mm-3499
===
Subject: Re: Minkowski Metric
 The metric associated with one particular choice of coordinate system
> is unique to that particular choice of coordinate system.
> Mathematically, you have
 ds^2 = g_ij dq^i dq^j
 Rewriting in matrix form, we have
 ds^2 = [g] * [dq^2] = g_ij dq^i dq^j
 Where
           [g_00, g_01, g_02, g_03]
> **  [g] = [g_10, g_11, g_12, g_13]
>           [g_20, g_21, g_22, g-23]
>           [g_30, g_31, g_32, g_33]
              [dq^0 dq^0, dq^0 dq^1, dq^0 dq^2, dq^0 dq^3]
> **  [dq^2] = [dq^1 dq^0, dq^1 dq^1, dq^1 dq^2, dq^1 dq^3]
>              [dq^2 dq^0, dq^2 dq^1, dq^2 dq^2, dq^2 dq^3]
>              [dq^3 dq^0, dq^3 dq^1, dq^3 dq^2, dq^3 dq^3]
 **  * = Dot product of 2 4-by-4 matrices
 I know what you are trying to do but the above is not the way to do it.
No, you still have no clue.
> One big problem is that if you define [g] and [dq^2] as you did - that
> is, as 4x4 matrices - then their product [g] * [dq^2] is also a 4x4
> matrix.
I did not define ([g] * [dq^2]) as a multiplication of two 4-by-4
matrices.  I defined them as DOT PRODUCT where the definition is also
explained as
[g] * [dq^2] = g_ij dq^i dq^j
> [...]
 Then you'll have this formula:
   ds^2 = [dq-transpose] * [g] * [dq]
[dq-transpose] x [g] x [dq] = [g] * [dq^2]
This is the same as my ones above.
Clearly, if the following is true,
    ds^2 = [g] * [dq^2] = g_ij dq^i dq^j
Then [g] must be only valid to the coordinate [dq^2] and no other
coordinates.  Therefore, [g] cannot be a tensor.
> Once again: on the RHS you have now the product of three matrices: 1x4
> times 4x4 times 4x1, which result in a 1x1 matrix ds^2.
 This is still quite misleading but it's much closer to be correct.
I am already several steps ahead of you.  When are you going to catch
up?
> If [dq^2] is different from choice of coordinate system to another, how
> can you justify [g] is the same from coordinate system to another?It's 
not. I don't know how many times I must repeat the same:
 1. [g] is a shorthand notation for the matrix of g_ij's, where each
> g_ij is a function on the manifold,
The above is correct.  However, the following is wrong still.  <shrug 2. matrix [g] defines a tensor (denoted by g or ds^2) which is a
> *function* on vectors (namely: the dot product function - a
> coordinate-independent thing).
If both ds^2 and [g] are tensors, then you are implying (ds^2 = [g])
which has been an utterly absurd conclusion on your part.
===
Subject: Re: Minkowski Metric
> The metric associated with one particular choice of coordinate system
>> is unique to that particular choice of coordinate system.
>> Mathematically, you have
>> ds^2 = g_ij dq^i dq^j
>> Rewriting in matrix form, we have
>> ds^2 = [g] * [dq^2] = g_ij dq^i dq^j
>> Where
>>           [g_00, g_01, g_02, g_03]
>> **  [g] = [g_10, g_11, g_12, g_13]
>>           [g_20, g_21, g_22, g-23]
>>           [g_30, g_31, g_32, g_33]
>>              [dq^0 dq^0, dq^0 dq^1, dq^0 dq^2, dq^0 dq^3]
>> **  [dq^2] = [dq^1 dq^0, dq^1 dq^1, dq^1 dq^2, dq^1 dq^3]
>>              [dq^2 dq^0, dq^2 dq^1, dq^2 dq^2, dq^2 dq^3]
>>              [dq^3 dq^0, dq^3 dq^1, dq^3 dq^2, dq^3 dq^3]
>> **  * = Dot product of 2 4-by-4 matrices
>> I know what you are trying to do but the above is not the way to do it.
 No, you still have no clue.
> One big problem is that if you define [g] and [dq^2] as you did - that
>> is, as 4x4 matrices - then their product [g] * [dq^2] is also a 4x4
>> matrix.
 I did not define ([g] * [dq^2]) as a multiplication of two 4-by-4
> matrices.  I defined them as DOT PRODUCT where the definition is also
> explained as
 [g] * [dq^2] = g_ij dq^i dq^j
> [...]
>> Then you'll have this formula:
>>   ds^2 = [dq-transpose] * [g] * [dq]
 [dq-transpose] x [g] x [dq] = [g] * [dq^2]
 This is the same as my ones above.
 Clearly, if the following is true,
    ds^2 = [g] * [dq^2] = g_ij dq^i dq^j
 Then [g] must be only valid to the coordinate [dq^2] and no other
> coordinates.  Therefore, [g] cannot be a tensor.
> Once again: on the RHS you have now the product of three matrices: 1x4
>> times 4x4 times 4x1, which result in a 1x1 matrix ds^2.
>> This is still quite misleading but it's much closer to be correct.
 I am already several steps ahead of you.  When are you going to catch
> up?
> If [dq^2] is different from choice of coordinate system to another, 
how
>> can you justify [g] is the same from coordinate system to another?It's 

>> not. I don't know how many times I must repeat the same:
>> 1. [g] is a shorthand notation for the matrix of g_ij's, where each
>> g_ij is a function on the manifold,
 The above is correct.  However, the following is wrong still.  <shrug
>> 2. matrix [g] defines a tensor (denoted by g or ds^2) which is a
>> *function* on vectors (namely: the dot product function - a
>> coordinate-independent thing).
 If both ds^2 and [g] are tensors, then you are implying (ds^2 = [g])
> which has been an utterly absurd conclusion on your part.
In this case the metric tensor g maps the spacetime displacement dX 
to a 
real number R given by
R = ds^2 = g(dX.dX)
That's how the metric tensor is defined.
Pete 
===
Subject: Re: JSH: Re-cap, when is unique also important?
> What is not in doubt is that back in 2002 after a problem solving
>> effort I found my own way to count prime numbers where last week I kept
>> talking about the sieve form of my prime counting function.
>> But that's not what I found years ago.
>> To understand the issues now you need to understand just a bit about
>> counting prime numbers, as in all of human history there have been only
>> two basic ways to count them:
>> 1.  Brute force, like from 1 to 10, you check that no natural numbers
>> below 2 divide it, except 1, so it is prime, as is 3, then you find
>> that 2 divides 4 so it's not, and then no naturals except 1 divide 5 so
>> it is, and so on...
>> 2.  Get smarter and notice that you can count the primes by first
>> finding primes.  That is, you find the primes below the positive square
>> root of the number you're counting up to, and get counts using those
>> primes.
>> Natural numbers are just the counting numbers, starting at 1, next 2,
>> then 3 and so forth.
>> The brute force method doesn't work that well, so people have figured
>> out smart ways with the second approach.  And with it you have sieve
>> methods, where again, you use primes to count more primes.
>> In thousands of years of human history there have been just these two
>> ways known to exactly count prime numbers, until I found my prime
>> counting function, where crucial here is in understanding what hasn't
>> been mentioned.
>> I did not find a sieve function.  Years ago when I posted I did not
>> post a sieve function.
>> What I found is a function that unlike any other previously known,
>> finds the primes it needs to do counting the smarter way, where you
>> just tell it that 2 is prime and it figures out the rest.
>> That is a P(x,y) function.  Where y is like x, just another natural
>> number.
>> In contrast with sieve functions like my prime counting function in its
>> sieve form P(x,n), there n is a count of prime numbers--the helper
>> primes you need to count primes.
>> Guess what?  I didn't first give the sieve form of my prime counting
>> function as it was a sci.math poster named Wim Benthem who did so,
>> years ago, after I put up something unlike anything mathematicians had
>> ever seen before--a function that did what people did, count primes by
>> first finding primes, on its own.
>> For days last week I argued with posters about whether or not there is
>> another known multi-variable prime counting function, as several
>> posters lied until cornered and then switched to saying my prime
>> counting function could be trivially related to a previously known phi
>> function.
>> But they lied again, as that's not the function I actually discovered.
>> And the one I actually discovered cannot be so related to anything else
>> known.  It cannot be directly related to any sieve function.
>> If I am wrong a poster should relate my P(x,y) function to something
>> previously known.
>> Those who followed the discussion may remember I asked posters if there
>> was anything else unique about my prime counting function besides being
>> multi-variable, and you may notice they did not give you the answer I
>> give to you now.
>> So you have the lies about there being another multi-variable prime
>> counting function, collapsing into lies that my prime counting function
>> was just trivially related to another multi-variable function, when in
>> fact what I actually discovered and posted about those years ago, can't
>> be related to anything else known, and it can do what no other known
>> mathematical function can do.
>> So what gives?
>> Why would mathematicians lie if it's so grand, and is there anything
>> else to this prime counting function of mine?
>> Well some of you know that modern mathematics can get a little
>> complicated.  You may also know that mathematicians can get a lot of
>> schooling to study difficult topics for years to gain expertise in
>> their subject areas.
>> It can be a difficult business with extreme complexity, difficult
>> arguments and years of effort just to understand the basics.
>> And I might have cut the Gordian knot in one of the most high profile
>> areas--prime numbers.
>> Imagine you are a Ph.D in mathematics specializing in prime numbers,
>> and you sent a grant proposal for federal funds totalling $500,000 US
>> to fund your research where most of that is your SALARY for five years
>> of research.  And then some nobody, from nowhere comes up with a simple
>> damn function that opens the door to a simple explanation and your
>> research is not needed.
>> Do you just give up on half a million dollars over the next five years?
>> Didn't think of it that way?  Don't understand how mathematical
>> research gets funded?
>> I give a simple answer, and mathematicians lose income--if they
>> acknowledge it.
>> If they don't, and notice, they didn't, those research grants keep
>> coming, the money keeps flowing, and that mathematics Ph.D is still
>> worth something.
>> Alternative explanations?
>> Give them please.
>> You have a unique function that does what no other in human history has
>> ever done in counting prime numbers.  Yet mathematicians have done
>> their best to completely ignore it for over four years.
>> Oh yeah, is there anything else to its uniqueness?
>> In its simplicity may be found the answers to the 'why' of prime
>> numbers.  Answers so simple, they might even make sense to
>> non-mathematicians.
>> A slashing breakdown of complexity, replaced with beautiful simplicity.
>> Yet most mathematicians completely ignore it, while you get Usenet
>> posters who lie about its details, continually getting caught in lies
>> and omissions.
>> Come on, deep down, you knew it had to be about money.
>> So simple answer is, they lie about it because of the money.
>> 
 James, you still do not understand why your work is not acknowledged,
> do you?
 M
>
of course he does, he hasn't got any work, he is pure troll. 
===
Subject: TI 84 Plus Silver calculator: Cheapest place to buy?
I'm a 48 yr old man. Laid off from job and now a full
time college student taking college level trigonometry
this year.
Where is best and cheapest place to buy a new Texas
Instruments 84 Plus Silver graphing calculator?
Local Walmart has it for $130. Can I do better via mail
order or online?
===
Subject: Re: Is continuum completely filled up?
   <4vvevuF1dt4kfU1@mid.individual.net>
   <engcs0$dkk$1@news1.stnet.ad.jp>
   <45a18eed@news2.lightlink.com>
   <virgil-F50F52.20450007012007@comcast.dca.giganews.com>
   <enst71$5cv$1@news1.stnet.ad.jp>
   <45a25f3d@news2.lightlink.com>
   <virgil-5DEE7D.18474108012007@comcast.dca.giganews.com>
   <45a3ab25@news2.lightlink.com
> What TO thinks and what mathematics shows rarely make contact.
>> You would argue if I said 2+2=4.
 Why? It is one of the few of TO's comments that conforms to anything
> mathematical.
> The successor operation, which is enough to imply an infinite set, 
is
>> not actually an order relation, though it it commonly used to 
generate
>> one.
>> So, this comes after that is not a statement of the order of this 
and
>> that? Hmm...sounds like one to me.

