Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

Writing a textbook that consolidates all the Adics and getting rid of base dependency; call them Infinite Integers and do they form a Field

8 views
Skip to first unread message

a_plu...@hotmail.com

unread,
Dec 19, 2005, 4:16:40 AM12/19/05
to
Dik will be happy to know I will define addition and multiplication and
they will have additive and multiplicative inverses. I will lump all
the Adics both p and n adics integers into one huge kettle. And I must
remark that at this very same time I am arguing in a different thread
that the definition of Field and Ring are ill-defined. I like
situations such as this because if one thread goes against me, the
other thread could go gloriously to my advantage. But I hope both
threads go for me.

Recently I bought two books on P-adics because Chris kept bringing up
the Hensel Lemma but on perusing these two books they have paid me a
huge dividend in insights. I expected no dividends and that I would end
up putting them on a shelve and rarely reading them. But today I
envisioned something quite remarkable. The two books are Gouvea P-ADIC
NUMBERS and Koblitz P-ADIC NUMBERS.

What I envisioned is how to get rid of the base dependency and I do it
this way.

...0000000, ....00001, .....000002, ....00003, .....000004, ....000005,
.....00006, ....00007, ....00008, ....00009, .....0000010, ....000011,
ad infinitum

I call it the first winding.

The second winding is this:

...11110, .....1111, ....1112, ....11113, .....1114, ....1115,
.....1116, ....1117, ....11118, ....111119, .....11111*10,
....11111*11, ad infinitum where * indicates distinction between say
the 1's and the numeral 10 or 11

The third winding is this:

...22220, .....22221, ....2222, ...22223, .....22224, ....22225,
.....2226, ....22227, ....22228, ....22229, .....2222*10, ....2222*11,
ad infinitum

All of them are INFINITE INTEGERS, the concept I came with when I first
came to the Internet in 1993 and never heard or knew of such a thing as
Adics and I was sadly surprized they were base dependent.

But tonight I wish to outline how to lump all the Adic Integers into
one set and toss out the base dependency.

Now every adic integer will be in that array above. The 10-adic
idempotents of ......2890625 and ....7109376 will be in the array of
....555555 winding and ...66666 winding respectively. Even though we
know that these idempotents will never end up looking like
555555555.....2890625

Addition and Multiplication of these numbers is the same as adding or
multiplying any two integers and they end up in the winding for which
the last number is such as the 10-adic idempotent is in the ....55555
winding.

Now the above array helps me with the nagging question of the mid 1990s
concerning my thesis that Natural-Numbers = Infinite Integers. The
question by dullards as to how the last finite-integer moves on to the
first infinite integer? So as one moves from 1, 2, 3, 4, 5, where does
that lead to the first number that looks like this....111111 or like
this ....454545. Where is the last finite integer that ends in 000000s
and the first infinite integer enter the sequence.

So the above array gives that answer by saying the first infinite
integer that enters the picture is ....1111110 as the first number in
the 2nd winding. But I cannot tell you the last number in the 1st
winding. But this should not trouble any of us because we can never
tell someone what is the last number in Rationals between 1 and 2 can
we. Nor can we tell someone what is the last number in Reals between 2
and 3, but we can say that the first number between 3 and 4 in Reals is
3.

So the above array dismisses the Adics with their base dependency.

And we should now be able to write a High School textbook that an
average High School student would be able to understand and work with
Infinite Integers or the Adics without base dependency.

Do these Infinite Integers form a Field? I think the answer is yes
because they have all the inverses and obey all the Field rules.

Now, do the above Infinite Integers answer the old problems of
mathematics of Fermat's Last Theorem, Goldbach Conjecture, Twin Primes,
Riemann Hypothesis? I think they do in that FLT is false and has
counterexamples in every power higher than squaring. As for the
Goldbach Conjecture there are numbers in that array that are even yet
are not the sum of two primes. The Riemann Hypothesis is shown false
because there are counterexamples such as ...9999999 or square roots of
-1.

Are there any new surprizes to Number theory? Yes, there are other
numbers besides 2 which are both even and yet prime such as the 10-adic
idempotent of ....7109376.

Do the Infinite Integers have anything new to say about what prime
numbers are and their infinitude? Yes, in that some primes are of a
different class than what we are familar with. Such as the even primes
besides 2 such as the 10-adic idempotent of ....7109376

A note about history of mathematics, in that the Counting numbers were
discovered in history first and later came base representation of
Counting numbers. As for Infinite Integers, their base representation
came first per Adics and only now has anyone offered the full scope of
these Infinite Integers that are baseless.

Now I called the above as "Windings" because I believe they form a
sphere like surface so starting with ....0000 and the first winding
traces out a circular path and the 2nd winding is another circular path
increasing the first until all the windings trace out a sphere surface.

Another note is that in p-adic integers such as 10-adics we had a
number such as ...99999 which was -1, but in the above arrays, the
number ....99999 is not -1 but just another successor number in the
10th winding. In fact there are no negative numbers in Infinite
Integers.

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

a_plu...@hotmail.com

unread,
Dec 19, 2005, 2:03:39 PM12/19/05
to
Now I used the term "winding" for the subset that ends in 0000s then
ends in 11111s then ends in 22222s etc etc. Perhaps a better term will
be that of "geodesic" because my intuition tells me that these Infinite
Integers (baseless All-Adics) are the Native Numbers of Riemann
geometry. Each winding is a geodesic such as a line of longitude. But
that is much later.

Today I want to discuss what I wish to pull out of these Infinite
Integers. Of course I want them to be base-free or baseless, unlike the
P-adics and N-adics which are confined to specific bases such as
2-adics to base 2, 3-adics to base 3. I want to dispense of that
nonsense of base. And these Infinite Integers does that.

However, a problem immediately creeps in, in that I have repeating
forms. But this is a problem of Reals and Rationals also in that
2.99999..... is the same as 3.00000... etc etc. Here in Infinite
Integers a number such as ......1111111115 on the 111s winding is the
same as .....111111115 on the Elevens winding. In fact it is
indistinguishable between 111s winding and elevens winding, the same
with 2 windings and the number twenty-two winding. But like in the
Reals we simply ignore the fact that 2.9999.... is the same as
3.0000....


Here is a partial list of things I want in this Array:
(1) Baselessness, like the Natural Numbers are baseless.
(2) I want to achieve in this ARRAY of windings is all-inclusiveness.
Do I have every possible *digit arrangement* somewhere in the Array. I
believe so.
(3) Easy operation of addition and multiplication and I believe that
was achieved for it is the same thing as Natural Numbers.
(4) Very Important-- Sequential Ordering to the maximum. Although I
cannot tell you the last number in each winding, I can tell you the
first number in the next higher up winding. Which is no discredit to me
because no-one can tell what the last Real is between the open-set of
(2,3) since there is no last Real but the first Real in the next
open-set (3,4) is 3. And the Reals are Ordered, and these Infinite
Integers are also Ordered.

Now I mentioned in my post last night that there are surprizes, in fact
huge big surprizes in that the prime numbers in Infinite Integers are
vastly larger and vastly different in complexity. Because every
irrational number in Reals is represented by a Infinite Integer such as
pi and e where e is in each winding as .....828172. And this Infinite
Integer e is even and yet prime because it is irrational. So the number
2 in Natural Numbers is not the only even prime number but that Natural
NUmbers have an infinitude of even prime numbers.

And of course my old thesis is true that Natural Numbers are the
Infinite Integers and that the old way of thinking that Natural Numbers
are confined to just finite integers as the 0000s winding is just
stupidity.

Gottfried Helms

unread,
Dec 19, 2005, 2:21:51 PM12/19/05
to
Am 19.12.2005 20:03 schrieb a_plu...@hotmail.com:
> And of course my old thesis is true that Natural Numbers are the
> Infinite Integers and that the old way of thinking that Natural Numbers
> are confined to just finite integers as the 0000s winding is just
> stupidity.
>
Stupidity... so I'd to struggle to be acknowlegded by you
at first..
I think, there'll be no way to get that acknowledgement
for a stupid moron like I seem to be...

Gottfried Helms

a_plu...@hotmail.com

unread,
Dec 19, 2005, 2:36:28 PM12/19/05
to
Gottfried Helms <h...@uni-kassel.de> wrote:

Gottfried Helms

A.P. writes:
Well I am glad Gottfried weighed in with his opinion this day in 2005
because he weighed in on one of my very earliest posts to the Internet
in 1993. So it is nice for me to know some people have kept up with my
posts starting 1993 and now today of 19 Dec 2005.

A.P.

a_plu...@hotmail.com

unread,
Dec 19, 2005, 9:45:44 PM12/19/05
to
Now I should be able to do even better than list the Arrays of Infinite
Integers. I should be able to given any Infinite Integer X be able to
say what is its predecessor and what is its successor. So given any
Infinite Integer X, say what is X -1 and what is X + 1. Then I will
have conquered all of this problem. So I have not given what is the
predecessor of ....1111110 in the 111s Winding from the preceding
Winding of the 0000s winding.

As a song goes "a little help from friends". I hope Dik is not on Xmas
holidays or vacation and can give that little help just about now.

I have three windings listed below

...0000000, ....00001, .....000002, ....00003, .....000004, ....000005,
.....00006, ....00007, ....00008, ....00009, .....0000010, ....000011,
ad infinitum

...11110, .....1111, ....1112, ....11113, .....1114, ....1115,


.....1116, ....1117, ....11118, ....111119, .....11111*10,
....11111*11, ad infinitum where * indicates distinction between say
the 1's and the numeral 10 or 11

...22220, .....22221, ....2222, ...22223, .....22224, ....22225,
.....2226, ....22227, ....22228, ....22229, .....2222*10, ....2222*11,
ad infinitum

Now if I can just provide the predecessors in each winding in going
from 000s to 111s or going from 1111s to 2222s, we will have this
conquered.

One idea that may pay off is to consider that in *all possible digit
arrangements* within each winding that the maximum is the 999s digits.
So the maximum in the 0000s winding would be a number that is
99999999...... Same for 1111s and 2222s windings.

So can I say that the Predecessor of ....111110 is 99999999......

Can I say the Predecessor of .....222220 is 999999.....

If I can say that then the predecessor of 999999..... is
99999.....999998 and so on down the line.

The trouble is that the meaning of the 1111s winding and 22222s winding
is that those 1s and 2s go to infinity and so how can I say they
somehow become 9s?

So on a very difficult problem like this, it is best that the world
puts more than just my one mind to the problem, that another human mind
can see something I may have neglected or not seen.

So if I can get that last piece of the puzzle, that given any Infinite
Integer, I can thence tell you its Successor and Predecessor, then this
dragon is slain.

a_plu...@hotmail.com

unread,
Dec 19, 2005, 10:48:00 PM12/19/05
to

Dik, can I say that in the 000s winding that those zeroes at infinity
become filled up with 9s and that the last number in that winding is
999999.....99999. A sort of convergence at infinity where 9 digit
replaces every 0 digit. Can I say that?

Can I say that in the 1111s winding that those 1s become replaced with
9s and thus at infinity the last number in the 111s winding is
99999....99999?

Gottfried Helms

unread,
Dec 19, 2005, 11:28:27 PM12/19/05
to
Am 19.12.2005 20:36 schrieb a_plu...@hotmail.com:
>
> A.P. writes:
> Well I am glad Gottfried weighed in with his opinion this day in 2005
> because he weighed in on one of my very earliest posts to the Internet
> in 1993. So it is nice for me to know some people have kept up with my
> posts starting 1993 and now today of 19 Dec 2005.
>
> A.P.
>
Assume it as a christmas surprise... :-)

Well, your articles in 1996 (I think) made me first time
to think about these things, and to write a small text about
it. It was an interesting experience...

Gottfried Helms

Proginoskes

unread,
Dec 20, 2005, 2:20:00 AM12/20/05
to

I found out more about the All-Adics in one hour than you did in a
month. I found the problems a whole lot quicker. You should read the
point-by-point analysis, because you've ignored so much that you should
have studied.

AP wrote:

> Dik will be happy to know I will define addition and multiplication and
> they will have additive and multiplicative inverses. I will lump all
> the Adics both p and n adics integers into one huge kettle. And I must
> remark that at this very same time I am arguing in a different thread
> that the definition of Field and Ring are ill-defined.

What you mean here is "badly defined"; "ill-defined" has a particular
meaning in mathematics, specifically, if you have two objects A and B
which are different but equivalent, and a function f such that
f(A) is not equivalent to f(B), then f is ill-defined.

Another example of a truly ill-defined concept is the winding number
you
present later on.

> I like
> situations such as this because if one thread goes against me, the
> other thread could go gloriously to my advantage. But I hope both
> threads go for me.

In this one, you'll be okay, since you're exploring new territory,
unless
you say something stupid like "Fermat really asked the question in the
context of the All-Adics", since the All-Adics (and not even the
p-adics)
were studied in Fermat's time.

> Recently I bought two books on P-adics because Chris kept bringing up
> the Hensel Lemma but on perusing these two books they have paid me a
> huge dividend in insights. I expected no dividends and that I would end
> up putting them on a shelve and rarely reading them. But today I
> envisioned something quite remarkable. The two books are Gouvea P-ADIC
> NUMBERS and Koblitz P-ADIC NUMBERS.
>
> What I envisioned is how to get rid of the base dependency and I do it
> this way.
>
> ...0000000, ....00001, .....000002, ....00003, .....000004, ....000005,
> .....00006, ....00007, ....00008, ....00009, .....0000010, ....000011,
> ad infinitum
>
> I call it the first winding.
>
> The second winding is this:
>
> ...11110, .....1111, ....1112, ....11113, .....1114, ....1115,
> .....1116, ....1117, ....11118, ....111119, .....11111*10,
> ....11111*11, ad infinitum where * indicates distinction between say
> the 1's and the numeral 10 or 11
>
> The third winding is this:
>
> ...22220, .....22221, ....2222, ...22223, .....22224, ....22225,
> .....2226, ....22227, ....22228, ....22229, .....2222*10, ....2222*11,
> ad infinitum
>
> All of them are INFINITE INTEGERS, the concept I came with when I first
> came to the Internet in 1993 and never heard or knew of such a thing as
> Adics and I was sadly surprized they were base dependent.

Actually, they are already known as "positive integer-valued vectors in

Z^infinity", and can also be represented by all functions from the
positive integers to the positive integers. (Note: I will use the term
"integers" as short for "finite integers", i.e., the standard usage,
and
refer to these new infinite integers as "All-Adics".)

Basically, you're representing the function f such that

f(1) = a
f(2) = b
f(3) = c
...

as the All-Adic (..., c,b,a). So they really aren't new.

> But tonight I wish to outline how to lump all the Adic Integers into
> one set and toss out the base dependency.
>
> Now every adic integer will be in that array above.

First mistake. Where does ...010101010101 go? It can't appear in any
of your windings, since it is not eventually constant.

More importantly, the number of All-Adics is uncountable; that is,
there
is no one-to-one correspondence with the set of integers, and
consequently
no one-to-one correspondence with an infinite array of integers. The
proof
of this, no matter what the array looks like, is the standard Cantor
argument.

> The 10-adic
> idempotents of ......2890625 and ....7109376 will be in the array of
> ....555555 winding and ...66666 winding respectively. Even though we
> know that these idempotents will never end up looking like
> 555555555.....2890625

Well, then WHY will they be in the ...555 winding? The way you've
described
windings above, the ...000 winding consists of all All-Adics which end
as
...0000a, for some nonnegative integer a. Therefore, the ...555 winding
should be all All-Adics that end in ...555b, for some nonnegative
integer
b. Since ...2890625 does not end this way, it can't be in the ...555
winding.

> Addition and Multiplication of these numbers is the same as adding or
> multiplying any two integers

So basically, ...1111111 + ...2221234 = ...3332345, and
...111013 * ...000104 = ...00[twelve].

> and they end up in the winding for which
> the last number is such as the 10-adic idempotent is in the ....55555
> winding.

So what you're saying HERE is that the All-Adic ...****a ends up in the
...aaaa winding (where a is a positive integer, and * represents any
arbitrary bunch of integers). This contradicts what you said above:

>> ...0000000, ....00001, .....000002, ....00003, .....000004, ....000005,
>> .....00006, ....00007, ....00008, ....00009, .....0000010, ....000011,
>> ad infinitum
>>
>> I call it the first winding.

Above, ...00001 would be put in the ...000 winding, but the paragraph
above states that you would put it in the ...111 winding. These are
different windings, so the "winding" of ...00001 is ill-defined. This
is your second mistake.

> Now the above array helps me with the nagging question of the mid 1990s
> concerning my thesis that Natural-Numbers = Infinite Integers. The
> question by dullards as to how the last finite-integer moves on to the
> first infinite integer?

No; it's a legitimate question. If you claim that you can get from a
All-Adic ...0000 (which is a finite integer) to the All-Adic ...111
(which is not a finite integer) by repeatedly adding 1's, then at some
point you must go from a finite integer to an infinite one, otherwise
all of your All-Adics would be finite integers.

Here you've cleverly avoided that question by making it impossible to
get to ...111 by repeatedly adding 1. (Adding 1 replaces an All-Adic
with the one to the right in your Array.) Specifically, to get to
...111
from ...000, you need to do some other operation.

> So as one moves from 1, 2, 3, 4, 5, where does
> that lead to the first number that looks like this....111111 or like
> this ....454545. Where is the last finite integer that ends in 000000s
> and the first infinite integer enter the sequence.
>
> So the above array gives that answer by saying the first infinite
> integer that enters the picture is ....1111110 as the first number in
> the 2nd winding.

What about ...1111010? Certainly ...1111010 < ...1111110

> But I cannot tell you the last number in the 1st
> winding. But this should not trouble any of us because we can never
> tell someone what is the last number in Rationals between 1 and 2 can
> we. Nor can we tell someone what is the last number in Reals between 2
> and 3, but we can say that the first number between 3 and 4 in Reals is
> 3.

And this analogy does carry over. Maybe your cranium hasn't frozen
over, after all.

> So the above array dismisses the Adics with their base dependency.
>
> And we should now be able to write a High School textbook that an
> average High School student would be able to understand and work with
> Infinite Integers or the Adics without base dependency.

And you, AP, should be able to do the homework problems I've assigned
starting at this point, to show that you understand what properties
the All-Adics have. None of them require anything beyond algebra and
an introduction to number theory (definition of prime, even numbers,
etc.) which, after all, every High School student should be able to
get through. Your statement is false if you can't. My comments should
convince you that you haven't through through the definitions you've
come up with.

> Do these Infinite Integers form a Field? I think the answer is yes
> because they have all the inverses and obey all the Field rules.

The answer here is no. (This is mistake #3.) The multiplicative
identity
I must have the property that I * A = A, where A is any All-Adic. Some
figuring shows that I = ...1111 fits the bill here. But now, there is
no inverse of ...11112, since no positive integer times 2 equals 1.
In fact, ...1111 is the ONLY All-Adic that has a multiplicative
inverse.
So you only get a "commutative ring with identity" here; you've missed
"field" by just one axiom.

(Later:) No, I take that back. You don't even have _additive_ inverses,
because there's no All-Adic such that ...0001 + A = ...0000, since
there
is no solution to 1 + n = 0 where n is in {0, 1, 2, 3, ...}. You can
get additive inverses if you let the "digits" be negative integers as
well.

> Now, do the above Infinite Integers answer the old problems of
> mathematics of Fermat's Last Theorem, Goldbach Conjecture, Twin Primes,
> Riemann Hypothesis?

Now you've done it. I have to flag this as mistake #4. The All-Adics
were not studied in Fermat's time, in Goldbach's time. (The p-adics
only
gained awareness in the early 20th century.)

Now, once the p-adics _were_ introduced, FLT was found to be false for
them. In particular, Hensel's Lemma constructs an example for any
p-adic,
where p is a prime number. In this case, Hensel's Lemma simplifies to:

Theorem. Let p be a prime, and n an integer which is at least 2.
If there is an integer k such that k^n = 1 (mod p), then there is
a p-adic c such that c^n = ...00011.

Since the existence of k is assured (k = 1), FLT is false in the
p-adics,
for any prime p, and for any n. The triple of p-adics is

a = ...00001
b = ...00010
c = (the p-adic which the Theorem above provides; it depends on n).

