mm-3259
===
Subject: Re: Matrix algebra: establishing identities
> Could anybody help me solve the following problem:
> [img]http://members.lycos.nl/fctikkieterug/identities.gif[/img]
> <a href=http://members.lycos.nl/fctikkieterug/identities.gif>Click
> here (link to picture</a Please if you can help me establish either 
identity 1,2, or 3 or all of
> them I would be very grateful. I've been working on it for hours but
> don't seem to get anything close.
If all matrices are square, then H * X = I implies that X * H = I,
which proves (1).
===
Subject: Re: Matrix algebra: establishing identities
>> Could anybody help me solve the following problem:
>> [img]http://members.lycos.nl/fctikkieterug/identities.gif[/img]
>> <a href=http://members.lycos.nl/fctikkieterug/identities.gif>Click
>> here (link to picture</a>
The identities were 
X^T V^(-1) X H = X^T V^(-1)                                            (1)
X^T V^(-1) = F^(-1) (F^(-1) + X^T Delta^(-1) X)^(-1) X^T Delta^(-1)    (2)
V H^T = X (F + (X^T Delta^(-1) X)^(-1))                                (3)
with definitions
V = X F X^T + Delta
H = (X^T Delta^(-1) X)^(-1) X^T Delta^(-1)
H X = I  (not a definition, but a consequence of the last one)
>> Please if you can help me establish either identity 1,2, or 3 or all of
>> them I would be very grateful. I've been working on it for hours but
>> don't seem to get anything close.
>If all matrices are square, then H * X = I implies that X * H = I,
>which proves (1).
It's still true if they aren't square.
If H is m x n and X is n x m, V and Delta must be n x n,
F must be m x m.  Now H X = I (which would be m x m) 
makes X H an n x n projection with Ker(X H) = Ker(H) 
and Ran(X H) = Ran(X).  Since X and H must have rank m,
m <= n. I - X H is the complementary projection with 
Ker(I - X H) = Ran(X H) = Ran(X) and 
Ran(I - X H) = Ker(X H) = Ker(H).
To show X^T V^(-1) X H = X^T V^(-1), it thus suffices 
to show X^T V^(-1) x = 0 when x is in Ker(H). 
Delta and X^T Delta^(-1) X are invertible if the definition
of H makes sense.
So x is in Ker(H) if and only if Delta^(-1)(x) is in Ker(X^T), i.e.
x = Delta y with X^T y = 0.  Then 
V y = X F X^T y + Delta y = x, i.e. 
X^T V^(-1) x = X^T y = 0.  This proves (1).
For (3),
V H^T = (X F X^T + Delta)(Delta^(-1)^T X (X^T Delta^(-1)^T X)^(-1))
      = X F (X^T Delta^(-1)^T X) (X^T Delta^(-1)^T X)^(-1) +
        X (X^T Delta^(-1)^T X)^(-1)
      = X F + X (X^T Delta^(-1)^T X)^(-1)
      = X (F + (X^T Delta^(-1)^T X)^(-1))
Assuming that all the inverses that are written in (2) exist,
multiply (2) on the left by (F^(-1) + X^T Delta^(-1) X) F and
on the right by V, so (2) is equivalent to
(F^(-1) + X^T Delta^(-1) X) F X^T = X^T Delta^(-1) V 
But the left side is 
  X^T + X^T Delta^(-1) X F X^T
  = X^T (I + Delta^(-1) X F X^T)
  = X^T Delta^(-1) (Delta + X F X^T)
  = X^T Delta^(-1) V 
as required.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
===
Subject: Re: Applied math: making the most of a poor voting system
>> In the following discussion, please let X be your end of the left/right,
>> liberal/conservative spectrum, and Y be the other end of the left/right,
>> liberal/conservative spectrum.  (I know it's not that simple in reality:
>> some folks are moderates, and many folks don't subscribe to the agenda 
of
>> their preferred party 100%, but for the sake of this discussion let's
>> assume everyone is at one end or the other, mkay?)
>> Imagine you are X, living in a very Y area.
>> There is one X candidate, who is showing very poorly in the straw polls
>> shortly before the election, and there are two Y candidates that are
>> showing much stronger in the straw polls shortly before the election.
>> In a scenario where you can only place one vote for the single candidate
>> you would like to win (IE, like in most of the USA, where you cannot say
>> I'd prefer candidate A, but if I cannot have them, then I want B, and
>> my last choice is C), how would you vote?
>> Would you:
>> 1) Vote for the unlikely-to-win X candidate hoping to get better 
campaign
>> funding in the next election?
>> ...or would you:
>> 2) Vote for the more moderate of the two Y candidates, hoping to keep 
the
>> area from being dragged even further in the direction of Y than it 
already
>> is, due to ending up with an extreme Y being put in office?
>> Are there other reasons you would vote one way or the other?
>Certainly with only two choices, the average person's new political
>slogan is not May the best man win!, it has become May the worst man
>lose!, and I suppose that the thinking man's strategy should be such
>as to accomplish that end.  On the other hand, there are situations in
>which good conscience must overcome cold logic.  Consider, for example,
>if you are a Jew, and you have in front of you a ballot with three
>candidates.  You know that one of the first two will be a sure winner,
>while the third doesn't stand a chance (something like a ballot in the
>U.S. elections).  The first two names are Adolph Hitler, and Herman
>Ghoering (sp.?).  For whom do you vote?
>cfe
Yes -- sometimes you have to vote on principle regardless of whether
or not the candidate has a chance to win. Votes can act as powerful
statements of support for alternative views. If no one is willing to
make such a statement, then one can expect no real alternatives.
The US system provides a clear example of this dilemma.
If voters can be convinced, using the logic that a 3rd party candidate
has essentially no chance to win, that there's no point voting for a
3rd party candidate, then we are pretty much locked into a 2-party
system. 
The big US corporations now realize this, so many corporations now
send large amounts of money to _both_ of the major parties. The idea
is simple -- if you own both parties, you can't lose.
Thus, the 2-party system has evolved into a 1-party system, with
potentially no escape.
quasi
===
Subject: Proof Without Words: d/dx (sin x) = cos x ?
I'm familiar with the standard limit of a difference quotient proof of 
d/dx (sin x) = cos x.
Does there exist a proof without words (a graphic with no commentary, a 
Behold illustration) of this fact?
L 
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
> I'm familiar with the standard limit of a difference quotient proof 
of
> d/dx (sin x) = cos x.
> Does there exist a proof without words (a graphic with no commentary, 
a
> Behold illustration) of this fact?
Prove both d/dx(cos(x))=-sin(x)  and d/dx(sin(x))=cos(x):
 Draw a unit circle oriented counterclockwise, pick a generic radius
(not horizontal, not vertical, not at pi/4 angles), then draw a unit
tangent vector (representing velocity vector, assuming unit speed)
attached to the enpoint of the radius.
 Mark the (x,y) coordinates of the endpoint of the radius, confronted with
the (x,y) coordinates of the velocity. Behold.
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
>I'm familiar with the standard limit of a difference quotient proof of 
>d/dx (sin x) = cos x.
>Does there exist a proof without words (a graphic with no commentary, a 
>Behold illustration) of this fact?
Take a look at <http://www.whim.org/nebula/math/images/ddxsinx.gif>
Rob Johnson <rob@trash.whim.org>
take out the trash before replying
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
>>I'm familiar with the standard limit of a difference quotient proof of 
>>d/dx (sin x) = cos x.
>>Does there exist a proof without words (a graphic with no commentary, 
a 
>>Behold illustration) of this fact?
>Take a look at <http://www.whim.org/nebula/math/images/ddxsinx.gifRob 
Johnson <rob@trash.whim.orgtake out the trash before replying
Similarly, plus with some interesting 'history', see
 http://www.ms.uky.edu/~carl/ma330/sin/sin1.html
and the large yellow right triangle at the bottom of the page.
 
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
>I'm familiar with the standard limit of a difference quotient proof of 
>d/dx (sin x) = cos x.
>Does there exist a proof without words (a graphic with no commentary, a 
>Behold illustration) of this fact?
Point A moving on unit cicle with unit speed.
Look at the velocity vector.
-- 
Rouben Rostamian
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
> I'm familiar with the standard limit of a difference quotient proof 
of
> d/dx (sin x) = cos x.
> Does there exist a proof without words (a graphic with no commentary, 
a
> Behold illustration) of this fact?
d/dx (sin x) = lim(h->0) (sin x+h - sin x)/h
        = lim (sin x cos h + cos x sin h - sin x)/h
        = lim ((sin x)(cos h - 1)/h + cos x (sin h)/h)
        = cos x
Since lim(h->0) (sin h)/h = 1
and for the other limit, start by noting
        (cos h - 1)/h = (cos^2 h - 1) / h(cos h + 1)
Riddle of the day:  is there love without poems?
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
> Riddle of the day:  is there love without poems?
A poem is a song without the music.
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
> I'm familiar with the standard limit of a difference quotient proof of 
> d/dx (sin x) = cos x.
> Does there exist a proof without words (a graphic with no commentary, 
a 
> Behold illustration) of this fact?
Look at the slope of the tangents to sin x as x varies from 0 to pi/2 to 
pi etc. You can see the slopes trace out a cosine curve.
Bob Kolker
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
> I'm familiar with the standard limit of a difference quotient proof of 
> d/dx (sin x) = cos x.
> Does there exist a proof without words (a graphic with no commentary, 
a 
> Behold illustration) of this fact?
Does behold the power series! count? 
===
Subject: Re: Proof Without Words: d/dx (sin x) = cos x ?
> Does behold the power series! count?
Well, to use this argument, you have to know that term-by-term 
differentiation is valid in the radius of convergence of the power series, 
and the radius of convergence of the series of sine is infinity. So... this 

shouldn't be that trivial. 
===
Subject: Re: vandermonde matrix question
<27556868.1130962933802.JavaMail.jakarta@nitrogen.mathforum.org>,
> Let T: P_n(F)->F^(n+1) be the linear transformation defined 
> by T(f)=(f(c_o), f(c_1), ... , f(c_n)), where 
> c_o, c_1, ..., c_n are distinct scalars in the infinite 
> field F.  Let B be the standard ordered basis for P_n(F) 
> and y be the standard ordered basis for F^(n+1).
> 1) How can I show that M=[T]y,B has the form:
> [1 c_o (c_o)^2 ...   (c_o)^n
>  1 c_1 (c_1)^2 ...   (c_1)^n
>  .
>  .
>  .
>  1 c_n (c_n)^2 ...   (c_n)^n]
I take it [T]y,B is notation for the matrix representing T 
with respect to the bases y and B. 
You say B is the standard ordered basis for P_n(F) - do you know 
what this means? 
You say y is the standard ordered basis for F^(n+1) - do you know 
what this means? 
Do you know how to find the matrix representing a linear 
transformation? 
> 2) How can I use the fact that I know that T is an
> isomorphism to prove that det(M)=/0? (doesn't equal 0)
Do you know what an isomorphism is? Do you know what you can say 
about a matrix if its determinant is not zero? 
> 3) How can I show that det(M)= [product] ((c_j)-(c_i))?
Now this is the only part of the problem that isn't trivial 
for someone who actually understands the definitions of the terms 
in the problem. Even so, if you know about the effects of row 
and column operations on determinants you might be able to put 
a proof together, especially if you keep one eye on the goal.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: vandermonde matrix question
Let T: P_n(F)->F^(n+1) be the linear transformation defined
> by T(f)=(f(c_o), f(c_1), ... , f(c_n)), where
> c_o, c_1, ..., c_n are distinct scalars in the infinite
> field F. Let B be the standard ordered basis for P_n(F)
> and y be the standard ordered basis for F^(n+1).
> 1) How can I show that M=[T]y,B has the form:
> [1 c_o (c_o)^2 ... (c_o)^n
> 1 c_1 (c_1)^2 ... (c_1)^n
> 1 c_n (c_n)^2 ... (c_n)^n]
I take it [T]y,B is notation for the matrix representing T
with respect to the bases y and B.
You say B is the standard ordered basis for P_n(F) - do you know
what this means?
You say y is the standard ordered basis for F^(n+1) - do you know
what this means?
Do you know how to find the matrix representing a linear
transformation?
A transformation is linear if T(cx+y)= cT(x)+ T(y).  
But I am not sure how to show this in this case.
> 2) How can I use the fact that I know that T is an
> isomorphism to prove that det(M)=/0? (doesn't equal 0)
Do you know what an isomorphism is? Do you know what you can say
about a matrix if its determinant is not zero?
If the det(M)=/0, then that matrix is invertible, if it
is square, which it is in this case.  
> 3) How can I show that det(M)= [product] ((c_j)-(c_i))?
Now this is the only part of the problem that isn't trivial
for someone who actually understands the definitions of the terms
in the problem. Even so, if you know about the effects of row
and column operations on determinants you might be able to put
a proof together, especially if you keep one eye on the goal.
If I change the row order, does the determinant change?
I am not sure how changing the rows would change anything...
I would think that I should perform row operations on
this matrix, hoping to arise at an upper triangular
matrix, and having the determinant be determined by
the entries on the diagonal...
===
Subject: Re: vandermonde matrix question
<23208101.1130990093114.JavaMail.jakarta@nitrogen.mathforum.org>,
>> Let T: P_n(F)->F^(n+1) be the linear transformation defined
>> by T(f)=(f(c_o), f(c_1), ... , f(c_n)), where
>> c_o, c_1, ..., c_n are distinct scalars in the infinite
>> field F. Let B be the standard ordered basis for P_n(F)
>> and y be the standard ordered basis for F^(n+1).
 1) How can I show that M=[T]y,B has the form:
>> [1 c_o (c_o)^2 ... (c_o)^n
>> 1 c_1 (c_1)^2 ... (c_1)^n
>> .
>> .
>> .
>> 1 c_n (c_n)^2 ... (c_n)^n]
>> I take it [T]y,B is notation for the matrix representing T
>> with respect to the bases y and B.
>> You say B is the standard ordered basis for P_n(F) - do you know
>> what this means?
>> You say y is the standard ordered basis for F^(n+1) - do you know
>> what this means?
>> Do you know how to find the matrix representing a linear
>> transformation?
> A transformation is linear if T(cx+y)= cT(x)+ T(y).  
> But I am not sure how to show this in this case.
A transformation is linear if T(c x + y) = c T(x) + T(y) 
for all c in F and all x and y in the domain, but it is 
neither necessary nor helpful to show that the given 
transformation is linear. The statement of the problem 
says that the transformation is linear. 
In any event, you didn't answer any of my questions. 
If you don't know the answers, say so - you don't have a prayer 
of solving the problem if you can't answer my questions, so 
that way we'll know where we have to start. 
>> 2) How can I use the fact that I know that T is an
>> isomorphism to prove that det(M)=/0? (doesn't equal 0)
>> Do you know what an isomorphism is? Do you know what you can say
>> about a matrix if its determinant is not zero?
> If the det(M)=/0, then that matrix is invertible, if it
> is square, which it is in this case.  
Yes, that's good. But what about the other question - do you know 
what an isomorphism is, in particular what distinguishes an 
isomorphism from any other kind of linear transformation? If not, 
say so - you have no hope of doing the problem until you find out. 
>> 3) How can I show that det(M)= [product] ((c_j)-(c_i))?
>> Now this is the only part of the problem that isn't trivial
>> for someone who actually understands the definitions of the terms
>> in the problem. Even so, if you know about the effects of row
>> and column operations on determinants you might be able to put
>> a proof together, especially if you keep one eye on the goal.
> If I change the row order, does the determinant change?
> I am not sure how changing the rows would change anything...
> I would think that I should perform row operations on
> this matrix, hoping to arise at an upper triangular
> matrix, and having the determinant be determined by
> the entries on the diagonal...
Sounds good - but be sure to do row operations that result in 
things like c_j - c_i showing up, so you have a shot at getting 
the intended answer.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: The Formula for The Eliptic to the Ellipse?
> I am trying to crack the elliptic curve cryptography system and the
> discrete step of the function as the elliptic are to make the same
> occur for the associated ellipse.  So all I need to do is take the
> formula for the elliptic and convert it to the ellipse.
> And it sounds all rather parametric, which is ok.  I just want the
> basic x, y  with the associated constants. A search function is to then
> be searched for.
> Is that easy for anybody?
I doubt it. There are some pretty clever regulars in this newsgroup, 
but I think even they will be hard-pressed to make any sense out of 
your question. 
To begin with, the relation between elliptic curves and ellipses 
is so distant, especially when we're talking about elliptic curves 
modulo some integer, that I'm not sure it even makes sense to ask for 
the associated ellipse to a given elliptic curve. 
Maybe if you showed us an example of how, in a very simple case, 
you manage to associate an ellipse to an elliptic curve, some of us 
will have a better idea of what you are talking about.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: central subgroup
> A very simple question (I hope the answer is too ;-)):
> What's the definition of a central subgroup?
> I searched on the internet and in my own 'library', but couldn't find it. 

Is there maybe an other equivalent def.?
Likely you mean the center of a group
        Z(G) = { x in G | for all g in G, xg = gx }
which is an Abelian subgroup of G.
===
Subject: Re: prove it if you can
> 
>>[...]
>>Let _s_ be the square root of 2. I don't know whether s^s is rational 
or
>>not. If it is, then you're done. Otherwise, consider
    (s^s)^s = s^{s^2} = s^2 = 2.
So, if s^s is irrational, you have the example that you want.
> For the record, s^s is not only irrational, but transcendental. This is
> a consequence of Lindemann's Theorem.
> How so? All I know about Lindemann's Theorem is what I
> how it implies that. The theorem tells us that if
> a is algebraic, then exp(a) is transcendental. But
> s^s is exp(ln(s)s), and surely ln(s)s isn't algebraic.
I think it needs Gelfond-Schneider, which came after Lindemann. 
If a is algebraic and not 0 or 1, and b is algebraic and irrational, 
then a^b is transcendental. 
http://mathworld.wolfram.com/GelfondsTheorem.html
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: prove it if you can
[s is sqrt(2)]
>For the record, s^s is not only irrational, but transcendental. This is
>a consequence of Lindemann's Theorem.
>> ...
> I think it needs Gelfond-Schneider, which came after Lindemann. 
> If a is algebraic and not 0 or 1, and b is algebraic and irrational, 
> then a^b is transcendental. 
> http://mathworld.wolfram.com/GelfondsTheorem.html
proof about s^s and (s^s)^s I was told that whether s^s was
irrational or not was open, and that was much more recently
than 1934. I guess that it's just not as well-known as
it should be.
===
Subject: Re: prove it if you can
   <3sro10Fpmu31U1@individual.net>
   <3sspfmFpsc6sU1@individual.net>
   <gerry-33C705.13584503112005@sunb.ocs.mq.edu.au
>>[...]
>>Let  s  be the square root of 2. I don't know whether s^s is rational 
or
>>not. If it is, then you're done. Otherwise, consider
    (s^s)^s = s^{s^2} = s^2 = 2.
So, if s^s is irrational, you have the example that you want.
> For the record, s^s is not only irrational, but transcendental. This 
is
> a consequence of Lindemann's Theorem.
> How so? All I know about Lindemann's Theorem is what I
> how it implies that. The theorem tells us that if
> a is algebraic, then exp(a) is transcendental. But
> s^s is exp(ln(s)s), and surely ln(s)s isn't algebraic.
> I think it needs Gelfond-Schneider, which came after Lindemann.
> If a is algebraic and not 0 or 1, and b is algebraic and irrational,
> then a^b is transcendental.
> http://mathworld.wolfram.com/GelfondsTheorem.html
Yes. I got those two mixed up.
     --- Christopher Heckman
===
Subject: free group
How to show that two commuting elements of a free group are powers of a 
third?
===
Subject: Re: free group
>How to show that two commuting elements of a free group are powers of a 
third?
It is very easy if you assume the theorem that a subgroup of a free group
is free.
It is also possible to do it by brute force, but messy, because you have to
take into account the possibility that parts of words cancel.
It goes roughly as follows. Let the two commuting reduced words be x, y.
First handle the case when there is no cancellation, so xy = yx as words.
If l(x) = l(y) then x = y and result is clear.
If l(x) < l(y), y has x as a prefix, so you can replace y by
the shorter word x^-1 y and use induction on word length.
If there is cancellation, then x = ua^-1, y = av  for some generator (or
inverse) a, and we have xy = uv, yx = avua^-1, so u has a as its first
letter, and hence x = ata^-1, y = asa^-1 for some s,t, and the result
follows by induction applied to s and t.
Derek Holt.
===
Subject: Re: free group
> How to show that two commuting elements of a free group are powers of a 
third?
< x,y | xyx^-1 y^-1 > is isomorphic to additive ZxZ.
What c in ZxZ is there for which
        n.c = (1,0), m.c = (0,1)
for some n,m in Z
Notation:  3.c = c + c + c, etc.
===
Subject: Re: free group
:> How to show that two commuting elements of a free group are powers of a 
third?
: < x,y | xyx^-1 y^-1 > is isomorphic to additive ZxZ.
That's a free abelian group, but not a free group ...
: What c in ZxZ is there for which
:       n.c = (1,0), m.c = (0,1)
: for some n,m in Z
: Notation:  3.c = c + c + c, etc.
===
Subject: Re: free group
 <Pine.BSI.4.58.0511021929150.9186@vista.hevanet.com> 
<dkca9r$fbl$1@news.Stanford.EDU :> How to show that two commuting elements 
of a free group are powers of
> a third?
> : < x,y | xyx^-1 y^-1 > is isomorphic to additive ZxZ.
> That's a free Abelian group, but not a free group ...
How so?  G = < x,y > is a free group and if x and y commute, then
G is the a fore mentioned group and the counter example persists.
> : What c in ZxZ is there for which
> :     n.c = (1,0), m.c = (0,1)
> : for some n,m in Z
> : Notation:  3.c = c + c + c, etc.
===
Subject: Re: free group
            days. My association with the Department is that of an alumnus.
>> :> How to show that two commuting elements of a free group are powers of
>> a third?
>> : < x,y | xyx^-1 y^-1 > is isomorphic to additive ZxZ.
>> That's a free Abelian group, but not a free group ...
>How so?  G = < x,y > is a free group and if x and y commute, then
>G is the a fore mentioned group
No. F is a free group. If a and b are commuting elements of F, then
<a,b> is a QUOTIENT of the aforementioned group. In that quotient, it
is possible for a and b to be powers of the same element.
> and the counter example persists.
>> : What c in ZxZ is there for which
>> :    n.c = (1,0), m.c = (0,1)
>> : for some n,m in Z
>> : Notation:  3.c = c + c + c, etc.
Not applicable. In point of fact, a free group cannot contain the free
abelian group of rank 2 as a subgroup (all subgroups of a free group
are [absolutely] free, so if a and b commute, the subgroup they
generate must be free of rank 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
===
Subject: Re: free group
            days. My association with the Department is that of an alumnus.
>> :> How to show that two commuting elements of a free group are powers of
>> a third?
>> : < x,y | xyx^-1 y^-1 > is isomorphic to additive ZxZ.
>> That's a free Abelian group, but not a free group ...
>How so?  G = < x,y > is a free group and if x and y commute, then
>G is the a fore mentioned group and the counter example persists.
>> : What c in ZxZ is there for which
>> :    n.c = (1,0), m.c = (0,1)
>> : for some n,m in Z
No, it does not. You misunderstand the question. The third element
does not have to lie in the subgroup generated by x and y. If you
map your free abelian group into the FREE group,
then you will find that any embedding will send x and y to elements
that are in fact powers of the third. The free group does not contain
copies of Z x Z (since every subgroup of a free group is free).
-- 
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
===
Subject: Re: free group
   <Pine.BSI.4.58.0511021929150.9186@vista.hevanet.com>
   <dkca9r$fbl$1@news.Stanford.EDU>
   <Pine.BSI.4.58.0511030242390.12256@vista.hevanet.com :> How to show that 
two commuting elements of a free group are powers 
of
> a third?
> : < x,y | xyx^-1 y^-1 > is isomorphic to additive ZxZ.
> That's a free Abelian group, but not a free group ...
> How so?  G = < x,y > is a free group and if x and y commute, then
> G is the a fore mentioned group and the counter example persists.
You haven't given a counterexample. You've proved that if there is a
counterexample, then the subgroup of G generated by x and y will be the
aforementioned group. That doesn't show that there is a counterexample.
> : What c in ZxZ is there for which
> :   n.c = (1,0), m.c = (0,1)
> : for some n,m in Z
> : Notation:  3.c = c + c + c, etc.
===
Subject: Re: a math question
   <dk8fca$5dj$1@nntp.itservices.ubc.ca The answer is wrong.  It should be
> 1 + ((i-1)*exp(pai*j*i/2)*(b1+i*b2-b*cot(3*pai*j/2)) + 2*b2)/
> (2*sqrt(a)) + ...
In fact, I do agree the answer is wrong,but I cannot deduce your answer
too. My answer is
1 +
((i-1)*exp(pi*j*i/2)*[b1+i*b2-b*cot(3*pai*j/2))-(i+1)*b2*exp(-pi*j*i/2)]/(2*

sqrt(a))
+ ...
===
Subject: Re: Help with an elliptic curve?
   <021120051115075565%chenrich@monmouth.com Hello all,
> I need to solve the equation,
> y^2 = 20x^3-19x^2+14x-11
> where x,y are rationals.
> (P.S. It has small integer solutions at x=1,3 but any other rational
> point?)
> You have 5 known rational points.  In homogeneous coordinates they are
> x:y:1 =
> 0, 1, 0
> 1, 2, 1
> 1, -2, 1
> 3, 20,1
> 3, -20,1
> The line through two of these points will intersect the curve in
> another rational point.  (Bring scratch paper, and several sharp
> pencils with good erasers.)
> --
> Chris Henrich
> http://www.mathinteract.com
> God just doesn't fit inside a single religion.
The method somehow slipped my mind. Seriously, this elliptic curve
arises naturally in a parametric equal sums of sixth powers.  I need to
know another explicit rational solution other then x=1,3.
-Titus
===
Subject: Re: Help with an elliptic curve?
>Hello all,
>I need to solve the equation,
>y^2 = 20x^3-19x^2+14x-11
>where x,y are rationals.
>(P.S. It has small integer solutions at x=1,3 but any other rational
>point?)
>>You have 5 known rational points.  In homogeneous coordinates they are
>>x:y:1 =
>>0, 1, 0
>>1, 2, 1
>>1, -2, 1
>>3, 20,1
>>3, -20,1
>>The line through two of these points will intersect the curve in
>>another rational point.  (Bring scratch paper, and several sharp
>>pencils with good erasers.)
>>--
>>Chris Henrich
>>http://www.mathinteract.com
>>God just doesn't fit inside a single religion.
> The method somehow slipped my mind. Seriously, this elliptic curve
> arises naturally in a parametric equal sums of sixth powers.  I need to
> know another explicit rational solution other then x=1,3.
> -Titus
Using Apecs with Maple we get:
Gcub(20,0,0,0,-19,0,-1,14,0,-11);
`The equation of the curve is`
                     20*U^3-19*U^2-V^2+14*U-11 = 0
`The point at infinity with homogeneous coordinates [0, 1, 0]`
`will be taken as the group O.`
`P0=Q is a flex.`
`We now invoke the apecs command ein(0,-76,0,4480,-281600)`
`to determine the minimal Weierstrass form.`
`The coordinates U,V of the original equation are related to the`
`coordinates X,Y of the Weierstrass equation by`
                                    3      2
                                 4 X  + 5 X  + 42 X - 179
              U = X/5 + 2/5, V = ------------------------
                                      10 Y + 5 X + 5
                                      5 U         5 V
                   X = 5 U - 2, Y = - --- + 1/2 + ---
                                       2           2
`To transfer points between the original curve and the Weierstrass form`
`use the commands Trcw(u,v) and Trwc(x,y).`
`These commands remain available during this apecs session.`
              present curve is , C110 = [1, 1, 1, 10, -45]
 > Tor();
`Tor already done with the following results.`
             The order of the torsion subgroup of E(Q) =, 5
`The list of torsion points is`
          O, [3, 3, 5], [13, -57, 5], [13, 43, 5], [3, -7, 5]
`The non-O point(s) are in the notation [x,y,d], d=order; they form`
`the sequence PP, while in the notat. [x,y] they form the list ouP.`
Maybe this helps.
Jaap Spies
===
Subject: Re: Help with an elliptic curve?
   <021120051115075565%chenrich@monmouth.com>
   <4369F3B9.2020207@hccnet.nl>
<snip>
What is the RANK???
===
Subject: Re: Help with an elliptic curve?
> <snip What is the RANK???
Rk();
`Values of rc,r4,r1 for curve D110 were 0 0 2`
`now updated to 2 0 2`
`Rank already determined to be 0`
`using (possibly all of) the T, B-SD, RH conjectures.`
`Consider using Seek, Bas or Crem ---`
`RkNC can do no more (2-descent involving the cubic field Q(x) where`
`[x,y] is a point of order 2 is not programmed in apecs).`
===
Subject: Re: bump function
Bump function: B(x)=(e^2)*e*(1-x)*e(x+1)=e^4(1-x)(1+x)
 e:R->R by e(x)= e^(-1/x) if x>0
0 if x<=0
How is this function positive inside the interval (-1,1)?
===
Subject: Re: bump function
>Bump function: B(x)=(e^2)*e*(1-x)*e(x+1)=e^4(1-x)(1+x)
> e:R->R by e(x)= e^(-1/x) if x>0
>0 if x<=0
>How is this function positive inside the interval (-1,1)?
Good grief!! I explained to you early on that the line B(x) is a bunch
of nonsense, and even gave you a clear statement of how your problem
should be stated. Why are you posting this now? If you don't
understand the previous posts, why don't you respond to them with
proper attribution and context?
--Lynn
===
Subject: Re: Pointwise Convergence (Re: Marble problem--put in 10 marbles, 
then remove 1)
>>Using your notation, you compute lim_{k->oo} f_k(j), and you assert
>>without proof that f, the indicator function of the marbles in the bag at
>>the end, is equal to lim_{k->oo} f_k.  It's the very same erroneous
>>conclusion.
> One of the purposes of modelling a physical process with mathematics is
> to use that mathematical model to make a prediction of what will happen
> in the physical process.
>| A: if a marble is taken out of the bag at some finite step and never
>|    returned, then it is outside the bag at the end
>| B: if a marble is placed into the bag at some finite step and never
>|    removed, then it is inside the bag at the end
  [ snip yet another repetition of points that have never been in dispute ]
>>Topology has nothing to do with it.  For one thing, when we are
>>discussing convergence of indicator functions, it makes absolutely no
>>difference whether we use the discrete topology or the metric topology.
>>They are the same on {0,1}.  The reason limits are irrelevant here has
>>nothing to do with whether they actually look like limits or not.  The
>>reason limits are irrelevant is simply that the stated problem does not
>>ask for the evaluation of a limit.  
> The stated problem asks for the contents of the bag at the end.
> Whether the word limit is actually used or not, the fact that the
> state at the end of an infinite process is being queried implicitly
> involves a limit.
Most definitely not, because limits sometimes give an indisputably wrong
answer.  Limits are quite simply the wrong approach to the problem.
> Suppose we are asked the following problem: given an empty bag, and
> for k = 1, 2, 3, ..., 1/k^2 ounces of sand are added to the bag at time
> 1-1/k.  How much sand is in the bag at time 1?  This problem does not
> ask explicitly for the evaluation of a limit; however, the problem is
> asking for the limit as n->oo of
>      n
>     ---  1
>     >   ---
>     --- k^2
>     k=1
This problem involves an infinite sum.  The marble problem does not.
Each marble is moved only finitely many times.
> I am not saying that this is the same as the original problem, since
> we are not taking anything out, but as far as limits are concerned,
> both problems involve them.
We are not told of the life history of individual grains of sand, and
this would be impossible, since there can only be finitely many of them
and therefore the 1/k^2 terms would soon reduce to a fraction of a grain.
>>The problem asks for f (in your notation), not for lim_{k->oo} f_k.  You
>>are assuming without proof that the two coincide.  Except when they
>>don't.
> I think my notation was f_oo, but f is more concise.  The problem does
> not tell how to determine anything but the values of f_k(j) for finite
> k.  Determining f with the LCM is modelled by taking the pointwise
> limit of the f_k.  Saying that marble j is not in the bag 'at the end'
> because it has been taken out at some step and never returned, is an
> application of the LCM.
The LCM allows us to determine f, not lim_{k->oo} f_k.
> I am trying to show that to model mathematically the physical 
situation
> of the marbles in the bag, one way is to use the topology of 
pointwise
> convergence and define the state at the end to be the limit of 
the
> finite states.  This model agrees with the hypothesized physical
> situation.  The same added premises are equivalent to those used in 
the
> arguments that supposedly don't use limits, to arrive at the state 
at
> the end.
I don't consider the state of the model when determining the final
position of ball k.  I only consider the position of ball k, which
changes only finitely many times and ends with ball k outside the bag for
each k.  You don't need limits when tracing a finite number of
transitions.  The LCM takes care of that.
>>I have understood from the beginning that you are trying to show exactly
>>that.  My point is that your efforts are doomed to failure, because what
>>you are trying to show simply is not so.
> Pointwise convergence of indicator functions does model the behavior of
> the marbles in the bag.
But only at times *before* the end.  We are asked about what happens *at*
the end.
>>Oh, you you don't just assume that f_k(t0) = lim_{t->t0-} f_k(t).  
You
>>define it, in the same way that David Ullrich defined the answer 
to
>>be 42.  I see.
> I do not.  What I am saying is that there is no a priori way, from the
> statement of the problem, to determine f_k(t0) mathematically without
> using some extra premises.  Simply because lim_{t->t0-} f_k(t) = 0 does
> not mean that f_k(t0) = 0.  Premises A and B allow us to say that the
> limit equals the value.
>>We have been over that.  Your premises A and B are acceptable.  The
>>trouble for your argument is, premises A and B do not address lim_{k->oo}
>>f_k at all.  They address f instead.  You are simply conflating the two,
>>when in fact we know that they are not always the same.
> Premises A and B imply that f(j) = lim_{k->oo}f_k(j) since f(j) tells
> which marbles are in/out of the bag at the end and lim_{k->oo}f_k(j)
> tells which marbles are put in/taken out and never removed/returned.
> Premises A and B say these are the same.
Except when they aren't.  If a marble is moved at the end, then premises
A and B say explicitly that the limit differs from the final value.  What
the problem asks for is the final value.  That's why the limits are
irrelevant.
-- 
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
===
Subject: Re: help on limits of continous functions
>>If you'd like a formula, consider something like f(x) = 
>>|cos(x)|^|x|.
> But that doesn't work!
But something like it does!
===
Subject: Re: help on limits of continous functions
On Wed, 02 Nov 2005 21:11:31 -0800, The World Wide Wade
>If you'd like a formula, consider something like f(x) = 
>|cos(x)|^|x|.
>> But that doesn't work!
>But something like it does!
I was actually making a little joke... never mind.
***
===
Subject: Re: Analysis
> Let f be a nonnegative measurable function, how do i show that there is
> an increasing sequence a_n of nonnegative simple functions each  of
> which vanishes outside the set of finite measure such that f = lim a_n.
> I need help please.
Let f be a nonnegative extended-real valued function.
Define f_n so that
f_n(x) = min{ 2^(-n) floor( 2^n f(x) ) , n  }for every x in [-n,n]
and f_n(x) = 0 for every x outside [-n,n].
Then f_n is an increasing sequence of nonnegative finitely-valued functions 

that converges pointwise to f.
If f is measurable, f_n are simple functions. 
===
Subject: Re: How many books does Alice have?
> I created this problem from x=639, so 639 is also a possible solution,
> I generalized
> that you may get lot of possible solutions with:
> x_i = x + nM, with n= ...,-2,-1,0,1,2,... and M=m_1*m_2*...._m_i=420
> so x_i = ..., 219, 639, 1059, 1479, 1899, ....
Yes, and all these solutions are congruent to 219 modulo 420.
That's what people mean when they said it is the only solution mod 420.
> I am not sure if this generalization is correct and I had doubts about
> why you said
> that 219 is the only solution mod 420, that last part I dont see clear.
Well, just like what I've said, it is implied by the CRT. 
===
Subject: Re: help on math notation
> and so in this case what is the meaning of u_0 ? it is the  function u
> with t =0 , is that ?
Yes.
===
Subject: proof is the inverse process of inference
you heard it here 1st!
proof(z)
(x^y->z) ^ (axiom(x) v proof(x)) ^ (axiom(y) v proof(y))
where modus ponens is a special case of x^y->z
x ^ (x->z) -> z
Like inference, proof too must lie outside the scope of theorems.
Herc
--
Pot ~ the intellectual drug :-)-~
how dangerous is 1 cigarette per week?
===
Subject: Re: proof is the inverse process of inference
Herc --
Proof being the inverse process of inference is interesting, but I
confess that I cannot follow your notation since I have not studied
symbolic logic other than very little.
As I understand modus ponens, it very simply says that if A then B, and
A exists, so B exists.
As I understand inference, it means that we cannot (or will not)
determine every case, but the pattern is towards a certain conclusion.
As I understand proof, it means that there can be no other conclusion.
Since inference falls short of a conclusive proof, is this all you mean
when you say that one is the inverse of the other?  Or, do you see some
deeper connection?  Can you explain it in English?
Very Respectfully,
Ray
===
Subject: Re: Help -- Similar to Fatou's Lemma
>The following is similar to Fatou's Lemma but different.  I'm unclear
>about what Integral f_n and Integral f have a meaning is supposed to
>be saying -- integrable is my guess but others have thought
>differently.  I could use some help with this proof.
>Let h be an integrable function and <f_n> a sequence of measurable
>functions with f_n >= -h and lim f_n = f.  Show that Integral f_n and
>Integral f have a meaning and that Integral f <= lower limit of the
>integral of f_n.
I'd say integral f has a meaning means either f is integrable or the 
integral is +infinity (or -infinity in general, but not in these cases).  
Certainly there's no reason to think that f_n is integrable here, 
because you only have a lower bound for it, not an upper bound.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
===
Subject: Re: Help -- Similar to Fatou's Lemma
   <dkchhb$lf9$1@nntp.itservices.ubc.caThe following is similar to Fatou's 
Lemma but different.  I'm unclear
>about what Integral f_n and Integral f have a meaning is supposed to
>be saying -- integrable is my guess but others have thought
>differently.  I could use some help with this proof.
>Let h be an integrable function and <f_n> a sequence of measurable
>functions with f_n >= -h and lim f_n = f.  Show that Integral f_n and
>Integral f have a meaning and that Integral f <= lower limit of the
>integral of f_n.
> I'd say integral f has a meaning means either f is integrable or the
> integral is +infinity (or -infinity in general, but not in these cases).
> Certainly there's no reason to think that f_n is integrable here,
> because you only have a lower bound for it, not an upper bound.
Couldn't f_n be bounded by h and -h since h >= - f_n  and -h <= f_n?
Since h is integrable then -h is also integrable and f_n is integrable.
 Perhaps I'm wrong.
Any ideas on how to prove that Integral f <= lower limit of the
integral of f_n?   I was thinking of the Lebesgue Dominated Convergence
theorem but it seems to follow almost directly from that if I'm not
mistaken.
TMH
===
Subject: joint&conditional prob
Good morning, i am a phd student in applied mathematics and i am
developing models for wireless networks;
actually i am facing the following probability problem which i show
you, i did not find any reasonable way to solve this since
unfortunately all references i found deal with conditional and joint
probability separately. Please, maybe you can give me some hint to
solve it or address me to someone who has already approached it. Please
note that in the following R' and R'' are two tx/rx nodes in a wireless
network and they all have a fixed radio range. Two nodes can hear each
other radio signals iff their radio coverage areas has not empty
intersection, of course.
Please refer to the following explanation:
1. given a certain curve C
1. given the height h' over C of the point R'
2. calculate the following probability:
P{D(R'')<=d, h(R'')<=h'' | h(R')=h'}
that is, we want to calculate the probability to find a point R'', such
that its orthogonal projection onto curve C has distance D(R'') less
than a certain value d, joint with the probability that its height
h(R'') over curve C is less than a certain value h''. This probability
is conditioned by the requirement that h(R')=h'.
I hope my explanation of the problem is clear enough. Otherwise, fell
free to ask me.
===
Subject: joint&conditional prob
Good morning, i am a phd student in applied mathematics and i am
developing models for wireless networks;
actually i am facing the following probability problem which i show
you, i did not find any reasonable way to solve this since
unfortunately all references i found deal with conditional and joint
probability separately. Please, maybe you can give me some hint to
solve it or address me to someone who has already approached it. Please
note that in the following R' and R'' are two tx/rx nodes in a wireless
network and they all have a fixed radio range. Two nodes can hear each
other radio signals iff their radio coverage areas has not empty
intersection, of course.
Please refer to the following explanation:
1. given a certain curve C
1. given the height h' over C of the point R'
2. calculate the following probability:
P{D(R'')<=d, h(R'')<=h'' | h(R')=h'}
that is, we want to calculate the probability to find a point R'', such
that its orthogonal projection onto curve C has distance D(R'') less
than a certain value d, joint with the probability that its height
h(R'') over curve C is less than a certain value h''. This probability
is conditioned by the requirement that h(R')=h'.
I hope my explanation of the problem is clear enough. Otherwise, fell
free to ask me.
===
Subject: joint&conditional prob
Good morning, i am a phd student in applied mathematics and i am
developing models for wireless networks;
actually i am facing the following probability problem which i show
you, i did not find any reasonable way to solve this since
unfortunately all references i found deal with conditional and joint
probability separately. Please, maybe you can give me some hint to
solve it or address me to someone who has already approached it. Please
note that in the following R' and R'' are two tx/rx nodes in a wireless
network and they all have a fixed radio range. Two nodes can hear each
other radio signals iff their radio coverage areas has not empty
intersection, of course.
Please refer to the following explanation:
1. given a certain curve C
1. given the height h' over C of the point R'
2. calculate the following probability:
P{D(R'')<=d, h(R'')<=h'' | h(R')=h'}
that is, we want to calculate the probability to find a point R'', such
that its orthogonal projection onto curve C has distance D(R'') less
than a certain value d, joint with the probability that its height
h(R'') over curve C is less than a certain value h''. This probability
is conditioned by the requirement that h(R')=h'.
I hope my explanation of the problem is clear enough. Otherwise, fell
free to ask me.
===
Subject: Re: sigma-algebra question
===
Subject: sigma-algebra question
> Let C subset R^n be a class of sets.
What do you mean?
that C is a collection of subsets of R^n as you
explained, namely C subset P(R^n).
> Is it true that  sigma(C times R) =  sigma(C) times R where
>  sigma(C times R) is the sigma-algebra generated by the rectangles
> A times R (A in C, R = real numbers) and  sigma(C) times R
> is the class of sets of the form A times R where A is in the sigma
> algebra generated by C? the left to right inclusion is clear because
>  sigma(C) times R is a sigma-algebra. It appears to me that the
> other inclusion should be obvious, but I can't prove it.
I'm confused what you mean by A times R means.
Usually A times R, ie AxR = { (x,y) | x in A, y in R }
Is that what you mean?  Or do you mean
        { AxB | A in C, B in P(R) }
ie
        { AxB | A in C, B subset R }
--
In general can one say for R subset P(X), S subset P(Y)
{ AxB | A in sigma R, B in sigma S } = sigma { AxB | A in R, B in S }
that the product of a sigma algebra over X and a sigma algebra over Y 
is
a sigma algebra over XxY ?
sigma A times sigma B = sigma (A times B)
--
BTW, you can omit all those  from TeX.
----
===
Subject: vector analysis
I have attempted a question and would like to know if what I've done is 
correct.
I have to evaluate the integral of F. dr (c lower limit) where F = (yz)i + 
(2y)j - (x^2)k and
i) C is the curve x=t, y=t^2, z=t^3   0<t<1 (less than or equal!)
My answer is as follows;
r= (x,y,z)
 = (t, t^2, t^3)
dr = (1,2t,3t^2) dt
F= ( (t^2), (2t), (-t^2))
so the integral of F. dr (limit 0 to 1)
= the integral (5t^2 -3t^4) dt
= (10/3)t - (3/5)t (lim 0 to 1)
therefore the answer equals 41/15?
Any help will be much appreciated if I have made a mistake on this 
question.
===
Subject: Re: vector analysis
F = (t^5, t^2, -t^2) ?
===
Subject: Re: vector analysis
> I have attempted a question and would like to know if what I've done is
> correct. I have to evaluate the integral of F. dr (c lower limit) where
> F = (yz)i + (2y)j - (x^2)k and i) C is the curve x=t, y=t^2, z=t^3 0<t<1
> (less than or equal!)
> My answer is as follows;
> r= (x,y,z)
>  = (t, t^2, t^3)
> dr = (1,2t,3t^2) dt
> F= ( (t^2), (2t), (-t^2))
Your evaulation of F is wrong.
> so the integral of F. dr (limit 0 to 1)
> = the integral (5t^2 -3t^4) dt
> = (10/3)t - (3/5)t (lim 0 to 1)
> therefore the answer equals 41/15?
> Any help will be much appreciated if I have made a mistake on this 
question.
===
Subject: Re: vector analysis
> I have attempted a question and would like to know if what I've done is 
correct.
> I have to evaluate the integral of F. dr (c lower limit) where F = (yz)i + 