> TO does not seem to know what being an 'order relation' involves.
 A strong order relation, <,  on a set S is a subset of SxS which is
>    assymetric (if x<y then not y< x),
>    transitive (if x < y and y < z then z < z) and
>    connected ( if x =|= y then either x < y or y < x).
 Which of those do you think does not apply to the naturals as a result
> of successor()?
Is your thinking really that fuzzy or are you just pretending?
None of those apply to the naturals. This is about a relation
we write <. The naturals are not a relation. Successor() is
not a relation. All of those properties apply to <, which is
a relation.
                     - Randy
===
Subject: Re: Is continuum completely filled up?
> TO does not seem to know what being an 'order relation' involves.
>  > A strong order relation, <,  on a set S is a subset of SxS which 
is
>    assymetric (if x<y then not y< x),
>    transitive (if x < y and y < z then z < z) and
>    connected ( if x =|= y then either x < y or y < x).
 Which of those do you think does not apply to the naturals as a result
> of successor()?
Is your thinking really that fuzzy or are you just pretending?
That's a good question. But, I doubt Tony will answer it.
-- 

===
Subject: Integer or Irrational
Consider the following equation under the given conditions:
y = (ab)^1/2[(AB)^1/k(UV)^1/2k]                   (1)
Conditions:        1. (a, b) = 1, but none is a perfect square
                        2. A, B are rational but none is a k-th power.
                        3. U, V are rational but none is a perfect
square
                        4. prime k > 5
                5. Both a, b > k
                        6. Each of A, B, U, V < 1.
To determine a set of conditions(if exist) under which y will always
remain an
 integer > k. Otherwise, y will remain irrational.
Any help to determine the conditions will be highly appreciated.
===
Subject: Re: JSH: Idea for TV comedy series
   <45a2e0f3$0$313$426a74cc@news.free.fr>
   <50hebsF1fqt9rU1@mid.individual.net
> The Last Danish Pastry a .8ecrit :
>> In the TV series Numb3rs an FBI agent has a brother who is a math
>> professor. This professor applies his mathematical knowledge to
>> solve various crimes.
>> Consider a series (provisionally titled The Hamm3r) in which an
>> FBI agent has a brother who is . James applies his
>> mathematical knowledge to try to solve various crimes with
>> hilarious results.
>
> Yes, and the FBI agent could be the (adoptive) son of Maxwell Smart,
> also known as Agent 86.
 Ah yes. Nice tie-in.
 I always liked the opening sequence to that show: the doors closing.
 And Agent 99 was very lovely...
>
Here's my idea for the series pilot.  The brother's are introduced
as the FBI agent is buying a life insurance policy on someone
and getting his oh-so-smart sibling to help pick out which one
has the best expected return on investment.
Next scene, the FBI agent is telling his bosses that he thinks
the whole terrorist war can be won in like a month, if they can
just get a knowledgeable guy on the inside to figure out how
they train to be able to transmit those powerful terrorist secret
messages through media interviews with lawyers, etc.
A guy in short who can talk that al Gebra talk like he's just
finished a class in it.
Hilariously the FBI's plans go awry when the CIA wiretaps
the brothers' cellphone and hotmail accounts, mistaking
him for real-al Gebra, gives the boy genius an extraordinary
rendition to an undisclosed location where he enjoys loud
music and waterboarding.
I think the second hour of the pilot can be like a reality TV
show, where we goofily bring in a bunch of people off the
street with no idea where they are being detained and
kind of compete to see who can get themselves voted
out of the house first.
At the end fake-al Gebra brings it all home as terrorists
succumb to his endless ranting about the Hammer, lying
newsgroup correspondants, and the journal editors who
let the math results of the millennium slip through their
fingers. They gladly confess to anything the CIA wants
just to be free of his paranoid megalomania.
The denouement has the brothers tearfully reunited. Our
hero has a medal, a laptop, and everything for saving the
country. In the parting shot we see the FBI agent weeping
as he pulls the life insurance policy out of his desk and
tears it to shreds just as we notice it was made out on...
his brother.
--c
===
Subject: Re: JSH: Idea for TV comedy series
> In the TV series Numb3rs an FBI agent has a brother who is a math
> professor. This professor applies his mathematical knowledge to solve
> various crimes.
 Consider a series (provisionally titled The Hamm3r) in which an FBI
> agent has a brother who is . James applies his
> mathematical knowledge to try to solve various crimes with hilarious
> results.
 --
> Clive Tooth
> http://www.shutterstock.com/cat.mhtml?gallery_id=61771
Hi Clive,
I recommend writing a pilot immediately.
All the best,
Jason
===
Subject: Re: Reference
Nntp-Posting-Host: apps.cwi.nl
...
 > sin(pi/7) and cos(pi/7) can be given explicitly, but, as they are
 > solutions of a cubic equation, their expression is pretty complicate
 > and involves intermediate complex numbers even if the result is real.
Fairly complicated, but it can be done.  Let me give it a try:
Set z = exp(2.pi/7), now we have:  z^7 - 1 = 0 and so
    z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 = 0. 
Now set y = z + 1/z (= 2c = 2.cos(2.pi/7)):
    y^3 + y^2 - 2.y - 1 = 0
or
    8.c^3 + 4.c^2 - 4.c - 1 = 0
To solve the cubic use the following method for the cubic
    x^3 + a.x^2 + b.x + c = 0
set the following:
    Q = (12.b - 4.a^2)
    R = (36.a.b - 108.c - 8.a^3)
    K1 = cbrt(R + sqrt(Q^3 + R^2))/2
    K2 = cbrt(R - sqrt(Q^3 + R^2))/2
    W = (-1 + sqrt(-3))/2
now the roots are:
    X_i = (-2.a + W^i.K1 + W^(2i).K2)/6.
(maple is not able to verify this, at least I do not find a way.)
which I expanded to the above generic solution for cubics.)
Filling in (and if I do the calculations correctly), I get
    K1 = cbrt((7 + 21.sqrt(-3))/2)
and with that and a = 1/2, follow the three solutions for the cosine.
There are three because all of cos(2.pi/7), cos(4.pi/7) and cos(6.pi/7)
share the initial equations.  Now we know:
    -1 < cos(6.pi/7) < cos(4.pi/7) < cos(2.pi/7) < 1
and so (with numerical verification) we get:
    c_1 = cos(6.pi/7)
    c_2 = cos(4.pi/7)
    c_3 = cos(2.pi/7).
As cos(pi/7) = - cos(6.pi/7) we need some backtracking.  Using K1 above
and K2 = cbrt((7 - 21.sqrt(-3))/2) we get:
    cos(pi/7) = (1 - W1.K1 - W2.K2)/6
(which conforms remarkably with other sources) and you can fill in the
numbers for W1, W2, K1 and K2.  And so:
    cos(pi/7) = (1 -
        - (-1 + sqrt(-3))/2 * cbrt((7 + 21.sqrt(-3))/2)
        - (-1 - sqrt(-3))/2 * cbrt((7 - 21.sqrt(-3))/2))/6.
which may be simplified to:
    cos(pi/7) = (4 +
        + (1 - sqrt(-3)) * cbrt(28 + 84.sqrt(-3))
        + (1 + sqrt(-3)) * cbrt(28 - 84.sqrt(-3)))/24.
Which is not even a very complicated expression.  Getting sin(pi/7) is
trivial from this.
-- 
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: Uncomputable numbers are all in your head
> > > > > > > > > > 
Define a real number as the output of a program, we can use a Universal 
Turing Machine
> > > > > > > > > > 
and a natural index for every possible program.
> > > > > > > > >  > > > > > > > > > 
real_n <=> UTM(n:N)
> > > > > > > > >  > > > > > > > > > 
e.g. UTM(5) is the 5th number
> > > > > > > > > > 
UTM(100) is the 100th number.
> > > > > > > > >  > > > > > > > > > 
To calculate the digits of the numbers give an input parameter DIGIT
> > > > > > > > > > 
to the number-program, and convert the output into MOD 10.
> > > > > > > > >  > > > > > > > > > 
UTM(n)(digit) mod 10
> > > > > > > > >  > > > > > > > > > 
e.g. say the 1000th number happens to be pi/10, then
> > > > > > > > > > 
UTM(1000) (1) mod 10 = 3
> > > > > > > > > > 
UTM(1000) (2) mod 10 = 1
> > > > > > > > > > 
UTM(1000) (3) mod 10 = 4
> > > > > > > > >  > > > > > > > > > 
The mod 10 is not required but it ensures digits output are from 0 - 9, and 
makes the
> > > > > > > > > > 
set of programs much richer in real numbers.
> > > > > > > > >  > > > > > > > > > 
How does Cantors proof hold up against such a definition of numbers?
> > > > > > > > >  > > > > > > > > >  > > > > > > > > > 
Perfectly well. If you require that reals be computable you should
> > > > > > > > > > 
require that bijections be computable as well. Then Cantor's proof
> > > > > > > > > > 
shows that there is no effective enumeration of the set of computable
> > > > > > > > > > 
numbers.
> > > > > > > > >  > > > > > > > > > How 
so?
> > > > > > > > >  > > > > > > > > > 
Herc
> > > > > > > > >  > > > > > > > > > Let 
f(m,n) be a total recursive function in two variables whose range
> > > > > > > > > is 
contained in the set of decimal digits. If we think of the function
> > > > > > > > > 
n->f(m,n) as a computable sequence of digits, then f may be viewed as
> > > > > > > > > an 
effective enumeration of a set of such sequences. (We may get a
> > > > > > > > > 
repetition of the same sequence, but we don't need to worry about
> > > > > > > > > that). 
There is an effective procedure for constructing a computable
> > > > > > > > > 
sequence which differs from the sequence n->f(m,n) for all m. Just
> > > > > > > > > 
construct the anti-diagonal number. This is guaranteed to be computable
> > > > > > > > > since 
f is total recursive.
> > > > > > > >  > > > > > > > > It may 
be computable, but its not indexable, so it doesn't qualify as an indexed 
real.
> > > > > > > > > Cantors 
proof is invalid if you require that all reals are indexed by a UTM, the
> > > > > > > > > antidiag 
isn't.
> > > > > > > >  > > > > > > > >  > > > > > > > > It is if 
it's the antidiag for an effective enumeration. In that case,
> > > > > > > > > there will 
be some Turing machine that computes it.
> > > > > > > >  > > > > > > > > If you 
take the enumeration whereby n corresponds to the computable
> > > > > > > > > real 
computed by Turing machine number n, that's not an effective
> > > > > > > > > 
enumeration, because it's not decidable whether a given number actually
> > > > > > > > > 
corresponds to a real. In that case the argument doesn't apply.
> > > > > > > >  > > > > > > >  > > > > > > > > If a TM 
computes the antidiagonal, call it TM_a, then there must exist some number 
a,
> > > > > > > > UTM(a) = 
TM_a.  Are you saying TM_a exists because there is no a where UTM(a)
> > > > > > > > computes the 
antidiagonal.
> > > > > > > >  > > > > > > > > Herc
> > > > > > > >  > > > > > > > > The 
antidiagonal for what?
> > > > > > > >  > > > > > > > > The 
antidiagonal for an effective enumeration is computable. That shows
> > > > > > > > that an 
effective enumeration cannot include every computable real.
> > > > > > > >  > > > > > > > > As I said, if 
you let the n-th real in the list be the one computed by
> > > > > > > > the Turing 
machine with number n, that's not an effective enumeration,
> > > > > > > > because not 
every Turing machine computes a real,
> > > > > > >  > > > > > > > Why not?
> > > > > > >  > > > > > > >  > > > > > > > There are various 
issues that can be got around by changing the
> > > > > > > > definition of 
Turing machine. But the one issue that can't be got
> > > > > > > > around is that not 
every Turing machine halts.
> > > > > > >  > > > > > > > It does not need to. 
3.14 is a real number just as
> > > > > > > 
3.140000000000........
> > > > > > >  > > > > > >  > > > > > > > The problem is that 
there's no effective way of deciding whether the
> > > > > > > Turing machine is 
going to go on forever without producing another
> > > > > > > digit or whether it 
will eventually produce another digit. So the point
> > > > > > > stands, it is not an 
effective enumeration.
> > > > > >  > > > > > > OK, let's say it is not 
an effective enumeration. But the Turing
> > > > > > > machine will either halt 
or not. If it does not halt we get 3.14, if
> > > > > > > it does halt we get 
3.140000.... (Some of them will produce the
> > > > > > > actual pi.) In any case 
all the Turing machines will produce nothing
> > > > > > > but real numbers. 
Whether it halts or not, whatever it will produce
> > > > > > > will be a real number.
> > > > > >  > > > > > > The antidiagonal will 
not be computable because at some point it will
> > > > > > > get stuck and stop 
producing digits. That is, there is no Turing
> > > > > > > machine that will 
produce the anti-diagonal. I think this is what Herc
> > > > > > > was trying to 
demonstrate.
> > > > > >  > > > > > >  > > > > > > That's right, there is an 
enumeration, not an effective one, and the
> > > > > > > anti-diagonal for it is 
not computable. If that is Herc's claim then I
> > > > > > > agree with him. My point 
that there is no effective enumeration of the
> > > > > > > computable reals still 
stands. If you believe in non-effective
> > > > > > > enumerations of the 
computable reals then you should also believe in
> > > > > > > non-computable reals.
> > > > > >  > > > > > > Non-effective enumeration 
means that some reals will be represented by
> > > > > > finite strings. A 
non-computable real means that no algorithm for its
> > > > > > computation exists.
> > > > > >  > > > > > > Case in point is the 
scenario above. There is no algorithm that will
> > > > > > compute the anti-diagonal. 
(An attempt to compute it will result in an
> > > > > > infinite loop and thus in a 
finite string that will not be different
> > > > > > from all the strings in the 
list.) By definition no algorithm exists to
> > > > > > compute infinite, 
incompressible strings that carry infinite amount of
> > > > > > information.
> > > > > >  > > > > > > In any case it does not 
matter. You can require that the function be
> > > > > > total. Then you will find 
out that the reals are not r.e. OK. But the
> > > > > > computable reals are still 
countable. You make a list indexed by
> > > > > > integers and the 
anti-diagonal is not copmutable. Again, I believe this
> > > > > > ,or something to that 
effect, was Herc's point.
> > > > > >  > > > > >  > > > > > > Well, my point is that if you 
only believe in computable reals you
> > > > > > should also only believe in 
computable bijections, in which case you
> > > > > > should still accept Cantor's 
theorem.
> > > > >  > > > > > Yes, I do believe only in 
computable reals. That includes all the
> > > > > > rational numbers, plus pi, e, 
the solutions of all algebraic equations
> > > > > > and much more. That is enough 
for me.
> > > > >  > > > > > How does the belief in 
computable bijection imply the belief in
> > > > > > Cantor's theorem? Which part of 
the theorem? That the anti-diagonal is
> > > > > > not on the list or that the 
anti-diagonal exists?
> > > > >  > > > > >  > > > > > Cantor's argument proves that 
there is no effective enumeration of the
> > > > > > computable reals.
> > > > >  > > > > > I accept it. But I accept that the 
anti-diagonal exists only if it is
> > > > > computable. If you list all the 
computable reals then the anti-diagonal
> > > > > is not computable, and effective 
enumeration is not required to
> > > > > establish this.
> > > > >  > > > >  > > > > > Please pay attention. There is no 
effective enumeration of the
> > > > > computable reals. Given an effective 
enumeration of some of the
> > > > > computable reals, the anti-diagonal 
exists, is computable, and is not
> > > > > in the set enumerated. If you list all 
the computable reals, that is
> > > > > not an effective enumeration. (And, as 
you correctly point out, the
> > > > > anti-diagonal is not computable). You 
are welcome to believe only in
> > > > > computable reals if you want. But if 
so then you should also believe
> > > > > only in effective enumerations,
> > > >  > > > > Why so?
> > > >  > > > >  > > > > Well, I can't think of a justification for 
refusing to accept
> > > > > non-computable reals that wouldn't also 
entail not accepting
> > > > > non-effective enumerations. They're pretty 
much the same kind of thing.
> > > > > Maybe you've thought of such a 
justification. Please tell me all about
> > > > > it.
> > > >  > > > > I simply don't see why the machines 
absolutely have to terminate.
> > > >  > > > >  > > > >  so you shouldn't talk about the
> > > > > possibility of listing all the 
computable reals. That's not a
> > > > > possibility in your world.
> > > >  > > > > Maybe not. But I do not need it as I am 
not attempting the diagonal
> > > > > argument.
> > > >  > > > > There is a list of modified Turing 
machines, which output every symbol
> > > > > they write on the tape. You can order 
tem lexicographically and by the
> > > > > input. (I think Herc was attempting a 
similar construct a little
> > > > > differently.) They will compute all the 
reals. Many of them will get
> > > > > stuck in infinite loops but so what? 
There are no more reals than those
> > > > > computed by the machines in the list. 
The anti-diagonal is not
> > > > > computable. There is an universal 
machine somewhere in the list that
> > > > > will try to emulate all the other 
machines in the list and add 1 mod b.
> > > > > It will be on line i. When it comes to 
computing the i-th digit of the
> > > > > diagonal it will call itself and go into 
an infinite regression.
> > > >  > > > >  > > > > If you only believe in computable reals, I 
don't think you should say
> > > > > There are only countably many reals 
unless you can find an effective
> > > > > enumeration of the computable reals. As 
Cantor's argument proves, there
> > > > > is no such enumeration.
> > > >  > > > > I think you have it all backwards. As I have 
shown above the diagonal
> > > > argument does not go through if you restrict 
yourself to computable
> > > > reals. The totality of the function is 
required for the argument to go
> > > > through. It serves no other purpose.
> > > >  > > >  > > > > If you require your reals to be computable, 
you should also require
> > > > your enumerations to be effective.
> > >  > > > I do not see the connection. All the the Turing 
machines are countable
> > > > and so are their inputs N. Effective enumeration 
is the range of one of
> > > > them.
> > >  > > >  > > > And a non-effective enumeration isn't. And I can't 
think of any good
> > > > reason for rejecting non-computable reals which 
wouldn't also be a
> > > > reason for rejecting non-effective enumerations. 
They are pretty much
> > > > the same kind of thing. An enumeration of the 
reals can be coded by a
> > > > real, and the real will be computable if and only 
if the enumeration is
> > > > an effective enumeration of a subset of the 
computable reals. If you
> > > > don't believe in non-computable reals, you 
shouldn't believe in any
> > > > enumerations that aren't effective enumerations of 
a subset of the
> > > > computable reals.
> > >  > > > I think what you are trying to tell ma that you want 
the diagonal
> > > argument to stand. For that you need a total 
function. Therefore I
> > > should also accept the requirement that the function 
be total. But it
> > > is a contrived case considering that for a given 
general recursive
> > > function you cannot even tell if it is total or 
not.
> > >  > >  > > > I don't think you should call it an enumeration of the 
reals if the
> > > function is not total. Because I don't think a Turing 
machine really
> > > computes a real if it won't ever tell you what the 
value of a
> > > particular digit is. If we happen to know that the 
Turing machine
> > > doesn't halt on this particular digit, then we can 
come up with a
> > > different Turing machine which does halt on every 
digit and say this
> > > is the real that the first Turing machine computes. 
But there's no
> > > effective procedure for coming up with such a Turing 
machine. You're
> > > saying, take the list where the n-th real is the real 
computed by the
> > > n-th Turing machine, where in the case where the 
Turing machine doesn't
> > > halt for some particular digit you're appealling to 
the fact that we
> > > can find another Turing machine which goes on 
producing zeroes forever
> > > after the point where the first Turing machine doesn't 
halt. But this
> > > is cheating, you're evading the fact that there's no 
effective
> > > procedure for finding such a Turing machine. What 
you've got doesn't
> > > deserve to be called an effective enumeration of all 
the reals, because
> > > there's no effective procedure for finding the m-th 
digit of the n-th
> > > real.
> >  > > Correct. That is what's wrong with the diagonal 
argument. The matrix
> > > will in fact be full of holes. Some of them will be on 
the diagonal, so
> > > the anti-diagonal is not computable.
> >  > >  > > There's nothing wrong with the diagonal argument as long 
as you're
> > > clear about what it purports to show. There is a 
non-effective
> > > enumeration of the computable reals, yes, this in no way 
contradicts
> > > the diagonal argument, and does not deserve to be 
paraphrased as The
> > > reals are countable. You're assuming the matrix with its 
holes in it
> > > actually exists. There's no effective way of telling 
whether a given
> > > entry in the matrix will have a hole in it or not, so to 
assume that
> > > the matrix somehow exists as a determinate object is 
on a par with
> > > assuming the existence of a non-computable real.
> >  > >  > > > The diagonal argument does show that there can be no 
effective
> > > enumeration of the set of all computable reals. What 
you've got, I
> > > claim, should be called a non-effective enumeration.
> >  > > It is still an enumeration.
> >  > > > And since I can't
> > > see any good reason to reject non-computable reals 
that doesn't also
> > > lead to rejecting non-effective enumerations, I don't 
accept that
> > > you've given me a reasonable stance which justifies 
the statement The
> > > reals are countable.
> >  > > I will give you a couple of reasons:
> > > 1) What happens in infinity is beyond the realm not only 
of observable
> > > but even imaginable phenomena. Hence any statements 
about infinity
> > > (such as one infinity is greater than another) are 
meaningless.
> >  > > Is the statement The computable reals are not 
effectively countable
> > > meaningless?
> >  > > Maybe not, but P(N) > N is.
> >  >  > > Why? It says there's an injection from N into P(N) and no 
bijection
> > from N onto P(N). Seems to me I understand it perfectly well.
>  > Can you picture one infinity greater than another?
>  >  > Can you picture four-dimensional space?
>  > No. Can you? That is I cannot picture a four-dimensional gemetric 
space
> as oppose to algebraic space defines as R x R X R x R.
>   > No, I can't, but that obviously has no bearing on what mathematical
> practice should be. So I was wondering why you thought it was 
relevant
> to ask whether I can picture one infinity greater than another,
> whatever that means.
>  >  I don't know what this question
> > means. It's irrelevant. I understand the meaning of the statement
> > perfectly well.
>  > If you cannot imagine what |R| > |N| means nothing.
>   > Well, that's an ill-formed sentence and it's hard to tell what you
> meant to say. I know perfectly well what |R|>|N| means.
>  >  >  > >  > > 2) Actual infinity (that is finished infinity) is a 
contradiction in
> > > terms
> > > 3) Infinite, incompressible strings carrying infinite 
amount of
> > > information cannot exist for reasons 1) and 2).
> >  > > So why do you believe your enumeration, which you conceded 
was
> > > non-effective, exists?
> >  > > I never said I believed in any enumeration.
> >  >  > > Good. So presumably you won't claim that the reals are 
countable.
>  > I do not believe that that there are more reals than natural 
numbers.
>  >  > So you either reject the claim that the naturals can be injected 
into
> > the reals, or you claim that the naturals can be bijected onto 
the
> > reals. Somehow I doubt the former. So how does this square with 
your
> > saying I never said I believed in any enumeration?
>  > I said effective enumeration of real numbers was not necessary.
>  > Well, that's just as well because it's impossible. So what exactly 
are
> you trying to prove?
>  > I also
> said that all the Turing machines that compute real numbers can be
> effectively enumerated.
>  > Well, that's actually not true.
>  > I also said that I did not see any reason why
> we needed a subset of the machines that compute infinite strings as
> 3.14 is a real number just as 3.14000000......
>   > Because if we allow the Turing machines which don't halt, there will 
be
> no effective procedure for determining the real number they compute 
to
> an arbitrary level of accuracy.
>  >  >  > >  > > 4) in a finite binary tree the ratio of edges to paths 
is (2*2^n -
> > > 2)/2^n. In an infinite tree the ratio is lim{n-->oo} 
(2*2^n - 2)/2^n =
> > > 2.
> >  > > Wrong, in an infinite tree it's not defined.
> >  > > Is it true that lim{n-->oo}(2*2^n - 2)/2^n = 2?
> >  > > Yes.
> >  > > is it true that we indexed all the layers by integers?
> >  > > Don't understand this one. You seem to be missing my point.
>  > Here is the point. lim{n-->oo}(2*2^n - 2)/2^n = 2 applies to 
the
> > INFINITE tree because there are infinitely many integers.
>  >  > Look, this is just drivel. You've given no reason for thinking 
this
> > limit has anything to do with anything.
>  > Hm ... So do you agree that (2*2^n - 2)/2^n is the ratio of edges 
to
> paths in a finite binary tree? What then is your interpretation of
> lim{n-->oo}(2*2^n - 2)/2^n?
>   > It's just an expression which you came up with. It's mathematically
> well-formed but you have given me no reason to think it has anything 
to
> do with anything.
 What then is your interpretation of lim{n-->oo}(2*2^n - 2)/2^n?

