MM-100
==
So when I say factors of 5, Im NOT saying
multiples of 5. For> instance, 2 and 3 are factors of 6. So
when I say factors of 6 it> doesnt mean multiples of 6. And
I emphasize that there is a> mathematical term multiples of
which applies.[...]In the integers, 5 has the property that if
m*n = 5, where m and nare integers, then either m is a unit,
or n is a unit.In the algebraic integers, 5 can be further
broken downas 5= sqrt(5)*sqrt(5), oras 5 = (2+i)*(2-i) .None
of sqrt(5), 2+i or 2-i are units, right?Id like to know your
take on this:Can 5 be broken down into a productof non-unit
factors, each of whichcannot be further broken down?If so,
how? If not, why? Is therea paradox here? Etc.This is an
open-ended question...David Bernier
Suppose that r_1 = 1
+ 2i and r_2 = 1 - 2i. Now sqrt(5) is a factor> of 5. Does it
go with r_1 or r_2? <deleted> See what I mean sci.skeptic?
The issue here is that Im actually> using standard
mathematical usage, while several posters apparently> keep
reading factors of 5 as multiples of 5. The problem has
to do with their misuse of mathematical terminology.Do you
have a problem with that? There is nothing non-standard
here.
Im putting this bit rst to highlight it:>>Note:
Keith Ramsay is pointing out that he has edited my post
in>>terms of placement of text, as he has moved a
section.[...]> | Consider that in the ring of algebraic
integers, 5 has algebraic> | integer factors, and given
algebraic integers r_1 and r_2, where their> | product is 5,
why are you acting as if its so difcult to comprehend> |
that there must be some distribution of factors of 5?> Im
not acting as if Im having any kind of difculty. Im
saying> that *youre* having difculty in realizing that
there must be some> distribution of factors of 5 is a
meaninglessly vague phrase. Youre> saying this as if to
say, Shouldnt this be true?, and not as if> you actually
had proven it was true.>>Given that r_1 r_2 = 5, it makes
sense that factors of 5 distribute in>>some way between r_1
and r_2.>>> It makes sense, but (assuming I know what you
mean - yes you>> _are_ using the language in nonstandard
ways) its simply>> _not_ _true_.>>> Suppose that r_1 = 1 +
2i and r_2 = 1 - 2i. Now sqrt(5) is a factor>> of 5. Does it
go with r_1 or r_2?<deleted>See what I mean sci.skeptic?
The issue here is that Im actually>using standard
mathematical usage, while several posters apparently>keep
reading factors of 5 as multiples of 5.Huh? You said that
factors of 5 distribute in some way between r_1and r_2. Take
r_1 and r_2 as above and tell me how sqrt(5)distributes
among them.>The problem has to do with their misuse of
mathematical terminology.Thats certainly one of the problems
here. But its _your_ misuseof standard terminology thats the
problem. You convenientlyignored the rest of my post where I
explained this.>For instance 2 is a factor of 6, but so is 3,
and in fact, so is 6.When you say, factor of, it means
something that is a factor of the>given number. And similarly
factors of would mean factors of the>given number.Nobodys
disputed that. It does not follow that for instance9 has a
factor of 12 - the way the language _is_ used,9 has a
factor of 12 does not mean that there is a factorof 12 which
divides 9, which is what you seem to mean,it means that 12 is
a factor of 9, which is clearly false.>But you have these
posters, like David Ullrich who is a math professor>at
Oklahoma State University, who are lost with rather
basic>mathematical language to the point that they *keep*
arguing *after*>Ive corrected them. You see their *belief*
system apparently is that>as mathematicians they cant be the
ones with the error, so amazingly,>they simply keep almost
mindlessly repeating it.You may guess that they say factors
of when they mean multiples>of, but Im using the proper
terminology, in the correct way.So when I say factors of 5,
Im NOT saying multiples of 5. For>instance, 2 and 3 are
factors of 6. So when I say factors of 6 it>doesnt mean
multiples of 6. And I emphasize that there is a>mathematical
term multiples of which applies.And in fact, using factors
of when you mean multiples of while>common, is technically
incorrect.I have never seen anyone, here or elsewhere, use
factors ofto mean multiples of. Except for that you have
a good point.>But people like Ramsay and Ullrich are unlikely
to reply to *correct*>their mistakes because the society of
sci.math lets people it>considers part of the society get
away with the dumbest mistakes.Thats how mathematicians
operate as demonstrated before your eyes.And thats how they
can have errors in their discipline for years, and>years
because they seek to by denition have a society that
is>perfect, when in fact, mathematicians are just people, and
people make>mistakes.>James
Harris
Nope.
There is no claim that y is a factor as its *provably* a>
factor.> *sigh* Could you give me your denition of a
factor, then? I suspect> its different from what
mathematicians usually mean, because the way to> prove y is a
factor of x is to nd some z such that y*z=x, with suitable>
properties for all three of them (i.e. all in the algebraic
integers,> say). If you havent done this, how can you prove
its a factor?Ive switched from saying provably a factor,
as that is problematicto saying that its implied to be a
factor, and an analogy is in thering of evens where you have
2 and 6, but 6 does not have 2 as afactor, though 6=2(3), in
the ring of integers.Here the problem is that members that
should be in the ring ofalgebraic integers are left out. So
imagine one such member m, andthe algebraic integers c
and a, such that c = amand youll nd that in the ring of
algebraic integers, a is coprimeto c, which is the
problem, as by the denition of factor, m asits not an
algebraic integer is NOT a factor of c in the ring
ofalgebraic integers, just like 2 is NOT a factor of 6 in the
ring ofevens. > Rather than deal with the term factor as
people get confused, in my> paper Advanced Polynomial
Factorization, I have three numbers a_1,> a_2, and a_3, and I
prove that a_3 is coprime to a factor I call f.> Argh, you
profess to want to get away from the term factor, and then>
you go and use it right again. Please dene your terms
rigorously.Im using the standard denition of factor. Here
probably I couldhave said that I prove that a_3 is coprime to
an algebraic integer Icall f.In the past, Ive done the
equivalent of saying that 6 in the ring ofevens has 2 as a
factor, which is incorrect, as the implication isthat Im
talking about the ring of evens, when to be correct I have
tohave switched rings.While 6 does have 2 as a factor in the
ring of integers, it does NOTin the ring of evens.What Im
doing now is correcting that mistaken usage, and Im
notsurprised that its taken some time and that many of you
may havebecome confused.The situation, however, is analogous
to someone talking about 6 in thering of evens, saying that 6
has 2 as a factor, when theyve had toswitch rings.The short
answer is that in the ring of evens saying 6 has 2 as afactor
is wrong. > Consider c=ab, where c is an algebraic integer,
and a is an> algebraic integer, but b is not.> Now if
you include fractions or move to a eld like algebraic
numbers> thats ok. But Im talking about b thats not in
any way a fraction> or fractional. Its like with the ring of
evens, and given 6 = 2(3), you have that 3> is outside the
ring, as in the ring of evens, 2 is NOT a factor of 6.> The
situation is analogous.> Do you understand?> No, I dont.
You seem to be operating under the assumption that all> are
you claiming that the set of evens is incomplete, because
2> should be a factor of 6, but isnt?Nope.Ok, so you had
reason to be confused by my switching around beforewhen I did
the equivalent of talk about 2 as a factor of 6 in the ringof
evens, when its not.The problem with algebraic integers is
that ring operations lead to anumber which *should* be an
algebraic integer, but its not.Heres basically my approach.
I get an expression like g = r + fcwhere g, r, f, and c are
algebraic integers.Now I nd out that g is not coprime to f,
and I can separate off somefactor of f, which gives me h = s
+ cwhere h appears to be a factor of g, and s appears to be a
factor ofr.you are very *strict* in your mathematics, and I
didnt help with whatI explained above in terms of that f
word.Still for those of you who prize mathematical knowledge,
its afascinating little problem, and can even be a challenge
to understand,which should only intrigue you more.James
Harris
something that is a factor of the> given number. And
similarly factors of would mean factors of the> given
number.Nice of you to clear this up with such an unambiguous
statement. We sure dont need circular denitions.--What
goes around, comes around. -- Unknown--Democracy: The triumph
of popularity over principle.--http://www.crbond.com
In
the past, Ive done the equivalent of saying that 6 in the
ring of> evens has 2 as a factor, which is incorrect, as the
implication is> that Im talking about the ring of evens,
when to be correct I have to> have switched rings.> While 6
does have 2 as a factor in the ring of integers, it does NOT>
in the ring of evens.> What Im doing now is correcting
that mistaken usage, and Im not> surprised that its taken
some time and that many of you may have> become confused.Ah,
good. Ill look forward to a cleaned-up version of the proof,
then.> Heres basically my approach. I get an expression like g = r + fc> \
where g, r, f, and c are algebraic integers. Now I nd out that g is not \
coprime to f, and I can
separate off some> factor of f, which gives me> h = s + c
where h appears to be a factor of g, and s appears to be a
factor of> r. factor of f that is f, was separated off.Im a
bit confused by this step... as I understand it, if g is
notcoprime to f, that means that they share *some* factor,
call it e,not that g is a multiple of f. In other words you
should only be ableto say: g/e = r/e + (f/e)c h = s +
(f/e)crather than dividing by the whole of f as you seem to
have done above.> Now to the appears part, as in checking
r, I nd that r is NOT an> algebraic integer!This seems
slightly weird, since r = g - fc, and if g, f, and c are
allalgebraic integers, r must be too. One of the other three
must alsoturn out not be an algebraic integer.
Im
putting this bit rst to highlight it:>>Note: Keith Ramsay is
pointing out that he has edited my post in>>terms of placement
of text, as he has moved a section.[...]> | Consider that in
the ring of algebraic integers, 5 has algebraic> | integer
factors, and given algebraic integers r_1 and r_2, where
their> | product is 5, why are you acting as if its so
difcult to comprehend> | that there must be some
distribution of factors of 5?> Im not acting as if Im
having any kind of difculty. Im saying> that *youre*
having difculty in realizing that there must be some>
distribution of factors of 5 is a meaninglessly vague
phrase. Youre> saying this as if to say, Shouldnt this be
true?, and not as if> you actually had proven it was
true.>>Given that r_1 r_2 = 5, it makes sense that factors of
5 distribute in>>some way between r_1 and r_2.>>> It makes
sense, but (assuming I know what you mean - yes you>> _are_
using the language in nonstandard ways) its simply>> _not_
_true_.>>> Suppose that r_1 = 1 + 2i and r_2 = 1 - 2i. Now
sqrt(5) is a factor>> of 5. Does it go with r_1 or
r_2?<deleted>See what I mean sci.skeptic? The issue here is
that Im actually>using standard mathematical usage, while
several posters apparently>keep reading factors of 5 as
multiples of 5.The problem has to do with their misuse of
mathematical terminology.It is about time you realize that
you are the one who is misusingterminology.>For instance 2 is
a factor of 6, but so is 3, and in fact, so is 6.When you say,
factor of, it means something that is a factor of the>given
number. And similarly factors of would mean factors of
the>given number.The problem is when you say x has a factor
of y. This is, instandard mathematical terminology, the same
as saying y is a factorof x; which is the same, in standard
mathematical terminology, as yis a divisor of x and as x
is a multiple of y.You do NOT mean x has a factor of y;
what you mean is x has[non-unit] factors in common with
y.
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 mans 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
gures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
indenite 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
*provably* a>> factor.>>> *sigh* Could you give me your
denition of a factor, then? I suspect>> its different from
what mathematicians usually mean, because the way to>> prove
y is a factor of x is to nd some z such that y*z=x, with
suitable>> properties for all three of them (i.e. all in the
algebraic integers,>> say). If you havent done this, how can
you prove its a factor?Ive switched from saying provably a
factor, as that is problematic>to saying that its implied to
be a factor, and an analogy is in the>ring of evens where you
have 2 and 6, but 6 does not have 2 as a>factor, though
6=2(3), in the ring of integers.Here the problem is that
members that should be in the ring of>algebraic integers are
left out. So imagine one such member m, and>the algebraic
integers c and a, such that c = amand youll nd that in
the ring of algebraic integers, a is coprime>to c, which
is the problem, as by the denition of factor, m as>its
not an algebraic integer is NOT a factor of c in the ring
of>algebraic integers, just like 2 is NOT a factor of 6 in
the ring of>evens.And, amazingly, you failed to answer the
question. Here it is again,in capital letters so you dont
miss it again:COULD YOU PLEASE GIVE ME YOUR DEFINITION OF A
FACTOR, THEN?You know, something like:Let R be a ring, and x
and y elements of R; then x IS A FACTOR OF y(in R) if and
only if ....orLet R be a ring, x and y elements of R; then
x HAS A FACTOR OF y (inR) if and only if ...No examples, no
analogies, no talking about what you used to do andwhat you
are doing now, no talking about why you changed the way
yousay things.Just give the damned denition. Dont give
analogies, dont giveexamples, dont explain what is wrong
with the usual, dont explainwhat is wrong with the
unusual.Just give the damned denition. That is a fundamental
part ofmathematics (precise language, carefully dened terms).
Until youlearn that, you are not doing mathematics, you care
doingcrypto-mathematics and
quackery.
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 mans 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
gures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
indenite 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
=[cut]>And in fact, using
factors of when you mean multiples of while>common, is
technically incorrect.> I have never seen anyone, here or
elsewhere, use factors of> to mean multiples of. Except
for that you have a good point.Ramsay said something like 6
has a factor of 3.Likewise, 15 has a factor of 3.That is,
has a factor of is the same as is a multiple of.When
Ramsay was discussing this, JSH conviently changed thephrase
from has a factor of to the phrase is a factor ofwhen he
tried to show Ramsay he was not using standard terminology.If
of,he would nd Ramsays usage to be the standard usage. I
did thesearch, but looked at only the rst couple of hits.
James wouldneed to nd a hit that said something like 15 has
a factor of 10.I doubt that he would nd any such usage like
that.-- Bill Hale
[cut]>And in fact, using factors
of when you mean multiples of while>common, is technically
incorrect.> I have never seen anyone, here or elsewhere, use
factors of> to mean multiples of. Except for that you have
a good point.> Ramsay said something like 6 has a factor of
3.> Likewise, 15 has a factor of 3.> That is, has a factor
of is the same as is a multiple of.However, it is not.
Common usage may use it that way, but expandedout, saying 15
has a factor of 3, is equivalent to saying 3 is afactor of
15, and as 3 is a factor of 3, 15 has a factor of 3.It can be
a multiple of whats given but thats not forced. If
thediscussion were with integers then it wouldnt be a big
deal for me touse what I see as slang; however, what
shouldnt be lost here is thatits not that simple.A better
example is, in the ring of algebraic integers, g has a
factorof 5, that is 1+2i.So the context is important
here.Remember that a *proof* is being discusssed while people
get excitedabout whether or not has factors of can mean
something other than isa multiple of, which should tell you
something.Theyre trying to deceive you rather than get to
the bottom of things. > When Ramsay was discussing this, JSH
conviently changed the> phrase from has a factor of to the
phrase is a factor of> when he tried to show Ramsay he was
not using standard terminology.Really? Where? In any event a
number can be said to have a factor of5, without that meaning
the number has 5 as that factor. Forinstance, 21 has a factor
of 12, in that 3 is a factor of both 21 and12.Now many may
read that as 21 has a multiple of 12, but thats notwhats
stated.And in fact, you can say that 12 has a factor of 3, as
3 is itself afactor of itself, but to be precise you can say
12 is a multiple of 3.And again, if it were *integers* being
discussed then I wouldnt havea problem with using 12 has a
factor of 3 as meaning 12 is a multipleof 3, as that shortcut
probably would be ok.However, in context, my usage ts the
situation, and it seems to methat posters have a problem with
my correct usage because they cantnd anything wrong with the
search on the phrase has a factor of,> he would nd
Ramsays usage to be the standard usage. I did the> search,
but looked at only the rst couple of hits. James would> need
to nd a hit that said something like 15 has a factor of
10.> I doubt that he would nd any such usage like that.>
-- Bill HaleSee what I mean? How many of you thought better
of mathematiciansbefore you saw the tricks they play?My usage
is correct as any of you can demonstrate for yourselves
bynoticing that 12 has a factor of 21, as 3 is a factor of
both. Butyou may also think to yourself that I should just
give in to themathematicians and posters, and play along if I
want to convince them.But you see, at least some of them are
mathematicians, so they aremath experts! I use precision;
they get upset.Rather than admit the truth--that my proofs
arecorrect--mathematicians play word games and debate me over
use offactor of because theyre deceitful.James Harris
> [cut]>>And in fact, using factors of when you mean
multiples of while>>common, is technically incorrect.>>>
I have never seen anyone, here or elsewhere, use factors
of>> to mean multiples of. Except for that you have a good
point.>>> Ramsay said something like 6 has a factor of 3.>>
Likewise, 15 has a factor of 3.>> That is, has a factor of
is the same as is a multiple of.However, it is not. Common
usage may use it that way, but expanded>out, saying 15 has a
factor of 3, is equivalent to saying 3 is a>factor of 15, and
as 3 is a factor of 3, 15 has a factor of 3.It can be a
multiple of whats given but thats not forced. If
the>discussion were with integers then it wouldnt be a big
deal for me to>use what I see as slang; however, what
shouldnt be lost here is that>its not that simple.A better
example is, in the ring of algebraic integers, g has a
factor>of 5, that is 1+2i.So the context is important
here.The context is that you INSIST on using terminology in
an incorrect ornon-standard manner.When you say g has a
factor of 5 to mean that 1+2i divides g, youare MISUSING the
terminology. The CORRECT and STANDARD way of saying what you
mean is to say that gand 5 have common [non-trivial]
factors.
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 mans 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
gures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
indenite 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
==
multiples of while>>common, is technically incorrect.>>>
I have never seen anyone, here or elsewhere, use factors
of>> to mean multiples of. Except for that you have a good
point.>>> Ramsay said something like 6 has a factor of 3.>>
Likewise, 15 has a factor of 3.>> That is, has a factor of
is the same as is a multiple of.However, it is not. Common
usage may use it that way, but expanded>out, saying 15 has a
factor of 3, is equivalent to saying 3 is a>factor of 15, and
as 3 is a factor of 3, 15 has a factor of 3.It can be a
multiple of whats given but thats not forced. If
the>discussion were with integers then it wouldnt be a big
deal for me to>use what I see as slang; however, what
shouldnt be lost here is that>its not that simple.A better
example is, in the ring of algebraic integers, g has a
factor>of 5, that is 1+2i.This is indeed a better example. It
proves, beyond a shadow of adoubt, that you are misusing
standard terminology; or else that youare using nonstandard
terminology.If when you say g has a factor of 5, you really
mean not that 5divides g, but rather that there is an
algebraic integer which dividesboth 5 and g, then the
STANDARD AND CORRECT way of saying it isg and 5 have a
common factor [in the ring of all algebraicintegers].(the
bracketed part may be omitted if it is clear from context
oralready agreed on).If you want to further specify that this
common factor is not a unit,then the STANDARD AND CORRECT way
of saying it is to say:g and 5 have a common
non-trivial/non-unit factor [in the ring of allalgebraic
integers].In the specic case of the ring of all algebraic
integers, this isequivalent to g and 5 are not coprime [in
the ring of all algebraic integers](equivalent left as an
exercise exercise to the competent reader;James, you should
skip it).
[.snip.]
such a reasoner as Mr. Smith? I answer as a deceased friend
of mine used to answer on like occasions - A mans 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
gures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
indenite 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
For instance, 21 has a factor of 12, in that 3 is a factor of
both 21 and 12.> But Magidin, 9 does not have a factor of
12.Now that makes a lot of sense... or maybe not.
My
usage is correct as any of you can demonstrate for yourselves
by> noticing that 12 has a factor of 21, as 3 is a factor of
both. That is some funny-ass shit, man. Youre a humorist,
right?
If when you say g has a factor of 5, you really
mean not that 5> divides g, but rather that there is an
algebraic integer which divides> both 5 and g, then the
STANDARD AND CORRECT way of saying it is> g and 5 have a
common factor [in the ring of all algebraic> integers].Oh,
for crying out loud, is *THAT* what James has been
meaning?This just shows what everyone has been saying all
along, that precisionand rigor in dening and using
mathematical language really is important.
The problem
is when you say x has a factor of y. This is, in> standard
mathematical terminology, the same as saying y is a factor>
of x; which is the same, in standard mathematical
terminology, as y> is a divisor of x and as x is a
multiple of y.I seem to have lost track of exactly what the
original statements thatprompted all this discussion were,
but I do recall seeing things like:x has a factor of 5 which
is sqrt(5).Perhaps people are just parsing this differently.
One way of looking atit is as x has a value which is a
factor of 5, that value being sqrt(5).Its still too
ambiguous to say for certain, but thats one
possibleinterpretation which would at least make sense and
could explain whyeveryone insists theyre using the correct
denition.Of course that just emphasizes the need for precise
terminology and usage...
See what I mean? How many of
you thought better of mathematicians> before you saw the
tricks they play?The more I read of this sort of thing, the
better I think ofmathematicians, and the worse I think of
you.> Rather than admit the truth--that my proofs are>
correct--mathematicians play word games and debate me over
use of> factor of because theyre deceitful.Their usage is
correct; yours is incorrect.-- Wayne Brown | When your
tails in a crack, you improvisefwbrown@bellsouth.net | if
youre good enough. Otherwise you give | your pelt to the
trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
> If when you say g has a factor of 5,
you really mean not that 5>> divides g, but rather that there
is an algebraic integer which divides>> both 5 and g, then the
STANDARD AND CORRECT way of saying it is>>> g and 5 have a
common factor [in the ring of all algebraic>> integers].>
Oh, for crying out loud, is *THAT* what James has been
meaning?> This just shows what everyone has been saying all
along, that precision> and rigor in dening and using
mathematical language really is important.When James says
something like x has a factor of y he *sometimes*means y
is a factor of x and *other* times he means x and y sharea
common factor. The problem is that he doesnt specify
*which*meaning hes using in a given instance; indeed he
appears to use theminterchangeably, as if he doesnt see any
difference. Sometimes he usesboth meanings in the same
argument, ip-opping between them wheneverit suits his
convenience. Thats why he doesnt like strict denitions:are
clear (if anywhere) only in his own head.-- Wayne Brown |
When your tails in a crack, you
improvisefwbrown@bellsouth.net | if youre good enough.
Otherwise you give | your pelt to the trapper.e^(i*pi) =
-1 -- Euler | -- John Myers Myers, Silverlock
=[...]|
See what I mean sci.skeptic? The issue here is that Im
actually| using standard mathematical usage, while several
posters apparently| keep reading factors of 5 as multiples
of 5.No. The problem was with the use of the phrase has a
factor of. Well, if r_1 has a factor of 5 that is 1+2i, and
r_2 has a factor of 5 that is 1-2i, then Id say there it is
clear there is a distribution of factors of 5. Ive noted
before how I talk of factors of and when I mean a specic
factor I say that is, for instance, a factor of 5 that is
1+2i.You seem to think that once we agree on the meaning of
is a factorof, the meaning of has a factor of is
automatic. But r_1 has 1+2idoesnt mean anything. Saying
r_1 has 1+2i, which is a factor of 5is not better. r_1 has
a factor of 5, which is 1+2i is equallynonstandard.It is like
saying I have two parents of Bruce. If I wanted to saythat
we were siblings, I would need to say something like I have
twoparents of Bruce *as my parents*. That at least would be
logical,although eccentric. You need to say something like
is divisible by1+2i, which is a factor of 5.It only makes
it somewhat worse that theres already an
establishedconventional meaning of has a factor of. [*] I
have no problem withyour not liking this usage. In fact
people tend to avoid it when theyare writing mathematics.
Its kind of informal. But please dont tryto replace it with
another LESS standard usage.| The problem has to do with their
misuse of mathematical terminology.The problem has to do with
your confusing use of terminology.| For instance 2 is a
factor of 6, but so is 3, and in fact, so is 6.| | When you
say, factor of, it means something that is a factor of the|
given number. And similarly factors of would mean factors of
the| given number.Yes.| But you have these posters, like David
Ullrich who is a math professor| at Oklahoma State University,
who are lost with rather basic| mathematical language to the
point that they *keep* arguing *after*| Ive corrected them.
