mm-268
===
Subject: need help ...Let X is a T_1 Space and A is a
subset of X.Prove that Derived set of A is closed
set.__sol> Let A'
means the derived set of A. & Let x in (A')^c ,complement of
derived set of A. since x not in A', there exist G ,a open
neighborhood of x s.t. (G{x}) intersectionA=empty. Now, for
showing (A')^c is open, i think i must show G is a subset
of(A')^c. maybe for showing this, T_1 space condition must be
used. but, unfortunately i can't any idea for using T_1
condition. If anyone give me help, I will really
===
unknown by support1.mathforum.org (8.11.6/8.11.6/The Math
greater than God, More evil than the devil, The poor have
it,The rich need it, And if you eat it, you'll die?
===
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
devil, The poor have it,>he rich need it, And if you eat it,
you'll die? Is there an epidemic of this? There have been at
least threethreads on the (completely non-mathematical) riddle
in the last fewdays.
Seehttp://mathforum.org/epigone/alt.math.undergrad/homzurskol/
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
===
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
some seeds like mittens? (Hint:There is a homonym in
theanswer.)Because they are sown (sewn)? [Naw, that's pushing
it ...]>What word has one pronunciation when it is
capitalized, and anotherpronunciation when it is not
capitalized?I think it is Polish and polish.>Why is the letter
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
numerals . . .> 500 IN THE BEGINNINGD = 500.We have: D _ _ _
_>500 IN THE ENDanother DWe have: D _ _ _ D>5 IN THE MIDDLE IS
SEENV = 5We have: D _ V _ D>FIRST OF ALL LETTERS,First letter =
AWe have: D A V _ D>FIRST OF ALL NUMBELIE IN BETWEEN.First
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
sin 2A = 2 tan A / 1 + tan^2 Aok...sin 2A = 2 sinA cosA = 2
tanA cosA cosA = 2 tanA cos^2 A =tan A (cos 2A + 1)I don't
think I am going in the right direction any
===
formulae by support1.mathforum.org (8.11.6/8.11.6/The Math
stuck...>Prove sin 2A = 2 tan A / 1 + tan^2 A>ok...>sin 2A = 2
sinA cosA = 2 tanA cosA cosA = 2 tanA cos^2 A =>tan A (cos 2A +
1)>I don't think I am going in the right direction any
longer... It's not so much the right direction as the right
side! convert 2 tan A/(1+ tan^2 A) to sin and cos: 2(sin A/cos
A)/(1+ sin^2 A/cos^2 A) Now multiply both numerator and
===
angle and half angle formulae by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
===
have problems with Mathematics ? by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
years old.The oldest had problems with her mathematics and,
just looking I founda marvellous site explaining lots of
things about maths, try it out
:http://www.outwar.com/page.php?x=1920394 . It can seem
===
calculus by support1.mathforum.org (8.11.6/8.11.6/The Math
prove [(R -> S) -> (R -> R)]can I write:1 [(R -> S) -> (R ->
R)]2 R -> S hyp13 R -> R mp,2,14 [(R -> S) -> (R -> R)]
discharge hyp1R -> R is always true, can i discharge hyp1 and
===
Quadratic Function Question by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
I am unable to figureout, and was hoping someone could offer
some hints or a point in theright direction:I've been given
three points (30,1500); (40,3600); and (50,6300), andI have
been asked to determine the equation for this quadratic. Ihave
graphed it, but it is not incredibly accurate, and I'm unsure
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
question on a math assignment that I am unable to figure>out,
and was hoping someone could offer some hints or a point in
the>right direction:>I've been given three points (30,1500);
(40,3600); and (50,6300), and>I have been asked to determine
the equation for this quadratic. I>have graphed it, but it is
not incredibly accurate, and I'm unsure of>how to approach
Greg,the easiest way to determine the quadratic is in the
formf(x) = ((x-40)*(x-50))/((30-40)*(30-50)) * 1500 +
((x-30)*(x-50))/((40-30)*(40-50)) * 3600 +
((x-30)*(x-40))/((50-30)*(50-40)) * 6300.Do you see the way
the function is defined by the three pointsyou gave ?Best
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
three points (30,1500); (40,3600); and (50,6300),>and I have
been asked to determine the equation for this quadratic.The
quadratic function has the form: y = ax^2 + bx + cWe know
three points and they tell us that:when x = 30, y = 1500when x
= 40, y = 3600when x = 50, y = 6300Plug those values into the
general form and we get:900a + 30b + c = 15001600a + 40b + c =
36002500a + 50b + c = 6300Solve this system of equations and
get: a = 3, b = 0, c = -1200Therefore, the function is: y =
3x^2 - 1200~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~To check:
substitute those points into this equation and
===
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
to solve the following equation, but not using>trial and
error>any ideas?>(1.2)^n > 1.6 + 0.4n First what do you mean
by trial and error? Secondly, since youused n instead of x do
you intend that n be an integer? There is no way to give a
formula for the solution. You could useany of a number of
numerical techniques (I certainly would not callNewton's
method trial and error) to solve the equation. I get x<-2.39
or x> 9.08.(-2.39< x< 9.08 has the inequality the other
===
coefficients of A are unknown. by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
A2) x = bwhere A1 and b are known but A2 is composed of
coefficents whosevalues are unknown and what we want is the
estimates of x. Anybody hasidea how to solve this
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
<360.bc cos (A - 45.bc) = 1/sqrt3 cosAwell...cos (A - 45.bc) =
tan30 cos Aand now I'm stuck because I'm not too sure of
basics...tan30 cos A = sin30 cos A / cos 30It would be nice if
===
Stuck again!! by support1.mathforum.org (8.11.6/8.11.6/The
for 0 < A <360.bc >cos (A - 45.bc) = 1/sqrt3 cosAIt doesn't
come out very neat . . .I used the Difference Formula for
Cosines:cos(A)cos(45) + sin(A)sin(45) =
cos(A)/sqrt3cos(A)/sqrt2 + sin(A)/sqrt2 = cos(A)sqrt3Multiply
through by sqrt6:sqrt3*cos(A) + sqrt3*sin(A) =
sqrt2*cos(A)sqrt3*sin(A) = sqrt2*cos(A) - sqrt3*cos(A) =
[sqrt2 - sqrt3]cos(A)sin(A)/cos(A) = [sqrt2 -
sqrt3]/sqrt3tan(A) = (sqrt2 - sqrt3)/sqrt3Then: A =
arctan[(sqrt2 - sqrt3)/sqrt3] = -10.39828583 degreesTherefore:
===
problems. by support1.mathforum.org (8.11.6/8.11.6/The Math
had gross profit percentage of 35% last year on sales
of$1,000,000. What was your gross profit in dollars? If you
reduced yourselling price by 10%, what increase in sales would
you need to makethe same gross profit?An manufacturer
guarantees their generators remain within 1.5% of theoptimum
specs for 10 years. Once in use, the buyer finds the
generatorproduces 34,754 watts per hr when it was ra 35,000
watts. Shouldthe buyer be concerned? Why/why not?I am just
having some trouble with these I am not sure how to startthem
out. If anyone could walk me through these I would really be
===
percent word problems.> Your company had gross profit
percentage of 35% last year on sales of> $1,000,000. What was
your gross profit in dollars? If you reduced your> selling
price by 10%, what increase in sales would you need to make>
the same gross profit?> An manufacturer guarantees their
generators remain within 1.5% of the> optimum specs for 10
years. Once in use, the buyer finds the generator> produces
34,754 watts per hr when it was ra 35,000 watts. Should> the
buyer be concerned? Why/why not?> I am just having some
trouble with these I am not sure how to start> them out. If
anyone could walk me through these I would really be most>
with 35% of $1M. On the second partof the first, make up an
example - assume the selling price is $10, forexample.On the
second one, you should be able to figure that out. What is
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
===
Could it be half-angle Formula? by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
cos^2 A - 2 sin A - 2 = 0I guess I should put everything in
terms of either cos or sin, butcan't really see hw, unless I
use the half angle formula, whichtotally defeats
===
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
the original message.I got it as soon as I sent the
===
formula? by support1.mathforum.org (8.11.6/8.11.6/The Math
simpler than that . . .Use: cos^2 A = 1 - sin^2 A3(1 - sin^2
A) - 2 sin A - 2 = 0which gives us: 3 sin^2 A + 2 sin A - 1 =
===
calculus text?>Back then I had algebra, geometry , and trig.
Not much else and no>math since..... You're not ready for
Calculus but you can get ready for it. Do this:Restudy
elementary and Intermediate algebra on your own for about 6 to
10months; you will need much of it for studying Calculus. Then,
don't worry muchabout Geometry, but go on to review
trigonometry for maybe 2 or 3 months, allon your own. Now,
find any old Calculus used textbook written in the 1970's
or1980's. Many of them are very good. If you have retained
your math learning from high school very well, and it hasonly
been a couple of year(...........no. You said you left school
at theend of Eisenhower's administration), well wait: If you
have retained yourconcepts and skills well for the last 45
yeathen MAYBE, you could studyfrom a College
AlgebraPreCalculus textbook which includes sections
onTrigonometry; and study this for 6 months. THEN, find any
old Calculus bookwritten in the 1970's or 1980's. Anyway, a
couple of good authors to watch for whose names would be on
someuseful calculus textbooks are Larson & Hostetler, and
Salas & Hill. Otherpeople will have other suggestions. You can
find such used books at Librarybooksales, both from college
===
Stewart's Precalculus:Mathematics for Calculus seems like a
pretty good bet for starters....>>Back then I had algebra,
geometry , and trig. Not much else and no>>math since.....
>You're not ready for Calculus but you can get ready for it.
Do this:>Restudy elementary and Intermediate algebra on your
own for about 6 to 10>months; you will need much of it for
studying Calculus. Then, don't worry much>about Geometry, but
go on to review trigonometry for maybe 2 or 3 months, all>on
your own. Now, find any old Calculus used textbook written in
the 1970's or>1980's. Many of them are very good. >If you have
retained your math learning from high school very well, and it
has>only been a couple of year(...........no. You said you
left school at the>end of Eisenhower's administration), well
wait: If you have retained your>concepts and skills well for
the last 45 yeathen MAYBE, you could study>from a College
AlgebraPreCalculus textbook which includes sections
on>Trigonometry; and study this for 6 months. THEN, find any
old Calculus book>written in the 1970's or 1980's. >Anyway, a
couple of good authors to watch for whose names would be on
some>useful calculus textbooks are Larson & Hostetler, and
Salas & Hill. Other>people will have other suggestions. You
can find such used books at Library>booksales, both from
===
Re: Factoring Conjecture: Triplets> People belong to groups .
Mathematicians tend to hang out> with other mathematicians.If
Nora Baron is a mathematician, then it's quite likely that>
someone on the newsgroup knows who that poster is.> I think
nobody in this newsgroup knows who I am. At least, except for>
Bob Silverman (who appears not to post anymore), I have met
nobody from> this group. And Bob Silverman only shortly when
he visi my institute> (I think not more than 5 minutes).> Yup,
I'm trying to figure out what type of group Nora Baron is part>
of, because I'm somewhat concerned here.I take what Nora says,
about her identity or anything else, at face> value because
she's demonstra herself to be trustworthy in the past.> You
should not. Nora also makes errors. Dik is of course exactly
right here (though he might have softened the blow a bit!).
