| Here m varies and you question that possibility for the a's to be
| continuous everywhere.  Now what do you see in those equations Keith
| Ramsay?
You say whether a1, a2, and a3 are continuous is irrelevant, so if you
want to you, feel free to skip the following explanation for why they
can't all three be continuous and defined everywhere. The explanation
can be made more precise if you don't believe me.
It's called monodromy. It's related to Galois theory! But it's
more geometric.
At m=0, two of the a's are the same. For m near 0, but not exactly at
0, two of them are close together, while the third is farther away,
staying close to the value it has at 0.
If we take m on a path around 0, going in small steps, because the a's
are continuous they would have also to change in small steps. What is
interesting is what happens to them when m is brought back to where it
started: those two a's are forced to exchange places with each other.
But that would contradict the fact that they have the values they do
at the original m close to 0, and that they're continuous, and defined
for all algebraic integers m. (It's possible to step along such a path
stepping only on algebraic integers, because there are algebraic
integers close to any point on the complex plane.)
It's easier to see in the case of x^2-m = (x-b1)(x-b2), so that b1 and
b2 are the two complex square roots of m (b2=-b1). If we take m in a
loop around m=0, then its two square roots exchange places with each
other. They also go around 0 in the complex plane, but half as fast as
m does. So for b1 and b2 both to be continuous is impossible. The usual
fix is to decide on a place to cut the plane, and make them switch as
they cross the cut.
I would have to do a more careful calculation to be sure that the third
a in the example from the paper can't be continuous either, because
on a different path it changes place with one of the first two, but I
think that's also true. One would have to look carefully at the way the
a's change going to one of the other values of m where two a's coincide,
and I don't see any way to make the calculation simple yet.
[...]
| Hmmm...maybe it'd help to consider the other g's now that you bring
| them 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^2
| 
| and I'm merely noting the consequence of P(0)/f^2 being coprime to f.
Namely, that the factor of f in g_1 must go away. Really, the main
question I had at this point was what you meant by go away. Dividing
g_1 g_2 g_3 by f^2 doesn't necessarily correspond to any operation on
g_1, g_2, and g_3 separately. This concept of what factor of f is
taken away from g_1 when the product of g_1, g_2, and g_3 is divided
by f^2 hasn't been defined.
It sounds like (this is a guess) you mean that you want to write a
common refinement of the two factorizations of P, namely f^2*(P/f^2)
and g_1*g_2*g_3. This would be a way to write P as a product ABCDEF
where AB=g_1, CD=g_2, and EF=g_3, and f^2=ACE, and P/f^2=BDF. It
further sounds like by the factor of f in g_1 going away you mean to
refer to the factor A in it, which is a common factor of f^2 and g_1.
If this is what you mean, you would be considering the product as being
divided by f^2 by having the factors A,C,E taken out of g_1, g_2, and
g_3 separately.
If this is so, then one of the main questions is what values this
common factor has when m is not 0. I don't see a problem with saying
it's f when m=0, but I also don't see why it shouldn't be a varying
factor of f as m changes.
[...]
| > It's 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, I'm figuring out how that
| factor f^2 goes through the g's.
| 
| What I find is that it only makes sense if only two of the g's have a
| factor of f that is f, but then you're forced out of the ring of
| algebraic integers, though it doesn't appear to have happened from
| within the argument, which is why that ring is flawed.
I was thinking we knew that two of the a's were 0 at m=0, so that the
corresponding g's were uf there, and hence divisible by f. I don't see
how you plan to apply this to the situation when m is not 0.
| > Perhaps it would help if you clarified and/or confirmed 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 you're saying the 
value
| > of g1 for each allowed value of m is divisible by f? 
| 
| It depends on context.  It's like if you have y=2x+1, and consider
| various values.  Now I see pressure to write that as f(x)=2x+1, but I
| don't think it's worth the potential confusion.
| 
| The problem is that though you see m as the key variable, it may in
| fact be a *factor* of m that is the true independent variable, which
| I'm sure has confused quite a few people.  Now it's a measure of your
| mathematical ability if you know what I mean there, and let's just say
| that I'm not interested in expanding out in a way that leaves *more*
| room for confusion.
Unfortunately, if the only way that the argument actually works for
you is that in the background, you're implicitly permitting yourself
this kind of reparametrization, then the thing you have written down
isn't a complete proof.
Yes, there are clever techniques involving substituting parameters,
sometimes used to circumvent the presence of a Galois group or
monodromy, but if you plan to go that route you still need to say what
you're doing.
| >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?
| 
| They're not necessarily functions of m.
At some point you need actually to say what kind of thing a1, a2, and
a3 are, without leaving any special wriggle room. If treating them as
functions of m is not quite good enough, then you really need to say
what they are.
| Think about it Keith Ramsay.  Sit down, go over what I've given and
| think very carefully over just that one point, and maybe, if you're
| honestly confused, it'll help you to break through to an
| understanding.
[...]
| Posts should get *shorter* and not longer.
Keith Ramsay
====
> 
> Here I'm right, and you're wrong, as the *definition* of algebraic
> integer leads to an incomplete ring, which I prove mathematically.
> 
> The trouble is, I'm still not following your logic, just to be able to
> see 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
Nope.  There is no claim that y is a factor as it's *provably* a
factor.
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.
You also have that a_1 a_2 a_3 have f^2 as a factor, so proving that
a_3 is coprime to f, means that a_1 and a_2 shouldn't be.
But there's a contradiction because they don't technically have *any*
factors of f, so that means that the definition of algebraic integers,
with a strict interpretation of the word factor would mean that
neither a_1, a_2, nor a_3 have f as a factor when a_1 a_2 a_3 has f^2
as a factor.
 
> (c) But there is no algebraic integer z (=x/y) such that y*z = x
> 
> This leads inexorably to the conclusion that the claim in (b) is simply
> false.  There are *lots* of possible algebraic integers which I can
> multiply by *something* to get x, but there are only a few choices that
> I can multiply by another algebraic integer to get x, and those are the
> factors.
You're running in circles.
First consider that the definition of algebraic integers as the
*roots* of monic polynomials with integer coefficients does not
specifically give you any protection from contradiction.
The definition is over the field of rationals, and it's not clear in
just looking at it, if it will give a complete ring.
Mathematicians apparently *assumed* it did, and have gone on that
assumption for some time.
People seem fixed on the *definition* with the assumption, when
there's no proof based on the definition that the ring is complete.
And there can't be such a proof because I've *proven* it's not
complete, as that definition leads to a contradiction.
> Basically, I do not see in what sense you can claim that y is a factor
> of x without demonstrating the properties of x/y... just being able to
> multiply y by something to get x isn't enough to be able to claim that.
> And without that claim, the conclusion about incompleteness isn't there.
Go to the following link, find and read my paper Advanced Polynomial
Factorization:
  http://groups.msn.com/AmateurMath
and you will see how you can prove that x has y as a factor.
> Perhaps it could help if you could give me a rigorous definition of
> what you mean by incomplete ring, or conversely complete ring.  
Is
> this a property which is distinct from being, say, closed under
> multiplication?
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 field like algebraic numbers
that's ok.  But I'm talking about 'b' that's not in any way a fraction
or fractional.
It's 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?
James Harris
====
> [...]
> | Here m varies and you question that possibility for the a's to be
> | continuous everywhere.  Now what do you see in those equations Keith
> | Ramsay?
> 
> You say whether a1, a2, and a3 are continuous is irrelevant, so if you
> want to you, feel free to skip the following explanation for why they
> can't all three be continuous and defined everywhere. The explanation
> can be made more precise if you don't believe me.
If I have y=mx+b in the ring of algebraic integers, is y continuous?
The ring is algebraic integers, but you're trying to wander off into
something unrelated to the paper, which I guess is some knowledge
you're proud to have memorized.
But do you *understand* it?
 
> It's called monodromy. It's related to Galois theory! But it's
> more geometric.
> 
> At m=0, two of the a's are the same. For m near 0, but not exactly at
> 0, two of them are close together, while the third is farther away,
> staying close to the value it has at 0.
> 
> If we take m on a path around 0, going in small steps, because the a's
> are continuous they would have also to change in small steps. What is
> interesting is what happens to them when m is brought back to where it
> started: those two a's are forced to exchange places with each other.
My guess is that you have some way to expand on that, and if you do,
and it depends on some p-adic interpretation then I will rip you apart
by forcing you down to the rigorous and EXACT numbers.
You see Mathematics is an infinite world where your brain can get
quite lost, but be proud of itself anyway as it can simply tell itself
that it's not lost.
> But that would contradict the fact that they have the values they do
> at the original m close to 0, and that they're continuous, and defined
> for all algebraic integers m. (It's possible to step along such a path
> stepping only on algebraic integers, because there are algebraic
> integers close to any point on the complex plane.)
Why don't you just quit babbling, solve for the a's, and with that
solution try to prove your claim?
And that's what I mean readers about people who think mathematics is a
fashion show.
Mathematics is an infinite subject, which forces perfection.  Human
beings, however, have mathematics, which is a finite subject made up
of what people can discover in a fascinatingly short period of time,
and unfortunately, people can also screw up and put errors in
mathematics.
> It's easier to see in the case of x^2-m = (x-b1)(x-b2), so that b1 and
> b2 are the two complex square roots of m (b2=-b1). If we take m in a
> loop around m=0, then its two square roots exchange places with each
> other. They also go around 0 in the complex plane, but half as fast as
> m does. So for b1 and b2 both to be continuous is impossible. The usual
> fix is to decide on a place to cut the plane, and make them switch as
> they cross the cut.
That's an operator problem.  That is, in trying to evaluate a value
for b1 or b2, you use the sqrt() operator, and get lost on its
inherent ambiguity.
However for the math there is no such problem as it simply deals with
the value of b1 and b2 at each point.  It doesn't think to itself,
take the square, figure out if it's negative or positive.
You, however, can get confused.
If you wish to see it in some sense the way the math sees it, use only
gaussian integers for m in your loop that are squares in the ring of
gaussian integers.
Try that and report back Keith Ramsay.
> I would have to do a more careful calculation to be sure that the third
> a in the example from the paper can't be continuous either, because
> on a different path it changes place with one of the first two, but I
> think that's also true. One would have to look carefully at the way the
> a's change going to one of the other values of m where two a's coincide,
> and I don't see any way to make the calculation simple yet.
And you've gone off on a tangent.  But as you mention it, I suggest
you go ahead and be *careful* instead of possibly hoping that you can
get away with casting doubt.
It's not a political discussion Keith Ramsay.  You're not on Fox News.
> 
> [...]
> | Hmmm...maybe it'd help to consider the other g's now that you bring
> | them 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^2
> | 
> | and I'm merely noting the consequence of P(0)/f^2 being coprime to f.
> 
> Namely, that the factor of f in g_1 must go away. Really, the main
> question I had at this point was what you meant by go away. Dividing
> g_1 g_2 g_3 by f^2 doesn't necessarily correspond to any operation on
> g_1, g_2, and g_3 separately. This concept of what factor of f is
> taken away from g_1 when the product of g_1, g_2, and g_3 is divided
> by f^2 hasn't been defined.
That's nonsensical.  Now I guess you're questioning the assumption
that given that g_1 g_2 g_3 have f^2 as a factor, some of the g's must
have some non unit factor in common with f.
Well gee, it turns out that *in the ring of algebraic integers* except
for at m=0, NONE of the g's share any non unit factors in common with
f.
It's bizarre; it's wacky; and it shows that the ring of algebraic
integers is incomplete.
> It sounds like (this is a guess) you mean that you want to write a
> common refinement of the two factorizations of P, namely f^2*(P/f^2)
> and g_1*g_2*g_3. This would be a way to write P as a product ABCDEF
> where AB=g_1, CD=g_2, and EF=g_3, and f^2=ACE, and P/f^2=BDF. It
> further sounds like by the factor of f in g_1 going away you mean to
> refer to the factor A in it, which is a common factor of f^2 and g_1.
> If this is what you mean, you would be considering the product as being
> divided by f^2 by having the factors A,C,E taken out of g_1, g_2, and
> g_3 separately.
> 
> If this is so, then one of the main questions is what values this
> common factor has when m is not 0. I don't see a problem with saying
> it's f when m=0, but I also don't see why it shouldn't be a varying
> factor of f as m changes.
Yeah, and I gave an example yesterday which should tell you why.
  P(x) = 2x^2 + 4x + 2
  (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2
  a_1 a_2 = 2x^2,
and I note that when x=0, at *least* one of the a's must equal 0, and
further note that the constant term of the factor a_1 + 1, which I'll
call g_1 is 1 at that point.  Further the factor a_2 + 2 has a factor
that is 2 at that point, and I'll call it g_2.
Now then Keith Ramsay, do you question what g_2 will do when x does
not equal 0?
So why do you do it above?
Ultimately, you may try and claim that irreducibility over Q is what's
pertinent, if so, explain in detail.
> [...]
> | > It's 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, I'm figuring out how that
> | factor f^2 goes through the g's.
> | 
> | What I find is that it only makes sense if only two of the g's have a
> | factor of f that is f, but then you're forced out of the ring of
> | algebraic integers, though it doesn't appear to have happened from
> | within the argument, which is why that ring is flawed.
> 
> I was thinking we knew that two of the a's were 0 at m=0, so that the
> corresponding g's were uf there, and hence divisible by f. I don't see
> how you plan to apply this to the situation when m is not 0.
I've generalized that factors of polynomials can be written as r+c,
where r=0, or varies while the polynomial factor varies, and c is
constant and is a factor of the constant term of the polynomial.
That's easy to see with something like x+2 as a factor of x^2+3x+2.
In the paper I have that as a lemma, with a proof.
If anyone had an actual error in the paper, they could simply show
fault with the lemma, as it's the key linchpin.
Instead I get overlong posts where people bring up tangents, and keep
questioning things that are obvious.
Mathematicians are such dweebs.
> | > Perhaps it would help if you clarified and/or confirmed 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 you're saying the 
value
> | > of g1 for each allowed value of m is divisible by f? 
> | 
> | It depends on context.  It's like if you have y=2x+1, and consider
> | various values.  Now I see pressure to write that as f(x)=2x+1, but I
> | don't think it's worth the potential confusion.
> | 
> | The problem is that though you see m as the key variable, it may in
> | fact be a *factor* of m that is the true independent variable, which
> | I'm sure has confused quite a few people.  Now it's a measure of your
> | mathematical ability if you know what I mean there, and let's just say
> | that I'm not interested in expanding out in a way that leaves *more*
> | room for confusion.
> 
> Unfortunately, if the only way that the argument actually works for
> you is that in the background, you're implicitly permitting yourself
> this kind of reparametrization, then the thing you have written down
> isn't a complete proof.
Then prove it.
There are lots of things that I think about that aren't needed to show
a proof, and I don't put them into discussions unless it suits me.
Remember, a proof begins with a truth and proceeds by logical steps to
a conclusion which then must be true.
I only need give that truth in the beginning and proceed by logical
steps.
That you can move deeper into portions in a way that isn't material to
the argument makes the math just a little more fun...well it does for
me at least.
> Yes, there are clever techniques involving substituting parameters,
> sometimes used to circumvent the presence of a Galois group or
> monodromy, but if you plan to go that route you still need to say what
> you're doing.
That is irrelevant.  See?  I knew that bringing the subject up might
just distract you.
Why don't you focus on the important path of the proof, and not worry
about sideroads?
> | >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?
> | 
> | They're not necessarily functions of m.
> 
> At some point you need actually to say what kind of thing a1, a2, and
> a3 are, without leaving any special wriggle room. If treating them as
> functions of m is not quite good enough, then you really need to say
> what they are.
You didn't connect back with your own statement, fascinating.
I'm repeating myself with that statement, but before you talked of
reparametrization in reply.
I fear you are a parrot.  You look for key words and then regurgitate
based on something you think is relevant, but you need to THINK.
> | Think about it Keith Ramsay.  Sit down, go over what I've given and
> | think very carefully over just that one point, and maybe, if you're
> | honestly confused, it'll help you to break through to an
> | understanding.
> 
> [...]
> | Posts should get *shorter* and not longer.
> 
> Keith Ramsay
And I'll reiterate that posts should get shorter and not longer.
It's called cutting to the chase, or getting to the nitty-gritty.
James Harris
====
| > [...]
| > | Here m varies and you question that possibility for the a's to be
| > | continuous everywhere.  Now what do you see in those equations Keith
| > | Ramsay?
| > 
| > You say whether a1, a2, and a3 are continuous is irrelevant, so if you
| > want to you, feel free to skip the following explanation for why they
| > can't all three be continuous and defined everywhere. The explanation
| > can be made more precise if you don't believe me.
| 
| If I have y=mx+b in the ring of algebraic integers, is y continuous?
Assuming you mean that m and b are algebraic integers, and y is being
given as a function of x, yes.
Remember (unlike e.g. the Gaussian integers) the algebraic integers
are dense in the complex plane, meaning that any circle, no matter
how small, has algebraic integers inside it.
 
| The ring is algebraic integers, but you're trying to wander off into
| something unrelated to the paper, which I guess is some knowledge
| you're proud to have memorized.
My memory is decent, but nothing to be proud of.
No, my main point in bringing this stuff up was to try to convince
you that defining a1, a2, and a3 in such a way that their values at
one point are related to their values at other points in a useful way.
If my explanation isn't working, fine, someone will eventually find
a better way of explaining it to you.
| > If we take m on a path around 0, going in small steps, because the a's
| > are continuous they would have also to change in small steps. What is
| > interesting is what happens to them when m is brought back to where it
| > started: those two a's are forced to exchange places with each other.
| 
| My guess is that you have some way to expand on that, and if you do,
| and it depends on some p-adic interpretation then I will rip you apart
| by forcing you down to the rigorous and EXACT numbers.
You have a way of being melodramatic when it doesn't make sense.
First of all, don't try to hint that I'm prone to pretending that I'm
dealing with complex numbers when I'm actually sneakily dealing with
p-adic numbers. You're the one who's been dealing with a1, a2, and
a3 having indeterminate type.
Second, p-adic numbers ARE exact. They're just a different system.
It's easier in a lot of ways to be rigorous with them than it is
to be rigorous with the real numbers or the complex numbers.
Third, you're surely aware that nobody will be ripping anybody
apart or forcing anybody to do anything. People have tried hard
to force you down to rigor, for example by telling you what,
allegedly, the common factors between 5 and the roots of that cubic
are. Doesn't phase you a bit! And you've shown us all how easy it is
to be impervious to demands for rigor.
| You see Mathematics is an infinite world where your brain can get
| quite lost, but be proud of itself anyway as it can simply tell itself
| that it's not lost.
on the algebraic integers, when the property I actually needed was
continuity not only at the algebraic integers, but at all complex
numbers. My mistake was to think that the fact that the algebraic
integers are dense in the plane would be good enough, but it isn't.
You might say that the discontinuities can be hidden at
non-algebraic-integer points. There is, for example, a unique
function f(z) defined on the complex numbers a+bi such that
b <> pi*a, satisfying f(z)^2=z, f(1)=1, and f(-1)=i. It can't
be extended to the line b=pi*a because it approaches different
limits on either side of that line (it jumps there) on the
side where a,b < 0. If we restrict it to the algebraic integers,
then it's continuous (because continuous would mean only continuous
on algebraic integers).
I'm still saying that the situation with a1, a2 and a3 is analogous
to the situation with the two so-called branches of the square
root multiple-valued function. It's not all that hard to give a
rigorous proof of what I said about continuous square root functions:
Suppose that f(z) is a function on the complex plane having complex
values, and satisfying f(z)^2=z for all z. The goal is to show that
this leads to a contradiction. Actually, it's enough to consider it
on the circle, which consists of complex numbers of the form e^{it}
where t is a real number.
Because for each real t, f(e^{it})^2 - e^{it}
= (f(e^{it})-e^{it/2})(f(e^{it})+e^{it/2}), we also have that
either f(e^{it}) = e^{it/2} or f(e^{it}) = -e^{it/2}. Denote by A
the set of t satisfying f(e^{it}) = e^{it/2} and by B the set of
t satisfying f(e^{it}) = -e^{it/2}. Since e^{it/2} <> 0 for any
real t, the two sets A and B are disjoint, and I've just shown
that their union is the whole real line.
The next thing is to prove that they are both closed sets. A closed
set is a set like A which has as an element every complex number z
having the property that for every epsilon>0, there is an element
of A within epsilon of z. But A and B are the sets of zeros of the
continuous functions f(e^{it})-e^{it/2} and f(e^{it})+e^{it/2}
respectively, and the set of zeros of a continuous function is always
closed. The proof of that is that if g is continuous, and g(z) <> 0,
then there exists by the definition of continuity a delta > 0 such
that |g(z')-g(z)|<|g(z)| when |z'-z|<delta, and |g(z')-g(z)|<|g(z)|
implies g(z')<>0 as well. So if g is continuous and z isn't a zero,
then there's a delta>0 such that no other zeros of g come within
delta of z. Conversely, if there are zeros within any delta>0 of
z, then z must also be a zero.
Finally, we use a standard result that the real line can't be
broken into two disjoint nonempty sets which are both closed.
The standard proof is by the least upper bound principle. If A
and B are the two nonempty closed sets, suppose a is an element
of A and b is an element of B. Without loss of generality, a < b;
the proof for b < a is essentially the same. Let a' be the least
upper bound of the elements of A less than b. Because a' is an
upper bound, there are no elements of A between a' and b. So for
every delta>0, delta<b-a', the element a'+delta is in B. So by
the fact that B is closed, a' is an element of B. On the other
hand, by the fact that A is closed, this means there is a delta>0
such that every number within delta of a' is also in B, including
the open segment from max(a, a'-delta) to a'. That means
max(a, a'-delta) is a lesser upper bound on A, contradicting the
fact that a' was the least upper bound.
It follows that one of A or B is empty, and the other is the whole
real line. But that's impossible, since if f(1) = f(e^{0*i}) =
e^{0*i/2} = 1 then f(1) = f(e^{2*pi*i}) = -(-1) = -e^{pi*i}
= -e^{2*pi*i/2}, and if f(e^{0*i})=-e^{0*i/2} then f(e^{2*pi*i})
= e^{2*pi*i/2}. This is where the wraparound comes in; if 0 is
in A then 2*pi is in B and vice-versa.
So in fact the original assumption that there was such an f was
false, Q.E.D.
Now, a1 and a2 are not square roots, but they are similar enough
to a constant times the square root of m to make the same type of
argument apply.
Keith Ramsay
====
> | > [...]
> | > | Here m varies and you question that possibility for the a's to be
> | > | continuous everywhere.  Now what do you see in those equations 
Keith
> | > | Ramsay?
> | > 
> | > You say whether a1, a2, and a3 are continuous is irrelevant, so if 
you
> | > want to you, feel free to skip the following explanation for why they
> | > can't all three be continuous and defined everywhere. The explanation
> | > can be made more precise if you don't believe me.
> | 
> | If I have y=mx+b in the ring of algebraic integers, is y continuous?
> 
> Assuming you mean that m and b are algebraic integers, and y is being
> given as a function of x, yes.
> 
> Remember (unlike e.g. the Gaussian integers) the algebraic integers
> are dense in the complex plane, meaning that any circle, no matter
> how small, has algebraic integers inside it.
Hmmm...interesting assertion, but what about singularities?
  
> | The ring is algebraic integers, but you're trying to wander off into
> | something unrelated to the paper, which I guess is some knowledge
> | you're proud to have memorized.
> 
> My memory is decent, but nothing to be proud of.
That's not what I said, as I assume that no matter what your level of
pride in your memory might be you do actually remember some things.
> No, my main point in bringing this stuff up was to try to convince
> you that defining a1, a2, and a3 in such a way that their values at
> one point are related to their values at other points in a useful way.
Um, but that is an insinuation that they aren't already rigorously
defined, when in fact they are.
> If my explanation isn't working, fine, someone will eventually find
> a better way of explaining it to you.
That is an irrational statement.  Where do you suppose this someone
is?
Possibly your memory tells you they're in some potential field?
> | > If we take m on a path around 0, going in small steps, because the 
a's
> | > are continuous they would have also to change in small steps. What is
> | > interesting is what happens to them when m is brought back to where 
it
> | > started: those two a's are forced to exchange places with each other.
> | 
> | My guess is that you have some way to expand on that, and if you do,
> | and it depends on some p-adic interpretation then I will rip you apart
> | by forcing you down to the rigorous and EXACT numbers.
> 
> You have a way of being melodramatic when it doesn't make sense.
> 
> First of all, don't try to hint that I'm prone to pretending that I'm
> dealing with complex numbers when I'm actually sneakily dealing with
> p-adic numbers. You're the one who's been dealing with a1, a2, and
> a3 having indeterminate type.
That is false as the a's are determined both in form and type.
> Second, p-adic numbers ARE exact. They're just a different system.
> It's easier in a lot of ways to be rigorous with them than it is
> to be rigorous with the real numbers or the complex numbers.
Is that something you have from memory Keith Ramsay?
> Third, you're surely aware that nobody will be ripping anybody
> apart or forcing anybody to do anything. People have tried hard
> to force you down to rigor, for example by telling you what,
> allegedly, the common factors between 5 and the roots of that cubic
> are. Doesn't phase you a bit! And you've shown us all how easy it is
> to be impervious to demands for rigor.
That's an appeal to the gallery, which is a logical fallacy.
> | You see Mathematics is an infinite world where your brain can get
> | quite lost, but be proud of itself anyway as it can simply tell itself
> | that it's not lost.
> 
> on the algebraic integers, when the property I actually needed was
> continuity not only at the algebraic integers, but at all complex
> numbers. My mistake was to think that the fact that the algebraic
> integers are dense in the plane would be good enough, but it isn't.
> 
> You might say that the discontinuities can be hidden at
> non-algebraic-integer points. There is, for example, a unique
> function f(z) defined on the complex numbers a+bi such that
> b <> pi*a, satisfying f(z)^2=z, f(1)=1, and f(-1)=i. It can't
> be extended to the line b=pi*a because it approaches different
> limits on either side of that line (it jumps there) on the
> side where a,b < 0. If we restrict it to the algebraic integers,
> then it's continuous (because continuous would mean only continuous
> on algebraic integers).
 I'm still saying that the situation with a1, a2 and a3 is analogous
> to the situation with the two so-called branches of the square
> root multiple-valued function. It's not all that hard to give a
> rigorous proof of what I said about continuous square root functions:
> 
> Suppose that f(z) is a function on the complex plane having complex
> values, and satisfying f(z)^2=z for all z. The goal is to show that
> this leads to a contradiction. Actually, it's enough to consider it
> on the circle, which consists of complex numbers of the form e^{it}
> where t is a real number.
> 
> Because for each real t, f(e^{it})^2 - e^{it}
> = (f(e^{it})-e^{it/2})(f(e^{it})+e^{it/2}), we also have that
> either f(e^{it}) = e^{it/2} or f(e^{it}) = -e^{it/2}. Denote by A
> the set of t satisfying f(e^{it}) = e^{it/2} and by B the set of
> t satisfying f(e^{it}) = -e^{it/2}. Since e^{it/2} <> 0 for any
> real t, the two sets A and B are disjoint, and I've just shown
> that their union is the whole real line.
Your limitation is in needing the sqrt() operator, but being unable to
simply give the result in general.  The math doesn't have that problem
as it doesn't need operators, but simply sees the result as it is.
For instance, consider (1+sqrt(-3))/2, which is in fact a unit in the
ring of algebraic integers, but it *looks* nothing like 1, or i,
because you have the two operators sqrt() and /.
(Purists may note that this algebraic integer cannot be expressed
using only ring operators.)
You have no meaningful means of further resolution.
But the math sees a number.
 
> The next thing is to prove that they are both closed sets. A closed
> set is a set like A which has as an element every complex number z
> having the property that for every epsilon>0, there is an element
> of A within epsilon of z. But A and B are the sets of zeros of the
> continuous functions f(e^{it})-e^{it/2} and f(e^{it})+e^{it/2}
> respectively, and the set of zeros of a continuous function is always
> closed. The proof of that is that if g is continuous, and g(z) <> 0,
> then there exists by the definition of continuity a delta > 0 such
> that |g(z')-g(z)|<|g(z)| when |z'-z|<delta, and |g(z')-g(z)|<|g(z)|
> implies g(z')<>0 as well. So if g is continuous and z isn't a zero,
> then there's a delta>0 such that no other zeros of g come within
> delta of z. Conversely, if there are zeros within any delta>0 of
> z, then z must also be a zero.
> 
> Finally, we use a standard result that the real line can't be
> broken into two disjoint nonempty sets which are both closed.
> The standard proof is by the least upper bound principle. If A
> and B are the two nonempty closed sets, suppose a is an element
> of A and b is an element of B. Without loss of generality, a < b;
> the proof for b < a is essentially the same. Let a' be the least
> upper bound of the elements of A less than b. Because a' is an
> upper bound, there are no elements of A between a' and b. So for
> every delta>0, delta<b-a', the element a'+delta is in B. So by
> the fact that B is closed, a' is an element of B. On the other
> hand, by the fact that A is closed, this means there is a delta>0
> such that every number within delta of a' is also in B, including
> the open segment from max(a, a'-delta) to a'. That means
> max(a, a'-delta) is a lesser upper bound on A, contradicting the
> fact that a' was the least upper bound.
> 
> It follows that one of A or B is empty, and the other is the whole
> real line. But that's impossible, since if f(1) = f(e^{0*i}) =
> e^{0*i/2} = 1 then f(1) = f(e^{2*pi*i}) = -(-1) = -e^{pi*i}
> = -e^{2*pi*i/2}, and if f(e^{0*i})=-e^{0*i/2} then f(e^{2*pi*i})
> = e^{2*pi*i/2}. This is where the wraparound comes in; if 0 is
> in A then 2*pi is in B and vice-versa.
> 
> So in fact the original assumption that there was such an f was
> false, Q.E.D.
Can you switch to congruences?  What about something like 
  f(e^{it})^2 - e^{it} = 0(mod 35)
in the ring of algebraic integers?
 
