Local: Fri, Jan 23 2009 3:35 pm Subject: #165 Chapter 7, The two large flaws in the Peano Axioms and how to fix them; new book 2nd edition: New True Mathematics

[plutonium....@gmail.com]

Here are the Peano Axioms for the Natural Numbers as given by
Wikipedia:
--- Quoting Wikipedia on Peano Axioms ---

The axioms

When Peano formulated his axioms, the language of mathematical logic
was in its infancy. The system of logical notation he created to
present the axioms did not prove to be popular, although it was the
genesis of the modern notation for set membership (∈ from Peano's ε)
and implication (⊃ from Peano's reversed 'C'). Peano maintained a
clear distinction between mathematical and logical symbols, which was
not yet common in mathematics; such a separation had first been
introduced in the Begriffsschrift by Gottlob Frege, published in 1879.
[4] Peano was unaware of Frege's work and independently recreated his
logical apparatus based on the work of Boole and Schröder.[5]

The Peano axioms define the properties of natural numbers, usually
represented as a set N or \mathbb{N}. The first four axioms describe
the equality relation.[6]

1. For every natural number x, x = x. That is, equality is
reflexive.
2. For all natural numbers x and y, if x = y, then y = x. That is,
equality is symmetric.
3. For all natural numbers x, y and z, if x = y and y = z, then x =
z. That is, equality is transitive.
4. For all a and b, if a is a natural number and a = b, then b is
also a natural number. That is, the natural numbers are closed under
equality.

The remaining axioms define the properties of the natural numbers. The
constant 0 is assumed to be a natural number, and the naturals are
assumed to be closed under a "successor" function S.

5. 0 is a natural number.
6. For every natural number n, S(n) is a natural number.

Peano's original formulation of the axioms used 1 instead of 0 as the
"first" natural number. This choice is arbitrary, as axiom 5 does not
endow the constant 0 with any additional properties. However, because
0 is the additive identity in arithmetic, most modern formulations of
the Peano axioms start from 0. Axioms 5 and 6 define a unary
representation of the natural numbers: the number 1 is S(0), 2 is S(S
(0)) (= S(1)), and, in general, any natural number n is Sn(0). The
next two axioms define the properties of this representation.

7. For every natural number n, S(n) ≠ 0. That is, there is no
natural number whose successor is 0.
8. For all natural numbers m and n, if S(m) = S(n), then m = n.
That is, S is an injection.

These two axioms together imply that the set of natural numbers is
infinite, because it contains at least the infinite subset { 0, S(0),
S
(S(0)), … }, each element of which differs from the rest. The final
axiom, sometimes called the axiom of induction, is a method of
reasoning about all natural numbers; it is the only second order
axiom.

9. If K is a set such that:
* 0 is in K, and
* for every natural number n, if n is in K, then S(n) is in
K,
then K contains every natural number.

--- end quoting Wikipedia on Peano Axioms ---

Now there are two errors in those Peano Axioms above and those two
are:

(1) failed to define "finite" versus "infinite" which thus, the Peano
axioms
end up describing this set 0000....0000, 0000....00001, 0000....00002,
0000....00003, ..... then much further on 5555....5555,
5555.....55556,
and still much further on we have 9999...99997, 9999...99998
and finally the last and largest integer in the world as
99999....99999

So because Peano failed to define Finite from Infinite, his above 9
axioms
delivers the AP-adic Integers or the Infinite Integers. They obey the
Peano
axioms (except for maybe 9999...9999 since it has no successor)

(2) the second error is more serious because it destroys the integers
that
Peano is trying to create. It is a mistake which can be called a "lack
of
defining a Ruler". In a previous post I said it was a Masonry or
Carpenter
mistake that Peano never labored on building a brick wall or
carpenter's
building walls. Well in order to have a Ruler, a ruler must have two
ends.
There are no one ended rulers in the world.
So in Peano's 5th axiom he says only one number is brought into
existence
and further on he calls up a Successor function S(0). Well, the
problem
and error is that Peano cannot create a Natural Number system without
creating
at least two numbers the 0 and then the 1, or the 1 and then the
number 2.
So Peano's 5th axiom should have created two new numbers and
thenceforth
Peano could create the Successor Function of N+1 because the "+1" was
created in the 5th axiom when creating two numbers.

As it stands Peano created a one-ended-ruler, and there is no such
thing. Peano
needed to create two new numbers for the Natural Numbers in order to
make a
Metric, a Parameter of the Natural Numbers.

As it stands at the moment with Peano's 5th axiom where he creates
only one
Natural Number, then this set are the Natural Numbers:

0, 1/2, 1, 3/2, 2, 5/2, etc etc

or this set is the Set of Peano Natural Numbers:

0, 1/3, 2/3, 1, 4/3, 5/3, 2, etc etc

Now some may say that those reduce to this set 0,1,2,3,4, etc etc
but that is not true because when Peano creates only a 0 or 1 but not
two new
numbers depends on his Induction Axiom of Successor, in that the N, N
+1
is not served by N, N+1/2, nor served by N, N+1/3

The only way to avoid that error of Peano was by creating two new
Natural Numbers
and thus creating a "Ruler" in the 5th axiom.

P.S. the way to solve Peano's error of "finite versus infinite" is to
introduce a 10th
axiom that basically says that "Finite" is synonymous with physics
measurement.
Where Physics is halted in measuring of about 10^500 is the end of
finiteness and
the beginning of Infinity.

Thus the Finite Natural Numbers are from 0 to 10^500 and the Infinite
Integers
or Infinite Natural Numbers are from 10^500 to 9999.....99999

Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies

[Archimedes Plutonium]

making sure it gets posted properly

[plutonium....@gmail.com]

Now in my next post I am going to build the Reals from the Peano
Axioms
without ever using rationals and irrationals and Dedekind Cut. I am
going
to build the Reals by focusing on a Successor Function of the Real
Number
0d0000....0001. So where Peano creates 0 and 1 and then applies a
Successor
function to build the Natural Numbers. I am going to create 0 and
0d000...0001
and create All the Reals without ever using Dedekind Cut.

And when I do that creating of the Reals from scratch without ever
using Dedekind
Cut, I will show why so important it was for Peano to create two new
numbers in
his 5th axiom, because in order to create the Reals, I need two new
Reals to
create the Successor Function.

And much later on in this thread I shall review the old unproven
conjectures of
Riemann Hypothesis and Fermat's Last Theorem and various other
unproven
Number theory problems. Because, is FLT true for Finite-Natural-
Numbers but
false for Infinite-Natural-Numbers. Here we get into the lovely
situation that a proof
on Finite Natural Numbers amounts to proving whether a statement is
true or false
for the first 10^500 Integers. So is Fermat's Last Theorem true for
the first
10^500 Integers. The idempotents in P-adics of 10-adics have only one
integer that
is a finite-integer and the other two integers (idempotents) are
Infinite-integers.

So is Fermat's Last Theorem true for the first 10^500 Integers, and I
believe that is
a fact of truth. But that FLT is false for All Natural Numbers because
beyond 10^500
we have boatloads (as a Princeton math student described it in 1993),
we have
boatloads of counterexamples.

So the mathematics of the 21st century has a divided house in Number
theory since
the Natural NUmbers as a whole are Infinite Integers and where we have
a segment
of them called Finite Natural NUmbers or Finite Integers from 0 to
10^500.

Noone should be frowning or unpleased that Physics comes into math and
forces
the number 10^500 down the throats of mathematicians. Noone should be
displeased
because Physics is superior to mathematics and physics creates math.

So it is only natural (sorry for the pun), natural that Physics has
the final word or
say as to what Finite means versus Infinite. If we can no longer count
or measure
or test or experiment with a number beyond 10^500, then, that is
FINITE. And beyond
10^500 is infinite.

[plutonium....@gmail.com]

As promised I am going to build the Reals in two different means, one
from All Possible
Digit Arrangements and second from the Peano Axioms with a Successor
function.

First let me do it the easy way from All Possible Digit Arrangements.
This is one
of my earlier chapters in this book where I simply took the decimal
ten digits and
said, let them go to infinity. Now all possible digit arrangments for
rightward
infinite strings gives us a smallest Real and a largest Real in the
open interval
0 to 1. The largest Real is 0d99999....9999 and the smallest Real is
0000...00001. Because I have a largest Real in that interval allows me
to
define exponentiation of place value and where the Real of
0d9999....9995
the "5" digit is going to be in the 10^(-)9999....99999 place-value.
Now that
may look funny to some that the exponentiation is seemingly larger
than the
number itself but remember these are infinite strings and that the
even
integers are infinitely the same as all the integers.

Then define multiplication on these All Possible Digit Arrangements
and we
encounter the question of 1d0000...000 x 0d0000....00001 =
0d000...00100...000
And we ask what the place value of the "1" digit is for that Real
Number.
And we thence define that Real Number as having the 10^(-500) place-
value.
The number is important because it is the limit of Physics measuring
and
testing and counting and experimenting in the World. So we define
10^500
or 10^(-)500 as the limit of finite to infinite. Numbers larger or
smaller than
10^500 or 10^(-)500 are infinite.

So we have all the Infinite strings of Reals by simply saying apply
All Possible
Digit Arrangements and that yields a special number of
0d000...010...000 which
we ascribe to as 10^(-)500 as the bridge from which finite goes to
infinite.
Now we use that number of 10^500 to assign the Finite portion leftward
string
of Reals. So to the left of the decimal point "d" in Reals can only go
to
10^500 and in between two Reals is an infinite supply of Reals.

----- second way of building the Reals ----

Here again is the Peano Axioms as given by Wikipedia

> The axioms

The old math used Dedekind Cuts to define the Reals. I need not go
that far
because the Peano Axioms refurbished define the Reals. Here is how
they
define the Positive Reals.

I replace the words "Natural Numbers" above with "Reals"
And the key number instead of "1" is the number 0d0000....00001 which
acts and behaves like a 1 for Reals. So make that replacement above in
the Peano Axioms.

Now when Peano does his 5th axiom which is flawed, we have to add two
elements not just one element of 0. We include both 0d0000....000
and 0d000....00001 so that the metric of distance of 0d0000....00001
serves
as the Successor function.

So now when we reach axiom 9th of Peano we see how the Successor of
endlessly adding 0d000....0001 to 0 ends up giving us every Real
number between
0 and 1 in the open interval (0,1) So the Peano axioms provides us
with all the
infinite rightward strings of Reals, but we must take care of the
finite leftward
portion on Reals. So here again we append a tenth new axiom to Peano
saying
in effect that Physics limit of measure is a number of 10^500 and
beyond that
number is no longer able to do Physics of any sort of meaning. So we
define
10^500 as the endpoint of Finite mathematics and the next number as
the beginning
of the infinite. Now we run the same Peano Axioms above with these
numbers from
0 to 10^500 using the integers 0 and 1 as the metric ruler that
supplies the Successor
function to craft every one of those numbers from 0 to 10^500.

So we end up with the Reals as being such examples:

1d00000....000008888

10^500 d33333....3333334

But where 10^501d 0000....0000 is not a Real Number because it
violates the
definition that a Real Number is only finite portion leftwards.

As anyone can see, the building of the Reals is easiest with All
Possible Digit
Arrangements. The building of the Reals can be done from the Peano
Axioms
also but with alot more definitions involved.

And in effect the building of the Reals are the numbers that are the
native numbers
of Euclidean Geometry. So the fact that there is an infinitude of
Reals between
two finite consecutive Reals such as between 0 and 1 gives continuity,
because
continuity only means an infinite supply of numbers between two
numbers. And
the fact that the Reals end at 10^500 means that every time a
Euclidean Geometry
book says the lines go to infinity are wrong. Lines in Euclidean
Geometry can only
go to 10^500 from 0 in a finite measure of distance since the Reals
leftward string
is defined finite. So how can Geometry books say that parallel lines
never meet
when no lines in Euclidean geometry go beyond 10^500? The answer is
that lines
are parallel in Euclidean Geometry by the test of any Reals on those
lines have
an equal distance spacing apart.

Obviously the Reals have only one infinity. And the concept of All
Possible Digit
Arrangements allows for the world of mathematics to have only one type
of infinity.
So much of mathematics of the past century with its Cantor hocus pocus
phony
baloney about infinity was fakery math.

And the Dedekind method of building the Reals was a phony method for
it started
its program with a delusional concept of "absolute continuity" where
given any two
Reals there is an infinity of Reals between them. If only Dedekind and
Cantor had
studied the concept of All Possible Digit Arrangements may have or
would have
then realized that there cannot be a third Real between that of
0d3333....33333 and 0d3333....333334

The Reals as constructed above are infinite in the rightward string
and are consecutive
in that they are countable with a successor and predecessor and that
they are
discrete and have gap-spacing between two consecutive Reals such as
0d5555...55557 and 0d5555....55558

Now in the next post I intend to build the Finite Natural Numbers and
Infinite
Natural Numbers and the AP-adics by the two methods of building using
All Possible Digit Arrangements and using Peano Axioms Revised

[plutonium....@gmail.com]

Alright, as promised I built the Reals from the Peano Axioms without
any
Dedekind Cut used. And I built the Reals from the easier construction
of
simply All Possible Digit Arrangements. And now I am going to build
the
AP-adics and the Natural-Numbers using both methods. Now the Natural
Numbers come in two classes or types. There is the Finite Natural
Numbers
and the Infinite Natural-Numbers and the two put together are the AP-
adic
Integers.

And there needs to be a wholescale revision and relooking at the
alleged
proofs and theorems of the old math. Fermat's Last Theorem is probably
true only for the Finite-Natural-Numbers of 0 to 10^500. I suspect a
computer
has checked most of the numbers from 0 to 10^500 whether any satisfy
the Fermat's Last Theorem. And the Riemann Hypothesis and the Goldbach
Conjecture and the Twin Primes Conjecture have to be checked for
numbers from 0 to 10^500. I say checked because the Peano Axioms
of the 19th century actually gave the Infinite Integers, not the
Finite Natural
Numbers. So that when Andrew Wiles announced a proof of Fermat's Last
Theorem, his work could only be a confirmation that none of the
Natural Numbers
from 0 to 10^500 satisfies the Fermat condition. Because the Infinite
Integers
of idempotents shows us that Fermat's Last Theorem is false for All-
Natural-Numbers.
Has the Riemann Hypothesis been checked for Natural Numbers from 0 to
10^500?
And what is the latest status of the prime number from 0 to 10^500? Do
we know
many of those prime numbers between 0 and 10^500?

Now as for the Goldbach Conjecture that every even number is the sum
of two primes.
Well, that conjecture was a terribly difficult math conjecture to try
to prove because
it is pitting addition up against multiplication and when you pitt
something as fundamental
as raw addition against raw multiplication (primes) then you do not
have much to
work with to prove or disprove the conjecture. But now that we have
the full Natural
Numbers as the Infinite Integers from 0 to 99999....99999, a
conjecture like
Goldbach becomes easy prey for proving. When the full Natural Numbers
from
0 to 9999...99999 are used on Goldbach Conjecture we immediately have
to ask
whether the concept of "prime" still exists over all the integers from
0 to 9999...999
and I would have to say at this moment that the concept of prime is
not a universal
concept but that there are "hybrids" in the numbers from 0 to
9999....99999 which are
hybrids from being prime or composite. For instance the number
10000....00001
whether it is prime or composite or a hybrid. Hybridization is a
problem for the
concepts of rational and irrational for example the number
07777...6666
is neither rational nor irrational for it does repeat in 6's and
repeats in 7's but there
is that lone "0" digit. If not for the 0 digit we could call it
rational but is it irrational
or a hybrid. So in All Natural Numbers we tend to lose concepts that
we had in
Finite-Natural-Numbers.

As for Goldbach Conjecture, if we can establish a concept of Prime for
all numbers
from 0 to 9999....99999 then it is likely that the Natural Numbers
from
9999....99999 to 99999....9999899, which are the last 100 numbers of
the Natural
Numbers that there are just as many primes of about 25% prime numbers
in the
last 100 Natural Numbers as there are primes from 0 to 100 which if
memory serves
me is about 25% prime numbers in that interval. So we see that the
Prime Distribution
theorem of the old math is false, for although the primes seem to thin
out as we go
beyond 100, that at some point they start to become more dense and in
the final
stretch of the last 100 Natural Numbers are as dense as the first 100
Natural Numbers.
Also, I may add, that we have situations where there are Triplet
Primes such as
3,5,7 are triplet primes. So as we go into the Infinite Natural
Numbers we run into
alot of triplet primes, and broken up by numbers ending in digits of
"5", but where
digits of 1,3,7,9 are fair game candidates for having a string of
primes interrupted only
by the 5 ending digit. Call that string a quadruple string primes.

So is the Goldbach Conjecture true or false? Well probably on Finite
Natural
Numbers, those from 0 to 10^500, the Goldbach Conjecture is probably
true,
and someone may have already found an argument that holds up to
10^500.
But as for All the Natural Numbers from 0 to 99999....99999 when we
consider
that the density of primes increases, provided that the concept of
prime still
is a useable and workable concept, and because the density increases,
that
we have to ask this question. Since in the first 100 numbers we have
25 primes
for which we have to permutate those 25 to deliver 50 even numbers,
that the
probability of Goldbach Conjecture being true over All the Natural
Numbers is
extremely high probability.

So, let me construct the Finite Natural Numbers and the Infinite
Natural Numbers
which the two put together are the All Natural Numbers.

Using All Possible Digit Arrangements is the easiest construction for
you just
wave a magic probability wand and you have all possible arrangements.
You
have 00000....00000 and 0000....00001 and all the way up to
99999....99999.

The second way to construct is to use the Peano Axioms, only we have
to
strengthen the 5th axiom to read "there exists both a 0 and 1" And the
Peano
Axioms delivers all the Natural Numbers from 0 to 99999....9999

So both construction methods delivers from 0 to 9999....99999. So
where are the
Finite Natural Numbers?

In All Possible Digit Arrangements we can force them to deliver a
special
number of 10000....0000 x 0000....00001 and ask what the answer is of
0000....0001000....00000 which I call the smallest infinity. Since
Physics is
the superior science over mathematics, physics places the definition
of
Finite as the number 10^500 where the measurement and testing in
Physics
grinds to a halt. So the number 10^500 is the end of Finite Integers
and add 1 more
is the beginning of Infinite Integers.

Now in Peano Axioms to construct the Natural Numbers we have to
include a tenth
new axiom saying that Finite is defined by Physics as 10^500.

And, there you have it.

[plutonium....@gmail.com]

plutonium.archime...@gmail.com wrote:
(snipped)

>
> And there needs to be a wholescale revision and relooking at the
> alleged
> proofs and theorems of the old math. Fermat's Last Theorem is probably
> true only for the Finite-Natural-Numbers of 0 to 10^500. I suspect a
> computer
> has checked most of the numbers from 0 to 10^500 whether any satisfy
> the Fermat's Last Theorem. And the Riemann Hypothesis and the Goldbach
> Conjecture and the Twin Primes Conjecture have to be checked for
> numbers from 0 to 10^500. I say checked because the Peano Axioms
> of the 19th century actually gave the Infinite Integers, not the
> Finite Natural
> Numbers. So that when Andrew Wiles announced a proof of Fermat's Last
> Theorem, his work could only be a confirmation that none of the
> Natural Numbers
> from 0 to 10^500 satisfies the Fermat condition. Because the Infinite
> Integers
> of idempotents shows us that Fermat's Last Theorem is false for All-
> Natural-Numbers.
> Has the Riemann Hypothesis been checked for Natural Numbers from 0 to
> 10^500?
> And what is the latest status of the prime number from 0 to 10^500? Do
> we know
> many of those prime numbers between 0 and 10^500?

I really should not be asking those type of questions because I know
that the
number 10^500 is more numbers to count, even by a supercomputer than
there
is time in the Universe to count them. And that is why I say the
definition of
Finite should be provided by Physics to where there is no longer any
Physics
to be measured or tested.

So I had a look a the efforts applied to the Riemann Hypothesis for
looking for
counterexamples and here is what Wikipedia says about the search
efforts:

--- quoting Wikipedia ---

http://en.wikipedia.org/wiki/Riemann_hypothesis

Searching for ζ-function zeroes
Absolute value of the ζ-function

There is a long history of computational attempts to explore as many
zeroes of the ζ-function as possible. One notable such attempt was
ZetaGrid, a distributed computing project, which checked over a
billion zeros a day when it was running. The project was shut down in
November 2005. As of 2008[update], no computational project has
succeeded in finding a counterexample to the Riemann hypothesis.

In 2004, Xavier Gourdon and Patrick Demichel verified the Riemann
hypothesis through the first ten trillion non-trivial zeros using the
Odlyzko-Schönhage algorithm.

Michael Rubinstein has made public an algorithm for generating the
zeros.

--- end quoting Wikipedia ---

So that computers have looked at the first ten trillion zeros or 10^13
of the zeta function. So 10^13 is
far short of 10^500 where Physics gives out on measuring.

So is there a way to show that the Counting Numbers of 10^500 and
beyond of the
10^500 zeroes of the Zeta Function that the first counterexample of
the RH becomes
evident?

Here I propose that we simply apply the 10-adics as if they were
Natural Numbers.
Can anyone offer a set of 10-adics, such as the idempotents and
pretend as if they
are Natural Numbers and then say, those are counterexamples of the
Riemann Hypothesis.

I offered the 10-adic idempotents as counterexamples to the Riemann
Hypothesis
in the 1990s, but noone wanted to listen.

Why the Riemann Hypothesis looks to be true is because as the Counting
Numbers
get ever so much larger, they bend and curve and become circular and
come back around
to their starting point of 0 then 1. Why is this? It is because
infinity at large is Elliptic Geometry
that bends and comes back around. So with small numbers like 10^13 we
cannot
sense the bending that is going on and we "feel" as though the Riemann
Hypothesis
is true, but when you get to actual large numbers such as beyond
10^500 such as the
10-adic idempotents, then the bending and curving of counting numbers
is unavoidable
and unmistakeable. And when we use the ultimate 10-adics of
9999....99991 which is
(-9) and 9999....99992 which is (-8) on up to 99999....99999 which is
(-1) and put those
into the Riemann Hypothesis we quickly realize they are
counterexamples.

But do the counterexamples to the Riemann Hypothesize, are they
apparent at the
10^500 scale? Or do we have to jump to the 9999....99991 scale to see
the obvious
destruction of the Riemann Hypothesis? I am hoping that we can see it
at the
10^500 level because that would tie together the end of Physics
measuring with the
end of mathematics in keeping a Euclidean straightline as "sort of
straight".

[plutonium....@gmail.com]

plutonium.archime...@gmail.com wrote:
(snipped)

Now here are the 10-adic Idempotents

.......000001
.......9977392256259918212890625
.......0022607743740081787109376

In the 1990s I did not have the concept of FrontView and BackView.

We clearly can see the BackView of those three Infinite Integers of
Natural-Numbers
or of AP-adics Integers.

Can we guess the FrontView and thus estimate what percentage of the
semicircle
of longitude, that those numbers are from 0 or from the 9999...99999
near the South Pole?

Well if the FrontView of one of them is the "9" digit then that number
is at least
90% of the way to the South Pole but making the other Idempotent as
less than
10% of the way to the South Pole.

In fact, can we determine if the Idempotents of 10-adics ended up
looking like this:

99999......9977392256259918212890625

and this

00000.....0022607743740081787109376

If that idempotent ended up looking like that then can that second one
end
up looking like this:

00000.....002260..376 where the ".." indicates a finite span. So that
the front
most "2" digit is in the 10^500 place value.

If so, then the Idempotents would be the Counterexamples of the
Riemann
Hypothesis.

[plutonium....@gmail.com]

So that we define finite as anything below 10^500 from the Physics
that such a number
is at the limit of measurement or testing or experimenting in Physics.
IF we cannot do something
and beyond the realm of "doing" then it is in the infinite realm. So
we define finite
as below 10^500.

Now I went to see what the largest Mersenne prime reckoning was and it
is of the order
of 10^43,000,000 but those numbers are pokes at large numbers and not
a counting.
Because no machine, no supercomputer can count to 10^500 since there
is not enough
time in the Universe to count that high.

And I checked to see what the largest prime number verification of
Goldbach Conjecture
was and it is 10^18 or thereabouts.

But I want to comment on another perspective of the Goldbach
Conjecture. Something
that may interest me in looking for a new angle to prove or disprove
it. For the Goldbach
Conjecture is the most stubborn conjecture that I ever encountered. I
found Fermat's
Last Theorem and Riemann Hypothesis a breeze compared to the Goldbach
Conjecture.
I have not been able to find a counterexample in the Infinite Integers
from 0 to
9999....9999. So that suggests the Goldbach Conjecture is a true
statement.

So I set up a grid of where count the number of primes from
0 to 100 there are 25 primes
0 to 200 there are 46 primes
0 to 300 there are 62 primes
0 to 400 there are 78 primes
0 to 500 there are 95 primes

Now I ask myself how many even numbers can 25 primes generate to cover
50 primes in the interval 0 to 100 and the answer is 25 x 25 = 625.
Then
how many even numbers can 46 primes generate to cover 100 even numbers
in the interval 0 to 200 and the answer is 46 x 46 = 2116.

So the question I begin to ask is does that number of generated even
numbers
ever come close to how many even numbers I need to cover?

Now according to Finite Natural Numbers, the primes thin out the
further out we go.
So we would think that the further out we go that the number of
generated even
numbers catches up with the number of even numbers we need to cover.

But in Infinite Natural Numbers, as I noted earlier that the number of
primes in the
last interval from 99999....99999 to 9999...9999899 is about 25 primes
in that interval

So Goldbach Conjecture would be a cinch to be true because in Infinite
Integers
the primes never really thin out but become denser episodically. They
become thinner
in layers but then start to become denser and thinner then denser and
cycling between
thin and dense.

But in Finite Natural Numbers from say 0 to 10^500, the primes beome
increasingly
thin. So how many primes in the interval 0 to 10^500 and would that
number be able
to cover the even numbers in that interval?

So if the Peano Axioms were held to a stricture that only Finite
Integers exist as
Natural Numbers then the thinning out would seem to almost certainly
force the
Goldbach Conjecture to be false. But if the Peano Axioms had a barrier
point of
10^500 as Finite Integers and the rest of the Natural Numbers are
Infinite Integers
then a Goldbach Conjecture would be true.

So here we have a conjecture of mathematics that is dependent on what
we call
Natural Numbers. If we call the Natural Numbers as only "finite
integers" then
the Goldbach Conjecture would certainly be false because the primes
would thin
out so badly that they fail to cover some very large even integers. On
the other
hand, if we see the Natural Numbers as finite up to a point such as
10^500 and
beyond that are Infinite-integers and thus recognize that the primes
beyond
10^500 cycle between thin and dense pockets of primes, such that the
primes
always can generate more even numbers than what is needed.

