Speak up,
lurkers!Hmmm... Within the last 24 Hours, James incorrectly
claimed that KeithRamsay was doing an appeal to the gallery,
and correctly noted thatappealing to the gallery is a logical
fallacy. Perhaps James does notrealize that what he is doing
here is a ->classic<- example ofappealing to the gallery?
After all, he does not know what appeal toauthority, ad
hominem, and any of the other logical fallacies heclaims
others are committing
are...
===


===
Why do you take so much trouble to expose
such a reasoner as Mr. Smith? I answer as a deceased friend
of mine used to answer on like occasions - A man's capacity
is no measure of his power to do mischief. Mr. Smith has
untiring energy, which does something; self-evident honesty
of conviction, which does more; and a long purse, which does
most of all. He has made at least ten publications, full of
?ures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
inde?ite something in the mysterious all this. They are
brought to the point of suspicion that the mathematicians
ought not to treat all this with such undisguised contempt,
at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by
Augustus de
Morgan
===


===
Arturo Magidinmagidin@math.berkeley.edu>
> <deleted> Let's go back to your original
polynomial,>> P(m) = 25(5000m^3 - 600 m^2 - 126m +
11),>>and assume it is factored *as you propose* in the
form>>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5).>> Now instead of m = 0, let's try m =
1:>Ok.> P(m) = 25 * (5000 - 600 - 126 +
11) = 25 * 4285.>> This can be factored in the form [*]
by letting>> a_1 = -5, a_2 = -5, and a_3 =
2140.>Um, Nora Baron, now you willy-nilly give the
a's values. Doesn't that>seem to be just a tad bit strange
to you?>>You haven't provided a way to de?e what the
a_i(m) are, so of course >>she can simply select values that
work.> That's false as the a's are given by the
factorization> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x
+ y)(a_3 x + y)> where v=-1+mf^2, and y=uf.> For my
example I used x=2, f=5, u=1. And now letting m=1, to test>
your claim I have> 13825x^3 - 72xy^2 + y^3 = (a_1 x +
y)(a_2 x + y)(a_3 x + y)> so the a's are the roots of the
cubic> a^3 + 72a^2 - 13825 = 0.> And checking with a=-5,
gives -12150, and not 0.> So Nora Baron's claim and yours
is refuted.I thought you said the a's vary with m. If they
are roots of the cubic, then they are constants. Note that in
your paper you use the a's in two different ways. Which way
are they used in THIS case? I believe Dale Hall is the one
who showed that if you view the a's as constants, they still
don't have the properties you want, but I could easily be
mistaken.>> Note that EACH ONE of these is divisible by 5:>
> That was Nora Baron's comment for her picked values for
the a's.>Ok, you *pick* values from the a's as if
they're not de?ed by a>cubic, which they are, and your
picked values are supposed to prove>something?>>They
AREN'T de?ed by a cubic. They are constrained by a cubic. If
>>you want them to be de?ed by a cubic, you'll have to say
more about >>how they are supposed to be constructed.>
That statement is false as I've shown by giving the
cubic.What you've shown is that you are inconsistent in the
use of your notation. Perhaps if you used a_1(m) when dealing
with functions and a_1 when dealing with constants this
confusion wouldn't arise.>> NONE ARE COPRIME TO
5.>So are your claiming that at m=1, a_1 = -5, a_2 =
-5, and a_3 = 2140?> That was my question to Nora Baron.>
>Remember you simply tossed those numbers out there, but
there *is* a>way to calculate the a's so your assertion can
be tested, as they are>roots to a cubic.>>Not that you've
provided. If you have a way in mind, please provide it >>to
us and include it in your paper. You've only claimed this
things >>exist and have certain properties. If you want to
change them, so be >>it. But that may cause more problems for
you.> I simplify and people like you come in and try to
confuse, and sadly,> I think that the readership of sci.math
wants to be confused by you.I didn't use the same symbols to
represent two different types of quantities.>Now if you can
do that Nora Baron then clearly I made some kind of>mistake
and there's no more room for me to argue.>>room for you to
argue.> Well now that I've given the cubic, I challenge
you to admit *you*> made a mistake.Challenge declined. You
want the a's to both vary with m and be constants. You'll
have to make up your mind which it is.>So why don't you go
back to the cubic which de?es the a's, stick in>all the
values and see if the a's come out as you have above, and
then>it's over.>>For each value of m, she can de?e
a_i(m) to be whatever three values >>makes both sides equal.