> Now, a1 and a2 are not square roots, but they are similar enough
> to a constant times the square root of m to make the same type of
> argument apply.
> 
> Keith Ramsay
Why don't you just give the a's explicitly and prove it then?
James Harris
====
[...]
|And you've gone off on a tangent.  But as you mention it, I suggest
|you go ahead and be *careful* instead of possibly hoping that you can
|get away with casting doubt.
|
|It's not a political discussion Keith Ramsay.  You're not on Fox News.
You're hardly in any position to complain, either about glibly
expressing doubts about things you don't understand, or about making
political statements, or about being careful either!
[...]
| > If this is so, then one of the main questions is what values this
| > common factor has when m is not 0. I don't see a problem with saying
| > it's f when m=0, but I also don't see why it shouldn't be a varying
| > factor of f as m changes.
| 
| Yeah, and I gave an example yesterday which should tell you why.
| 
|   P(x) = 2x^2 + 4x + 2
| 
|   (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2
| 
|   a_1 a_2 = 2x^2,
Is this last equation another assumption? Otherwise, what's to rule
out a_1=1, a_2 = x^2+2x-1, for instance?
 
| and I note that when x=0, at *least* one of the a's must equal 0, and
| further note that the constant term of the factor a_1 + 1, which I'll
| call g_1 is 1 at that point.  Further the factor a_2 + 2 has a factor
| that is 2 at that point, and I'll call it g_2.
|
| Now then Keith Ramsay, do you question what g_2 will do when x does
| not equal 0?
| 
| So why do you do it above?
Continually expressing shock that I don't agree with you, when you
haven't indicated a principle to justify your claim, is just a bunch
of posturing. What's the general principle?
| Ultimately, you may try and claim that irreducibility over Q is what's
| pertinent, if so, explain in detail.
The two cases aren't parallel. You really should get out of the
habit of thinking that analogy and cajolery add up to a logical
argument. Giving a bunch of examples which work the way you want
another example to work is not a proof.
If I were to give an explanation for why the examples that work,
work, I'd give it in terms of the following principle:
 (*)  If a divides bc, then there exist m and n such that a=mn,
      m divides b and n divides c.
You give examples where the product of two polynomials is divisible
by an integer. Given that (*) is true in these cases, we can then
try to figure out what m and n would work by considering the constant
terms and so on as you do. All of your simplified examples are ones
where (*) works.
But (*) is not generally true for functions having algebraic integer
values.
[...]
| > | >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?
| > | 
| > | They're not necessarily functions of m.
| > 
| > At some point you need actually to say what kind of thing a1, a2, and
| > a3 are, without leaving any special wriggle room. If treating them as
| > functions of m is not quite good enough, then you really need to say
| > what they are.
| 
| You didn't connect back with your own statement, fascinating.
| 
| I'm repeating myself with that statement, but before you talked of
| reparametrization in reply.
| 
| I fear you are a parrot.  You look for key words and then regurgitate
| based on something you think is relevant, but you need to THINK.
I didn't think I would need to put in the previous quoted text in
order for you to remember what you had just written:
#> Perhaps it would help if you clarified and/or confirmed 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 you're saying the value
#> of g1 for each allowed value of m is divisible by f? 
#
#It depends on context.  It's like if you have y=2x+1, and consider
#various values.  Now I see pressure to write that as f(x)=2x+1, but I
#don't think it's worth the potential confusion.
#
#The problem is that though you see m as the key variable, it may in
#fact be a *factor* of m that is the true independent variable, which
#I'm sure has confused quite a few people.  Now it's a measure of your
#mathematical ability if you know what I mean there, and let's just say
#that I'm not interested in expanding out in a way that leaves *more*
#room for confusion.
So you're telling us that m might not be the true independent
variable. Do you realize that switching from one independent
variable to another one is known as reparametrizing?
A reasonable reader, who had read the paper but not this discussion,
would naturally tend to suspect that a1, a2 and a3, which are
described as having values at given values of m, would tend to
suspect that they are simply functions of m. If this is not
necessarily so, it's because you've left out any explanation for
what else you intend for them to be.
This stuff about how being more specific is bad because the rest of
us (being inferior to you in mathematical ability of course) would
be liable to be confused is just an excuse. You are, because of
your lack of practice in writing actual proofs, failing to be
specific, and as a result getting confused. I think you're just trying
to bluff your way past the fact that you don't know how to define
them in such a way as to get the argument in the paper to work.
Keith Ramsay
====
[...]
|And you've gone off on a tangent.  But as you mention it, I suggest
|you go ahead and be *careful* instead of possibly hoping that you can
|get away with casting doubt.
|
|It's not a political discussion Keith Ramsay.  You're not on Fox News.
You're hardly in any position to complain, either about glibly
expressing doubts about things you don't understand, or about making
political statements, or about being careful either!
[...]
| > If this is so, then one of the main questions is what values this
| > common factor has when m is not 0. I don't see a problem with saying
| > it's f when m=0, but I also don't see why it shouldn't be a varying
| > factor of f as m changes.
| 
| Yeah, and I gave an example yesterday which should tell you why.
| 
|   P(x) = 2x^2 + 4x + 2
| 
|   (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2
| 
|   a_1 a_2 = 2x^2,
Is this last equation another assumption? Otherwise, what's to rule
out a_1=1, a_2 = x^2+2x-1, for instance?
 
| and I note that when x=0, at *least* one of the a's must equal 0, and
| further note that the constant term of the factor a_1 + 1, which I'll
| call g_1 is 1 at that point.  Further the factor a_2 + 2 has a factor
| that is 2 at that point, and I'll call it g_2.
|
| Now then Keith Ramsay, do you question what g_2 will do when x does
| not equal 0?
| 
| So why do you do it above?
Continually expressing shock that I don't agree with you, when you
haven't indicated a principle to justify your claim, is just a bunch
of posturing. What's the general principle?
| Ultimately, you may try and claim that irreducibility over Q is what's
| pertinent, if so, explain in detail.
The two cases aren't parallel. You really should get out of the
habit of thinking that analogy and cajolery add up to a logical
argument. Giving a bunch of examples which work the way you want
another example to work is not a proof.
If I were to give an explanation for why the examples that work,
work, I'd give it in terms of the following principle:
 (*)  If a divides bc, then there exist m and n such that a=mn,
      m divides b and n divides c.
You give examples where the product of two polynomials is divisible
by an integer. Given that (*) is true in these cases, we can then
try to figure out what m and n would work by considering the constant
terms and so on as you do. All of your simplified examples are ones
where (*) works.
But (*) is not generally true for functions having algebraic integer
values.
[...]
| > | >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?
| > | 
| > | They're not necessarily functions of m.
| > 
| > At some point you need actually to say what kind of thing a1, a2, and
| > a3 are, without leaving any special wriggle room. If treating them as
| > functions of m is not quite good enough, then you really need to say
| > what they are.
| 
| You didn't connect back with your own statement, fascinating.
| 
| I'm repeating myself with that statement, but before you talked of
| reparametrization in reply.
| 
| I fear you are a parrot.  You look for key words and then regurgitate
| based on something you think is relevant, but you need to THINK.
I didn't think I would need to put in the previous quoted text in
order for you to remember what you had just written:
#> Perhaps it would help if you clarified and/or confirmed 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 you're saying the value
#> of g1 for each allowed value of m is divisible by f? 
#
#It depends on context.  It's like if you have y=2x+1, and consider
#various values.  Now I see pressure to write that as f(x)=2x+1, but I
#don't think it's worth the potential confusion.
#
#The problem is that though you see m as the key variable, it may in
#fact be a *factor* of m that is the true independent variable, which
#I'm sure has confused quite a few people.  Now it's a measure of your
#mathematical ability if you know what I mean there, and let's just say
#that I'm not interested in expanding out in a way that leaves *more*
#room for confusion.
So you're telling us that m might not be the true independent
variable. Do you realize that switching from one independent
variable to another one is known as reparametrizing?
A reasonable reader, who had read the paper but not this discussion,
would naturally tend to suspect that a1, a2 and a3, which are
described as having values at given values of m, would tend to
suspect that they are simply functions of m. If this is not
necessarily so, it's because you've left out any explanation for
what else you intend for them to be.
This stuff about how being more specific is bad because the rest of
us (being inferior to you in mathematical ability of course) would
be liable to be confused is just an excuse. You are, because of
your lack of practice in writing actual proofs, failing to be
specific, and as a result getting confused. I think you're just trying
to bluff your way past the fact that you don't know how to define
them in such a way as to get the argument in the paper to work.
Keith Ramsay
====
[...]
|And you've gone off on a tangent.  But as you mention it, I suggest
|you go ahead and be *careful* instead of possibly hoping that you can
|get away with casting doubt.
|
|It's not a political discussion Keith Ramsay.  You're not on Fox News.
You're hardly in any position to complain, either about glibly
expressing doubts about things you don't understand, or about making
political statements, or about being careful either!
[...]
| > If this is so, then one of the main questions is what values this
| > common factor has when m is not 0. I don't see a problem with saying
| > it's f when m=0, but I also don't see why it shouldn't be a varying
| > factor of f as m changes.
| 
| Yeah, and I gave an example yesterday which should tell you why.
| 
|   P(x) = 2x^2 + 4x + 2
| 
|   (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2
| 
|   a_1 a_2 = 2x^2,
Is this last equation another assumption? Otherwise, what's to rule
out a_1=1, a_2 = x^2+2x-1, for instance?
 
| and I note that when x=0, at *least* one of the a's must equal 0, and
| further note that the constant term of the factor a_1 + 1, which I'll
| call g_1 is 1 at that point.  Further the factor a_2 + 2 has a factor
| that is 2 at that point, and I'll call it g_2.
|
| Now then Keith Ramsay, do you question what g_2 will do when x does
| not equal 0?
| 
| So why do you do it above?
Continually expressing shock that I don't agree with you, when you
haven't indicated a principle to justify your claim, is just a bunch
of posturing. What's the general principle?
| Ultimately, you may try and claim that irreducibility over Q is what's
| pertinent, if so, explain in detail.
The two cases aren't parallel. You really should get out of the
habit of thinking that analogy and cajolery add up to a logical
argument. Giving a bunch of examples which work the way you want
another example to work is not a proof.
If I were to give an explanation for why the examples that work,
work, I'd give it in terms of the following principle:
 (*)  If a divides bc, then there exist m and n such that a=mn,
      m divides b and n divides c.
You give examples where the product of two polynomials is divisible
by an integer. Given that (*) is true in these cases, we can then
try to figure out what m and n would work by considering the constant
terms and so on as you do. All of your simplified examples are ones
where (*) works.
But (*) is not generally true for functions having algebraic integer
values.
[...]
| > | >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?
| > | 
| > | They're not necessarily functions of m.
| > 
| > At some point you need actually to say what kind of thing a1, a2, and
| > a3 are, without leaving any special wriggle room. If treating them as
| > functions of m is not quite good enough, then you really need to say
| > what they are.
| 
| You didn't connect back with your own statement, fascinating.
| 
| I'm repeating myself with that statement, but before you talked of
| reparametrization in reply.
| 
| I fear you are a parrot.  You look for key words and then regurgitate
| based on something you think is relevant, but you need to THINK.
I didn't think I would need to put in the previous quoted text in
order for you to remember what you had just written:
#> Perhaps it would help if you clarified and/or confirmed 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 you're saying the value
#> of g1 for each allowed value of m is divisible by f? 
#
#It depends on context.  It's like if you have y=2x+1, and consider
#various values.  Now I see pressure to write that as f(x)=2x+1, but I
#don't think it's worth the potential confusion.
#
#The problem is that though you see m as the key variable, it may in
#fact be a *factor* of m that is the true independent variable, which
#I'm sure has confused quite a few people.  Now it's a measure of your
#mathematical ability if you know what I mean there, and let's just say
#that I'm not interested in expanding out in a way that leaves *more*
#room for confusion.
So you're telling us that m might not be the true independent
variable. Do you realize that switching from one independent
variable to another one is known as reparametrizing?
A reasonable reader, who had read the paper but not this discussion,
would naturally tend to suspect that a1, a2 and a3, which are
described as having values at given values of m, would tend to
suspect that they are simply functions of m. If this is not
necessarily so, it's because you've left out any explanation for
what else you intend for them to be.
This stuff about how being more specific is bad because the rest of
us (being inferior to you in mathematical ability of course) would
be liable to be confused is just an excuse. You are, because of
your lack of practice in writing actual proofs, failing to be
specific, and as a result getting confused. I think you're just trying
to bluff your way past the fact that you don't know how to define
them in such a way as to get the argument in the paper to work.
Keith Ramsay
====
[...]
|And you've gone off on a tangent.  But as you mention it, I suggest
|you go ahead and be *careful* instead of possibly hoping that you can
|get away with casting doubt.
|
|It's not a political discussion Keith Ramsay.  You're not on Fox News.
You're hardly in any position to complain, either about glibly
expressing doubts about things you don't understand, or about making
political statements, or about being careful either!
[...]
| > If this is so, then one of the main questions is what values this
| > common factor has when m is not 0. I don't see a problem with saying
| > it's f when m=0, but I also don't see why it shouldn't be a varying
| > factor of f as m changes.
| 
| Yeah, and I gave an example yesterday which should tell you why.
| 
|   P(x) = 2x^2 + 4x + 2
| 
|   (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2
| 
|   a_1 a_2 = 2x^2,
Is this last equation another assumption? Otherwise, what's to rule
out a_1=1, a_2 = x^2+2x-1, for instance?
 
| and I note that when x=0, at *least* one of the a's must equal 0, and
| further note that the constant term of the factor a_1 + 1, which I'll
| call g_1 is 1 at that point.  Further the factor a_2 + 2 has a factor
| that is 2 at that point, and I'll call it g_2.
|
| Now then Keith Ramsay, do you question what g_2 will do when x does
| not equal 0?
| 
| So why do you do it above?
Continually expressing shock that I don't agree with you, when you
haven't indicated what principle it is that justifies your claim,
is just a bunch of posturing. What's the general principle?
| Ultimately, you may try and claim that irreducibility over Q is what's
| pertinent, if so, explain in detail.
The two cases aren't parallel. You really should get out of the
habit of thinking that analogy and cajolery add up to a logical
argument. Giving a bunch of examples which work the way you want
another example to work is not a proof.
If I were to give an explanation for why the examples that work,
work, I'd give it in terms of the following principle:
 (*)  If a divides bc, then there exist m and n such that a=mn,
      m divides b and n divides c.
You give examples where the product of two polynomials is divisible
by an integer. Given that (*) is true in these cases, we can then
try to figure out what m and n would work by considering the constant
terms and so on as you do. All of your simplified examples are ones
where (*) works.
But (*) is not generally true for functions having algebraic integer
values.
Another way to look at it is that in order to show that a1 is divisible
by f (say), you have either to show that all three of them are divisible
by f, or you have to make some additional assumption which somehow forces 
a1
at each value of m not to be the value or one of the values of the three
(a1, a2, and a3) that isn't divisible by f. If a1, a2 and a3 were simply
three functions of m, which at each specific m satisfy the equation 
defining
them, then the order in which they're assigned to those three values is 
still
arbitrary, leaving nothing to keep a1 from being the wrong value.
[...]
| > | >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?
| > | 
| > | They're not necessarily functions of m.
| > 
| > At some point you need actually to say what kind of thing a1, a2, and
| > a3 are, without leaving any special wriggle room. If treating them as
| > functions of m is not quite good enough, then you really need to say
| > what they are.
| 
| You didn't connect back with your own statement, fascinating.
| 
| I'm repeating myself with that statement, but before you talked of
| reparametrization in reply.
| 
| I fear you are a parrot.  You look for key words and then regurgitate
| based on something you think is relevant, but you need to THINK.
I didn't think I would need to put in the previous quoted text in
order for you to remember what you had just written:
|> Perhaps it would help if you clarified and/or confirmed 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 you're saying the value
|> of g1 for each allowed value of m is divisible by f? 
|
|It depends on context.  It's like if you have y=2x+1, and consider
|various values.  Now I see pressure to write that as f(x)=2x+1, but I
|don't think it's worth the potential confusion.
|
|The problem is that though you see m as the key variable, it may in
|fact be a *factor* of m that is the true independent variable, which
|I'm sure has confused quite a few people.  Now it's a measure of your
|mathematical ability if you know what I mean there, and let's just say
|that I'm not interested in expanding out in a way that leaves *more*
|room for confusion.
So you're telling us that m might not be the true independent
variable. Do you realize that switching from one independent
variable to another one is known as reparametrizing?
A reasonable person reading the proof (and not this discussion)
would look at that formula supposedly defining a1, a2, and a3,
and since you don't say specifically what they are, they'd have to
make a bit of a guess. That they are at least functions of m would
be quite a reasonable interpolation. Otherwise, you're leaving them
in the dark.
This stuff about how being more specific is bad because the rest of
us (being inferior to you in mathematical ability of course) would
be liable to be confused is just an excuse. You are, because of
your lack of practice in writing actual proofs, failing to be
specific, and as a result getting confused.
I think you're just trying to bluff your way past the problem. Anybody
who actually had a proof would be able to answer such simple questions
as these much more specifically.
Keith Ramsay
====
> If my explanation isn't working, fine, someone will eventually find
> a better way of explaining it to you.
> 
> That is an irrational statement.  Where do you suppose this someone
> is?
The someone who can bring JSH to understanding does not exist.
====
>> Remember (unlike e.g. the Gaussian integers) the algebraic integers
>> are dense in the complex plane, meaning that any circle, no matter
>> how small, has algebraic integers inside it.
Hmmm...interesting assertion, but what about singularities?
I'll bite. What about singularities?
We aren't talking about a function here, just about two sets, the
complex numbers and the algebraic integers. There aren't any
singularities. And if there were, what does that have to do with
whether there are infinitely many algebraic integers everywhere you
look?
                - Randy
====
Sorry about the duplicate postings; I got caught by this thing where
I get a connection timed out page while trying to post with Google,
but apparently the posting goes out anyway.
[...]
| > Remember (unlike e.g. the Gaussian integers) the algebraic integers
| > are dense in the complex plane, meaning that any circle, no matter
| > how small, has algebraic integers inside it.
| 
| Hmmm...interesting assertion, but what about singularities?
As usual by a circle I mean one having positive radius, which doesn't
have singularities.
[...]
| > No, my main point in bringing this stuff up was to try to convince
| > you that defining a1, a2, and a3 in such a way that their values at
| > one point are related to their values at other points in a useful way.
| 
| Um, but that is an insinuation that they aren't already rigorously
| defined, when in fact they are.
I don't see much sign of your knowing what it means to give a rigorous
But an equation is only meaningful if one knows somehow what kind of
things the terms in it represent. The paper refers to the value one
has at m=0, so there has to be some sense in which one can substitute
m=0 into one of them. It's natural then to assume you mean for them
to be functions of m. But then you say that they're not necessarily
functions of m, without saying specifically what else they might be.
(You hint that expanding on this point might cause confusion.)
Obviously you're not giving a rigorous definition.
| > If my explanation isn't working, fine, someone will eventually find
| > a better way of explaining it to you.
| 
| That is an irrational statement.  Where do you suppose this someone
| is?
I see at least a couple of other people who also understand this point
about how the values of a1, a2, and a3 at one point relate to their
values at another point, and have been trying to explain it to you.
I'm just assuming that eventually one of us will come up with a better
way of explaining it to you.
| Possibly your memory tells you they're in some potential field?
??
 
[...]
| That is false as the a's are determined both in form and type.
As in, sort of like functions of m but not necessarily, and you
wouldn't want to confuse poor confused us by going into further
details about it.
| > Second, p-adic numbers ARE exact. They're just a different system.
| > It's easier in a lot of ways to be rigorous with them than it is
| > to be rigorous with the real numbers or the complex numbers.
| 
| Is that something you have from memory Keith Ramsay?
No, from my own experience.
| > Third, you're surely aware that nobody will be ripping anybody
| > apart or forcing anybody to do anything. People have tried hard
| > to force you down to rigor, for example by telling you what,
| > allegedly, the common factors between 5 and the roots of that cubic
| > are. Doesn't phase you a bit! And you've shown us all how easy it is
| > to be impervious to demands for rigor.
| 
| That's an appeal to the gallery, which is a logical fallacy.
I'm just answering this suggestion you made that if I had a further
explanation in mind using p-adic numbers, you were going to rip me
apart and force me to be rigorous. I'll be less likely to remind you
of how poor your rigor has been up until now, if you refrain from
bold predictions about what mine is going to be like in the future.
[...]
| > So in fact the original assumption that there was such an f was
| > false, Q.E.D.
| 
| Can you switch to congruences?  What about something like 
| 
|   f(e^{it})^2 - e^{it} = 0(mod 35)
| 
| in the ring of algebraic integers?
That's an interesting question. The multiples of 35 in the algebraic
integers are also dense, which means that we have a lot of freedom
to adjust the values of f. I would guess that this makes it possible
to have a function f which is continuous on the whole complex plane
and satisfies f(z)^2 = z (mod 35) for all algebraic integers z, but
I would have to think some more to prove it.
e^{it} is an algebraic integer only for certain selected t, of course. 
| > Now, a1 and a2 are not square roots, but they are similar enough
| > to a constant times the square root of m to make the same type of
| > argument apply.
| > 
| > Keith Ramsay
| 
| Why don't you just give the a's explicitly and prove it then?
If you mean the a1, a2 and a3 as I would define them, i.e. as
functions on the complex integers with complex integer values,
it's just the time required. Applying the cubic formula, as you
seem to be suggesting I do, is pretty tedious when the coefficients
are expressions involving m and another constant or two. We'll see
whether I have enough spare time this week.
If it's a1, a2 and a3 that might have a different independent
variable than m, it would of course depend upon what that variable
is.
Keith Ramsay
====
[...]
>>Remember (unlike e.g. the Gaussian integers) the algebraic integers
>>are dense in the complex plane, meaning that any circle, no matter
>>how small, has algebraic integers inside it.
> 
> 
> Hmmm...interesting assertion, but what about singularities?
[...]
        2> sqrt(2) > 1  so,
        1> sqrt(2) - 1 >0
Also, sqrt(2) - 1 is an algebraic integer because sqrt(2) and 1 are.
In fact, 1/2 > sqrt(2) - 1 >0  .
So for any n>=1:
  (1/2)^n > (sqrt(2) - 1)^n > 0  .
Note that, by repeated multiplication, (sqrt(2) - 1)^n is also an
algebraic integer.
Given any positive real number d, there exists an n such that:
     d> (sqrt(2) - 1)^n > 0.
Letting M and N vary through all ordinary integers, we get a lattice
or square grid from the collection of algebraic integers
   M* (sqrt(2)-1)^n  +  N*i*(sqrt(2)-1)^n  .
The basic , small squares in the grid have a side of
   (sqrt(2) - 1)^n < d   [ the n depends on d...]
So we can construct grids with d=0.1, 0.01, 0.001, ....
  d= 10^(-100) ... , and all the numbers in all the grids
are algebraic integers.
David Bernier
====
> 
>> Remember (unlike e.g. the Gaussian integers) the algebraic integers
>> are dense in the complex plane, meaning that any circle, no matter
>> how small, has algebraic integers inside it.
Hmmm...interesting assertion, but what about singularities?
> 
> I'll bite. What about singularities?
> 
> We aren't talking about a function here, just about two sets, the
> complex numbers and the algebraic integers. There aren't any
> singularities. And if there were, what does that have to do with
> whether there are infinitely many algebraic integers everywhere you
> look?
> 
>                 - Randy
A singularity is a circle of 0 radius, and in the complex plane I can
pick an infinite number of fractions.
Fascinating how far the discussion is from my paper, but maybe not so
far from the subject line, eh?
James Harris
====
> [...]
> |And you've gone off on a tangent.  But as you mention it, I suggest
> |you go ahead and be *careful* instead of possibly hoping that you can
> |get away with casting doubt.
> |
> |It's not a political discussion Keith Ramsay.  You're not on Fox News.
> 
> You're hardly in any position to complain, either about glibly
> expressing doubts about things you don't understand, or about making
> political statements, or about being careful either!
Support what you say mathematically.  It's not a political discussion
and you're not on Fox News!
> [...]
> | > If this is so, then one of the main questions is what values this
> | > common factor has when m is not 0. I don't see a problem with saying
> | > it's f when m=0, but I also don't see why it shouldn't be a varying
> | > factor of f as m changes.
> | 
> | Yeah, and I gave an example yesterday which should tell you why.
> | 
> |   P(x) = 2x^2 + 4x + 2
> | 
> |   (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2
> | 
> |   a_1 a_2 = 2x^2,
> 
> Is this last equation another assumption? Otherwise, what's to rule
> out a_1=1, a_2 = x^2+2x-1, for instance?
The last equation rules it out, which is why I gave it.
> | and I note that when x=0, at *least* one of the a's must equal 0, and
> | further note that the constant term of the factor a_1 + 1, which I'll
> | call g_1 is 1 at that point.  Further the factor a_2 + 2 has a factor
> | that is 2 at that point, and I'll call it g_2.
> |
> | Now then Keith Ramsay, do you question what g_2 will do when x does
> | not equal 0?
> | 
> | So why do you do it above?
> 
> Continually expressing shock that I don't agree with you, when you
> haven't indicated a principle to justify your claim, is just a bunch
> of posturing. What's the general principle?
The general principle is that factors of polynomials can be written as
r+c, where r varies or is 0, while c is constant.
That's it.  Like I've said before that simple thing is the linchpin of
my position.
To refute my mathematical argument, you have to refute it.
However, posters keep making long-winded posts that manage to avoid
it.
> | Ultimately, you may try and claim that irreducibility over Q is what's
> | pertinent, if so, explain in detail.
> 
> The two cases aren't parallel. You really should get out of the
> habit of thinking that analogy and cajolery add up to a logical
> argument. Giving a bunch of examples which work the way you want
> another example to work is not a proof.
I've given the proof.
 
> If I were to give an explanation for why the examples that work,
> work, I'd give it in terms of the following principle:
> 
>  (*)  If a divides bc, then there exist m and n such that a=mn,
>       m divides b and n divides c.
> 
> You give examples where the product of two polynomials is divisible
> by an integer. Given that (*) is true in these cases, we can then
> try to figure out what m and n would work by considering the constant
> terms and so on as you do. All of your simplified examples are ones
> where (*) works.
 But (*) is not generally true for functions having algebraic integer
> values.
Irrelevant.
> [...]
> | > | >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?
> | > | 
> | > | They're not necessarily functions of m.
> | > 
> | > At some point you need actually to say what kind of thing a1, a2, and
> | > a3 are, without leaving any special wriggle room. If treating them as
> | > functions of m is not quite good enough, then you really need to say
> | > what they are.
> | 
> | You didn't connect back with your own statement, fascinating.
> | 
> | I'm repeating myself with that statement, but before you talked of
> | reparametrization in reply.
> | 
> | I fear you are a parrot.  You look for key words and then regurgitate
> | based on something you think is relevant, but you need to THINK.
> 
> I didn't think I would need to put in the previous quoted text in
> order for you to remember what you had just written:
It's not necessary that they be functions of m.  The question of
whether or not the a's for the specific example I gave are functions
of m, keeps coming up, but it's not really relevant.
In general, there is no necessity that given P(x), a factor of P(x)
can be written as a function of x plus a constant.
> #> Perhaps it would help if you clarified and/or confirmed 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 you're saying the 
value
> #> of g1 for each allowed value of m is divisible by f? 
> #
> #It depends on context.  It's like if you have y=2x+1, and consider
> #various values.  Now I see pressure to write that as f(x)=2x+1, but I
> #don't think it's worth the potential confusion.
> #
> #The problem is that though you see m as the key variable, it may in
> #fact be a *factor* of m that is the true independent variable, which
> #I'm sure has confused quite a few people.  Now it's a measure of your
> #mathematical ability if you know what I mean there, and let's just say
> #that I'm not interested in expanding out in a way that leaves *more*
> #room for confusion.
> 
> So you're telling us that m might not be the true independent
> variable. Do you realize that switching from one independent
> variable to another one is known as reparametrizing?
That's not relevant.
> A reasonable reader, who had read the paper but not this discussion,
> would naturally tend to suspect that a1, a2 and a3, which are
> described as having values at given values of m, would tend to
> suspect that they are simply functions of m. If this is not
> necessarily so, it's because you've left out any explanation for
> what else you intend for them to be.
Questions about whether or not they're functions of m are not
relevant.
> This stuff about how being more specific is bad because the rest of
> us (being inferior to you in mathematical ability of course) would
> be liable to be confused is just an excuse. You are, because of
> your lack of practice in writing actual proofs, failing to be
> specific, and as a result getting confused. I think you're just trying
> to bluff your way past the fact that you don't know how to define
> them in such a way as to get the argument in the paper to work.
The paper has a lemma.  Given that lemma its conclusion follows easily
enough.
If you have a problem with the paper, you need to show it
mathematically, instead of talking on all these extraneous subjects.
For instance, the question of whether or not the a's are functions of
m, is irrelevant.  The lemma is in the paper for a reason.  In dodging
the lemma and in continually bringing up irrelevant issues, it seems
to me that you are deliberately being intellectually dishonest, with a
perverse need to try and influence others to believe falsehoods.
James Harris
====
[snip]
> The paper has a lemma.  Given that lemma its conclusion follows easily
> enough.
The lemma is wrong.
> If you have a problem with the paper, you need to show it
> mathematically, instead of talking on all these extraneous subjects.
The errors have been shown mathematically already. You have chosen to ignore 
the truth.
> For instance, the question of whether or not the a's are functions of
> m, is irrelevant.  The lemma is in the paper for a reason.  In dodging
> the lemma and in continually bringing up irrelevant issues, it seems
> to me that you are deliberately being intellectually dishonest, with a
> perverse need to try and influence others to believe falsehoods.
The arguments you present in your paper have been conclusively refuted. It 
is you who are being
intellectually dishonest, with a perverse need to try and influence others 
to believe falsehoods.
You have consistently revealed yourself to be a defender of lies and an 
enemy of truth. When confronted with
specific errors you ignore the facts and continue your delirium.
James Harris, you should be ashamed of your behavior -- especially since you 
persist in document it in public
records.
--
A man with integrity will identify, acknowledge and correct his errors. A 
man without integrity will ignore,
deny or defend them.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
> Sorry about the duplicate postings; I got caught by this thing where
> I get a connection timed out page while trying to post with Google,
> but apparently the posting goes out anyway.
> 
> [...]
> | > Remember (unlike e.g. the Gaussian integers) the algebraic integers
> | > are dense in the complex plane, meaning that any circle, no matter
> | > how small, has algebraic integers inside it.
> | 
> | Hmmm...interesting assertion, but what about singularities?
> 
> As usual by a circle I mean one having positive radius, which doesn't
> have singularities.
0 is a positive number.
 
> [...]
> | > No, my main point in bringing this stuff up was to try to convince
> | > you that defining a1, a2, and a3 in such a way that their values at
> | > one point are related to their values at other points in a useful 
way.
> | 
> | Um, but that is an insinuation that they aren't already rigorously
> | defined, when in fact they are.
> 
> I don't see much sign of your knowing what it means to give a rigorous
> But an equation is only meaningful if one knows somehow what kind of
> things the terms in it represent. The paper refers to the value one
> has at m=0, so there has to be some sense in which one can substitute
> m=0 into one of them. It's natural then to assume you mean for them
> to be functions of m. But then you say that they're not necessarily
> functions of m, without saying specifically what else they might be.
> (You hint that expanding on this point might cause confusion.)
> 
> Obviously you're not giving a rigorous definition.
Obviously?
The a's are defined by
 (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),
where v=-1+mf^2.
claims with mathematics.
 