Now since the number of primes has been charted out to 10^18 and that
10^500
is vastly far away. I just wonder if that number 10^500 is where the
number of
prime generated even numbers catches up with the amount of even
numbers that
requires a Goldbach covering?

[plutonium....@gmail.com]

Oops, made a big mistake above. My vision of the primes was flawed.
I had seen them as thinning out. But the thinning out was never to be
anything noteworthy. Because the density of primes follows
the formula of x / Ln(x) and gives this chart:

10 is 4
10^2 is 25
10^3 is 168
10^4 is 1,229

So my understanding of the primes was flawed, in that I thought they
thinned
out drastically. I suppose I read too many schlock math books which
said
the primes thin out badly. But with the Prime Counting Function, we
see that
really that is not the case. They do thin out but it is a thinning out
that is so
gradual as to be not even noteworthy. It is almost a straightline
graph.

It is similar to the Planet Earth surface that the planet is so big
relative to us
that we feel it is flat although the curvature is there and is
gradual. Same thing
with the distribution of primes.

So that Goldbach Conjecture should be true regardless of my spiel
about finite
or infinite integers. That the number of primes in an interval is
always going to
generate more Goldbach even numbers than what are needed for a
covering
of the even numbers in that interval.

And although a proof is not begot from saying the supply of even
numbers generated
by primes is vastly larger than the even numbers needed, nonetheless
it is a
stark indicator that the Goldbach Conjecture is true.

So has anyone tabulated that the Goldbach prime generated even numbers
in
the interval 0 to 100 is 25 x 25 = 625 even numbers when only 50 are
needed?
How many of those even numbers are covered twice or three times? How
many have
only one Goldbach covering?

So if I can find some sort of way to link the generated even numbers
with the needed
even number coverings, I would have a proof of Goldbach because there
is always a
larger supply of Goldbach generated even numbers than the even numbers
within that
specific interval. So some fact or feature as to why addition of
primes should traverse
over every even number since there are vastly more generated even
numbers.

Perhaps what we should do to solve Goldbach is devise a Prime Counting
Function
to where it just gives us say 3 primes in the first 10 and only 24 in
the 10^2 or 167 in
the 10^3 and see if there is a even number not covered by Goldbach. If
not, devise
a Prime Counting Function that delivers only 2 primes in the first 10
and only 23 in
the first 10^2 until we hit upon a formula where a even number is not
Goldbach covered.

Suppose 3 was not prime then we would miss 6 and 8. Would we miss any
others?

Or suppose 97 were not prime, then what even number would be missing
in a Goldbach
covering? Such questions may unlock the inner connections.

[plutonium....@gmail.com]

Here is an outline of the Goldbach Conjecture proven to be true.

I did this in the 1990s but had a flaw in my reasoning. But not all is
thrown out the
window. The deep connection between adding two primes to make an even
integer,
is the connection that we multiply those two primes. So there is a
unique connection
between addition and multiplication.

For example, in Goldbach we have for the case of 12 that it is covered
by that of 5 + 7, or in the case of 14, it is covered by either 7+7 or
11 +3.

So here is the deep connection that proves Goldbach true. In that 12 =
5 + 7, we focus on what 5 x 7 is and it is 35. In the case of 14 = 7 +
7
we focus on 7 x 7 = 49 and in the case of 14 = 11 + 3 we focus on
11 x 3 =33.

So now we ask the important question of the link? The link in that
every
even number has a primes sum has to be true so that there exists no
holes
in multiplication so that every Integer that is the product of two-
primes exists.

So the gist of the arguement or proof is that if a even number say 12
is not the
sum of two primes then the product of two primes does not exist or
that the
number 35 is nonexistant.

Now in the 1990s I tried to prove Goldbach with the Chebyschev theorem
that between
N and 2N exists a prime. I was on the correct track to link 35 = 7 x 5
and
12 = 7 + 5.

So the key is to proving Goldbach is that if there exists a even
integer that is not
Goldbach covered by the sum of two primes, then there exists a hole in
the Natural
Numbers, a hole of a number which is not the product of two primes.

So what I have to do is develop a formula such as a Prime Counting
Function, but
instead of counting primes that it counts integers that are the
product of two primes.

In the first 100 numbers I have this list
2 yields 15 two-prime composites
3 yields 12 two-prime composites
5 yields 7 two-prime composites
7 yields 4 two-prime composites
11
13
17
19
23
29
31
37
41
43
47

Now if I did not make any mistakes up there, that list should give
me 38 numbers that are two-prime-composites in the interval 0 to 100

So now the Prime Counting Function is x/Ln(x) which gives 25 primes in

the interval 0 to 100

So I have to find what the Function is for two-prime composites?
Is it similar to x/Ln(x). I suspect it is, only if the intervals were
adjusted.

So if I can show that Prime Counting Function is the same as
Two-Prime-Composites, then, I can show that if there exists a even
number
that has no Goldbach addition covering then there exists a hole in the
integers
where there is a missing integer of a two prime composite.

[plutonium....@gmail.com]

One minor mistake above. I have to figure the function for the number
of
Goldbach addition coverings in the intervals because some
even numbers such as 14 have two Goldbach coverings of 7 +7 and
that of 11 + 3.
Now whether the Prime Counting Function can help in discovery of that
formula for the number of Goldbach coverings, is questionable.

If it can, then I weigh the Goldbach coverings against the Two-Prime-
Composites.
Those two must agree, for if they do not agree, then the Goldbach
Conjecture
is false.

[plutonium....@gmail.com]

Some good news and some bad news. The good news is I found a quick and
easy
proof of the Goldbach Conjecture. It takes probably only one page to
prove it, after a
modification of the Chebychev Theorem of primes. The bad news is that
this has kept
me up all night long. But I rather do this and miss some sleep.

Here is the proof, and it pretty much follows my "slightly marred
proof attempt of the
1990s" I was talked out of that proof attempt and failed to make this
minor adjustment.
This minor adjustment now makes the Goldbach Conjecture a simple
proof. Here it is.

First I need to set up partitions of additions such as this:

0 4
1 3
2 2
3 1
4 0

that above was the 4 partition and we see where 4 = 2 + 2 to satisfy
Goldbach

this is the 6 partition:

0 6
1 5
2 4
3 3
4 2
5 1
6 0

and we see from the 6-partition that Goldbach is satisfied by the 3 +
3

Now I write the 8-partition as such:

0 8
1 7
2 6
3 5
4 4
5 3
6 2
7 1
8 0

And we see Goldbach satisfied in the fact that 8 = 3 + 5

Now we write the Chebychev Theorem that there exists a prime number
between N and 2N

Now we reformulate the Chebychev Theorem as a Corollary: Between N
and 2N there exists at least one Composite number which is the product
of two primes. And for the case of 8 = 3 + 5 we have 3 x 5 = 15

That is easy to prove of that Chebychev Corollary that between N and
2N exists at least one Composite composed of just two prime numbers.

Now we are almost at home.

Given any even number. Set up a partition of it as above for 10 then
12, then
14, ad infinitum. Then apply the Chebychev Corollary. So there has to
exist
a Composite of two-primes in all even numbered partition. That Two
Primed
Composite within that particular partition proves that all even
numbers
are the addition of two primes.

Glory be thee 231Pu!!

[plutonium....@gmail.com]

I noticed in Wikipedia there is another Goldbach Conjecture called the
Weak
Goldbach.

--- quoting Wikipedia ---

* Strong Goldbach conjecture: Every even integer greater than 2 can be
written as a sum of two primes.
* Weak Goldbach conjecture: Every odd integer greater than 5 can be
written as a sum of three primes.

--- end quoting Wikipedia ---

Well, if my method is true and accurate for proving the Strong
Goldbach Conjecture,
it should be valid to apply to the Weak form of Goldbach. All I would
have to do
is rearrange the partitions to accomodate odd integers and to have a
three sum
and finally to remake a second Corollary to Chebychev theorem stating
that between
N and 2N has to exist at least one Composite-of-Three-Primes

So let me see if I can rearrange the partitions

>
> First I need to set up partitions of additions such as this:
>
> 0 4
> 1 3
> 2 2
> 3 1
> 4 0
>

0 7
1 6
2 5
3 4
4 3
5 2
6 1
7 0

The above was the 7-partition, and I am having some difficulty in
trying to
make a 3 sided partition. Whether I can improve on the above is open
for
question. And it is already late at night.

I can easily make a second Corollary to Chebychev saying that between
N and 2N exists a Composite that is composed of Three Primes and in
the
above case it would be 7 = 2 + 2 + 3 to satisfy Weak-Goldbach
Conjecture.

So the Composite of Three Primes I am looking for with respect to 7 is
that
of 2 x 2 x 3 = 12

So the proof of Weak Goldbach Conjecture would say that build any odd
number
partition and there must exist a Three Prime Composite within that
partition.
Thus Weak Goldbach is proven true.

[plutonium....@gmail.com]

> 0 7
> 1 6
> 2 5
> 3 4
> 4 3
> 5 2
> 6 1
> 7 0
>

0 7
1 3 + 3
2 2 + 3
3 2 + 2
4 1 + 2
5 1 + 1
6 1 + 0
7 0

This would be the 9 partition

0 9
1 5 + 3
2 5 + 2
3 5 + 1
4 3 + 2
5 2 + 2
6 2 + 1
7 1 +1
8 1 + 0
9 0

I am having some trouble in keeping some order to the odd-partitions.
Some way of a all inclusive form for all possible arrangements. In the
even-partitions, every possible combination is vouched for, but with
odd partitions, I have yet to find a way of all inclusive combinations
without repeating the one column. For example the 1 5 + 3 could just
as easily have been 1 4 + 4 or could have been 1 6 + 2

In the even partitions they were all-inclusive of combinations, so I
need
that same aspect on odd-partitions. Maybe I should strive for only the
primes breakdown which would have eliminated the 1 4 + 4

[plutonium....@gmail.com]

Losing some of last night's sleep is always worth doing, when proving
the Goldbach
Conjecture. The outline of the proof is the same as what I did way
back in 1993.
The idea in proving the Goldbach Conjecture is that the Prime numbers
are
scattered amoungst all the Natural Numbers, but that scattering is not
chaotic
but rather superdetermined so that you have a prime in a certain
location so that
all the Natural Numbers exist. If in a certain spot where a number is
supposed
to exist and it is superdetermined that a prime number exists in that
spot, that is
how it must be in order for all the Natural Numbers to exist. The
Prime Counting
Function is x/ Ln(x), meaning that a composite or prime has to be
precisely fixed
at each point along the Natural Numbers for all of them to exist. That
is the
rock bottom foundation for such Conjectures as the Goldbach where we
have
addition comingled with multiplication.

So the proof of Goldbach asks whether every even number from 4 and
beyond is
the sum of two primes. So to prove that, we have to ask whether the
special-Composite
numbers-- those Composites that have only two prime factors, are so
well-distributed.
So the Goldbach Conjecture reduces to the question that 2+2 = 4 is
because
2 x 2 = 4, and next 3 + 3 = 6 because 3 x 3 = 9, and next because 3 +5
= 8
because 3 x 5 = 15.

So in other words, if there is a even number call it M such that there
are no two primes
when added together equals that M, means there exists a hole in the
Natural Numbers
because there is a hole of g x h = K. The formula for the Counting of
Primes is
x /Ln(x) and the formula for Counting the Two-Prime-Composites is
parallel to x/Ln(x)
So that if there is a M which is not covered by a Goldbach addition,
then there is
a hole or missing number K.

Now here is the 4-Partition

0 4
1 3
2 2
3 1
4 0

Now there is excess in that partition because addition and
multiplication are
commutative and so we shorten or truncate the 4-Partition to be this:

0 4
1 3
2 2

Now this is the 4-Partition to addition

0 + 4
1 + 3
2 + 2

And this is the 4-Partition to multiplication

0 x 4
1 x 3
2 x 2

Now Goldbach is satisfied in the 4-Partition because 2 + 2 is a
Goldbach covering
of the even number 4, but notice also that the 4-Partition has a
Goldbach covering
because 4 is the multiplication of 2 x 2

Now for the Even Number 6 we end up with the 6-Partition to add and
multiply as such:

0 + 6
1 + 5
2 + 4
3 + 3

0 x 6
1 x 5
2 x 4
3 x 3

So in proving Goldbach there must be a Two-Prime-Composite in that
Partition
or else a hole exists in the Natural Numbers where a number g x h = K
does
not exist. For the even number 6, if not for 3 + 3 there would not be
a 9 as a
Natural Number.

Moving on to the Even Number 8 provides this:

0 + 8
1 + 7
2 + 6
3 + 5
4 + 4

0 x 8
1 x 7
2 x 6
3 x 5
4 x 4

And so for the Even Number 8, if 3 + 5 did not exist as a Goldbach
covering
then the Natural Number 3 x 5 = 15 would not exist.

So the proof of Goldbach, whether the Strong Goldbach with Two-Prime-
Composites
or the Weak Goldbach Conjecture involving Three-Prime-Composites, are
proofs
the rely on the insight that if a number is not Goldbach covered by
addition then there
exists a Composite which is missing and thus the entire Natural
Numbers are gone.

Now my proof attempt in the 1990s was similar to the above with its
partitions
and the idea of special Composite numbers and I applied the Chebychev
Theorem
that between N and 2N must exist a prime number. Now what I did in the
1990s
that did not finish this proof was that I was trying to make the
Chebychev
Theorem go down one side of the partition such as in the 8-Partition
to go down the left hand side:

0 + 8
1 + 7
2 + 6
3 + 5
4 + 4

with only the left hand side as:

0
1
2
3
4

And to have the Chebychev Theorem dictate that since there is a prime
there
of 2 and then 3, that applying the Chebychev on the other side of the
8-partition
that the Chebychev would force or dictate a second prime to have to
exist
across from the 2 or 3 prime on the other side of the partition.

So in the 1990s I was trying to finagle the Chebychev Theorem so that
it forced
a prime on the left side of the partition to pair up with the prime on
the right
side of the partition. And if I did that, instantly Goldbach
Conjecture is proven.

So what have I got now that I did not have in the 1990s? What I have
now
is that I am going to rig the Chebychev Theorem as a Partition
Analysis.

Notice that in the 4-Partition we have these Goldbach multiplications

1x3 = 3
2x2 =4

Notice in the 6-Partition we have these Goldbach multiplications

2x4 =8
3x3 =9

Notice in the 8-Partition we have these Goldbach multiplications

3x5=15
4x4=16

Notice in the 10-Partition we have these Goldbach multiplications

4x6 =24
5x5 =25

So that in the Goldbach multiplications the number (1/2N)^2 where N is
the even
number partition, is important. For the even number 4 we have 2^2 = 4
for the even number 6 we have 3^2 = 9 for the even number 8 we have
4^2
= 16 for the even number 10 we have 5^2 = 25. So the application of
the
Chebychev Theorem in Goldbach multiplication is focused on 1/2N
squared.

[plutonium....@gmail.com]

Alright, this time it looks as though Goldbach Conjecture is solved.
The problem
I had in the past was not adjusting the Chebychev Theorem to lead the
proof. I
tried to use the Chebychev theorem on both the left and rightsides of
the partitions
to line up a prime on the leftside with the rightside. But this time I
have a trick
to tease the Chebychev theorem into obeying the proof. But it is these
Partitions
that describe the whole situation of Goldbach Conjecture. A picture
tells a
thousand words. Well Goldbach proof is best done with just showing the
Partitions and here I have assembled the first Even Partitions from 4
to 20
with the multiplication off to the side:

the 4-Partition is:

0 4
1 3 multi = 3
2 2 multi = 4

the 6-Partition is:

0 6
1 5
2 4 multi = 8
3 3 multi = 9

the 8-Partition is:

0 8
1 7
2 6 multi = 12
3 5 multi = 15
4 4 multi = 16

the 10-Partition is:

0 10
1 9
2 8
3 7 multi = 21
4 6 multi = 24
5 5 multi = 25

the 12-Partition is:

0 12
1 11
2 10
3 9 multi = 27
4 8 multi = 32
5 7 multi = 35
6 6 multi = 36

the 14-Partition is:

0 14
1 13
2 12
3 11
4 10 multi = 40
5 9 multi = 45
6 8 multi = 48
7 7 multi = 49

the 16-Partition is:

0 16
1 15
2 14
3 13
4 12
5 11 multi = 55
6 10 multi = 60
7 9 multi = 63
8 8 multi = 64

the 18-Partition is:

0 18
1 17
2 16
3 15
4 14
5 13 multi = 65
6 12 multi = 72
7 11 multi = 77
8 10 multi = 80
9 9 multi = 81

the 20-Partition is:

0 20
1 19
2 18
3 17
4 16
5 15
6 14 multi = 84
7 13 multi = 91
8 12 multi = 96
9 11 multi = 99
10 10 multi = 100

So how is the Goldbach Conjecture conquered? Both the Strong and Weak
forms
of the Conjecture. Well in the 1990s I tried using the Chebychev
Theorem that says
between N and 2N always exists a prime. I tried using that to show
that a prime
was on the leftside of a individual partition and paired with a prime
on the rightside
so that in the example of 8, that the 3 would be paired to the 5. But
Chebychev
Theorem was not going to guarantee the 3 paired up with the 5.

But I did not thence try to adjust the Chebychev theorem so that I
could be
guaranteed the matching up or pairing up of the primes on the leftside
with
rightside of partition. At least one prime pairing guaranteed in each
partition
is the proof of Goldbach.

But I now know how to make the Chebychev Theorem guarantee that each
Partition
has a paired set of primes. I do that by adjusting the Chebychev
Theorem so that
it speaks of Composites that have only two prime factors such as 3 x 5
= 15 and
where the 3 + 5 is the prime pairing for Goldbach 8 integer.

In the above Partitions off to the sidelines I have indicated what the
multiplication
of some of those pairings are. And notice there is a pattern for the
(1/2)(N) squared
such as in the 12 Partition that we have 36 and 35 and 32 and 27. The
pattern is that
the numbers are consecutive for at least two of the multiplications --
example
in the 20 Partition we have 100 and then 99.

So I alter the Chebychev Theorem to incorporate multiplication and I
alter it so that
between N and (1/2N)^2 of each partition must exist at least one Two-
Primed-Composite.

For the 4 it is 4, for 6 it is 9, for 8 it is 15, for 10 it is either
21 or 25, for 12 it is 35.

So whereas in the past of my efforts on Goldbach was to make the
Chebychev guarantee
the leftside prime paired with the rightside prime. I alter the
Chebychev and force
it to give me at least one Two-Primed-Composite within each partition.
So that
the forced handover of a Two-Primed-Composite out of each and every
Even Numbered
Partition satisfies the Strong Goldbach Conjecture. And I easily
adjust the Partitions
for Odd Numbers with addition of three numbers to satisfy the Weak
Goldbach
and adjusting the Chebychev to handle Three-Primed-Composites.

Now some are going to raise a stink over the fact that there are gaps
such as in
the 10 Partition there is 21, then 24, then 25. So some incomplete-
thinkers are
going to worry about the fact that 22 and 23 are not represented in
the 10-Partition.
And my answer is that they are not complete-thinkers. What I have done
with the
Chebychev alteration is said that the primes in Chebychev are existing
in
N to 2N. That measure is still true for two-primed-composites. So that
instead
of looking at the N-Partition and wondering where is there a prime on
the left column
paired to a prime on the right column, I simply have shifted the
Chebychev theorem
to tell me that there is a Two-Prime-Composite within that specific
Partition, even though
there are gaps in the multiplication. So if you think long enough, you
will understand
that I have replaced the hunt for a single prime in the leftside
column hoping it
pairs up with the prime in the rightside column, and simply
turned the hunt into finding just one Two Primed Composite.

[plutonium....@gmail.com]

Marvellous, I would say, that the proof of the Goldbach Conjecture is
unfolding here on
the Internet in sci.math, sci.physics and sci.logic. Not in some ivory
tower, but here
on the Internet where millions of viewers can see it unfold. Where in
1990s I had the
Partitions and had Chebychev Theorem, but was caught on a snap in the
2000s
because I could not guarantee the prime on the rightside of the
partition paired with
the prime on the leftside of the partition. All I needed was that
guarantee. Well, several
years later, January 2009, here I would find that guarantee, by simply
making an
adjustment to the Chebychev. By saying that not a "solo prime" was
between
N and 2N but that a solo-two-primed-composite had to exist between N
and 2N.
That instead of worrying that 3 would pair up with 5 in the 8-
Partition where the
3 is on the leftside of the Partition and the 5 is on the rightside of
the Partition.
Why worry about whether they pair up with the Chebychev since the 5 is
between
4 and 8. Why worry if the Chebychev pairs them up? Why worry whether
the Chebychev pairs up any Even Number.

Here is the 8-Partition:

0 8
1 7
2 6

3 5
4 4

So why worry that the Chebychev pairs up the 3 on the leftside with
the 5
on the rightside? If I can make the Chebychev guarantee the prime on
the
leftside has a pairing up with a Chebychev prime on the rightside,
then
I will have proven Goldbach.

So what I discovered is that adjust the Chebychev. The distribution of
primes
is the Prime Counting Function of x/Ln(x).

Drop the prime 2 because it is never going to be a Goldbach prime
beyond 4.
So just drop the prime 2 from further consideration.

Since "Solo Primes" obey x/Ln(x) which is factored into the Chebychev
Theorem
then what would the distribution of "Two-Primed-Composites" be? Well
we dropped
2 in our consideration, but the distribution of Two Primed Composites
will also
be that of x/Ln(x). And the distribution of Three Primed Composites
for the Weak
Goldbach will be that of x/Ln(x).

So, now, we dropped 2 from the Goldbach Conjecture because 2 is only
needed
for the 4-Partition.

Now you can see or anticipate, if you are math inclined, you can
anticipate how
the Goldbach proof is proven. Instead of using the basic most
primitive Chebychev
Theorem that between N and 2N in the case of the 8-Partition I can
only say that
a prime is on the rightside but I cannot guarantee it to pair up with
the 3.

So now we bring in the refurbished Chebychev Theorem, that says in the
8-Partition
there exists a Two Primed Composite instead of the 5 on the rightside
there exists

3 x 5 = 15.

So you see, there exists this two primed composite because two primed
composites
have to obey the Chebychev Theorem whether it is a solo-prime or a Two-
Primed
Composite since both follow x/Ln(x)

So, some have a hard time of grasping how I can make that guarantee
that 3 pairs
with 5 in the 8 Partition. I make that guarantee because instead of
using the
most basic Chebychev Theorem that tells me that a 5 exists on the
rightside,
the refurbished Chebychev tells me that between N and 2N there exists
a
Two Primed Composite.

Now I can draw nonmath people a simple analogy that they can picture
the above.

Think of a very long building that has doors on the rightside and
leftside.

Here is the 8-Partition:

0 8
1 7
2 6

3 5
4 4

Now that 8-Partition is a long building with five doors on the
leftside
and five doors on the rightside. And only where there are primes
paired
up is there a walkthrough or alley that goes through. Where the
numbers
are not paired primes there is a internal wall so the doors enter a
enclosed
room. Only where the doors are primes and they pair up is it an
alleyway.

So what the basic Chebychev Theorem does is tell me there is a prime
door on the rightside, but it does not guarantee that the prime pairs
and
creates an alleyway.

So what I do is refurbish the Chebychev and consider not solo-primes
but those Composite numbers which have two-primes such as 3 x 5 = 15

And I look at a generalized building as a 2M-Partition

0 M
1 M-1
2 M-2
3 M-3
. .
.
.

If I used the basic Chebychev I know only that there is at least on
prime
on the leftside and one Chebychev prime on the rightside, but the
basic
Chebychev cannot guarantee that the prime will pair and make an
alleyway
in the general even number 2M-Partition.

So I bring in the Refurbished Chebychev. I do not need to know there
is a
prime on the leftside. I do not care. Because the Refurbished
Chebychev
is not concerned with "solo primes". It is concerned with Composites
of
two-primes (where 2 is not in the work). And this Refurbished
Chebychev
is as good as the basic Chebychev because both operate on x/Ln(x).
They are like a ruler that has been picked up and moved to make a
different
measure, but the ruler is still the same with the same ratcheted marks
telling
the distance.

So in this analogy of a long building, I no longer care about the
prime on the
leftside of the building. I have the Refurbished Chebychev at work
here. I simply
know from the Refurbished Chebychev that there must exist a Two-Primed-
Composite
within that Partition, whether the partition is 8 or 10 or 12 or 2M.
The Refurbished Chebychev tells me that a Two Primed Composite must
exist in that
partition, and for the case of 8-Partition it is 3 x 5 = 15.

In the case of any even number beyond 4, that the Refurbished
Chebychev guarantees
at least one Two-Primed-Composite in any given Even Numbered
Partition.

Thus, the reader is witnessing the proof of the Goldbach Conjecture,
both Strong and
Weak forms.

[plutonium....@gmail.com]

I never intended to find nor offer a proof of the Goldbach Conjecture
as part of this book
and in the future will probably write a separate book on the Goldbach
Conjecture proof.

Why did I take this detour of Goldbach? Because if the Natural Numbers
as per Peano
Axioms are really 0, 1, 2, 3, ...., 99999....9999 and if the Goldbach
Conjecture is really
true then there should not be counterexamples within the Natural
Numbers from
0 to 9999....9999. So I spent a few moments trying to think of a
counterexample, and
could not. And the largest Goldbach covering would be for 9999...99998
which is
satisfied by 9999...99991 + 00000....000007

And since I could not find a counterexample, I sort of remembered
where I last
left off on Goldbach some 2 or 3 years ago, a snag in the Chebychev
Theorem.
And would you know it, a new attempt becomes successful. Most math
proofs
are some sort of "trick in logic". The trick for Infinitude of Primes
is to construct
"multiply the lot and add 1". The trick for Goldbach is to Recalibrate
the Chebychev
Theorem in that there is a Two-Primed-Composite between N and 2N.

So it appears that the concept of Prime extends throughout the
Naturals as
from 0 to 9999...9999. The concept of Even and Odd extends through the
entire
Naturals from 0 to 9999...9999 and now it seems as though the concept
of
Prime extends throughout and that 9999...99997 is the world's largest
prime number.

But does the Riemann Hypothesis hold true for the Naturals from 0 to
9999....9999

I believe the answer is no, that the Riemann Hypothesis is false.

Now that would not conflict with the Goldbach Conjecture. It may
conflict with
the Chebychev Theorem which is dependent on x/Ln(x). That formula on
the
distribution of primes may be false in general. But the point of logic
is that there
is a formula, even though it is not that of x/Ln(x) in general.

I could not find counterexamples for Goldbach in the Infinite Integers
and so that
lead to a "normal proof".