You've given nothing more speci? about the >>values to
require more.> That is false as I've repeatedly given the
expression de?ing the> a's.If the a's are constants. If they
are, then your expression above fails to be true when you
vary m.>> Now, what's your explanation?>> Nora
B.>Well I've explained above, and again, of course,
you can't just *pick*>values for the a's for a particular
m, as their values are set>rigorously. What I'm curious
about are the people who believe Nora>Baron had a valid
point.>>Where have you established how to rigorously
compute them. What are they?> For m=1, f=5, the a's are
the roots of the cubic> a^3 + 72a^2 - 13825 = 0.There's
nothing in the nature of the problem that (to me) makes such
a de?ition either obvious or necessary.>Speak up, and
don't be shy as I suspect you may believe she still has>a
valid point and I need to understand why she's so effective
in>convincing people.>>Because you have not stated all of
the conditions you want to exist. >>Once you do that, you will
lose the results you are trying to achieve.> I've stated
those conditions repeatedly. Are you now claiming that> you
have never seen the factorization> (v^3+1)x^3 -3vxy^2 + y^3
= (a_1 x + y)(a_2 x + y)(a_3 x + y)?Is THIS where it's
de?ed??? Ok... and do you want the choice of your a_i(m) to
hold for all x and y? Your simpli?ation does not make that
obvious. Worse, there is no good way of assigning the values
generated for each choice of m to each a_i(m).Note: when you
CHOSE values for x and y, you changed the nature of the
problem being looked at. The observations made by myself and
(I believe) Nora are based on the special case of x=2, y=5.
The irony is, you've been defending a leap from speci? to
general in your arguments, but won't let us look at the
speci? while ignoring the general.Have you considered
keeping everything in one thread so it's easier to follow
things?>Come on, speak up, do you think I answered her
objections? Do you>believe they were valid in the ?st
place?>>I think you didn't understand her objections.>
I think you're lying.That is your perogative.> James Harris--
Will Twentyman> Um, Nora Baron, now you willy-nilly give
the a's values. Doesn't that> seem to be just a tad bit
strange to you?For strangeness, nothing equals JSH's attempts
at mathematics.> I think you didn't understand her
objections.> I think you're lying.> James HarrisWe
doubt that you are thinking and we know you're lying!>
> <deleted> Let's go back to your
original polynomial,>> P(m) = 25(5000m^3 - 600 m^2
- 126m + 11),>>and assume it is factored *as you
propose* in the form>>[*] P(m) = (2 a_1 + 5)(2 a_2
+ 5)(2 a_3 + 5).>> Now instead of m = 0, let's try
m = 1:>Ok.> P(m) = 25 *
(5000 - 600 - 126 + 11) = 25 * 4285.>> This can be
factored in the form [*] by letting>> a_1 = -5, a_2
= -5, and a_3 = 2140.>Um, Nora Baron, now you
willy-nilly give the a's values. Doesn't that>seem to be
just a tad bit strange to you?>>You haven't provided a
way to de?e what the a_i(m) are, so of course >>she can
simply select values that work.> That's false as
the a's are given by the factorization> (v^3+1)x^3
-3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)>
where v=-1+mf^2, and y=uf.> For my example I used x=2,
f=5, u=1. And now letting m=1, to test> your claim I have>
> 13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x
+ y)> so the a's are the roots of the cubic> a^3
+ 72a^2 - 13825 = 0.> And checking with a=-5, gives
-12150, and not 0.> So Nora Baron's claim and yours is
refuted.> I thought you said the a's vary with m. If they
are roots of the cubic, > then they are constants. Note that
in your paper you use the a's in two > different ways. Which
way are they used in THIS case? I believe Dale > Hall is the
one who showed that if you view the a's as constants, they >
still don't have the properties you want, but I could easily
be mistaken.Are you being deliberately obtuse? Nora Baron's
assertion is withm=1. >> Note that EACH ONE of these is
divisible by 5:> That was Nora Baron's comment for
her picked values for the a's.>Ok, you *pick*
values from the a's as if they're not de?ed by a>cubic,
which they are, and your picked values are supposed to prove>
>something?>>They AREN'T de?ed by a cubic. They are
constrained by a cubic. If >>you want them to be de?ed by
a cubic, you'll have to say more about >>how they are
supposed to be constructed.> That statement is
false as I've shown by giving the cubic.> What you've shown
is that you are inconsistent in the use of your > notation.