There is no reason to trust amathematician. Everyone makes
errors. The math however shouldstand on its own. I try to
write things as simply as I can so that people can understand
what I am saying and verify itfor themselves and not just
accept it because it sounds likeI know what I am talking
about. Dik I think does much thesame.> I also make errors. Not
often!> I think> everyone in this newsgroup makes the
occasional error. In most cases> such an error is spot by
another reader (Arturo is pretty good at> spotting errors),
and life goes on. But you'd better not spot an error>
irrelevant.> I distrust virtually everything you say, because
you've demonstra> yourself *repealy* to be *un*trustworthy.>
That is too much. In general the algebraic manipulations
(except for> occasionally quite a few false starts) are
correct. It is just the> conclusions where James fails. With
respect to his math, Harris should be trea like everyoneelse:
don't believe it unless you can verify it for yourself. > Nora
clearly has shown that she's a mathematician with the quality>
of the work she's pos here.> She has shown that she is educa
with quite a bit of mathematics,> I do not know whether she is
a mathematician or not (what makes one> a mathematician?), but
again, that is not so very interesting.> Fermat also was not a
===
Re: Factoring Conjecture: TripletsI take what Nora says, about
her identity or anything else, at face> value because she's
demonstra herself to be trustworthy in the past.> You should
not. Nora also makes errors. I also make errors. I think>
everyone in this newsgroup makes the occasional error. In most
cases> such an error is spot by another reader (Arturo is
pretty good at> spotting errors), and life goes on. But you'd
better not spot an error> irrelevant.By taking it at face
value I don't mean I assume everything Nora saysis without
error. I mean that I assume she is being honest in whatevershe
says, because she never has given me a reason to think
otherwise.If she makes an error, I believe it to be an honest
error. I feel thebehavior, I distrust him even when he
*appears* to be telling the truth;I always assume a hidden
agenda or an ulterior motive. IOW, I considerhim to have
forfei the right to be given the benefit of any doubt;most
people are innocent until proven guilty but I view James as
guiltyuntil proven innocent, and even then it would take
===
TripletsI take what Nora says, about her identity or anything
else, at face> value because she's demonstra herself to be
trustworthy in the past.> You should not. Nora also makes
errors. I also make errors. I think> everyone in this
newsgroup makes the occasional error. In most cases> such an
error is spot by another reader (Arturo is pretty good at>
spotting errors), and life goes on. But you'd better not spot
an error> irrelevant.> By taking it at face value I don't mean
I assume everything Nora says> is without error. I mean that I
assume she is being honest in whatever> she says, because she
never has given me a reason to think otherwise.> If she makes
an error, I believe it to be an honest error. I feel the>
behavior, I distrust him even when he *appears* to be telling
the truth;> I always assume a hidden agenda or an ulterior
motive. IOW, I consider> him to have forfei the right to be
given the benefit of any doubt;> most people are innocent
until proven guilty but I view James as guilty> until proven
innocent, and even then it would take *iron-clad* proof.All a
clear demonstration that Wayne Brown doesn't
understandmathematics at all.Mathematics isn't about trust;
mathematics is about truth determinableby logical methods.Sure
I play fast and loose with conjectures, which are just
guesses,and have even thought I had proofs quite a few times
only to be wrong,but the point that seems to escape Wayne
Brown, and people like WayneBrown out there is that
mathematically it doesn't matter.The mathematics doesn't
change.It's not about reputations or of democracy. Mathematics
is NOT ademocracy.If something is true mathematically then it's
just true, and people bedamned.Opinions don't matter in
mathematics. So Wayne Brown and all thepeople like him can
just off, as opinions will NEVER rulemathematical
===
what Nora says, about her identity or anything else, at face>
value because she's demonstra herself to be trustworthy in the
> past.You should not. Nora also makes errors. I also make
errors. I think> everyone in this newsgroup makes the
occasional error. In most cases> such an error is spot by
another reader (Arturo is pretty good at> spotting errors),
and life goes on. But you'd better not spot an error>
irrelevant.> By taking it at face value I don't mean I assume
everything Nora says> is without error. I mean that I assume
she is being honest in whatever> she says, because she never
has given me a reason to think otherwise.> If she makes an
error, I believe it to be an honest error. I feel the>
behavior, I distrust him even when he *appears* to be telling
the truth;> I always assume a hidden agenda or an ulterior
motive. IOW, I consider> him to have forfei the right to be
given the benefit of any doubt;> most people are innocent
until proven guilty but I view James as guilty> until proven
innocent, and even then it would take *iron-clad* proof.> All
a clear demonstration that Wayne Brown doesn't understand>
mathematics at all.it is his undersatnding of JSH in question
here, and that appears to be superb.> Mathematics isn't about
trust; mathematics is about truth determinable> by logical
methods.No one can prove everything for himself (herself). So
even in mathematics, one must accept most things on trust. So
one's trustworthiness is an important mathematical
consideration. One usually determines another's
trustworthiness largely by their track record of correctness
versus error.On that bsis, Wayne Brown is judging Nora Baron
and JSH exactly > Sure I play fast and loose with conjectures,
which are just guesses,> and have even thought I had proofs
quite a few times only to be wrong,> but the point that seems
to escape Wayne Brown, and people like Wayne> Brown out there
is that mathematically it doesn't matter.You sound like the
boy who cried wolf.> The mathematics doesn't change.But you
don't produce any.> It's not about reputations or of
democracy. Mathematics is NOT a> democracy.If it were, you
would long ago have been vo out.> If something is true
mathematically then it's just true, and people be> damned.Very
well, then, damn you for not producing any of that true
mathematics.> Opinions don't matter in mathematics. So Wayne
Brown and all the> people like him can just off, as opinions
will NEVER rule> mathematical truth.Your opinions least of
===
taking it at face value I don't mean I assume everything Nora
says>> is without error. I mean that I assume she is being
honest in whatever>> she says, because she never has given me
a reason to think otherwise.>> If she makes an error, I
believe it to be an honest error. I feel the>> behavior, I
distrust him even when he *appears* to be telling the truth;>>
I always assume a hidden agenda or an ulterior motive. IOW, I
consider>> him to have forfei the right to be given the
benefit of any doubt;>> most people are innocent until proven
guilty but I view James as guilty>> until proven innocent, and
even then it would take *iron-clad* proof.> All a clear
demonstration that Wayne Brown doesn't understand> mathematics
at all.> Mathematics isn't about trust; mathematics is about
truth determinable> by logical methods.> Sure I play fast and
loose with conjectures, which are just guesses,> and have even
thought I had proofs quite a few times only to be wrong,> but
the point that seems to escape Wayne Brown, and people like
Wayne> Brown out there is that mathematically it doesn't
matter.The point that seems to escape is that I wasn't talking
aboutmathematical truth; I'm talking about basic decency and
trustworthiness.(This whole discussion began, remember, with
James questioning whetherNora really is who she claims to be,
not with a question of mathematicaltruth.) When Nora Baron
says something, whether it's right or wrong,I believe *she*
believes it's right; and if she discovers she's wrong,she'll
be honest enough to admit it. James, I'm not criticizing
yourmath here (others are much more capable of doing that).
I'm saying thatyou yourself are not a decent and trustworthy
person. I'm not sayingyou're wrong; I'm saying you yourself
DON'T CARE if you're wrong or not,and you'll keep claiming to
be right EVEN IF YOU KNOW YOU ARE WRONG --because you have
demonstra time and again that you're a *liar.*> The
mathematics doesn't change.> It's not about reputations or of
democracy. Mathematics is NOT a> democracy.> If something is
true mathematically then it's just true, and people be>
damned.> Opinions don't matter in mathematics. So Wayne Brown
and all the> people like him can just **** off, as opinions
will NEVER rule> mathematical truth.Truth, mathematical or any
other kind, means *nothing* to you, and I'mreally tired of
===
Conjecture: Triplets>> In short, I trust Nora, and I don't
trust you. If you took one of>> your idiotic newsgroup polls
on this issue I'm certain you'd find an>> overwhelming
majority who feel the same way. (Not that it means her math>>
is infallible, of course; it just means I trust her not to lie
about it,>> whereas I *expect* you to lie at every
opportunity.)> If someone doesn't know the truth, is it
possible for him to lie?There are differences between not
knowing the truth, refusing to face thetruth, and knowing the
truth but lying about it. James has provided uswith examples
of all three on many occasions -- often all three within--
===
trust Nora, and I don't trust you. If you took one of>your
idiotic newsgroup polls on this issue I'm certain you'd find
an>overwhelming majority who feel the same way. (Not that it
means her math>is infallible, of course; it just means I trust
her not to lie about it,>whereas I *expect* you to lie at every
opportunity.)>>If someone doesn't know the truth, is it
possible for him to lie?> There are differences between not
knowing the truth, refusing to face the> truth, and knowing
the truth but lying about it. James has provided us> with
examples of all three on many occasions -- often all three
withinI think in the case of JSH his need to believe in
himself is so powerful that it clouds the picture. His state
of delusion is such that he doesn't know the truth, about
himself, about otheor about mathematics. We are dealing with a
mentally disturbed person here, and I think the word lie does
===
Factoring Conjecture: Triplets> In short, I trust Nora, and I
don't trust you. If you took one of> your idiotic newsgroup
polls on this issue I'm certain you'd find an> overwhelming
majority who feel the same way. (Not that it means her math>
is infallible, of course; it just means I trust her not to lie
about it,> whereas I *expect* you to lie at every
opportunity.)> If someone doesn't know the truth, is it
possible for him to lie?> GibYes. If someone makes a statement
and knows that it isn't true, butnevertheless claims that it is
true, then that person is lying. Thelying is not about the
contents of the statement itself, but about theclaim that it
is the truth. Knowledge of the truth is not necessary inthis
case. I guess James suspects a lot of mathematicians in this
groupdoing this.For example, if I have zero mean normal
probability density functionpdf(x) with sigma=1, then I just
don't know right now whatint_sqrt(3)^oo pdf(x) dx is. But if I
claim that the answer is 22, thenI'm lying for sure(or call me
===
Conjecture: TripletsPeople belong to groups . Mathematicians
tend to hang out> with other mathematicians.If Nora Baron is a
mathematician, then it's quite likely that> someone on the
newsgroup knows who that poster is.I'm not asking anyone to
reveal the poster, just to verify that the> poster isn't
totally anonymous.Yup, I'm trying to figure out what type of
group Nora Baron is part> of, because I'm somewhat concerned
here.> I take what Nora says, about her identity or anything
else, at face> value because she's demonstra herself to be
trustworthy in the past.> I distrust virtually everything you
say, because you've demonstra> yourself *repealy* to be
*un*trustworthy.> Nora clearly has shown that she's a
mathematician with the quality> of the work she's pos here.