This was known a while ago. (AP cited an example in the 10-adics in his
1990 posts. Hensel's Lemma may not be true if p is not prime.) Of
course,
you are using a completely different addition and multiplication here.

> I think they do in that FLT is false and has
> counterexamples in every power higher than squaring. As for the
> Goldbach Conjecture there are numbers in that array that are even yet
> are not the sum of two primes.

Homework assignment: Which All-Adics are prime? (Hint:
....11a1 * ....111b = ....11ab, for any positive integers a, b. Also,
an All-Adic A is prime if A is not ...0001, and there do not exist
B and C (not not equal to ...0001) such that A = B * C.)

> The Riemann Hypothesis is shown false
> because there are counterexamples such as ...9999999 or square roots of
> -1.

Homework assignment: Prove this. What assumptions do you need to make?
(Hint: How is zeta(A) defined if A is an All-Adic?)

> Are there any new surprises to Number theory? Yes, there are other


> numbers besides 2 which are both even and yet prime such as the 10-adic
> idempotent of ....7109376.

Homework assignment: Prove this statement is false. (Hint: the only
prime
All-Adics are of the form A = ...111p111...111, where p is a prime
number.
An even All-Adic is one of the form 2 * B; hence prove that there is no
All-Adic B which makes 2 * B = ...111p111...111.)

> Do the Infinite Integers have anything new to say about what prime
> numbers are and their infinitude? Yes, in that some primes are of a

> different class than what we are familiar with. Such as the even primes


> besides 2 such as the 10-adic idempotent of ....7109376

Homework assignment: Prove that all idempotents in the All-Adics are of
the form ...a(3)a(2)a(1)a(0), where a(i) = 0 or 1, for all i.

> A note about history of mathematics, in that the Counting numbers were
> discovered in history first and later came base representation of
> Counting numbers. As for Infinite Integers, their base representation
> came first per Adics and only now has anyone offered the full scope of
> these Infinite Integers that are baseless.
>
> Now I called the above as "Windings" because I believe they form a
> sphere like surface so starting with ....0000 and the first winding
> traces out a circular path and the 2nd winding is another circular path
> increasing the first until all the windings trace out a sphere surface.
>
> Another note is that in p-adic integers such as 10-adics we had a
> number such as ...99999 which was -1, but in the above arrays, the
> number ....99999 is not -1 but just another successor number in the
> 10th winding. In fact there are no negative numbers in Infinite
> Integers.

This past statement, if nothing else, is full of cutzpah.

The All-Addics don't have additive inverses, so the All-Adics cannot be
a
field.

Back to the drawing board with you!

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 20, 2005, 2:24:32 AM12/20/05
to
And can you share with us what that text was about.

a_plu...@hotmail.com

unread,
Dec 20, 2005, 2:45:19 AM12/20/05
to
Chris Heckman wrote:
First mistake. Where does ...010101010101 go? It can't appear in any
of your windings, since it is not eventually constant.

More importantly, the number of All-Adics is uncountable; that is,
there
is no one-to-one correspondence with the set of integers, and
consequently
no one-to-one correspondence with an infinite array of integers. The
proof
of this, no matter what the array looks like, is the standard Cantor
argument.

A.P. writes:
Your ....101010101 goes with the 11th winding focused on the number 10.
Each number has its own winding 0 winding, 1 winding, 2 winding, ...10
winding
where we repeat 10 so we have .......10101010 or we could say the 101
winding.

Pick any interval of repetition you want, say you want a repeating of
....7327173271
we find that in the 73271 winding

Now suppose you want a clever string such as pi or e or
....121110987654321 which you recognize as the counting numbers in
succession. Well those are found in each and every winding.

So every number has its repetition in a winding and thus every possible
digit arrangement exists in one or several windings.

Yes, I made a mistake about these Infinite Integers for they do not
constitute a field nor a ring, but neither does the Natural Numbers and
so I am happy because the native numbers of Riem geom do not form a
field or ring.

As for much of your other comments Chris, you seem to be unaware of my
major theme that the Natural Numbers are these Infinite Integers. They
are one and the same set. You pretend as if Natural Numbers are a
distinct and different set from these Infinite Integers.

So when a Conjecture like the Riemann Hypothesis comes along, you are
sitting there thinking that 0,1,2,3,..... is the universe of the
conjecture, whereas I put all of these infinite integers beckoning
Riemann Hypothesis.

a_plu...@hotmail.com

unread,
Dec 20, 2005, 3:01:44 AM12/20/05
to
I wrote earlier tonight:

Dik, can I say that in the 000s winding that those zeroes at infinity
become filled up with 9s and that the last number in that winding is
999999.....99999. A sort of convergence at infinity where 9 digit
replaces every 0 digit. Can I say that?

Can I say that in the 1111s winding that those 1s become replaced with
9s and thus at infinity the last number in the 111s winding is
99999....99999?

A.P. writes:
I have a form of justification that is very logically satisfying for
why the last number in each winding is 99999....99999 and in a sense
each winding converges at infinity to 9999....99999. The justification
is that each infinite-integer has an infinite number of place values,
call them slots where one places a digit from amoung
0,1,2,3,4,5,6,7,8,and 9. Now the Success Axiom in Peano axioms makes
the Natural-Numbers be Infinite Integers due to the endless adding of
1. Thus "Finite Integers" are a badly-defined concept and Finite
Integers is shadow set.

So in the first three windings:


...0000000, ....00001, .....000002, ....00003, .....000004, ....000005,
.....00006, ....00007, ....00008, ....00009, .....0000010, ....000011,
ad infinitum

...11110, .....1111, ....1112, ....11113, .....1114, ....1115,
.....1116, ....1117, ....11118, ....111119, .....11111*10,
....11111*11, ad infinitum where * indicates distinction between say
the 1's and the numeral 10 or 11

...22220, .....22221, ....2222, ...22223, .....22224, ....22225,
.....2226, ....22227, ....22228, ....22229, .....2222*10, ....2222*11,
ad infinitum

all of those 0s in the 0 winding eventually become all 9s. The same in
the 1s winding where they eventually turn into 9 and once a place value
turns into a 9, the endless adding of 1 turns the successive place
values into 9.

So the argument justification is that the Peano Axioms is an endless
adding to 1 fills those place values so that each winding ends up with
a last number of 9999.....99999.

You see the Successor Axiom is like one arm and the other arm is that
9s eventually fill every place value and these two arms wash the hands
of both one another.

So I think I am there at the point to conclude that given any
Infinite-Integer call it X that I can tell you what X+1 is and I can
tell you what X-1 is. That is a great feat because that is a proof that
Infinite Integers are the Natural Numbers themselves.

Proginoskes

unread,
Dec 20, 2005, 3:39:46 AM12/20/05
to

The most important part of mathematics is that it be free of
contradictions. Your definition of the All-Adics fails this test.

a_plu...@hotmail.com wrote:
> Chris Heckman wrote:
> First mistake. Where does ...010101010101 go? It can't appear in any
> of your windings, since it is not eventually constant.
>
> More importantly, the number of All-Adics is uncountable; that is,
> there
> is no one-to-one correspondence with the set of integers, and
> consequently
> no one-to-one correspondence with an infinite array of integers. The
> proof
> of this, no matter what the array looks like, is the standard Cantor
> argument.
>
> A.P. writes:
> Your ....101010101 goes with the 11th winding focused on the number 10.

Nope. The 10th winding is either all All-Adics that are eventually ten,
or the All-Adics that end with the number ten. Neither one applies to
...1010101. Try again.

> Each number has its own winding 0 winding, 1 winding, 2 winding, ...10
> winding
> where we repeat 10 so we have .......10101010 or we could say the 101
> winding.

No, because the 101 winding would have All-Adics like ...101101101101.
But 0101 would also be the winding for ...101010101. Or I should say
"a" winding, because ...1111 is now in an infinite number of windings:

1
11
111
1111
...

> Pick any interval of repetition you want, say you want a repeating of
> ....7327173271
> we find that in the 73271 winding
>
> Now suppose you want a clever string such as pi or e or
> ....121110987654321 which you recognize as the counting numbers in
> succession. Well those are found in each and every winding.

Think again. The only Adics which are in a winding are those which are
eventually periodic. ...121110987654321 (which can be interpeted in two
different ways, BTW) is not eventually periodic, in either
interpretation. (...121110987654321 can be thought of as
(...,12,11,10,9,8,7,6,5,4,3,2,1) or (..., 1, 2, 1, 1, 1, 0, 9, 8, 7, 6,
5, 4, 3, 2, 1).)

> So every number has its repetition in a winding and thus every possible
> digit arrangement exists in one or several windings.
>
> Yes, I made a mistake about these Infinite Integers for they do not
> constitute a field nor a ring, but neither does the Natural Numbers

The Natural numbers form a ring. There is a field that contains the
natural numbers (namely the rational numbers) with operations based on
+ and * for natural numbers, but there is no field that contains the
All-Adics, with your definitions of + and *.

> and
> so I am happy because the native numbers of Riem geom do not form a
> field or ring.
>
> As for much of your other comments Chris, you seem to be unaware of my
> major theme that the Natural Numbers are these Infinite Integers. They
> are one and the same set. You pretend as if Natural Numbers are a
> distinct and different set from these Infinite Integers.

But that's how you defined the All-Adics in your original post,
(essentially) as sequences of positive integers, not as sequences of
All-Adics!

Let's take a look at what you had a few paragraphs above:

>> Now suppose you want a clever string such as pi or e or

>> ....121110987654321 which you recognize as the COUNTING NUMBERS in
>> succession.

All I'm doing is saying that what you call "counting numbers" are the
same as the (standard) natural numbers.

All I'm asking for (and a requirement of mathematics) is consistency.

The p-adics bottom out at the integers modulo p, and for a good reason
(to avoid an infinite regression).

> So when a Conjecture like the Riemann Hypothesis comes along, you are
> sitting there thinking that 0,1,2,3,..... is the universe of the
> conjecture, whereas I put all of these infinite integers beckoning
> Riemann Hypothesis.

Which means you haven't read all of my post. The Riemann hypothesis
says:

] If zeta(X) = 0, then X = -1, -2, -3, ..., or Re(X) = 1/2.

(And no, the universe of the conjecture is not the nonnegative
integers; it's the complex numbers.)

You haven't defined zeta(X) where X is an All-Adic, so the RH is
impossible to state. It's like a language where there is no
word/expression for "color" (or any of the vocabulary associated with
color); there's no way to say what a rainbow is. Similarly, if the RH
can't be stated in the All-Adics, no one can even consider the question
of whether it's true.

So what is beckoning here is: How is zeta(X) defined, if X is an
All-Adic?

For instance, what is zeta(...1111) ?

--- Christopher Heckman

Proginoskes

unread,
Dec 20, 2005, 3:55:01 AM12/20/05
to

a_plu...@hotmail.com wrote:
> Dik will be happy to know I will define addition and multiplication and
> they will have additive and multiplicative inverses. [...]

One thing that's been bothering me is: Do you even bother to check out
the "facts" that you include in your post? For instance, did you even
try to find an additive inverse for, say, ...111, with your definition
of addition of Adics, _before_ posting it? Or do you just type your
reponse as you read my (and other people's) posts?

An important part of mathematics (arguably the most important part) is
that you should check the consequences of various results. In physics,
this is called using the scientific method. (I couldn't help but notice
that you posted this thread to sci.physics for some unfathomable
reason.)

(Once someone becomes familiar enough with mathematics, AFTER DOING
ENOUGH OF IT, they are able to respond "on the fly" and provide an
example which is typical and which has all the properties desired. But
a large amount of practice is needed, and (frankly) you are nowhere
near being able to do it. I can do it, but only in some contexts.)

If you don't check the results of your assumptions and definitions,
you're not doing math or physics. You're not even doing science. You're
writing fiction, something that belongs more in alt.religion than in
sci.math or sci.physics.

So, to avoid embarassment to yourself again, PLEASE read my posts in
their entirety and DO THE CALCULATIONS. Just ignoring them makes you
look like a flake.

--- Christopher Heckman

Gottfried Helms

unread,
Dec 20, 2005, 5:28:43 AM12/20/05
to
Am 20.12.2005 08:24 schrieb a_plu...@hotmail.com:
> And can you share with us what that text was about.
>
Sicher. Aber bitte keine zu großen Erwartung: es war
sozusagen mein Jungfernflug in die Welt des mathematischen
Argumentierens...
http://141.51.96.22/divers/irrationalzahlen/index.htm

Ich hatte das dann noch ein bißchen weitergetrieben,
aber dann beiseite gelegt.

merry christmas -

Gottfried Helms

denis feldmann

unread,
Dec 20, 2005, 7:28:11 AM12/20/05
to
Proginoskes a écrit :

Surely it is 1+1/2^...1111+1/3^...1111+..., where n^...1111 is
n*n^10*n^100*... , no?


>
> --- Christopher Heckman
>

a_plu...@hotmail.com

unread,
Dec 20, 2005, 12:51:04 PM12/20/05
to
Gottfried Helms wrote:

merry christmas -

Gottfried Helms


A.P. writes:
Sure enough I searched your website and found this reference. Sorry
Gottfried I can no longer speak or understand German. Although I was
born in Germany, Arzberg, my parents moved to the USA when I was 4-5
years old and although I was speaking German very well, having to learn
English, subdued my German.

--- quoting from that website listed ---
//// Anmerkung 6'2001:... ab hier wird es etwas durcheinander, da ich
in den Zusammenhang mit den ADICS-Zahlen und den Konstruktionen dieses
Archimedes Plutonium gekommen bin

////


Ich will das deshalb hier auch noch nicht weiterverfolgen, sondern
einen anderen Aspekt vorstellen, der sich ergibt, wenn man konsequent
das Konstruktionsprinzip einfordert und dies von bestimmten
Anforderungen an es befreit, wie sie durch die üblicherweise
Beschäftigung mit aktualen oder auch irrationalen Zahlen implizit
auftreten. Eine dieser unausgesprochenenen Beschränkungen verbirgt
sich
--- end quoting ---

Gottfried, I have moved on from Adics. Trouble with Adics is that they
are base dependent and thus unable to see them "fully". It is like
walking around in the world with always some colored glasses on and
never seeing the world with uncolored glasses. At one moment blue
glasses, another red glasses, another yellow glasses and never any
clear glasses.

But now I have made the Natural-Numbers the Infinite Integers and can
see them fully. With Adics, I was always having to be restricted to
some play-pen confinement and never able to see them All at once, but
now I can.

And there are many surprizing results. There are Natural Numbers that
repeat as we count. This is amazing because we thought that the Peano
Axioms would never allow a number to occur more than once when we
count. Numbers like ....99999 and ......999998 occur infinitely many
times as we start to count 0,1,2,.... Another surprize is that there
are infinitely many even primes and that 2 is not the only even prime
for example the square root of 5 when digitized is a even Infinite
Integer that is prime. The number e when digitized is an even prime
Infinite Integer.

So the Goldbach conjecture is false and counterexamples would be e
digitized of .....828172. Since it is irrational and transcendental in
Reals, it carries over some of those qualities as a Infinite Integer
where it cannot be a prime x + prime y = ....828172. So Goldbach
Conjecture falls apart and is false.

The Riemann Hypothesis falls apart in Natural Numbers also because as
we count them, numbers like ...99999 and ....999998 and ....999997 etc
etc occur an infinite number of times as we count and that is a shock
to the conjecture of Riemann Hypothesis for it cannot accomodate a
Natural Number that is infinitely occurring as a successor.

As for FLT, well, the Infinite Integers no longer need special Adic
numbers to counterexample. I pick any two Infinite Integers I want and
I can deliver to you a number that satisfies FLT to any power. In Reals
pick 2 and cube it is 8 and pick 3 and cube it is 27. Add them is 35.
Take the cube root of 35 in Reals satisfies FLT.
But now transpose that result to Infinite Integers and we have the
triple ....00002 and ....00003 and ....cube root of 35 digitized as a
Counterexample of FLT.

We live in amazing times in the sciences and mathematics.

a_plu...@hotmail.com

unread,
Dec 20, 2005, 1:03:15 PM12/20/05
to
Chris Heckman wrote:

No, because the 101 winding would have All-Adics like ...101101101101.
But 0101 would also be the winding for ...101010101. Or I should say
"a" winding, because ...1111 is now in an infinite number of windings:

A.P. writes: Chris you are a professor of mathematics and being such I
would expect you to be able to pick up loose ends and loose threads. I
made a typing error because I typed too fast that should be 1,010
winding for it is the same as the 10 winding and the same as the future
winding of the number 101,010. Now the loose end I expect you to have
picked up on is that, yes, your example starts with a *1* whereas the
10 winding would look like this .....1010101010 starting with a *0*.
But you being a professor of mathematics you should have been able to
remedy that because you know you can simply add a number to the winding
for the winding looks like this

.......1010101010 , .....101010101, ......101010102, .....1010103 ad
infinitum

I make alot of typing errors and I leave alot of loose ends, but expect
those that read my posts to have the ability to fill in those loose
spots.

a_plu...@hotmail.com

unread,
Dec 20, 2005, 1:45:56 PM12/20/05
to
First I need to define a *digitizing* of a Real number into a Infinite
Integer. Take for example the Reals of pi and e and sqrt 5. When
digitizing e and sqrt 5 they become Infinite Integers of .....828172
and ....606322 respectively.

Now the Infinite Integers are the Natural Numbers as given by the Peano
Axioms. Mathematicians of past centuries, let us call it the olden
days, were too stupid to realize that if you have a postulate that is
open ended, where you have the endless adding of 1 that you cannot keep
those numbers confined to ending in just 000s but that they eventually
spill over into being Infinite Integers.

The first three windings (or perhaps call them geodesics) are these:

...0000000, ....00001, .....000002, ....00003, .....000004,
....000005,
.....00006, ....00007, ....00008, ....00009, .....0000010, ....000011,
ad infinitum

...11110, .....1111, ....1112, ....11113, .....1114, ....1115,
.....1116, ....1117, ....11118, ....111119, .....11111*10,
....11111*11, ad infinitum where * indicates distinction between say
the 1's and the numeral 10 or 11

...22220, .....22221, ....2222, ...22223, .....22224, ....22225,
.....2226, ....22227, ....22228, ....22229, .....2222*10, ....2222*11,
ad infinitum

Now the Riemann Hypothesis is false because there are an infinitude of
numbers as we start with 0 and count to 1 then 2, then 3 etc etc we
eventually come to the number such as ....9999 and its predecessor of
....99998. But that these two numbers occur in every winding and bridge
the winding of the 000s to that of 1111s. So the predecessor of
....111110 is ....99999 and the predecessor of ....22220 is also
...99999

The Infinite Integers obey every one of the Peano Axiom postulates
because, given any Infinite Integer X, I can then tell you what X-1 is
and what X+1 is.

Now, let us conquer a few long outstanding unsolved math problems
starting with the Riemann Hypothesis: It is false because poor Mr.
Riemann never knew that the Natural-Numbers would end up looking like
what I am posting today. Never in his wildest imagination would have
realized that as you count from 0 to 1 then 2 then 3, that if you go on
counting infinitely many times that you would find a number that seems
to come up again and again and again such as ....99999 and ....999998.
Riemann and Peano had imagined that the Naturals would count only once.

So the Riemann Hypothesis is false and counterexamples are ....9999 and
....99998 in fact there are an infinite number of counterexamples.

Now to Goldbach Conjecture. It is patently false because poor Mr.
Goldbach, not in his wildest imagination would have realized that there
are an infinite number of even primes and what that would do to his
"simplistic conjecture". A few counterexamples to Goldbach is digitized
e and digitized sqrt 5. They are .....828172 and ....606322 and both
are "irrational Infinite Integers" although I have to define what
irrational infinite integer is. Anyway, there does not exist a sum of
two other Infinite Integers that are prime and add up to digitized e.
Another counterexample is the digitized 10-adic idempotents.

By the way, the adics are not Infinite Integers but just a cute and
fancy way to rewrite Rational numbers of the Reals.

Moving on to Infinitude of Even Primes for Natural Numbers. Digitized e
and digitized sqrt 5 are two examples of even Natural Numbers yet
prime.

Proof: There are an infinitude of irrationals that end in an even
number such as the "2" ending of sqrt5 digitized and e digitized. Now
are they prime as Infinite Integers? Yes, because as a Real they were
irrational and some transcendental thus we cannot divide them once they
are digitized.

a_plu...@hotmail.com

unread,
Dec 20, 2005, 2:03:38 PM12/20/05
to
Now I forgot to mention FLT in this post but had discussed it in the
post to Gottfried.