But I believe the Riemann Hypothesis has boatloads of Counterexamples.
With numbers like the 10-adic Idempotents or just plain old numbers
such as

99999....999999 or 50000....000000

The reasoning for the counterexamples is that the Riemann Hypothesis
goes
to infinity and that is in Euclidean Geometry, but in Euclidean
Geometry we
cannot go to infinity since the leftward string of a Real Number is
always a FINITE
string. To go to infinity then you have to have Elliptic Geometry
where the
leftward string goes to infinity from 0 to 9999....999999. So can we
put the Riemann
Hypothesis on a sphere surface and expect a 1/2 Real line to go to
infinity? Of
course not because no lines remain straight and uncurved and unbent in
Elliptic Geometry. So the question for the Riemann Hypothesis becomes
at what
number is the Riemann Zeta Function obviously bending and curving? Is
it bending
and curving at about 10^500 where the end of Physics measuring occurs?
I believe
so.

In the 1990s I offered a proof of the Riemann Hypothesis using the
Rectangle of
Whirling Squares.

The Fibonacci sequence of 1, 1, 2, 3, 5, 8, 13, 21, ...... is a model
of the
Riemann Hypothesis, where if the RH is true then the logarithmic
spiral always
remains enclosed inside the Rectangle of Whirling Squares.

If the Riemann Hypothesis is false, then the logarithmic spiral gets
out of the
Whirling Squares.

So, the intelligent logical question to ask, is whether the Rectangle
of Whirling
Squares when placed on a sphere surface, then at what point is it
obvious that
the logarithmic spiral trespasses outside of the Whirling Squares? Now
that is
hard to picture because the surface of a sphere does not have squares,
much
less whirling squares.

I am going to show the counterexamples of the Riemann Hypothesis.

I believe what is going to happen is that we have a convergence of
Physics
with Pure-Mathematics in that the Riemann Hypothesis breaks down in
a big way at about 10^500. The 10-adic Idempotents are obvious
counterexamples.

[David R Tribble]

Archimedes Plutonium wrote:
> Here are the Peano Axioms for the Natural Numbers as given by
> Wikipedia:

> 6. For every natural number n, S(n) is a natural number.
>

> [...]

> So because Peano failed to define Finite from Infinite, his above 9
> axioms delivers the AP-adic Integers or the Infinite Integers. They obey the
> Peano axioms (except for maybe 9999...9999 since it has no successor)

Maybe? Either it obeys the axioms or it doesn't. Which is it?

If 999...999 has no successor, then it does not obey axiom 6,
therefore it is not a natural.

[angryb...@gmail.com]

It's even better than that:

If 999...999 has no successor, then 999...999 is not a natural number.
If 999...999 is not a natural number, then a number whose successor is
999...999 is also not a natural number.
More generally, if S(n) is not a natural number, then n is not a
natural number.

Therefore, none of Archie's proposed infinite numbers are natural
numbers.

-o

[LudovicoVan]

Isn't, for (A)P-adics, 0 the successor of 999...999? Then it would
rather be axiom 7 that is not satisfied:

"7. For every natural number n, S(n) ≠ 0. That is, there is no natural
number whose successor is 0."

Then I can glimpse some problems, where 0, the smallest (in value)
element, is not a least element to found a well-order...

-LV

[plutonium....@gmail.com]

Yes, thanks LV for clarifying to Tribble.

I would have replied that in any axiom system such as Peano or the
Euclidean
for Geometry, that in any axiom system you have what I consider the
humdrum
low priority axioms which are "adjustable" and then you have the pride
of the
pack axiom, which is like the engine of a airplane or the engine of a
moving
vehicle. The other axioms are the frame and body whereas the most
interesting
axiom is the engine or motor that makes it all come together and do
its purpose.

So for the Euclidean Geometry, the motor axiom is the parallel line
axiom, the one
in which you can get the NonEuclidean Geometries.

For Peano Axiom, the only real essential one is the endless adding of
1 which is
the motor of that axiom system.

The body or framework axioms are easily adjustable.

And as I pointed out of the flaws of Peano is that he needed to call
into existence
two initial Counting Numbers of both 0 and 1, not just "0" alone.
Because you
need a "ruler" for the Natural Numbers and no ruler is possible
without two ratchet
marks to give a distance.

But the "motor axiom" of Peano is the Successor axiom of endless
adding of 1.
It is this axiom that makes 1001 a Natural Number and also
9999...99991
a Natural Number.

LV, the way I set up the AP-adics is that 9999....99999 is a longitude
that
is one unit distance short of the South Pole but in Hensel P-adics
9999....9999
is -1 and one unit short of 0. So it looks as though in P-adics that 0
does have
a predecessor but not in AP-adics. And in AP-adics the successor of
999...9999
is the South Pole, an imaginary-AP-adic which I think is pi or maybe
"e" or maybe
something else.

But tinkering with the framework-axioms or body-axioms is minor work,
or work
of no real importance. The importance of the Peano axioms or the
Euclidean
Axioms is the Motor-axiom. What drives them is important. We can
constantly
make adjustments to the peripheral axioms but we need to keep our eyes
focused
on the motor axiom.

[LudovicoVan]

On 29 Jan, 21:07, plutonium.archime...@gmail.com wrote:
> LudovicoVan wrote:
> > On 29 Jan, 17:16, David R Tribble <da...@tribble.com> wrote:
> > > Archimedes Plutonium wrote:
> > > > Here are the Peano Axioms for the Natural Numbers as given by
> > > > Wikipedia:
> > > > 6. For every natural number n, S(n) is a natural number.
>
> > > > [...]
> > > > So because Peano failed to define Finite from Infinite, his above 9
> > > > axioms delivers the AP-adic Integers or the Infinite Integers. They obey the
> > > > Peano axioms (except for maybe 9999...9999 since it has no successor)
>
> > > Maybe? Either it obeys the axioms or it doesn't. Which is it?
>
> > > If 999...999 has no successor, then it does not obey axiom 6,
> > > therefore it is not a natural.
>
> > Isn't, for (A)P-adics, 0 the successor of 999...999? Then it would
> > rather be axiom 7 that is not satisfied:
>
> > "7. For every natural number n, S(n) ≠ 0. That is, there is no natural
> > number whose successor is 0."
>
> > Then I can glimpse some problems, where 0, the smallest (in value)
> > element, is not a least element to found a well-order...
>

> Yes, thanks LV for clarifying to Tribble.

To be true, I didn't mean to clarify anything to Mr tribble :). The
question mark was not rethorical.

> And as I pointed out of the flaws of Peano is that he needed to call
> into existence
> two initial Counting Numbers of both 0 and 1, not just "0" alone.
> Because you
> need a "ruler" for the Natural Numbers and no ruler is possible
> without two ratchet
> marks to give a distance.

At a merely intuitive level, I can glimpse what you are talking about,
as I have encountered something similar in my own wanderings through
number theory. The problem is that at the moment I cannot even
remember where that was, so I am afraid I cannot be any more precise
than this. Anyway this might come back to my mind later.

> LV, the way I set up the AP-adics is that 9999....99999 is a longitude
> that
> is one unit distance short of the South Pole but in Hensel P-adics
> 9999....9999
> is -1 and one unit short of 0.

Yes, that is what I was talking about: the P-adics are "cyclic".

> So it looks as though in P-adics that 0
> does have
> a predecessor but not in AP-adics. And in AP-adics the successor of
> 999...9999
> is the South Pole, an imaginary-AP-adic which I think is pi or maybe
> "e" or maybe
> something else.

What about an "omega"? Why would you think of one of the known
irrationals?

> But tinkering with the framework-axioms or body-axioms is minor work,
> or work
> of no real importance. The importance of the Peano axioms or the
> Euclidean
> Axioms is the Motor-axiom. What drives them is important. We can
> constantly
> make adjustments to the peripheral axioms but we need to keep our eyes
> focused
> on the motor axiom.

I mostly agree, I do believe in playing with different constructions.
On the other hand, and speaking in general, I do also believe in
extreme rigour (although, said like this, doesn't sound very
exciting...), that is, every construction is legitimate as long as it
is indeed a "construction": intuition and "vision" are fundamental
ingredients for sure, but, on the abstract, rigour is as important to
make the vision something really workable.

-LV

[plutonium....@gmail.com]

LudovicoVan wrote:

>
> Yes, that is what I was talking about: the P-adics are "cyclic".
>
> > So it looks as though in P-adics that 0
> > does have
> > a predecessor but not in AP-adics. And in AP-adics the successor of
> > 999...9999
> > is the South Pole, an imaginary-AP-adic which I think is pi or maybe
> > "e" or maybe
> > something else.
>
> What about an "omega"? Why would you think of one of the known
> irrationals?

Well, the concept of All Possible Digit Arrangements is the easiest
way
of building either the Reals or AP-adics. That means there is one and
only
one type of infinity. So infinity is not a number and infinity is
cheapened
by alot when the definition of Finite is 10^500 or less. So not only
did the
past history of mathematics inflate the importance of Algebra but it
inflated the importance of infinity for it means little in the face of
Finite =
10^500 or less and that All Possible Digit Arrangements makes infinity
and omega sort of childishness.

Elliptic Geometry is all positive curvature and that means all
positive
numbers. So that Zero or the North Pole must be some imaginary number
for that Elliptic Geometry cannot have zero and thus must be
substituted for.
Now what to make of the South Pole as a number? It cannot be
9999....9999
+ 1 = 10000....00000 because 9999....99999 is the last and largest
number
as the last possible digit arrangement given a preset-place-value. If
we say
that there are only two place values then 99 is the largest number
there. If three
place values is all that is possible than 999 is the largest number.
So the
South Pole and North Pole have to have imaginary appended numbers.
Same
with Hyperbolic Geometry and Negative AP-adics in that zero is not a
negative
number so it is imaginary.

I do not know what the final numbers for the North and South Pole are.
With
time I expect to resolve that.

>
> > But tinkering with the framework-axioms or body-axioms is minor work,
> > or work
> > of no real importance. The importance of the Peano axioms or the
> > Euclidean
> > Axioms is the Motor-axiom. What drives them is important. We can
> > constantly
> > make adjustments to the peripheral axioms but we need to keep our eyes
> > focused
> > on the motor axiom.
>
> I mostly agree, I do believe in playing with different constructions.
> On the other hand, and speaking in general, I do also believe in
> extreme rigour (although, said like this, doesn't sound very
> exciting...), that is, every construction is legitimate as long as it
> is indeed a "construction": intuition and "vision" are fundamental
> ingredients for sure, but, on the abstract, rigour is as important to
> make the vision something really workable.
>
> -LV

I am striving for rigor. Much of the AP-adics is not finalized. So in
this state
of having pieces of the puzzle and only partial assemblage, I have to
postpone
the rigor until near the end. Algebra is the subject in math that well
defines the
operations on numbers. So I have alot of the Algebra still cloudy and
tentative.

I do not know what to make of Irrational Integers such as
9999.....151413121110987654321
or this
99999......0000099990000999000990090
or this
00000.....9999900009999000999009909

When you are at the beginning or middle of a project, you do not
tighten the
screws and put in all the finishing nails of rigor but rather have
things so that
they can be switched or changed, and you want the natural order of
sequence
so that the final tightened of the screws and finishing boards come at
the end.

[Dan Christensen]

> axioms […]

I don't see why this is a problem.

> end up describing this set 0000....0000, 0000....00001, 0000....00002,
> 0000....00003, ..... then much further on 5555....5555,
> 5555.....55556,

Actually, it is more like 0, S(0), S(S(0))), …

> and still much further on we have 9999...99997, 9999...99998
> and finally the last and largest integer in the world as
> 99999....99999
>

99999....99999 is not a well-defined number. In PA, there is a first
number (0 here), but no last number. Informally, every number has a
successor that is not any previous number. There is no "looping back,"
if you will.

> So because Peano failed to define Finite from Infinite, his above 9
> axioms
> delivers the AP-adic Integers or the Infinite Integers. They obey the
> Peano
> axioms (except for maybe 9999...9999 since it has no successor)
>
> (2) the second error is more serious because it destroys the integers
> that
> Peano is trying to create. It is a mistake which can be called a "lack
> of
> defining a Ruler". In a previous post I said it was a Masonry or
> Carpenter
> mistake that Peano never labored on building a brick wall or
> carpenter's
> building walls. Well in order to have a Ruler, a ruler must have two
> ends.
> There are no one ended rulers in the world.

There is no notion of distance in PA.

> So in Peano's 5th axiom he says only one number is brought into
> existence
> and further on he calls up a Successor function S(0). Well, the
> problem
> and error is that Peano cannot create a Natural Number system without
> creating
> at least two numbers the 0 and then the 1, or the 1 and then the
> number 2.
> So Peano's 5th axiom should have created two new numbers and
> thenceforth

[snip]

Again, this is not necessary. PA (or the modern, set-theoretic version
of it) works perfectly well as a basis for number theory. If you can
show some inconsistency within number theory, then we would have to
rethink these axioms. Otherwise, probably not.

Dan
Download my DC Proof software at http://www.dcproof.com

[hagman]

Helloooo???
Since the AP-adic integer 9999...9999 has no successor we immediately
see that this is *NOT* what Peano defined.
In a set of natureal numbers as described by the Peano axioms
*EVERY* natural number n has a successor.
Otherwise you might just as well claim that the set of naturals is
{0,1,2,3} because all axioms are fulfilled except that 3 has no
successor.

>
> (2) the second error is more serious because it destroys the integers
> that
> Peano is trying to create. It is a mistake which can be called a "lack
> of
> defining a Ruler". In a previous post I said it was a Masonry or
> Carpenter
> mistake that Peano never labored on building a brick wall or
> carpenter's
> building walls. Well in order to have a Ruler, a ruler must have two
> ends.
> There are no one ended rulers in the world.
> So in Peano's 5th axiom he says only one number is brought into
> existence
> and further on he calls up a Successor function S(0). Well, the
> problem
> and error is that Peano cannot create a Natural Number system without
> creating
> at least two numbers the 0 and then the 1, or the 1 and then the
> number 2.
> So Peano's 5th axiom should have created two new numbers and
> thenceforth
> Peano could create the Successor Function of N+1 because the "+1" was
> created in the 5th axiom when creating two numbers.

+1 is only defined later (i.e. both '+' and '1' are defined later).
We define 1 as the natural number S(0) the existence and uniqueness of
which
follows from the axioms.
We define n+0=n and n+S(m) = S(n+m) - which needs the concept of
recursive
definition, which I'll skip.
It is then a *theorem* that S(n) = n+1 for all n.
Proof:
n + 1
= n + S(0) by definition of 1
= S(n+0) by recursive definition of addition, i.e. use n+S(m)=S(n
+m) with m=0
= S(n) by the definition of +

>
> As it stands Peano created a one-ended-ruler, and there is no such
> thing. Peano
> needed to create two new numbers for the Natural Numbers in order to
> make a
> Metric, a Parameter of the Natural Numbers.
>
> As it stands at the moment with Peano's 5th axiom where he creates
> only one
> Natural Number, then this set are the Natural Numbers:
>
> 0, 1/2, 1, 3/2, 2, 5/2, etc etc

WIth a properly defined successor function, yes, this is a model
of the Peano axioms, e.g. 0 is 0 and S is x|->x+1/2

>
> or this set is the Set of Peano Natural Numbers:
>
> 0, 1/3, 2/3, 1, 4/3, 5/3, 2, etc etc
>
> Now some may say that those reduce to this set 0,1,2,3,4, etc etc
> but that is not true because when Peano creates only a 0 or 1 but not
> two new
> numbers depends on his Induction Axiom of Successor, in that the N, N
> +1
> is not served by N, N+1/2, nor served by N, N+1/3
>
> The only way to avoid that error of Peano was by creating two new
> Natural Numbers
> and thus creating a "Ruler" in the 5th axiom.
>
> P.S. the way to solve Peano's error of "finite versus infinite" is to
> introduce a 10th
> axiom that basically says that "Finite" is synonymous with physics
> measurement.

The error was on your side.
The term finite is not defined by the axioms but we may define
the property.

[David R Tribble]

Archimedes Plutonium wrote:
> Well, the concept of All Possible Digit Arrangements is the easiest
> way of building either the Reals or AP-adics. That means there is one and
> only one type of infinity. So infinity is not a number and infinity is
> cheapened by alot when the definition of Finite is 10^500 or less. So not only
> did the past history of mathematics inflate the importance of Algebra but it
> inflated the importance of infinity for it means little in the face of
> Finite = 10^500 or less and that All Possible Digit Arrangements makes infinity
> and omega sort of childishness.

Archimedes Plutonium wrote [earlier]:
>> So we define Reals as their leftward finite portion goes to 10^500 and
>> we define the Natural Numbers from 0 to 10^500 as Finite Integers and beyond
>> that are Infinite Integers
>

David R Tribble wrote:
>> So 10^500+1 is infinite?
>

Archimedes Plutonium wrote:
> Yes, for none of us can count to 10^500 nor is anything "Physically
> Measureable" at 10^500 or 10^(-)500
> So that 10^500 +1 is just as out of reach as is 55555....55555
>
> I have looked for a definition of "Finite" purely from a mathematical
> view, and there is none, [...]

Obviously you have not read enough math books then.
"Finite" and "infinite" are defined in several ways in mathematics.

In set theory, a finite set is any set that can be placed in a
one-to-one correspondence (bijected with) a natural.
An infinite set can't.

A Dedekind-infinite set can be bijected with one or more
of its proper subsets. A Dedekind-finite set can't.

In analysis, an infinite limit is defined by ever-decreasing
epsilon/delta pairs.

And so forth.

Your arbitrary choice of 10^500 as the divding line between finite
and infinite has no rigorous mathematical basis.

[plutonium....@gmail.com]

David R Tribble wrote:

>
> Archimedes Plutonium wrote:
> > Yes, for none of us can count to 10^500 nor is anything "Physically
> > Measureable" at 10^500 or 10^(-)500
> > So that 10^500 +1 is just as out of reach as is 55555....55555
> >
> > I have looked for a definition of "Finite" purely from a mathematical
> > view, and there is none, [...]
>
> Obviously you have not read enough math books then.
> "Finite" and "infinite" are defined in several ways in mathematics.
>

Obviously you have not thought long enough but are leaning on your
assumptions.

> In set theory, a finite set is any set that can be placed in a
> one-to-one correspondence (bijected with) a natural.
> An infinite set can't.
>

You need to go back to whether Naturals distinguish themselves apart
whether 00091111....1111 is finite since it ends in zeroes to the
left.
You never did this, nor did Peano and that is why his axioms are
defunct and incomplete. You admit that 00000.....911 is finite. You
admit
this only because you were spoon fed this in school, but you never
were able to tell whether 00000000091111......1111 was finite or
infinite
because noone spoon fed you on that.

And in your list here, you forgot to mention the definition that your
mind
works with. Your mind's definition of "finite" is that the leftward
string
ends in zeroes such as 00000....00005 and you conveniently write 5
and Peano thinks of it as 5. But what about 0000050000......000000
It ends leftward in zeroes also and is a number in All Possible Digit
Arrangements. And none of the definitions of finite is able to handle
that
number, nor Peano Axioms is dumbfounded and floored with that number.

> A Dedekind-infinite set can be bijected with one or more
> of its proper subsets. A Dedekind-finite set can't.
>

What do you say to the bijection of the Reals between 0 and 1
and the AP-adic Integers where you take any of those Reals and
flip it over so that 0d3333.....3333 is 3333....3333r (d for decimal
point and r for radix) So here we have a bijection, yet the math
community is unwilling to understand that the infinity of Reals
is the same infinity as the Counting Numbers as AP-adic Integers.

> In analysis, an infinite limit is defined by ever-decreasing
> epsilon/delta pairs.
>

When the Reals are built out of All Possible Digit Arrangements, we
do not even need the Limit concept and see it as a silly contraption.
Like asking Olympic atheletes in the 100 meter dash to use
crutches in their performance.

> And so forth.
>
> Your arbitrary choice of 10^500 as the divding line between finite
> and infinite has no rigorous mathematical basis.

Peano never defined Finite from Infinite and so his axioms ends up
delivering this set of numbers:

0, 1, 2, 3, 4, ....., 55555....55555, .... , 7373737373....737373, and
all the
way up to 9999....999997, 9999....999998, 99999....999999

Because Peano never defined finite from infinite, and that the math
community
never bothered nor recognized the problem, they thus, still, have
pseudo problems
such as the Fermat's Last Theorem or Riemann Hypothesis since they do
not
know that Natural Numbers as given by Peano is this set:
0,1,2,..... , 99999....99999

Once the math community defines finite from infinite then all those
Number theory
problems melt away.

[plutonium....@gmail.com]

hagman wrote:

>
> Helloooo???
> Since the AP-adic integer 9999...9999 has no successor we immediately
> see that this is *NOT* what Peano defined.
> In a set of natureal numbers as described by the Peano axioms
> *EVERY* natural number n has a successor.
> Otherwise you might just as well claim that the set of naturals is
> {0,1,2,3} because all axioms are fulfilled except that 3 has no
> successor.
>

The successor or 9999....99999 is the South Pole, of one unit distance
from
9999...99999.

In Elliptic Geometry the famous Euler Identity of e^(2pi x i) = 1
becomes in Elliptic Geometry that of the default equation
0 ^(0 x i) = 1 In Elliptic Geometry all the numbers have to be
positive so a Zero cannot exist. As we know in Reals sqrt -1
cannot exist and so we define it as i imaginary appended.
Well, in Elliptic Geometry the Poles exist but they cannot be
any zero number, yet they must be a zero number to satisfy
algebra. Sqrt -1 is appended to Reals to satisfy algebra
and so is Zero appended to Elliptic Geometry to satisfy
algebra.

Thus, the North Pole is not zero but appended 2pi, but analytically
or algebraically 2pi is zero as the North Pole. The South Pole is
"e" but another form of zero.

So the successor of 9999....99999 is "e". Just yesterday someone
wanted me to say that the South Pole is Omega. But omega is not
a algebraic appended need, like that of sqrt-1 or 2pi or e. Those are
actual needed numbers. Omega is merely a concept, not a number.

[plutonium....@gmail.com]

The Peano Axioms delivers this set 0,1,2,3,4, ......, 99999....99999
with its Successor axiom of endless adding of 1.

The problem is the schizophrenia of the mathematics community that
they
do not want to recognize that the Peano axioms delivers that set. They
want
to say that a finite integer is like this 0000.....00009 is equal to 9
and write
9 without its infinite string of leftward zeros. But when you have
endless adding
of 1 you end up with 99999....99999

So, now you can see the problem. And then when schizophrenic
mathematicians
define finite they usually assume it means a leftward ending in zeros
so that

0000.....00000747 is the finite 747

trouble here is that what is

07474747.....747

It ends leftward in a 0 in the 10^9999....99998 place value

This number

000066666....66666 ends in zeros leftward in the 10^9999....99995
place value

So according to Peano or Christensen that the number 00006666...6666
is finite as is 000000.....000066 is finite 66

The problem with the Peano Axioms is that they fail to provide the
difference
between what it means to be finite versus infinite and the way to
solve that
is for a tenth axiom that simply says Physics gives out at 10^500 and
so
10^500 +1 is infinity and below is finite.

That makes sense, because Physics is the Master science and math is
but the servant science. So where Physics is unable to measure is as
good as infinity. Where Physics is able to go in and measure is
finite.

>
> > end up describing this set 0000....0000, 0000....00001, 0000....00002,
> > 0000....00003, ..... then much further on 5555....5555,
> > 5555.....55556,
>
> Actually, it is more like 0, S(0), S(S(0))), …
>
>
> > and still much further on we have 9999...99997, 9999...99998
> > and finally the last and largest integer in the world as
> > 99999....99999
> >
>
> 99999....99999 is not a well-defined number. In PA, there is a first
> number (0 here), but no last number. Informally, every number has a
> successor that is not any previous number. There is no "looping back,"
> if you will.
>

You made several mistakes there. In All Possible Digit Arrangements,
9999...999 is just as valid as 0000....0009 or 0000....00001 or even
these bizarre combinations such as 9999....151413121110987654321

So in All Possible Digit Arrangements the numbers are automatically
defined
and thrust upon us. So a number in mathematics is not left up to some
error prone human to want to decide "well, I like that number but not
that one
and so I will accept and reject willy nilly"

The Hensel P-adics loop around. The AP-adics also loop around once you
factor in the other hemisphere.

It is still somewhat a mystery to me as to why mathematics has the
characteristic
that as the numbers grow larger that the numbers have a natural bend
and arc
and they come back around. My justification is that Physics is king
and the Universe
is just a one big atom which is Elliptical and so big numbers have to
be Elliptic geometry.

>
> > So because Peano failed to define Finite from Infinite, his above 9
> > axioms
> > delivers the AP-adic Integers or the Infinite Integers. They obey the
> > Peano
> > axioms (except for maybe 9999...9999 since it has no successor)
> >
> > (2) the second error is more serious because it destroys the integers
> > that
> > Peano is trying to create. It is a mistake which can be called a "lack
> > of
> > defining a Ruler". In a previous post I said it was a Masonry or
> > Carpenter
> > mistake that Peano never labored on building a brick wall or
> > carpenter's
> > building walls. Well in order to have a Ruler, a ruler must have two
> > ends.
> > There are no one ended rulers in the world.
>
> There is no notion of distance in PA.
>

Wrong, the Successor axiom is a distance axiom, whether you want to
call it adding one more unit distance to a prior number, or whether
you want
to call it successor. It means a distance function for the successor
has
embodied the addition operation of an algebra. Call it the Unit Vector
of Addition.

>
> > So in Peano's 5th axiom he says only one number is brought into
> > existence
> > and further on he calls up a Successor function S(0). Well, the
> > problem
> > and error is that Peano cannot create a Natural Number system without
> > creating
> > at least two numbers the 0 and then the 1, or the 1 and then the
> > number 2.
> > So Peano's 5th axiom should have created two new numbers and
> > thenceforth
>
> [snip]
>
> Again, this is not necessary. PA (or the modern, set-theoretic version
> of it) works perfectly well as a basis for number theory. If you can
> show some inconsistency within number theory, then we would have to
> rethink these axioms. Otherwise, probably not.
>
> Dan

Well the error of the Peano axioms with finite and infinite is easily
seen in
the fact that Number theory conjectures piled up and are still
unprovable some
thousands of years later, not because they are tough problems but
because
the Counting Numbers are not what Peano axioms deliver.

As for a problem in the error of Peano not creating or saying that two
numbers
exist of 0 and 1 at the outset. Well the problem here is that the
Peano Axioms
cannot tell whether 0, 1/2, 1, 3/2, 2, ....... are the Counting
Numbers.

Your Peano Axioms is unable to tell whether the Counting Numbers are
0, 1/2, 1, 3/2, 2, .......

or are

0,1,2,3,.......

Every ruler that is going to measure, say a meter, must have two marks
on the
ruler to tell you the start of the meter distance and then the
endpoint of the meter
distance. No ruler exists that has one mark. So the Peano Axioms that
calls
into existence just the number 0 and not 0 and 1 is not a ruler. So
when you
get to the Successor Axiom with its endless adding of 1, well, you
never built
1 into the system in the first place and so you cannot have a
consistent successor.