You see their *belief* system apparently is that| as
mathematicians they cant be the ones with the error, so
amazingly,| they simply keep almost mindlessly repeating
it.It seems to me that its actually you who obstinately
continues toargue after having been corrected.Even supposing
one or both of us were arrogant, this idea that werearrogant
based on a belief that as mathematicians we are infallible
isabsurd. Nobody says mathematicians are infallible. They
just dontagree that youre doing a very good job at nding
our mistakes. I wasnever sure whether Pertti Lounesto was
especially good at catching ourmistakes either, but you have
some ways to go before youre as good ashe was.| You may
guess that they say factors of when they mean multiples|
of, but Im using the proper terminology, in the correct
way.We dont say factors of to mean multiples of. We just
happen toknow that what people besides yourself mean if they
do say x _HAS_ afactor of Y (in it) is that x _IS_ a
multiple of Y.| So when I say factors of 5, Im NOT saying
multiples of 5. For| instance, 2 and 3 are factors of 6. So
when I say factors of 6 it| doesnt mean multiples of 6. And
I emphasize that there is a| mathematical term multiples of
which applies.Ill note that youre assuming were making a
mistake which is dumberthan the kind of mistake I personally
tend to expect you to make, andI think its silly. Surely you
can see by now that this is not whatwere thinkinig.| And in
fact, using factors of when you mean multiples of while|
common, is technically incorrect.It isnt common at all!| But
people like Ramsay and Ullrich are unlikely to reply to
*correct*| their mistakes because the society of sci.math
lets people it| considers part of the society get away with
the dumbest mistakes.Were not posting corrections and
apologies now because we donthave good reason to believe
that were wrong.Its not about punishment. After hundreds of
fallacious proofs ofFermat, its pretty clear that you dont
care very much about gettingpunished for crying wolf either,
do you? Its fairly difcult foranyone to get punished for
dumb mistakes on usenet, regardless ofstatus.| Thats how
mathematicians operate as demonstrated before your eyes.| |
And thats how they can have errors in their discipline for
years, and| years because they seek to by denition have a
society that is| perfect,Huh? What denition?| when in fact,
mathematicians are just people, and people make|
mistakes.Keith RamsayP.S. [*] It depends on the meaning of
of! ;-)Factor of can mean divisor of. But it is also
used in phrases likeMy measurement of Plancks constant was
off by a factor of 2, wherefactor of 2 does mean a number
that divides 2. Its referring to afactor, 2. Factor still
means factor, but the way in which itsbeing associated
with 2 is different. I have some sympathy with yourthinking
that the phrase factor *that is* 2 would be a good way
tosay this, but thats just not how anyone else says it.I
think most often its used when talking about formulas, to
refer tofactors appearing in the formula. You dropped a
factor of 2 on theleft hand side in this step.
=[...]|
Given that r_1 r_2 = 5, it makes sense that factors of 5
distribute in| some way between r_1 and r_2.I would guess
that you mean that there exist algebraic integersdividing 5,
which also divide r_1 and r_2, except that thats apeculiar
thing to ask. The obvious choice of divisors of 5 to useas
divisors of r_1 and r_2 are just r_1 and r_2.So is that what
do you actually mean by it?| For instance, if r_1 = r_2 =
sqrt(5), you have the same factor of 5 in| both r_1 and r_2,
while with r_1=5, r_2 = 1, you have all the non unit| factors
of 5 with r_1.|| It hardly seems like rocket science.|| Can
you explain why you as a mathematician are having difculty
in| the area Keith Ramsay?One day as I was boarding a bus, a
woman asked the driver a questionwhich sounded to me exactly
like where is the outside wall?To the driver this was a bit
of a problem. It wasnt a problem withthe drivers knowledge
of geography. He actually knew the answer tothe womans
question. The question wasnt complicated. It was just
alanguage problem. She wanted to know where the pedestrian
mall indowntown was, and once the driver gured out that
thats what shemeant, there was no problem answering her
question.Thats what my problem with you is like. As soon
as you can make amathematical question clear, the answer is
probably not going to behard to nd. There are tools for
working this sort of thing out.[...]| The mistake is in
assuming that denition of the ring of algebraic| integers
does not lead to contradictions.Thats absurd. Theres no
possible way that the *denition* ofalgebraic integer or
ring of algebraic integers could lead to acontradiction.
Ive seen you clutch at straws before, but this takesthe
cake.[...]| Ive also noted that its odd that mathematicians
would rather try to| duel with proofs than simply nd an
error in my proof. But, of| course, as it is a proof, there
is no error, which is why I think they| try to nd other
means.|| Thats not mathematics. Thats cheating.I can
demonstrate what happens if one of us simply nds an error
inyour proof.When you consider a factorization P(m) = (a1x +
uf)(a2x + uf)(a3x + uf),you dont correctly describe what a1,
a2, and a3 are. You claim theyare algebraic integers, but
since they vary with m this cannot beprecisely correct. You
talk about what values they have when m=0 andwhen m=1, for
example. In order to satisfy such an equation, they haveto be
something like algebraic functions of m.People have done a
fair bit of work to make good sense of the notionof
algebraic function, in such a way that one can talk about
thevalues at individual points. As m assumes different
values, we cantalk about the unordered triple of values of
a which could be usedin a factorization like that. But if
we want to label one of the threeas being a1, another as a2,
and a third as a3, we have to make achoice somehow. Its a
bit like deciding which of the roots of t^2=xis going to be
called sqrt(x) and which is going to be called-sqrt(x). Part
of the trouble is that theres no way to make thechoice which
makes it dened for all x and continuous. On the otherhand,
you need for there to be some kind of relationship between
thevalues of a1 at m=0 and its values elsewhere in your
argument.This mistake helps to make the rest of the argument
confusing. Yourefer to one of the as not being coprime to m.
In what ring? The ringhas to contain the as, but they appear
to be functions of m.At the bottom, and I assume someone else
has pointed this out, youwrite out P(m) but dont distribute
the factor of f^2 ;-) in all ofthe terms. You have (Im
taking this from the posting quoting yourpaper): (m^3 f^6 -
3m^2 f^4 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 + u^3but it should be
(m^3 f^6 - 3m^2 f^4 + 3mf^2)x^3 - 3( - 1 + mf^2 )xu^2f^2 +
u^3f^3(a factor of f ;-) was left out in the last term
too). This does notevaluate to 65x^3 - 12x + 1 if you
substitute f=sqrt(5), u=1, m=1. Itwould have to be
u=1/sqrt(5). But you assumed that u was an algebraicinteger
when you introduced it. You use that to say that f is a
factorof uf, for example. You write that the values of a1,
a2, and a3 dontdepend on u, but that doesnt explain how
its valid to assume u is analgebraic integer and then try to
apply the result to a polynomial,that you get by substituting
a non-integral algebraic number for u.---Ok, now lets see
what happens.I think frustration with you over the way you
respond to people whond mistakes of yours is one of the main
reasons why people sometimesdecide to try something different,
like writing a proof that theconclusion is false. But also I
would say that its often more fun,even when dealing with our
favorite posters, to go and write up my ownproof of a result
related to what they are writing, than it is to justreact to
the way theyve done it.| > | If you disagree with that
assessment, can you please explain why you| > | believe there
should be something simple and short enough for me to| > |
quote?| >| > In mathematics, we try to make each
*mathematical* statement have its| > own well-dened meaning.
(There are of course nonmathematical claims| > that we
sometimes make, which arent always precise.) If a sequence
of| > well-dened mathematical statements is wrong, its
wrong because (at| > least) one of the statements is wrong.|
>| > The key is making sure all the denitions are sound.||
As Ive stated before a proof begins with a truth and
proceeds by| logical statements to a conclusion that then
must be true.|| Therefore it follows that denitions must be
sound, or the steps will| not be logical.Well, nice to see we
agree. And now that youve identied thedenition of ring of
algebraic integers as the culprit, reallyyouve done
essentially what Ive asked-- we can quote this
denitionspecically. Unfortunately (or fortunately), we
still have no goodreason to think that the denition could
create a contradiction.| > Mathematics that is vaguely enough
written that it can be wrong| > without any individual
statements in it being denitely wrong, is| > considered very
bad. Its bad because an author hasnt dened their| > terms
well enough.|| The term bad is a human term inapplicable in
context as mathematics| is about truth.|| That is, a proof
is neither bad nor good, it is true or false.This is naive.
Realistically, one has to be concerned with the qualityof a
proof as well as with whether its simply true or
false.Getting rid of all errors is quite hard. So in the
middle of workingon a proof, the usual situation is to be
working on a faulty sketch ofa proof. There is a huge
difference between a proof that is false,but is written
*well*, and has good parts in it, so that correcting itis
easier, and one that is false and written *badly*, so that
itshard even to see where mistakes might be.[...]| > The
standard usage is also inconsistent with the rest of the
paper.| > You write, for example, ...proving that two of the
as have a factor| > that is f at a point where you plainly
have NOT shown that f is a| > factor of any of the as, in
any of the usual ways factor is| > dened.|| Hmmm...thats
an interesting point. What has happened at that point| in the
paper is that Ive considered the constant term P(0), and the|
constant term with f^2 separated off, which is P(0)/f^2 = 3x
u^2 + u^3| f, and noted that it is coprime to f.Just to be
clear, why is 3x+u coprime to f?| Now I then consider g_1 at
m=0, where c=g_1, and notice it has a| factor of f,Assuming u
is an algebraic integer.| and then based on P(0)/f^2 being
coprime to f, I have| that r + c, must have a factor of f,
which proves that r must have a| factor of f that is f.I
guess you mean g_1=r+c as in the lemma. What makes you think
thatf divides r+c?| > Elsewhere, you refer to an algebraic
integer f coprime to x. This| > has no standard meaning in
the context you use it, where x is a| > variable. If you
meant coprime in some ring of polynomials, which is| >
closest to a standard meaning, it would mean that there
exists| > polynomials P and Q such that P*f+Q*x=1. But if f
is an algebraic| > integer, thats possible (if and) only if
f is a unit, with Q=0, and| > you also assume in the same
sentence that f is a non unit.|| But f while a variable is
constant, and so is x. So it is not in any| ring of
polynomials.|| Unfortunately, you seem to be xated on the
*letters* as in seeing an| x you may assume that its
varying. Nope. Its constant.Says who? Taking it to be a
constant is not only confusing, butinconsistent with the rest
of the paper.When you introduce P(m), you remark that the new
variables provide*additional* degrees of freedom. If x is a
constant, along witha1, a2, and a3, whats degree of freedom
did you start out with?When you refer to the factorization
of P(m) into linear terms(a1x+uf)(a2x+uf)(a3x+uf), thats
consistent with x being a variable.If x is a constant,
theres no uniqueness in such a factorization (andnot much
that you can prove about the values of a1, a2, and a3). Inthe
next step you infer that the coefcient of x^3 on the right
handside (a1a2a3) is the same as the coefcient of x^3 on the
left handside. This also is consistent with x being a
variable. If twopolynomials are equal for all values of x,
then their correspondingcoefcients have to be equal. But if
x is a constant, that step isinvalid.Referring to f being
coprime to x isnt a tipoff that x isnt avariable, as you
also consider whether something is coprime to m,and m is
denitely being varied.In fact, I have no reason to believe,
except your say-so, that youintended for x to be a constant
until you saw my question.And its still unclear why two of
the as are supposed to bedivisible by f.| Thats troubling
Keith Ramsay as thats basic algebra.The problem is not with
the algebra. The problem is with the writing.| The symbols|
have to be understood as dened, not based on your experience
of how| you may be used to seeing them.But you didnt dene x.
Nearly all the context implies that x isbeing used as a
polynomial variable. What do *you* think we should doin a
case like that?We can, on the one hand, try to make
reasonable assumptions aboutthings which typically are done
in a certain way. Based on all thecontext I mention above,
normally this would be because youretreating x as a
polynomial variable.It seems, however, as though you want to
say now that whenever I makesuch an assumption (which you
decide to declare incorrect) its myfault. (And a fault in
algebra, too.) So really, I should assumenothing.But if I did
this, you have not in your wildest dreams imagined theamount
of nitpicking clarication Id be asking you to do. You
wouldat the very least have to say what kind of variable each
variable youuse is. Meanwhile, you also complain when people
keep pestering you to clarifydetails like that. You imply
theyre doing it only as a distractionfrom the real
mathematical content.You cant have it both ways. Either you
agree to some reasonablereading between the lines, or you
write your paper in such a waythat reading between the lines
isnt needed.| > | > I dont know what you think is a fair
way for things to work, but the| > | > way things actually
work, this kind of problem with terminology will| > | >
absolutely prevent you from making any headway.| > || > |
What will prevent me from making headway is if
mathematicians| > | continually lie.| >| > Mathematicians
dont continually lie. What you keep deciding are| > lies
are just disagreements with you.|| Thats not true. What Ive
done is explain clearly and in detail.No, its far from clear,
and the detail is still poor at just theplace in the argument
where it needs to be best, right near the end.| In reply I
nd people making specious issues, like your claims about|
factor when you apparently are sticking in multiple.Which
again was merely a misunderstanding.| > Perhaps the biggest
problem of all is that youve reached a point| > where you
dont have anybody you are willing to trust, who could help|
> you sort out which of the claims people are making are
valid and which| > are baloney. So you tend automatically to
assume that the things| > people say that seem wrong to you
are baloney of some kind or another.|| How can I take people
like you seriously when you have trouble with| such simple
things as factor of?Because I dont have trouble with
this, except with people whovedecided to play Humpty Dumpty
(and pay their words extra to mean whatthey want them to
mean). If I said I had a teacher of children, and Iexpected
people to understand me as saying that there was someone
whotaught both me and children, they would have a problem
with that too.The problem would not be with what has a
teacher means.| > This also leaves you without any very good
way to learn how to improve| > your mathematical tinkering. We
keep telling you different things| > which _we_ claim would
help you avoid pitfalls like you keep falling| > into. But
(a) you dont trust us enough to believe weve identied a| >
problem with what youre doing, and (b) you dont trust us to
give you| > straight advice on how to avoid it; you suspect
us of just trying to| > waste your time. So youre stuck with
only your own inklings of whats| > a good way to work with
proofs.|| Yet Ive caught you in a contradiction with
yourself, where you| apparently have been inserting the word
multiple when I say| something like factor of 5, so that
you read multiples of 5. If| thats not what youre doing
then, who knows whats going through that| brain of yours.I
believe I can explain well enough.| Readers should consider
what follows if they think thats too harsh.|| > | Now then
in what way is it improper terminology to talk about| > |
algebraic integer factors of 5?| >| > I didnt say it was
improper to talk about algebraic integer factors| > of 5.
That has a perfectly well-dened meaning. An algebraic
integer| > r is a factor of 5 if it divides 5 in the
algebraic integers, meaning| > that 5/r is also an algebraic
integer.|| <deleted>|| Youre contradicting yourself.No, as I
explained elsewhere. You made the leap of assuming that hasa
factor of should mean has a common factor with. I have a
teacherof children does not mean I share a teacher with
some children.6 has a factor of 3 does not mean 6 and 3
share a factor.| Go back and read over what you said at| the
top of this post. Im tiring of this exercise in pointing out|
your errors.*You* think *youre* getting tired of correcting
*my* errors? Comingfrom you thats a laugh.Keith Ramsay
> [cut]>>And in fact, using factors of when you mean
multiples of while>>common, is technically incorrect.>>>
I have never seen anyone, here or elsewhere, use factors
of>> to mean multiples of. Except for that you have a good
point.>>> Ramsay said something like 6 has a factor of 3.>>
Likewise, 15 has a factor of 3.>> That is, has a factor of
is the same as is a multiple of.However, it is not. Yes it
is.>Common usage may use it that way, but expanded>out,
saying 15 has a factor of 3, is equivalent to saying 3 is
a>factor of 15, and as 3 is a factor of 3, 15 has a factor of
3.Huh? This is typical - your answers always seem to miss
thepoint. We all agree that 15 has a factor of 3, 3 is a
factorof 15 are both true, and say the same thing. The
problemis that this is not the only way you use has a factor
of -the way you use the language 15 has a factor of 25would
be true, because 5 is a factor of 25, and 15 has5 as a
factor. In _fact_ 15 has a factor of 25 is _false_,if
youre using words to mean what everyone else meansby
them.>It can be a multiple of whats given but thats not
forced. If the>discussion were with integers then it wouldnt
be a big deal for me to>use what I see as slang; however, what
shouldnt be lost here is that>its not that simple.A better
example is, in the ring of algebraic integers, g has a
factor>of 5, that is 1+2i.So the context is important
here.Uh, what seems to me to be important is the question
that yousimply ignored a _second_ time: You say that if r_1
r_2 = 5then the factors of 5 distribute among r_1 and r_2.
Letr_1 = 1+2i, r_2 = 1-2i. Note that sqrt(5) is a factor of 5
(inthe algebraic integers of course). Does that mean
thatsqrt(5) is a factor of r_1 or of r_2?>Remember that a
*proof* is being discusssed while people get excited>about
whether or not has factors of can mean something other than
is>a multiple of, which should tell you something.Theyre
trying to deceive you rather than get to the bottom of
things.>> When Ramsay was discussing this, JSH conviently
changed the>> phrase from has a factor of to the phrase is
a factor of>> when he tried to show Ramsay he was not using
standard terminology.Really? Where? In any event a number can
be said to have a factor of>5, without that meaning the number
has 5 as that factor. For>instance, 21 has a factor of 12, in
that 3 is a factor of both 21 and>12.No. You are _revising_
termimology here. Saying 21 has a factorof 12 and then
castigating _us_ for our sloppy usage is _exactly_the same as
saying that the integers are irrational and thenlaughing at
all the people who dont realize thats so.>Now many may read
that as 21 has a multiple of 12, but thats not>whats
stated.And in fact, you can say that 12 has a factor of 3, as
3 is itself a>factor of itself, but to be precise you can say
12 is a multiple of 3.And again, if it were *integers* being
discussed then I wouldnt have>a problem with using 12 has a
factor of 3 as meaning 12 is a multiple>of 3, as that
shortcut probably would be ok.However, in context, my usage
ts the situation, and it seems to me>that posters have a
problem with my correct usage because they cant>nd anything
wrong with the math--so they argue semantics.You really think
that nobodys noticed that you _continue_ to ignorehalf of my
posts here, answering (incorrectly) the part about thelanguage
but simply not replying to the part about the math?Youve
laughed at people for not following this, saying it
doesntseem like rocket science. So answer the
question.Ignoring questions about the math and then saying
thatthere are no questions about the math would be
funnyexcept its much too transparent. You can do much
has a factor of,>> he would nd Ramsays usage to be the
standard usage. I did the>> search, but looked at only the
rst couple of hits. James would>> need to nd a hit that
said something like 15 has a factor of 10.>> I doubt that
he would nd any such usage like that.>>> -- Bill HaleSee
what I mean? How many of you thought better of
mathematicians>before you saw the tricks they play?My usage
is correct as any of you can demonstrate for yourselves
by>noticing that 12 has a factor of 21, as 3 is a factor of
both. But>you may also think to yourself that I should just
give in to the>mathematicians and posters, and play along if
I want to convince them.But you see, at least some of them
are mathematicians, so they are>math experts! I use
precision; they get upset.No, you use language in totally
nonstandard ways, people try toexplain that what youre
saying doesnt mean what you think itdoes and you refuse to
believe them.When you insist on speaking your own private
language youshouldnt be surprised when people dont believe
you - theyreassuming you mean what you _say_. (Its exactly
this problemwith has a factor of that prevented Ramsay from
understandingwhat you meant a few posts up in this very
thread. Honest.)>Rather than admit the truth--that my proofs
are>correct--mathematicians play word games and debate me
over use of>factor of because theyre deceitful.>James
Harris
[...]> |
Given that r_1 r_2 = 5, it makes sense that factors of 5
distribute in> | some way between r_1 and r_2.> I would
guess that you mean that there exist algebraic integers>
dividing 5, which also divide r_1 and r_2, except that thats
a> peculiar thing to ask. The obvious choice of divisors of 5
to use> as divisors of r_1 and r_2 are just r_1 and r_2.>
So is that what do you actually mean by it?The issue that is
raised at the start of the paper is a question abouthow one
would know that given x^2 + x - 5 = (x - r_1)(x - r_2)
wherefactors of 5 might be in r_1 and r_2, in terms of the
question ofwhether or not either could be a unit factor of
5.For those who wonder what a unit factor is, its like how
1 is afactor of 5, but in the ring of algebraic integers there
are aninnity of numbers that are factors of 1 that are
themselves not 1. For instance (1+sqrt(-3))/2 is a unit
factor because it multipliestimes (1-sqrt(-3))/2 to give 1,
and both are algebraic integers.It turns out that
mathematicians dont know how to prove whether ornot r_1 and
r_2 in the example I gave are unit factors. However, ifyou
tell them that they may get upset and start showering you
with alot of mathematical statements, which if you look
carefully, dontprove it, or disprove it.But mathematicians
have apparently *decided* that neither r_1 nor r_2can be unit
factors of 5, though they dont know how to prove it.
Unfortunately, they refuse to admit the reality, and argue
usingreasoning that has been shown by me to be awed in order
to claimthat they can.So the preamble in my paper is
highlighting that problem. Imbasically hitting them in the
face with it at the start. And youllnotice how much arguing
has gone on about the *preamble* and not thethe main part of
the paper. > | For instance, if r_1 = r_2 = sqrt(5), you have
the same factor of 5 in> | both r_1 and r_2, while with r_1=5,
r_2 = 1, you have all the non unit> | factors of 5 with r_1.>
|> | It hardly seems like rocket science.> |> | Can you
explain why you as a mathematician are having difculty in> |
the area Keith Ramsay?> One day as I was boarding a bus, a
woman asked the driver a question> which sounded to me
exactly like where is the outside wall?> To the driver
this was a bit of a problem. It wasnt a problem with> the
drivers knowledge of geography. He actually knew the answer
to> the womans question. The question wasnt complicated. It
was just a> language problem. She wanted to know where the
pedestrian mall in> downtown was, and once the driver gured
out that thats what she> meant, there was no problem
answering her question.> Thats what my problem with you
is like. As soon as you can make a> mathematical question
clear, the answer is probably not going to be> hard to nd.
There are tools for working this sort of thing out.Yeah and
how *long* did it take the driver?My problem with
mathematicians as highlighted by the subject line ofthis
thread is that Im seeing sickening evidence that they
areworking to *hide* the truth.So Keith Ramsay, lets see if
this post of yours changes thatassessment in any way.You see
Keith Ramsay, in the real world, people tend to try to get
tothe bottom of things quickly because they have a job to do,
and valuetheir time. Now I think mathematicians value their
time like otherpeople, but the twist is that I see you as
working to lengthen thetime in which the world is left
ignorant of the truth, and thatswrong.> [...]> | The mistake
is in assuming that denition of the ring of algebraic> |
integers does not lead to contradictions.> Thats absurd.
Theres no possible way that the *denition* of> algebraic
integer or ring of algebraic integers could lead to a>
contradiction. Ive seen you clutch at straws before, but
this takes> the cake.Yet Ive proven my assertions
mathematically, and Ive presented thatmathematics
repeatedly.> [...]> | Ive also noted that its odd that
mathematicians would rather try to> | duel with proofs than
simply nd an error in my proof. But, of> | course, as it is a
proof, there is no error, which is why I think they> | try to
nd other means.> |> | Thats not mathematics. Thats
cheating.> I can demonstrate what happens if one of us
simply nds an error in> your proof.> When you consider a
factorization P(m) = (a1x + uf)(a2x + uf)(a3x + uf),> you
dont correctly describe what a1, a2, and a3 are. You claim
they> are algebraic integers, but since they vary with m this
cannot be> precisely correct. You talk about what values they
have when m=0 and> when m=1, for example. In order to satisfy
such an equation, they have> to be something like algebraic
functions of m.Ive talked on this issue using P(x) = x+1, in
the ring of integers.Now then, your objection is like saying
that its wrong for me to saythat P(x) is an integer, though
the ring is declared to be integers,and x+1 will give an
integer for any integer x.Now notice that P(x) = x+1 is an
algebraic function of x, but it*still* gives integers for any
integer x. > People have done a fair bit of work to make good
sense of the notion> of algebraic function, in such a way
that one can talk about the> values at individual points. As
m assumes different values, we can> talk about the unordered
triple of values of a which could be used> in a
factorization like that. But if we want to label one of the
three> as being a1, another as a2, and a third as a3, we have
to make a> choice somehow. Its a bit like deciding which of
the roots of t^2=x> is going to be called sqrt(x) and which
is going to be called> -sqrt(x). Part of the trouble is that
theres no way to make the> choice which makes it dened for
all x and continuous. On the other> hand, you need for there
to be some kind of relationship between the> values of a1 at
m=0 and its values elsewhere in your argument.Your lengthy
statement is sophistry. I can give a simple example:Consider
the set {1,2,3}, and take a member m from that set.It is true
that *one* and only one member of the set is even.Notice that
I can get that logical conclusion without labeling.However, I
can *arbitrarily* label m_1=1, m_2=2, and m_3=3, and itdoesnt
change that conclusion.Now I can just as easily label m_1=2,
m_2=3, m_3=1, and it doesntchange that conclusion.That is,
its *still* true that only one of the ms is even!!!What I
want readers to understand is that the most
reasonableconclusion to draw is that mathematicians are
*deliberately* lying tothem. > This mistake helps to make the
rest of the argument confusing. You> refer to one of the as
not being coprime to m. In what ring? The ring> has to
contain the as, but they appear to be functions of m.For a
nonzero value of m, it is *proven* that *one* of the as
iscoprime to f.Ive explained above simply how that is
possible without there being amistake. > At the bottom, and I
assume someone else has pointed this out, you> write out P(m)
but dont distribute the factor of f^2 ;-) in all of> the
terms. You have (Im taking this from the posting quoting
your> paper):> (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1 + mf^2
)xu^2 + u^3And for the readers, after that youd have 65x^3 -
12x + 1.> but it should be> (m^3 f^6 - 3m^2 f^4 + 3mf^2)x^3
- 3( - 1 + mf^2 )xu^2f^2 + u^3f^3Which is a false statement.