> It is the unique numer x, if it exists, such that for all positive real
> numbers epsilon, there exists a positive integer N, such that for all
> n>N, |x-((2*2^n-2)/2^n)|<epsilon. It happens to be equal to 2.
Is it true that An((2*2^n - 2)/2^n) >= 1)? Is it true that
An(#eges/#path >= 1)?
>  >  >  > > Is it true that
> > N is infinite?
> >  >  > > Yes.
> >  > >  > > It contradicts Cantor's result that the cardinality of 
the paths is
> > > greater than the number of the edges.
> > > 5) The diagonal argument shows that for any matrix of 
size n there is a
> > > string of size n that is not in the matrix. Yet for any 
n such a string
> > > can be generated and added to the list. You could as 
well have counters
> > > A and B with values v(B) and v(A) such that vhen v(A) = 
n then v(B) =
> > > n+1. You could keep incrementing these counters till 
infinity and find
> > > that for any n v(B) > v(A). Yet it would be absurd to 
argue that in
> > > infinity B > A. But the Cantorists do precisely that 
arguing that they
> > > can always find one more antidiagonal that is not on the 
list yet.
> >  > > We say that one infinite set is of larger size than 
another when the
> > > latter can be injected into the former but not bijected 
onto the
> > > former. Cantor proves this is the case for the set of real 
numbers and
> > > the set of natural numbers.
> >  > > You missed the point. I am saying that any time you find 
the
> > anti-diagonal I can put back on the list. What leads you to 
believe
> > that once we reach infinity there will be one outstanding 
anti-diagonal
> > left before I have a chance to put it in the list.
> >  >  > > Given any countable set of real numbers, there is a real 
number not in
> > the set. That shows that the set of real numbers is not 
countable.
>  > Any real that is not on the list can be added to the list.
>  >  > So what?
>  > So all the reals will be on the list.
>   > Wrong.
>  >  >  > >  > > When
> > > you observe a dog chasing its tail you do not conclude 
that one
> > > infinity is greater than another.
> >  > >  > > > Secondly procedures that are not total are still 
considered effective.
> > > Case in point is my example of a list of modified 
Turing machines. It
> > > does not matter one iota if any given machine will 
terminate or not
> > > because in either case it will compute a real 
number. And there no more
> > > computable real numbers than there are modified 
Turing machines.
> > >  > >  > > > You are not justified in saying the real numbes are 
countable from your
> > > standpoint unless you can give an effective 
enumeration of the
> > > computable reals, which gives an effective procedure 
for finding the
> > > m-th digit of the n-th real.
> >  > > You added this arbitrary condition to preserve the 
diagonal argument.
> >  > >  > > It's no more arbitrary than the requirement that reals be 
computable.
> >  > >  If the enumeration you've given is good
> > > enough to be called an enumeration, why shouldn't I 
take the
> > > anti-diagonal for it and say that's good enough to be 
called a real
> > > number?
> >  > > Because the diagonal will have holes.
> >  > >  > > When I objected to you that the enumeration will have 
holes, you said,
> > > Fine, get rid of the holes
> >  > > I do not recall saying this.
> >  >  > I could not find the quote there.
>  >  >  > `It does not need to. 3.14 is a real number just as
> > 3.140000000000........'
>  > You're saying the fact that there are holes doesn't matter. You
> > advocate replacing 3.14 followed by an infinite string of holes 
with
> > 3.1400000000......
>  > No, I am not advocating that. Quite the contrary. I am saying that 
3.14
> stands on its own.
>   > Well, it doesn't, because you have know way of knowing what it is. 
You
> know it starts with 3.14, and that's all you know. You don't know
> whether there are any more digits.
 Why do I need to know how many more digits? All i want to know is if it
> is a real, and a real it is.

> But you don't know which real it is. If you're saying you can have a
> list of reals where there's no effective procedure for telling what the
> n-th real is, then why don't you allow a real where there's no
> effective procedure for telling what the n-th digit is?
I do.
  >  >  > > without worrying about the fact that
> > > there's no effective procedure for doing so. You still 
thought this was
> > > a good enough reason to call it an enumeration. Why 
shouldn't I do the
> > > same thing here and say that the anti-diagonal with the 
holes removed
> > > deserves to be called a real, even though it's not 
computable?
> >  > >  > > > Whatever. It doesn't really matter. As long as you 
translate your claim
> > > into my language I'll agree with you. Just say The 
computable reals
> > > are non-effectively countable.
> >  > > OK, they are not effectiveky countable.
> >  > >  > > Good. Also, there is nothing wrong with the diagonal 
argument, and you
> > > cannot justify saying The reals are countable without 
twisting the
> > > meanings of words in totally arbitrary ways.
> >  > >  That's fine. I think that to paraphrase
> > > this as The real numbers are countable is silly, 
that's what I object
> > > to.
> >  > > Countable means the same cardinality as N.
> >  > >  > > And same cardinality means there exists a bijection. 
Why am I
> > > supposed to accept the existence of the bijection but not 
the
> > > anti-diagonal?
> >  > > I thought the anti-diagonal was supposed to prove that there 
was no
> > bijection.
> >  >  > > It does. You are telling me the enumeration exists but not 
the
> > anti-diagonal, therefore the enumeration is a bijection. I see 
no good
> > reason for accepting the existence of the enumeration but not 
the
> > anti-diagonal.
> >  > >  It's not good enough to say the anti-diagonal will have
> > > holes. You can only justify your claim that the bijection 
exists by
> > > getting rid of the holes (in a necessarily non-effective 
way).
> >  > > Obviously there is bijection between Turing machines and N, 
holes or
> > not.
> >  >  > > Well, yes, I think it's obvious, but I also think it's obvious 
that the
> > anti-diagonal exists. I don't think you can have one without 
the other.
>  > If some of the (countable many) Turing machines go into infinite 
loop
> > or infinite recursion then the diagonal will not be computable.
>  >  > That's right, the enumeration will not be effective, and the 
diagonal
> > will not be computable. I don't understand why you believe in the
> > enumeration but not the diagonal.
>  > I believe that the set of machines that compute infinitely long 
real
> numbers is smaller than the set of machines that compute all the 
real
> numbers. The set that computes all the real numbers is definitely
> denumerable, even effectively enumerable.
>   > But this is no good to you because for some of these machines you 
don't
> know which real number they compute.
 You just said which real number. So you know it is a real number
> after all.

> Yes, *I* know it's a real number. But I believe in non-computable reals
> as well. I don't think you should allow yourself to speak of a list of
> real numbers where you don't know what the n-th real number is if you
> won't allow yourself to speak of a real where you don't know what the
> n-th digit is.
I am a bit lost here. Are you saying that if you do not know whether
the machine will compute the next digit then the real is not
computable?
  That doesn't deserve to be called
> an effective enumeration of the computable reals.
 If you do not like to call it an effective enumeration of the
> computable reals I am OK with that.

> Good. Now, if you don't believe in non-computable reals, why do you
> believe in non-effective enumerations? Or, if you don't believe in
> non-effective enumerations, why do you say the reals are countable?
Because I have no reason to believe they are uncountable.
  > Regarding a given
> Turing machine such that we don't know whether it halts for every 
digit
> or not, I don't see how you could justify the claim This Turing
> machine definitely computes some real number
 What do you think it could possibly compute. Could it compute something
> that is not a real number?