Perhaps if you used a_1(m) when dealing with functions and >
a_1 when dealing with constants this confusion wouldn't
arise.The following should not be new to you: (v^3+1)x^3
-3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where
v=-1+mf^2, and y=uf.And what I did to *simplify* was to let
x=2, f=5, u=1. Then NoraBaron made an assertion about the a's
for m=1, and I gave the de?ingcubic.>> NONE ARE COPRIME
TO 5.>So are your claiming that at m=1, a_1 =
-5, a_2 = -5, and a_3 = 2140?> That was my question
to Nora Baron.>Remember you simply tossed those
numbers out there, but there *is* a>way to calculate the
a's so your assertion can be tested, as they are>roots to
a cubic.>>Not that you've provided. If you have a way in
mind, please provide it >>to us and include it in your
paper. You've only claimed this things >>exist and have
certain properties. If you want to change them, so be >>it.
But that may cause more problems for you.> I
simplify and people like you come in and try to confuse, and
sadly,> I think that the readership of sci.math wants to be
confused by you.> I didn't use the same symbols to represent
two different types of > quantities.If I have y=mx+b, and then
a little later I have y=x+1, and then latertalk of with x=1,
y=2, is that using the same symbols to representdifferent
types of quantities?If you're having trouble keeping up, then
you can work through it*yourself* putting in whatever symbols
help you.However, I've yet to be convinced that others would
be more helped bymy adding in extra than hurt by the
additional level of complexity.For instance, if I have P(m)=
(v^3+1)x^3 -3vxy^2 + y^3 = (a_1(m) x + y)(a_2(m) x +
y)(a_3(m)x + y)does that help more than it hurts?And it's
even busier with the FLT argument where there are even
MOREsymbols.Are there people who will now get even more
confused?What kind of arguments might I face then from people
who want to dragme into arguments about functions? >Now if
you can do that Nora Baron then clearly I made some kind of>
>mistake and there's no more room for me to argue.>
>>room for you to argue.> Well now that I've given
the cubic, I challenge you to admit *you*> made a mistake.>
> Challenge declined. You want the a's to both vary with m and
be > constants. You'll have to make up your mind which it
is.You lack intellectual honesty.>So why don't you go
back to the cubic which de?es the a's, stick in>all the
values and see if the a's come out as you have above, and
then>it's over.>>For each value of m, she can de?e
a_i(m) to be whatever three values >>makes both sides equal.