You have shown that you are *not*> a mathematician in exactly
the same way.> Yeah right. What else has Nora Baron pos? In
any NG with math in its name what more than math need anyone
post?JSH cerrtainly posts a lot of OT material for math NGs.>
Where else has that> poster made comments besides my
threads?If the posts Nora has made are relevant to the threads
in which she pos them, which they uniformly have been, the
existence of other posts by her in other threads is not
relevant.> Do you know Wayne Brown?Irrelevant.> Or are you
just babbling here?It seems to be JSH who is babbling
===
that I'm relying on rather basic principles of algebra,and I'm
going to see if explaining that might help.I'm still using the
example pos by a Rick Decker, a professor atHamilton College,
and as usual here's some identifying headers so youcan find
Decker's post, which also shows that he is indeed at
===
HamiltonCollege:Subject: Re: Mathematical consistency,
courageIn his post Decker claimed to mirror my argument using
a quadraticinstead of a cubic, where he has(5a_1(x) +
7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where I'll now make a
slight change to have(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 +
30x) + 7(2) and his a's are roots of a^2 - (x - 1)a + 7(x^2 +
x).My emphasis has been on the actual constant terms as while
it appearsfomr that factorization that both are 7 on the left
that doesn't addup with what's on the right, so I set x=0 to
find that one of the a'sequals 0, then, asa^2 + a = 0and
picking a_1(0) = 0, then a_2(1) = -1, so it makes sense I
think now to introduce b_2(x), where a_2(x) = b_2(x) - 1so
that at x=0, b_2(0) = 0, and making that substitution
gives(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x) + 7(2) and
now the constant terms match up. Now I've seen posters make a
bigdeal out of even the phrase constant terms so I've now
moved togetting rid of them!!!That's easy enough as all I have
to do is multiply out to get25 a_1(x) b_2(x) + 10 a_1(x) + 35
b_2(x) + 14 = 7(25x^2 + 30x) + 14and not surprisingly the
constant terms match up, so let's delete themoff!!!That
gives25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 +
30x).The point here, and the rather basic algebraic principle
is that in afactorization you have contributions from
different portions to theproduct.It's like you can
separate7(25x^2 + 30x) + 7(2) into A + B + Cwhere the various
pieces of(5a_1(x) + 7)(5b_2(x) + 2) each have their place and
contribute to each part.So, with A = 7(25x^2) I have that
5a_1(x) and 5b_2(x) have acontribution to it, and they also,
because they are somewhatcomplica, contribute to B = 7(30x),
but, and this is important,they have NO contribution to C =
7(2).If you can understand that simple bit of algebra, then
you canunderstand how it all works.That's because my
construction deliberately focuses things so that
thecontributions to C are isola in the factorization, which
you cansee, since7 times 2 equals 7(2), and I have(5a_1(x) +
7)(5b_2(x) + 2) = 7(25x^2 + 30x) + 7(2).Some posters seem to
have an extraordinary amount of difficulty here,so I'll use
c_1 = 7 and c_2 = 2, so that you have(5a_1(x) + c_1)(5b_2(x) +
c_2) = 7(25x^2 + 30x) + 7(2)to *emphasize* that the part of the
factors(5a_1(x) + 7) and (5b_2(x) + 2)which produces 7(2) in
7(25x^2 + 30x) + 7(2)is easily seen.It's algebra. The way it
works is that you multiply in a rigid way toget from(5a_1(x) +
7)(5b_2(x) + 2) to7(25x^2 + 30x) + 7(2).And that *rigid* format
only allows for 7 and 2 in(5a_1(x) + 7)(5b_2(x) + 2) to
contribute to the constant term, which not surprisingly
is7(2).That's algebra.So, then what happens if you divide both
sides by 7?Let's say you try to have both factors on the left
have sqrt(7) as afactor, then(5a_1(x)/sqrt(7) +
sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x) + 2and
notice that now the parts that produce 2 on the right
aresqrt(7) and 2/sqrt(7)which may not seem to be a big deal to
you, but you see, 2/sqrt(7) isNOT in the ring of algebraic
integers as it is not the root of a monicpolynomial with
integer coefficients.Now then, remember, in algebra, when you
multiply out a factorizationlike(5a_1(x) + 7)(5b_2(x) + 2)
each part contributes in a rigid way to its product
like7(25x^2 + 30x) + 7(2)and if it's really hard to understand
that, you can split up 7(25x^2 + 30x) + 7(2) into the parts A,
B, and C, whereA = 7(25x^2), B = 7(30x), and C= 7(2)to see
that the contribution to the C part comes *only* from the 7
and2 in(5a_1(x) + 7)(5b_2(x) + 2).The algebra is fascinating
because some of you may have never realizedthat's how a
factorization multiplies because you never had to thinkthat
way!!!So, let me repeat, as it may be a hard effort for some
of you:Different parts of a factorization contribute in a
rigid way to theproduct.With(5a_1(x) + 7)(5b_2(x) + 2) =
7(25x^2 + 30x) + 7(2)you can see the 7 and 2 on the left,
which multiply to give the 7(2)on the right.If you divide both
sides by 7, then if you divide each factor on theleft by
sqrt(7), you getsqrt(7) and 2/sqrt(7) on the left, which
rightly gives 2 on the right,but now you're NOT in the ring of
algebraic integers because 2/sqrt(7)is NOT in that ring.If you
can understand how parts of a factorization like(5a_1(x) +
7)(5b_2(x) + 2)contribute to the product7(25x^2 + 30x) +
7(2)then you can understand why dividing by 7 *in the ring of
algebraicintegers* requires that you have(5a_1(x)/7 +
1)(5b_2(x) + 2) = 25x^2 + 30x + 2.Now, it IS true that when
you do that you find out something weird!!! It turns out that
5a_1(x)/7 is not in general in the ring of
algebraicintegers!!!Some people when faced with that necessity
of algebra, have decided toquestion algebra, and how different
parts of a factorizationcontribute to the product.These people
have been very dedica in their questioning of algebra,and I
doubt that they will stop now, but I have a suggestion for
thoseof you who believe in algebra and understand it:Conclude
that dividing boths sides of(5a_1(x) + 7)(5b_2(x) + 2) =
7(25x^2 + 30x) + 7(2)by 7 gives factors on the left side that
are not in general in thering of algebraic
===
turns out that I'm relying on rather basic principles of
algebra,>and I'm going to see if explaining that might
help.You _really_ look like an idiot when you say this sort of
thing,implying that the reason not one of the many
professionalmathematicians you've contac here and
elsewhereagrees with you is that _none_ of them understand
thebasic principles of
===
Understanding the algebra> It turns out that I'm relying on
rather basic principles of algebra,> and I'm going to see if
explaining that might help.> I'm still using the example pos
by a Rick Decker, a professor at> Hamilton College, and as
usual here's some identifying headers so you> can find
Decker's post, which also shows that he is indeed at Hamilton>
===
College:> Subject: Re: Mathematical consistency, courage> In
his post Decker claimed to mirror my argument using a
quadratic> instead of a cubic, where he has> (5a_1(x) +
7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where I'll now make a
slight change to have> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 +
30x) + 7(2) > and his a's are roots of > a^2 - (x - 1)a +
7(x^2 + x).> My emphasis has been on the actual constant terms
as while it appears> fomr that factorization that both are 7 on
the left that doesn't add> up with what's on the right, so I
set x=0 to find that one of the a's> equals 0, then, as> a^2 +
a = 0> and picking a_1(0) = 0, then a_2(1) = -1, > so it makes
sense I think now to introduce b_2(x), where > a_2(x) = b_2(x)
- 1> so that at x=0, b_2(0) = 0, and making that substitution
gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x) + 7(2) >
and now the constant terms match up. Now I've seen posters
make a big> deal out of even the phrase constant terms so I've
now moved to> getting rid of them!!!> That's easy enough as all
I have to do is multiply out to get> 25 a_1(x) b_2(x) + 10
a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x) + 14> and not
surprisingly the constant terms match up, so let's delete
them> off!!!> That gives> 25 a_1(x) b_2(x) + 10 a_1(x) + 35
b_2(x) = 7(25x^2 + 30x).> The point here, and the rather basic
algebraic principle is that in a> factorization you have
contributions from different portions to the> product.> It's
like you can separate> 7(25x^2 + 30x) + 7(2) into A + B + C>
where the various pieces of> (5a_1(x) + 7)(5b_2(x) + 2) > each
have their place and contribute to each part.> So, with A =
7(25x^2) I have that 5a_1(x) and 5b_2(x) have a> contribution
to it, and they also, because they are somewhat> complica,
contribute to B = 7(30x), but, and this is important,> they
have NO contribution to C = 7(2).> If you can understand that
simple bit of algebra, then you can> understand how it all
works.> That's because my construction deliberately focuses
things so that the> contributions to C are isola in the
factorization, which you can> see, since> 7 times 2 equals
7(2), and I have> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x)
+ 7(2).> Some posters seem to have an extraordinary amount of
difficulty here,> so I'll use c_1 = 7 and c_2 = 2, so that you
have> (5a_1(x) + c_1)(5b_2(x) + c_2) = 7(25x^2 + 30x) + 7(2)>
to *emphasize* that the part of the factors> (5a_1(x) + 7) and
(5b_2(x) + 2)> which produces 7(2) in > 7(25x^2 + 30x) + 7(2)>
is easily seen.> It's algebra. The way it works is that you
multiply in a rigid way to> get from> (5a_1(x) + 7)(5b_2(x) +
2) > to> 7(25x^2 + 30x) + 7(2).> And that *rigid* format only
allows for 7 and 2 in> (5a_1(x) + 7)(5b_2(x) + 2) > to
contribute to the constant term, which not surprisingly is>
7(2).> That's algebra. Is this algebra also? Say w1 and w2 are
algebraic integers and w1*w2 = 7. Say a_1(x)/w1 and a_2(x)/w2
are algebraic integers. Since b_2(x) = a_2(x) + 1, 5*b_2(x) +
2 = 5*a_2(x) + 7 and (5*b_2(x) + 2)/w2 = 5*a_2(x)/w2 +
7/w2,which is the sum of two algebraic integers. Therefore
(5*a_1(x)/w1 + 7/w1)*(5*a_2(x)/w2 + 7/w2) = (25*x^2 + 30*x +
2),as required. All the coeffcients on the left side of
thisequation are algebraic integers. So the *algebra* looks
fine. Where you must disagree with this is in the implied
assertion that w1 and w2 satisfying the above *exist*. If you
can prove they cannot exist then you are rightand I am wrong.
I say they exist. Specifically, let w1 = GCD(a_1(x), 7) and w2
= GCD(a_2(x), 7). These are GCDs in the algebraic integers.