> | > If my explanation isn't working, fine, someone will eventually find
> | > a better way of explaining it to you.
> | 
> | That is an irrational statement.  Where do you suppose this someone
> | is?
> 
> I see at least a couple of other people who also understand this point
> about how the values of a1, a2, and a3 at one point relate to their
> values at another point, and have been trying to explain it to you.
> I'm just assuming that eventually one of us will come up with a better
> way of explaining it to you.
Well I should have helped you and these hypothetical others out with
the information above.  Solve that cubic and see what happens.
> | Possibly your memory tells you they're in some potential field?
> 
> ??
Don't worry about it.
> [...]
> | That is false as the a's are determined both in form and type.
> 
> As in, sort of like functions of m but not necessarily, and you
> wouldn't want to confuse poor confused us by going into further
> details about it.
The a's are defined by
 (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),
where v=-1+mf^2.
And how do you know?  Wouldn't it be from your *memory* Keith Ramsay?
That's why it's called experience, as in it happened in the past, so
you have to use memory unless possibly you're relying on writings?
 
> | > Third, you're surely aware that nobody will be ripping anybody
> | > apart or forcing anybody to do anything. People have tried hard
> | > to force you down to rigor, for example by telling you what,
> | > allegedly, the common factors between 5 and the roots of that cubic
> | > are. Doesn't phase you a bit! And you've shown us all how easy it is
> | > to be impervious to demands for rigor.
> | 
> | That's an appeal to the gallery, which is a logical fallacy.
> 
> I'm just answering this suggestion you made that if I had a further
> explanation in mind using p-adic numbers, you were going to rip me
> apart and force me to be rigorous. I'll be less likely to remind you
> of how poor your rigor has been up until now, if you refrain from
> bold predictions about what mine is going to be like in the future.
Your mathematical arguments should stand on their own Keith Ramsay. 
And there's no need to shrink back if you're correct.  It's about the
math, not your fears.
 
> [...]
> | > So in fact the original assumption that there was such an f was
> | > false, Q.E.D.
> | 
> | Can you switch to congruences?  What about something like 
> | 
> |   f(e^{it})^2 - e^{it} = 0(mod 35)
> | 
> | in the ring of algebraic integers?
> 
> That's an interesting question. The multiples of 35 in the algebraic
> integers are also dense, which means that we have a lot of freedom
> to adjust the values of f. I would guess that this makes it possible
> to have a function f which is continuous on the whole complex plane
> and satisfies f(z)^2 = z (mod 35) for all algebraic integers z, but
> I would have to think some more to prove it.
> 
> e^{it} is an algebraic integer only for certain selected t, of course. 
Of course.  Feel free to think further on it.
> | > Now, a1 and a2 are not square roots, but they are similar enough
> | > to a constant times the square root of m to make the same type of
> | > argument apply.
> | > 
> | > Keith Ramsay
> | 
> | Why don't you just give the a's explicitly and prove it then?
> 
> If you mean the a1, a2 and a3 as I would define them, i.e. as
> functions on the complex integers with complex integer values,
> it's just the time required. Applying the cubic formula, as you
> seem to be suggesting I do, is pretty tedious when the coefficients
> are expressions involving m and another constant or two. We'll see
> whether I have enough spare time this week.
> 
> If it's a1, a2 and a3 that might have a different independent
> variable than m, it would of course depend upon what that variable
> is.
> 
> Keith Ramsay
So you're admitting to bringing up what I've called irrelevant issues
without having actually checked them, while refusing to fully
acknowledge the mathematical argument in the paper including a rather
basic lemma that you simply don't mention.
James Harris
====
        ... stuff deleted ...
> The paper has a lemma.  Given that lemma its conclusion follows easily
> enough.
> 
> If you have a problem with the paper, you need to show it
> mathematically, instead of talking on all these extraneous subjects.
> 
I have shown that your Primary Argument reaches a false conclusion,
by showing (what I'm sure you must have seen by now, but are choosing
to ignore) explicit formulas for factors that each of your a's have
in common with 5.
Surely that cannot be considered extraneous, as it has to do with
what one must assume is the primary result in the paper. That
primary result is false, thus your argument must be incorrect.
in threads that you have started, and keep participating in. The
arithmetic is simple, if tedious, and the argument, unlike yours,
can be verified, line by line.
Yet, you persist in this denial.
the argument. Just say the word.
> For instance, the question of whether or not the a's are functions of
> m, is irrelevant.  The lemma is in the paper for a reason.  In dodging
> the lemma and in continually bringing up irrelevant issues, it seems
> to me that you are deliberately being intellectually dishonest, with a
> perverse need to try and influence others to believe falsehoods.
> 
Is it irrelevant to point out that your conclusion (regarding the a's
being coprime to 5) is false? Do you deny that I have done this?
Is it perverse to have demonstrated the falsity of your conclusion?
Is it perverse to mention that fact?
Is it a falsehood that I am perpetrating?
If so, then why?
If not, then why do you insist on asserting that others
are lying, when you knowingly insist that an incorrect
result is true?
> 
> James Harris
Dale.
====
> 
>>[...]
>>| > If this is so, then one of the main questions is what values this
>>| > common factor has when m is not 0. I don't see a problem with saying
>>| > it's f when m=0, but I also don't see why it shouldn't be a varying
>>| > factor of f as m changes.
>>| 
>>| Yeah, and I gave an example yesterday which should tell you why.
>>| 
>>|   P(x) = 2x^2 + 4x + 2
>>| 
>>|   (a_1 + 1)(a_2 + 2) = 2x^2 + 4x + 2
>>| 
>>|   a_1 a_2 = 2x^2,
>>Is this last equation another assumption? Otherwise, what's to rule
>>out a_1=1, a_2 = x^2+2x-1, for instance?
> 
> 
> The last equation rules it out, which is why I gave it.
The last equation is not necessarily true.  What must be true is that 
a_1(x) a_2(x) + a_2(x) + 2 a_1(x) = 2x^2 + 4x.  Because we do not know 
the a_i(x) are polynomials, we cannot assume that they will behave 
exactly the way terms of a polynomial would.
>>| and I note that when x=0, at *least* one of the a's must equal 0, and
>>| further note that the constant term of the factor a_1 + 1, which I'll
>>| call g_1 is 1 at that point.  Further the factor a_2 + 2 has a factor
>>| that is 2 at that point, and I'll call it g_2.
>>|
>>| Now then Keith Ramsay, do you question what g_2 will do when x does
>>| not equal 0?
>>| 
>>| So why do you do it above?
>>Continually expressing shock that I don't agree with you, when you
>>haven't indicated a principle to justify your claim, is just a bunch
>>of posturing. What's the general principle?
> 
> 
> The general principle is that factors of polynomials can be written as
> r+c, where r varies or is 0, while c is constant.
This is trivially true, but because you have failed to say anything 
further about the nature of r, it also communicates virtually nothing.
> That's it.  Like I've said before that simple thing is the linchpin of
> my position.
It's trivially true, but you've done nothing rigorous with it.
> To refute my mathematical argument, you have to refute it.
No one is trying to refute that fact.  It's the conclusions you leap to 
from it that are the problem.  If you say something a little more 
precise about r, then you may be able to fill in the holes in your lemma.
> However, posters keep making long-winded posts that manage to avoid
> it.
No, it's the only part of the paper that is true.  It's the *rest* of 
your paper that has problems.
>>| Ultimately, you may try and claim that irreducibility over Q is what's
>>| pertinent, if so, explain in detail.
>>The two cases aren't parallel. You really should get out of the
>>habit of thinking that analogy and cajolery add up to a logical
>>argument. Giving a bunch of examples which work the way you want
>>another example to work is not a proof.
> 
> 
> I've given the proof.
No, you've given a claim that you have failed to defend.  You also tend 
to change the subject or start new threads rather than deal with the 
problems that have been shown.
>>If I were to give an explanation for why the examples that work,
>>work, I'd give it in terms of the following principle:
>> (*)  If a divides bc, then there exist m and n such that a=mn,
>>      m divides b and n divides c.
>>You give examples where the product of two polynomials is divisible
>>by an integer. Given that (*) is true in these cases, we can then
>>try to figure out what m and n would work by considering the constant
>>terms and so on as you do. All of your simplified examples are ones
>>where (*) works.
>>But (*) is not generally true for functions having algebraic integer
>>values.
> 
> Irrelevant.
> 
What you are trying to prove, and the reason your proof doesn't work is 
irrelevant?  How does that work?
> 
>>[...]
>>| > | >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?
>>| > | 
>>| > | They're not necessarily functions of m.
>>| > 
>>| > At some point you need actually to say what kind of thing a1, a2, and
>>| > a3 are, without leaving any special wriggle room. If treating them as
>>| > functions of m is not quite good enough, then you really need to say
>>| > what they are.
>>| 
>>| You didn't connect back with your own statement, fascinating.
>>| 
>>| I'm repeating myself with that statement, but before you talked of
>>| reparametrization in reply.
>>| 
>>| I fear you are a parrot.  You look for key words and then regurgitate
>>| based on something you think is relevant, but you need to THINK.
>>I didn't think I would need to put in the previous quoted text in
>>order for you to remember what you had just written:
> 
> 
> It's not necessary that they be functions of m.  The question of
> whether or not the a's for the specific example I gave are functions
> of m, keeps coming up, but it's not really relevant.
> 
> In general, there is no necessity that given P(x), a factor of P(x)
> can be written as a function of x plus a constant.
They can be constant functions of x.  Then you have a constant (written 
as a function) plus a constant.  Also note that you've just said it's ok 
to do what you objected to at the beginning of this post.  Keith made 
a_1(x) = 1, a_2(x)=x^2+2x-1.  Here, a_1(x)+1 is in the form constant + 
constant.  Is this ok or not?
> 
>>#> Perhaps it would help if you clarified and/or confirmed 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 you're saying the 
value
>>#> of g1 for each allowed value of m is divisible by f? 
>>#
>>#It depends on context.  It's like if you have y=2x+1, and consider
>>#various values.  Now I see pressure to write that as f(x)=2x+1, but I
>>#don't think it's worth the potential confusion.
>>#
>>#The problem is that though you see m as the key variable, it may in
>>#fact be a *factor* of m that is the true independent variable, which
>>#I'm sure has confused quite a few people.  Now it's a measure of your
>>#mathematical ability if you know what I mean there, and let's just say
>>#that I'm not interested in expanding out in a way that leaves *more*
>>#room for confusion.
>>So you're telling us that m might not be the true independent
>>variable. Do you realize that switching from one independent
>>variable to another one is known as reparametrizing?
> 
> That's not relevant.
Are you saying it doesn't matter what your independent variables are? 
If so, then you aren't saying anything and can draw no conclusions.  It 
may be inconvenient, but is highly relevant.
>>A reasonable reader, who had read the paper but not this discussion,
>>would naturally tend to suspect that a1, a2 and a3, which are
>>described as having values at given values of m, would tend to
>>suspect that they are simply functions of m. If this is not
>>necessarily so, it's because you've left out any explanation for
>>what else you intend for them to be.
> 
> Questions about whether or not they're functions of m are not
> relevant.
If they cannot vary with m, they are constants.  If they are constants, 
they can be computed and we can easily verify whether your paper is 
true.  If they do vary with m, they are functions of m.  In that case, 
different types of analysis can be done on them.  If we don't know what 
is being talked about, then there is no way to conclude anything.  You 
are refusing to address the key issue of defining what you are talking 
about.  When you do that, you are not saying anything meaningful. 
Instead, you are babbling.
>>This stuff about how being more specific is bad because the rest of
>>us (being inferior to you in mathematical ability of course) would
>>be liable to be confused is just an excuse. You are, because of
>>your lack of practice in writing actual proofs, failing to be
>>specific, and as a result getting confused. I think you're just trying
>>to bluff your way past the fact that you don't know how to define
>>them in such a way as to get the argument in the paper to work.
> 
> 
> The paper has a lemma.  Given that lemma its conclusion follows easily
> enough.
Then start by defending your lemma.  There have been several objections 
(and disproofs) that you have completely failed to address.
> If you have a problem with the paper, you need to show it
> mathematically, instead of talking on all these extraneous subjects.
> 
What do you think everyone's been doing?  You are the one who brings up 
Fox News.  You are the one who brings up politics.  You are the one who 
brings up moral character and the supposed flaws in our characters. 
Please deal with the math as we have been.
> For instance, the question of whether or not the a's are functions of
> m, is irrelevant.  The lemma is in the paper for a reason.  In dodging
> the lemma and in continually bringing up irrelevant issues, it seems
> to me that you are deliberately being intellectually dishonest, with a
> perverse need to try and influence others to believe falsehoods.
> 
The lemma has been shown to be highly questionable at best.  You have 
failed to defend it.  Who is dodging issues about the lemma?
> James Harris
-- 
Will Twentyman
====
>  0 is a positive number.
What is your definition of a positive number?
Can a number be both positive and negative?
If you have a non-standard notion of positive,
then  the following argument may prove zero also
is negative.
Suppose +x = -x.
Then 2x = 0
and x = 0.
So, -0 = 0 = +0.
====
>> Sorry about the duplicate postings; I got caught by this thing where
>> I get a connection timed out page while trying to post with Google,
>> but apparently the posting goes out anyway.
>> 
>> [...]
>> | > Remember (unlike e.g. the Gaussian integers) the algebraic integers
>> | > are dense in the complex plane, meaning that any circle, no matter
>> | > how small, has algebraic integers inside it.
>> | 
>> | Hmmm...interesting assertion, but what about singularities?
>> 
>> As usual by a circle I mean one having positive radius, which doesn't
>> have singularities.
0 is a positive number.
Good one; about equivalent to your infamous integers are
irrationals.
No, 0 is NOT a positive number. Positive numbers are strictly larger
than 0. 0 is a non-negative number, and a non-positive number; it is
not a positive number.
======================================================================
Why do you take so much trouble to expose such a reasoner as
 Mr. Smith? I answer as a deceased friend of mine used to answer
 on like occasions - A man's capacity is no measure of his power
 to do mischief. Mr. Smith has untiring energy, which does 
 something; self-evident honesty of conviction, which does more;
 and a long purse, which does most of all. He has made at least
 ten publications, full of figures few readers can critize. A great
 many people are staggered to this extend, that they imagine there
 must be the indefinite 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 Magidin
magidin@math.berkeley.edu
====
[snip]
Give it up, James Harris. Your so-called proof of FLT has gone down in 
flames. Your pitiful attempts to
rescue it with diversions and side shows is laughable. You have covered the 
territory from gadfly to troll to
loon to failure to fool to clown. Take down your idiotic and failed attempt 
at a proof. It's polluting the
internet.
Recently you seem focussed on the notion that mathematicians are liars. (See 
topic of this thread.) Why not
post a list of those lies so others can observe them?--or are *you* the 
liar?
--
There are two things you must never attempt to prove: the unprovable -- and 
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
> 
>  0 is a positive number.
> 
> What is your definition of a positive number?
> Can a number be both positive and negative?
> If you have a non-standard notion of positive,
> then  the following argument may prove zero also
> is negative.
> 
> Suppose +x = -x.
> Then 2x = 0
> and x = 0.
> So, -0 = 0 = +0.
Oh, ok, if you all agree that 0 isn't a positive number that's fine with 
me.
James Harris
====
[snip....
>> As usual by a circle I mean one having positive radius, which doesn't
>> have singularities.
0 is a positive number.
> 
I think I have boiled all of JHs problems down.....
[snip...
====
 > 
 >  0 is a positive number.
 > 
 > What is your definition of a positive number?
James must be a follower of Bourbaki.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> 
>  0 is a positive number.
> 
> What is your definition of a positive number?
> Can a number be both positive and negative?
> If you have a non-standard notion of positive,
> then  the following argument may prove zero also
> is negative.
> 
> Suppose +x = -x.
> Then 2x = 0
> and x = 0.
> So, -0 = 0 = +0.
> 
> Oh, ok, if you all agree that 0 isn't a positive number that's fine with 
me.
> 
> 
> James Harris
Just to keep JSH guessing, to the Bourbaki set, 0 is both positive 
and negative, but not strictly either.
====
[...]
| > Obviously you're not giving a rigorous definition.
| 
| Obviously?
Yes.
| The a's are defined by
| 
|  (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),
| 
| where v=-1+mf^2.
You seem often to have the attitude that equations speak for
themselves. But if you had just left this equation by itself,
it could have been referring to quaternion-valued functions
a_1(m), a_2(m) and a_3(m). Just writing an equation doesn't tell
us what the equation means.
But then you denied that guess. m is not necessarily the independent
variable. So evidently not only does the definition not quite say
what type of functions or values the a_i are, it misleads the reader
into thinking that they're something more obvious than they are.
Don't try to just skim past this issue of what the independent
variable is in these functions.
 
| 
|   a^3 + 3va^2 -(v^3+1) = 0, where again v=-1+mf^2, and the a's are its
| roots.
| 
| So you can solve for the a's and get some result, instead of endlessly
| talking.
The usual solution of the cubic starts by changing the variable to
get rid of the square term. So let b = a+v, and we get
b^3 - 3v^2b + (v^3-1) = 0. The discriminant is
   D = (v^3-1)^2/4 - v^6
     = -(3/4)v^6 - (1/2)v^3 + (1/4)
     = -(3v^3 - 1)(v^3 + 1)/4.
(although in number theory we use a different constant factor).
This shows there are six values of v for which two of the roots
coincide.
By Cardan's formula:
  b = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3},
or
  a = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3} - v,
| It might be especially interesting to see the result and see what
| happens at v=-1.
Because D = (v+1)(1-3v^3)(1+v+v^2)/4, near v=-1, D is approximately
v+1. The two values of sqrt(D) trade places with each other as we
sqrt(D) with -sqrt(D) wouldn't change the value of a. But the two
cube roots are chosen so that at v=-1, for example, they are either
both 1, or one is (-1+sqrt(3)*i)/2 and the other is (-1-sqrt(3)*i)/2,
or the first is (-1-sqrt(3)*i)/2 and the other is (-1+sqrt(3)*i)/2.
So changing the sign on the square root of D is equivalent to leaving
the first root (where both cube roots are taken to be 1) alone, while
exchanging the other two roots. As I said before, if you take a loop
around v=-1, the value of one of the roots comes back to the value of
the other one.
 
| Reminder to readers, Keith Ramsay has brought up a continuity
| question, which actually is irrelevant, but what the hell?
I was trying to convince you that it's unsafe to assume that the
values of a_1, a_2, and a_3 extrapolate in a natural way from one
value of m to another. This would be needed, for example, for a_2
to be always divisible by f or anything like that. Otherwise, how
could it possibly be true that it's one particular one of the three
that has a certain divisibility property, if the manner in which
we've decided which of the three roots to assign to each of the
three variables in an arbitrary way depending on the value of m?
I'm sorry if the relevance doesn't seem clear to you.
The one obvious way to attempt to extrapolate values of a_1, a_2
and a_3 from one value of m to another is by continuity. Given a
path in the complex plane which avoids all the points where two
of the roots coincide, we can see what happens to the a_i if we
continuously change m along that path, while letting the a_i also
change continuously as we go. Unfortunately (or fortunately,
actually) which a_i is which when we reach the end of the path
will depend upon which path we took from one endpoint to the other.
So if I have a given value of a at m=0 (say a=0), the question of
which of the three values at m=1 corresponds to it is meaningless
unless I make an arbitrary convention.
That's essentially what prevents them from being continuous
everywhere.
I don't have the time today, but what one can do is to show that
for each epsilon>0, there exists a delta>0 having the property that
if |v+1| < delta, then one of the roots is within epsilon of the
nonzero root at v=-1 (what was it, a=1?), while the other two satisfy
|a^2/(v+1) - C| < epsilon where C is some constant (probably just 1).
Then we can consider the values v = -1+e^{it}. For small epsilon,
the inequality forces a/e^{it/2} to be close to one of the square
roots of C. Then we can carry out the same argument as last time
but with the two sets being the t for which |a/e^{it/2} - sqrt(C)|
is smaller and the t for which |a/e^{it/2} + sqrt(C)| is smaller.
[...]
| So you're admitting to bringing up what I've called irrelevant issues
| without having actually checked them, while refusing to fully
| acknowledge the mathematical argument in the paper including a rather
| basic lemma that you simply don't mention.
I find the lemma very uninteresting. I don't understand how you
manage to think of it as important.
Similar to a lot of arguments written by undergraduate math majors
early in their studies, this argument plods its way slowly through
the routine parts of the argument, but condenses into a rapid sprint
the part which is most in need of explanation, and also hardest to
believe.
Keith Ramsay
====
> [...]
> | > Obviously you're not giving a rigorous definition.
> | 
> | Obviously?
> 
> Yes.
Let's see if you support your accusation.
> | The a's are defined by
> | 
> |  (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),
> | 
> | where v=-1+mf^2.
> 
> You seem often to have the attitude that equations speak for
> themselves. But if you had just left this equation by itself,
> it could have been referring to quaternion-valued functions
> a_1(m), a_2(m) and a_3(m). Just writing an equation doesn't tell
> us what the equation means.
On the contrary, the equation clearly shows that there is a
factorization.
> a_1, a_2, and a_3 to be functions defined for algebraic integers m,
> with algebraic integer values, having the property that for each m
> they make the equation above hold for all x and y.
Of course the ring is algebraic integers as that's been the ring
throughout the thread and is the ring in the paper Advanced Polynomial
Factorization.
The factorization should hold for all f, m, x and y, in the ring.
There is no need to mention functions.
 
> If that guess had been correct, then I would have said that the
> definition was at least close to rigorous, just leaving a couple of
> things implicit that should be explicit. All that it would take to
> make it rigorous would be a few clarifying words. (As I tried to
> explain to you before, you also couldn't be taking x to be a constant,
> for example.)
That's false.  There is no requirement on the factorization that x not
be constant.
Factorizations don't work that way.
> But then you denied that guess. m is not necessarily the independent
> variable. So evidently not only does the definition not quite say
> what type of functions or values the a_i are, it misleads the reader
> into thinking that they're something more obvious than they are.
Factorizations are not dependent on the dependency of the variables.
> Don't try to just skim past this issue of what the independent
> variable is in these functions.
Factorizations are not dependent on functions.
> | 
> |   a^3 + 3va^2 -(v^3+1) = 0, where again v=-1+mf^2, and the a's are its
> | roots.
> | 
> | So you can solve for the a's and get some result, instead of endlessly
> | talking.
> 
> The usual solution of the cubic starts by changing the variable to
> get rid of the square term. So let b = a+v, and we get
> b^3 - 3v^2b + (v^3-1) = 0. The discriminant is
> 
>    D = (v^3-1)^2/4 - v^6
>      = -(3/4)v^6 - (1/2)v^3 + (1/4)
>      = -(3v^3 - 1)(v^3 + 1)/4.
> 
> (although in number theory we use a different constant factor).
> This shows there are six values of v for which two of the roots
> coincide.
> 
> By Cardan's formula:
> 
>   b = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3},
> 
> or
> 
>   a = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3} - v,
> 
> | It might be especially interesting to see the result and see what
> | happens at v=-1.
Readers should note that assuming the above is correct at v=-1, D=0,
so you get
  a = 1^{1/3} + 1^{1/3} + 1
which gives you the three a's, a_1=0, a_2=0, a_3=3.
Now consider Keith Ramsay's reply.
 
> Because D = (v+1)(1-3v^3)(1+v+v^2)/4, near v=-1, D is approximately
> v+1. The two values of sqrt(D) trade places with each other as we
> sqrt(D) with -sqrt(D) wouldn't change the value of a. But the two
> cube roots are chosen so that at v=-1, for example, they are either
> both 1, or one is (-1+sqrt(3)*i)/2 and the other is (-1-sqrt(3)*i)/2,
> or the first is (-1-sqrt(3)*i)/2 and the other is (-1+sqrt(3)*i)/2.
> 
> So changing the sign on the square root of D is equivalent to leaving
> the first root (where both cube roots are taken to be 1) alone, while
> exchanging the other two roots. As I said before, if you take a loop
> around v=-1, the value of one of the roots comes back to the value of
> the other one.
Um, did *anyone* see an answer in there?
> | Reminder to readers, Keith Ramsay has brought up a continuity
> | question, which actually is irrelevant, but what the hell?
> 
> I was trying to convince you that it's unsafe to assume that the
> values of a_1, a_2, and a_3 extrapolate in a natural way from one
> value of m to another. This would be needed, for example, for a_2
> to be always divisible by f or anything like that. Otherwise, how
> could it possibly be true that it's one particular one of the three
> that has a certain divisibility property, if the manner in which
> we've decided which of the three roots to assign to each of the
> three variables in an arbitrary way depending on the value of m?
> I'm sorry if the relevance doesn't seem clear to you.
There is no relevance.  In polynomial equations you *can* have it
where certain roots have a factor while others do not.
> The one obvious way to attempt to extrapolate values of a_1, a_2
> and a_3 from one value of m to another is by continuity. Given a
> path in the complex plane which avoids all the points where two
> of the roots coincide, we can see what happens to the a_i if we
> continuously change m along that path, while letting the a_i also
> change continuously as we go. Unfortunately (or fortunately,
> actually) which a_i is which when we reach the end of the path
> will depend upon which path we took from one endpoint to the other.
> So if I have a given value of a at m=0 (say a=0), the question of
> which of the three values at m=1 corresponds to it is meaningless
> unless I make an arbitrary convention.
Actually you simply take the three values of the cuberoot operator,
just like with square roots you take the two values of the square root
operator.
For those who are confused simply remember that x^3 - 1 = 0, gives you
three values for x.
> That's essentially what prevents them from being continuous
> everywhere.
Looks to me like you put down a lot of words Keith Ramsay, yet somehow
managed to not even refer back to
 a = ((1-v^3)/2 + sqrt(D))^{1/3} + ((1-v^3)/2 - sqrt(D))^{1/3} - v
from your work before.
> I don't have the time today, but what one can do is to show that
> for each epsilon>0, there exists a delta>0 having the property that
> if |v+1| < delta, then one of the roots is within epsilon of the
> nonzero root at v=-1 (what was it, a=1?), while the other two satisfy
> |a^2/(v+1) - C| < epsilon where C is some constant (probably just 1).
> Then we can consider the values v = -1+e^{it}. For small epsilon,
> the inequality forces a/e^{it/2} to be close to one of the square
> roots of C. Then we can carry out the same argument as last time
> but with the two sets being the t for which |a/e^{it/2} - sqrt(C)|
> is smaller and the t for which |a/e^{it/2} + sqrt(C)| is smaller.
Well, if you don't have the time it's not really worth mentioning, now
is it?
After all, the discussion isn't a political one, and you're NOT on Fox
News!!!
> [...]
> | So you're admitting to bringing up what I've called irrelevant issues
> | without having actually checked them, while refusing to fully
> | acknowledge the mathematical argument in the paper including a rather
> | basic lemma that you simply don't mention.
> 
> I find the lemma very uninteresting. I don't understand how you
> manage to think of it as important.
That lemma is the linchpin of the paper!!!
> Similar to a lot of arguments written by undergraduate math majors
> early in their studies, this argument plods its way slowly through
> the routine parts of the argument, but condenses into a rapid sprint
> the part which is most in need of explanation, and also hardest to
> believe.
Your belief here is irrelevant as a proof begins with a truth then
proceeds by logical steps to a conclusion which then must be true. 
Challenging a paper requires that you find that it did not begin with
a truth, or that you find a break in the logical chain.
James Harris
====
>> 
>>> Remember (unlike e.g. the Gaussian integers) the algebraic integers
>>> are dense in the complex plane, meaning that any circle, no matter
>>> how small, has algebraic integers inside it.
>>Hmmm...interesting assertion, but what about singularities?
>> 
>> I'll bite. What about singularities?
>> 
>> We aren't talking about a function here, just about two sets, the
>> complex numbers and the algebraic integers. There aren't any
>> singularities. And if there were, what does that have to do with
>> whether there are infinitely many algebraic integers everywhere you
>> look?
>> 
A singularity is a circle of 0 radius,
No, it's not. A singularity is a property of a function. And in
topological statements like a circle, no matter how small, it is
understood that the term circle refers to a finite radius.
> and in the complex plane I can
>pick an infinite number of fractions.
What does that have to do with the statement? The statement is this:
Pick a complex value z. Pick a radius epsilon >0. The neighborhood {w
in C: |z - w| < epsilon} has infinitely many algebraic integers in it.
What the heck does I can pick an infinite number of fractions have
to do with the density of algebraic integers?
Within a radius of 0.01 of 1 + i there are an infinite number of
algebraic integers.
Within a radius of 0.00002 of -sqrt(3) + i*pi there are an infinite
number of algebraic integers.
What does I can pick an infinite number of fractions have to do with
statements like those?
                  - Randy
====
yeah, but a circle of no radius would have itself
as a singularity.  did we uncover a new property/axiom/pedagogical twist?
otherwise, Buddy,
a ring is what I say it is, now; capiche?
 