With only the existence of 0 given and no other "measure" to the next
Natural Number
you have no ruler in the Natural Numbers and so this set, according to
Peano
Axioms:

0, 1/10, 2/10, 3/10, ...... is the set of Natural Numbers.

If Peano had created into existence both 0 and 1, he would have
eliminated all sets
except for:

0,1,2,3,4,5,........

Archimedes Plutonium

[Jan Burse]

> trouble here is that what is
>
> 07474747.....747

What you are defining are not numbers, they are strings.
Thus mappings from M to {0,..,9}, i.e. from an index
to a digit where M is either a finite set or an infinite
sets.

Digit strings are okay, they exist in set theory for
example. Its also possible to define number like
operations on them. Assume the most right index equals
to 0. So we can go on and define addition (a+b) of
two strings a and b as follows:

(a+b)[i]:=(a[i]+b[i]+c[i]) mod 10,
c[i]:=(a[i]+b[i]+c[i]) div 10.

c is the carry string.

It can be easily seen that for this addition we have:

x+0 = x
x+(y+z) = (x+y)+z

Multiplication is also possible to define, namely
we can define:

(a*b)[i]=(sum_j+k=i (a[j]+b[k])+c[i]) mod 10,
c[i]=(sum_j+k=i (a[j]+b[k])+c[i]) div 10.

c is again the carry string.

(If the string is truely infinite, then the carry
will grow and grow, but this doesn't matter, each
digit in the infinite resulting string is defined)
It can also be seen that:

x*0=0
x*1=x
x*(y*z)=(x*y)*z
x*(y+z)=x*y+x*z

Via S(x):=x+1 we can relate them to the peano axioms.
So to some extend they are a non-standard model of
peano. It is an interesting question to see which
axiom of peano is capable to exclude them.

Actually in peano each subset should have a minimal
element. This is equivalent to the induction axiom,
see for example Machovers book on Set Theory.

Lets take the following subset of infinite strings
as a counter example for the claim that infinite
strings are a model of peano.

s0 ...222221
s1 ...222212
s2 ...222122
...

Or more generally:

sk[i] := if k=i then 1 else 2

According to the infinite string arithmetic s1 < s0,
because s1 + 9 = s0. Also s2 < s1, because s2 + 90 = s1,
and so on. Thus we have:

.. < s2 < s1 < s0

Therefore we have a subset without a minimal element,
and therefore infinite strings ARE NOT (NEVER EVER)
a model of peano arithmetic.

Sorry

Bye

[David R Tribble]

Archimedes Plutonium wrote:
>> trouble here is that what is
>> 07474747.....747
>

Jan Burse wrote:
> What you are defining are not numbers, they are strings.
> Thus mappings from M to {0,..,9}, i.e. from an index
> to a digit where M is either a finite set or an infinite
> sets.

> [...]

> Via S(x):=x+1 we can relate them to the peano axioms.
> So to some extend they are a non-standard model of
> peano. It is an interesting question to see which
> axiom of peano is capable to exclude them.

> [...]

> Therefore we have a subset without a minimal element,
> and therefore infinite strings ARE NOT (NEVER EVER)
> a model of peano arithmetic.
> Sorry

It's worse than that.

AP's "infinite integers" are something different from simple infinite
strings. At least with your infinite strings extension we have
something
resembling arithmetic numbers. AP's "infinite integers" don't even get
that far.

For example, he defines the number 012...789 as having a "Front View"
composed of the digits '012', and a "Back View" composed of the digits
'789', and an "infinite number of digits" in between the two, whatever
that means.

He refuses to define exactly what the '...' is supposed to mean beyond
the vacuous explanation of "an infinite number of digits", offering no
mathematically sound basis or explanation of it. According to his
beliefs,
"you get such a number from the Peano Axioms by continuously adding
1 to 0", which is quite obviously impossible.

[Jan Burse]

David R Tribble schrieb:

> For example, he defines the number 012...789 as having a "Front View"
> composed of the digits '012', and a "Back View" composed of the digits
> '789', and an "infinite number of digits" in between the two, whatever
> that means.

Ok, obviously there more infinite strings possible in set
theory as those that I have just defined. I defined them
as a mapping of

M -> {0,..,9}

Where M was finite or the infinite omega. But M can
be any ordinal. For example omega+omega. Then we
would have strings like:

s = ... 111 ... 222

Where:

s[<i,0>]=2
s[<i,1>]=1

(I am coding omega+omega by omega x 2)

Addition is not so much a problem to define. But multiplication
is not straight forward. But we already get a contradiction
with addition, my example is sufficient, just prefix by 0,
i.e.:

...00 ..221
...00 ..212
...00 ..122
...

We might further discuss n+omega instead omega+omage as
index set, or n+n etc.. But this doesn't give anything
new.

Bye

[plutonium....@gmail.com]

Jan Burse wrote:
> > trouble here is that what is
> >
> > 07474747.....747
>
> What you are defining are not numbers, they are strings.
> Thus mappings from M to {0,..,9}, i.e. from an index
> to a digit where M is either a finite set or an infinite
> sets.
>

Post of Jan Burse is so full of mistakes and bs that I do not
have the time to correct them.

Karl Heuer said in 1993, where even Jan Burse can respect
what Karl Heuer says, even though Jan knows next to nothing
about math.

But Karl defined the Reals and P-adics in 1993, or was it 1994 where
he defined the Reals as leftward finite strings with an infinite
rightward
string. And the P-adics as opposite with a radix instead of decimal.

So I mean, here is Karl Heuer, respected throughout mathematics,
defining the Reals and P-adics as strings.

And here we have what, Jan Burse, some clumsy, opinionated person
who should have never been given a degree in mathematics.

He can't even follow the logic of a conversation. Where the logic at
this
moment is where does the Peano Axioms say that
00000.....00009 is a Counting Number yet 90000.....00000 is not.

So, Jan, as miserable as you are with mathematics, could you ever set
your mind to telling us how the Peano Axioms says that 0000....0009
is a Natural Number yet 90000....0000 is not? I doubt it, you probably
will
ramble on about some irrelevant thing about set theory.

Tribble has become irritated because I insult him. Well, I insult
anyone who
cannot think logically and enters a discussion with illogic.

Burse, be thankful that I snipped your remainder of opinionated crap.

So answer the question, --

How does your beloved Peano Axioms tell you that 0000...00009 is a
Natural Number and not 90000....0000. Do you have at least one
stitch of logic to answer that. You probably cannot answer that
because, well, your defeated, your just full of opinion and no logic.

If you cannot show how the Peano Axioms accepts 0000....0009
but turns away 90000....00000 then your are defeated.

P.S. by the way, any of these posts where I get off track of math and
have
to point out the illogic of responders will not be in
the publish form of this book. But sadly that is what most readers are
attracted to. And most readers can pick up and understand the larger
story from these clashes between the highly logical myself versus the
opinionated rabble-rouser who lack logic.

[plutonium....@gmail.com]

David R Tribble wrote:

>
> It's worse than that.
>
> AP's "infinite integers" are something different from simple infinite
> strings. At least with your infinite strings extension we have
> something
> resembling arithmetic numbers. AP's "infinite integers" don't even get
> that far.
>

The old way of doing math. Everything new is extensions of old.
Imagine
if Quantum Mechanics had to be some extension of Newtonian Mechanics.

And we see that why Physics is the king of science whereas math is a
tiny
subset of physics. Those in mathematics have never really cleaned out
there
dirty house where the floors and foundation have cracked and rotted
away,
yet they still build extensions here and there.

They can never admit that Peano Axioms are wrong, and that they need
total replacement. But they easily accept any old ivory tower crackpot
who offers some extension of the failed Peano Axioms.

Why noone in mathematics ever raised his/her voice and said, what is
the
Validity of having Extensions? If in physics we had extensions to
other things
pretty soon those extensions and the entire building is going unstable
and
falls due to gravity. But there in mathematics, they just keep
plugging along
with ever more extensions and never question that their foundation is
false
or cracked.

> For example, he defines the number 012...789 as having a "Front View"
> composed of the digits '012', and a "Back View" composed of the digits
> '789', and an "infinite number of digits" in between the two, whatever
> that means.
>

One of the reasons I insult you is because you keep making up these
liarings
about my definitions and progress. I said numerous times that the
"0" in 012....789 is in the 10^9999....99999 place value and that the
"1"
is in the 10^9999....9998 place value.

And that whereever you see the "infinitculum as these four dots .... "
That they mean the rest of the 10^999.....999999 place values for
which
the digits are not shown.

> He refuses to define exactly what the '...' is supposed to mean beyond
> the vacuous explanation of "an infinite number of digits", offering no
> mathematically sound basis or explanation of it. According to his
> beliefs,

I defined it many times and frequently posted what the "infinitculum
of
four dots means.

This is one of the reasons I insult you because you keep liaring over
definitions
and concepts that I have made clear yet you pretend as if I had not.
So a
liar deserves to be insulted.

> "you get such a number from the Peano Axioms by continuously adding
> 1 to 0", which is quite obviously impossible.

Then show me and everyone else, including Jan Burse who minutes ago I
put the same question to him.

Show me how the Peano Axioms tells you that 00000.....000008 is a
Natural Number which you conveniently abbreviate to 8, but that the
Peano
Axioms do not include 800000.....00000. Show me how the endless adding
of
1 to 0 gets you 8 but never gets you 800000.....000000.

So the burden is on you and when you say "quite obviously impossible"
is because
you are not intelligent enough to understand. It has been your track
record to never
directly reply to question put to you.

So how does Peano Axioms tell you that 0000....0008 is a Natural
Number
but that 80000.....00000 is not??

Your answer needs more than your usual Tribbling of "quite obviously
impossible"

[Jan Burse]

plutonium....@gmail.com schrieb:

> So how does Peano Axioms tell you that 0000....0008 is a Natural
> Number
> but that 80000.....00000 is not??

Actually the peano axiom system is not about the decision
whether individual objects are numbers or not. It is a
predication of a whole object system, whether this object
system corresponds to what peano had in mind as a number
system.

That another object system is not a peano number system
is a predication that is neither evil nor good. Its just
an observation in the current case, that depending on
what your "...." mean, that you might end up in an object
system that is not a peano system.

And nobody, at least not me, does dispute the fact, that
one can find object systems with arithmetic properties
that are not peano systems. Mathematical logic does not
impose any restrictions on the theory that one wants to
deal with.

Feel free to work out your alternative object system, but
don't be unfair to those that correctly point out that
your object system is not a peano system.

Bye

[Jan Burse]

plutonium....@gmail.com schrieb:

> And here we have what, Jan Burse, some clumsy, opinionated person
> who should have never been given a degree in mathematics.

Correctly speaking I don't have a degree in mathematics.
I come from a different inheritance.

Bye

[plutonium....@gmail.com]

Jan Burse wrote:
> plutonium....@gmail.com schrieb:
> > So how does Peano Axioms tell you that 0000....0008 is a Natural
> > Number
> > but that 80000.....00000 is not??
>
> Actually the peano axiom system is not about the decision
> whether individual objects are numbers or not. It is a
> predication of a whole object system, whether this object
> system corresponds to what peano had in mind as a number
> system.

Since you are not a mathematician, you really should not be in this
thread, talking, you should be only listening to what is said. For
almost
everything that comes out of your mouth is wrong.

Here are the Peano Axioms as listed by Wikipedia:

The Peano axioms define the properties of natural numbers, usually

represented as a set N. The first four axioms describe the equality
relation.[6]

1. For every natural number x, x = x. That is, equality is
reflexive.
2. For all natural numbers x and y, if x = y, then y = x. That is,
equality is symmetric.
3. For all natural numbers x, y and z, if x = y and y = z, then x =
z. That is, equality is transitive.
4. For all a and b, if a is a natural number and a = b, then b is
also a natural number. That is, the natural numbers are closed under
equality.

The remaining axioms define the properties of the natural numbers. The
constant 0 is assumed to be a natural number, and the naturals are
assumed to be closed under a "successor" function S.

5. 0 is a natural number.
6. For every natural number n, S(n) is a natural number.

Peano's original formulation of the axioms used 1 instead of 0 as the
"first" natural number. This choice is arbitrary, as axiom 5 does not
endow the constant 0 with any additional properties. However, because
0 is the additive identity in arithmetic, most modern formulations of
the Peano axioms start from 0. Axioms 5 and 6 define a unary
representation of the natural numbers: the number 1 is S(0), 2 is S(S
(0)) (= S(1)), and, in general, any natural number n is Sn(0). The
next two axioms define the properties of this representation.

7. For every natural number n, S(n) does not equal 0. That is,

there is no natural number whose successor is 0.
8. For all natural numbers m and n, if S(m) = S(n), then m = n.
That is, S is an injection.

These two axioms together imply that the set of natural numbers is
infinite, because it contains at least the infinite subset { 0, S(0), S
(S(0)), … }, each element of which differs from the rest. The final
axiom, sometimes called the axiom of induction, is a method of
reasoning about all natural numbers; it is the only second order
axiom.

9. If K is a set such that:

(i) 0 is in K, and
(ii) for every natural number n, if n is in K, then S(n) is

in K,
then K contains every natural number.

And what Jan Burse should have done when asked why he thought that
0000....00009 was a number in Peano Axioms, specifically 9, and why
900000.....00000 would not be a Peano Axiom number, is to list the
Peano axioms and go through them one by one to see where any of those
axioms excludes 90000....00000 yet retains 0000....00009

>
> That another object system is not a peano number system
> is a predication that is neither evil nor good. Its just
> an observation in the current case, that depending on
> what your "...." mean, that you might end up in an object
> system that is not a peano system.
>

Oh, please do not take up the silly excuse that Tribble uses
saying he does not understand "....". I have explained this many
a times. The Finitculum of a series of two dots, if two dots means a
finite stretch of digits one is too lazy to list or for convenience
sake
or to save space. The fact that Tribble spends so much time on
dots is only an indication of how trivial he has resorted to and
because
he is defeated in the overall discussion. Tribble is like so many in
math, who can never admit themselves wrong, and pull up more excuses.

The Infinitculum of three or more dots simply means the number is too
long to list all the digits.

In the case of 00000.....0000009 it means an infinity of zeros
In the case of 90000......0000000 it means an infinity of zeros,
and where the "9" digit is in the 10^999.....99999 place-value and
where 9999.....9999999 infinitculum means infinite string of 9s.

In the case of 070000....0000060 the infinitculum means an infinity of
zeroes where the "7" is in the 10^9999....99998 place value and the
"6" is in the tens place value.

Noone can write out an infinity of digits so we use an Infinitculum
and
for you Burse or Tribble to end up basing your argument around
something
as trivial as the three or four dots, shows how far defeated you are.

So there is no more excuses remaining for you to go through one by one
of
the Peano Axioms and to say why the Peano Axioms includes 0000...0009
but not 90000....00000

> And nobody, at least not me, does dispute the fact, that
> one can find object systems with arithmetic properties
> that are not peano systems. Mathematical logic does not
> impose any restrictions on the theory that one wants to
> deal with.
>

Okay, some more yappity yap on your part, but get down to the business
that you entered this thread on-- you implied that 0000...0009 was a
Natural Number but that 90000....00000 or say 9999....99999 was not.
So tell us how you get such an opinion by actually going through the
Peano
Axioms and telling us how they separate out 0000....00009 from
90000....00000

> Feel free to work out your alternative object system, but
> don't be unfair to those that correctly point out that
> your object system is not a peano system.
>
> Bye

You pointed out nothing but your own self conceited ignorance of the
subject discussed.

So I listed the Peano Axioms again for you. Now go through them
and tell us why you think that 0000....00009 which is 9, why it is a
Natural Number but that 90000....00000 is not a Natural Number from
the Peano Axioms.

P.S. Jan, or Tribble since I pressed both to answer the above. If
neither one
of you can go through the Peano Axioms, then we all can say that both
of
you lost this argument. Both of you do not know what you are talking
about.
Both of you are in the wilderness. Both of you insist that 9000...0000
or 9999....999999 or 22222....22222 is not a Peano Natural Number, but
neither of you is willing or capable of pointing out where the Peano
axioms
actually eliminates them. The reason neither Tribble or Burse can do
that
is because those are Peano Axiom Natural Numbers.

Pretty pitiful that Tribble and Burse say that 90000.....0000 is not a
Natural
Number from Peano Axioms because of "......"

Goodness gracious, one of the most pitiful excuses I ever heard of.

At least Jan Burse admits that he does not have a College degree in
mathematics
but is Tribble also a nonmathematician? Is he a computer graduate who
dabbles
in math? Probably, because he cannot even support his allegation about
9000....0000

[David R Tribble]

David R Tribble wrote:
>> It's worse than that.

>> For example, he defines the number 012...789 as having a "Front View"
>> composed of the digits '012', and a "Back View" composed of the digits
>> '789', and an "infinite number of digits" in between the two, whatever
>> that means.
>

Archimedes Plutonium wrote:
> One of the reasons I insult you is because you keep making up these

> liarings [sic]

> about my definitions and progress. I said numerous times that the
> "0" in 012....789 is in the 10^9999....99999 place value and that the
> "1" is in the 10^9999....9998 place value.
>

> And that whereever you see the "infinitculum [sic] as these four dots .... "

> That they mean the rest of the 10^999.....999999 place values for
> which the digits are not shown.

Which is obviously a circular definition. What is the "..." in
10^999...999 supposed to mean then? Your circular definition
fails to explain anything.

But keep the insults coming. They make you look smart.

[David R Tribble]

Archimedes Plutonium wrote:
> But Karl Heuer defined the Reals and P-adics in 1993, or was it 1994 where

> he defined the Reals as leftward finite strings with an infinite
> rightward string. And the P-adics as opposite with a radix instead of decimal.

The p-adics were invented by by Kurt Hensel in 1897.
http://en.wikipedia.org/wiki/P-adic

Or did you mean something different by "P-adics"?

[David R Tribble]

Archimedes Plutonium wrote:
> How does your beloved Peano Axioms tell you that 0000...00009 is a
> Natural Number and not 90000....0000. Do you have at least one
> stitch of logic to answer that. You probably cannot answer that

> because, well, your [sic] defeated, your just full of opinion and no logic.

> If you cannot show how the Peano Axioms accepts 0000....0009

> but turns away 90000....00000 then your [sic] are defeated.

It's incumbent on you to show us why you think it's true, not the
other way around.

It's easy to show that 900...000 cannot be a Peano natural.
Starting with the natural 0, we find it successor, 1, then the
successor to 1, 2, and so on. The Peano axioms tell us that
for any finite natural, it has a finite natural successor.

Which, obviously, means that every natural is a successor
to some other finite natural, except for 0, which has is not
the successor of any natural.

Your so-called number 999...999 has an infinite sequence of
non-zero digits, so tell us at what point in finding a successor
to a finite natural (with a finite sequence of digits) do you get
a finite natural with an infinite number of digits? Because we
don't see where that happens using the Peano axioms.

> P.S. by the way, any of these posts where I get off track of math and
> have to point out the illogic of responders will not be in

> the publish [sic] form of this book. But sadly that is what most readers are

> attracted to. And most readers can pick up and understand the larger

> story from these clashes between the highly logical myself [sic] versus the
> opinionated rabble-rouser who lack [sic] logic.

Oh, look, another insult.

[plutonium....@gmail.com]

That is delightfully funny. Sometimes we have just funny and then we
have
low funny and occasionally high funny. But that is sweet delightful
funny.

I hope Karl Heuer is reading this to sip that funniness, that someone
actually thought Karl was Kurt.

You see Tribble, that is why I must insult you alot because you live
in a fantasy
and not reality. A fantasy of your own making, because everything I
say you
look for some darkness in it, something to undermine everything I say.
And it is
sad that you are in science or math for you belong in neither. A
scientist does
not have time for playing games, for psychology, or chips on the
shoulder, and believe me, you
have more than one chip on your shoulder.

Tell us where you got your degree Tribble and in what subject? I would
guess Computer
Science, where you never had any Physics or Chemistry but a few
dabbling math
courses. Probably from a big state school like Univ of Texas.

I don't expect that colleges and universities teach etiquette and
morals but they should
and the one that Tribble needs to understand is the one that says:
(A) A conversation or discussion between two persons has to be one of
respect
for the other person. If you do not respect the other person then stop
talking with them.

Not worth saving in my archive, but a pointing out of bad behaviour

[plutonium....@gmail.com]

David R Tribble wrote:
> Archimedes Plutonium wrote:
> > How does your beloved Peano Axioms tell you that 0000...00009 is a
> > Natural Number and not 90000....0000. Do you have at least one
> > stitch of logic to answer that. You probably cannot answer that
> > because, well, your [sic] defeated, your just full of opinion and no logic.
> > If you cannot show how the Peano Axioms accepts 0000....0009
> > but turns away 90000....00000 then your [sic] are defeated.
>
> It's incumbent on you to show us why you think it's true, not the
> other way around.

The number of times I showed you why 90000....00000 is a Natural
Number is more than four times. Four times directly did I explain to
you why Peano Natural Numbers is 0,1,2,...., 9999....99999

And recently I asked you how you could differentiate whether 77 was
a Natural but that 777....7777 was not Natural and you refused to give
me any reasoning other than to say you did not understand "...."

>
> It's easy to show that 900...000 cannot be a Peano natural.
> Starting with the natural 0, we find it successor, 1, then the
> successor to 1, 2, and so on. The Peano axioms tell us that
> for any finite natural, it has a finite natural successor.

Just because you say it is "easy" does not mean it is easy.
Because what Peano Axiom tells you that 1 is finite, yet
11111....11111 is not finite? There are no Peano axioms that
even mentions "finite and infinite". The Peano Axioms are silent
as to the concept of "finite" or the concept of "infinite"

So, David, you made that up above. You said "Peano axioms tell
us that for any finite natural". So David, which axiom is that
specifically?

>
> Which, obviously, means that every natural is a successor
> to some other finite natural, except for 0, which has is not
> the successor of any natural.
>
> Your so-called number 999...999 has an infinite sequence of
> non-zero digits, so tell us at what point in finding a successor
> to a finite natural (with a finite sequence of digits) do you get
> a finite natural with an infinite number of digits? Because we
> don't see where that happens using the Peano axioms.
>
>

Okay, this is at least the fifth time I went over this with Tribble
about
why the Peano Axioms goes from 0 to 9999....9999.

All of the Peano Axioms are obeyed by infinite-integers, where 0 has
no predecessor because 9999...9999 falls 1 unit short of the South
Pole
and that 9999...9999 itself has a Successor in the South Pole, and
whether
that successor is Omega or "e", whatever it is imaginary and also is
the
North Pole as imaginary of 2pi or 0.

So, none of the Peano Axioms talks about finite or infinite. So the
reader can
make up their own definition of finite or infinite.

Now as for the hardship that most readers have and which troubles
Tribble is
their minds cannot see how the Naturals transition from zeroes on the
leftward
into a number which has nonzeroes on the leftward. So how does the
Natural
Numbers go from 00000....888 into say 111111.....888 where they easily
picture 888 and call it finite but how does endless adding of one
eventually
get to 11111....888 or 22222....22222 ??

And I answer Tribble for the fifth time the same as the fourth and
third etc etc.

Look at the Peano Axioms of the Successor or Math Induction. If you
know
Math Induction, and know it well, you know it is equivalent to the
concept that
given any Natural Number N, it has a N+1 and a N-1. Those are
equivalent
formulations of the Math Induction in Peano Axioms.

So, give me any Infinite Integer say, give me 90000....00000 then its
predecessor
was 899999....999999 and its successor is 900000....00001

Or, give me 11111.....888 and its predecessor was 11111....887 and its
successor
is 11111....889

So you see Mr. David Tribble and Jan Burse. The Peano Axioms do not
give any
word of guidance as to what is finite or infinite. And the Peano
Axioms of its Math
Induction demand and force that the Natural Numbers are from 0 to 1 to
2 all the
way up to 99999....9999

And it is only the ignorance of Tribble, Burse and anyone else who
opines that
90000...0000 is infinite and different and not a Peano Natural Number.

That is why the Peano Natural Numbers need a 10th axiom which says
that Finite
is 10^500 or below and anything beyond 10^500 is not measurable or
testable by
Physics. And since Physics is King over the servant of mathematics,
that beyond
10^500 is infinite.

So, Mr, David Tribble, from your past behaviour, you will deny the
above and say some
excuse and then cause me to call you a liar and then you call me an
insultor.
But for me there will not be a sixth time for you because you will be
killfiled and that is the
end of my time on you. A fifth time on you is the last.

[Jan Burse]

plutonium....@gmail.com schrieb:

> And what Jan Burse should have done when asked why he thought that
> 0000....00009 was a number in Peano Axioms, specifically 9, and why
> 900000.....00000 would not be a Peano Axiom number, is to list the
> Peano axioms and go through them one by one to see where any of those
> axioms excludes 90000....00000 yet retains 0000....00009

For you its still Sir Jan Burse. And I am not interested in working
out your example. You can do that by yourself.

I gave another counter example for some idea of infinite string numbers.
And that was enough for me on this annoying topic, so that I can again
divert from it and delve into more interesting stuff.

Be

[angryb...@gmail.com]

You are operating under the assumption that the Peano axioms only
define one set. This is not true; they define infinitely many "Peano
sets": one for each choice of '0' and successor function. For example,
the subset of the rationals starting with 1, closed under the
successor function S(x) = x / 2, is a model of the Peano axioms. So is
the subset set of real numbers, starting with π, closed under the
successor function S(x) = x * π. So is the subset of the natural
numbers, starting with 0, closed under the successor function S(x) = x
+ 2. The natural numbers under S(x) = x + 1 are only the most
intuitive model of the axioms, and inferred definitions of addition,
subtraction, and multiplication from the Peano axioms map well to
intuitive definitions of addition, subtraction, and multiplication
over this model/

Furthermore, even if two sets satisfy the Peano axioms, each under its
own successor function, the union of the two sets does not necessarily
satisfy the Peano axioms under either successor function. To go back
to the π example, the union of the Peano set (π, F(x) = x * π) and the
Peano set (0, G(x) = x + 1) is not a Peano set under either F or G,
and elements of either set will not be generated by repeated
application of the other set's successor function to members of that
set.

So even if {0, 1, 2, ...} and {777...700, 777...701, 777...702, ...}
both satisfy the Peano axioms under S(x) = x + 1, that does not imply
that {0, 1, 2, ..., 777...700, 777...701, 777...702, ...} satisfies
them under S(x) = x + 1. You haven't shown this yet.

Some fairly easy questions:

- What is the smallest AP-adic that is leftwardly zero (000...0x)
that has as an AP-adic that is not leftwardly zero under the successor
function S(x) = x + 000...001?
- Is this the only AP-adic that is leftwardly zero whose successor
under S(x) = x + 000...001 is not leftwardly zero?
- Given S(x) = x + 000...001, what is S(999...999)? What is S(S
(999...999))? What is S(S(S(999...999)))? More generally, what is S^n
(999...999), where n > 0? (These must exist, or the set containing
999...999 is not a model of the Peano axioms under S(x) = x +
000...001 but may be a model of them under some other successor
function.)