I said:
As for FLT, well, the Infinite Integers no longer need special Adic
numbers to counterexample. I pick any two Infinite Integers I want and
I can deliver to you a number that satisfies FLT to any power. In Reals
pick 2 and cube it is 8 and pick 3 and cube it is 27. Add them is 35.
Take the cube root of 35 in Reals satisfies FLT.
But now transpose that result to Infinite Integers and we have the
triple ....00002 and ....00003 and ....cube root of 35 digitized as a
Counterexample of FLT.


What I am banking on is the idea that the cube root of 35 digitized as
a Infinite Integer when cubed ends up being .....0000000035 itself. I
no longer need Adic idempotents or Adics to give counterexamples.

a_plu...@hotmail.com

unread,
Dec 20, 2005, 2:12:56 PM12/20/05
to
I guess I did not need Dik to help me bridge the gap of predecessor and
successor of windings in Infinite Integers. But I would like for Dik to
tell me why were the Adics discovered before the Infinite-Integers? I
mean why would history find base-dependent integers of infinite expanse
before history discovers baseless Infinite-Integers. That timetable of
discovery is most baffling to me. I would have thought Hensel would
discover baseless Infinite Integers easier then finding adics that
depend on base.

Perhaps I am picking on Dik too much, anyone else familar with the
history of mathematics who can offer a reason as to why history found
base dependent adics before history found baseless Infinite Integers??

a_plu...@hotmail.com

unread,
Dec 20, 2005, 2:27:46 PM12/20/05
to
Chris Heckman wrote:
I found out more about the All-Adics in one hour than you did in a
month. I found the problems a whole lot quicker. You should read the
point-by-point analysis, because you've ignored so much that you should
have studied.

A.P. writes:
But Chris, how can that be true when it is I who invented the All-adics
and invented Infinite Integers. Tell me of any author such as Koblitz
or Gouvea or anyone who has used or delved into All-adics? Over the
years of 1993 to 2005, Dik Winter has often mentioned to newcomers of
the Internet that what I have waiting for them is a nice package
surprize of consolidating all the adics into one lump set which I
called All-adics. And that was the first time Dik ever saw this concept
of All-adics. So, Chris, are you telling me that someone else has
thought of and developed All-adics before me? If it is true, I would
not be saddened because it is the Infinite Integers that I care about
and the adics were just a means of getting the Infinite Integers.

Minus XVII

unread,
Dec 20, 2005, 4:58:08 PM12/20/05
to
there's a special place for you;
you'll get the most Plutonium Stars, for sure,
in the Math Special Olympics.

--les Protocols de George Elder chez Kyoto!
(emmissions-trading scheme online in USA as of Feb.12)
http://tarpley.net/bush8.htm
http://larouchepub.com/other/2002/2903_chapter_11.html
http://www.rwgrayprojects.com/synergetics/plates/plates.html

a_plu...@hotmail.com

unread,
Dec 20, 2005, 11:06:29 PM12/20/05
to

With the Infinite Integers in hand, I will have to rethink the Odd
Perfect Numbers conjecture. Some define perfect number so that 1 is odd
perfect, others do not. But with the Natural Numbers equal to the
Infinite-Integers a candidate for being odd perfect is digitized pi. of
.....562951413. Perhaps a better term then digitized may occur, and
perhaps numeralized or something else but for the moment I call it
digitized. Digitized pi appears in every Infinite Integer winding, ie,
the 00s winding, the 11s winding etc etc.

Now the Real pi is both transcendental and irrational but the reason it
maybe a odd perfect number candidate is the fact that when we draw a
circle we can inscribe regular polygons in the circle and thus
divisible by say 2, by 3, by 4 etc etc and infinitely many divisibles
of regular polygons. So one is tempted to conclude that pi is evenly
divisible by an infinite number of factors and adding up those factors
would equal digitized pi itself.

Arguments against digitized pi being odd perfect is that since it is
irrational it has no factors at all. Just as we assumed or concluded
that digitized e and digitized sqrt5 and 10-adic idempotents were
irrational and thus prime.

So is pi perhaps special over transcendental e and all other irrational
numbers?

Has anyone ever researched whether Real e and Real pi have different
characteristics concerning the inscribing of regular polygons?

But I have opened up a gigantic new world to mathematics with Infinite
Integers = Natural Numbers. And there is so much work that I can only
do a very little.

Proginoskes

unread,
Dec 21, 2005, 12:50:53 AM12/21/05
to

denis feldmann wrote:
> Proginoskes a écrit :
> > The most important part of mathematics is that it be free of
> > contradictions. Your definition of the All-Adics fails this test.
> >
> > a_plu...@hotmail.com wrote:
> > [...]

> >>So when a Conjecture like the Riemann Hypothesis comes along, you are
> >>sitting there thinking that 0,1,2,3,..... is the universe of the
> >>conjecture, whereas I put all of these infinite integers beckoning
> >>Riemann Hypothesis.
> >
> >
> > Which means you haven't read all of my post. The Riemann hypothesis
> > says:
> >
> > ] If zeta(X) = 0, then X = -1, -2, -3, ..., or Re(X) = 1/2.
> > [...]

> > You haven't defined zeta(X) where X is an All-Adic, so the RH is
> > impossible to state. It's like a language where there is no
> > word/expression for "color" (or any of the vocabulary associated with
> > color); there's no way to say what a rainbow is. Similarly, if the RH
> > can't be stated in the All-Adics, no one can even consider the question
> > of whether it's true.
> >
> > So what is beckoning here is: How is zeta(X) defined, if X is an
> > All-Adic?
> >
> > For instance, what is zeta(...1111) ?
>
> Surely it is 1+1/2^...1111+1/3^...1111+..., where n^...1111 is
> n*n^10*n^100*... , no?

(1) I think you're referring to the definition

zeta(s) = 1 + 1/2^s + 1/3^s + 1/4^s + ...,

but this is only true if the real part of s is > 1. See
http://mathworld.wolfram.com/RiemannZetaFunction.html for the actual
definition, right before formula (12).

(2) Even if the formula is correct, what is 2^...1111 ? AP didn't
define exponentiation of the All-Adics.

--- Christopher Heckman


--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 1:04:21 AM12/21/05
to

a_plu...@hotmail.com wrote:
> Gottfried Helms wrote:
> Am 20.12.2005 08:24 schrieb a_pluton...@hotmail.com:
> > And can you share with us what that text was about.
>
> Sicher. Aber bitte keine zu großen Erwartung: es war
> sozusagen mein Jungfernflug in die Welt des mathematischen
> Argumentierens...
> http://141.51.96.22/divers/irrationalzahlen/index.htm
>
> Ich hatte das dann noch ein bißchen weitergetrieben,
> aber dann beiseite gelegt.
>
> merry christmas -
>
> Gottfried Helms
>
>
> A.P. writes:
> [...]

> And there are many surprizing results.

As opposed to a "surprising" result.

> There are Natural Numbers that
> repeat as we count. This is amazing because we thought that the Peano
> Axioms would never allow a number to occur more than once when we
> count.

Why is AP saying "we" here? Is he pregnant?

The Peano Axioms _directly imply_ that no numbers repeat. (This is due
to S being one-to-one, and S(x) nonzero for all x.)

> Numbers like ....99999 and ......999998 occur infinitely many
> times as we start to count 0,1,2,.... Another surprize is that there
> are infinitely many even primes

No, there are zero even primes. AP hasn't bothered to read my outline
of this.

Basically, All-Adics are functions from W to W, where W = {0,1,2,...}.
+ and * are defined componentwise. (This is bad because there no
All-Adics have additive or multiplicative inverses, except for 0 and 1.
AP thinks this is a good thing, for some ungottliche reason.)

Using standard definitions and playing around (two things that AP hates
to do) show that
an All-Adic is a prime if it is of the form
....1,1,1,p,1,1...,1, for some natural number prime p.

None of these All-Adics equal 2 = ...0002 times another Adic; hence no
prime All-Adics are even. Nicht wahr? (Please don't respond entirely in
German; I lost a lot of the vocabulary when I had my nervous
breakdown.)

> and that 2 is not the only even prime
> for example the square root of 5 when digitized is a even Infinite
> Integer that is prime. The number e when digitized is an even prime
> Infinite Integer.
>
> So the Goldbach conjecture is false and counterexamples would be e

> digitized of .....828172. [...]

Actually, this isn't surprising, because even with the standard
definitions, ...0006 can't be written as the sum of two prime
All-Adics, either. Of course, this sheds no light on the real Goldbach
Conjecture, which was formulated centures before p-adics.

> The Riemann Hypothesis falls apart in Natural Numbers also because as
> we count them, numbers like ...99999 and ....999998 and ....999997 etc
> etc occur an infinite number of times as we count and that is a shock
> to the conjecture of Riemann Hypothesis for it cannot accomodate a
> Natural Number that is infinitely occurring as a successor.

And AP is too dumm to realize that a contradiction means you did
something wrong, not that you did something right.

> As for FLT, well, the Infinite Integers no longer need special Adic
> numbers to counterexample. I pick any two Infinite Integers I want and
> I can deliver to you a number that satisfies FLT to any power. In Reals
> pick 2 and cube it is 8 and pick 3 and cube it is 27. Add them is 35.
> Take the cube root of 35 in Reals satisfies FLT.

Everyone who is surprised by this, lobotomize yourself.

> But now transpose that result to Infinite Integers and we have the
> triple ....00002 and ....00003 and ....cube root of 35 digitized as a
> Counterexample of FLT.

AP thinks he has a version of Hensel's Lemma in the All-Adics. Poor
fool.

--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 1:13:20 AM12/21/05
to

a_plu...@hotmail.com wrote:
> Chris Heckman wrote:
>
> No, because the 101 winding would have All-Adics like ...101101101101.
> But 0101 would also be the winding for ...101010101. Or I should say
> "a" winding, because ...1111 is now in an infinite number of windings:
>
> A.P. writes: Chris you are a professor of mathematics and being such I
> would expect you to be able to pick up loose ends and loose threads.

That's what I did when I said "0101 would also be the winding for
...101010101." Pay attention, Ludwig.

But I am not sure whether mistakes like these are typos, or whether you
really intended to say what you're saying, and you believe it to be
true.

> I made a typing error because I typed too fast that should be 1,010
> winding for it is the same as the 10 winding and the same as the future
> winding of the number 101,010. Now the loose end I expect you to have
> picked up on is that, yes, your example starts with a *1* whereas the
> 10 winding would look like this .....1010101010 starting with a *0*.

All-Adics don't "start" anywhere; they "end" at a particular place,
though.
...1010101 is the same All-Adic as ...010101, just like 111 = 0111 in
the natural numbers.

> But you being a professor of mathematics you should have been able to
> remedy that because you know you can simply add a number to the winding
> for the winding looks like this
>
> .......1010101010 , .....101010101, ......101010102, .....1010103 ad
> infinitum

Which means that ...1010101 is in both winding 10 AND winding 01. Thus
your concept of "winding of an All-Adic" is ill-defined (in the literal
sense of the word: ambiguous).

> I make alot of typing errors and I leave alot of loose ends, but expect
> those that read my posts to have the ability to fill in those loose
> spots.

Of course, reading over your posts one last time would also cut down on
these typing mistakes. And actually working out the examples I've
posted would do the same for the math errors.

For instance, there are NO even prime All-Adics, contrary to what you
say. Using the standard definition of primes and your definition of
multiplication, every prime All-Adic looks like

....1,1,1,p,1,1,1,...,1 (where I've put in commas to separate the
individual digits)

for some prime p. (...,0,0,0,2 isn't prime because ...,0,0,2 =
...,1,0,1,2 * ...,0,1,0,1.)

And being a multiple of 2 means every digit is an even natural number.

--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 1:15:29 AM12/21/05
to

a_plu...@hotmail.com wrote:
> First I need to define a *digitizing* of a Real number into a Infinite
> Integer.

Nope, FIRST you need to go back and redo your definitions of Infinite
Integers (All-Adics), since there are half a dozen problems with them.

> [irrelevant derivations snipped]

--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 1:19:54 AM12/21/05
to

a_plu...@hotmail.com wrote:
> [...]

> What I am banking on is the idea that the cube root of 35 digitized as
> a Infinite Integer when cubed ends up being .....0000000035 itself.

Too bad, pay up.

THEOREM. There is no Infinite Integer whose cube is ...00035, whether
you interpret it as ....,0,0,35 (the correct way) or ...,0,0,3,5 (the
incorrect way; this is base-dependent).

Proof: Assume there is such an Infinite Integer A, and its last "digit"
is a. Then we must have the equation

A^3 = ....00035, so

(....,a)^3 = (...,b), where b = 5 or 35.
(..., a^3) = (...,b),

which implies a^3 = b, where a and b are FINITE INTEGERS, and b is one
of 5 or 35. But in either case, there is no "a" that makes the equation
true. Reducto ad absurdum, QED.

You can send your $1000 wager to me, care of the Department of
Mathematics and Statistics, at Arizona State University.

--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 1:26:58 AM12/21/05
to

a_plu...@hotmail.com wrote:
> I guess I did not need Dik to help me bridge the gap of predecessor and
> successor of windings in Infinite Integers.

And, to no surprise, you got it wrong. There is NO predecessor for
....1110 (more properly, ...,1,1,1,0.)

> But I would like for Dik to

> tell me why were the Adics discovered before the Infinite-Integers? [...]

They weren't. What you're calling infinite integers really are the set
F of functions from W to W, where W = {0,1,2,3,...} of finite integers.
They were discovered (and abandoned) long ago, and no one cares about
them because the "obvious" definitions of + and * don't give you
additive or multiplicative inverses. The semi-group people might be
interested in them, but anyone who demands that (F,+) be a group will
never consider this example.

http://mathworld.wolfram.com/Semigroup.html

(F,+) is a monoid, since there is an additive identity: ...000, and
addition is associative.

http://mathworld.wolfram.com/Monoid.html

--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 1:47:23 AM12/21/05
to

a_plu...@hotmail.com wrote:
> Chris Heckman wrote:
> I found out more about the All-Adics in one hour than you did in a
> month. I found the problems a whole lot quicker. You should read the
> point-by-point analysis, because you've ignored so much that you should
> have studied.
>
> A.P. writes:
> But Chris, how can that be true when it is I who invented the All-adics
> and invented Infinite Integers.

It's time for you to face some unpleasant, but true, facts.

If you had taken a course in modern algebra, you would have the tools
to realize that the All-Adics fail to be a field. If you took a course
in number theory, you would realize that there are in fact no even
prime numbers in the All-Adics. A real mathematician would
automatically consider the "homework assignments" that I gave to you in
my post; this separates the professionals from the putterer, the
dedicated from from the dabblers.

This is further proof that you haven't read it and studied my post. At
times you act like a ten-year old, with Attention Deficit Disorder. You
don't bother checking details, even when they are pointed out to you
and outlined. My parents' dog has a better chance at making a
mathematical breakthrough than you, because when when I aim his head in
the right direction, he can see what you're talking about.

> Tell me of any author such as Koblitz
> or Gouvea or anyone who has used or delved into All-adics?

The All-Adics, as you've defined them, lack properties of a field, and
are thus uninteresting to 99% of mathematicians. In fact, you don't
have any kind of inverses, additive or multiplicative, which means
you've found two examples of monoids, _possibly_ providing them with a
nice example of the things that "monoid people" study. (A "monoid" is a
set S with an operation that is closed, associative, and has an
identity. Inverses need not exist.)

> Over the
> years of 1993 to 2005, Dik Winter has often mentioned to newcomers of
> the Internet that what I have waiting for them is a nice package
> surprize of consolidating all the adics into one lump set which I
> called All-adics.

And I have been waiting for the same thing. But it simply didn't live
up to its promise. If you had spent some time checking the field axioms
instead of ranting and raving about other subjects, you would have
saved yourself a lot of time and embarassment.

> And that was the first time Dik ever saw this concept
> of All-adics. So, Chris, are you telling me that someone else has
> thought of and developed All-adics before me?

Yes: thought of, developed, and abandoned as "esoterica".

But any professional mathematician is able to work with any
definition/assumptions that he/she comes across, and to find
consequences of those assumptions. Courses like Number Theory and
Modern Algebra given him/her a chance to practice and figure out how to
find the "important" things to try. That's what all the graduate work
in mathematics is for.

In this case, it only took a few minutes, because the intuition I've
developed over the years led to results. If it hadn't, you wouldn't
have heard me say anything, and I would have worked on the problem
until it was resolved one way or another.

To a professional, it doesn't matter whether the idea is five minutes
old or five hundred years old; the results are obtained the same way.

> If it is true, I would
> not be saddened because it is the Infinite Integers that I care about
> and the adics were just a means of getting the Infinite Integers.

--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 2:01:12 AM12/21/05
to

a_plu...@hotmail.com wrote:
> With the Infinite Integers in hand, I will have to rethink the Odd
> Perfect Numbers conjecture. Some define perfect number so that 1 is odd
> perfect, others do not. [...]

You're starting down the "crank" road again. Here's a question that you
should consider:

Q1. What would it mean for an All-Adic A to be perfect?

For natural numbers, this means that the sum of all the (positive)
divisors of n, except for n, equals n. This also means the sum of the
(positive) divisors of n is 2*n. "Sum" is a concept we have already,
but we need to consider what the (positive) divisors of an All-Adic
are. So, before we can answer question Q1, we need to answer:

Q2. What does it mean for B to be a divisor of A?

Again, using the defintions, this would suggest that if we can find an
All-Adic C such that
B*C = A, then B is a divisor of A. So now the problem becomes:

Q3. Given an All-Adic A, how do we find all divisors of A?

In the natural numbers, we have the prime factorization tool to help
us. Assuming that we have unique prime factorization (which hasn't been
proven yet; that would be a topic for another day), it would look like
an answer to Q3 is:

A3. Write A = P(1)^n(1) * P(2)^n(2) * ... * P(k)^n(k), where P(i) are
all prime All-Adics, and n(i) is a positive integer for all i. (You
haven't defined A^B where A and B are both All-Adics, so let's stick to
what we can define.) Any factor of A "should look like"

P(1)^m(1) * P(2)^m(2) * ... * P(k)^m(k),

where m(i) are nonnegative (finite) integers which are <= n(i), for all
i. This gives us a finite list of All-Adics, so we add them together,
and compare the sum with 2*A.

It looks like we have answered Q1, but we're on thin ice, because we
assumed that prime factorization is possible in the All-Adics. That is
another topic to study, and it involves asking what the prime All-Adics
are, and whether every All-Adic has a prime factorization. It also
involves more work.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 21, 2005, 2:48:51 AM12/21/05
to

Proginoskes wrote:
> a_plu...@hotmail.com wrote:
> > I guess I did not need Dik to help me bridge the gap of predecessor and
> > successor of windings in Infinite Integers.
>
> And, to no surprise, you got it wrong. There is NO predecessor for
> ....1110 (more properly, ...,1,1,1,0.)
>

Chris, please stop calling these things All-adics. The
Infinite-Integers are not the adics for the adics are not ordered. I
have ordered the Infinite-Integers and they are base-independent. And
the Infinite-Integers are never any rational number, they are all
integers. And they are never negative numbers such as ....99999 = -1 in
10-adics. In short, the Infinite Integers are wholly different from
adics. I used the adics for the past several months from listening to
what Dik and you wrote about them to finally come to this point where I
invent the Infinite Integers. Whether I invent them or discovered them
is immaterial.

I was never happy with adics when I first heard about them in late
1993. I could never think in them, nor ever be confident about any
results or ever visualize anything about them. They seemed purely
mechanical. And I always had the impression that they were like colored
glasses when I want to see the world as it is.

The major difference between adics and Infinite-Integers is that adics
have no sequential order, whereas these Infinite-Integers given any X I
can tell you what X-1 is and what X+1 is, thus they are the Natural
Numbers themselves.

So kindly dispense of All-adics and call them Infinite Integers as set
up in the windings array. Understandably features of that array my
change and improvements are inevitable.

The predecessor or X-1 of the first number in the 111s winding of
.....1111110 is the number ......999999 which is the last number in the
0000s winding. How did I get that?
Because in the Successor postulate of Peano Axioms is an endless adding
of 1. So as a person goes farther and farther out there in the 000s
winding those 0s eventually become filled with the digit 9.