Notice that Keith Ramsay made a change,and below he talks
about *his* change as not working as if its myproblem.> (a
factor of f ;-) was left out in the last term too). This
does not> evaluate to 65x^3 - 12x + 1 if you substitute
f=sqrt(5), u=1, m=1. It> would have to be u=1/sqrt(5). But
you assumed that u was an algebraic> integer when you
introduced it. You use that to say that f is a factor> of uf,
for example. You write that the values of a1, a2, and a3
dont> depend on u, but that doesnt explain how its valid
to assume u is an> algebraic integer and then try to apply
the result to a polynomial,> that you get by substituting a
non-integral algebraic number for u.The gist of it is that
the factorization for 65x^3 - 12x + 1is independent of x,
which is a trivial enough thing usually, but VERYimportant
here.For instance, you have x^2 + 4x + 4 = (x+2)(x+2) without
worrying thatthe factorizaton changes as x changes, as thats
the point--itdoesnt.What I do in my paper is use a larger
more complicated expressionwhich *includes* 65x^3 - 12x + 1,
in what you might call a superset. I nd a limitation, which
then automatically applies.The equivalent is noting that with
x=1, you have x^2 + 4x + 4 = 9.With that you *still* have the
factorization, you also now have thatx+2 is a factor, and
indeed it is.Notice though that with the symbols gone, just
looking at 9, you have*less* information, than you have with
x^2 + 4x + 4.Isnt algebra grand? > ---> Ok, now lets see
what happens.Indeed.> I think frustration with you over the
way you respond to people who> nd mistakes of yours is one
of the main reasons why people sometimes> decide to try
something different, like writing a proof that the>
conclusion is false. But also I would say that its often
more fun,> even when dealing with our favorite posters, to go
and write up my own> proof of a result related to what they
are writing, than it is to just> react to the way theyve
done it.I want readers to consider the negative implication
and condescendingtone.> | > | If you disagree with that
assessment, can you please explain why you> | > | believe
there should be something simple and short enough for me to>
| > | quote?> | | > In mathematics, we try to make each
*mathematical* statement have its> | > own well-dened
meaning. (There are of course nonmathematical claims> | >
that we sometimes make, which arent always precise.) If a
sequence of> | > well-dened mathematical statements is
wrong, its wrong because (at> | > least) one of the
statements is wrong.> | | > The key is making sure all the
denitions are sound.> |> | As Ive stated before a proof
begins with a truth and proceeds by> | logical statements to
a conclusion that then must be true.> |> | Therefore it
follows that denitions must be sound, or the steps will> |
not be logical.> Well, nice to see we agree. And now that
youve identied the> denition of ring of algebraic
integers as the culprit, really> youve done essentially
what Ive asked-- we can quote this denition> specically.
Unfortunately (or fortunately), we still have no good> reason
to think that the denition could create a
contradiction.Stated without proof.> | > Mathematics that is
vaguely enough written that it can be wrong> | > without any
individual statements in it being denitely wrong, is> | >
considered very bad. Its bad because an author hasnt dened
their> | > terms well enough.> |> | The term bad is a human
term inapplicable in context as mathematics> | is about
truth.> |> | That is, a proof is neither bad nor good, it
is true or false.> This is naive. Realistically, one has to
be concerned with the quality> of a proof as well as with
whether its simply true or false.Stated without proof.>
Getting rid of all errors is quite hard. So in the middle of
working> on a proof, the usual situation is to be working on
a faulty sketch of> a proof. There is a huge difference
between a proof that is false,> but is written *well*, and
has good parts in it, so that correcting it> is easier, and
one that is false and written *badly*, so that its> hard
even to see where mistakes might be.A proof cannot be false,
by denition.What you might be trying to say is that when
looking for a proof, aperson might see the outlines of what
they think is a proof. As theypursue that outline they may
nd pieces of it which are faulty i.e.that are false.If there
is a proof there, however, it does not change.Its up to the
discoverer to nd it, if they can.> [...]> | > The standard
usage is also inconsistent with the rest of the paper.> | >
You write, for example, ...proving that two of the as have
a factor> | > that is f at a point where you plainly have
NOT shown that f is a> | > factor of any of the as, in any
of the usual ways factor is> | > dened.> |> |
Hmmm...thats an interesting point. What has happened at that
point> | in the paper is that Ive considered the constant
term P(0), and the> | constant term with f^2 separated off,
which is P(0)/f^2 = 3x u^2 + u^3> | f, and noted that it is
coprime to f.> Just to be clear, why is 3x+u coprime to
f?Thats just bizarre Keith Ramsay. The full expression is 3x
u^2 + u^3 fand f is coprime to 3, x and u, so its coprime to
that expression,which is what I said.So why did you ask about
3x + u?> | Now I then consider g_1 at m=0, where c=g_1, and
notice it has a> | factor of f,> Assuming u is an algebraic
integer.The ring has already been declared to be the ring of
algebraicintegers.For other readers, you may be wondering why
mathematicians are soclearly lying.Good question.However,
consider that if they tell the truth, it makes headlines.That
is, its news big enough to dominate the pages of
newspapersaround the world.Now if you found you could just
lie, and the world trusted you so thatthey let you lie for
quite some time, might you not hope that youcould just keep
lying endlessly, and never be held accountable?> | and then
based on P(0)/f^2 being coprime to f, I have> | that r + c,
must have a factor of f, which proves that r must have a> |
factor of f that is f.> I guess you mean g_1=r+c as in the
lemma. What makes you think that> f divides r+c?Its shown in
the paper. Specically, c is constant, so as g_1changes, there
must exist r = g_1 - c, and r must change.Its determined that
g_1 has a factor of f that is f at m=0. Itsfurther determined
that when m does not equal 0, separating off f^2from P(m)
leaves a constant term that is coprime to f; therefore,
afactor that is f must separate off from c, and the result is
coprimeto f.And yes, other readers, its a lot easier to look
over that section inthe paper where you have a lot more
information in front of you, thanguring it out from that
explanation.> | > Elsewhere, you refer to an algebraic
integer f coprime to x. This> | > has no standard meaning
in the context you use it, where x is a> | > variable. If you
meant coprime in some ring of polynomials, which is> | >
closest to a standard meaning, it would mean that there
exists> | > polynomials P and Q such that P*f+Q*x=1. But if f
is an algebraic> | > integer, thats possible (if and) only if
f is a unit, with Q=0, and> | > you also assume in the same
sentence that f is a non unit.> |> | But f while a variable
is constant, and so is x. So it is not in any> | ring of
polynomials.> |> | Unfortunately, you seem to be xated on
the *letters* as in seeing an> | x you may assume that its
varying. Nope. Its constant.> Says who? Taking it to be a
constant is not only confusing, but> inconsistent with the
rest of the paper.That is false. The key variable that
changes in the paper is m.> When you introduce P(m), you
remark that the new variables provide> *additional* degrees
of freedom. If x is a constant, along with> a1, a2, and a3,
whats degree of freedom did you start out with?The pertinent
example is x^2 + 4x + 4, like from before, as you get
anadditional degree of freedom from the use of x here, which
allows youto talk about a family of values, versus just 9.The
point of algebra is that using variables allows for
conclusions tobe drawn about large classes of numbers rather
than forcing you tocheck each number.Without algebra, a
person can gure out that 3(3)=9. With algebra,and an
additional degree of freedom, you have
that(x+2)(x+2)=x^2+4x+4, which also tell you about 16, 49,
and 64 alongwith an innity of other numbers.> When you refer
to the factorization of P(m) into linear terms>
(a1x+uf)(a2x+uf)(a3x+uf), thats consistent with x being a
variable.> If x is a constant, theres no uniqueness in such
a factorization (and> not much that you can prove about the
values of a1, a2, and a3). In> the next step you infer that
the coefcient of x^3 on the right hand> side (a1a2a3) is the
same as the coefcient of x^3 on the left hand> side. This
also is consistent with x being a variable. If two>
polynomials are equal for all values of x, then their
corresponding> coefcients have to be equal. But if x is a
constant, that step is> invalid.Thats false. Consider again
x^2 + 4x + 4 = (x+2)(x+2), as that istrue whether or not you
consider x variable or not.The *factorization* is whats
important, and it exists without regardto whether or not you
call P(x) = x^2 + 4x + 4 a polynomial, or justhave x^2 + 4x +
4 with x=2.For readers that particular falsehood is an
important one to considercarefully because its one that I
suggest to you is hard to explain asanything other than a
willful falsehood--that is, a lie.> Referring to f being
coprime to x isnt a tipoff that x isnt a> variable, as you
also consider whether something is coprime to m,> and m is
denitely being varied.While it is true that m is varied,
its also true that you canconsider a family of values for m
that are each coprime to f.That condition states that any
particular m is coprime to f.> In fact, I have no reason to
believe, except your say-so, that you> intended for x to be a
constant until you saw my question.Thats irrelevant. Remember
a proof begins with a truth, and proceedsby logical steps to a
conclusion which then must be true.If you have an issue it
should be with the beginning, as to whether ornot it is a
truth, or with a logical step. > And its still unclear why
two of the as are supposed to be> divisible by f.The
conclusion is that one of the as is coprime to f.> | Thats
troubling Keith Ramsay as thats basic algebra.> The
problem is not with the algebra. The problem is with the
writing.Stated without proof.> | The symbols> | have to be
understood as dened, not based on your experience of how> |
you may be used to seeing them.> But you didnt dene x.
Nearly all the context implies that x is> being used as a
polynomial variable. What do *you* think we should do> in a
case like that?You should stick to whats in the paper, and
with logic.> We can, on the one hand, try to make reasonable
assumptions about> things which typically are done in a
certain way. Based on all the> context I mention above,
normally this would be because youre> treating x as a
polynomial variable.Nope. I use a factorization, and as Ive
pointed out, a factorizationexists independent of what you
mention, like with x^2 + 4x + 4 =(x+2)(x+2), as explained
above.> It seems, however, as though you want to say now that
whenever I make> such an assumption (which you decide to
declare incorrect) its my> fault. (And a fault in algebra,
too.) So really, I should assume> nothing.You go with whats
in the paper.> But if I did this, you have not in your
wildest dreams imagined the> amount of nitpicking
clarication Id be asking you to do. You would> at the very
least have to say what kind of variable each variable you>
use is. Sounds like a lawyer tactic.> Meanwhile, you also
complain when people keep pestering you to clarify> details
like that. You imply theyre doing it only as a distraction>
from the real mathematical content.Posters have asked
questions answered in the paper.Stick with the paper.> You
cant have it both ways. Either you agree to some reasonable>
reading between the lines, or you write your paper in such a
way> that reading between the lines isnt needed.Which begs
the question of your status as a math expert, as thedenition
of mathematician is math expert.> | > | > I dont know what
you think is a fair way for things to work, but the> | > |
> way things actually work, this kind of problem with
terminology will> | > | > absolutely prevent you from making
any headway.> | > |> | > | What will prevent me from making
headway is if mathematicians> | > | continually lie.> | | >
Mathematicians dont continually lie. What you keep deciding
are> | > lies are just disagreements with you.> |> | Thats
not true. What Ive done is explain clearly and in detail.>
No, its far from clear, and the detail is still poor at just
the> place in the argument where it needs to be best, right
near the end.Then why didnt you ask more questions about
that section in thispost?> | In reply I nd people making
specious issues, like your claims about> | factor when you
apparently are sticking in multiple.> Which again was
merely a misunderstanding.Yet several posters argued about
it, and I suggest to readers that itis strong evidence of a
tendency from mathematicians to care lessabout the truth than
their own beliefs, and egos.> | > Perhaps the biggest problem
of all is that youve reached a point> | > where you dont
have anybody you are willing to trust, who could help> | >
you sort out which of the claims people are making are valid
and which> | > are baloney. So you tend automatically to
assume that the things> | > people say that seem wrong to you
are baloney of some kind or another.> |> | How can I take
people like you seriously when you have trouble with> | such
simple things as factor of?> Because I dont have
trouble with this, except with people whove> decided to
play Humpty Dumpty (and pay their words extra to mean what>
they want them to mean). If I said I had a teacher of
children, and I> expected people to understand me as saying
that there was someone who> taught both me and children, they
would have a problem with that too.> The problem would not
be with what has a teacher means.My usage resolves
ambiguity, and its telling that youre *still*trying to nd
a way to argue about it. For readers who didnt catchthe
fascinating exchange Ive had with several posters, a
highlight ismy pointing out that you can say 12 has a factor
of 21 as 3 is afactor of both 12 and 21. But posters have
used has a factor of tomean is a multiple of, so theyd
say 21 has a factor of 3, which iscorrect, but can lead to
ambiguity.For instance, consider the statement: x has a
factor of 12.That factor could be 2, 3, or 4. Or if I move to
another ring, itcould be 1+i.Simply because *certain* people
choose to read that as, x is amultiple of 12, does not change
the reality.> | > This also leaves you without any very good
way to learn how to improve> | > your mathematical
tinkering. We keep telling you different things> | > which
_we_ claim would help you avoid pitfalls like you keep
falling> | > into. But (a) you dont trust us enough to
believe weve identied a> | > problem with what youre
doing, and (b) you dont trust us to give you> | > straight
advice on how to avoid it; you suspect us of just trying to>
| > waste your time. So youre stuck with only your own
inklings of whats> | > a good way to work with proofs.> |>
| Yet Ive caught you in a contradiction with yourself, where
you> | apparently have been inserting the word multiple when
I say> | something like factor of 5, so that you read
multiples of 5. If> | thats not what youre doing then, who
knows whats going through that> | brain of yours.> I
believe I can explain well enough.Then do so.> | Readers
should consider what follows if they think thats too harsh.>
|> | > | Now then in what way is it improper terminology to
talk about> | > | algebraic integer factors of 5?> | | > I
didnt say it was improper to talk about algebraic integer
factors> | > of 5. That has a perfectly well-dened meaning.
An algebraic integer> | > r is a factor of 5 if it divides 5
in the algebraic integers, meaning> | > that 5/r is also an
algebraic integer.> |> | <deleted |> | Youre contradicting
yourself.> No, as I explained elsewhere. You made the leap
of assuming that has> a factor of should mean has a common
factor with. I have a teacher> of children does not mean
I share a teacher with some children.> 6 has a factor of
3 does not mean 6 and 3 share a factor.And for other
readers, consider dealing with several people like
KeithRamsay, who go on and on, refuse to work at getting to
the bottom ofthings, and when corrected, they simply go off
on a tangent.Im am *tired* of having to deal with
mathematicians.Theyre so damn irrational!!!> | Go back and
read over what you said at> | the top of this post. Im
tiring of this exercise in pointing out> | your errors.>
*You* think *youre* getting tired of correcting *my* errors?
Coming> from you thats a laugh.> Keith RamsayIsnt it
ironic, dont you think?A little too ironic, I really do
think.My tidbit from Alanis.James Harris
=James, go back
and learn the denitions used in mathematics. It is
evidentthat you are the only one not playing by the rules.
You have to dene yourterms CLEARLY and stick to those
denitions. Mathematicians are NOT at fault,its you. Your
ego is so big that you dont realize this.David Moran
>
[...]>> | Given that r_1 r_2 = 5, it makes sense that factors
of 5 distribute in>> | some way between r_1 and r_2.>>> I
would guess that you mean that there exist algebraic
integers>> dividing 5, which also divide r_1 and r_2, except
that thats a>> peculiar thing to ask. The obvious choice of
divisors of 5 to use>> as divisors of r_1 and r_2 are just
r_1 and r_2.>>> So is that what do you actually mean by
it?The issue that is raised at the start of the paper is a
question about>how one would know that given x^2 + x - 5 = (x
- r_1)(x - r_2) where>factors of 5 might be in r_1 and r_2, In
general a factor of 5 does not have to be in r_1 _or_ in
r_2.>in terms of the question of>whether or not either could
be a unit factor of 5.For those who wonder what a unit
factor is, its like how 1 is a>factor of 5, but in the ring
of algebraic integers there are an>innity of numbers that are
factors of 1 that are themselves not 1. >For instance
(1+sqrt(-3))/2 is a unit factor because it multiplies>times
(1-sqrt(-3))/2 to give 1, and both are algebraic integers.For
anyone still wondering, a unit factor is a factor which
isalso a unit. _In_ the present context unit means unit in
thealgebraic integers, which means algebraIc integer
whosereciprocal is also an algebraic integer.>It turns out
that mathematicians dont know how to prove whether or>not
r_1 and r_2 in the example I gave are unit factors. Thats
nonsense. Its _obvious_ to _me_, someone who knowsnothing
about such things, how to prove whether or not r_1and r_2 in
that example are unit factors! In fact they are _not_.Proof:
Since r_1 and r_2 are roots of x^2 + x - 5 its clear
that1/r_1 and 1/r_2 are roots of 1 + x - 5x^2. Thats an
irreduciblenon-monic polynomial, so 1/r_1 and 1/r_2 are not
algebraicintegers.This is a special case of the incredibly
obvious resultthat an algebraic integer is a unit if and only
if the_constant_ term in its minimial polynomial is plus
orminus 1.> However, if>you tell them that they may get upset
and start showering you with a>lot of mathematical statements,
which if you look carefully, dont>prove it, or disprove
it.Really? Actually if you look carefully you see that the
simple argument above _does_ prove it.>[...]For instance,
consider the statement: x has a factor of 12.That factor
could be 2, 3, or 4. Or if I move to another ring, it>could
be 1+i.Simply because *certain* people choose to read that
as, x is a>multiple of 12, does not change the
reality._exactly_ like the fact that certain people say
that integers arerational does not change the reality that
theyre irrational.agrees that youre using the language
incorrectly then you _are_using it incorrectly - what words
and phrases mean _is_ decidedby majority vote. And _nobody_
but you thinks that the statement15 has a factor of 12 is
true.What you _mean_ when you say 15 has a factor of 12 is
ofcourse true. But when you say 15 has a factor of 12
peopleare _not_ going to know what you mean, theyre going
toassume you mean what anyone else would mean by thesame
words (making the statement obviously false.)What do you
think you accomplish by inventing your ownprivate language
this way?>> | > [...]>>> No, as I explained elsewhere. You
made the leap of assuming that has>> a factor of should
mean has a common factor with. I have a teacher>> of
children does not mean I share a teacher with some
children.>> 6 has a factor of 3 does not mean 6 and 3
share a factor.And for other readers, consider dealing with
several people like Keith>Ramsay, who go on and on, refuse to
work at getting to the bottom of>things, and when corrected,
they simply go off on a tangent.Im am *tired* of having to
deal with mathematicians.Theyre so damn irrational!!!Then
why do you keep coming back for more dealings
withmathematicians?>> | Go back and read over what you said
at>> | the top of this post. Im tiring of this exercise in
pointing out>> | your errors.>>> *You* think *youre*
getting tired of correcting *my* errors? Coming>> from you
thats a laugh.>>> Keith RamsayIsnt it ironic, dont you
think?A little too ironic, I really do think.My tidbit
from Alanis.>James Harris************************David C.
Ullrich
In the past, Ive done the equivalent of
saying that 6 in the ring of> evens has 2 as a factor, which
is incorrect, as the implication is> that Im talking about
the ring of evens, when to be correct I have to> have
switched rings.> While 6 does have 2 as a factor in the
ring of integers, it does NOT> in the ring of evens.> What
Im doing now is correcting that mistaken usage, and Im not>
surprised that its taken some time and that many of you may
have> become confused.> Ah, good. Ill look forward to a
cleaned-up version of the proof, then.I assume you mean my
paper Advanced Polynomial Factorization, but asIve noted I
typically used coprime so that I didnt have to
usefactor.When I do use factor in the paper it follows
logically.> Heres basically my approach. I get an expression
like> g = r + fc> where g, r, f, and c are algebraic
integers.> Now I nd out that g is not coprime to f, and I
can separate off some> factor of f, which gives me> h = s +
c where h appears to be a factor of g, and s appears to be a
factor of> r. factor of f that is f, was separated off.>
Im a bit confused by this step... as I understand it, if g
is not> coprime to f, that means that they share *some*
factor, call it e,> not that g is a multiple of f. In other
words you should only be able> to say:> g/e = r/e + (f/e)c>
h = s + (f/e)c> rather than dividing by the whole of f as
you seem to have done above.But then its forced that f/e be
a unit, and I explain in detailbelow.> Now to the appears
part, as in checking r, I nd that r is NOT an> algebraic
integer!> This seems slightly weird, since r = g - fc, and
if g, f, and c are all> algebraic integers, r must be too.
One of the other three must also> turn out not be an
algebraic integer.But r, g, f and c are all algebraic
integers.It might help to expand out.I have a polynomial P(m)
where P(m) = g_1 g_2 g_3, and P(0) is a multiple of f, as it
has as afactor f^2, and also P(0)/f^2 is coprime to f.Further
I have that P(m) has a factor that is f^2, which is true
forall m.And I nd that when m=0, g_1 has a factor that is f.
Now myselection of indices is arbitrary, but I still nd that
there is oneand only one other of the gs that has a factor
that is f, when m=0,which means Ive now covered the factor
f^2. Notice here thatbasically Im saying that two and only
two of the gs can have nonunit factors of f, and each has a
factor that is f. The actualindices I use are arbitrary.Next
I consider P(m)/f^2, with P(m)/f^2 = h_1 h_2 h_3, where the
hsare factors of the gs.Now I also have for each h,
something like h_1 = s_1 + d_1, where d_1doesnt change as m
changes, which is just use of my lemma, and I knowthat d_1
must be coprime to f, as I know that P(0)/f^2 is coprime tof,
and d_1 is a factor of P(0).THAT is the basic irrefutable
point. Notice that with what you haveabove you end up with h
= s + (f/e)c, and unless f/e is a unit, youhave a
contradiction.James Harris
=[lots snipped]> Now I nd out
that g is not coprime to f, and I can separate off some>
factor of f, which gives me Im a bit confused by this
step... as I understand it, if g is not> coprime to f, that
means that they share *some* factor, call it e,> not that g
is a multiple of f. And I nd that when m=0, g_1 has a factor
that is f.Ah, theres the trouble. In the post I was
originally replying to, youindicated that g was not coprime
to f, but in the followup, you clariedthat g was an actual
multiple of f as well, so it is reasonable to divideoff the
whole of f.(Note, of course, that I have no idea whether any
of that is actuallytrue, but at least the particular quibble
I was replying to is settled.)
See what I mean? How many
of you thought better of mathematicians>> before you saw the
tricks they play?The more I read of this sort of thing, the
better I think of>mathematicians, and the worse I think of
you.> Rather than admit the truth--that my proofs are>>
correct--mathematicians play word games and debate me over
use of>> factor of because theyre deceitful.Their usage is
correct; yours is incorrect.consistently; and (b) give a
precise denition at the beginning, orat least when
confusions arise; then there would be littleproblems.
However, as has been noted elsewhere, he does not use
itconsistently either:In this thread, he said:My usage is
correct as any of you can demonstrate for yourselves by
noticing that 12 has a factor of 21, as 3 is a factor of
earlier, he had said: But Magidin, 9 does not have a factor
of
nothing inherently wrong with non-standard usage. It
isJamess insistence on UNDEFINED and EQUIVOCAL usage which
is
damning.
such a reasoner as Mr. Smith? I answer as a deceased friend
of mine used to answer on like occasions - A mans 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
gures few readers can critize. A great many people are
staggered to this extend, that they imagine there must be the
indenite 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
==
 
The issue that is raised at the start of the paper is a question
about> how one would know that given x^2 + x - 5 = (x -
r_1)(x - r_2) where> factors of 5 might be in r_1 and r_2, in
terms of the question of> whether or not either could be a
unit factor of 5.> For those who wonder what a unit
factor is, its like how 1 is a> factor of 5, but in the
ring of algebraic integers there are an> innity of numbers
that are factors of 1 that are themselves not 1.> For
instance (1+sqrt(-3))/2 is a unit factor because it
multiplies> times (1-sqrt(-3))/2 to give 1, and both are
algebraic integers.> It turns out that mathematicians dont
know how to prove whether or> not r_1 and r_2 in the example I
gave are unit factors. However, if> you tell them that they
may get upset and start showering you with a> lot of
mathematical statements, which if you look carefully, dont>
prove it, or disprove it.Depends on who is doing the
looking.If it is JSH, there are lots of things that he may
overlook that would beeasily seen by anyone who wants to see
them.For each algebraic integer there is a monic polynomial
with integercoefcients of minimal degree which is satised
by that algebraic integer,and that algebraic integer is a
unit in the ring of algebraic integers ifand only if the
constant term of that minimal polynomial is a unit in thering
of rational integers.