> No, it's not. A singularity is a property of a function. And in
> topological statements like a circle, no matter how small, it is
> understood that the term circle refers to a finite radius.
--A church-school McCrusade (Blair's ideals?):
Harry-the-Mad-Potter want's US to kill Iraqis?...
 http://www.tarpley.net/bush25.htm (Thyroid Storm ch.)
  http://www.rwgrayprojects.com/synergetics/plates/plates.html
   http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX
====
I'm trying to figure out how to correctly form the logical negation of
this sentence:
For all real numbers satisfying a<b, there exists a natural number n,
such that a + 1/n < b.
Converting this sentence into symbolic form gives something like:
(forall a,b in R ):( a < b),(exists n in N):( a+1/n <b )
and then forming the negation, I have
(exists a,b in R):(a>= b),(forall n in N):(a+1/n >= b)
i.e. There exist real numbers satisfying a>=b, such that for all
natural numbers n, a+1/n >= b.
However, I am obviously doing something wrong in the negation, since
BOTH of the above sentences appear to be true, and by the law of
excluded middle, P and not P cannot BOTH be true.
So, what am I doing wrong?
thanks.
====
> I'm trying to figure out how to correctly form the logical negation of
> this sentence:
> For all real numbers satisfying a<b, there exists a natural number n,
> such that a + 1/n < b.
> 
> Converting this sentence into symbolic form gives something like:
> (forall a,b in R ):( a < b),(exists n in N):( a+1/n <b )
> 
> and then forming the negation, I have
> 
> (exists a,b in R):(a>= b),(forall n in N):(a+1/n >= b)
> 
> i.e. There exist real numbers satisfying a>=b, such that for all
> natural numbers n, a+1/n >= b.
> 
> However, I am obviously doing something wrong in the negation, since
> BOTH of the above sentences appear to be true, and by the law of
> excluded middle, P and not P cannot BOTH be true.
> 
> So, what am I doing wrong?
Your original statement says that if there are two reals, a and b with a 
< b, there exists n such that a + 1/n < b.
The negation of that would be that there exist two reals, a and b, with 
a < b, such that for all n, a + 1/n >= b.
I think you got into trouble when you swapped a < b for a >= b in your 
attempted negation.
====
> For all real numbers satisfying a<b, there exists a natural number n,
> such that a + 1/n < b.
 Converting this sentence into symbolic form gives something like:
> (forall a,b in R ):( a < b),(exists n in N):( a+1/n <b )
>
for all a,b in R, (a < b ==> some n in N with a+1/n < b)
> and then forming the negation, I have
> (exists a,b in R):(a>= b),(forall n in N):(a+1/n >= b)
>
some a,b in R with (a < b and for all n in N, not a+1/n < b)
> i.e. There exist real numbers satisfying a>=b, such that for all
> natural numbers n, a+1/n >= b.
>
Archimedean principle for real numbers:
        for all r in R, some n in N with r < n.
if a < b, then b-a /= 0
        some n in N with 1/(b-a) < n
        1/n < b-a as 0 < b-a
        a + 1/n < a + b-a = b
> So, what am I doing wrong?
You converted wrong and you may have used
        ~(p&q) <-> (~p&~q)
instead of
        ~(p&q) <-> (~p v ~q)
when forming the negation.
====
> I'm trying to figure out how to correctly form the logical negation of
> this sentence:
> For all real numbers satisfying a<b, there exists a natural number n,
> such that a + 1/n < b.
The tricky part here is the phrase satisfying a < b.
Actually, the same problem applies to the phrase real also,
except that real is a unary relation, while a < b is a
binary relation.
That is, strictly, you can use only for all x and not
for all real x, etc. That is, the phrase for all real x
is an abreviation for the phrase for all x,... (if x is real,
then ...).
Note that you can swap (for all real a)(for all real b) to
get (for all real b)(for all real a) with no problem.
But, you can not swap (for all real a)(for all real b with a < b)
to get (for all real b with a < b)(for all real a) since
the first a is then left unquantified.
> Converting this sentence into symbolic form gives something like:
> (forall a,b in R ):( a < b),(exists n in N):( a+1/n <b )
This is the first time I have seen the notation :(a < b).
This is what is causing you a problem. You don't have a
correct rule on how to handle the negation through it.
I convert that sentence into symbolic form as follows:
  (for all a in R)(for all b in R)(for some n in N)
     (if a < b then a + 1/n < b)
Since I am going to form the negation, I transform the
if ... then ... into and and or and not terms
as follows:
  (for all a in R)(for all b in R)(for some n in N)
     (not(a < b) or (a + 1/n < b))
> and then forming the negation, I have
> 
> (exists a,b in R):(a>= b),(forall n in N):(a+1/n >= b)
  (for some a in R)(for some b in R)(for all n in N)
     ((a < b) and (a + 1/n >= b))
> i.e. There exist real numbers satisfying a>=b, such that for all
> natural numbers n, a+1/n >= b.
You should not have negated a<b, just like you did not
negate a in R.
> However, I am obviously doing something wrong in the negation, since
> BOTH of the above sentences appear to be true, and by the law of
> excluded middle, P and not P cannot BOTH be true.
> 
> So, what am I doing wrong?
Eliminating a in R etc in the phrase for a in R, I have:
  (for all a)(for all b)(for some n)
     (if a in R and b in R and n in N and a < b then a + 1/n < b)
Try negating that and see that a in R remains and does not
become a not in R.
Or, even better, since I don't like the phrase a < b unless
I definitely already know that a and b are real numbers, I suppose
that the above should be written:
  (for all a)(for all b)(for some n)
     (if a in R and b in R and n in N, then if a < b then a + 1/n < b)
You could try and negate that also. I didn't do it myself.
I hope it works out ok.
-- Bill Hale
====
>I'm trying to figure out how to correctly form the logical negation of
>this sentence:
>For all real numbers satisfying a<b, there exists a natural number n,
>such that a + 1/n < b.
>Converting this sentence into symbolic form gives something like:
>(forall a,b in R ):( a < b),(exists n in N):( a+1/n <b )
  (forall a,b in R) (a < b --> exists n in N (a + 1/n < b))
>and then forming the negation, I have
>(exists a,b in R):(a>= b),(forall n in N):(a+1/n >= b)
Nope.  The negation is 
  (exists a,b in R) ~(a < b --> exists n in N (a + 1/n < b)).
Since ~(p --> q) is p & ~q, the desired negation is
  (exists a,b in R) (a < b & ~exists n in N (a + 1/n < b)),
or, finally,
  (exists a,b in R) (a < b & forall n in N (a + 1/n >= b)).
[...]
Brian
====
[cut]
> Eliminating a in R etc in the phrase for a in R, I have:
> 
>   (for all a)(for all b)(for some n)
>      (if a in R and b in R and n in N and a < b then a + 1/n < b)
I eliminated n in N from the phrase for some n in N wrong.
It should be:
   (for all a)(for all b)(for some n)
      (n in N and (if a in R and b in R and a < b then a + 1/n < b))
-- Bill Hale
====
>I'm trying to figure out how to correctly form the logical negation of
>this sentence:
>For all real numbers satisfying a<b, there exists a natural number n,
>such that a + 1/n < b.
For some real numbers a<b, there is no natural number n such that 
a+1/n < b.
The negation of For all X, Y is For some X, not Y. (Or: There 
exists at least one X such that Y.)
>i.e. There exist real numbers satisfying a>=b, such that for all
>natural numbers n, a+1/n >= b.
This is For some not X, Z. :-)
-- 
Stan Brown, Oak Road Systems, Cortland County, New York, USA
                                  http://OakRoadSystems.com/
You find yourself amusing, Blackadder.
I try not to fly in the face of public opinion.
Mime-version: 1.0
====
> I'm trying to figure out how to correctly form the logical negation of
> this sentence:
> For all real numbers satisfying a<b, there exists a natural number n,
> such that a + 1/n < b.
> 
> Converting this sentence into symbolic form gives something like:
> (forall a,b in R ):( a < b),(exists n in N):( a+1/n <b )
> 
> and then forming the negation, I have
> 
> (exists a,b in R):(a>= b),(forall n in N):(a+1/n >= b)
> 
> i.e. There exist real numbers satisfying a>=b, such that for all
> natural numbers n, a+1/n >= b.
> 
> However, I am obviously doing something wrong in the negation, since
> BOTH of the above sentences appear to be true, and by the law of
> excluded middle, P and not P cannot BOTH be true.
> 
> So, what am I doing wrong?
> 
> thanks.
Assuming that your N does not contain 0, consider:
(exists a,b in R)(forall n in N):if(a >= b) then (a+1/n >= b)
Which negates to:
(forall a,b in R)(exists n in N):(a>= b) and (a+1/n < b)
====
Determine a>0 so that the equationsystem:
        x^2+y^2=18
{
        y=-x+a
has exactly one solution.
How Do I solve a problem of this kind?
====
>Determine a>0 so that the equationsystem:
>        x^2+y^2=18
>        y=-x+a
>has exactly one solution.
How Do I solve a problem of this kind?
Solve the equations simultaneously. Since one is a quadratic, you 
will end up with some +/- radical expression, and the radicand will 
probably contain a. Determine a so that the choice of + or - makes 
no difference, i.e. so that the radicand is 0.
-- 
Stan Brown, Oak Road Systems, Cortland County, New York, USA
                                  http://OakRoadSystems.com/
You find yourself amusing, Blackadder.
I try not to fly in the face of public opinion.
====
> Determine a>0 so that the equationsystem:
> 
>         x^2+y^2=18
> {
>         y=-x+a
> 
> has exactly one solution.
> 
> How Do I solve a problem of this kind?
> 
> 
To rephrase another answer, substitute a - x into x^2 + y^2 - 18 = 0.
Square the a - x and collect things up and you'll have a quadratic.
Think of the quadratic formula. The discriminant, the part under the
square root (b^2 - 4*a*c - this a is not _your_ a) determines 
how
many solutions there are: if it is zero there is only one so set it
equal to zero and solve for a. I get a = 6 (and a = -6.)
-- 
Paul Sperry
Columbia, SC (USA)
====
> Determine a>0 so that the equationsystem:
> 
>         x^2+y^2=18
> {
>         y=-x+a
> 
> has exactly one solution.
> 
> How Do I solve a problem of this kind?
> 
> 
> To rephrase another answer, substitute a - x into x^2 + y^2 - 18 = 0.
> 
> Square the a - x and collect things up and you'll have a quadratic.
[snip]
That'll all work, but another approach is a little geometry
and some trig.
x^2 + y^2 = 18 is a circle centered on (0,0) with
radius 3sqrt(2).  y=-x+a with a>0 is a line with slope -1 that
passes through quadrants I, II, and IV.  Having one solution
means the line intersects the circle at only one point, which
means it must be tangent to the circle.  By the constraints
of the problem, we'll be in quadrant I.  In quadrant I, a
circle only has slope -1 at the point where x=y.  That
immediately gives us the point (3,3).
So the desired tangent line has slope -1 and passes through
(3, 3):
   y - 3 = (-1)(x - 3)
   y - 3 = -x + 3
   y = -x + 6
Thus a=6.
-- 
Rich Carreiro                            rlcarr@animato.arlington.ma.us
====
>Determine a>0 so that the equationsystem:
        x^2+y^2=18
>{
>        y=-x+a
has exactly one solution.
How Do I solve a problem of this kind?
>To rephrase another answer, substitute a - x into x^2 + y^2 - 18 = 0.
>>Square the a - x and collect things up and you'll have a quadratic.
>>    
>[snip]
That'll all work, but another approach is a little geometry
>and some trig.
x^2 + y^2 = 18 is a circle centered on (0,0) with
>radius 3sqrt(2).  y=-x+a with a>0 is a line with slope -1 that
>passes through quadrants I, II, and IV.  Having one solution
>means the line intersects the circle at only one point, which
>means it must be tangent to the circle.  By the constraints
>of the problem, we'll be in quadrant I.  In quadrant I, a
>circle only has slope -1 at the point where x=y.  That
>immediately gives us the point (3,3).
So the desired tangent line has slope -1 and passes through
>(3, 3):
>   y - 3 = (-1)(x - 3)
>   y - 3 = -x + 3
>   y = -x + 6
Thus a=6.
>
And avoids the spurious a=-6 solution.
This reminds me of one definition of a mathematician as someone who will 
spend two weeks looking for the way to reduce two hours of calculation 
to 10 minutes.  (Actually, I first heard that as trick rather than 
way, but many of the tricks are actually rather useful, more so 
than  a method is a device, used twice would lead you to believe.)
My experience has been that very few students appreciate the elegance of 
this and most view it as a trick.  In other words, it's totally 
inappropriate for classroom use.
I wonder what we'll find out about Mr. Aspen's mathematical sense.
Jon Miller
====
> Determine a>0 so that the equationsystem:
         x^2+y^2=18
> {
>         y=-x+a
 has exactly one solution.
 How Do I solve a problem of this kind?

I dont know what more to say! =)
Said Aspen
====
[skip]
> Since I explained it the implication that I didn't is false.
Since you didn't actually explained anything, the implication is correct.
I asked you to explain what *kind* of thing factor g is, and your 
response was -
Not even close. Read the paper.  That's not an explanation.
====
Does James Harris have a degree in math?
====
> Does James Harris have a degree in math?
No.  He claims to have an undergraduate degree in physics.
-- 
Wayne Brown                | When your tail's in a crack, you improvise
fwbrown@bellsouth.net      |  if you're good enough.  Otherwise you give
                           |  your pelt to the trapper.
e^(i*pi) = -1  -- Euler  |           -- John Myers Myers, 
Silverlock
====
>  
> I was going back over my work as I often do when I finally realized
> that my lemma in my paper
> 
> Advanced Polynomial Factorization
> http://groups.msn.com/AmateurMath/Documents/FactProof.pdf
> 
> was wrong.
> 
> I had a factor g of the polynomial P(x) where g = r+c, and r is
> supposedly proven to not be coprime to x, when in fact it can be
> coprime to x.
> 
> What I was actually using for the paper is the fact that r varies 
with
> x, while c remains constant, so I made the simplest change.
> 
> It is possible to prove that r is not coprime to P(x)-P(0), but not
> worth the effort here.
> 
> My apologies for the error.
> 
> In the first paragraph you give the wrong roots for x^2+x-5=0. You
> might want to correct them, too.
> 
> 
> James, I am new to your work. You say: As an example consider
> sqrt(x+1) which is a non polynomial factor of x+1. What do you mean
> by saying that sqrt(x+1) is a factor of x+1?
> 
> 
> John Peters
There are two ways to look at it.  Since the ring is algebraic
integers, you can see it as that given x, which is going to be an
algebraic integer as that's the ring, sqrt(x+1) is a factor of x+1.
For instance with x=2, which is an algebraic integer, sqrt(3) is a
factor of 3.
The other way to look at it is that x+1 can be considered the result
of the multiplication sqrt(x+1) times sqrt(x+1) equals x+1; therefore,
sqrt(x+1) is a factor of x+1, where ring operations allow that
expression.  But they don't allow (x+1)/(y+1) in general as the / is
not allowed in general as division is not a ring operation.
So I can also say that given the polynomial P(x)=x+1, sqrt(x+1) is a
factor of the polynomial, though you may wonder what to now call
sqrt(x+1).  I don't care what you call it.
Now sure, if you qualify by saying y=x, then y+1 is a factor of x+1 in
the ring, but not, in general, otherwise, if all you know is that y+1
is in the ring.
Some may see that I don't have to talk specifically about polynomial
rings or something wackier to handle sqrt(x+1) by taking this
approach.
If you find that makes your head swim, you can consider values of x+1
from the first point of view.
James Harris
====
> 
> ...
> sqrt(x+1) is a factor of x+1, where ring operations allow that
> ...
In _what_ is sqrt(x+1) a factor of x+1?
GC
====
>> 
>> ...
>> sqrt(x+1) is a factor of x+1, where ring operations allow that
>> ...

>In _what_ is sqrt(x+1) a factor of x+1?
I believe this is what JSH means: 
If x is an algebraic integer, then so is x+1 and sqrt(x+1). Moreover, there 

is an algebraic integer r (viz. r = sqrt(x+1)) such that r*sqrt(x+1) = x+1. 

Therefore sqrt(x+1) is a factor of x+1 in the ring of algebraic integers.
If this IS what he means, then he is, of course, quite correct. (Oh, GOD! 
It's five o'clock in the morning of midsummer's eve and JSH is starting to 
make sense!)
The problem is that he categorically denies that 2 is a factor of x^2 + x 
in the ring of integers, so my interpretation may be wrong.
-- 
Thomas Wasell     | A great many people think they are thinking when they
wasell@bahnhof.se | are merely rearranging their prejudices. 
                  |     -- William James
====
>> ...
>> sqrt(x+1) is a factor of x+1, where ring operations allow that
>> ...
In _what_ is sqrt(x+1) a factor of x+1?
 I believe this is what JSH means:
 If x is an algebraic integer, then so is x+1 and sqrt(x+1).
Moreover, there
> is an algebraic integer r (viz. r = sqrt(x+1)) such that r*sqrt(x+1)
= x+1.
> Therefore sqrt(x+1) is a factor of x+1 in the ring of algebraic
integers.
 If this IS what he means, then he is, of course, quite correct. (Oh,
GOD!
> It's five o'clock in the morning of midsummer's eve and JSH is
starting to
> make sense!)
 The problem is that he categorically denies that 2 is a factor of
x^2 + x
> in the ring of integers, so my interpretation may be wrong.
James,
Building on the insight provided by Thomas, let me frame a possible
definition:
1) Let S be a set.
2) Let f, g and h be [single-valued] functions defined on S and taking
values in S. That is, their domain and codomain are S.
3) Let * be a [single-valued] binary operator whose operands are
elements of S and which takes values in S.
4) Suppose, for all x in S, f(x)*g(x)=h(x).
Definition: In this case we say that f is a factor of h. Or
(equivalently) that h is divisible by f.
In the case you are interested in, S is the set of algebraic integers,
and * is (complex) multiplication.
(Note that our sqrt(x+1) example might have some problems because
sqrt(x+1) might not be single-valued.)
Is the above definition correct?
John Peters
====
>> 
>> ...
>> sqrt(x+1) is a factor of x+1, where ring operations allow that
>> ...

>In _what_ is sqrt(x+1) a factor of x+1?
> 
> I believe this is what JSH means: 
> 
> If x is an algebraic integer, then so is x+1 and sqrt(x+1). Moreover, 
there 
> is an algebraic integer r (viz. r = sqrt(x+1)) such that r*sqrt(x+1) = 
x+1. 
> Therefore sqrt(x+1) is a factor of x+1 in the ring of algebraic integers.
> 
> If this IS what he means, then he is, of course, quite correct. (Oh, GOD! 

> It's five o'clock in the morning of midsummer's eve and JSH is starting to 

> make sense!)
> 
> The problem is that he categorically denies that 2 is a factor of x^2 + x 

> in the ring of integers, so my interpretation may be wrong.
That's not true.  The original post had something about that being in
the same sense if I remember correctly.  So you left off key
information in your claim.
After all, in the ring of algebraic integers, 2 is NOT in general a
factor of x^2 + x.
For instance, with x=(1+sqrt(-3))/2, x^2 + x = sqrt(-3).
That it is in the ring of integers is, in my mind, trivia, which is
not in the same sense as something like x being a factor of x^2 + x,
which IS true in the ring of algebraic integers, so it's NOT in the
same sense.
What I've seen time and time again is a grasping at any little comment
of mine to try and see someway to make it not make sense.
Here when I talk of factorizations in what I think is a rather obvious
way, bringing up a trivial result in the ring of integers as if it has
great signficance just makes me wonder.
Don't ANY of you care about what the truth is?  
Don't bother me with trivia. 
James Harris
====
nevermind.  let's start another item!
>  > I had a factor g of the polynomial P(x) where g = r+c, and r is
>  > supposedly proven to not be coprime to x, when in fact it can be
>  > coprime to x.
>  ...
>  > Or r can be constant and 0.
> 
> What you fail to state anywhere is how g is restricted.  Or is it
> unrestricted?  If unrestricted:
> 
>    g(x) = def: 1 (if x = 0), x + 5 (elsewhere)
> g(x) is a factor of P(x) = x + 5 in the sense you state.  What are r and 
c?
--les ducs d'Enron!
 http://quincy4board.homestead.com/
 Funny.html  (schoolboard stuffin')
====
> The problem is that he categorically denies that 2 is a factor of x^2 + 
x 
> in the ring of integers, so my interpretation may be wrong.
> 
> That's not true.  The original post had something about that being in
> the same sense if I remember correctly.  So you left off key
> information in your claim.
> 
> After all, in the ring of algebraic integers, 2 is NOT in general a
> factor of x^2 + x.
> 
> For instance, with x=(1+sqrt(-3))/2, x^2 + x = sqrt(-3).
> 
But, 2 IS a factor of sqrt(-3), the other factor being sqrt(-3)/2,
which is a root of 4x^2+3, hence algebraic.
====
[snip]
>But, 2 IS a factor of sqrt(-3), the other factor being sqrt(-3)/2,
>which is a root of 4x^2+3, hence algebraic.
But it's not an algebraic *integer*. I.e. it's not the root of a *monic* 
polynomial.
-- 
Thomas Wasell     | For a man to truly understand rejection, he must first
wasell@bahnhof.se | be ignored by a cat. 
====
> [snip]
>But, 2 IS a factor of sqrt(-3), the other factor being sqrt(-3)/2,
>which is a root of 4x^2+3, hence algebraic.
> 
> But it's not an algebraic *integer*. I.e. it's not the root of a *monic* 
> polynomial.
I'm just playing by the same rules that James does --
a polynomial is
an algebraic integer,
a function,
an integer,
a square root,
a ring of polynomials,
a real number,
a ring of algebraic integers,
a monic polynomial,
a complex number,
a transcendental function,
a Laplace Transform, a ...
Hell, no wonder math is full of contradictions!
====
        ... a buncha crap ...
> What I've seen time and time again is a grasping at any little comment
> of mine to try and see someway to make it not make sense.
> 
> Here when I talk of factorizations in what I think is a rather obvious
> way, bringing up a trivial result in the ring of integers as if it has
> great signficance just makes me wonder.
> 
> Don't ANY of you care about what the truth is?  
> 
Apparently more than you do. Note this:
Define the polynomials
        q(x) = 8 x^2 - 76 x -185
        r(x) = 8 x^2 - 4 x - 45
        s(x) = 4 x^2 - 37 x - 104
Then for z any root of the polynomial
        p(x) = x^3 - 12 x^2 + 65,
we have
        q(z)*r(z) = 5
        r(z)*s(z) = z.
By virtue of the polynomials q,r,s having integer coefficients, and
of z being a root of p(x), we know that z, q(z), r(z), and s(z) are all
algebraic integers.
Finally, the minimal polynomial for r(z) is
        MinPoly(r(z)) = x^3 - 969 x^2 + 315 x + 5
The divisibility properties were verified in Maxima; the minimal
polynomial was provided by KASH, and verified in KASH.
Your claim regarding the properties of the coefficients ai in
the factorization:
        65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
namely, that ONE of the ai's is coprime to 5, is clearly
incorrect. Therefore, your argument is flawed.
Once again, your methods are shown to produce incorrect
results, despite your confidence in your own ability to
spot correct logic. Your logic, your methods, your grasp
of mathematics are *all* unsound.
> Don't bother me with trivia. 
> 
How trivial was that?
Score:
        sci.math        omega
        JSH             nullset
> 
> James Harris
Dale.
====
John Peters <tafelnder@yahoo.com> skrev i melding
> ...
>> sqrt(x+1) is a factor of x+1, where ring operations allow that
>> ...
In _what_ is sqrt(x+1) a factor of x+1?
 I believe this is what JSH means:
 If x is an algebraic integer, then so is x+1 and sqrt(x+1).
> Moreover, there
> is an algebraic integer r (viz. r = sqrt(x+1)) such that r*sqrt(x+1)
> = x+1.
> Therefore sqrt(x+1) is a factor of x+1 in the ring of algebraic
> integers.
 If this IS what he means, then he is, of course, quite correct. (Oh,
> GOD!
> It's five o'clock in the morning of midsummer's eve and JSH is
> starting to
> make sense!)
 The problem is that he categorically denies that 2 is a factor of
> x^2 + x
> in the ring of integers, so my interpretation may be wrong.
 James,
 Building on the insight provided by Thomas, let me frame a possible
> definition:
 1) Let S be a set.
 2) Let f, g and h be [single-valued] functions defined on S and taking
> values in S. That is, their domain and codomain are S.
 3) Let * be a [single-valued] binary operator whose operands are
> elements of S and which takes values in S.
 4) Suppose, for all x in S, f(x)*g(x)=h(x).
 Definition: In this case we say that f is a factor of h. Or
> (equivalently) that h is divisible by f.
 In the case you are interested in, S is the set of algebraic integers,
> and * is (complex) multiplication.
 (Note that our sqrt(x+1) example might have some problems because
> sqrt(x+1) might not be single-valued.)
Isn't sqrt(a) single-valued? Don't mistake it to be +/- ,because it isn't.
Sqrt is allways positive! It follows from the definition of Sqrt.
 Is the above definition correct?
 John Peters

====
don't bother me with mathematics, jerk.
    now, let's start another item!
 
> By virtue of the polynomials q,r,s having integer coefficients, and
> of z being a root of p(x), we know that z, q(z), r(z), and s(z) are all
> algebraic integers.
> 
> Finally, the minimal polynomial for r(z) is
> 
>       MinPoly(r(z)) = x^3 - 969 x^2 + 315 x + 5
> 
> The divisibility properties were verified in Maxima; the minimal
> polynomial was provided by KASH, and verified in KASH.
> 
> Your claim regarding the properties of the coefficients ai in
> the factorization:
> 
>       65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
> 
> namely, that ONE of the ai's is coprime to 5, is clearly
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected
to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
====
I'm living in HK. We had one of the biggest rally against the HK gov't since 
the June 4th '89 incident. The press reported that there are 500K people
went to the street for the protest.
I'm very interested with the reported figures. Is there really a 
'reasonable' method based on statistical model to approximate the total no. 
of
protesters with acceptable degree of confidence? Can anyone show me the 
concept.
====
>I'm living in HK. We had one of the biggest rally against the HK gov't 
since the June 4th '89 incident. The press reported that there are 500K 
people
>went to the street for the protest.
I'm very interested with the reported figures. Is there really a 
'reasonable' method based on statistical model to approximate the total no. 
of
>protesters with acceptable degree of confidence? Can anyone show me the 
concept.
>
I Have Been Told (tm) that no one estimating crowd sizes uses any 
reasonable method to estimate them.  I don't remember the reference.
The way to estimate crowd size is the same as the one astronomers use to 
estimate the number of visible stars.  You pick a small area of 
surveillance, find the volume of space that area views, actually count 
the number of people/stars/whatever.  Then pick another area and repeat. 
 You need to use statistical sampling methodology to assure that your 
samples are representative of the total population.  You can get quite 
sophisticated with this, so that your efforts are concentrated in the 
areas with the most importance/difficulty/whatever.  So for crowds, 
you'd probably concentrate your efforts at dense points, rather than 
along the edges.  Then you average (possibly with weights) your 
observations to get an average density, and then multiply that by the 
volume to get the total number.  Of course, for crowds you'd use surface 
area instead of volume, but you get the same idea.
I have done this, but if your interest isn't in the results (that is, if 
BigNum crowds in headlines sell newspapers/support your political point 
of view/etc. and accuracy isn't really relevant) it's probably not worth 
the effort.
At some time, someone will probably add a fairly simple routine to the 
camera analysis software to have crowd size printed out in the 
automatically generated report, but that's at least a couple of years 
down the road.  http://www.ohio.com/mld/ohio/news/6211959.htm  I'm 
guessing that deployment of the CTS software/hardware will take a couple 
of years to implement, and crowd size estimation will be low enough 
priority that they just don't get around to it.
So they won't be able to tell you how many people are in the crowd, but 
they'll be able to give you all their names.  Well, there's another 
method:  just count the number of names in the list of attendees.  Maybe 
it won't take a couple of years.
Jon Miller
====
Jon/Mary,
The methodology you mentioned, does it make a assumption that the crowd in 
question is stationary just like the stars in the clear sky? My scenario is
the crowd to be measured has varied mobility (e.g. the crowd is moving at 
non-constant speed at one direction). When the crowd moves forward, new
in-coming crowd comes in and it took 7 hours to travel from Point A to Point 
B.
I'm just thinking of differential equations to solve it. Does it make 
sense?
James
>I'm living in HK. We had one of the biggest rally against the HK gov't 
since the June 4th '89 incident. The press reported that there are 500K 
people
>>went to the street for the protest.
>>I'm very interested with the reported figures. Is there really a 
'reasonable' method based on statistical model to approximate the total no. 
of
>>protesters with acceptable degree of confidence? Can anyone show me the 
concept.
>I Have Been Told (tm) that no one estimating crowd sizes uses any 
>reasonable method to estimate them.  I don't remember the reference.
The way to estimate crowd size is the same as the one astronomers use to 
>estimate the number of visible stars.  You pick a small area of 
>surveillance, find the volume of space that area views, actually count 
>the number of people/stars/whatever.  Then pick another area and repeat. 
> You need to use statistical sampling methodology to assure that your 
>samples are representative of the total population.  You can get quite 
>sophisticated with this, so that your efforts are concentrated in the 
>areas with the most importance/difficulty/whatever.  So for crowds, 
>you'd probably concentrate your efforts at dense points, rather than 
>along the edges.  Then you average (possibly with weights) your 
>observations to get an average density, and then multiply that by the 
>volume to get the total number.  Of course, for crowds you'd use surface 
>area instead of volume, but you get the same idea.
I have done this, but if your interest isn't in the results (that is, if 
>BigNum crowds in headlines sell newspapers/support your political point 
>of view/etc. and accuracy isn't really relevant) it's probably not worth 
>the effort.
At some time, someone will probably add a fairly simple routine to the 
>camera analysis software to have crowd size printed out in the 
>automatically generated report, but that's at least a couple of years 
>down the road.  http://www.ohio.com/mld/ohio/news/6211959.htm  I'm 
>guessing that deployment of the CTS software/hardware will take a couple 
>of years to implement, and crowd size estimation will be low enough 
>priority that they just don't get around to it.
So they won't be able to tell you how many people are in the crowd, but 
>they'll be able to give you all their names.  Well, there's another 
>method:  just count the number of names in the list of attendees.  Maybe 
>it won't take a couple of years.
Jon Miller
====
> The methodology you mentioned, does it make a assumption that the crowd
> in question is stationary just like the stars in the clear sky? My
> scenario is the crowd to be measured has varied mobility (e.g. the crowd
> is moving at non-constant speed at one direction). When the crowd moves
> forward, new in-coming crowd comes in and it took 7 hours to travel from
> Point A to Point B.
>
No, it makes sense to photograph crowd from satellite, plane or
hellicopter perhaps several times to assure estimation of peak event.
However, these times, as protests are ignored, crowd estimates aren't
needed nor event and attendance deemed news wirthy.
> I'm just thinking of differential equations to solve it. Does it make 
sense?
No, renting a plane and an aerial survey camara does.
====
>In alt.math.undergrad you write:
>  
>At some time, someone will probably add a fairly simple routine to the 
>>camera analysis software to have crowd size printed out in the 
>>automatically generated report, but that's at least a couple of years 
>>down the road.
>>    
>
>I recall some press release from a company that I think was dealing with
>designing places for selling tickets, handling crowds, that sort of thing, 