> As a matter of fact, if it doesn't halt for every digit I don't think
> it computes anything.
If it computes 3.14 then it looks like a perfectly good number to me.
> You say even in that case it still computes a
> real number, but you can only justify that by implicitly appealing to
> an oracle machine that can see for which digits the Turing machine
> won't halt.
Not really. I am saying that whether it halts or not whatever it will
compute is a real. No oracle is required.
 If you accept such a real number as an entry in your list,
> saying in principle we could get rid of the holes,
I am definitely not saying that. The holes are there to stay.
> then I think you
> should accept the anti-diagonal for the list as well. If you say ah,
> but we can't form the anti-diagonal because of the holes then I say
> we can't even form the list in the first place because of the holes.
Let's say we cannot form a list. But we do not have any reason to say
that the reals are uncountable either.
 > from a constructivist
> standpoint. You're trying to sneak in non-constructivist assumptions
> when it suits you, but you won't allow non-computable reals.
 Are you saying that Turing machines that do not halt are not
> constructivist?
 If non-computable reals are not constructivist, then non-effective
> lists of reals are not constructivist either. (Actually, we can justify
> talking about non-computable reals perfectly well from a constructivist
> standpoint, but you and |-|erc are ignoring that fact so I'm ignoring
> it too for the sake of argument).
 Well maybe, it is a question of terminology. I am not
> necessarily advocating one label or another. All I am saying is that
> you can enumerate all the machines that will compute real numbers
 But you can't. If it deserves to be called an enumeration, then the
> anti-diagonal deserves to be called a real.
Is it your position that if the machine does not compute infinitely
many digits then it is not a real number?
 and
> that I do not care if any of those machines halts or not because
> whatever it computes can be nothing but a real.
  >  >  > > So I
> > > should be allowed to do the same with the anti-diagonal 
and say Here
> > > is a non-computable real number.
> >  > > What justifies saying that it exists?
> >  >  > > What justifies saying that the enumeration exists? If it does, 
the
> > anti-diagonal should exist too.
>  > If the enumeration does not exist then there is no diagonal.
>  >  > True. The nonexistence of the enumeration is the point of 
Cantor's
> > argument.
>  >  > >  > >  > > >  > > > > And in that case the argument works
> > > > fine.
> > >  > > > Are you suggesting that there is some 
contradiction in recognizing only
> > > > countable reals?
> > >  > > > No, countable reals doesn't mean anything. If 
you substitute
> > > > computable reals, I suggested no such thing.
> > >  > > > Hardly. You only need to drop the axiom of 
extent
> > >  > > > There is no axiom of extent.
> > >  > > > I meant axiom of comprehension.
> > >  > > >  > > > or
> > > > the the power set axiom and the diagonal 
argument will not go through.
> > >  > > > Yes, it will, it will show that there is no 
countable set containing
> > > > all the reals.
> > >  > > > No, it will not. If there is no power sat then you 
cannot show that it
> > > is greater than N.
> > >  > > >  > > > And since you believe that ZFC is consistent 
then a weaker system is
> > > > even more consistent.
> > >  > > >  > > > There are subtheories of ZFC in which the diagonal 
argument doesn't go
> > > > through. That is no objection to the diagonal 
argument. Your job is to
> > > > find a consistent theory in which the reals are 
countable (which
> > > > necessarily won't be a subtheory of ZFC), and 
convince me that I should
> > > > accept this theory. I expect you could do the 
former, but I doubt you
> > > > can do the latter.
> > >  > > > I already told how to construct a consistent theory 
where the diagonal
> > > argument will not go through. Drop the axiom of 
comprehension or drop
> > > the power set axiom.
> > >  > > > That of course is not entirely satisfactory theory. 
So I propose adding
> > > the axiom: A class x is set only if x is definable. 
This will result in
> > > proving that
> > >       B = {x e N| x ~e phi(x)}, phi:N --> P(N)
> > > is not a set. After all B is just Russell's set in 
blue.
> > >  > > >  > > > And not knowing which Turing machine will halt 
how can you specify a
> > > > total function?
> > >  > > >  > > > Don't understand this point.
> > >  > > >  > > >  > > > >  > > > >  > > > > >  > > > > >  > > > > > >  > > > > > > > > and it's not 
a
> > > > > > > > computable 
matter to decide which do and which don't.
===
Subject: Re: Uncomputable numbers are all in your head
   <MPG.200b54dcdf772400989a9e@news.rcn.com>
   <eo01mf$nmv$1@mailhub227.itcs.purdue.edu
>> > Given any countable set of real numbers, there is a real number 
not in
>> > the set. That shows that the set of real numbers is not 
countable.
>> > Any real that is not on the list can be added to the list.
>> > So what?
>> So all the reals will be on the list.
>> Are you joking or are you serious? You prove that the reals are
>> uncountable. You take a list. You find a real not on it. So, you add 
the
>> new real to the list to create another list. Of course, the proof 
still
>> works, so there is a real not on the second list.
 But I can add the second anti-diagonal immediately to the second list.
  How many reals do you
>> think you need to add to the list until they are all there?
 Infinitely many.
 Yes, but only a countable infinity.  There are uncountably many numbers to 
be
> added.
Are you saying that you can generate uncountably many anti-diagonals,
but I can put only countably many back on the list?

> --
> Dave Seaman
> U.S. Court of Appeals to review three issues
> concerning case of Mumia Abu-Jamal.
> <http://www.mumia2000.org/>
===
Subject: Re: Uncomputable numbers are all in your head
> Yes, but only a countable infinity.  There are uncountably many numbers 
to be
> added.
Are you saying that you can generate uncountably many anti-diagonals,
> but I can put only countably many back on the list?
Depends what you mean by can generate. If you give us uncountably many 
lists, we can certainly generate an antidiagonal for each one. 
However, the main thing that we are saying is that there are uncountably 
many reals, but a list can only contain a countable number of items (the 
latter is pretty much by definition of what a list is).
-- 

===
Subject: Re: Uncomputable numbers are all in your head
   <MPG.200b54dcdf772400989a9e@news.rcn.com
> > Any real that is not on the list can be added to the list.
>  > So what?
>  > So all the reals will be on the list.
>  > Are you joking or are you serious? You prove that the reals are
> uncountable. You take a list. You find a real not on it. So, you add 
the
> new real to the list to create another list. Of course, the proof 
still
> works, so there is a real not on the second list.
 But I can add the second anti-diagonal immediately to the second list.
 Ye gods. Of course you can re-iterate the diagonal construction, at
> each point constructing a new countable list, and then diagonalizing to
> define a new number not on that list, we which can add to get an
> extended list. But you will never get *all* real numbers that, only a
> countably infinite subset of them.
I am proposing to put ALL the anti-diagonals on the list one by one.
===
Subject: Re: Uncomputable numbers are all in your head
> 
 > > Any real that is not on the list can be added to the list.
>  > So what?
>  > So all the reals will be on the list.
>  > Are you joking or are you serious? You prove that the reals are
> uncountable. You take a list. You find a real not on it. So, you add 
the
> new real to the list to create another list. Of course, the proof 
still
> works, so there is a real not on the second list.
>  > But I can add the second anti-diagonal immediately to the second 
list.
 Ye gods. Of course you can re-iterate the diagonal construction, at
> each point constructing a new countable list, and then diagonalizing to
> define a new number not on that list, we which can add to get an
> extended list. But you will never get *all* real numbers that, only a
> countably infinite subset of them.
I am proposing to put ALL the anti-diagonals on the list one by one.
What makes you think you can do that? 
Aren't you just assuming what you want to show? You want to show the 
reals are countable. This means you want to show that you can put them 
on a list. And, you are proposing to do this one by one, which seems 
to be just a different way of saying that the reals can be put on a 
list. But, you haven't yet shown that the reals can be put on a list. 
And, in fact, we have proved that the reals can't be put on a list. So, 
this seems to boil down to the following:
Theorem. The reals are countable.
Proof: The reals can be put on a list. QED
-- 