In the past, Ive done the
equivalent of saying that 6 in the ring of> evens has 2 as a
factor, which is incorrect, as the implication is> that Im
talking about the ring of evens, when to be correct I have
to> have switched rings.> While 6 does have 2 as a factor
in the ring of integers, it does NOT> in the ring of evens. What Im doing \
now is correcting that mistaken usage, and
Im not> surprised that its taken some time and that many of
you may have> become confused.> Ah, good. Ill look forward
to a cleaned-up version of the proof, then.> I assume you
mean my paper Advanced Polynomial Factorization, but as> Ive
noted I typically used coprime so that I didnt have to use>
factor.> When I do use factor in the paper it follows
logically.I emphasize that important point with this minor
correction to myearlier post, and the correction is below. >
Heres basically my approach. I get an expression like> g =
r + fc> where g, r, f, and c are algebraic integers.> Now
I nd out that g is not coprime to f, and I can separate off
some> factor of f, which gives me> h = s + c where h
appears to be a factor of g, and s appears to be a factor of>
r. factor of f that is f, was separated off.> Im a bit
confused by this step... as I understand it, if g is not>
coprime to f, that means that they share *some* factor, call
it e,> not that g is a multiple of f. In other words you
should only be able> to say:> g/e = r/e + (f/e)c> h = s +
(f/e)c> rather than dividing by the whole of f as you seem
to have done above.> But then its forced that f/e be a
unit, and I explain in detail> below.> Now to the appears
part, as in checking r, I nd that r is NOT an> algebraic
integer!> This seems slightly weird, since r = g - fc, and
if g, f, and c are all> algebraic integers, r must be too.
One of the other three must also> turn out not be an
algebraic integer.> But r, g, f and c are all algebraic
integers.> It might help to expand out.> I have a
polynomial P(m) where> P(m) = g_1 g_2 g_3, and P(0) is a
multiple of f, as it has as a> factor f^2, and also P(0)/f^2
is coprime to f.> Further I have that P(m) has a factor
that is f^2, which is true for> all m.> And I nd that when
m=0, g_1 has a factor that is f. Now my> selection of indices
is arbitrary, but I still nd that there is one> and only one
other of the gs that has a factor that is f, when m=0,> which
means Ive now covered the factor f^2. Notice here that>
basically Im saying that two and only two of the gs can
have non> unit factors of f, and each has a factor that is f.
The actual> indices I use are arbitrary.Notice everything is
still ok at m=0, but things get bizarre fornonzero m. > Next
I consider P(m)/f^2, with P(m)/f^2 = h_1 h_2 h_3, where the
hs> are factors of the gs.And thats where I have again
accidentally used the erroneous usagebecause only one of the
hs is an algebraic integer. That is,*reasonably* youd
suppose that the hs are all factors of the gs butonly one
of them can be in the ring of algebraic integers. Its
abizarre and fascinating result.That in the ring of
algebraic integers is very important as theproblem comes
because the denition of the ring leaves this hole Ivebeen
talking about which represents an error in taught
mathematics.> Now I also have for each h, something like h_1
= s_1 + d_1, where d_1> doesnt change as m changes, which is
just use of my lemma, and I know> that d_1 must be coprime to
f, as I know that P(0)/f^2 is coprime to> f, and d_1 is a
factor of P(0).Thats still correct.> THAT is the basic
irrefutable point. Notice that with what you have> above you
end up with h = s + (f/e)c, and unless f/e is a unit, you>
have a contradiction.And that is the irrefutable point, and
no amount of talking willchange that fact. My demonstration
of the hole is mathematicallycorrect and therefore
irrefutable, so mathematicians should quitplaying word games
and work to understand rather than confuse.Remember,
mathematics *currently* has this problem. So ifmathematicians
get to dance rather than x it, then students will belearning
bogus math that has clearly been shown to have a problem.So
why would mathematicians ght the truth? I suspect its
becausetheyve claimed perfection from errors like the one
Ive found. Ithink theyre ghting for that delusion of
perfection in theirdiscipline which has now been
shattered.James Harris
| Go back and read over what you
said at> | the top of this post. Im tiring of this exercise
in pointing out> | your errors.> *You* think *youre* getting
tired of correcting *my* errors? Coming> from you thats a
laugh.> Keith Ramsay> Isnt it ironic, dont you think?> A
little too ironic, I really do think.> My tidbit from
Alanis.As we all know, 90% of the examples in her song
arent actually irony,in the traditional sense. So in the
song, it seems that Ms Morissetteis using the word ironic
in a way that differs from its conventionaldenition.Now
*thats* ironic :)-- Dave TaylorGrrr! If anyone wants me,
Remember, mathematics *currently* has this problem.The
problem is named James S Harris.> So if> mathematicians get
to dance rather than x it, then students will be> learning
bogus math that has clearly been shown to have a problem.If
the students are learning Harrisonian math, it is mostly
bogus, and willhamper their futures in any eld requiring
correctness and accuract.But most mathematicians here would
rather laugh at Harris that attempt tox him.
<3ef47aaa$1_3@newsfeed>
My usage
resolves ambiguityAs an other reader (who JSH constantly
appeals to) may I commend thisas .sig. fodder?-- MinSo where
What I want
readers to understand is that the most reasonable> conclusion
to draw is that mathematicians are *deliberately* lying to>
them.And what everyone else wants the readers to understand
is how contumaciousis the character of the one calling
mathemeticians names.Fortunately, the great mass of readers,
even in the most non-mathematicallyoriented of newsgroups in
which Harris posts, quickly get the measure ofJames Steven
Harris, and thereafter read him for amusement rather
thanedication.
I have a polynomial P(m) where> P(m) =
g_1 g_2 g_3, and P(0) is a multiple of f, as it has as a>
factor f^2, and also P(0)/f^2 is coprime to f.> Further I
have that P(m) has a factor that is f^2, which is true for>
all m.> And I nd that when m=0, g_1 has a factor that is
f. Now my> selection of indices is arbitrary, but I still nd
that there is one> and only one other of the gs that has a
factor that is f, when m=0,> which means Ive now covered the
factor f^2. Notice here that> basically Im saying that two
and only two of the gs can have non> unit factors of f, and
each has a factor that is f. The actual> indices I use are
arbitrary.> Notice everything is still ok at m=0, but
things get bizarre for> nonzero m.> Next I consider
P(m)/f^2, with P(m)/f^2 = h_1 h_2 h_3, where the hs> are
factors of the gs.> And thats where I have again
accidentally used the erroneous usage> because only one of
the hs is an algebraic integer. That is,> *reasonably* youd
suppose that the hs are all factors of the gs but> only one
of them can be in the ring of algebraic integers. Its a>
bizarre and fascinating result.Youve lost me again.
According to the above, both g_1 and, say,g_2 are multiples
of f. So, dividing through by f^2, you can take: h_1 = g_1/f
h_2 = g_2/f h_3 = g_3and all three hs should thus be
algebraic integers by what youve said.
= > The issue that
is raised at the start of the paper is a question about > how
one would know that given x^2 + x - 5 = (x - r_1)(x - r_2)
where > factors of 5 might be in r_1 and r_2, in terms of the
question of > whether or not either could be a unit factor of
5.a unit factor. Pray look up that proof and respond. To
recap: When r_1 is a unit, its inverse (1/r_1) is an
algebraic integer, so it is a root of a monic irreducible
polynomial. It is the root of the irreducible polynomial 5x^2
- x - 1, so is not an algebraic integer.So, r1 is not a unit.
> It turns out that mathematicians dont know how to prove
whether or > not r_1 and r_2 in the example I gave are unit
factors. However, if > you tell them that they may get upset
and start showering you with a > lot of mathematical
statements, which if you look carefully, dont > prove it, or
disprove it.Eh? See above...-- dik t. winter, cwi, kruislaan
413, 1098 sj amsterdam, nederland, +31205924131home:
bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
>The problem is when you say x
has a factor of y. This is, in>>standard mathematical
terminology, the same as saying y is a factor>>of x; which
is the same, in standard mathematical terminology, as y>>is
a divisor of x and as x is a multiple of y.> I seem to
have lost track of exactly what the original statements that>
prompted all this discussion were, but I do recall seeing
things like:> x has a factor of 5 which is sqrt(5).>
Perhaps people are just parsing this differently. One way of
looking at> it is as x has a value which is a factor of 5,
that value being sqrt(5).> Its still too ambiguous to
say for certain, but thats one possible> interpretation
which would at least make sense and could explain why>
everyone insists theyre using the correct denition.> Of
course that just emphasizes the need for precise terminology
and usage...The problem is that JSH does not care to have a
constant meaning for histerminology; that would require him
to take responsibility for what hethe standard JSH meaning is
x is a multiple of sqrt(5). The reason heresists all these
attempts (at getting him to bring his terminology intoline
with standard usage) so assiduously is that it would
necessitatehis careful reading of every one of his
divisibility statements anddetermine which way the
divisibility goes, whats a multiple of what,and so forth.
JSH doesnt have the fortitude to subject his own work tosuch
scrutiny.Dale
=| As we all know, 90% of the examples in
her song arent actually irony,| in the traditional sense. So
in the song, it seems that Ms Morissette| is using the word
ironic in a way that differs from its conventional|
denition.People seem to be using ironic nowadays as if it
hada much broader meaning.Keith Ramsay
| As we all know,
90% of the examples in her song arent actually irony,> |
in the traditional sense. So in the song, it seems that Ms
Morissette> | is using the word ironic in a way that
differs from its conventional> | denition.> People seem to
be using ironic nowadays as if it had> a much broader
meaning.Check out
http://www.guardian.co.uk/weekend/story/0,3605,985375,00.
htmlfor all your ironic needs. > Keith Ramsay
The
issue that is raised at the start of the paper is a question
about> how one would know that given x^2 + x - 5 = (x -
r_1)(x - r_2) where> factors of 5 might be in r_1 and r_2,
in terms of the question of> whether or not either could be
a unit factor of 5.> a unit factor. Pray look up that proof
and respond. To recap:> When r_1 is a unit, its inverse
(1/r_1) is an algebraic integer, so> it is a root of a monic
irreducible polynomial. It is the root of> the irreducible
polynomial 5x^2 - x - 1, so is not an algebraic> integer.>
So, r1 is not a unit.Yes, if r_1 is a unit, but what if r_1
is coprime to 5 and NOT a unit?The problem is that the
denition of algebraic integers leaves thatpossibility open,
as you cant *prove* that neither r_1 nor r_2 can becoprime
to 5 with Galois Theory.So yeah, r_1 cant be a unit, but
that doesnt mean it cant becoprime to 5, do you understand?
> It turns out that mathematicians dont know how to prove
whether or> not r_1 and r_2 in the example I gave are unit
factors. However, if> you tell them that they may get upset
and start showering you with a> lot of mathematical
statements, which if you look carefully, dont> prove it,
or disprove it.> Eh? See above...OOPS!!! Its been difcult
for me to switch from the confusingusage, which Ive explained
is like talking about 2 and 6 in the ringof evens where 2 is
NOT a factor of 6, and later talking of 2 being afactor of 6,
as you think of the ring of integers.You are correct though
here that my statement is wrong in context asyou can
reasonably assume that Im talking about the ring of
algebraicintegers, as in the ring of algebraic integers its
true that neithercan be a unit. But that doesnt mean that
neither can be coprime to 5in the ring of algebraic
integers.Its a fascinating problem created by the denition
of algebraicintegers, rather neat in a way.James Harris
> I have a polynomial P(m) where> P(m) = g_1 g_2 g_3, and
P(0) is a multiple of f, as it has as a> factor f^2, and also
P(0)/f^2 is coprime to f.> Further I have that P(m) has a
factor that is f^2, which is true for> all m.> And I nd
that when m=0, g_1 has a factor that is f. Now my> selection
of indices is arbitrary, but I still nd that there is one>
and only one other of the gs that has a factor that is f,
when m=0,> which means Ive now covered the factor f^2.
Notice here that> basically Im saying that two and only two
of the gs can have non> unit factors of f, and each has a
factor that is f. The actual> indices I use are arbitrary.>
Notice everything is still ok at m=0, but things get bizarre
for> nonzero m.> Next I consider P(m)/f^2, with P(m)/f^2 =
h_1 h_2 h_3, where the hs> are factors of the gs.> And
thats where I have again accidentally used the erroneous
usage> because only one of the hs is an algebraic integer.
That is,> *reasonably* youd suppose that the hs are all
factors of the gs but> only one of them can be in the ring
of algebraic integers. Its a> bizarre and fascinating
result.> Youve lost me again. According to the above, both
g_1 and, say,> g_2 are multiples of f. So, dividing through by
f^2, you can take:> h_1 = g_1/f> h_2 = g_2/f> h_3 = g_3>
and all three hs should thus be algebraic integers by what
youve said.Yes, they all apparently *should* be but provably
arent, which is whytheres a problem.Notice that the
conclusion is forced by the argument that proves thatone of
the gs must be coprime to f as two of the gs, with m
notequal 0, apparently have f as a factor. I say apparently
because theydont in the ring of algebraic integers, so
youre suddenly forced outof the ring by the denition of
algebraic integer.However, despite that fact, its true that
one of the gs *must* becoprime to f in the ring of algebraic
integers, and the other factwith it means that *all* are
forced to be coprime to f in the ring ofalgebraic
integers.However, within my paper its reasonable to suppose
at that point thatyoure still in the ring, but it turns out
that using the result thatone of the gs and therefore one of
the as is coprime to f leads to acontradiction as described
often enough by posters.And thats a aw in mathematics
thats been there for over a hundredyears.James
Harris
=Note: issues of coprime have been dealt with. I
used an older denition which does not apply, so all those
issues will be deleted. My only issues at the moment are the
proof of the lemma and the notion of incomplete ring.
This is to focus on the math, not style.>Proof. Let x=0,
then g must be a factor of P(0), so at that point c g.>(1)
If when x does not equal 0, g=c, r=0.>(2) If when x does not
equal 0, g =/= c there must exist r which>>varies>with x, and
as r equals 0 when x equals 0 it is not coprime to>x.>Im not
sure step 2 makes sense. I dont see why it makes r
coprime>>to x.The argument isnt complicated but I simplied
that section rather>than deal with it, after I realized it
wasnt correct>That is, r is not coprime to x or contains a
unit factor of x.>With that unit factor things get messy as
Im sure some would jump on>that and claim that since r would
always have a unit factor of x, its>meaningless.>>If it gets
messy, then you should present it rather than force the
>>reader to try to slog through all of that on their own.
Skip simple >>things that are clear cut. Messy means not
obvious a lot of the time.Huh? What I found was that I
could use something simpler than I had>that kept it from
getting messy. I decided to go with simpler.>>You left a gap
in your proof that you didnt address.> That is a lie.>
The lemma asserts that any factor of a polynomial can be
written as > r + c> where r is always 0, or a changing
value, while c is constant.> Since for any polynomial
factor g, a value at 0 exists, which gives c,> then
obviously,> r = g - c> so it hardly need proving.> But
you assert a gap. It hardly seems possible that you could be>
confused on something so simple, so you are lying.Rather than get into an \
involved explanation with more room
for>confusion, I realized that I really was just using the
fact that any>factor of a polynomial can be split up between
whats constant and>whats varying.That is, r changes, but c
does not, as x varies.The proof of that is actually trivial
as all you do is take g at 0,>which gives you c, then r =
g-c.Given that c is constant, while g is not, obviously r is
not either.>>This I agree with.>Excellent! Then you just got
past the lemma!!!>Wrong, I just got past the statement of the
lemma. Not the proof.> You need a proof of that lemma?
Again you dont sound like a math> expert.> For polynomial
factors of a polynomial, like x+1 is a factor of x^2 +> 2x +
1, its easy to *look* at it, as you have x, which changes,
while> 1 does not.> I merely generalized beyond polynomial
factors, and you claimed a gap> in the proof, and now claim
to not be able to get past the proof of> the lemma!!!> Now
for readers, how can a person keep posting against me making
such> statements, getting continually refuted, but keep
coming?> Because mathematicians are *social* creatures. As
long as he feels> hes doing a job for the math society, then
hell probably keep at his> hatchet job, no matter how stupid
his statements, nor how many times I> show how stupid or
deceitful they are.> My point for all of you is that
mathematicians *will* work to confuse> against the truth, in
order to ght for their society.>Are you sure youre a
mathematician?>>Only if a masters degree counts.>Forget that
question as Ill admit I was quick on the gun.The point is
that the lemma is there so that I can talk
about>non-polynomial factors of a polynomial later. Thats
the point of it.>Then nish prooving the lemma.> So
youre asking me to prove that a factor of a polynomial can
be> split up as r+c, where r equals 0 and changes with the
polynomial> variable, for instance, for P(x) that variable is
x, while c is> constant?> I assert that the request shows
that you are either lying or are> incompetent.> If you are
incompetent then you fail the denition of mathematician,> as
a mathematician is dened to be a math expert.> If you are
not a mathematician I have little interest in trying to>
teach you basic mathematics.> What you have shown: r=g-cNo
problem.What you have claimed without adequate proof: r is
not coprime to x when x is not 0, g is not c.Until you prove
this claim, your lemma does not stand and you have
accomplished nothing. I dont care how easy you claim it is.
I dont see it. If its easy, do it.-- Will Twentyman
Youve lost me again. According to the above, both g_1 and,
say,> g_2 are multiples of f. So, dividing through by f^2,
you can take:> h_1 = g_1/f> h_2 = g_2/f> h_3 = g_3> and
all three hs should thus be algebraic integers by what
youve said.> Yes, they all apparently *should* be but
provably arent, which is why> theres a problem.> Notice
that the conclusion is forced by the argument that proves
that> one of the gs must be coprime to f as two of the gs,
with m not> equal 0, apparently have f as a factor. I say
apparently because they> dont in the ring of algebraic
integers, so youre suddenly forced out> of the ring by the
denition of algebraic integer.Wait a sec, so they *dont*
have f as a factor after all? Why did yousay they did,
then?This whole exercise strikes me as just magnifying a
subtle error inyour proof into a blatant error and then
claiming a contradiction.
=it seems that he has unique
denitions of to be or not to be. > Wait a sec, so they
*dont* have f as a factor after all? Why did you> say they
did, then?> This whole exercise strikes me as just
magnifying a subtle error in> your proof into a blatant error
and then claiming a contradiction.--Dec.2000 WAND Chairman
Paul ONeill, reelectedto Board.
Newsish?http://www.rand.org/publications/randreview/issues/rr
.12.00/http://members.tripod.com/~american_almanac
=If it
helps any, for a long time James has used (i) a has a factor
of bto mean (ii) a shares a factor with b (one suspects
that (i) comes to mean (ii) via somethinglike a has a factor
which is a factor of b.) People haveexplained many times that
(i) actually means that b isa factor of a, so he shouldnt say
(i) when he means (ii).Why he insists on using the terminology
the way he doesis as mysterious as various other aspects of
all this.Im putting this bit rst to highlight it:[...]>|
Consider that in the ring of algebraic integers, 5 has
algebraic>| integer factors, and given algebraic integers r_1
and r_2, where their>| product is 5, why are you acting as if
its so difcult to comprehend>| that there must be some
distribution of factors of 5?Im not acting as if Im having
any kind of difculty. Im saying>that *youre* having
difculty in realizing that there must be some>distribution
of factors of 5 is a meaninglessly vague phrase.
Youre>saying this as if to say, Shouldnt this be true?,
and not as if>you actually had proven it was true.By the
standard meanings of the terms, saying an algebraic integer
r_1>has a factor of 5 in the algebraic integers is just the
same as>saying its divisible by 5 in the algebraic integers.
Thats the same>as saying that theres an algebraic integer t
for which r_1=5t.There isnt, and I think you know there
isnt. Since neither r_1 nor>r_2 is divisible by 5, theres
no distribution of factors of 5>between them.I think you
use ill-dened terms because on some level youre aware>that
using only well-dened terms spoils your fun. Its
disillusioning.>It clear away the fog of obfuscation.
Suddenly youre back to reality,>and its the rather ordinary
reality that people keep telling you that>youre in, not the
amazing story you wish it was.Back to the top.| > [...]>| >
| What I can do is take out the use of the term in the paper,
and use>| > | factor in the ring of algebraic integers, and
note that will lead to>| > | a contradiction as the point of
the paper is that the ring is>| > | incomplete.>| > |>| > |
What I want to impress upon readers is that Im quite willing
to work>| > | with mathematicians to explain the mistake that
theyre teaching.>| | > If you want to show that a mistake is
being taught, you should quote it>| > specically.>|>|
Thats not very sensible if you expect that for a mistake to
be taught>| for any length of time itd have some subtlety,
unless youre also>| questioning the competence of
mathematicians.Quite to the contrary-- its because
mathematicians are generally>competent that one can
ordinarily say very specically what is wrong>when they make
mistakes.What makes a mathematical mistake subtle is not what
it takes to point>it out once its been recognized; its the
difculty in recognizing>it in the rst place.| It seems to
me that rather than deal with the important question,>| which
is whether or not Im correct, people often try to introduce>|
weird ad hoc conditions.You created an ad hoc rule that in
order to refute a claim of yours,>its not good enough for
someone else to prove that the conclusion is>wrong. You think
a proof of yours can show that a proof of a>mathematician is
wrong, but you appear unwilling to accept the>possibility
that a proof from a mathematician can just as well show>that
one of yours is wrong (i.e., not really a proof).| If you
disagree with that assessment, can you please explain why
you>| believe there should be something simple and short
enough for me to>| quote?In mathematics, we try to make each
*mathematical* statement have its>own well-dened meaning.
(There are of course nonmathematical claims>that we sometimes
make, which arent always precise.) If a sequence
of>well-dened mathematical statements is wrong, its wrong
because (at>least) one of the statements is wrong.The key is
making sure all the denitions are sound.Mathematics that is
vaguely enough written that it can be wrong>without any
individual statements in it being denitely wrong,
is>considered very bad. Its bad because an author hasnt
dened their>terms well enough.| > | The ring of algebraic
integers is incomplete, its easy to show, and>| > | Ive
shown it with my paper.>| > |>| > | If mathematicians are
having trouble understanding any part of it, Im>| > | able
to explain further.>| | > You have a big problem with
terminology. You use terms that arent>| > used by others in
the context in which you use them, such as counting>| >
factors of 5 in algebraic integers. I havent seen anything I
would>| > consider an adequate denition of those terms. You
dont explain>| > why your terms should be considered
relevant to the ones mathematicians>| > are using.>|>| That
statement possibly has something to do with what I now call
the>| preamble in the paper as it is there to help explain
context, but>| isnt part of the actual argument.>|>| Heres
what I say, copied from the paper (some editing for
format):>|>| To highlight the standard belief consider the
algebraic integers>| (-1-sqrt(21))/2 and (-1+sqrt(21))/2
which are roots of x^2 + x - 5.>|>| While you know that the
algebraic integer factors are themselves>| factors of 5, in
what way is each a factor of 5? Can either not have>| non
unit factors of 5? How do you know?>|>| There Im talking
about algebraic integers, so one can assume Im>| talking
about algebraic integer factors. Given that 5 has algebraic>|
integer factors, how is what I say nonstandard Keith
Ramsay?>|>| Id like you to carefully explain your assertion.
The question Im>| raising is the possibility that you
lied.No, I havent lied. You are not using have non unit
factors of 5 in>a standard way.If at this point in the
paper, you meant it in a completely standard>way, heres what
the meaning would be. Have a factor of 5 is>synonymous with
have 5 as a factor, or be divisible by 5. Also 5>is a
non-unit (so that part is redundant).Now obviously we all
know you dont mean that. Its possible you mean>something
like, shares non-unit factors with 5, meaning that
there>are non-unit algebraic integers that are factors of
both numbers. But>this is just guesswork. Any serious writer
of mathematics takes>responsibility for taking out the
guesswork, NOT forcing the reader to>keep inferring which of
various possible meanings is the intended one.The standard
usage is also inconsistent with the rest of the paper.>You
write, for example, ...proving that two of the as have a
factor>that is f at a point where you plainly have NOT shown
that f is a>factor of any of the as, in any of the usual ways
factor is>dened.Elsewhere, you refer to an algebraic
integer f coprime to x. This>has no standard meaning in the
context you use it, where x is a>variable. If you meant
coprime in some ring of polynomials, which is>closest to a
standard meaning, it would mean that there exists>polynomials
P and Q such that P*f+Q*x=1. But if f is an algebraic>integer,
thats possible (if and) only if f is a unit, with Q=0,
and>you also assume in the same sentence that f is a non
unit.| > I dont know what you think is a fair way for
things to work, but the>| > way things actually work, this
kind of problem with terminology will>| > absolutely prevent
you from making any headway.>|>| What will prevent me from
making headway is if mathematicians>| continually
lie.Mathematicians dont continually lie. What you keep
deciding are>lies are just disagreements with you.Perhaps
the biggest problem of all is that youve reached a
point>where you dont have anybody you are willing to trust,
who could help>you sort out which of the claims people are
making are valid and which>are baloney. So you tend
automatically to assume that the things>people say that seem
wrong to you are baloney of some kind or another.This also
leaves you without any very good way to learn how to
improve>your mathematical tinkering. We keep telling you
different things>which _we_ claim would help you avoid
pitfalls like you keep falling>into. But (a) you dont trust
us enough to believe weve identied a>problem with what
youre doing, and (b) you dont trust us to give you>straight
advice on how to avoid it; you suspect us of just trying
to>waste your time. So youre stuck with only your own
inklings of whats>a good way to work with proofs.| Now then
in what way is it improper terminology to talk about>|
algebraic integer factors of 5?I didnt say it was improper
to talk about algebraic integer factors>of 5. That has a
perfectly well-dened meaning. An algebraic integer>r is a
factor of 5 if it divides 5 in the algebraic integers,
meaning>that 5/r is also an algebraic integer.| > In an
experiment in communication, I once tried elaborating on such
a>| > concept for you. You usually write as though the number
of factors of>| > 5 in an algebraic number satised certain
axioms which are familiar>| > to me: being a rational number
v(x)>=0 associated with each algebraic>| > integer x <> 0,
where v(5)=1, v(xy)=v(x)+v(y) for any algebraic>| > integers
x<>0 and y<>0, and v(x+y)>=min{v(x),v(y)} for algebriac>| >
integers x<>0, y<>0, and x+y<>0. Assuming that we have such a
function>| > v, then we can show that v(r1)=0 and
v(r2)=v(r3)=1/2 for the roots>| > r1, r2, and r3 of your
cubic, if we take them in the right order.>| > I think this
is the best way to try to make sense of your argument>| > in
your advanced polynomial factorization thing.>| | >
Unfortunately, this known concept is not so directly related
to>| > divisibility in the algebraic integers, so even if you
did manage to>| > produce a denition on these lines, it
wouldnt show mathematicians>| > are doing anything wrong
when they make claims about divisibility in>| > the algebraic
integers as they teach about it in classrooms. Theyre>| >
just apples and oranges.>|>| There is no divisibility
argument within the paper, and I only even>| mention roots
when the roots are algebraic integers.I guess you think of
factorization as thoroughly unrelated to>divisibility.