> > Which, obviously, means that every natural is a successor
> > to some other finite natural, except for 0, which has is not
> > the successor of any natural.
>
> > Your so-called number 999...999 has an infinite sequence of
> > non-zero digits, so tell us at what point in finding a successor
> > to a finite natural (with a finite sequence of digits) do you get
> > a finite natural with an infinite number of digits? Because we
> > don't see where that happens using the Peano axioms.
>
> Okay, this is at least the fifth time I went over this with Tribble
> about
> why the Peano Axioms goes from 0 to 9999....9999.
>
> All of the Peano Axioms are obeyed by infinite-integers, where 0 has
> no predecessor because 9999...9999 falls 1 unit short of the South
> Pole
> and that 9999...9999 itself has a Successor in the South Pole, and
> whether
> that successor is Omega or "e", whatever it is imaginary and also is
> the
> North Pole as imaginary of 2pi or 0.
>
> So, none of the Peano Axioms talks about finite or infinite. So the
> reader can
> make up their own definition of finite or infinite.

Axioms 6 through 8 collectively imply that for any "last" element of a
Peano set under some function S, you can always apply S to it and
generate a new "last" element that has not yet appeared. No further
definition of "infinite" is required within the context of the Peano
axioms (but may be required for other definitions of other
properties).

> Now as for the hardship that most readers have and which troubles
> Tribble is
> their minds cannot see how the Naturals transition from zeroes on the
> leftward
> into a number which has nonzeroes on the leftward. So how does the
> Natural
> Numbers go from 00000....888 into say 111111.....888 where they easily
> picture 888 and call it finite but how does endless adding of one
> eventually
> get to 11111....888 or 22222....22222 ??

- What is the smallest AP-adic that is leftwardly zero (000...0x,
where all the elided digits are 0) that has as an AP-adic that is not
leftwardly zero under the successor function S(x) = x + 000...001?
- Is this the only AP-adic that is leftwardly zero whose successor
under S(x) = x + 000...001 is not leftwardly zero?

> Or, give me 11111.....888 and its predecessor was 11111....887 and its
> successor
> is 11111....889

For clarity, is 11111...888 rewritable as 111118...888 or as
11111...1888?

-o

[plutonium....@gmail.com]

plutonium.archime...@gmail.com wrote:
> David R Tribble wrote:

> >
> > It's easy to show that 900...000 cannot be a Peano natural.
> > Starting with the natural 0, we find it successor, 1, then the
> > successor to 1, 2, and so on. The Peano axioms tell us that
> > for any finite natural, it has a finite natural successor.
>

(big snip here)

> Look at the Peano Axioms of the Successor or Math Induction. If you
> know
> Math Induction, and know it well, you know it is equivalent to the
> concept that
> given any Natural Number N, it has a N+1 and a N-1. Those are
> equivalent
> formulations of the Math Induction in Peano Axioms.
>
> So, give me any Infinite Integer say, give me 90000....00000 then its
> predecessor
> was 899999....999999 and its successor is 900000....00001
>
> Or, give me 11111.....888 and its predecessor was 11111....887 and its
> successor
> is 11111....889
>

I need to talk a little more about the misconception or delusion of
the Peano
Axioms.

For I can actually point to the very number that Tribble wants to know
where
finite integers goes into infinite integers. But this point is
arbitrary for mathematics.

The point in which Tribble wants to know where finite enters infinite
is the point
0999999....99999 and 10000....00000

I could not do this in the 1990s because I did not have AP-adics nor
FrontView
with BackView.

The common misconception of finite and infinite is that finite to them
means
ending in a leftward string of zeroes. So to Tribble and the rest of
the math
community 88 is finite because it is 0000....0000088 and 10^400 is
finite because
it is 0000.... 10^400.

So to old math and their conceptions of finite and infinite is that a
Finite Number
ends leftwards with zeroes.

But there is some trouble with that because what is 001299999....99999
where
the "2" is in the 10^9999...9996 place value and the "1" in the
10^9999...9997
place value and the zeroes occupy the 10^9999....9998 and
10^9999...9999
place values.

So in the old math 00129999...99999 would be finite.

So, now, where is the transition point where finite becomes infinite?
Simple. Since every Peano Axiom Nonzero Number has both a predecessor
and successor then the "Last Finite Integer" that Tribble wants so
badly
is the integer 099999.....999999 for it is the last and final Finite
Integer as
what the Old Math defines as finite. So adding 1 to that last and
final
Finite Integer we have 100000....00000 and its successor is easily
10000....00001.

So any reader can see. That the concept of Finite in old math was
arbitrary because
it was defined as ending in zeroes leftward. So that 9999...9999 or
7777....7777 was
not Finite to Tribble or to anyone of the Old Math because they did
not end leftwards
in zeroes such as 00000....0000947 or 00000....0001000000000000 which
is 947
and one trillion respectively.
But the endless adding of 1 easily gets to 09999....999999 which would
be Tribble's
last Finite Integer, the Last of the Mohicans and add one more and we
have
what the Old Math calls Infinite Integers.

So you can see, that the definition of Finite by the math community
was nothing but
an arbitrary and artifical contrivance and was hidden as a error of
the Peano Axioms.
The Peano Axioms once corrected of its 5th axiom that 0 and 1 exists.
And then
append a 10th new axiom saying that Finite is 10^500 and below and
infinity is beyond
10^500.

So I have been accused of arbitrariness and artificiality with 10^500,
but take a look
at the artificiality and arbitrariness of the definition that Finite
means ending leftward
in zeroes. What about the Integer of 00000.....00001000.....000000
where the
"1" digit is in the 10^100000.....000000 place-value ??

So, you can see, clearly that the Peano Axioms do not and cannot
distinguish between
finite and infinite. Peano never mentions finite or infinite. And that
the Peano Axioms
delivers this set of numbers called Natural Numbers from 0 to 1 to 2
to eventually
99999....999999

It does not come any clearer than that

[plutonium....@gmail.com]

If you can post with a real name I will answer those questions

> the subset set of real numbers, starting with ð, closed under the
> successor function S(x) = x * ð. So is the subset of the natural

> numbers, starting with 0, closed under the successor function S(x) = x
> + 2. The natural numbers under S(x) = x + 1 are only the most
> intuitive model of the axioms, and inferred definitions of addition,
> subtraction, and multiplication from the Peano axioms map well to
> intuitive definitions of addition, subtraction, and multiplication
> over this model/
>
> Furthermore, even if two sets satisfy the Peano axioms, each under its
> own successor function, the union of the two sets does not necessarily
> satisfy the Peano axioms under either successor function. To go back

> to the ð example, the union of the Peano set (ð, F(x) = x * ð) and the

[angryb...@gmail.com]

On Jan 31, 8:23 pm, plutonium.archime...@gmail.com wrote:
> If you can post with a real name I will answer those questions

If you can convince Google Groups to show my profile name rather than
my email address, be my guest. In the interim, my name is Owen
Jacobson.

Ignoring questions to focus on the asker is the basis of the "ad
hominem" fallacy. Do be careful.

-o

[David R Tribble]

Archimedes Plutonium wrote:
> You see Tribble, that is why I must insult you alot because you live
> in a fantasy
> and not reality. A fantasy of your own making, because everything I
> say you
> look for some darkness in it, something to undermine everything I say.
> And it is
> sad that you are in science or math for you belong in neither. A
> scientist does
> not have time for playing games, for psychology, or chips on the
> shoulder, and believe me, you
> have more than one chip on your shoulder.
>

> I don't expect that colleges and universities teach etiquette and
> morals but they should
> and the one that Tribble needs to understand is the one that says:
> (A) A conversation or discussion between two persons has to be one of
> respect
> for the other person. If you do not respect the other person then stop
> talking with them.

I guess in future posts we should use your text above as an
example of the proper way of showing respect, huh?

[plutonium....@gmail.com]

> For clarity, is 11111...888 rewritable as 111118...888 or as
> 11111...1888?

> -o

So, sign your post as your full name. I do not read or reply to
posters
with fake names, for they are not serious enough for me.

The first thing going on is ALL Possible Digit Arrangements.

And the definition of the Infinitculum (those dots) is omission of an
infinite stretch of digits of the same pattern.

So is this number 11111.....888 the same digit arrangement as
111118....888 or as 11111....1888.

Well 11111.....888 is either one or the other but not both.

It goes to show that we can write numbers that are ambiguous.

But in All Possible Digit Arrangements each number is unique for it
has
a unique digit arrangement.

Suppose I wanted to write the number that has a 7 in the middle of
the infinite string of zeroes with a "1" digit at both ends.

So I could write it as 100000......7.....000001 but that would be
ambiguous. So I improve on it as 10000.....00700.....00001
and best of all I write 10000....00700....00001 and say in words
that the "7" digit is in the 10^49999....9999 place value.

I should write in some length about binary representation of AP-adics.
I have
avoided this because the Reals are known in decimal representation.
The important issue of Binary versus Decimal is that it has some
profound
implications because of infinity rather than finiteness. And the
implication for
Peano Axioms is that half of the Natural Numbers in binary are finite
as per the
definition that ending leftwards in zeros. So in Binary, half of the
numbers
are ending leftwards in zeroes and the other half in 1s digit. So in
binary one
can say that Finite has half the quantity of numbers and infinite
integers are
the other half quantity. Whereas in Decimal representation we can say
that
10% of the Peano Natural Numbers is Finite and 90% Infinite Integers
using
the definition of finite as leftward ending in zero.

Now the reason I use Decimal is because Reals uses Decimal. And the
proofs
and theorems do not depend on representation. However, because the
Natural
Numbers are Infinite Integers, the representation matters more so than
in the
old way of doing mathematics.

In Decimal, All Possible Digit Arrangements makes 9999....99999 the
largest number
but in Binary 11111.....11111 becomes the largest number. In Decimal
the leftmost
place-value is 10^9999.....99999 and in Binary the leftmost place-
value is
2^1111....111111. So representation does matter.

[plutonium....@gmail.com]

Owen Jacobson wrote:
> On Jan 31, 6:19 pm, plutonium.archime...@gmail.com wrote:

(snipped)

>
> You are operating under the assumption that the Peano axioms only
> define one set. This is not true; they define infinitely many "Peano
> sets": one for each choice of '0' and successor function. For example,

Not so. And you are at fault in that you assume Peano Axioms as the
ultimate foundation, but in reality, comparing All Possible Digit
Arrangements
as the foundation is a far firmer and is the "ultimate foundation"

So if you want to understand mathematics, you must embrace
All Possible Digit Arrangements and not some cobbled together axioms.

For the Natural Numbers, you see, I can build them two ways. Simply
All Possible Digit Arrangements. or, the harder way of 9 or more
axioms.

As I said in many previous posts, that Reals are modeled by the Peano
Axioms where 0d0000....00001 acts as 1 and 0d0000....00002 acts as
2 and that all the Reals between 0 and 1 refect the same as integers
from 0 to 99999....99999

So Owen fails to recognize or understand All Possible Digit
Arrangements.
He is still back in the weeds thinking that Peano Axioms is the best
way,
whereas it is the worst compared to All Possible Digit Arrangements.

So the question that Owen should ask himself is how many number
systems
are there possible with a All Possible Digit Arrangement as
foundation? And the
answer is two systems the Reals and AP-adics.

> the subset of the rationals starting with 1, closed under the
> successor function S(x) = x / 2, is a model of the Peano axioms. So is

> the subset set of real numbers, starting with ð, closed under the
> successor function S(x) = x * ð. So is the subset of the natural

> numbers, starting with 0, closed under the successor function S(x) = x
> + 2. The natural numbers under S(x) = x + 1 are only the most
> intuitive model of the axioms, and inferred definitions of addition,
> subtraction, and multiplication from the Peano axioms map well to
> intuitive definitions of addition, subtraction, and multiplication
> over this model/
>

When you start with All Possible Digit Arrangements, then the Reals
have the same Peano mechanisms of Successor. You may have to
restrict the interval of Reals such as between 0 and 1 or between 3
and
4. But essentially the same sort of mechanisms that exist in Counting
Numbers or Natural Numbers exists in Reals because both are
systems that are created from one concept--- All Possible Digit
Arrangements.

There are only two types of geometry--- Euclidean and NonEuclidean
where NonEuclidean is Elliptic and Hyperbolic and when you break
the symmetry of Euclidean you end up with Elliptic + Hyperbolic.

> Furthermore, even if two sets satisfy the Peano axioms, each under its
> own successor function, the union of the two sets does not necessarily
> satisfy the Peano axioms under either successor function. To go back

> to the ð example, the union of the Peano set (ð, F(x) = x * ð) and the

> Peano set (0, G(x) = x + 1) is not a Peano set under either F or G,
> and elements of either set will not be generated by repeated
> application of the other set's successor function to members of that
> set.
>
> So even if {0, 1, 2, ...} and {777...700, 777...701, 777...702, ...}
> both satisfy the Peano axioms under S(x) = x + 1, that does not imply
> that {0, 1, 2, ..., 777...700, 777...701, 777...702, ...} satisfies
> them under S(x) = x + 1. You haven't shown this yet.
>

That is where Owen is really muddled up for that would not have even
been
a question if Owen had started with All Possible Digit Arrangements
rather
than a gaggle of cobbled axioms.

> Some fairly easy questions:
>
> - What is the smallest AP-adic that is leftwardly zero (000...0x)

Here is a question that makes sense in All Possible Digit Arrangements
but most of Owen's questions are nonsense since he still does not
start at a "true foundation"

From what I can understand there is a smallest infinity demanded by
the Reals in the interval 0 to 1 where the multiplication of

1d0000....0000 x 0d0000.....00001 equals a number and that number
is 0d000000.....0001000.....00000000 where the "1" digit is in the 10^
(-)
50000....00000 place value

And in AP-adics I have the same symmetric number when

1000000....00000 x 000000....00001 = 00000....00100....000000 where
the 1 digit is in the 10^50000....00000 place value

I call them the smallest infinity

As for the remainder of Owen's questions, they are not questions under
All Possible Digit Arrangements but are confusions on the part of
Owen.

[plutonium....@gmail.com]

I believe Binary rather than Decimal representation can shed light on
Prime definition
and Prime distribution. In Decimal the Sphere surface is divided into
ten bands
where the Equator becomes the 50000....0000 line but in Binary there
are only
two bands, the 0-band and the 1-band depending on what the leftmost
digit is.
And the Equator in binary is the 10000....00000 line. So in Binary we
only
have to visualize the distribution of primes within two bands, the 0-
band and 1-band.

Now we can do the same for Reals between 0 and 1 in that in binary
there are
two types of Reals between 0 and 1 as those ending in the 2^(-)
1111....1111
place value as either 0 or 1 digit. And in Binary the Real
0d50000...0000 or
1/2 becomes 0b00000....00100......00000 where the 1 digit is in the 2^
(-)1000...000
place value.

So to what extent do the two bands in Binary obey the Prime
Distribution
formula of x/Ln(x)?

I think it obeys that formula up to the smallest infinity of
0000....00001000....00000 which is 1000....0000 x 0000....00001 in
Decimal. But I maybe wrong in that it only obeys up to the definition
of
Finite as that of 10^500 or less. So there are alot of visualizations
that
need to be made for more progress.

[Dan Christensen]

On Jan 30, 5:41 pm, plutonium.archime...@gmail.com wrote:

[snip]

> > > Now there are two errors in those Peano Axioms above and those two
> > > are:
>
> > > (1) failed to define "finite" versus "infinite" which thus, the Peano
> > > axioms […]
>
> > I don't see why this is a problem.
>
> The Peano Axioms delivers this set 0,1,2,3,4, ......, 99999....99999

There is nothing in PA about strings of digits representing numbers --
finite or otherwise. It "delivers" only the number 0 (the "first"
number) and a successor function S:

0, S(0), S(S(0)), ....

> with its Successor axiom of endless adding of 1.
>
> The problem is the schizophrenia of the mathematics community that
> they
> do not want to recognize that the Peano axioms delivers that set.
> They
> want
> to say that a finite integer is like this 0000.....00009 is equal to 9
> and write
> 9 without its infinite string of leftward zeros. But when you have
> endless adding
> of 1 you end up with 99999....99999
>
> So, now you can see the problem.

[snip]

Actually, no. I guess what I would like to see from you is a number-
theoretic statement that you think should be true, but isn't under PA.
Or one that can be shown to be both true and false under PA. IMHO, if
you cannot do so, you are probably wasting your time. Sorry.

Dan
Download my DC Proof software at http://www.dcproof.com

[Owen Jacobson]

On 2009-02-01 14:32:34 -0500, plutonium....@gmail.com said:

>
> Owen Jacobson wrote:
>> On Jan 31, 6:19 pm, plutonium.archime...@gmail.com wrote:
>
> (snipped)
>>
>> You are operating under the assumption that the Peano axioms only
>> define one set. This is not true; they define infinitely many "Peano
>> sets": one for each choice of '0' and successor function. For example,
>
> Not so.

I gave you examples of several distinct sets that satisfy the Peano
axioms without being the set of all natural numbers. If you'd care to
identify which axiom rejects the set {1, 1/2, 1/4, 1/8, 1/32, ...} but
accepts the set {0, 1, 2, 3, 4, ...}, by all means, be my guest - it
would be rather educational.

The "zero" element of the first set is the rational number 1; the
successor function is F(x) = x / 2.

The "zero" element of the second set is the natural number 0; the
successor function is G(x) = x + 1.

This should be sufficient for you to show the axiom or set of axioms
that rejects the first but not the second. I quite honestly can't see
it, myself.

> And you are at fault in that you assume Peano Axioms as the
> ultimate foundation, but in reality, comparing All Possible Digit
> Arrangements
> as the foundation is a far firmer and is the "ultimate foundation"

I assume nothing of the sort. Axioms are neither true nor false in any
absolute sense. The closest thing to a "false" set of axioms is a set
that proves both P and ¬P for some statement P, which doesn't make that
set wrong, just inconsistent.

Theorems of an axiom set A are always conditional on the axioms used to
prove them. It is always possible to create another set of axioms B in
which some or all of the theorems of A are either not provable or are
provably false.

Peano's axioms prove theorems that we accept as "intuitively" true: 2 +
2 = 4 is a theorem of the Peano axioms, for suitable definitions of
"2", "+", "=", and "4". However, there are many axiom sets that prove
that 2 + 2 ≠ 4. Neither set "wins"; both Peano arithmetic and other
axiom sets prove interesting theorems.

You appear to be claiming either
a) that the AP-adics also satisfy the Peano axioms, under some
successor function,
or
b) that the AP-adics satisfy an axiom set extending the Peano axioms.
If neither of these match your claim you might save both of us some
time by clarifying what it is you're actually claiming about the
relationship between your AP-adics and the Peano axioms.

> So if you want to understand mathematics, you must embrace
> All Possible Digit Arrangements and not some cobbled together axioms.
> For the Natural Numbers, you see, I can build them two ways. Simply
> All Possible Digit Arrangements. or, the harder way of 9 or more
> axioms.

What are the axioms of "all possible digit arrangements"?

Are fewer axioms required for "all possible digit arrangements"?

Are theorems provable from those axioms compatible or incompatible with
theorems provable from other axiom sets?

>> Some fairly easy questions:
>>
>> - What is the smallest AP-adic that is leftwardly zero (000...0x)
>
> Here is a question that makes sense in All Possible Digit Arrangements
> but most of Owen's questions are nonsense since he still does not
> start at a "true foundation"
>
> From what I can understand there is a smallest infinity demanded by
> the Reals in the interval 0 to 1 where the multiplication of
>
> 1d0000....0000 x 0d0000.....00001 equals a number and that number
> is 0d000000.....0001000.....00000000 where the "1" digit is in the 10^
> (-)
> 50000....00000 place value
>
> And in AP-adics I have the same symmetric number when
>
> 1000000....00000 x 000000....00001 = 00000....00100....000000 where
> the 1 digit is in the 10^50000....00000 place value
>
> I call them the smallest infinity

That's very interesting, but that doesn't answer the question I asked.
The question as asked spanned multiple lines, so I forgive you for not
noticing the rest; here it is in its entirety:

- What is the smallest AP-adic that is leftwardly zero (000...0x)

that has as an AP-adic that is not leftwardly zero under the successor
function S(x) = x + 000...001?

I'd like to clarify "leftwardly zero" - an AP-adic is leftwardly zero
if and only if it has a finite number of non-zero digits, and the most
signifigant non-zero digit is only finitely far from the right-hand end
of the number. To the left of the last non-zero digit, all further
digits are zero. 000...010..0 is leftwardly zero (given that .. in your
notation denotes a finite span of digits); 000...010...0 is not
leftwardly zero, as the last non-zero digit is more than finitely far
from the right end of the number. Clear enough?

I mention "leftwardly zero" AP-adics because, if the AP-adics are
compatible with the natural numbers, then there exists a mapping from
the natural numbers to the AP-adics such that the image of every
natural number is leftwardly zero. One such mapping treats each natural
as a string of digits: 102 is the string "102". Then we can simply
prepend a string of infinitely many "0" digits to the number to arrive
at the string "000...0102", which represents the AP-adic 000...0102.
All of the digits in the ... are also 0.

This appears compatible with your notion of AP-adic numbers and implies
some useful things.

-o

[plutonium....@gmail.com]

Dan Christensen wrote:

>
> There is nothing in PA about strings of digits representing numbers --
> finite or otherwise. It "delivers" only the number 0 (the "first"
> number) and a successor function S:
>
> 0, S(0), S(S(0)), ....
>

Funny how you complain about "strings" when it is in front of your
eyes.

The Successor is another word for "string". And String is another word
for Series.

Series of digits. S(S(0)) is just another way of writing the string
0000...0002
or the Series 0000....000002 or 1+1.

From your comments I don't think you have a degree in math. Probably
you dabble in philosophy. A philosopher is more of a language warrior
than
a scientist. And a philosopher is mostly argumentative with never
wanting
a resolution that a mathematician seeks.

Look at every one of your lines of comment. They belch and bellow in
just plain argumentative nonsense. Dan, you simply do not belong in
this
thread. Your mind is not open for anything new, but is loaded with
your rigid old
brainwash.

> > with its Successor axiom of endless adding of 1.
> >
> > The problem is the schizophrenia of the mathematics community that
> > they
> > do not want to recognize that the Peano axioms delivers that set.
> > They
> > want
> > to say that a finite integer is like this 0000.....00009 is equal to 9
> > and write
> > 9 without its infinite string of leftward zeros. But when you have
> > endless adding
> > of 1 you end up with 99999....99999
> >
> > So, now you can see the problem.
>
> [snip]
>
> Actually, no. I guess what I would like to see from you is a number-
> theoretic statement that you think should be true, but isn't under PA.
> Or one that can be shown to be both true and false under PA. IMHO, if
> you cannot do so, you are probably wasting your time. Sorry.
>
> Dan

Tell me, is there any post of yours where agree with someone? Probably
not.
The first sign of a intelligent person is not crab crab crab, like
you, but the first
sign of intelligence is that two people moving forward to solve some
problem.

Everyone of your lines is never an intersection with attempts to move
forward.
Everyone of your lines is seeking to turn everything I do backwards.

So, Dan, you do not belong in this thread for you are far too stupid
on any of this
subject matter.

You never once mentioned All Possible Digit Arrangements which would
counter
every one of your silly complaints.

Are you simply too stupid to understand or comprehend All Possible
Digit Arrangements.
Probably. I think it was Charles Sanders Pierce, the pragmatist that
said a
"philosopher" was just a ball of spun up twine or words to that
effect.

So go quietly and leave this thread, for your nonsense takes up too
much time.

I think there is a disqualifer for a person to be in science, for that
person has
psychological chips on their shoulders--- entering a discussion where
you simply
deny and reverse everything the other person says. Well, maybe the
Canadian
winters provide Dan with nothing but complaining.

So get you out of this thread. I rather noone replies than to read
garbage like Dan's

As David Tribble would say "Obviously clear"

[plutonium....@gmail.com]

Owen Jacobson wrote:
> On 2009-02-01 14:32:34 -0500, plutonium....@gmail.com said:
>
> >
> > Owen Jacobson wrote:
> >> On Jan 31, 6:19 pm, plutonium.archime...@gmail.com wrote:
> >
> > (snipped)
> >>
> >> You are operating under the assumption that the Peano axioms only
> >> define one set. This is not true; they define infinitely many "Peano
> >> sets": one for each choice of '0' and successor function. For example,
> >
> > Not so.
>
> I gave you examples of several distinct sets that satisfy the Peano
> axioms without being the set of all natural numbers. If you'd care to
> identify which axiom rejects the set {1, 1/2, 1/4, 1/8, 1/32, ...} but
> accepts the set {0, 1, 2, 3, 4, ...}, by all means, be my guest - it
> would be rather educational.
>

That is why Peano axioms are deficient and need a fixing by the
axiom that creates 0 should also create 1 along with it. For when
Peano gets to the Successor axiom that his "spanner" so to speak
or ruler is a span of 1 unit distance.

> The "zero" element of the first set is the rational number 1; the
> successor function is F(x) = x / 2.
>
> The "zero" element of the second set is the natural number 0; the
> successor function is G(x) = x + 1.
>
> This should be sufficient for you to show the axiom or set of axioms
> that rejects the first but not the second. I quite honestly can't see
> it, myself.
>

No, the tone of your questions is that you think you are bright and
smart
but in fact a ignorant klutz, and you fill some of your sentences with

"honestly can't see it, myself".

> > And you are at fault in that you assume Peano Axioms as the
> > ultimate foundation, but in reality, comparing All Possible Digit
> > Arrangements
> > as the foundation is a far firmer and is the "ultimate foundation"
>
> I assume nothing of the sort. Axioms are neither true nor false in any
> absolute sense. The closest thing to a "false" set of axioms is a set
> that proves both P and ¬P for some statement P, which doesn't make that
> set wrong, just inconsistent.

No, the point is Owen, this entire book is based on the firm
foundation
of All Possible Digit Arrangements and how that replaces axiom systems
of cobbled together thousands of definitions and hundreds of axioms.

So on the one hand, there is Archimedes Plutonium who brings to Math
just one object ---- All Possible Digit Arrangements and builds both
Reals and AP-adics (containing the Natural Numbers)

And then there is Owen who brings to the Convention Center of Math,
thousands of definitions and hundreds of gaggly cobbled together
axioms.

I build the Reals and Natural Numbers in 5 seconds.

Owen has to take 2 years to go through his mess to achieve the same
thing.

So why do I not answer your questions? Because you are not a
mathematician
but a argumentative dufus.

You do not even know what All Possible Digit Arrangements is. You
never mentioned
them and you fail to understand how they replace your entire stupid
and silly understanding
of Peano Axioms.

You do not belong in this conversation because you are a waste of
time. Your
mind belongs back in the 19th century with Peano where your silly
beliefs would be
a warm comfort to Peano, but totally out of date in this 21st century.