===
Subject: JSH    All Your Base Belong To Me.     All Your Ass Belong To Me 
Too.
You guys are a bunch of sissies. None of you can successfully argue
against Harris's mathematical Theory of Triviality and QG.
You are all a bunch of pussies.
Harris's Theorem 
The Existence of a Trivial is Indeterminate.
===
Subject: Re: Kant - why knucmo and i really agree (but argue because he is 
wrong)
> Knucmo,
   I think we both agree that the fundamental-most categories
>   we use to interpret our world are strongly inherent to our
>   being.  I have said that in many of my posts on the topic,
>   and you have argued that is Kant's position.  You have also
>   allowed Kant's position the possibility of corporeal being,
>   and contingency.
   Sir Frederick and other regular posters, as well as numerous
>   other modern philosophers, have covered theories of truth in
>   the brain.  How we pattern recognise, the process of
>   categorisation, what we remember and why.
   If you are a realist, mind is corporeal and ontogenic and
>   phylogenic learning are the only forms of knowledge.
>   Phylogenic learning, or evolution, gives us our mostly-
>   common categories, the basis on which language, logic, and
>   our mathematics are built.  They make the game possible.
   Where we differ, Knucmo, is in the role Kant plays in this
>   description.  You believe that he intends a priori to
>   allow for contingent facts, that the synthetic a priori that
>   he founds his metaphysics on is not necessarily absolute.
   You have stated in other leaves that you do believe there
>   is some absolute truth.  1 + 1 = 2.  That is the type of
>   thing Kant believed, and I will agree with you if you
>   characterise him so.  But he also believed a lot more was
>   absolute.
   I have given many quotations of Kant where he details the
>   extent of the necessity he felt his foundations provided.
>   They were necessary because they involved his fundamental
>   questions, which he (again as I have quoted) are to be
>   made absolute.
   Mathematics?  Absolute.
>   Science and a foundation of absolute, to interpret the contingent.
>   Metaphysics from transcendant truth.
>
Kant speaks for himself, here;
V. IN ALL THEORETICAL SCIENCES OF REASON SYNTHETIC
A PRIORI JUDGMENTS ARE CONTAINED AS PRINCIPLES
1. All mathematical judgments, without exception, are synthetic. This
fact, though incontestably certain and in its consequences very
important, has hitherto escaped the notice of those who are engaged in
the analysis of human reason, and is, indeed, directly opposed to all
their conjectures. For as it was found that all mathematical inferences
proceed in accordance with the principle of contradiction (which the
nature of all apodeictic certainty requires), it was supposed that the
fundamental propositions of the science can themselves be known to be
true through that principle. This is an erroneous view. For though a
synthetic proposition can indeed be discerned in accordance with the
principle of contradiction, this can only be if another synthetic
proposition is presupposed, and if it can then be apprehended as
following from this other proposition; it can never be so discerned in
and by itself.
First of all, it has to be noted that mathematical propositions,
strictly so called, are always judgments a priori, not empirical;
because they carry with them necessity, which cannot be derived from
experience. If this be demurred to, I am willing to limit my statement
to pure mathematics, the very concept of which implies that it does not
contain empirical, but only pure a priori knowledge.
We might, indeed, at first suppose that the proposition 7 & 5 = 12 is a
merely analytic proposition, and follows by the principle of
contradiction from the concept of a sum of 7 and 5. But if we look more
closely we find that the concept of the sum of 7 and 5 contains nothing
save the union of the two numbers into one, and in this no thought is
being taken as to what that single number may be which combines both.
The concept of 12 is by no means already thought in merely thinking
this union of 7 and 5; and I may analyse my concept of such a possible
sum as long as I please, still I shall never find the 12 in it. We have
to go outside these concepts, and call in the aid of the intuition
which corresponds to one of them, our five fingers, for instance, or,
as Segner does in his Arithmetic, five points, adding to the concept of
7, unit by unit, the five given in intuition. For starting with the
number 7, and for the concept of 5 calling in the aid of the fingers of
my hand as intuition, I now add one by one to the number 7 the units
which I previously took together to form the number, and with the aid
of that figure [the hand] see the number 12 come into being. That 5
should be added to 7, I have indeed already thought in the concept of a
sum = 7 & 5, but not that this sum is equivalent to the number
12. Arithmetical propositions are therefore always synthetic. This is
still more evident if we take larger numbers. For it is then obvious
that, however we might turn and twist our concepts, we could never, by
the mere analysis of them, and without the aid of intuition, discover
what [the number is that] is the sum.
Just as little is any fundamental proposition of pure geometry
analytic. That the straight line between two points is the shortest, is
a synthetic proposition. For my concept of straight contains nothing of
quantity, but only of quality. The concept of the shortest is wholly an
addition, and cannot be derived, through any process of analysis, from
the concept of the straight line. Intuition, therefore, must here be
called in; only by its aid is the synthesis possible. What here causes
us commonly to believe that the predicate of such apodeictic judgments
is already contained in our concept, and that the judgment is therefore
analytic, is merely the ambiguous character of the terms used. We are
required to join in thought a certain predicate to a given concept, and
this necessity is inherent in the concepts themselves. But the question
is not what we ought to join in thought to the given concept, but what
we actually think in it, even if only obscurely; and it is then
manifest that, while the predicate is indeed attached necessarily to
the concept, it is so in virtue of an intuition which must be added to
the concept, not as thought in the concept itself.
Some few fundamental propositions, presupposed by the geometrician,
are, indeed, really analytic, and rest on the principle of
contradiction. But, as identical propositions, they serve only as links
in the chain of method and not as principles; for instance, a = a; the
whole is equal to itself; or (a & b) a, that is, the whole is greater
than its part. And even these propositions, though they are valid
according to pure concepts, are only admitted in mathematics because
they can be exhibited in intuition.
2. Natural science (physics) contains a priori synthetic judgments as
principles. I need cite only two such judgments: that in all changes of
the material world the quantity of matter remains unchanged; and that
in all communication of motion, action and reaction must always be
equal. Both propositions, it is evident, are not only necessary, and
therefore in their origin a priori, but also synthetic. For in the
concept of matter I do not think its permanence, but only its presence
in the space which it occupies. I go outside and beyond the concept of
matter, joining to it a priori in thought something which I have not
thought in it. The proposition is not, therefore, analytic, but
synthetic, and yet is thought a priori; and so likewise are the other
propositions of the pure part of natural science.
3. Metaphysics, even if we look upon it as having hitherto failed in
all its endeavours, is yet, owing to the nature of human reason, a
quite indispensable science, and ought to contain a priori synthetic
knowledge. For its business is not merely to analyse concepts which we
make for ourselves a - priori of things, and thereby to clarify them
analytically, but to extend our a priori knowledge. And for this
purpose we must employ principles which add to the given concept
something that was not contained in it, and through a priori synthetic
judgments venture out so far that experience is quite unable to follow
us, as, for instance, in the proposition, that the world must have a
first beginning, and such like. Thus metaphysics consists, at least in
intention, entirely of a priori synthetic propositions.
http://www.arts.cuhk.edu.hk/Philosophy/Kant/cpr/
http://www.bright.net/~jclarke/kant/
http://en.wikipedia.org/wiki/Critique_of_Pure_Reason
>   I have given quotations like:
>    Here is a great and established branch of knowledge, encompassing
>     even now a wonderfully large domain and promissing an unlimited
>     extension in the future, yet carrying with it thoroughly apodictic
>     certainty, that is, absolute necessity, and therefore resting upon
>     no empirical grounds. 
   Kant was monovalent.  He was an apologist for monotheism.  This is
>   why I have such a negative view of Kant, and why I think others
>   may have a more positive view.  This belief in one truth.
   When I quote him waxing on the geometrically absolute qualities
>   of space and implying that the inverse square law is necessary,
>   as I have done, it is because it illustrates the error.
   if (KantsProgrammeSuccessful) then (SomePhenomenologicalStatement)
>   ~(SomePhenomenologicalStatement)
>   ----
>   ~(KantsProgrammeSuccessful)
   You, Knucmo, have claimed this syllogism invalid.  You seem to
>   claim the error lies in the first step, that no implication is
>   possible.  But you do not answer why the quotes I have given
>   claim otherwise, in Kant's own words.
   You see, I dislike Kant's programme because I believe he is
>   strongly monovalent.  I believe he claims this in a number of
>   places, and uses it to found his theory of morality.  He makes
>   numerous claims that the universality of morality is founded
>   in the absolute necessity of the categorical imperative.
   So, as with your 1 + 1 = 2, he believes there is an absolutely
>   unquestionable core to rational thought that determines not
>   only what we perceive but what we ought decide to do.  It is
>   a strange interpretation of Kant in my eyes to believe he
>   accepts an evolutionary a priori that is contingent but still
>   reveals absolute imperatives to the intuition, and one that
>   Kant himself denies.
   Kant is not to be credited for simply saying that we order our
>   perceptions and give structure to the senses.  That is old and
>   common to many traditions, to very old ones (Buddhism and
>   Stoicism) which I have mentioned and you have yet to respond to,
>   Knucmo.  In fact, in a response to Fernando Revilla in the
>   kant thread, I mentioned in passing Ratnakirti, who even
>   specifically pointed to antinomies in designating boundaries
>   for concepts, hundreds of years prior to Kant.  Chryssipus
>   was
   But the nominalist schools of Buddhism and the Stoic schools of
>   classic philosophy both have extensive theories of human error.
>   Kant was often quite explicit that his intensions were to explain
>   the basis of human certainty.  When pressed, many with Kantian
>   tendencies will claim an absolute, like your 1 + 1 = 2.  My
>   objection is when the absolute is denied and Kant's own
>   statements ignored.
   You could answer my quotes, which I entered in by hand from
>   source material, Knucmo.
   Or you could read some of the available information out there,
>   like
   http://post.queensu.ca/~miller/Papers/Innate%20ideas.pdf
   and other professional journal content.
   But instead of confronting the issues raised and avoiding
>   acknowledging the quotes, you have decided to post some juvenile
>   remarks and belittle me by your Hes.
   I do not know if I can get any resolution.  Perhaps my mind is
>   too much made up.  I had thought someone could explain the
>   quotes, which seem explicit to me.  Particularly:
    If we consider the properties of the circle, by which this figure
>     combines in itself so many arbitrary determinations of space
>     in a universal rule, we cannot avoid attributing a constitution
>     to this geometrical thing.  Two straight lines, for example, which
>     intersect each other and the circle, howsoever they may be
>     drawn, are always divided so that the rectangle constructed
>     with the segments of the one is equal to that constructed with
>     the segments of the other.  The question now is: Does this
>     law lie in the circle or the understanding?  That is, does this
>     figure, independently of the understanding, contain in itself the
>     ground of the law; or does the understanding, having constructed
>     according to its concepts (of the equality of the radii) the figure
>     itself, introduce into it this law of the chords intersecting in
>     geometrical proportion?  When we follow the proofs of this law,
>     we soon perceive that it can only be derived from the condition
>     on which the understanding founds the construction of this
>     figure, namely, the concept of the equality of radii.  But if we
>     enlarge this concept to pursue further the unity of various
>     properties of geometrical figures under the common laws and
>     consider the circle as a conic section, which of course is
>     subject to the same fundamental conditions of construction as
>     other conic sections, we shall find that all the chords which
>     intersect within the ellipse, parabola, and hyperbola always
>     intersect so that the rectangles of segments are not indeed
>     equal but always bear a constant ratio to one another.  If we
>     proceed still farther to the fundamental teachings of astronomy,
>     we find a physical law of reciprocal attraction applicable to
>     all material nature, the rule of whichis that it decreases
>     inversely as the square of its distance from each attracting
>     point, that is, as the spherical surfaces increases over which
>     this force spreads, which law seems to be necessarily inherent
>     in the very nature of things, and hence is usually propounded as
>     knowable a priori.  Simple as the sources of this law are,
>     merely resting on the relation of spherical surfaces of different
>     radii, its consequences are so valuable with regard to the
>     variety and simplicity of their agreement that not only are all
>     possible orbits of the celestial bodies conic sections, but
>     such a relation of these orbits to one another results that no
>     other law of attraction than that of inverse square of the distance
>     can be imagined as fit for a cosmical system.
     Here accordingly is nature, which rests upon laws the understanding
>     knows a priori, and chiefly from the universal principles of the
>     determination of space.
 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
> galathaea: prankster, fablist, magician, liar
===
Subject: Re: Cantor Confusion
Nntp-Posting-Host: apps.cwi.nl
 > No, that does AC (although the statement is indirectly).  V=L is the 
axiom
 > of constructibility and states that every set is constructible.  Most
 > mathematicians think it is false, but it can not be disproven and is
 > consistent with ZF.  It *gives* a construction.
 > 
 > But, alas,  this construction cannot be communicated. It is and remains
 > a top secret. Otherwise most mathematicians could easily be proved
 > wrong by simply constructing a well-ordering of R.
It can be communicated.  But you are not even willing to look at the
axiom and its consequences.
 >  > Eh?  As I understand it, there is only one union.  But, what is 
the
 >  > relevance?
 >  >
 >  > There is a union of paths for every path.
 >
 > How do you define a union of paths?
 > 
 > For example: the infinite union of {0.01} and {0.010} and {0.0101} ...
 > yields the path corresponding to 1/3.
I asked for a definition, not an example.  How do you define the union
of paths?  What (for instance) is the union of {0.1111} and {0.1011}?
 >  > Every node which is placed on a finite level (= every digit with a
 >  > finite index) is in the union of all finite trees. Therefore every 
path