Every winding has a bridge or gap to be connected from the winding
below and the winding above and this gap is filled by ....99999.

The end goal is that *every possible digit arrangement* exists in at
least one of the windings. And every number has a predecessor and
successor. Thus the Infinite Integers are the Natural Numbers of the
Peano Axioms.

> > But I would like for Dik to
> > tell me why were the Adics discovered before the Infinite-Integers? [...]
>
> They weren't. What you're calling infinite integers really are the set
> F of functions from W to W, where W = {0,1,2,3,...} of finite integers.
> They were discovered (and abandoned) long ago, and no one cares about
> them because the "obvious" definitions of + and * don't give you
> additive or multiplicative inverses. The semi-group people might be
> interested in them, but anyone who demands that (F,+) be a group will
> never consider this example.
>
> http://mathworld.wolfram.com/Semigroup.html
>
> (F,+) is a monoid, since there is an additive identity: ...000, and
> addition is associative.
>
> http://mathworld.wolfram.com/Monoid.html
>
> --- Christopher Heckman

F of functions from W to W. Did they call it the Desperation Function
back then or now? Sounds like you are making things up here, maybe you
had too much Tequila before Xmas.

a_plu...@hotmail.com

unread,
Dec 21, 2005, 3:07:57 AM12/21/05
to
Chris Heckman wrote:

Too bad, pay up.

A^3 = ....00035, so

--- Christopher Heckman

A.P. writes:
Well thank goodness I see no word All-adics. But you still seem to use
the Adic definitions for multiplication.

What I meant by "banking" is that face it, we will never be able to
multiply out an irrational number such as cube root of 35 because those
digits never end nor repeat in blocks. Since it is impossible for us to
fully see this number and fully multiply it in a cubing. That I give
myself a free license or freedom to say that if I could multiply the
cube root of 35 digitized that magically in the multiplication the end
result will be ....00000035.

So what I am saying is that since we can never do the cubing of
digitized 35 to see if it truly becomes all zeroes except for 35, that
we can thus say or define it as its Real counterpart that it does end
up as ....00000035.

We all know that it has some end result but none of us can tell what it
is and so I am banking on that the Real result is the same as the
Infinite Integer result.

a_plu...@hotmail.com

unread,
Dec 21, 2005, 3:19:44 AM12/21/05
to
There is another supporting piece of evidence in favor of calling
....99999 as the last number in each winding. Our numbers are decimal
based numbers and in Reals before we go from one Whole Real to the next
Whole Real such as say from going from 3.00... to 4.0000.... that the
number 3.99999..... is one and the same as 4.000.....

So in a sense I would be justified in saying that the first number of
the 111s winding is ......11111110 and it predecessor in the 00s
winding is .....999999.

Setting up these windings in a Array for the Infinite Integers makes me
think that these windings are like the Whole Reals themselves where
.....11111110 is a Whole Infinite Integer and the next Whole Infinite
Integer is ....2222220 preceded by .....99999

So these Windings force these Infinite Integers into a Sequential
Ordering.

Proginoskes

unread,
Dec 21, 2005, 4:32:48 AM12/21/05
to

a_plu...@hotmail.com wrote:
> There is another supporting piece of evidence in favor of calling
> ....99999 as the last number in each winding.

But if you do, you're using base 10, and you don't have a baseless
representation.

--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 4:40:33 AM12/21/05
to
a_plu...@hotmail.com wrote:
> Proginoskes wrote:
> > a_plu...@hotmail.com wrote:
> > > I guess I did not need Dik to help me bridge the gap of predecessor and
> > > successor of windings in Infinite Integers.
> >
> > And, to no surprise, you got it wrong. There is NO predecessor for
> > ....1110 (more properly, ...,1,1,1,0.)
> >
>
> Chris, please stop calling these things All-adics.
> [...] In short, the Infinite Integers are wholly different from
> adics. [...]

Okay; I won't call them that. But my question still is valid: What do
you claim is the predecessor for ....1110?

More importantly, what exactly is an Infinite Integer?

> The major difference between adics and Infinite-Integers is that adics
> have no sequential order, whereas these Infinite-Integers given any X I
> can tell you what X-1 is and what X+1 is, thus they are the Natural
> Numbers themselves.

What is ....11110 - 1?

> So kindly dispense of All-adics and call them Infinite Integers as set
> up in the windings array. Understandably features of that array my
> change and improvements are inevitable.

Will you post updates as you do so? You don't seem to mind changing
definitions and not telling anyone about it.

> The predecessor or X-1 of the first number in the 111s winding of
> .....1111110 is the number ......999999 which is the last number in the
> 0000s winding.

Clearly, ...111110 < ...9999 (since each digit of ...999 is greater
than each digit of ...1110), so you can't have ...9999 + 1 = ...11110

Besides, you're using base 10 here. I could just as easily argue that
...8888 is the proper "predecessor" by using base 9.

> How did I get that?
> Because in the Successor postulate of Peano Axioms is an endless adding
> of 1. So as a person goes farther and farther out there in the 000s
> winding those 0s eventually become filled with the digit 9.

... and then with the digit ten, then the digit eleven, etc. After all,
these things are supposed to be base-independent, right?

> > > But I would like for Dik to
> > > tell me why were the Adics discovered before the Infinite-Integers? [...]
> >
> > They weren't. What you're calling infinite integers really are the set
> > F of functions from W to W, where W = {0,1,2,3,...} of finite integers.
> > They were discovered (and abandoned) long ago, and no one cares about
> > them because the "obvious" definitions of + and * don't give you
> > additive or multiplicative inverses. The semi-group people might be
> > interested in them, but anyone who demands that (F,+) be a group will
> > never consider this example.
> >
> > http://mathworld.wolfram.com/Semigroup.html
> >
> > (F,+) is a monoid, since there is an additive identity: ...000, and
> > addition is associative.
> >
> > http://mathworld.wolfram.com/Monoid.html
>

> F of functions from W to W. Did they call it the Desperation Function
> back then or now?

No; I just gave it a name so I could talk about it. You call the same
things "Infinite Integers" now. And F does not refer to a function; it
refers to a SET of functions.

> Sounds like you are making things up here, maybe you
> had too much Tequila before Xmas.

"Monoid" and "Semigroup" are recognized by the mathematics community.
(How could I have gotten those pages posted if I had made them up? Or
given them MSC numbers?)

http://www.ams.org/msc/20Mxx.html (Semigroup areas)
http://www.ams.org/msc/19Dxx.html (Monoids)

OTOH, "Infinite Integer" isn't standard terminology. You can try to
search for it at

http://www.ams.org/msc/

Therefore, YOU are the one making things up here. QED.

--- Christopher Heckman

Proginoskes

unread,
Dec 21, 2005, 4:49:23 AM12/21/05
to

a_plu...@hotmail.com wrote:
> Chris Heckman wrote:
> a_pluton...@hotmail.com wrote:
> > [...]
> > What I am banking on is the idea that the cube root of 35 digitized as
> > a Infinite Integer when cubed ends up being .....0000000035 itself.
>
> Too bad, pay up.
>
> THEOREM. There is no Infinite Integer whose cube is ...00035, whether
> you interpret it as ....,0,0,35 (the correct way) or ...,0,0,3,5 (the
> incorrect way; this is base-dependent).
>
> Proof: Assume there is such an Infinite Integer A, and its last "digit"
> is a. Then we must have the equation
>
> A^3 = ....00035, so
>
> (....,a)^3 = (...,b), where b = 5 or 35.
> (..., a^3) = (...,b),
>
> which implies a^3 = b, where a and b are FINITE INTEGERS, and b is one
> of 5 or 35. But in either case, there is no "a" that makes the equation
> true. Reducto ad absurdum, QED.
>
> A.P. writes:
> Well thank goodness I see no word All-adics. But you still seem to use
> the Adic definitions for multiplication.

Nope; I'm doing it component-wise, just like you said.

...,0,1,2 * ...,2,3,4 = ...,0,3,8 and
...,0,1,2 + ...,2,3,4 = ...,2,4,6.

The Adic definitions are base-dependent; these aren't.

However, you seem to have slipped in the word "digitized" without
defining it.

> What I meant by "banking" is that face it, we will never be able to
> multiply out an irrational number such as cube root of 35 because those
> digits never end nor repeat in blocks. Since it is impossible for us to
> fully see this number and fully multiply it in a cubing.

However, it could be possible to prove it doesn't exist, just like it's
possible to prove that the square root of 2 is irrational (without
generating all of its digits).

> That I give
> myself a free license or freedom to say that if I could multiply the
> cube root of 35 digitized that magically in the multiplication the end
> result will be ....00000035.

Yes, and the proof is whether N * N * N = ....000035

> So what I am saying is that since we can never do the cubing of
> digitized 35 to see if it truly becomes all zeroes except for 35, that
> we can thus say or define it as its Real counterpart that it does end
> up as ....00000035.
>
> We all know that it has some end result but none of us can tell what it
> is and so I am banking on that the Real result is the same as the
> Infinite Integer result.

On the contrary, anyone who works hard enough can tell what it is,
provided you explain what all of your terms are and don't change them
in the middle of a thread.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 21, 2005, 1:49:00 PM12/21/05
to

Baseless in the sense that the Natural-Numbers written in decimal form
are baseless.

Now tell me something Chris about these FLT counterexamples in exponent
3.

cube root(35) = 3.27106631......

digitized cube root 35 is ......136601723

and this Infinite Integer would be found in every Winding.

....000002^3 = ....000008
....000003^3 = ....000027

add the two gives ....0000035

Now would you not agree Chris that ....136601723^3 is equal to
....0000035

And would I have to prove that every digitized Real yields the same
result for Infinite Integers as it does yield in Reals.

Does that require a proof or is it so obvious that I just say so.

Infinite Integers are *not* adics, not p-adics not all-adics not
anything adics. I used you and Dik to get a sense of what Infinite
Integers are by discussing just adics for the past 3 months.

Infinite-Integers are the Natural Numbers themselves for they obey the
Peano Axioms and especially the Successor Postulate of endless adding
of 1.

Up until 1993, the world of Mathematics was using a partial deck of
cards, they were using only a tiny fraction of what the
cards-of-all-the-Natural-Numbers were. Mathematicians before the
Infinite-Integers were playing card games not with the full deck of
cards but with only 3 or 4 cards, and what kind of bridge or poker or
solitaire or hearts can anyone play with only 4 cards with all the
remaining cards missing.

Now I need someone good in geometry to tell me if there is something
different about the Real number e and the number pi. Pi on a circle can
be inscribed with *regular polygons*. Does that mean that pi is evenly
divisible by 5 by 6 by 8 etc etc? Does it mean that digitized pi is
evenly divisible by 5, 6,8

Yet e digitized seems to be not evenly divisible. Perhaps the log
spiral is the key.

So I need a good geometer to tell me if there is this difference
between pi and e

a_plu...@hotmail.com

unread,
Dec 21, 2005, 2:10:21 PM12/21/05
to
Gottried Helms-- can I enlist you to write a textbook in German.
Titled: Infinite-Integers equals Natural-Numbers.

It will show the Arrays of Windings, the 000s winding, the 111s winding
etc etc

Any average High School student (over in Germany and much of Europe
call it a different name than High School here in the USA of grades
10,11,12, in Australia it would be called Form)

Anyway, the text has to be accessible to an average student of age 15
to 18 so that he/she can add, multiply Infinite Integers as easily as
with Reals.

Gottfried, this is the most important Mathematics text since Euclid's
Elements because for the first time in human history we are playing
with the full deck of Natural Numbers and not just 3 or 4 cards and the
entire rest of the deck missing.

So we need a person in each language to write a elementary text on
Infinite Integers accessible and understandable by an average girl or
boy of age 15.

Carbon in us,
Carbon of plutonium,
Make us wholly thine,
Take us to the Nucleus
Nucleosynthesis Divine

Minus XVII

unread,
Dec 21, 2005, 9:11:48 PM12/21/05
to
sorry; I meant, the Math Special Olympiad.

I've been a dishwasher at a university dorm, as well, but
I don't share your problem with English. I believe that
that is your main problem, as far as comprehension goes;
i.e. you can tell that Dik Winter is not a native speaker, but
it doesn't stop him from "following" what is **.

thus quoth:


Yes: thought of, developed, and abandoned as "esoterica".

thus:
mister MacGunken (canned-spam-guy),
time is always a "dimension" in a phase-diagram;
the rest is crappola. if you don't like that analogy,
try the astronomer royal's flipbook-universe. (yes,
they ka-night these people .-)

thus:
http://www.woodcraftarts.com/jacob.htm
you must have shown the Wolframites some thing,
that made them put it on their site, aside
from what is evident, thereat. (or,
it's sir David's compulsion to cover
every thing in Universe, that may be cellular automata ...
like you & me !-)

thus quoth:
<deletives impleted>

Proginoskes

unread,
Dec 22, 2005, 12:54:19 AM12/22/05
to

a_plu...@hotmail.com wrote:
> Proginoskes wrote:
> > a_plu...@hotmail.com wrote:
> > > There is another supporting piece of evidence in favor of calling
> > > ....99999 as the last number in each winding.
> >
> > But if you do, you're using base 10, and you don't have a baseless
> > representation.
>
> Baseless in the sense that the Natural-Numbers written in decimal form
> are baseless.

I'm confused here. It looks like you are just using ten-adics, since
you have the following equations below:

...0003^3 = ...00027
...00027 + ...0008 = ...00035

and these are also true for the 10-adics. Can you provide a calculation
in the Infinite Integers which is different from one in the 10-adics?
For instance, tell me how to choose A and B so that A+B in the 10-adics
gives me a different answer from A+B in the Infinite Integers.

Maybe you should check to see if I'm using the right definitions then.
(If you want to publish this as a mathematical book, you need to
include definitions like these.)

DEFINITION 1. An Infinite Integer is a sequence of digits; that is,
each of a(0), a(1), a(2), a(3), ... is a digit (an integer between 0
and 9), and the Infinite Integer is written as ...a(3)a(2)a(1)a(0).

DEFINITION 2. If A and B are Infinite Integers, then the sum A+B is
calcuated as follows:

(1) Let c(i) = a(i) + b(i), for all i.
(2) For each i, if c(i) > 9, replace c(i) with c(i) - 10 and c(i+1)
with c(i+1) + 1
(3) Then A+B = C.

(This was generalized from your equation ...00027 + ...0008 =
...00035.)

DEFINITION 3. If A and B are Infinite Integers, then the product A*B is
calculated similarly. (Follow the elementary school method of
multiplying numbers with > 1 digit.)

DEFINITION 4. If A is an Infinite Integer, and n is a positive integer,
then A^n is the product A*A*...*A, where there are n A's.

(This was generalized from your equation ...0003^3 = ...00027.)

> Now tell me something Chris about these FLT counterexamples in exponent
> 3.
>
> cube root(35) = 3.27106631......

Only in base ten, of course.

> digitized cube root 35 is ......136601723

Would the digitized cube root of 35000 be ...136601723 as well, since
cube root (35000) = 32.7106631 ... ?

Would the digitized square root of 2 be ...4124141 ? (I'm pretty sure
about this one.)

Since the cube root of 35 can also be written as 03.27106631...,
wouldn't that mean that the digitized cube root of 35 could also be
...1366017230 ? (I suspect not ...)

DEFINITION 5. If R is a real number of the form
a(-2)a(-1)a(0).a(1)a(2)a(3)..., with a(-2) not zero (decimal form),
then the digitized version of R is the Infinite Integer
...a(3)a(2)a(1)a(0)a(-1)a(-2).

(This was generalized from your equation digitized (cube root 35) is
......136601723.)

> and this Infinite Integer would be found in every Winding.

Once again, this is confusing, because you've used "winding" in two
different senses:

(1) ...111112 is in winding ...111, because the digits are eventually
...111.
(2) ...111112 is in winding ...222, because it ends in ...222.

Clearly, no Infinite Integer can be in the ...111 and ...222 windings,
if definition (2) is used, since no Infinite Integer ends in 1 and 2
(at the same time); but if you mean (1), then the digits of A are
eventually all 1's (in order to be in the ...111 winding), but then A
can't be in the ...222 winding, since the digits will not eventually be
all 2's.

So neither one of these definitions actually works. So I have a
question:

QUESTION. If A and B are Infinite Integers, then when is A in Winding
#B?

And this is one you _must_ answer.

> ....000002^3 = ....000008
> ....000003^3 = ....000027
>
> add the two gives ....0000035
>
> Now would you not agree Chris that ....136601723^3 is equal to
> ....0000035

I would not agree. That's because anything that ends in 3, cubed, ends
in 7. In fact,
...136601723^3 = ...259320067, from DEFINITION 3 and DEFINITION 4.

Besides, if your method were correct, you could also say that
twenty-seven digitized is ....00072, and since ...0003^3 = ....00027,
then ...0003^3 is not ...00072.

> And would I have to prove that every digitized Real yields the same
> result for Infinite Integers as it does yield in Reals.
>
> Does that require a proof or is it so obvious that I just say so.

It looks like you're proposing that if a^3 = b in the real numbers,
then A^3 = B in the Infinite Integers, where A is the digitization of
a, and B is the digitization of b.

If so, no such proof exists. See the counterexample above.

> Infinite Integers are *not* adics, not p-adics not all-adics not
> anything adics.

Then how do they differ?

> [rant snipped]


>
> Now I need someone good in geometry to tell me if there is something
> different about the Real number e and the number pi. Pi on a circle can
> be inscribed with *regular polygons*.

By some of them, yes. I don't think all of them can be, if you limit
yourself to compass and straightedge. (If you're allowed to use
trigonometry, then they all can be.)

> Does that mean that pi is evenly
> divisible by 5 by 6 by 8 etc etc? Does it mean that digitized pi is

> evenly divisible by 5, 6,8 [...]

Digitized pi ends in 3, so it can't be divisible by 5 (the last digit
would have to be 5 or 0), 6 or 8 (3 isn't even). And of course, I think
you meant to ask:

"Does it mean that digitized pi is evenly divisible by ...0005,
...0006, ...0008?"

--- Christopher Heckman

Proginoskes

unread,
Dec 22, 2005, 12:58:37 AM12/22/05
to

Minus XVII wrote:
> sorry; I meant, the Math Special Olympiad.
>
> I've been a dishwasher at a university dorm, as well, but
> I don't share your problem with English. I believe that
> that is your main problem, as far as comprehension goes;
> i.e. you can tell that Dik Winter is not a native speaker, but
> it doesn't stop him from "following" what is **.

Dishwashing in itself should not be looked down upon; watching the
water pour into various dishes and bowls, I gained instant
enlightenment on why a particular max-flow-min-cut algorithm works.

--- Christopher Heckman

Proginoskes

unread,
Dec 22, 2005, 1:00:38 AM12/22/05
to

a_plu...@hotmail.com wrote:
> Gottried Helms-- can I enlist you to write a textbook in German.
> Titled: Infinite-Integers equals Natural-Numbers.

You need to get the math right first, otherwise it can only go in the
fiction section.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 22, 2005, 4:05:54 AM12/22/05
to

Proginoskes wrote:
> a_plu...@hotmail.com wrote:
> > Proginoskes wrote:
> > > a_plu...@hotmail.com wrote:
> > > > There is another supporting piece of evidence in favor of calling
> > > > ....99999 as the last number in each winding.
> > >
> > > But if you do, you're using base 10, and you don't have a baseless
> > > representation.
> >
> > Baseless in the sense that the Natural-Numbers written in decimal form
> > are baseless.
>
> I'm confused here. It looks like you are just using ten-adics, since
> you have the following equations below:
>
> ...0003^3 = ...00027
> ...00027 + ...0008 = ...00035
>
> and these are also true for the 10-adics. Can you provide a calculation
> in the Infinite Integers which is different from one in the 10-adics?
> For instance, tell me how to choose A and B so that A+B in the 10-adics
> gives me a different answer from A+B in the Infinite Integers.

I guess it is hard for you to stop thinking and using adics since you
were buried in them for the past several months. I guess the title of
this thread is misleading also. Adics have nothing to do with Infinite
Integers. Reals have more to do with Infinite Integers.

Infinite Integers are base-free, just as the Natural Numbers are base
free. All results are the same no matter what base you have the Natural
Numbers in, same for Reals.