So kindly leave this thread. And do not bother

As David Tribble likes to say " good gosh golly, no insults"

[plutonium....@gmail.com]

Sorry, I am going to have to walk or tread lightly here. I have only a
few glimpses
or vague images of Irrational-Integers and their distribution as per x/
Ln(x). I have
a better picture of the Prime Reals, but sadly, the Irrational-Reals
has become
more vague, ever since All Possible Digit Arrangments.

So I am not in a position to offer a definition of Irrational Integer
or Prime-Real. I
can speak about them and given some time, perhaps can settle on some
sort of definition.

There are some very troubling thoughts about this. Because we all know
from
old-math that it was thought that between any two Rational Reals there
existed
an Irrational Real. And you can already anticipate and appreciate the
problem.
These are the first three Reals starting with 0 then 0d000...001 and
then
0d0000....0002 So there is no Irrational Real between any one of those
first
three Real Numbers starting at 0. So the question, bizarre feature, of
the Reals
is where do we encounter the first Irrational Real because there are
an infinity
of well ordered Reals starting with 0 then 0d000....00001 and like the
Counting
Numbers going to infinity. So, do you see the problem? The first
infinite Reals
starting at 0 has no Irrational Real Number.

Now let me try to make a preliminary stab at a definition of
Irrational Real as
that which nothing repeats in blocks or strings of digits and let me
offer this
Real Number as perhaps the smallest manufactured Irrational Real.

If 0d0000....00001 is the first nonzero Real then I offer this as the
first
Irrational Real

0d0000......00001000100101

I do some bicycle riding and bicycle mechanics on my bikes and there
is something
called "spacer" on the hub bolt. So I like to think of that Irrational
Number as a
Spacer Number where there is one 0 then two 0s then three 0s. Now I
could have
made the spacings larger.

So that Irrational Real is not going to be one of the Reals out to 10^
(-)500.

So the vague picture I am having is that the Irrational Reals are
nonexistant for
infinitely many Reals close to 0.

And then for Irrational Integers, a similar image emerges where you go
infinitely
far out such as in Decimal Integers 00000....0001000......0000000
which is
far beyond 10^500 or even 10^500^500^500

I do not know what to think of this emerging image, that
"irrationality" is nonexistent
for infinitely many numbers near 0.

And for the concept of Prime-Reals which mirror reflects the Primes of
Integers
so that 0d000...001 is like 1 and 0d0000....002 is like the prime 2
for Reals between
0 and 1.

So is there some link or connection with the distribution of
Irrational Reals and that
of Prime Reals?

Since Integer Primes are distributed by x/Ln(x) for the first 10^500
Natural Numbers
can we expect that when the Irrational Integers start up such as

00000......00001000100101, are they distributed by x/Ln(x)

And since the Prime Reals are distributed by x/Ln(x) can we say that
the Irrational Reals when they start up, would they have the same
formula?

I do not think so, because given a Irrational Integer such as

00000.....00001000100101 I can build Prime Integers from that as root
stock
and have 40 primes in a one-hundred number interval.

So that tells me that the distribution of Primes is far denser near
Irrational Integers
then the density of Irrational Integers.

The same sort of inference on Reals, that the Prime Reals get denser
near Irrational
Reals then the density of Irrational Reals.

Taking a Irrational Integer and using it as "root stock" I can graft
40 Prime Integers
in a hundred number interval. I can do the same for Irrational Reals
and Prime
Reals.

So, somewhere in the counting of Integers or the counting of Reals
starting at
zero, somewhere the Primes stop following the formula x/Ln(x) and
become
more dense. And where they become more dense is where there are
Irrational Integers and Irrational Reals.

Now I spoke of Binary representation as an aid in visualization since
the Binary
has only two bands, whereas Decimal has ten bands to think about.

Maybe what I should do is go further down -- backwards starting with
9999....9999 and going backwards. In those last hundred number
interval
where 9999....99997 is the world's largest prime number, I counted 24
primes
where there are 25 primes in the interval 0 to 100. So if I went
backwards from
9999....9999 and see what happens.

[Dan Christensen]

On Feb 2, 1:53 am, plutonium.archime...@gmail.com wrote:
> Dan Christensen wrote:
>
> > There is nothing in PA about strings of digits representing numbers --
> > finite or otherwise. It "delivers" only the number 0 (the "first"
> > number) and a successor function S:
>
> > 0, S(0), S(S(0)), ....
>
> Funny how you complain about "strings" when it is in front of your
> eyes.
>
> The Successor is another word for "string". And String is another word
> for Series.
>
> Series of digits. S(S(0)) is just another way of writing the string
> 0000...0002
> or the Series 0000....000002 or 1+1.

[snip]

A technical point: Neither 0000....000002 nor addition is defined in
PA. However, 0, S(0), S(S(0)), ... are all defined in PA.

> > > So, now you can see the problem.
>
> > [snip]
>
> > Actually, no. I guess what I would like to see from you is a number-
> > theoretic statement that you think should be true, but isn't under PA.
> > Or one that can be shown to be both true and false under PA. IMHO, if
> > you cannot do so, you are probably wasting your time. Sorry.
>
> > Dan
>
> Tell me, is there any post of yours where agree with someone? Probably
> not.

[snip]

You are bothered by some seemingly counter-intuitive aspects of PA
that, it is true, I cannot fathom. Maybe, as you suggest, I am just
not clever enough. By the same token, it seems you have not identified
any real problems with PA -- not in a number-theoretic sense, anyway.
Sorry, but I find it difficult to take you seriously.

[plutonium....@gmail.com]

Dan Christensen wrote:

>
> You are bothered by some seemingly counter-intuitive aspects of PA
> that, it is true, I cannot fathom. Maybe, as you suggest, I am just
> not clever enough. By the same token, it seems you have not identified
> any real problems with PA -- not in a number-theoretic sense, anyway.
> Sorry, but I find it difficult to take you seriously.
>
> Dan

Well I have been on the Internet since 1993, Dan.

And there is a huge swell of people whose psychology is utterly
baffling and
illogical such as you.

(1) you hate the guts of the person you reply to and you think they
are wrong
in everything they say

YET, (2) you still reply to them as if, and pretending to have a
discussion or
conversation.

So, by Logic, if a person is bright, smart and intelligent, they would
know whether
their actions and behaviour are as a result of hatred and loathing of
another person
or, whether their action and behaviour is a result of a desire for
learning and
exchange of ideas.

To put it simply, Dan, why do you persist in posting to my threads
when you
know that you hate me and think that nothing I offer is true.

The only answer can be is that you, Dan are not intelligent and that
you post
because you have a psychological hole in your head.

Logical and rational and intelligent people do not engage with people
who they deem
"wrong". They avoid them. So that can only mean, Dan, that you are not
intelligent.

And I am not going to waste time on your lousy questions because you
will never
agree to anything I say.

You do not belong in sci.math, probably no science newsgroup. You
belong in some
Monty Python sketch where you see the Department of Arguments

You hate my guts, and you think everything I say is wrong, and yet you
still reply
and pretend to ask questions of me. Spells only one thing-- you lack
intelligence.

Go fly a kite.

[plutonium....@gmail.com]

plutonium.archime...@gmail.com wrote:
> Dan Christensen wrote:
>
> >
> > You are bothered by some seemingly counter-intuitive aspects of PA
> > that, it is true, I cannot fathom. Maybe, as you suggest, I am just
> > not clever enough. By the same token, it seems you have not identified
> > any real problems with PA -- not in a number-theoretic sense, anyway.
> > Sorry, but I find it difficult to take you seriously.
> >

Alright, I can learn from all posters. And perhaps was overboard rude
to Dan.

So I will see if this new pattern of mine works better. Whether Dan
wants
to try or not is up to him.

Dan, --

Do you know what I mean when I say the concept All Possible Digit
Arrangements
of leftward strings out to infinity as Natural Numbers and in the
rightward direction
as Reals.

Do you know what is meant by All Possible Digit Arrangements?

(So, instead of me getting riled, maybe I should take posters like Dan
and
do a Socratic method on them)

[plutonium....@gmail.com]

Owen Jacobson wrote:

> I mention "leftwardly zero" AP-adics because, if the AP-adics are
> compatible with the natural numbers, then there exists a mapping from
> the natural numbers to the AP-adics such that the image of every
> natural number is leftwardly zero. One such mapping treats each natural
> as a string of digits: 102 is the string "102". Then we can simply
> prepend a string of infinitely many "0" digits to the number to arrive
> at the string "000...0102", which represents the AP-adic 000...0102.
> All of the digits in the ... are also 0.
>
> This appears compatible with your notion of AP-adic numbers and implies
> some useful things.
>
> -o

And probably, I was rude to Owen. So I am willing to give this a
second try, if
Owen is willing. On a Socratic type of dialogue. But that means I
cannot handle
a post where you ask zillions of questions. I feel if someone is
really into science
then the way they ask a question shows they are serious.

So Owen, do you know what I mean when I say the concept All Possible
Digit
Arrangements leftward of a radix point? Do you know what All Possible
Digit Arrangements
to a place value of 10^99999....99999 means?

[Dan Christensen]

On Feb 2, 4:50 am, plutonium.archime...@gmail.com wrote:
> plutonium.archime...@gmail.com wrote:
> > Dan Christensen wrote:
>
> > > You are bothered by some seemingly counter-intuitive aspects of PA
> > > that, it is true, I cannot fathom. Maybe, as you suggest, I am just
> > > not clever enough. By the same token, it seems you have not identified
> > > any real problems with PA -- not in a number-theoretic sense, anyway.
> > > Sorry, but I find it difficult to take you seriously.
>
> Alright, I can learn from all posters. And perhaps was overboard rude
> to Dan.
>
> So I will see if this new pattern of mine works better. Whether Dan
> wants
> to try or not is up to him.
>
> Dan, --
>
> Do you know what I mean when I say the concept All Possible Digit
> Arrangements
> of leftward strings out to infinity as Natural Numbers and in the
> rightward direction
> as Reals.
>
> Do you know what is meant by All Possible Digit Arrangements?
>

I only know what you have posted about them here. When you write
999999....999999, I assume you mean an infinite (leftward) string of
9's. Intuitively, we know that not all infinite leftward strings of
digits, as you call them, represent a natural number. Why do you think
your string 999999....999999 represents a natural number? And why do
you think it has anything to do with PA? PA says nothing about the
representation of numbers using strings of digits. Did you consider
that perhaps it is your representation theory, and not PA that is
faulty?

I don't mean to "rile" you. I myself have spent years of my spare time
pondering these things, and have come to many dead ends. For years I
tried, for example, to invent a non-set-theoretic number theory. It
was fun and I think I learned a lot, but it ultimately led nowhere.

Dan
Download my DC Proof software at http://www.dcproof.com

> (So, instead of me getting riled, maybe I should take posters like Dan

[plutonium....@gmail.com]

Hello Dan, let us try this over again.

Let me ask you to write All Possible Digit Arrangements for two
digits, 0,1
to two place values. Please write them here:

Secondly please write a matrix of ten digits in two-place-value, here,
next

Once you have done that, I can easily answer the above question you
have
and more. So a dialogue, if you want one, needs a common ground
starting
point.

[Dan Christensen]

Note: Google is saying it does not currently support sci.math. I will
post to the other groups.

On Feb 2, 1:28 pm, plutonium.archime...@gmail.com wrote:
> Dan Christensen wrote:
> > On Feb 2, 4:50 am, plutonium.archime...@gmail.com wrote:
>
> > I only know what you have posted about them here. When you write
> > 999999....999999, I assume you mean an infinite (leftward) string of
> > 9's. Intuitively, we know that not all infinite leftward strings of
> > digits, as you call them, represent a natural number. Why do you think
> > your string 999999....999999 represents a natural number? And why do
> > you think it has anything to do with PA? PA says nothing about the
> > representation of numbers using strings of digits. Did you consider
> > that perhaps it is your representation theory, and not PA that is
> > faulty?
>
> > I don't mean to "rile" you. I myself have spent years of my spare time
> > pondering these things, and have come to many dead ends. For years I
> > tried, for example, to invent a non-set-theoretic number theory. It
> > was fun and I think I learned a lot, but it ultimately led nowhere.
>
> > Dan
>
> Hello Dan, let us try this over again.
>
> Let me ask you to write All Possible Digit Arrangements for two
> digits, 0,1
> to two place values. Please write them here:
>

If I understand you correctly:

0 0
0 1
1 0
1 1

> Secondly please write a matrix of ten digits in two-place-value, here,
> next
>

0 0
0 1
0 2
...
9 8
9 9

> Once you have done that, I can easily answer the above question you
> have
> and more. So a dialogue, if you want one, needs a common ground
> starting
> point.
>

I hope you will get around to explaining why you think your infinite
string ...99999 represents a natural number. And how PA fits into all
this.

[Owen Jacobson]

On 2009-02-02 04:55:45 -0500, plutonium....@gmail.com said:

> So Owen, do you know what I mean when I say the concept All Possible
> Digit Arrangements leftward of a radix point?

For any natural number, N, I have a formal definition of "all possible
digit arrangements" of length N. Under this formalism, there are B^N
possible N-digit arrangements for digits in base B.

For example, for N = 2, B = 2, we can enumerate them: {"00", "01", "11", "10"}

I also have a formal definition of infinitely long "digit arrangements"
based on the limit of the definition of finite arrangements as N grows
without bound. Under this formalism, there (probably) are uncountably
many possible infinitely-long digit arrangements. As a consequence of
the construction, the resulting strings of digits have only one end:
either a "first digit" or a "last digit", but not both.

It's pretty clear to me that while you probably accept the first
formalism, you also probably reject with the second.

> Do you know what All Possible Digit Arrangements to a place value of
> 10^99999....99999 means?

In the absence of an axiomatic definition of 999...999, no, I don't. I
can make an intuitive guess at it, but such a guess is useless to me
for mathematics.

Hence the request for axioms or some other rigorous basis. My questions
have all been intended to "dig out" the fundamentals of your number
system; they're not intended to be difficult or offensive questions.

-o

[plutonium....@gmail.com]

Dan Christensen wrote:

> >
>
> If I understand you correctly:
>
> 0 0
> 0 1
> 1 0
> 1 1
>
> > Secondly please write a matrix of ten digits in two-place-value, here,
> > next
> >
>
> 0 0
> 0 1
> 0 2
> ...
> 9 8
> 9 9
>

Yes, very good, exactly what I wanted.

We need a tad more though to see the progression.

As you listed the above matrix of ten digits for the two-place-value

I need you to list the ten-digits for the three-place-value and the
next matrix of four-place-value

And then I need you to list a matrix for an infinity-place-value

Then we can get down to business

Try to list these matrices all on one page for future reference so I
do not
have to cut and paste from different posts.

[Archimedes Plutonium]

Owen Jacobson wrote:
> On 2009-02-02 04:55:45 -0500, plutonium....@gmail.com said:
>

Hello Owen, glad you are willing to continue, even after I threw a tirade.

Owen Jacobson wrote:
> On 2009-02-02 04:55:45 -0500, plutonium....@gmail.com said:
>
> > So Owen, do you know what I mean when I say the concept All Possible
> > Digit Arrangements leftward of a radix point?
>
> For any natural number, N, I have a formal definition of "all possible
> digit arrangements" of length N. Under this formalism, there are B^N
> possible N-digit arrangements for digits in base B.
>
> For example, for N = 2, B = 2, we can enumerate them: {"00", "01",
"11", "10"}
>

Okay that is a good start for it shows you are bringing in Probability
combinations
Personally I forgotten whether they are called permutations in
probability theory

But it is important that you bring up ideas of Probability theory
because it is more
important than Set theory and I made a claim recently that all of Set
theory fits inside
of Probability theory. But that is a topic for a whole new book.

> I also have a formal definition of infinitely long "digit arrangements"
> based on the limit of the definition of finite arrangements as N grows
> without bound. Under this formalism, there (probably) are uncountably
> many possible infinitely-long digit arrangements. As a consequence of
> the construction, the resulting strings of digits have only one end:
> either a "first digit" or a "last digit", but not both.
>

But I don't like that detour, for it does not appreciate the true value
of All Possible Digit Arrangements as a concept more powerful than
all the Peano Axioms and definitions to build the Natural Numbers and
Reals. So I need to ask some favor of you.

So Owen, I am asking a favor of you to list the matrix of ten items:
apple (a for short), banana (b for short), cucumber, dachshund, energy,
fire, geography, history, intelligence, juice

Owen, please list the entire matrix for two place value of those ten
objects,

Then based on your two place matrix, list what the first four and last four
"entries (which will be numbers)" are for the three place value matrix.

Finally, list, based on your combinations in two and three place value
what an Infinity place value matrix would look like by listing the first
four
and last four entries.

[Dan Christensen]

On Feb 3, 2:31 am, plutonium.archime...@gmail.com wrote:
> Dan Christensen wrote:
>
> > If I understand you correctly:
>
> > 0 0
> > 0 1
> > 1 0
> > 1 1
>
> > > Secondly please write a matrix of ten digits in two-place-value, here,
> > > next
>
> > 0 0
> > 0 1
> > 0 2
> > ...
> > 9 8
> > 9 9
>
> Yes, very good, exactly what I wanted.
>
> We need a tad more though to see the progression.
>
> As you listed the above matrix of ten digits for the two-place-value
>
> I need you to list the ten-digits for the three-place-value and the
> next matrix of four-place-value
>

I get the idea.

> And then I need you to list a matrix for an infinity-place-value
>

Aye, there's the rub! Not all entries will correspond to natural
numbers. In particular, your string ...999999 (all 9's) is not a
natural number. Neither is ...11111 (all 1's), or ...22222 (all 2's),
etc.

And I'm quite sure you cannot prove (from PA plus suitable + and ^
functions) that

...99999

or

9 + 90 + 900 + 9000 + ...

is a well-defined number. Do you not agree?

[plutonium....@gmail.com]

I disagree about you "getting the idea" and so I want to pursue this
further
with you. For not only will All Possible Digit Arrangements yield
*all* the Natural
Numbers but will yield *all* the Reals and Dan, this is important, in
that the
single concept of All Possible Digit Arrangements as the creator of
the Natural
Numbers and Reals, creates the axioms of Peano and Dedekind.

I have not displayed this power of the concept All Possible Digit
Arrangements
and that is why both you and Owen will display it.

Please continue but let me drop the concept All Possible Digit
Arrangements down
further to its most primitive state. Let me drop it down so that it
not only creates
all the Natural Numbers and all the Reals but creates the very own
Peano Axioms.

So please continue Dan. Please list All the Possible Arrangements,
notice I deleted
the word "digit" Please list all the Possible Arrangements of
these ten objects of K,L, M, N, P,R,S,T, U, W to that of Two-Slots
__ __

As you pointed out earlier, you will have a hundred arrangements

Please list them, because I hope you list them chaotically, as I hope
that
Owen lists his chaotically. Because in the chaos of your listing, if
you
list chaotically. The only thing that matters is that you list all the
possibilities.

So I do not care how chaotic or controlled your listing, so long as
you list them
all.

Then I will come in and then assign a few control measures, such as
the relisting
will have the first entry as the Additive Identity, and the second
entry will be
a Multiplicative Identity, then the third entry must have to be the
second plus second
entry. Then the fourth entry is a continuation of what math calls Math
Induction.

So, in other words, Dan or Owen, be as sloppy as you want in listing
All Possible
Arrangements. I will tack on a few requirements which would be called
in math
axioms that forces your listing, whether chaotic or not chaotic,
forces the listing
to become highly controlled and thus the All Natural Numbers:
0,1,2,....,
9999...9999 and the All Reals between 0 and 1 starting out as
0d0000....0000 then 0d0000....00001, then 0d00000....00002

So I hope Dan and Owen are not tired yet. Not tired before anything
exciting has
taken place.

For what I am showing is that we start out all of Numbers in
Mathematics as
All Possible Arrangements of items. Give me ten items and I require of
you to
arrange them in all possible two combinations or place-value or two
slots. Then
I ask you for those ten items in three slots or three place value.

So you provide the list of those arrangements, and all that is
demanded of you is that
you provide the corrent total outlay of those arrangements and that
you have no missing
member or element of the arrangements. Then I step in and demand a few
*ordering
contraints* These contraints force you to list the elements in a
particular unique
pattern. For example, you may have been utterly chaotic in your
listing as such
if the ten objects were numbers 0,1,2,3,4,5,6,7,8,9

28
68
40
11
93
so on so forth on up to one hundred listed elements whether they were
ten numbers
or ten letters as asked of Dan and Owen.

So then I look at your list and force upon it a few constraints. I
force the number
of constraints that would make the list totally patterned so that the
first element
that Dan or Owen listed would be not 28, but have to be 00 and the
second
element that they list would not have been 68 but had to be 01.

So please Dan and Owen oblige me further with listing. For I hope you
list chaotically
and from the chaos, I force a few contraints, which would be axioms in
modern
day speak. And those axioms that I force upon your chaotic listing
would be
the Peano Axioms if we are doing Natural Numbers, or they would be
Dedekind Program
if we are doing Reals.

Either one, the Reals or Natural Numbers are built from All Possible
Arrangements.

I sense Dan is getting impatient, but he should not because we have
not hit the
fun part of where the All Possible Arrangements builds the very own
Peano Axioms.

So please Dan, let us go further. Please list the All Possible
Arrangements to two
slots for those letters I gave above. And please list them to however
is easiest for
you to list so that it takes the least amount of time for you to list
them all, and list
them so that you do not forget a member.

Because in the concept of All Possible Arrangements, there seems to be
a
Counting function of a counting so that you list them all in the
shortest period of
time. Now the listing in the shortest period of time may be such as
99, 10, 20, 30, .. out to one hundred listed items

So the burden on me is to give you the smallest number of enforcement
demands
such as the first element has to be the additive identity, and the
second element
has to be a multiplicative identity and the third element is the
second plus the second

Maybe it takes only three enforcements to turn any chaotic listing of
those hundred
elements into the listing of the first hundred elements of the Natural
Numbers and
the first hundred elements of the Reals to infinity-place-value of
0d0000....00000 then 0d0000....00001 then 0d00000....00002

So I hope Dan and Owen are not frustrated and impatient because the
good things
are still to come.

[David R Tribble]

Archimedes Plutonium wrote:
> So Owen, do you know what I mean when I say the concept All Possible
> Digit
> Arrangements leftward of a radix point? Do you know what All Possible
> Digit Arrangements
> to a place value of 10^99999....99999 means?

You are saying that a number like 123...789 has an infinite
number of digits, and that this particular string of decimal
digits it is only one of all the possible combinations of infinitely
long decimal digit strings. We get that.

What we don't get is how the '1' digit has a value of
1 x 10^999...999, because this is a circular definition.
Specifically, the leftmost '9' of the exponent of 10^999...999 is
also in the 10^999...999 position, right? Correct me if I'm wrong,
but that's what you seem to be saying.

If that is indeed the case, then you have to explain how 999...999
is equal to 9 x 10^999...999 + ..., which in turn means that it
is equal to 9 x (9 x 10^999...999 + ...) + ..., which in turn means it
is equal to 9 x (9 x (9 x 10^999...999 + ...) + ...) + ..., and so on.
How is this circular definition supposed to work?

[Dan Christensen]

OK, but some infinite arrangements (leftward) will not correspond to
any natural numbers; e.g. ...11111 (all 1's), ...22222 (all 2's), etc.

> but will yield *all* the Reals and Dan, this is important, in
> that the
> single concept of All Possible Digit Arrangements as the creator of
> the Natural
> Numbers and Reals, creates the axioms of Peano and Dedekind.
>

[snip]

Don't you have it backwards? I think they usually develop number
theory STARTING with PA. (PA themselves may be derived from set
theoretic axioms, as in ZFC.) And part of this development of number
theory will be a way to represent numbers using strings of digits. Do
you really propose to reverse this process -- to start with a
representation system and then derive PA? Sorry, I don't see how this
can possibly succeed. Without something like PA as a starting point,
it seems to me that it would be impossible to formalize any
representation system. Best of luck in your indeavour, however. :^)

[plutonium....@gmail.com]

David R Tribble wrote:
> Archimedes Plutonium wrote:
> > So Owen, do you know what I mean when I say the concept All Possible
> > Digit
> > Arrangements leftward of a radix point? Do you know what All Possible
> > Digit Arrangements
> > to a place value of 10^99999....99999 means?
>
> You are saying that a number like 123...789 has an infinite
> number of digits, and that this particular string of decimal
> digits it is only one of all the possible combinations of infinitely
> long decimal digit strings. We get that.
>
> What we don't get is how the '1' digit has a value of
> 1 x 10^999...999, because this is a circular definition.
> Specifically, the leftmost '9' of the exponent of 10^999...999 is
> also in the 10^999...999 position, right? Correct me if I'm wrong,
> but that's what you seem to be saying.

Well, I used to think that the leftmost 9 in 9999....99999 was in the
10^9999...9998 place value since the place value had to be smaller
than the number. Then I thought since these are infinite that we can
have 10^9999....9999

But with my most recent conversation on All Possible Digit
Arrangements
that perhaps the largest place value of say a 5 in 53333.....333333
has
a unique number in that All Possible Digit Arrangements.

For instance, in ten digits to the 2 place value is the formula 10^2
for
ten digits arranged in two ways for 3 place value of 10 digits is
10^3 possible arrangements. And those numbers of 10 and 2 and 3
are amoung the arrangements.

So the biggest arrangement possible is 99999...9999 and there are
10 digits and so, what is the place-value of that largest number. So
the
exponent for 10 is one of those numbers within all possible
arrangements.

So now we see this progression:

99 is 10^2
999 is 10^3
9999 is 10^4

so what is 999999.....999999 is 10^?

Is it a smaller infinite number where 10000.....00000 is smaller than
9999....999999?

Well if we define finite as that of 10^500 and below and everything
above as
infinite. Then by logic should not the 8 in 899999.....999999 be in
the
10^501 place-value?

>
> If that is indeed the case, then you have to explain how 999...999
> is equal to 9 x 10^999...999 + ..., which in turn means that it
> is equal to 9 x (9 x 10^999...999 + ...) + ..., which in turn means it
> is equal to 9 x (9 x (9 x 10^999...999 + ...) + ...) + ..., and so on.
> How is this circular definition supposed to work?