 >  > containing nodes on a finite level (= every sequence of digits at
 >  > places with finite indexes) is in the union of finite trees. There 
is
 >  > nothing remaining!
 >
 > But that does *not* mean that the set of paths is the union of the sets 
of
 > paths in the finite trees.
 > 
 > If the tree is complete with respect to edges, then it cannot be
 > completed any further.
Makes no sense.  It does not even refute my statement.
 > You are again, *not* talking about the union of sets of paths.  So what 
is
 > the relevance?
 > 
 > If the tree is complete with respect to edges, then it cannot be
 > completed any further.
And again, *not* talking about the union of sets of paths.  So what is the
relevance?
 >  > Nevertheless, the representations in the union of all finite trees 
are
 >  > countable as the countable union of finite sets.
 >
 > A new term again.  What are the representations?
 > 
 > The paths are representations of real numbers.
Oh.  Perhaps.  Pray define.
-- 
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: Cantor Confusion
Nntp-Posting-Host: apps.cwi.nl
 >
 > What is a ghost path?
 > 
 > A ghost path is a path which does not exist in a tree which contains
 > the paths of all real numbers the bits of which can be indexed by
 > natural numbers.
That is something I do not understand.  Please elaborate.
 >  > The union of all finite trees is also the union of all last levels.
 >  > That is the union of all levels. If some path is missing, then it 
must
 >  > stretch over more than all levels indexed by natural numbers.
 >
 > Ah, again about that diagonal.  No, you are wrong.
 > 
 > All path the bits of which can be indexed by natural numbers are in the
 > union of all finite trees.
But *not* in the union of the sets of paths in the finite tree.
 > You know it. You cannot point to a bit of any number of [0, 1] which is
 > missing. But you claim there were further numbers.
 > 
 > Try to complete the union of all finite trees to the complete infinite
 > tree. Which level, node or edge would you add?
Pray, properly define the union of trees.  In my opinion a tree consists
of three sets.  First the set E of edges, next the set N of nodes and
finally the set P of paths.  What is the union of two trees?
 > Why was it not a proof?
 > 
 > Because you used crippled plants which are not under discussion in a
 > forest of well grown trees.
My proof was about the set of paths in the union of finite trees not being
the union of the sets of paths in the finite trees.  What was wrong about
that proof?  Please, once come up with a proper definition of the union
of two trees.  Before you come up with such a definition it is impossible
to even entertain a discussion.
 > I am talking about the union of *sets* of paths, not the union of 
paths.
 > Do you not see the difference?  How do you define the union of paths?
 > And how do you define the union of the *sets* of paths that are in the
 > finite trees?
 > 
 > Which edge of any infinite path is missing? Which bit could be added to
 > one of the numbers represented?
What is the relevance?  If you cna state that there is some infinite path
in the union of the *sets* of paths in finite trees, you should also be
able to point to a finite tree that contains that infinite path.  There is
no edge missing, it is only that your infinite path is not a path in any
of the finite trees, so it is not in the union of the sets of paths in
finite trees.
 > Ok, so you do *not* use the union of the sets of paths but something 
else.
 > 
 > The material from which the paths are constructed.
Makes no sense.
 > Indeed, you actually do not use those sets at all in forming your 
union.
 > Nevertheless you want to draw conclusions about the cardinality of the
 > *set* of paths in that union from the cardinalities of the *sets* of
 > paths in the finite trees.
 > 
 > The completeness of U{n e N} {1,2,3,...n} = N is the same as the
 > completeness of the union of all levels of the infinite tree, and that
 > is the same as the union of all finite trees. And all these unions are
 > counatble.
As you still fail to provide proper definitions of unions of trees, I 
wonder.
 >  How do you *define* the union of two trees?
 > I saw a tree as consisting of three sets: E, N and P for edges, nodes
 > and paths.
 > 
 > You can also unify the levels.
Eh.  Pray expand.
 >  And I defined the union of two trees T1 and T2 as:
 >    [ {E1, E2}, {N1, N2}, {P1, P2} ]
 > (where those unions are sets, so duplicated elements can be elided).
 > 
 > It is sufficient to build the union of all finite trees, or of all
 > levels, or of all edges, or of all nodes. This union is an infinite
 > tree. An infinite tree cannot be surpassed (by another tree with
 > naturally indexed levels). Therefore the union is *the* infinite tree
 > and, hence, it contains all possible paths.
Again, no definition, only assertions.
-- 
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: Cantor Confusion
Nntp-Posting-Host: apps.cwi.nl
...
 >  > Ok? Every infinite tree, which contains all levels enumerated with
 >  > natural numbers, contains all possible paths and, therefore, 
contains
 >  > the representations of all real numbers of [0, 1].
 >
 > In that case the set of paths in the infinite tree is not the union of 
the
 > sets of paths in the finite trees.
 >
 > The union of all finite trees (= union of all levels) of the tree
 > automatically contains all paths, because there is no path or part of a
 > path outside of this union.
And again, in that case the set of paths in the union is not the union of
the sets of paths.  I did show that with a finite graph.  So I wonder
why you think it holds for an infinite graph (which a tree is)?
 >  > The union of all finite initial segments {1,2,3,...n} is the 
infinite
 >  > initial segment N. So the infinite tree is the union of all finite
 >  > trees.
 >
 > Yes, so what?  I repeat that I am talking about the union of the sets 
of
 > paths.
 > 
 > The union of all finite trees (= union of all levels) of the tree
 > automatically contains all paths, because there is no path or part of a
 > path outside of this union.
Yes, but that does *not* make the *set* of paths the union of the *sets*
of paths in the finite trees.
 > No, because I do not state that.  I only state (I repeat):
 >     there is no infinite number in the union of initial segments
 > and that means that N does not contain an infinite number.  It 
contains
 > only finite numbers (but infinitely many).  And the union of all 
finite
 > initial segments *is* N.
 > 
 > Correct. As the union of all finite paths turning always right is
 > 0.111... .
Pray, define the union of paths.  But I am talking about *sets* of paths.
 > U{n is natural} {n} = N.  Because every k in N is in one of the sets 
used
 > in the union, namely {k}.  {0.111...} is *not* in
 > U[ {0.1}, {0.11}, {0.111}, ...} ] because it is in *none* of the sets
 > used in the union.
 > 
 > But {0.111...} is  {{0.1}, {0.11}, {0.111}, ...} (see below).
Oh.
 > The union of {0.1}, {0.11}, {0.111}, ..., is *not* {0.111...}, it is:
 > {0.1, 0.11, 0.111, 0.1111, ...}.  Remember: unions are defined for 
*sets*.
 > I do not know any definition of union for numbers as these.
 >
 > And what is the set {0.1, 0.11, 0.111, 0.1111, ...}, or, translated to
 > digits, the set {1,2,3,...}? Is it not {0.111...} or, translated to
 > digits, N (or omega)?
Perhaps.   But what is the relevance?  How about the set {0.1, 0.10, 0.101,
0.1010, ...}?  How do you define the *union* of paths?  In a level three
tree I can encounter the paths 0.111 and 9.101.  What is their union?
 > So we go a bit abstract already.
 > 
 > So consider the circles below. Their meaning should be unique.
For you, apparently.
 >  >  > They will understand, at least by experiment,
 >  >  >
 >  >  > oo
 >  >  > ooo
 >  >  > _____
 >  >  > ooooo
 >  >
 >  > You apparently have not done the experiment.  No, they will not
 >  > understand, and such experiments have been done.
 >  >
 >  > Depends on their intelligence, which I don't know.
 >
 > It has nothing to do with intelligence.  In their culture they do not 
use
 > *any* abstraction.  They do everything with concrete things only.
 > 
 > In above circles there is no abstraction.
You think so.  I see (in my non-abstracted view) a line with circles, 
beneath
it another line of circles, a dash and a third line of circles.  I have no
idea what the meaning of that is.  A picture, apparently.  Looks nice.  But
what is the meaning?
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: (i) = 1000...0000, (e)= -1, pi = 3 solves the old beautiful 
equation e^(pi*i)= -1
   <JBLvKs.GsC@cwi.nl ...
>  > So if ....88888 has a square root then they could be ....2xyz or
>  > ....9xyz but apparently as you point out above they can be neither.
>    > Natural Numbers may or may not have square roots but they are between
>  > perfect squares.
>    > The number .....999999 does have a square root and it either ends in 7
>  > or 3
>    > Apparently it is the number ....31xyz....7 or ....31xyz....3
 It is neither.
>   ...03 * ...03 = ...09    ...07 * ...07 = ...49
>   ...13 * ...13 = ...69    ...17 * ...17 = ...89
>   ...23 * ...23 = ...29    ...27 * ...27 = ...29
>   ...33 * ...33 = ...89    ...37 * ...37 = ...69
>   ...43 * ...43 = ...49    ...47 * ...47 = ...09
>   ...53 * ...53 = ...09    ...57 * ...57 = ...49
>   ...63 * ...63 = ...69    ...67 * ...67 = ...89
>   ...73 * ...73 = ...29    ...77 * ...77 = ...29
>   ...83 * ...83 = ...89    ...87 * ...87 = ...69
>   ...93 * ...93 = ...49    ...97 * ...97 = ...09
> none of them ends with 99.
At first I thought a carryover would save this number 31...3 but there
is no more carryover.
So I have to resort to (i) as either 1000....0000 or -1000....00000
I typed this post earlier today but Google was having troubles so it
may appear later.
Now the biggest number in Natural Numbers is ....999999 and it
resembles the number -1. It is not -1 but can substitute for -1.
Now the number 1000....00000 has an automatic square root identical to
itself as 100....0000 but can this number be -1????
It can if we consider and hold true that 999....9999 is the largest
number to exist. And thus if we demand 9999....999 the largest number
when we subtract 10000....00000 from 9999....99999 we have -1
So for our equation that we wish to solve of e^(i*pi) = -1
we can rewrite as:
e^ (100...000 x pi) =  -1
Now I am still searching to find what the variable of e ranges over.
The variable of pi, (pi not in Euclidean geometry but pi in Riemann
geometry varies over 2 to 3.14159.....) that I will use pi as
......000003 or simply 3
So the above becomes this equation:
e^ (100...000 x 3) =  -1
So, now all I have to take care of is the number (e). A geometrical
interpretation of (e) is that it is a number for which a geometrical
object is growing in a Eucl. or Riem or Loba space for which the object
remains in a state of constant proportionality. Analogy: is the
seashell as it grows looks the same when small as when it grows to full
maturity. Another example is the onion which remains in a constant
proportionality of its parts as it grows from small to large.
So, what is the range of values that (e) has in Euclidean geometry and
the answer is that it has only the value of 2.71828..... But the value
of (e) in Riemannian geometry I think ranges from +1 to that of
+2.71828..... And most important of all, what is the range of value of
(e) in Lobachevskian geometry and my intuition tells me it is -1 to
-2.71828....... So I need to verify these claims. But am pretty
confident they are true. And I would be astounded if I were the first
mathematician in history to realize that e is a variable that has the
range of the above listed. Then the mathematics community was asleep at
the helm.
So, now, I have all the values to solve for the ages old beautiful
equation of e^(pi*i)=-1
When I replace the number (e) with -1 I get this:
(-1)^(3* 1000....0000) = -1
And that does it, that solves the beautiful ages old equation of
e^(pi*i) = -1
It solves it because the exponent of pi x i is a odd number of
30000....00000 which since 1000....0000 is -1 that 3000....0000 is -3
and so -1^ -3 is still -1.
Now I am a bit uneasy about that and not extremely satisfied since I
have changed the values repeatedly over the past few days. But it is
nonetheless the world's first valued filled equation which is probably
the world's most beautiful and mysterious of equations. And even though
there were many changes to reach this point, it must be seriously
examined as the first time anyone has satisfied that equation with
numeric values.
Now what is the geometrical interpretation of the above. That is easy
for me with the Atom Totality theory. It says the Universe itself is
one big atom and that humanity is viewing the last six electrons of the
Atom Totality where galaxies are tiny pieces of the last 6 electrons.
So we are inside a huge Lob of the 5f6 of the Atom Totality. We are in
a Lobachevskian geometry as the inside of an electron-lob. And the
value for e would be -1 and the value for pi would be 3 and the value
for i would be 1000....00000.
P.S. If 1000....0000 does not work out, well, I have only one last
straw up my sleeve--- -1000.....000000, at least for the time being.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
===
Here is a short course, enjoyment and fun:
Audio speech by Benjamin Freedman in his own voice you can hear here :
Free Science History Ebook: THE MANUFACTURE AND SALE of Saint Einstein
The following passage by Benjamin Disraeli (who has very interesting
quotes in the searcheable free pdf book) at
http://jewishracism.com/SaintEinstein.htm
After Hechler came David Ben-Gurion, who stated,
The First World War brought us the Balfour Declaration. The Second
ought to bring us the Jewish State.1772 .......... read the book
===
Subject: Why some mathematical ancestry in U.S. trace back to Germany?
Recently I have been searching for mathematical ancestries of all
mathematicians that I person know on the Mathematics Genealogy website
(which tells you who's whose doctoral adviser). These mathematicians
that I searched for are mostly in U.S., but it seems trace back several
generation (in student-adviser sense) it always end up in Germany.
===
Subject: Re: Can ZFC prove Addition is Associative?
>> 
>On Tue, 09 Jan 2007 10:41:23 +0100, Han de Bruijn
>> 
> schreef:
>Uh, that's correct. In an informal sense the probability of a random
>integer being even is 1/2, but if we're talking carefully about
>probability the way it is formulated mathematically then there
>is no such probability (because there is no uniform probability
>measure on the integers).
>>Uh, and THAT is only because you are able to talk about THE 
integers.
>>With any finite set of integers N = {1,2,3, .. ,n} there is no problem.
>> 
>The question I was talking about was whether it's correct to say
>that the probability that a random natural number is even is 1/2.
>You appeared to be saying something about that question as well,
>in particular you seemed to be suggesting that it was _correct_
>to say that, since the density of the evens is 1/2.
>> 
>You're talking about the set of all natural numbers as well.
>With any finite set {1,...n} we're talking about a different question.
>> 
>> This is an example of the absurdity of Han's position.  He
>> does not believe in the set {1, 2, 3, .... },  yet he objects
>> to the fact that according to the standard definitions
>> there does not exist a uniform probability measure on {1, 2, 3, .... }.
> The explanation is in the fact that my set {1, 2, 3, .... } is quite
> different from yours. It isn't there as a whole (: actual infinity).
> But we have gone through all this, haven't we?
Yes, your set {1, 2, 3, ... } is quite different.  So why do
you insist that what is true of your set {1, 2, 3, ... } should
be true of the standard set {1, 2, 3, ... }?  They are different
sets.   You know that, yet you try to pretend that they are
the same set when it suits your rants.   Your contradiction
is simply a result of switching which set your are talking
about in the middle of your argument.
Stephen
===
Subject: Re: Size of equivalence class of Cauchy sequences
   <45A5E7DD.4020507@netscape.net
>Given a rational number, r, how many Cauchy sequences
>converge to r?