Note P(m) has a factor that is f^2 has nothing to do>with
divisibility of P(m) by f^2, then. This just highlights how
far>your terminology departs from standard. Yes, I
automatically take []>is a factor of [] as a synonym for
[] divides [], because thats>the way it is with any normal
usage of the terms. Its only a mistake>if one is dealing with
someone like you who departs so far from>standard usage.| Ive
talked>| about x/y in discussing the error in taught
mathematics, but did make>| a post explaining that was when
an argument is considered in the eld>| of rationals.>|>|
Your statement falls at Keith Ramsay, and I think youre
just trying>| to sound good enough to fool people, rather
than trying to get to the>| truth.People have seldom had a
hard time getting at the mathematical truth>in these
discussions, whenever the actual mathematical question
has>been well-dened. (There was *once* a question about
polynomial>factorization over the algebraic integers which
required some effort,>but that was a rare exception.)The
problem is with getting well-dene claims. Take your
term>incomplete ring for example. Your denition of
incomplete ring>for example amounts essentially to a ring
in which one element doesnt>divide another one, but it
*should* divide it. You only imagine that>the difculties
created by using this meaninglessly vague term are>someone
elses fault.| What is clear is that the paper does NOT
operate over any elds, and>| depends only on ring
operations.You seem to have an odd idea of what one is
allowed to say after>having declared an argument to be in a
ring. The whole point is to>make all of your mathematical
statements be well-dened, not to obey>arbitrary
conventions.When you say that you are working in the
algebraic integers, that>provides us with a certain kind of
context, which allows us to>understand the meaning of certain
terms. For example, after youve>said that youre working in
the algebraic integers, if you say y>divides x, it means y
divides x in the algebraic integers.>Otherwise it might be
ambiguous what it meant, or mean something>different from
what you intended.Saying youre doing this does NOT mean that
it magically becomes>forbidden to refer to x/y. Certainly it
would be incorrect to assume>that x/y is dened in the
algebraic integers. But theres an obvious>meaning to the
expression x/y, so its legitimate to refer to it (just>so
long as one realizes that it isnt necessarily an algebraic
integer>itself). If were trying to gure out whether there
exists an>algebraic integer z with the property that x=yz, it
makes perfect>sense for us to consider rst the fact that
there exists one and only>one z that satises x=yz, and THEN
ask whether that z (which exists>in the algebraic numbers) is
a member of the algebraic integers (which>is the ring were
considering primarily). Saying that x/y is an>algebraic
integer means just the same thing as saying that there>exists
an algebraic integer z such that yz=x.| > Say we dene the
number of factors of 5 in an algebraic integer x to be>| >
the highest rational number r such that 5^r divides x in the
algebraic>| > integers, i.e. such that there exists an
algebraic integer y such that>| > 5^r*y = x. I think this
denition works (i.e., I think there is such a>| > highest
rational power for each algebraic number, although I
havent>| > tried to write out a proof). If we dene it that
way, though, then it>| > simply doesnt have one of the
properties I listed above.>|>| Your post isnt coherent
mathematically given what I say in the paper.>| Consider that
in the ring of algebraic integers, 5 has algebraic>| integer
factors, and given algebraic integers r_1 and r_2, where
their>| product is 5, why are you acting as if its so
difcult to comprehend>| that there must be some distribution
of factors of 5?Just to repeat, its not a difculty I have;
its your difculty in>saying what you actually mean by that.
And I dont think you have a>clear idea of what you mean by
that yourself.| For instance, both could have a factor that
is sqrt(5), or one could>| have a factor of 1+2i, while the
other had a factor of 1-2i.>|>| The question Im trying to
get the reader to explore is, given that>| they are roots of
this particular polynomial, could one of them only>| have
unit factors of 5?>|>| How do you know?If you were using the
terms in the standard way, this question would>be, How can
you tell when one algebraic integer divides another>algebraic
integer?If you actually dened your question, it would become
clear either>that its the same as the standard one (and that
the answer>mathematicians give is correct), or that its a
different question>from the standard one (so that even if you
get a different answer to>your question, its just irrelevant
to whether mathematicians are>being truthful). I think this
has a lot to do with why you simply keep>the question vague;
this kind of clarity would take the drama out of it.Keith
Ramsay
Youve
lost me again. According to the above, both g_1 and, say,>
g_2 are multiples of f. So, dividing through by f^2, you can
take:> h_1 = g_1/f> h_2 = g_2/f> h_3 = g_3> and all three
hs should thus be algebraic integers by what youve said.>
Yes, they all apparently *should* be but provably arent,
which is why> theres a problem.> Notice that the
conclusion is forced by the argument that proves that> one of
the gs must be coprime to f as two of the gs, with m not>
equal 0, apparently have f as a factor. I say apparently
because they> dont in the ring of algebraic integers, so
youre suddenly forced out> of the ring by the denition of
algebraic integer.> Wait a sec, so they *dont* have f as a
factor after all? Why did you> say they did, then?As Ive
explained before its like if youre talking about 2 and 6
inthe ring of evens, and later say that 6 has 2 as a factor.
Clearly,youve moved outside the ring. > This whole exercise
strikes me as just magnifying a subtle error in> your proof
into a blatant error and then claiming a contradiction.Then
prove that assertion.What I ask for should be simple which is
that mathematicians orposters *prove* their assertions
mathematically.If you cant then admit it. If you can, then
> and all three hs
should thus be algebraic integers by what youve said.>
Yes, they all apparently *should* be but provably arent,
which is why> theres a problem.Unless JSH has again made an
error, in which familiar case the problem isone that
everyone but JSH knows how to deal with.
Wait a sec, so
they *dont* have f as a factor after all? Why did you> say
they did, then?> As Ive explained before its like if
youre talking about 2 and 6 in> the ring of evens, and later
say that 6 has 2 as a factor. Clearly,> youve moved outside
the ring.Mathematically, you need to specify such things with
rigor when youbring them into a proof. The difference between
a factor in thealgebraic integers and a factor in some other
ring or eld is veryvital to keep straight, especially if the
whole POINT of your proos to highlight a hypothetical
deciency in the way the algebraicintegers are dened or
used.In other words, if you divide by something that isnt a
factor, youshouldnt act surprised when you suddenly drop out
of the ring. Thatsnot a problem with mathematics, its a
problem with what you just did.> This whole exercise strikes
me as just magnifying a subtle error in> your proof into a
blatant error and then claiming a contradiction.> Then
prove that assertion.> What I ask for should be simple
which is that mathematicians or> posters *prove* their
assertions mathematically.> If you cant then admit it. If
you can, then do so.Your proof is not well-specied. It is
unclear and ambiguous, asexemplied by your constant misuse
and reuse of terms like factor,coprime, polynomial,
variable, and constant. These are thesubtle errors to
which I refer.Whenever I or another poster points out a
problem with it, a longdiscussion ensues in which it is
ultimately revealed, as in thiscase, that what you *said* in
the proof and what you *meant* in theproof are two completely
different things.So, at this point, I am completely in the
dark as to what you reallymean by your proof, because I cant
really believe what you say in itaccurately reects what you
mean by it.If and when you learn to specify your ideas with
mathematical rigor,Ill have another look at it. Another
poster requested a proof and/orclarication of one of your
lemmas: that sounds like a good start.
I think Ive
gured out a way to show basically all of you, including>
people who think they dont know any math that mathematicians
have> been lying about my work. Its so trivial you *should*
wonder why> they thought they could get away with it.And its
worth reminding people of the start of this thread which
hasmore than a hundred posts in it.What I want to emphasize
to the sci.skeptic and alt.math.undergradreaders is that its
hard to see how mathematicians could not be lyinghere. > Here
goes.> My paper Advanced Polynomial Factorization depends
on considering a> factor of a polynomials that I call g.>
(Paper linked to at http://groups.msn.com/AmateurMath as
usual.)> 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 polynomials value changes,> while c does not.And why
would anyone argue?> Now you can consider all factors of a
given polynomial using gs, with> something like> g_1...g_k
= P(x)> where you have k factors. 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 Im 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.> For my example up above its easy, as with
g_1 = x + 1, x varies as x> varies, while 1 does not.> Now
thats enough that the proof in the paper is straightforward,
but> posters have argued with me anyway, with some trying to
argue over the> denition of polynomial, amazingly enough.
However, consider that the gs have an important feature,
which is> that when x=0, I have> g_1...g_k = P(0).Thats
very important. > 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 gs must> multiply to give the constant
term.> So then, maybe you still want to believe the
mathematicians and> question that I can write g = r + c.>
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), > which means that if you believe the mathematicians
then theyve> 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.Which is what should be sickening to *some* of you as
mathematicianschallenging me are actually challenging algebra
itself.But theyre stuck. The math is simple enough that if
they just followalgebra, then they have to admit Im right,
or at least not claim Imwrong.So as Ive said mathematicians
have been shown to be a social groupthat like many others
tries to manipulate truth and here I can showyou easily
because it IS math.Which is why the subject line is what it
is.> So why would mathematicians argue against such a simple
result?> 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, they either believe that
you will go along, if you knowtheyre lying, or that you are
incapable of catching them lying, evenwith basic algebra.>
That is, MOST of you will doubt algebra itself rather than
consider> that mathematicians, whom you probably dont even
personally know,> would lie.> So where does this 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.>
For instance, you may be saying, HEY, whats with the m
when you> had x before??!!!> Well, theres 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.Now think about it, if
mathematicians gure that they can confuse youon that point,
what does that tell you?> Finally, the weirder thing is that
one poster in particular got a lot> of mileage out of
questioning my nding 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> dened at that
point.Consider that passage carefully, as what Im
emphasizing to you isthat there are people who are lying to
you, and the evidence is rightin front of you.> Well thats
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, its of degree 1, so if you solve for the> roots,
youll get funky stuff.> Now when I was nding the proof of
FLT...remember the process took> some years...at times Id
talk of polynomials with respect to other> than m, but I
rened my discourse as my understanding improved.> However,
people arguing with me did not.And why wouldnt they? I gure
its because if they concedeanything, theyre afraid of being
forced to give the full truth--andthey hate the truth in this
case.> You may *choose* to believe that they did not because
they dont know> enough mathematics to follow, but were
talking about actual> mathematicians here.> Whats more
rational?> I say its more rational to suppose that they
*did* gure 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 wasnt so great.Which continues to
explain. And remember, *look* at this thread withover a
hundred posts in it.> For most of you, theres probably the
belief that theres some funky> higher math involved that
your pitiful brain cant follow or you> dont know about, as
you may suppose that mathematicians either> wouldnt lie, or
they wouldnt lie in such a dumb way where I could> catch
them so easily.> But consider whats in the balance:> 1.
I discovered a proof of Fermats Last Theorem thats more>
available to people in general than most of what
mathematicians have> been producing lately.> 2. Worse I did
so having said Id nd 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 Fermats Last Theorem.> And those are just
highlights as theres 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.And it
gets worse with each passing day.> EXCEPT, it looks like all
they have to do is either stay quiet, or> *claim* Im wrong. Many of you \
simply believe them, and question algebra
itself, which is> rather sad. Id think at least some of you
valued your educations.Yet I remember talking to one fellow
about this in a privatecommunication, and he basically told
me that he didnt trust what hedbeen taught at university.He
wasnt surprised at the treatment Ive faced from
mathematicians.> Others of you may gure it doesnt matter,
maybe because Western> civilization seems to be based on
lying anyway, and maybe you gure I> should just grow up,
accept that everybody lies and move on. And I> dont have to
talk about Enron or pedophile Catholic priests or things> in
that vein.> I mean, look at George W. Bush and Iraq. If
people can be *killed*> over lies, without consequences to
the liars, then whats with some> freaking stupid math?And
the shortsightedness of that is whats sickening. Many of
youmight *claim* care about life and liberty, as long as its
your own.But youll ght for the right to kill even old men
and pregnantwomen, if it suits you.How many of you care or
paid attention to reports from Iraq like thatof a doddering
old man on a cane, gunned down in the streets ofBaghdad by a
heavily armored American soldier?How many of you gave a damn
about the bombs that dropped on a wing ofa hospital in Iraq
for pregnant women?No, youd just care if it happened in your
town, at a local hospital. And you call that the American
way?But you still pray, now dont you? You still pray to a
God that youexpect to protect you and your own, while you
kill women, children andold men. > Good point.>
Mathematicians *are* a part of society after all. Why should
they> tell the truth now? Itd 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.> After all, thats so easy, now isnt it?>
Which is why you need to understand why I talk about
mathematicians> potentially being prosecuted. Liars dont
just stop because the gig> is up, as then, they wouldnt
necessarily be liars, then eh?> Itd be against their
*true* natures.But life is about truth and consequences.
Think of a world thatfaults America for its wrongs as like
seeing a boulder rolling downhill. Its the laws of physics
at work. Some of you seem to thinkthat the boulder can roll
up against gravity--like praying to God toprotect you despite
your lies--but you see, if you ever see a boulderrolling up a
mountain on this earth, youre upside down.James Harris
=  The issue that is raised at the start of the paper is a
question about > how one would know that given x^2 + x - 5
= (x - r_1)(x - r_2) where > factors of 5 might be in r_1
and r_2, in terms of the question of > whether or not
either could be a unit factor of 5. > a unit factor. Pray
look up that proof and respond. To recap: > When r_1 is a
unit, its inverse (1/r_1) is an algebraic integer, so > it
is a root of a monic irreducible polynomial. It is the root
of > the irreducible polynomial 5x^2 - x - 1, so is not an
algebraic > integer. > So, r1 is not a unit. > Yes, if r_1
is a unit, but what if r_1 is coprime to 5 and NOT a unit?If
r_1 is coprime to 5, there are elements x and y of the ring
in questionsuch that x.r_1 + y.5 = 1. (That is the denition
of being coprime.)When we multiply this equation by r_2, we
get: 5.x + 5.y.r2 = r2, or5(x + y.r2) = r2. So 5 is a divisor
of r_2. On the other hand, r_2 is adivisor of 5. This is only
possible if r_2 is a conjugate of 5, i.e. thereis a unit u
such that r_2 = u.5. But r_2.r_1 = 5, so u = 1/r_1,
meaningr_1 is a unit. Which is not possible. So r_1 is not
coprime to 5. > The problem is that the denition of
algebraic integers leaves that > possibility open, as you
cant *prove* that neither r_1 nor r_2 can be > coprime to 5
with Galois Theory.See the proof above. Galois Theory is not
needed for this. It is asimple application of the denitions.
> So yeah, r_1 cant be a unit, but that doesnt mean it cant
be > coprime to 5, do you understand?See above. > You are
correct though here that my statement is wrong in context as
> you can reasonably assume that Im talking about the ring
of algebraic > integers, as in the ring of algebraic integers
its true that neither > can be a unit. But that doesnt mean
that neither can be coprime to 5 > in the ring of algebraic
integers.See above. > Its a fascinating problem created by
the denition of algebraic > integers, rather neat in a way.I
see no problem.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025
1.0
What I want to emphasize to the sci.skeptic and
alt.math.undergrad> readers is that its hard to see how
mathematicians could not be lying> here.When something is so
obvious to others but is so hard to see for ones self,perhaps
one needs new glasses.
I think Ive gured out a way to
show basically all of you, including> people who think they
dont know any math that mathematicians have> been lying
about my work. Its so trivial you *should* wonder why> they
thought they could get away with it.> And its worth
reminding people of the start of this thread which has> more
than a hundred posts in it.Argumentum ad populum> What I want
to emphasize to the sci.skeptic and alt.math.undergrad>
readers is that its hard to see how mathematicians could not
be lying> here.> Here goes.> My paper Advanced Polynomial
Factorization depends on considering a> factor of a
polynomials that I call g.> (Paper linked to at
http://groups.msn.com/AmateurMath as usual.)> 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 polynomials value
changes,> while c does not.> And why would anyone argue?>
Now you can consider all factors of a given polynomial using
gs, with> something like> g_1...g_k = P(x)> where you
have k factors. 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 Im 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.> For my
example up above its easy, as with g_1 = x + 1, x varies as
x> varies, while 1 does not.> Now thats enough that the
proof in the paper is straightforward, but> posters have
argued with me anyway, with some trying to argue over the>
denition of polynomial, amazingly enough. However,
consider that the gs have an important feature, which is>
that when x=0, I have> g_1...g_k = P(0).> Thats very
important.> 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 gs must> multiply to give the constant term. So then, maybe \
you still want to believe the mathematicians
and> question that I can write g = r + c.> 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), > which means that
if you believe the mathematicians then theyve> 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.Argumentum ad nauseam>
Which is what should be sickening to *some* of you as
mathematicians> challenging me are actually challenging
algebra itself.> But theyre stuck. The math is simple
enough that if they just follow> algebra, then they have to
admit Im right, or at least not claim Im> wrong.> So as
Ive said mathematicians have been shown to be a social
group> that like many others tries to manipulate truth and
here I can show> you easily because it IS math.> Which is
why the subject line is what it is.> So why would
mathematicians argue against such a simple result?> 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, they either believe that you will
go along, if you know> theyre lying, or that you are
incapable of catching them lying, even> with basic algebra. That is, MOST of \
you will doubt algebra itself rather than
consider> that mathematicians, whom you probably dont even
personally know,> would lie.> So where does this 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.>
For instance, you may be saying, HEY, whats with the m
when you> had x before??!!!> Well, theres 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.> Now think about it, if
mathematicians gure that they can confuse you> on that
point, what does that tell you?> Finally, the weirder thing
is that one poster in particular got a lot> of mileage out of
questioning my nding 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> dened at that point. Consider that \
passage carefully, as what Im emphasizing to
you is> that there are people who are lying to you, and the
evidence is right> in front of you.> Well thats 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, its of
degree 1, so if you solve for the> roots, youll get funky
stuff.> Now when I was nding the proof of FLT...remember
the process took> some years...at times Id talk of
polynomials with respect to other> than m, but I rened my
discourse as my understanding improved.> However, people
arguing with me did not.> And why wouldnt they? I gure
its because if they concede> anything, theyre afraid of
being forced to give the full truth--and> they hate the truth
in this case.> You may *choose* to believe that they did not
because they dont know> enough mathematics to follow, but
were talking about actual> mathematicians here.> Whats
more rational?> I say its more rational to suppose that
they *did* gure 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 wasnt so great.> Which continues to
explain. And remember, *look* at this thread with> over a
hundred posts in it.> For most of you, theres probably the
belief that theres some funky> higher math involved that
your pitiful brain cant follow or you> dont know about, as
you may suppose that mathematicians either> wouldnt lie, or
they wouldnt lie in such a dumb way where I could> catch
them so easily.> But consider whats in the balance:> 1.
I discovered a proof of Fermats Last Theorem thats more>
available to people in general than most of what
mathematicians have> been producing lately.Argumentum ad
novitatem> 2. Worse I did so having said Id nd 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 Fermats Last Theorem.> And
those are just highlights as theres 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.> And it gets worse with each passing day.>
EXCEPT, it looks like all they have to do is either stay
quiet, or> *claim* Im wrong.> Many of you simply believe
them, and question algebra itself, which is> rather sad. Id
think at least some of you valued your educations.> Yet I
remember talking to one fellow about this in a private>
communication, and he basically told me that he didnt trust
what hed> been taught at university.> He wasnt surprised
at the treatment Ive faced from mathematicians.> Others of
you may gure it doesnt matter, maybe because Western>
civilization seems to be based on lying anyway, and maybe you
gure I> should just grow up, accept that everybody lies and
move on. And I> dont have to talk about Enron or pedophile
Catholic priests or things> in that vein.> I mean, look at
George W. Bush and Iraq. If people can be *killed*> over
lies, without consequences to the liars, then whats with
some> freaking stupid math?> And the shortsightedness of that
is whats sickening. Many of you> might *claim* care about
life and liberty, as long as its your own.> But youll
ght for the right to kill even old men and pregnant> women,
if it suits you.> How many of you care or paid attention to
reports from Iraq like that> of a doddering old man on a cane,
gunned down in the streets of> Baghdad by a heavily armored
American soldier?> How many of you gave a damn about the
bombs that dropped on a wing of> a hospital in Iraq for
pregnant women?> No, youd just care if it happened in your
town, at a local hospital. > And you call that the American
way?> But you still pray, now dont you? You still pray to
a God that you> expect to protect you and your own, while you
kill women, children and> old men.Ignoratio elenchi> Good
point.> Mathematicians *are* a part of society after all.