Try to get it through your head, Chris, that Infinite Integers are the
Natural Numbers themselves, for every axiom of Peano is satisfied by
these Infinite Integers and the old finite integers were only a few
members. Most of the cards of the deck that is Natural Numbers were
missing by playing only with finite-integers.

Now *most* everything that happens for Reals in addition, in
multiplication happens with these Infinite Integers. In a sense the
Infinite Integers are exactly like the Reals once you drop or delete
every decimal point. So the square root of a Real is the square root of
a Infinite Integer.

The Reals are continuous, the Infinite Integers are discrete and
separated by 1 unit.

Affirmative. But there are alot of repetitions in Infinite Integers.

I could spend time defining much of what you ask above. But that would
cut into my time of discovery of new and better things. I leave
definition-pushing to those not on a track of discovery.

> Would the digitized square root of 2 be ...4124141 ? (I'm pretty sure
> about this one.)

In essence the Infinite Integers do and behave almost identically to
the operations performed on the Reals. None of this is Adics and forget
Adics.

These Infinite Integers are identical to Reals which have their decimal
point removed. And where Reals are continuous, Infinite Integers are
discrete points.


>
> Since the cube root of 35 can also be written as 03.27106631...,
> wouldn't that mean that the digitized cube root of 35 could also be
> ...1366017230 ? (I suspect not ...)
>

No, the Infinite Integers do not recognize zeroes to the left of Reals
that are vacuous zeroes. The predecessor of 136601723 is 136601722 and
the successor is 136601724.


> DEFINITION 5. If R is a real number of the form
> a(-2)a(-1)a(0).a(1)a(2)a(3)..., with a(-2) not zero (decimal form),
> then the digitized version of R is the Infinite Integer
> ...a(3)a(2)a(1)a(0)a(-1)a(-2).
>
> (This was generalized from your equation digitized (cube root 35) is
> ......136601723.)
>
> > and this Infinite Integer would be found in every Winding.
>
> Once again, this is confusing, because you've used "winding" in two
> different senses:
>
> (1) ...111112 is in winding ...111, because the digits are eventually
> ...111.
> (2) ...111112 is in winding ...222, because it ends in ...222.

Every counting number starting with 0 forms a winding. 000s winding
then comes 111s winding then comes 222s winding. As you count every
number it forms a winding.

The reason I created Windings is to Order the Infinite Integers so that
given any Infinite Integer call it X, I can tell you what X-1 is and
tell you what X+1 is. The reason I need that is because these numbers
thence satisfy the Peano Axioms and thus are the Natural Numbers
themselves. You cannot do that with the Adics and I ditched them.

When I created the Infinite Integers in 1993, I kept getting nagging
questions from a trio out of Princeton University that given any
Infinite Integer X, I was unable to tell that trio what X-1 and X+1
was. And because I was unable to do that those three math graduate
students of Princeton never believed that the Infinite Integers were
the Natural Numbers. Here I am 12 years later and now able to do that
task.

The windings simply order these Infinite Integers so that I can give
you X-1 and X+1 whenever you insist on giving me X. Thus Natural
Numbers = Infinite Integers.

>
> Clearly, no Infinite Integer can be in the ...111 and ...222 windings,

Yes they can. There is a Real Number of 0.11111.....
and 0.222222.... for which you have no qualms about.

I need to know if regular polygons can be inscribed in logarthmic
spiral and whether a hexagon, for instance divides the spiral in 6 even
spots. I would guess it is impossible.

So I would guess that pi and e are fundamentally different in that pi
is evenly divisible by an infinitude of numbers and that e is a prime
irrational, whereas pi is a composite irrational. Both pi and e are
transcendental and that allows pi to be composite irrational. But other
irrationals that are not transcendental are all prime irrationals.

Jonathan Hoyle

unread,
Dec 22, 2005, 10:16:30 AM12/22/05
to
>> Now, do the above Infinite Integers answer the old
>> problems of mathematics of Fermat's Last Theorem,
>> Goldbach Conjecture, Twin Primes, Riemann
>> Hypothesis? I think they do in that FLT is false and
>> has counterexamples in every power higher than
>> squaring.
<Crankish ranting snipped>

None of those problems' domains are the p-adic spaces. FLT's domain,
for example, is the natural numbers. Hell, I could "solve" FLT by
changing the domain to reals, and observing that z = (a^n+b^n)^(1/n),
and thus "FLT is false" everywhere. But that isn't the challenge. The
challenge is to solve over the natural numbers.

a_plu...@hotmail.com

unread,
Dec 22, 2005, 3:32:49 PM12/22/05
to

I was asked to respond to this since Hoyle is in my killfile. People in
my killfile are mostly incurable loons (sorry to normal people for such
strong language but the Internet allows to many of these bully type
loons to enter a thread for which they were never competent to enter).

Hoyle would have had a valid argument above in 1993 and 1994 and up
till this week. But after this week, Hoyle is utterly wrong.

The change this week is that given any Infinite Integer (not adics but
infinite integers) call it X, I can provide you with X-1 and X+1. Those
Windings provide me with the predecessor and successor to any given
Infinite Integer. I did not have that ability in 1993 but in December
of 2005, the last and final objection to Infinite Integers =
Natural-Numbers is now lifted. Every Peano Axiom, or postulate is
satisfied by these Infinite Integers.

So Hoyle and loons like him could have validly objected that I was
shifting domain from 1993 to 2005. But that objection is no longer
valid. The tables are turned against Hoyle and anyone who thinks like
him.

Now I accuse Hoyle and any like minded person, that your
Natural-Numbers = Finite Integers is only a partial deck of cards and
you are not playing with the full deck of cards that is the Natural
Numbers. Say I take out of a deck of cards all of the cards beyond the
number 7 and put them in a different room or threw them in the trash.
And so any card game you were playing was not a legitimate game of
cards.

Before Dec 2005 the accusation that I shifted the domain of Number
theory could be held as a valid accusation, because I could not provide
X-1 and X+1 to any specified Natural Number X. Now I can and thus the
Infinite Integers are the full set of Natural Numbers and Number Theory
played with only Finite Integers (those of the 000s Winding) is a
corrupted Number theory.

Jonathan Hoyle

unread,
Dec 22, 2005, 5:18:43 PM12/22/05
to
>> The change this week is that given any Infinite Integer
>> (not adics but infinite integers) call it X
<More crankery snipped>

There are no infinite integers, all integers are finite. I suppose you
could extend the question of FLT into Non-Standard Analysis, and ask if
this applies in the domain of hyper-integers, but we would still be
talking about a different domain than the natural numbers. I am not
saying that Non-Standard Number Theory is unmeaningful. I'm sure it
has its own fascinations and interesting resulting. Let's just
remember that Fermat was not concerned it, since Fermat dealt only with
the standard (finite) integers.

>> Now I accuse Hoyle and any like minded person, that
>> your Natural-Numbers = Finite Integers is only a partial
>> deck of cards and you are not playing with the full deck
>> of cards that is the Natural Numbers.

Again, we appear to be playing the "definition" game. Like it or not,
the natural numbers are finite by definition. Archie's "integers" are
not the integers that FLT (and similar number theory problems) are
considering. Calling these infinite things "integers" does not change
the fact that the original problem looks at only finite solutions.

Assuming for the moment that Archie's "infinite integers" are really
the hyper-integers of NSA (which I am doubtful of), then I believe FLT
would still remain true by NSA's Transfer Principle. Irrespective,
this obsession with changing a definition just seems silly. If you
must translate into Archie-speak, just do the following:

1. From English to Archie-Speak: Place the adjective "finite" in front
of each instance of "integer" or "natural number". (It is redundant
for English speakers.)

2. From Archie-speak to English: Place the prefix "hyper" in front of
each instance of "integer" or "natural number". (It is redundant for
Archie speakers.)


On a completely unrelated note, I have solved the Twin Prime Theorem!
I note that limiting the term "prime" to only those numbers divisible
by themselves and 1 is a corrupted Number Theory. (After all, aren't
we as a society supposed to be above discrimination?) Therefore, I
have promoted the definition of "prime" to refer to any integer. Now
we can easily prove that there are an infinite number of twin primes!
:-)

Jonathan Hoyle
Eastman Kodak

a_plu...@hotmail.com

unread,
Dec 22, 2005, 10:57:46 PM12/22/05
to

You know Kodak as a company has done poorly over the past 2 decades and
one of the major reasons is because its management fails to see a
realistic picture of their business environment, constantly clinging to
their old ways of their market niche, unable to see that digital was
coming on strong.

Similarly, a person like Hoyle, unrealistically, thinks that if a group
of men comes together in the 1800s and defines Natural Numbers in Peano
Axioms that those definitions are air tight waterproof and consistent
forever. That once somebody defines something, silly people like Hoyle,
think they are everlastingly perfect definitions, flawless forever.

No wonder Kodak was on the verge of bankruptcy by the early 2000s when
it has people like Hoyle who live in the "past" and have a mind not
able to bend, change or learn.

When a problem such as Fermat's Last Theorem or Riemann Hypothesis or
Goldbach come along and ask that "every Natural Number has this
property". Those conjectures do not say "stop with only the Numbers
ending in 0000s", for they want to know if every Natural Number obeys
the conjecture. The Natural Numbers are defined by Peano Axioms and
those axioms do not stop with finite integers but spill over into the
Infinite Integers. Hoyle does not possess a good enough mind to
recognize this. His is an old dogmatic mind, unable to learn, unable to
grow, unable to see things clearly.

What Peano Axioms define with its endless adding of 1 is that the
Natural Numbers cruise into the Infinite Integers and that no-one can
cut out a set that is purely finite. Either you toss out endless adding
of 1 or you toss out Finite-Integers.

It is good to know that Hoyle is not in the teaching profession because
dogmatists who ceased to learn do not belong in education.

Proginoskes

unread,
Dec 23, 2005, 1:47:43 AM12/23/05
to

a_plu...@hotmail.com wrote:
> Proginoskes wrote:
> > [...]

> > I'm confused here. It looks like you are just using ten-adics, since
> > you have the following equations below:
> >
> > ...0003^3 = ...00027
> > ...00027 + ...0008 = ...00035
> >
> > and these are also true for the 10-adics. Can you provide a calculation
> > in the Infinite Integers which is different from one in the 10-adics?
> > For instance, tell me how to choose A and B so that A+B in the 10-adics
> > gives me a different answer from A+B in the Infinite Integers.
>
> I guess it is hard for you to stop thinking and using adics since you
> were buried in them for the past several months. [...]

No, it isn't. I got rid of the idea of 10-adics until your calculations
reminded me of them. Never assume that things that are hard for you are
hard for other people.

Now I'll repeat what I actually asked, and which you didn't answer:
Computationally, how are Infinite Integers different from 10-adics?

When you say ...0003^3 = ...00027 and ...00027 + ...0008 =
...00035, that suggests 10-adic addition. You should be able to give me
an explicit example of two numbers A and B where A+B is one thing if
you think of them as 10-adics, and another thing if you think of them
as Infinite Integers.

Does this mean all of my post (up to here) is correct, or just the part
about the digitized cube root of 35000 being ...136601723 ?

> But there are alot of repetitions in Infinite Integers.
>
> I could spend time defining much of what you ask above. But that would
> cut into my time of discovery of new and better things. I leave
> definition-pushing to those not on a track of discovery.

Au contraire. You need the definitions to formalize what you're talking
about, without getting contradictions. Once I applied actual
definitions for the (now damned for eternity) "All-Adics", your
analysis fell apart. (I.e., your statement that you thought the AA's
were a field simply failed, on a basic level.)

Creativity in mathematics _can_ be done while remaining formal. Hell,
_I_'ve done it myself.

> > Would the digitized square root of 2 be ...4124141 ? (I'm pretty sure
> > about this one.)
>
> In essence the Infinite Integers do and behave almost identically to
> the operations performed on the Reals. None of this is Adics and forget
> Adics.
>
> These Infinite Integers are identical to Reals which have their decimal
> point removed.

But now I'm stumped by your equation that says ...0003^3 = ...00027.
Since the number 27 equals the real number 27.0000..., and the number 3
equals the real number 3.0000...., that should make 27 be ...000072,
and your equation is really

...00003^3 = ...00072

(This is what I'm talking about when I mention consistency.)

But now we have a definite problem, because 1 + 10 = 11, but if we
digitize all the numbers, we get

digitized(1) + digitized(10) = ...0001 + ...0001 = ...0002

which is not digitized(11). So it is not true that

digitized(x) + digitized(y) = digitized (x + y)

in general.

> > Since the cube root of 35 can also be written as 03.27106631...,
> > wouldn't that mean that the digitized cube root of 35 could also be
> > ...1366017230 ? (I suspect not ...)
> >
>
> No, the Infinite Integers do not recognize zeroes to the left of Reals
> that are vacuous zeroes. The predecessor of 136601723 is 136601722 and
> the successor is 136601724.

What is the successor of ...9999? What is the predecessor of ...000?

> > DEFINITION 5. If R is a real number of the form
> > a(-2)a(-1)a(0).a(1)a(2)a(3)..., with a(-2) not zero (decimal form),
> > then the digitized version of R is the Infinite Integer
> > ...a(3)a(2)a(1)a(0)a(-1)a(-2).
> >
> > (This was generalized from your equation digitized (cube root 35) is
> > ......136601723.)
> >
> > > and this Infinite Integer would be found in every Winding.
> >
> > Once again, this is confusing, because you've used "winding" in two
> > different senses:
> >
> > (1) ...111112 is in winding ...111, because the digits are eventually ...111.
> > (2) ...111112 is in winding ...222, because it ends in ...222.
>
> Every counting number starting with 0 forms a winding. 000s winding
> then comes 111s winding then comes 222s winding. As you count every
> number it forms a winding.

(1) What numbers are on the ...000 winding?

(2) Doesn't 001 have a winding?

> The reason I created Windings is to Order the Infinite Integers so that
> given any Infinite Integer call it X, I can tell you what X-1 is and
> tell you what X+1 is. The reason I need that is because these numbers
> thence satisfy the Peano Axioms and thus are the Natural Numbers
> themselves. You cannot do that with the Adics and I ditched them.

But now you're stuck with the old question: "How do you get from a
finite number of nonzero digits to an infinite number by adding 1?"
again.

> When I created the Infinite Integers in 1993, I kept getting nagging
> questions from a trio out of Princeton University that given any
> Infinite Integer X, I was unable to tell that trio what X-1 and X+1
> was. And because I was unable to do that those three math graduate
> students of Princeton never believed that the Infinite Integers were
> the Natural Numbers. Here I am 12 years later and now able to do that
> task.

And now you think that the Peano Axioms apply to your Infinite
Integers, but they don't. That's because if you consider the successor
to ...999, you run into a contradiction:

Let's let the successor of ...999 be A = ...a(2)a(1)a(0). The
successor of ...999 cannot be zero; this violates a Peano Axiom, so at
least one number a(i) must be nonzero. Let k be the smallest number for
which this is true.

If k = 0, then let B be the Infinite Integer ...a(2)a(1)[a(0)-1]; i.e.,
the predecessor of A. If k > 0, then a(k-1), a(k-2), ..., a(0) are all
zeros. The predecessor of A is thus
B = ...a(k+1)[a(k)-1]999...999.

Now, notice that not all the digits of B are 9's, since one of the
digits is a(k)-1<= 9-1=8.
Thus B and ...999 are two different Infinite Integers with the same
successor. This also violates a Peano Axiom.

Therefore Infinite Integers are not Natural Numbers.

> > Clearly, no Infinite Integer can be in the ...111 and ...222 windings,
>
> Yes they can. There is a Real Number of 0.11111.....
> and 0.222222.... for which you have no qualms about.

I'm talking about a particular Infinite Integer A being in two
windings, not two Infinite Integers in two windings.

> > if definition (2) is used, since no Infinite Integer ends in 1 and 2
> > (at the same time); but if you mean (1), then the digits of A are
> > eventually all 1's (in order to be in the ...111 winding), but then A
> > can't be in the ...222 winding, since the digits will not eventually be
> > all 2's.
> >
> > So neither one of these definitions actually works. So I have a
> > question:
> >
> > QUESTION. If A and B are Infinite Integers, then when is A in Winding
> > #B?
> >

> > And this is one you _must_ answer. [...]

And I repeat: You MUST answer this one.

And you clearly didn't read any more.

--- Christopher Heckman

Proginoskes

unread,
Dec 23, 2005, 1:49:03 AM12/23/05
to

You evidently haven't heard of the "AP way of solving problems", where
you change definitions until the result becomes trivial to prove, then
claim success. He also did it for the Map-Coloring problem.

--- Christopher Heckman

Proginoskes

unread,
Dec 23, 2005, 1:55:19 AM12/23/05
to

a_plu...@hotmail.com wrote:
> Jonathan Hoyle wrote:
> > >> Now, do the above Infinite Integers answer the old
> > >> problems of mathematics of Fermat's Last Theorem,
> > >> Goldbach Conjecture, Twin Primes, Riemann
> > >> Hypothesis? I think they do in that FLT is false and
> > >> has counterexamples in every power higher than
> > >> squaring.
> > <Crankish ranting snipped>
> >
> > None of those problems' domains are the p-adic spaces. FLT's domain,
> > for example, is the natural numbers. Hell, I could "solve" FLT by
> > changing the domain to reals, and observing that z = (a^n+b^n)^(1/n),
> > and thus "FLT is false" everywhere. But that isn't the challenge. The
> > challenge is to solve over the natural numbers.
>
> Hoyle would have had a valid argument above in 1993 and 1994 and up
> till this week. But after this week, Hoyle is utterly wrong. [...]

No; AP is utterly wrong. Fermat framed his problem BEFORE 1993, IN THE
TERMINOLOGY OF HIS DAY, which meant that natural numbers did not mean
Infinite Integers, since Fermat didn't live to see them. If Fermat
could be sent to the year 2005 (or 2006, or whenever it can be shown
that Infinite Integers are contradiction-free), and if he postulated
his "theorem" after exposure to the Infinite Integers, THEN AND ONLY
THEN can you claim that the problem belongs in the domain of the
Infinite Integers.

You're "changing the essential elements" of the problem, something you
claimed other people (who work on the 4CT and other difficult problems)
did.

> Now I accuse Hoyle and any like minded person, that your
> Natural-Numbers = Finite Integers is only a partial deck of cards and
> you are not playing with the full deck of cards that is the Natural
> Numbers. Say I take out of a deck of cards all of the cards beyond the
> number 7 and put them in a different room or threw them in the trash.
> And so any card game you were playing was not a legitimate game of

> cards. [...]

Well, it could be argued that the Infinite Integers themselves aren't
the right definition, only another step towards the correct
definition. That would mean that you aren't playing with a full deck,
either.

--- Christopher Heckman

P.S. You left yourself open to this one, AP.

Proginoskes

unread,
Dec 23, 2005, 1:57:56 AM12/23/05
to

Jonathan Hoyle wrote:
> [...]

> On a completely unrelated note, I have solved the Twin Prime Theorem!
> I note that limiting the term "prime" to only those numbers divisible
> by themselves and 1 is a corrupted Number Theory. (After all, aren't
> we as a society supposed to be above discrimination?) Therefore, I
> have promoted the definition of "prime" to refer to any integer. Now
> we can easily prove that there are an infinite number of twin primes!
> :-)

You might think you're being clever, but AP simply does not believe in
using analogies. Hence your point will be lost on him.

--- Christopher Heckman

Proginoskes

unread,
Dec 23, 2005, 2:14:47 AM12/23/05
to

a_plu...@hotmail.com wrote:
> Gottried Helms-- can I enlist you to write a textbook in German.
> Titled: Infinite-Integers equals Natural-Numbers.