(2y)j - (x^2)k and
> i) C is the curve x=t, y=t^2, z=t^3   0<t<1 (less than or equal!)
> My answer is as follows;
> r= (x,y,z)
>  = (t, t^2, t^3)
> dr = (1,2t,3t^2) dt
OK.
> F= ( (t^2), (2t), (-t^2))
Check your expressions for the components of F. For example,
the first component is yz where y=t^2 and z=t^3. How
did you get t^2 there?
===
Subject: Re: Analysis 2
On 2 Nov 2005 10:38:35 -0800, jennifer <scrilla_12_1999@yahoo.coma. I am 
trying to show that it is possible to have strict inequality in
>Fatou's Lemma.
>I started off by considering  a sequence f_n and defined it as
>follows:-
>f_n (x) =1 if n is less than or equal x, and x is less than  n+1,
>with f_n (x) = 0 otherwise
>This is what i have so far and i know fatou's lemma.
>Please help, where do i go from here?
The best place to go from here would be to the realization
that you should at least work on your homework a _little_
bit on your own, instead of immediately posting every
question to sci.math!
Really. The book or someone _gave_ you those f_n as a hint,
which means that you really haven't done _anything_ on
this problem. The reason this problem is especially
appropriate for a little lecture on how you need to
learn to do these things yourself is that it's _obvious_
where to go from here! Look:
You want to show that there can be strict inequality
in FL. So calculate both sides of the inequality for
these f_n, and note that the two things turn out to
be unequal.
(In calculating int(lim inf f_n) you first calculate
lim inf f_n and then integrate. In calculating
lim inf(int f_n) you first calculate int f_n and
then take the lim inf of the result...)
>b. I want to show also that the Monotone convergence theorem need not
>hold for decreasing sequences of functions.
>Do i start by defining a sequence of functions by f_n (x) = 0 if x<n
>and  f_n (x)= 1 for x>=n.
>If so, where do i go from here?
Find lim int f_n and int lim f_n?
***
===
Subject: Re: Paradox
   <dk87v4$5sbi@odds.stat.purdue.edu I hope Herman agrees that the paradox 
doesn't actually show we can get
> a contradiction in ZFC. I've shown why that doesn't go through. In that
> sense, the paradox is resolved.
Isn't your argument is circular? To show that the set S cannot be
defined you seem to presume that one cannot derive a condradiction in
ZFC.
===
Subject: Re: Paradox
>> I hope Herman agrees that the paradox doesn't actually show we can get
>> a contradiction in ZFC. I've shown why that doesn't go through. In that
>> sense, the paradox is resolved.
>Isn't your argument is circular? To show that the set S cannot be
>defined you seem to presume that one cannot derive a condradiction in
>ZFC.
If ZFC is not consistent, anything can be proved, so there is
no problem.  If is only if ZFC is consistent that there is a
problem.  Since any definition of the set yields a contradiction,
there cannot be such a definition.
Now if you can provide a definition within the language 
of ZFC, this would prove that ZFC is inconsistent.  The
claimed definition is not a definition in ZFC.
Goedel's result is that this type of self-reference cannot
be made in any language adequate for Peano arithmetic.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
===
Subject: derivatives and dependent variables
Consider the function
f(x + y, y + z) = exp(x + y) + exp(y + z)
where x, y, and z are independent. How do I compute
partial(f)/partial(x + y)? I obviously know how to
partial-differentiate the first summand, but how do I do the second
one:
partial exp(y + z)/partial (x + y)  ??
===
Subject: Re: derivatives and dependent variables
I'm not familiar with a standard defiition of partial/partial(x+y).  If
it's just the directional derivative, then its just partial/partial
x+partial/partial y.
The usual definition of convexity for multivariate functions is
f(cx+(1-c)y)<=cf(x)+(1-c)f(y) for vectors x and y and a real c with
0<=c<=1.
martin dowd
===
Subject: Re: derivatives and dependent variables
And as a follow-up question, does it ever make sense to talk about
convexity of such a function?
===
Subject: mathematical analysis
Question:
    lim(n*(exp(1/n)-1))^n = ?
I have spend much time on this problem, but even now I don't solve it.
Can you help me to solve it?
===
Subject: Re: mathematical analysis
> Question:
> lim(n*(exp(1/n)-1))^n = ?
> I have spend much time on this problem,
> but even now I don't solve it. Can you
> help me to solve it?
I'll assume you want the limit as n --> oo.
Let y = [ n*( exp(1/n) - 1 ) ]^n. We'll evaluate
limit as n --> oo of ln(y) and then use continuity
of the logarithm function to conclude that the
limit you want is exp[limit as n --> oo of ln(y)].
ln(y) = n*[ln(n) + ln( exp(1/n) - 1 )]
      = [ln(n) + ln( exp(1/n) - 1 )] / (1/n)
Now make the variable change m = 1/n, and note that
limit as n --> oo of ln(y) is equivalent to
limit as m -->0+  of ln(y). Then we have
ln(y) = [-ln(m) + ln( e^m - 1 )] / m
At this point, apply L'Hopital's rule by differentiating
in the appropriate manner. You'll get
        (e^m)/(e^m - 1) - (1/m)
      = [m*e^m - e^m + 1] / [m*(e^m - 1)].
Now use the standard series expansion for e^m,
do some algebra, keep only the dominate terms
in the numerator and the denominator, and
you'll be left with limit as m --> 0+ of
[m^2 - (1/2)*m^2] / m^2, which is 1/2. Hence,
the limit you want is exp(1/2).
A few years ago I posted several elementary ways
of evaluating limits in sci.math, and you might
find that post to be a useful reference. The post
but it's in the Math Forum sci.math archive at
http://mathforum.org/kb/message.jspa?messageID=225974
Dave L. Renfro
===
Subject: Re: mathematical analysis
>Question:
>lim(n*(exp(1/n)-1))^n = ?
>I have spend much time on this problem, but even now I >don't solve it.
>Can you help me to solve it?
(e^x-(1+x))/x^2 -> 1/2   (x->0) ---->
(e^(1/x)-(1+1/x))*x^2 = x^2*(e^(1/x)-1)-x -> 1/2 (x->oo)
---->
n^2*(e^(1/n)-1)-n -> 1/2 (n->oo)  ---->
(1+(n^2*(e^(1/n)-1)-n)/n)^n -> e^(1/2) (n->oo)
The last step uses the fact that if a_n -> a (n->oo),
then (1+a_n/n)^n -> e^a (n->oo).
Best wishes
Torsten.
===
Subject: Re: mathematical analysis
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
 VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
> Question:
>     lim(n*(exp(1/n)-1))^n = ?
> I have spend much time on this problem, but even now I don't solve it.
> Can you help me to solve it?
Taylor series.  exp(x) = 1 + x + x^2/2 + O(x^3).  You can probably
prove the result with other methods, but getting it in the first place
will be most convenient using Taylor.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: mathematical analysis
 <858xw5eucr.fsf@lola.goethe.zz Question:
>     lim(n*(exp(1/n)-1))^n = ?
> Taylor series.  exp(x) = 1 + x + x^2/2 + O(x^3).  You can probably
> prove the result with other methods, but getting it in the first place
> will be most convenient using Taylor.
sqr e ?
n.log(n(e^(1/n) - 1))
[e^(1/n) - 1] / (1/n)
e^(1/n) (-1/n^2) / (-1/n^2) = e^(1/n) -> 1
log(n(e^(1/n) - 1) / (1/n)
1/(n(e^(1/n) - 1) * [e^(1/n) - 1 + n.e^(1/n) (-1/n^2)] / (-1/n^2)
-n + e^(1/n) / (e^(1/n) - 1)
((1 - n)e^(1/n) + n) / (e^(1/n) - 1)
Yicks, does numerator -> 0 ?
===
Subject: Re: mathematical analysis
William Elliot <marsh@hevanet.remove.com> escribi.97:
> Question:
>     lim(n*(exp(1/n)-1))^n = ?
>> Taylor series.  exp(x) = 1 + x + x^2/2 + O(x^3).  You can probably
>> prove the result with other methods, but getting it in the first
>> place will be most convenient using Taylor.
> sqr e ?
e^(1/n) = 1 + 1/n + 1/(2n^2) + O(1/n^3)
L = Lim(n(e^(1/n) - 1)^n, n, inf)
Ln(L)  = Lim(n*Ln(n(1/n + 1/(2n^2) + O(1/n^2))), n, inf)
          = Lim(n*Ln(1 + 1/(2n) + O(1/n^2)), n, inf)
(As L(1 + x) ~ x, when  x --> 0)
          = Lim(n*(1/(2n) + O(1/n^2)), n, inf)
          = Lim(1/2 + O(1/n)), n, inf) = 1/2
Then, L = e^(1/2)
-- 
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com 
===
Subject: Re: Information on Nguyen-Widrow equation
   <dkaeo7$fvf$1@newsg3.svr.pol.co.uk>
Hi All:
Strange, tried again and found it,maybe  I should stick to slide rules
===
Subject: 4 points in plane with only integer distances
iNo 3 on a line, of course.
Evident solutions are a rectangle (e.g. 3x4) or rhombus,
but does a general solution exist? (It's just *another*
diophantine nightmare :-)
-- 
Hauke Reddmann <:-EX8    fc3a501@uni-hamburg.de
His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn
===
Subject: Re: 4 points in plane with only integer distances
> iNo 3 on a line, of course.
> Evident solutions are a rectangle (e.g. 3x4) or rhombus,
> but does a general solution exist? (It's just *another*
> diophantine nightmare :-)
problem is famous but unsolved, and it doesn't help
either as I also need all AREAS to be integer either
(which holds for the rectangle or rhombus example).
-- 
Hauke Reddmann <:-EX8    fc3a501@uni-hamburg.de
His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn
===
Subject: Re: Wronskian, linear algebra question
> you'll get an equation
> that should look like something in the determinant
> chapter of your linear algebra book...
But what if his linear algebra book was Linear Algebra Done Right and
has almost no mention of determinants?
-- 
G. A. Edgar                              
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Wronskian, linear algebra question
   <031120050805061074%edgar@math.ohio-state.edu.invalid you'll get an 
equation
> that should look like something in the determinant
> chapter of your linear algebra book...
> But what if his linear algebra book was Linear Algebra Done Right and
> has almost no mention of determinants?
Heh, maybe I should have said your multilinear algebra book.
(Is it possible to do *that* right?  I call dibs on the title.)
===
Subject: Re: Wronskian, linear algebra question
   <031120050805061074%edgar@math.ohio-state.edu.invalid you'll get an 
equation
> that should look like something in the determinant
> chapter of your linear algebra book...
> But what if his linear algebra book was Linear Algebra Done Right and
> has almost no mention of determinants?
> --
> G. A. Edgar                              
http://www.math.ohio-state.edu/~edgar/
Then they can use Proffesor Strang's lectures at the MIT OpenCourseWare
site.
http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/index.h

tm
===
Subject: Re: Wronskian, linear algebra question
   <12487993.1130982394716.JavaMail.jakarta@nitrogen.mathforum.org Let y_1, 
y_2, y_n be linearly independent functions in C^oo.
> For each y in C^oo, define T(y) in C^oo by
> [T(y)](t)=
> det [y(t) y_1(t) ... y_n(t)
>    y'(t) (y_1)'(t) ... (y_n)'(t)
>      .
>      .
>      .
>    (y^(n)(t) ((y_1)^n)(t) ((y_n)^n)(t)]
> Where y^n represents the nth derivative of y(t).
> 1) Prove that T:C^oo -> C^oo is a linear transformation.
> Now really, you expect us to do *all* your work
> for you? This problem may look daunting, but
> really it's practically trivial if you just write
> a few things down.
> Write down the definition of linear transformation,
> and then write down the equation (or equations) that
> T must satisfy to agree with that definition. If
> you even bother to do this, you'll get an equation
> that should look like something in the determinant
> chapter of your linear algebra book...
> To be a linear transformation, T must satisfy
> T(cx+y)=cT(x)+T(y).  But I don't understand what my T(x)
> and T(y) are... I guess I don't understand what my function is.
Heh, well y and x are themselves functions, maybe
that's what is confusing you.  T is a function
(as we will see, a linear transformation) taking
the C^oo function y to a different C^oo function T(y),
and similarly it takes x to T(x).
So, your job is to prove that the LHS of your eqn
T(cx+y) is the *same* function as the RHS cT(x)+T(y).
How do you prove that two functions are the same?
You show that no matter real number t is, the two
functions take it to the same value, i.e. you show
that [T(cx+y)](t) (a real number) has the same value
as [cT(x)+T(y)](t).  And by the way, the latter is
by definition the same as c*[T(x)](t) + [T(y)](t) so
of course it is a real number too.
In other words, the way to proceed is just to write
it all out with cx+y as the argument to T, and dink
around with it until it rearranges to a sum of the 2
determinants c*[T(x)](t) and [T(y)](t).
   [T(cx+y)](t) = det [(cx+y)(t) y_1(t) ... y_n(t)
                       (cx+y)'(t) y_1'(t) ...
                          ...
                       (cx+y)^(n)(t) y_1(n)(t) ...]
          = det [c*x(t)+y(t) y_1(t) ... y_n(t)
                 c*x'(t)+y'(t) y_1'(t) ... y_n'(t)
                      ...
                 c*x^(n)(t)+y^(n)(t) y_1^(n)(t) ...]
                 ^^^^^^^^^^^^^^^^^^^
                 note all this is the 1st column
where I have used the chain rule and other properties
of derivatives to expand the first column -- you need
to verify this.  And from here we can just use the
property of determinants that they are linear in each
column (in particular the first column).
> 2) Prove that N(T) contains span{y_1, y_2, ... , y_n}.
> This is also easy. You can easily prove that
> T takes each of those vectors to the zero function;
> just write down [T(y_1)](t) for example, and look
> at what's special about the first two columns on
> the RHS.
> Once again, I really don't get this... I know that the
> N(T)= {x in C^oo: T(x)=0}.  But after this I don't
> understand how to show something is in the null space, or
> that is spans something else.
You are given that y_1, ..., y_n spans a certain
subspace of C^oo, so you don't have to prove it;
your job is to prove that T maps any vector in
that subspace to the zero function.
If you prove that each of the y_1, ..., y_n get
mapped to the zero function, then any linear
combination of them must map to zero as well,
right?  If you can't see this, write down some
arbitrary v = v1y_1 + ... + vny_n and see what
T maps it to.
===
Subject: Re: Wronskian, linear algebra question
   <12487993.1130982394716.JavaMail.jakarta@nitrogen.mathforum.org Heh, well 
y and x are themselves functions,
In other words, your vectors here are functions;
the vector space C^oo is a space of functions.
Hopefully I did not confuse you by sometimes
calling x,y, y_1, etc. functions and at other
It's good to remember that vector spaces sometimes
are more abstract than perhaps we are used to
dealing with in Physics 101.  But all the rules
are the same -- so, if you're confused, pretend
that you are not and everything will be OK.  ;-)
===
Subject: Re: Wronskian, linear algebra question
   <12487993.1130982394716.JavaMail.jakarta@nitrogen.mathforum.org>
define T(y) in C^oo by
> [T(y)](t)=
> det [y(t) y_1(t) ... y_n(t)
>    y'(t) (y_1)'(t) ... (y_n)'(t)
>      .
>      .
>      .
>    (y^(n)(t) ((y_1)^n)(t) ((y_n)^n)(t)]
> Where y^n represents the nth derivative of y(t).
Please take note of this definition for [T(y)].
> To be a linear transformation, T must satisfy
> T(cx+y)=cT(x)+T(y).  But I don't understand what my T(x)
> and T(y) are... I guess I don't understand what my function is.
Which should answer your question on what the function is.
===
Subject: Re: Ancient Remainder Arithmetic
Hi all,
Since David and his mythical clone have disappeared,
no longer wishing to defend their classical Greek positions, may I return to 

reading and reporting the Egyptian texts that were created before and after 

1500 BCE. All created exact remainder when dividing weights and measures 
units - with ease. 
That is, the mythical classical position that Egyptian
scribes used only an additive form of multiplicaiton,
and scribal division was an inverse process thereto
is 100% WRONG. Reading over 40 examples of volume
unit division (per a hekat) recorded in two easy to
find texts, The Rhind Mathematical Papyrus - with 
at least 35 examples, and the parent Akhmim Wooden
Tablet - 2000 BCE - directly explain every step to the
interested reader.  Scribes did the following:
(64/64)/n = Q/64 + R/(n*64)
when n was less than 64. 
In addition, a constant remainder common divisor, 1/320,
was created for all n less than 64, for ease of granary
and other workers.
Q/64 as a Horus-Eye, binary series - that was
uneffected by the 1/320 common divisor. For example
n = 3 crated Q = 21 or (16 + 4 + 1)/64 = 1/4 1/16 1/64
with the remainder term being altered to with the
1/64 term being replaced by its equal 5/320, or
(5*R/n)*1/320
with (5*R/n) being written as an Egyptian fraction
series, and with 1/320 being written as a word = ro.
Example, continuing with n = 3, R = 1 or the scribe
(5*1/3)* 1/320 =(5/3)*1/3 = (1 2/3)*1/320
Finally, only in the AWT did the scribe prove his
work, that step consisting of taking his final
Q/64 + (5*R/n)*ro (answer)
and multiplying both parts of the answer by n,
such that
(Q/64)*n returned to 63/64 when n = 3
and,
(5/3)*1/320 times 3 = 5/320 = 1/64 or
63/64 + 1/64 = 64/64, or the beginning hekat unity form.
Interesting? Try another n, say 13, where
Q = 4 and R = 12 and see the beauty of the 
method. (hint 4/64 = 1/4 + 
        (5*R/n) = 60/13 = 4 + 8/13 (yes vulgar fractions
were easily handed
                 = 4 + 1/2 + (2 + 1)/(2*13)
                 = 4 + 1/2 + 1/13 + 1/26
or (64/64)/13 = 1/4 + (4 1/2 1/13 1/26)*1/320)
and so forth for any n less than 64.
    
All positive comments will be responded to in a
timely manner.
Milo Gardner
===
Subject: Calculator with mod function?
remainder of a division).
Could you please suggest me one?
===
Subject: Re: Calculator with mod function?
> remainder of a division).
> Could you please suggest me one?
TI-89 & Voyage 200 have it, as mod(n1,n2). They also have fpart(n) and
ipart(n) for the integer and fractional parts of a number.
--
john
===
Subject: Periodic homeomorphisms in Hilbert space
Let H be a Hilbert space. Does there exist a homeomorphism f:H-->H of period 

3 (i.e. f(f(f(x)))=x for all x in H) without fixed points?
===
Subject: Re: Periodic homeomorphisms in Hilbert space
On Thu, 03 Nov 2005 08:51:57 EST, Asikhit Pshenavan <funcan@mail.ruLet H be 
a Hilbert space. Does there exist a homeomorphism f:H-->H of 
period 3 (i.e. f(f(f(x)))=x for all x in H) without fixed points?
Yes, assuming that a curious fact I heard from a usually reliable
source here on sci.math is actually true:
If X is an infinite-dimensional Banach space then X is
homeomorphic to its unit _sphere_ S = {x in X : ||x|| = 1}.
So consider f:S -> S defined by f(x) = wx where w is a third
root of unity.
===
Subject: Re: Real Tetration Solution
> What I mean by order is the y in x^^y, and what I mean by
> non-oscillating (for the n-th derivative), is that
> (the n-th derivative of x^^y with respect to y) > 0 for all x > 1
> and all y > 0. The (x > 1) restriction is there because even x^^y
> itself is oscillating for 0 < x < 1.
> By saying that x^^y is infinitely differentiable with respect to y,
> I mean that even a piecewise-defined function (where each piece is
> defined for when k < y <= k+1 for integer k), will have continuous
> n-th derivatives with respect to y for all real y, all natural number n.
OK, I see what you are saying. Yeah, the second construct in my paper is
oscillating. (I haven't checked the first). That's only natural, 
because
I've used bump functions which are C^{oo}, and then defined the tetration
function as Int(sum(bump),0,x). Therefore, the derivative would be exactly
the underlying constructor bump function at the specified point, which is
C^{oo}. The derivative is actually 0 at the naturals. So
d^p/dy^p[(e^z)^^y]|_{n=0}, if n is a natural.
> When these two conditions are combined, (when x^^y is defined by an
> infinitely-differentiable piecewise-defined function with non-oscillating
> n-th derivatives with respect to y for all natural number n),
> I beleive that this describes a unique extension of tetration.
Actually, constructing a non-oscillating tetration extension is not such 
a
big deal. I will give you one to play with, and compare with your results:
First of all, table 1 on page 12 of my paper, gives all the correct
expansions of (e^z)^^y for natural y=m in N. This is independent of any
additional results or extensions you or I might have, so whatever extension
you construct, has to agree with the values obtained when you evaluate the
extension for y=m in N using the series in table 1. If it doesn't agree,
don't waste your time.
Now, you can choose a plethora of methods for example, to interpolate
between those values. In my second example in the aforementioned paper, I
used a bump function interpolation, thus the result. You can instead choose
simple Lagrangian interpolation.
Fix n in the series expansion of (e^z)^^m, m in N in table I. Take the 
first
m pairs {[j, a_{j,n}]}, j in {0,1,2,...m-1} from the vertical column for
that n in that table and get a Lagrange polynomial P_{m,n}(x) of degree m,
which passes through these points.
Trivially these polynomials are C^{oo}, (polynomials are always C^{oo}),
P_{m,n}(x) are all bounded in [0,m-1] (since for fixed n, a_{m,n} is
eventually constant for all m in N) and additionally, their derivatives 
will
be non-oscillating if you increase m (and thus the degree) enough and if 
you
additionally supply one or two points to the left of the first pair
{0,a_{0,n}}, to flatten any possible oscillations at the start.
Again, for fixed n in N, the coefficients in this table eventually become
constant, so for such n, if you increase m, the sequence of the polynomials
{P_{0,n}(x),P_{1,n}(x),...,P_{m-1,n}(x)}, will converge in [0,m-1] to a
bounded series S_n(x) in [0,m-1].
The series S_n(x) interpolates in a C^{oo} and in a non-oscillating way
between all the coefficients a_{m,n}, for all m in N in table 1, in the
interval [0,m-1].
Now define for y > 0, (e^z)^^y = Sum(S_n(y)*z^n,n=0..oo).
When y = m in N, then the previous reduces to (e^z)^^m =
Sum(S_n(m)*z^n,n=0..oo) = Sum(a_{m,n}*z^n,n=0..oo), which is the correct
expansion for (e^z)^^m, in table 1, for y = m.
So there's another non-oscillating tetration extension. Clearly this is
C^{oo} and additionally, the derivative will be monotone increasing for all
y>0, since there exists m in N (sufficiently large), where
d^p/dy^p[(e^z)^^y] > 0.
By using the above extension with sufficient accuracy, you can extract
another value of e^^Pi, for example.
> I'm now trying to find if there is one and only one
> non-oscillating extension, and if there is one and only one,
Hmmm. What do you think are the chances you've got the only 
non-oscillating
extension, with me giving you this one, above, just right off the top of 
my
head? :-) The above extension is just *one* way of doing it. You could have
used for example a Spline interpolation or a Pade approximation on the
coefficients a_{m,n}, m in N. If you manage to prove that any/all these
possible extensions (via interpolation or otherwise), converge to something
unique, and if you additionally prove that *your* extension is 
equivalent
to *that* unique extension, then you might be on to something.
> Andrew Robbins
-- 
I. N. Galidakis
http://users.forthnet.gr/ath/jgal/
Eventually, _everything_ is understandable
===
Subject: Re: Real Tetration Solution
I've tried playing around with series expansions in z of (e^z)^^y, and I was 

very impressed by your recurrence relation (a and alpha in your paper: 
Extensions.pdf). Once I understood the recurrence relation, I found that 
instead of defining the recurence relation 'a' and the real number version 
of 
the recurence relation 'alpha' seperately, you can define them all at the 
same time, and this allows for the simplest termwise extension I've 
found, expressed by:
(e^x)^^y == sum[k=0..00] x^k*alpha_k(y)
alpha_0(y) == 1 if y >= 0
alpha_k(y) == y^k/k! if 0 <= y <= 1
alpha_k(y) == sum[j=1..k] if y > 1
    (j/k)*alpha_(k-j)(y)*alpha_(j-1)(y-1)
(I might have forgotten part of the definition, but I think thats correct) 
Anyways, by using the recurrence relation, you can ensure that the bivariate 

function you get by defining is this way f(x, y) == f(e^z, y) still has the 

property f(x, y+1) == x^f(x, y). But from my research, if I remember 
correctly, If you just interpolate the terms of this expansion at a fixed k 

for real y, any way you want, then its possible to come up with a solution 
where f(x, n) == x^^n for integer n, but the property f(x, y+1) == x^f(x, y) 

DOESNT hold for all real y. And in my mind, in order for a continuous 
extension of tetration to be valid, it should still obey the property f(x, 
y+1) == x^f(x, y), not just coinside with integer tetration.
Have you determined whether your series extension obeys that property for 
real y?
Another thing of note: using Lagrange interpolating polynomials for 
generating 'alpha' for fixed k, is extremely oscilating if you try and 
connect terms where y > k, because they go from changing alot, to no change 

at all. And I have a feeling this oscillation in the terms will produce an 
oscillation in the series.
Andrew Robbins
===
Subject: Re: Real Tetration Solution
> I've tried playing around with series expansions in z of (e^z)^^y,
> and I was very impressed by your recurrence relation
> (a and alpha in your paper: Extensions.pdf).
> Once I understood the recurrence relation, I found that instead
> of defining the recurence relation 'a' and the real number version
> of the recurence relation 'alpha' seperately, you can define them
> all at the same time, and this allows for the simplest termwise
extension
> I've found, expressed by:
> (e^x)^^y == sum[k=0..00] x^k*alpha_k(y)
> alpha_0(y) == 1 if y >= 0
> alpha_k(y) == y^k/k! if 0 <= y <= 1
> alpha_k(y) == sum[j=1..k] if y > 1
>     (j/k)*alpha_(k-j)(y)*alpha_(j-1)(y-1)
> (I might have forgotten part of the definition, but I think thats 
correct)
> Anyways, by using the recurrence relation, you can ensure that the
> bivariate function you get by defining is this way f(x, y) == f(e^z, y)
> still has the property f(x, y+1) == x^f(x, y). But from my research,
> if I remember correctly, If you just interpolate the terms of this
expansion at
> a fixed k for real y, any way you want, then its possible to come up with
a
> solution where f(x, n) == x^^n for integer n, but the property f(x, y+1)
== x^f(x, y)
> DOESNT hold for all real y.
Correct.
> And in my mind, in order for a continuous extension of tetration to be
valid,
> it should still obey the property f(x, y+1) == x^f(x, y),
> not just coinside with integer tetration.
Well, there's no standard definition about what _exactly_ constitutes a
tetration extension, so I'd think that both versions (extensions satisfying
this functional equation vs not at the non-naturals) could be considered. 
Is
an extension that satisfies this functional equation at all positive points
more interesting than one which doesn't? Then the first construct in 
the
paper is more interesting than the second C^{oo} one. So in order to
improve upon the situation, you'd have to find something which satisfies
this functional equation for all y>0, which is ALSO C^{oo} and which is 
ALSO
non-oscilating. Have you got anything like that perhaps?
> Have you determined whether your series extension obeys that property for
real y?
The first construct does. The second obeys it only at natural
hyperexponents, but see my reply to Gerald.
> Another thing of note: using Lagrange interpolating polynomials for
generating 'alpha'
> for fixed k, is extremely oscilating if you try and connect terms where y
> k, because
> they go from changing alot, to no change at all. And I have a feeling 
this
oscillation
>  in the terms will produce an oscillation in the series.
Well, if you are trying to find e^^Pi, for example, I don't see why it
matters what the approximating series S_n(x) does OUTSIDE the interval
[0,m-1]. The functional variation of S_n(x) outside [0,m-1] would affect 
the
value S_n(Pi) very little, as your m gets larger.
So, if you can add points left of the first pair to flatten the initial
oscillation so that the initial pair is dealt with, the rest of the
interpolants will be very smooth, because as I said, as m increases, the
coefficients eventually stabilize to the same value. So you'd be ok in the
interval [0,m-1], m in N, which is what we are interested in.
The approximating resultant series S_n(x) will be very smooth throughout
[0,m-1]. Again, I don't see why we should care about what S_n(x) does
outside [0,m-1], because we are interested for what happens at y=Pi, for
example, which is where S_n(x) will be extremely smooth, when m gets large
enough.
> Andrew Robbins
-- 
I. N. Galidakis
http://users.forthnet.gr/ath/jgal/
Eventually, _everything_ is understandable
===
Subject: Re: Real Tetration Solution
I think, if you want to call it tetration, then more is required
than merely interpolating the integer values.  I think
you want something like this:
A nice function F(s,a,x) such that:
  F(0,a,x) = x
  F(1,a,x) = a^x
  F(s+t,a,x) = F(s,a,F(t,a,x))
Then define a^^b = F(b,a,1)
Intuitively, F(s,a,x) is the iterate of order s of the function a^x,
so if s is an integer it is a^a^a^...^a^x with s copies of a.
Of course, merely interpolating F(s,a,x) from integer s to real s
will not usually preserve the crucial functional equation
F(s+t,a,x) = F(s,a,F(t,a,x)) .
-- 
G. A. Edgar                              
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Real Tetration Solution
Actually, I found a really interesting formula to connect that function with 

2-argument tetration, and it involves the tetralog function 'slog'. First of 

all using your notation: x^^y == F(y,x,1) yes? the 1 being the top-most 
exponent in the powertower/tetration. There are other notations online that 

use x^^y@1 for this, or in general F(y, x, z) == x^^y@z. I call this 'z' the 

auxilliary, but its possible to do without it. First we must know that we 
can add different y's with composed functions of F. so:
F(u, x, F(v, x, z)) == F(u+v, x, z)
or equivalently:
x^^u@(x^^v@z) == x^^(u+v)@z
OK, now we need to know the value of slog_x(z), which I can do now with my 
method of solving for 'slog'. The slog function can be defined by the 
relation:
x^^(slog_x(z)) == z == F(slog_x(z), x, 1)
Using this relationship, your iterated exponential function F can now be 
written in a different form:
F(y, x, z) == F(y, x, F(slog_x(z), x, 1))
F(y, x, z) == F(y + slog_x(z), x, 1)
F(y, x, z) == x^^[y + slog_x(z)]
Using the additive property from above, and converting to 2-argument 
tetration from 3-argument tetration, we can write this as:
x^^y@z == x^^[y + slog_x(z)]
Andrew Robbins
===
Subject: Re: Real Tetration Solution
> I think, if you want to call it tetration, then more is required
> than merely interpolating the integer values.  I think
> you want something like this:
> A nice function F(s,a,x) such that:
>   F(0,a,x) = x
>   F(1,a,x) = a^x
>   F(s+t,a,x) = F(s,a,F(t,a,x))
> Then define a^^b = F(b,a,1)
> Intuitively, F(s,a,x) is the iterate of order s of the function a^x,
> so if s is an integer it is a^a^a^...^a^x with s copies of a.
Well, the first construct, in my paper, satisfies all that and preserves
this functional equation for all positive hyperexponents. It is not C^{oo},
however, so I don't see how your comment applies. If it did, it would imply
that a simple linear fit extension such as this extension, is more
interesting than a C^{oo} one.
> Of course, merely interpolating F(s,a,x) from integer s to real s
> will not usually preserve the crucial functional equation
> F(s+t,a,x) = F(s,a,F(t,a,x)) .
The construct I gave to Mr. Robbins in the previous post, preserves this
functional equation at the naturals, since there, it preserves the correct
expansions for natural hyperexponents. For the inbetween values, I haven't
checked, but one could conceivably add constraints to the interpolation so
that the construct interpolates the values at the naturals AND satisfies
this functional equation at selected points inbetween. I expect this to be 
a
little messy, but I cannot see why it cannot be done in principle.
Now, if Mr. Robbins has found something that preserves this functional
equation for ALL positive hyperexponents AND is C^{oo} AND is
non-oscillating, that'd be interesting. But I don't quite see how one
could prove that this construct would satisfy this functional equation for
all positive hyperexponents, if one is working only with numerical
approximations.
Perhaps he can show us some more details?
> -- 
> G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
-- 
I. N. Galidakis
http://users.forthnet.gr/ath/jgal/
Eventually, _everything_ is understandable
===
Subject: Re: y'=yz/x^2, z'=2(1-y)
> I HAVE NOT solved this:
> y'=yz/x^2
> z'=2(1-y)
> I have proved that if
> y=f(x) , z=g(x)
> is a soloution, then
> y=f(x/b) z=b.g(x/b)
> is a soloution.
> AND y=1 , z=0 is a soloution.
Does the dash refer to differentiation WRT x or t ?
(If it is WRT x then differentiating the first WRT y and the second WRT
z
and plugging the originals back in the results eliminates all
derivatives
and should give a solution in principle.)
===
Subject: RE: y'=yz/x^2, z'=2(1-y)
y AND z are functions of x and y'=dy/dx , z'=dz/dx
THIS equation come from a physical problem.
> y'=yz/x^2
> z'=2(1-y)
Does the dash refer to differentiation WRT x or t ?
(If it is WRT x then differentiating the first WRT y and the second WRT
z
and plugging the originals back in the results eliminates all
derivatives
and should give a solution in principle.)
===
Subject: Re: y'=yz/x^2, z'=2(1-y)
   <1Asaf.149876$dP1.507673@newsc.telia.net>
[Nils, it is easier for readers to follow the discussion
if you put your replies after the text you are replying to.
I've changed the following to how it should look]
> y'=yz/x^2
> z'=2(1-y)
 Does the dash refer to differentiation WRT x or t ?
> y AND z are functions of x and y'=dy/dx , z'=dz/dx
> THIS equation come from a physical problem.
Drat - I found what I think is a general solution for
dy/dt = yz/x^2 and dz/dt = 2(1-y). But your version
is tougher. Here's the solution to the dt version:
Given dy/dt = yz/x^2 and dz/dt = 2(1-y)
The first is equivalent to the following for some f(t):
  y/x,  z/x  =  f.dy/dt,  1/f
and differentiating the second of these WRT t gives:
  1 - y  =  d/dt(x/f) / 2
         =  (f.dx/dt - x.df/dt) / 2f^2
to which adding y = x.f.dy/dt gives:
  2.f^2  =  f.dx/dt - x.df/dt + 2.x.f^3.dy/dt
         =  f.dx/dt + x.(2.f^3.dy/df - 1).df/dt
So taking:
     2.f^3.dy/df - 1  =  1
<=>  y = C - 1/(2.f^2)                  [1.1]
we obtain:
   d(xf)/dt  =  2.f^2
<=>  x = 2.(int(f^2).dt + D) / f        [1.2]
and from z = x/f:
     z = 2.(inf(f^2).dt + D) / f^2      [1.3]
Plugging these back in y/x = f.dy/dt gives:
  f.(C - 1/(2.f^2))/(int(f^2).dt + D) = 2.f.(1/f^3).df/dt
<=> (C.f^3  - f/2) = 2.(int(f^2).dt + D).df/dt
                       = 2.f^3/3 + 2D.df/dt
which gives finally:
   t = 2D.int( 1 / ((C-2).f^3 - f/2)).df + E
(Note that C, D, E are integration constants.)
I suppose you could try equating the resulting
expression for t to x = 2.(int((f^2.(dt/df)).df + D) / f
noting that dt/df can be expressed as a function
of f via C.f^3 - f/2) = ... above.
John R Ramsden  (jhnrmsdn@yahoo.com.uk)
 * Remove m from com to reply
===
Subject: Re: y'=yz/x^2, z'=2(1-y)
>I HAVE NOT solved this:
>y'=yz/x^2
>z'=2(1-y)
>I have proved that if
>y=f(x) , z=g(x)
>is a soloution, then
>y=f(x/b) z=b.g(x/b)
>is a soloution.
>AND y=1 , z=0 is a soloution.
So are you asking us to solve it?
I doubt that you'll get a closed-form solution, though Maple
does give an implicit solution involving the integral of
the solution of a first-order differential equation:
if b'(t) = (2 t - t^2) b(t)^3 + (1 - t) b(t)^2,
B(t) = int b(t) dt,  ln(x) = B(z/x), and y = 1 - z'/2, 
then y and z are a solution.  
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
  
===
Subject: Re: Standard Deviation of PISA
>According to YOUR lexicon, NOT ours, our very own Founding Forefathers,
> Yep.
>The white, European character of the United States was enshrined in
>law. 
> Note that white in American law has always included Jews, Arabs,
> Persians, and even Asian Indians.  Nothing about Europe in it.
>A few localities recognized
>free blacks as citizens of states, but the Supreme Court ruled in 1857
>that no black, slave or free, could be a citizen of the United States.
>Blacks did gain U.S. citizenship under the post-Civil War amendments,
>but other races did not.
> The 14th amendment gave citizenship to anyone born here regardless of
> race.
>The ban on immigration and naturalization of Chinese, established in
>1882, continued until 1943. It was only when the United States found
>itself allied with China in the Second World War that Congress repealed
>the Chinese exclusion laws-but not by much. It set an annual quota of
>105 Chinese. Needless to say, it permitted no immigration from Japan.
> All those Japanese-Americans who were interned during WW II can
> testify to the contrary.  
>Until 1965, the United States had a national origins immigration
>policy designed explicitly to keep the country white.
> Then we grew up.
> The evolutionary trend has been more or less constant against your
> racist xenophobia.  You've lost, loser.
John obviously did not write the body of that post.  Not only 
is it stylistically not his, there are too many accurate historical
facts.
-- cary
===
Subject: Re: Standard Deviation of PISA
> DURN!
> You know what we forgot to put on the nigger poll, Gray?
Sure: only white trash uses words like nigger.
> http://christianparty.net/pollblacks.htm
> jews too
Oh, okay.
White trash tends to be collectivist in nature with little actual 
difference between the lunatic fringe of the so-called left and 
the lunatic fringe of the  so-called right.
Someone who has such terrible ego problems and the knowledge [or 
suspicion] of his own inferiority should seek help because he 
has a poor - or no - grip on reality.
> He forgot his email address, but remembered what we forgot, which is
> that there should have been at LEAST one question about you despicable
> jews.
You're still speaking the language of your native planet of Gibber. 
You don't do a very good job with the English language
 