Well, if the tenth Peano Axiom should be that finite is 10^500, and
infinite
is above that, then the 9 in 96666....66666 is in the 10^501 place
value

[plutonium....@gmail.com]

Dan Christensen wrote:

>
> Don't you have it backwards? I think they usually develop number
> theory STARTING with PA. (PA themselves may be derived from set
> theoretic axioms, as in ZFC.) And part of this development of number
> theory will be a way to represent numbers using strings of digits. Do
> you really propose to reverse this process -- to start with a
> representation system and then derive PA? Sorry, I don't see how this
> can possibly succeed. Without something like PA as a starting point,
> it seems to me that it would be impossible to formalize any
> representation system. Best of luck in your indeavour, however. :^)
>

Looks as though you ended the conversation, for I asked you to list
the matrices

Goodbye

[Owen Jacobson]

On 2009-02-03 02:57:02 -0500, Archimedes Plutonium
<a_plu...@hotmail.com> said:

>
>
> Owen Jacobson wrote:
> > On 2009-02-02 04:55:45 -0500, plutonium....@gmail.com said:
> >
> > > So Owen, do you know what I mean when I say the concept All Possible
> > > Digit Arrangements leftward of a radix point?
> >
> > For any natural number, N, I have a formal definition of "all possible
> > digit arrangements" of length N. Under this formalism, there are B^N
> > possible N-digit arrangements for digits in base B.
> >
> > For example, for N = 2, B = 2, we can enumerate them: {"00", "01",
> "11", "10"}
> >
>
> Okay that is a good start for it shows you are bringing in Probability
> combinations
> Personally I forgotten whether they are called permutations in
> probability theory

Ah... no, actually. While I'm aware of the relationship, I was actually
thinking of formal languages (an extension of set theory and formal
logic) when I wrote that. Read ^ as the exponentiation operator. No
matter; the conclusion is the same.

> > I also have a formal definition of infinitely long "digit arrangements"
> > based on the limit of the definition of finite arrangements as N grows
> > without bound. Under this formalism, there (probably) are uncountably
> > many possible infinitely-long digit arrangements. As a consequence of
> > the construction, the resulting strings of digits have only one end:
> > either a "first digit" or a "last digit", but not both.
> >
>
> But I don't like that detour, for it does not appreciate the true value
> of All Possible Digit Arrangements as a concept more powerful than
> all the Peano Axioms and definitions to build the Natural Numbers and
> Reals. So I need to ask some favor of you.
>
> So Owen, I am asking a favor of you to list the matrix of ten items:
> apple (a for short), banana (b for short), cucumber, dachshund, energy,
> fire, geography, history, intelligence, juice
>
> Owen, please list the entire matrix for two place value of those ten objects,

'ge', 'ga', 'gj', 'gd', 'gb', 'gg', 'gc', 'gi', 'gh', 'gf',
'ie', 'ia', 'ij', 'id', 'ib', 'ig', 'ic', 'ii', 'ih', 'if',
'je', 'ja', 'jj', 'jd', 'jb', 'jg', 'jc', 'ji', 'jh', 'jf',
'de', 'da', 'dj', 'dd', 'db', 'dg', 'dc', 'di', 'dh', 'df',
'ce', 'ca', 'cj', 'cd', 'cb', 'cg', 'cc', 'ci', 'ch', 'cf',
'ae', 'aa', 'aj', 'ad', 'ab', 'ag', 'ac', 'ai', 'ah', 'af',
'fe', 'fa', 'fj', 'fd', 'fb', 'fg', 'fc', 'fi', 'fh', 'ff',
'he', 'ha', 'hj', 'hd', 'hb', 'hg', 'hc', 'hi', 'hh', 'hf',
'be', 'ba', 'bj', 'bd', 'bb', 'bg', 'bc', 'bi', 'bh', 'bf',
'ee', 'ea', 'ej', 'ed', 'eb', 'eg', 'ec', 'ei', 'eh', 'ef'

Without an axiom for describing a total order, this enumeration is as
valid as any other. The moment you stepped outside of numbers, I
stopped assuming *any* order - if you'd like to use one, say so; it
doesn't change my own work at all, but I can accomodate.

> Then based on your two place matrix, list what the first four and last four
> "entries (which will be numbers)" are for the three place value matrix.

'gec', 'gej', 'gei', 'ged', 'geb', 'gef', 'gea', 'geg', 'geh', 'gee',
'gac', 'gaj', 'gai', 'gad', ..., 'efa', 'efg', 'efh', 'efe'

I've listed the first fourteen, rather than the first four, so that you
can see that there is a pattern. I'm generating each new dimension by
appending one more character to each number.

Aside: what's with this fixation on base 10? Any number that's
describable in base 10 is also describable in base 2 and vice versa,
and it'd make these lists much shorter and easier to work with. Once
again, it makes no difference to me, and I can accomodate ten-digit
demonstrations if it makes you more comfortable, but it's rather
tedious to generate digit groups from ten digits, even with mechanical
help.

> Finally, list, based on your combinations in two and three place value
> what an Infinity place value matrix would look like by listing the first four
> and last four entries.

One enumeration begins 'gec...', 'gec...', 'gec...', 'gec...' for four
different sets of trailing digits. Before you object that I haven't
shown the last place of these enumerations, read to the end before you
start explaining why I'm wrong (under your formalism, I likely am wrong
- but I do not have your formalism available to me, so I must use my
own).

The digit sequences were generated using a total recursive function
from the domain of digit strings to the range of sets of digit strings
that reveals new numbers from any given number (assuming the axiom of
choice):

f (x, s) = (s = {}) | {}
(s ≠ {}) | {x + g (x)} ∪ (f (x, s - {g (x)}))

where g (x) is any function that choses one element of x (for example,
g = min would be the function that choses the least element of x), + is
the string concatenation operator ("a" + "b" = "ab") and - is the set
difference operator ({1, 2} - {1} = {2}).

(I actually used a different definition of g at each step, but it
doesn't matter as any definition of g generates all numbers, albiet in
different orders.)

Starting with a zero-digit string (the null string of digits), we can
demonstrate that this function generates all possible digit
arrangements by repeated application:

D = {"0", "1"},

f applied to all zero-digit numbers...
f ("", D) = {"0", "1"}
...gives all of the one-digit numbers without repetition, and applied
to all one-digit numbers...
f ("0", D) = {"00", "01"}
f ("1", D) = {"10", "11"}
...gives all two-digit numbers without repetition, and so on. It covers
all possible arrangements.

Under this construction, I hold that the concept of a "last place" in
an infinite string of digits is meaningless, which is why it wasn't
presented in the list of numbers above. Here's why:

Assume there is a last digit in an infinitely-long number. We know from
the way we enumerate numbers that there is also a first digit.
Therefore, we can write a specific infinitely-long number as a...b.

Since this number must be generatable through f, above, and since every
element of a result of f is a valid argument to f (proofs left as an
exercise; they're not very hard), we must be able to generate new
numbers from this infinitely long number by passing it through f, giving

D' = {0, 1, ..., 9},
f ("a...b", D') = {"a...b0", "a...b1", ..., "a...b9"}

all of which are also infinitely long.

Since we can add another digit, 'b' is not the last digit. We can't
rule out these numbers, because their existence follows naturally from
the formalism I'm using to enumerate numbers in the first place, but
their existence contradicts a consequence of the assumption that there
is a last digit of an infinitely-long number, so either there must not
be a last digit or the number a...b is not infinitely-long.

I actually urge you to *disagree* with this post! It is not intended as
an argument that you're wrong, but as an illustration of the thoughts
that occur to me when you ask for "all possible digit arrangements" -
my mind creates a (computable!) formalism for enumerating digit
arrangements, then makes some conclusions from this formalism that
contradict the point you're trying to make. Therefore my formalism
disagrees with yours.

Neither is more right than the other! I'm merely hoping to better
understand yours.

-o

[Owen Jacobson]

Depending on whether you accept the concept of a limit, there is a
formalism that we can use to describe double-ended numbers with either
even or odd numbers of digits which grow without bound. Consider the
case of even digit counts, since it's slightly simpler. We partition
each number into its "high half" and its "low half":

99 = (90) + (9)
= (9 * 10^1) + (9 * 10^0)

9999 = (9000 + 900) + (90 + 9)
= (9 * 10^3 + 9 * 10^2) + (9 * 10^1 + 9 * 10+0)

lo(x, k) = ∑ (n from 0 to k) x * 10^n
lo(9, 0) = 9
lo(9, 1) = 99

hi(x, k) = 10^(k + 1) * ∑ (n from 0 to k) x * 10^(2k - n)
hi(9, 0) = 90
hi(9, 1) = 9900

num(x, k) = hi(x, k) + lo(x, k)
num(9, 0) = 99
num(9, 1) = 9999
num(9, 2) = 999999
...

We can see that this generates every number made of 9s with an even
number of digits. The limit of num(x, k) as k grows without bound could
be read as the number "xxx... ...xxx". Standard mathematics would
instead say that the limit also grows without bound, and since this
function is only defined for integer values of k (for which the total
number of digits is 2k), additional axioms are needed to pin down the
meaning of num (x, k) "for infinitely-large k".

-o

[David R Tribble]

Archimedes Plutonium wrote:
>> So Owen, do you know what I mean when I say the concept All Possible
>> Digit
>> Arrangements leftward of a radix point? Do you know what All Possible
>> Digit Arrangements
>> to a place value of 10^99999....99999 means?
>

David R Tribble wrote:
>> You are saying that a number like 123...789 has an infinite
>> number of digits, and that this particular string of decimal
>> digits it is only one of all the possible combinations of infinitely
>> long decimal digit strings. We get that.
>>
>> What we don't get is how the '1' digit has a value of
>> 1 x 10^999...999, because this is a circular definition.
>> Specifically, the leftmost '9' of the exponent of 10^999...999 is
>> also in the 10^999...999 position, right? Correct me if I'm wrong,
>> but that's what you seem to be saying.
>

Archimedes Plutonium wrote:
> Well, I used to think that the leftmost 9 in 9999....99999 was in the
> 10^9999...9998 place value since the place value had to be smaller
> than the number. Then I thought since these are infinite that we can
> have 10^9999....9999
>
> But with my most recent conversation on All Possible Digit Arrangements
> that perhaps the largest place value of say a 5 in 53333.....333333

> has a unique number in that All Possible Digit Arrangements. [...]

>
> Well if we define finite as that of 10^500 and below and everything
> above as
> infinite. Then by logic should not the 8 in 899999.....999999 be in
> the 10^501 place-value?

Maybe, but I still don't see why 10^500 is finite and 10^500+1 is
infinite. Perhaps you should use a different term, like "AP-infinite",
to distinguish your concept of "infinite" from the usual meaning.

In the usual meaning, infinity+1 = infinity. But obviously 10^500+1 ≠
10^500,
so your "AP-infinite numbers" do not follow the usual rules of
infinity.

David R Tribble wrote:
>> If that is indeed the case, then you have to explain how 999...999
>> is equal to 9 x 10^999...999 + ..., which in turn means that it
>> is equal to 9 x (9 x 10^999...999 + ...) + ..., which in turn means it
>> is equal to 9 x (9 x (9 x 10^999...999 + ...) + ...) + ..., and so on.
>> How is this circular definition supposed to work?
>

Archimedes Plutonium wrote:
> Well, if the tenth Peano Axiom should be that finite is 10^500, and infinite
> is above that, then the 9 in 96666....66666 is in the 10^501 place value

Then that means that the "..." in your 96666...66666 is a series of
500-10
decimal digits, all of them '6'. It also means that 999...999 is a
sequence
of 500-6 digits (or is the total number of digits 500?).

So when you say that "..." indicates an "infinite number of digits",
you
really mean that it indicates an "AP-infinite number of digits", which
is
about 500 digits, right?

[plutonium....@gmail.com]

Owen Jacobson wrote:
> On 2009-02-03 02:57:02 -0500, Archimedes Plutonium

(snipped to the important stuff)

> >
> > Owen, please list the entire matrix for two place value of those ten objects,
>
> 'ge', 'ga', 'gj', 'gd', 'gb', 'gg', 'gc', 'gi', 'gh', 'gf',
> 'ie', 'ia', 'ij', 'id', 'ib', 'ig', 'ic', 'ii', 'ih', 'if',
> 'je', 'ja', 'jj', 'jd', 'jb', 'jg', 'jc', 'ji', 'jh', 'jf',
> 'de', 'da', 'dj', 'dd', 'db', 'dg', 'dc', 'di', 'dh', 'df',
> 'ce', 'ca', 'cj', 'cd', 'cb', 'cg', 'cc', 'ci', 'ch', 'cf',
> 'ae', 'aa', 'aj', 'ad', 'ab', 'ag', 'ac', 'ai', 'ah', 'af',
> 'fe', 'fa', 'fj', 'fd', 'fb', 'fg', 'fc', 'fi', 'fh', 'ff',
> 'he', 'ha', 'hj', 'hd', 'hb', 'hg', 'hc', 'hi', 'hh', 'hf',
> 'be', 'ba', 'bj', 'bd', 'bb', 'bg', 'bc', 'bi', 'bh', 'bf',
> 'ee', 'ea', 'ej', 'ed', 'eb', 'eg', 'ec', 'ei', 'eh', 'ef'
>

Excellent, excellent, all is excellent here. And I hope Owen can
help me extract the Peano axioms out of this array. Not only
the original Peano Axioms, because they are probably inflated
with excess baggage, but even a more streamlined packet
of Peano Axioms.

The point of this exercise is to show that ALL Possible Arrangements
whether digits or anything else is far more powerful than the Peano
Axioms and the Dedekind Program. Where the All Possible Arrangements
can deliver the Peano Axioms and the Dedekind Program because they
are minor subsets of All Possible Arrangements.

Prior to this posting I have called it All Possible Digit Arrangements
but I need
to be more Universal and call it All Possible Arrangements. It comes
from
Probability Theory as its Space of All Possible Outcomes. And where
Set theory
is a minor subset of Probability Theory.

So I am hoping that Owen can help me get this streamlined Peano
Axioms.

> Without an axiom for describing a total order, this enumeration is as
> valid as any other. The moment you stepped outside of numbers, I
> stopped assuming *any* order - if you'd like to use one, say so; it
> doesn't change my own work at all, but I can accomodate.

Good, I kind of wished you were more chaotic in your listing to make
it
even a bigger challenge.

But it says something to me, metamathematically. That Counting is
deeper than counting numbers or counting objects. That the Universe
puts itself in an "ordering"

Now suppose we gave every human the challenge of writing out all
possible
combinations of those ten letters in two-slots as what Owen did.

Now given that challenge to every human alive that can do that
challenge
how many of those inputs would have what can be described as no
pattern
whatsoever, and those that have a pattern such as Owen's above. A
pattern
that can facilitate the challengee so that they can get every
combination
in a short time period.

Of course there will be some inputs that look as though they thought
randomly
and chaotically and by the time they reached several rows they have to
go back
through them to make sure they are not repeating a combination.

Now if I were given that challenge of a,b,c,d,e,f,g,h,i,j in two slots
all possible
arrangements I would have given this as my input:

aa
ab
ac
ad
ae
af
ag
ah
ai
aj
ba
bb
bc
bd
be
bf
bg
bh
bi
bj
ca
cb
cc
cd
ce
cf
cg
ch
ci
cj
.
.
.
ja
jb
jc
jd
je
jf
jg
jh
ji
jj

And the reason I would have given that is because I was given
a,b,c,d,e,f,g,h,i,j
in that order and I wanted to do the list as fast as possible without
having to go
back and forth constantly to see if I was repeating a combination
earlier.
By listing in the way I have done above I eliminate that worry of
repeating.

>
> > Then based on your two place matrix, list what the first four and last four
> > "entries (which will be numbers)" are for the three place value matrix.
>
> 'gec', 'gej', 'gei', 'ged', 'geb', 'gef', 'gea', 'geg', 'geh', 'gee',
> 'gac', 'gaj', 'gai', 'gad', ..., 'efa', 'efg', 'efh', 'efe'
>

Okay, good, in my input the first four and last four would have looked
like this:

aaa
aab
aac
aad
.
.
.

jjg
jjh
jji
jjj

> I've listed the first fourteen, rather than the first four, so that you
> can see that there is a pattern. I'm generating each new dimension by
> appending one more character to each number.
>

Okay, as I said before, there seems to be a metamath of a Counting
before we even get to Numbers and count those numbers. That counting
is contained inside the concept of All Possible Arrangements and that
is what I now want to fetch out of the listing of All Possible
Arrangements.

> Aside: what's with this fixation on base 10? Any number that's
> describable in base 10 is also describable in base 2 and vice versa,
> and it'd make these lists much shorter and easier to work with. Once
> again, it makes no difference to me, and I can accomodate ten-digit
> demonstrations if it makes you more comfortable, but it's rather
> tedious to generate digit groups from ten digits, even with mechanical
> help.
>

Not a fixation but a habit of modern society. Much like the English
versus the Metric, once a country uses the English measures it is
out of habit that they do not use metric.

I prefer Binary, but we all use Decimal.

Okay, Owen. Let us now try to see how we get the Peano Axioms out of
All Possible Arrangements.

Picture yourself at this huge math convention where every human being
who
can possibly fill out a paper page with the task of All Possible
Arrangements
of ten items in two slots. And you are the judge at this Convention
and has
before him 3 billion sheets of paper of submitted All Possible
Arrangements
and it is your task to make up some Rules (which would be axioms) that
forces
every one of those inputs to end up looking like this, which would
mean also
that your own input above has to look like this:

aa
ab
ac
ad
ae
af
ag
ah
ai
aj
ba
bb
bc
bd
be
bf
bg
bh
bi
bj
ca
cb
cc
cd
ce
cf
cg
ch
ci
cj
.
.
.
ja
jb
jc
jd
je
jf
jg
jh
ji
jj

Rules:
1) given a,b,c,d,e,f,g,h,i,j we take that as the primary order where
"a" is first and
"j" is last
2) there is an algebra over this with addition and multiplication
3) there is an additive identity and it is the first element
4) there is a multiplicative identity and it is to be the second
element
5) The succeeding element is to be the previous element plus the
multiplicative identity

Owen, I believe my five Rules above are the Peano Axioms in a more
distilled
and refined form. And it shows that All Possible Arrangements is
deeply, more
basic, more fundamental than even the axiom way of doing mathematics.

So when you are the Math Convention Judge with a 3 billion sheets of
paper submitted
by 3 billion persons who have listed All Possible Arrangements of ten
letters
in two slots, that the above five Rules will force every submitted
paper to end up
looking like this:

aa
ab
ac
ad
ae
af
ag
ah
ai
aj
ba
bb
bc
bd
be
bf
bg
bh
bi
bj
ca
cb
cc
cd
ce
cf
cg
ch
ci
cj
.
.
.
ja
jb
jc
jd
je
jf
jg
jh
ji
jj

Comment: what is fascinating about those five Rules as the Peano
Axioms is the
fact that you need to state a initial ordering as Rule one. Peano
Axioms never did that.
And so, the Peano Axioms could deliver the negative integers rather
than the positive
integers. So according to the Peano Axioms, they could be 0, -1, -2,
-3,..... Peano
has only the existence of 0. By not giving both 0 and 1. The Peano
Axioms are the
Negative Integers. But my first rule eliminates that Peano flaw.

[plutonium....@gmail.com]

Simultaneous to this conversation is the one with Owen where we
look at how many elements in a probability space of ten items arranged
in 2 slots and we have the formula items^slots. So that ten items in 2
slots
is 10^2 , ten items in three slots is 10^3 and that makes sense in
that
we have 99 as the largest number for 2 slots and 999 as the largest
number
for 3 slots. So slots become Place-Value.

So, now, let me finally solve this Place-Value problem. This is the
second
edition and I do not want to go into the 3rd edition without solving
Place Value.

I am almost there, David Tribble, as to solving place value.

I need the idea of defining finite from infinite and that is the
Physics cutoff
of 10^500 as no more measurable Physics allowed. So that is as good as
infinity region or realm.

I have another cutoff that is mathematical. This is a special math
number
of Algebra where I have 10000....00000 x 00000....00001 = ?
The product is not 10000.....00000 but is 0000....000100....000000
and the question is what place-value is that "1" digit in that number?

Similarly in Reals we have the same situation where we make the cutoff
of
finite to infinite as 10^(-)500. Then we have the special Algebraic
Number
of

1d0000.....00000 x 0d0000....00001 = ?

The answer is not 0d0000....00001 but rather is
0d0000.....000100....0000

So here I can assign the place value of that special Infinite Integer
above
or that special Real Number as that of 10^501 and 10(-)501
respectively.

But I am not finished in defining Exponentiation. I need to define it
so that
All Possible Digit Arrangements allows such numbers:
01234567891011121314.........9999....999989999.....99999

What that number is is the encapsulation of every number into making
one number.

Suppose that the only numbers that exists are 1,2,and 3 then that
encapsulated would
be a new number that is 123. So instead of the number 1, number 2, and
number 3
we have the new number of 123.

So in All Possible Digit Arrangements there has to exist these
Encapsulated Numbers
inside that Array, not just 0000....00001 and 22222....22222 but every
number
encapsulated as a new number.

So Exponentiation is fine and dandy up to 10^500, as what we think of
as finite
exponentiation. But beyond 10^500 we defined as infinite.

And infinity behaves differently from finite as we well know that
there are as many
Even Numbers 2,4,6,8, ..... as there are all-numbers 1,2,3,4,5,.... in
that we can
have a 1 to 1 correspondence or bijection. So exponention for place-
value on infinite
integers is going to be different from finite integers.

And we all know that in finite integers, the mulitiplication delivers
a larger number than
its multipliers. But in Infinite Integer multiplication the product is
often smaller than
either multipliers. So the Place-Value Exponentiation has to be
different from finite
versus infinite.

So what I have thought of is that Exponentiation of Place Value
becomes only
meaningful in terms of "location" of a digit. So that in Finite
Integers the location of
the "5" in 9995777 is the 10^3 place value. The location of the "5" in
58888....88888 is all that Place Value function that is needed. Place
Value is not
needed as a measure of size or quantity as is in finite integers. So
Place-Value
has only one function that is required for both Infinite and Finite
numbers and that
function is location. For finite integers, the place-value takes on
expanded functions
of telling you not only location but quantity and size. The set of
even integers is the
same quantity and same size as the total integers. So that means that
Place Value
in mathematics has only one true and important function and that is to
tell location.

So for Finite Numbers of 10^500 or less we use the ordinary Place-
value. But for numbers
that are Infinite we use Place-Value purely to tell the location of a
digit within a string
of digits.

So, for the number 0123456.....99999....999989999....99999 which is
the
encapsulation of every Integer as one Integer, and by the way that
number would
be considered a Finite Integer in the Old Way of doing mathematics
because it
ended leftwards in zero digit. So for that number what is the Place
Value of the
"0" digit above? Well, here I am going to use a "trick" because Place
Value is no
measure of size or quantity of a number as it is for Finite numbers
but only a measure
of location. So the location of the "0" in the above number is
10^0012345.....9999

What is the place value of the "7" in this number?
799995999.....999999
So the place-value of the "7" is 10^0799995999....99999 and the place
value of the
"5" is going to be 10^0000005999....9999

In other words for Place Value of Finite Integers the Place Value
serves two functions
of size and location of a digit in a digit arrangement. But once we
get into Infinite
Realm the Place Value has only the one function of location and so I
define
Place Value on an infinite string as purely a means of communication
as to where
a digit is located.

[plutonium....@gmail.com]

plutonium.archime...@gmail.com wrote:

Also, some other interesting observations for which the Algebra and
Peano Axioms of the past never really highlighted or even realized.

That Numbers Systems, whether they be Natural Numbers or
Reals, that they have to begin with the Additive Identity as the first
member and the Second Member has to be the Multiplicative Identity.

So the Peano system never realized that important feature.

As for Reals since they are infinite between the Additive and
Multiplicative Identity of 0 to 1, that we have a special Real Number
of 0d0000....00001 which is the multiplicative identity in the open
interval of 0 to 1 in Reals. So the Reals actually have two
multiplicative
identities one of 1d0000....00000 and one of 0d00000....00001.

[David R Tribble]

Archimedes Plutonium wrote:
> But I am not finished in defining Exponentiation. I need to define it so that
> All Possible Digit Arrangements allows such numbers:
> 01234567891011121314.........9999....999989999.....99999
> What that number is is the encapsulation of every number into making
> one number.

> [...]

>
> So, for the number 0123456.....99999....999989999....99999 which is

> the encapsulation of every Integer as one Integer, [...]

If that number is an Integer, and it contains all Integers, does that
number contain itself?

You are saying that it contains the Integer 0123...9998, since it
contains all the Integers, so I'm guessing that the next Integer
to the right of 0123...9998 is the Integer 0123...9999 itself,
right?

[Dan Christensen]

Yes. Sorry.

Just some advice, for what it is worth, on how you might proceed in
developing your new number theory: If you want to start from what I
have been calling your representation system, I think you would have
to first invent a formalism in the language of the set theory of your
choice to model this representation system of yours. (Yes, I am
probably using the word "model" incorrectly here.) You would probably
not want to use the standard ZFC axioms since PA can be derived
directly from them -- but not in a very pleasing way, IMHO. My guess
is that you will find this exceedingly difficult, if not impossible.
But what do I know? ;^) Best of luck.

Dan
Download my DC Proof software a http://www.dcproof.com

PS: You might want to play around with this DC Proof software of mine
to try out your ideas. It is based on my own simplified set theory --
not ZFC. (Yes, I am one of those awful math cranks that you see
here!!!)

[plutonium....@gmail.com]

David R Tribble wrote:
> Archimedes Plutonium wrote:
> > But I am not finished in defining Exponentiation. I need to define it so that
> > All Possible Digit Arrangements allows such numbers:
> > 01234567891011121314.........9999....999989999.....99999
> > What that number is is the encapsulation of every number into making
> > one number.
> > [...]
> >
> > So, for the number 0123456.....99999....999989999....99999 which is
> > the encapsulation of every Integer as one Integer, [...]
>
> If that number is an Integer, and it contains all Integers, does that
> number contain itself?
>

In the old math where things are half-way defined, with the aircorp
defining
things that were foggy, then you have paradoxes. Not because the math
warrants paradoxes, but because you defined halfway. In physics we
have
the same situation with Big Bang and Black Holes. When you get into
fiction
within science then you have paradoxes arise. Not because the paradox
is
a part of the science, no, but because the science is ill-defined.

So now the All Possible Arrangements of ten digits in 1 slot is
0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Now if the Probability Space is just ten digits in 1 slot then a
number such
as 0123456789 is not a number of that Space, but is a number for ten
digits
in 10-slots.

So at infinity-slots, there no longer is the pitiful so called Paradox
of set of all
sets. It was never a paradox within Mathematics. It was those who were
confused
about infinity. If you have infinity, you no longer have a Set of All
Sets Paradox.
If you no longer have gravity as angels driving planets, you no longer
have Angel
Paradox of how many can fit on the head of a needle.

So in the Old Math with there lack of understanding of the concept of
infinity, sure,
they would glum onto a piece of idiocy of set of all set paradox.

Infinity such as the even numbers are bijection with all numbers,
means you can
not have a set of all set paradox. So that at Infinity such as
99999.....999999
there are numbers as the above within 0 to 1 to 99999....999999.

The set of all set paradox arises when people do not properly
understand infinity.

Getting back on track to place-value.