 c,  the cardinality of the reals.
So, let me see if I understand correctly: If we use the equivalence
classes of Cauchy sequences method to define real numbers, then each
real number is itself a set with cardinality that of the set of real
numbers. If we use the Dedekind cut method to define real numbers, then
each real number is a denumerable set. (?)
Moeblee
===
Subject: Re: Could you help me with complex numbers?
David Ullrich gave great answers to your questions, so I won't repeat them 
other than to second his suggestion that you roundly ignore whoever told 
you that 15-year-old girls have no business being interested in complex 
numbers (or anything else that piques your intellectual curiosity).
On second thought I have given a partial answer to your question about 
what complex numbers measure, see below.
> math and try to study a bit more thn required by my school. The complex
> numbers have always made me curious, so I tried to understand them to
> the extent it's possible for someone my age.
Formal education will try to distill this natural and healthy impulse to 
learn out of you. Don't let it!
> At first, I was introduced to the misteryous i = sqrt(-1), something
> really kabbalistic to me. Then, I got to know what mathematicians did
> was extend those algebraic laws of the real field (I've read something
> about groups, rings and fields) to the vector space R^2. Not sure if
> this is correct, but that's what I think. In R^2 we can add and
> multiply by a scalar, but we can't multiply, for example, (2,3) and
> (3,7). But then it was defined that, in the complex plane, (Argand
> Gauss plane, right?) (a,b) * (c,d) = (ac - bd , ad + bc). So, it seems
> to me that, just like R^2, the complex plane C is formed by  orderd
> pairs of elements extracted from the real line. And the only difference
> between them is that R^2 doesn't have all of those algebraic laws and C
> does. In other words, C is a field and R^2 is not. It can't be that
> simple, of course there's something  I don't gather, I don't
> understand. But thats my perception. If I'm right, and probably I'm
> wrong, if we see R^2 and C just like sets and dont take into account
> those field operations, then apparently they are exactly the same set.
As David suggested elsewhere in this thread, the distinction is entirely a 
matter of point of view. You are entirely correct. Most mathematicians use 
R^2 to denote a two dimensional real vector space and C to denote the 
field, but one can set everything up so that they have the same 
underlying set.
You may be interested to know that there are theorems about R and C that 
guarantee that even if you have a different construction of C (or R) in 
mind than I do, we are still talking about fundamentally the same objects: 
all of the a priori different things that get called C are isomorphic to 
each other.
> So, whast's the difference between, for example (2,3) and 2 + 3i?   2 +
> 3i suggests a vector notation, as though we could see the complex
> numbers as vectors on the plane, like those we study in Physics, like
> forces, velocities, etc. Are the complex numbers actual numbers or are
> they vectors?  That is, does it make sense to measure something in
> complex numbers. Sorry for my stupid question, but does it make any
> sense to say you bought something for (30  + 50 i)  US$ ? If instead of
> money it was say, distance, I could understand, since the real axis may
> be seen as the horizontal axis and the imaginary axis stands for the
> vertical axis.
There is one thing that comes to mind, for what it's worth: in the theory 
of AC electrical circuits, the quantity of _impedance_ is represented as 
a complex number.
http://en.wikipedia.org/wiki/Electrical_impedance
This may not be very enlightening but at least it is a partial answer to 
the question can complex numbers measure things? Perhaps not in the 
sense of rulers or voltmeters, but it is very useful to be able to imagine 
them doing so in a way consistent with the theory.
> It seems to me that the complex numbers are somwhat artificial,
> because from what I could read they were say constructed. I mean, fisrt
> you have the natural numbers {1,2,3....} and you always could add but
> subtraction was impossible if n < m. Then the negative integers were
> created and solved this problem, the set Z of all the integers became
> available. But then division was not always possible and the set of the
> rationals Q was created. But then you couldn't compute things like
> sqrt(2) and so came the irrationalls. That is, so far you solved the
> problems, but with the complexes, didn't  matematician just construct
> something rather artificial? They got out of the real line.
They did get out of the real line, but in a natural extension of the 
process you describe. The desire for solutions to
k = m - n (or, if you like, m + n = 0)
for arbitrary positive integer m and n leads to the negative integers in a 
way similar to the way that the desire for solutions to
z^2 + x = 0
for arbitrary real numbers x leads to the complex numbers.
It is a bit surprising when first meeting these objects when one realizes 
that the expansion of the system of the real numbers that allows for 
solutions to x^2 + 1 = 0 actually allows you to solve ALL polynomial 
equations
a_n x^n + ... + a_1 x + a_0 = 0,
even when the a's are complex! See Fundamental Theorem of Algebra.
You may be interested in further progress in this direction. Google for 
quaternions, octonions, normed division algebra, and, if you're 