> But, don't worry, Gray.  Since you jews are one third nigger,
There are those who say that Imitation is the sincerest form of 
flattery but, as far as I'm concerned, you're just Rex Currying 
me.
you repeat it, you're just reinforcing the opinion of most of us 
that you don't have a sense of inferiority but that you're 
actually inferior.
 > this ought to cover you, too.
Oh, peckerhead, you're out of your league. You should go back to 
your fellow alcoholics and drunkards on the primrose path. You're 
not sharp enough to play in the grown-up areas of UseNet.
Let's see if I can remember the groups you try and convince people 
that you are superior because you trash groups that you're scared 
to death are superior to you.
Women, blacks, Jews, Muslims, Christians (I'll be happy to strike 
that as soon as you can produce a couple of quotes from your 
psychopathetic babblings that show you to emulate Jesus), men 
[rather than poor, pathetic boys like you who have never grownup], 
politicians who you don't like, the corner grocer and any other 
decent person.
> John Knight
> ps--THERE ARE A LOT OF FOLKS OUT THERE WHO AGREE 100% WITH THIS POLL,
> GRAY!!!
President Bush43 got re-elected so I'm not too strong on you 
drooling  with A LOT OF FOLKS OUT THERE WHO AGREE 100%.
You seem to be very impressed with numbers that you can't provide 
data on. So, now you state a POLL has all these answers but you 
don't provide the questions (you really are an idiot).
Was this a poll taken on your website hosted by the one-a-month 
club and the Richard Nixon hot dog stand, airline discount ticket 
weirdo and public latrine?
The reason most people laugh at you is that you're a joke and a 
funny little almost man.
Not even a good try, El Flako,
Gray Shockley
--------------------------
Swinehood hath no remedy.
                  - Sidney Lanier
> race:              White
> sex:               Man
> age:               49
===
> Subject:           race
> SubjectOther:
> Username:
> UserEmail:
> UserTel:
> UserFAX:
> ContactRequested:
> question1:         Strongly Agree
> question2:         Strongly Agree
> onepercent:        Strongly Agree
> zimbabwe:          Strongly Agree
> la:                Strongly Agree
> justicestats:      Strongly Agree
> hispanics:         Strongly Agree
> nd:                Strongly Agree
> povertyb:          Strongly Agree
> povertyh:          Strongly Agree
> twentiethcentury:  Strongly Agree
> twentyfirst:       Strongly Agree
> clearance:         Strongly Agree
> drugs:             Strongly Agree
> wardrugs:          Strongly Agree
> guncontrol:        Strongly Agree
> second:            Strongly Agree
> stp:               Strongly Agree
> taxes:             Strongly Agree
> taxesh:            Strongly Agree
> savings:           Strongly Agree
> slaves:            Strongly Agree
> prison:            Strongly Agree
> potential:         Strongly Agree
> desire:            Strongly Agree
> loss:              Strongly Agree
> aids:              Agree
> life:              Strongly Agree
> income:            Strongly Agree
> math:              Strongly Agree
> dc:                Strongly Agree
> education:         Strongly Agree
> ancestor:          Strongly Agree
> solomon:           Strongly Agree
> sin:               Strongly Agree
> beasts:            Strongly Agree
> liberia:           Strongly Agree
> ps:                Strongly Agree
> cuttaxes:          Strongly Agree
> reducehomicide:    Strongly Agree
> increaseclearance: Strongly Agree
> raisescores:       Strongly Agree
> speech:            Strongly Agree
> repealgunlaws:     Strongly Agree
> reducedrugs:       Strongly Agree
> autotheft:         Strongly Agree
> rape:              Strongly Agree
> assault:           Strongly Agree
> robbery:           Strongly Agree
> drugabuse:         Strongly Agree
> dui:               Strongly Agree
> theft:             Strongly Agree
> lawyers:           Strongly Agree
> divorce:           Strongly Agree
> illegitimacy:      Strongly Agree
> illegitmacy:
> inmates:           Strongly Agree
> garden:            Strongly Agree
> discriminated:     Strongly Agree
> repatriation:      Strongly Agree
> betterenvironment: Strongly Agree
> twohundredk:       Strongly Agree
> thirtyfivek:       Strongly Agree
> exile:             Strongly Agree
> Time:              12:05:33 PM
> Remote Name:       ....164.198.19
> Remote User:
> HTTP User Agent:   Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1;
> SV1; .NET CLR 1.1.4322)
> Comments:
> jews too
===
Subject: Re: Standard Deviation of PISA
> Country avg IQ
You just kinda forgot your citations. Were you drunk again (still)?
Not very well qualified in arithmetic, are ya, BubbleButt?
It's pretty obvious where Ken Who Is Not a Ten is picking up bad 
habits. [I just hope that's all he's picking up from you.]
Oh, by the way, ScatBoy, I haven't seen the examples I'm sure that 
you've collected that prove your statement that:
           > Just like jews like Gray Shockley 
           > who can't write three sentences 
           > without engaging in scatology, 
Now, JohnnieBoy, you just need to produce some instances of this.
Heck, SonnyBoy, you can just start with one example although to 
prove your point, you're going to have to prove I do it in every 
message I write.
           > Just like jews like Gray Shockley 
           > who can't write three sentences 
           > without engaging in scatology, 
Mebbe you can get your new best friend, Kenneth CLifton, to help 
you research UseNet (that's what this is called, SonnyBoy, I 
realize you have certain problems which limit your understanding 
of reality and sanity but, gee whiz, if we can't make fun of white 
trash like you and Clifton and Tommy Tupper Topaz, the place 
could get boring.
So, Jonnie It's going to be a bad Knight, let's have your first 
installment of my posts where:
           > Just like jews like Gray Shockley 
           > who can't write three sentences 
           > without engaging in scatology, 
I realize that I shouldn't make fun of lower animals like you; 
after all, back then at your birth, no one even knew what a crack 
baby     was.
But you're the porker who stated:
           > Just like jews like Gray Shockley 
           > who can't write three sentences 
           > without engaging in scatology, 
and, now, just about everybody (with the exceptions of those 
closest to you - both of them) thinks of you as a sociopathetic 
liar and a worthless failure.
Now you hear, JohnnieBoy, now doncha forget to show all those 
examples of:
           > Just like jews like Gray Shockley 
           > who can't write three sentences 
           > without engaging in scatology, 
or even those two will turn on you and you'll be saying, Drop 
kick me, Jesus, through the goalposts of life.
Gray Shockley
------------------
Bring on the brothers, who've gone on before
And all of the sisters, who've knocked on your door
All the departed, dear, loved ones of mine
Stick 'em up front in the offensive line.
 - Paul Craft
Sung by: Bobby Bare
You say you want more?
I saw a garage sale 
Pulled up in the yard 
Found a statue of Jesus 
It was eight feet tall 
He held out his arms 
And he seemed all alone 
So I loaded him up 
And I drove him home 
Out by my driveway he 
Looks down the street 
With long hair and sandals made 
Of rebar and concrete 
I painted him white with a long purple robe 
He's a rock of ages on our gravel road 
Chorus: 
He's an eight hundred pound Jesus 
Standing taller than a tree 
He's an eight hundred pound Jesus 
A bigger man than you or me 
I thought losin' my job was 
The end of the world 
'Til my best pal ran off with my best girl 
I felt suicidal with no real friends 
So I walked outside with a rope in my hand 
Out by that statue there's a big oak tree 
So I stood on his shoulders 
And I counted to three 
I had every intention of buying the farm 
But when I jumped off 
He caught me in his arms
I wanted to return the favor to him 
'Cause I never had a more solid friend 
So I planted some flowers 
All around his feet 
And I bought him a flock 
Of ceramic sheep.
He's a bigger man than you or me . 
There! [Sawyer Brown lyrics, btw]
> Hong Kong 107
> South Korea 106
> Japan 105
> Taiwan 104
> Singapore 103
> Austria 103
> Germany 102
> Netherlands 102
> Italy 101
> New Zealand 101
> Sweden 101
> Switzerland 101
> United Kingdom 101
> Belgium 100
> China 100
> Quote:
> Denmark 100
> Lichtenstein 100
> Norway 100
> Finland 99
> Hungary 99
> Poland 99
> Australia 99
> France 99
> Ireland 99
> Canada 98
> Slovakia 98
> Czech Republic 97
> Romania 97
> Slovenia 97
> Spain 97
> United States 97
> Argentina 96
> Russia 96
> Uruguay 96
> Portugal 95
> Israel 94
> Bulgaria 93
> Georgia 92
> Greece 92
> Malaysia 92
> Thailand 91
> Croatia. 90
> Peru 90
> Turkey 90
> Armenia 89
> Colombia 89
> Indonesia 89
> Azerbaijan 88
> Brazil 87
> Iraq 87
> Mexico 87
> Western Samoa 87
> Tonga 87
> Lebanon 86
> Philippines 86
> Cuba 85
> Morocco 85
> Fiji 84
> Iran 84
> Puerto Rico 84
> Egypt 83
> India 81
> Ecuador 80
> Guatemala 79
> Barbados 78
> Nepal 78
> Qatar 78
> Zambia 77
> Congo 73
> Uganda 73
> Jamaica 72
> Kenya. 72
> South Africa 72
> Sudan 72
> Tanzania 72
> Ghana. 71
> Nigeria 67
> Guinea 66
> Zimbabwe 66
> Congo 65
> Sierra Leone 64
> Ethiopia 63
> Equatorial Guinea 59
> The above is actually good news for American Whites, because there's
> zero evidence that I'm aware of that American blacks who're 13% of
> our population score even a point higher than their racial cousins back
> in Ghana, or that American Mexicans who're 10% of the population
> score any higher than their racial cousins back in Guatemala. We really
> can't compare the Mexicans here to the ones in Mexico, because all the
> good ones stayed behind in Mexico.
> (.77 x X) + (.13 x 71) + (.10 x 79) = 97
> .77X = 97 - 9.23 - 7.9 = 79.87
> X = 104
> What this list says is (not taking into account the 1% Indians, the 2%
> Arabs, the 2% STUPID jews, and the 3% white Hispanics) that American
> Whites have an IQ of 104, on par with Singapore, Taiwan, Japan, and
> Korea, and only 3 points lower than Hong Kong.
> hmm, better re-calculate this taking into account those other muds, eh?
> Let's do this assuming that white Hispanics, who're racial cousins to
> the jews who were expelled from Spain in 1492, Arabs, Indians, and
> jews, who're 8% of the population, have a collective IQ of 80:
> (.69 x X) + (.08 x 80) + (.13 x 71) + (.10 x 79) = 97
> .69X + 6.4 + 9.23 + 7.9 = 97
> .69X = 73.47
> X = 106.5
> There, that's better!
> John Knight
> ps--cary, do you think American niggers score closer to Guinea than to
> Zimbabwe?  Which do you think most accurately reflects nigger IQ?
===
Subject: Re: Standard Deviation of PISA
   <0h5em19o8h7raieldeg31it78acuc451v5@4ax.com>
   <dk8hne$f5t$1@onion.ccit.arizona.edu>
   <0001HW.BF8EECA900447EF6F0284550@news.giganews.com Country avg IQ
> You just kinda forgot your citations. Were you drunk again (still)?
> Not very well qualified in arithmetic, are ya, BubbleButt?
He still thinks the average of log10 and log100 is log55.
===
Subject: Re: Standard Deviation of PISA
   <dk8hed$e9e$1@onion.ccit.arizona.edu>
   <dkal94$25h$1@onion.ccit.arizona.edu>
Nobody commented on the following post from the woman who voted to
exile the niggers, and it deserves to be posted again, because it's
dead nuts 100% correct.
Why does she have such a view?  Because she READ, and UNDERSTOOD, the
Holy Bible.
Can just ANYBODY who reads the Holy Bible UNDERSTAND it?  Absolutely
not.  As she aptly points out in this post, evangelizing to niggers is
worse than talking to trees.  There ain't a nigger, a latrino, an
Indian, a Hispanic, a Mexican, a Puerto Rican, a Cuban, nor any other
muds or mamzers, who can EVER understand it like she does:
I believe that the caucasian race has been the only race to respond
consistently to the gospel message and salvation by the millions for
centuries.
I also believe that our race is intellectually superior than any other
races on earth. The white race has contributed
medically,technologically,and economically to the benefit of MANKIND
and the BEASTS of various nationalities(referenced towards other races
throughout scripture).No other race can make this claim(if they do WE
have contributed to their knowledge in various areas).
I believe that the caucasian race are Yahweh Gods chosen Israelite
peoples. Furthermore I believe that we have fullfilled all prophecy and
continue to do so in a positive way.
I know the European nations have evangelized worldwide to all the other
racial lands to no avail.There is not even the hint of a christian
nation(other than white) in spite of numerous efforts throughout the
centuries. Most of the martyrs are of caucasians descent.
Yahshua(Jesus) came ONLY for the Israelites. The Israelites were the
only ones made in Yahwehs image.
The bible hints at the idea of there being other pre Adamic races on
earth before Adam was created in Yahwehs image(no other race was).
The OT and NT message and salvation was ONLY given to the House of
Judah and the house of Israel.
There are ONLY 12 gates into the Kingdom of heaven and the New
Jerusalem holy city promise land of the new covenant.NOONE other than
the israelites that accepted Yahshuas message and salvation will be
able to enter in through the gates labeled with the names of the tribes
of Israel. No other gates exist.
I know and Yahweh has been revealing His truth to me for several months
now.
The ancient meaning of gentile is mistranslated. The gentile are the
heathen israelites of the house of Israel that were separated from the
House of Judah in dispersion and captivities. Judah continued in the
law but the house of Israel practiced heathen ways and disregarded it.
The Messiah Yahshua(Jesus) came to earth to reconcile the House of
Israel and Judah back to Himself and break down the middle wall of
partition between them and destroying the enmity between them. Gentiles
are NOT non israelites. The word gentile never really existed in the
hebrew(which was goyim) and is mistranslated(in my belief for reasons
of control).
Noone but an israelite in the FLESH has an opportunity for salvation.
1st Adam was made in Yahwehs image and so was the 2nd come to us to
reddeem those that were of the first Adam.
Keep up the good work speaking the truth in spite of opposition. he
time for reality and TRUTH is NOW!!
Love and hugs in Yahshua,
Her post is 100% correct.  But there's one ADDITION which needs to be
made to it:
<<<I know the European nations have evangelized worldwide to all the
other
racial lands to no avail.There is not even the hint of a christian
nation(other than white) in spite of numerous efforts throughout the
centuries. Most of the martyrs are of caucasians descent.>
This paragraph is 100% correct, but it's not complete, because those
who evangelize anyone other than an Israelite OF THE HOUSE OF ISRAEL is
in violation of God's Law, Jesus' commandment, and all common sense.
They do not have the authority to toss out the following:
These twelve Jesus sent forth, and commanded them, saying, Go not into
the way of the Gentiles, and into any city of the Samaritans enter ye
not: But go rather to the lost sheep of the house of Israel. And  as ye
go, preach, saying, The kingdom of heaven is at hand.  Matthew 10:5-9
John Knight
===
Subject: Re: Standard Deviation of PISA
> Nobody commented on the following post from the woman who voted to
> exile the niggers, and it deserves to be posted again, because it's
> dead nuts 100% correct.
> Why does she have such a view?  Because she READ, and UNDERSTOOD, the
> Holy Bible.
> Can just ANYBODY who reads the Holy Bible UNDERSTAND it?  Absolutely
> not.  
Ah, I love the sound of self-reference in the morning...
-- cary
===
Subject: Re: Standard Deviation of PISA
>> Nobody commented on the following post from the woman who voted to
>> exile the niggers, and it deserves to be posted again, because it's
>> dead nuts 100% correct.
>> Why does she have such a view?  Because she READ, and UNDERSTOOD, the
>> Holy Bible.
>> Can just ANYBODY who reads the Holy Bible UNDERSTAND it?  Absolutely
>> not.  
> Ah, I love the sound of self-reference in the morning...
> -- cary
I was more enthused with Jake's, because 
it's dead nuts 100% correct.
The planet of Gibber sure turns out some wackos and the Border 
Patrol has a rough time coping with /all/ the illegal aliens, not 
just Jake the Wack and Kenneth, the apparent psychopath.
Gray Shockley
--------------------------
Now my own suspicion is that the 
Universe is not only queerer than 
we suppose, but queerer than 
we can suppose.
                       - J. B. S. Haldane
===
Subject: Re: Standard Deviation of PISA
>Nobody commented on the following post from the woman who voted to
>exile the niggers,
Nobody but nincompoops like you care what other racist nincompoops
write.
And no it was not a post from her - it was a post from you quoting
her, probably without her permission.
>Why does she have such a view?
Because she, if she really exists, is a racist nincompoop just like
you.
-- 
lojbab                                             lojbab@lojban.org
Bob LeChevalier, Founder, The Logical Language Group
(Opinions are my own; I do not speak for the organization.)
Artificial language Loglan/Lojban:                 http://www.lojban.org 
===
Subject: Re: Standard Deviation of PISA
   <dk8hed$e9e$1@onion.ccit.arizona.edu>
   <dkal94$25h$1@onion.ccit.arizona.edu>
In the New Testament, however, we find a refreshing breath
of anti-racism:
    Where there is neither Greek nor Jew, circumcision nor
    uncircumcision, Barbarian, Scythian, {...}
Of course that's from Paul, and we are all to aware of your
strong ambivalence about this self-proclaimed Jew of Tarsus.
Paul didn't have the authority to change God's everlasting covenant.
Wherever on Earth do you even THINK he would try to do that?  Why would
you HATE Paul so much that you would claim that he would even WANT to
change God's covenant?
If you KNEW that Paul was referring ONLY to the House of Israel [read:
Paul referred to the ISRAELITES of the House of Israel living in Judaea
as Judaeans, or jews, and he referred to ISRAELITES of the House of
Israel living in Greece as Greeks.
If you persist in your belief that Paul would change God's Law or God's
covenant like this--YOU MUST TOSS PAUL OUT.
http://christianparty.net/weiland.htm
But he did NOT do that.
If you think Paul had the authority to ignore what JESUS said about the
jews, that they're children of the devil, and march off by himself
and just invite these SCUMBAGS into the covenant, or the Israelite
community, or the church, then you think that what you think Paul
said has more authority than what you KNOW Jesus said.
John Knight
===
Subject: Re: Standard Deviation of PISA
> In the New Testament, however, we find a refreshing breath
> of anti-racism:
>     Where there is neither Greek nor Jew, circumcision nor
>     uncircumcision, Barbarian, Scythian, {...}
> Of course that's from Paul, and we are all to aware of your
> strong ambivalence about this self-proclaimed Jew of Tarsus.
> Paul didn't have the authority to change God's everlasting covenant.
> Wherever on Earth do you even THINK he would try to do that?  Why would
> you HATE Paul so much that you would claim that he would even WANT to
> change God's covenant?
Hey, it's not like God, your version of, doesn't change His mind
all the time anyhow.  One day eating oysters is an ABOMINATION,
the next day the only thing you have to watch out for is months 
with `R's in them.  One day blood sacrifices are absolutely
required, the next day they're out.  God creates Man.
Then He regrets creating Man, so he wipes Man out, nearly.  Next 
thing you know He's all regretful He did THAT too, so He promises
He'll never do it again.  Well, not in exactly the same
manner, anyhow.
He the Lord your God is a fickle God.
It's all moot anyhow: if you're an Israelite, then I'm a Seraphim.
-- cary
===
Subject: Re: Standard Deviation of PISA
>Paul didn't have the authority to change God's everlasting covenant.
So what?
>Wherever on Earth do you even THINK he would try to do that?
Why not?
>Why would
>you HATE Paul so much that you would claim that he would even WANT to
>change God's covenant?
I don't think that he much cared.
>If you KNEW that Paul was referring ONLY to the House of Israel
He wasn't, so I could not know such a thing.
>[read:
>Paul referred to the ISRAELITES of the House of Israel living in Judaea
>as Judaeans, or jews, and he referred to ISRAELITES of the House of
>Israel living in Greece as Greeks.
In other words, he was talking about Jews and Greeks, and only in your
Cowardly New World Newspeak was he talking about ISRAELITES of the
House of Israel living anywhere.
>If you persist in your belief that Paul would change God's Law or God's
>covenant like this--YOU MUST TOSS PAUL OUT.
I have no problem with tossing out that which is wrong.  Nor did
Thomas Jefferson.
lojbab
-- 
lojbab                                             lojbab@lojban.org
Bob LeChevalier, Founder, The Logical Language Group
(Opinions are my own; I do not speak for the organization.)
Artificial language Loglan/Lojban:                 http://www.lojban.org 
===
Subject: Re: Standard Deviation of PISA
>Paul didn't have the authority to change God's everlasting covenant.
> So what?
>Wherever on Earth do you even THINK he would try to do that?
> Why not?
>Why would
>you HATE Paul so much that you would claim that he would even WANT to
>change God's covenant?
> I don't think that he much cared.
>If you KNEW that Paul was referring ONLY to the House of Israel
> He wasn't, so I could not know such a thing.
>[read:
>Paul referred to the ISRAELITES of the House of Israel living in Judaea
>as Judaeans, or jews, and he referred to ISRAELITES of the House of
>Israel living in Greece as Greeks.
> In other words, he was talking about Jews and Greeks, and only in your
> Cowardly New World Newspeak was he talking about ISRAELITES of the
> House of Israel living anywhere.
>If you persist in your belief that Paul would change God's Law or God's
>covenant like this--YOU MUST TOSS PAUL OUT.
> I have no problem with tossing out that which is wrong.  Nor did
> Thomas Jefferson.
Do they still have Teejs' razor, I wonder?  It would make a catchy
museum exhibit:
    This is the razor with which Thomas Jefferson (1743-1826,
    Governor of Virginia, third Preident of the United States),
    edited his Bible, removing the miracles, many of the sayings 
    attributed to Jesus, all of the writings of Paul, and the 
    entire Old Testament.  Copies of the resulting book, sometimes
    known as The Jefferson Bible, may be purchased in
    our gift shop.
    
    
    
-- cary
===
Subject: Re: Standard Deviation of PISA
> Do they still have Teejs' razor, I wonder?  It would make a catchy
> museum exhibit:
>     This is the razor with which Thomas Jefferson (1743-1826,
>     Governor of Virginia, third Preident of the United States),
>     edited his Bible, removing the miracles, many of the sayings 
>     attributed to Jesus, all of the writings of Paul, and the 
>     entire Old Testament.  Copies of the resulting book, sometimes
>     known as The Jefferson Bible, may be purchased in
>     our gift shop.
>     
>     
>     
> -- cary
The Gift Shop is not open during lunchtime but you are cordially 
invited to dine in the Sally Hemmings Dining Room until the Gift 
Shop opens.
  ++ gray
===
Subject: Re: Standard Deviation of PISA
> Do they still have Teejs' razor, I wonder?  It would make a catchy
> museum exhibit:
> 
>     This is the razor with which Thomas Jefferson (1743-1826,
>     Governor of Virginia, third Preident of the United States),
>     edited his Bible, removing the miracles, many of the sayings 
>     attributed to Jesus, all of the writings of Paul, and the 
>     entire Old Testament.  Copies of the resulting book, sometimes
>     known as The Jefferson Bible, may be purchased in
>     our gift shop.
>     
>     
>     
> -- cary
> The Gift Shop is not open during lunchtime but you are cordially 
> invited to dine in the Sally Hemmings Dining Room until the Gift 
> Shop opens.
I thought that was the Sally Hemmings Boutique?  Formerly
known as Sally's Secret?
-- cary
===
Subject: Re: Standard Deviation of PISA
> Do they still have Teejs' razor, I wonder?  It would make a catchy
> museum exhibit:
> 
> This is the razor with which Thomas Jefferson (1743-1826,
> Governor of Virginia, third Preident of the United States),
> edited his Bible, removing the miracles, many of the sayings 
> attributed to Jesus, all of the writings of Paul, and the 
> entire Old Testament.  Copies of the resulting book, sometimes
> known as The Jefferson Bible, may be purchased in
> our gift shop.
> 
> 
> 
> -- cary
>> The Gift Shop is not open during lunchtime but you are cordially 
>> invited to dine in the Sally Hemmings Dining Room until the Gift 
>> Shop opens.
> I thought that was the Sally Hemmings Boutique?  Formerly
> known as Sally's Secret?
> -- cary
Microsoft bough out the whole thing.
Fortunately, Microsoft opens windows, 
Gray Shockley
-------------------
And Apple Opens Doors
===
Subject: Re: Standard Deviation of PISA
   <dkd6ao$e0k$1@onion.ccit.arizona.edu>
   <0001HW.BF8F8A2A0005071AF0284550@news.giganews.com>
   <dkdboi$1pr$1@onion.ccit.arizona.edu>
   <0001HW.BF8F91F40006DA40F0284550@news.giganews.com> The Gift Shop is not 
open during lunchtime but you are cordially
>> invited to dine in the Sally Hemmings Dining Room until the Gift
>> Shop opens.
> I thought that was the Sally Hemmings Boutique?  Formerly
> known as Sally's Secret?
> -- cary
Microsoft bough out the whole thing.
Fortunately, Microsoft opens windows,
Gray Shockley
You jews have been trying to discredit Mr. Jefferson for almost three
centuries, claiming that he would not have found sex with a nigger any
more repugnant than James Garfield did:
<<<Andrew Johnson felt the same way: This is a country for white men,
government for white men . . . . James Garfield certainly agreed.
repugnance when I think of the negro being made our political equal and
I would be glad if they could be colonized, sent to heaven, or got rid
of in any decent way . . . .
To a White man like Mr. Jefferson, about the only thought more
repugnant than sex with a nigger would be sex with a jewess, most of
whom had sex with faggots like Dirshowitz (if Dirshowitz isn't too
feeble to have sex, that is).
lobmamzer claims that the percentage of the population of DC who're
niggers is only 60%.  The percentage of nigger children who took NAEP
was 84%.  So to calculate the murderous tendencies of niggers, we can
use both figures to figure out how high the homicide rate of niggers
has to be for DC to have attained the coveted spot as the Murder
Capitol of the World in 1991, with a record high 80.6 homicides per
100k population:
SIXTY PERCENT NIGGERS
.6X x (.1 x 20 homicide rate of latrinos) + (.04 x .05 homicide rate of
chinks) + (.26 x 1 homicide rate of honkies) = 80.6
.6X = 78.338
X = 130.6 = homicide rate of niggers IF niggers are only 60% of DC's
population
EIGHTY THREE PERCENT NIGGERS
.83X + (.1 x 20 homicide rate of latrinos) + (.03 x .05 homicide rate
of chinks) + (.04 x 1 homicide rate of honkies) = 80.6
.83X = 78.558
X = 94.6 = homicide rate of niggers IF niggers are 83% of DC's
population
Which figure do you jews think is correct, Gray?  Do  you think
American niggers murder each other and lots of others at a rate of
only 95 per 100,000 population (about four TIMES the rate at which
South African niggers murder each other and lots of others), or do you
think  they murder each other (and lots of others) at a rate of 131, on
par with Swaziland?
http://christianparty.net/murdrateworld.htm
Or is today not your day to think, Gray?
John Knight
http://christianparty.net/pollblacks.htm is going extremely well, btw.
===
Subject: Re: Standard Deviation of PISA
>> The Gift Shop is not open during lunchtime but you are cordially
>> invited to dine in the Sally Hemmings Dining Room until the Gift
>> Shop opens.
> I thought that was the Sally Hemmings Boutique?  Formerly
> known as Sally's Secret?
> -- cary
> Microsoft bough out the whole thing.
> Fortunately, Microsoft opens windows,
> Gray Shockley
> You jews have been trying to discredit Mr. Jefferson for almost three
> centuries, 
Whoa, hey, finally ... f i n a l l y ... I get to be a Jew too.
`bout damn time John.  I was trired of dusting the space I'd set aside for 
that honor in my trophy case of Unearned Yet Cherished Internet 
Accolades.  But the wait was worth all, and I shall wear
the badge proudly, if undeservedly.
And my college girlfriend would be delighted.  
-- cary
===
Subject: Re: Standard Deviation of PISA
   <0h5em19o8h7raieldeg31it78acuc451v5@4ax.com>
   <dk8hed$e9e$1@onion.ccit.arizona.edu>
   <69tgm15eqo5sqcu9sf7iav9lqj2drklqhh@4ax.com>
For 6,000 years, the CONCEPT of racist, or anti-Semite, or 
sexist
didn't exist in ANY dictionary.  For a millennia, it didn't exist in
the English language until it was ADDED by RACISTS who were mere
GUESTS, whose ticket has been PUNCHED, who, according to EIGHTY FIVE
PERCENT (85%) of those who've taken the poll demand that they must now
LEAVE.
http://christianparty.net/liberia.htm
According to YOUR lexicon, NOT ours, our very own Founding Forefathers,
Jesus, and God were racists:
Abraham Lincoln also favored colonization. He was the first President
ever to invite a delegation of blacks officially to visit the White
House; he held the meeting to ask them to persuade their people to
leave. Even in the midst of a desperate war with the Confederacy,
Lincoln found time to study the problem of black colonization, and to
appoint Rev. James Mitchell as Commissioner of Emigration.
His successor Andrew Johnson felt the same way: This is a country for
be a government for white men . . . . James Garfield certainly agreed.
repugnance when I think of the negro being made our political equal and
I would be glad if they could be colonized, sent to heaven, or got rid
of in any decent way . . . .
What of 20th century Presidents? Theodore Roosevelt thought blacks were
a perfectly stupid race, and blamed Southerners for bringing them to
solution to the terrible problem offered by the presence of the Negro
on this continent . . . he is here and can neither be killed nor driven
away . . . . As for Indians, he once said, I don't go so far as to
think that the only good Indians are the dead Indians, but I believe
nine out of ten are, and I shouldn't inquire too closely into the
health of the tenth.
Woodrow Wilson was a confirmed segregationist, and as president of
Princeton prevented blacks from enrolling. He enforced segregation in
government offices and was supported in this by Charles Eliot,
president of Harvard, who argued that civilized white men could not
be expected to work with barbarous black men. During the Presidential
campaign of 1912, Wilson campaigned to keep Asians out of the country:
I stand for the national policy of exclusion. . . . We cannot make a
homogeneous population of a people who do not blend with the Caucasian
race. . . . Oriental coolieism will give us another race problem to
solve and surely we have had our lesson.
Henry Cabot Lodge took the view that there is a limit to the capacity
of any race for assimilating and elevating an inferior race, and when
you begin to pour in unlimited numbers of people of alien or lower
races of less social efficiency and less moral force, you are running
the most frightful risk that any people can run.
Harry Truman is remembered for integrating the armed services by
executive order, but in his private correspondence was as much a
separatist as Jefferson: I am strongly of the opinion Negroes ought to
be in Africa, yellow men in Asia and white men in Europe and America.
As recent a President as Dwight Eisenhower argued that although it
might be necessary to grant blacks certain political rights, this did
not mean social equality or that a Negro should court my daughter. It
is only with John Kennedy that we finally find a President whose public
pronouncements on race begin to be acceptable by today's standards
(although he made virtually no effort to end segregation).
I have quoted politicians because they are cautious people who
recirculate the bromides of their times. Mark Twain, who never sought
desirable subject for extermination if ever there was one. Jack London
explained that part of the appeal of socialism was that it was devised
so as to give more strength to these certain kindred favored races so
that they may survive and inherit the earth to the extinction of the
lesser, weaker races.
Samuel Gompers, probably the most famous labor leader in American
believe in unrestricted immigration want this country Chinaized. But I
firmly believe that there are too many right-thinking people in our
country to permit such an evil. He went on to add, It must be clear
to every thinking man and woman that while there is hardly a single
reason for the admission of Asiatics, there are hundreds of good and
strong reasons for their absolute exclusion.
The white, European character of the United States was enshrined in
law. The first naturalization bill, passed in 1790, made citizenship
available only to free white persons. A few localities recognized
free blacks as citizens of states, but the Supreme Court ruled in 1857
that no black, slave or free, could be a citizen of the United States.
Blacks did gain U.S. citizenship under the post-Civil War amendments,
but other races did not. State and federal laws excluded Asians, and in
1914 the Supreme Court upheld the principle that citizenship could be
denied to foreign-born Asians.
The ban on immigration and naturalization of Chinese, established in
1882, continued until 1943. It was only when the United States found
itself allied with China in the Second World War that Congress repealed
the Chinese exclusion laws-but not by much. It set an annual quota of
105 Chinese. Needless to say, it permitted no immigration from Japan.
Until 1965, the United States had a national origins immigration
policy designed explicitly to keep the country white.
Their desires WILL be fulfilled.
John Knight
===
Subject: Re: Standard Deviation of PISA
>According to YOUR lexicon, NOT ours, our very own Founding Forefathers,
Yep.
>The white, European character of the United States was enshrined in
>law. 
Note that white in American law has always included Jews, Arabs,
Persians, and even Asian Indians.  Nothing about Europe in it.
>A few localities recognized
>free blacks as citizens of states, but the Supreme Court ruled in 1857
>that no black, slave or free, could be a citizen of the United States.
>Blacks did gain U.S. citizenship under the post-Civil War amendments,
>but other races did not.
The 14th amendment gave citizenship to anyone born here regardless of
race.
>The ban on immigration and naturalization of Chinese, established in
>1882, continued until 1943. It was only when the United States found
>itself allied with China in the Second World War that Congress repealed
>the Chinese exclusion laws-but not by much. It set an annual quota of
>105 Chinese. Needless to say, it permitted no immigration from Japan.
All those Japanese-Americans who were interned during WW II can
testify to the contrary.  
>Until 1965, the United States had a national origins immigration
>policy designed explicitly to keep the country white.
Then we grew up.
The evolutionary trend has been more or less constant against your
racist xenophobia.  You've lost, loser.
lojbab
-- 
lojbab                                             lojbab@lojban.org
Bob LeChevalier, Founder, The Logical Language Group
(Opinions are my own; I do not speak for the organization.)
Artificial language Loglan/Lojban:                 http://www.lojban.org 
===
Subject: Re: Help with Fibonacci & Co
   <3105785.1130692275805.JavaMail.jakarta@nitrogen.mathforum.org But when 
speaking of the 2000th digit, it is inappropriate to round.
> What about your other big numbers? Are they rounded also?
David C Ullrich had the right idea. It is all a bit of fun, and with
his comments he had me chuckling all day.
Yes, I make big numbers - but only for fun. My various researches (such
as into data-compression, fractional factorials &c.) often deliver a
natural constant.
When the page has been written, I often relax with a little bit of
machine-code, knowing that it is of no great importance. The root-two
program is given below.
As I produce the natural constant, I also ponder whether there is any
aspect of the work that I have omitted. If not, I add the big number as
a joke, and declare the work to be done.
For the number Phi, one simply takes the square-root program, and in
place of 2 one uses one-and-a-quarter.
When the program is redirected to a text file, the result will be
1.118033989. A spot of text-editing turns it into 1.618033989, which is
Phi. I then crop the thing to two thousand places, and round up the
last if the 2001st was 5 or more.
But it's no big deal.
So those who speculated that it was done by Newton's method were
correct.
Regarding Pi, my page on floating point describes much of what one
needs to know, for the arctan.
Pi divided by 4 is 1 - 1/3 + 1/5 - 1/7 and so on.
The convergence is poor.
However, if one finds the arctan of 22.5 degrees it is easier. Tan 22.5
is root-two minus 1.
Root-two minus 1 can be translated into 0.198912367 (Tan 11.25
degrees).
That can be halved again.
So it is possible to use Pi/ (2 to the power 42) or so, and get twenty
or more decimal places at a time.
Charles Douglas Wehner
****************
offset    equ    256
buffer    equ    2048
buffsize  equ    833
ascii0    equ    48
printchar equ    512
dot       equ    46
places    equ    2002
top1      equ    buffer+buffsize-1
buffer2   equ    buffer+buffsize
top2      equ    buffer2+buffsize-1
buffer3   equ    buffer2+buffsize
top3      equ    buffer3+buffsize-1
call firstguess
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call newrap
call readout
int 32
newrap:
call octave
call division
call addguess
call halves
call transfer
ret
firstguess:
   mov di,buffer
   mov cx,buffsize-2
   cld
   xor al,al
   repz
   stosb
   mov al,6ah
   stosb
   mov al,1
   stosb
ret
octave:
   mov di,buffer2
   mov cx,buffsize+buffsize-1
   cld
   xor al,al
   repz
   stosb
   mov al,2
   stosb
ret
transfer:
   mov si,buffer2
   mov di,buffer
   mov cx,buffsize
   cld
   transloop:
      lodsb
      stosb
   loop transloop
ret
moveup:
   mov si,buffer2
   mov di,buffer3
   mov cx,buffsize
   cld
   moveloop:
      lodsb
      stosb
   loop moveloop
ret
addguess:
   mov si,buffer
   mov di,buffer2
   mov dx,buffer2
   xor ax,ax
   xor bx,bx
   mov cx,buffsize
   cld
   addloop:
      xor ah,ah
      lodsb
      add bx,ax
      push si
      mov si,dx
      lodsb
      mov dx,si
      pop si
      add ax,bx
      stosb
      mov bl,ah
      xor bh,bh
   loop addloop
ret
halves:
   mov si,top2
   mov di,top2
   xor ax,ax
   mov bx,2
   mov cx,buffsize
   std
   halfloop:
      lodsb
      div bl
      stosb
   loop halfloop
ret
double:
   mov si,buffer2
   mov di,buffer2
   mov cx,buffsize+buffsize
   cld
   doublloop:
      lodsb
      adc al,al
      stosb
   loop doublloop
ret
subtract?:
   mov si,buffer
   mov di,buffer3
   mov dx,buffer3
   xor ax,ax
   mov cx,buffsize
   cld
   subloop:
      lodsb
      mov bl,al
      push si
      mov si,dx
      lodsb
      mov dx,si
      pop si
      sbb al,bl
      stosb
   loop subloop
   jc payback
stc
ret
   payback:
   mov si,buffer
   mov di,buffer3
   mov dx,buffer3
   xor ax,ax
   mov cx,buffsize
   cld
   payloop:
      lodsb
      mov bl,al
      push si
      mov si,dx
      lodsb
      mov dx,si
      pop si
      adc al,bl
      stosb
   loop payloop
clc
ret
divbit:
   call subtract?
   call double
ret
division:
   xor al,al
   mov [buffer2],al
   mov cx,buffsize-1
   divloop:
      push cx
      call divbit
      call divbit
      call divbit
      call divbit
      call divbit
      call divbit
      call divbit
      call divbit
      pop cx
   loop divloop
      call divbit
ret
readout:
   mov si,top1
   cld
   lodsb
   add al,ascii0
   mov dl,al
   mov ax,printchar
   int 33
   mov dl,dot
   mov ax,printchar
   int 33
   mov di,top1
   xor al,al
   cld
   stosb
   mov cx,places
   numloop:
      push cx
      mov si,buffer
      mov di,buffer
      xor ax,ax
      xor bx,bx
      mov cx,buffsize
      xor dx,dx
      cld
      tensloop:
          lodsb
          mov bx,ax
          add ax,ax
          add ax,ax
          add ax,bx
          add ax,ax
          add ax,dx
          stosb
          mov dl,ah
          xor dh,dh
          xor ah,ah
      loop tensloop
      mov si,top1
      lodsb
      add al,ascii0
      mov dl,al
      mov ax,printchar
      int 33
      mov di,top1
      cld
      xor al,al
      stosb
      pop cx
   loop numloop
ret
===
Subject: Graph Theory Question
Given a graph of order n with [n^2/4]+2 edges. Show that it contains a
bowtie, ie two triangles with exactly one vertex in common.
===
Subject: Re: Graph Theory Question
As stated it's not true.  K_4 has no such subgraph and it has exactly
[4^2/4]+2 = 6 edges
--  
===
Subject: Re: Graph Theory Question
As stated it's not true.  K_4 has no such subgraph and it has exactly
[4^2/4]+2 = 6 edges
--  
===
Subject: Re: River meandering.
As far as I know there isn't.
If there was it would probably be a Mandelbrot type function
as Per Bak has written some about drainage basin's being fractal  ( How 
Nature Works) and the Nile river studies by and hydrologist gave us the 
Hurst exponent. The coast line of Britain is one of those
fractal objects as well...
> Is there a parametric equation for rivers meandering, where t is the
> time it took for you to get from some set point to a point on the
> river? I know there's an integral form of the equation, but I'd like to
> see if there's also a parametric form of the equation.
===
Subject: Re: A tiny puzzle in number theory
   <18502326.1130972362243.JavaMail.jakarta@nitrogen.mathforum.org Proof or 
disproof:
> Every positive integer >0 has the same irrational
> factor.
> Tapio
> PROOF
> For every m,n positive integers
> n=e*(1/e)*n
> m=e*(1/e)*m
> e=lim(1+(1/x))^x (x -> inf.)
SIMPLER:
Theorem:
Every positive integer N has the same irrational factor sqr(2).
PROOF
N = m.n
m = sqr(2)
n  = N/sqr(2)
===
Subject: Re: A tiny puzzle in number theory
>> Proof or disproof:
>> Every positive integer >0 has the same irrational
>> factor.
>> Tapio
>> PROOF
>> For every m,n positive integers
>> n=e*(1/e)*n
>> m=e*(1/e)*m
>> e=lim(1+(1/x))^x (x -> inf.)
> SIMPLER:
> Theorem:
> Every positive integer N has the same irrational factor sqr(2).
> PROOF
> N = m.n
> m = sqr(2)
> n  = N/sqr(2)
Well, I can accept the both answers as the problem was badly formulated. 
Anyway, I had an idea that inverse factors like sqrt(2) and 1/sqrt(2) 
and 
also logarithmic ones are not allowed, but still you can find a common 
irrational factor. Can you find the answer now as the problem is still 
tiny?
Tapio 
===
Subject: Re: A tiny puzzle in number theory
> Proof or disproof:
> Every positive integer >0 has the same irrational
> factor.
> Tapio
> PROOF
> For every m,n positive integers
> n=e*(1/e)*n
> m=e*(1/e)*m
> e=lim(1+(1/x))^x (x -> inf.)
>> SIMPLER:
>> Theorem:
>> Every positive integer N has the same irrational factor sqr(2).
>> PROOF
>> N = m.n
>> m = sqr(2)
>> n  = N/sqr(2)
>Well, I can accept the both answers as the problem was badly formulated. 
>Anyway, I had an idea that inverse factors like sqrt(2) and 1/sqrt(2) 
and 
>also logarithmic ones are not allowed, but still you can find a common 
>irrational factor. Can you find the answer now as the problem is still 
>tiny?
Maybe if you can tell us exactly what you consider to be an allowed 
factor, you will have a well-defined problem.  As it stands, we can
only guess what you mean.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
===
Subject: Re: A tiny puzzle in number theory
>> Proof or disproof:
>> Every positive integer >0 has the same irrational
>> factor.
>> Tapio
 PROOF
>> For every m,n positive integers
>> n=e*(1/e)*n
>> m=e*(1/e)*m
>> e=lim(1+(1/x))^x (x -> inf.)
> SIMPLER:
> Theorem:
> Every positive integer N has the same irrational factor sqr(2).
> PROOF
> N = m.n
> m = sqr(2)
> n  = N/sqr(2)
>>Well, I can accept the both answers as the problem was badly formulated.
>>Anyway, I had an idea that inverse factors like sqrt(2) and 1/sqrt(2) 
>>and
>>also logarithmic ones are not allowed, but still you can find a common
>>irrational factor. Can you find the answer now as the problem is 
still
>>tiny?
> Maybe if you can tell us exactly what you consider to be an allowed
> factor, you will have a well-defined problem.  As it stands, we can
> only guess what you mean.
You are right. :-)
The problem is still badly formulated - I do agree with you, because you can 

always divide both sides of the equation by some irrational - for example, 
but you do not know easy what are the other irrationals in the product of 
irrational factors.
I wish the problem solver finds the new sofisticated but simple answer (that 

I already assume to know) where any positive integer can be presented as a 
product of distinct irrationals so that every positive rational including 
any positive integers has the same irrational in the simple product of 
distinct irrationals.
My background idea is that knowledge of such kind of  method would convert 
diophantine problems to reduced problems of irrationals as the other common 

divisors are also irrational. I have a common idea to shift the attention of 

number theory problems from Gaussian complex numbers to handle only 
irrationals, that looks somehow more interesting than complex numbers - only 

from my limited point of view (See for example Harvey Cohn: Advanced Nuber 
Theory, Dover Publications, 1962).
I proposed the problem in pure hope somebody else has also found the method. 