>and if I'm not confused, I think they had done substantial work on 
counting
>automatically with cameras.  But it has been a while and I'm not able to
>think of any of the keywords that might turn this up.
>
Could be.  Now that I think about it, I think the discussion I read was 
about the size of the crowd at the Million Man March, so my info is 
certainly dated.
I guess I'll post this to alt.math.undergrad since it's relevant, and I 
don't mind appearing stupid (I have children, so it happens all the time 
Jon Miller
====
Market research for a new mobile phone suggests that the potential market
has a ceiling of 40 million units and that 0 sales will be achieved without
marketing. After spending 1 million on marketing 20 million sales were
recorded.
Can anyone model for sales (S) in terms of advertising expenditure (x) of
the form:
S = a+(be)POWER-cx
====
> Market research for a new mobile phone suggests that the potential market
> has a ceiling of 40 million units and that 0 sales will be achieved 
without
> marketing. After spending 1 million on marketing 20 million sales were
> recorded.
> 
> Can anyone model for sales (S) in terms of advertising expenditure (x) of
> the form:
> 
> S = a+(be)POWER-cx
I assume you mean
  S = a + b e^(-cx)
Well, you have some known values, so use them:
  (0, 0)
  (+oo, 40mil)
  (1mil, 20 mil)
Plug those (x, S) pairs into the equation for S and see what they
imply about constants a, b, and c.
-- 
Rich Carreiro                            rlcarr@animato.arlington.ma.us
====
> 
> 
>>Market research for a new mobile phone suggests that the potential market
>>has a ceiling of 40 million units and that 0 sales will be achieved 
without
>>marketing. After spending 1 million on marketing 20 million sales were
>>recorded.
>>Can anyone model for sales (S) in terms of advertising expenditure (x) of
>>the form:
>>S = a+(be)POWER-cx
> 
> 
> I assume you mean
>   S = a + b e^(-cx)
> 
> Well, you have some known values, so use them:
>   (0, 0)
>   (+oo, 40mil)
>   (1mil, 20 mil)
> 
> Plug those (x, S) pairs into the equation for S and see what they
> imply about constants a, b, and c.
> 
====
Let's say I have a function, f(x), whose inverse for some reason I can't
calculate.  Can I use the formula:
g(x) = f^(-1)(x)
g'(x) = 1 / [ f'(g(x)) ]
so find f^(-1)(x) explicitly?  If so, can I always find it?
Also, does every function whose inverse can be expressed in terms of a
finite number of elementary operations be expressed in terms of a finite
number of elementary operations?  Also, is it possible to have a function
that can't be expressed in terms of a finite number of elementary 
operations
have an inverse that can be?
Jeremy
====
> Let's say I have a function, f(x), whose inverse for some reason I can't
> calculate.  Can I use the formula:
 g(x) = f^(-1)(x)
> g'(x) = 1 / [ f'(g(x)) ]
>
g(x) = integral g'(x) dx = integral 1/f'(g(x)) dx
let u = g(x)
        = integral du/f'(u)g'(x) = integral du = u = g(x)
> so find f^(-1)(x) explicitly?  If so, can I always find it?
>
Let's test it out.  For n > 5 let
f(x) = sum(i=0..n) x^i restricted to some interval where it's 1-1
g'(x) = 1/sum(i=1..n) i.g(x)^(i-1)
What's used for g(x) to calculate this sum?
> Also, does every function whose inverse can be expressed in terms of a
> finite number of elementary operations be expressed in terms of a finite
> number of elementary operations?
> Also, is it possible to have a function that can't be expressed in terms
> of a finite number of elementary operations have an inverse that can be?
Basically these two questions are the same.
Let g be the inverse of f as given above.
Can g be expressed in a finite number of elementary operations?
====
> Let's say I have a function, f(x), whose inverse for some reason I can't
> calculate.  Can I use the formula:
> 
> g(x) = f^(-1)(x)
> 
> g'(x) = 1 / [ f'(g(x)) ]
> 
> so find f^(-1)(x) explicitly?  If so, can I always find it? [...]
Do you mean, by integrating 1 / (f'(g(x)) explicitly? I don't see how
you can do the integration without knowing what g(x) is, unless you want
to solve an integral equation. (And the way to solve that particular
integral equation will probably involve finding the inverse of f(x) first.)
     -- Christopher Heckman
====
> Do you mean, by integrating 1 / (f'(g(x)) explicitly? I don't see how
> you can do the integration without knowing what g(x) is, unless you want
> to solve an integral equation. (And the way to solve that particular
> integral equation will probably involve finding the inverse of f(x)
first.)
Not necessarily by doing it explicitly, but maybe by doing something more
clever than I can think of, or by finding a method to solve the 
differential
equation, or at least do something good to aid in the finding of f^(-1)(x).
- Jeremy
====
> so find f^(-1)(x) explicitly?  If so, can I always find it?
 Let's test it out.  For n > 5 let
> f(x) = sum(i=0..n) x^i restricted to some interval where it's 1-1
> g'(x) = 1/sum(i=1..n) i.g(x)^(i-1)
> What's used for g(x) to calculate this sum?
I'm confused on what you're doing here.
> Also, does every function whose inverse can be expressed in terms of a
> finite number of elementary operations be expressed in terms of a 
finite
> number of elementary operations?
 Also, is it possible to have a function that can't be expressed in 
terms
> of a finite number of elementary operations have an inverse that can 
be?
 Basically these two questions are the same.
> Let g be the inverse of f as given above.
> Can g be expressed in a finite number of elementary operations?
I would guess so, but I'm confused as how to prove it.
- Jeremy
====
> Let's say I have a function, f(x), whose inverse for some reason I can't
> calculate.  Can I use the formula:
 g(x) = f^(-1)(x)
 g'(x) = 1 / [ f'(g(x)) ]
 so find f^(-1)(x) explicitly?  If so, can I always find it?
 Also, does every function whose inverse can be expressed in terms of a
> finite number of elementary operations be expressed in terms of a finite
> number of elementary operations?  Also, is it possible to have a function
> that can't be expressed in terms of a finite number of elementary
operations
> have an inverse that can be?
 Jeremy

Suppose I want to find the inverse of
f(x) = e^x + x - 1
x = e^y + y - 1    swap variables
1 = (e^y)y' + y'   take the derivative of both sides
y' = 1/(e^y + 1)   solve for y'
y'' = ...
.
.
y^(p) = ...
note that when y=1, x = e  so y(e) = 1
(if y=g(x), then y(e)= g(e) = 1)
The Taylor expression becomes,
y = 1 + y'(e) (x-e)+y''(e) (x-e)^2 /2! + ...
swapping back variables leads to the inverse,
so yes, implicit differentation can be used.
====
 so find f^(-1)(x) explicitly?  If so, can I always find it?
 Let's test it out.  For n > 5 let
> f(x) = sum(i=0..n) x^i restricted to some interval where it's 1-1
> g'(x) = 1/sum(i=1..n) i.g(x)^(i-1)
> What's used for g(x) to calculate this sum?
> I'm confused on what you're doing here.
As you clipped the orginal problem, I'm limited to general description.
How would you like it if I clipped down to 'What's used ... '
I'm taking the function f(x) as I described and applying your clipped
formula to it to point out the circularity of your attempt and to give
example for later.
> Also, does every function whose inverse can be expressed in terms of 
a
> finite number of elementary operations be expressed in terms of a 
finite
> number of elementary operations?
 Also, is it possible to have a function that can't be expressed in 
terms
> of a finite number of elementary operations have an inverse that can 
be?
 Basically these two questions are the same.
> Let g be the inverse of f as given above.
> Can g be expressed in a finite number of elementary operations?
 I would guess so, but I'm confused as how to prove it.
>
It depends upon what you mean as elementary.  For starters there's Abel's
famous theorem that a polynomial of degree greater than 4 can't be solved
by radicals.  Indeed prior to his time, the solution for quartics was the
big ado.
====
> Let's say I have a function, f(x), whose inverse for some reason I can't
> calculate.  Can I use the formula:
> 
> g(x) = f^(-1)(x)
> 
> g'(x) = 1 / [ f'(g(x)) ]
> 
> so find f^(-1)(x) explicitly?  If so, can I always find it?
> 
> Also, does every function whose inverse can be expressed in terms of a
> finite number of elementary operations be expressed in terms of a finite
> number of elementary operations?  Also, is it possible to have a function
> that can't be expressed in terms of a finite number of elementary 
operations
> have an inverse that can be?
> 
> Jeremy
More on the subject that you want to know is contained is the book by
famous British mathematician G.H. Hardy The integration of functions
of a single variable Hafner, New York, 1971
A formula due to J. L. Liouville uses inverse function to find a
primitive function of a function.
Say that you want to integrate f(x)
 Int[f(x)]dx = Int[x'*f(x)]dx {Since x« is 1}  (A)
Now since f^-1(f(x)) = x, then x' = (f^-1(f(x)))'
 then (A) is equal to:
 Int[f(x)]dx = Int[(f^-1(f(x)))'*f(x)]dx  {using partial integration}
             = f^-1(f(x))*f(x) - Int[f^-1(f(x))*f'(x)]dx
             = x*f(x) - Int[(F^-1(f(x)))'dx]
                                             
             = x*f(x) - F^-1(f(x))
where F^-1 is primitive of f^-1
This is because f^-1(f(x))*f'(x) = [F^-1(f(x))]' f^-1(f(x))*f'(x)
For example if f(x) = ln(x), x > 0, the above forumula gives:
Int[ln(x)]dx = x*ln(x) - exp(ln(x)) + C
             = x[ln(x) - 1] + C
Since primitive of exp(x) is exp(x) + C
/Marko
====
Wow, thanks, I'll have to look into that further!
====
A quick question: Is every finite 2-generated abelian group, i.e. G =
< x, y >, isomorphic to Z_a x Z_b for some integers a and b? It seems
so (since such a G can be displayed in a lattice just like one would
use to visually represent Z_a x Z_b, and there must be relations of
the form mx=0 for some m, ny=0 for some n), but I'm not 100% certain
one way or the other.
BDG
====
> A quick question: Is every finite 2-generated abelian group, i.e. G =
> < x, y >, isomorphic to Z_a x Z_b for some integers a and b? It seems
> so (since such a G can be displayed in a lattice just like one would
> use to visually represent Z_a x Z_b, and there must be relations of
> the form mx=0 for some m, ny=0 for some n), but I'm not 100% certain
> one way or the other.
> 
> BDG
If you mean G is isomorphic to the direct sum of at most two
non-trivial cyclics the answer is yes.
If you mean G is isomorphic to the direct sum of _exactly_ two
non-trivial cyclics then the answer is no: Z_5 = < 2, 3 >.
One way to prove the former is to note that G is the homorphic image of
Z x Z and the kernel is of the form K x H with K <= Z x {0} and 
H <= {0} x Z.
-- 
Paul Sperry
Columbia, SC (USA)
====
> 
> A quick question: Is every finite 2-generated abelian group, i.e. G =
> < x, y >, isomorphic to Z_a x Z_b for some integers a and b? It seems
> so (since such a G can be displayed in a lattice just like one would
> use to visually represent Z_a x Z_b, and there must be relations of
> the form mx=0 for some m, ny=0 for some n), but I'm not 100% certain
> one way or the other.
> 
> BDG
> 
> If you mean G is isomorphic to the direct sum of at most two
> non-trivial cyclics the answer is yes.
> 
> If you mean G is isomorphic to the direct sum of _exactly_ two
> non-trivial cyclics then the answer is no: Z_5 = < 2, 3 >.
> 
> One way to prove the former is to note that G is the homorphic image of
> Z x Z and the kernel is of the form K x H with K <= Z x {0} and 
> H <= {0} x Z.
If we adopt Z_1={0} then apparently everything is OK, Isn't it?
Alireza Abdollahi
====
> A quick question: Is every finite 2-generated abelian group, i.e. G =
>> < x, y >, isomorphic to Z_a x Z_b for some integers a and b? It seems
>> so (since such a G can be displayed in a lattice just like one would
>> use to visually represent Z_a x Z_b, and there must be relations of
>> the form mx=0 for some m, ny=0 for some n), but I'm not 100% certain
>> one way or the other.
>> 
>> BDG
If you mean G is isomorphic to the direct sum of at most two
>non-trivial cyclics the answer is yes.
If you mean G is isomorphic to the direct sum of _exactly_ two
>non-trivial cyclics then the answer is no: Z_5 = < 2, 3 >.
But Z_5 = Z_5 x Z_1, so the answer to his question is still yes.
An even more trivial special case is < 1, 1 > = Z_1 x Z_1.
Derek Holt.
>One way to prove the former is to note that G is the homorphic image of
>Z x Z and the kernel is of the form K x H with K <= Z x {0} and 
>H <= {0} x Z.
-- 
>Paul Sperry
>Columbia, SC (USA)
====
> A quick question: Is every finite 2-generated abelian group, i.e. G =
>> < x, y >, isomorphic to Z_a x Z_b for some integers a and b? It seems
>> so (since such a G can be displayed in a lattice just like one would
>> use to visually represent Z_a x Z_b, and there must be relations of
>> the form mx=0 for some m, ny=0 for some n), but I'm not 100% certain
>> one way or the other.
>> 
>> BDG
If you mean G is isomorphic to the direct sum of at most two
>non-trivial cyclics the answer is yes.
If you mean G is isomorphic to the direct sum of _exactly_ two
>non-trivial cyclics then the answer is no: Z_5 = < 2, 3 >.
> 
> But Z_5 = Z_5 x Z_1, so the answer to his question is still yes.
Or no. (As far as I'm concerned, Z_1 is trivial.)
[...]
-- 
Paul Sperry
Columbia, SC (USA)
====
>>> A quick question: Is every finite 2-generated abelian group, i.e. G =
>>> < x, y >, isomorphic to Z_a x Z_b for some integers a and b? It seems
>>> so (since such a G can be displayed in a lattice just like one would
>>> use to visually represent Z_a x Z_b, and there must be relations of
>>> the form mx=0 for some m, ny=0 for some n), but I'm not 100% certain
>>> one way or the other.
>>If you mean G is isomorphic to the direct sum of at most two
>>non-trivial cyclics the answer is yes.
>>If you mean G is isomorphic to the direct sum of _exactly_ two
>>non-trivial cyclics then the answer is no: Z_5 = < 2, 3 >.
>> But Z_5 = Z_5 x Z_1, so the answer to his question is still yes.
>Or no. (As far as I'm concerned, Z_1 is trivial.)
The answer to his question as he asked it is 'yes'; you're adding
a condition that wasn't part of the question.
Brian
====
>> A quick question: Is every finite 2-generated abelian group, i.e. G =
>>> < x, y >, isomorphic to Z_a x Z_b for some integers a and b? It seems
>>> so (since such a G can be displayed in a lattice just like one would
>>> use to visually represent Z_a x Z_b, and there must be relations of
>>> the form mx=0 for some m, ny=0 for some n), but I'm not 100% certain
>>> one way or the other.
>>> 
>>> BDG
>>If you mean G is isomorphic to the direct sum of at most two
>>non-trivial cyclics the answer is yes.
>>If you mean G is isomorphic to the direct sum of _exactly_ two
>>non-trivial cyclics then the answer is no: Z_5 = < 2, 3 >.
>> 
>> But Z_5 = Z_5 x Z_1, so the answer to his question is still yes.
>Or no. (As far as I'm concerned, Z_1 is trivial.)
But the answer to *his* question is yes (since 1 is an integer).
Derek Holt.
====
Let A and B be nonempty subsets of R which are bounded above.  Also
define the following sets of upper bounds for A and B:
S_A : = {s in R : forall a in A, a <= s}
S_B : = {s in R : forall b in B, b <= s}
I want to find formal (rigorous) justification for the following:
            B subseteq A  --> S_A subseteq S_B
In words:  If B is a subset of A, then the set of upper bounds for A
is a subset of the set of upper bounds for B.
I can easily see that this is true by drawing a picture, but I would
like to have formal justification, starting from the defintions of
subset and upper bound.  Any takers?  thanks.
====
> Let A and B be nonempty subsets of R which are bounded above.  Also
> define the following sets of upper bounds for A and B:
> S_A : = {s in R : forall a in A, a <= s}
> S_B : = {s in R : forall b in B, b <= s}
> 
> I want to find formal (rigorous) justification for the following:
> 
>             B subseteq A  --> S_A subseteq S_B
> 
> In words:  If B is a subset of A, then the set of upper bounds for A
> is a subset of the set of upper bounds for B.
> 
> I can easily see that this is true by drawing a picture, but I would
> like to have formal justification, starting from the defintions of
> subset and upper bound.  Any takers?  thanks.
Let x be a member of S_A. By definition of S_A we have a <= x for all 
elements a of A. As B is a subset of A, every element b of B is also an 
element of A, and therefore b <= x. So for every element b of B we have 
b <= x, which means by definition of S_B that x is an element of S_B. As 
x was an arbitrary member of S_A, we have shown that every element of 
S_A is also an element of S_B, which means that S_A is a subset of S_B. 
By the way, the upper bound is irrelevant. Every relation between a 
and s in the definitions of S_A and S_B would work exactly in the same 
way.
====
I have a question concerning finite-state machines (FSM). I need to
show that no FSM with input and output set {0, 1} outputs a 1 whenever
the number of ones and zeros input are equal and 0 otherwise.
In my attempt of constructing such an FSM, I ran into a problem. It
seems that the number of states required for a particular input string
is not the same for all input strings. I then figured the proof would
be by contradiction involving the size of the set of states of the
FSM. However, I haven't had luck in developing any rigorous argument
to convince myself of such. Any clues or suggestions?
Bernd
====
> I have a question concerning finite-state machines (FSM). I need to
> show that no FSM with input and output set {0, 1} outputs a 1 whenever
> the number of ones and zeros input are equal and 0 otherwise.
> 
> In my attempt of constructing such an FSM, I ran into a problem. It
> seems that the number of states required for a particular input string
> is not the same for all input strings. I then figured the proof would
> be by contradiction involving the size of the set of states of the
> FSM. However, I haven't had luck in developing any rigorous argument
> to convince myself of such. Any clues or suggestions?
The important thing to know is how many ahead (or behind) the 1s are
on the 0s.  At every stage of reading a string, this is some number
n.  Can the machine be in the same state for two different values of
n?
/olov
-- 
Supplicant:  I don't know if I'm alive or dead.
Oracle: So much for Descartes.
                        -- Internet Oracularity #1151-05
====
> The important thing to know is how many ahead (or behind) the 1s are
> on the 0s.  At every stage of reading a string, this is some number
> n.  Can the machine be in the same state for two different values of
> n?
> 
> /olov
Apparently not since each value of n requires a seperate state.
However, I can't entirely convince myself of this. I think the
argument that surges is that the number of states changes for
different values of n. Am I thinking in the right direction?
Bernd
====
> 
> 
>>The important thing to know is how many ahead (or behind) the 1s are
>>on the 0s.  At every stage of reading a string, this is some number
>>n.  Can the machine be in the same state for two different values of
>>n?
>>/olov
> 
> 
> Apparently not since each value of n requires a seperate state.
> However, I can't entirely convince myself of this. I think the
> argument that surges is that the number of states changes for
> different values of n. Am I thinking in the right direction?
> 
> Bernd
This post is relatively old, so nobody may still be interested in it. 
You are indeed thinking in the right direction. The problem can be 
proved using the Pumping Lemma for regular languages. A google search is 
fruitful, especially since the first entry (for me) was a delightful 
poem-turned-lemma!
Robert
====
> 
> ...
> 
> The gist of it is that I have a *proof* that an algebraic integer
> which I called x in the original post has another algebraic integer
> factor that I called y, but x/y is NOT an algebraic integer.
> 
> That's it.  That's the contradiction which I say proves that the ring
> of algebraic integers is incomplete as x/y should be included.
> 
> ...
> 
> 
> Why?  Why if  x  and  y  are in a ring, should  x/y  be in it as well? 
> Consider the ring of rational integers.
> 
> GC
> 
> In a special case, I prove that a number I'll call x has a factor I'll
> here call y, where both are in the ring of algebraic integers.
> 
> Given that y is a factor x, by definition x/y is in the ring of
> algebraic integers, but in the special case it provably is not.
> 
> That's the contradiction.
<deleted>
It's been hard to explain.
Later I realized that y can't be a factor of x in the ring of
algebraic integers, so I'd been going to a higher ring, like talking
of 2 and 6 being coprime in the ring of evens, and then saying that 2
is a factor of 6, which it is, but in the ring of integers.
Thinking about it this afternoon it seems to me that the gist of the
argument is that given xyz=w, where w is an integer, you can find x, y
and z, algebraic integers, such that neither xy, xz, nor yz, is an
algebraic integer.
More specifically considering the expressions I've often used, you
have
  a_1 a_2 a_3 = 65
where neither a_1 a_2, a_2 a_3, nor a_1 a_3 is an algebraic integer.
The a's here given by, from the expression used in my paper Advanced
Polynomial Factorization:
  a^3 + 12a^2 - 65 = 0.
Given that you have the product of two algebraic integers NOT being an
algebraic integer, the ring is incomplete.
Note:  Any proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an algebraic
integer would definitely cause me problems.  And I'm not saying
proof but a proof, which means that the mathematical argument is
true, not just that there's some yahoo claiming they have a proof,
when actually they just have a flawed argument.
James Harris
====
> Thinking about it this afternoon it seems to me that the gist of the
> argument is that given xyz=w, where w is an integer, you can find x, y
> and z, algebraic integers, such that neither xy, xz, nor yz, is an
> algebraic integer.
You may want to check
   http://www.jmilne.org/math/CourseNotes/math594f.pdf
J.S.Milne, Fields and Galois Theory, Page 9, Lemma 1.15: The algebraic 
integers form a subring of the complex numbers. 
====
   [.snip.]
>Thinking about it this afternoon it seems to me that the gist of the
>argument is that given xyz=w, where w is an integer, you can find x, y
>and z, algebraic integers, such that neither xy, xz, nor yz, is an
>algebraic integer.
In short, you are saying that the algebraic integers do not form a
ring, as they are not closed under multiplication.
That will come as a great surprise to Eisenstein, Gauss's favorite
pupil, who proved that if x and y are algebraic integers, then so are
x+y and x*y...
>Note:  Any proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an algebraic
>integer would definitely cause me problems.  And I'm not saying
>proof but a proof, which means that the mathematical argument is
>true, not just that there's some yahoo claiming they have a proof,
>when actually they just have a flawed argument.
Amazing. You have just described perfectly most of your arguments
(that's some yahoo claiming they have a proof, when actually they
just have a flawed argument).
Recall: a complex number c is an ALGEBRAIC INTEGER if and only if
there exists a monic polynomial f(x) with integer coefficients such
that f(c)=0.
Recall also that by subring of C we mean a nonempty set R of complex
numbers with the following properties:
(i) 0 is in R.
(ii) 1 is in R.
(iii) If a and b are in R, then a-b is in R (usual complex
      subtraction).
(iv) If a and b are in R, then a*b is in R (usual complex
     multiplication).
Finally, let R be a subring of C. We say that the additive group of R
is finitely generated to mean that there exist a finite number of
elements of R, a_1,...,a_m, such that every element of R may be
written in at least one way as
c_1*a_1 + ... + c_m*a_m, where c_1,...,c_m are (rational) integers.
THEOREM. The following are equivalent for a complex number a (that is,
if one of them is true, then they are all true; if one of them is
false, then they are all false):
  (i) a is an algebraic integer.
 (ii) The additive group of the ring Z[a] (the smallest ring contained
      in the complex numbers which contains all integers and contains a; it
      consists of all polynomial expressions in a with coefficients in Z)
      is finitely generated.
(iii) a is a member of some subring of the complex numbers whose
      additive group is finitely generated.
 (iv) There exists a finitely generated additive subgroup of C, called
       A, with the property that aA is contained in A.
Proof.  To prove the equivalence, we show that if (i) holds, then (ii)
holds; that if (ii) holds, then (iii) holds; that if (iii) holds, then
(iv) holds; and that if (iv) holds, then (i) holds. That is, we prove
that (i)->(ii)->(iii)->(iv)->(i).
(i)->(ii) Assume that a is an algebraic integer. Then there
exists a monic polynomial with integer coefficients, f(x), such that
f(a)=0; let
f(x) = x^n + a_{n-1}*x^{n-1} + ... + a_1*x + a_0,
with a_i an integer. 
By substituting a^n = -(a_{n-1}*a^{n-1} + ... + a_1*a + a_0) into any
expression of the form
b_m*a^m + ... b_1*a + b_0, with b_i integers, we see that every
element of Z[a] may be written as a polynomial expression in a of
degree strictly less than n and integer coefficients. That is,
Z[a] = {c_{n-1}*a^{n-1} + ... + c_1*a + c_0   : c_i integers}.
As a group, Z[a] is generated by 1, a, a^2, ... , a^{n-1}. Therefore,
Z[a] is finitely generated, proving (ii).
(ii) -> (iii). Assume that Z[a] has finitely generated additive
group. Then Z[a] is a subring of C which contains a, and with the
property that its additive group is finitely generated. Thus, if (ii)
holds, then (iii) must hold as well.
(iii)->(iv) Assume there is a subring R of the complex numbers, with
the properties that a is in R, and the additive group of R is finitely
generated. 
Since R is a ring, the product of any two elements of R must be in
R. Therefore, aR = {a*r : r in R} is a subset of R (since both a and r
are in R); thus, A=R shows that (iv) is true.
Finally, we prove that if (iv) holds, then (i) holds.
Assume that (iv) holds. Let A be a finitely generated additive
subgroup of C such that a*A is contained in A. Let x_1,...,x_m be
elements of A such that every element of A may be written as
c_1*x_1 + ... + c_m*x_m, with c_1,...,c_m integers.
Furthermore, we may assume that none of the x_i are equal to 0.
Our assumption is that for every choice of integers c_1,...,c_m,
(c_1*x_1 + ... + c_m*x_m)*a is in A.
For each i, a*x_i lies in A; therefore, a*x_i may be written as
above. We write:
a*x_1 = b_{11}*x_1 + b_{21}*x_2 + ... + b_{m1}*x_m
a*x_2 = b_{12}*x_1 + b_{22}*x_2 + ... + b_{m2}*x_m
      .
      .
      .
a*x_m = b_{1m}*x_1 + b_{2m}*x_2 + ... + b_{mm}*x_m.
Let M be the matrix
    ( b_{11}   b_{21}   ...  b_{m1} )
    ( b_{12}   b_{22}   ...  b_{m2} )
    (  .        .       .      .    )
 M= (  .        .        .     .    )
    (  .        .         .    .    )
    ( b_{1m}   b_{2m}   ...  b_{mm} )
this is an m x m matrix with integer coefficients, such that
  (x_1)     (x_1)
  (x_2)     (x_2)
  ( . )     ( . )
a*( . ) =  M( . )
  ( . )     ( . )
  (x_m)     (x_m)
If I is the m x m identity matrix, this means that
         (x_1)
         (x_2)
         ( . )
(a*I - M)( . ) 
         ( . )
         (x_m)
is the zero vector. Since none of the x_i are equal to 0, it follows
that a is an eigenvalue of M and the vector of the x_i is an
eigenvector of M. That means that det(aI-M) = 0. Equivalently, the
characteristic polynomial of M, det(xI-M), has a as a root. Since M
has integer coefficients, the characteristic polynomial of M is monic
and has integer coefficients. Therefore, a is the root of a monic
polynomial with integer coefficients. Therefore, a is an algebraic
integer, proving that (i) holds whenever (iv) holds. QED
Corollary. Let a and b be algebraic integers. Then a+b, a-b, and a*b
are all  algebraic integers.
Proof. Consider the ring Z[a,b], the smallest subring of C which
contains a, b, and all the integers. Note that it contains a+b, a-b,
and a*b, by definition of subring.
By clause (iii) of the Theorem, in order to prove that a+b, a-b, and
a*b are algebraic integers it is enough to prove that Z[a,b] has
finitely generated additive group.
Since a is an algebraic integer, we know that a satisfies a monic
polynomial with integer coefficients, say:
f(x) = x^n + a_{n-1}*x^{n-1} + ... + a_1*x + a_0;
likewise, b satisfies a monic polynomial with integer coefficients,
say:
g(x) = x^m + b_{m-1}*x^{m-1} + ... + b_1*x + b_0.
CLAIM: Z[a,b] is generated, as an additive group, by
1, a, a^2, ..., a^{n-1}, b, ba, ba^2, ..., ba^{n-1}, b^2,....,
b^{m-1}a^{n-1}.
Note that establishing the claim will prove the corollary.
Every element of Z[a,b] may be written as a polynomial expression in a
and b with integer coefficients. Since
a^n = -(a_{n-1}*a^{n-1} + ... + a_1*a + a_0)
b^m = -(b_{m-1}*b^{m-1} + ... + b_1*b + b_0),
we may substitute any occurance of any power of a greater than
a^{n-1}, and any power of b greater than b^{m-1}, by an expression
involving only lesser powers. Therefore, by repeating the
substitution, we may write every element of Z[a,b] as:
c_{n-1,m-1}*a^{n-1}*b^{m-1} + c_{n-1,m-2}*a^{n-1}*b^{m-2} + ... +
   + c_{n-1,0}*a^{n-1} + c_{n-2,m-1}*a^{n-2}b^{m-1} + ... + a_{0,0}*1.
This proves the claim. This proves the Corollary. QED
This proves that if a and b are algebraic integers, then so is a*b.
This definitely cause[s you] problems.
======================================================================
Why do you take so much trouble to expose such a reasoner as
 Mr. Smith? I answer as a deceased friend of mine used to answer
 on like occasions - A man's capacity is no measure of his power
 to do mischief. Mr. Smith has untiring energy, which does 
 something; self-evident honesty of conviction, which does more;
 and a long purse, which does most of all. He has made at least
 ten publications, full of figures few readers can critize. A great
 many people are staggered to this extend, that they imagine there
 must be the indefinite 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 Magidin
magidin@math.berkeley.edu
====
[snip latest sideshow]
> Note:  Any proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an algebraic
> integer would definitely cause me problems.  And I'm not saying
> proof but a proof, which means that the mathematical argument is
> true, not just that there's some yahoo claiming they have a proof,
> when actually they just have a flawed argument.
It's already been done by several posters (not yahoos). Wake up and smell 
the coffee you moron. You are
hereby banned from the sandbox. Go climb back under the rock whence you 
came.
--
There are two things you must never attempt to prove: the unprovable -- and 
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
> 
> ...
> 
> The gist of it is that I have a *proof* that an algebraic integer
> which I called x in the original post has another algebraic integer
> factor that I called y, but x/y is NOT an algebraic integer.
> 
> That's it.  That's the contradiction which I say proves that the 
ring
> of algebraic integers is incomplete as x/y should be included.
> 
> ...
> 
> 
> Why?  Why if  x  and  y  are in a ring, should  x/y  be in it as well? 

> Consider the ring of rational integers.
> 
> GC
> 
> In a special case, I prove that a number I'll call x has a factor I'll
> here call y, where both are in the ring of algebraic integers.
> 
> Given that y is a factor x, by definition x/y is in the ring of
> algebraic integers, but in the special case it provably is not.
> 
> That's the contradiction.
> 
> <deleted 
> It's been hard to explain.
> 
> Later I realized that y can't be a factor of x in the ring of
> algebraic integers, so I'd been going to a higher ring, like talking
> of 2 and 6 being coprime in the ring of evens, and then saying that 2
> is a factor of 6, which it is, but in the ring of integers.
> 
> Thinking about it this afternoon it seems to me that the gist of the
> argument is that given xyz=w, where w is an integer, you can find x, y
> and z, algebraic integers, such that neither xy, xz, nor yz, is an
> algebraic integer.
Nope.  That's not it.
 
> More specifically considering the expressions I've often used, you
> have
> 
>   a_1 a_2 a_3 = 65
> 
> where neither a_1 a_2, a_2 a_3, nor a_1 a_3 is an algebraic integer.
They are as is easily shown.
> The a's here given by, from the expression used in my paper Advanced
> Polynomial Factorization:
> 
>   a^3 + 12a^2 - 65 = 0.
> 
> Given that you have the product of two algebraic integers NOT being an
> algebraic integer, the ring is incomplete.
> 
> Note:  Any proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an algebraic
> integer would definitely cause me problems.  And I'm not saying
> proof but a proof, which means that the mathematical argument is
> true, not just that there's some yahoo claiming they have a proof,
> when actually they just have a flawed argument.
Oh well, that was wrong.
However, I find it interesting that my earlier post received some replies.  