They *exist* becauseof an old (& difficult) theorem of
Dedekind. Here a_1 and a_2 are the roots of a^2 - (x - 1)*a +
7*(x^2 + x). You can easily check that w1 and w2 as just
defined have these properties: 1. Both are algebraic integers
2. Both are divisors of 7. That is, 7/w1 and 7/w2 are
algebraic integers. 3. Both a_1(x)/w1 and a_2(x)/w2 are
algebraic integers. So I claim that establishes that w1 and w2
both *exist*. Notethat both of them are dependent on x, so it
would make senseto write them as w1(x) and w2(x). Note also
that for MOST integer values of x, w1 and w2 are not equalto
either 7 or 1. This happens only when x = 0. Note thatx = 0 is
really a special case, because when x = 0, a^2 - (x - 1)*a +
7*(x^2 + x)is reducible. In general it is not. If, for x = 1,
you think that w1 = 7, you will arrive at a contradiction to
one of the statements above: a_1(1)/7 isnot an algebraic
integer. as you note below.> So, then what happens if you
divide both sides by 7? Depends on how you do it.> Let's say
you try to have both factors on the left have sqrt(7) as a>
factor, then> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) +
2/sqrt(7)) = > 25x^2 + 30x) + 2> and notice that now the parts
that produce 2 on the right are> sqrt(7) and 2/sqrt(7) sqrt(7)
is the right factor only when x = 1. In generalit is more
complex. showed that for x = 2, the right factor is a divisor
of 7 which is the root of an11th degree monic polynomial.>
which may not seem to be a big deal to you, but you see,
2/sqrt(7) is> NOT in the ring of algebraic integers as it is
not the root of a monic> polynomial with integer coefficients.
There is no requirement that 2/sqrt(7) be in the ring
ofalgebraic integers. The key requirement is that the
wholeexpression (5 b_2(x) + 2)/sqrt(7) be an algebraic
integer.> Now then, remember, in algebra, when you multiply
out a factorization> like> (5a_1(x) + 7)(5b_2(x) + 2) > each
part contributes in a rigid way to its product like> 7(25x^2 +
30x) + 7(2)> and if it's really hard to understand that, you
can split up > 7(25x^2 + 30x) + 7(2) into the parts A, B, and
C, where> A = 7(25x^2), B = 7(30x), and C= 7(2)> to see that
the contribution to the C part comes *only* from the 7 and> 2
in> (5a_1(x) + 7)(5b_2(x) + 2).> The algebra is fascinating
because some of you may have never realized> that's how a
factorization multiplies because you never had to think> that
way!!!> So, let me repeat, as it may be a hard effort for some
of you:> Different parts of a factorization contribute in a
rigid way to the> product. No, it's nearly as rigid as you
think. Note that (5 b_2(x) + 2) = (5 a_2(x) + 7),so even
though 2/sqrt(7) is not an algebraic integer,7/sqrt(7) very
definitely is. > With> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 +
30x) + 7(2)> you can see the 7 and 2 on the left, which
multiply to give the 7(2)> on the right.> If you divide both
sides by 7, then if you divide each factor on the> left by
sqrt(7), you get> sqrt(7) and 2/sqrt(7) on the left, which
rightly gives 2 on the right,> but now you're NOT in the ring
of algebraic integers because 2/sqrt(7)> is NOT in that ring.>
If you can understand how parts of a factorization like>
(5a_1(x) + 7)(5b_2(x) + 2)> contribute to the product> 7(25x^2
+ 30x) + 7(2)> then you can understand why dividing by 7 *in
the ring of algebraic> integers* requires that you have>
(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2.> Now, it IS
true that when you do that you find out something weird!!! >
It turns out that 5a_1(x)/7 is not in general in the ring of
algebraic> integers!!! Exactly right, for x <> 0. That is
because you should not have factored 7 as 7 * 1.You should
have factored it as 7 = w1(x) * w2(x), as I definedabove. If
you had down that you end up with algebraic integer
coefficients everywhere, and no contradiction of thekind you
just encountered. Two facts: 1. Your factorization does not
work. We all agree on that. 2. There is another factorization,
7 = w1(x)*w2(x), as defined above by the GCD expressions, which
DOES work. It is only on the latter point that we disagree.
Instead ofcontinuing to ignore our objections and start a
blizzard of newthreads on the same topic, why don't you try
showing what is wrong with w1(x)*w2(x) ?> Some people when
faced with that necessity of algebra, have decided to>
question algebra, and how different parts of a factorization>
contribute to the product. No, no one questions algebra. We
simply disagree with your belief that yours is the only way to
factor 7 out of your polynomial factors. We have shown there is
another way whichdoes not incur the problem that a_1(x)/w1(x)
is not an algebraicinteger. We all agree that there is a
problem with your proposedfactorization. Let's agree to quit
beating on that. What you now need to do, if you can, is show
what is wrongwith OUR proposed factorization, 7 = w1(x)*w2(x),
with w1(x)and w2(x) as defined above. And please, try answering
this instead of starting yetanother new thread. > These people
have been very dedica in their questioning of algebra,> and I
doubt that they will stop now, but I have a suggestion for
those> of you who believe in algebra and understand it:>
Conclude that dividing boths sides of> (5a_1(x) + 7)(5b_2(x) +
2) = 7(25x^2 + 30x) + 7(2)> by 7 gives factors on the left side
that are not in general in the> ring of algebraic integers.>
===
that dividing boths sides of> (5a_1(x) + 7)(5b_2(x) + 2) =
7(25x^2 + 30x) + 7(2)> by 7 gives factors on the left side
that are not in general in the> ring of algebraic integers.So
what? Dividing in number rings is generally going to create
quantitiesnot in the ring. I'd find it extremely odd if you
could run arounddividing algebraic integers by other
quantities and still produce nothingbut algebraic
integers.Look at: (3+1) = 4 in the ring of integers or
algebraic integers. Nowdivide by 2, so that you have (3/2+1/2)
= 4/2 = 2. How many integers oralgebraic integers do you
see?--There are two things you must never attempt to prove:
the unprovable --and the
===
Understanding the algebra> With> (5a_1(x) + 7)(5b_2(x) + 2) =
7(25x^2 + 30x) + 7(2)> you can see the 7 and 2 on the left,
which multiply to give the 7(2)> on the right.> If you divide
both sides by 7, then if you divide each factor on the> left
by sqrt(7), you get> sqrt(7) and 2/sqrt(7) on the left, which
rightly gives 2 on the right,> but now you're NOT in the ring
of algebraic integers because 2/sqrt(7)> is NOT in that
ring.Quite true, 2/sqrt(7) isn't an algebraic integer, but
that is of noimportance here.With x = 1 as an example, we
havea_1(1) = sqrt(-14) and a_2(1) = -sqrt(-14), so b_2(1) = 1
- sqrt(-14)Now we have(5a_1(1) + 7)(5b_2(1) + 2) = 7(25(1)^2 +
30(1) + 2) = 7(57)and it's not hard to see that neither5b_2(1)
= 5(1 - sqrt(-14)) nor 2 are divisible by sqrt(7).However,
*both* of5a_1(1) + 7 = 5sqrt(-14) + 7and5b_2(1) + 2 =
-5sqrt(-14) + 7are indeed divisible in the algebraic integers
===
that at x=0, b_2(0) = 0, and making that substitution gives>
(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x) + 7(2)> and now
the constant terms match up.And your point is that 5a_1(x) has
to be divisible by 7 for some reason.Go back to your original
equation:> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x) + 7(2)
#1You don't like the '7' in the second part of the LHS so you
introduce b_2(x)to manipulate the 'constant term'. You want
the first term is divisible by7 and the second is divisible by
2 so you do your substitution to align theconstants with your
expectation.Why couldn't I do this:Introduce b_1(x) = a_1(x)