Therefore I'm kean to see what kind of answers are offered 

and therefore you 
have to guess as always in puzzles.  :-)
Unfortunately I'm not a good formulator of the problem - 
sorry.
A kind of background problem of this tiny puzzle consider Adrica's 
conjecture that has an important link to Riemann hypothesis - as I see. 
Honest question Robert!. I wish my answer was also honest and good enough. 
:-)
Tapio
> Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia            Vancouver, BC, Canada 
===
ultimately i was not seeing significance that a piece of the derivation 
tree
could be taken out and duplicated (that is why) and indefinitely producing
(pumping) strings in L
R
===
Subject: Zermelo-Fraenkel Theory Of Sets
I'm reading a book called The Theory Of Sets And Transfinite
Arithmetic by Alexander Abian (1965; no ISBN; Library of Congress
catalog card number 65-23086). Looks like a pretty good book.
The first chapter covers propositional logic, predicate logic and set
membership. So far so good.
Now, I arrived to a part that discusses equality. I'm having trouble
accepting some of the text, because it doesn't make sense to me.
Hopefully, someone can clear up my misunderstanding. BTW, this is
self-study.
The following two entries are from the book. Here, I use A (rather
than -/...) as the universal quantifier, e as the membership
predicate and == as the equivalence predicate which looks like an
equals sign (=) but written with three horizontal lines.
*Definition 1: We say that a set x is equal to a set y if x and y have
the same elements.
Formally,
(x = y) == (Az)((z e x) <--> (z e y))
*The Axiom Of Extensionality: Equal sets are elements of the same sets.
Formally,
(Ax)(Ay)((x = y) --> (Az)((x e z) --> (y e z)))
Definition 1 makes all the sense in the world. However, the axiom of
extensionality appears to me as incredibly counterintuitive.
The following is an example I came up with:
Sally's Classes = {Math, History, Art}
Tom's Classes = {Math, History, Art}
Age 8 Students = {Sally, Melissa, Joey}
Age 9 Students = {Martha, Tom}
So obviously, equal sets need not be elements of the same sets! :(
The question then is, what have I not understood correctly???
===
Subject: Re: Zermelo-Fraenkel Theory Of Sets
>*The Axiom Of Extensionality: Equal sets are elements of the same sets.
>Formally,
>(Ax)(Ay)((x = y) --> (Az)((x e z) --> (y e z)))
>Definition 1 makes all the sense in the world. However, the axiom of
>extensionality appears to me as incredibly counterintuitive.
>The following is an example I came up with:
>Sally's Classes = {Math, History, Art}
>Tom's Classes = {Math, History, Art}
Let 
x = Sally's Classes 
y = Tom's Classes 
z = the set of non-conflicting schedules of classes
(x e z) --> (y e z)
--Keith Lewis              klewis {at} mitre.org
===
Subject: Re: Zermelo-Fraenkel Theory Of Sets
>I'm reading a book called The Theory Of Sets And Transfinite
>Arithmetic by Alexander Abian (1965; no ISBN; Library of Congress
>catalog card number 65-23086). Looks like a pretty good book.
The excerpts you have presented suggest otherwise.  Are you sure you 
quoted them correctly?
>[...]
>The following two entries are from the book. Here, I use A (rather
>than -/...) as the universal quantifier, e as the membership
>predicate and == as the equivalence predicate which looks like an
>equals sign (=) but written with three horizontal lines.
>*Definition 1: We say that a set x is equal to a set y if x and y have
>the same elements.
>Formally,
>(x = y) == (Az)((z e x) <--> (z e y))
>*The Axiom Of Extensionality: Equal sets are elements of the same sets.
>Formally,
>(Ax)(Ay)((x = y) --> (Az)((x e z) --> (y e z)))
As  JFH has pointed out, this is *not* ZF.  Equality is a standard 
concept without need of defintion for each axiom system.  This concept 
implies what Abian identifies as his axiom.
The standard definition Axiom of Extensionality is
Ax Ay  {[Az  (z e x  <-->  z e y)]  -->  x = y}.
(Some authors use  <-->  in place of  -->.)
I suggest you get hold of a different book.  Halmos is my favorite, 
though it has its detractors.  Suppes, available in Dover, is widely 
recommended.
-- 
Stephen J. Herschkorn                        sjherschko@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan
===
Subject: Re: Zermelo-Fraenkel Theory Of Sets
> I'm reading a book called The Theory Of Sets And Transfinite
> Arithmetic by Alexander Abian (1965; no ISBN; Library of Congress
> catalog card number 65-23086). Looks like a pretty good book.
> The first chapter covers propositional logic, predicate logic and set
> membership. So far so good.
> Now, I arrived to a part that discusses equality. I'm having trouble
> accepting some of the text, because it doesn't make sense to me.
> Hopefully, someone can clear up my misunderstanding. BTW, this is
> self-study.
> The following two entries are from the book. Here, I use A (rather
> than -/...) as the universal quantifier, e as the membership
> predicate and == as the equivalence predicate which looks like an
> equals sign (=) but written with three horizontal lines.
> *Definition 1: We say that a set x is equal to a set y if x and y have
> the same elements.
> Formally,
> (x = y) == (Az)((z e x) <--> (z e y))
> *The Axiom Of Extensionality: Equal sets are elements of the same sets.
> Formally,
> (Ax)(Ay)((x = y) --> (Az)((x e z) --> (y e z)))
> Definition 1 makes all the sense in the world. However, the axiom of
> extensionality appears to me as incredibly counterintuitive.
All it is saying is that if x = y, and x is an element of z,
then y is an element of z.  That seems quite intuitive.
> The following is an example I came up with:
> Sally's Classes = {Math, History, Art}
> Tom's Classes = {Math, History, Art}
> Age 8 Students = {Sally, Melissa, Joey}
> Age 9 Students = {Martha, Tom}
> So obviously, equal sets need not be elements of the same sets! :(
> The question then is, what have I not understood correctly???
Your example seems to have nothing to do with this,
as it does not contain any sets that are members of other sets.
In your example, I am assuming that x=Sally's Classes,
and y=Tom's Classes, but what is z?
You seem to have some confused idea about set membership.
Just because Sally is an element of Age 8 Students
it does not mean that {Math, History, Art} is an element
of Age 8 Students.  Sally is not a set, is she?
Well I know everything can be considered a set, but is
that what you had in mind?
If you are claiming that Sally = {Math, History, Art}
then Age 8 Students = { { Math, History, Art}, Melissa, Joey },
and if Tom = { Math, History, Art }, then Tom is an element
of Age 8 Students.  If Tom is not 8, it just means you
have chosen an incorrect model of the world.
Stephen
===
Subject: Re: Zermelo-Fraenkel Theory Of Sets
 Discussion, linux)
>> I'm reading a book called The Theory Of Sets And Transfinite
>> Arithmetic by Alexander Abian (1965; no ISBN; Library of Congress
>> catalog card number 65-23086). Looks like a pretty good book.
>> The first chapter covers propositional logic, predicate logic and set
>> membership. So far so good.
>> Now, I arrived to a part that discusses equality. I'm having trouble
>> accepting some of the text, because it doesn't make sense to me.
>> Hopefully, someone can clear up my misunderstanding. BTW, this is
>> self-study.
>> The following two entries are from the book. Here, I use A (rather
>> than -/...) as the universal quantifier, e as the membership
>> predicate and == as the equivalence predicate which looks like an
>> equals sign (=) but written with three horizontal lines.
>> *Definition 1: We say that a set x is equal to a set y if x and y have
>> the same elements.
>> Formally,
>> (x = y) == (Az)((z e x) <--> (z e y))
>> *The Axiom Of Extensionality: Equal sets are elements of the same sets.
>> Formally,
>> (Ax)(Ay)((x = y) --> (Az)((x e z) --> (y e z)))
>> Definition 1 makes all the sense in the world. However, the axiom of
>> extensionality appears to me as incredibly counterintuitive.
> All it is saying is that if x = y, and x is an element of z,
> then y is an element of z.  That seems quite intuitive.
Perhaps it's intuitive, but it's not the Axiom of Extensionality.
Definition 1 is the Axiom of Extensionality, and this second thing is
a consequence of a standard property of equality (so is one of the
directions of Definition 1).
Namely, for all propositions P with one free variable,
x = y -> P(x) <-> P(y).
His text seems to have odd and misleading terminology.
-- 
Jesse F. Hughes
You may not realize it but THOUSANDS of people read my posts.
You are putting your stupidity on wide display.
  -- James S. Harris knows about wide displays of stupidity.
===
Subject: Re: Zermelo-Fraenkel Theory Of Sets
   <dkdcrm$k43$1@news.msu.edu I'm reading a book called The Theory Of Sets 
And Transfinite
> Arithmetic by Alexander Abian (1965; no ISBN; Library of Congress
> catalog card number 65-23086). Looks like a pretty good book.
> The first chapter covers propositional logic, predicate logic and set
> membership. So far so good.
> Now, I arrived to a part that discusses equality. I'm having trouble
> accepting some of the text, because it doesn't make sense to me.
> Hopefully, someone can clear up my misunderstanding. BTW, this is
> self-study.
> The following two entries are from the book. Here, I use A (rather
> than -/...) as the universal quantifier, e as the membership
> predicate and == as the equivalence predicate which looks like an
> equals sign (=) but written with three horizontal lines.
> *Definition 1: We say that a set x is equal to a set y if x and y have
> the same elements.
> Formally,
> (x = y) == (Az)((z e x) <--> (z e y))
> *The Axiom Of Extensionality: Equal sets are elements of the same sets.
> Formally,
> (Ax)(Ay)((x = y) --> (Az)((x e z) --> (y e z)))
> Definition 1 makes all the sense in the world. However, the axiom of
> extensionality appears to me as incredibly counterintuitive.
> All it is saying is that if x = y, and x is an element of z,
> then y is an element of z.  That seems quite intuitive.
> The following is an example I came up with:
> Sally's Classes = {Math, History, Art}
> Tom's Classes = {Math, History, Art}
> Age 8 Students = {Sally, Melissa, Joey}
> Age 9 Students = {Martha, Tom}
> So obviously, equal sets need not be elements of the same sets! :(
> The question then is, what have I not understood correctly???
> Your example seems to have nothing to do with this,
> as it does not contain any sets that are members of other sets.
> In your example, I am assuming that x=Sally's Classes,
> and y=Tom's Classes, but what is z?
> You seem to have some confused idea about set membership.
> Just because Sally is an element of Age 8 Students
> it does not mean that {Math, History, Art} is an element
> of Age 8 Students.  Sally is not a set, is she?
> Well I know everything can be considered a set, but is
> that what you had in mind?
> If you are claiming that Sally = {Math, History, Art}
> then Age 8 Students = { { Math, History, Art}, Melissa, Joey },
> and if Tom = { Math, History, Art }, then Tom is an element
> of Age 8 Students.  If Tom is not 8, it just means you
> have chosen an incorrect model of the world.
> Stephen
As said, your example has nothing to do with your question.  Just
because Tom's classes are the same set as Sally's does not mean Tom =
Sally.
There are several comments to be made.  My first is that Abian was an
one.  But leaving that all aside, you might say that Tom's classes are
not the same set as Sally's because you intended them to be different.
That is the difference between intentional and extentional equality.
Actually, in ZF set theory, they aren't sets at all.  A set consists of
elements or is empty and each element is a set that consists of
elements or is empty and each of those elements ...  and the axiom of
well-foundedness says that if you take an element and an element of
that element and so on, you come, after finitely many steps, to the
empty set.
Is this really what a set is?  Well, each person has to decide for
himself, but I am not a fan of ZF set theory.  I would rather view this
as a model for set theory that demonstrates a kind of intuitive
consistency.  And there are other approaches, not all of which satisfy
extentionality, or well-foundedness, for that matter.  The categorical
approach doesn't even start with elements.
===
Subject: Re: Zermelo-Fraenkel Theory Of Sets
   <dkdcrm$k43$1@news.msu.edu There are several comments to be made. My 
first
> is that Abian was an abyssmal teacher and I amazed
I had always been under the impression that Abian
was considered an excellent teacher, although this
isn't based on anything I know for sure. Instead,
it just seems to be a reasonable observation based
on the large number of elementary and expository
type papers he published. A former colleague of
mine was one of Abian's Ph.D. students, but I
don't recall now if he said anything about Abian's
teaching.
I also don't understand why being an abysmal
teacher would lead you to think Abian had never
written a book, because writing books and papers
takes time away from grading papers, extra office
hours, time spent preparing lectures, etc. I realize
that plenty of college professors find their research
stimulates their teaching, but my perception has
been that the vast majority of students, if they
have any opinion at all on the matter, think that
writing books and papers takes away from teaching.
I find myself in the strange position of defending
Abian, despite some negative things I've posted
about Abian's publishing record in the past and
despite my thinking that he was quite crankish in
many of his non-math science viewpoints (which
long time sci.math posters are no doubt aware of,
because Abian used to post quite a bit in sci.math).
However, I'm motivated to respond because I can't
believe that anyone who knows the least bit about
Abian would not also know about his very lengthy
publishing record. In fact, Abian is almost the
mathematical poster child of the fact that one
can publish a huge amount (3 books and 254 papers)
and not be considered a very strong mathematician.
To be fair, I find many of Abian's papers useful
precisely *because* they're not pitched at such
a rarified level that only a few specialists in
the field can read them. Unfortunately, Abian
also had a bad tendency of re-proving results
that can be found easily in the literature, and
for giving a bibliography that suggests he didn't
look much beyond a few standard textbooks before
writing the paper.
Dave L. Renfro
===
Subject: Re: Zermelo-Fraenkel Theory Of Sets
> *The Axiom Of Extensionality: Equal sets are elements of the same sets.
> Formally,
> (Ax)(Ay)((x = y) --> (Az)((x e z) --> (y e z)))
> Definition 1 makes all the sense in the world. However, the axiom of
> extensionality appears to me as incredibly counterintuitive.
> The following is an example I came up with:
> Sally's Classes = {Math, History, Art}
> Tom's Classes = {Math, History, Art}
> Age 8 Students = {Sally, Melissa, Joey}
> Age 9 Students = {Martha, Tom}
> So obviously, equal sets need not be elements of the same sets! :(
I don't understand what your example has to do with
the stated axiom.
===
Subject: ideal gas theory
How did the scientists like Boyle and Avagadro did experiments and still got 

results, which supported the ideal gas theory? It is actually wrong and 
still 
how did they get results that volume is inversely proportional to pressure 
while temperature directly to pressure. I think the only chance of 
disproving 
ideal theory is experiment. There is no theoretical reason for such a 
behavior of real gases. Scientists who are considered to be great developed 

the ideal theory. So I doubt their Excellency if they made mistakes in 
observations.
===
Subject: quest
How many dimensions exist in universe? In logic we can say the universe to 
be three-dimensional. But I have heard that Einstein called the universe to 

be four-dimensional or space-time universe which also contains one time-like 

dimension. But now I am hearing that there are more than the four 
dimensions. 
Can anyone here help me
===
Subject: magnetism
I was hearing headphone and I noticed something. The headphone is not 
properly stereophonic. But I shook my head when someone asked me something 
and then I noticed that the sound is better as well as it becomes 
stereophonic in a particular direction. This happens when it is held in 
North-South direction. Has it something to do with magnetism? 
                    I have seen the degauss button in the computer monitor 
and this setting has to be restored when one changes its direction 
considerably. I also know that such settings are even applicable to 
television but the only difference is that for television such settings are 

inside the cabinet.
                    Are both of these properties interrelated with each 
other? Can you explain the science behind this?
===
Subject: Re: Smooth Interpolations
> On Wed, 02 Nov 2005 04:28:55 EST, Maury Barbato
>> On Fri, 28 Oct 2005 04:57:17 EDT, Maury Barbato
>> 
> On Thu, 27 Oct 2005 02:55:16 EDT, Maury Barbato
> 
>I'm considering the following problem.
>>Let {(X_n, Y_n)} be a sequence of points of
> R^2
>> with
> the following
>properties:
>(I) X_n>0 for every n>=0;
>(II) X_{n+1}<(X_n)/2 for every n>=0;
>(III) {|Y_n|} is strictly decreasing and it
> tends
>> to
> 0;
>(IV) |Y_{n}|/X_n is strictly decreasing and it
>tends to 0.
>>Problem. Does there exist a subsequence {(x_n,
>> y_n)}
>of {(X_n, Y_n)} and a smooth function
>> f:[0,x_0]->R 
>such that f(x_n)=y_n for every n>=0?
>>If we apply Whitney's Extension Theorem we
>> conclude 
>that there exists a C^1 function g:[0,X_0]->R
>> whose 
>graph {(x, g(x)): x in [0,X_0]} passes through
>> all
> the
>points of {(X_n,Y_n)}. 
>If you don't know Whitney's Extension Theorem,
>> you
> can
>find a constructive proof of this result in
> the
> paper of
>Ring, Schwaiger, Schoepf (see proof of Theorem
> 4)
>> at
> the
>page 
>>http://math.uni-graz.at/ring/papers/zorzitto.pdf
>>  . 
>>Maybe the problem has a positive answer if we
> require
>that f is a C^k function, but in the stronger 
>formulation above I think the answer is no.
> 
> No, you can't even get a C^2 function.
> 
> The function would certainly have f(0) = f'(0)
>> 0.
> If it's C^2 this implies |f(x)| <= c x^2 for
> small
>> x.
> Let X_n = 3^{-n}, Y_n = (X_n)^(3/2).
> 
>Maury
> 
> 
> ***
> 
> 
How couldn't I think it! I sometimes surprise
> myself
>>for my stupidity ...
Anyway, I'm now thinking about what happens if we
>>try to find a function f:(0,X_0]->R such that
>>such that f(X_n)=Y_n for every n>=0.
The following piecewise construction shows that 
>>we always find a smooth function with this 
>>property.
[...]
However, I don't what happens if we require that
>>f is analytic: does it exists?
>> 
>> Certainly you can find an analytic f if there's
>> no condition on the behavior of f(t) as t -> 0:
>> 
>> If O is an open set in the plane, z_n is a
> sequence
>> of distinct points of O that tend to the boundary
>> of O, and c_n is any sequence of complex numbers
>> then there exists a function f holomorphic in O
>> with f(z_n) = c_n for all n. (I don't know if
>> this theorem has a name; it's often proved using
>> the Mittag-Leffler theorem, so you can look in
>> complex books shortly after the M-L theorem for
>> a proof.)
>> 
Who assures us that the restriction of this function to
(0,X_0) is real?
>> What seems more interesting to me is the 
>> question of what condition you need to add
>> to get f in C^2([0,1]). I conjecture that
>> a necessary and sufficient additional condition
>> is
>> 
>> (*) (y_n/x_n - y_{n+1}/x_{n+1})/(x_n - x_{n+1})
>>     has a limit as n -> infinity.
>> 
>> (Writing x_n in place of X_n for convenience,
>> similarly for y_n.)
>> 
>> Let's see. As above we have f(0) = f'(0) = 0.
>> So
>> 
>>   y_n/x_n = int_0^1 f'(x_n t) dt.
>> 
>> And in general
>> 
>>   f'(p) - f'(q) = int_p^q f''(t) dt
>> 
>>           = (q-p) int_0^1 f''(p + t(q-p)) dt,
>> 
>> so we get
>> 
>>   (y_n/x_n - y_{n+1}/x_{n+1})/(x_n - x_{n+1})
>> 
>>     = int_0^1 int_0^1 t 
>>         f''(x_{n+1} t + s(x_n t - x_{n+1} t)) ds
> dt.
>> 
>> If f is C^2 on [0,1] then this converges to
>> 
>>     f''(0) int_0^1 int_0^1 t ds dt = f''(0)/2,
>> 
>> so (*) is in fact necessary. I bet it's sufficient
>> as well.
>What makes you think that? 
> I didn't say I was certain, just seemed to me
> that under those conditions it shouldn't be
> too hard to construct an interpolating
> function directly.
>I tried to define f'(x_n)
>and f''(x_n) in a way such that the hypotheses of
>Whitney's Extension Theorem are satisfied, but
> without
>success. So, if your conjecture is true, surely it
> has
>no trivial proof!
> I see.
>I managed to prove your conjecture only when
>(y_n/x_n - y_{n+1}/x_{n+1})/(x_n - x_{n+1})
>tends to zero as n -> infinity.
> Well it _almost_ seems like the general
> case should follow, considering
> z_n = y_n - c x_n^2 in place of y_n.
> Of course that's not quite a proof, since
> the z_n will not satisfy (iii) and (iv)
> at the top, but they come pretty close to
> satisfying (iii) and (iv)...
This proves that there's a subsequence (x_n, y_n)
and a C^2 function f:[0,X_0]->R such that f(x_n)=y_n,
but this isn't yet your conjecture. I'll try to gain
something.
>> 
>>Maury
>> 
>> 
>> ***
>> 
>> 
>Maury Barbato
> ***
> 
Maury
===
Subject: Re: Smooth Interpolations
On Thu, 03 Nov 2005 11:17:33 EST, Maury Barbato
>> On Wed, 02 Nov 2005 04:28:55 EST, Maury Barbato
> On Fri, 28 Oct 2005 04:57:17 EDT, Maury Barbato
> 
> On Thu, 27 Oct 2005 02:55:16 EDT, Maury Barbato
>> 
>>I'm considering the following problem.
>>Let {(X_n, Y_n)} be a sequence of points of
>> R^2
> with
>> the following
>>properties:
>>(I) X_n>0 for every n>=0;
>>(II) X_{n+1}<(X_n)/2 for every n>=0;
>>(III) {|Y_n|} is strictly decreasing and it
>> tends
> to
>> 0;
>>(IV) |Y_{n}|/X_n is strictly decreasing and it
>>tends to 0.
>>Problem. Does there exist a subsequence {(x_n,
> y_n)}
>>of {(X_n, Y_n)} and a smooth function
> f:[0,x_0]->R 
>>such that f(x_n)=y_n for every n>=0?
>>If we apply Whitney's Extension Theorem we
> conclude 
>>that there exists a C^1 function g:[0,X_0]->R
> whose 
>>graph {(x, g(x)): x in [0,X_0]} passes through
> all
>> the
>>points of {(X_n,Y_n)}. 
>>If you don't know Whitney's Extension Theorem,
> you
>> can
>>find a constructive proof of this result in
>> the
>> paper of
>>Ring, Schwaiger, Schoepf (see proof of Theorem
>> 4)
> at
>> the
>>page 
>>http://math.uni-graz.at/ring/papers/zorzitto.pdf
>  . 
>>Maybe the problem has a positive answer if we
>> require
>>that f is a C^k function, but in the stronger 
>>formulation above I think the answer is no.
>> 
>> No, you can't even get a C^2 function.
>> 
>> The function would certainly have f(0) = f'(0)
>> =
> 0.
>> If it's C^2 this implies |f(x)| <= c x^2 for
>> small
> x.
>> Let X_n = 3^{-n}, Y_n = (X_n)^(3/2).
>> 
>>Maury
>> 
>> 
>> ***
>> 
>> 
>>How couldn't I think it! I sometimes surprise
>> myself
>for my stupidity ...
>>Anyway, I'm now thinking about what happens if we
>try to find a function f:(0,X_0]->R such that
>such that f(X_n)=Y_n for every n>=0.
>>The following piecewise construction shows that 
>we always find a smooth function with this 
>property.
>>[...]
>>However, I don't what happens if we require that
>f is analytic: does it exists?
> 
> Certainly you can find an analytic f if there's
> no condition on the behavior of f(t) as t -> 0:
> 
> If O is an open set in the plane, z_n is a
>> sequence
> of distinct points of O that tend to the boundary
> of O, and c_n is any sequence of complex numbers
> then there exists a function f holomorphic in O
> with f(z_n) = c_n for all n. (I don't know if
> this theorem has a name; it's often proved using
> the Mittag-Leffler theorem, so you can look in
> complex books shortly after the M-L theorem for
> a proof.)
> 
>Who assures us that the restriction of this function to
>(0,X_0) is real?
We want a real-analytic function, right? The
real part of a holomorphic function is real-analytic.
> What seems more interesting to me is the 
> question of what condition you need to add
> to get f in C^2([0,1]). I conjecture that
> a necessary and sufficient additional condition
> is
> 
> (*) (y_n/x_n - y_{n+1}/x_{n+1})/(x_n - x_{n+1})
>     has a limit as n -> infinity.
> 
> (Writing x_n in place of X_n for convenience,
> similarly for y_n.)
> 
> Let's see. As above we have f(0) = f'(0) = 0.
> So
> 
>   y_n/x_n = int_0^1 f'(x_n t) dt.
> 
> And in general
> 
>   f'(p) - f'(q) = int_p^q f''(t) dt
> 
>           = (q-p) int_0^1 f''(p + t(q-p)) dt,
> 
> so we get
> 
>   (y_n/x_n - y_{n+1}/x_{n+1})/(x_n - x_{n+1})
> 
>     = int_0^1 int_0^1 t 
>         f''(x_{n+1} t + s(x_n t - x_{n+1} t)) ds
>> dt.
> 
> If f is C^2 on [0,1] then this converges to
> 
>     f''(0) int_0^1 int_0^1 t ds dt = f''(0)/2,
> 
> so (*) is in fact necessary. I bet it's sufficient
> as well.
What makes you think that? 
>> I didn't say I was certain, just seemed to me
>> that under those conditions it shouldn't be
>> too hard to construct an interpolating
>> function directly.
>>I tried to define f'(x_n)
>>and f''(x_n) in a way such that the hypotheses of
>>Whitney's Extension Theorem are satisfied, but
>> without
>>success. So, if your conjecture is true, surely it
>> has
>>no trivial proof!
>> I see.
>>I managed to prove your conjecture only when
>>(y_n/x_n - y_{n+1}/x_{n+1})/(x_n - x_{n+1})
>>tends to zero as n -> infinity.
>> Well it _almost_ seems like the general
>> case should follow, considering
>> z_n = y_n - c x_n^2 in place of y_n.
>> Of course that's not quite a proof, since
>> the z_n will not satisfy (iii) and (iv)
>> at the top, but they come pretty close to
>> satisfying (iii) and (iv)...
>This proves that there's a subsequence (x_n, y_n)
>and a C^2 function f:[0,X_0]->R such that f(x_n)=y_n,
>but this isn't yet your conjecture. I'll try to gain
>something.
===
Subject: Re: Stuck with an integral
> Int((x-2)*e^x/x^3 dx). 
> What do you want to do with this?  It is not an elementary function.
> It needs to be expressed in terms of Ei(x),  the exponential integral.
False, in this special case.
 By hand, split the integrand into e^x/x^2 - 2*e^x/x^3, and integrate the
second term by parts.
 (There are two evident possibilities, one will fail, the other will
succeed.)
===
Subject: Re: abs of matrix entries
> I have some programming code that takes the absolute value of all the
> entries in a linear transformation matrix (representing rotations and
> scalings in the plane).  I can't figure out why the absolute values were
> taken and what this does to the geometry.
> Is there any theory on this?
(Just a guess): This may have something to do with improved, or optimal
scaling. The theory says:
 Given a matrix A (invertible, and most of the time irreducible, as your
context suggests), let B be its inverse. Then we can find two (positive)
diagonal matrices R (for rows)  and C (for columns) so that the
condition number of R^(-1)*A*C is minimal. The norm for this condition
number is the maximum absolute row sum.
 The entries of C are the respective entries of the Perron eigenvector of
M = abs(B)*abs(A), hence the occurrence of absolute values.
 The entries of R are the absolute row sums of A*C.
 The background theory is called Perron-Frobenius Theory:
http://mathworld.wolfram.com/Perron-FrobeniusTheorem.html
http://www.prenhall.com/divisions/esm/app/ph-linear/leon/html/perron.html
and Google showed about 101,000 other references.
What may optimal scaling do to the geometry? Perhaps the scaled matrix
transforms the unit cube into a solid whose shape is as close as possible
to the unit cube.
===
Subject: Re: Marble problem--put in 10 marbles, then remove 1
   <Ip2GxD.1BE@cwi.nl>
   <ITSnetNOTcom#virgil-DC4A3B.17173129102005@comcast.dca.giganews.com>
   <85y84b9700.fsf@lola.goethe.zz   How about you
> take for example a course in probability theory (which I would argue is
> applied very commonly) where the distinction between countable and
> uncountable is very VERY important.
> I take it there are examples in the real world you can show me which
> deal with infinity?
Not in the way that you think.  You wish to see an example of an
infinite (say countable) set in the tangible universe.  As you cannot
find such an example you conclude that mathematics with the infinite is
pointless to the tangible world.
What is commonly done is to model the real world which may be finite
and perhaps quantized or whatnot with mathematics that is infinite and
continuous.  Despite the primary intuition that finite is simpler, it
is often much easier to model with the infinite and continuous.
> The closest thing I got to an answer was physics. But where has an
> example similar to the marble counting problem appeared in physics?
That is a thought exercise, not a real world problem.  The exercise is
in set theory which is what modern mathematics is based on.
> Note, at this point I have retracted my insistence that the bag is
> full, but I still wish for proof in the real world before jumping to
> ANY conclusions.
proof in the real world?  You must not have been listening when
people say that it is not a real world problem.
===
Subject: REPOST: Re: Pomerance/Crandall Prime Numbers book: inequality 
question
>The second edition of Pomerance and Crandall's Prime Numbers book is 
>out, and before I buy it I figured I should read the first edition, 
>which I already own (I bought it a few years ago and only read a few 
>sections that were of interest to me at the time).
>Sadly, I can't figure out the following inequality, which occurs -- even 
>more sadly -- on page 6:
>x <= PROD( floor( 1 + (ln x) / (ln p) ) ), where x is a positive integer 
>and the product is taken over all primes p <= x.
>Why is this inequality true?
Call the right side P(x) = product_p f(x,p).
f(x,p) = floor(1 + ln(x)/ln(p) = n iff p^(n-1) <= x < p^n.  Thus f(x,p) 
is the number of powers of p that are <= x.  Consider selecting 
one of those powers of p for each prime p <= x and multiplying them 
together.  This gives us P(x) distinct positive integers, and each 
of the x positive integers <= x can be obtained in this way, so 
x <= P(x).
 
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
===
Subject: Re: Pomerance/Crandall Prime Numbers book: inequality question
Bytes: 570
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===
Subject: REPOST: Re: Pomerance/Crandall Prime Numbers book: inequality 
question
> The second edition of Pomerance and Crandall's Prime Numbers book is 
> out, and before I buy it I figured I should read the first edition, 
> which I already own (I bought it a few years ago and only read a few 
> sections that were of interest to me at the time).
> Sadly, I can't figure out the following inequality, which occurs -- even 
> more sadly -- on page 6:
> x <= PROD( floor( 1 + (ln x) / (ln p) ) ), where x is a positive integer 
> and the product is taken over all primes p <= x.
> Why is this inequality true?
Go back up the paragraph where it says 
       x = the number of solutions of { product (p_i)^(a_i) le x }
In any solution, (p_i)^(a_i) le x for each i, so a_i is at most 
(log x) / (log p_i), so there are at most 1 + ( (log x) / (log p_i) ) 
possible values of a_i for each i. So, the number of solutions, 
which is x, is at most the product of the numbers 
1 + ( (log x) / (log p_i) ). 
I've been sloppy, leaving out the floor symbols, but you can put 
them back in if you like.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Pomerance/Crandall Prime Numbers book: inequality question
Bytes: 616
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===
Subject: REPOST: Re: Pomerance/Crandall Prime Numbers book: inequality 
question
> The second edition of Pomerance and Crandall's Prime Numbers book is 
> out, and before I buy it I figured I should read the first edition, 
> which I already own (I bought it a few years ago and only read a few 
> sections that were of interest to me at the time).
> Sadly, I can't figure out the following inequality, which occurs -- even 
> more sadly -- on page 6:
> x <= PROD( floor( 1 + (ln x) / (ln p) ) ), where x is a positive integer 
> and the product is taken over all primes p <= x.
> Why is this inequality true?
===
Subject: Re: Pomerance/Crandall Prime Numbers book: inequality question
Control: cancel <11mgggj4pl0drfe@corp.supernews.com>
===
Subject: REPOST: Pomerance/Crandall Prime Numbers book: inequality 
question
The second edition of Pomerance and Crandall's Prime Numbers book is 
out, and before I buy it I figured I should read the first edition, 
which I already own (I bought it a few years ago and only read a few 
sections that were of interest to me at the time).
Sadly, I can't figure out the following inequality, which occurs -- even 
more sadly -- on page 6:
x <= PROD( floor( 1 + (ln x) / (ln p) ) ), where x is a positive integer 
and the product is taken over all primes p <= x.
Why is this inequality true?
===
Subject: Pomerance/Crandall Prime Numbers book: inequality question
Bytes: 560
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===
Subject: Re: How do you say it in English?
[answers inserted]
> how do you call the binomial coefficient in English, say
>   5       5!
>  ( ) = -------
>   3    3! * 2!
> Is it 5 above 3 or 5 over 3 or something else?
 five choose three
> And how do you call a fraction, say 5/3?
> Is it 5 by 3 or something else?
 five over three
> And how do you call the simplification of a fraction
> when we devide both the numerator and the denominator
> by their greatest common divisior, say 10/15 = 2/3.
> Is it to shorten the fraction or something else?
 reduce the fraction -- that does not require greatest common divisor,
      so that 40/60 is reduced to 10/15, or to 20/30, or to 2/3.
 reduce to lowest terms or write in lowest terms uses gcd, so 2/3
      would be 40/60 in lowest terms.
Rado se stalo, ZVK(Slavek)
> --
> Pavel Pokorny
> Math Dept, Prague Institute of Chemical Technology
> http://www.vscht.cz/mat/Pavel.Pokorny
===
Subject: Re: How do you say it in English?
 And how do you call a fraction, say 5/3?
> Is it 5 by 3 or something else?
 5 over 3 or (less often) 5 divided by 3
> In the UK and its cultural dependencies, you may hear 5 on 3.
> --
> Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
> In the UK I've only ever heard the first referred to as
> 5 choose 3.
> I must say I've never heard of the second being called
> 5 on 3. Much more common is 5 over 3 or 5 by 3.
Well, the first person I ever heard say a on b for what I 
would have called a over b was an Australian mathematician, 
lecturing in the US. Later, I was in England for a year, 
and I'm sure that a on b was the common usage. So, 
1. you are a lot younger than I am, and in the years since I 
was in England the American usage has taken over, or 
2. you and I spent time in different parts of the UK, where 
different usages predominated, or 
3. one of us is nuts.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: How do you say it in English?
 And how do you call a fraction, say 5/3?
>> Is it 5 by 3 or something else?
>> 
>> 5 over 3 or (less often) 5 divided by 3
> In the UK and its cultural dependencies, you may hear 5 on 3.
> You may hear 5 on 3 in Canada but it has a totally different 
> meaning. :)
This is a break away from the thread proper.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Geometric R.V
>> Also is Z geometric? why?
>> and MUI. Let Z = min(X,Y).
>I think you mean that X and Y are EXPONENTIAL, not geometric.
>(Geometric random variables are discrete, with Prob{X = n} =(1-p)
>if you wanted to, but I very much doubt that this is what you are
>doing.) Anyway, assuming you mean exponential rather than geometric,
>your stated result is false. The random variable Z would have rate
It depends on how you define the parameters, even for the
exponential.  The same proof works for both.  For simplicity,
start the geometric at 0.
Now just compute P(Z >= n).
One could do the same for the exponential, replacing n 
by x in the above, for non-negative real x.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
===
Subject: Geometric R.V
> Also is Z geometric? why?
As RG Vickson said, it should be  lambda + mu, and it works
for both Geometric and Exponential, and for the same reason.
RGV gave a formal reason, so here is an informal one to go with it.
Random variable X represents the time you have to wait for
an incident to occur, where they occur completely at random
with an average rate of  Lambda  per second.
Ditto for Y and Mu.
Now, if these two processes run independently and in parallel,
then min(X,Y) is the time you wait before EITHER type of
incident occurs; and these super-incidents are still at random
but now with rate (lambda + mu), as you may easily observe
by regarding how many averagely occur in a fixed time interval.
Q.E.D.  (quod easily done)
-------------------------------------------------------------------------
        Bill Taylor             W.Taylor@math.canterbury.ac.nz
-------------------------------------------------------------------------
             Have you paid your random road tax recently?
             If not - then visit your local speed camera!
-------------------------------------------------------------------------
===
Subject: Re: Geometric R.V---possible correction
> Also is Z geometric? why?
> and MUI. Let Z = min(X,Y).
It may be that you really do mean geometric, unlike my assumption in
my previous response. (I might have been thrown off by the weird
common p or q or similar.) Anyway, assuming geometric X and Y, your
that the distsributions are P{X=n} = lambda (1-lambda)^(n-1) and P{Y=n}
= mu (1-mu)^(n-1)). I won't give the details, but will try to guide you
to discovery of the method yourself. (1) Is it clear to you that if you
can get the cumulative distribution P{Z <= n} or complementary
cumulative P{Z > n} for each n = 1,2,3, ... , you can get the
probability mass function? (2) Looking at the complementary cumulative,
for instance, how would you relate the event {Z > n} to events
involving X and Y? (Are you using a textbook? Does it have relevant
material?) (3) What happens if you very carefully examine the
conditions for independence of X and Y? (Again: are you using a
textbook, and does it have relevant material?) If you write things down
and think a bit, you should be able to get the result.
Good luck
R.G. Vickson
===
Subject: Re: Well Ordering the Reals
   <Ip8Cy8.8pu@cwi.nl>
   <MPG.1dd0180a70f3a7bc98a5a5@newsstand.cit.cornell.edu>
   <Ip90rJ.BA5@cwi.nl>
   <MPG.1dd1622bc3553e7398a5ce@newsstand.cit.cornell.edu>
   <dk86eg02vr8@drn.newsguy.com>
   <MPG.1dd16f5d868a36cf98a5d8@newsstand.cit.cornell.edu>
   <ITSnetNOTcom#virgil-1F1039.16494501112005@comcast.dca.giganews.com>
   <MPG.1dd2a9fa77839ca298a601@newsstand.cit.cornell.edu>
   <MPG.1dd2c395a19d17b898a615@newsstand.cit.cornell.edu>
   <3sscabFpuriqU1@individual.net>
   <MPG.1dd302f1c360e08898a628@newsstand.cit.cornell.edu>
   <3sstiqFpfjifU3@individual.net>
   <3ssvb2FpgffaU1@individual.net The universe is infinite, infinite sets 
are equivalent.
You've been asked this many times, but if all infinite sets are
equivalent (by which I assume you mean that they all have the
same number of elements), can you please show us a bijection
between N and R, or between N and the real interval [0,1]?
Such a mapping could, of course, be used to well-order the reals.
===
Subject: Re: Well Ordering the Reals
   <Ip90rJ.BA5@cwi.nl>
   <MPG.1dd1622bc3553e7398a5ce@newsstand.cit.cornell.edu>
   <dk86eg02vr8@drn.newsguy.com>
   <MPG.1dd16f5d868a36cf98a5d8@newsstand.cit.cornell.edu>
   <ITSnetNOTcom#virgil-1F1039.16494501112005@comcast.dca.giganews.com>
   <MPG.1dd2a9fa77839ca298a601@newsstand.cit.cornell.edu>
   <MPG.1dd2c395a19d17b898a615@newsstand.cit.cornell.edu>
   <3sscabFpuriqU1@individual.net>
   <MPG.1dd302f1c360e08898a628@newsstand.cit.cornell.edu>
   <3sstiqFpfjifU3@individual.net>
   <3ssvb2FpgffaU1@individual.net The universe is infinite, infinite sets 
are equivalent.
> You've been asked this many times, but if all infinite sets are
> equivalent (by which I assume you mean that they all have the
> same number of elements), can you please show us a bijection
> between N and R, or between N and the real interval [0,1]?
> Such a mapping could, of course, be used to well-order the reals.
Have you heard of the natural/unit equivalency function, or EF?
Basically EF = n/d, as d->oo.  EF(0)=0, EF(n+1)>EF(n), lim n->oo
EF(n)=1, because x/x=1.  It's monotonic increasing, the naturals scaled
to the unit interval.
As sets, they're sets of numbers.  There is basically the notion of
iota as a least positive real, and I'm aware of the fact that the reals
are the complete ordered field, and say, non-standardly, that they
are as well contiguous sequence of points.  By Cantor's first there are
no infinite sequences or nested intervals, the antidiagonal is off the
range or dually represented.  In terms of powerset and
Cantor/Bernstein, that gets into technical philosophy and projective
extensions.
Standard arguments against it include that for any n, EF(n) is zero,
and that the reals are a complete ordered field so (EF(n)+EF(n+1)) / 2
would be a real number.  Basically the response is that n is non-zero
and infinity is very large, so it's recriprocal is very small and some
infinitesimal defined as the least.
Now the unit impulse function is not a real function, but it's useful.
If the reals are well-orderable, then there exist adjacent points in
the normal ordering of the reals, and EF is a bijection from N to
R[0,1], and other bijections from N to continuous segments of R are
piecewise, or transformed, compositions of EF.
Yes:  well-order the reals.  That's a knotty problem that for decades
and a century has resisted calls for action from the mathematical
community writ at large.  What chagrin:  they were already
well-ordered.
So, well-order the reals.
Ross
===
Subject: Re: Well Ordering the Reals
   <Ip90rJ.BA5@cwi.nl>
   <MPG.1dd1622bc3553e7398a5ce@newsstand.cit.cornell.edu>
   <dk86eg02vr8@drn.newsguy.com>
   <MPG.1dd16f5d868a36cf98a5d8@newsstand.cit.cornell.edu>
   <ITSnetNOTcom#virgil-1F1039.16494501112005@comcast.dca.giganews.com>
   <MPG.1dd2a9fa77839ca298a601@newsstand.cit.cornell.edu>
   <MPG.1dd2c395a19d17b898a615@newsstand.cit.cornell.edu>
   <3sscabFpuriqU1@individual.net>
   <MPG.1dd302f1c360e08898a628@newsstand.cit.cornell.edu>
   <3sstiqFpfjifU3@individual.net>
   <3ssvb2FpgffaU1@individual.net The universe is infinite, infinite sets 
are equivalent.
> You've been asked this many times, but if all infinite sets are
> equivalent (by which I assume you mean that they all have the
> same number of elements), can you please show us a bijection
> between N and R, or between N and the real interval [0,1]?
> Such a mapping could, of course, be used to well-order the reals.
> Have you heard of the natural/unit equivalency function, or EF?
Isn't this just a piece of nonsensical babble that you have invented?
>By Cantor's first there are
> no infinite sequences or nested intervals, ...
Isn't this just your total misunderstanding of the implications of the
claim that there is a well-ordering of the reals. You somehow think it
must be the natural order, even though plainly it isn't.
> and infinity is very large, so it's recriprocal is very small and some
I suppose this is somewhat ridiculous, telling someone who hasn't
mastered basic grammar, but it's is a misspelling above. (So is
recriprocal...)
And btw, infinity is not very large, it is really really really
really huge. (Joke)
> If the reals are well-orderable, then there exist adjacent points in
> the normal ordering of the reals, ...
No there don't. But babble on...
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Well Ordering the Reals
David R Tribble said:
> The universe is infinite, infinite sets are equivalent.
> You've been asked this many times, but if all infinite sets are
> equivalent (by which I assume you mean that they all have the
> same number of elements), can you please show us a bijection
> between N and R, or between N and the real interval [0,1]?
> Such a mapping could, of course, be used to well-order the reals.
log oo(x) certainly works. If I can prove that my enumeration includes all 
reals, then that also provides a bijection between the naturals and the 
reals 
through the binary strings, but any effort in this area will simply be met 
with 
Cantor's two proofs. I'll go read up on Cauchy sequences and Dedekind 
cuts, 
and see if I can't prove the second, since the first is surely abhorrent to 