James Harris
====
>> [...]
<deleted
>It's been hard to explain.
[...]
Thinking about it this afternoon it seems to me that the gist of the
>argument is that given xyz=w, where w is an integer, you can find x, y
>and z, algebraic integers, such that neither xy, xz, nor yz, is an
>algebraic integer.
Yes, that _is_ hard to explain. 
Why? Because it's simply _false_. Easily shown to be false, also
well-known to be false. When you say this is hard to explain it's
_exactly_ like if you'd said you'd had a hard time explaining your
theory, then you realized the gist of it was that 2 + 2 = 5.
(Well, not exactly like that - the difference is that _you_ know
enough math to see that 2 + 2 = 5 is false. But to someone
who actually knows something about this stuff it's exactly
the same. You talk a lot about how we all lie about these
things. When you do that the impression you make is
exactly like someone complaining about people lying,
saying that 2 + 2 = 4. Honest.)
>More specifically considering the expressions I've often used, you
>have
  a_1 a_2 a_3 = 65
where neither a_1 a_2, a_2 a_3, nor a_1 a_3 is an algebraic integer.
The a's here given by, from the expression used in my paper Advanced
>Polynomial Factorization:
  a^3 + 12a^2 - 65 = 0.
Given that you have the product of two algebraic integers NOT being an
>algebraic integer, the ring is incomplete.
Note:  Any proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an algebraic
>integer would definitely cause me problems.  And I'm not saying
>proof but a proof, which means that the mathematical argument is
>true, not just that there's some yahoo claiming they have a proof,
>when actually they just have a flawed argument.

>James Harris
************************
David C. Ullrich
====
with this informal admission,
I hereby (again) declare my solemn intention
not to participate in heckling monsieur Harris, any more, if
I can possibly help it.  even if he can't help instigating it.
    of course, there could be an ulterior motive to all of this,
as it's so hard to believe!
 
> Note:  Any proof that a_1 a_2, a_2 a_3, or a_1 a_3 is an algebraic
> integer would definitely cause me problems.  And I'm not saying
> proof but a proof, which means that the mathematical argument is
> true, not just that there's some yahoo claiming they have a proof,
> when actually they just have a flawed argument.
> 
> Oh well, that was wrong.
--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...
La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto:
(FOSSILISATION [McCainanites?] (TM/sic))/
BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.
Http://www.tarpley.net/bushb.htm (content partiale, below):
  17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81
====
> And I'm not saying
> proof but a proof, which means that the mathematical argument is
> true, not just that there's some yahoo claiming they have a proof,
> when actually they just have a flawed argument.
> 
> 
> James Harris
JSH has lots of experience in this area.
====
I was leading my class through related rates yesterday, and one of 
the students stunned me by asking for a good qualitative description 
of what a derivative means. (He asked it in a more basic form, which 
is why I was stunned.)
Our textbook, Larson/Hostetler/Edwards, really doesn't do a very 
good job of explaining what a derivative _means_, now that I look at 
it with fresh eyes.
I pointed him at /Calculus Made Easy/, which I think is pretty good 
at that. But is there a Web resource I can give to all my students? 
Once again, the issue is not how to find derivatives but how to 
understand them.
-- 
Stan Brown, Oak Road Systems, Cortland County, New York, USA
                                  http://OakRoadSystems.com/
You find yourself amusing, Blackadder.
I try not to fly in the face of public opinion.
====
A derivative is the rate of change of a function. If you graph the function
the derivative is the slope of the graph.
A good example is that the derivative of position is velocity, the
derivative of velocity is acceleration.
 I was leading my class through related rates yesterday, and one of
> the students stunned me by asking for a good qualitative description
> of what a derivative means. (He asked it in a more basic form, which
> is why I was stunned.)
 Our textbook, Larson/Hostetler/Edwards, really doesn't do a very
> good job of explaining what a derivative _means_, now that I look at
> it with fresh eyes.
 I pointed him at /Calculus Made Easy/, which I think is pretty good
> at that. But is there a Web resource I can give to all my students?
 Once again, the issue is not how to find derivatives but how to
> understand them.
 -- 
> Stan Brown, Oak Road Systems, Cortland County, New York, USA
>                                   http://OakRoadSystems.com/
> You find yourself amusing, Blackadder.
> I try not to fly in the face of public opinion.
====
> A derivative is the rate of change of a function. If you graph the
function
> the derivative is the slope of the graph.
> A good example is that the derivative of position is velocity, the
> derivative of velocity is acceleration.
derivative is often interpreted as a rate of change.  You mean well, we
know, and I may be wrong but I think what Stan has in mind is he is having
trouble explaining, satisfactorilly, and at a certain intuitive/qualitative
level, what a derivative _means_.  Textbooks are good at giving very
unambiguous definitions, but not so good many times at explaining these
things at a more basic level.
Although I don't have the most current edition of Larson, et al, I do have
the 5th ed. handy and if the one Stan has is anything like it, I understand
exactly what he is talking about.  In my edition, the way the authors
attempt to put a derivative in layman's terms, so to speak, is fairly
confusing.  They begin this attempt somewhere in ch. 1 with a section
entitled the tangent line problem then they drop that discussion 
entirely,
moving on to other things, and later in chapter 2 I believe, they pick back
up with that discussion.  At face value, this seems not that unreasonable
since the problem of finding a derivitive *is* essentially the same as
findling the slope of a tangent line (hence, the tangent line itself) but 
if
you have had to experience reading the actual material, in the manner
presented, you may agree that the authors were not very effective in this
attempt to explain a derivative.  Yeah, they do a good job of explaing 
how
the secant line approaches the tangent line, etc, etc, and how the limiting
position of the secant line is the tangent line, etc. but this really
doesn't address what a derivative _means_.  It just addresses what one 
_is_.
so why _are_ we interested in tangent lines?  What do _they_ mean?
All that said, it's not the easiest thing in the world to explain what a
derivative _means_ as Stan has acknowledged (and Larson, et al.
demonstrates).  But here are a couple of places you may want to look, if 
you
haven't already:
http://karlscalculus.org/calculus.html
Karl goes into extreme vivid detailed explanations, in layman's terms,
(borderline children's talk, in some places, if that kind of stuff doesn't
bother you or your students).  All the rigor is there, but buried deep
within lengthy, informal explanations in story format.  It may be necessary
to start his story with the limit chapter.
http://www.sosmath.com/calculus/diff/der00/der00.html
This is an excellent site that is much more concise than karl's, geared to
more of a grown up audience, if you will.  They discuss both common
interpretations, that of a physical rate of change and that of the
(geometrical) slope of a tangent line.  but they do a much better job at 
it,
IMO, than larson, et. al.
HTH,
-- 
Darrell

> I was leading my class through related rates yesterday, and one of
> the students stunned me by asking for a good qualitative description
> of what a derivative means. (He asked it in a more basic form, which
> is why I was stunned.)
 Our textbook, Larson/Hostetler/Edwards, really doesn't do a very
> good job of explaining what a derivative _means_, now that I look at
> it with fresh eyes.
 I pointed him at /Calculus Made Easy/, which I think is pretty good
> at that. But is there a Web resource I can give to all my students?
 Once again, the issue is not how to find derivatives but how to
> understand them.
 -- 
> Stan Brown, Oak Road Systems, Cortland County, New York, USA
>                                   http://OakRoadSystems.com/
> You find yourself amusing, Blackadder.
> I try not to fly in the face of public opinion.

====
>> A derivative is the rate of change of a function. If you graph the
>function
>> the derivative is the slope of the graph.
>> A good example is that the derivative of position is velocity, the
>> derivative of velocity is acceleration.
derivative is often interpreted as a rate of change.  You mean well, we
>know, and I may be wrong but I think what Stan has in mind is he is having
>trouble explaining, satisfactorilly, and at a certain 
intuitive/qualitative
>level, what a derivative _means_. 
You interpret rightly. Since I'm teaching calculus, I should 
certainly hope I know what a derivative is! :-)
>http://karlscalculus.org/calculus.html
>Karl goes into extreme vivid detailed explanations, in layman's terms,
>(borderline children's talk, in some places, if that kind of stuff doesn't
>bother you or your students).
Nope. When the student was explaining his difficulty, he said, Talk 
to me like I'm 4.
>http://www.sosmath.com/calculus/diff/der00/der00.html
>This is an excellent site that is much more concise than karl's, geared to
>more of a grown up audience, if you will. 
Every time I look at sosmath I block it out mentally because I don't 
like the ad for calc101.com, which is essentially Pay us and we'll 
do your homework. But the page itself does look good.
-- 
Stan Brown, Oak Road Systems, Cortland County, New York, USA
                                  http://OakRoadSystems.com/
You find yourself amusing, Blackadder.
I try not to fly in the face of public opinion.
====
I'm trying to derive the properties of determinants, as opposed to just 
proving
them. Hence I'm pretending ignorance of determinants, and trying to compute
eigenvalues by primitive methods. In case that kind of thing doesn't 
interest
you, you've been warned. This is where I'm at:
Suppose A is a square matrix and I is the identity matrix of the same order. 
Let
X_0 be (A-xI), where x is a variable. Assuming X_0[1,1]<>0, there are row
operations that convert X_0 into a matrix X_1 with X_1[1,j]=0 for j>1.
Continuing in the same way you eventually obtain a matrix X_n that can be
converted into I provided none of the entries on its main diagonal is zero.
Hence every eigenvalue of A is a root of 
numer(X_n[1,1])*...*numer(X_n[n,n]),
where numer(r) means the polynomial p with r=p/q for p and q relatively 
prime.
Now, it turns out that numer(X_n[n,n]) is the characteristic polynomial of 
A
(recall that I don't officially know about the characteristic polynomial at 
this
point), so that none of the numer(X_n[i,i])'s with i<n has an eigenvalue of 
A
for a root.
Can anybody explain this fact without invoking a bunch of involved theory?
--
Peace,
EJ
====
> I'm trying to derive the properties of determinants, as opposed to just 
proving
> them. Hence I'm pretending ignorance of determinants, and trying to 
compute
> eigenvalues by primitive methods.
I don't think computing eigenvalues is a property of determinants.
So this might not be the best way to investigate determinants.
I think of the properties of a determinant as being multilinear and
skew symmetric.
One approach at viewing determinants is to consider it as the
volume of the parallelpiped formed by the row vectors of the matrix.
Another approach is to consider it as a differential n-form.
A third approach is to see how determinants pop-up when trying
to solve n equations in n unknowns using Kramer's rule. I like
to use Kramer's rule in the form D * x_i = c_i since it is
true even when D is zero (here D is the determinant).
> In case that kind of thing doesn't interest
> you, you've been warned. This is where I'm at:
> 
> Suppose A is a square matrix and I is the identity matrix of the same 
order. Let
> X_0 be (A-xI), where x is a variable. Assuming X_0[1,1]<>0, there are row
> operations that convert X_0 into a matrix X_1 with X_1[1,j]=0 for j>1.
> Continuing in the same way you eventually obtain a matrix X_n that can be
> converted into I provided none of the entries on its main diagonal is 
zero.
> Hence every eigenvalue of A is a root of 
numer(X_n[1,1])*...*numer(X_n[n,n]),
> where numer(r) means the polynomial p with r=p/q for p and q relatively 
prime.
> 
> Now, it turns out that numer(X_n[n,n]) is the characteristic polynomial of 
A
> (recall that I don't officially know about the characteristic polynomial 
at this
> point), so that none of the numer(X_n[i,i])'s with i<n has an eigenvalue 
of A
> for a root.
> 
> Can anybody explain this fact without invoking a bunch of involved 
theory?
Let A be the square matrix [a,b;c,d].
Let u = (x,y) be an eigenvector with eigenvalue k.
Then, A*u = k*u.
Thus,
    a*x + b*y = k*x
    c*x + d*y = k*y.
That is,
    (a-k)*x + b*y     = 0
      c*x   + (d-k)*y = 0
Note that (0,0) is not considered an eigenvector.
Thus, either x or y is non-zero.
I have two equations in two unknowns x and y, with paramenter k.
Solve for the non-zero unknown, say x, using the method you
learned in high school.
You get:
   (a-k)*(d-k)*x - b*c*x = 0.
Since x is not zero, you get (a-k)*(d-k) - b*c = 0.
This is just the determinant since the determinant
pops up when you solve linear equations for the unknowns.
This doesn't seem to be what you are looking for though.
-- Bill Hale
====
> I'm trying to derive the properties of determinants, as opposed to just 
proving
> them. Hence I'm pretending ignorance of determinants, and trying to 
compute
> eigenvalues by primitive methods. In case that kind of thing doesn't 
interest
> you, you've been warned. This is where I'm at:
> 
> Suppose A is a square matrix and I is the identity matrix of the same 
order. Let
> X_0 be (A-xI), where x is a variable. Assuming X_0[1,1]<>0, there are row
> operations that convert X_0 into a matrix X_1 with X_1[1,j]=0 for j>1.
> Continuing in the same way you eventually obtain a matrix X_n that can be
> converted into I provided none of the entries on its main diagonal is 
zero.
> Hence every eigenvalue of A is a root of 
numer(X_n[1,1])*...*numer(X_n[n,n]),
> where numer(r) means the polynomial p with r=p/q for p and q relatively 
prime.
> 
> Now, it turns out that numer(X_n[n,n]) is the characteristic polynomial of 
A
> (recall that I don't officially know about the characteristic polynomial 
at this
> point), so that none of the numer(X_n[i,i])'s with i<n has an eigenvalue 
of A
> for a root.
> 
> Can anybody explain this fact without invoking a bunch of involved 
theory?
Have a look at the book Linear Algebra Done Right by Sheldon Axler,
for a consistently determinant-free approach to basic linear algebra. 
The book's homepage is at:
http://math.sfsu.edu/axler/LADR.html
HTH,
Felix.
P.S: Note that Axler allows the zero vector to be an eigenvector,
which is quite non-standard usage.  Apart from that, the book is an
excellent presentation of classic material in an elegant way.
P.P.S[Full Disclousre]: Personally, I think that determinants are just
ok.
====
>URGENT PLEASE READ
>Our friend recently suffered a heart attack at 46 years old. 
>He is a regularly working contracted employee (self employed) who will be 
unable to work for 6 or 7 weeks from the time of the heart attack with 
absolutely no income.
>He has paid taxes all of his life but is not eligible for government 
assistance
>of any type as his wife is working earning $1400.00 monthly. This obviously 
does not support their family including their 2 kids but does exclude them 
from any agency help. 
>Mrs. Sheffer was even told that if she were a crack head or unemployed, 
they could then offer assistance.
>They have now missed their rent, had to give up their car, had to add $300 
monthly in medications and the quality of their life has sunken to new lows. 

>This is a kind and considerate hard working family who has fallen on 
temporary hard times and could sure use a kindly hand before it's too late.
>We are trying to prove that people DO still care even if our government 
doesn't.
>We have set up a donation PO Box at the address below and could sure use 
your help!
Donna Sheffer, 3216 Eglinton ave. east, Scarborough, Ontario, Canada, M1J 
2H6
>Your generous donation $1, $2, $5, $10, $20 whatever would be greatly 
appreciated.
>Help restore a little faith and save a real family.
>God Bless
>See the angioplasty image here.  http://www.otwebdesign.com/apimage.jpg
This is no scam
>Please help!!
 I faxed you a $100 bill !
  Do you really think we're all idiots?
====
[snip by Stan]
>>This is no scam
>>Please help!!

> I faxed you a $100 bill !
  Do you really think we're all idiots?
Some of we apparently are. You found it necessary to repost the 
whole scam, including the Web site and the postal address, thus 
giving the spammer another shot at our eyes.
-- 
Stan Brown, Oak Road Systems, Cortland County, New York, USA
                                  http://OakRoadSystems.com/
You find yourself amusing, Blackadder.
I try not to fly in the face of public opinion.
====
 [snip by Stan]
>>This is no scam
>>Please help!!

> I faxed you a $100 bill !
  Do you really think we're all idiots?
 Some of we apparently are. You found it necessary to repost the
> whole scam, including the Web site and the postal address, thus
> giving the spammer another shot at our eyes.
However, the tragedy may be real, and the arrogant
replies thus all the more hurtful. But we cannot
know for sure: hence paranoia prevails...
Christian
--
ÈIn the paranoid-schizoid position ... there is no concern
for the consequences of actions, and certainly not for their
effects on others. At most there is the operation of a talion
morality, a cold exchange of goods and evils.Ç
- Eugene Victor Wolfenstein: 'Psychoanalytic-Marxism, Groundwork'
====
I need help with proving the following statement:
    When the vertex of an angle is in the exterior of a circle, the measure
of the angle is one-half the difference of the measures of the intercepted
arcs.
Any hints or directions to look in would be appreciated.
====
>I need help with proving the following statement:
    When the vertex of an angle is in the exterior of a circle, the 
measure
>of the angle is one-half the difference of the measures of the intercepted
>arcs.
Any hints or directions to look in would be appreciated.
Are we to assume that the sides of the angle are tangent to the 
circle?
If so, try drawing radii to the two points of tangency. You will 
then have two right angles; you will also have an interior angle of 
the circle (which has a simple relationship to both arc lengths). If 
you draw the diagonal of the quadrilateral from the center of the 
circle to the vertex of the external angle, you will have two 
congruent right triangles.
I think if you try that and do some playing around you should see a 
method of attack. But I should disclose that I haven't completed the 
proof myself, so I'm telling you how I _would_ start and not how I 
_did_ start.
-- 
Stan Brown, Oak Road Systems, Cortland County, New York, USA
                                  http://OakRoadSystems.com/
Walrus meat as a diet is less repulsive than seal.
====
>     When the vertex of an angle is in the exterior of a circle, the 
measure
> of the angle is one-half the difference of the measures of the 
intercepted
> arcs.
 Any hints or directions to look in would be appreciated.
>
What happens if the angle's line miss the circle or just one passes thru
or touches the circle?
====
In a particular retail store the average price of a typical two-seater sofa
has been £200 and average sales were 1000 per year. This year the 
store has
run a half price sale on such sofas and found that sales increased by 50%.
Question: Find a hyperbolic model for demand in terms of price.
====
Suppose that in a triangle ABC  a,b,c are the sides ,
2s=a+b+c , r= radius of incircle ,  R= radius of circumcircle.
Denote  
        D= sqrt{s^2-3r(4R+r)} .
Prove that
(1)           D =< max{a,b,c} - min{a,b,c}  =< 2D/sqrt{3} .
If  a =< b =< c ,  prove or disprove the inequalities:
      
(2)   min{b-a,c-b} =< D/sqrt{3} ,
(3)  D =< 2c-(a+b) =< 2D , |2b-(a+c)| =< D ,  D=< (b+c)-2a =< 2D .
=====
To: geometry-college@moderators.isc.org
====
Suppose that in a triangle ABC  a,b,c are the sides ,
 r= radius of incircle ,  R= radius of circumcircle.
Denote  
        D= sqrt{s^2-3r(4R+r)} .
Prove that
(1)           D =< max{a,b,c} - min{a,b,c}  =< 2D/sqrt{3} .
If  a =< b =< c ,  prove or disprove the inequalities:
      
(2)   min{b-a,c-b} =< D/sqrt{3} ,
(3)  D =< 2c-(a+b) =< 2D , |2b-(a+c)| =< D ,  D=< (b+c)-2a =< 2D .
=====
====
Let G be an infinite abelian group such that every proper subgroup of
G is finite. Show that G is isomorphic to Z(p^infty), where
Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }
I've been looking at this one for a week and nothing. Can somebody
help! It's driving me crazy.
====
>Let G be an infinite abelian group such that every proper subgroup of
>G is finite. Show that G is isomorphic to Z(p^infty), where
Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }
I've been looking at this one for a week and nothing. Can somebody
>help! It's driving me crazy.
You say you have gotten nothing, so some comments, which may help
getting you started; I haven't thought it through in detail, so I
could be wrong in some things. But it seems right to me:
First, note that G is a torsion group; as such, it can be decomposed
into a direct sum of its p-parts, that is
G = G_2 oplus G_3 oplus G_5 oplus ...
where G_p = {g in G;  g^{p^m}=e for some m>0}.
Now, let's consider only the non-trivial p-parts; so we have
G = G_{p_1} oplus G_{p_2} oplus ...
Note that there cannot be an infinite number of non-trivial p-parts;
if there were, then the subgroup G_{p_2}oplus ... would be a proper
infinite subgroup. So
G= G_{p_1} oplus G_{p_2} oplus ... oplus G_{p_m};
since G is infinite, at least one of the G_{p_i} is infinite. But
then, G_{p_i} itself is an infinite subgroup of G, nontrivial. Hence
G=G_{p_i}. That is, G is an infinite p-group.
Now consider the quotients G/pG,  pG/p^2G, p^2G/p^3G,...
these quotients are vector spaces over F_p, hence they have a
well-defined dimension. Prove that the dimensions are non-increasing,
able to conclude (a) that there exists m>0 such that p^mG/p^{m+1}/G is
cyclic; and then that (b) all of them are cyclic; using the fact that
G does not have infinite proper subgroups.
Once you have that all of them are cyclic, it should not be hard to
write down an isomorphism.
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
Hmmm...I think I follow you, but I still don't see the actual 
isomorphism...
>Let G be an infinite abelian group such that every proper subgroup of
>G is finite. Show that G is isomorphic to Z(p^infty), where
Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }
I've been looking at this one for a week and nothing. Can somebody
>help! It's driving me crazy.
> 
> You say you have gotten nothing, so some comments, which may help
> getting you started; I haven't thought it through in detail, so I
> could be wrong in some things. But it seems right to me:
> 
> First, note that G is a torsion group; as such, it can be decomposed
> into a direct sum of its p-parts, that is
> 
> G = G_2 oplus G_3 oplus G_5 oplus ...
> 
> where G_p = {g in G;  g^{p^m}=e for some m>0}.
> 
> Now, let's consider only the non-trivial p-parts; so we have
> 
> G = G_{p_1} oplus G_{p_2} oplus ...
> 
> Note that there cannot be an infinite number of non-trivial p-parts;
> if there were, then the subgroup G_{p_2}oplus ... would be a proper
> infinite subgroup. So
> 
> G= G_{p_1} oplus G_{p_2} oplus ... oplus G_{p_m};
> 
> since G is infinite, at least one of the G_{p_i} is infinite. But
> then, G_{p_i} itself is an infinite subgroup of G, nontrivial. Hence
> G=G_{p_i}. That is, G is an infinite p-group.
> 
> Now consider the quotients G/pG,  pG/p^2G, p^2G/p^3G,...
> these quotients are vector spaces over F_p, hence they have a
> well-defined dimension. Prove that the dimensions are non-increasing,
> able to conclude (a) that there exists m>0 such that p^mG/p^{m+1}/G is
> cyclic; and then that (b) all of them are cyclic; using the fact that
> G does not have infinite proper subgroups.
> 
> Once you have that all of them are cyclic, it should not be hard to
> write down an isomorphism.
> 
> ======================================================================
> It's not denial. I'm just very selective about
>  what I accept as reality.
>     --- Calvin (Calvin and Hobbes)
> ======================================================================
> 
> Arturo Magidin
> magidin@math.berkeley.edu
====
> Let G be an infinite abelian group such that every proper subgroup of
> G is finite. Show that G is isomorphic to Z(p^infty), where
> 
> Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }
> 
> I've been looking at this one for a week and nothing. Can somebody
> help! It's driving me crazy.
This may involve some machinery that you don't have available. 
You can easily see that G must be a p-group. There is a theorem
[Kulikov] that says G has a direct summand of the form Z_(p^n), 
1 <= n <= oo. That summand can't be finite or else its complementary
summand would also be finite and hence G would be finite. So, the
summand is Z_(p^oo). The complementary summand must be 0 or else G has
an infinite proper subgroup. Hence G is (isomorphic to) Z_(p^oo).
-- 
Paul Sperry
Columbia, SC (USA)
====
> Let G be an infinite abelian group such that every proper subgroup of
> G is finite. Show that G is isomorphic to Z(p^infty), where
> 
> Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }
But is every proper subgroup of this finite?
Perhaps the question was: every proper subgroup of finite index?
-- 
Timothy Murphy  
tel: +353-86-233 6090
====
>>Let G be an infinite abelian group such that every proper subgroup of
>>G is finite. Show that G is isomorphic to Z(p^infty), where
>>Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }
>>I've been looking at this one for a week and nothing. Can somebody
>>help! It's driving me crazy.
>> 
>> You say you have gotten nothing, so some comments, which may help
>> getting you started; I haven't thought it through in detail, so I
>> could be wrong in some things. But it seems right to me:
>> 
>> First, note that G is a torsion group; as such, it can be decomposed
>> into a direct sum of its p-parts, that is
>> 
>> G = G_2 oplus G_3 oplus G_5 oplus ...
>> 
>> where G_p = {g in G;  g^{p^m}=e for some m>0}.
>> 
>> Now, let's consider only the non-trivial p-parts; so we have
>> 
>> G = G_{p_1} oplus G_{p_2} oplus ...
>> 
>> Note that there cannot be an infinite number of non-trivial p-parts;
>> if there were, then the subgroup G_{p_2}oplus ... would be a proper
>> infinite subgroup. So
>> 
>> G= G_{p_1} oplus G_{p_2} oplus ... oplus G_{p_m};
>> 
>> since G is infinite, at least one of the G_{p_i} is infinite. But
>> then, G_{p_i} itself is an infinite subgroup of G, nontrivial. Hence
>> G=G_{p_i}. That is, G is an infinite p-group.
>> 
>> Now consider the quotients G/pG,  pG/p^2G, p^2G/p^3G,...
>> these quotients are vector spaces over F_p, hence they have a
>> well-defined dimension. Prove that the dimensions are non-increasing,
>> able to conclude (a) that there exists m>0 such that p^mG/p^{m+1}/G is
>> cyclic; and then that (b) all of them are cyclic; using the fact that
>> G does not have infinite proper subgroups.
>> 
>> Once you have that all of them are cyclic, it should not be hard to
>> write down an isomorphism.
>Hmmm...I think I follow you, but I still don't see the actual 
isomorphism...
Well, assuming everything I said was correct, and you have shown that
G is a p-group, and moreover, that each of p^iG/p^{i+1}G is cyclic,
what does that mean? Can you show that G/p^iG is cyclic for all i>0?
Now, take G/pG. This is cyclic of order p, say with generator
x+pG. Since G/p^2G is also cyclic, you should be able to see that it
is generated by an element which, when multiplied by p, is equal to
x. Call it x/p. Then you need to find one which when multiplied by p
is x/p; call it x/p^2. Etc.
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
>Let G be an infinite abelian group such that every proper subgroup of
>>G is finite. Show that G is isomorphic to Z(p^infty), where
>>Z(p^infty) = { [m/p^n] in Q/Z | m in Z, n in N }
>>I've been looking at this one for a week and nothing. Can somebody
>>help! It's driving me crazy.
>> 
>> You say you have gotten nothing, so some comments, which may help
>> getting you started; I haven't thought it through in detail, so I
>> could be wrong in some things. But it seems right to me:
>> 
>> First, note that G is a torsion group; as such, it can be decomposed
>> into a direct sum of its p-parts, that is
>> 
>> G = G_2 oplus G_3 oplus G_5 oplus ...
>> 
>> where G_p = {g in G;  g^{p^m}=e for some m>0}.
>> 
>> Now, let's consider only the non-trivial p-parts; so we have
>> 
>> G = G_{p_1} oplus G_{p_2} oplus ...
>> 
>> Note that there cannot be an infinite number of non-trivial p-parts;
>> if there were, then the subgroup G_{p_2}oplus ... would be a proper
>> infinite subgroup. So
>> 
>> G= G_{p_1} oplus G_{p_2} oplus ... oplus G_{p_m};
>> 
>> since G is infinite, at least one of the G_{p_i} is infinite. But
>> then, G_{p_i} itself is an infinite subgroup of G, nontrivial. Hence
>> G=G_{p_i}. That is, G is an infinite p-group.
>> 
>> Now consider the quotients G/pG,  pG/p^2G, p^2G/p^3G,...
>> these quotients are vector spaces over F_p, hence they have a
>> well-defined dimension. Prove that the dimensions are non-increasing,
>> able to conclude (a) that there exists m>0 such that p^mG/p^{m+1}/G is
>> cyclic; and then that (b) all of them are cyclic; using the fact that
>> G does not have infinite proper subgroups.
>> 
>> Once you have that all of them are cyclic, it should not be hard to
>> write down an isomorphism.
>Hmmm...I think I follow you, but I still don't see the actual 
isomorphism...
Probably because the final two paragraphs are complete bollocks... All
the quotients given will be trivial.
I screwed up; I used the p-powers subgroups instead of the subgroups
of elements of exponent the p-powers...
Here's a better way to proceed; we know that G is an infinite
group. First, consider, pG. If pG were trivial, then G would be of
exponent p, hence an infinite dimensional vector space over F_p, which
clearly has infinite proper subgroups; so pG is nontrivial. If pG
is finite, then G/pG is an infinite dimensional vector space over F_p,
and you can lift an infinite proper subgroup to G. So pG is infinite,
hence pG=G. So G is divisible (being a p-group and p-divisible).
At this point you could invoke classification theorems for divisible 
abelian
groups, which will tell you that G must be a bunch of copies of
Z_p^{infty}, and thereby conclude that it must be equal to one of
them. But here's a way to proceed without invoking those
classification theorems:
For each n>0, let G[p^n] = {g in G : g^{p^n}=e}, the subgroup of
elements of epxonent p^n. 
G[p] is nontrivial; let x in G[p] be different from e1. Since G is
p-divisible, there is an element x_2 such that px_2 = x; an element
x_3 such that px_3=x_2; ...; an element x_{n+1} such that px_{n+1} =
x_n.
Let H = < x, x_2, x_3,...>
It is now easy to verify that H is nontrivial and infinite, so H=G. It
is now also easy to given an isomorphism from Z_{p^{infty}} to G: sent
1/p + Q/Z to x, and 1/p^n +Q/Z to x_n,  n>1.
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
Can someone explain why the set of all integers is not a field?
====
> Can someone explain why the set of all integers is not a field?
>
What's the multiplicative inverse of 2?  Is it an integer?
====
>Can someone explain why the set of all integers is not a field?
A field is a set F with two binary operations, usually denoted + and *
(sum and multiplication, respectively). The fact that they are
binary operations means they are functions from FxF to F (pairs of
elements to elements).  However, it is usual in most places to use
infix notation, and write x+y instead of +(x,y).
The set F, together with the two operations + and *, must satisfy
certain properties before it is called a field:
F must be an abelian group under the operation +. That means:
(1) + must be associative: for all x,y,z in F,  
        (x+y)+z = x+ (y+z).
(2) There exists an element 0 in F such that for all x in F,
        x+0 = 0+x = x.
(3) For every x in F there exists y in F such that x+y = y+x = 0. y is
        called the inverse of x and is often denoted -x; in that
        case, x-y means x+(-y).
(4) + is commutative: for all x,y in F,  
       x+y = y+x.
In addition to these properties, F must be a commutative ring under
the operations + and *. That means:
(5) * must be associative: for all x,y,z in F,
       (x*y)*z = z*(y*z).
(6) * must distribute over +: for all x,y,z in F,
      x*(y+z) = (x*y) + (x*z)
      (y+z)*x = (y*x) + (z*x).
(7) * must be commutative: for all x,y, in F,
     x*y = y*x.
In addition to these properties, F must have a one:
(8) There exists an element 1 in F such that for all x in F,
     1*x = x*1 = x.
And finally, in order to be a field, every nonzero element of F must
have a multiplicative inverse:
(9) For every x in F, if x is not equal to 0 (the same 0 in property
    (2) above), then there exists y in F such that
      x*y = y*x = 1.
Now, are the integers a field? First you must specify which two
operations you are considering. If you are using the usual addition
and multiplication, then the answer is no because the integers fail
property (9): for example, if x=2, then x is not equal to 0, but there
does not exist any integer y such that 2*y = 1.
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
alt.math.undergrad, Cieslak Family <cieslakr002@hawaii.rr.com> 
>Can someone explain why the set of all integers is not a field?
http://mathworld.wolfram.com/FieldAxioms.html
Can you see which field axiom is _not_ satisfied by the integers?
(spoiler below)
What is the multiplicative inverse of 6? Is it a member of the 
integers?
-- 
Stan Brown, Oak Road Systems, Cortland County, New York, USA
                                  http://OakRoadSystems.com/
Walrus meat as a diet is less repulsive than seal.
====
> Can someone explain why the set of all integers is not a field?