You've given nothing more speci? about the >>values to
require more.> That is false as I've repeatedly
given the expression de?ing the> a's.> If the a's are
constants. If they are, then your expression above fails > to
be true when you vary m.You can't be that lost. I simply ?d
it hard to believe that youdon't know more as I think you do
but are attempting to confuseothers.>> Now, what's your
explanation?>> Nora B.>Well I've
explained above, and again, of course, you can't just *pick*>
>values for the a's for a particular m, as their values are
set>rigorously. What I'm curious about are the people who
believe Nora>Baron had a valid point.>>Where have
you established how to rigorously compute them. What are
they?> For m=1, f=5, the a's are the roots of the
cubic> a^3 + 72a^2 - 13825 = 0.> There's nothing in
the nature of the problem that (to me) makes such a >
de?ition either obvious or necessary.You mean like
expressions that actually de?e the a's?You seem to think you
don't need to bother with the math!>Speak up, and don't be
shy as I suspect you may believe she still has>a valid
point and I need to understand why she's so effective in>
>convincing people.>>Because you have not stated all
of the conditions you want to exist. >>Once you do that,
you will lose the results you are trying to achieve.>
> I've stated those conditions repeatedly. Are you now
claiming that> you have never seen the factorization>
(v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)?>
> Is THIS where it's de?ed??? Ok... and do you want the
choice of your > a_i(m) to hold for all x and y? Your
simpli?ation does not make > that obvious. Worse, there is
no good way of assigning the values > generated for each
choice of m to each a_i(m).That is b.s. as you can easily
enough solve for the a's, even with theexpression in that
form by creating a cubic as I have done, and thensolving
it.There'll be a LOT of symbols, but it can be done.> Note:
when you CHOSE values for x and y, you changed the nature of
the > problem being looked at. The observations made by
myself and (I > believe) Nora are based on the special case
of x=2, y=5. The irony is, > you've been defending a leap
from speci? to general in your arguments, > but won't let us
look at the speci? while ignoring the general.Actually my
work has always covered a family of values, which is why Ican
stick in values as I did in this thread. Previously, with FLT,
ofcourse, I couldn't stick in values for x, y and z.> Have you
considered keeping everything in one thread so it's easier to
> follow things?The threads become HUGE with hundreds of
posts, and then it's harderto follow things.>Come on,
speak up, do you think I answered her objections? Do you>
>believe they were valid in the ?st place?>>I think
you didn't understand her objections.> I think
you're lying.> That is your perogative.You've either been
making a horrendous number of mistakes in replyingto me while
continually ignoring information that has been given, oryou're
lying.James Harris> <deleted>>
> a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)>
> For any integer m, a_1 a_2 a_3 is a
multiple of 25. i.e. not coprime to 5.>
> Perhaps you mean that none of the a's are coprime to 5?>
> Which is trivially obvious and uninteresting after
all.> Sorry to have bothered you.>
> It's actually fascinating as it's a bizarre result, and I
think you> realize that I'm correct as you *again*
deleted out the following> wonderful simpli?ation
which shows clearly that I'm right.> Yuck.
Unfortunately in my haste I neglected to refute the claim,
as> it is in fact NOT true that none of the a's are
coprime to 5, as in> fact the fascinating as it's a
bizarre result is that they ALL are> coprime to 5 in
the ring of algebraic integers.> The problem
is with the ring of algebraic integers as I've explained.>
> Consider the following which you deleted yet
again from my previous> post:>
Hmmm...how about this explanation?>
Remember that the polynomial is> P(m) =
25(5000m^3 - 600 m^2 - 126m + 11),> which gives
the reader a chance to see P(m) with only the m left as>
> a symbol> Yes! I like this example. See
below.> Fascinating.> and you see that
factor 25. Now imagine that I have the polynomial>
Q(m), where> P(m) = 25 Q(m)>
> so you have as a factorization> Q(m) =
(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25> and
the question is, how does that factor of 25 divide out? >
> I want readers to consider how many posters have
apparently attempted> to make them believe that the
factor of f^2, here 25, was in some> sense welded into
the expression, when in fact it's a factor of P(m)>
such that I can write> P(m) = 25 Q(m)>
> as I've done here, or P(m) = f^2 Q(m), in general.>
> Well, checking at Q(0), gives 11, so how can those
factors of 25> divide through Q(m) in such a way as to
give 11, at m=0?> There's only one way,
which gives you> Q(m) = (2 a_1/5 + 1)(2
a_2/5 + 1)(2 a_3 + 5).> If that bothers
you, remember that at m=0, two of the a's equal 0.>
> Proving what, exactly? Clearly a_1, a_2 and a_3 are
functions> of m. When m = 0, two of them are 0. However
when m is not zero,> none of them are zero. So how do
their values when m = 0 > determine what they will be
when m <> 0 ?> Well you have P(m) = 25 Q(m) and that
factor 25, which has been the> focus of all the arguing.