I need a "best way" of Place-value location for Infinite Integers.
Someone said
Omega may do the trick for me. Since it is only location that I need
place value
for infinity that we call the 7 in 799999.....99999 as that of
Omega^-1 and the
2 in 8288888.....88888 as Omega^-2 place value so that we can easily
navigate
the location of digits at infinity. Of course, at the opposite end we
have the normal
place value of ones, tens 10^2, 10^3 etc etc. At the other end we have
a Omega
count down. So here, Lwalk comes in handy with his love of omega, in
that all of its
use is simply just one tiny use-- place value. Where Omega is simply a
fancy word
for Infinity. And there is only one type and one kind of Infinity, not
the Cantor nonsense.

But I think I have a better definition for infinity place value that
would be the most
useful. Remember, it is only location of a digit that place value
serves for Infinity strings.
So what if I said that the place value of 5 in 54444....444444 was at
the 10^544...43
place value. In other words, the place value of a digit is going to be
a predecessor
number. But will that work. Suppose the question is the place value
of
3 in 111111311111.....11111. Now the Omega scheme would quickly give
that as Omega^-6 but the predecessor scheme causes me too much labor
in trying
to arrive at 0000003111...11110 as the exponent for 10

At this moment it looks as though the Omega scheme or if Omega has too
much
"bad history" and bad connotation as to false ideas that it fostered
in the past, well
then we just use the word Infinity^-6 example above.

Because, all we need for Place Value is the leftmost portion of the
string as a location
finder and nothing more. Whereas in the rightmost portion of the
infinite string up to
10^500 the Place Value has not only a location function but a size
function and quantity
function but after 10^500 it has only a location function.

[lwa...@lausd.net]

On Jan 30, 3:54 pm, David R Tribble <da...@tribble.com> wrote:
> He refuses to define exactly what the '...' is supposed to mean beyond
> the vacuous explanation of "an infinite number of digits", offering no
> mathematically sound basis or explanation of it.

So once again, Tribble rejects the AP-reals just because
some of the require an ellipsis, the three dots '...', in
order to specify their digits.

In that case, one can reject the standard reals for the
exact same reason! There are some _standard_ reals whose
decimal digits can't be given without an ellpsis! One
well-known example is the uncomputable Chaitin's Omega.

The following link:

http://www.mathrix.org/liquid/archives/the-history-of-the-chaitin-leibniz-medallion

gives the first few binary digits of Omega as:

"000000100000010000011000100001101000111111..."

including at the end, that darned ellipsis! In
decimal, this converts to:

.0078749972... (decimal).

Tribble, if you could please define exactly what the
ellipsis above is supposed to mean beyond the vacuous
explanation of "an infinite number of digits," and
specify the digits of Chaitin's Omega -- or for that
matter, _any_ uncomputable standard real -- without
using those three dots. Until he does so, Tribble has
_no_ right to demand that AP do the same with the
ellipses in his AP-reals.

[lwa...@lausd.net]

On Jan 30, 11:59 am, David R Tribble <da...@tribble.com> wrote:
> Your arbitrary choice of 10^500 as the divding line between finite
> and infinite has no rigorous mathematical basis.

The problem is that AP wants there to exist a largest
finite number and a smallest infinite number.

Notice that with other attempts to extend the reals --
hyperreals, surreals, even Tribble's own suprareals --
there is neither a largest finite (limited, H_0, etc.)
number nor a smallest infinite (unlimited, H_1) number.

Since AP wants a largest finite, he chose what he
considers to be the largest meaningful number in
physics, 10^500. If Tribble doesn't consider googol^5
to be the largest physically meaningful natural, then
maybe one can achieve the same effect by letting N be
what Tribble considers to be the largest useful natural
number in physics. Then let N be the largest finite
number, and N+1 be the smallest infinite number.

Notice that this is similar to what WM and some of the
other ultrafinitists are proposing. Let the largest
useful natural in physics be the largest finite. The
only difference is that to WM, there are no infinite
numbers at all -- certainly not N+1.

I don't fully agree with AP's plan to call 10^500 the
largest finite number, but I feel that he was coerced
into naming such a number after Tribble kept insisting
that AP define the ellipses '..." in his AP-adics. I
feel that no largest finite AP-adic is necessary.

[lwa...@lausd.net]

On Feb 4, 6:28 pm, David R Tribble <da...@tribble.com> wrote:

> Archimedes Plutonium wrote:
> > So, for the number 0123456.....99999....999989999....99999 which is
> > the  encapsulation of every Integer as one Integer, [...]
> If that number is an Integer, and it contains all Integers, does that
> number contain itself?
> You are saying that it contains the Integer 0123...9998, since it
> contains all the Integers, so I'm guessing that the next Integer
> to the right of 0123...9998 is the Integer 0123...9999 itself,
> right?

It wouldn't necessarily be a contradiction if it did. There's no
reason that a string of digits can't contain itself as a proper
substring of digits. All that means is that the string of digits
must be Dedekind infinite.

Once again, if Tribble criticizes 0123...9999 because it must
apparently contain itself as a proper substring, then one might
point out that the exact same thing happens with the standard
reals -- in particular, the some standard rationals, as well!

Let's consider 1/p, for p a prime coprime to the base 10 of the
decimal system. Say p = 7, giving the decimal expansion:

0. 142857142857142857...

Then let's consider the proper substring of the string defined
by the standard real 1/7, given by removing the first six
decimal digits of the real. We obtain:

142857142857142857142857...

the exact same string of digits!

Thus the standard real 1/7 has the exact same problem that the
AP-adic 0123...9999 has! Tribble's objections to 0123...9999
apply _equally_ to 1/7 as well!

So it's possible for an AP-adic such as 0123...9999 to be
the concatenation of all the AP-adic strings, even including
0123...9999 itself. One can show that the digits therefore
must not be wellordered -- but that doesn't stop them from
being totally ordered or well-defined.

So consider a relation alpha such that:

1. alpha is a total order.
2. Every object save the minimal object has a predecessor,
and every object save the maximal object has a successor.
3. alpha is not isomorphic to 0 or 1.
4. alpha x alpha is isomorphic to alpha itself.
(x is the multiplication of order-types)
5. alpha is not a wellorder.
(redundant, but mentioned for emphasis)

Not until someone proves that no such alpha having all
five properties above can exist, can anyone claim that
AP's notions are contradictory.

Otherwise let the digits of any AP-adic, as well as
the AP-adics themselves, be order-isomorphic to alpha.

[plutonium....@gmail.com]

lwal...@lausd.net wrote:

(some snipping)

Thanks for the term "ellipses" for the three dots to indicate
infinity.
I much rather prefer a new term I made up as "infinitculum" and
"finitculum".

Lwal, do you have an opinion as to my calling the place-value of
say "1" in 137999.....99999 as the Omega place value and the
"3" as the Omega^-1 place value and the "7" as the Omega^-2 place-
value
and so on and so forth? I like the idea of using the negative numbers
going
backwards.

Or should I use the word "Infinity" rather than the word "Omega"

>
> So it's possible for an AP-adic such as 0123...9999 to be
> the concatenation of all the AP-adic strings, even including
> 0123...9999 itself. One can show that the digits therefore
> must not be wellordered -- but that doesn't stop them from
> being totally ordered or well-defined.
>

The arguement of whether 0123....99999 is a submember as well
as the Set of All Sets Paradox is to me a no-brainer. That in
infinity,
infinity itself is the Set of All Sets and thus there is no paradox.
As well as there are numbers that encapsulate all the other numbers
since they are one of all possible digit arrangements. As you said
yourself, the Reals have these encapsulating numbers.

So I simply say that Set of All Sets was never a paradox because
they just forgot that these are in the infinite realm, not the finite
realm
where they tripped up and became confused and thus pinned it as
a paradox when it was just their lack of understanding of infinity.

> So consider a relation alpha such that:
>
> 1. alpha is a total order.
> 2. Every object save the minimal object has a predecessor,
> and every object save the maximal object has a successor.
> 3. alpha is not isomorphic to 0 or 1.
> 4. alpha x alpha is isomorphic to alpha itself.
> (x is the multiplication of order-types)
> 5. alpha is not a wellorder.
> (redundant, but mentioned for emphasis)
>
> Not until someone proves that no such alpha having all
> five properties above can exist, can anyone claim that
> AP's notions are contradictory.
>
> Otherwise let the digits of any AP-adic, as well as
> the AP-adics themselves, be order-isomorphic to alpha.

I do not know what you are calling "total order" from that of "well
order".
The AP-adics are All Possible Digit Arrangements and that gives
"for any given number N, there is an N-1 and a N+1"

To me, given any number N there is a predecessor and successor
means it is Well Ordered and Total Ordered.

So, how, Lwal, are you defining Well Order as different from Total
Order.
And does not Well Order mean N has both a predecessor and successor.

What I think is that you have vestiges of Cantor and Reals and is
thinking
that Cantor said the Reals are not well-ordered.

So does Well Ordered mean N has both predecessor and successor? I
think it
does.

And I think that All Possible Arrangements of Digits creates
inescapably a matrix
that is Well Ordered as N with both predecessor and successor.

So give me any definition of Well Order, and I bet it always comes
back around to
simply meaning N has a predecessor and successor.

As for Total Order, that is the first time I seen that one.

[plutonium....@gmail.com]

plutonium.archime...@gmail.com wrote:
> lwal...@lausd.net wrote:
>

> > So consider a relation alpha such that:
> >
> > 1. alpha is a total order.
> > 2. Every object save the minimal object has a predecessor,
> > and every object save the maximal object has a successor.
> > 3. alpha is not isomorphic to 0 or 1.
> > 4. alpha x alpha is isomorphic to alpha itself.
> > (x is the multiplication of order-types)
> > 5. alpha is not a wellorder.
> > (redundant, but mentioned for emphasis)
> >

Took a look at definitions of Total Order, Well Order, Axiom of
Choice.

--- quoting Wikipedia on those three ---

Wikipedia on Total Order and Well Order and Axiom of Choice

In mathematics and set theory, a total order, linear order, simple
order, or (non-strict) ordering is a binary relation (here denoted by
infix ≤) on some set X. The relation is transitive, antisymmetric, and
total. A set paired with a total order is called a totally ordered
set, a linearly ordered set, a simply ordered set, or a chain.

If X is totally ordered under ≤, then the following statements hold
for all a, b and c in X:

If a ≤ b and b ≤ a then a = b (antisymmetry);If a ≤ b and b ≤ c then a
≤ c (transitivity);a ≤ b or b ≤ a (totality).

In mathematics, a well-order relation (or well-ordering) on a set S is
a total order on S with the property that every non-empty subset of S
has a least element in this ordering. Equivalently, a well-ordering is
a well-founded total order. The set S together with the well-order
relation is then called a well-ordered set.

In mathematics, the axiom of choice, or AC, is an axiom of set theory.
Informally put, the axiom of choice says that given any collection of
bins, each containing at least one object, it is possible to make a
selection of exactly one object from each bin, even if there are
infinitely many bins and there is no "rule" for which object to pick
from each. The axiom of choice is not required if the number of bins
is finite or if such a selection "rule" is available.
--- end quoting Wikipedia ---

I am too tired tonight, but hope by tomorrow to show that since the AP-
adics
have the feature of given any N it has both a predecessor N-1 and a
successor
N+1, and that confers Well Ordering onto the AP-adics as also the
Reals are
Well Ordered.

Now I wonder if I can prove it purely from Probability theory of its
Universal Space
of All Possible Outcomes? Lwal, would it not be logical that a
Probability Space
must be Well Ordered? It seems to me that Probability cannot operate
with its
Universal Space of Outcomes as being anything less than Well Ordered,
for you
would not have a subject of Probability theory without it being Well
Ordered.

I did read some of the equivalencies of Axiom of Choice. But I bet
that since
Peano Natural Numbers as finite numbers, since they are well-ordered
that
Well-Ordering and Axiom of Choice are nothing but confused definitions
of
this concept--- given any N, there is a predecessor N-1 and a
successor N+1.

I would almost bet that those are equivalent.

[Jesse F. Hughes]

lwa...@lausd.net writes:

> I don't fully agree with AP's plan to call 10^500 the
> largest finite number, but I feel that he was coerced
> into naming such a number after Tribble kept insisting
> that AP define the ellipses '..." in his AP-adics. I
> feel that no largest finite AP-adic is necessary.

Yes, that bastard should stop forcing AP to make himself clear. When
he does so, he coerces AP into saying stupid things. It's a damn
shame.

--
Jesse F. Hughes

"You know that view most people have of mathematicians as brilliant
people? What if they're not?" -- James S. Harris

[Jesse F. Hughes]

lwa...@lausd.net writes:

> On Jan 30, 3:54 pm, David R Tribble <da...@tribble.com> wrote:
>> He refuses to define exactly what the '...' is supposed to mean beyond
>> the vacuous explanation of "an infinite number of digits", offering no
>> mathematically sound basis or explanation of it.
>
> So once again, Tribble rejects the AP-reals just because
> some of the require an ellipsis, the three dots '...', in
> order to specify their digits.
>
> In that case, one can reject the standard reals for the
> exact same reason! There are some _standard_ reals whose
> decimal digits can't be given without an ellpsis! One
> well-known example is the uncomputable Chaitin's Omega.

How deceitful.

In Chaitin's Omega, it is perfectly clear what is intended to follow
the three dots: an omega sequence of digits. Which digits may be
unknown, but we know what type fills in the dots. In AP's stuff, it
is perfectly unclear. Consider

999...999

Is there an omega sequence of 9's followed by three 9's? (Probably
not.) Is there an omega sequence of 9's, followed by an omega^op
sequence of 9's? Maybe, but if I recall correctly, he sometimes
writes something like

987...000...789,

and that can't be described thus.

Tribble's complaint is not that he doesn't know what digits go in the
dots, but rather that he has no idea what sort of thing the dots
indicate.

> Tribble, if you could please define exactly what the
> ellipsis above is supposed to mean beyond the vacuous
> explanation of "an infinite number of digits," and
> specify the digits of Chaitin's Omega -- or for that
> matter, _any_ uncomputable standard real -- without
> using those three dots. Until he does so, Tribble has
> _no_ right to demand that AP do the same with the
> ellipses in his AP-reals.

Oh, bullshit.

Your response is utterly disingenuous. It's a swell and noble calling
you have here, this crank defense organization, but please don't sink
to their ignorance in your responses.

--
"I'm the theory guy.
Other people are the experimental people.
If you push me on details I get annoyed, as I'm the theory guy.
I'm the theoretical amateur mathematician." --James S. Harris, poet

[Ken Quirici]

Neat!

> Thus the standard real 1/7 has the exact same problem that the
> AP-adic 0123...9999 has! Tribble's objections to 0123...9999
> apply _equally_ to 1/7 as well!
>

However your example was decimal representation which was
a repeating finite decimal, so I'm not sure it covers the
case of an infinite sequence of digits that occurs within
itself.

If that's not the case, then at least 'constructively' the
decimal string never gets to it's end so it can start over
again.

Is that because the number of digits in a decimal
representation is countable?

So it would seem necessary to find an irrational between
0 and 1 whose decimal representation contains itself - or
prove it's possible anyway. I'm not sure what condition
on a decimal representation guarantees it represents an
irrational - that it's last part is not a repeating
[finite?] string of digits?

e.g.

.089220..(non-repeating stuff)......458458458..(repeating)..

(this of course is an example of a decimal representation that
doesn't contain itself).

[Ken Quirici]

I think what I was (barely articulately) groping for was
this:

Suppose A is the decimal representation which
contains itself as a proper subset.

Suppose B is a proper subset of A which = A.

Then in your case of the repeating decimal,
A-B will be finite for any proper subset B
which = A (I think).

What about the case where A-B is not finite?

Regardless of whether A-B, or B comes 'first'
in enumerating the digits of A, you'll never
get to it's end so you can start the other.

[plutonium....@gmail.com]

Alright, I was too tired last night to prove that AP-adics and Reals
are Well Ordered. And
I had to refresh myself on what Total-Order means.

It is known that Axiom of Choice is equivalent to Well-Order. It is
known that the
Peano Axioms are WellOrdered. And why is that? It is because of
Mathematical-Induction.

So we have this known, that Axiom of Choice == WellOrder == Peano
Natural Numbers ==
Mathematical Induction (where == means equivalent)

So all I have to do is show that for any AP-adic that given any number
N (except 0)
that N has both a N-1 and a N+1. Given any number N (except 0) has
both a
predecessor and a successor. That means AP-adics have Mathematical
Induction.
Now for the Reals, they also have Mathematical Induction for given any
Real Number
call it R there is a R- 0d0000....00001 and a R+ 0d0000.....00001 (and
including the
Real 0) So the Reals have Math Induction and are WellOrdered.

Now the real exciting aspect that I raised the issue on last night was
that I said
since Set theory is only a minor subset of Probability theory that
Probability theory
can never function as a theory because All Possible Digit Arrangements
or All
Possible Arrangements comes from Probability theory as its Outcome
Space.
So you cannot have Probability theory if it was not WellOrdered. And
now here I can
make sense of that statement. What is Mathematical Induction in its
strongest form?
The strongest form of Math Induction is what I gave above that given
any N
you have both a predecessor and successor (except for 0). What is that
in terms of
Probability theory? It is simply Causality. Causality backwards for
the past history
and causality forwards for the future.

So Probability theory has to be wellordered because that means it has
Causality
where such and such an action leads to such and such a result.

So now I have these Equivalencies where == means equivalent:

Math Induction == All Possible Arrangements == mainspring of
Probability theory == Causality
== Axiom of Choice == WellOrdering == Peano Natural Numbers == AP-
adics
== Reals

Math Induction is Causality is WellOrdering for if we want to know
history is a
WellOrdering backwards in time as the predecessor and continuing down
the
line of predecessors. And science experimentation is WellOrdering of
Successor
and up the line of future successors.

So in the history of mathematics with much of its time spent on Axiom
of Choice
and WellOrdering was time wasted on too much over nothing. For the
Axiom of
Choice and WellOrdering were simply diverse forms of one and the same
concept--
you have Math Induction and that means for every N there is a
predecessor and
successor (except 0).

So I leave this with a question for Lwal. Please tell me if you know
of any set
that does not have the feature of given any N there is a predecessor
and successor.
I know the Reals and AP-adics and the Peano Natural-Numbers have that
feature.
Does Hensel P-adics have that feature? I suspect they do and so the
Hensel P-adics
are a morphed form of the Reals. Do the surreals have that feature--
they do not and
so they are fiction. Do the hyperreals have that feature -- they do
not and again
they are fiction.

So it boils down to only two sets in all of mathematics that has the
feature of
WellOrdered and those two are AP-adics and Reals and everything else
is a
morphed (error prone) subset of either AP-adics or Reals.

[LudovicoVan]

On 6 Feb, 12:53, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> lwal...@lausd.net writes:
> > On Jan 30, 3:54 pm, David R Tribble <da...@tribble.com> wrote:
> >> He refuses to define exactly what the '...' is supposed to mean beyond
> >> the vacuous explanation of "an infinite number of digits", offering no
> >> mathematically sound basis or explanation of it.
>
> > So once again, Tribble rejects the AP-reals just because
> > some of the require an ellipsis, the three dots '...', in
> > order to specify their digits.
>
> > In that case, one can reject the standard reals for the
> > exact same reason! There are some _standard_ reals whose
> > decimal digits can't be given without an ellpsis! One
> > well-known example is the uncomputable Chaitin's Omega.
>
> How deceitful.
>
> In Chaitin's Omega, it is perfectly clear what is intended to follow
> the three dots: an omega sequence of digits.  Which digits may be
> unknown, but we know what type fills in the dots.  In AP's stuff, it
> is perfectly unclear.  Consider
>
>   999...999
>
> Is there an omega sequence of 9's followed by three 9's?  (Probably
> not.)  Is there an omega sequence of 9's, followed by an omega^op
> sequence of 9's?

As far as I have seen, that's simply meant as an omega sequence (of
9's, in this case), from start to end. (Although, in analogy to the P-
adics, I am not 100% sure calling it an omega sequence is completely
correct: my own lack of proper knowledge, in any case. Just think P-
adics, and you won't get it too wrong.)

> Maybe, but if I recall correctly, he sometimes
> writes something like
>
>   987...000...789,
>
> and that can't be described thus.

I've seen AP consistently writing the same digit before and after the
ellipsis, so that no ambiguity is ever possible. In fact, if I am not
mistaken, he has more than once restated the concept in this thread
too.

You are just playing dumb: no big news, of course.

-LV

[plutonium....@gmail.com]

plutonium.archime...@gmail.com wrote:

> So I leave this with a question for Lwal. Please tell me if you know
> of any set
> that does not have the feature of given any N there is a predecessor
> and successor.
> I know the Reals and AP-adics and the Peano Natural-Numbers have that
> feature.
> Does Hensel P-adics have that feature? I suspect they do and so the
> Hensel P-adics
> are a morphed form of the Reals. Do the surreals have that feature--
> they do not and
> so they are fiction. Do the hyperreals have that feature -- they do
> not and again
> they are fiction.
>
> So it boils down to only two sets in all of mathematics that has the
> feature of
> WellOrdered and those two are AP-adics and Reals and everything else
> is a
> morphed (error prone) subset of either AP-adics or Reals.
>

Sorry, I made a mistake there for I should have thought quicker on my
feet.
I should have remembered what makes the Reals wellordered is the fact
that
you can use 0d0000.....0001 as a multiplicative identity and endless
add that
number to collect every Real Numbers.

But with Hensel P-adics the radix must be finite and so the Hensel P-
adics
could not be a morphed Reals. So the question is whether the Hensel P-
adics
are another form of AP-adics? And I think they are except they need
cleaning
up of the dirt encrusted silly definitions and restrictions that the
Hensel P-adics
have associated. For example, they believe 9999....9999 was -1. But
then
Hensel P-adic followers never had FrontView and BackView and they
never
had All Possible Digit Arrangements. So there was alot they never had
in
Hensel P-adics.

[Ken Quirici]

And what are the start and end of the omega sequence
3.14159....., the digits of the decimal representation of pi?

In fact, if you start

9999....9999999 from the left, when do you meet the sequence coming
from
the right?

Given any 987234.....09234, and given that this indicates you
have a calculable sequence of digits from both left and right,
how can you not have the two sequences fly by each other - and
note there's no particular nth digit from the left or
mth digit from the right such that it represents a meeting place
of the two sequences, since they would then both be finite. So
they both grope towards each other
never meeting.

So to define any abc......xyz, you really can
only claim this represents two distinct sequences that you
choose to represent as one. What advantage do you gain
by this? I'm not saying there isn't any, I just can't
see it at the moment.

[herbzet]

plutonium....@gmail.com wrote:
> lwal...@lausd.net wrote:
>
> (some snipping)
>
> Thanks for the term "ellipses" for the three dots to indicate
> infinity.

That's "ellipsis". It has several meanings.

--
hz

[Jesse F. Hughes]

LudovicoVan <ju...@diegidio.name> writes:

>> In Chaitin's Omega, it is perfectly clear what is intended to follow
>> the three dots: an omega sequence of digits.  Which digits may be
>> unknown, but we know what type fills in the dots.  In AP's stuff, it
>> is perfectly unclear.  Consider
>>
>>   999...999
>>
>> Is there an omega sequence of 9's followed by three 9's?  (Probably
>> not.)  Is there an omega sequence of 9's, followed by an omega^op
>> sequence of 9's?
>
> As far as I have seen, that's simply meant as an omega sequence (of
> 9's, in this case), from start to end.

Can't be!

Omega sequences have no end. It makes no sense to write 999...999 to
stand for an omega sequence of 9s.

--
Jesse F. Hughes

"Do not click any hyperlinks that you do not trust. Type them in the
Address bar yourself." -- Microsoft gives security advice.

[Jesse F. Hughes]

LudovicoVan <ju...@diegidio.name> writes:

>> Maybe, but if I recall correctly, he sometimes
>> writes something like
>>
>>   987...000...789,
>>
>> and that can't be described thus.
>
> I've seen AP consistently writing the same digit before and after the
> ellipsis, so that no ambiguity is ever possible. In fact, if I am not
> mistaken, he has more than once restated the concept in this thread
> too.
>
> You are just playing dumb: no big news, of course.

Again, you're simply wrong. Even with the same digit before and after
the ellipsis, notation like 0000....00100....00000 is very ambiguous.
What does it mean? In what position is the "1"?

Furthermore, he discusses all possible digit sequences, so *obviously*
the notation is ambiguous! Anything can go in the ellipsis.

Finally, I don't believe that he is so consistent as you say (that the
last digit before and after the ellipsis are the same), but I confess
I haven't found a counterexample. Seems they came up various times
that he's tried to define multiplication, but I didn't find such an
example.

--
"When you go to class today, if your professor talks about algebraic
number theory, or misuses Galois Theory[,] I want you to carefully
notice how you feel. Hold on to that feeling so that you never forget
it." --James S. Harris, on channeling rage via Galois theory.

[LudovicoVan]

On 7 Feb, 00:03, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> >> In Chaitin's Omega, it is perfectly clear what is intended to follow
> >> the three dots: an omega sequence of digits.  Which digits may be
> >> unknown, but we know what type fills in the dots.  In AP's stuff, it
> >> is perfectly unclear.  Consider
>
> >>   999...999
>
> >> Is there an omega sequence of 9's followed by three 9's?  (Probably
> >> not.)  Is there an omega sequence of 9's, followed by an omega^op
> >> sequence of 9's?
>
> > As far as I have seen, that's simply meant as an omega sequence (of
> > 9's, in this case), from start to end.
>
> Can't be!
>
> Omega sequences have no end.  It makes no sense to write 999...999 to
> stand for an omega sequence of 9s.  

You (I don't mean you only) seem to have problems with notation: I
suppose you would have no objection in reading (9), but you cannot
accept 9...9.

-LV

[LudovicoVan]

On 7 Feb, 00:12, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> >> Maybe, but if I recall correctly, he sometimes
> >> writes something like
>
> >>   987...000...789,
>
> >> and that can't be described thus.
>
> > I've seen AP consistently writing the same digit before and after the
> > ellipsis, so that no ambiguity is ever possible. In fact, if I am not
> > mistaken, he has more than once restated the concept in this thread
> > too.
>
> > You are just playing dumb: no big news, of course.
>
> Again, you're simply wrong.  Even with the same digit before and after
> the ellipsis, notation like 0000....00100....00000 is very ambiguous.
> What does it mean? In what position is the "1"?

Nobody has been using that notation: no more than one ellipsis is
allowed, otherwise of course it is ambiguous. You keep injecting
confusion.

> Furthermore, he discusses all possible digit sequences, so *obviously*
> the notation is ambiguous!  Anything can go in the ellipsis.

Nope. Talking about all possible digit arrangements and the ambiguous
notation you inject have nothing to do together. Again, think P-adics:
at least that's something you should have respect for, I suppose.

> Finally, I don't believe that he is so consistent as you say (that the
> last digit before and after the ellipsis are the same), but I confess
> I haven't found a counterexample.

I appreciate you honesty.

-LV