mathematicians. 
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
   <Ip8Cy8.8pu@cwi.nl>
   <MPG.1dd0180a70f3a7bc98a5a5@newsstand.cit.cornell.edu>
   <Ip90rJ.BA5@cwi.nl>
   <MPG.1dd1622bc3553e7398a5ce@newsstand.cit.cornell.edu>
   <dk86eg02vr8@drn.newsguy.com>
   <MPG.1dd16f5d868a36cf98a5d8@newsstand.cit.cornell.edu>
   <ITSnetNOTcom#virgil-1F1039.16494501112005@comcast.dca.giganews.com>
   <MPG.1dd2a9fa77839ca298a601@newsstand.cit.cornell.edu>
   <MPG.1dd2c395a19d17b898a615@newsstand.cit.cornell.edu>
   <3sscabFpuriqU1@individual.net>
   <MPG.1dd302f1c360e08898a628@newsstand.cit.cornell.edu>
   <dkbfag0r2a@drn.newsguy.com>
Tony Orlow says...
>> No, I believe it was Daryl (I might be wrong, maybe Tribble?) who was
>> explaining how the set size is the first ordinal after the largest 
element,
> I didn't say any such thing.
And likewise I never said anything like that either.
> What I said was that omega is the
> smallest ordinal that is larger than *every* finite ordinal. There
> is no largest finite ordinal, but there is a smallest infinite
> ordinal.
I said something similar.  I also provided a trivial proof that
since any given finite k in N is followed by another larger
finite k+1, there can be no finite k that represents the size
of N and also be a member of N.  We are forced to conclude
that N has a size that is not a finite natural number.  Hence
the definitions of Aleph_0 and omega.
===
Subject: Re: Well Ordering the Reals
David R Tribble said:
> Tony Orlow says...
>> No, I believe it was Daryl (I might be wrong, maybe Tribble?) who was
>> explaining how the set size is the first ordinal after the largest 
element,
> I didn't say any such thing.
> And likewise I never said anything like that either.
> What I said was that omega is the
> smallest ordinal that is larger than *every* finite ordinal. There
> is no largest finite ordinal, but there is a smallest infinite
> ordinal.
> I said something similar.  I also provided a trivial proof that
> since any given finite k in N is followed by another larger
> finite k+1, there can be no finite k that represents the size
> of N and also be a member of N.  We are forced to conclude
> that N has a size that is not a finite natural number.  Hence
> the definitions of Aleph_0 and omega.
Right, and yet, if we start with 1, which you said was stupid, then 
there is 
an equivalence between set size and largest element, and it becomes clear 
that, 
if all elements are finite, then the set size is too. That's better than 
dragging hebrew and greek letters into the number system.
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
 
>Right, and yet, if we start with 1, which you said was stupid,
>then there is an equivalence between set size and largest element
The empty set has size 0, but its largest element isn't 0. So
your convention works for some sets, but not for others. In
particular, it doesn't work for sets of size 0 or sets of size
omega.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>imaginatorium@despammed.com said:
>> Or are there really
>> more representations of numbers in C than there are numbers in C. Or 
it
>> the relation between binary N and whatever N is the size of C 
just
>> so as to make them match up after all?
>It depends whether you are interpeting the strings as quantities, or just 
as 
>raw strings. See? You can give a reasonable answer, even if the question 
didn't
>expect it.
> So you *are* saying that whether set A is bigger than set B is a
> *subjective* judgement; it depends on how you *interpret* the elements
> of A?
> --
> Daryl McCullough
> Ithaca, NY
I wouldn't say it is subjective, but dependent on context. It depends 
what 
question you are asking, whether you are talking about the set of strings 
constructed from some alphabet, or the set of quantities that those strings 

represent. Maybe it's too hard for standard set theorists to distinguish 
strings from that which they represent. Oh well.
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Daryl McCullough said:
> Tony Orlow says...
imaginatorium@despammed.com said:
> 
>> Or are there really
>> more representations of numbers in C than there are numbers in C. Or 
it
>> the relation between binary N and whatever N is the size of C 
just
>> so as to make them match up after all?
>It depends whether you are interpeting the strings as quantities, or 
just 
>as 
>raw strings. See? You can give a reasonable answer, even if the 
question 
>didn't
>expect it.
> 
> So you *are* saying that whether set A is bigger than set B is a
> *subjective* judgement; it depends on how you *interpret* the elements
> of A?
> 
> --
> Daryl McCullough
> Ithaca, NY
> 
> 
> I wouldn't say it is subjective, but dependent on context.
Which makes it subjective.
In mathematics, a set is entirely determined by what objects are members 
and what are non-members of that set. The context in which that set 
occurs is irrelevant to the set's identity. 
If two sets have the same members in any context then they are 
identically equal sets in every context.
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>Daryl McCullough said:
>> So you *are* saying that whether set A is bigger than set B is a
>> *subjective* judgement; it depends on how you *interpret* the elements
>> of A?
>I wouldn't say it is subjective, but dependent on context.
If it is context-dependent, that means that you really *don't*
have a notion of the relative sizes of two sets.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>Daryl McCullough said:
>> So you *are* saying that whether set A is bigger than set B is a
>> *subjective* judgement; it depends on how you *interpret* the elements
>> of A?
>I wouldn't say it is subjective, but dependent on context.
> If it is context-dependent, that means that you really *don't*
> have a notion of the relative sizes of two sets.
> --
> Daryl McCullough
> Ithaca, NY
No, it means it depends on the definition of the elements.
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>Daryl McCullough said:
>> Tony Orlow says...
Daryl McCullough said:
>> If it is context-dependent, that means that you really *don't*
>> have a notion of the relative sizes of two sets.
>No, it means it depends on the definition of the elements.
Once again, let A be the set of all finite strings that only
use characters '0' and '1' and such that 0 is the only string
starting with the character '0'.
So A = { 0, 1, 10, 11, ... }
Let B = the set of all natural numbers { 0, 1, 2, 3, ... }
Let C = the set of all natural numbers that can be written
in base-10 using only '0' and '1': { 0, 1, 10, 11, 100, 101, ... }
Does A have the same size as B, or does it have the same size as C?
--
Daryl McCullough
Ithaca, NY
  
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>No, I believe it was Daryl (I might be wrong, maybe Tribble?) who was 
>explaining how the set size is the first ordinal after the largest 
element,
> I didn't say any such thing. What I said was that omega is the
> smallest ordinal that is larger than *every* finite ordinal. There
> is no largest finite ordinal, but there is a smallest infinite
> ordinal.
> --
> Daryl McCullough
> Ithaca, NY
You gave finite examples with a largest element, and then said that since 
there 
was no largest to the infinite set, one uses the first limit ordinal, which 

is 
the next after all the successor ordinals. Isn't that what you said?
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Daryl McCullough said:
> Tony Orlow says...
>  
>No, I believe it was Daryl (I might be wrong, maybe Tribble?) who 
>was explaining how the set size is the first ordinal after the 
>largest element,
> 
> I didn't say any such thing. What I said was that omega is the 
> smallest ordinal that is larger than *every* finite ordinal. There 
> is no largest finite ordinal, but there is a smallest infinite 
> ordinal.
> 
> -- Daryl McCullough Ithaca, NY
> 
> 
> You gave finite examples with a largest element, and then said that 
> since there was no largest to the infinite set, one uses the first 
> limit ordinal, which is the next after all the successor ordinals. 
> Isn't that what you said?
It is correct whether it was said or not, and it differs noticeably from 
what TO said was said. TO is still being delusional about 'largest 
naturals'.
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>Daryl McCullough said:
>> Tony Orlow says...
>>  
>>No, I believe it was Daryl (I might be wrong, maybe Tribble?) who was 
>>explaining how the set size is the first ordinal after the largest 
element,
>> I didn't say any such thing. What I said was that omega is the
>> smallest ordinal that is larger than *every* finite ordinal. There
>> is no largest finite ordinal, but there is a smallest infinite
>> ordinal.
>You gave finite examples with a largest element, and then said that since 
there
>was no largest to the infinite set, one uses the first limit ordinal, which 

is 
>the next after all the successor ordinals. Isn't that what you said?
What I didn't say was the set size is the first ordinal after the
largest element.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>Fine, if you say so, but I can certainly entertain notions of 
infinite-tuples, 
>and am not the only one. Perhaps they aren't usually called n-tuples, but 
for 
>infinite n, there is no difference. What was your point, anyway?
> Going back in the thread, I said that in mathematics, a list refers
> to a set of elements indexed by finite natural numbers, and you disputed
> that. You provided a bunch of URLs that all agreed with my terminology.
> --
> Daryl McCullough
> Ithaca, NY
And this was in the context of a well ordering as a list? Well, then, no 
well 
ordering could ever be performed on an uncountable set, then, could it?
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Daryl McCullough said:
>  I said that in mathematics, a list refers
> to a set of elements indexed by finite natural numbers, and you 
> disputed that. You provided a bunch of URLs that all agreed with my 
> terminology.
> 
> And this was in the context of a well ordering as a list? Well, then, no 
well 
> ordering could ever be performed on an uncountable set, then, could it?
If one accepts the axiom of choice as being true, then, at least 
theoretically, every set must be well orderable, even if we do not know 
exactly how to do it for every set.
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>And this was in the context of a well ordering as a list? Well, then, no 
well 
>ordering could ever be performed on an uncountable set, then, could it?
On the contrary, if we let omega_1 = the set of all countable ordinals,
then omega_1 is uncountable, and it is well-ordered.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>And this was in the context of a well ordering as a list? Well, then, no 
well 
>ordering could ever be performed on an uncountable set, then, could it?
> On the contrary, if we let omega_1 = the set of all countable ordinals,
> then omega_1 is uncountable, and it is well-ordered.
> --
> Daryl McCullough
> Ithaca, NY
Are you saying there is no infinite descending chain from omega? I would 
imagine there is. Hw do you get around that?
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>Are you saying there is no infinite descending chain from omega? I would 
>imagine there is.
Well, there clearly isn't. If alpha_1, alpha_2, alpha_3, etc. is a
descending chain from omega, then by definition of descending chain,
alpha_1 = omega,
alpha_2 = some ordinal less than omega.
By definition, omega is the smallest infinite ordinal. So if
alpha_2 < omega, then alpha_2 is *finite*. Therefore, the chain
alpha_2,alpha_3, ... must be a finite chain. The chain alpha_1, alpha_2, 
...
must be finite, as well.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>Are you saying there is no infinite descending chain from omega? I would 
>imagine there is.
> Well, there clearly isn't. If alpha_1, alpha_2, alpha_3, etc. is a
> descending chain from omega, then by definition of descending chain,
> alpha_1 = omega,
> alpha_2 = some ordinal less than omega.
> By definition, omega is the smallest infinite ordinal. So if
> alpha_2 < omega, then alpha_2 is *finite*. Therefore, the chain
> alpha_2,alpha_3, ... must be a finite chain. The chain alpha_1, alpha_2, 
...
> must be finite, as well.
> --
> Daryl McCullough
> Ithaca, NY
Oh!! So omega is a finite number of steps from 0, then? That makes 
sense......
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Are you saying there is no infinite descending chain from omega? I would 
> imagine there is.
Do more. Describe it.
> Hw do you get around that?
By observing that your failure of understanding does not
constitute a proof.
===
Subject: Re: Well Ordering the Reals
Robert Low said:
> Are you saying there is no infinite descending chain from omega? I would 
> imagine there is.
> Do more. Describe it.
The set of elements less than omega is infinite, right? Is that not an 
infinite 
descending chain? Perhaps you get around this by noting that omega is not 
successor to any particular element, so that you cannot define ANY chain 
descending from it without naming a finite ordinal, and therefore making the 

chain finite. Is that about right? 
> Hw do you get around that?
> By observing that your failure of understanding does not
> constitute a proof.
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Robert Low said:
>Are you saying there is no infinite descending chain from omega? I would 
>imagine there is.
>>Do more. Describe it.
> The set of elements less than omega is infinite, right? Is that not an 
infinite 
> descending chain? Perhaps you get around this by noting that omega is not 

> successor to any particular element, so that you cannot define ANY chain 
> descending from it without naming a finite ordinal, and therefore making 
the 
> chain finite. Is that about right? 
Fairly close, though this is true for *any* ordinal, even
if it's uncountable. An infinite descending chain would
violate well-foundedness.
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>Robert Low said:
>> Are you saying there is no infinite descending chain from omega? I 
would 
>> imagine there is.
>> Do more. Describe it.
>The set of elements less than omega is infinite, right? Is that not an
>infinite descending chain?
No. The definition of an infinite descending chain is a totally ordered
set with no smallest element. The set of all elements less than omega
has a smallest element, namely 0.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
   <Ip8Cy8.8pu@cwi.nl>
   <MPG.1dd0180a70f3a7bc98a5a5@newsstand.cit.cornell.edu>
   <Ip90rJ.BA5@cwi.nl>
   <MPG.1dd1622bc3553e7398a5ce@newsstand.cit.cornell.edu>
   <dk86eg02vr8@drn.newsguy.com>
   <MPG.1dd16f5d868a36cf98a5d8@newsstand.cit.cornell.edu>
   <dk8d8g0pop@drn.newsguy.com>
   <MPG.1dd19d3ad68c592798a5ef@newsstand.cit.cornell.edu>
   <dk8mu101mv6@drn.newsguy.com>
   <MPG.1dd297c4141de8498a5fa@newsstand.cit.cornell.edu>
   <dkan0001k5a@drn.newsguy.com>
   <MPG.1dd2c048d47bda4a98a613@newsstand.cit.cornell.edu>
   <dkaue602anh@drn.newsguy.com>
   <MPG.1dd304921ac8154098a62a@newsstand.cit.cornell.edu>
   <dkbf070q67@drn.newsguy.com>
   <MPG.1dd3f357f3049e8498a635@newsstand.cit.cornell.edu Daryl McCullough 
said:
> Tony Orlow says...
>Fine, if you say so, but I can certainly entertain notions of 
infinite-tuples,
>and am not the only one. Perhaps they aren't usually called n-tuples, 
but for
>infinite n, there is no difference. What was your point, anyway?
> Going back in the thread, I said that in mathematics, a list refers
> to a set of elements indexed by finite natural numbers, and you 
disputed
> that. You provided a bunch of URLs that all agreed with my terminology.
> And this was in the context of a well ordering as a list? Well, then, no 
well
> ordering could ever be performed on an uncountable set, then, could it?
Tony, I wonder if it will occur to you to look up the definition of
well order in a book? It is not the same as enumeration: in
particular (we're told) the axiom of choice implies that every set has
a well-ordering, but a very simple proof shows that there cannot be an
enumeration of the (standard) reals. You are making things rather hard
for yourself by trying to prove the one that is impossible, instead of
the one that should be possible.
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Well Ordering the Reals
imaginatorium@despammed.com said:
> Daryl McCullough said:
> Tony Orlow says...
Fine, if you say so, but I can certainly entertain notions of 
infinite-tuples,
>and am not the only one. Perhaps they aren't usually called n-tuples, 
but for
>infinite n, there is no difference. What was your point, anyway?
 Going back in the thread, I said that in mathematics, a list 
refers
> to a set of elements indexed by finite natural numbers, and you 
disputed
> that. You provided a bunch of URLs that all agreed with my 
terminology.
 And this was in the context of a well ordering as a list? Well, then, no 
well
> ordering could ever be performed on an uncountable set, then, could it?
> Tony, I wonder if it will occur to you to look up the definition of
> well order in a book? It is not the same as enumeration: in
> particular (we're told) the axiom of choice implies that every set has
> a well-ordering, but a very simple proof shows that there cannot be an
> enumeration of the (standard) reals. You are making things rather hard
> for yourself by trying to prove the one that is impossible, instead of
> the one that should be possible.
I have looked it up on mathworld, which is nice because it has links to 
explanations of the terms used. It has no definition on the Enumeration 
page. 
I take it you mean Cantor's diagonal proof of the uncountability of the 
reals? 
It proves there are more strings than digits, ala N=S^L. Cantor's first? It 

shows we need infinite digits to represent the elements of the infinite set. 

Nothing says to me there cannot be an enumeration. Wouldn't any explicit 
well 
ordering really require an enumeration?
> Brian Chandler
> http://imaginatorium.org
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
   <Ip8Cy8.8pu@cwi.nl>
   <MPG.1dd0180a70f3a7bc98a5a5@newsstand.cit.cornell.edu>
   <Ip90rJ.BA5@cwi.nl>
   <MPG.1dd1622bc3553e7398a5ce@newsstand.cit.cornell.edu>
   <dk86eg02vr8@drn.newsguy.com>
   <MPG.1dd16f5d868a36cf98a5d8@newsstand.cit.cornell.edu>
   <dk8d8g0pop@drn.newsguy.com>
   <MPG.1dd19d3ad68c592798a5ef@newsstand.cit.cornell.edu>
   <dk8mu101mv6@drn.newsguy.com>
   <MPG.1dd297c4141de8498a5fa@newsstand.cit.cornell.edu>
   <dkan0001k5a@drn.newsguy.com>
   <MPG.1dd2c048d47bda4a98a613@newsstand.cit.cornell.edu>
   <dkaue602anh@drn.newsguy.com>
   <MPG.1dd304921ac8154098a62a@newsstand.cit.cornell.edu>
   <dkbf070q67@drn.newsguy.com>
   <MPG.1dd3f357f3049e8498a635@newsstand.cit.cornell.edu Daryl McCullough 
said:
> Tony Orlow says...
>Fine, if you say so, but I can certainly entertain notions of 
infinite-tuples,
>and am not the only one. Perhaps they aren't usually called n-tuples, 
but for
>infinite n, there is no difference. What was your point, anyway?
> Going back in the thread, I said that in mathematics, a list refers
> to a set of elements indexed by finite natural numbers, and you 
disputed
> that. You provided a bunch of URLs that all agreed with my terminology.
> And this was in the context of a well ordering as a list? Well, then, no 
well
> ordering could ever be performed on an uncountable set, then, could it?
Incorrect. There is no requirement for a well-ordering to be in the
form
of a list.
                 - Randy
===
Subject: Re: Well Ordering the Reals
Virgil said:
> David Kastrup said:
> So in your opinion the set
> 
> {..., 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 0, 2, 4, 6, 8, 10, 12, 14 
> ...}
> 
> where every member happens to have an infinite number of 
> predecessors and an infinite number of successors would be an 
> infinite set?
> 
> 
> Do you actually allow infinite values for these elements which are 
> finitely spaced on the number line? 
> TO keeps claiming that his version of the set of naturals, which is well 
> ordered, contains a nonempty set with no first member.
> But then, TO-matics (TO's version of mathematics) only exists in a 
> looking glass world where one has to believe in 8 impossible things 
> before breakfast to get by.
Look at the name of the thread. Do you know the difference between naturals 

and 
reals?
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Virgil said:
> 
> David Kastrup said:
> 
> So in your opinion the set
> 
> {..., 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 0, 2, 4, 6, 8, 10, 12, 
> 14 ...}
> 
> where every member happens to have an infinite number of 
> predecessors and an infinite number of successors would be an 
> infinite set?
> 
> 
> Do you actually allow infinite values for these elements which 
> are finitely spaced on the number line? 
>  
> TO keeps claiming that his version of the set of naturals, which is 
> well ordered, contains a nonempty set with no first member.
> 
> But then, TO-matics (TO's version of mathematics) only exists in a 
> looking glass world where one has to believe in 8 impossible things 
> before breakfast to get by.
> 
> Look at the name of the thread. Do you know the difference between 
> naturals and reals?
Since TO was commenting on David's set of naturals, I had supposed that 
not every set under discussion had to be the set of all reals.
Obviously in TO's world, TO gets to talk about anything but everyone 
else gets to talk about only what TO allows.
Well, this isn't TO's world.
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>How about you not post any more regarding the well ordering
>unless you can decide whether you are using standard or non-standard
>math?
> I used your notation to show that your ordering was not a well-ordering
> for your reals. I used standard notation to show that the standard 
ordering
> on the reals is not a well-ordering. What else do you want? The fact
> is that you did *not* come up with a well-ordering of either the
> standard reals, nor your nonstandard reals. So your claim to have
> a well-ordering of the reals is false.
> --
> Daryl McCullough
> Ithaca,  NY
I want a standard refutation of the well ordering of the standard reals 
which 
is put in standard terms. Consider -oo and oo not to be members of the set, 

and 
give me a standard objection to the well ordering. I assume a countably 
infinite number of bits is sufficient for enumerating the reals? Let's see 
what 
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>I want a standard refutation of the well ordering of the standard reals 
which 
>is put in standard terms. Consider -oo and oo not to be members of the set, 

and
>give me a standard objection to the well ordering. I assume a countably 
>infinite number of bits is sufficient for enumerating the reals? Let's see 

what
You describe a mapping from bitstrings to reals.
The problem, though, is that if you only use finite
bitstrings, then your mapping doesn't include all the
reals. On the other hand, if you include the *infinite*
bit strings, then there is no (known) well-ordering of
the infinite bitstrings. Your ordering of the reals
relies on the lexicographic ordering of the
set of bitstrings, and that is not a well-ordering.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>I want a standard refutation of the well ordering of the standard reals 
which 
>is put in standard terms. Consider -oo and oo not to be members of the 
set, and
>give me a standard objection to the well ordering. I assume a countably 
>infinite number of bits is sufficient for enumerating the reals? Let's 
see what
> You describe a mapping from bitstrings to reals.
> The problem, though, is that if you only use finite
> bitstrings, then your mapping doesn't include all the
> reals. 
Now, wait a minute. What about Cantor's First? Is he allowing an infinite 
number of terms in his sequences or not? If he is, then it becomes obvious, 

at 
least in the context of my well ordering, that c is the point at n=oo where 

sequences A and B meet. If not, then how can he claim that he has 
determined, 
in a finite number of subdivisions, that c is not in R?
> On the other hand, if you include the *infinite*
> bit strings, then there is no (known) well-ordering of
> the infinite bitstrings. Your ordering of the reals
> relies on the lexicographic ordering of the
> set of bitstrings, and that is not a well-ordering.
Why is that? The dictionary certainly nevers seems to be confused about what 

comes after what, or what comes first. What condition of well ordering does 

that ordering violate?
> --
> Daryl McCullough
> Ithaca, NY
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
   <85oe54p94x.fsf@lola.goethe.zz>
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   <MPG.1dd415227eb3836098a63e@newsstand.cit.cornell.edu Daryl McCullough 
said:
<snip On the other hand, if you include the *infinite*
> bit strings, then there is no (known) well-ordering of
> the infinite bitstrings. Your ordering of the reals
> relies on the lexicographic ordering of the
> set of bitstrings, and that is not a well-ordering.
> Why is that? The dictionary certainly nevers seems to be confused about 
what
> comes after what, or what comes first. What condition of well ordering 
does
> that ordering violate?
You know, the one truly astonishing thing about all this is that as you
spew out your endless nonsense, people like me are actually learning
something from the process. OK, a slim volume I have here is the
Penguin Dict. of Math., and it says A set is well-ordered if it is
ordered [I suppose that means a total order] and every subset has a
first element.
Well, in the lexicographic ordering of infinite bit strings, the set
{1000..., 0100..., 0010..., ...} has no first element. Ergo it isn't
well-ordered.
Go on - now babble a bit. <g>
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Well Ordering the Reals
imaginatorium@despammed.com said:
> Daryl McCullough said:
> <snip On the other hand, if you include the *infinite*
> bit strings, then there is no (known) well-ordering of
> the infinite bitstrings. Your ordering of the reals
> relies on the lexicographic ordering of the
> set of bitstrings, and that is not a well-ordering.
> Why is that? The dictionary certainly nevers seems to be confused about 
what
> comes after what, or what comes first. What condition of well ordering 
does
> that ordering violate?
> You know, the one truly astonishing thing about all this is that as you
> spew out your endless nonsense, people like me are actually learning
> something from the process. OK, a slim volume I have here is the
> Penguin Dict. of Math., and it says A set is well-ordered if it is
> ordered [I suppose that means a total order] and every subset has a
> first element.
> Well, in the lexicographic ordering of infinite bit strings, the set
> {1000..., 0100..., 0010..., ...} has no first element. Ergo it isn't
> well-ordered.
> Go on - now babble a bit. <g>
If you are working with the infinite bit strings, then it is only your 
restriction of finiteness on the number of intermediate zeroes which makes 
that 
sequence not end. If you allow the intermediate zeroes to be infinite in 
number, then the bottom of that chain is ...00001, wouldn't you say?
> Brian Chandler
> http://imaginatorium.org
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
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   <dk8b900iii@drn.newsguy.com>
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imaginatorium@despammed.com said:
> Daryl McCullough said:
> <snip On the other hand, if you include the *infinite*
> bit strings, then there is no (known) well-ordering of
> the infinite bitstrings. Your ordering of the reals
> relies on the lexicographic ordering of the
> set of bitstrings, and that is not a well-ordering.
> Why is that? The dictionary certainly nevers seems to be confused 
about what
> comes after what, or what comes first. What condition of well ordering 
does
> that ordering violate?
> You know, the one truly astonishing thing about all this is that as you
> spew out your endless nonsense, people like me are actually learning
> something from the process. OK, a slim volume I have here is the
> Penguin Dict. of Math., and it says A set is well-ordered if it is
> ordered [I suppose that means a total order] and every subset has a
> first element.
> Well, in the lexicographic ordering of infinite bit strings, the set
> {1000..., 0100..., 0010..., ...} has no first element. Ergo it isn't
> well-ordered.
> Go on - now babble a bit. <g If you are working with the infinite bit 
strings, then it is only your
> restriction of finiteness on the number of intermediate zeroes which makes 

that
> sequence not end. If you allow the intermediate zeroes to be infinite in
> number, then the bottom of that chain is ...00001, wouldn't you say?
Yes, I suppose if I were at the point at which infinity ends, I woud
say that, but I would also say everything else, and I would be the King
of Peru, because there is no such place.
Yawn oh yawn, you can now do your Bongo trick, where you say No
largest natural yahooo!, No end to the unending.... Spliftjhgy! Or
whatever. You simply cannot grasp the concept of an infinite set. You
can understand a big set, or a verrrry big set, or a very very really
hugely enormous set, but you cannot understand the idea of considering
an unending sequence in its totality. Of course not, because you have
this simplistic idea that totality must mean you got to the last one.
Well, it doesn't. Never mind.
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>imaginatorium@despammed.com said:
>> Well, in the lexicographic ordering of infinite bit strings, the set
>> {1000..., 0100..., 0010..., ...} has no first element. Ergo it isn't
>> well-ordered.
>If you are working with the infinite bit strings, then it is only your 
>restriction of finiteness on the number of intermediate zeroes which makes 

that
>sequence not end. If you allow the intermediate zeroes to be infinite in 
>number, then the bottom of that chain is ...00001, wouldn't you say?
So what? If a set is well-ordered then *every* subset has a smallest
element. So even if you want to allow a nonstandard number of bits,
that's fine. We can still talk about the set of all bit strings such
there is exactly one 1, and its index is finite (if you start counting
from the left).
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
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   <dkdlop01fid@drn.newsguy.com Tony Orlow says...
>imaginatorium@despammed.com said:
>> Well, in the lexicographic ordering of infinite bit strings, the set
>> {1000..., 0100..., 0010..., ...} has no first element. Ergo it isn't
>> well-ordered.
>If you are working with the infinite bit strings, ...
By force of habit, when I see the words infinite bit strings I think
of what we call infinite ones, that go on with no end. I forget that
you only grasp imponderably enormous ones, so you can always get to the
end if need be. Well, tough.
> ... then it is only your
>restriction of finiteness on the number of intermediate zeroes which 
makes that
>sequence not end.
I'm not sure which are supposed to be intermediate, but no, even
though normally every one of the sequence
100000....
010000....
001000....
000100....
000010....
...
would have a finite index to reach the '1',... for you I'll happily
allow an infinite string of zeros, followed by a 1, followed by more
zeros, followed by many more zeros, followed by many many more zeros,
followed by just hugely enormously many zeros, followed by, er, an
unending string of zeros. The pattern is some zeros, a 1, some more
zeros.
> If you allow the intermediate zeroes to be infinite in
>number, then the bottom of that chain is ...00001, wouldn't you say?
So no, of course, under no circumstances would I say that P is the end
of an unending sequence R, for any R or P.
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Well Ordering the Reals
David Kastrup said:
> David Kastrup said:
>> 
>> Virgil said:
> 
> So how does one tell when a number of elements is infinite, 
rather 
> than finite? 
>> When a set is recursively defined with no terminating state, and
>> there is no restriction of finiteness on the number of recursive
>> iterations.
>> 
>> Well, like
>> 10 is a member of X
>> if k is a member of X, so is floor(k/2)
>> a set containing 10, and for each of its members k containing
>> floor(k/2), contains all members of X.
>> 
>> Then it happens that X = {0,1,2,5,10}, even though recursively defined
>> with no terminating state and no restriction of finiteness on the
>> number of recursive iterations.  And you would like to claim that X is
>> an infinite set?
> I didn't seem to say it above,
> Oh, you made a mistake then?  How extraordinary!  How in the world was
> that possible?
> but I said elsewhere that elements would generate at least one
> UNIQUE successor.
> You said wagonloads of incoherent and contradictory crap elsewhere, so
> you can't expect us to choose from that heap, as it would come
> tumbling down on our heads.
> 1 and 0 have the same successor. Oh well. Better luck next time.
> Ah, so the set defined with
>     0 is a member of X
>     if k is a member of X, so is ((k+1) mod 5)
>     a set containing 0, and for each of its members k containing ((k+1)
>     mod 5) contains all members of X
> is an infinite set, since all elements have unique successors.  No
> doubt you'll come up with another piece of hogwash that you think you
> might have uttered elsewhere that will cater for this case.
Very well. An infinite set is a set defined recursively so that each element 

has at least one unique successor WHICH IS NOT ITSELF, with no limitation on 

the number of successor operations. Happy?
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> David Kastrup said:
> Ah, so the set defined with
> 
>     0 is a member of X
>     if k is a member of X, so is ((k+1) mod 5)
>     a set containing 0, and for each of its members k containing ((k+1)
>     mod 5) contains all members of X
> 
> is an infinite set, since all elements have unique successors.  No
> doubt you'll come up with another piece of hogwash that you think you
> might have uttered elsewhere that will cater for this case.
> 
> 
> Very well. An infinite set is a set defined recursively so that each 
element 
> has at least one unique successor WHICH IS NOT ITSELF, with no limitation 

on 
> the number of successor operations. Happy?
None of the elements in David's example has itself as its successor.
===
Subject: Re: Well Ordering the Reals
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
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 VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
> David Kastrup said:
>> David Kastrup said:
> 
> Virgil said:
>> 
>> So how does one tell when a number of elements is infinite, 
rather 
>> than finite? 
> When a set is recursively defined with no terminating state, and
> there is no restriction of finiteness on the number of recursive
> iterations.
> 
> Well, like
> 10 is a member of X
> if k is a member of X, so is floor(k/2)
> a set containing 10, and for each of its members k containing
> floor(k/2), contains all members of X.
> 
> Then it happens that X = {0,1,2,5,10}, even though recursively 
defined
> with no terminating state and no restriction of finiteness on the
> number of recursive iterations.  And you would like to claim that X 
is
> an infinite set?
 I didn't seem to say it above,
>> Oh, you made a mistake then?  How extraordinary!  How in the world was
>> that possible?
>> but I said elsewhere that elements would generate at least one
>> UNIQUE successor.
>> You said wagonloads of incoherent and contradictory crap elsewhere, so
>> you can't expect us to choose from that heap, as it would come
>> tumbling down on our heads.
>> 1 and 0 have the same successor. Oh well. Better luck next time.
>> Ah, so the set defined with
>>     0 is a member of X
>>     if k is a member of X, so is ((k+1) mod 5)
>>     a set containing 0, and for each of its members k containing ((k+1)
>>     mod 5) contains all members of X
>> is an infinite set, since all elements have unique successors.  No
>> doubt you'll come up with another piece of hogwash that you think you
>> might have uttered elsewhere that will cater for this case.
> Very well. An infinite set is a set defined recursively so that each 
element 
> has at least one unique successor WHICH IS NOT ITSELF, with no limitation 

on 
> the number of successor operations. Happy?
The above circular set (sets don't actually have an order)
{0,1,2,3,4} still would be infinite according to that definition,
since each element has a unique successor which is not itself.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: Well Ordering the Reals
David Kastrup said:
> David Kastrup said:
>> 
>> David Kastrup said:
> 
> Virgil said:
>> 
>> So how does one tell when a number of elements is infinite, 
rather 
>> than finite? 
> When a set is recursively defined with no terminating state, and
> there is no restriction of finiteness on the number of recursive
> iterations.
> 
> Well, like
> 10 is a member of X
> if k is a member of X, so is floor(k/2)
> a set containing 10, and for each of its members k containing
> floor(k/2), contains all members of X.
> 
> Then it happens that X = {0,1,2,5,10}, even though recursively 
defined
> with no terminating state and no restriction of finiteness on the
> number of recursive iterations.  And you would like to claim that X 
is
> an infinite set?
 I didn't seem to say it above,
>> 
>> Oh, you made a mistake then?  How extraordinary!  How in the world was
>> that possible?
>> 
>> but I said elsewhere that elements would generate at least one
>> UNIQUE successor.
>> 
>> You said wagonloads of incoherent and contradictory crap elsewhere, so
>> you can't expect us to choose from that heap, as it would come
>> tumbling down on our heads.
>> 
>> 1 and 0 have the same successor. Oh well. Better luck next time.
>> 
>> Ah, so the set defined with
>> 
>>     0 is a member of X
>>     if k is a member of X, so is ((k+1) mod 5)
>>     a set containing 0, and for each of its members k containing 
((k+1)
>>     mod 5) contains all members of X
>> 
>> is an infinite set, since all elements have unique successors.  No
>> doubt you'll come up with another piece of hogwash that you think you
>> might have uttered elsewhere that will cater for this case.
>> 
>> 
> Very well. An infinite set is a set defined recursively so that each 
element 
> has at least one unique successor WHICH IS NOT ITSELF, with no 
limitation on 
> the number of successor operations. Happy?
> The above circular set (sets don't actually have an order)
> {0,1,2,3,4} still would be infinite according to that definition,
> since each element has a unique successor which is not itself.
Lovely! A circular succession in ways can be considered infinite, but your 
point is well taken. It would seem that one would have to add the constraint 

of 
partial order. How are we doing now, David?
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
Robert Low said:
> Robert Low said:
>Now, is the only objection to my well ordering Daryl's infinite 
descending 
>chain? 
>>Since that proves that it isn't a well ordering, it's enough.
> Sure, except that set definition isn't allowed in standard mathematics, 
so 
> within the context that seeks a well ordering, it is not a valid 
infinite 
> descending chain. Next.....
> Your argument that your well-ordering is OK is that proofs that it
> isn't use your own incorrect ideas, and hence are invalid? 5/5 for
> chutzpah. 0/5 for anything resembling reason.
If I am playing your game now, in offering the well ordering of the reals 
that 
standard mathematics seeks, then to discredit it, you must stick to the 
standard rules. I am not doing anything non-standard in my well ordering, so 

your statement is vacuous.
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> http://www.people.cornell.edu/pages/aeo6/WellOrder/
Where does TO present a rule for testing whether one binary string 
precedes or follows another in TO's ordering? Whithout such a criterion, 
TO does not have an ordering, muchless a well ordering. 
Where is TO's proof that every real is represented in his ordering?
Can TO show, for example the representstion for e, or pi, or sqrt(2), or 
the real roots of an arbitrary  polynomial?
Can TO show that the set of real numbers he represents has the least 
upper bound property?
There are massive enough holes in TO's presentation to allow us to 
conclude that his claim is NOT PROVEN.
===
Subject: Re: Well Ordering the Reals
Virgil said:
> You know, it suddenly occurs to me that, while it was some source of 
> delight to see you using my notation for infinite numbers, that is 
> not standard mathematical notation, and therefore is not available to 
> you in proving my well-ordering false.
> If TO is going to require standard notation in the analysis of his 
> system, he is constrained to use only standard notation to present his 
> system. And there is no way that standard notation will allow a string 
> of digits with a firs digit and a last one and infinitely many in 
> between.
> Thus if TO decalres a representational system, anyone is justified in 
> using  that very system to analyse it.
You speak from ignorance. Tell me where in my well ordering I used any non-
standard notation. You haven't even looked at it, and you deign to comment. 