Now considering Q(m) it turns out that> looking at the
factorization> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 +
5)/25> the a's are in the way, and it's not clear how
you divide out.> Now you *know* that the 25 divides
through some way, and it's clear> that whatever way that is
would be true for any m, just like with> 2(x^2 + 2x + 1)
= (x+1)(2x+2)> when you divide off 2, you do it for all
x.> For readers, people like Nora Baron have basically
been arguing that> you can have something like>
Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 +
h_1 h_2)> where h_1 h_2 h_3 = 5, and each is not a
unit.> But notice then that setting m=0, gives (h_1 h_2
h_3)^2 = 25, when in> fact Q(0)=11.> Given that
rigorous fact they claim that m=0 is a some kind of special>
> case and try to cast doubt on the result, which is an
attack on> algebra that I call voodoo math.>
Surprisingly they've been quite successful from what I've
gathered in> convincing people, which makes you wonder
about people's understanding> of algebra.> Because
from algebra, it turns out that you can just set m=0, to>
?ure out how that 25 divides out, as I've done.> No
voodoo at all. What you would like to have here is that > the
same a_1, a_2, and a_3 work when m = 0 as when m <> 0. > But
you know, and everyone else knows, that that is totally >
impossible. Why? Because when m = 0, two of the a's must be >
zero. When m <> 0, NONE of the a's can be zero. That everyone
else knows is an appeal to the gallery which is alogical
fallacy.The a's are de?ed by (v^3+1)x^3 -3vxy^2 + y^3 = (a_1
x + y)(a_2 x + y)(a_3 x + y),where v=-1+mf^2, and I guessed
that with so many symbols people wereconfused, so I stuck in
some values to lessen the symbol load.> It's real simple.>
The a's are functions of m. And I don't mean CONSTANT
functions.> Got that? > Different m's give different
a's.> You believe that because a_1 is divisible by 5 when m
= 0,> it must be divisible by 5 for other values of m also. It
is> some kind of hunch that you have, based on intuition. It
seems> like a nice parallel construction. You think it's
obvious. > It isn't. In fact it is false. Your intuition this
time is > wrong. Your hunch has been disproved.<deleted>Here
you claim that I'm acting on a hunch about the a's.But,
when I used x=2, f=5, y=5, I have P(m) = 25(5000m^3 - 600 m^2
- 126m + 11), and noticing that 25 I used P(m) = 25 Q(m), so I
have Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25where the
question is, how does that 25 divide through?But since I know
that Q(0)=11, I know that it must be Q(m) = (2 a_1/5 + 1)(2
a_2/5 + 1)(2 a_3 + 5).At Q(0) that is Q(0) = (2 (0) + 1)(2
(0) + 1)(2 (3) + 5) = 11.So in fact it's algebra and not a
hunch.It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here
you can see howthe 2 divides out, but if you couldn't see,
and couldn't guess, likeif you had P(x)= 2(x^2+2x+1) = (a_1 +
1)(a_2 + 2), with a_1 a_2 = 2x^2, a_1 + a_2 = 4x,you could
have P(x) = 2Q(x), and set x=0, with Q(0) to get 1,
showingthat Q(x) = (a_1 + 1)(a_2/2 + 1), is correct.Of
course, you can also just see that is the only way that
works.But regardless there is only ONE way that will work,
and it doesn'tchange as x changes, and it doesn't change
above as m changes.Assaulting algebra is a rather disturbing
step taken by posterscontinuing to argue with me. What is
also troubling is how uncaringthe newsgroup seems to be about
their denial of basic algebra. But itis telling.James
Harris[snip]> Assaulting algebra is a rather disturbing
step taken by posters> continuing to argue with me. What is
also troubling is how uncaring> the newsgroup seems to be