+1Than after substitution your original equation (#1) becomes
(5b_1(x) + 2)(5a_2(x) + 7) = 7(25x^2 + 30x) + 7(2)Now what
does that tell you? Is the first term now divisible by 2 and
thesecond by 7? But they are the same terms as you had
originally... Takingpart of the 'constant term' and
rearranging things so that you hide a partin the 'functional
part' (e.g. b_2(x)) doesn't change the divisibility ofthe
entire term.Does this give you any insight?--Stan
===
up in a discussion with a poster who seemed to believe that I>
don't consider the possibility that I'm wrong. I've seen that>
position put up quite a few times, and I think it's worth
addressing,> so I'll go back yet again to an example pos by a
Rick Decker, a> professor at Hamilton College, to go over what
I say happens and> consider how I might be wrong.> Here's some
identifying to find Decker's post, which also shows that> he
===
is indeed at Hamilton College:> Subject: Re: Mathematical
consistency, courage> In his post Decker claimed to mirror my
argument using a quadratic> instead of a cubic, where he has>
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where I'll
now make a slight change to have> (5a_1(x) + 7)(5a_2(x) + 7) =
7(25x^2 + 30x) + 7(2) > and his a's are roots of > a^2 - (x -
1)a + 7(x^2 + x).> My emphasis has been on the actual constant
terms as while it appears> fomr that factorization that both
are 7 on the left that doesn't add> up with what's on the
right, so I set x=0 to find that one of the a's> equals 0,
then, as> a^2 + a = 0> and picking a_1(0) = 0, then a_2(1) =
-1, > so it makes sense I think now to introduce b_2(x), where
> a_2(x) = b_2(x) - 1> so that at x=0, b_2(0) = 0, and making
that substitution gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2
+ 30x) + 7(2) > and now the constant terms match up. Now I've
seen posters make a big> deal out of even the phrase constant
terms so I've now moved to> getting rid of them!!!> That's
easy enough as all I have to do is multiply out to get> 25
a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x) +
14> and not surprisingly the constant terms match up, so let's
delete them> off!!!> That gives> 25 a_1(x) b_2(x) + 10 a_1(x) +
35 b_2(x) = 7(25x^2 + 30x).> Now nothing spectacular so far, so
let's move to where there's been> EXTREME disagreement as my
position is that from the constant terms it> only makes sense
that if you divide both sides of> (5a_1(x) + 7)(5b_2(x) + 2) =
7(25x^2 + 30x + 2) > by 7, you end up with> (5a_1(x)/7 +
1)(5b_2(x) + 2) = 25x^2 + 30x + 2> if you're to remain in the
ring of algebraic integers.> That's an important point.>
Multiplying out as before gives me> 25 a_1(x) b_2(x)/7 + 10
a_2(x)/7 + 5 b_2(x) + 2 = 25x^2 + 30x + 2> and again getting
rid of the constant terms, I have> 25 a_1(x) b_2(x)/7 + 10
a_2(x)/7 + 5 b_2(x) = 25x^2 + 30x> and from before I had> 25
a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)> and
dividing both sides of it gives> 25 a_1(x) b_2(x)/7 + 10
a_1(x)/7 + 5 b_2(x) = 25x^2 + 30x,> so the answers the same.>
Now that might not impress some of you, so I'm going to let
x=3, and> consider what would happen if somehow there was a
*different* way to> divide both sides of> (5a_1(x) +
7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> by 7, so let x=3, then>
(5a_1(3) + 7)(5b_2(3) + 2) = 7(25(3)^2 + 30(3) + 2)> and even
if you believe that 7 splits up as a *function* of x to>
divide the left side, you should also believe that at x=3 its
factors> actually have some definite value.> So let's use
w_1(3), and w_2(3) for those factors where w_1(3) w_2(3)> = 7,
and dividing both sides gives> (5a_1(3)/w_1(3) +
w_2(3))(5b_2(3)/w_2(3) + 2/w_2(3)) = 25(3)^2 + 30(3)> + 2> but
now there's a problem as 2/w_2(3) isn't in the ring of
algebraic> integers if w_2(3) isn't a unit, as 2 and 7 are
coprime.> Now the way I see it w_2(3) and 2/w_2(3) are factors
of 2 on the right> side, so there's no way to argue with
2/w_2(3) being forbidden in the> ring of algebraic integebut
some of you might wish to claim that> you can escape having to
have such a factor by concluding that> (5b_2(3) + 2)/w_2(3) is
in the ring.> you can simply mush all your numbers together
here and have it all> work out in the end.> That's actually
kind of funny as a math position. I'll call it the> mushing
position.> So then from my analysis, for me to be wrong, it
must be true for> 2/w_2(3) to be in the ring of algebraic
integer when w_2(3) is not a> unit, which is a contradiction.>
I just don't accept that mushing position as it's not
mathematics. See my previous reply on this. The mushing
position is correct. I am glad you have chosen to focus on x =
3. When you do that,we are down to the heart of the matter.
Everything in the equationsis just numbers and we can think
about factorizations of numbersinstead of the perhaps
confusing background of functions and polynomials. Let A = (5
a_1(3) + 7), B = (5 a_2(3) + 7), and C = (25*3^2 + 30*3 + 2) =
257.Note that A * B = 7 * C,and that 7 is coprime to C.Theorem:
If A, B, and C are algebraic integeand p is a prime which is
coprime to C, then there exist w1 and w2 such that: 1. w1*w2 =
p 2. A/w1 and B/w2 are algebraic integers 3. (A/w1) * (B/w2) =
C.Proof: Define w1 = GCD(A, p) and w2 = GCD(B, p). The restis
immediate, QED.Here is how this applies to your factorization:
p = 7 and A = 5 a_1(3) + 7, B = 5 a_2(3) + 7, C = 257.Use w1
and w2 as defined in the theorem. Note that 7/w1 and 7/w2 are
algebraic integers. Clearly 7 = w1*w2.Finally, note that by
the theorem, it must also betrue that 5 a_1(3)/w1 and 5
a_2(3)/w2are algebraic integers. Therefore (5 a_1(3)/w1 +
7/w1)*(5 a_2(3)/w2 + 7/w2) = C = 257is a factorization of the
desired form in which***all the coefficients are algebraic
integers***,which is what you need. Note that in general w1
and w2 are not equal to 7 or 1.(Perhaps *only* when x = 0).
For x = 1, 2, 3, etc.,assuming that either w1 = 7 or w2 = 7
leads to acontradiction because a_1(3)/7 is not an
algebraicinteger (as you know). The only non-explicit part of
this is the GCDfunction, which is however guaranteed to exist
bya deep theorem of Dedekind. > As a sidenote, when x=1, it IS
the case that you're pushed out of the> ring of algebraic
integeas w_1(1) = w_2(1) = sqrt(7).> That looks like>
(5a_1(1)/sqrt(7) + sqrt(7))(5b_2(1)/sqrt(7) + 2/sqrt(7)) =
25(1)^2 +> 30(1) + 2> and it turns out that Rick Decker's
example is a special case where> that happens with an integer
x.> Notice that everything follows from figuring out the
factors of the> constant term with> (5a_1(3) + 7)(5b_2(3) + 2)
= 7(25(3)^2 + 30(3) + 7(2)> as once they're identified, you can
track what happens to them when> you divide both sides by 7.>
The argument is simple enough that I find myself looking for>
*emotional* reasons to explain why people keep arguing with me
about> it.> What's especially frustrating to me is that the
argument is rather> basic, but in response I see people who
seem intent on bulldozing> through with VERY LONG replies,
where they usually try to claim> there's some alternate
argument supporting their position.> Somehow I doubt it'll be
different this time, but I like the exercise> of going over
===
conclusions> I ended up in a discussion with a poster who
seemed to believe that I> don't consider the possibility that
I'm wrong. I've seen that> position put up quite a few times,
and I think it's worth addressing,> so I'll go back yet again
to an example pos by a Rick Decker, a> professor at Hamilton
College, to go over what I say happens and> consider how I
might be wrong.> Here's some identifying to find Decker's
post, which also shows that> he is indeed at Hamilton
===
College:> Subject: Re: Mathematical consistency, courage> In
his post Decker claimed to mirror my argument using a
quadratic> instead of a cubic, where he has> (5a_1(x) +
7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where I'll now make a
slight change to have> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 +
30x) + 7(2) > and his a's are roots of > a^2 - (x - 1)a +
7(x^2 + x).> My emphasis has been on the actual constant terms
as while it appears> fomr that factorization that both are 7 on
the left that doesn't add> up with what's on the right, so I
set x=0 to find that one of the a's> equals 0, then, as> a^2 +
a = 0> and picking a_1(0) = 0, then a_2(1) = -1, > so it makes
sense I think now to introduce b_2(x), where > a_2(x) = b_2(x)
- 1> so that at x=0, b_2(0) = 0, and making that substitution
gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x) + 7(2) >
and now the constant terms match up. Now I've seen posters
make a big> deal out of even the phrase constant terms so I've
now moved to> getting rid of them!!!> That's easy enough as all
I have to do is multiply out to get> 25 a_1(x) b_2(x) + 10
a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x) + 14> and not
surprisingly the constant terms match up, so let's delete
them> off!!!> That gives> 25 a_1(x) b_2(x) + 10 a_1(x) + 35
b_2(x) = 7(25x^2 + 30x).In my previous post I went on a long,
involved and unnecessarycalculation which I then managed to
screw up evaluating at x=-1. Ohwell.It turns out, of course,
that it's easy enough to calculate b_2(x),froma_2(x) = b_2(x)
- 1anda_2(x)^2 - (x - 1)a_2(x) + 7(x^2 + x) = 0,as
substituting givesb_2(x)^2 - 2b_2(x) + 1 -x b_2(x) + x +
b_2(x) -1 + 7(x^2 + x) = 0which isb_2(x)^2 -(x+1) b_2(x) +
7x^2 + 8x = 0which givesb_2(x) = ((x+1)+/- sqrt(x^2 + 2x + 1 -
28 x^2 - 32 x))/2which isb_2(x) = ((x+1) +/- sqrt(-27x^2 - 30x
===
previous post I went on a long, involved and unnecessary>
calculation which I then managed to screw up evaluating at
x=-1. Oh> well.Oops!--There are two things you must never
attempt to prove: the unprovable -- and the
===
Questioning my conclusions> I ended up in a discussion with a
poster who seemed to believe that I> don't consider the
possibility that I'm wrong. I've seen that> position put up
quite a few times, and I think it's worth addressing,> so I'll
go back yet again to an example pos by a Rick Decker, a>
professor at Hamilton College, to go over what I say happens
and> consider how I might be wrong.> Here's some identifying
to find Decker's post, which also shows that> he is indeed at
===
Hamilton College:> Subject: Re: Mathematical consistency,
courage> In his post Decker claimed to mirror my argument
using a quadratic> instead of a cubic, where he has> (5a_1(x)
+ 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where I'll now make a
slight change to have> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 +
30x) + 7(2) > and his a's are roots of > a^2 - (x - 1)a +
7(x^2 + x).> My emphasis has been on the actual constant terms
as while it appears> fomr that factorization that both are 7 on
the left that doesn't add> up with what's on the right, so I
set x=0 to find that one of the a's> equals 0, then, as> a^2 +
a = 0> and picking a_1(0) = 0, then a_2(1) = -1, > so it makes
sense I think now to introduce b_2(x), where > a_2(x) = b_2(x)
- 1> so that at x=0, b_2(0) = 0, and making that substitution
gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x) + 7(2) >
and now the constant terms match up. Now I've seen posters
make a big> deal out of even the phrase constant terms so I've
now moved to> getting rid of them!!!> That's easy enough as all
I have to do is multiply out to get> 25 a_1(x) b_2(x) + 10
a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x) + 14> and not
surprisingly the constant terms match up, so let's delete
them> off!!!> That gives> 25 a_1(x) b_2(x) + 10 a_1(x) + 35
b_2(x) = 7(25x^2 + 30x).I can divide both sides by 5, which
gives5 a_1(x) b_2(x) + 2 a_1(x) + 7 b_2(x) = 7(5x^2 + 6x)and
group to geta_1(x)(5 b_2(x) + 2) = 7(5x^2 + 6x) - 7b_2(x)and
now solve for a_1(x) to geta_1(x) = [7(5x^2 + 6x) -
7b_2(x)]/(5b_2(x) + 2).a^2 - (x - 1)a + 7(x^2 + x), soa_1(x)^2
- (x-1)a_1(x) + 7(x^2 + x) = 0,and substituting gives[7(5x^2 +
6x) - 7b_2(x)]^2/(5b_2(x) + 2)^2 - (x-1)[7(5x^2 + 6x)
-7b_2(x)]/(5b_2(x) + 2) + 7(x^2 + x) = 0and multiplying both
sides by (5b_2(x) + 2)^2 gives[7(5x^2 + 6x) - 7b_2(x)]^2 -
(x-1)[7(5x^2 + 6x) - 7b_2(x)](5b_2(x) +2) + 7(x^2 + x)(5b_2(x)
+ 2)^2 = 0Expanding a bit49[(5x^2 + 6x)^2 -2(5x^2 + 6x)b_2(x) +
b_2(x)^2] - 7(x-1)[5(5x^2 +6x)b_2(x) - 5b_2(x)^2 + 2(5x^2 + 6x)
- 2b_2(x)] + 7(x^2+ x)(25b_2(x)^2+ 20b_2(x) + 4) = 0simplifying
a bit49[(5x^2 + 6x)^2 -2(5x^2 + 6x)b_2(x) + b_2(x)^2] -
7(x-1)[(25x^2 + 30x- 2)b_2(x) - 5b_2(x)^2 + 2(5x^2 + 6x) ] +
7(x^2+ x)(25b_2(x)^2 +20b_2(x) + 4) = 0Simplifying and
grouping gives7[7 + 5(x-1) + 25(x^2+x)]b_2(x)^2 +[49(-2)(5x^2
+ 6x) -7(x-1)(25x^2 +30x - 2) + 7(20(x^2 + x)]b_2(x) + 49(5x^2
+ 6x)^2 - 14(x-1)(5x^2 + 6x)+ 28(x^2 + x) = 0.Dividing by 7
gives[7 + 5(x-1) + 25(x^2+x)]b_2(x)^2 +[7(-2)(5x^2 + 6x)
-(x-1)(25x^2 + 30x- 2) + (20(x^2 + x)]b_2(x) + 7(5x^2 + 6x)^2
- 2(x-1)(5x^2 + 6x) +4(x^2 + x) = 0and that's intimidating
enough that I'll use x=-1, to get somethingless complica:[7 -
10 + 25(1 - 1)]b_2(-1)^2 +[7(-2)(5 - 6) -(-2)(25 - 30 - 2)
+(20(1 - 1)]b_2(-1) + 7(5 - 6)^2 - 2(-2)(5 - 6) + 4(1 - 1) =
0which is[-3 ]b_2(-1)^2 +[14 +(2)(7)]b_2(-1) + 7 - 4 = 0which
is-3 b_2(-1)^2 + 28b_2(-1) + 3 = 0which verifies that it's
non-monic and irreducible over Q.Now then, the point is that
*in general* b_2(x) is NOT an algebraicinteger function!!!Get
===
you're at it again... > Now nothing spectacular so far, so
let's move to where there's been > EXTREME disagreement as my
position is that from the constant terms it > only makes sense
that if you divide both sides of > (5a_1(x) + 7)(5b_2(x) + 2)
= 7(25x^2 + 30x + 2) > by 7, you end up with > (5a_1(x)/7 +
1)(5b_2(x) + 2) = 25x^2 + 30x + 2 > if you're to remain in
the ring of algebraic integers.No, James, and you *know* that.