Grow up.
> Can you provide some infinite 
> descending chain using standard notation and concepts? 
> Since there is reasonable doubt that TO's system even exists outside of 
> the dream world of TOmatics, TO must prove that his system even exists 
> in standard mathematics before any such constrint can be put on analysis 
> of it.
Ignorant.
> If, as I suspect, it only exists in that odd world of TOmatics, then 
> TOmatics is the only place, and TOmatic methods are the only methods, by 
> which it can be analysed. 
Ignorant.
>If I am to use 
> your standard definition of well-ordering...
> Since TO does not use the standard definitions of anything else, why 
> does TO bother to stick with the standard here?
Because now I am playing your game. So, now you want to play mine? My game 
doesn't have well orderings persay.
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Virgil said:
> 
> 
> You know, it suddenly occurs to me that, while it was some source of 
> delight to see you using my notation for infinite numbers, that is 
> not standard mathematical notation, and therefore is not available to 
> you in proving my well-ordering false.
> 
> If TO is going to require standard notation in the analysis of his 
> system, he is constrained to use only standard notation to present his 
> system. And there is no way that standard notation will allow a string 
> of digits with a first digit and a last one and infinitely many in 
> between.
> 
> Thus if TO declares a representational system, anyone is justified in 
> using  that very system to analyse it.
> You speak from ignorance.
How is it ignorance to say that TO is reaponsible for his own arguments, 
and that if they should be faulty, they can be used against him?
===
Subject: Re: Well Ordering the Reals
Virgil said:
> Virgil said:
> 
> Respond specifically to my comments on Cantor's First. Tell me it 
> that, in the context of my well ordering, the real number c in the 
> proof does not boil down to the last element, after you read my 
> response.
> 
> What sequence does TO alleged 'c' is to be the last of? 
> The cmbination of A and B, or either one.
> The whole point is that c is NOT in either or in both sequences at all!
It is in both, if you allow infinite iterations, as I showed. I know what 
the 
proof purports to say. I demonstrated that, given my well ordering, which I 

am 
sure you have still failed to read, that very c is the limit of the elements 

in 
the sequence as the iterations approach oo. That only says to me that any 
repreentation of such a number would conceptually have an infinite number of 

symbols, but we already know that reals can have infinite bits, so that 
means 
nothing.
> 
> It is certainly neither the A_n's nor the B_n's.
> It is, after an infintie number of iterations, with an infinite number 
of 
> bits/exponentiations.
> Since these sequences are mappings from the infinite set of finite 
> naturals to the reals, there is no such after.
Use your standard concept of limit, as above.
> 
> One has an arbitrary injective function from the infinite set of 
finite 
> naturals to the even more infinite set of reals.
> Yes, well, if you only have a finite number of iterations, then you 
really 
> aren't including every real int he set anyway.
> WE are including the infinitely many finite naturals, which is all 
> anyone needs to establish the theorem.
So, it is considered sufficient to enumerate the reals using COUNTABLY many 

bits? Fine, then the infinite descending chain fails to exist, since any bit 

is 
finitely far from the start of the string. Tada!
> 
> Then n -> A_n, and n -> B_n are subsequences of that original function 
> chosen so that n -> A_n is strictly order-preserving , n -> B_n is 
> strictly order reversing, each B_n is an upper bound on the set {A_n}, 
> and each A_n is a lower bound on the set {B_n}.
> Yes, and that is the way it remains for every finite iteration. 
>  But we have iterations for the infinite set of finite naturals, so we 
> are not so limited.
> I t requires an infinite number of bits to represent the values in a 
> truly infinite set.
> Not in actual mathematics, only in the delusional world of TOmatics!
In reality, it does, by N=S^L.
> 
> real number c that cannot be in the original injection, but what 'c' 
is 
> supposed to be last element of is not evident. If one means the 
set of 
> reals given by the original mapping, it is not a member of that set at 
> all, so it cannot be a last with respect to that set.
> Read the damn web page. I am relating Cantor's First to my ordering of 
the 
> reals.
> I have no interest in reading that which is damned.
Then you're a fool.
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Virgil said:
> 
> Virgil said:
> 
> Respond specifically to my comments on Cantor's First. Tell me it 
> that, in the context of my well ordering, the real number c in the 
> proof does not boil down to the last element, after you read my 
> response.
> 
> What sequence does TO alleged 'c' is to be the last of? 
> The cmbination of A and B, or either one.
> 
> The whole point is that c is NOT in either or in both sequences at all!
> It is in both, if you allow infinite iterations, as I showed.
But the proof, being done in standard mathematics and not in the 
dreamworld of TOmatics, does not either require nor allow iteration 
beyond those indexes found in the infinite set of finite naturals.
> Since these sequences are mappings from the infinite set of finite 
> naturals to the reals, there is no such after.
> Use your standard concept of limit, as above.
The standard concept of limit does not have any such after.
Definition: 
For N being the the infinite set of finite naturals, 
and R being the set of finite reals, 
   'lim_{n -> oo} f(n) = L' 
is defined to mean
   'for every positive real epsilon in R 
   there exists an n in N such that  
   whenever m in N and m larger than n, 
   then |L-f(m)| is smaller than epsilon.'
Where in that definition does it mention after?
> WE are including the infinitely many finite naturals, which is all 
> anyone needs to establish the theorem.
> So, it is considered sufficient to enumerate the reals using COUNTABLY 
many 
> bits?
The whole point, which TO seems to be missing, is that one cannot 
ennumerate the reals with only the infinite set of finite naturals.
 
One wonders what TO thinks the Cantor theorem is all about if he does 
not understand even that much of it.
> 
> Then n -> A_n, and n -> B_n are subsequences of that original 
function 
> chosen so that n -> A_n is strictly order-preserving , n -> B_n is 
> strictly order reversing, each B_n is an upper bound on the set 
{A_n}, 
> and each A_n is a lower bound on the set {B_n}.
> 
> Yes, and that is the way it remains for every finite iteration. 
>  But we have iterations for the infinite set of finite naturals, so we 
> are not so limited.
> 
> I t requires an infinite number of bits to represent the values in a 
> truly infinite set.
> 
> Not in actual mathematics, only in the delusional world of TOmatics!
> In reality, it does, by N=S^L.
The reality of TOmatics is quite unreal outside TOmatics.
 And TO's N=S^L argument is sufficient evidence of that delusion.
> 
> Read the damn web page. I am relating Cantor's First to my ordering of 
> the 
> reals.
> 
> I have no interest in reading that which is damned.
> Then you're a fool.
 
Perhaps, but at least not a damned fool like TO!
===
Subject: Re: Well Ordering the Reals
 > Dik T. Winter said:
...
 > That the interval [0, 1] with length 1 can be covered by a collection 
of
 > intervals with a total length less than 1.
 > 
 > Which may be some kind of contradiction in measure theory, but in set
 > theory, 
Why is it a contradiction in measure theory?  It comes because the 
assumption
is made that the reals are countable.
 > do you not consider the set of reals in [0,1] to be equal to the set in
 > [0,2]? 
Not in measure theory.  In set theory they are also not equal but
equi-numerous.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>Daryl McCullough said:
>> Why do you keep bringing up the end of a set which you agree
>> has no end?
>Because in order to measure it, you have to conceptualize some end to it
> So, you are saying that although you agree that the naturals have no
> largest element, you prefer to reason as if it *did* have a largest
> element. You prefer to reason using false premises.
> --
> Daryl McCullough
> Ithaca, NY
No, I prefer to reason inductively and formulaically using variable value 
ranges, rather than declare the existence of things that aren't there, like 

an 
exact set size for a set that has no end. Do you know what a variable is? 
They 
are used in formulas, and can take on any real value. Some, like myself, 
even 
consider the possiblity that variables may take on infinite values. It's 
like a 
limit concept, except that, instead of approaching oo but being too scared 
to 
touch it, in Bigulosity we pet it, and pick it up, and cradle it in our 
arms. 
After all, infinity is not a dangerous beast. It will not gore you with its 

horn. Go ahead and touch the little unicorn.
I do like getting poetic about such things. Visuals are very nice. Of 
course, 
it probably doesn't make you think I am NOT a crackpot, but I really 
couldn't 
give a flying, um, spitball. 
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Daryl McCullough said:
> Tony Orlow says...
Daryl McCullough said:
> 
>> Why do you keep bringing up the end of a set which you agree
>> has no end?
> 
>Because in order to measure it, you have to conceptualize some end to 
it
What TO has to do to measure it, is something only TO has to do. The 
rest of us have better methods. Though TO doesn't seem able to 
understand them.
> 
> So, you are saying that although you agree that the naturals have no
> largest element, you prefer to reason as if it *did* have a largest
> element. You prefer to reason using false premises.
> 
> --
> Daryl McCullough
> Ithaca, NY
> 
> 
> No, I prefer to reason inductively
 That's because deductive reasoning is totally beyond TO's capabilities, 
as evidenced by his many failed attempts at it.
> and formulaically using variable value 
> ranges, rather than declare the existence of things that aren't there, 
Like value ranges for sets which do not have them?
> I do like getting poetic about such things. Visuals are very nice. 
TO may be able to wax poetic about things poetic, and learn simple 
things with the aid of sufficiently well designed visuals, but he is 
totally incapable when faced with the formalities of hard logic, as 
required by any serious mathematics.
===
Subject: Re: Well Ordering the Reals
imaginatorium@despammed.com said:
> Daryl McCullough said:
> Tony Orlow says...
David Kastrup said:
> How is that a basic property?  Adding 1 to each element of a 
set
>> that has numbers in a ring is a reversible operation, so it can 
hardly
>> imply smaller set size.  And yet adding 1 to each element of N 
is
>> _exactly_ equivalent to removing 0 from the set, without adding 
any
>> element to it.
>Yes, because you do not consider what happens to the set at the other 
end.
 On the one hand, you claim that there is no largest natural number.
> That implies that there is no end. So what happens at the end
> is a nonsensical notion.
 Why do you keep bringing up the end of a set which you agree
> has no end?
 --
> Daryl McCullough
> Ithaca, NY
> Because in order to measure it, you have to conceptualize some end to 
it, as a
> variable which can assume infinite values. I thought I said that.
> Yeah, you arranged words in roughly that sequence many times. But what
> does it mean, we wonder? To determine the size of an infinite set,
> first conceptualise that it is _not_ an infinite set, then write down
> the answer quick before it deconceptualises back to being infinite
> again. Do you know how to catch a pink elephant?
Don't be obnoxious. Do you know what a variable is? Do you know what a limit 

is? As x goes to oo, which is larger, 2x or x^2? Do some thinking, and not 
just 
about how you can refute the concept. As they said once on Car Talk, embrace 

the smell.
> A: nug tnahpele knip a htiw ti toohS.
> And how do you catch a grey elephant then?
> A: nug tnahpele knip a htiw ti toohs dna knip seog ti litnu eson sti
> dloH.
> Seems to me this technique might be applicable to measuring infinite
> sets, somehow.
> But meanwhile, we are agog to hear quite how you measure these sets of
> strings that Daryl is giving you. Until you can do that, well-ordering
> the reals is taking on rather a lot, I fancy.
The well ordering of the reals has happened, Brian, and it didn't require 
any 
pink guns.
> Brian Chandler
> http://imaginatorium.org
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> imaginatorium@despammed.com said:
> Yeah, you arranged words in roughly that sequence many times. But what
> does it mean, we wonder? To determine the size of an infinite set,
> first conceptualise that it is _not_ an infinite set, then write down
> the answer quick before it deconceptualises back to being infinite
> again. Do you know how to catch a pink elephant?
> Don't be obnoxious. 
Does To claim a patent on being obnoxious? He is certainly more praticed 
at it that anyone else currently posting.
> Do you know what a variable is? Do you know what a limit 
> is? As x goes to oo, which is larger, 2x or x^2?
For most purposes it is quite irrelevant which is larger, since they are 
going where x goes, beyond any finite number.
> Do some thinking
To is deeply into giving advice he does not follow himself.
> The well ordering of the reals has happened, Brian, and it didn't require 

any 
> pink guns.
If it has happeneed, it wasn't TO that did it.
===
Subject: Re: Well Ordering the Reals
   <MPG.1dd15a2eedc716aa98a5c8@newsstand.cit.cornell.edu>
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<snip The well ordering of the reals has happened, Brian, and it didn't 
require 
any
> pink guns.
You mean the stuff on your web page? Well, you know what to do then.
Print it out and take a copy or two round to the Cornell math dept,
where I'm sure a claim to have produced a well-ordering of the reals
will be received with more interest than any mention of infinite
integers, angle trisections, or any of the more usual crank obsessions.
Don't feel obliged to tell us what happens if you don't want to, but do
not, I repeat, do not hold your breath. If you had a clearer conception
of what well-ordering means to mathematicians you might be even
better off; since they will notice fairly quickly that you are (I
suppose still) claiming an enumeration of them, interest may drop off
fairly quickly. Particularly as I don't think there's even a hint of a
proof that your list actually contains all of anything.
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Well Ordering the Reals
imaginatorium@despammed.com said:
> <snip The well ordering of the reals has happened, Brian, and it didn't 
require any
> pink guns.
> You mean the stuff on your web page? Well, you know what to do then.
> Print it out and take a copy or two round to the Cornell math dept,
> where I'm sure a claim to have produced a well-ordering of the reals
> will be received with more interest than any mention of infinite
> integers, angle trisections, or any of the more usual crank obsessions.
Yes, well, the thing I had approached Professor Shore with the first time 
was a 
bit of refutation of transfinite set theory, which evoke an immediate 
response 
of disdain. This subject is likely to be more palatable. This discussion has 

help.
> Don't feel obliged to tell us what happens if you don't want to, but do
> not, I repeat, do not hold your breath. If you had a clearer conception
> of what well-ordering means to mathematicians you might be even
> better off; since they will notice fairly quickly that you are (I
> suppose still) claiming an enumeration of them, interest may drop off
> fairly quickly. Particularly as I don't think there's even a hint of a
> proof that your list actually contains all of anything.
How would one prove that it contained all reals? It is certainly a set dense 

in 
the reals, since between any two values is an intermediate value, but this 
is 
also true of the rationals. The digital representations - are they 
considered 
to enumerate all real values? If so, why? If not, what other criterion does 

one 
need? Any ideas?
> Brian Chandler
> http://imaginatorium.org
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
   <MPG.1dd16dcef6a2e5bc98a5d7@newsstand.cit.cornell.edu>
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imaginatorium@despammed.com said:
> <snip The well ordering of the reals has happened, Brian, and it didn't 
require any
> pink guns.
> You mean the stuff on your web page? Well, you know what to do then.
> Print it out and take a copy or two round to the Cornell math dept,
> where I'm sure a claim to have produced a well-ordering of the reals
> will be received with more interest than any mention of infinite
> integers, angle trisections, or any of the more usual crank obsessions.
> Yes, well, the thing I had approached Professor Shore with the first time 

was a
> bit of refutation of transfinite set theory, which evoke an immediate 
response
> of disdain. This subject is likely to be more palatable. This discussion 
has
> help.
> Don't feel obliged to tell us what happens if you don't want to, but do
> not, I repeat, do not hold your breath. If you had a clearer conception
> of what well-ordering means to mathematicians you might be even
> better off; since they will notice fairly quickly that you are (I
> suppose still) claiming an enumeration of them, interest may drop off
> fairly quickly. Particularly as I don't think there's even a hint of a
> proof that your list actually contains all of anything.
> How would one prove that it contained all reals?
Well, that's your problem, isn't it? It's your claim. But in normal
mathematics it would be vastly easier (I think) to prove the contrary,
since it would probably only involve clarifying terminology enough to
see what you actually mean.
> It is certainly a set dense in
> the reals, since between any two values is an intermediate value, but this 

is
> also true of the rationals.
Exactly. So it gets you nowhere.
> The digital representations - are they considered
> to enumerate all real values?
What digital representations? The standard decimal representations of
the standard reals can be shown not to be enumerable, by the diagonal
argument. (I know, you can't really understand this; tough. I'm not
going through it all again.) Anyway, no, they are not considered to
enumerate, but anyway in real mathematics it's not a business of
considering this or that to be something, but of proving that it is
or isn't.
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Well Ordering the Reals
Tony Orlow says...
>How would one prove that it contained all reals?
To prove that a set contains all real numbers, you need to
show that it contains all rational numbers, and then show
that your set is complete. To be complete means that for
every Cauchy sequence of reals in your set converges to a
real in your set.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Well Ordering the Reals
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 Discussion, linux)
> Tony Orlow says...
>>How would one prove that it contained all reals?
Daryl, with all due respect, I found your response odd.
> To prove that a set contains all real numbers, you need to
  ^^ One way to...[1]
> show that it contains all rational numbers, and then show
> that your set is complete. To be complete means that for
> every Cauchy sequence of reals in your set converges to a
                           ^^^^^ rationals
> real in your set.
  ^^^ element (which is necessarily real)
Better?
Footnotes: 
[1]  There are other ways.  Certainly, I don't need to follow this
path to prove that R u R contains R.
-- 
Jesse F. Hughes
It's easy folks.  Just talk about my approach to your favorite
mathematician.  If they can't be interested in it, they've
demonstrated a lack of mathematical skill. -- James Harris
===
Subject: Re: Well Ordering the Reals
Daryl McCullough said:
> Tony Orlow says...
>How would one prove that it contained all reals?
> To prove that a set contains all real numbers, you need to
> show that it contains all rational numbers, and then show
> that your set is complete. To be complete means that for
> every Cauchy sequence of reals in your set converges to a
> real in your set.
> --
> Daryl McCullough
> Ithaca, NY
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
Robert Low said:
> Robert Low said:
>Tony Orlow says...
>>If one can, as you did, select an infinite element, can form a set by 
>>considering elements finitely less than that element, then any 
uncountable 
>>set has subsets which are infinite descending chains. 
>So you are saying that no uncountable set can be well-ordered.
>So you *don't* claim to have a well-ordering for the reals?
>>Wouldn't that admission require him to admit that the
>>reals aren't countable? (I can't tell any more...)
> IF Daryl's infinite descending chain is valid in standard mathematics, 
which I 
> guess it's not, since it uses MY non-standard infinite bitstrings. 
Hmmm...
> IF you were capable of understanding Daryl's argument, you'd
> notice that all the strings he lists are well-defined infinite
> binary strings. Your strings (whatever they might be) certainly
> include all the ones Daryl used. Hence, whatever your set
> might be, it contains an infinite descending chain. Therefore
> it is not well-ordered.
Well, the binary representation is one way to represent the set, but the set 

does not consist of binary strings. Any ordered set can be bijected with a 
set 
of binary strings, so if this is a valid objection, then any uncountably 
infinite set would suffer from this drawback, and not be well orderable. Did 

Hilbert and Goedel miss this fact? Or, is it even allowed for me to have 
uncountably many bits? If a countably infinite number of bits is sufficient 

for 
the enumeration, then we can consider, for the purposes of satisfying the 
idiosyncracies of set theory, that all such bit strings only contain bits a 

finite distance from the beginning of the string, in which case there is no 

infinite descending chain. So, what is it? Do I NEED uncountably many bits, 

in 
which case a well ordering on any uncountable set is simply impossible, or 
can 
I say I have countably many bits, and still maintain, in your system, that 
this 
ordering covers the set? Since the power set of a countably infinite set is 

considered to be uncountable, it would seem that a countable number N of 
bits 
would suffice, since that would give 2^N element representations, equivalent 

to 
the uncountable power set. So, consider the bits to be countably infinite, 
and 
find an objection under those circumstances.
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Robert Low said:
> Well, the binary representation is one way to represent the set, but 
> the set does not consist of binary strings. Any ordered set can be 
> bijected with a set of binary strings
Another of the wild claims TO cannot prove!
> Or, is it even allowed for me to have uncountably many bits?
 
Not in a string! At least until TO can prove that uncountable is only 
countable.
===
Subject: Re: Well Ordering the Reals
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   <3snennFp7fmoU1@individual.net>
   <dk622p02m49@drn.newsguy.com>
   <MPG.1dd15aa99a3e4e3998a5c9@newsstand.cit.cornell.edu>
   <dk848702mnf@drn.newsguy.com>
   <MPG.1dd16867c47639098a5d5@newsstand.cit.cornell.edu>
   <dk87jb03af@drn.newsguy.com>
   <MPG.1dd17620b963865598a5d9@newsstand.cit.cornell.edu>
   <dk8ccb0mfp@drn.newsguy.com>
   <MPG.1dd185d2f1bc36ee98a5e6@newsstand.cit.cornell.edu>
   <dk8fi9010nu@drn.newsguy.com>
   <MPG.1dd1a1132eb94ae498a5f3@newsstand.cit.cornell.edu>
   <dk8nqg01pdd@drn.newsguy.com>
   <MPG.1dd29d6abfccf45198a5fc@newsstand.cit.cornell.edu>
   <dkampc01jl6@drn.newsguy.com>
   <3ss7sfFpng8eU1@individual.net>
   <MPG.1dd2d0eb6bcfdf6b98a61b@newsstand.cit.cornell.edu>
   <3ssfhnFpqrk6U1@individual.net>
   <MPG.1dd3e03a2d695a8c98a62c@newsstand.cit.cornell.edu Any ordered set can 
be bijected with a set
> of binary strings, so if this is a valid objection, then any uncountably
> infinite set would suffer from this drawback, and not be well orderable.
> [...]
> Do I NEED uncountably many bits, in
> which case a well ordering on any uncountable set is simply impossible, or 

can
> I say I have countably many bits, and still maintain, in your system, that 

this
> ordering covers the set? Since the power set of a countably infinite set 
is
> considered to be uncountable, it would seem that a countable number N of 
bits
> would suffice, since that would give 2^N element representations, 
equivalent to
> the uncountable power set. So, consider the bits to be countably infinite, 

and
> find an objection under those circumstances.
You've just described a commonly used mapping of binary real fractions
to the powerset of N.  Each real r in [0,1] represented as a binary
fraction maps to a subset S of N, where each 1 digit in r indicates
the membership of a natural k in subset S.  Since each real is an
infinitely long binary string, composed of the sum of a countably
infinite number of powers of 2, such numbers/bitstrings are capable
of mapping to all the 2^Aleph_0 members of P(N).
This same mapping also forms the basis for Cantor's diagonal proof,
which proves that the members of N cannot map to all the members
of P(N).  Part of the reason for this is that real fractions are
infinitely long bitstrings but naturals are not, which means that there
are more real fractions than naturals.  Thus the naturals are countable
but the real fractions are not.
So while any given real has a representation consisting of a countably
infinite number of bits, there are an uncountably infinite number of
such bitstrings.
Any well-ordering of the reals has to provide ordering and successor
operations for an uncountably infinite amount of numbers (or subsets).
===
Subject: Re: Well Ordering the Reals
David R Tribble said:
> Any ordered set can be bijected with a set
> of binary strings, so if this is a valid objection, then any 
uncountably
> infinite set would suffer from this drawback, and not be well 
orderable.
> [...]
> Do I NEED uncountably many bits, in
> which case a well ordering on any uncountable set is simply impossible, 
or can
> I say I have countably many bits, and still maintain, in your system, 
that this
> ordering covers the set? Since the power set of a countably infinite set 
is
> considered to be uncountable, it would seem that a countable number N of 
bits
> would suffice, since that would give 2^N element representations, 
equivalent to
> the uncountable power set. So, consider the bits to be countably 
infinite, and
> find an objection under those circumstances.
> You've just described a commonly used mapping of binary real fractions
> to the powerset of N.  Each real r in [0,1] represented as a binary
> fraction maps to a subset S of N, where each 1 digit in r indicates
> the membership of a natural k in subset S.  Since each real is an
> infinitely long binary string, composed of the sum of a countably
> infinite number of powers of 2, such numbers/bitstrings are capable
> of mapping to all the 2^Aleph_0 members of P(N).
So, you are saying that a countably infinite number of bits, where every bit 

is 
in a finite position, would be sufficient to cover the reals? I was not 
speaking exactly of the power set representation, but comparing it to the 
set 
of countably long bitstrings, which would also have the size of 2^N, 
considered 
uncountable.
> This same mapping also forms the basis for Cantor's diagonal proof,
> which proves that the members of N cannot map to all the members
> of P(N).  Part of the reason for this is that real fractions are
> infinitely long bitstrings but naturals are not, which means that there
> are more real fractions than naturals.  Thus the naturals are countable
> but the real fractions are not.
Well, you know that I advocate the use of infinitely long bit strings on 
both 
sides of the digital point, contrary to most mathematicians. Personally, I 
don't see anything especially wrong with Ross ER function of inverting the 
bits 
and bijecting the naturals with the reals in [0,1]. It's not a bad idea to 
consider N naturals and N reals in [0,1].
> So while any given real has a representation consisting of a countably
> infinite number of bits, there are an uncountably infinite number of
> such bitstrings.
Right, because 2^n marks the boundary, somehow, between countable and 
uncountable, by standard theory. That makes no sense to me, but I am trying 

to 
understand the precepts of your system.
> Any well-ordering of the reals has to provide ordering and successor
> operations for an uncountably infinite amount of numbers (or subsets).
Does a well ordering need to have a successor to every element? What element 

is 
omega successor to in the well ordering of the ordinals? 
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> Robert Low said:
>>IF you were capable of understanding Daryl's argument, you'd
>>notice that all the strings he lists are well-defined infinite
>>binary strings. Your strings (whatever they might be) certainly
>>include all the ones Daryl used. Hence, whatever your set
>>might be, it contains an infinite descending chain. Therefore
>>it is not well-ordered.
 > wibble wibble wibble thpppt.
(OK, I paraphrased it slightly.)
===
Subject: Re: Well Ordering the Reals
   <MPG.1dcfee5958a7e4cb98a597@newsstand.cit.cornell.edu Proginoskes said:
> Hi All -
 I am told it is impossible to well order the real numbers such that 
there is a
> first and a successor operation which defines each other. [...]
> I suspect you're using the wrong definition.
> A well-ordering of a set S is a total ordering < such that every
> non-empty subset T of S has a first (minimum) element. A well-ordering
> of the positive integers is
> 1, 2, 3, 4, 5, 6, ...
> because if you have a set T which is a nonempty subset of the positive
> integers, it contains some positive integer n -- not necessarily the
> smallest. There are only a finite number of elements of T less than n,
> and you just check cases until you find the smallest one. (If 1 is in
> T, you're done; otherwise if 2 is in T, you're done; etc., up to n,
> where the checking procedure is guaranteed to stop, after a finite
> number of checks.)
> By less than, you mean preceding in the order, correct?
Yes.
> A well-ordering of the integers is
> 0, 1, -1, 2, -2, 3, -3, 4, -4, ...
> Certainly 1 is not less than -1, so less than is not exactly the 
word you
> want to use, is it?
It's not the standard less than, but then again, I'm not using the
standard ordering. In this ordering, 1 _does_ precede -1, so it's okay
to say that 1 is less than -1 in this ordering.
      --- Christopher Heckman
===
Subject: Re: Well Ordering the Reals
   <MPG.1dd15a2eedc716aa98a5c8@newsstand.cit.cornell.edu>
   <85oe54p94x.fsf@lola.goethe.zz>
   <MPG.1dd16dcef6a2e5bc98a5d7@newsstand.cit.cornell.edu>
   <853bmgp6lb.fsf@lola.goethe.zz>
   <MPG.1dd17762fe459c9798a5da@newsstand.cit.cornell.edu>
   <85pspknomu.fsf@lola.goethe.zz>
   <MPG.1dd184f96dd3ad4098a5e5@newsstand.cit.cornell.edu>
   <dk8fl201103@drn.newsguy.com>
   <MPG.1dd1a18a4b65878798a5f4@newsstand.cit.cornell.edu>
   <85ll08m345.fsf@lola.goethe.zz>
   <MPG.1dd1a7cffcdc969998a5f7@newsstand.cit.cornell.edu>
   <dk8pbg01v36@drn.newsguy.com>
   <MPG.1dd29f66a260f6a398a5fe@newsstand.cit.cornell.edu>
   <dkamka01j63@drn.newsguy.com>
   <MPG.1dd2bc83afed89498a610@newsstand.cit.cornell.edu>
   <MPG.1dd301e3edcdbdb98a627@newsstand.cit.cornell.edu 
imaginatorium@despammed.com said:
> Daryl McCullough said:
> Well, then consider the following three sets:
     A = { 0, 1, 10, 11, 100, 101, 110, 
111 ... }
>     B = { 0, 1, 2, 3, 4, 5, 6, 7, ... }
>     C = { 0, 1, 10, 11, 100, 101, 110, 111, ... }
 A = the set of all bit strings such that the only string
> starting with the character '0' is 0.
> B = the set of all natural numbers.
> C = the set of all natural numbers that can be written in
> base-10 using only '1' and '0'.
 <Orlovian wisdom begins here Option 3 is the standard, simplistic answer, 
without considering the 
properties
> of the elements. This is a very good example for illustrating the role 
of N=S^L
> in the comparison of symbolic sets.
 A is the set of all bit strings, with leading and trailing zeroes 
removed. So,
> the size of this set is binary N. String lengths will be considered to 
extend
> to log2(N).
 B is the set of all natural numbers, whose size is pure quantitative 
N.
> Can you expand slightly on the relation between binary N (this is
> some sort of Tinfinity, no?) and pure quantitative N (some sort 
of
> food additive?)
> Sure. Quantitative N is the length of the real number line in units, and 
given
> the identity function between value and count for naturals, also the count 

of
> the set of naturals on that line. Binary and decimal N are considered to 
be
> equal to quantitative N, so that the set of binary or decimal naturals 
will be
> the same size as the set of naturals. But, this means that the strings in
> binary are log2(N) in length, while the strings in decimal are log10(N) 
in
> length. Given that, we can say the set of binary integers is the same as 
the
> set of decimal integers. However, when comparing the sets without any
> quantitative interpretation of the elements, we can only consider the size 

of
> the alphabet, S, and the length of the strings, L, in which case we must 
use
> N=S^L to compare the sets of strings, assuming some common variable 
maximum
> length.
> C, as you say, is a subset of B, and should be smaller. How can we 
measure the
> size of this set, relative to B? We use N=S^L, as follows. The entire 
set of
> decimal numbers is decimal N. The number of bits in each is therefore 
log10(N).
> Since each digit is either a 0 or 1, we have 2^log10(N) decimal 
strings with
> only 0's and 1's. You will notice that if you plug in any finite N for 
a
> maximum value (powers of 10, or it starts to require floor() and such) 
that
> this function is correct.
> So what does this make the Tsize of C? Is it 2^log10(N) ? What sort
> of N is this one? Pure decimal N? (Genuine question, insofar as 
a
> question with no meaningful answer can be genuine.)
> It is the length of the real line, and the count of the naturals, which is 

of
> course not an absolute number.
> And when we compare sets A and C, we find that notice that A is the set
> of strings representing the elements in C, yet (modulo different
> varieties of N), there seem to be more elements in A than in C:
> A = { 0, 1, 10, 11, 100, 101, 110, 111 
... }
> C = {  0,   1,   10,   11,   100,   101,   110,   111, ... }
> Why do you say there are more elements in A than in C.
Because above, *you* said (I quote):
A is the set of all bit strings, with leading and trailing zeroes
removed. So,
the size of this set is binary N.
B is the set of all natural numbers, whose size is pure quantitative N.
Apart from the different flavours of N, this appears to say that A
and B are the same size. Is this correct? I mean, you're not surely
going to agree that Bigulosity can't compare them after all?
Next you say:
C, as you say, is a subset of B, and should be smaller.
Well, is it smaller or not? If should be means it is, then *you* are
saying that C is smaller than B, which is the same size as A. This
usually implies that C is smaller than A. Despite the fact that A
clearly consists of the string representations of elements of C.
Tony, take a step back. You might even remember that (goodness, 6000?
8000? posts ago) I said yes, of course the subset relationship provides
a partial ordering on sets. I also bemoaned the fact that, tired of
arguing with cranks, yet contradictorily unable to resist them, people
tend to make inaccurate and sweeping statements, like Cardinality is
the unique measure of set size. Well, of course cardinality is a blunt
instrument, but it is a universal one. Given any two sets there either
is or is not a bijection between them; the existence of a bijection is
an equivalent relation, and this puts sets in equivalence classes. The
surprising thing is not that this is a blunt tool, but that it does
make distinctions between sets which are unending (which can't be
distinguished by a Does the ditty stop? argument).
Of course it is possible to make much finer distinctions between
infinite sets in some other cases - some infinite subsets of natural
numbers have a limiting density, which can be used to compare these
subsets. But your attempts to extend this to all sets plainly fail,
as we see as soon as we consider sets of strings, *as sets of strings*.
Anyway, you've been round in huge circles, and along the way have
decided that not only is Bigulosity a useful and interesting tool (your
view, I presume: actually I don't think it is), but that somehow just
about everything in maths is Wrong, and can be corrected by you in a
weekend's cogitation. So you flail here and there, arguing with
everything that comes with a proof, and generally making a bit of a
fool of yourself.
In a curious way, it seems to me, mathematics is the art of discovering
possible. Otto Von Bismarck, remark, Aug. 11, 1867. Right, this is the
difference: politics (and most soft- and pseudo-science) proceeds by
deciding where you want to get, and bending and forcing a set of
ill-defined guidelines into allowing one to get there. Mathematics is
about choosing a set of utterly inflexible rules, and seeing where it
leads. I suppose the attraction of mathematics to cranks is that they
know only the PSS (politics-and-soft-science) approach, and told that
there is no ruler-and-compass construction for trisecting an angle, see
this as a challenge to be overcome. Unfortunately, the result can never
be mathematics. If there is no bijection between a set and its power
set - a fact with a crystal-clear one-line proof - then you should not
really be surprised if an announcement that you have found one is
treated with hilarity. The only real issue is whether you have written
anything clear enough that we can see if you have just genuinely
misunderstood the definitions concerned, or have simply made a blunder
in the demonstration.
> Pretty amazing stuff this, Tony, if you can pull it off. Have you tried
> Cornell maths dept?
> I emailed a professor who does set theory here a while back, who at first
> dismissed the very idea, and then said his father had just died. I haven't 

been
> back. I don't want to bother him right now. Maybe over winter break. But, 

you
> just want to hear about how I made a fool of myself in person on my own 
turf,
> don't you? Heh.
It should be fairly clear to you that the chances that suddenly the
non-cranks on sci.math start accepting declaring ends to unending
sequences, and similar contradictions, is zero, for all practical
purposes. You may get various cranks to nod in your direction from time
to time, but the trouble is that in the end every crank has his* own
scheme to push, and is not likely to join in helping you. So it's going
to be a lonely road.
* The fairer sex is in a tiny minority on sci.math: but are there, have
there ever been any crankesses?
Anyway, carry on. I reckon you could be entertaining us for a while to
come...
Brian Chandler
http://imaginatorium.org
===
Subject: Re: Well Ordering the Reals
imaginatorium@despammed.com said:
> imaginatorium@despammed.com said:
> Daryl McCullough said:
> Well, then consider the following three sets:
     A = { 0, 1, 10, 11, 100, 101, 110, 
111 ... }
>     B = { 0, 1, 2, 3, 4, 5, 6, 7, ... }
>     C = { 0, 1, 10, 11, 100, 101, 110, 111, ... }
 A = the set of all bit strings such that the only string
> starting with the character '0' is 0.
> B = the set of all natural numbers.
> C = the set of all natural numbers that can be written in
> base-10 using only '1' and '0'.
> <Orlovian wisdom begins here
> Option 3 is the standard, simplistic answer, without considering the 
properties
> of the elements. This is a very good example for illustrating the 
role of N=S^L
> in the comparison of symbolic sets.
 A is the set of all bit strings, with leading and trailing zeroes 
removed. So,
> the size of this set is binary N. String lengths will be considered 
to extend
> to log2(N).
 B is the set of all natural numbers, whose size is pure quantitative 
N.
 Can you expand slightly on the relation between binary N (this is
> some sort of Tinfinity, no?) and pure quantitative N (some 
sort of
> food additive?)
> Sure. Quantitative N is the length of the real number line in units, and 
given
> the identity function between value and count for naturals, also the 
count of
> the set of naturals on that line. Binary and decimal N are considered to 
be
> equal to quantitative N, so that the set of binary or decimal naturals 
will be
> the same size as the set of naturals. But, this means that the strings 
in
> binary are log2(N) in length, while the strings in decimal are log10(N) 
in
> length. Given that, we can say the set of binary integers is the same as 
the
> set of decimal integers. However, when comparing the sets without any
> quantitative interpretation of the elements, we can only consider the 
size of
> the alphabet, S, and the length of the strings, L, in which case we must 
use
> N=S^L to compare the sets of strings, assuming some common variable 
maximum
> length.
 C, as you say, is a subset of B, and should be smaller. How can we 
measure the
> size of this set, relative to B? We use N=S^L, as follows. The 
entire set of
> decimal numbers is decimal N. The number of bits in each is 
therefore log10(N).
> Since each digit is either a 0 or 1, we have 2^log10(N) decimal 
strings with
> only 0's and 1's. You will notice that if you plug in any finite N 
for a
> maximum value (powers of 10, or it starts to require floor() and 
such) that
> this function is correct.
 So what does this make the Tsize of C? Is it 2^log10(N) ? What 
sort
> of N is this one? Pure decimal N? (Genuine question, insofar 
as a
> question with no meaningful answer can be genuine.)
> It is the length of the real line, and the count of the naturals, which 
is of
> course not an absolute number.
 And when we compare sets A and C, we find that notice that A is the 
set
> of strings representing the elements in C, yet (modulo different
> varieties of N), there seem to be more elements in A than in C:
 A = { 0, 1, 10, 11, 100, 101, 110, 111 
... }
> C = {  0,   1,   10,   11,   100,   101,   110,   111, ... }
> Why do you say there are more elements in A than in C.
> Because above, *you* said (I quote):
> A is the set of all bit strings, with leading and trailing zeroes
> removed. So,
> the size of this set is binary N.
> B is the set of all natural numbers, whose size is pure quantitative N.
> Apart from the different flavours of N, this appears to say that A
> and B are the same size. Is this correct? I mean, you're not surely
> going to agree that Bigulosity can't compare them after all?
> Next you say:
> C, as you say, is a subset of B, and should be smaller.
Yes, sorry, that makes sense. If C is decimal, then it only contains 
2^log10(N) 
of the N naturals. This depends on how you interpret A. If it is the set of 

all 
binary strings with leading zeroes removed, then its size is N. If we 
interpret 
those strings as quantities, then the size of the set depends on the base we 

are using. If we use base 2, then it covers all naturals, but if we use base 

10, then it doesn't.
> Well, is it smaller or not? If should be means it is, then *you* are
> saying that C is smaller than B, which is the same size as A. This
> usually implies that C is smaller than A. Despite the fact that A
> clearly consists of the string representations of elements of C.
That is not at all clear. You defined A as the set of bit strings, without 
any 
reference their quantitative interpretation. In absence of such 
interpretation, 
they are treated as strings constructed from an alphabet. You need to be 
able 
to distinguish between a string and what it represents. Is apple an 
apple?
> Tony, take a step back. You might even remember that (goodness, 6000?
> 8000? posts ago) I said yes, of course the subset relationship provides
> a partial ordering on sets. I also bemoaned the fact that, tired of
> arguing with cranks, yet contradictorily unable to resist them, people
> tend to make inaccurate and sweeping statements, like Cardinality is
> the unique measure of set size. Well, of course cardinality is a blunt
> instrument, but it is a universal one. Given any two sets there either
> is or is not a bijection between them; the existence of a bijection is
> an equivalent relation, and this puts sets in equivalence classes. The
> surprising thing is not that this is a blunt tool, but that it does
> make distinctions between sets which are unending (which can't be
> distinguished by a Does the ditty stop? argument).
above.
> Of course it is possible to make much finer distinctions between
> infinite sets in some other cases - some infinite subsets of natural
> numbers have a limiting density, which can be used to compare these
> subsets. But your attempts to extend this to all sets plainly fail,
> as we see as soon as we consider sets of strings, *as sets of strings*.
No it really doesn't. For lack of quantitative interpretation of the 
strings, 
we apply N=S^L. Consider the set A, where we actually have 10 characaters in 

our alphabet, but use only two. In this case, we have 2^log10(N) of the N 
strings, just like C has 2^log10(N) of the N decimal naturals.
> Anyway, you've been round in huge circles, and along the way have
> decided that not only is Bigulosity a useful and interesting tool (your
> view, I presume: actually I don't think it is), but that somehow just
> about everything in maths is Wrong, and can be corrected by you in a
> weekend's cogitation. So you flail here and there, arguing with
> everything that comes with a proof, and generally making a bit of a
> fool of yourself.
I analyze your proofs, and almost always, discover a last element implied 
which 
creates your contradiction. I offer methods which satisfy intuitive notions. 