Why should they> tell the truth now? Itd 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.> After all, thats so easy, now
isnt it?> Which is why you need to understand why I talk
about mathematicians> potentially being prosecuted. Liars
dont just stop because the gig> is up, as then, they
wouldnt necessarily be liars, then eh?> Itd be against
their *true* natures.> But life is about truth and
consequences. Think of a world that> faults America for its
wrongs as like seeing a boulder rolling down> hill. Its the
laws of physics at work. Some of you seem to think> that the
boulder can roll up against gravity--like praying to God to>
protect you despite your lies--but you see, if you ever see a
boulder> rolling up a mountain on this earth, youre upside
down.> James Harris
=I doubt Ill have the time to
answer all of this at once. Im willingto return to other
parts later on, but it may be a few days before Ihave time
again.[...]| It turns out that mathematicians dont know how
to prove whether or| not r_1 and r_2 in the example I gave
are unit factors. However, if| you tell them that they may
get upset and start showering you with a| lot of mathematical
statements, which if you look carefully, dont| prove it, or
disprove it.On usenet, we can excuse people for accidentally
incorrectlyparaphrasing what other people have said, once in
a while, but this isridiculous.[...]| > One day as I was
boarding a bus, a woman asked the driver a question| > which
sounded to me exactly like where is the outside wall?| >| >
To the driver this was a bit of a problem. It wasnt a problem
with| > the drivers knowledge of geography. He actually knew
the answer to| > the womans question. The question wasnt
complicated. It was just a| > language problem. She wanted to
know where the pedestrian mall in| > downtown was, and once
the driver gured out that thats what she| > meant, there
was no problem answering her question.| >| > Thats what my
problem with you is like. As soon as you can make a| >
mathematical question clear, the answer is probably not going
to be| > hard to nd. There are tools for working this sort of
thing out.|| Yeah and how *long* did it take the driver?Not
very long.However, the woman wasnt refusing to answer
questions about what shemeant. When asked whether she meant
one thing or another thing, shedidnt declare it to be
irrelevant hair-splitting, designed to delayher arrival. She
repeated the question in different phrasings, ratherthan keep
reiterating the same phrasing with exasperated cries ofyoure
just not listening!. She didnt assert that her way of
sayingthings was correct, even if the rest of us who were
more familiar withthe language might all be in a habit of
saying it another way.She wasnt attempting to prove the
incompetence of the driver byshowing that he was incapable of
driving there. She didnt call him apathological liar bent on
concealing the true location of herdestination, and liable to
ruin civilization with his lies. She wasnot attempting to
introduce novel geographical concepts that shethought the
driver needed to know before he could *really* go there.She
wasnt trying to prove she was a superior driver to the
driver.She didnt say that we needed to take a different
route to get therethan the usual one. If she had done very
many of those things, wewould have had to leave before
answering the question.[...]| You see Keith Ramsay, in the
real world, people tend to try to get to| the bottom of
things quickly because they have a job to do, and value|
their time.Wow, thats insightful.| Now I think
mathematicians value their time like other| people, but the
twist is that I see you as working to lengthen the| time in
which the world is left ignorant of the truth, and thats|
wrong.I agree that that would be immoral. Youre assuming
again that notonly are you right, but that we also know
youre right, however.Are you sure?[...]| > When you consider
a factorization P(m) = (a1x + uf)(a2x + uf)(a3x + uf),| > you
dont correctly describe what a1, a2, and a3 are. You claim
they| > are algebraic integers, but since they vary with m
this cannot be| > precisely correct. You talk about what
values they have when m=0 and| > when m=1, for example. In
order to satisfy such an equation, they have| > to be
something like algebraic functions of m.|| Ive talked on
this issue using P(x) = x+1, in the ring of integers.|| Now
then, your objection is like saying that its wrong for me to
say| that P(x) is an integer, though the ring is declared to
be integers,| and x+1 will give an integer for any integer
x.|| Now notice that P(x) = x+1 is an algebraic function of
x, but it| *still* gives integers for any integer x.First, it
*is* incorrect to use P(x) is an integer to mean that
forevery x, that P(x) is an integer. Theres no practical
reason why youshould say the rst instead of the
second.Second, saying that each value of a1 is an algebraic
integer doesnttell me what specic type of thing having
algebraic integer valuesit is.I know you consider this
needlessly picky, but it is relevantto the manner in which
a1, a2, and a3 vary from point to point. Withoutmaking any
assumptions on them at all, the order in which the
threevalues are assigned to a1, a2, and a3 is completely
independent ateach point. Knowing that a1 is the one of the
three thatYou make an analogy with the following
situation:[...]| Consider the set {1,2,3}, and take a member
m from that set.|| It is true that *one* and only one member
of the set is even.|| Notice that I can get that logical
conclusion without labeling.|| However, I can *arbitrarily*
label m_1=1, m_2=2, and m_3=3, and it| doesnt change that
conclusion.|| Now I can just as easily label m_1=2, m_2=3,
m_3=1, and it doesnt| change that conclusion.|| That is,
its *still* true that only one of the ms is even!!!But the
cases arent parallel. Its always hard to know how to
respondwhen youre attempted to explain something by
analogy.Say we assume that a1, a2, and a3 are some kind of
function on thealgebraic integers, with values in the
algebraic integers. (Youhavent said this is so, but its
hard to see what else you might havein mind.) In a typical
such set of three functions, each of the threefunctions is
discontinuous everywhere. Im doubtful that this is thesort
of thing you have in mind.Later, you say that that g1 must
have that same factor (of f) ingeneral, where g1 is
associated with a1. This does not representany kind of plain
fact, if g1 is discontinuous everywhere.| What I want readers
to understand is that the most reasonable| conclusion to draw
is that mathematicians are *deliberately* lying to| them.I
propose that the most reasonable conclusion to draw is that
yourwriting is unclear and hard to read, and that youre just
egotisticalenough to assume that it needs no improvement, but
it clear as itstands.| > This mistake helps to make the rest
of the argument confusing. You| > refer to one of the as not
being coprime to m. In what ring? The ring| > has to contain
the as, but they appear to be functions of m.|| For a
nonzero value of m, it is *proven* that *one* of the as is|
coprime to f.|| Ive explained above simply how that is
possible without there being a| mistake.The usual meaning
associated with a is not coprime to m stillremains there
do not exist u and v (in the same ring as a and m) forwhich
au+vm=1. My question is not have you proven this or not,
butwhat do you mean by it.| > At the bottom, and I assume
someone else has pointed this out, you| > write out P(m) but
dont distribute the factor of f^2 ;-) in all of| > the
terms. You have (Im taking this from the posting quoting
your| > paper):| >| > (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1 +
mf^2 )xu^2 + u^3|| And for the readers, after that youd
have|| 65x^3 - 12x + 1.|| > but it should be| >| > (m^3 f^6 -
3m^2 f^4 + 3mf^2)x^3 - 3( - 1 + mf^2 )xu^2f^2 + u^3f^3|| Which
is a false statement. Notice that Keith Ramsay made a change,|
and below he talks about *his* change as not working as if
its my| problem.I thought that you were rewriting P, and I
think thats a fairlynatural misunderstanding on my part,
given the context. You have anargument allegedly showing
something about P, and then at the end youcome out with a
polynomial thats looks quite a bit like P, and youwrite as
if the result applied equally well to it too. Now letting
m=1, f=sqrt(5), where I can let u=1 as its value doesnt
change the as, I have (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1
+ mf^2 )xu^2 + u^3 = 65x^3 - 12x + 1But with those
substitutions m^3 f^6 - 3m^2 f^4 + 3m = 1*5^3-3*5^2+3*1= 53
<> 65. So something is wrong. It sure looks like you
didntdistribute the factor of f^2 everywhere it belongs. I
think you couldstate more clearly what the connection between
P and this otherexpression is supposed to be, and why the
result about P implies theresult about this expression.[...]|
> I think frustration with you over the way you respond to
people who| > nd mistakes of yours is one of the main
reasons why people sometimes| > decide to try something
different, like writing a proof that the| > conclusion is
false. But also I would say that its often more fun,| > even
when dealing with our favorite posters, to go and write up my
own| > proof of a result related to what they are writing,
than it is to just| > react to the way theyve done it.|| I
want readers to consider the negative implication and
condescending| tone.I dont think youre in any position to
complain about negative andcondescending language, given
how much you produce yourself.[...]| > Well, nice to see we
agree. And now that youve identied the| > denition of
ring of algebraic integers as the culprit, really| > youve
done essentially what Ive asked-- we can quote this
denition| > specically. Unfortunately (or fortunately), we
still have no good| > reason to think that the denition
could create a contradiction.|| Stated without proof.I dont
need a proof. Im just stating the obvious absence of good
reasonto believe it.The reason you offer for thinking this is
true is poor, for starters,because it requires that we assume
the contradiction you reach is dueto the culprit youve
identied, and not something else. Why should Iconsider it
more likely that the contradiction has come from
thatdenition, rather than something wrong with the proof of
thefundamental theorem of algebra, for example? Perhaps the
rules oogic being applied are invalid under some conditions.
Obviously Idont think this is the problem, but compared to
the possibility ofthe denition being the problem, these are
not so much less likely.Giving a new (not previously used)
name to a property doesnt createa contradiction. Thats all
the denition of algebraic integerdoes. It allow us to use
that name instead of using the wholedenition in each place
where we say algebraic integer. If acontradiction can arise
while using the term, substitute thedenition for the term in
each place where it appears, and you get away to get the same
contradiction without using the term itself.| > | >
Mathematics that is vaguely enough written that it can be
wrong| > | > without any individual statements in it being
denitely wrong, is| > | > considered very bad. Its bad
because an author hasnt dened their| > | > terms well
enough.| > || > | The term bad is a human term inapplicable
in context as mathematics| > | is about truth.| > || > | That
is, a proof is neither bad nor good, it is true or false.|
>| > This is naive. Realistically, one has to be concerned
with the quality| > of a proof as well as with whether its
simply true or false.|| Stated without proof.Who cares?
Im simply pointing out the obvious.| > Getting rid of all
errors is quite hard. So in the middle of working| > on a
proof, the usual situation is to be working on a faulty
sketch of| > a proof. There is a huge difference between a
proof that is false,| > but is written *well*, and has good
parts in it, so that correcting it| > is easier, and one that
is false and written *badly*, so that its| > hard even to
see where mistakes might be.|| A proof cannot be false, by
denition.So when you said it [a proof] is true or false
you were just gettingready to play word games with me,
eh?[...]| > | Hmmm...thats an interesting point. What has
happened at that point| > | in the paper is that Ive
considered the constant term P(0), and the| > | constant term
with f^2 separated off, which is P(0)/f^2 = 3x u^2 + u^3| > |
f, and noted that it is coprime to f.| >| > Just to be clear,
why is 3x+u coprime to f?|| Thats just bizarre Keith Ramsay.
The full expression is|| 3x u^2 + u^3 f|| and f is coprime to
3, x and u, so its coprime to that expression,| which is
what I said.|| So why did you ask about 3x + u?Sorry, I
simply missed the f which had wrapped to the next line.
Itlooked like 3xu^2+u^3 = u^2(3x+u).Of course, given the way
youve been, I expect you to conclude that itwasnt an honest
mistake, but rather that I was trying to lie aboutyour
wonderful result again.| > | Now I then consider g_1 at m=0,
where c=g_1, and notice it has a| > | factor of f,| >| >
Assuming u is an algebraic integer.|| The ring has already
been declared to be the ring of algebraic| integers.So you
claim Im speaking redundantly. I thought it deserved
emphasis.| For other readers, you may be wondering why
mathematicians are so| clearly lying.|| Good question.Notice
that what just happened is I said something that you
ndredundant, but you decided to consider it clearly to be a
lie.This is pretty typical of your discoveries of
lies.[...]| > | and then based on P(0)/f^2 being coprime to
f, I have| > | that r + c, must have a factor of f, which
proves that r must have a| > | factor of f that is f.| >| > I
guess you mean g_1=r+c as in the lemma. What makes you think
that| > f divides r+c?|| Its shown in the paper.
Specically, c is constant, so as g_1| changes, there must
exist r = g_1 - c, and r must change.Im looking at the copy
quoted in that other guys posting. I dontsee how in what
follows the fact that r changes (without saying muchabout
*how* r changes) will help.| Its determined that g_1 has a
factor of f that is f at m=0.Because its uf when m=0.| Its|
further determined that when m does not equal 0, separating
off f^2| from P(m) leaves a constant term that is coprime to
f;Usually by constant term of a polynomial, one means the
same thingas the value of the polynomial when its variable is
set to 0. Sowhats the signicance of saying m is not equal to
0?| therefore, a| factor that is f must separate off from c,
and the result is coprime| to f.I thought you had c=uf, and u
was coprime to f just by assumption.| And yes, other readers,
its a lot easier to look over that section in| the paper
where you have a lot more information in front of you, than|
guring it out from that explanation.Indeed, youre not being
much less cryptic here than you are in thepaper. It would be
nice to get to the truth faster, but if youregoing to
consider this kind of thing to be an adequate
explanationfor a step.| > | > Elsewhere, you refer to an
algebraic integer f coprime to x. This| > | > has no
standard meaning in the context you use it, where x is a| > |
> variable. If you meant coprime in some ring of polynomials,
which is| > | > closest to a standard meaning, it would mean
that there exists| > | > polynomials P and Q such that
P*f+Q*x=1. But if f is an algebraic| > | > integer, thats
possible (if and) only if f is a unit, with Q=0, and| > | >
you also assume in the same sentence that f is a non unit.|
> || > | But f while a variable is constant, and so is x. So
it is not in any| > | ring of polynomials.| > || > |
Unfortunately, you seem to be xated on the *letters* as in
seeing an| > | x you may assume that its varying. Nope.
Its constant.| >| > Says who? Taking it to be a constant is
not only confusing, but| > inconsistent with the rest of the
paper.|| That is false. The key variable that changes in the
paper is m.|| > When you introduce P(m), you remark that the
new variables provide| > *additional* degrees of freedom. If
x is a constant, along with| > a1, a2, and a3, whats degree
of freedom did you start out with?|| The pertinent example is
x^2 + 4x + 4, like from before, as you get an| additional
degree of freedom from the use of x here, which allows you|
to talk about a family of values, versus just 9.|| The point
of algebra is that using variables allows for conclusions to|
be drawn about large classes of numbers rather than forcing
you to| check each number.Okay, I see why you call it a
degree of freedom.| Without algebra, a person can gure out
that 3(3)=9. With algebra,| and an additional degree of
freedom, you have that| (x+2)(x+2)=x^2+4x+4, which also tell
you about 16, 49, and 64 along| with an innity of other
numbers.|| > When you refer to the factorization of P(m)
into linear terms| > (a1x+uf)(a2x+uf)(a3x+uf), thats
consistent with x being a variable.| > If x is a constant,
theres no uniqueness in such a factorization (and| > not
much that you can prove about the values of a1, a2, and a3).
In| > the next step you infer that the coefcient of x^3 on
the right hand| > side (a1a2a3) is the same as the coefcient
of x^3 on the left hand| > side. This also is consistent with
x being a variable. If two| > polynomials are equal for all
values of x, then their corresponding| > coefcients have to
be equal. But if x is a constant, that step is| > invalid.||
Thats false.What you write in the following doesnt show
that the step is valid(with x a constant). Youre not really
defending the step at all--youre just describing what sort
of confusion you assume Im sufferingfrom to cause me to
disagree with you. Well stay stuck on this pointso long as
this remains your attitude.| Consider again x^2 + 4x + 4 =
(x+2)(x+2), as that is| true whether or not you consider x
variable or not.|| The *factorization* is whats important,
and it exists without regard| to whether or not you call P(x)
= x^2 + 4x + 4 a polynomial, or just| have x^2 + 4x + 4 with
x=2.This doesnt explain why the factorization can be said to
beunique. Obviously youve picked this particular
factorization outbecause its the one which holds *for all
values of x*. But if x isconstant, an equation such as
x^2+4x+4 = (x+a)(x+b) means only thatthe values of the two
sides for that one specic value of x are thesame. Thats not
a strong enough condition to force the factorizationto be
unique. For each value of x, there are many
differentfactorizations of this form.The fact that you called
the factorization above unique indicatesthat youre
interested in the specic one which holds not just forone
value of x, but the one which holds for all values of x.| For
readers that particular falsehood is an important one to
consider| carefully because its one that I suggest to you is
hard to explain as| anything other than a willful
falsehood--that is, a lie.Not hard for anybody but you.
Frankly, it seems youve given upany attempt to explain our
disagreements with you with any otherexplanation than that
were lying. It used to be that you had anassortment of
self-serving explanations of other kinds, which hadat least a
smidgeon more plausibility to them.In this particular case, it
would be a lie for me to claim that fora *constant* x, the
factorization (x+a)(x+b) = x^2+4x+4 is unique.For a constant
x, each way of factoring the right hand side (whichis also a
constant) gives another pair of a and b which satisfy
theequation for that x.| > Referring to f being coprime to x
isnt a tipoff that x isnt a| > variable, as you also
consider whether something is coprime to m,| > and m is
denitely being varied.|| While it is true that m is varied,
its also true that you can| consider a family of values for
m that are each coprime to f.|| That condition states that
any particular m is coprime to f.Then the domain of m is not
the whole ring you started with, but asubset of it? If this
is what you mean to say, then youve used amisleading way to
indicate it in the paper.[...]| > And its still unclear why
two of the as are supposed to be| > divisible by f.|| The
conclusion is that one of the as is coprime to f.In the last
paragraph, ...proving that two of the as have a factorthat
is f.| > | Thats troubling Keith Ramsay as thats basic
algebra.| >| > The problem is not with the algebra. The
problem is with the writing.|| Stated without proof.But true
nonetheless.[...]| > It seems, however, as though you want to
say now that whenever I make| > such an assumption (which you
decide to declare incorrect) its my| > fault. (And a fault
in algebra, too.) So really, I should assume| > nothing.||
You go with whats in the paper.But the paper leaves out
things like saying that x is constant.| > But if I did this,
you have not in your wildest dreams imagined the| > amount of
nitpicking clarication Id be asking you to do. You would| >
at the very least have to say what kind of variable each
variable you| > use is.|| Sounds like a lawyer tactic.If the
paper were able to stand completely on its own, without
anyimplicit assumptions at all, it would be able to withstand
lawyerlyscrutiny. But it cant. You complain both when I
make a reasonableassumption, like that x is being used as a
variable, and complain whenI refrain from doing so but
instead ask that you state clearly whichyou mean. You dont
get to have it both ways.| > Meanwhile, you also complain
when people keep pestering you to clarify| > details like
that. You imply theyre doing it only as a distraction| >
from the real mathematical content.|| Posters have asked
questions answered in the paper.So what? There are also
questions that are incorrectly answered in thepaper.| Stick
with the paper.It hardly answers all reasonable questions.
The most important part ot is terribly cryptic: Now as noted
before in general P(m) has a factor that is f^2 , and
separating that factor off, gives a constant term coprime to
f; therefore, given g1 = a1x + uf > where with m = 0, g1
gives a factor of f it must have that same factor > in
general, proving that two of the as have a factor that is
f.As usual, you cram the part of the reasoning where your
story divergesfrom what is usually believed into a short and
very confusing spot.Expand the explanation of this spot, or
nobody will get what you mean.| > You cant have it both
ways. Either you agree to some reasonable| > reading between
the lines, or you write your paper in such a way| > that
reading between the lines isnt needed.|| Which begs the
question of your status as a math expert, as the| denition
of mathematician is math expert.I wonder whether you know the
meaning of begs the question. Gettinga PhD in mathematics
doesnt make a person able to read minds. Theresno valid way
to just realize that in your paper you intended for xto be a
constant.[...]| > No, its far from clear, and the detail is
still poor at just the| > place in the argument where it
needs to be best, right near the end.|| Then why didnt you
ask more questions about that section in this| post?Well, as
usual Im in a habit of preferring to x up the earliersteps
in an argument before going on to the ones depending on
them.Partly, its hard to see what question to ask besides
why?.[...]| > Because I dont have trouble with this,
except with people whove| > decided to play Humpty Dumpty
(and pay their words extra to mean what| > they want them to
mean). If I said I had a teacher of children, and I| >
expected people to understand me as saying that there was
someone who| > taught both me and children, they would have
a problem with that too.| > The problem would not be with
what has a teacher means.|| My usage resolves ambiguity, and
its telling that youre *still*| trying to nd a way to argue
about it. For readers who didnt catch| the fascinating
exchange Ive had with several posters, a highlight is| my
pointing out that you can say 12 has a factor of 21 as 3 is
a| factor of both 12 and 21.Your usage is confusing to
everybody but you. It would help if youadmitted it. Its not
consistent with the way other phrases are used.| But posters
have used has a factor of to| mean is a multiple of,Not
just posters. This is just what it normally means.| so theyd
say 21 has a factor of 3, which is| correct, but can lead to
ambiguity.No, it doesnt. It could lead to an ambiguity if a
person were likeyou, and already in a habit of thinking like
this:| For instance, consider the statement: x has a factor
of 12.|| That factor could be 2, 3, or 4. Or if I move to
another ring, it| could be 1+i.Saying x has 2 *as a factor*
is ne, and so would be saying that xhas a factor of 12 as a
factor. But saying x has 2, without statingthat it has it as
a factor, is meaningless, and so would x has afactor of 12
except that factor of 12 is used to mean either anumber
dividing 12, or 12 itself used as a factor, depending on
thecontext.| Simply because *certain* people choose to read
that as, x is a| multiple of 12, does not change the
reality.I dont know what you think determines linguistic
reality, but itscertainly not just your own intuitions as
to what things should mean.[...]| And for other readers,
consider dealing with several people like Keith| Ramsay, who
go on and on, refuse to work at getting to the bottom of|
things, and when corrected, they simply go off on a
tangent.Bullshit. You just imagine that things I say that you
dont understandare irrelevant.| Im am *tired* of having to
deal with mathematicians.|| Theyre so damn irrational!!!Well
at least youre coming up with an alternative explanation to
theidea that Im just completely immoral.If you would take
the trouble to write your proof in the way thatevery
undergraduate math major learns how to write proofs, wed
havegotten to the bottom of this already. In the many years
youve beenposting, youve had more than plenty of time to
learn how. But youthink youre above doing it that way, which
is why we end up withthese curiously prolonged
discussions.These undergraduates would know, that to be sure
of what were doingwe have to build our way up from simple
stuff. For instance, beforeusing things like if x and y are
both coprime to z, then xy is coprime to zin the algebraic
integers, we should either prove that its true, ornd a
proof someone else has done, or maybe nd a place where
itsstated that its true, by someone we can trust. Now, do
you actuallyknow that this is true? Did you make or see a
proof? You use it. Youshouldnt use it based on guesses.Now
if you knew that the hairiest steps of your argument followed
fromprinciples like that, and knew how to prove them if
necessary, thenyou could unroll the proof around there, and
wed get somewhere.Keith Ramsay
I think Ive gured out a
way to show basically all of you, including> people who think
they dont know any math that mathematicians have> been lying
about my work. Its so trivial you *should* wonder why> they
thought they could get away with it.> And its worth
reminding people of the start of this thread which has> more
than a hundred posts in it.> Argumentum ad populumThe
assertion is that Im making a fallacious appeal to the
gallery.That is, Ive concluded that its worth reminding
people of how Istarted the thread, which now has more than a
hundred posts in it.So I guess Im to understand that the
poster Brett disagrees with myconclusion that its worth so
reminding, seeing it as being based on afallacious appeal to
the crowd.I disagree. > What I want to emphasize to the
sci.skeptic and alt.math.undergrad> readers is that its hard
to see how mathematicians could not be lying> here. Here
goes.> My paper Advanced Polynomial Factorization depends
on considering a> factor of a polynomials that I call g.>
(Paper linked to at http://groups.msn.com/AmateurMath as
usual.)> 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 polynomials value changes,> while c does not.> And
why would anyone argue?> Now you can consider all factors
of a given polynomial using gs, with> something like>
g_1...g_k = P(x)> where you have k factors. 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 Im
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.> For my example up above
its easy, as with g_1 = x + 1, x varies as x> varies, while
1 does not.> Now thats enough that the proof in the paper
is straightforward, but> posters have argued with me anyway,
with some trying to argue over the> denition of
polynomial, amazingly enough. However, consider that the
gs have an important feature, which is> that when x=0, I
have> g_1...g_k = P(0).> Thats very important.> 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 gs must> multiply to give the constant term.> So then,
maybe you still want to believe the mathematicians and>
question that I can write g = r + c.> 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), > which means that
if you believe the mathematicians then theyve> 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.> Argumentum ad
nauseamThe assertion here is that Im simply being repetitive
which impliesthat I dont support my argument. However,
readers can see themathematics itself.Theres no support for
the claim that my conclusion is based onrepetition as the
details above havent been given before, and mostimportantly,
the mathematics is correct.That is, its fallacious to claim a
true conclusion is wrong, even ifthe argument for it *has*
been repeated.After all, if youve explained something to a
child, and they havedifculty understanding, its hardly a
logical fallacy to explainagain.> Which is what should be
sickening to *some* of you as mathematicians> challenging me
are actually challenging algebra itself.> But theyre
stuck. The math is simple enough that if they just follow>
algebra, then they have to admit Im right, or at least not
claim Im> wrong.> So as Ive said mathematicians have been
shown to be a social group> that like many others tries to
manipulate truth and here I can show> you easily because it
IS math.> Which is why the subject line is what it is.> So
why would mathematicians argue against such a simple result? 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, they either believe that you
will go along, if you know> theyre lying, or that you are
incapable of catching them lying, even> with basic algebra. That is, MOST of \
you will doubt algebra itself rather than
consider> that mathematicians, whom you probably dont even
personally know,> would lie.> So where does this 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.>
For instance, you may be saying, HEY, whats with the m
when you> had x before??!!!> Well, theres 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.> Now think about it, if
mathematicians gure that they can confuse you> on that
point, what does that tell you?> Finally, the weirder thing
is that one poster in particular got a lot> of mileage out of
questioning my nding 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> dened at that point.>
Consider that passage carefully, as what Im emphasizing to
you is> that there are people who are lying to you, and the
evidence is right> in front of you.> Well thats 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, its of
degree 1, so if you solve for the> roots, youll get funky
stuff.> Now when I was nding the proof of FLT...remember
the process took> some years...at times Id talk of
polynomials with respect to other> than m, but I rened my
discourse as my understanding improved.> However, people
arguing with me did not.> And why wouldnt they? I gure
its because if they concede> anything, theyre afraid of
being forced to give the full truth--and> they hate the truth
in this case.> You may *choose* to believe that they did not
because they dont know> enough mathematics to follow, but
were talking about actual> mathematicians here.> Whats
more rational?> I say its more rational to suppose that
they *did* gure 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 wasnt so great.> Which continues to
explain. And remember, *look* at this thread with> over a
hundred posts in it.> For most of you, theres probably the
belief that theres some funky> higher math involved that
your pitiful brain cant follow or you> dont know about, as
you may suppose that mathematicians either> wouldnt lie, or
they wouldnt lie in such a dumb way where I could> catch
them so easily.> But consider whats in the balance:> 1.
I discovered a proof of Fermats Last Theorem thats more>
available to people in general than most of what
mathematicians have> been producing lately.> Argumentum ad
novitatemHere the assertion is that Im claiming that the
newness of my work iswhats important. But there cant be a
logical fallacy as theres noconclusion at this point as Im
building an argument.Besides based on whats stated its not
newness, but accessibilitythat Im stressing.That is, the
argument being given isnt based on the newness of theresult,
but at least partly on the assertion of greater availabilityto
people in general.> 2. Worse I did so having said Id nd 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 Fermats Last Theorem.>
And those are just highlights as theres 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.> And it gets worse with each
passing day.> EXCEPT, it looks like all they have to do is
either stay quiet, or> *claim* Im wrong.> Many of you
simply believe them, and question algebra itself, which is>
rather sad. Id think at least some of you valued your
educations.> Yet I remember talking to one fellow about
this in a private> communication, and he basically told me
that he didnt trust what hed> been taught at university.>
He wasnt surprised at the treatment Ive faced from
mathematicians.> Others of you may gure it doesnt matter,
maybe because Western> civilization seems to be based on lying
anyway, and maybe you gure I> should just grow up, accept
that everybody lies and move on. And I> dont have to talk
about Enron or pedophile Catholic priests or things> in that
vein.> I mean, look at George W. Bush and Iraq. If people
can be *killed*> over lies, without consequences to the
liars, then whats with some> freaking stupid math?> And
the shortsightedness of that is whats sickening. Many of
you> might *claim* care about life and liberty, as long as
its your own.> But youll ght for the right to kill even
old men and pregnant> women, if it suits you.> How many of
you care or paid attention to reports from Iraq like that> of
a doddering old man on a cane, gunned down in the streets of>
Baghdad by a heavily armored American soldier?> How many of
you gave a damn about the bombs that dropped on a wing of> a
hospital in Iraq for pregnant women?> No, youd just care
if it happened in your town, at a local hospital. > And you
call that the American way?> But you still pray, now dont
you? You still pray to a God that you> expect to protect you
and your own, while you kill women, children and> old men.>
Ignoratio elenchiThe assertion is that my conclusion is
irrelevant.However, George W. Bush and many Americans have
made a point ofemphasizing religion in politics, and
specically--prayer.Here Im pointing out a fundamental
contradiction in people claimingto be religious and claiming
God is on their side, or requesting aidfrom a Supreme Being,
when what theyre doing is antithetical to mostpeoples idea
of God.Its like a crook praying to God to help him rob
banks.That follows from whats given as I note various things
that havehappened in Iraq, like the old man doddering on a
cane gunned down bya heavily armed American soldier on the
streets of Baghdad.I read that a while back from an
embedded reporter who almostnonchalantly reported it. There
was a sense of the bizarre, and afeeling that my country had
fallen quite far, but strangely didntseem to realize that
people like George W. Bush cant stand for good,when what
theyre causing to be done is so reprehensible. > Good
point.> Mathematicians *are* a part of society after all.