about their denial of basic algebra. But it> is telling.No
one is assaulting algebra. They are denying that you have
constructed a valid proof. And they are correct. Youare
wrong. What is telling here is your monumental
incompetence.--There are two things you must never attempt to
prove: the unprovable -- and the obvious.--Democracy: The
triumph of popularity over
principle.--http://www.crbond.com>>Let's go back to
your original polynomial,>> P(m) = 25(5000m^3 - 600
m^2 - 126m + 11),>>and assume it is factored *as you
propose* in the form>>[*] P(m) = (2 a_1 + 5)(2 a_2 +
5)(2 a_3 + 5).>>Now instead of m = 0, let's try m =
1:>> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 *
4285.>>This can be factored in the form [*] by
letting>> a_1 = -5, a_2 = -5, and a_3 =
2140.>Um, Nora Baron, now you willy-nilly give
the a's values. Doesn't that>seem to be just a tad bit
strange to you?>>You haven't provided a way to de?e
what the a_i(m) are, so of course >>she can simply select
values that work.>That's false as the a's are given
by the factorization>> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x
+ y)(a_2 x + y)(a_3 x + y)>>where v=-1+mf^2, and
y=uf.>>For my example I used x=2, f=5, u=1. And now
letting m=1, to test>your claim I have>> 13825x^3 -
72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)>>so the
a's are the roots of the cubic>> a^3 + 72a^2 - 13825 =
0.>>And checking with a=-5, gives -12150, and not
0.>>So Nora Baron's claim and yours is refuted.>>I
thought you said the a's vary with m. If they are roots of
the cubic, >>then they are constants. Note that in your paper
you use the a's in two >>different ways. Which way are they
used in THIS case? I believe Dale >>Hall is the one who
showed that if you view the a's as constants, they >>still
don't have the properties you want, but I could easily be
mistaken.> Are you being deliberately obtuse? Nora
Baron's assertion is with> m=1.Correct. She was DEFINING the
values of the a_i(1). The discussion had dealt with what the
a_i(0) were, but she simply showed that when m changes from 0
to 1, it is possible for the a_i(m) to change from having the
properties you wanted to ones that you did NOT want. Nothing
obtuse. Now, do you understand what a function is? Do you
perhaps see the importance of de?ing the characteristics of
these functions at all values of m if you want them to have
certain properties?>>Note that EACH ONE of these is
divisible by 5:>That was Nora Baron's comment for her
picked values for the a's.>Ok, you *pick* values
from the a's as if they're not de?ed by a>cubic, which
they are, and your picked values are supposed to
prove>something?>>They AREN'T de?ed by a cubic.
They are constrained by a cubic. If >>you want them to be
de?ed by a cubic, you'll have to say more about >>how they
are supposed to be constructed.>That statement is
false as I've shown by giving the cubic.>>What you've shown
is that you are inconsistent in the use of your >>notation.
Perhaps if you used a_1(m) when dealing with functions and
>>a_1 when dealing with constants this confusion wouldn't
arise.> The following should not be new to you:>
(v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)>
> where v=-1+mf^2, and y=uf.> And what I did to *simplify*
was to let x=2, f=5, u=1. Then Nora> Baron made an assertion
about the a's for m=1, and I gave the de?ing> cubic.When you
set x=2,f=5,u=1, then you change the nature of the problem. If
you would prefer to think of the a_i as varying over m,x,f,u,
we can do that but then Nora is still able to choose the
values she did. If you don't want people de?ing the a_i's in
inconvenient ways, you will have to specify what is legal.