Why bother with him? You can do it yourself, using a resource like
AltaVista ( http://babelfish.altavista.com/translate.dyn ). For
instance, the paragraph

The major difference between adics and Infinite-Integers is that adics
have no sequential order, whereas these Infinite-Integers given any X I

can tell you what X-1 is and what X+1 is, thus they are the Natural
Numbers themselves.

turns out as the following in German:

Der Hauptunterschied zwischen adics und Endlos-Ganzzahlen ist, daß
adics keinen aufeinanderfolgenden Auftrag haben, während diese
Endlos-Ganzzahlen, die jedem möglichem X mich gegeben werden, Ihnen
erklären können, was X-1 ist und welches X+1 ist, so sind sie sich
numerieren das natürliche.

and the following in Russian:

Основная отличия между adics и
Инфинитн-Intejerami что adics не имеет никакой
последовательный заказ, тогда как эти
Инфинитн-Intejery, котор дали любому х меня
могут сказать вам X-1 и что X+1, таким
образом они естественными нумерует.

and Swedish:

Zee mäyør difference between ädicee änd Infinite-Integeree ädicee
noo sequentiäl ørdy, vheræ Infinite-Integeree given äny X Me cän
tell vhät X-1 änd vhät Xplus1 , thsky they Näturäl Numberee
themselvy.

But why stop there? Here it is in Pig Latin (via
http://www.snowcrest.net/donnelly/piglatin.html ):

Ethay ajormay ifferenceday etweenbay adicsway andway
Infiniteway-Integersway isway atthay adicsway avehay onay equentialsay
orderway, ereaswhay esethay Infiniteway-Integersway ivengay anyway XAY
Iway ancay elltay ouyay atwhay XAY-1 isway andway atwhay XAY+1 isway,
usthay eythay areway ethay Aturalnay Umbersnay emselvesthay.

Or let B1FF ghost-write it ( http://www.shortbus.net/dialect.html ):

THE MAJOR DIFFURENCE BETWEEN ADICZ + MENF1NITE-INTEGURZ 1Z THAT ADICZ
GOT NO SEQUENTIAL ORDUR. WHEARAZ THESE 1NFINITE-1NTEGURZ GIVEN ANY X I
CAN TEL U WUT X-1 1Z AND WUT X+1 IS. THUZ THEY R THE NATURAL NUMBURZ
THUMSELVES!!!!1!!1!

Or reach inner city kids with the following translation:

De majo' difference between adics and Infinite-Integers be dat adics
gots' no sequential o'der, whereas dese Infinite-Integers given any X
ah' can tell ya' whut X-1 be and whut X+1 is, dus dey are da damn
Natural Numbers demselves.

Or even reach other species:

The major difference between adics and Infinite-Integers is like, ya
know, that adics have no sequential order, like, wow, whereas these
Infinite-Integers given any X I can tell you what X-1 is and what X+1
is, like, wow, thus they are thuh Natural Numbers themselves.

But if I can be serious again ...

If you do succeed in finding a publisher to print your book, I'd like
to buy the first copy.

--- Christopher Heckman

Proginoskes

unread,
Dec 23, 2005, 2:19:33 AM12/23/05
to

a_plu...@hotmail.com wrote:
> [...] a person like Hoyle, unrealistically, thinks that if a group

> of men comes together in the 1800s and defines Natural Numbers in Peano
> Axioms that those definitions are air tight waterproof and consistent
> forever. That once somebody defines something, silly people like Hoyle,
> think they are everlastingly perfect definitions, flawless forever.

And of course A.P. is a realistic person who comes along, changes
"blue" to mean "red", and then goes on to say -- and "prove" -- that
all apples are blue -- and claim that this is new knowledge.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 23, 2005, 2:32:11 AM12/23/05
to
Chris Heckman wrote:
What is the successor of ...9999? What is the predecessor of ...000?

A.P. writes: 0000 is easy to answer in that the Peano Axioms tell us
that O has no predecessor.

As for the ....99999 that is a tougher one, because it has the
potential of sinking this entire Array. The trouble I have in answering
it is because the number ....99999 appears as a bridge in every
successive winding. So if I cannot answer this decently, I may be
sinking.

You see, the bridge between ....0000 winding has ...99999 as its last
number bridging to the next winding of ...1111 and then ....99999
appears again as the last number in the ....1111 winding to link over
into the first number of the ....2222 winding which is ....2222220.

So now Chris has asked me what is the successor of ....99999.

This is critical, because it could destroy this entire Array.

But let me ask Chris a question. Where in the Peano Axioms does it say
that a number must have a unique successor? Consider the number that
has a billion digits of 9. We do not expect such a number to arise
again further out once we reach this number. We all know that as we
count numbers we count them only once and never again meet up with them
the further out we count. So the question is Chris, where in the Peano
Axioms does it forbid a number to recurr. I know there was one
postulate that said 0 has no predecessor, but I do not remember a
postulate that said for all n, n is never recurring in the count. I
know you posted the Peano Axioms but I do not remember if there was
such a postulate.

Now if there is such a postulate, then it is inconsistent with the
postulate of endless adding of 1. Inconsistent because it denies "all
possible digit arrangements".

I am hoping the Peano Axioms are a consistent set of axioms and thus
yield all possible digit arrangements and thus the Infinite Integers.

So to answer the question what is the successor of ....99999 it is
....1111110. But it is also ....222220 and also .....333330.

So this is a grave challenge to my Array. Does the Peano Axioms state
that a number cannot be repeated in the count?

Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom

where dots of the electron-dot-cloud are galaxies.

a_plu...@hotmail.com

unread,
Dec 23, 2005, 2:43:32 AM12/23/05
to
Chris Heckman wrote:
No; AP is utterly wrong. Fermat framed his problem BEFORE 1993, IN THE
TERMINOLOGY OF HIS DAY, which meant that natural numbers did not mean
Infinite Integers, since Fermat didn't live to see them. If Fermat
could be sent to the year 2005 (or 2006, or whenever it can be shown
that Infinite Integers are contradiction-free), and if he postulated
his "theorem" after exposure to the Infinite Integers, THEN AND ONLY
THEN can you claim that the problem belongs in the domain of the
Infinite Integers.

A.P. writes:
Convoluted logic of Chris Heckman would then say that since the angels
moved the planets and the Sun in orbit during the Ptolemy century,
since it was defined as such, that our modern telescopes should reveal
those angels flying and keeping the planets in their perfect circle
orbits.

It is astounding that any person has a degree in mathematics and argues
that because a problem is defined and formulated in the 1600s can only
be answered by 1600 ideas and cannot be challenged with the new facts
and ideas that arose in future centuries.

The reason Heckman and Hoyle are writing so irrationally is because
they will say anything rather than admit that I have found a new truth.
They are poor sports and their game is anything but admit that I am
correct.

Proginoskes

unread,
Dec 23, 2005, 2:43:43 AM12/23/05
to

a_pluton...@hotmail.com wrote:
> Chris Heckman wrote:
> What is the successor of ...9999? What is the predecessor of ...000?
>
> A.P. writes: 0000 is easy to answer in that the Peano Axioms tell us
> that O has no predecessor.

What does the "A.P. definition of Infinite Integers" say, not refering
to the Peano Axioms?

> As for the ....99999 that is a tougher one, because it has the

> potential of sinking this entire Array. [...]


> But let me ask Chris a question. Where in the Peano Axioms does it say
> that a number must have a unique successor?

The successor operation must be a function, which means there's a
unique successor for each number. (#2 on MathWorld's Peano Axioms page;
they talk about "the" successor, so it must be unique. If the idea of a
successor weren't unique, you could only talk about "a" successor.)

> [...] So the question is Chris, where in the Peano


> Axioms does it forbid a number to recurr.

> [...]


> I know there was one
> postulate that said 0 has no predecessor, but I do not remember a
> postulate that said for all n, n is never recurring in the count. I
> know you posted the Peano Axioms but I do not remember if there was
> such a postulate.

The axiom that the successor function is one-to-one. (#4 on MathWorld's
Peano Axioms page)

Check out MathWorld's definition. (This is a good link to have for the
basic definition of any math concept.)
http://mathworld.wolfram.com/PeanosAxioms.html

> Now if there is such a postulate, then it is inconsistent with the
> postulate of endless adding of 1. Inconsistent because it denies "all
> possible digit arrangements".

And which also implies that the Infinite Integers are a bigger set than
the Natural Numbers.

Also, the standard Cantorian diagonalization proof can be used to show
that the Natural Numbers and the Infinite Integers have different
cardinalities, which also implies the two sets can't be the same.

> I am hoping the Peano Axioms are a consistent set of axioms

They are consistent in the sense that you cannot prove something that
is false.

> and thus yield all possible digit arrangements and thus the Infinite Integers.
>
> So to answer the question what is the successor of ....99999 it is
> ....1111110. But it is also ....222220 and also .....333330.

If you take the predecessor of the successor of a number N, you should
get N back. The predecessor of ...1111110 is ...11111109, which is not
...999.

> So this is a grave challenge to my Array. Does the Peano Axioms state
> that a number cannot be repeated in the count?

Yep.

--- Christopher Heckman

Proginoskes

unread,
Dec 23, 2005, 2:56:59 AM12/23/05
to

a_pluton...@hotmail.com wrote:
> Chris Heckman wrote:
> No; AP is utterly wrong. Fermat framed his problem BEFORE 1993, IN THE
> TERMINOLOGY OF HIS DAY, which meant that natural numbers did not mean
> Infinite Integers, since Fermat didn't live to see them. If Fermat
> could be sent to the year 2005 (or 2006, or whenever it can be shown
> that Infinite Integers are contradiction-free), and if he postulated
> his "theorem" after exposure to the Infinite Integers, THEN AND ONLY
> THEN can you claim that the problem belongs in the domain of the
> Infinite Integers.
>
> A.P. writes:
> Convoluted logic of Chris Heckman would then say that since the angels
> moved the planets and the Sun in orbit during the Ptolemy century,

They did? It wasn't gravity, way back then? Wow. I never knew that. I
never even suspected it. Is this part of your revolution in Physics?
What other revelations are out there? That luminous aether exists after
all? And do I need to go back to my undergraduate college to take a few
more classes, so that I'll still have a minor in Physics?

Don't say that I am assuming things that I'm not.

> since it was defined as such,

Oh, now I see his point. No, it wasn't defined as such; it was ASSUMED
to be such. The thing responsible for moving the planets is simply the
thing that causes the planets to move. Whether it's angels or gravity
or oxen or skyhooks is not a question of definition, but of conjecture
(or an axiom).

The two things may be the same in physics, but they're not the same in
mathematics. Which AP would know if he had ever bothered to take a
basic logic course.

> It is astounding that any person has a degree in mathematics and argues
> that because a problem is defined and formulated in the 1600s can only
> be answered by 1600 ideas and cannot be challenged with the new facts
> and ideas that arose in future centuries.

It is astounding that any sane person argues that he can change a
"basic problem element" and still be solving the same problem. Unless
he/she is a crank, or some such thing.

> The reason Heckman and Hoyle are writing so irrationally is because
> they will say anything rather than admit that I have found a new truth.
> They are poor sports and their game is anything but admit that I am
> correct.

Because AP never HAS been. He refuses to read the proofs of why he's
wrong, so in his mind, he's batting 1.000.

And speaking of sports, maybe he should take up professional baseball.
Then he can say that any time he swings the bat counts as a grand slam,
and he'll be the greatest baseball player in history.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 23, 2005, 3:21:51 AM12/23/05
to
Okay I looked up the Peano Axioms in Wikipedia:
1 There is a natural number 0.
2 Every natural number a has a successor, denoted by S(a) or x'.
3 There is no natural number whose successor is 0.
4 Distinct natural numbers have distinct successors: a = b if and
only if S(a) = S(b).
5 If a property is possessed by 0 and also by the successor of
every natural number which possesses it, then it is possessed by all
natural numbers. (This axiom, also known as axiom of induction, ensures
that the proof technique of mathematical induction is valid.)

And the #4 postulate challenges the idea that a number can recurr in
the counting.

So I have two routes to take from here. I can say there is a different
Array to patch up that messy spot of ...9999 recurring with nonunique
successors, but that would be fruitless because other numbers recurr in
the counting. The number ....1010103 is in the 10 winding but also in
the 1,010 winding. So there are infinite repeats and #4 prohibits
repeats.

One option is to say the Peano Axioms are steadfast and true, like
everyone else does. But I am not like everyone else.

And I choose the third option. I say that postulate #4 is inconsistent
with #2, the endless adding of 1. That you cannot have these two
postulates and have a consistent set of axioms. You cannot have endless
adding of 1 and never have repeating of a number when they are counted.

Endless adding of 1 produces all possible digit arrangements. And it is
impossible to count that set of all possible digit arrangements without
repeating some number in the count.

So I say the Peano Axioms are inconsistent, internally inconsistent.

Proginoskes

unread,
Dec 23, 2005, 3:43:27 AM12/23/05
to

a_plu...@hotmail.com wrote:
> Okay I looked up the Peano Axioms in Wikipedia:
> 1 There is a natural number 0.
> 2 Every natural number a has a successor, denoted by S(a) or x'.
> 3 There is no natural number whose successor is 0.
> 4 Distinct natural numbers have distinct successors: a = b if and
> only if S(a) = S(b).
> 5 If a property is possessed by 0 and also by the successor of
> every natural number which possesses it, then it is possessed by all
> natural numbers. (This axiom, also known as axiom of induction, ensures
> that the proof technique of mathematical induction is valid.)
>
> And the #4 postulate challenges the idea that a number can recurr in
> the counting.

Not only challenge it, it forbids it.

> So I have two routes to take from here. I can say there is a different
> Array to patch up that messy spot of ...9999 recurring with nonunique
> successors, but that would be fruitless because other numbers recurr in
> the counting. The number ....1010103 is in the 10 winding but also in
> the 1,010 winding. So there are infinite repeats and #4 prohibits
> repeats.
>
> One option is to say the Peano Axioms are steadfast and true, like
> everyone else does. But I am not like everyone else.
>
> And I choose the third option. I say that postulate #4 is inconsistent
> with #2, the endless adding of 1. That you cannot have these two
> postulates and have a consistent set of axioms. You cannot have endless
> adding of 1 and never have repeating of a number when they are counted.

Why not? That would mean there is something wrong with the function
f(n) = n+1.

Axiom #2 says this function is defined for all nonnegative integers n.
(In fact, it's even defined on bigger sets, like the real numbers, the
complex numbers, the quaternions, so that's not the problem.)

Axiom #4 says that f(x) = f(y) iff x = y. So your claim is that you can
have
x + 1 = y + 1
and at the same time, x and y are different.

> Endless adding of 1 produces all possible digit arrangements. And it is
> impossible to count that set of all possible digit arrangements without
> repeating some number in the count.
>
> So I say the Peano Axioms are inconsistent, internally inconsistent.

I say that you are a woman. That doesn't make it true.

If you could _prove_ the Peano Axioms are inconsistent, that would be a
solution. But there is no inconsistency in the Peano Axioms.

You have a fourth option: Look up Gerhard Gentzen's proof and find a
mistake.

You have a fifth option: Deny that Infinite Integers really are Natural
Numbers. Another point in favor of this option is that Cantor's
diagonal argument also applies to Infinite Integers. The Wikipedia link
for this proof is:
http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 23, 2005, 5:16:18 AM12/23/05
to

a_plu...@hotmail.com wrote:
> Okay I looked up the Peano Axioms in Wikipedia:
> 1 There is a natural number 0.
> 2 Every natural number a has a successor, denoted by S(a) or x'.
> 3 There is no natural number whose successor is 0.
> 4 Distinct natural numbers have distinct successors: a = b if and
> only if S(a) = S(b).
> 5 If a property is possessed by 0 and also by the successor of
> every natural number which possesses it, then it is possessed by all
> natural numbers. (This axiom, also known as axiom of induction, ensures
> that the proof technique of mathematical induction is valid.)
>

Okay, what Chris Heckman set in motion tonight with his question of
what is the Successor of ....9999 set a grave and severe challenge to
my Array of the Infinite Integers in that it could not deliver a unique
successor and thus violates #4.

I had a important post to add to the other current thread of Riem geom
+ Loba geom = Eucl geom but this question riled me so much that I
decided to take a shower and go to bed. But in the midst of the shower
I found a resolution answer.

The thought kept going through my mind of DeBroglie pilot wave, the
wave that guides light waves. So I needed a pilot wave to sequentally
order the Infinite-integers and obey all the Peano axioms.

New Array:
0, ....000000, 1, ....11111, 2, .....22222, 3, .....333333, 4,
....44444, 5, ....55555, 6, ....66666, 7, ....777777, 8, .....88888, 9,
....99999, 10, ....10101010, 11, .....111111, 12 ad infinitum

That is the New Array and I went to 11 so as to show the ....11111
repetition with that of ....1111 for 11 and for 1.

Now I obey all of the Peano Axioms because when Chris asks for what is
the successor of ....99999, he has to ask what is the Successor of 9,
and .....9999 and then my answer is 10. If he asked for what is the
successor of 99, and ....9999 then it is 100. You see the pilot? And
this obeys Peano's #4 because the uniqueness is not a solitary
uniqueness but a dualistic uniqueness. Pairs of numbers yield a unique
successor.

Now one may bulk at the idea that adding 1 to 2 is not ....22222 but 3,
and my rejoinder is that adding 1 to 2 has to consider all
possible-place-values before reaching 3 and the ....22222 represents
all possible place values. So this satisfies #2 of the Peano axioms and
#4 and so all of the Peano axioms are satisfied.

So give me any X-1 and X, I can give you X+1 that is unique. Give me
any X+1 and X and I can give you X-1. That is not as strong as give me
any X and I can give you X-1 and X+1, but strong enough or weak enough
to satisfy the Peano Axioms.

Question: can we define Reals in most compact form as a string of
digits that has a finite leftward portion and an infinite rightward
portion. And which possesses all possible digit arrangements? Is that
true? The most compact definition of Reals? I think so.

Now, the most compact definition of Peano Axiom Integers is an infinite
leftward string which has all possible digit arrangements contained. I
believe this is the most compact definition of the counting numbers.
The Peano Axioms are an elaboration of this compact definition. The
endless adding of 1 forces all possible digit arrangements to emerge
and so the alleged finite-integers could not fulfill the Peano Axioms.

Now many may jokingly comment that every time they count to ten they
missed ten other numbers in their count and my answer would be like the
Pilot wave or the dualism that the endless adding of 1 is really the
endless adding of 2, not 1. So we start with 0 and add two new terms of
....00000 and 1, then to 1 we add two new terms of ....1111 and 2, etc
etc.

Thus I keep all the Infinite Integers, and I keep every possible digit
arrangement and I keep a sequential ordering, and I preserve every one
of the Peano Axioms.

Thus, Natural Numbers = Infinite Integers.

I still have time to make that geometry post.

Proginoskes

unread,
Dec 23, 2005, 5:56:15 AM12/23/05
to

a_plu...@hotmail.com wrote:
> a_plu...@hotmail.com wrote:
> > Okay I looked up the Peano Axioms in Wikipedia:
> > 1 There is a natural number 0.
> > 2 Every natural number a has a successor, denoted by S(a) or x'.
> > 3 There is no natural number whose successor is 0.
> > 4 Distinct natural numbers have distinct successors: a = b if and
> > only if S(a) = S(b).
> > 5 If a property is possessed by 0 and also by the successor of
> > every natural number which possesses it, then it is possessed by all
> > natural numbers. (This axiom, also known as axiom of induction, ensures
> > that the proof technique of mathematical induction is valid.)
> >
>
> Okay, what Chris Heckman set in motion tonight with his question of
> what is the Successor of ....9999 set a grave and severe challenge to
> my Array of the Infinite Integers in that it could not deliver a unique
> successor and thus violates #4.
> [...]

> New Array:
> 0, ....000000, 1, ....11111, 2, .....22222, 3, .....333333, 4,
> ....44444, 5, ....55555, 6, ....66666, 7, ....777777, 8, .....88888, 9,
> ....99999, 10, ....10101010, 11, .....111111, 12 ad infinitum
>
> That is the New Array and I went to 11 so as to show the ....11111
> repetition with that of ....1111 for 11 and for 1.

This doesn't work either; and #4 is the property that fails. This is
because 1 and 11 have the same successor, and 1 and 11 are different.

> Now I obey all of the Peano Axioms because when Chris asks for what is
> the successor of ....99999, he has to ask what is the Successor of 9,
> and .....9999 and then my answer is 10. If he asked for what is the
> successor of 99, and ....9999 then it is 100. You see the pilot? And
> this obeys Peano's #4 because the uniqueness is not a solitary
> uniqueness but a dualistic uniqueness. Pairs of numbers yield a unique
> successor.

Well, then you should be listing _pairs_ of Infinite Integers.
(Besides, the successor to 111, 1111, 11111, etc., will all be
...1111.)