If you do that you do *not* remainin the ring algebraic
integers. It is your wish that you *should*remain in that
ring, but there is nothing that makes it necessary.Because
there are other ways that you can do the division suchthat you
remain in the ring of algebraic integers. > Now that might not
impress some of you, so I'm going to let x=3, and > consider
what would happen if somehow there was a *different* way to >
divide both sides of > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2
+ 30x + 2) > by 7, so let x=3, then > (5a_1(3) +
7)(5b_2(3) + 2) = 7(25(3)^2 + 30(3) + 2) > and even if you
believe that 7 splits up as a *function* of x to > divide the
left side, you should also believe that at x=3 its factors >
actually have some definite value. > So let's use w_1(3),
and w_2(3) for those factors where w_1(3) w_2(3) > = 7, and
dividing both sides gives > (5a_1(3)/w_1(3) +
w_2(3))(5b_2(3)/w_2(3) + 2/w_2(3)) = 25(3)^2 + 30(3) > + 2No
James. It does *not* give that if you wish to stay in the
ringof algebraic integers. It gives: [(5a_1(3) +
7)/w_1(3)][(5b_2(3) + 2)/w_2(3)] = 25(3)^2 + 30(3) + 2The two
are different. While with this formulation you stay in thering
of algebraic integewith your formulation you do not.
Especially,while [(5b_2(3) + 2)/w_2(3)] is an algebraic
integer, this does not implythat both 5b_2(3)/w_2(3) and
2/w_2(3) are (or should be) algebraic integers.You are using
some sort of an inverse form of the distributive propertywhich
does not apply when you are not in a field. > but now there's a
problem as 2/w_2(3) isn't in the ring of algebraic > integers
if w_2(3) isn't a unit, as 2 and 7 are coprime.There is no
need, because in the correct formulation that term doesnot
occur. > Now the way I see it w_2(3) and 2/w_2(3) are factors
of 2 on the right > side, so there's no way to argue with
2/w_2(3) being forbidden in the > ring of algebraic integebut
some of you might wish to claim that > you can escape having
to have such a factor by concluding that > (5b_2(3) +
2)/w_2(3) is in the ring. > you can simply mush all your
numbers together here and have it all > work out in the
end.Yup, indeed. > That's actually kind of funny as a math
position. I'll call it the > mushing position.Oh. > So then
from my analysis, for me to be wrong, it must be true for >
2/w_2(3) to be in the ring of algebraic integer when w_2(3) is
not a > unit, which is a contradiction.No. You're analysis is
wrong. You say that with algebraic integerfunctions p(x) and
q(x) and algebraic integer r, that when(p(x) + r)/q(x) is an
algebraic integer, r/q(x) should be an algebraicinteger.
Something that does not even hold in the ring of integers.--
===
a discussion with a poster who seemed to believe that I>don't
consider the possibility that I'm wrong. Who cares whether you
consider the possibility?You _are_ wrong - the fact that you
don't realizethat after so many careful explanations is a
lotmore significant than the question of whether youconsider
the possibilty that you might be
wrong.**************************As far as I'm concerend you're
trying to wait until I die, so I figuremaybe you should die
instead. How about that, eh? Wouldn't that be abetter
twist?You refuse to follow the math, so the great Powers that
controlreality and *speak* in mathematics decide to kill you
instead of me.So what do you think about that, eh? Oh, can't
hear Them talking?Well, I guess that's because you don't
really understand Mathematics,the true language, which is THE
language.They're talking about you now, and They agree with my
assessment, andwill not penalize me as They allowed the others
like Galois and Abelto be penalized.They will kill you
===
Re: JSH: Questioning my conclusions> I ended up in a
discussion with a poster who seemed to believe that I> don't
consider the possibility that I'm wrong. I've seen that>
position put up quite a few times, and I think it's worth
addressing,> so I'll go back yet again to an example pos by a
Rick Decker, a> professor at Hamilton College, to go over what
I say happens and> consider how I might be wrong.> Here's some
identifying to find Decker's post, which also shows that> he
===
is indeed at Hamilton College:> Subject: Re: Mathematical
consistency, courage> In his post Decker claimed to mirror my
argument using a quadratic> instead of a cubic, where he has>
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where I'll
now make a slight change to have> (5a_1(x) + 7)(5a_2(x) + 7) =
7(25x^2 + 30x) + 7(2) > and his a's are roots of > a^2 - (x -
1)a + 7(x^2 + x).> My emphasis has been on the actual constant
terms as while it appears> fomr that factorization that both
are 7 on the left that doesn't add> up with what's on the
right, so I set x=0 to find that one of the a's> equals 0,
then, as> a^2 + a = 0> and picking a_1(0) = 0, then a_2(1) =
-1, > so it makes sense I think now to introduce b_2(x), where
> a_2(x) = b_2(x) - 1> so that at x=0, b_2(0) = 0, and making
that substitution gives> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2
+ 30x) + 7(2) > and now the constant terms match up. Now I've
seen posters make a big> deal out of even the phrase constant
terms so I've now moved to> getting rid of them!!!> That's
easy enough as all I have to do is multiply out to get> 25
a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x) +
14> and not surprisingly the constant terms match up, so let's
delete them> off!!!> That gives> 25 a_1(x) b_2(x) + 10 a_1(x) +
35 b_2(x) = 7(25x^2 + 30x).> Now nothing spectacular so far, so
let's move to where there's been> EXTREME disagreement as my
position is that from the constant terms it> only makes sense
that if you divide both sides of> (5a_1(x) + 7)(5b_2(x) + 2) =
7(25x^2 + 30x + 2) > by 7, you end up with> (5a_1(x)/7 +
1)(5b_2(x) + 2) = 25x^2 + 30x + 2> if you're to remain in the
ring of algebraic integers. ... and of course you know, and we
agree, that when you divide by 7 in this way you do NOT remain
in the algebraic integers: inparticular, a_1(x)/7 in general
is not an algebraic integer. > That's an important point.>
Multiplying out as before gives me> 25 a_1(x) b_2(x)/7 + 10
a_2(x)/7 + 5 b_2(x) + 2 = 25x^2 + 30x + 2> and again getting
rid of the constant terms, I have> 25 a_1(x) b_2(x)/7 + 10
a_2(x)/7 + 5 b_2(x) = 25x^2 + 30x> and from before I had> 25
a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)> and
dividing both sides of it gives> 25 a_1(x) b_2(x)/7 + 10
a_1(x)/7 + 5 b_2(x) = 25x^2 + 30x,> so the answers the same.>
Now that might not impress some of you, so I'm going to let
x=3, and> consider what would happen if somehow there was a
*different* way to> divide both sides of> (5a_1(x) +
7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> by 7, so let x=3, then>
(5a_1(3) + 7)(5b_2(3) + 2) = 7(25(3)^2 + 30(3) + 2)> and even
if you believe that 7 splits up as a *function* of x to>
divide the left side, you should also believe that at x=3 its
factors> actually have some definite value.> So let's use
w_1(3), and w_2(3) for those factors where w_1(3) w_2(3)> = 7,
and dividing both sides gives> (5a_1(3)/w_1(3) +
w_2(3))(5b_2(3)/w_2(3) + 2/w_2(3)) = 25(3)^2 + 30(3)> + 2 Yes.
This is exactly what happens!> but now there's a problem as
2/w_2(3) isn't in the ring of algebraic> integers if w_2(3)
isn't a unit, as 2 and 7 are coprime. Correct.> Now the way I
see it w_2(3) and 2/w_2(3) are factors of 2 on the right>
side, so there's no way to argue with 2/w_2(3) being forbidden
in the> ring of algebraic integebut some of you might wish to
claim that> you can escape having to have such a factor by
concluding that> (5b_2(3) + 2)/w_2(3) is in the ring. Also
correct. Remember that you can re-write (5b_2(3) + 2) as
(5a_2(3) + 7), and the quotient becomes 5a_2(3)/w_2(3) +
7/w_2(3).Note that since w_2(3) is a factor of 7, we must have
7/w_2(3) *is* an algebraic integer.With the *correct*
definition of w_2(3), it is also the case that
5a_2(3)/w_2(3)is an algebraic integer. Put all this together,
and it means that (5b_2(3) + 2)/w_2(3) *is also* an algebraic
integer: which is exactly what you need inthe factorization.>
you can simply mush all your numbers together here and have it
all> work out in the end. Exactly right. > That's actually kind
of funny as a math position. I'll call it the> mushing
position. Sure, it's kind of funny. It's unintuitive. That
doesn't mean it's wrong. Do you have a *proof* that it cannot
be true, or just a feeling thatit is kind of funny ?> So then
from my analysis, for me to be wrong, it must be true for>
2/w_2(3) to be in the ring of algebraic integer when w_2(3) is
not a> unit, which is a contradiction. No, that's where you're
wrong. The mushing position is right!2/w_2(3) is NOT an
algebraic integer. Remember: it is quite possible for the sum
of two things that are not algebraic integersto be an
algebraic integer. That is what happens here.> I just don't
accept that mushing position as it's not mathematics. That's
where you are wrong. And saying it's not mathematicsis *not* a
disproof. It's just an emotional statement.> As a sidenote,
when x=1, it IS the case that you're pushed out of the> ring
of algebraic integeas w_1(1) = w_2(1) = sqrt(7). By which you
mean that 2/w_2(1) is not an algebraic integer.Absolutely
correct. But look at what happens in that case to
thefactorization: a_1(1) = a_2(1) = sqrt(-14), and b_2(1) =
sqrt(-14) + 1 Therefore (5*b_2(1) + 2)/sqrt(7) = (5*sqrt(-14)
+ 5 + 2)/sqrt(7) = (5*sqrt(-2) + 7/sqrt(7)) = (5*sqrt(-2) +
sqrt(7)),which - lo and behold - IS an algebraic integer.This
is the mushing position!This is just simple arithmetic. The
whole point of Decker's examplewas, with this choice of w_1(1)
and w_2(1), the expression factorswith all the coefficients
being algebraic integers. This is exactly whatyou say cannot
happen. The mushing position is demonstrato be exactly
correct. It's just simple arithmetic. > That looks like>
(5a_1(1)/sqrt(7) + sqrt(7))(5b_2(1)/sqrt(7) + 2/sqrt(7)) =
25(1)^2 +> 30(1) + 2 You're back to the wrong factorization
again: you keep insisting on7 = 7 * 1 is the only possibility.