Don't be surprised if the speci?ations you provide limit you
to cases that don't give the broad result you want to
achieve.>>NONE ARE COPRIME TO 5.>So are
your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 =
2140?>That was my question to Nora
Baron.>>Remember you simply tossed those numbers
out there, but there *is* a>way to calculate the a's so
your assertion can be tested, as they are>roots to a
cubic.>>Not that you've provided. If you have a way in
mind, please provide it >>to us and include it in your
paper. You've only claimed this things >>exist and have
certain properties. If you want to change them, so be >>it.
But that may cause more problems for you.>I simplify
and people like you come in and try to confuse, and
sadly,>I think that the readership of sci.math wants to be
confused by you.>>I didn't use the same symbols to
represent two different types of >>quantities.> If I have
y=mx+b, and then a little later I have y=x+1, and then later>
talk of with x=1, y=2, is that using the same symbols to
represent> different types of quantities?I have no problem
with this.> If you're having trouble keeping up, then you can
work through it> *yourself* putting in whatever symbols help
you.I do. That's why you keeping seeing my posts with
a_1(m).> However, I've yet to be convinced that others would
be more helped by> my adding in extra than hurt by the
additional level of complexity.It's not for the others, it's
to clearly de?e what are the independent variables for the
functions a_i. It's so that you know what you're talking
about and can clearly communicate it to us.> For instance, if
I have> P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1(m) x +
y)(a_2(m) x + y)(a_3(m)> x + y)> does that help more than
it hurts?It helps a great deal. How does it hurt?> And it's
even busier with the FLT argument where there are even
MORE> symbols.> Are there people who will now get even more
confused?Unlikely. It may be busier but it is now precise. You
have clearly indicated that the a_i depend only on m, and that
when you change x or y, the a_i will not change. Of course, y
and v are both de?ed in terms of f, so there's probably a
little more going on. Your introductions of v and y is hiding
a relationship.> What kind of arguments might I face then from
people who want to drag> me into arguments about
functions?Ones that you can either clearly defend against or
concede to as everyone will be discussing the same things.
Right now there are no de?itions so people are interpreting
the same symbols differently.>Now if you can do that Nora
Baron then clearly I made some kind of>mistake and there's
no more room for me to argue.>>room for you to
argue.>Well now that I've given the cubic, I
challenge you to admit *you*>made a mistake.>>Challenge
declined. You want the a's to both vary with m and be
>>constants. You'll have to make up your mind which it is.>
You lack intellectual honesty.You are the one that is being
dragged into de?ing your symbols.>So why don't you go
back to the cubic which de?es the a's, stick in>all the
values and see if the a's come out as you have above, and
then>it's over.>>For each value of m, she can de?e
a_i(m) to be whatever three values >>makes both sides equal.
You've given nothing more speci? about the >>values to
require more.>That is false as I've repeatedly given
the expression de?ing the>a's.>>If the a's are
constants. If they are, then your expression above fails >>to
be true when you vary m.> You can't be that lost. I simply
?d it hard to believe that you> don't know more as I think
you do but are attempting to confuse> others.If the a's vary
with m, then there must be a way to de?e them for each
choice of m. You have NOT done that. You have given what
looks like a de?ition but would lead to the a's being
constants. Perhaps if you go back and POST your paper HERE we
can see if you've changed things enough to address the issues
that you feel are valid. I'm working with a copy that is
approximately a month old.>>Now, what's your
explanation?>>Nora B.>Well I've
explained above, and again, of course, you can't just
*pick*>values for the a's for a particular m, as their
values are set>rigorously. What I'm curious about are the
people who believe Nora>Baron had a valid
point.>>Where have you established how to rigorously
compute them. What are they?>For m=1, f=5, the a's
are the roots of the cubic>> a^3 + 72a^2 - 13825 =
0.>>There's nothing in the nature of the problem that (to
me) makes such a >>de?ition either obvious or necessary.>