Why should they> tell the truth now? Itd 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.> After all, thats so easy, now
isnt it?> Which is why you need to understand why I talk
about mathematicians> potentially being prosecuted. Liars
dont just stop because the gig> is up, as then, they
wouldnt necessarily be liars, then eh?> Itd be against
their *true* natures.> But life is about truth and
consequences. Think of a world that> faults America for its
wrongs as like seeing a boulder rolling down> hill. Its the
laws of physics at work. Some of you seem to think> that the
boulder can roll up against gravity--like praying to God to>
protect you despite your lies--but you see, if you ever see a
boulder> rolling up a mountain on this earth, youre upside
down.Ive thought about that for a while as I kept wondering
who mightreply back with some way that they say a boulder
could be seen rollingup a mountain without you having to be
upside down.Note: I used
http://users.tru.eastlink.ca/~brsears/reafault.htm
toreference the logical fallacies.James Harris
>I think
Ive gured out a way to show basically all of you,
including>>people who think they dont know any math that
mathematicians have>>been lying about my work. Its so
trivial you *should* wonder why>>they thought they could get
away with it.> And its worth reminding people of the
start of this thread which has> more than a hundred posts in
it.> What I want to emphasize to the sci.skeptic and
alt.math.undergrad> readers is that its hard to see how
mathematicians could not be lying> here.It might be
interesting to set up a website to conduct a poll. Each
respondant enters their math prociency and whether they
believe Mr. Harris. Results to be sorted accordingly. I
wonder if hes convinced anyone that hes right.>>Here
goes.>>My paper Advanced Polynomial Factorization depends on
considering a>>factor of a polynomials that I call g.>>(Paper
linked to at http://groups.msn.com/AmateurMath as usual.)>>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 polynomials
value changes,>>while c does not.> And why would anyone
argue?Only with the sloppy notation. (I know, you dont want
to discuss style issues)>>Now you can consider all factors of
a given polynomial using gs, with>>something like>> g_1...g_k
= P(x)>>where you have k factors. 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 Im 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.>>For my example up above its easy, as with g_1 = x + 1,
x varies as x>>varies, while 1 does not.>>Now thats enough
that the proof in the paper is straightforward, but>>posters
have argued with me anyway, with some trying to argue over
the>>denition of polynomial, amazingly enough.>>However,
consider that the gs have an important feature, which
is>>that when x=0, I have>> g_1...g_k = P(0).> Thats
very important.And very sloppy. g_1(0) g_2(0) ... g_k(0) =
P(0), and g_i(x) is not necessarily a polynomial. What ring
_is_ it in? Do you care?>>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 gs must>>multiply to give
the constant term.>>So then, maybe you still want to believe
the mathematicians and>>question that I can write g = r +
c.>>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),
>>which means that if you believe the mathematicians then
theyve>>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.>
Which is what should be sickening to *some* of you as
mathematicians> challenging me are actually challenging
algebra itself.Agreed, I dont know where g_i or r_i is being
evaluated in any of that. If P(x) is a constant, everything to
left is constant. If P(x) is a non-constant, so is everything
to the left. Theres nothing revolutionary here, just
bizarrely written... if you mean what you appear to mean.>>So
why would mathematicians argue against such a simple
result?>>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, they either believe
that you will go along, if you know> theyre lying, or that
you are incapable of catching them lying, even> with basic
algebra.Or perhaps it doesnt say anything new, and says it
poorly.>>That is, MOST of you will doubt algebra itself
rather than consider>>that mathematicians, whom you probably
dont even personally know,>>would lie.>>So where does this
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.>>For
instance, you may be saying, HEY, whats with the m when
you>>had x before??!!!>>Well, theres 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.> Now think about it, if
mathematicians gure that they can confuse you> on that
point, what does that tell you?I thought you were introducing
more room for error. The more complications you introduce, the
more you must account for in your conclusion.>>Finally, the
weirder thing is that one poster in particular got a lot>>of
mileage out of questioning my nding 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>>dened at that point.> Consider that passage
carefully, as what Im emphasizing to you is> that there are
people who are lying to you, and the evidence is right> in
front of you.After careful consideration, that should have
warned you that your work was likely to generate a lot of
skepticism if you had to argue for that result for years and
then want to use it as part of your proof. Did you ever bring
your skeptic to reason before using it?>>Well thats 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, its of
degree 1, so if you solve for the>>roots, youll get funky
stuff.That indicates the degree of the polynomial depends on
v and is _not_ necessarily 3. Perhaps you would do better to
think of it as a degree 6 polynomial over v,x,y?>>Now when I
was nding the proof of FLT...remember the process took>>some
years...at times Id talk of polynomials with respect to
other>>than m, but I rened my discourse as my understanding
improved.>>However, people arguing with me did not.> And
why wouldnt they? I gure its because if they concede>
anything, theyre afraid of being forced to give the full
truth--and> they hate the truth in this case.Or perhaps
because you never seem to post your revisions in a
location/format that everyone is able to access... like
here.>>You may *choose* to believe that they did not because
they dont know>>enough mathematics to follow, but were
talking about actual>>mathematicians here.>>Whats more
rational?>>I say its more rational to suppose that they
*did* gure 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 wasnt so great.> Which continues to
explain. And remember, *look* at this thread with> over a
hundred posts in it.You mean the thread where you wont
acknowledge or refute arguments that dispute signicant
portions of your argument? (Im not calling either a proof
right now since we cant seem to get either side to
acknowledge anything the other claims.)>>For most of you,
theres probably the belief that theres some funky>>higher
math involved that your pitiful brain cant follow or
you>>dont know about, as you may suppose that mathematicians
either>>wouldnt lie, or they wouldnt lie in such a dumb way
where I could>>catch them so easily.>>But consider whats in
the balance:>>1. I discovered a proof of Fermats Last
Theorem thats more>>available to people in general than most
of what mathematicians have>>been producing lately.Where is it
published?>>2. Worse I did so having said Id nd 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 Fermats Last Theorem.Where is
his error?>>And those are just highlights as theres 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.> And it gets worse with each
passing day.Yup. Youve been reduced to replying to your own
posts.>>EXCEPT, it looks like all they have to do is either
stay quiet, or>>*claim* Im wrong.>>Many of you simply
believe them, and question algebra itself, which is>>rather
sad. Id think at least some of you valued your
educations.Yet you are the one who is saying something is
wrong with math as taught. I thought youd want people to
question algebra.> Yet I remember talking to one fellow about
this in a private> communication, and he basically told me
that he didnt trust what hed> been taught at university.>
He wasnt surprised at the treatment Ive faced from
mathematicians.Do you believe that one person siding with you
is vindication?>>Others of you may gure it doesnt matter,
maybe because Western>>civilization seems to be based on
lying anyway, and maybe you gure I>>should just grow up,
accept that everybody lies and move on. And I>>dont have to
talk about Enron or pedophile Catholic priests or things>>in
that vein.>>I mean, look at George W. Bush and Iraq. If
people can be *killed*>>over lies, without consequences to
the liars, then whats with some>>freaking stupid math?>
And the shortsightedness of that is whats sickening. Many of
you> might *claim* care about life and liberty, as long as
its your own.Do you know what liberty is? It is not the
right to have everyone believe everything you say. Some
people think its the right to not be insulted, yet it isnt
so. How are you dening liberty? Also, how are you sure we
are interested only in our own?> But youll ght for the
right to kill even old men and pregnant> women, if it suits
you.Actually, I wont.> How many of you care or paid
attention to reports from Iraq like that> of a doddering old
man on a cane, gunned down in the streets of> Baghdad by a
heavily armored American soldier?Didnt even see the report.
Can you site a reference to it?> How many of you gave a damn
about the bombs that dropped on a wing of> a hospital in Iraq
for pregnant women?Who dropped the bombs? What caused them to
drop there?> No, youd just care if it happened in your town,
at a local hospital. > And you call that the American
way?Nope.> But you still pray, now dont you? You still pray
to a God that you> expect to protect you and your own, while
you kill women, children and> old men.Ignoring the personal
attack, I still pray. Now, what does ANY of this have to do
with math? Or are you so desperate that you can only resort
to character assassination? If you want to accuse me of being
scum, please learn a little more about me as a person rst. If
you want to discuss math, please leave my personal habits out
of it. If you dont mean to attack me personally, then
_which_ mathematicians are these terrible people?>>Good
point.>>Mathematicians *are* a part of society after all. Why
should they>>tell the truth now? Itd 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.>>After all, thats so easy, now isnt it?>>Which
is why you need to understand why I talk about
mathematicians>>potentially being prosecuted. Liars dont
just stop because the gig>>is up, as then, they wouldnt
necessarily be liars, then eh?>>Itd be against their *true*
natures.> But life is about truth and consequences. Think
of a world that> faults America for its wrongs as like seeing
a boulder rolling down> hill. Its the laws of physics at
work. Some of you seem to think> that the boulder can roll up
against gravity--like praying to God to> protect you despite
your lies--but you see, if you ever see a boulder> rolling up
a mountain on this earth, youre upside down.> James
HarrisJames, if you want to prove mathematicians lie. You
should be able to do so easily and then stop arguing.
Otherwise, drop the claim. The other items dont help your
case.-- Will Twentyman
I doubt Ill have the time to
answer all of this at once. Im willing> to return to other
parts later on, but it may be a few days before I> have time
always hard to know how to respond> when youre attempted to
explain something by analogy.> Say we assume that a1, a2,
and a3 are some kind of function on the> algebraic integers,
with values in the algebraic integers. (You> havent said
this is so, but its hard to see what else you might have> in
mind.) In a typical such set of three functions, each of the
three> functions is discontinuous everywhere. Im doubtful
that this is the> sort of thing you have in mind.> Later,
you say that that g1 must have that same factor (of f) in>
general, where g1 is associated with a1. This does not
represent> any kind of plain fact, if g1 is discontinuous
answers all reasonable questions. The most important part of>
it is terribly cryptic:> Now as noted before in general P(m)
has a factor that is f^2 , and> separating that factor off,
gives a constant term coprime to f; therefore,> given g1 =
a1x + uf> where with m = 0, g1 gives a factor of f it must
have that same factor> in general, proving that two of the
as have a factor that is f.> As usual, you cram the part of
the reasoning where your story diverges> from what is usually
believed into a short and very confusing spot.> Expand the
explanation of this spot, or nobody will get what you mean.>
It is interesting that Keith Ramsay here focusses on
exactlythe same troublesome passage in Advanced Polynomial
Factorizationas I have done in another thread: [Statement of
problem with ringof algebraic integers]. Keith rightly
describes it as the most important part. Two mathematicians
have arrived at exactlythe same problem independently. Why
would that be? Must be aMathematician Conspiracy, eh? That
phrase ... must have that same factor IN GENERAL (caps
added), wherein James Harris generalizes from a degenerate
1st-degree instance of his polynomial (when m = 0) to the
more general 3rd degree polynomial (m not zero) is the core
problem. No proof for it, no hint of justication, no
statement of an underlying principle. Worse yet, as shown by
both me and W. Dale Hall independently in the other thread,
the main conclusion of Advanced Polynomial Factorization
can be proven to be just plain false. Not only is the IN
GENERAL statement not true and not supported, there is no
way to x it! At least James Harris is an equal opportunity
insulter. He doesnt single out either of us for identifying
errors in hisargument. He calls both of us liars with equal
fervor. Nora B.
I doubt Ill have the time to answer all
of this at once. Im willing> to return to other parts later
on, but it may be a few days before I> have time again.>
hard to know how to respond> when youre attempted to explain
something by analogy.> Say we assume that a1, a2, and a3
are some kind of function on the> algebraic integers, with
values in the algebraic integers. (You> havent said this is
so, but its hard to see what else you might have> in mind.)
In a typical such set of three functions, each of the three>
functions is discontinuous everywhere. Im doubtful that this
is the> sort of thing you have in mind.> Later, you say that
that g1 must have that same factor (of f) in> general,
where g1 is associated with a1. This does not represent> any
kind of plain fact, if g1 is discontinuous
everywhere.Fascinating considering that he started by saying
Say we assume.It so happens that considered as functions
they probably arecontinuous.That is, Im fairly certain they
are continuous but as its not reallyrelevant to the paper, I
paper.> It hardly answers all reasonable questions. The
most important part of> it is terribly cryptic:> Now as
noted before in general P(m) has a factor that is f^2 , and>
separating that factor off, gives a constant term coprime to
f; therefore,> given g1 = a1x + uf> where with m = 0, g1
gives a factor of f it must have that same factor> in
general, proving that two of the as have a factor that is
f.> As usual, you cram the part of the reasoning where your
story diverges> from what is usually believed into a short and
very confusing spot.> Expand the explanation of this spot, or
nobody will get what you mean.I have no problem doing
so.Consider that 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 readers can *see* the factor
f^2 of P(m).Now for P(0), I have P(0) = f^2 (3xu^2 + u^3 f),
which is thatconstant term I talk about a lot. And now notice
that P(0)/f^2 =3xu^2 + u^3 f, which is coprime to f (yes, f is
coprime to 3, x, and uwhich is given in the paper).Heres
where it gets interesting as while the key variable is m,
Inotice that focusing on those xs it looks like another
cubic, so Iconsider the factorization P(m) = (a_1 x + uf)(a_2
x + uf)(a_3 x + uf)and specically focus on the factor a_1 x +
uf, which I call g.I notice that if a_1 is 0, when m=0, as at
least one of the as mustbe, then at that point g=uf, and has
a factor that is f.However, P(m)/f^2 has a factorization as
well, but its constant termis P(0)/f^2 = 3xu^2 + u^3 fwhich
is coprime to f, and considering g I see that factor of f
thatis visible with g = a_1 x + ufhas to go away, and in
going away it must go through a_1.Now posters in arguing with
me have claimed that some non-unit factorof f, is what goes
through, but lets call that j, where f = jk, andboth j and k
are non unit algebraic integers, then youd havesomething like
g/j = a_1/j x + ukand at m=0, youd have uk, which is not
coprime to f, so that cant betrue.Its actually rather easy
mathematics. But people keep arguing withme about it. > It is
interesting that Keith Ramsay here focusses on exactly> the
same troublesome passage in Advanced Polynomial
Factorization> as I have done in another thread: [Statement
of problem with ring> of algebraic integers]. Keith rightly
describes it as the most > important part. Two
mathematicians have arrived at exactly> the same problem
independently. Why would that be? Must be a> Mathematician
Conspiracy, eh? Not unless its a conspiracy of incompetence.
I think that a lot ofposters like Nora Baron and Keith Ramsay
simply arent that good atmathematics, and they lie a lot
when they dont exactly know whatshappening, to keep up the
delusion that theyre better than they are.They probably
memorized a lot of stuff, but here they have to *think*.Then
again they do often lie.> That phrase ... must have that
same factor IN GENERAL (caps added), > wherein James Harris
generalizes from a degenerate 1st-degree > instance of his
polynomial (when m = 0) to the more general 3rd > degree
polynomial (m not zero) is the core problem. No proof for >
it, no hint of justication, no statement of an underlying
principle. > Worse yet, as shown by both me and W. Dale Hall
independently in > the other thread, the main conclusion of
Advanced Polynomial > Factorization can be proven to be
just plain false. Not only is > the IN GENERAL statement
not true and not supported, there is > no way to x
it!Babbling. The problem is that if you *believe* Nora Baron,
you haveto disbelieve in algebra. Strangely enough some of you
seem quiteready to toss out algebra when posters like Nora
Baron, Keith Ramsay,or Arturo Magidin chatter.I nd it
fascinating, and yes, annoying.> At least James Harris is an
equal opportunity insulter. He > doesnt single out either of
us for identifying errors in his> argument. He calls both of
us liars with equal fervor.> Nora B.Well when youre a liar
then hey, youre a liar.James Harris
I doubt Ill have
the time to answer all of this at once. Im willing> to
return to other parts later on, but it may be a few days
cases arent parallel. Its always hard to know how to
respond> when youre attempted to explain something by
analogy.> Say we assume that a1, a2, and a3 are some kind
of function on the> algebraic integers, with values in the
algebraic integers. (You> havent said this is so, but its
hard to see what else you might have> in mind.) In a typical
such set of three functions, each of the three> functions is
discontinuous everywhere. Im doubtful that this is the> sort
of thing you have in mind.> Later, you say that that g1
must have that same factor (of f) in> general, where g1
is associated with a1. This does not represent> any kind of
plain fact, if g1 is discontinuous everywhere.> Fascinating
considering that he started by saying Say we assume.> It
so happens that considered as functions they probably are>
continuous.> That is, Im fairly certain they are
continuous but as its not really> relevant to the paper, I
the paper.> It hardly answers all reasonable questions. The
most important part of> it is terribly cryptic:> Now as
noted before in general P(m) has a factor that is f^2 , and>
separating that factor off, gives a constant term coprime to
f; therefore,> given g1 = a1x + uf> where with m = 0, g1
gives a factor of f it must have that same factor> in
general, proving that two of the as have a factor that is
f.> As usual, you cram the part of the reasoning where
your story diverges> from what is usually believed into a
short and very confusing spot.> Expand the explanation of
this spot, or nobody will get what you mean.> I have no
problem doing so.> Consider that> 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
readers can *see* the factor f^2 of P(m).> Now for P(0), I
have P(0) = f^2 (3xu^2 + u^3 f), which is that> constant term
I talk about a lot. And now notice that P(0)/f^2 3xu^2 +
u^3 f, which is coprime to f (yes, f is coprime to 3, x, and
u> which is given in the paper).> Heres where it gets
interesting as while the key variable is m, I> notice that
focusing on those xs it looks like another cubic, so I>
consider the factorization> P(m) = (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf)> and specically focus on the factor a_1 x
+ uf, which I call g.> I notice that if a_1 is 0, when m=0,
as at least one of the as must> be, then at that point g=uf,
and has a factor that is f. However, P(m)/f^2 has a
factorization as well, but its constant term> is> P(0)/f^2
= 3xu^2 + u^3 f> which is coprime to f, and considering g I
see that factor of f that> is visible with> g = a_1 x + uf has to go away, \
and in going away it must go through a_1.>
Here you are thinking of a_1 as a function of m, and youhave
said above that when m = 0, a_1 = 0. In general of coursewhen
m is nonzero, a_1 is nonzero. > Now posters in arguing with me
have claimed that some non-unit factor> of f, is what goes
through, but lets call that j, where f = jk, and> both j and
k are non unit algebraic integers, then youd have> something
like> g/j = a_1/j x + uk> and at m=0, youd have uk,
which is not coprime to f, so that cant be> true.> Here you
are essentially saying that there exist j, k, and
k,algebraic integers, such that a_1 = j*k and f = j*k. This
means that g/j can be written as g/j = k + u*k. Of course
since you are thinking of a_1 as a function of m,it must also
be assumed that j and k are functions of m. Now above you say
that when m = 0, u*k is not coprime to f.That is OK. However,
what you want is that when m is NOT 0, k + u*k is not
coprime to f. Just to remind: k cannot be assumed to be
constant. Whenm = 0, k = 0 = a_1/j, and certainly in that
case you get g/j is notcoprime to f. When m <> 0, there is no
reason to assumek is 0 or even a divisor of f. Therefore as
far as you know, when m <> 0, g/j = k + u*k may be coprime
to f. You do not obtain thecontradiction you want. You have
been mislead by your notation. You are leadto think that
a_1/j is a constant. If it were, thenwhatever holds when m =
0 would also hold when m <> 0. But it cannot be assumed that
that a_1/j = k is a constant.You must assume that it is a
function of m. The bottom line is, when m <> 0, you cannot
deduce anything about whether k + u*k is coprime to f from
its behavior when m = 0. > Its actually rather easy
mathematics. But people keep arguing with> me about it.> No,
its not easy mathematics. Its wrong mathematics.> It is
interesting that Keith Ramsay here focusses on exactly> the
same troublesome passage in Advanced Polynomial
Factorization> as I have done in another thread: [Statement
of problem with ring> of algebraic integers]. Keith rightly
describes it as the most > important part. Two
mathematicians have arrived at exactly> the same problem
independently. Why would that be? Must be a> Mathematician
Conspiracy, eh? > Not unless its a conspiracy of
incompetence. I think that a lot of> posters like Nora Baron
and Keith Ramsay simply arent that good at> mathematics, and
they lie a lot when they dont exactly know whats> happening,
to keep up the delusion that theyre better than they are.>
They probably memorized a lot of stuff, but here they have to
*think*.> Then again they do often lie.> Obviously I
disagree. Feel free to point out any lies in thetext above.>
That phrase ... must have that same factor IN GENERAL (caps
added), > wherein James Harris generalizes from a degenerate
1st-degree > instance of his polynomial (when m = 0) to the
more general 3rd > degree polynomial (m not zero) is the core
problem. No proof for > it, no hint of justication, no
statement of an underlying principle. > Worse yet, as shown
by both me and W. Dale Hall independently in > the other
thread, the main conclusion of Advanced Polynomial >
Factorization can be proven to be just plain false. Not only
is > the IN GENERAL statement not true and not supported,
there is > no way to x it!> Babbling. The problem is that
if you *believe* Nora Baron, you have> to disbelieve in
algebra. Not babbling. My proof and Dale Halls proof are
given in the other thread, Statement of problem with ring of
algebraic integers. James Harris has not found a valid
objection or errorin either of these proofs, which show that
the main conclusionof APF is wrong. As long as he maintains
that conclusion, hisactual argument (as given here, or as
vaguely implied in APF)is unimportant anyway. Nora B.>
Strangely enough some of you seem quite> ready to toss out
algebra when posters like Nora Baron, Keith Ramsay,> or
Arturo Magidin chatter.> I nd it fascinating, and yes,
annoying.> At least James Harris is an equal opportunity
insulter. He > doesnt single out either of us for
identifying errors in his> argument. He calls both of us
liars with equal fervor.> Nora B.> Well when youre a
liar then hey, youre a liar.> James Harris
=[...]| >
When you refer to the factorization of P(m) into linear
terms| > (a1x+uf)(a2x+uf)(a3x+uf), thats consistent with x
being a variable.| > If x is a constant, theres no
uniqueness in such a factorization (and| > not much that you
can prove about the values of a1, a2, and a3). In| > the next
step you infer that the coefcient of x^3 on the right hand| >
side (a1a2a3) is the same as the coefcient of x^3 on the left
hand| > side. This also is consistent with x being a variable.
If two| > polynomials are equal for all values of x, then
their corresponding| > coefcients have to be equal. But if x
is a constant, that step is| > invalid.| | Thats false.
Consider again x^2 + 4x + 4 = (x+2)(x+2), as that is| true
whether or not you consider x variable or not.| | The
*factorization* is whats important, and it exists without
regard| to whether or not you call P(x) = x^2 + 4x + 4 a
polynomial, or just| have x^2 + 4x + 4 with x=2.Im taking a
second shot at explaining this. (Perhaps I should pointout
that probably the most attractive feature of entering
intodiscussions like this is the chance to try out
explanations and seewhich of them make sense to you.)It has
to do with the difference between a constant and a variable.
Inyour example, whether we take x to be a constant or a
variable isirrelevant, because in either case theres just
one quantierinvolved, and its a universal quantier: the
claim is that thisequation holds for *all* values of x. But
this is misleading, becausein other cases the statement given
in a proof has more than onequantier in it, and they come in
a different order depending onwhether x is a constant or a
variable. Ill explain both situations.Suppose we have a
proof where x is introduced as a complex number, aconstant.