You should have left it as sqrt(7)*sqrt(7). It's hard to
believe you don't understand this.> and it turns out that Rick
Decker's example is a special case where> that happens with an
integer x. It happens with virtually all other integers x
also. The expressions aremore complica. But the principle is
exactly the same. It happens whenever the polynomial a^2 - (x
- 1)*a + 7*(x^2 + x)is irreducible. It is irreducible when x =
1, 2, 3, and most otherinteger values. Not, however, when x =
0. x = 0 is a special,singular case. You cannot generalize
from it to other values.> Notice that everything follows from
figuring out the factors of the> constant term with> (5a_1(3)
+ 7)(5b_2(3) + 2) = 7(25(3)^2 + 30(3) + 7(2)> as once they're
identified, you can track what happens to them when> you
divide both sides by 7.> The argument is simple enough that I
find myself looking for> *emotional* reasons to explain why
people keep arguing with me about> it. Not at all. Try looking
in an objective, unbiased way at the arithmetic above. There is
nothing wrong with it. If you findexplicitly where there is an
error, point it out. Whatever your emotional investment is in
this, if you are honest with yourself for even a few seconds,
you will at a minimum admit that: it's kind of funny it's the
mushing position it's not mathematics--- NONE of these is even
remotely a mathematical argument. They are just gut feelings.
If there is one thing you should have learned in the past 8
yeait is that gut feelings are a good guide to what is true -
especially not when it comes to the ring of algebraic
integers.> What's especially frustrating to me is that the
argument is rather> basic, but in response I see people who
seem intent on bulldozing> through with VERY LONG replies,
where they usually try to claim> there's some alternate
argument supporting their position.> Somehow I doubt it'll be
different this time, but I like the exercise> of going over
the argument yet again. It won't be different. You are still
===
Efficiency> I already answered this in alt.algebra.help. It is
not useful to ask the> same> question multiple times.Bill>
Sorry, I didn't know.Depends on the groups. It's a question of
the local mores. If you reallywant to troll, try crossposting
the same question to an evolution group anda fundamentalist
Christian group.But if a serious question is really relevant
to two or more groups, go aheadand crosspost. That way
===
JSH: Algebra is supremeThe algebra is simple. And algebra IS
supreme, whether this or thatposter wishes to challenge
it.Rick Decker, a professor at Hamilton College, gave this
example, whichI like because it's a quadratic, which is a lot
easier to play withthan the cubics I've used before. If you
want to see his originalpost here are some headers which also
===
show that he is indeed atHamilton College:Subject: Re:
Mathematical consistency, courageDecker put forward the
quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where
his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).I think that
is what leaves some of you guessing and hesitant becauseit's
not a factorization like you're used to seeing.Normally you
see something like(x+2)(x+1) = x^2 + 3x + 2and there's nothing
hidden, it's easy. There's no need to trust thealgebra, as you
can trace everything out *easily*, while the Deckerexample
forces you to accept algebra not because you can see
everydetail, but because it's true.For instance,
algebraically, factorizations multiply out in a certainway
demonstra by(a+b)(c+d) = ac + ad + bc + bdand that's just a
FACT which is not open to debate.Looking at(5a_1(x) +
7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) it IS true that25 a_1(x)
a_2(x) + 35 a_1(x) + 35 a_2(x) + 49 = 7(25x^2 + 30x + 2).Now
subtracting 14 from both sides gives25 a_1(x) a_2(x) + 35
a_1(x) + 35 a_2(x) + 35 = 7(25x^2 + 30x).Now notice that you
have 35 on the left side, but 0 on the right interms of
*constants* i.e. terms that don't vary as x varies.That is
just algebra. It's simple, and direct.What I do is to use
a_2(x) = b_2(x) - 1, so that I have(5a_1(x) + 7)(5b_2(x) + 2)
= 7(25x^2 + 30x + 2) which again is JUST ALGEBRA, so there
isn't anything weird.Now multiplying out I get25 a_1(x) b_2(x)
+ 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x + 2)and
subtracting 14 from both sides gives25 a_1(x) b_2(x) + 10
a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)which again is JUST
ALGEBRA, so there's NO ROOM FOR DEBATE.Now then, clearly
with(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)the 7 and 2
visible to the naked eye on the left, are factors of 7(2)on the
right, which is verified by multiplying out!!!It's just
algebra. There's no room for debate. There's no poll totake,
and no reason to go look for votes either way!The simple fact
is that 7 and 2 on the left are factors of 7(2) on theright,
and you can multiply out, and subtract 14 from both sides
toget25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 +
30x)and it's not magic that the *constants* are now gone from
bothsides!!!It's ALGEBRA.Now then, if 7 and 2 on the left are
the factors of 7(2) on the right,then what happens if you
divide by 7?Well you get 1 and 2 on the left as the factors of
2 on the right.That looks like(5a_1(x)/7 + 1)(5b_2(x) + 2) =
25x^2 + 30x + 2and here's where certain posters decide to
attack algebra, as if it'sa debate, as if it's a democracy, as
if mathematics gives a damn thatthey don't like a
conclusion!That's because the two factors (5a_1(x)/7 + 1) and
(5b_2(x) + 2) arenot necessarily algebraic integeand for some
values, like withx=1, or x=1 mod 7, you can *see* that they
aren't algebraic integers!The problem is that these posters
believe that there's SOME WAY youcan divide 7 from both sides
*in general* and get algebraic integerfactors on the left
side!That's it!!!That's what all the arguing is about because
the algebra say NO, andthey say YES, and I keep saying what
the algebra says, and they keepsaying what they say, and for
some reason most of you seem todisbelieve the algebra!!!The
problem is that if you divide both sides by 7, and decide that
onefactor has sqrt(7) as well as the other then you
get(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) =
25x^2 + 30x + 2and that's JUST ALGEBRA, there's NO ROOM FOR
DEBATE!!!No point in calling your congressman to promote your
right to have amathematical conclusion that fits your
needs!!!The FACT is that if both factors have sqrt(7) as a
factor for ANYVALUE of x, then what you have
is(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) =
25x^2 + 30x + 2and that gives the factors of 2 on the right
assqrt(7) and 2/sqrt(7)and if you argue with that ALGEBRAIC
FACT then you are debasingyourself from the realm of sentient
debate into an animal realm whereyou're showing a weak need to
push whatever the hell makes you happyover LOGIC!!!IF BOTH
FACTORS HAVE sqrt(7) AS A FACTOR THEN ALGEBRA TELLS YOU
THAT(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) =
25x^2 + 30x + 2which tells you that the factors of 2 on the
right are sqrt(7) and 2/sqrt(7)which works to give you 2, BUT
NOT IN THE RING OF ALGEBRAIC INTEGERS!!!The logical conclusion
which follows mathematically is that given(5a_1(x) + 7)(5b_2(x)
+ 2) = 7(25x^2 + 30x + 2)dividng both sides by 7 is not
possible within the ring of algebraicintegers in general.That
is, there is no decomposition of 7 into algebraic integer
factorsof(5a_1(x) + 7) and (5b_2(x) + 2)in general in the ring
of algebraic integers.That's it! Sorry if it hurts your
feelings. Sorry if it makes youwant to go cry to God or your
mother because it JUST IS THAT WAY!I didn't make it that
way!!!I'm not forcing algebra to be something else just to
ruin your life,hurt your feelings, make you mad, or do
anything at all because IT'SNOT POSSIBLE TO CHANGE THE
ALGEBRA!ALGEBRA IS SUPREME!It's the supreme authority
===
wave equation, in its differential form[- d^2/d^2 x + V(x)]
Psi = E PsiFor some reason, people expect this system to
simula for every possiblehistory, numerically. However, what
they fail to realizes is that one canwrite this equation, in
alegbraic form, it is simula for all possiblehistory in just 1
step.For example,X + Y = Zif restric for reals X,Y, Z, would be
true for all X , Y, Z, withoutactually simulating the system.>
The algebra is simple. And algebra IS supreme, whether this or
that> poster wishes to challenge it.> Rick Decker, a professor
at Hamilton College, gave this example, which> I like because
it's a quadratic, which is a lot easier to play with> than the
cubics I've used before. If you want to see his original> post
here are some headers which also show that he is indeed at>
===
Hamilton College:> Subject: Re: Mathematical consistency,
courage> Decker put forward the quadratic> (5a_1(x) +
7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)> where his a's are roots
of> a^2 - (x - 1)a + 7(x^2 + x).> I think that is what leaves
some of you guessing and hesitant because> it's not a
factorization like you're used to seeing.> Normally you see
something like> (x+2)(x+1) = x^2 + 3x + 2> and there's nothing
hidden, it's easy. There's no need to trust the> algebra, as
you can trace everything out *easily*, while the Decker>
example forces you to accept algebra not because you can see
every> detail, but because it's true.> For instance,
algebraically, factorizations multiply out in a certain> way
demonstra by> (a+b)(c+d) = ac + ad + bc + bd> and that's just
a FACT which is not open to debate.> Looking at> (5a_1(x) +
7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)> it IS true that> 25
a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 49 = 7(25x^2 + 30x +
2).> Now subtracting 14 from both sides gives> 25 a_1(x)
a_2(x) + 35 a_1(x) + 35 a_2(x) + 35 = 7(25x^2 + 30x).> Now
notice that you have 35 on the left side, but 0 on the right
in> terms of *constants* i.e. terms that don't vary as x
varies.> That is just algebra. It's simple, and direct.> What
I do is to use a_2(x) = b_2(x) - 1, so that I have> (5a_1(x) +
7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> which again is JUST
ALGEBRA, so there isn't anything weird.> Now multiplying out I
get> 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 +
30x + 2)> and subtracting 14 from both sides gives> 25 a_1(x)
b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)> which again
is JUST ALGEBRA, so there's NO ROOM FOR DEBATE.> Now then,
clearly with> (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)>
the 7 and 2 visible to the naked eye on the left, are factors
of 7(2)> on the right, which is verified by multiplying
out!!!> It's just algebra. There's no room for debate. There's
no poll to> take, and no reason to go look for votes either
way!> The simple fact is that 7 and 2 on the left are factors
of 7(2) on the> right, and you can multiply out, and subtract
14 from both sides to> get> 25 a_1(x) b_2(x) + 10 a_1(x) + 35
b_2(x) = 7(25x^2 + 30x)> and it's not magic that the
*constants* are now gone from both> sides!!!> It's ALGEBRA.>
Now then, if 7 and 2 on the left are the factors of 7(2) on
the right,> then what happens if you divide by 7?> Well you
get 1 and 2 on the left as the factors of 2 on the right.>
That looks like> (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x +
2> and here's where certain posters decide to attack algebra,
as if it's> a debate, as if it's a democracy, as if
mathematics gives a damn that> they don't like a conclusion!>
That's because the two factors (5a_1(x)/7 + 1) and (5b_2(x) +
2) are> not necessarily algebraic integeand for some values,
like with> x=1, or x=1 mod 7, you can *see* that they aren't
algebraic integers!> The problem is that these posters believe
that there's SOME WAY you> can divide 7 from both sides *in
general* and get algebraic integer> factors on the left side!>
That's it!!!> That's what all the arguing is about because the
algebra say NO, and> they say YES, and I keep saying what the
algebra says, and they keep> saying what they say, and for
some reason most of you seem to> disbelieve the algebra!!!>
The problem is that if you divide both sides by 7, and decide
that one> factor has sqrt(7) as well as the other then you
get> (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7))
=> 25x^2 + 30x + 2> and that's JUST ALGEBRA, there's NO ROOM
FOR DEBATE!!!> No point in calling your congressman to promote
your right to have a> mathematical conclusion that fits your
needs!!!> The FACT is that if both factors have sqrt(7) as a
factor for ANY> VALUE of x, then what you have is>
(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) =>
25x^2 + 30x + 2> and that gives the factors of 2 on the right
as> sqrt(7) and 2/sqrt(7)> and if you argue with that
ALGEBRAIC FACT then you are debasing> yourself from the realm
of sentient debate into an animal realm where> you're showing
a weak need to push whatever the hell makes you happy> over
LOGIC!!!> IF BOTH FACTORS HAVE sqrt(7) AS A FACTOR THEN
ALGEBRA TELLS YOU THAT> (5a_1(x)/sqrt(7) +
sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) => 25x^2 + 30x + 2>
which tells you that the factors of 2 on the right are>
sqrt(7) and 2/sqrt(7)> which works to give you 2,> BUT NOT IN
THE RING OF ALGEBRAIC INTEGERS!!!> The logical conclusion
which follows mathematically is that given> (5a_1(x) +
7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)> dividng both sides by 7
is not possible within the ring of algebraic> integers in
general.> That is, there is no decomposition of 7 into
algebraic integer factors> of> (5a_1(x) + 7) and (5b_2(x) +
2)> in general in the ring of algebraic integers.> That's it!