I 
provide an original ordering of the reals, the ramifications of which have 
not 
yet panned out. This isn't a weekend of cogitation. I heard a good quote on 

NPR's The Writer's Almanac this morning, from (Joe?) Queenan, a humorist 
with 
you don't know anything, so shut up. Hmmm.... How long have I been working 
on 
these things? It's time to write.
> In a curious way, it seems to me, mathematics is the art of discovering
> possible. Otto Von Bismarck, remark, Aug. 11, 1867. Right, this is the
> difference: politics (and most soft- and pseudo-science) proceeds by
> deciding where you want to get, and bending and forcing a set of
> ill-defined guidelines into allowing one to get there. Mathematics is
> about choosing a set of utterly inflexible rules, and seeing where it
> leads. I suppose the attraction of mathematics to cranks is that they
> know only the PSS (politics-and-soft-science) approach, and told that
> there is no ruler-and-compass construction for trisecting an angle, see
> this as a challenge to be overcome. Unfortunately, the result can never
> be mathematics. If there is no bijection between a set and its power
> set - a fact with a crystal-clear one-line proof - then you should not
> really be surprised if an announcement that you have found one is
> treated with hilarity. The only real issue is whether you have written
> anything clear enough that we can see if you have just genuinely
> misunderstood the definitions concerned, or have simply made a blunder
> in the demonstration.
The whole point with the bijection with the powerset was to get you actually 

thinking about the value range, and the different infinities of bits, and to 

consider whether those arguments concerning the power set also apply to 
other 
sets. We have a bijection between the power set and the set, with no 
identifiable point of breakdown. The set of all elements which are not in 
the 
set to which they map is the set of all elements, which never ends, and 
therefore would be mapped to an element which never ends. In this case, the 

endless nature of the bijection is not sufficient for you to claim the sets 

are 
equal, whereas in all other cases it is. In reality, the endlessness of the 

bijection is NOT sufficient to say two sets are of equal size. The NATURE of 

the bijection needs to be taken into account, to turn the hammer into an 
axe.
> Pretty amazing stuff this, Tony, if you can pull it off. Have you 
tried
> Cornell maths dept?
> I emailed a professor who does set theory here a while back, who at 
first
> dismissed the very idea, and then said his father had just died. I 
haven't been
> back. I don't want to bother him right now. Maybe over winter break. 
But, you
> just want to hear about how I made a fool of myself in person on my own 
turf,
> don't you? Heh.
> It should be fairly clear to you that the chances that suddenly the
> non-cranks on sci.math start accepting declaring ends to unending
> sequences, and similar contradictions, is zero, for all practical
> purposes. You may get various cranks to nod in your direction from time
> to time, but the trouble is that in the end every crank has his* own
> scheme to push, and is not likely to join in helping you. So it's going
> to be a lonely road.
No kidding. It's been a lonely road for 40 years. Big deal. It's true that 
everyone who objects to transfinite set theory has a different take on the 
matter and a different solution. For most, the solution is to disregard 
infinity entirely, and among the others, there is some overlap and some 
difference of opinion. That's a GOOD thing. If we all agreed, we'd learn 
nothing from each other. This is a part of evolution, which may not be 
considered at all related to math, but it is.
> * The fairer sex is in a tiny minority on sci.math: but are there, have
> there ever been any crankesses?
Women don't tend to divorce themselves from reality enough or obsess enough 

to 
come up with wacky new ideas in math very often. Men can focus on one insane 

idea to the detriment of their hygiene even. You may consider my ideas to be 

a 
dinosaur destined to extinction, and yet, it may be destined to fly, like 
the 
dinosaurs that survive today as birds.
> Anyway, carry on. I reckon you could be entertaining us for a while to
> come...
> Brian Chandler
> http://imaginatorium.org
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
> imaginatorium@despammed.com said:
> Well, is it smaller or not? If should be means it is, then *you* 
are
> saying that C is smaller than B, which is the same size as A. This
> usually implies that C is smaller than A. Despite the fact that A
> clearly consists of the string representations of elements of C.
> That is not at all clear. You defined A as the set of bit strings, 
> without any reference their quantitative interpretation.
So that now the size of a set of well defined objects depends on how one 
interprets those objects? 
Thus the size of a set depends on the context in which it occurs?
This is TO with his amBigulosity running on empty again.
In mathematics, sets are not variable objects, the way TO tries to make 
them.
> Bigulosity handles these differences well. 
TO's AmBigulosity, does not handle anything well.
> I analyze your proofs, and almost always, discover a last element 
> implied which creates your contradiction.
What TO discovers lies only in the eye of the beholder.
And in his subsequent posts.
> I offer methods which satisfy intuitive notions.
And contradict logical requirements in the process.
 
> I provide an original ordering of the reals, the ramifications of 
> which have not yet panned out.
It is not even yet clear that it includes all reals.
> The whole point with the bijection with the powerset was to get you 
actually 
> thinking about the value range
One can get more out of contemplating unicorns. They, at least, have a 
coherent description. 
> We have a bijection between the power set and the set, with no 
> identifiable point of breakdown. 
 
What is this We ? 
We, meaning everyone except TO, find that there is no such bijection 
form any set to its power set.
We, meaning everyone except TO, find convincing and irrefutable the 
proof every mapping from any set to its power set must fail to be a 
bijection by any mathematical standards.
 
The set of all elements which are not in the 
> set to which they map is the set of all elements, which never ends,
A set which nenver ends is not a set. Sequences can be never ending, 
and processes can be never ending, but sets are not works in progress, 
they just are.
> 
> 
> Pretty amazing stuff this, Tony, if you can pull it off. Have 
> you tried Cornell maths dept?
> I emailed a professor who does set theory here a while back, who 
> at first dismissed the very idea, and then said his father had 
> just died. I haven't been back. I don't want to bother him right 
> now. Maybe over winter break. But, you just want to hear about 
> how I made a fool of myself in person on my own turf, don't you? 
> Heh.
Since you regularly make a fool of yourself on this turf, what keeps you 
from doing it everywhere? 
The knowledge that you are making a fool of yourself?
===
Subject: Re: Well Ordering the Reals
> Tony Orlow says...
>Daryl McCullough said:
>     A = { 0, 1, 10, 11, 100, 101, 110, 
111 ... }
>     B = { 0, 1, 2, 3, 4, 5, 6, 7, ... }
>     C = { 0, 1, 10, 11, 100, 101, 110, 111, ... }
>> Let me introduce yet another set, D:
>> 
>>     D = { 0, 1, 2, 3, 4, 5, 6, 7, 
8, 9, 10, ... }
>> 
>> That is, D is the base-10 representations of the naturals. So D has
>> the same size as the naturals, right? You also say that A has the same
>> size as the naturals, right? But A is a subset of D. So the size of A
>> must be smaller than the size of D. Right? Isn't that a contradiction?
>Yes, it certainly depends on how you are considering the set.
> So whether one set is bigger, or smaller, or the same size as
> a second set is purely subjective? It depends on how one thinks
> about it.
> Your original objection to Cantorian cardinality was that
> set A could be a proper subset of set D, but that they had
> the same cardinality. It appears that your new, improved
> notion of size has this same flaw. But it's worse: you
> are basically saying that the size of a set depends on how
> you look at it. So your mathematics is as fuzzy and subjective
> as art appreciation.
> Well, your mathematics is ugly.
> --
> Daryl McCullough
> Ithaca, NY
Actually, it is not even mathematics. 
Mathematics, at least, is constrained by the rules of formal  logic, but 
what TO does violates all sorts of logical constraints.
===
Subject: Re: Well Ordering the Reals
> Daryl McCullough said:
> Tony Orlow says...
> 
>You know, it suddenly occurs to me that, while it was some source of 
>delight to
>see you using my notation for infinite numbers, that is not standard 
>mathematical notation, and therefore is not available to you in proving 
my 
>well-ordering false.
> 
> Are you saying that you are allowed to use your nonstandard
> notation to formulate your ordering, but I'm *not* allowed
> to use your notation to prove that it's not a well-ordering?
> That's inconsistent. Either we both use the standard notation,
> or we both use your notation. Either way is fine with me.
> I didn't use any non-standard notation. 
Lie 1
> I never used infinite bit strings. 
Lie 2.
> If a countable number of bits is sufficient (and required) by 
> standard set theory, then for the purposes of a standard 
> well-ordering, let's stick with that. 
Is that TO-countable or Dedekind-countable?
A Dedkind countable sequence of decreasing terms is suffient to disprove 
well ordering.
> Does that somehow limit the well ordering in some prohibited way? If 
> so, then a well ordering is probably entirely impossible for an 
> uncountable set, given standard definitions.
Given the axiom of choice, every set can be well ordered, including 
uncountably infinite ones. Because the axiom of choice say they can!
> 
>Can you provide some infinite descending chain using 
>standard notation and concepts?
Only if TO can produce his allegedly well-ordered uncountable set using 
only standard notation and concepts. Whatever notatain and concepts are 
used to produce the alleged well ordering may be fairly used in its 
refutation.
> The way a well-ordering of a set works is this: Some elements are
> successor elements, and other elements are limit elements. If you
> are going down in the ordering, and you get to a limit element, then
> the only way to go down further is to skip *infinitely* many
> elements.
> Uh, doesn't skipping elements automatically mean you have not enumerated 
or 
> ordered the entire set? 
One is not required to enumerate the set to disprove well ordering, one 
need only find a non-ending descending sequence, which is allowed to 
skip over as many elements as needed.
> 
> Here's an example: consider the set of ordered pairs (x,y) where
> x and y are naturals. We order this set lexicographically. What
> that means is
> 
>    1. If x < y, then (x,z) < (y,z).
>    2. If w < z, then (x,w) < (x,z).
> 
> So the ordering looks like this:
> 
>    (0,0) (0,1) (0,2) (0,3) ...
>    (1,0) (1,1) (1,2) (1,3) ...
>    (2,0) (2,1) (2,2) (2,3) ...
>    ...
> 
> where each element is greater than the element to the left of it,
> and each element is greater than every element that is on an earlier
> row.
> Huh! Kind of liek my enumeration of the rationals.....
> 
> The limit elements are (1,0), (2,0), ...
> 
> This is a well-ordering, in that if you keep going down, you
> will eventually (after a finite number of steps) reach (0,0).
> 
> Try it yourself: can you come up with an infinite (or, in your
> terms, finite but unbounded) descending chain? But I can
> come up with an infinite *ascending* chain:
> 
>     (0,0) (0,1) (0,2) ...
> If I do as you did above, and pretend up is down
While up is often down for TO, it is not because he is copying anyone 
else, but because he is not.
> then that's a descending 
> chain. But that's not allowed, now, is it?
Only if one claims reverse well-ordering.
> 
> A well-ordering makes a distinction between ascending chains
> and descending chains. The only well-ordering that does *not*
> make that distinction are finite well-orderings.
> Well, you should make that distinction as well, and note that your 
> descending chain is in ascending order in my well ordering. 
But there are endless ascending chains which in TO's ordering become 
endless descending chains, and so TO still fails to well order the reals.
> 
> --
> Daryl McCullough
> Ithaca, NY
> 
>  
> 
===
Subject: Re: Well Ordering the Reals
> Robert Low said:
> 
> Now, is the only objection to my well ordering Daryl's infinite 
> descending 
> chain? 
> 
> Since that proves that it isn't a well ordering, it's enough.
> 
> Sure, except that set definition isn't allowed in standard mathematics, so 

> within the context that seeks a well ordering, it is not a valid infinite 

> descending chain. 
If one  is required to use standard mathematics, none of TO's 
constructions are valid in the first place, so TO has nothing properly 
called a set to well order.
===
Subject: Re: Well Ordering the Reals
> David Kastrup said:
> 
> Virgil said:
>> 
>> So how does one tell when a number of elements is infinite, 
rather 
>> than finite? 
> When a set is recursively defined with no terminating state, and
> there is no restriction of finiteness on the number of recursive
> iterations.
Then by TO's definition above, the number of elements in the closed real 
interval, [0,1], must be finite.
===
Subject: Re: Well Ordering the Reals
> David Kastrup said:
> 
> David Kastrup said:
>> 
>> Daryl McCullough said:
> Tony Orlow says...
>> So in your opinion the set
>> 
>> {..., 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 0, 2, 4, 6, 8, 10, 12, 14 
...}
>> 
>> where every member happens to have an infinite number of 
predecessors
>> and an infinite number of successors would be an infinite set?
>> 
>> 
> Do you actually allow infinite values for these elements which are
> finitely spaced on the number line?
> 
> You are balking at your own words again.  Every of these elements has
> an infinite number of predecessors and successors, all of them being
> finite.
> No, if they are all finite, then they are all a finite distance from any 
> other. 
> Sorry.
Whether they are finite, or are finite distances from one aother, is 
irrelevant to whether they have infinitely many predecessors or 
successors. As usual, TO demonstrates a form of QD.
> 
> If so, it is infinite. If not, then whatever finite range you have
> can only contain a finite number of such spacings. It doesn't matter
> that it go in two directions, only that the number of iterations not
> be limited to finite values.
> 
> The number of iterations to left and right is not limited to any
> finite value.
> But it is not allowed to be infinite, as you just said, so therefore it is 

> limited to finite values. 
But infinitely many of them!
TO is perpetually perplexed by the distinction between the finiteness of 
naturals individually and the infiniteness of them collectively
===
Subject: Re: Well Ordering the Reals
   <dk61ub02llj@drn.newsguy.com>
   <MPG.1dd15a2eedc716aa98a5c8@newsstand.cit.cornell.edu>
   <85oe54p94x.fsf@lola.goethe.zz>
   <MPG.1dd16dcef6a2e5bc98a5d7@newsstand.cit.cornell.edu>
   <853bmgp6lb.fsf@lola.goethe.zz>
   <MPG.1dd17762fe459c9798a5da@newsstand.cit.cornell.edu>
   <dk8b900iii@drn.newsguy.com>
   <MPG.1dd180d0d9444a2998a5e3@newsstand.cit.cornell.edu>
   <dk8duh0rqu@drn.newsguy.com>
   <MPG.1dd19f952d1b026398a5f0@newsstand.cit.cornell.edu>
   <dk8mod01me0@drn.newsguy.com>
   <MPG.1dd294a67a1367298a5f9@newsstand.cit.cornell.edu>
   <853bmfjc2l.fsf@lola.goethe.zz>
   <MPG.1dd2b9129734161598a60f@newsstand.cit.cornell.edu>
   <85br13hqsx.fsf@lola.goethe.zz>
   <MPG.1dd2e2ae194b50da98a623@newsstand.cit.cornell.edu>
   <ITSnetNOTcom#virgil-109009.19521702112005@comcast.dca.giganews.com ...
They really should use some phrase other than top posters.  It wasn't
always, but in recent times top posting is seen as bad netiquette.
Virgil, we know already, you don't need to tell us.
So, well-order the reals.  That's what this is about, isn't it?
You have all this vitriol for Tony and none for me?  Damn, I thought
you had enough for everybody.
I think what mathematics needs is more exhaustive reasoning about sets
dense in the reals.  That means, sets dense in the reals who complement
is dense and their union is the reals, eg the rationals and
irrationals, algebraic and non-algebraics, basically natural, dense,
subsets of the reals, and their correlation to Banach-Tarski, points on
a line as they are and as well in the infinital geometric mutation,
etc.
So, again, well-order the reals.  Now, in application of extension of
Cantor's first, are there countably many nested intervals?  Keep in
mind you're talking about a well-ordering of the reals.
There can be only one axiomless theory.
Ross
===
Subject: Re: Well Ordering the Reals
Ross A. Finlayson said:
> ...
> They really should use some phrase other than top posters.  It wasn't
> always, but in recent times top posting is seen as bad netiquette.
> Virgil, we know already, you don't need to tell us.
> So, well-order the reals.  That's what this is about, isn't it?
> You have all this vitriol for Tony and none for me?  Damn, I thought
> you had enough for everybody.
Virgil has a love/hate thing for me. I am the cat on the fence to his 
yappy-ass 
dog. Are you jealous of the cross they want to nail me to? Try referring to 

things like Virgilogic, and you too can be the envy of all those 
standing on 
the ground. You're much too nice to hate. :D
> I think what mathematics needs is more exhaustive reasoning about sets
> dense in the reals.  That means, sets dense in the reals who complement
> is dense and their union is the reals, eg the rationals and
> irrationals, algebraic and non-algebraics, basically natural, dense,
> subsets of the reals, and their correlation to Banach-Tarski, points on
> a line as they are and as well in the infinital geometric mutation,
> etc.
There are probably some interesting things left to discover in that area, 
sure. 
I am not sure what direction to go with that, but you probably have some 
good 
ideas.
> So, again, well-order the reals.  Now, in application of extension of
> Cantor's first, are there countably many nested intervals?  Keep in
> mind you're talking about a well-ordering of the reals.
It would seem, based on Cantor's First, that only countably many nested 
intervals are considered, otherwise c would be considered in the range of R. 

Given that, let's use countably many bits and exponentiations in my well 
ordering. Then, there is no infinite descending chain. What do you think of 

my 
construction, Ross? Perhaps we should discuss this by phone?
> There can be only one axiomless theory.
Perhaps that depends what kind of logic you're using. Logic is somewhat 
based 
on axioms as well, but at least they are well justified and always work.
> Ross
-- 
Smiles,
Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
===
Subject: Re: Well Ordering the Reals
   <85oe54p94x.fsf@lola.goethe.zz>
   <MPG.1dd16dcef6a2e5bc98a5d7@newsstand.cit.cornell.edu>
   <853bmgp6lb.fsf@lola.goethe.zz>
   <MPG.1dd17762fe459c9798a5da@newsstand.cit.cornell.edu>
   <dk8b900iii@drn.newsguy.com>
   <MPG.1dd180d0d9444a2998a5e3@newsstand.cit.cornell.edu>
   <dk8duh0rqu@drn.newsguy.com>
   <MPG.1dd19f952d1b026398a5f0@newsstand.cit.cornell.edu>
   <dk8mod01me0@drn.newsguy.com>
   <MPG.1dd294a67a1367298a5f9@newsstand.cit.cornell.edu>
   <853bmfjc2l.fsf@lola.goethe.zz>
   <MPG.1dd2b9129734161598a60f@newsstand.cit.cornell.edu>
   <85br13hqsx.fsf@lola.goethe.zz>
   <MPG.1dd2e2ae194b50da98a623@newsstand.cit.cornell.edu>
   <ITSnetNOTcom#virgil-109009.19521702112005@comcast.dca.giganews.com>
   <MPG.1dd4017a6928834598a638@newsstand.cit.cornell.edu Ross A. Finlayson 
said:
> ...
> They really should use some phrase other than top posters.  It 
wasn't
> always, but in recent times top posting is seen as bad netiquette.
> Virgil, we know already, you don't need to tell us.
> So, well-order the reals.  That's what this is about, isn't it?
> You have all this vitriol for Tony and none for me?  Damn, I thought
> you had enough for everybody.
> Virgil has a love/hate thing for me. I am the cat on the fence to his 
yappy-ass
> dog. Are you jealous of the cross they want to nail me to? Try referring 
to
> things like Virgilogic, and you too can be the envy of all those 
standing on
> the ground. You're much too nice to hate. :D
> I think what mathematics needs is more exhaustive reasoning about sets
> dense in the reals.  That means, sets dense in the reals who complement
> is dense and their union is the reals, eg the rationals and
> irrationals, algebraic and non-algebraics, basically natural, dense,
> subsets of the reals, and their correlation to Banach-Tarski, points on
> a line as they are and as well in the infinital geometric mutation,
> etc.
> There are probably some interesting things left to discover in that area, 

sure.
> I am not sure what direction to go with that, but you probably have some 
good
> ideas.
> So, again, well-order the reals.  Now, in application of extension of
> Cantor's first, are there countably many nested intervals?  Keep in
> mind you're talking about a well-ordering of the reals.
> It would seem, based on Cantor's First, that only countably many nested
> intervals are considered, otherwise c would be considered in the range of 

R.
> Given that, let's use countably many bits and exponentiations in my well
> ordering. Then, there is no infinite descending chain. What do you think 
of my
> construction, Ross? Perhaps we should discuss this by phone?
> There can be only one axiomless theory.
> Perhaps that depends what kind of logic you're using. Logic is somewhat 
based
> on axioms as well, but at least they are well justified and always work.
> Ross
> --
> Smiles,
> Tony
> http://www.people.cornell.edu/pages/aeo6/WellOrder/
No, I intend no such thing, re-axiomatization.
I'm sincere, I think I'm right.
Tony, you remind me of myself, years and years ago we had voluminous
threads that covered many elementary, and advanced, notions of infinite
sets and numbers,  and sets and numbers.  Many regular posters and
lurkers on sci.math would know that.
Virgil is, kind of a pseudo-literary.  Where he calls you TO, he these
days assiduously does not refer to me if at all possible, for he has
imposed on himself an ignorance because I say things that would
otherwise demand his very own personal re-examination of tenets he
holds dear.  He's kind of like a frickin' zombie, but he's more of the
sort of a hydrophobic.  Ha ha ha.
Yeah, feel free to call, Ross Finlayson: 208-476-3831.  (Well, it's not
a toll-free number.)  That goes for other participants on this thread,
Tribble, Dave, you're welcome to telephone me, Virgil, I look forward
to speaking with you, Brian, I guess that would be an international
phone call.
Tony, I went off for years, I talked about things like your log_oo and
so on, scalar infinity, unit scalar infinity, half a unit scalar
infinity, etcetera.
Over that time, I've gained some education, and derived some novel
notions that have to do with these mathematics, logic, mathematical
logic, and in some fringe sense physics.  Now, it is generally
ignorant, and arrogant, to think that a contemporary individual can
contribute in a meaningful way to the art or science of mathematics in
its very foundations, because of the concerted, and presumed
exhaustive, effort of all those who came before, where each of those
was thus accordingly ignorant, and arrogant, in thinking they could do
so.
While that is so, it's not necessarily wrong.
Have you provided an example of the well-ordering of the reals?  Are
you sincere, do you honestly think you have?  Does each subset ordered
by that ordering have a least element with the comparator of your
ordering?  Does it satisfy in extension of Cantor's first that it is a
well-ordering of all the reals?
This is sci.math, humor is irrelevant.
The universe is infinite.
Well-order the reals.
Ross
===
Subject: Another cantor set question
Hi there.
If C is a cantor set, [0,1] - C is the complement,
Those intervals in the complement, do they have parwise disjoint closure.
Im trying to show, either. But the problem is i dont know if the claim is 
true 
===
Subject: Re: Another cantor set question
Try to prove that every element of the cantor set is either a
descending or ascending limit of other elements in the cantor set.
Once you do that, it shouldn't be difficult to prove your claim.
--  
===
Subject: Re: triangle with inscribed circle
>That being so, think of the line between two acute
>angles as forming the base of the construction:
>   /_______(     )___
>Now what can you say about the circle diameter parallel
>to this base, relative to the endpoints of the base?
I see only that the point on the circle diameter are in the distance
greater than 1 from endpoints of the base, but it doesn,t help me.
===
Subject: Re: triangle with inscribed circle
> That being so, think of the line between two acute
> angles as forming the base of the construction:
>   /_______(     )___
> Now what can you say about the circle diameter parallel
> to this base, relative to the endpoints of the base?
> I see only that the point on the circle diameter are in the distance
> greater than 1 from endpoints of the base, but it doesn,t help me.
OK, what can you say about the _length_ of the triangle base
compared to the length of the circle diameter parallel to it? ...
===
Subject: Re: Has anyone ever done this?
 <Ken.Pledger-981674.09195801112005@bats.mcs.vuw.ac.nz>
 <1bff9$4368875d$82a1e3ad$15309@news1.tudelft.nl>
In message <1bff9$4368875d$82a1e3ad$15309@news1.tudelft.nl>, Han de 
...
>Now listen, dude! There are _many_ things in elementary mathematics that
>are not known to professional mathematicians.
>Here is a challenge. Maybe you know that there exists a _simple_ Finite
>Element in two dimensions, called the linear triangle. Function values
>inside that triangle can be found by linear interpolation of the values
>at its vertices. Now give me at least _one_ other simple element shape
>in 2-D which has a linear interpolation inside. I take the risk that I
>am wrong (actually: I hope that I'm wrong and that this is well known).
>
How about a disc?
Where a vertex is defined to be any point on the perimeter...
-- 
Jeremy Boden
===
Subject: Re: Has anyone ever done this?
>>How can I be a crank if my mathematics are correct?
>By definition, a crank is someone who espouses incorrect mathematics
>and attempts to pass them off as proof.
>>Any mathematician can produce incorrect mathematics.  The crank's
>>distinguishing feature is refusal to listen to reason. 
Refusal to listen to disproofs, yes.  Refusal to listen
to reasoning based on opinion, at worst a maybe, unless
there is an insistence that an incorrect step is correct.
>>I think that Starb...@Earthlink qualifies as a crank because of his
>>repeated
>>failure to listen to those more knowledgeable than he is.  See the add
>>10 marbles,
>>remove 1 thread.
Who decides more knowledgeable here?  Here is a quote from Galileo:
In questions of science, the authority of a thousand is not worth
 the humble reasoning of a single individual.
>My statement that Starb doesn't qualify as a crank was based solely
>threads where he may or may not qualify.
>Robert Israel                                israel@math.ubc.ca
>Department of Mathematics        http://www.math.ubc.ca/~israel 
>University of British Columbia            Vancouver, BC, Canada
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
===
Subject: Re: infinity ...
   <MPG.1dc85a4d4881712598a56a@newsstand.cit.cornell.edu>
   <MPG.1dcfff7f104b00a498a59e@newsstand.cit.cornell.edu> If it doesn't 
contain any infinite members, it's not infinite. Those 
terms
>> differ by more than a constant finite amount, but rather a rapidly 
growing
>> amount greater than 1. There is no way you have an infinite number of 
them
>> without achieving infinite values within the set.
>> Yes, you and Albrecht keep saying that repeatedly.  Please demonstrate
>> why it must be so, because it's not.
>> Your argumentation is not fair, but I don't wonder about that.
>> _You_ has to show, that in the case of the whole set there is no
>> natural number as big as the whole set.
>> You argue: there is no infinite natural number since the peano axioms
>> don't allow an infinite natural number.
>> That's right. I agree with you.
>> Alas, Albrecht, it is not true. There is nothing in the Peano axioms 
that
>> states any such thing. The inductive proof of the finiteness of the 
naturals is
>> flawed in that it applies an increment to each successor, noting that 
adding 1
>> does not turn a finite into an infinite, but ignores the intrinsic nature 

of
>> inductive proof as a recursively defined infinite concatenation of 
logical
>> implications, over which an infinite number of increments does indeed 
produce
>> an infinite value as successor. Of course, this cannot be achhieved in 
any
>> finite or countably infinite (finite but unbounded) number of steps.
On one side we have Tony, who believes infinite naturals exist but
that the set of finite naturals is not infinite.  On the other side
we have Albrecht, who also does not believe the set of naturals is
infinite but also does not believe infinite naturals exist.
They are obviously both wrong, but for different reasons.
It's amusing to see this kind of discussion.
===
Subject: Re: infinity ...
   <MPG.1dc85a4d4881712598a56a@newsstand.cit.cornell.edu
> Consider the set of reals in the interval [0,1], that is, the set
> S = {x in R : 0 <= x <= 1}.  The elements of this set cannot be
> enumerated by the naturals (which is why it is called an 
uncountably
> infinite set).  But all sets have a size, so this set must have a
> size that is not a natural number.  It is meaningless (and just
> plain false) to say this set has no size or is not a set.
> I'm not shure if the reals build a set in spite of you and Cantor and
> others are shure.
> A set is defined by consisting of discrete, distinguishable, individual
> elements. Now tell me: what separates a point on a line from the very
> next point on the line to be discrete? What separates sqrt(2) from the
> very next real number to be discrete?
> If you look only on individual points, you may have a set. But if you
> look on all of them?
> So, your above argumentation has no relevance to me. Proof the reals to
> be a set, then let's talk again.
> What is your difficulty with the standard definitions?
> Certainly the Integers { ... -3,-2,-1,0,1,2,3 ... }
> form a set
Yes. If we accept infinite sets.
> Take the set of all pairs of integers (i,j) where
> the second integer is not 0.
> Take some equivalence classes of the above and we
> have the rationals.  (Note the rationals are not
> discrete).
> Take paris of sets of rationals (A,B) where all
> the rationals in A are less than those in B and
> where A union B is all the rationals.  Now
> we have the reals.
You think about Dedekind cuts?
> At which step do we fail to have a set?
>       -William Hughes
All this don't proof anything.
You know that the constructable real numbers are denumerable infinite.
If all real numbers are nondenumerable, there must be nondenumerable
infinite many real numbers between every pair of constructable reals.
Proof that this numbers are entities.
AS
===
Subject: Re: infinity ...
> You know that the constructable real numbers are denumerable infinite.
> If all real numbers are nondenumerable, there must be nondenumerable
> infinite many real numbers between every pair of constructable reals.
> Proof that this numbers are entities.
No numbers are entities without the assumption of their existence, 
since no 'number' exists in the physical world.
The easiest way to get started is to assume the naturals and a  little 
set theory as a basis, and construct the rest upon that basis.
===
Subject: Re: infinity ...
   <MPG.1dc85a4d4881712598a56a@newsstand.cit.cornell.edu
 Consider the set of reals in the interval [0,1], that is, the set
> S = {x in R : 0 <= x <= 1}.  The elements of this set cannot be
> enumerated by the naturals (which is why it is called an 
uncountably
> infinite set).  But all sets have a size, so this set must have a
> size that is not a natural number.  It is meaningless (and just
> plain false) to say this set has no size or is not a set.
> I'm not shure if the reals build a set in spite of you and Cantor and
> others are shure.
> A set is defined by consisting of discrete, distinguishable, 
individual
> elements. Now tell me: what separates a point on a line from the very
> next point on the line to be discrete? What separates sqrt(2) from 
the
> very next real number to be discrete?
> If you look only on individual points, you may have a set. But if you
> look on all of them?
 So, your above argumentation has no relevance to me. Proof the reals 
to
> be a set, then let's talk again.
> What is your difficulty with the standard definitions?
> Certainly the Integers { ... -3,-2,-1,0,1,2,3 ... }
> form a set
> Yes. If we accept infinite sets.
> Take the set of all pairs of integers (i,j) where
> the second integer is not 0.
> Take some equivalence classes of the above and we
> have the rationals.  (Note the rationals are not
> discrete).
> Take paris of sets of rationals (A,B) where all
> the rationals in A are less than those in B and
> where A union B is all the rationals.  Now
> we have the reals.
> You think about Dedekind cuts?
Yes, this is one of the standard constructions.
> At which step do we fail to have a set?
>       -William Hughes
> All this don't proof anything.
> You know that the constructable real numbers are denumerable infinite.
> If all real numbers are nondenumerable, there must be nondenumerable
> infinite many real numbers between every pair of constructable reals.
> Proof that this numbers are entities.
Since all of these numbers are of the form (A,B) where
A and B are sets of rationals they are clearly entities.
                            -William Hughes
===
Subject: Re: infinity ...
   <MPG.1dc85a4d4881712598a56a@newsstand.cit.cornell.edu>
   <MPG.1dcfff7f104b00a498a59e@newsstand.cit.cornell.edu The problem, math 
has to deal with, is, that it needs actual infinite
> objects like the decimal expansion of irreal numbers or transcendent
Pardon. Not irreal number but irrational number was meant
> numbers or just rational numbers like 1/3, or the numbers of points
> which lays on a straight line or in an intervall on an straight line or
> the set of natural numbers, etc, pp.
> But this problems has to be solved without inconsitent systems like the
> cantorian, which leads to nonsense like transfinite numbers which don't
> have any sense.
AS
===
Subject: Re: infinity
   <ITSnetNOTcom#virgil-FFCC1F.12173330102005@comcast.dca.giganews.com>
   <ITSnetNOTcom#virgil-6E36F8.14390930102005@comcast.dca.giganews.com Can 
you map Series S= (0,0),(1,0),(2,0)....... , 
(0,1),(1,1),(2,1)...}
> to N such that f(0)=(0,0) , f(1)=(1,0), f(2)=(2,0),......
> can their exist a bijection between N and S in the above order.
 If you say yes then please show me that.
> if you say No. then my answer is the cardinality of S or it's field 
is
> not the same as N.
 Failing to find a mapping from one series to another series for a
> particular ordering proves nothing.
> Yes you are right according to the standard way of thinking.
> However it happens that my ways(probably erronus) are different from
> the standard ways.
> In my simple way of thinking the order of a set shouldn't affect
> the ability of that set to biject with another set of the same
> cardinality.
> So if Card (set A)= Card (set B) , then any order set A is thrown into
> shouldn't prevent it from bijecting to set B , I mean ALL different
> orders of set A should biject to set B, becasue cardinality is not
> changed by rearranging the members of a set.
> Now in set F(S) defined above their is a particular order which
> prevents it from being bijected to set N.
> In my simple thinking this means that bijecting set F(S) with set N is
> order modulated or affected by order ,since some orders of set F(S)
> prevents bijecting it with set N. Therefore F(S) cannot have the same
> cardinality as set N.
> My argument is a simple as that .
> Ok, let F(S) have order A, so you can biject it with N, call
> this F_A.  Let F(S) have order B so you can't biject it with N, call
> this F_B.  Clearly you cannot biject F_A to F_B.
Right!
> Therefore F(S) does not have the same cardinality as itself!
No this conclusion is wrong and I didn't say that.
The problem with you mathematicians is that you do not spend enough
time to understand what others are saying.
I will use  your  symbology since it is good.
What I want to say here is that F_A if bijected to N , while F_B if not
bijected to N.
Then  F(S) is not equal in cardinality to N, and of caorse both
of its forms F_A and F_B are not equal in cardinality to N.
Now  what do the bijection between F_A and N  means.
My answer this bijection only means that the order of F_A is similar to
the order of N.
and consequently the ordinal number of F_A is the same as the ordinal
number of N.
But it doesn't mean that card F_A = card N.
To me only bijection between two set doesn't mean that these sets has
the same cardinality, it only means that they have the same ordinality.
To say that any two sets have the same cardinality you should
demonstrate that
ALL the arrangments of one of the sets can be bijected to the other
set.
Did you grasp what I am doing? I am changing the definition of
cardinality existing in your head dear mathematicians.
Formally speaking ( although I know it is clumsy )
if P means a property of
so P(set A) means a property of set A
P(set A) = v  , means that v is the value of that property of set A
for example Cardinality(set A) = 5  , here 5 is the value of
cardinality of set A
Similarily  P(set A) = v means that v is the value of that Property of
set A.
Now in your heads dear mathematicians exist the following rules:
If P(set A) = P( set B)  when a bijection can exist between set A and
set B .
and P (set A) <> P(set B) when a bijection cannot exist between set A
and set B.
Then P is  Cardinality.
To me that is not enough to define P as cardinality.
In order to say P is cardinality another requirement in addition to the
above should
be fullfilled , that is:
P ( set A _ a) = P ( set A_b) = P(set A_c)=......=P( set A_n)
where set A_( a xor b xor c xor,,,,xor n)  means a certain rearrangment
of set A.
For example :   {1,2} has bijection to {a,b} but that is not enough to
demontrate that
these two sets has the same cardinality.
Only if you say that { 1,2} has bijection to {b,a}  , then I can say
that their is equal cardinality between both sets.
Try to understand fully what others are saying before you critise them.
Zuhair
> Your definitions of cardinality is useless for infinite sets.
This is becasue you didn't understand them.
> (Note the standard explanation is simple.  A bijection either
> exists or does not exist.  The ordering you assign to a set
> makes no difference.
Yes that is exactly what I am asserting and you are violating.
 True, some orderings make it easier to
> see the bijection than others, but if a bijections exists, even
> if you use an ordering that makes it hard to see.  If a bijection
> exists for one ordering it exists for all orderings.)
This is only true if that bijection was between that ordering and a set
that has equal
cardinality to F(S).
>                   - William Hughes
>                       -William Hughes
> Zuhair
Try to understand read my posts carefully again.
Zuhair
===
Subject: Re: infinity
   <ITSnetNOTcom#virgil-FFCC1F.12173330102005@comcast.dca.giganews.com>
   <ITSnetNOTcom#virgil-6E36F8.14390930102005@comcast.dca.giganews.com Ok, 
let F(S) have order A, so you can biject it with N, call
> this F_A.  Let F(S) have order B so you can't biject it with N, call
> this F_B.  Clearly you cannot biject F_A to F_B.
> Right!
> To say that any two sets have the same cardinality you should
> demonstrate that
> ALL the arrangments of one of the sets can be bijected to the other
> set.
Since F can not be bijected to all of its own arrangements,
F does not have the same zu-cardinality as itself.
That is an elementary consequence of both of the above statements.
              - Randy
===
Subject: Re: infinity
   <ITSnetNOTcom#virgil-FFCC1F.12173330102005@comcast.dca.giganews.com>
   <ITSnetNOTcom#virgil-6E36F8.14390930102005@comcast.dca.giganews.com Can 
you map Series S= (0,0),(1,0),(2,0)....... , 
(0,1),(1,1),(2,1)...}
> to N such that f(0)=(0,0) , f(1)=(1,0), f(2)=(2,0),......
> can their exist a bijection between N and S in the above order.
 If you say yes then please show me that.
> if you say No. then my answer is the cardinality of S or it's 
field is
> not the same as N.
 Failing to find a mapping from one series to another series for a
> particular ordering proves nothing.
 Yes you are right according to the standard way of thinking.
 However it happens that my ways(probably erronus) are different from
 the standard ways.
 In my simple way of thinking the order of a set shouldn't affect
> the ability of that set to biject with another set of the same
> cardinality.
> So if Card (set A)= Card (set B) , then any order set A is thrown 
into
> shouldn't prevent it from bijecting to set B , I mean ALL different
> orders of set A should biject to set B, becasue cardinality is not
> changed by rearranging the members of a set.
 Now in set F(S) defined above their is a particular order which
> prevents it from being bijected to set N.
 In my simple thinking this means that bijecting set F(S) with set N 
is
> order modulated or affected by order ,since some orders of set F(S)
> prevents bijecting it with set N. Therefore F(S) cannot have the same
> cardinality as set N.
 My argument is a simple as that .
> Ok, let F(S) have order A, so you can biject it with N, call
> this F_A.  Let F(S) have order B so you can't biject it with N, call
> this F_B.  Clearly you cannot biject F_A to F_B.
> Right!
> Therefore F(S) does not have the same cardinality as itself!
> No this conclusion is wrong and I didn't say that.
> The problem with you mathematicians is that you do not spend enough
> time to understand what others are saying.
> I will use  your  symbology since it is good.
> What I want to say here is that F_A if bijected to N , while F_B if not
> bijected to N.
> Then  F(S) is not equal in cardinality to N, and of caorse both
> of its forms F_A and F_B are not equal in cardinality to N.
At this point we have
    There exists a bijection from F_A to N, call it g
    There does not exist a bijection from F_B to N
    Therefore there does not exist a bijection from F_B to F_A
    (assume one exists, call it f. then f followed by g is a bijection
     from F_B to N)
(note that here you are not using the standard meaning of bijection,
otherwise
the identity is a bijection between F_B and F_A)
So since there is no bijection (your definition) between F_B and F_A,
F_B and F_A do not have the same cardinality (your definition)
> Now  what do the bijection between F_A and N  means.
> My answer this bijection only means that the order of F_A is similar to
> the order of N.
> and consequently the ordinal number of F_A is the same as the ordinal
> number of N.
This is your definition of bijection, the standard definition makes
no reference to order.
> But it doesn't mean that card F_A = card N.
No but it does mean that there is a bijection (your definition) from
F_A to N
> To me only bijection between two set doesn't mean that these sets has
> the same cardinality, it only means that they have the same ordinality.
> To say that any two sets have the same cardinality you should
> demonstrate that
> ALL the arrangments of one of the sets can be bijected to the other
> set
So a set X has the same cardinality as a set Y iff all arrangements
of Y can be bijected (you definition) to X.
So a set X has the same cardinality as a set X iff all arrangements
of X can be bijected (your definition)  to X.
So F(S) has the same cardinality as a set F(S) iff all arrangements
of F(S) can be bijected (your definition)  to F(S).
But, F_B cannot be bijected (your definition) to F_A.
Therefore F(S) does not have the same cardinality as F(S)
>Did you grasp what I am doing? I am changing the definition of
> cardinality existing in your head dear mathematicians.
Since I make the claim that a set does not have the same
cardinality as itself,  and this is ridiculous using the standard
definitions, it is clear to me that you are changing the
definition of cardinality.
> Formally speaking ( although I know it is clumsy )
> if P means a property of
> so P(set A) means a property of set A
> P(set A) = v  , means that v is the value of that property of set A
> for example Cardinality(set A) = 5  , here 5 is the value of
> cardinality of set A
> Similarily  P(set A) = v means that v is the value of that Property of
> set A.
> Now in your heads dear mathematicians exist the following rules:
> If P(set A) = P( set B)  when a bijection can exist between set A and
> set B .
> and P (set A) <> P(set B) when a bijection cannot exist between set A
> and set B.
> Then P is  Cardinality.
> To me that is not enough to define P as cardinality.
> In order to say P is cardinality another requirement in addition to the
> above should
> be fullfilled , that is:
> P ( set A _ a) = P ( set A_b) = P(set A_c)=......=P( set A_n)
> where set A_( a xor b xor c xor,,,,xor n)  means a certain rearrangment
> of set A.
> For example :   {1,2} has bijection to {a,b} but that is not enough to
> demontrate that
> these two sets has the same cardinality.
> Only if you say that { 1,2} has bijection to {b,a}  , then I can say
> that their is equal cardinality between both sets.
> Try to understand fully what others are saying before you critise them.
Words to live by.
                                        -William Hughes
P.S.  I strongly suggest that you come up with a couple of new terms
         to describe your definiton of bijection, and your definition
of cardinality
         (e.g. zijection, zardinality, but if you don't like these make
up
         your own terms).  This would greatly reduce confusion.
===
Subject: Re: infinity
   <ITSnetNOTcom#virgil-FFCC1F.12173330102005@comcast.dca.giganews.com>
   <ITSnetNOTcom#virgil-6E36F8.14390930102005@comcast.dca.giganews.com Can 
you map Series S= (0,0),(1,0),(2,0)....... , 
(0,1),(1,1),(2,1)...}
> to N such that f(0)=(0,0) , f(1)=(1,0), f(2)=(2,0),......
> can their exist a bijection between N and S in the above order.
 If you say yes then please show me that.
> if you say No. then my answer is the cardinality of S or it's 
field is
> not the same as N.
 Failing to find a mapping from one series to another series for a
> particular ordering proves nothing.
 Yes you are right according to the standard way of thinking.
 However it happens that my ways(probably erronus) are different 
from
 the standard ways.
 In my simple way of thinking the order of a set shouldn't affect
> the ability of that set to biject with another set of the same
> cardinality.
> So if Card (set A)= Card (set B) , then any order set A is thrown 
into
> shouldn't prevent it from bijecting to set B , I mean ALL different
> orders of set A should biject to set B, becasue cardinality is not
> changed by rearranging the members of a set.
 Now in set F(S) defined above their is a particular order which
> prevents it from being bijected to set N.
 In my simple thinking this means that bijecting set F(S) with set N 
is
> order modulated or affected by order ,since some orders of set F(S)
> prevents bijecting it with set N. Therefore F(S) cannot have the 
same
> cardinality as set N.
 My argument is a simple as that .
 Ok, let F(S) have order A, so you can biject it with N, call
> this F_A.  Let F(S) have order B so you can't biject it with N, call
> this F_B.  Clearly you cannot biject F_A to F_B.
> Right!
> Therefore F(S) does not have the same cardinality as itself!
> No this conclusion is wrong and I didn't say that.
> The problem with you mathematicians is that you do not spend enough
> time to understand what others are saying.
> I will use  your  symbology since it is good.
> What I want to say here is that F_A if bijected to N , while F_B if not
> bijected to N.
> Then  F(S) is not equal in cardinality to N, and of caorse both
> of its forms F_A and F_B are not equal in cardinality to N.
> At this point we have
>     There exists a bijection from F_A to N, call it g
>     There does not exist a bijection from F_B to N
>     Therefore there does not exist a bijection from F_B to F_A
>     (assume one exists, call it f. then f followed by g is a bijection
>      from F_B to N)
No this conclusion is erronous, This is like saying since the absolute
value of the difference between 2 and 1 is 1 and for the difference
between -2 and 1 is 3 then
the absolute value of the difference between 2 and any number x is not
equal to the absolute value of the difference between -2 and x.Which is
wrong.
Now although F_A can be bijected to N ,while F_B cannot, this doesn't
mean that
their is no set were both F_A and F_B can be bijected to.
F_A can be bijected to the set
N*=1,2,3,4,.......,w,w+1,w+2,w+3,........
so does F_B . And accordingly we can find a bijection between F_A and
F_B.
Another set that both F_A and F_B can be bijected to is without doubt
F(S).
Of coarse without any doubt F(S) can be bijected to N*.
Also a set X can always be found to have bijections to all arrangments
of itself.
that's why it has One cardinality.
However if you insist that set X cannot always be found to have
bijections to all arrangments of itself , then this doesn't follow that
it has a cardinality that it not
equal to it's cardinality, since this is absured. The logical
conculsion would be that set X which cannot be shown to have bijections
to all arrangments of itself, is
to be called a set of UNDETERMINED CARDINALITY.
Now if you proove that a set has one of it's arrangments non bijectable
to it, then
we say that this set HAS NO CARDINALITY. ( ie it doesn't possess a
property
that is irrespective of the individual properties of its members and
their order).
Zuhair
===
Subject: Re: infinity
   <ITSnetNOTcom#virgil-FFCC1F.12173330102005@comcast.dca.giganews.com>
   <ITSnetNOTcom#virgil-6E36F8.14390930102005@comcast.dca.giganews.com Also 
a set X can always be found to have bijections to all arrangments
> of itself.
> that's why it has One cardinality.
Here is the set N.
N = {0,1,2,3,...}
Here is an arrangement of set N
N = {0,2,4,...  ,1,3,5,...}
Please illustrate the bijection from the first arrangement to the
second
arrangement.
> However if you insist that set X cannot always be found to have
> bijections to all arrangments of itself , then this doesn't follow that
> it has a cardinality that it not
> equal to it's cardinality, since this is absured. The logical
> conculsion would be that set X which cannot be shown to have bijections
> to all arrangments of itself, is
> to be called a set of UNDETERMINED CARDINALITY.
Let X be any Dedekind-infinite set. Then there exists Y a proper subset
of
X such that there is a bijection from X to Y. Consider an ordering
of X as:
X = {Y ,  X/Y }
Since we have a bijection from X to Y, there are no elements of X
left to map to X/Y in that mapping. Hence X can not be bijected to
this arrangement of itself.
Therefore in zuhair set theory, no infinite set X has a cardinality.
                  - Randy
===
Subject: Re: infinity
   <ITSnetNOTcom#virgil-FFCC1F.12173330102005@comcast.dca.giganews.com>
   <ITSnetNOTcom#virgil-6E36F8.14390930102005@comcast.dca.giganews.com Can 
you map Series S= (0,0),(1,0),(2,0)....... , 
(0,1),(1,1),(2,1)...}
> to N such that f(0)=(0,0) , f(1)=(1,0), f(2)=(2,0),......
> can their exist a bijection between N and S in the above order.
 If you say yes then please show me that.
> if you say No. then my answer is the cardinality of S or it's 
field is
> not the same as N.
 Failing to find a mapping from one series to another series for a
> particular ordering proves nothing.
 Yes you are right according to the standard way of thinking.
 However it happens that my ways(probably erronus) are different from
 the standard ways.
 In my simple way of thinking the order of a set shouldn't affect
> the ability of that set to biject with another set of the same
> cardinality.
> So if Card (set A)= Card (set B) , then any order set A is thrown 
into
> shouldn't prevent it from bijecting to set B , I mean ALL different
> orders of set A should biject to set B, becasue cardinality is not
> changed by rearranging the members of a set.
 Now in set F(S) defined above their is a particular order which
> prevents it from being bijected to set N.
 In my simple thinking this means that bijecting set F(S) with set N 
is
> order modulated or affected by order ,since some orders of set F(S)
> prevents bijecting it with set N. Therefore F(S) cannot have the same
> cardinality as set N.
 My argument is a simple as that .
> Ok, let F(S) have order A, so you can biject it with N, call
> this F_A.  Let F(S) have order B so you can't biject it with N, call
> this F_B.  Clearly you cannot biject F_A to F_B.
> Right!
> Therefore F(S) does not have the same cardinality as itself!
> No this conclusion is wrong and I didn't say that.
> The problem with you mathematicians is that you do not spend enough
> time to understand what others are saying.
> I will use  your  symbology since it is good.
> What I want to say here is that F_A if bijected to N , while F_B if not
> bijected to N.
> Then  F(S) is not equal in cardinality to N, and of caorse both
Correction: I meant The cardinality of F(S) is not equal to the
cardinality of N.
> of its forms F_A and F_B are not equal in cardinality to N.
Correction: I meant  cardinality of F_A and cardinality of F_B are not
equal to
cardinality of N.
> Now  what do the bijection between F_A and N  means.
> My answer this bijection only means that the order of F_A is similar to
> the order of N.
> and consequently the ordinal number of F_A is the same as the ordinal
> number of N.
> But it doesn't mean that card F_A = card N.
> To me only bijection between two set doesn't mean that these sets has
> the same cardinality, it only means that they have the same ordinality.
> To say that any two sets have the same cardinality you should
> demonstrate that
> ALL the arrangments of one of the sets can be bijected to the other
> set.
> Did you grasp what I am doing? I am changing the definition of
> cardinality existing in your head dear mathematicians.
> Formally speaking ( although I know it is clumsy )
> if P means a property of
> so P(set A) means a property of set A
> P(set A) = v  , means that v is the value of that property of set A
> for example Cardinality(set A) = 5  , here 5 is the value of
> cardinality of set A
> Similarily  P(set A) = v means that v is the value of that Property of
> set A.
> Now in your heads dear mathematicians exist the following rules:
> If P(set A) = P( set B)  when a bijection can exist between set A and
> set B .
> and P (set A) <> P(set B) when a bijection cannot exist between set A
> and set B.
> Then P is  Cardinality.
> To me that is not enough to define P as cardinality.
A simple addition: and if P(set A)= v , and P(set B)= x
then if set A can be bijected to set B  gives rise to
P(set A) = P(set B) = v=x
and when set A cannot be bijected to set B gives rise to
P(set A ) <> P ( set B) <==> v <> x
Then P is Cardinality( or a function of cardinality), and v and x are
the values of that cardinality in set A and set B respectively.
Of coarse I am against that and I stress that in addition to the above
a second
requirement should be fulfilled like below.
> In order to say P is cardinality another requirement in addition to the
> above should
> be fullfilled , that is:
> P ( set A _ a) = P ( set A_b) = P(set A_c)=......=P( set A_n)
Better writtin as P ( set A _ a) = P ( set A_b) = P(set A_c)=......=P(
set A_n) = v
> where set A_( a xor b xor c xor,,,,xor n)  means a certain rearrangment
> of set A.
> For example :   {1,2} has bijection to {a,b} but that is not enough to
> demontrate that
> these two sets has the same cardinality.
> Only if you say that { 1,2} has bijection to {b,a}  , then I can say
> that their is equal cardinality between both sets.
More precisely speaking would be:
Only if you say that { 1,2} ALSO has bijection to {b,a}  , then I can
say
> that their is equal cardinality between both sets.
So the bijections required to assure card {1,2} = card{a,b} are:
1,2
a,b
AND
1,2
b,a
Only if both are fullfiled then I can say that card {1,2} = card{a,b}
> Try to understand fully what others are saying before you critise them.
> Zuhair
> Your definitions of cardinality is useless for infinite sets.
> This is becasue you didn't understand them.
> (Note the standard explanation is simple.  A bijection either
> exists or does not exist.  The ordering you assign to a set
> makes no difference.
> Yes that is exactly what I am asserting and you are violating.
>  True, some orderings make it easier to
> see the bijection than others, but if a bijections exists, even
> if you use an ordering that makes it hard to see.  If a bijection
> exists for one ordering it exists for all orderings.)
> This is only true if that bijection was between that ordering and a set
> that has equal
> cardinality to F(S).
>                   - William Hughes
>                       -William Hughes
 Zuhair
> Try to understand read my posts carefully again.
> Zuhair
===
Subject: Re: infinity
   <ITSnetNOTcom#virgil-FFCC1F.12173330102005@comcast.dca.giganews.com>
   <ITSnetNOTcom#virgil-6E36F8.14390930102005@comcast.dca.giganews.com Can 
you map Series S= (0,0),(1,0),(2,0)....... , 
(0,1),(1,1),(2,1)...}
> to N such that f(0)=(0,0) , f(1)=(1,0), f(2)=(2,0),......
> can their exist a bijection between N and S in the above order.
 If you say yes then please show me that.
> if you say No. then my answer is the cardinality of S or it's field 
is
> not the same as N.
 Failing to find a mapping from one series to another series for a
> particular ordering proves nothing.
> Yes you are right according to the standard way of thinking.
> However it happens that my ways(probably erronus) are different from
> the standard ways.
> In my simple way of thinking the order of a set shouldn't affect
> the ability of that set to biject with another set of the same
> cardinality.
> So if Card (set A)= Card (set B) , then any order set A is thrown into
> shouldn't prevent it from bijecting to set B , I mean ALL different
> orders of set A should biject to set B, becasue cardinality is not
> changed by rearranging the members of a set.
> Now in set F(S) defined above their is a particular order which
> prevents it from being bijected to set N.
> In my simple thinking this means that bijecting set F(S) with set N is
> order modulated or affected by order ,since some orders of set F(S)
> prevents bijecting it with set N. Therefore F(S) cannot have the same
> cardinality as set N.
> My argument is a simple as that .
> Ok, let F(S) have order A, so you can biject it with N, call
> this F_A.  Let F(S) have order B so you can't biject it with N, call
> this F_B.  Clearly you cannot biject F_A to F_B.
Right!
> Therefore F(S) does not have the same cardinality as itself!
No this conclusion is wrong and I didn't say that.
The problem with you mathematicians is that you do not spend enough
time to understand what others are saying.
I will use  your  symbology since it is good.
What I want to say here is that F_A if bijected to N , while F_B if not
bijected to N.
Then  F(S) is not equal in cardinality to N, and of caorse both
of its forms F_A and F_B are not equal in cardinality to N.
Now  what do the bijection between F_A and N  means.
My answer this bijection only means that the order of F_A is similar to
the order of N.
and consequently the ordinal number of F_A is the same as the ordinal
number of N.
But it doesn't mean that card F_A = card N.
To me only bijection between two set doesn't mean that these sets has
the same cardinality, it only means that they have the same ordinality.
To say that any two sets have the same cardinality you should
demonstrate that
ALL the arrangments of one of the sets can be bijected to the other
set.
Did you grasp what I am doing? I am changing the definition of
cardinality existing in your head dear mathematicians.
Formally speaking ( although I know it is clumsy )
if P means a property of
so P(set A) means a property of set A
P(set A) = v  , means that v is the value of that property of set A
for example Cardinality(set A) = 5  , here 5 is the value of
cardinality of set A
Similarily  P(set A) = v means that v is the value of that Property of
set A.
Now in your heads dear mathematicians exist the following rules:
If P(set A) = P( set B)  when a bijection can exist between set A and
set B .
and P (set A) <> P(set B) when a bijection cannot exist between set A
and set B.
Then P is  Cardinality.
To me that is not enough to define P as cardinality.
In order to say P is cardinality another requirement in addition to the
above should
be fullfilled , that is:
P ( set A _ a) = P ( set A_b) = P(set A_c)=......=P( set A_n)
where set A_( a xor b xor c xor,,,,xor n)  means a certain rearrangment
of set A.
For example :   {1,2} has bijection to {a,b} but that is not enough to
demontrate that
these two sets has the same cardinality.
Only if you say that { 1,2} has bijection to {b,a}  , then I can say
that their is equal cardinality between both sets.
Try to understand fully what others are saying before you critise them.
Zuhair
> Your definitions of cardinality is useless for infinite sets.
This is becasue you didn't understand them.
> (Note the standard explanation is simple.  A bijection either
> exists or does not exist.  The ordering you assign to a set
> makes no difference.
Yes that is exactly what I am asserting and you are violating.
 True, some orderings make it easier to
> see the bijection than others, but if a bijections exists, even
> if you use an ordering that makes it hard to see.  If a bijection
> exists for one ordering it exists for all orderings.)
This is only true if that bijection was between that ordering and a set
that has equal
cardinality to it.
>                   - William Hughes
>                       -William Hughes
> 
Next time read my posts well !!!
Zuhair
===
Subject: Re: infinity
   <ITSnetNOTcom#virgil-FFCC1F.12173330102005@comcast.dca.giganews.com>
   <ITSnetNOTcom#virgil-6E36F8.14390930102005@comcast.dca.giganews.com Can 
you map Series S= (0,0),(1,0),(2,0)....... , 
(0,1),(1,1),(2,1)...}
> to N such that f(0)=(0,0) , f(1)=(1,0), f(2)=(2,0),......
> can their exist a bijection between N and S in the above order.
 If you say yes then please show me that.
> if you say No. then my answer is the cardinality of S or it's field 
is
> not the same as N.
 Failing to find a mapping from one series to another series for a
> particular ordering proves nothing.
> Yes you are right according to the standard way of thinking.
> However it happens that my ways(probably erronus) are different from
> the standard ways.
> In my simple way of thinking the order of a set shouldn't affect
> the ability of that set to biject with another set of the same
> cardinality.
> So if Card (set A)= Card (set B) , then any order set A is thrown into
> shouldn't prevent it from bijecting to set B , I mean ALL different
> orders of set A should biject to set B, becasue cardinality is not
> changed by rearranging the members of a set.
> Now in set F(S) defined above their is a particular order which
> prevents it from being bijected to set N.
> In my simple thinking this means that bijecting set F(S) with set N is
> order modulated or affected by order ,since some orders of set F(S)
> prevents bijecting it with set N. Therefore F(S) cannot have the same
> cardinality as set N.
> My argument is a simple as that .
> Ok, let F(S) have order A, so you can biject it with N, call
> this F_A.  Let F(S) have order B so you can't biject it with N, call
> this F_B.  Clearly you cannot biject F_A to F_B.
Right!
> Therefore F(S) does not have the same cardinality as itself!
No this conclusion is wrong and I didn't say that.
The problem with you mathematicians is that you do not spend enough
time to understand what others are saying.
I will use  your  symbology since it is good.
What I want to say here is that F_A if bijected to N , while F_B if not
bijected to N.
Then  F(S) is not equal in cardinality to N, and of caorse both
of its forms F_A and F_B are not equal in cardinality to N.
Now  what do the bijection between F_A and N  means.
My answer this bijection only means that the order of F_A is similar to
the order of N.
and consequently the ordinal number of F_A is the same as the ordinal
number of N.
But it doesn't mean that card F_A = card N.
To me only bijection between two set doesn't mean that these sets has
the same cardinality, it only means that they have the same ordinality.
To say that any two sets have the same cardinality you should
demonstrate that
ALL the arrangments of one of the sets can be bijected to the other
set.
Did you grasp what I am doing? I am changing the definition of
cardinality existing in your head dear mathematicians.
Formally speaking ( although I know it is clumsy )
if P means a property of
so P(set A) means a property of set A
P(set A) = v  , means that v is the value of that property of set A
for example Cardinality(set A) = 5  , here 5 is the value of
cardinality of set A
Similarily  P(set A) = v means that v is the value of that Property of
set A.
Now in your heads dear mathematicians exist the following rules:
If P(set A) = P( set B)  when a bijection can exist between set A and
set B .
and P (set A) <> P(set B) when a bijection cannot exist between set A
and set B.
Then P is  Cardinality.
To me that is not enough to define P as cardinality.
In order to say P is cardinality another requirement in addition to the
above should
be fullfilled , that is:
P ( set A _ a) = P ( set A_b) = P(set A_c)=......=P( set A_n)
where set A_( a xor b xor c xor,,,,xor n)  means a certain rearrangment
of set A.
For example :   {1,2} has bijection to {a,b} but that is not enough to
demontrate that
these two sets has the same cardinality.
Only if you say that { 1,2} has bijection to {b,a}  , then I can say
that their is equal cardinality between both sets.
Try to understand fully what others are saying before you critise them.
Zuhair
> Your definitions of cardinality is useless for infinite sets.
This is becasue you didn't understand them.
> (Note the standard explanation is simple.  A bijection either
> exists or does not exist.  The ordering you assign to a set
> makes no difference.
Yes that is exactly what I am asserting and you are violating.
 True, some orderings make it easier to
> see the bijection than others, but if a bijections exists, even
> if you use an ordering that makes it hard to see.  If a bijection
> exists for one ordering it exists for all orderings.)
This is only true if that bijection was between that ordering and a set
that has equal
cardinality to F(S).
>                   - William Hughes
>                       -William Hughes
> 
Next time read my posts well !!!
Zuhair
===
Subject: vector analysis correction
I have gone over my mistake from my earlier post and it would be helpful if 