We would have something like this: Suppose that x is a
complex number. . . . Clearly, x^2+4x+4 = (x+2)(x+2).If we
dont introduce any additional assumptions about x,
thestatement at that step is that this equation holds for an
arbitrarycomplex number x, in other words that for every
complex number x, wehave x^2+4x+4 = (x+2)(x+2). The every
is the universal quantier.It would be bad style to say at
this point that there is _some_ xsatisfying the equation,
because weve already assumed that we hadsome given value of
it, and we shouldnt go switching to talking aboutthis
(possibly different) value of x.If we introduce x as a
variable, we would have something like this: Let x be a
complex-valued variable. . . . We know that x^2+4x+4 =
(x+2)(x+2) holds for every x.Thats just another way of
making the same statement about theequation. In this case, we
could substitute holds for some x orholds for at least
three xs and so on, because we havent assumed aspecic
value of x is given yet.If more than one quantier is
involved (quantifying over more thanone variable), things get
more subtle. We might want to say, forexample that there exist
complex numbers a and b having the propertythat for every
complex number x, x^2+4x+4 = (x+a)(x+b) holds. That hasthe
universal quantier for every on x, but before it theres
alsoan existential quantier there exist on a and b. The
only advantageof having declared x to be a complex-valued
variable previously isthat we get to present this fact
without having to say again that x iscomplex: Let x be a
complex-valued variable. . . . We know that there exist
complex numbers a and b such that x^2+4x+4 = (x+a)(x+b) holds
for every x.We could even say that there are unique complex
numbers a and b suchthat x^2+4x+4 = (x+a)(x+b) holds for
every x, since only a=b=2 willdo.If, however, x is declared
to be a specic complex number, aconstant, then we cant make
this same statement at this step in theproof. When x is a
constant, it doesnt make sense to talk about aproperty
holding for every x later on, because that would imply that
xis being allowed to vary again. If we had this in the proof:
Let x be a given complex number. . . . We know that there
exist complex numbers a and b such that
x^2+4x+4=(x+a)(x+b).we would be making a much weaker
statement. This step would representthe statement that for
every complex number x, there exist complexnumbers a and b
such that x^2+4x+4=(x+a)(x+b). The difference betweenthe two
statements is the order of the quantiers. In this one,
theuniversal quantier on x comes before the existential
quantier on aand b.The statement with the quantier on a and
b coming rst says thattheres a way I could win a simple
game, where I have to give you aand b, and then youre
allowed to pick any x you like, and I win
ifx^2+4x+4=(x+a)(x+b) is true. I dont have any real choice
in how topick a and b if I want to win.The statement with the
quantier on x coming rst means that you canwin a game where
I start by picking x, and then youre allowed to picka and b,
and you win if x^2+4x+4=(x+a)(x+b) is true. In effect, the
aand b are allowed to depend upon the x.Now applying this to
your paper, if x is taken to be a constant, itmeans that the
factorization P(m)=(a1*x+uf)(a2*x+uf)(a3*x+uf) onlymeans that
for the given x, the values a1, a2, and a3 can be any
onnitely many solutions to that equation for the given x.
If youmean to say that a1, a2, and a3 are values that make
the equation holdfor all x, then it restricts a1, a2, and a3
much more tightly; forgiven values of m, u, and f, there are
just the six different ways ofpermuting the three factors. It
seems to be what would make the mostsense in relation to what
youre trying to do, but that would meanthat x is not a
constant at that point.Keith Ramsay
=I said that a typical
set of three functions a1, a2, and a3 satisfyingthe
factorization formula would be discontinuous everywhere.
Thiswould be like the function f(x) = 1 if x is rational and
0 if x isirrational. Theres a sense in which the typical
function whose valuesare 0 and 1 is discontinuous
everywhere.[...]| It so happens that considered as functions
they probably are| continuous.| | That is, Im fairly certain
they are continuous but as its not really| relevant to the
paper, I havent checked thoroughly.If its irrelevant, ne,
but for what its worth, Im reasonablycondent, without
having checked, that its not possible for a1, a2,and a3 to
be continuous everywhere, for the same sort of reason asits
not possible to have a complex-number valued function
f(z)dened for every complex number z which is continuous and
satisesf(z)^2 = z for every z.[...]| > Expand the explanation
of this spot, or nobody will get what you mean.| | I have no
problem doing so.Its good to see you take up the
challenge....| Consider that| | 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 readers can
*see* the factor f^2 of P(m).| | Now for P(0), I have P(0) =
f^2 (3xu^2 + u^3 f), which is that| constant term I talk
about a lot. And now notice that P(0)/f^2 =| 3xu^2 + u^3 f,
which is coprime to f (yes, f is coprime to 3, x, and u|
which is given in the paper).|| Heres where it gets
interesting as while the key variable is m, I| notice that
focusing on those xs it looks like another cubic, so I|
consider the factorization| | P(m) = (a_1 x + uf)(a_2 x +
uf)(a_3 x + uf)| | and specically focus on the factor a_1 x
+ uf, which I call g.| | I notice that if a_1 is 0, when m=0,
as at least one of the as must| be, then at that point g=uf,
and has a factor that is f.| | However, P(m)/f^2 has a
factorization as well, but its constant term| is| | P(0)/f^2
= 3xu^2 + u^3 f| | which is coprime to f,So far theres
nothing too strange-looking, and I doubt theres anyobstacle
to lling in details like what exactly a1, a2, and a3 are.|
and considering g I see that factor of f that| is visible
with| | g = a_1 x + uf| | has to go away, and in going away
it must go through a_1.Nora Baron has posted, sounding like
she thinks she knows what youmean by this, but I dont. By
the factor of f that is visible, itsounds like you mean
simply the f appearing in the formula. Perhapsyou mean (like
you said just above) just that the value of g at m=0
isdivisible by f.But what you mean then by has to go away
is very unclear to me.g1(0)g2(0)g3(0) is P(0), which you just
got through saying was aproduct of f^2 with something coprime
to f; and you also just saidthat g1(0) is divisible by f.
What has to go away? I think you couldput it much more
plainly.| Now posters in arguing with me have claimed that
some non-unit factor| of f, is what goes through, but lets
call that j, where f = jk, and| both j and k are non unit
algebraic integers, then youd have| something like| | g/j =
a_1/j x + uk| | and at m=0, youd have uk, which is not
coprime to f, so that cant be| true.Dont you mean uk is not
divisible by f? Its also true that its notcoprime to f,
since k divides f and uk both, and k isnt a unit, butthat
seems less relevant.Its looking like the factor going away
is supposed to be a factorof a_1 in some sense or another. If
so, what sense?Perhaps it would help if you claried and/or
conrmed a few things.When in the paper you write that g1
must have that same factor (off) in general, does in
general mean that youre saying the valueof g1 for each
allowed value of m is divisible by f? Or do you meansomething
else? Is it a fair paraphrase of the step where youintroduce
a1, a2, and a3, to say that they are three functions fromthe
algebraic integers to the algebraic integers, a1(m), a2(m),
anda3(m), having the property that P(m) =
(a1*x+uf)(a2*x+uf)(a3*x+uf) forevery x?| Its actually rather
easy mathematics. But people keep arguing with| me about it.I
still am curious at what it is that seems to you like a
cogentargument here. In addition to context, which also
appears in yourpaper, youve inserted just a few statements
here, expressed in termsof a factor going through (whatever
that means).When you claimed to be able to satisfy the right
(was it rightwing?) denition of mathematical proof, as
given by that well-knownmathematician/writer whose name I
forget, I think you probably failedto realize how stringent a
denition he had in mind. One stepfollowing from another in a
purely logical fashion would mean thateach step was like the
syllogism Aristotle is a man. All men are mortal. Therefore
Aristotle is mortal.Now if necessary, we could break down
proofs (when theyre reallyproofs) into smaller pieces, until
the pieces are nally much likethat syllogism (or similar
logical principle). The reason this is notordinarily done is
that there are too many pieces for it to beconvenient
(although theres been some of this done by people
usingautomated proof checking programs). If theres a trouble
spot, though,at least one can move in that direction until
its clearer why theauthor thought it was validIts clear
that what youve got is a good ways away from the
ultimatelevel of detail! So it should be possible to break
the steps down intomany smaller ones. You can look at Douglas
Hofstadters proof of thecommutativity of addition in the
natural numbers (in Goedel, Escher,Bach), which has a couple
dozen or so steps if I remember correctly.Even such mundane
things as if x is relatively prime to y, and xdivides yz,
then x divides z can be expanded upon.Weve both said
theres supposed to be a chain of true statements.Still, all
I can see are apparently true statements on one side andfalse
ones on the other, with the elementary statements we all
knoware true to start with on the one side, and the
conclusion of theargument on the other side, with steps in
the middle having no obviousconsequence looking like the next
step.| > It is interesting that Keith Ramsay here focusses on
exactly| > the same troublesome passage in Advanced
Polynomial Factorization| > as I have done in another
thread: [Statement of problem with ring| > of algebraic
integers]. Keith rightly describes it as the most | >
important part. Two mathematicians have arrived at exactly|
> the same problem independently. Why would that be? Must be
a| > Mathematician Conspiracy, eh? | | Not unless its a
conspiracy of incompetence. I think that a lot of| posters
like Nora Baron and Keith Ramsay simply arent that good at|
mathematics, and they lie a lot when they dont exactly know
whats| happening, to keep up the delusion that theyre
better than they are.| | They probably memorized a lot of
stuff, but here they have to *think*.Im pleased to see you
at least consider incompetence as anexplanation for some of
the things you nd puzzling, rather than onlycynical attempts
to suppress a truth. Consider also the possibilitythat a
contradiction in the mathematics weve been taught all
theseyears would be very hard for us to believe.| Then again
they do often lie.I dont lie very often. The things you
think are lies, are cases whereyou think youre obviously
right, but you at least appear very muchmistaken.I have read
that the average person lies about twice a day on average,as
a way of smoothing out social situations. The other day a
guystarted chatting with me on the bus, and eventually
suggested I callhim up, especially if a job opening turned up
where I work. He askedme if I would. I think this is the kind
of situation where peopleoften lie, but it didnt seem like a
good thing to do.Ive come to think that something, which at
rst glance is quitedifferent from lying, is not so much less
of a moral fault (or asdifferent, in essence) as is commonly
thought. Namely, even decentpeople often state falsehoods not
on purpose, but simply due to nothaving been concerned enough
with the truth to avoid them. Cliffordhas an essay where he
claims something on similar lines. Reaganmisquoting himself
on the percentage of a certain pollutant that comesfrom trees
would be an example. A person claiming denitely to have
aproof of a result, when theyve not been any more careful
than manyprevious occasions when this person mistakenly
thought they had one,would be an example. I dont claim to be
innocent of this kind offault, but I do try to avoid it, and I
hope youll try to avoid ittoo.| > That phrase ... must have
that same factor IN GENERAL (caps added), | > wherein James
Harris generalizes from a degenerate 1st-degree | > instance
of his polynomial (when m = 0) to the more general 3rd | >
degree polynomial (m not zero) is the core problem. No proof
for | > it, no hint of justication, no statement of an
underlying principle. | > Worse yet, as shown by both me and
W. Dale Hall independently in | > the other thread, the main
conclusion of Advanced Polynomial | > Factorization can be
proven to be just plain false. Not only is | > the IN
GENERAL statement not true and not supported, there is | >
no way to x it!| | Babbling. The problem is that if you
*believe* Nora Baron, you have| to disbelieve in algebra.
Strangely enough some of you seem quite| ready to toss out
algebra when posters like Nora Baron, Keith Ramsay,| or
Arturo Magidin chatter.| | I nd it fascinating, and yes,
annoying.Has to go away isnt a term from algebra.| > At
least James Harris is an equal opportunity insulter. He | >
doesnt single out either of us for identifying errors in
his| > argument. He calls both of us liars with equal
fervor.| > | > Nora B.| | Well when youre a liar then hey,
youre a liar.When the shoe doesnt t, I dont pretend to be
wearing it.Keith Ramsay
I said that a typical set of
three functions a1, a2, and a3 satisfying> the factorization
formula would be discontinuous everywhere. This> would be
like the function f(x) = 1 if x is rational and 0 if x is>
irrational. Theres a sense in which the typical function
whose values> are 0 and 1 is discontinuous
everywhere.Relevancy? > [...]> | It so happens that
considered as functions they probably are> | continuous.> | >
| That is, Im fairly certain they are continuous but as its
not really> | relevant to the paper, I havent checked
thoroughly.> If its irrelevant, ne, but for what its
worth, Im reasonably> condent, without having checked, that
its not possible for a1, a2,> and a3 to be continuous
everywhere, for the same sort of reason as> its not possible
to have a complex-number valued function f(z)> dened for
every complex number z which is continuous and satises>
f(z)^2 = z for every z.Thats naive. What I have is that a_1
a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m) a_1(a_2 + a_3) + a_2
a_3 = 0and a_1 + a_2 + a_3 = - 3(-1 + mf^2 )where f is a
constant and is a prime integer other than 3.Here m varies
and you question that possibility for the as to becontinuous
everywhere. Now what do you see in those equations
KeithRamsay?What I do know is that each of the as is an
algebraic integer for anym.I know that at m=0, two of the as
equal 0, and one equals 3. However, theres no indication that
thats a point of discontinuity,as you can approach from
either side by varying m appropriately. > [...]> | > Expand
the explanation of this spot, or nobody will get what you
mean.> | > | I have no problem doing so.> Its good to see
you take up the challenge....I enjoy discussing the argument.
Its nifty.> | Consider that> | > | 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
readers can *see* the factor f^2 of P(m).> | > | Now for
P(0), I have P(0) = f^2 (3xu^2 + u^3 f), which is that> |
constant term I talk about a lot. And now notice that
P(0)/f^2 | 3xu^2 + u^3 f, which is coprime to f (yes, f is
coprime to 3, x, and u> | which is given in the paper).> |> |
Heres where it gets interesting as while the key variable is
m, I> | notice that focusing on those xs it looks like
another cubic, so I> | consider the factorization> | > | P(m)
= (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)> | > | and specically
focus on the factor a_1 x + uf, which I call g.> | > | I
notice that if a_1 is 0, when m=0, as at least one of the as
must> | be, then at that point g=uf, and has a factor that is
f.> | > | However, P(m)/f^2 has a factorization as well, but
its constant term> | is> | > | P(0)/f^2 = 3xu^2 + u^3 f> | >
| which is coprime to f,> So far theres nothing too
strange-looking, and I doubt theres any> obstacle to lling
in details like what exactly a1, a2, and a3 are.For any given
algebraic integer m, the as are algebraic integers.> | and
considering g I see that factor of f that> | is visible with>
| > | g = a_1 x + uf> | > | has to go away, and in going away
it must go through a_1.> Nora Baron has posted, sounding
like she thinks she knows what you> mean by this, but I
dont. By the factor of f that is visible, it> sounds like
you mean simply the f appearing in the formula. Perhaps> you
mean (like you said just above) just that the value of g at
m=0 is> divisible by f.Well you didnt have a problem with
P(0)/f^2 being coprime to f.Here Im considering the constant
element of g, which must also becoprime to f, so that factor
of f that is f in uf must go away.Ive found that all the
symbols might confuse people, so how aboutusing f=7, u=1?Then
I have P(m) = 49((2401m^3 - 147 m^2 + 3m)x^3- 3(-1 + 49m)x +
7)and notice that P(0)= 49(3x + 7), and P(0)/49 = 3x + 7.So
looking at g = a_1 x + 7 I have that 7 in there that cant
bethere after I separate off (or divide off if you prefer) 49
from P(m).Its kind of tricky I know because you have to pay
careful attentionto whats happening.1. P(m) has that factor
of f^2, which with f=7, is 49.2. Separating that off, means
that the constant term P(0)/49 iscoprime to 7.3. But g = a_1
x + 7, with f=7 and u=1, so the 7 has to go away.Now you
questioned 3., but you have to remember that g is a factor
ofP(m).Lets say that the 7 remains, then at m=0, you have 7,
which is NOTcoprime to 7, so thats not possible.Its clear
the 7 must go.> But what you mean then by has to go away is
very unclear to me.> g1(0)g2(0)g3(0) is P(0), which you just
got through saying was a> product of f^2 with something
coprime to f; and you also just said> that g1(0) is divisible
by f. What has to go away? I think you could> put it much
more plainly.Hmmm...maybe itd help to consider the other gs
now that you bringthem up.You have to remember that I can
consider P(m)/f^2, and then you have P(m)/f^2 = g_1 g_2
g_3/f^2and Im merely noting the consequence of P(0)/f^2
being coprime to f. > | Now posters in arguing with me have
claimed that some non-unit factor> | of f, is what goes
through, but lets call that j, where f = jk, and> | both j
and k are non unit algebraic integers, then youd have> |
something like> | > | g/j = a_1/j x + uk> | > | and at m=0,
youd have uk, which is not coprime to f, so that cant be> |
true.> Dont you mean uk is not divisible by f? Its also
true that its not> coprime to f, since k divides f and uk
both, and k isnt a unit, but> that seems less relevant.I
mean that its not coprime to f, or if you prefer that it
shares acommon factor with f, and here that factor is k. >
Its looking like the factor going away is supposed to be a
factor> of a_1 in some sense or another. If so, what sense?In
considering P(m)/f^2 = g_1 g_2 g_3/f^2, Im guring out how
thatfactor f^2 goes through the gs.What I nd is that it
only makes sense if only two of the gs have afactor of f
that is f, but then youre forced out of the ring ofalgebraic
integers, though it doesnt appear to have happened fromwithin
the argument, which is why that ring is awed.> Perhaps it
would help if you claried and/or conrmed a few things.>
When in the paper you write that g1 must have that same
factor (of> f) in general, does in general mean that
youre saying the value> of g1 for each allowed value of m is
divisible by f? It depends on context. Its like if you have
y=2x+1, and considervarious values. Now I see pressure to
write that as f(x)=2x+1, but Idont think its worth the
potential confusion.The problem is that though you see m as
the key variable, it may infact be a *factor* of m that is
the true independent variable, whichIm sure has confused
quite a few people. Now its a measure of yourmathematical
ability if you know what I mean there, and lets just saythat
Im not interested in expanding out in a way that leaves
*more*room for confusion.>Or do you mean> something else? Is
it a fair paraphrase of the step where you> introduce a1, a2,
and a3, to say that they are three functions from> the
algebraic integers to the algebraic integers, a1(m), a2(m),
and> a3(m), having the property that P(m) =
(a1*x+uf)(a2*x+uf)(a3*x+uf) for> every x?Theyre not
necessarily functions of m.Think about it Keith Ramsay. Sit
down, go over what Ive given andthink very carefully over
just that one point, and maybe, if yourehonestly confused,
itll help you to break through to anunderstanding.> | Its
actually rather easy mathematics. But people keep arguing
with> | me about it.> I still am curious at what it is that
seems to you like a cogent> argument here. In addition to
context, which also appears in your> paper, youve inserted
just a few statements here, expressed in terms> of a factor
going through (whatever that means).The discussion is
informal and Im looking for ways to help peopleunderstand. >
When you claimed to be able to satisfy the right (was it
right> wing?) denition of mathematical proof, as given by
that well-known> mathematician/writer whose name I forget, I
think you probably failed> to realize how stringent a
denition he had in mind. One step> following from another in
a purely logical fashion would mean that> each step was like
the syllogism> Aristotle is a man.> All men are mortal.>
Therefore Aristotle is mortal.> Now if necessary, we could
break down proofs (when theyre really> proofs) into smaller
pieces, until the pieces are nally much like> that syllogism
(or similar logical principle). The reason this is not>
ordinarily done is that there are too many pieces for it to
be> convenient (although theres been some of this done by
people using> automated proof checking programs). If theres
a trouble spot, though,> at least one can move in that
direction until its clearer why the> author thought it was
valid> Its clear that what youve got is a good ways away
from the ultimate> level of detail! So it should be possible
to break the steps down into> many smaller ones. You can look
at Douglas Hofstadters proof of the> commutativity of
addition in the natural numbers (in Goedel, Escher,> Bach),
which has a couple dozen or so steps if I remember
correctly.> Even such mundane things as if x is relatively
prime to y, and x> divides yz, then x divides z can be
expanded upon.> Weve both said theres supposed to be a
chain of true statements.> Still, all I can see are
apparently true statements on one side and> false ones on the
other, with the elementary statements we all know> are true to
start with on the one side, and the conclusion of the>
argument on the other side, with steps in the middle having
no obvious> consequence looking like the next step.Well I
dont believe you. Time after time Ive lled in steps
andwhat makes me REALLY suspicious is that back when I did
have a gap,posters like yourself narrowed in on it.Then I
lled the gap and the games began in earnest.Now then how
about you consider P(m)/49 = (2401m^3 - 147 m^2 + 3m)x^3-
3(-1 + 49m)x + 7having a constant term P(0)/49 = 3x + 7,
which is coprime and thenrevisit those comments.Here Ive
lessened the number of symbols to try and helpunderstanding.
> | > It is interesting that Keith Ramsay here focusses on
exactly> | > the same troublesome passage in Advanced
Polynomial Factorization> | > as I have done in another
thread: [Statement of problem with ring> | > of algebraic
integers]. Keith rightly describes it as the most > | >
important part. Two mathematicians have arrived at exactly>
| > the same problem independently. Why would that be? Must
be a> | > Mathematician Conspiracy, eh? > | > | Not unless
its a conspiracy of incompetence. I think that a lot of> |
posters like Nora Baron and Keith Ramsay simply arent that
good at> | mathematics, and they lie a lot when they dont
exactly know whats> | happening, to keep up the delusion
that theyre better than they are.> | > | They probably
memorized a lot of stuff, but here they have to *think*.>
Im pleased to see you at least consider incompetence as an>
explanation for some of the things you nd puzzling, rather
than only> cynical attempts to suppress a truth. Consider
also the possibility> that a contradiction in the mathematics
weve been taught all these> years would be very hard for us
to believe.> | Then again they do often lie.> I dont lie
very often. The things you think are lies, are cases where>
you think youre obviously right, but you at least appear
very much> mistaken.Then prove it mathematically.In this
ENTIRE POST Keith Ramsay you do not have a
relevantmathematical statement which you prove.Now does that
say mathematical competence to you? > I have read that the
average person lies about twice a day on average,> as a way
of smoothing out social situations. The other day a guy>
started chatting with me on the bus, and eventually suggested
I call> him up, especially if a job opening turned up where I
work. He asked> me if I would. I think this is the kind of
situation where people> often lie, but it didnt seem like a
good thing to do.> Ive come to think that something, which
at rst glance is quite> different from lying, is not so much
less of a moral fault (or as> different, in essence) as is
commonly thought. Namely, even decent> people often state
falsehoods not on purpose, but simply due to not> having been
concerned enough with the truth to avoid them. Clifford> has
an essay where he claims something on similar lines. Reagan>
misquoting himself on the percentage of a certain pollutant
that comes> from trees would be an example. A person claiming
denitely to have a> proof of a result, when theyve not been
any more careful than many> previous occasions when this
person mistakenly thought they had one,> would be an example.
I dont claim to be innocent of this kind of> fault, but I do
try to avoid it, and I hope youll try to avoid it> too.Um,
havent you gured out that I skim through when I dont see
math?> | > That phrase ... must have that same factor IN
GENERAL (caps added), > | > wherein James Harris
generalizes from a degenerate 1st-degree > | > instance of
his polynomial (when m = 0) to the more general 3rd > | >
degree polynomial (m not zero) is the core problem. No proof
for > | > it, no hint of justication, no statement of an
underlying principle. > | > Worse yet, as shown by both me
and W. Dale Hall independently in > | > the other thread, the
main conclusion of Advanced Polynomial > | > Factorization
can be proven to be just plain false. Not only is > | > the
IN GENERAL statement not true and not supported, there is >
| > no way to x it!> | > | Babbling. The problem is that if
you *believe* Nora Baron, you have> | to disbelieve in
algebra. Strangely enough some of you seem quite> | ready to
toss out algebra when posters like Nora Baron, Keith Ramsay,>
| or Arturo Magidin chatter.> | > | I nd it fascinating, and
yes, annoying.> Has to go away isnt a term from
algebra.The discussion is informal.> | > At least James
Harris is an equal opportunity insulter. He > | > doesnt
single out either of us for identifying errors in his> | >
argument. He calls both of us liars with equal fervor.> | >
| > Nora B.> | > | Well when youre a liar then hey, youre a
liar.> When the shoe doesnt t, I dont pretend to be
wearing it.> Keith RamsayIm looking for posters who get to
the bottom of things.Posts should get *shorter* and not
longer.Being an expert means you can cut to the chase.James
Harris
=ah, so; mea culpa -- Im trying! > Um, havent you
gured out that I skim through when I dont see math? > The
discussion is informal. > Posts should get *shorter* and not
longer.> Being an expert means you can cut to the chase.--A
church-school McCrusade (Blairs ideals?):Harry-the-Mad-Potter
wants US to kill Iraqis?...For a 1000-year anglo-american
hegemony?HEY, JIMMY; LETS 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/les/curriculum/Cosmo.PCX==
= Here Im right, and youre wrong, as the *denition* of
algebraic> integer leads to an incomplete ring, which I prove
mathematically.The trouble is, Im still not following your
logic, just to be able tosee what the great contradiction
is.(a) I have two algebraic integers x and y(b) I claim that
y is a factor of x in the algebraic integers(c) But there is
no algebraic integer z (=x/y) such that y*z = xThis leads
inexorably to the conclusion that the claim in (b) is
simplyfalse. There are *lots* of possible algebraic integers
which I canmultiply by *something* to get x, but there are
only a few choices thatI can multiply by another algebraic
integer to get x, and those are thefactors.Basically, I do
not see in what sense you can claim that y is a factorof x
without demonstrating the properties of x/y... just being
able tomultiply y by something to get x isnt enough to be
able to claim that.And without that claim, the conclusion
about incompleteness isnt there.Perhaps it could help if you
could give me a rigorous denition ofwhat you mean by
incomplete ring, or conversely complete ring. Isthis a
property which is distinct from being, say, closed
undermultiplication?