I'm not so sure it satisfies #3, either, unless you distinguish the
natural number zero from the Infinite Integer zero, but then you have
to concede that one of your Infinite Integers (...000) isn't a natural
number.

> Now one may bulk at the idea that adding 1 to 2 is not ....22222 but 3,
> and my rejoinder is that adding 1 to 2 has to consider all
> possible-place-values before reaching 3 and the ....22222 represents

> all possible place values. [...]

This is true and correct. Given the successor function, addition is
defined recursively as

x + 0 = x, and x + S(y) = S(x + y), so, for instance,

1 + ...111 = S(S(0)) + S(S(S(0))) = S[S(S(0)) + S(S(0))] =
S[S[S(S(0))+S(0)]]
= S[S[S[S(S(0)) + 0]]] = S[S[S[S(S(0))]]] = ...222,

which just says "the second element plus the third element is the fifth
element." (0 is counted as the "zeroth element".)

> So give me any X-1 and X, I can give you X+1 that is unique. Give me
> any X+1 and X and I can give you X-1. That is not as strong as give me
> any X and I can give you X-1 and X+1, but strong enough or weak enough
> to satisfy the Peano Axioms.

No; for the Peano Axioms, every NUMBER has exactly one successor and at
most one predecessor.

I don't see ...111112 on the list, though, so you don't have every
possible digit arrangement.

> Question: can we define Reals in most compact form as a string of
> digits that has a finite leftward portion and an infinite rightward
> portion. And which possesses all possible digit arrangements? Is that
> true? The most compact definition of Reals? I think so.

It's not quite the best one, since 0.999... = 1.000... Dedekind Cuts
are the usual method when using the Peano Axioms, once you define
rational numbers. http://mathworld.wolfram.com/DedekindCut.html

> Now, the most compact definition of Peano Axiom Integers is an infinite
> leftward string which has all possible digit arrangements contained. I

> believe this is the most compact definition of the counting numbers. [...]

Nope, because the cardinalities of the Natural Numbers (as defined by
the Peano Axioms) and Infinite Integers are different.

THEOREM. (Cantor) There is no way to put the Infinite Integers into
one-to-one correspondence with the Natural Numbers.

Proof: Diagonalization.

--- Christopher Heckman

mathedman

unread,
Dec 23, 2005, 10:06:57 AM12/23/05
to
On 23 Dec 2005 02:16:18 -0800, a_plu...@hotmail.com wrote:

>
>Okay, what Chris Heckman set in motion tonight with his question of
>what is the Successor of ....9999 set a grave and severe challenge to
>my Array of the Infinite Integers in that it could not deliver a unique
>successor and thus violates #4.
>

This statement doesn't even make sense!
0.999... isn't a real number! It's a symbol for an infinite
series which is, in turn, a sequence.
To say that a sequence "has a successor" is meaningless
rambling.

Proginoskes

unread,
Dec 23, 2005, 11:19:40 PM12/23/05
to

The objects ("numbers") which are in the Peano Axioms don't have to be
actual numbers; just an infinite number of distinct elements.

What A.P. is doing here has very little to do with standard
mathematics. For instance, his Infinite Integers are really the
10-adics, without the addition or multiplication. (Don't tell him that,
though.)

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 24, 2005, 12:36:13 AM12/24/05
to

No, I overcome #4 by pairings. You have to stipulate a pair of
predecessors.
So to put the Infinite Integers into a sequental ordering so they can
be counted, apply the trick of pairing. With the Windings the
successors cannot be unique but by pairing the Successors can be
unique.

But another major problem crops up. With Windings I can get all
possible digit arrangements and especially the irrational Infinite
Integers such as digitized e and digitized pi and digitized square root
2. But can I get the irrational Infinite Integers by this New-Array?
Does it only produce rational Infinite Integers. If it cannot produce
the irrational Infinite Integers my work is not finished.

Can I say the New Array yields the irrational Infinite Integers in that
they converge at infinity?

You fail with your last comment Chris and should know better. There is
some metaphor analogy where the Hero of a myth pulls himself from his
bootstraps by pulling on them and thus lifting himself out of the
quagmire or quicksand or swamp. Of course we all know that we cannot
pull ourselves from a sinking pond by pulling up on the bootstraps.

Are the Natural Numbers the finite-integers or the infinite-integers is
the question? The consequence of finite-integers is a different
cardinality and the consequence of infinite-integers is equal
cardinality. So you cannot use cardinality to decide between whether
the Natural Numberse are finite or infinite integers.

But perhaps a good new path to explore is opened here in that since the
10-adics are of equal cardinality with the Reals, then can I make the
10-adics into a sequental ordering by some diagonal means and thus use
that as the Array for Infinite Integers.

Is Sequental Ordering contradictory to endless adding 1.

a_plu...@hotmail.com

unread,
Dec 24, 2005, 12:50:48 AM12/24/05
to
I wrote a few minutes ago:

Can I say the New Array yields the irrational Infinite Integers in that
they converge at infinity?

A.P. further writes:
Let us take digitized Irrational Infinite Integer pi .....562951413.

There is for example 413 and its pair ....413413413

Later on there is 1413 and its pair ....141314131413

Later on there is 51413 and its pair ....5141351413

But those pairs are all rational Infinite Integers. Can I claim that
irrational Infinite Integers enter the counting as convergences??

You know, this all reminds me of the Dedekind Cut of Rationals to
obtain Reals. Is there a Dedekind Cut on the Natural Numbers to get
irrational Infinite Integers from rational Infinite Integers?

Proginoskes

unread,
Dec 24, 2005, 1:12:44 AM12/24/05
to

That's not part of the rules. #2 says that every _natural number_ has a
successor, not a _pair of natural numbers_. Try again.

> But another major problem crops up. With Windings I can get all
> possible digit arrangements and especially the irrational Infinite
> Integers such as digitized e and digitized pi and digitized square root
> 2. But can I get the irrational Infinite Integers by this New-Array?
> Does it only produce rational Infinite Integers. If it cannot produce
> the irrational Infinite Integers my work is not finished.

You cannot get all possible digit arrangements; there will always be at
least one missing. This is equivalent to a result Cantor proved. (See
below.) In this particular list. ...1112 is missing, something that you
seem to have ignored.

> > > Now, the most compact definition of Peano Axiom Integers is an infinite
> > > leftward string which has all possible digit arrangements contained. I
> > > believe this is the most compact definition of the counting numbers. [...]
> >
> > Nope, because the cardinalities of the Natural Numbers (as defined by
> > the Peano Axioms) and Infinite Integers are different.
> >
> > THEOREM. (Cantor) There is no way to put the Infinite Integers into
> > one-to-one correspondence with the Natural Numbers.
> >
> > Proof: Diagonalization.
>

> You fail with your last comment Chris and should know better.

No, you simply haven't faced the following fact: If you look at the set
of natural numbers (as defined by Peano) and the set of Infinite
Integers (as defined by yourself), the two sets have different
cardinality. Therefore, they _cannot_ be the same set.

Just like: {1,3,4} has a different number of elements from {2,3,4,5},
so {1,3,4} and {2,3,4,5} are different sets: If you try to pair up the
elements of {1,3,4} with the elements of {2,3,4,5}, you'll run out of
elements in {1,3,4} before you run out of the ones in {2,3,4,5}.

Just like: The set of natural numbers is not equal to {4}, since the
set of natural numbers is infinite, and {4} is finite. Infinity =/= 1,
so the set of natural numbers is not the same as the set {4}.

> Are the Natural Numbers the finite-integers or the infinite-integers is
> the question? The consequence of finite-integers is a different
> cardinality and the consequence of infinite-integers is equal
> cardinality.

You can't say something is "a different cardinality". You have to say
it's "a different cardinality" FROM something else.

The Infinite Integers can be defined in the following way: Consider a
function "a" from N (the set of the natural numbers) to {0,1,2,3,..,9},
and create the Infinite Integer
...a(3)a(2)a(1)a(0). (Basically, fill in the digits one at a time.) For
every function from N to {0,1,2,3,...,9}, this will produce exactly one
Infinite Integer.

Now, if you look at the set of all functions above (and call that set
S), that set has the same cardinality as the set of Infinite Integers.
Cantor showed that N and S do not have the same size. (In fact, if you
limit a(0), a(1), ... to 0 or 1, the two sets still don't have the same
size.)

There are just TOO MANY INFINITE INTEGERS.

> But perhaps a good new path to explore is opened here in that since the
> 10-adics are of equal cardinality with the Reals, then can I make the
> 10-adics into a sequental ordering by some diagonal means and thus use
> that as the Array for Infinite Integers.

I don't know; can you wash dishes by using a screwdriver?

> Is Sequental Ordering contradictory to endless adding 1.

I've already answered this question, several times.

--- Christopher Heckman

Proginoskes

unread,
Dec 24, 2005, 1:23:41 AM12/24/05
to

a_plu...@hotmail.com wrote:
> I wrote a few minutes ago:
> Can I say the New Array yields the irrational Infinite Integers in that
> they converge at infinity?
>
> A.P. further writes:
> Let us take digitized Irrational Infinite Integer pi .....562951413.
>
> There is for example 413 and its pair ....413413413
>
> Later on there is 1413 and its pair ....141314131413
>
> Later on there is 51413 and its pair ....5141351413
>
> But those pairs are all rational Infinite Integers. Can I claim that
> irrational Infinite Integers enter the counting as convergences??

Yes, but they can't be "natural numbers". (This is what's known as "the
completion of the rational numbers by Cauchy sequences".)

> You know, this all reminds me of the Dedekind Cut of Rationals to
> obtain Reals. Is there a Dedekind Cut on the Natural Numbers to get
> irrational Infinite Integers from rational Infinite Integers?

Yes, but with a different definition of "rational Infinite Integers".

In the Peano Axioms, after + and * are defined (and some important
properties of + and * are proven), you can generate the "rational
numbers" by considering all ordered pairs (x,y) of natural numbers.

However, two ordered pairs (a,b) and (c,d) can represent the same
rational number, exactly in the same way as 1/2 = 2/4 = 3/6 = 4/8 = ...
. (In fact, (a,b) should be thought of as a/b.) So (back in the Peano
Axioms), (a,b) and (c,d) are the same rational number if a*d = b*c.

Now you can order the rational numbers. (This is what's needed before
using Dedekind Cuts.) It's not too difficult; order the natural numbers
in the "natural way":
0 < S(0) < S(S(0)) < S(S(S(0))) < ... , then say that
(a,b) < (c,d) if a*d < b*c.

Technically, you need to show that if (a,b) < (c,d), and (c,d) = (e,f),
then
(a,b) < (e,f); otherwise, you could conceivably have equations like
1/2 = 2/4 < 5/6 < 8/16 = 1/2. (This doesn't happen, of course, but it's
not obvious that it can't happen.) THEN come Dedekind cuts.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 24, 2005, 1:27:45 AM12/24/05
to
Chris Heckman wrote:
You cannot get all possible digit arrangements; there will always be at
least one missing. This is equivalent to a result Cantor proved. (See
below.) In this particular list. ...1112 is missing, something that you
seem to have ignored.

A.P. writes:
Okay, say you had an abbreviated Peano Axioms where you had just one
axiom of the Successor -- the endless adding of 1.

Now would that abbreviated Axioms yield all possible digit arrangements
of Infinite Integers?

Proginoskes

unread,
Dec 24, 2005, 1:43:38 AM12/24/05
to

a_plu...@hotmail.com wrote:
> Chris Heckman wrote:
> You cannot get all possible digit arrangements; there will always be at
> least one missing. This is equivalent to a result Cantor proved. (See
> below.) In this particular list. ...1112 is missing, something that you
> seem to have ignored.
>
> A.P. writes:
> Okay, say you had an abbreviated Peano Axioms where you had just one
> axiom of the Successor -- the endless adding of 1.

Then you can have all sorts of things which certainly are not the
natural numbers.

For instance, if you got rid of the axiom that said that S(x) is never
zero, you could have the following situation:

S(0) = 1
S(1) = 2
S(2) = 0

which is "modular 3 arithmetic".

If you remove essential elements of the description, you will get
something different.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 24, 2005, 3:30:27 AM12/24/05
to
These are the Peano Axioms as given by Wikipedia:

1 There is a natural number 0.
2 Every natural number a has a successor, denoted by S(a) or x'.
3 There is no natural number whose successor is 0.
4 Distinct natural numbers have distinct successors: a = b if and
only if S(a) = S(b).
5 If a property is possessed by 0 and also by the successor of
every natural number which possesses it, then it is possessed by all
natural numbers. (This axiom, also known as axiom of induction, ensures
that the proof technique of mathematical induction is valid.)

It is time to clean up these Axioms.

#1 is okay and it should be the first one as it creates 0.

However, #2 should create the number 1, because you cannot jump into a
successor axiom as #2 which is endless adding of 1 without there
existing the number 1. So the first two axioms of the Natural Numbers
should be this:

1. There is a natural number 0.
2. There is a natural number 1.

Now we can have the Successor Axiom as #3. The Successor Axiom does not
simultaneously create a natural number that is 1 and thence endlessly
add 1. So the old Peano Axioms is missing at least this creation of 1
axiom.

So the third axiom should be the Successor Axiom and it should be
stated as such

3. There is a successor series of endless adding of 1

Now, so far, these three axioms create all possible digit arrangements
of infinite leftward strings. Infinite rightward strings with a finite
leftward portion of all possible digit arrangements creates the Real
Numbers. So the best way to create the Reals is forget about the
evolution process of creating Naturals then Rationals and then Reals
out of Rationals. The more commonsense method is simply say infinite
rightward strings with finite portion leftward with all possible digit
arrangements is the Reals.

Now the above three axioms are the Adics for they are an endless adding
of 1 and they possess both 0 and 1 and they are all possible digit
arrangements leftward infinite strings. So the above three axioms are
also the Infinite Integers.

Now we come to a crucial question, if we can build the Adics or the
Infinite Integers with the endless adding of 1, then it makes
commonsense that by starting with 0 and 1 that all the other numbers
created are in a Sequential Ordering. I mean, given only 0 and 1 and
then endlessly adding 1 to build every Natural Number must be a
sequential building and thus have internal order to be Counted.

Now moving on to Peano's #3 and it is not okay because it is biased in
that it never considered that perhaps as we count out far enough that
Natural Numbers have a curvature and come back around. So Peano is not
justified in this axiom for it is in keeping with the previous axioms
that either the axiom or its denial conforms to the previous axioms,
and probably even the denial conforms to the Math Induction axiom.

4. The number 0 has no predecessor, or its denial, that 0 can have a
predecessor is of no conflict with the other three previous axioms.

Now it is Peano's #4 that I suspect is internally contradictory to the
Successor axiom. As I said above, if you invent a system where you
endless add 1 then there must be an order as you crank out each new
successor number, and that new natural numbers are not cranked out
simultaneously en masse, but cranked out individually and thus be
Sequentally Ordered and thus able to be Counted. That means the Adics
are countable and the Infinite Integers are countable. So #4 is at odds
with the Successor postulate.

Finally #5, and it appears to me as an oddball for it looks more like a
characteristic of the Natural Numbers for which it was unnecessary to
devote an axiom to it. That the Successor function or series should be
able to prove Induction as a theorem and unnecessary to make it a
full-fledged axiom itself.

I suspect in the history there was offered some set that seemed to obey
all the postulates except the Induction one and so Peano, because of
this example, decided he had to include Math Induction as a full
fledged axiom. In a sense, Math Induction is merely saying that given 0
and 1 and the Successor postulate is satisfied for all the numbers in
question that you have the Natural Numbers. So Math Induction is merely
a summation of three previous postulates into one more postulate that
is redundant.

I suspect that when Peano was formulating this axiom set that he faced
a example that he thought he needed to include Math Induction. But I
feel that math induction can be proven from the previous axioms and it
is redundant.

Now the important issue is that if you can create all the Adics or all
the Infinite Integers with simply one axiom-- Successor of endless
adding of 1, that this creation must have internal ordering that allows
you to Count, one by one each number and thus a Countable set.

Another thing that is very messy about the Peano Axioms. It creates all
of the Adics and the Infinite Integers in step #2 of endless adding of
1, but by step #4 where it tosses out all those numbers which lack a
unique predecessor or successor is a very messy situation. Because here
the Peano axioms create let us say the Adics or Infinite Integers and
then by step #4 is going to try to weed out or toss out all of those
adics or Infinite Integers that seem to disobey the uniqueness. A clean
wholesome Axiomatics would not be creating more numbers than wanted and
then inject a axiom that acts as a removal axiom.

a_plu...@hotmail.com

unread,
Dec 24, 2005, 12:00:33 PM12/24/05
to
It is best to do these major posts as brief as possible so the theme
sinks in.

The Peano axioms are flawed and I will correct them and clean them up.
But the major theme is embodied in one axiom-- the Successor axiom of
endless adding of 1. Let me call it a Endless Adding of 1 Machine.

If a axiom system has such a machine which the Natural Numbers do and
the Adics also have this very same machine, then the final product or
outcome of such a machine is an infinite set that is countable. The
Adics are countable, the Infinite Integers are countable and in fact
the two are the same set only one is base dependent and the other is
base independent.

The Reals are countable. I will show you and pinpoint where the Cantor
diagonal argument is a utter hoax and a fake. But here I must dwell on
this major theme.

Theme: If you construct numbers and you use Successor Axiom to build
those numbers then those numbers are Countable.

A machine that builds every number by the adding of 1 to the previous
number is a Countable set of numbers. The Adics are countable, and the
Infinite Integers are countable and the Reals are countable.

a_plu...@hotmail.com

unread,
Dec 24, 2005, 12:21:31 PM12/24/05
to
I need to keep each one of these as brief as possible so the main
message shines through.

The only reason people of the past believed there are different
cardinalities of infinite sets is because of the Cantor diagonal
argument which comes about from a flawed set of axioms of Peano.
Infinity is endlessness and there is only one kind of infinity. So let
me straighten out the diagonal argument.

Cantor claims that the Reals are of a different infinity from counting
numbers. He sets up any given number of Reals, say 2 Reals, say a 1,000
Reals and he draws a diagonal line down those list of Reals and he
changes one digit in each of those listed Reals (purpose of the
diagonal). Cantor thus claims he has a new Real not on any given list
and further claims that the Reals are thus uncountable.

The trouble here is that the Peano Axioms are flawed and not true to
what they are because they miss the negative counting numbers. The
Successor Axiom of Peano creates the negative numbers and they are
easily seen in the Adics as ....9999, ....99998 etc.

So, now, let us do the Cantor diagonal again with the Reals, and give
me _any list_ of Reals, and now draw your diagonal and craft your new
Real not on the original list, and now I assign it -1. Draw a second
list, craft your new Real and I assign it -2. Thus the Reals have the
same cardinality as the Counting Numbers.

But to see this even better, let me in the next post tell you what the
axioms of Reals, of Counting Numbers should really look like instead of
the Flawed Peano and Real constructions.

P.S. also in the above I did not mention that the Reals have the
problem in Cantor Diagonal of not able to eliminate all of those Reals
that are repeated such as 1.999.... is 2.0000.... and so you have
repeated Reals in Cantor's diagonal but no repeated counting numbers.

a_plu...@hotmail.com

unread,
Dec 24, 2005, 12:39:04 PM12/24/05
to

a_plu...@hotmail.com wrote:
> These are the Peano Axioms as given by Wikipedia:
> 1 There is a natural number 0.
> 2 Every natural number a has a successor, denoted by S(a) or x'.
> 3 There is no natural number whose successor is 0.
> 4 Distinct natural numbers have distinct successors: a = b if and
> only if S(a) = S(b).
> 5 If a property is possessed by 0 and also by the successor of
> every natural number which possesses it, then it is possessed by all
> natural numbers. (This axiom, also known as axiom of induction, ensures
> that the proof technique of mathematical induction is valid.)
>

These are what the Peano Axioms should be and they should only be these
four.


>
> 1. There is a natural number 0.
> 2. There is a natural number 1.

> 3. There is a successor series of endless adding of 1

4. A number is an infinite series and the Counting Numbers are
infinite leftward of all possible digit arrangements.

Why only these four? Because these four create the Adics and the
Infinite Integers. And they are symmetric with the axioms of the Reals
for which #4 is replaced with infinite string rightwards with finite
portion leftwards from a "decimal point".