> You mean like expressions that actually de?e the a's?>
You seem to think you don't need to bother with the math!Ok,
I just noticed the below... That connection would have been
helpful. Why do you berate me for not connecting something
below with that above?>Speak up, and don't be shy as I
suspect you may believe she still has>a valid point and I
need to understand why she's so effective in>convincing
people.>>Because you have not stated all of the
conditions you want to exist. >>Once you do that, you will
lose the results you are trying to achieve.>I've
stated those conditions repeatedly. Are you now claiming
that>you have never seen the factorization>> (v^3+1)x^3
-3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)?>>Is THIS
where it's de?ed??? Ok... and do you want the choice of your
>>a_i(m) to hold for all x and y? Your simpli?ation does
not make >>that obvious. Worse, there is no good way of
assigning the values >>generated for each choice of m to each
a_i(m).> That is b.s. as you can easily enough solve for the
a's, even with the> expression in that form by creating a
cubic as I have done, and then> solving it.> There'll be a
LOT of symbols, but it can be done.Here's the funny thing:
Nora did it. It worked. You just don't like it. If she failed
to do it correctly, point out the *speci?* ?Note: when
you CHOSE values for x and y, you changed the nature of the
>>problem being looked at. The observations made by myself
and (I >>believe) Nora are based on the special case of x=2,
y=5. The irony is, >>you've been defending a leap from
speci? to general in your arguments, >>but won't let us look
at the speci? while ignoring the general.> Actually my work
has always covered a family of values, which is why I> can
stick in values as I did in this thread. Previously, with
FLT, of> course, I couldn't stick in values for x, y and
z.What family of values? It might help if you informed people
about that in advance.>>Have you considered keeping everything
in one thread so it's easier to >>follow things?> The
threads become HUGE with hundreds of posts, and then it's
harder> to follow things.It's easier than trying to keep
track of 20 short threads. I can mark a thread for future
reference. It's harder to mark a lot of threads, and harder
to follow the logic between the threads. A concession in one
thread is not apparent in another thread.> James Harris--
Will Twentyman>>[deleted]> It's like with 2(x^2+2x+1)
= (x+1)(2x+2), where here you can see how> the 2 divides out,
but if you couldn't see, and couldn't guess, like> if you
had> P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2), > with a_1
a_2 = 2x^2, a_1 + a_2 = 4x,Correction 1: 2 a_1(x) + a_2(x) =
4x.Correction 2: only if a_1(x) = x, a_2(x) = 2x.If you want
to allow anything else, your assertion is incorrect. And you
consistently want to allow the a_i(x) to be
non-polynomials.-- Will Twentyman>
<deleted>> a_1 a_2 a_3 =
25 (625 m^3 - 75 m^2 + 3m)> For any
integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime
to 5.> Perhaps you mean that none
of the a's are coprime to 5?> Which is trivially
obvious and uninteresting after all.> Sorry to
have bothered you.> It's actually
fascinating as it's a bizarre result, and I think you>
> realize that I'm correct as you *again* deleted out the
following> wonderful simpli?ation which shows
clearly that I'm right.> Yuck.
Unfortunately in my haste I neglected to refute the claim,
as> it is in fact NOT true that none of the a's are
coprime to 5, as in> fact the fascinating as it's a
bizarre result is that they ALL are> coprime to 5 in
the ring of algebraic integers.> The
problem is with the ring of algebraic integers as I've
explained.> Consider the following
which you deleted yet again from my previous>
post:> Hmmm...how about this
explanation?> Remember that the
polynomial is> P(m) = 25(5000m^3 - 600
m^2 - 126m + 11),> which gives the reader a
chance to see P(m) with only the m left as> a
symbol> Yes! I like this example. See
below.> Fascinating.> and you see
that factor 25. Now imagine that I have the polynomial>
> Q(m), where> P(m) = 25 Q(m)>
> so you have as a factorization>
> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25>
> and the question is, how does that factor of 25
divide out? > I want readers to consider
how many posters have apparently attempted> to make
them believe that the factor of f^2, here 25, was in some>
> sense welded into the expression, when in fact it's a
factor of P(m)> such that I can write>
> P(m) = 25 Q(m)> as I've done here, or
P(m) = f^2 Q(m), in general.> Well,