====
>Can someone explain why the set of all integers is not a field?
 A field is a set F with two binary operations, usually denoted + and *
> (sum and multiplication, respectively). The fact that they are
> binary operations means they are functions from FxF to F (pairs of
> elements to elements).  However, it is usual in most places to use
> infix notation, and write x+y instead of +(x,y).
 The set F, together with the two operations + and *, must satisfy
> certain properties before it is called a field:
 F must be an abelian group under the operation +. That means:
 (1) + must be associative: for all x,y,z in F,
>         (x+y)+z = x+ (y+z).
> (2) There exists an element 0 in F such that for all x in F,
>         x+0 = 0+x = x.
> (3) For every x in F there exists y in F such that x+y = y+x = 0. y is
>         called the inverse of x and is often denoted -x; in that
>         case, x-y means x+(-y).
> (4) + is commutative: for all x,y in F,
>        x+y = y+x.
 In addition to these properties, F must be a commutative ring under
> the operations + and *. That means:
 (5) * must be associative: for all x,y,z in F,
>        (x*y)*z = z*(y*z).
 (6) * must distribute over +: for all x,y,z in F,
       x*(y+z) = (x*y) + (x*z)
>       (y+z)*x = (y*x) + (z*x).
 (7) * must be commutative: for all x,y, in F,
      x*y = y*x.
 In addition to these properties, F must have a one:
 (8) There exists an element 1 in F such that for all x in F,
      1*x = x*1 = x.
 And finally, in order to be a field, every nonzero element of F must
> have a multiplicative inverse:
 (9) For every x in F, if x is not equal to 0 (the same 0 in property
>     (2) above), then there exists y in F such that
       x*y = y*x = 1.

> Now, are the integers a field? First you must specify which two
> operations you are considering. If you are using the usual addition
> and multiplication, then the answer is no because the integers fail
> property (9): for example, if x=2, then x is not equal to 0, but there
> does not exist any integer y such that 2*y = 1.
 ======================================================================
> It's not denial. I'm just very selective about
>  what I accept as reality.
>     --- Calvin (Calvin and Hobbes)
> ======================================================================
 Arturo Magidin
> magidin@math.berkeley.edu
>
====
I am currently learning to do integration by parts. I'm having
difficulty on one problem, for which I get an answer that seems to be
right to me, but which disagrees with the answer in the book. I've
looked for the errata for the book, but can't find it. Since I can't
do the integral sign, I've made an image which can be found at
http://www.geocities.com/venture_free/Integration.jpg which displays
my problem. I'll wait here while you go look at it.
Now, the answer I keep coming up with is:
xln2x - 1/2x - ln2 +1/2
The answer in the back of the book, however, is:
xln2x - x - ln2 + 1
I think that I am assuming something different than the book is, and I
don't know if I'm wrong, or the book is. Basically, all else being
equal, I get my answer if:
d/dt ln2t = 1/2t
and I get the books answer if:
d/dt ln2t = 1/t. 
Which of us is right, and why?
Also, since I have your attention, is there some easier way to find
book errata than by just looking on Google?
-----------------------------
====
[...]
>  I get my answer if:
> 
> d/dt ln2t = 1/2t
> 
> and I get the books answer if:
> 
> d/dt ln2t = 1/t. 
> 
> Which of us is right, and why?
Sorry, your book is right: d/dt(ln(2t)) = 1/(2t)*d/dt(2t) = 
2/(2t) = 1/t.
Or: ln(2t) = ln(2) + ln(t) so d/dt(ln(2t)) = d/dt(ln(t) = 1/t.
For future reference, your problem was int(ln(2t), t=1..x).
> Also, since I have your attention, is there some easier way to find
> book errata than by just looking on Google?
It never would have occurred to me to look _there_.
-- 
Paul Sperry
Columbia, SC (USA)
====
> I am currently learning to do integration by parts. I'm having
> difficulty on one problem, for which I get an answer that seems to be
> right to me, but which disagrees with the answer in the book. I've
> looked for the errata for the book, but can't find it. Since I can't
> do the integral sign, I've made an image which can be found at
> http://www.geocities.com/venture_free/Integration.jpg which displays
> my problem. I'll wait here while you go look at it.
Bah, boo.
> Now, the answer I keep coming up with is:
> xln2x - 1/2x - ln2 +1/2
integeral log x dx = x log x - x
> The answer in the back of the book, however, is:
> xln2x - x - ln2 + 1
 I get my answer if:
> d/dt ln2t = 1/2t
>
d/dt log 2t = 1/2t * 2 = 1/t
> and I get the books answer if:
> d/dt ln2t = 1/t.
> Which of us is right, and why?
>
The book, because you made a mistake.
> Also, since I have your attention, is there some easier way to find
> book errata than by just looking on Google?
>
Go googling by title, publisher and/or author.
====
I'm trying to do a Fourier Series, but it's been too long since I took
Calculus.  
I'm starting with:
Integral [limits 0 to pi]  x*sin(nx)dx
Integrating by parts:
u = x 
du = dx
dv = sin(nx)dx
v = -cos(nx)  
Integral  x*sin(nx)dx = -x*cos(nx) - [integral] (-1)cos(nx)dx
     = -x*cos(nx) + [integral]  cos(nx)dx
====
> I'm trying to do a Fourier Series, but it's been too long since I took
> Calculus.  
> 
> I'm starting with:
> 
> Integral [limits 0 to pi]  x*sin(nx)dx
> 
> Integrating by parts:
> 
> u = x 
> du = dx
> dv = sin(nx)dx
> v = -cos(nx)  
> 
> Integral  x*sin(nx)dx = -x*cos(nx) - [integral] (-1)cos(nx)dx
>      = -x*cos(nx) + [integral]  cos(nx)dx
> 
> constant (n). 
Odd, since you presumably (incorrectly) integrated sin(nx) to get v.
In both cases substitute for nx.
-- 
Paul Sperry
Columbia, SC (USA)
====
>I'm trying to do a Fourier Series, but it's been too long since I took
>Calculus.  
I'm starting with:
Integral [limits 0 to pi]  x*sin(nx)dx
Integrating by parts:
u = x 
>du = dx
>dv = sin(nx)dx
>v = -cos(nx)  
Integral  x*sin(nx)dx = -x*cos(nx) - [integral] (-1)cos(nx)dx
>     = -x*cos(nx) + [integral]  cos(nx)dx
constant (n).  Is this part going to be equal to sin(nx) or something
>else?
>
 DUH! THINK! what's the derivative of sin(nx)  ????
====
I am trying to model two different savings plans here in the US, and
seem to be running into contradictory results.
There are two plans, an unmatched plan and a matched plan.
You can add money to the unmatched plan like so:
2004: 16k
2005: 18k (it remains at 18k from here on)
Then there is a matched plan (the employer adds 4k)
2004: 20k
2005: 22k (it remains at 22k from here on)
When you withdraw your money, the unmatched plan is taxed at 15%,
while the matched plan is taxed at 35% (this tax is only applied once,
when withdrawing all the money). Assume 10% interest compounded
anually.
It initially appears that after several years, the extra 4k added over
and over again will make the matched plan more attractive. Apparently
the latest BusinessWeek mentioned something to this effect.
me contradictory results.
Approaching the problem mathematically, I solved for U_n and M_n (for
n > 2)
U_n = unmatched funds after n years
M_n = matched funds after n years
I determined the formulas to be:
U_n = 14000 * (1.1)^n + 16000 * (1.1)^(n-1) + 18000* (1.1^(n-1) -
1.1)/(.1)
M_n = 18000 * (1.1)^n + 20000 * (1.1)^(n-1) + 22000* (1.1^(n-1) -
1.1)/(.1)
The last terms of each equation are a simplified geometric series
I want n, such that  .85 * U_n = .65 * M_n. How do I solve for n?
If you see any obvious problems with my program, feel free to point
them out.
If the funds never equal each other (which is what my program hints
at), how could I go about proving this?
This problem has been bugging me. I would greatly appreciate any help.
Chris
Begin Python Program
---------------
rate = .1
years = 50
Over50 = [14000, 16000, 18000]
Over50M = [18000, 20000, 22000]
def GetSum(funds):
        result = []
        total = 0.0
        for x in range(years):
                if x < 2:
                        total += funds[x]
                else:
                        total += funds[-1]
                total *= (1.0 + rate)
                result.append(total)
        return result
def PrintResults(a, b):
        print 'Year', 't', 'Unmatched', 't', 'Matched', 't', 
'Unmatched -
matched'
        for x in range(years):
                unmat = a[x] * .85
                mat = b[x] * .65
                print (x+1), 't', unmat, 't', mat, 't', unmat - mat
resultOver50 = GetSum(Over50)
resultOver50M = GetSum(Over50M)
PrintResults(resultOver50, resultOver50M)
====
> I am trying to model two different savings plans here in the US, and
> seem to be running into contradictory results.
> 
> There are two plans, an unmatched plan and a matched plan.
> 
> You can add money to the unmatched plan like so:
> 2004: 16k
> 2005: 18k (it remains at 18k from here on)
> 
> Then there is a matched plan (the employer adds 4k)
> 2004: 20k
> 2005: 22k (it remains at 22k from here on)
> 
> When you withdraw your money, the unmatched plan is taxed at 15%,
> while the matched plan is taxed at 35% (this tax is only applied once,
> when withdrawing all the money). Assume 10% interest compounded
> anually.
> 
> It initially appears that after several years, the extra 4k added over
> and over again will make the matched plan more attractive. Apparently
> the latest BusinessWeek mentioned something to this effect.
> 
> me contradictory results.
> 
> Approaching the problem mathematically, I solved for U_n and M_n (for
> n > 2)
> 
> U_n = unmatched funds after n years
> M_n = matched funds after n years
> 
> I determined the formulas to be:
> 
> U_n = 14000 * (1.1)^n + 16000 * (1.1)^(n-1) + 18000* (1.1^(n-1) -
> 1.1)/(.1)
> M_n = 18000 * (1.1)^n + 20000 * (1.1)^(n-1) + 22000* (1.1^(n-1) -
> 1.1)/(.1)
> 
> The last terms of each equation are a simplified geometric series
> 
> I want n, such that  .85 * U_n = .65 * M_n. How do I solve for n?
You don't. As you observed, they are never equal.
[...]
U_n = 14000*A + 16000*B + 18000*K;
M_n = 18000*A + 20000*B + 22000*K;
0.85*U_n = 11900*A + 13600*B + 15300*K;
0.65*M_n = 11700*A + 13000*B + 14300*K.
And A, B and K are positive (I don't agree with your K but it doesn't
matter).
So unmatched, taxed at 15% is always better that matched at 35%.
-- 
Paul Sperry
Columbia, SC (USA)
====
>I am trying to model two different savings plans here in the US, and
>seem to be running into contradictory results.
There are two plans, an unmatched plan and a matched plan.
You can add money to the unmatched plan like so:
>2004: 16k
>2005: 18k (it remains at 18k from here on)
Then there is a matched plan (the employer adds 4k)
>2004: 20k
>2005: 22k (it remains at 22k from here on)
When you withdraw your money, the unmatched plan is taxed at 15%,
>while the matched plan is taxed at 35% (this tax is only applied once,
>when withdrawing all the money). Assume 10% interest compounded
>anually.
>
Where do you get these tax rates?  I'll bet the difference isn't the 
calculations, it's the assumptions.
Is it possible that the unmatched plan paid with after-tax dollars, so 
that some percentage is going to taxes (or you're really making bigger 
contributions)?  Maybe the best thing to do is give us the generic names 
of the plans, like Roth IRA vs. employer-sponsered 401(k) plan.
The rates also depend on income.  You need to use your (foreknown) 
marginal tax rate for the years you're going to be withdrawing the money.
Keeping the rest because downthread comments may refer to them.
Jon Miller
>It initially appears that after several years, the extra 4k added over
>and over again will make the matched plan more attractive. Apparently
>the latest BusinessWeek mentioned something to this effect.
me contradictory results.
Approaching the problem mathematically, I solved for U_n and M_n (for
>n > 2)
U_n = unmatched funds after n years
>M_n = matched funds after n years
I determined the formulas to be:
U_n = 14000 * (1.1)^n + 16000 * (1.1)^(n-1) + 18000* (1.1^(n-1) -
>1.1)/(.1)
>M_n = 18000 * (1.1)^n + 20000 * (1.1)^(n-1) + 22000* (1.1^(n-1) -
>1.1)/(.1)
The last terms of each equation are a simplified geometric series
I want n, such that  .85 * U_n = .65 * M_n. How do I solve for n?
If you see any obvious problems with my program, feel free to point
>them out.
If the funds never equal each other (which is what my program hints
>at), how could I go about proving this?
This problem has been bugging me. I would greatly appreciate any help.
Chris
Begin Python Program
>---------------
rate = .1
>years = 50
Over50 = [14000, 16000, 18000]
>Over50M = [18000, 20000, 22000]
def GetSum(funds):
>       result = []
>       total = 0.0
>       for x in range(years):
>               if x < 2:
>                       total += funds[x]
>               else:
>                       total += funds[-1]
>               total *= (1.0 + rate)
>               result.append(total)
>       return result
def PrintResults(a, b):
>       print 'Year', 't', 'Unmatched', 't', 'Matched', 't', 
'Unmatched -
>matched'
>       for x in range(years):
>               unmat = a[x] * .85
>               mat = b[x] * .65
>               print (x+1), 't', unmat, 't', mat, 't', unmat - mat
resultOver50 = GetSum(Over50)
>resultOver50M = GetSum(Over50M)
PrintResults(resultOver50, resultOver50M)
>
====
Can anyone help with the following....
I've been given a function h(x) = 1/1600 x, x>=0
and told that the function Q(x) = exp (- integral between x and 0 h(u) du)
I've evaluated this to e ^ -1/3200x^2
Is this right - my integration is VERY rusty.
I've then worked out that the CDF of X is
F(x) = 1 -Q(x)
       = 1 - e ^ -1/3200x^2
Hence the pdf is
f(x) = 1/3200 e ^-1/3200x^2
Which has given me an exponential distribution with lambda = square root
1/3200 ( I've square rooted cos of the ^2 term in e above) giving a mean of
56.56854249 (1/lambda)
Can someone just check my calculations
====
> Can anyone help with the following....
 I've been given a function h(x) = 1/1600 x, x>=0
> and told that the function Q(x) = exp (- integral between x and 0 h(u) 
du)
 I've evaluated this to e ^ -1/3200x^2
> Is this right - my integration is VERY rusty.
>
integral(0,x) 1/1600u du = (log u)/1600 |0,x
        which gives problem at u = 0
integral(0,x) (1/1600)u du = (1/2)(1/1600)u^2 |0,x = x^2 / 3200
        = (x^2)/3200
Becareful about 1/2 u, 1/2u, (1/2)u, 1/(2u)
> I've then worked out that the CDF of X is
> F(x) = 1 -Q(x)
>        = 1 - e ^ -1/3200x^2
1 - e^[(-1/3200)x^2]
It appeared you said
        1 - e^[-1/(3200x^2)]
Beware e^a+b, e^a + b, e^(a+b), (e^a) + b.
[clip]
 Can someone just check my calculations
>
The problem I see is lack of clarity writing formulas in one line ascii.
====
Debbie Sewell <debbie.sewell@virgin.net>:
> I've been given a function h(x) = 1/1600 x, x>=0
> 
> and told that the function Q(x) = exp (- integral between x and 0 h(u) 
du)
> 
> I've evaluated this to e ^ -1/3200x^2
> Is this right - my integration is VERY rusty.
> 
> I've then worked out that the CDF of X is
> F(x) = 1 -Q(x)
>        = 1 - e ^ -1/3200x^2
> Hence the pdf is
> f(x) = 1/3200 e ^-1/3200x^2
> 
> Which has given me an exponential distribution with lambda = square root
> 1/3200 ( I've square rooted cos of the ^2 term in e above) giving a mean 
of
> 56.56854249 (1/lambda)
Close, but be careful with the details. 
This problem is formally identical to finding the event density given
a hazard function in survival analysis. That doesn't help solve integrals,
It looks to me like you're OK up to and including the calculation of
F(x) = 1 - exp(-(1/3200) x^2). Recall that the derivative of exp(y(x))
is y'(x) exp(y(x)) where y(x) is some function of x. So we have
  f(x) = -(d/dx) (-(1/3200) x^2) exp(-(1/3200) x^2)
       = (x/1600) exp(-(1/3200) x^2)
definitions given, this is always the case: the hazard function 
is the ratio of the event density to the survival function.
Note that this density f(x) is not an exponential density; an 
exponential event density has a constant (not linear) hazard function.
The expected value of the event density is integral(0,inf) x f(x) dx,
which you can solve by integration by parts. Recall that (d/dx) exp(x^2)
is just 2 x exp(x^2).
Hope this helps,
Robert Dodier
--
If I have not seen as far as others, it is because giants were
standing on my shoulders. -- Hal Abelson
====
> Debbie Sewell <debbie.sewell@virgin.net>:
 I've been given a function h(x) = 1/1600 x, x>=0
 and told that the function Q(x) = exp (- integral between x and 0 h(u)
du)
 I've evaluated this to e ^ -1/3200x^2
> Is this right - my integration is VERY rusty.
 I've then worked out that the CDF of X is
> F(x) = 1 -Q(x)
>        = 1 - e ^ -1/3200x^2
> Hence the pdf is
> f(x) = 1/3200 e ^-1/3200x^2
 Which has given me an exponential distribution with lambda = square 
root
> 1/3200 ( I've square rooted cos of the ^2 term in e above) giving a 
mean
of
> 56.56854249 (1/lambda)
 Close, but be careful with the details.
 This problem is formally identical to finding the event density given
> a hazard function in survival analysis. That doesn't help solve 
integrals,
 It looks to me like you're OK up to and including the calculation of
> F(x) = 1 - exp(-(1/3200) x^2). Recall that the derivative of exp(y(x))
> is y'(x) exp(y(x)) where y(x) is some function of x. So we have
   f(x) = -(d/dx) (-(1/3200) x^2) exp(-(1/3200) x^2)
>        = (x/1600) exp(-(1/3200) x^2)
 definitions given, this is always the case: the hazard function
> is the ratio of the event density to the survival function.
 Note that this density f(x) is not an exponential density; an
> exponential event density has a constant (not linear) hazard function.
 The expected value of the event density is integral(0,inf) x f(x) dx,
> which you can solve by integration by parts. Recall that (d/dx) exp(x^2)
> is just 2 x exp(x^2).
 Hope this helps,
> Robert Dodier
> --
> If I have not seen as far as others, it is because giants were
> standing on my shoulders. -- Hal Abelson
integration above x^2 etc.. !!
So, can you help me with the step by step way of using integration by parts
for integral(0,inf) exp(-(1/3200) x^2). ??
Debbie
====
Is there a way to completely describe the isomorphisms from Z_p x Z_q
(where p and q are any integers, not necessarily relatively prime) to
itself? What are these automorphisms?
====
>Is there a way to completely describe the isomorphisms from Z_p x Z_q
>(where p and q are any integers, not necessarily relatively prime) to
>itself? What are these automorphisms?
Well, since any finite abelian group G is the direct product of its Sylow
subgroups P_1,...,P_r, and Aut(G) = Aut(P_1) x ... x Aut(P_r), this problem
immediately reduces to the case when p amnd q are both powers of the same
prime - so let's change notation, and consider  A = Aut(G), with
G = Z_{p^a} x Z_{p^b}, where p is prime and 0 <= a <= b.
In all cases, A has a normal p-subgroup N which acts trivially on G/G^p,
and A/N is the induced action of A on G/G^p. I did a quick calculation,
(and may have made mistakes!), and got
Case 1. a = 0, G is cyclic. This case is well known.
        |N| = p^(b-1) and |A/N| = p-1.
Case 2. 0 < a < b. |N| = p^(3a+b-2), |A/N| = (p-1)^2.
Case 3. 0 < a = b. |N| = p^(4(a-1)), |A/N| = |GL(2,p)| = (p^2-1)(p^2-p).
I am sure the structure could be  described more precisely if necessary.
I believe that I have seen it in the literature.
Derek Holt.
====
> Is there a way to completely describe the isomorphisms from Z_p x Z_q
> (where p and q are any integers, not necessarily relatively prime) to
> itself? What are these automorphisms?
I know of no nice description for arbitrary p and q.
Hom(Z_p x Z_q, Z_p x Z_q) is isomorphic to
Hom(Z_p, Z_p) x Hom(Z_p, Z_q) x Hom(Z_q, Z_p) x Hom(Z_q, Z_q).
The first and last Hom's are isomorphic to Z_p and Z_q respectively.The
two mixed Hom's depend strongly on how p and q are related. Moreover,
it is not obvious (to me, anyway) how to identify the images of the
automorphisms of Z_p x Z_q in the product of the Hom's.
-- 
Paul Sperry
Columbia, SC (USA)
====
> Is there a way to completely describe the isomorphisms from Z_p x Z_q
> (where p and q are any integers, not necessarily relatively prime) to
> itself? What are these automorphisms?
(I shall use the notation Z/p and Z/q instead of Z_p and Z_q)
Further, and in contrast to Derek Holt's more enlightened approach, I
have an elementary prime-free description of the situation.
Let g = gcd(p, q), and p' = p/g, q' = q/g.
Let e_1 and e_2 be generators of Z/p x Z/q, with orders p and q
respectively.
Then one can represent the endomorphism ring of Z/p x Z/q as the ring
of 2x2 matrices
a b   =   A
c d
such that
a belongs to Z/p,
b belongs to Z/p and is a multiple of p'
c belongs to Z/q and is a multiple of q'
d belongs to Z/q.
These matrices act on the group in a natural way, and also multiply
naturally to represent composition.
Necessary conditions for such a matrix A to represent an automorphism
(i.e. to be invertible) are
(1) gcd(det A, p, q) = 1.
(2) gcd(a, p') = 1
(3) gcd(d, q') = 1
Conversely, suppose that A satisfies all three conditions.  Let A' be
the reduction of A mod g.  Then (1) tells us that A' is invertible,
with inverse B', say.  We want to construct a matrix B of the above
form which reduces to B' mod g.  Well, the non-diagonal entries of B'
determine those of B.  Then apply the Chinese remainder theorem to get
the diagonal entries of B, and hopefully a simple calculation should
show that the matrix we construct is indeed the inverse of A.
====
T  H  O  M  A  S
20 8 15 13  1 19 = 76
   I met Joshua in Bessborough Park beside the drinking fountain.
76+ Mom 16 7 53 197/168 +1312
289 Joshua 4 8 81 216/149 8934
Joshua 74 Christopher 139 Thomas 76
   Joshua is a 6 lettered name, Joshua is Bible Book 6, Joshua adds to
74 (a factor of 666). Joshua begins with the 10th letter of the
alphabet (6th non-prime). The vowels and consonants in Joshua both add
non-prime). Joshua was born on the 216th (6x6x6th) day of the year, it
is the 13x13th non-prime (while 13 in turn is the 6th prime). Joshua
(Book 6) contains 24 (6+6+6+6) chapters, I am 24 years older than
Joshua.
1-50 - Genesis
51-90 - Exodus
91-117 - Leviticus
118-153 - Numbers
154-187 - Deuteronomy
188-211 - Joshua
930-957 - Matthew
958-973 - Mark
974-997 - Luke
998-1018 - John
1019-1046 - Acts
1047-1062 - Romans
 188 <-the opening chapter of Book 6 is 6x6x6 short
       of the 404 verses of Bible Book 66, it is the
       6th prime squared (13x13) short of the 357
       verses of Daniel (also in part about 666)
 193 <-Book 6 chapter 6 is the 44th prime,
       while 44 is in turn 66.666...% of 66
 211 <-the terminating chapter of Book 6 is
       approximately 66.6% of the 66th prime (317)
 357 <-the opening chapter of Book 6 plus the 6th
       prime squared is the 357 verses of Daniel
       (in part about 666)
 404 <-the 6th prime squared (13x13) plus the 6th
       prime squared (13x13) plus 66 adds to the
       404 verses of Bible Book 66
1062 <-666 plus 6x66 is a combination of the 658
       verses of Bible Book 6 plus the 404 verses
       of Bible Book 66, and is the terminating
       chapter of New Testament Book 6
1070 <-666 plus the 404 verses of Book 66 is the
       1070 verses of Job (Book 6+6+6)
1213 <-Exodus terminates at chapter 90 (66th non-
       prime) with 1213 verses (the 198th or the
       66+66+66th prime)
1292 <-the 658 verses of Book 6 plus twice the
       66th prime (317) is the 1292 verses of
       Isaiah (the Book contains 66 chapters)
   Mom was born on the 16th day of the month, perhaps in 53 (16th
prime), and she gave birth on the 216th day of the year. She gave
birth with 149 days remaining in the year, it's the length of Bible
Book 48 (16+16+16). Joshua's name adds to 289, corresponding to the
(166th non-prime).
Primes    Non-Primes
   2           1
   3           4
   5           6
   7           8
  11           9
  13          10
  17          12
  19          14
  23          15
  29          16
  31          18
  37          20
  41          21
  43          22
  47          24
  53          25
  59          26
  61          27
  67          28
  71          30
  73          32
  79          33
  83          34
  89          35
  97          36
 101          38
 103          39
 107          40
 109          42
 113          44
 127          45
 131          46
 137          48
 139          49
 149          50
 151          51
 157          52
 163          54
 167          55
 173          56
 179          57
 181          58
 191          60
 193          62
 197          63
 199          64
 211          65
 223 <-48th-> 66
----         ---
4661        1710
   Joshua was born with 149 days remaining in the year, it's the
the 48th prime (223) and the 48th non-prime (66).
                Primes 
               In Prime
    Primes     Positions
 1     2
 2     3 <-       3
 3     5 <-       5
 4     7
 5    11 <-      11
 6    13
 7    17 <-      17
 8    19
 9    23
10    29
11    31 <-      31
12    37
13    41 <-      41
                ---
                108
   Mom and Joshua were born on days of the year adding to 413, it is
exactly 59 weeks (the 17th prime). Joshua was born in 81 (59th
non-prime while 59 is the 17th prime). He was born a multiple of 17
days into the century (29801=17x1753). Mom and Joshua were together
born a multiple of 108 days into the century (49356=108x457), there
are 108 verses in Bible Book 59 (the 17th prime, it's the 17th prime
day, month and year of birth adds to 93 (Leviticus 3 with 17 verses).
plus the 17th prime, chapter 76 is Exodus 26 (17th non-prime). Today
mom and Joshua are together 71.77 years old.
                Primes 
               In Prime
    Primes     Positions
 1     2
 2     3 <-       3
 3     5 <-       5
 4     7
 5    11 <-      11
 6    13
 7    17 <-      17
 8    19
 9    23
10    29
11    31 <-      31
12    37
13    41 <-      41
14    43
15    47
16    53
17    59 <-      59
                ---
                167
              Esther
              Book 17
   Joshua was born 19 days after mom's birthday. Mom was born on day
197, it's the number of verses in Bible Book 28 (19th non-prime).
Mom's day, month and year of birth adds to 76 and she married Thomas
(76=4x19). Joshua was born 67 days closer to the end of the year than
to the beginning of the year (19th prime), and keep in mind that the
2460 verses of Bible Book 19 is
19x19+19x19+19x19+19x19+19x19+19x19+19x19 minus the 19th prime (67).
unrepeated letters add to 77 (the primes up to 19). Mom was born on
the 19555th day of the century and married Thomas (76, the 55th
non-prime). The Samuels are Books 9 and 10 (for 19) with 55 chapters
and 1505 verses.
   I criticized churches (I said they had Egyptian penises on their
roofs, and this was in opposition to God's Second Commandment),
Protestants and Catholics lobbied my abusive parents to have me
treated, my abusive parents quickly complied and I was repeatedly
Protestants and Catholics then sat on psychiatric appeal panel
detain and torture me. I begged and begged for assistance to get out
of the country to no avail, people would tell me that I was there (in
psychiatric facilities) because I was supposed to be there. And so it
is with Joshua, he now finds himself posted on the usenet becuase he
is supposed to be here. I showed Joshua gems, and like everybody else,
he was so cheap and ignorant that he did not even have the decency to
time out of my life to do so. But come December, he will please his
parents (or his girlfriend's parents) and attend their church, he will
comment on the beauty of their decorated trees or their steeple (a
represntation of a penis), and will give them money (the churches that
teach you people to turn pagan fertility symbols into decorated idols
are supposedly worthy of your support and respect). In my effort to
get out of the brutal cycle of psychiatric abuuse, I told the
Protestants and Catholics seated in judgment against me that the Bible
repeatedly condemns turning trees into idols, and that they in fact
bow to their decorated trees via the placement and retieval of
presents at the base of their trees. At one psychiatric appeal panel
hearing Dr. Gene Marcoux heard me say this, he responded by saying
that I was religiously deluded to believe this, and then all the
Protestants and Catholics nodded in agreement and gave him permisson
to detain and torture me for the following three weeks. Year after
year you people collectively spend billions of dollars on turning
trees into decorated idols, you have spent millions of dollars having
me tortured in an attempt to make me shut up about this and other
traditions being taught by your churches, and you are so cheap and
ignorant that you can't even offer to pay me minimum wage for my work
so I could attempt to get out of this country on my own dime. I was
frantic to leave the country as I lost summer after summer after
summer after summer after summer after summer after summer to
psychiatric torture, and the subsequent summers were effectly lost as
well as I remained frantic to leave the country (I expected to be
arrested and tortured in the years that I was not), and there isn't a
single person who has the compassion to assist me to get out of this
country for one single winter!!! You are the shit of the earth Joshua
Christopher Thomas, you are an incompassionate turd, and now I beg God
to honor Exodus 20:5 and Hosea 4:6 as promises so that He would
terminate your life, or at least turn away from you in your hour of
need. If I hear of your death I will cheer with utter glee, it will be
in accordance to Scripture (Psalm 137:9), and I will post your stats
again. God is about to spread you people out like dung over the
surface of the earth, and in this I rejoice. You have billions of
dollars to spend on turning trees into idols, you have millions of
and so ignorant that you can't even buy a postage stamp in order to
send me a cheap letter expressing thanx for showing you evidence that
your very name is a gift from God.
Daryl Shawn Kabatoff
Box 7134
Saskatoon Saskatchewan
Canada
S7K 4J1
Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!
====
T  H  O  M  A  S
20 8 15 13  1 19 = 76
   I met Joshua in Bessborough Park beside the drinking fountain.
76+ Mom 16 7 53 197/168 +1312
289 Joshua 4 8 81 216/149 8934
Joshua 74 Christopher 139 Thomas 76
   Joshua is a 6 lettered name, Joshua is Bible Book 6, Joshua adds to
74 (a factor of 666). Joshua begins with the 10th letter of the
alphabet (6th non-prime). The vowels and consonants in Joshua both add
non-prime). Joshua was born on the 216th (6x6x6th) day of the year, it
is the 13x13th non-prime (while 13 in turn is the 6th prime). Joshua
(Book 6) contains 24 (6+6+6+6) chapters, I am 24 years older than
Joshua.
1-50 - Genesis
51-90 - Exodus
91-117 - Leviticus
118-153 - Numbers
154-187 - Deuteronomy
188-211 - Joshua
930-957 - Matthew
958-973 - Mark
974-997 - Luke
998-1018 - John
1019-1046 - Acts
1047-1062 - Romans
 188 <-the opening chapter of Book 6 is 6x6x6 short
       of the 404 verses of Bible Book 66, it is the
       6th prime squared (13x13) short of the 357
       verses of Daniel (also in part about 666)
 193 <-Book 6 chapter 6 is the 44th prime,
       while 44 is in turn 66.666...% of 66
 211 <-the terminating chapter of Book 6 is
       approximately 66.6% of the 66th prime (317)
 357 <-the opening chapter of Book 6 plus the 6th
       prime squared is the 357 verses of Daniel
       (in part about 666)
 404 <-the 6th prime squared (13x13) plus the 6th
       prime squared (13x13) plus 66 adds to the
       404 verses of Bible Book 66
1062 <-666 plus 6x66 is a combination of the 658
       verses of Bible Book 6 plus the 404 verses
       of Bible Book 66, and is the terminating
       chapter of New Testament Book 6
1070 <-666 plus the 404 verses of Book 66 is the
       1070 verses of Job (Book 6+6+6)
1213 <-Exodus terminates at chapter 90 (66th non-
       prime) with 1213 verses (the 198th or the
       66+66+66th prime)
1292 <-the 658 verses of Book 6 plus twice the
       66th prime (317) is the 1292 verses of
       Isaiah (the Book contains 66 chapters)
   Mom was born on the 16th day of the month, perhaps in 53 (16th
prime), and she gave birth on the 216th day of the year. She gave
birth with 149 days remaining in the year, it's the length of Bible
Book 48 (16+16+16). Joshua's name adds to 289, corresponding to the
(166th non-prime).
Primes    Non-Primes
   2           1
   3           4
   5           6
   7           8
  11           9
  13          10
  17          12
  19          14
  23          15
  29          16
  31          18
  37          20
  41          21
  43          22
  47          24
  53          25
  59          26
  61          27
  67          28
  71          30
  73          32
  79          33
  83          34
  89          35
  97          36
 101          38
 103          39
 107          40
 109          42
 113          44
 127          45
 131          46
 137          48
 139          49
 149          50
 151          51
 157          52
 163          54
 167          55
 173          56
 179          57
 181          58
 191          60
 193          62
 197          63
 199          64
 211          65
 223 <-48th-> 66
----         ---
4661        1710
   Joshua was born with 149 days remaining in the year, it's the
the 48th prime (223) and the 48th non-prime (66).
                Primes 
               In Prime
    Primes     Positions
 1     2
 2     3 <-       3
 3     5 <-       5
 4     7
 5    11 <-      11
 6    13
 7    17 <-      17
 8    19
 9    23
10    29
11    31 <-      31
12    37
13    41 <-      41
                ---
                108
   Mom and Joshua were born on days of the year adding to 413, it is
exactly 59 weeks (the 17th prime). Joshua was born in 81 (59th
non-prime while 59 is the 17th prime). He was born a multiple of 17
days into the century (29801=17x1753). Mom and Joshua were together
born a multiple of 108 days into the century (49356=108x457), there
are 108 verses in Bible Book 59 (the 17th prime, it's the 17th prime
day, month and year of birth adds to 93 (Leviticus 3 with 17 verses).
plus the 17th prime, chapter 76 is Exodus 26 (17th non-prime). Today
mom and Joshua are together 71.77 years old.
                Primes 
               In Prime
    Primes     Positions
 1     2
 2     3 <-       3
 3     5 <-       5
 4     7
 5    11 <-      11
 6    13
 7    17 <-      17
 8    19
 9    23
10    29
11    31 <-      31
12    37
13    41 <-      41
14    43
15    47
16    53
17    59 <-      59
                ---
                167
              Esther
              Book 17
   Joshua was born 19 days after mom's birthday. Mom was born on day
197, it's the number of verses in Bible Book 28 (19th non-prime).
Mom's day, month and year of birth adds to 76 and she married Thomas
(76=4x19). Joshua was born 67 days closer to the end of the year than
to the beginning of the year (19th prime), and keep in mind that the
2460 verses of Bible Book 19 is
19x19+19x19+19x19+19x19+19x19+19x19+19x19 minus the 19th prime (67).
unrepeated letters add to 77 (the primes up to 19). Mom was born on
the 19555th day of the century and married Thomas (76, the 55th
non-prime). The Samuels are Books 9 and 10 (for 19) with 55 chapters
and 1505 verses.
   I criticized churches (I said they had Egyptian penises on their
roofs, and this was in opposition to God's Second Commandment),
Protestants and Catholics lobbied my abusive parents to have me
treated, my abusive parents quickly complied and I was repeatedly
Protestants and Catholics then sat on psychiatric appeal panel
detain and torture me. I begged and begged for assistance to get out
of the country to no avail, people would tell me that I was there (in
psychiatric facilities) because I was supposed to be there. And so it
is with Joshua, he now finds himself posted on the usenet because he
is supposed to be here. I showed Joshua gems, and like everybody else,
he was so cheap and ignorant that he did not even have the decency to
time out of my life to do so. But come December, he will please his
parents (or his girlfriend's parents) and attend their church, he will
comment on the beauty of their decorated trees or their steeple (a
representation of a penis), and will give them money (the churches
that teach you people to turn pagan fertility symbols into decorated
idols are supposedly worthy of your support and respect). In my effort
to get out of the brutal cycle of psychiatric abuse, I told the
Protestants and Catholics seated in judgment against me that the Bible
repeatedly condemns turning trees into idols, and that they in fact
bow to their decorated trees via the placement and retrieval of
presents at the base of their trees. At one psychiatric appeal panel
hearing Dr. Gene Marcoux heard me say this, he responded by saying
that I was religiously deluded to believe this, and then all the
Protestants and Catholics nodded in agreement and gave him permission
to detain and torture me for the following three weeks. Year after
year you people collectively spend billions of dollars on turning
trees into decorated idols, you have spent millions of dollars having
me tortured in an attempt to make me shut up about this and other
traditions being taught by your churches, and you are so cheap and
ignorant that you can't even offer to pay me minimum wage for my work
so I could attempt to get out of this country on my own dime. I was
frantic to leave the country as I lost summer after summer after
summer after summer after summer after summer after summer to
psychiatric torture, and the subsequent summers were effectively lost
as well as I remained frantic to leave the country (I expected to be
arrested and tortured in the years that I was not), and there isn't a
single person who has the compassion to assist me to get out of this
country for one single winter!!! You are the shit of the earth Joshua
Christopher Thomas, you are an incompassionate turd, and now I beg God
to honor Exodus 20:5 and Hosea 4:6 as promises so that He would
terminate your life, or at least turn away from you in your hour of
need. If I hear of your death I will cheer with utter glee, it will be
in accordance to Scripture (Psalm 137:9), and I will post your stats
again. God is about to spread you people out like dung over the
surface of the earth, and in this I rejoice. You have billions of
dollars to spend on turning trees into idols, you have millions of
and so ignorant that you can't even buy a postage stamp in order to
send me a cheap letter expressing thanx for showing you evidence that
your very name is a gift from God.
Daryl Shawn Kabatoff
Box 7134
Saskatoon Saskatchewan
Canada
S7K 4J1
Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!
====
Lately I've been explaining how I found an error in taught mathematics
as I prove that in the ring of algebraic integers a number I'll here
call x has a factor I'll here call y, but x/y is not an algebraic
integer.
Now at least *some* of you should be wondering how that works given
that division is not a ring operation.
Part of the problem for those of you who wonder about that is that the
mathematicians among you probably know how that all plays out as I've
been arguing for a while now, but a poster named Randy Poe posted some
things that lead me to consider that all of you aren't necessarily
math experts i.e. mathematicians.
Ok, here it goes.  I'm going to go kind of fast but you should be able
to keep up as it's simple.
First thing is my paper Advanced Polynomial Factorization which is as
usual found at
  http://groups.msn.com/AmateurMath
shows that given these three numbers two of them have this factor I
call f and the ring is algebraic integers.
That's all you get there, and there's no mention of anything like
x/y, which I've talked about a lot.
And there's no mention because the ring operations are addition and
multiplication.
Division is not a ring operation.  Division is defined for fields, not
rings.
So then, it's *provable* that a number I'll call x has a factor I'll
call y in the ring of algebraic integers, is what the paper covers. 
There you have the a's like before and two of the a's have a factor f,
while the third is coprime to f, which is proven in the paper.
That is, given the definition for algebraic integer I can prove that
there's this number I'll call x, which is itself an algebraic integer,
that has this number I'll call y, which is an algebraic integer, as a
factor.
That's it for the paper.
Now then, taking that result in that ring, I can now move to the field
of algebraic numbers.
Now I have division, and in that field, I take x/y and notice that
it's NOT an algebraic integer, which is a contradiction.
That's how it's done.
Now for the mathematicians, what I just said shouldn't be news, but a
lot of you are basically, um, what's the right word?  Well there
probably isn't a right word so I'll leave it there.
Notice I'm only posting to sci.math and alt.math.undergrad here, as I
don't see the point in bothering anyone else with such trivial
details.
James Harris
====
[...]
> Now then, taking that result in that ring, I can now move to the field
> of algebraic numbers.
> 
> Now I have division, and in that field, I take x/y and notice that
> it's NOT an algebraic integer, which is a contradiction.
> 
> That's how it's done.
The field of algebraic numbers over the rationals Q contains the ring
of algebraic integers over Q, but the two are not equal.
So if x and y are algebraic integers over Q, and y is not zero,
then surely x/y is an algebraic number.  By no means does
x, y algebraic integers imply x/y is an algebraic integer.
David Bernier
====
> Lately I've been explaining how I found an error in taught mathematics
> as I prove that in the ring of algebraic integers a number I'll here
> call x has a factor I'll here call y, but x/y is not an algebraic
> integer.
> 
Can you explain what you mean by factor in this instance?
For example, 3 and 5 are algebraic integers, but 3/5 is not an algebraic 
integer.  That's because the algebraic integers are a ring but not a 
field.  
What do you mean that y is a factor of x?  If x and y are algebraic 
integers, the accepted definition of factor is that there is some 
algebraic integer z such that yz = x.  If so, then we say that y is a 
factor of x.
If we consider the example above in the ring of the usual integers, 5 is 
not a factor of 3.  But if we extend the ring of integers to the field 
of rationals, it turns out that there is a rational, namely 3/5, with 
the property that 5 * 3/5 = 3.  
The same example works in the ring of algebraic integers.
So all you seem to be saying is that you've determined that the 
algebraic integers are a ring but not a field.  This is already well 
known.
Is this what you're talking about?
====
>Lately I've been explaining how I found an error in taught mathematics
>as I prove that in the ring of algebraic integers a number I'll here
>call x has a factor I'll here call y, but x/y is not an algebraic
>integer.
>
[...]
Now then, taking that result in that ring, I can now move to the field
>of algebraic numbers.
Now I have division, and in that field, I take x/y and notice that
>it's NOT an algebraic integer, which is a contradiction.
No, that's not a contradiction. Because you get a factor in the
field of algebraic numbers, _not_ in the algebraic integers.
The following looks like one of those nasty parodies people
sometimes post, but actually it's an illustration of _exactly_
the error you're making, translated to a more familiar context:
Lately I've been explaining how I found an error in taught
mathematics as I prove that in the ring of integers a number 
I'll here call x has a factor I'll here call y, but x/y is not an
integer.
Let x = 2, y = 3.
Now then, taking that result in that ring, I can now move to the field
of rational numbers.
Now I have division, and in that field, I take x/y and notice that
it's NOT an integer, which is a contradiction.
Honest. What you're saying here makes _exactly_ as much
sense as saying that 3 is a factor of 2 (because 2/3 exists
in the rationals) although 2/3 is not an integer, contradiction.
There's no contradiction, just two true facts:
(i) 3 is _not_ a factor of 2, _in_ the integers.
(ii) 3 _is_ a factor of 2 _in_ the rationals (and luckily
  2/3 is rational.)
************************
David C. Ullrich
====
> 
> Lately I've been explaining how I found an error in taught mathematics
> as I prove that in the ring of algebraic integers a number I'll here
> call x has a factor I'll here call y, but x/y is not an algebraic
> integer.
> 
> 
> Can you explain what you mean by factor in this instance?
> 
> For example, 3 and 5 are algebraic integers, but 3/5 is not an algebraic 
> integer.  That's because the algebraic integers are a ring but not a 
> field.  
> 
> What do you mean that y is a factor of x?  If x and y are algebraic 
> integers, the accepted definition of factor is that there is some 
> algebraic integer z such that yz = x.  If so, then we say that y is a 
> factor of x.
> 
> If we consider the example above in the ring of the usual integers, 5 is 
> not a factor of 3.  But if we extend the ring of integers to the field 
> of rationals, it turns out that there is a rational, namely 3/5, with 
> the property that 5 * 3/5 = 3.  
> 
> The same example works in the ring of algebraic integers.
> 
> So all you seem to be saying is that you've determined that the 
> algebraic integers are a ring but not a field.  This is already well 
> known.
> 
> Is this what you're talking about?
This is going to seem rather odd.  There are times when I think
we put too much stress on form and not enough on 
substance.  Allowing for the fact that James Harris actively
resists the use of standard terminology and wants to remain
ignorant of conventions and definitions, is there any way
that his statement about x and y could make sense?
I think so.  If x and y are algebraic integers and y is
a factor of x in the algebraic integers, anyone would
naturally conclude that x/y must be an algebraic 
integer simply from the definition of factor.  But 
there is still the possibility that x/y could be a root 
of a non-monic polynomial with integer coefficients, 
primitive and irreducible over the rationals: that is, 
that possibility would exist if it were not for the 
following known theorem:
     Theorem.  If r is a root of a non-monic polynomial
     with integer coefficients that is primitive and 
     irreducible over the rationals, then r cannot be 
     an algebraic integer.
Now if Harris could show that x/y is such a root, one
would have to conclude that either (1) the theorem above
is wrong, or (2) something else is wrong with core
mathematics, i.e., the mathematics that is currently
taught.
Harris argued for many months that the Theorem is wrong.
He tried over and over again to come up with a counter-
example.  He finally consented to read a proof of the
theorem, *only after* someone that he regarded as an
authority, Robert Israel, stated that the theorem was
valid.  He then said that he accepted the theorem.  
Whether he ever actually read the proof I don't know.
However it appears now that he has slid back to grinding this
old axe in a new form: he no longer says the theorem
is wrong; instead he says there is something wrong with
core mathematics and in particular something wrong 
with the definition of algebraic integers, or that the
ring of algebraic integers is incomplete.
The point here is, there is possible content in what
he says.  I totally disagree with his conclusion, but
it could make a loony kind of sense.  If he actually did
find x and y as above, and a non-monic polynomial 
with integer coefficients, primitive and irreducible
over the rationals with the algebraic integer x/y as a 
root, he would indeed have found something wrong with 
mathematics.  Of course he hasn't, but theoretically it 
could make sense.
Thus: quit taking him to task for the superficial problem
of not knowing the definition of 'factor'.  Take him to 
task for yet again trying to show that the Theorem above, 
which he has accepted, is false. Ask him to specify the 
non-monic polynomial of which x/y is a root.
Andrzej
====
> What do you mean that y is a factor of x?  If x and y are algebraic 
> integers, the accepted definition of factor is that there is some 
> algebraic integer z such that yz = x.  If so, then we say that y is a 
> factor of x.
[snip]
> This is going to seem rather odd.  There are times when I think
> we put too much stress on form and not enough on 
> substance.  Allowing for the fact that James Harris actively
> resists the use of standard terminology and wants to remain
> ignorant of conventions and definitions, is there any way
> that his statement about x and y could make sense?
> 
> I think so.  If x and y are algebraic integers and y is
> a factor of x in the algebraic integers, anyone would
> naturally conclude that x/y must be an algebraic 
> integer simply from the definition of factor.
Uh, yeah.
>  But 
> there is still the possibility that x/y could be a root 
> of a non-monic polynomial with integer coefficients, 
> primitive and irreducible over the rationals: that is, 
> that possibility would exist if it were not for the 
> following known theorem:
> 
>      Theorem.  If r is a root of a non-monic polynomial
>      with integer coefficients that is primitive and 
>      irreducible over the rationals, then r cannot be 
>      an algebraic integer.
Snip. You haven't clarified the problem as far as I can
see. If the relationship between x and y is not
x/y is an algebraic integer, then what is it?
You seem to be saying that the only thing known about
x/y is that it is the root of such a polynomial. That's
easy.
> The point here is, there is possible content in what
> he says.
And what is that? I don't get what you're trying to
say. It seems to be that:
   (1) There exists x and y algebraic integers such that
   (2) r = x/y is not an algebraic integer, and
   (3) r is the root of a non-monic polynomial with
    integer coefficients, that is primitive and irreducible
    over the rationals.
Is that it? If so, isn't that just a trivial restatement
of the theorem?
            - Randy
====
>  
> What do you mean that y is a factor of x?  If x and y are algebraic 
> integers, the accepted definition of factor is that there is some 