Sorry if it hurts your feelings. Sorry if it makes you> want
to go cry to God or your mother because it JUST IS THAT WAY!>
I didn't make it that way!!!> I'm not forcing algebra to be
something else just to ruin your life,> hurt your feelings,
make you mad, or do anything at all because IT'S> NOT POSSIBLE
TO CHANGE THE ALGEBRA!> ALGEBRA IS SUPREME!> It's the supreme
===
funnierTake the differential equation of the harmonic
osscilator(d^2/dt^2 + 1/2 m w^2 x^2 ) Psi = E PsiEvery kid who
scored 100% in an introductory quantum mechanics class
incollege, has been able to find an exact solution in
algebraic form. Doesthat mean that, he implicitly simula all
possible such systems, in just afew steps? Or is there, just
one history because it would be economical ?-sureshThe Lord of
the Rain( Suresh __NoJunkMail kumar)> It's funny.> Take the
wave equation, in its differential form> [- d^2/d^2 x + V(x)]
Psi = E Psi> For some reason, people expect this system to
simula for every possible> history, numerically. However, what
they fail to realizes is that one can> write this equation, in
alegbraic form, it is simula for all possible> history in just
1 step.> For example,> X + Y = Z> if restric for reals X,Y, Z,
would be true for all X , Y, Z, without> actually simulating
the system.> The algebra is simple. And algebra IS supreme,
whether this or that> poster wishes to challenge it.Rick
Decker, a professor at Hamilton College, gave this example,
which> I like because it's a quadratic, which is a lot easier
to play with> than the cubics I've used before. If you want to
see his original> post here are some headers which also show
===
that he is indeed at> Hamilton College:Subject: Re:
Mathematical consistency, courageDecker put forward the
quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)where
his a's are roots ofa^2 - (x - 1)a + 7(x^2 + x).I think that
is what leaves some of you guessing and hesitant because> it's
not a factorization like you're used to seeing.Normally you see
something like(x+2)(x+1) = x^2 + 3x + 2and there's nothing
hidden, it's easy. There's no need to trust the> algebra, as
you can trace everything out *easily*, while the Decker>
example forces you to accept algebra not because you can see
every> detail, but because it's true.For instance,
algebraically, factorizations multiply out in a certain> way
demonstra by(a+b)(c+d) = ac + ad + bc + bdand that's just a
FACT which is not open to debate.Looking at(5a_1(x) +
7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)it IS true that25 a_1(x)
a_2(x) + 35 a_1(x) + 35 a_2(x) + 49 = 7(25x^2 + 30x + 2).Now
subtracting 14 from both sides gives25 a_1(x) a_2(x) + 35
a_1(x) + 35 a_2(x) + 35 = 7(25x^2 + 30x).Now notice that you
have 35 on the left side, but 0 on the right in> terms of
*constants* i.e. terms that don't vary as x varies.That is
just algebra. It's simple, and direct.What I do is to use
a_2(x) = b_2(x) - 1, so that I have(5a_1(x) + 7)(5b_2(x) + 2)
= 7(25x^2 + 30x + 2)which again is JUST ALGEBRA, so there
isn't anything weird.Now multiplying out I get25 a_1(x) b_2(x)
+ 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x + 2)and
subtracting 14 from both sides gives25 a_1(x) b_2(x) + 10
a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)which again is JUST
ALGEBRA, so there's NO ROOM FOR DEBATE.Now then, clearly
with(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)the 7 and 2
visible to the naked eye on the left, are factors of 7(2)> on
the right, which is verified by multiplying out!!!It's just
algebra. There's no room for debate. There's no poll to> take,
and no reason to go look for votes either way!The simple fact
is that 7 and 2 on the left are factors of 7(2) on the> right,
and you can multiply out, and subtract 14 from both sides to>
get25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 +
30x)and it's not magic that the *constants* are now gone from
both> sides!!!It's ALGEBRA.Now then, if 7 and 2 on the left
are the factors of 7(2) on the right,> then what happens if
you divide by 7?Well you get 1 and 2 on the left as the
factors of 2 on the right.That looks like(5a_1(x)/7 +
1)(5b_2(x) + 2) = 25x^2 + 30x + 2and here's where certain
posters decide to attack algebra, as if it's> a debate, as if
it's a democracy, as if mathematics gives a damn that> they
don't like a conclusion!That's because the two factors
(5a_1(x)/7 + 1) and (5b_2(x) + 2) are> not necessarily
algebraic integeand for some values, like with> x=1, or x=1
mod 7, you can *see* that they aren't algebraic integers!The
problem is that these posters believe that there's SOME WAY
you> can divide 7 from both sides *in general* and get
algebraic integer> factors on the left side!That's it!!!That's
what all the arguing is about because the algebra say NO, and>
they say YES, and I keep saying what the algebra says, and
they keep> saying what they say, and for some reason most of
you seem to> disbelieve the algebra!!!The problem is that if
you divide both sides by 7, and decide that one> factor has
sqrt(7) as well as the other then you get(5a_1(x)/sqrt(7) +
sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2and
that's JUST ALGEBRA, there's NO ROOM FOR DEBATE!!!No point in
calling your congressman to promote your right to have a>
mathematical conclusion that fits your needs!!!The FACT is
that if both factors have sqrt(7) as a factor for ANY> VALUE
of x, then what you have is(5a_1(x)/sqrt(7) +
sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = 25x^2 + 30x + 2and
that gives the factors of 2 on the right assqrt(7) and
2/sqrt(7)and if you argue with that ALGEBRAIC FACT then you
are debasing> yourself from the realm of sentient debate into
an animal realm where> you're showing a weak need to push
whatever the hell makes you happy> over LOGIC!!!IF BOTH
FACTORS HAVE sqrt(7) AS A FACTOR THEN ALGEBRA TELLS YOU
THAT(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) =
25x^2 + 30x + 2which tells you that the factors of 2 on the
right aresqrt(7) and 2/sqrt(7)which works to give you 2,BUT
NOT IN THE RING OF ALGEBRAIC INTEGERS!!!The logical conclusion
which follows mathematically is that given(5a_1(x) + 7)(5b_2(x)
+ 2) = 7(25x^2 + 30x + 2)dividng both sides by 7 is not
possible within the ring of algebraic> integers in
general.That is, there is no decomposition of 7 into algebraic
integer factors> of(5a_1(x) + 7) and (5b_2(x) + 2)in general in
the ring of algebraic integers.That's it! Sorry if it hurts
your feelings. Sorry if it makes you> want to go cry to God or
your mother because it JUST IS THAT WAY!I didn't make it that
way!!!I'm not forcing algebra to be something else just to
ruin your life,> hurt your feelings, make you mad, or do
anything at all because IT'S> NOT POSSIBLE TO CHANGE THE
ALGEBRA!ALGEBRA IS SUPREME!It's the supreme authority
===
simple.Then maybe you'd better start using the algebra,
instead of whatever thehell it is that you've been using the
===
algebra is simple. And algebra IS supreme, whether this or
that>poster wishes to challenge it.Nobody's been challenging
===
is supreme[snip boring nonsense full of gloating over
discovering the obvious]> The logical conclusion which follows
mathematically is that given> (5a_1(x) + 7)(5b_2(x) + 2) =
7(25x^2 + 30x + 2)> dividng both sides by 7 is not possible
within the ring of algebraic> integers in general.> That is,
there is no decomposition of 7 into algebraic integer factors>
of> (5a_1(x) + 7) and (5b_2(x) + 2)> in general in the ring of
algebraic integers.You showed two candidate decompositions
which did not work. It DOES NOTFOLLOW that there is no
decomposition which will work. That is anunsuppor conclusion.
Logic says that you have found two cases whichfail -- where
does it say that ALL cases will fail?> That's it! Sorry if it
hurts your feelings. Sorry if it makes you> want to go cry to
God or your mother because it JUST IS THAT WAY!> I didn't make
it that way!!!> I'm not forcing algebra to be something else
just to ruin your life,> hurt your feelings, make you mad, or
do anything at all because IT'S> NOT POSSIBLE TO CHANGE THE
ALGEBRA!No, but it *is* possible to leap triumphantly to false
conclusions. Yourconclusion may be true, but you didn't prove
it, you simply jumped to it.> ALGEBRA IS SUPREME!> It's the
supreme authority here.> James (Not-the-supreme-authority)
Harris--There are two things you must never attempt to prove:
the unprovable --and the
===
result, test REVISION 2 Adjunct Assistant Professor at the
University of Montana.>So the actual theorem reads:> Let's
have M = f1.f2, with f1 and f2 distinctive primes.> Primes
that wear really loud ties? (-: Oh, distinct. (-;>Reminds me
to make English my mother tongue ;-).You and me both...>I
*think* that when j is a divisor of both (f1-1) and of (f2-1),
!= 1,>there are counterexamples, but I do not have an off-hand
James Dolan also>mentioned the Chinese remainder theorem.It's
the obvious thing to try: working modulo an odd prime power
isrelatively easy, since the multiplicative group of units is
cyclic; somake sure you can make things work the way you want
them modulo f1,make sure you can make things work the way you
want them modulo f2,and then use the Chinese Remainder Theorem
to make sure things workthe way you want them modulo f1*f2.--