feeling particularly enthusiastic, Clifford algebra. Given what you say 

in your post, I should not think an understanding of the basics of any of 
these ideas to be beyond you (with the possible exception of the Clifford 
algebras, at least without some more reading). At the worst it will be 
confusing; at best it will introduce you to some of the most beautiful and 
fascinating mathematics around.
http://en.wikipedia.org/wiki/Quaternion
http://en.wikipedia.org/wiki/Octonion
http://en.wikipedia.org/wiki/Normed_divison_algebra
http://en.wikipedia.org/wiki/Clifford_algebra
> Well, thank you all and sorry, I know a 15 yo girl is not supposed to
> ask such silly questions to the top dogs of Math, but I'm really
> curious.
Ha. Ask whatever you like, we're happy to help. We like math so much that 
many of us do it for a living and all of us kill time on sci.math. At 
least in part we are here because we (sadly) talk to relatively few people 
in real life who express such interest in the things that also interest 
us.
> (Dont get me wrong, though I like Math I'm a normal 15 yo girl)
I venture to suggest the two ought not to be thought mutually exclusive.
Best,
-dave
***********************************************************
        Papa, can you multiply triples?
        No, I can only add and subtract them.
-an (apocryphal) conversation between W.R. Hamilton and
his young son
***********************************************************
===
Subject: Re: Irrational numbers questions
   <JBIy99.DJH@cwi.nl>
   <JBJ2q7.1tF@cwi.nl>
   <636a1$45a4e0bf$82a1e228$26741@news1.tudelft.nl>
   <878xgbj99y.fsf@phiwumbda.org>
   <87irfeimw0.fsf@phiwumbda.org>
   <fajcq2h5h75j1eor1t1i1pf2n0vs1dlpsr@4ax.com>
   <64114$45a65bf9$82a1e228$1014@news1.tudelft.nl>
   <87d55lh7ta.fsf@phiwumbda.org>
Jesse F. Hughes schreef:

>> On Wed, 10 Jan 2007 16:37:35 -0500, Jesse F. Hughes
> meant irrational?
>> No, if we assume his post was in good faith he didn't mock me for
>> the typo, he was too stupid to realize it _was_ a typo.
 False accusation.
 You mean he *shouldn't* have assumed that you were being honest when
> you mistook a typo for a more egregious error?
Uhm, right, silly word games to win a silly debate. 
Han de Bruijn
===
Subject: Re: Irrational numbers questions
Not a proof that gamma is irrational
The Euler-Mascheroni constant called 'gamma' is defined by:
     lim (n -> oo) [ sum(k = 1..n) 1/k - ln(n) ]
It's value is, approximately, given by:
         0.577215664901532860606512090082402431042
It is hitherto unknown whether 'gamma' is an irrational number or not.
The following is NOT a proof of a conjecture that the Euler-Mascheroni
constant, called 'gamma', would be an _irrational_ number.
A method is implemented by the following algorithm: start walking down
through the Stern-Brocot tree in the first place. Which is implemented
in the accompanying program as the 'while true do' loop. See:
   http://en.wikipedia.org/wiki/Stern-Brocot_tree
At each stage in the tree, consider the approximation which is made
possible by R.M. Young's theorem, as has been explained in:
Namely: L(n) < gamma < R(n) .
Where L(n) is monotonically increasing towards gamma,
while R(n) is monotonically decreasing towards gamma.
The width of the interval which delimits the approximation for gamma
is:
         G(n)-1/(2(n+1)) - [ G(n)-1/(2.n) ] = 1/(2.n(n+1))
Where:  G(n) = [ sum(k = 1..n) 1/k - ln(n) ] .
Make certain that lower bound L(n) as well as the upper bound R(n) are
well _within_ the intervals, as defined by subsequent fractions in the
Stern-Brocot tree. This can always be done, because the abovementioned
width of R.M. Young's interval can be made as small as desired: being
implemented in the accompanying program as the 'while not OK do' loop.
The bounds L,R can never be _equal_ to these fractions, because there
is a natural logarithm ln(N) in them. And this logaritm is irrational,
hence the bounds L,R are also irrational themselves.
This proves that any bounds to gamma are never fractions in the tree.
But the tree exhausts all fractions. And gamma is within these bounds.
But, uhm .. suppose that gamma actually _is_ a fraction F . Then, at a
certain place in the Stern-Brocot tree, L(n) < F < R(n). And this will
will remain so, no matter how the interval [ L(n) , R(n) ] is squeezed
to zero. Meaning that the 'while not OK do' loop is never ending. That
is why the above is NOT a proof that gamma itself would be irrational.
But perhaps we can not _prove_ that the Euler-Mascheroni constant gamma
is irrational, but we _can_ determine approximations for its Simplicity
for sure. This is done in the following program, where it is determined
that the simplest simplicity of gamma is less than: 1/13 . Therefore it
cannot be concluded that gamma is quite close to irrational, mainly due
to software limitations: _insufficient precision_. Software limitations
can be overcome however, so I expect that better estimates can be found.
----------------------------------------------------------------------
program Macaroni;
{
   Gamma as a Path in the Stern-Brocot Tree
   http://en.wikipedia.org/wiki/Stern-Brocot_tree
}
const
   tol : double = 1.E-11;
   gamma : double = 0.577215664901532860606512090082402431042;
procedure Young(eps : double);
{
   Walk through the tree
   all along until gamma
}
var
   m1,m2,n1,n2,N : Longint;
   L,U,som,G : double;
   tel,boven,onder,min : integer;
   function OK : boolean;
   begin
     OK := (((n1+n2)*U < m1+m2) and (n1*L > m1))
        or (((n1+n2)*L > m1+m2) and (n2*U < m2));
   end;
begin
   Writeln;
{ Initialize tree }
   m1 := 0; n1 := 1;
   m2 := 1; n2 := 0;
{ # iterates }
   tel := 0;
   N := 1;
   som := 1;
   G := som - ln(N);
   L := G - 1/(2*n);
   U := G - 1/(2*(n+1));
   boven := 0;
   onder := 0;
   while true do
   begin
     tel := tel + 1;
   { if gamma not in between
     approximating fractions }
     while not OK do
     begin
       N := N + 1;
       som := som + 1/N;
       G := som - ln(N);
       L := G - 1/(2*n);
       U := G - 1/(2*(n+1));
     end;
     if (n1+n2)*U < (m1+m2) then
   { Tightening }
     begin
       Write('U');
     { Upper Bound }
       m2 := m1+m2;
       n2 := n1+n2;
       boven := boven + 1;
     end else begin
       Write('L');
     { Lower Bound }
       m1 := m1+m2;
       n1 := n1+n2;
       onder := onder + 1;
     end;
   { Think about this: }
     if eps*n1*n2 > 1 then Break;
   end;
   Writeln;
   Writeln('   ',G,'   ');
   Writeln('   ',m1/n1,' < ',L);
   Writeln(' < ',gamma,' < ');
   Writeln('   ',U,' < ',m2/n2);
   Writeln(' +/- ',eps:1);
   Writeln(' ',m1,' / ',n1,' < gamma < ',m2,' / ',n2);
   Writeln(' ',tel,' iterations < ',Round(ln(eps)/ln(gamma)));
   min := boven; if onder < boven then min := onder;
   Writeln(' Simplest Rational Simplicity = 1/',min);
end;
begin
   Young(tol);
end.
----------------------------------------------------------------------
Output:
ULUULUULUUUULLLUUUUUUUUUUUUULLLLLULUUUUUUUUL
     5.77216947247995E-0001
     5.77215664900508E-0001 <  5.77215664900787E-0001
  <  5.77215664901533E-0001 <
     5.77215664904075E-0001 <  5.77215664904076E-0001
  +/-  1.0E-0011
  323007 / 559595 < gamma < 289166 / 500967
  44 iterations < 46
  Simplest Rational Simplicity = 1/13
----------------------------------------------------------------------
Disclaimer: the program defines the algorithm, illustrates the proof,
but is not itself evidence. Note however that there are long sequences
of U(pper) bound adaptions while executing the program. This behaviour
is typical for fractions. Meaning that gamma is quite alike a fraction
- while being found eventually being irrational. This might be kind of
an explanation why the irrationality of gamma is so hard to prove. But
oh well, the same kind of behaviour is found for Pi ...
For what it is worth.
Han de Bruijn