> algebraic integer z such that yz = x.  If so, then we say that y is a 

> factor of x.
> [snip]
> 
> This is going to seem rather odd.  There are times when I think
> we put too much stress on form and not enough on 
> substance.  Allowing for the fact that James Harris actively
> resists the use of standard terminology and wants to remain
> ignorant of conventions and definitions, is there any way
> that his statement about x and y could make sense?
> 
> I think so.  If x and y are algebraic integers and y is
> a factor of x in the algebraic integers, anyone would
> naturally conclude that x/y must be an algebraic 
> integer simply from the definition of factor.
> 
> Uh, yeah.
> 
>  But 
> there is still the possibility that x/y could be a root 
> of a non-monic polynomial with integer coefficients, 
> primitive and irreducible over the rationals: that is, 
> that possibility would exist if it were not for the 
> following known theorem:
> 
>      Theorem.  If r is a root of a non-monic polynomial
>      with integer coefficients that is primitive and 
>      irreducible over the rationals, then r cannot be 
>      an algebraic integer.
> 
> Snip. You haven't clarified the problem as far as I can
> see. If the relationship between x and y is not
> x/y is an algebraic integer, then what is it?
>
  Obviously it is.  What Harris may be trying to say is,
I have found an algebraic integer which is a root of a
*non-monic* polynomial etc., thus defying a well-known
theorem.  Therefore there is something wrong with
mathematics: either the theorem is wrong (as he has
long contended) or the definition of algebraic integer
is wrong, or something else.
  I am not sure why this needs explaining.  I am very
definitely not saying that I agree with Harris: simply
that instead of continuing to criticize him for refusing
to recognize a tautology, consider that he may actually
have had something more in mind - something equally wrong,
to be sure, but at a different level.
  I actually don't know how he goes from his main Advanced
Polynomial Factorization result (which has now been
proved false by two other posters) to obtaining x and
y as he describes.  I am not sure he remembers how 
either - not that it matters at all anyway.
 
> You seem to be saying that the only thing known about
> x/y is that it is the root of such a polynomial. That's
> easy.
> 
  What??? Basically the monic criterion for algebraic
integers is if and only if - more precisely, an algebraic
number is an algebraic integer if and only if its minimal
polynomial with integer coefficients is monic.
> The point here is, there is possible content in what
> he says.
> 
> And what is that? I don't get what you're trying to
> say. 
  Yes, I see that.
> It seems to be that:
>    (1) There exists x and y algebraic integers such that
>    (2) r = x/y is not an algebraic integer, and
     No - just the opposite - r = x/y IS an algebraic integer.
>    (3) r is the root of a non-monic polynomial with
>     integer coefficients, that is primitive and irreducible
>     over the rationals.
> 
> Is that it? If so, isn't that just a trivial restatement
> of the theorem?
> 
  Essentially he may be saying yet again, for the nth time,
that he has found a counterexample to the theorem.  The 
obvious question he should answer is: what is the polynomial?
  Andrzej
  
>             - Randy
====
>   Obviously it is.  What Harris may be trying to say is,
> I have found an algebraic integer which is a root of a
> *non-monic* polynomial etc., thus defying a well-known
> theorem.  Therefore there is something wrong with
> mathematics: either the theorem is wrong (as he has
> long contended) or the definition of algebraic integer
> is wrong, or something else.
OK, if he's made that statement. I thought you were
interpreting the remarks about I have proved that x is a
factor of y in the algebraic integers, but x/y is not
an algebraic integer.
Or perhaps you are. Is that his proof that x is a
factor of y? I've never yet been able to figure out
which part of James' writings are supposed to be the
proof part, but every once in a while he starts
talking as if a proof went by when nobody was looking.
           - Randy
====
>   Obviously it is.  What Harris may be trying to say is,
> I have found an algebraic integer which is a root of a
> *non-monic* polynomial etc., thus defying a well-known
> theorem.  Therefore there is something wrong with
> mathematics: either the theorem is wrong (as he has
> long contended) or the definition of algebraic integer
> is wrong, or something else.
> 
> OK, if he's made that statement. I thought you were
> interpreting the remarks about I have proved that x is a
> factor of y in the algebraic integers, but x/y is not
> an algebraic integer.
> 
> Or perhaps you are. Is that his proof that x is a
> factor of y? I've never yet been able to figure out
> which part of James' writings are supposed to be the
> proof part, but every once in a while he starts
> talking as if a proof went by when nobody was looking.
> 
>            - Randy
  In connection of the x/y thing, he says he has found
problems in core mathematics and taught mathematics -
he has said the algebraic integers are incomplete
(presumably as opposed to objects, where he has a
circular definition and he has yet to identify even one 
object which is not an integer).
  My reply was none too clear, partly because I am not
certain of what he is implying.  I <think> he could be saying,
I can prove that x/y is an algebraic integer but also
the theorem implies that it's *not* an algebraic integer.
Therefore something must be wrong with math.  With regard
to where the proof is in this case, as I said before, I
don't know and I am not sure he now knows.  If it depends
on his Advanced Polynomial Factorization paper, it is
nowhere, per Hall and Baron.
  Andrzej
====
Ullrich, you're busted;
you have a *lot* of explaining to do,
to the Order of the Undead Mathematicians!
> Lately I've been explaining how I found an error in taught
> mathematics as I prove that in the ring of integers a number 
> I'll here call x has a factor I'll here call y, but x/y is not an
> integer.
> 
> Let x = 2, y = 3.
> 
> Now then, taking that result in that ring, I can now move to the field
> of rational numbers.
> 
> Now I have division, and in that field, I take x/y and notice that
> it's NOT an integer, which is a contradiction.
--A church-school McCrusade (Blair's ideals?):
Harry-the-Mad-Potter want's US to kill Iraqis?...
For a 1000-year anglo-american hegemony?
HEY, JIMMY; LET'S US and SU FIGHT -then-PM of England & Zbiggy
http://www.tarpley.net/bush25.htm (Thyroid Storm ch.)
  http://www.rwgrayprojects.com/synergetics/plates/plates.html
   http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX
====
Q. There is a Chinese steel company that produces steel rods for local 
construction company. The plant manager knew from past statistics that the 
rods
produced are normally distributed with a mean length of 300cm. The standard 
deviation is 8cm. If 22% of the sample means are more than specific length
K, what is the value of K?
Solution:
Given:  u = 300cm,  sigma = 8cm,  n = 14,  p(>K)= 0.22
Let K = be the length of the steel rod
Since  u (population) = u (sample) = 300, sigma (sample) = sigma 
/squareroot(14) = 2.138099
P(bar X > K) = [K - u(sample)]/2.138099 = Zvalue(0.22)
(K - 300) / 2.138089935 =  Zvalue(0.22)
(K - 300) / 2.138089935 =  0.58
K = 2.138089935(0.58) + 300
K = 301.2401cm
Therefore, if 22% of the sample means are more than a specific length K, 
then K is = 301.2401cm.
Is my answer correct?
====
I am a cabinetmaker and a novice programmer. I have a interest in writing
a plywood optimization application. This app takes a number of pieces
of plywood and finds the best way to orient the pieces on a standard
sheet size i.e. 4 x 8. I`m looking for a mathematical algorithm that deals
with this type of problem. I`m wondering if you might able to guide me
in the right direction.
Here is a link of a commercial product that provides a optimization of
plywood. http://www.sheetlayout.com/
Best Wishes, Barry Golash
====
I tried to post this message before. I hope I'm not just being impatient 
and posting the same thing twice. Anyway, this is my question. 
Let«s say 
  I have y=f[x1(t),x2(t)] and I want to plot y(t) after removing the 
effect of x2 so that, in reality, I can have an idea of what x1(t) looks 
like. The data I have is y(t) (in which x1 and x2 are both taken into 
account) and x2(t). Ok, I hope I have made myself clear.
====
>   I have y=f[x1(t),x2(t)] and I want to plot y(t) after removing the
> effect of x2 so that, in reality, I can have an idea of what x1(t) looks
> like. The data I have is y(t) (in which x1 and x2 are both taken into
> account) and x2(t). Ok, I hope I have made myself clear.
No, hardly at all.  You want to plot, y = f(x1(t), x2(a)) to see what
x1(t) looks like?  Oh, you have y(t) = f(x1(t), x2(t)) and want to find
x1(t) from y(t) and x2(t) ?
What if f(x,y) = g(y) ?  Then y(t) = f(x1(t), x2(t)) = g(x2(t))
and x1(t) could be anything and do anything it wanted as along
as y(t) and x2(t) were arbitrarly related as y(t) = g(x2(t)).
====
> I tried to post this message before. I hope I'm not just being impatient 
> and posting the same thing twice. Anyway, this is my question. 
Let«s say 
>   I have y=f[x1(t),x2(t)] and I want to plot y(t) after removing the 
> effect of x2 so that, in reality, I can have an idea of what x1(t) looks 
> like. The data I have is y(t) (in which x1 and x2 are both taken into 
> account) and x2(t). Ok, I hope I have made myself clear.
In order to solve such a problem, you need to assume something about
the relationship between y, x1 and x2 (e.g. that x1 and x2 are
independent, and that y is dependent on both variables).  In other
words, you need to specify the problem more precisely.  In particular,
you must make an assumption about the functional relationship.  For
example, a common problem in linear system theory is that of
deconvolution. i.e. assume y = x1 * x2, where * denotes convolution.
 The goal in that case would be to solve for x1 given x2 and y. 
Generally, this sort of problem falls under the category of an
inverse problem.  This is a very complicated subject with many
possible approaches.  The chosen approach depends very much on your
particular problem, and what you may assume about the relationship
between your variables x1, x2, and y.  I hope this helps.
====
> 
>>I tried to post this message before. I hope I'm not just being impatient 
>>and posting the same thing twice. Anyway, this is my question. 
Let«s say 
>>  I have y=f[x1(t),x2(t)] and I want to plot y(t) after removing the 
>>effect of x2 so that, in reality, I can have an idea of what x1(t) looks 
>>like. The data I have is y(t) (in which x1 and x2 are both taken into 
>>account) and x2(t). Ok, I hope I have made myself clear.
> 
> 
> In order to solve such a problem, you need to assume something about
> the relationship between y, x1 and x2 (e.g. that x1 and x2 are
> independent, and that y is dependent on both variables).  In other
> words, you need to specify the problem more precisely.  In particular,
> you must make an assumption about the functional relationship.  For
> example, a common problem in linear system theory is that of
> deconvolution. i.e. assume y = x1 * x2, where * denotes convolution.
>  The goal in that case would be to solve for x1 given x2 and y. 
> Generally, this sort of problem falls under the category of an
> inverse problem.  This is a very complicated subject with many
> possible approaches.  The chosen approach depends very much on your
> particular problem, and what you may assume about the relationship
> between your variables x1, x2, and y.  I hope this helps.
It does help. If nothing else, it lets me know that it wasn't as simple 
====
By the way, I do know that y is dependent on both variables and that x1 
and x2 are independent but that's about it. Oh, and I also have the 
following data: y vs t and x2 vs t. Anyway, like I said, I'll have to 
look into it some more.
> 
>>I tried to post this message before. I hope I'm not just being impatient 
>>and posting the same thing twice. Anyway, this is my question. 
Let«s say 
>>  I have y=f[x1(t),x2(t)] and I want to plot y(t) after removing the 
>>effect of x2 so that, in reality, I can have an idea of what x1(t) looks 
>>like. The data I have is y(t) (in which x1 and x2 are both taken into 
>>account) and x2(t). Ok, I hope I have made myself clear.
> 
> 
> In order to solve such a problem, you need to assume something about
> the relationship between y, x1 and x2 (e.g. that x1 and x2 are
> independent, and that y is dependent on both variables).  In other
> words, you need to specify the problem more precisely.  In particular,
> you must make an assumption about the functional relationship.  For
> example, a common problem in linear system theory is that of
> deconvolution. i.e. assume y = x1 * x2, where * denotes convolution.
>  The goal in that case would be to solve for x1 given x2 and y. 
> Generally, this sort of problem falls under the category of an
> inverse problem.  This is a very complicated subject with many
> possible approaches.  The chosen approach depends very much on your
> particular problem, and what you may assume about the relationship
> between your variables x1, x2, and y.  I hope this helps.
====
> Lately I've been explaining how I found an error in taught
mathematics
> as I prove that in the ring of algebraic integers a number I'll here
> call x has a factor I'll here call y, but x/y is not an algebraic
> integer.
James,
I tried to frame a definition of 'factor' in the thread 'Error in my
paper Advanced Polynomial Factorization'. I asked if my proposed
definition was correct, but you did not respond. Is that definition
correct?
John Peters
====
 Lately I've been explaining how I found an error in taught
> mathematics
> as I prove that in the ring of algebraic integers a number I'll
here
> call x has a factor I'll here call y, but x/y is not an algebraic
> integer.
 James,
 I tried to frame a definition of 'factor' in the thread 'Error in my
> paper Advanced Polynomial Factorization'. I asked if my proposed
> definition was correct, but you did not respond. Is that definition
> correct?
 John Peters
Oops, posting under wrong pseudonym.
Clive Tooth
====
<3c65f87.0306171917.10e891a3@posting.google.com>,
> 
> Oh yeah, it'd help if you know mathematics.
> 
> 
> James Harris
If who knew mathematics?