anyone can tell me that I have done it right.
I have to evaluate the integral of F. dr (c lower limit) where F = (yz)i + 
(2y)j - (x^2)k and
i) C is the curve x=t, y=t^2, z=t^3 0<t<1 (less than or equal!)
So, F = (t^5, 2t^2, -t^2)which I hope is right.
and the answer I get is 17/30.
===
Subject: Re: vector analysis correction
On Thu, 03 Nov 2005 13:40:56 EST, jaspal <jsdhariwal@yahoo.co.ukI have gone 
over my mistake from my earlier post and it would be helpful if 
anyone can tell me that I have done it right.
>I have to evaluate the integral of F. dr (c lower limit) where F = (yz)i + 

(2y)j - (x^2)k and
>i) C is the curve x=t, y=t^2, z=t^3 0<t<1 (less than or equal!)
>So, F = (t^5, 2t^2, -t^2)which I hope is right.
>and the answer I get is 17/30.
Correct.
===
Subject: Re: Please help me with radicals simplification, i partly get it, 
but need help...
> my textbook has an answer of
> sqrt (6) / 2 for this question, and i have problems reaching it by
> using ur way, can u finish the problem, and explain  the last part
> more.
I gave you a start, since there was no evidence that you had started
it. I'm not going to finish without evidence that you did some
work first.
If this is what the book gave as an answer, you probably copied
the problem wrong.
> By the way, after
> 3rd-rt(2)*sqrt(3) - 3rd-rt(2)*sqrt(2) + sqrt(3)*sqrt(3) ,
> I got 3rd-rt(6) - 3rd-rt(4) + sqrt(9) - sqrt(6). And how did you get
> -2^(1/3)*2^(1/2) for the second term. thnx man
You have 3rd-rt(2) in your original question. That is the same as
2^(1/3).
sqrt(2) is the same as 2^(1/2).
                - Randy
===
Subject: Re: Please help me with radicals simplification, i partly get it, 
but need help...
> my textbook has an answer of
> sqrt (6) / 2 for this question, and i have problems reaching it by
> using ur way, can u finish the problem, and explain  the last part
> more.
If this is what the book gave as an answer, you probably copied
the problem wrong.
> By the way, after
> 3rd-rt(2)*sqrt(3) - 3rd-rt(2)*sqrt(2) + sqrt(3)*sqrt(3) ,
> I got 3rd-rt(6) - 3rd-rt(4)
If by 3rd-rt(2) you really mean the cube root of 2, then you
can't multiply a cube root by a square root and pretend they
are both square roots.
> + sqrt(9) - sqrt(6). And how did you get
> -2^(1/3)*2^(1/2) for the second term. thnx man
You have 3rd-rt(2) in your original question. That is the same as
2^(1/3).
sqrt(2) is the same as 2^(1/2).
                - Randy
===
Subject: Matrix square root, sherman morrison?
Hi all,
Can anybody help me with a numerical linear algebra homework problem?
(the apostrophe denotes transposition)
Given a vector u, I want to show how to compute a matrix X satisfying
X*(I + k*uu')*X = I
so that X can be thought of as the square root of
( I + k*uu' )^-1
in class, we did the Sherman-Morrison formula, which says that
(I + k*u*u')^-1  = I - [ k / (1+k*u'u) ]*uu'
but I've been playing with the algebra for days and haven't worked out
such an X.  Does anybody have suggestions?
KH
===
Subject: Re: Matrix square root, sherman morrison?
> Hi all,
> Can anybody help me with a numerical linear algebra homework problem?
> (the apostrophe denotes transposition)
> Given a vector u, I want to show how to compute a matrix X satisfying
> X*(I + k*uu')*X = I
> so that X can be thought of as the square root of
> ( I + k*uu' )^-1
> in class, we did the Sherman-Morrison formula, which says that
> (I + k*u*u')^-1  = I - [ k / (1+k*u'u) ]*uu'
> but I've been playing with the algebra for days and haven't worked out
> such an X.  Does anybody have suggestions?
> KH
Yes. This is equivalent to solving a Fredholm integral equation with a 
rank-1
(that is, separable) kernel. Have a look at Mathews & Walker, or Riley, et 
al.
books on mathematical methods of physics.
M = I + |u>k<u| is the unit matrix plus a rank-1 matrix. Thus its resolvent
satisfies
        RM = MR = I
If you define <v| = <u|R you can easilty find R in closed form.
Try the ansaatz
        X = I + |u>a<u|
and obtain an equation for a. You can assume |u> is normalized, i.e.
        <u|u> = 1
because that just amounts to a redefinition of k.
Note, BTW, how much simpler it looks if you use Dirac's bra-ket notation
for vectors and their adjoints, as I have done. It's why he invented it!
-- 
Julian V. Noble
Professor Emeritus of Physics
http://galileo.phys.virginia.edu/~jvn/
   For there was never yet philosopher that could endure the 
    toothache patiently.  
        -- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1.
===
Subject: Re: Matrix square root, sherman morrison?
>Hi all,
>Can anybody help me with a numerical linear algebra homework problem?
>(the apostrophe denotes transposition)
>Given a vector u, I want to show how to compute a matrix X satisfying
>X*(I + k*uu')*X = I
>so that X can be thought of as the square root of
>( I + k*uu' )^-1
>in class, we did the Sherman-Morrison formula, which says that
>(I + k*u*u')^-1  = I - [ k / (1+k*u'u) ]*uu'
>but I've been playing with the algebra for days and haven't worked out
>such an X.  Does anybody have suggestions?
Hint: For convenience write u u' = V and u' u = w, and note that
V^2 = w V.  An answer (maybe not the only one) can be written as
X = I + t V for some scalar t.  Expand out X (I + k V) X and
see what t has to be to make this I.  
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada
===
Subject: Re: Matrix square root, sherman morrison?
On 2005-11-03 15:04:30 -0400, Ken Honda <Honda_Kiai@hotmail.com> said:
> Hi all,
> Can anybody help me with a numerical linear algebra homework problem?
> (the apostrophe denotes transposition)
> Given a vector u, I want to show how to compute a matrix X satisfying
> X*(I + k*uu')*X = I
> so that X can be thought of as the square root of
> ( I + k*uu' )^-1
> in class, we did the Sherman-Morrison formula, which says that
> (I + k*u*u')^-1  = I - [ k / (1+k*u'u) ]*uu'
> but I've been playing with the algebra for days and haven't worked out
> such an X.  Does anybody have suggestions?
> KH
Try looking at the eigenvalues and eigenvectors of the symmetric matrix
   M = I + k*uu'
and thinking about square roots and inverses of the eienvalues.
Assume u'u = 1 as otherwise all you need is a revised value for k.
Sherman-Morrison has lots of uses. And also with the Woodbury 
generalization.
But not all class exercises want you to show you understood the most
complicated thing that has been mentioned in class. So go back to simpler
things.
===
Subject: Excellence and international dating
C.S. Lewis described false humility as smart men thinking themselves
stupid and pretty women thinking themselves ugly. This does not result
in humble people; it results in broken, malicious, bitter, poisonous
people. And the true humility as he supposed it is this: Rejoicing in
the good of the next person as much as one would in one's own.
At the incentive level, this is also what makes possible the best
world. When Christian ethics are misused to entrap and diminish, the
result is hideousness for everyone. When ethics are used instead to
motivate and enhance, the result is an improvement in life and in the
living (and the world bequeathed to the yet-to-exist).
At the incentive level, furthermore, it is important to recognize that
which makes excellence and make those qualities assets within the
civilization, in order that excellence be pursued and rewarded. And it
is for this reason that beauty and intellect, both of which are aspects
of excellence, be cultivated and rewarded. The reason, again? To
achieve the most excellent humanity, and the most excellent life for
humanity. Which is true for all aspects of human existence, and
includes such things as: Style; quality and tastefulness of
architecture and technology; elegance, richness and class in all human
works and in all human affairs; and beauty as value in and of itself.
Why is that important? Because that is what makes the highest quality
existence, and the highest quality civilization. The moral
consideration, being used to enhance and improve rather than degrade
and diminish, becomes used as a tool of good instead of a tool of
wrong. And that results in life bettered rather than worsened through
its utilization.
The psychological approach is flawed for one and simple reason: It
fails to differentiate the processes that it studies from its own
method in fathoming them. That is, the scientific method that it
believes it uses gets projected upon the person as vision of mental
health, and all mental processes - most of which have nothing at all to
do with scientific methods but nevertheless have their own logic - get
judged rather than studied. I have found, I believe, a way to
understand mindsets - and cultures, religions, experiences and
civilizations - that sees it in its external effects and experiences it
as it is experienced by the participants. I call it the integrative
cognition: Combining objective observation with subjective experience,
and arriving at a complete three-dimensional impressionistic assayal of
the mindset being viewed.
Now there is a technique of argumentation I refer to as Gordian knot.
It is a combination of two half-truths in such a manner as to create a
complete lie, entangling the soul and making its truth inaccessible.
One such Gordian knot that I've seen is used by American feminists. The
first half-truth is that beauty is relative. As I have stated and
continue to state, with my source of evidence being scientific studies,
there is both absolute beauty and relative beauty. The absolute beauty
is manifest in the facial proportions that, when shown to people from
all cultures by Judith Langlois, were recognized as being beautiful.
The relative beauty has been found in an experiment that showed 500
faces to 20000 people, and each face was found the most beautiful by at
least one person (though some by many more). The truthful way to fathom
the subject is therefore this: There is absolute beauty; and there is
relative beauty. The latter of which, in fact, should be found
encouraging: It means that even if people in your school or whatever
hideous locus of socialization found you ugly, there is someone for
you. And if that - not being seen as hopeless and not being attacked -
is your sole goal, then I am with you all the way.
But that's not what we are seeing. We are seeing something much worse,
and that is: Savage attack on both the absolutely beautiful women and
those catering more to the relative tastes of the place at hand.
Half-truth # 1: Beauty is relative. Half-truth #2: Beautiful women
are shallow and arrogant. Gordian knot: Let's attack the beautiful
women while thinking ourselves righteous and honest and serving the
good of woman.
How do I untie this Gordian knot? By seeing it objectively and
experiencing it within and seeing it completely and in parts,
completing the central half-truths and taking the half-lies apart.
Which is as follows: If beauty is solely relative, then it makes no
sense to attack beautiful women, as all women are equally beautiful or
ugly; and the fact that one is attacking the beautiful women while
being good to the ugly women means that one knows, like everyone else
does, what is beauty - and the central contention of beauty being
relative is therefore refuted by one's actions. And if beautiful women
are shallow and arrogant, then that shows some kind of inverse
correlation between being one's best and being good, with those who
seek to be excellent being worse than those who don't - which of course
is an absurd argument; as inner beauty and inner goodness will motivate
the person to be one's best, and that includes among other things being
one's most beautiful. What results instead is a kind of reverse
bigotry, in which a woman who seeks to be her best gets treated worse
than does a woman who doesn't; and that results in a wrong set of
incentives, a set of incentives that results in degradation of human
experience and of humanity as such.
Which means that it becomes necessary to intervene and destroy such
reverse-bigotry institutions and say that beautiful women are human
beings as well, even if the world's harpies don't like them. And who,
besides being human like them, also seek to be their best; which gives
them a greater EARNED set of rights. The argument being pushed down
people's throats - that inner beauty and outer beauty are somehow
incompatible - is based on envy and hatred and marks those who use it
as possessing neither beauty outer nor inner. And to use that as a
supposed conscience or heart of a country, moving men who have liberal
sympathies to power-hungry harpies while denying them to women who seek
to be their best, is indeed an outrage.
Now there is something else that is a part of this as well. Social
movements create a kind of cartel, in which competition is hobbled in
order that those who perpetrate the cartel mindset have control over
people's choices. Thus, unions used to punch men who worked too hard
for their tastes; and harpies attack and abuse women who are beautiful
inside and out. The goal is of course to make people think that they
have only one choice; and anything that gives promise otherwise must be
demolished. Which is another factor behind the hideousness with which
ugly women treat beautiful ones.
So in response to the aforementioned movement, I say: You will get
precisely what other cartel movements, anti-competitive movements, mob
movements, in America have gotten.
We'll see an increase in couplings between long-suffering American men
and magnificent women from other civilizations.
Which set of matches - between the people who have the most to offer
each other - will realize in both American men and foreign women
leading far better lives than they otherwise could have hoped for.
In the same way as for example we've seen the same done with American
business and Chinese labor.
And which as in the aforementioned case will give people - men and
women - better lives than they could have dreamed about, while creating
real-world incentive for people to excel and to treat their partners
well.
Be well NAFTA, be well WTO, and be well international relationships.
Ilya Shambat.
===
Subject: Re: Excellence and international dating
Well it started out really good but then I noticed the massive 
cross-posting.
Too bad.
Lost this reader with all those cross-posts.
Doug
> C.S. Lewis described false humility as smart men thinking themselves
> stupid and pretty women thinking themselves ugly. This does not result
> in humble people; it results in broken, malicious, bitter, poisonous
> people. And the true humility as he supposed it is this: Rejoicing in
> the good of the next person as much as one would in one's own.
> At the incentive level, this is also what makes possible the best
> world. When Christian ethics are misused to entrap and diminish, the
> result is hideousness for everyone. When ethics are used instead to
> motivate and enhance, the result is an improvement in life and in the
> living (and the world bequeathed to the yet-to-exist).
> At the incentive level, furthermore, it is important to recognize that
> which makes excellence and make those qualities assets within the
> civilization, in order that excellence be pursued and rewarded. And it
> is for this reason that beauty and intellect, both of which are
> aspects of excellence, be cultivated and rewarded. The reason, again?
> To achieve the most excellent humanity, and the most excellent life
> for humanity. Which is true for all aspects of human existence, and
> includes such things as: Style; quality and tastefulness of
> architecture and technology; elegance, richness and class in all human
> works and in all human affairs; and beauty as value in and of itself.
> Why is that important? Because that is what makes the highest quality
> existence, and the highest quality civilization. The moral
> consideration, being used to enhance and improve rather than degrade
> and diminish, becomes used as a tool of good instead of a tool of
> wrong. And that results in life bettered rather than worsened through
> its utilization.
> The psychological approach is flawed for one and simple reason: It
> fails to differentiate the processes that it studies from its own
> method in fathoming them. That is, the scientific method that it
> believes it uses gets projected upon the person as vision of mental
> health, and all mental processes - most of which have nothing at all
> to do with scientific methods but nevertheless have their own logic -
> get judged rather than studied. I have found, I believe, a way to
> understand mindsets - and cultures, religions, experiences and
> civilizations - that sees it in its external effects and experiences
> it as it is experienced by the participants. I call it the integrative
> cognition: Combining objective observation with subjective experience,
> and arriving at a complete three-dimensional impressionistic assayal
> of the mindset being viewed.
> Now there is a technique of argumentation I refer to as Gordian knot.
> It is a combination of two half-truths in such a manner as to create a
> complete lie, entangling the soul and making its truth inaccessible.
> One such Gordian knot that I've seen is used by American feminists.
> The first half-truth is that beauty is relative. As I have stated and
> continue to state, with my source of evidence being scientific
> studies, there is both absolute beauty and relative beauty. The
> absolute beauty is manifest in the facial proportions that, when
> shown to people from all cultures by Judith Langlois, were recognized
> as being beautiful. The relative beauty has been found in an
> experiment that showed 500 faces to 20000 people, and each face was
> found the most beautiful by at least one person (though some by many
> more). The truthful way to fathom the subject is therefore this:
> There is absolute beauty; and there is relative beauty. The latter of
> which, in fact, should be found encouraging: It means that even if
> people in your school or whatever hideous locus of socialization
> found you ugly, there is someone for you. And if that - not being
> seen as hopeless and not being attacked - is your sole goal, then I
> am with you all the way.
> But that's not what we are seeing. We are seeing something much worse,
> and that is: Savage attack on both the absolutely beautiful women and
> those catering more to the relative tastes of the place at hand.
> Half-truth # 1: Beauty is relative. Half-truth #2: Beautiful women
> are shallow and arrogant. Gordian knot: Let's attack the beautiful
> women while thinking ourselves righteous and honest and serving the
> good of woman.
> How do I untie this Gordian knot? By seeing it objectively and
> experiencing it within and seeing it completely and in parts,
> completing the central half-truths and taking the half-lies apart.
> Which is as follows: If beauty is solely relative, then it makes no
> sense to attack beautiful women, as all women are equally beautiful or
> ugly; and the fact that one is attacking the beautiful women while
> being good to the ugly women means that one knows, like everyone else
> does, what is beauty - and the central contention of beauty being
> relative is therefore refuted by one's actions. And if beautiful women
> are shallow and arrogant, then that shows some kind of inverse
> correlation between being one's best and being good, with those who
> seek to be excellent being worse than those who don't - which of
> course is an absurd argument; as inner beauty and inner goodness will
> motivate the person to be one's best, and that includes among other
> things being one's most beautiful. What results instead is a kind of
> reverse bigotry, in which a woman who seeks to be her best gets
> treated worse than does a woman who doesn't; and that results in a
> wrong set of incentives, a set of incentives that results in
> degradation of human experience and of humanity as such.
> Which means that it becomes necessary to intervene and destroy such
> reverse-bigotry institutions and say that beautiful women are human
> beings as well, even if the world's harpies don't like them. And who,
> besides being human like them, also seek to be their best; which gives
> them a greater EARNED set of rights. The argument being pushed down
> people's throats - that inner beauty and outer beauty are somehow
> incompatible - is based on envy and hatred and marks those who use it
> as possessing neither beauty outer nor inner. And to use that as a
> supposed conscience or heart of a country, moving men who have liberal
> sympathies to power-hungry harpies while denying them to women who
> seek to be their best, is indeed an outrage.
> Now there is something else that is a part of this as well. Social
> movements create a kind of cartel, in which competition is hobbled in
> order that those who perpetrate the cartel mindset have control over
> people's choices. Thus, unions used to punch men who worked too hard
> for their tastes; and harpies attack and abuse women who are beautiful
> inside and out. The goal is of course to make people think that they
> have only one choice; and anything that gives promise otherwise must
> be demolished. Which is another factor behind the hideousness with
> which ugly women treat beautiful ones.
> So in response to the aforementioned movement, I say: You will get
> precisely what other cartel movements, anti-competitive movements, mob
> movements, in America have gotten.
> We'll see an increase in couplings between long-suffering American men
> and magnificent women from other civilizations.
> Which set of matches - between the people who have the most to offer
> each other - will realize in both American men and foreign women
> leading far better lives than they otherwise could have hoped for.
> In the same way as for example we've seen the same done with American
> business and Chinese labor.
> And which as in the aforementioned case will give people - men and
> women - better lives than they could have dreamed about, while
> creating real-world incentive for people to excel and to treat their
> partners well.
> Be well NAFTA, be well WTO, and be well international relationships.
> Ilya Shambat. 
===
Subject: Re: two functions
> Not necessarily.
> f(f(x)) = f (x(a+b)) = ax(a+b) + b | x(a+b) |
> = ax(a+b) + bx | a+b |
But in this case x:=x(a+b) and x>0 => a+b>0
if that above is correct, then 
f(f(x))=x(a+b)^2
g(g(x))=x(a-b)^2
x(a+b)^2=x
f(f(x))=g(g(x))=x;
|a+b|=|a-b|=1 
reply for:
> Well, you could also have a = 1/2 and b = 1/2, right? >  Then x(a-b)^2 is
> not x.
yes it is not, but we're searching for such a, b that the
equation |a+b|=|a-b| is correct...Only in this case 
f(f(x))=g(g(x))=x
===
Subject: BOYLES law UNDER GRAVITATION
IN an ideal gas with constant TEMPERATURE
PV=RT gives BOYLES law.
Now LET THE gas bee spherical symmetric.
AND LET there bee gravity.
I'M looking for static soloutions.
IN aproppiate UNITS:
P(r)=PRESSURE AT distance r from ORIGO
M(r)=MASS in a BALL with radius r. with center IN ORIGO.
 THEN:
P'=-PM/r^2
M'=P.r^2
P'=-PM/r^2
M'=P.r^2
ONE SOLOUTION:
P=2.r^-2
M=2.r
What is the pressure in ORIGO if the total mass
in all space is m?
I'M want m to bee finite.
THIS might give a formula for calculating the pressure in THE STARS!
===
Subject: Re: New packings of unit squares in squares
   <dk3i1t$upq$1@news7.svr.pol.co.uk>
   <20051030191327.315$X1@newsreader.com message 
<20051025004458.819$BG@newsreader.com>:
> Packing 29 unit squares in a square of side length s = 5.9465+
 <http://img480.imageshack.us/img480/1738/29sqs7ic.gif The link throws a 403 
forbidden error when I try it
> I noted earlier in the thread that I would probably be deleting that 
image
> because it's now available on Erich's web page and in his dynamic survey
> For comparison, the previous best known packing of 29 squares, due
> to Bidwell in 1997, having s = 5.965 approx, can be seen at Erich's
> Packing Center <http://www.stetson.edu/~efriedma/squinsqu/>.
> AFAIK, no figure of the Bidwell packing is still available on the web.
As nearly always in those cases the wayback  machine helps:
Hugo Pfoertner
> David
===
Subject: Re: New packings of unit squares in squares
   <20051102174732.068$up@newsreader.com>
 > I'd been thinking about such an algorithm myself. A few months ago,
I came  up
>with what I thought would work well, but had not yet implemented it.  Based 

on my
>first glance at your paper, I see that my idea was based on  what you call 

maximal
>inflation. In any event, you had the idea and  implemented it before I 
even had the
>idea. Congratulations!
 maximal inflation of two squares (on line and free in the electronic
journal Forum geometricorum) in which we construct with ruler and
compass the percussion point between two inflated squares,
see :   http://forumgeom.fau.edu/FG2005volume5/FG200504index.html
> As we have run it on a slow computer- the computations are heavy- we
> have just succeed to find good packings up to 37 squares.   Then run it on 

a faster computer!  >
ok.
> This leads us to the next topic:
> -corrected the bound s17   It's not just _that_
> upper bound. I haven't done an exhaustive check, but  it seems that 
whenever the
> dynamic survey gives an upper bound in decimal  form, it should actually 
be taken as
> having ellipsis after the last digit.  For example, when it says s(17) 
=< 4.6755, it
> should actually be  understood as s(17) =< 4.6755... (or, in the form 
which Erich
> uses on his  web page, 4.6755+).
If it is the case we are very sory...(on the marvelous website of Erich
Friedman we see 4.675+)
>   It somewhat resembles the packing for n = 29 which I found only very  
recently.
 >The improvement is very messy. As you said, It appears to us
impossible to  find
>such a packing without a computer-aided method!  > and for n=37: 
s37<6.603...,
>but >you have improved this packing again,   I wonder why your program 
didn't find
 >my packing. I guess you just needed  to let the program run longer.
You are right and we will.
>BTW, I'm sending a copy of this message to Erich Friedman. AFAIK, he's  
unaware >of your paper.
box was in trouble.
 Thierry Gensane
===
Subject: Re: New packings of unit squares in squares
   <3ss8krFp5n8jU1@individual.netIs it running on Windows PC?
>Can we download it from somewhere?
We would like to write a convivial version, but you need to waite a
moment...
><http://www.stetson.edu/~efriedma/squinsqu/says s11 = 3.877+ and names 
Walter Trump as author
>of this solution dated 1979.
>What's your improvement and how does it look?
David Cantrell is right, it is the same. But, if Erich F. says
s11<3.8772 that is because
W. Trump did not succed to modelize exactly the packing and the titled
angle has not been optimized enough. Perhaps I am wrong.
>This is s17 = 4.675+ from John Bidwell, 1979 again.
>What's the corrected value? Could please inform
>Erich Friedman too?
See my answer to David.
<David's solution can be seen as s37 = 6.598+ and is
<three years old. When did you make your s37? And where
<can we have a look at it (online, I mean.)?
It is online on the website of Discrete and Computational Geometry,
marvelous packing but definitely dead.
I will send you the picture.
Thierry Gensane
===
Subject: Re: New packings of unit squares in squares
I will let the friends of de.sci.mathematik know
about your findings. I already did so with David's
findings. And there devoloped a little discussion
in which Hugo Pf.9artner gave some interesting remarks
and encouraged others to become busy packers :-)
It would be great to play with a web-version in the
near future. And it would be nice to find someone with
a homepage for hosting your 29-packing. I will ask
Hugo in de.sci.mathematik right now ;-)
Rainer Rosenthal
r.rosenthal@web.de
===
Subject: any point in a trainge
I'm finishing  solving bigger problem. But first I must prove one thing.
There is  a point H (any) inside a trangle ABC. In this trangle we can 
inscribe a circle (r). How to prove that form this point H to the farest 
vertex is at least 2r??