Peano Axioms have too many flaws. He forgets to create "1". He has two
axioms that build his system of numbers and the other three are a
removal technique on his two axioms of construction. This is not
logical to devise a system where you construct on one side and tear
down and throw away on the other side.

According to Peano, negative counting numbers are simply defined, not
constructed, and this is a major flaw. According to Peano, the
Rationals and then Reals are also defined and not constructed which
increases the flaws.

You cannot define Reals from Counting Numbers when they are so vastly
different fundamentally since Counting Numbers are infinite leftward
strings and Reals are infinite rightward strings with a finite leftward
portion.

But the overall major flaw of Peano is to build a system of numbers for
which he has a construction process by two axioms and then the other
three axioms are removal and tossing out of constructions. Mathematics
of such importance as the Counting Numbers would not have removal and
deciding axioms but be purely construction axioms.

a_plu...@hotmail.com

unread,
Dec 24, 2005, 1:04:04 PM12/24/05
to
I am going to give a list of logical flaws of most every practicing
mathematician today. Flaws that they believe in and that are fakeries.
Flaws of the Peano axioms and the flaw of most of Cantor's work. I will
just list them below:

(1) The flaw that mathematicians can accept and believe that infinity
has various orders or levels of infinity, or cardinalities of infinity.
This is a utter flight from commonsense reason. Because infinity is
simply "endlessness" and you do not have species or variety of
endlessness. Endlessness comes in only one type. A nice proof of this
uniqueness of infinity is that zero concept is the inverse of infinity
and since zero is unique, infinity is unique. Cantor cardinalities is a
silly crackpot faddish phenomenon and most every mathematician has
fallen for it.

(2) Peano axiom system has a construction axiom which is the Successor
axiom of endless adding of 1. The commonsense argument here is that
this same axiom builds the Adics and Infinite Integers. So if it builds
the Finite Integers and also builds the Adics and Infinite Integers
then all three sets must be countable. So the process of endless adding
of 1 is a process that orders those numbers and makes them countable.
So Successor axiom is equal to the ability to Count the set produced.

(3) There is a flaw to a system that axiomatizes the Natural Numbers
and goes from that axiom system to defining everything else up to the
Reals, when the Reals are so vastly different from the starting point
of Natural Numbers. So the system is flawed when it relies on defining
structure. And when a much simplier process of creating the Reals and
Natural Numbers exists that uses few if any definitions.

Creating the Natural Numbers with Successor axiom and infinite leftward
strings with all possible digit arrangement and then creating Reals as
infinite rightward strings with all possible digit arrangements is not
only an economy of brevity but is logically superior. This method has
the intrinsic symmetry between Natural Numbers and Reals whereas the
old method of Peano is longer and cumbersome and one can never know if
definitions are badly formed or well formed.

(4) The Peano Axiom method has two construction axioms and the other
three axioms are decision making axioms as to whether to remove numbers
already created by the successor axiom. This is a untidy and flawed way
of creating Natural Numbers. The better method is to have only
constructive axioms, not destructive axioms in the mix.

The Ghost In The Machine

unread,
Dec 24, 2005, 4:00:11 PM12/24/05
to
In sci.logic, a_plu...@hotmail.com
<a_plu...@hotmail.com>
wrote
on 24 Dec 2005 00:30:27 -0800
<1135413027....@g49g2000cwa.googlegroups.com>:

> These are the Peano Axioms as given by Wikipedia:
> 1 There is a natural number 0.
> 2 Every natural number a has a successor, denoted by S(a) or x'.
> 3 There is no natural number whose successor is 0.
> 4 Distinct natural numbers have distinct successors: a = b if and
> only if S(a) = S(b).
> 5 If a property is possessed by 0 and also by the successor of
> every natural number which possesses it, then it is possessed by all
> natural numbers. (This axiom, also known as axiom of induction, ensures
> that the proof technique of mathematical induction is valid.)
>
> It is time to clean up these Axioms.
>
> #1 is okay and it should be the first one as it creates 0.
>
> However, #2 should create the number 1, because you cannot jump into a
> successor axiom as #2 which is endless adding of 1 without there
> existing the number 1. So the first two axioms of the Natural Numbers
> should be this:
>
> 1. There is a natural number 0.
> 2. There is a natural number 1.

There is no requirement for this, as 1 = successor(0). "Adding 1"
is an interpretation done later on in the system; in fact, one can
define addition as follows:

a. 0+0 = 0.
b. a+b' = (a+b)'.
c. a'+b = (a+b)'.

therefore one can prove that a+1 = a+0' = (a+0)' = a', and
1+a = 0'+a = (0+a)' = a'.

Of course one might define the two numbers in a single axiom:

1. There are natural numbers 0 and 1, with 1 = 0'

though that might be construed as cheating. :-) But from the looks
of it thus far you're attacking a non-problem.

Note also that many formulations define 1 as the first natural
number, and define addition slightly differently, starting
out with 1+1 = 2 (1'). In such a formulation the standard additive
identity comes later, when subtraction requires an answer to the
question 'a - a = ?'.

Such actually happens in various finite groups. For example, the
finite additive group with 5 elements has the following tables:

+ 0 1 2 3 4 * 0 1 2 3 4
0 0 1 2 3 4 0 0 0 0 0 0
1 1 2 3 4 0 1 0 1 2 3 4
2 2 3 4 0 1 2 0 2 4 1 3
3 3 4 0 1 2 3 0 3 1 4 2
4 4 0 1 2 3 4 0 4 3 2 1

This is a true field, as far as I know; questions such as 1/4 make
perfect sense (the answer, of course, is 1/4 = 4). Note that 1/5
is division by 0 in this field, since 0 = 5 (mod 5).

Another possibility is a group defined by rotations, which has
period 2pi. (An isomorphism exists between this group with
addition, and the unit circle on C with multiplication.)


> So Peano is not
> justified in this axiom for it is in keeping with the previous axioms
> that either the axiom or its denial conforms to the previous axioms,
> and probably even the denial conforms to the Math Induction axiom.
>
> 4. The number 0 has no predecessor, or its denial, that 0 can have a
> predecessor is of no conflict with the other three previous axioms.

Note that 4' = 0 (mod 5), contradicting #3.

>
> Now it is Peano's #4 that I suspect is internally contradictory to the
> Successor axiom.
> As I said above, if you invent a system where you
> endless add 1 then there must be an order as you crank out each new
> successor number, and that new natural numbers are not cranked out
> simultaneously en masse, but cranked out individually and thus be
> Sequentally Ordered and thus able to be Counted. That means the Adics
> are countable and the Infinite Integers are countable. So #4 is at odds
> with the Successor postulate.

Your logic looks rather suspect. If a < b, then 0 < S(a) < S(b);
if S(a) < (b), then a < b.

>
> Finally #5, and it appears to me as an oddball for it looks more like a
> characteristic of the Natural Numbers for which it was unnecessary to
> devote an axiom to it. That the Successor function or series should be
> able to prove Induction as a theorem and unnecessary to make it a
> full-fledged axiom itself.

#5 is a funny axiom in any event. It's almost a meta-axiom, as,
rather than defining an axiom about a number, it defines an axiom
about *predicates* involving a number -- or, in some formulations,
an axiom about sets of numbers.

The rest are easily expressible in an RS(1) logic system:

1. 0 in N.
2. (a)(a in N => a' in N).
3. (a)(a in N => (a' != 0) ).
4. (a)(b)(a = b <=> a' = b' ).

Now 5 requires RS(2), if my notation is correct:

5. (S)((S in P(N) . 0 in S . (a)(a in S => a' in S)) => S = N)

or

5'. (P)((P in P(N) . P(0) . (a)(P(a) => P(a')) ) => P = T(N))

where T(N) is the true predicate for all a in N, and P(N) is
either the power set of N or the set of all single-argument
predicates whose argument is in N.

or

5". (P)((P in P(N x {t,f}) . (a)(~((a,t) in P, (a,f) in P))
. (0,t) in P . (a)((a,t) in P => (a',t) in P) ) => P = N x {t})

where the second clause is a formal requirement that P(a) makes sense.

>
> I suspect in the history there was offered some set that seemed to obey
> all the postulates except the Induction one and so Peano, because of
> this example, decided he had to include Math Induction as a full
> fledged axiom. In a sense, Math Induction is merely saying that given 0
> and 1 and the Successor postulate is satisfied for all the numbers in
> question that you have the Natural Numbers. So Math Induction is merely
> a summation of three previous postulates into one more postulate that
> is redundant.

Highly debatable. In particular, all of these are true,
if one defines a' = (a+1) with + being standard real addition:

0 in R+0 (R+0 = the set of all nonnegative reals)
a in R+0 => a' in R+0.
a = b <=> a' = b' in R+0.
No a in R+0 is such that a' = 0.

but one cannot conclude that

(S)((S in P(R+0) . 0 in S . (a)(a in S => a' in S)) => S = R+0)

because R+0 also contains elements such as 0.5, pi, and e.

Either one has to insert an additional axiom along the lines of

(a)(a in N . a != 0 => (Eb)(b in N . b' = a))

("for every non-zero natural number there exists another natural number
which is a predecessor")

or take the 5th axiom in its original form.

Another possibility is the set C1, where C1 is the continuum
half-open interval [0,2*pi) (or, if one prefers, the
unit circle) and the operation of succession is

a' = a+1 if a+1 in C1
a+1-2*pi if a+1 is not in C1

This meets axioms 1-4 but C1 also contains elements such as 0.5,
pi, and e, which cannot be generated.

>
> I suspect that when Peano was formulating this axiom set that he faced
> a example that he thought he needed to include Math Induction. But I
> feel that math induction can be proven from the previous axioms and it
> is redundant.

So prove it then, and show that R+0 and C1 do *not* meet these
requirements. The fact that R+0 and C1 appear to meet these
requirements (sans axiom 5) makes your logic highly suspect,
as they are counterexamples to a working 4-axiom system.

>
> Now the important issue is that if you can create all the Adics or all
> the Infinite Integers with simply one axiom-- Successor of endless
> adding of 1, that this creation must have internal ordering that allows
> you to Count, one by one each number and thus a Countable set.
>
> Another thing that is very messy about the Peano Axioms. It creates all
> of the Adics and the Infinite Integers in step #2 of endless adding of
> 1, but by step #4 where it tosses out all those numbers which lack a
> unique predecessor or successor is a very messy situation. Because here
> the Peano axioms create let us say the Adics or Infinite Integers and
> then by step #4 is going to try to weed out or toss out all of those
> adics or Infinite Integers that seem to disobey the uniqueness. A clean
> wholesome Axiomatics would not be creating more numbers than wanted and
> then inject a axiom that acts as a removal axiom.
>
> Archimedes Plutonium
> www.iw.net/~a_plutonium
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies
>


--
#191, ewi...@earthlink.net
It's still legal to go .sigless.

ken.q...@excite.com

unread,
Dec 24, 2005, 8:57:36 PM12/24/05
to
a_plu...@hotmail.com wrote:
> I need to keep each one of these as brief as possible so the main
> message shines through.
>
> The only reason people of the past believed there are different
> cardinalities of infinite sets is because of the Cantor diagonal
> argument which comes about from a flawed set of axioms of Peano.
> Infinity is endlessness and there is only one kind of infinity. So let
> me straighten out the diagonal argument.
>
> Cantor claims that the Reals are of a different infinity from counting
> numbers. He sets up any given number of Reals, say 2 Reals, say a 1,000
> Reals and he draws a diagonal line down those list of Reals and he
> changes one digit in each of those listed Reals (purpose of the
> diagonal). Cantor thus claims he has a new Real not on any given list
> and further claims that the Reals are thus uncountable.
>
> The trouble here is that the Peano Axioms are flawed and not true to
> what they are because they miss the negative counting numbers. The
> Successor Axiom of Peano creates the negative numbers and they are
> easily seen in the Adics as ....9999, ....99998 etc.
>
> So, now, let us do the Cantor diagonal again with the Reals, and give
> me _any list_ of Reals, and now draw your diagonal and craft your new
> Real not on the original list, and now I assign it -1. Draw a second
> list, craft your new Real and I assign it -2. Thus the Reals have the
> same cardinality as the Counting Numbers.
>

There's a subtle point here that Cantor's argument is making that I
think you're missing. Yes, it's true that if you apply Cantor's
'construction' to get a new real, and do that countably many times,
then of course, if the reals were countable to start with, they'd still
be countable after countably many additions to the list.

The point however is that Cantor's argument is not I believe intended
to construct the reals, one -at-a-time (which could only result, as I
said above, in countably many). It is rather to show that the
assumption that the reals are countable leads to a contradiction,
and hence must be false. Take our favorite reals in [0,1) and
list them - put them in a 1-1 correspondence with the natural
numbers. Examine their decimal expansions. You stated the
argument perfectly well. We can find a decimal expansion that
ISN'T in the list, but, as a decimal expansion, must be a real.
Contradiction. Therefore the reals aren't countable.

A separate point - you say somewhere in your chain of notes that
the infinite adics are countable. My understanding is that they
are uncountable, and this point has been made before.


Thanks.

Ken

a_plu...@hotmail.com

unread,
Dec 24, 2005, 10:52:51 PM12/24/05
to

No, there is a requirement for an axiom to state the existence of 1,
setting the stage for the Successor axiom which uses the 1 as a endless
adding.

To give you a metaphor, a caliper cannot be a calipre if it has only
one arm-- 0 for it must have the other arm to be of any use-- the 1.
One cannot start a Successor function with only 0 in existence for one
needs a 1 to set the gauge of the adding to 0.
Likewise another metaphor is a scissors with only one blade and the
other blade missing and so it is no longer a scissors. So one cannot
start the succession process until two numbers are specified of 0 and 1
and thence the Successor Axiom can start.


>
> a. 0+0 = 0.
> b. a+b' = (a+b)'.
> c. a'+b = (a+b)'.
>
> therefore one can prove that a+1 = a+0' = (a+0)' = a', and
> 1+a = 0'+a = (0+a)' = a'.
>
> Of course one might define the two numbers in a single axiom:
>

It is probably a good idea to define the two numbers into one single
postulate because they are in a sense paired--- nothing and something.


> 1. There are natural numbers 0 and 1, with 1 = 0'
>
> though that might be construed as cheating. :-) But from the looks
> of it thus far you're attacking a non-problem.

No, I am revealing the gaps of logic. Without stating the existence of
1, we cannot make a Successor Axiom which uses 1. We know 0 exists but
we have no clue that 1 exists and yet it is assumed in the Successor
axiom. So that is a flaw of the Peano axioms and must be fixed.

Well I am wondering about the set of negative integers for they would
obey all of the Peano axioms except for 0 has no predecessor. So it
seems like a artifical contrivance imposed by human mind rather than
something intrinsic and innate in the structure of mathematics.

>
> > So Peano is not
> > justified in this axiom for it is in keeping with the previous axioms
> > that either the axiom or its denial conforms to the previous axioms,
> > and probably even the denial conforms to the Math Induction axiom.
> >
> > 4. The number 0 has no predecessor, or its denial, that 0 can have a
> > predecessor is of no conflict with the other three previous axioms.
>
> Note that 4' = 0 (mod 5), contradicting #3.
>

The negative integers are a more valuable example.

> >
> > Now it is Peano's #4 that I suspect is internally contradictory to the
> > Successor axiom.
> > As I said above, if you invent a system where you
> > endless add 1 then there must be an order as you crank out each new
> > successor number, and that new natural numbers are not cranked out
> > simultaneously en masse, but cranked out individually and thus be
> > Sequentally Ordered and thus able to be Counted. That means the Adics
> > are countable and the Infinite Integers are countable. So #4 is at odds
> > with the Successor postulate.
>
> Your logic looks rather suspect. If a < b, then 0 < S(a) < S(b);
> if S(a) < (b), then a < b.
>

No, the logic is simple and straightforward. The Adics are built from
the same Successor axiom as the Naturals of endless adding of 1, so the
logic then goes that if endless adding of 1 builds the Natural Numbers
and also build the Adics, call it the Endless adding of 1 Machine. So
if this same machine builds Adics and builds the Counting Numbers then
the Adics must be countable. Even a layman can understand that if you
add 1 to a number to get a new number and to get all the numbers of
that set then:

endless adding of 1 Machine is equal to the ability to order and count
those numbers.

Since Adics are 1-1 correspondence with Reals then Reals are also
countable.

> >
> > Finally #5, and it appears to me as an oddball for it looks more like a
> > characteristic of the Natural Numbers for which it was unnecessary to
> > devote an axiom to it. That the Successor function or series should be
> > able to prove Induction as a theorem and unnecessary to make it a
> > full-fledged axiom itself.
>
> #5 is a funny axiom in any event. It's almost a meta-axiom, as,
> rather than defining an axiom about a number, it defines an axiom
> about *predicates* involving a number -- or, in some formulations,
> an axiom about sets of numbers.

I really believe it is a summary of the axioms of successor and
existence of 0 (and the missing existence axiom of 1). It should not be
an axiom but proven as a theorem.


>
> The rest are easily expressible in an RS(1) logic system:
>
> 1. 0 in N.
> 2. (a)(a in N => a' in N).
> 3. (a)(a in N => (a' != 0) ).
> 4. (a)(b)(a = b <=> a' = b' ).
>
> Now 5 requires RS(2), if my notation is correct:
>
> 5. (S)((S in P(N) . 0 in S . (a)(a in S => a' in S)) => S = N)
>
> or
>
> 5'. (P)((P in P(N) . P(0) . (a)(P(a) => P(a')) ) => P = T(N))
>
> where T(N) is the true predicate for all a in N, and P(N) is
> either the power set of N or the set of all single-argument
> predicates whose argument is in N.
>
> or
>
> 5". (P)((P in P(N x {t,f}) . (a)(~((a,t) in P, (a,f) in P))
> . (0,t) in P . (a)((a,t) in P => (a',t) in P) ) => P = N x {t})
>
> where the second clause is a formal requirement that P(a) makes sense.
>

It would be interesting to revisit the history of Peano when he was
writing these things and to see what prompted him to think that Math
Induction was needed.

I do not see your operation as succession; but as a correspondence
mapping and not a Succession function. It is easy to fool oneself in
thinking that a dreamed up function is a Succession function. One must
keep in mind that a Succession function on par with endless adding of 1
goes to infinity and so one can be fooled into thinking that some
"cycling type function" such as a trig function is a succession when it
is not. If it were a true Succession, you would be able to give the
first three numbers and the reader would then be able to generate every
number thereafter without ever recourse to your prescription. But since
you prescription is vital for obtaining every new number, then it is
not a Succession.

As for a proof that Math Induction is redundant is quite simple. Given
the existence of 0 and 1 we have the initial step. Given the Successor
function we have it true for 0,1,2. Given any X, a generalized X and if
we can show for X+1, is a recursion of the Successor Axiom. In other
words, Math Induction is merely a Summarization of the axioms of 0
exists and 1 exists and the Successor Function. So Math Induction is
not a axiom for it can be proven by the other postulates.

a_plu...@hotmail.com

unread,
Dec 24, 2005, 11:16:35 PM12/24/05
to
Ken Quirici writes:

The point however is that Cantor's argument is not I believe intended
to construct the reals, one -at-a-time (which could only result, as I
said above, in countably many). It is rather to show that the
assumption that the reals are countable leads to a contradiction,
and hence must be false.

A separate point - you say somewhere in your chain of notes that


the infinite adics are countable. My understanding is that they
are uncountable, and this point has been made before.

A.P. writes:
Finite Integers is only a partial deck of cards. Say the Reals were 52
cards. Now when you have a partial deck of say only the numeral cards
then of course when you try to pair up the Reals there will be extras
in the Reals. The Infinite Integers and the Adics makes a full deck of
Natural Numbers and they are 52, just as many as the Reals.

The commonsense argument is that to construct the Natural Numbers, Ken,
you use what I call the Endless adding of 1 Machine. As you add one you
also count that number. The Adics are built from only this Machine,
hence the Adics are Countable and since the Adics are equinumerous with
the Reals, then hence again, the Reals are Countable.

The only debate here is whether this contention is true of false. The
Endless adding of 1 Machine is another way of saying it is a Countable
Machine. For the Natural Numbers have only one creator function and it
is this Endless adding machine.

The Adics have only this one creator function. So it all boils down to
whether Endless adding of 1 Machine is the same as a Counting Machine.

How I empathize with Copernicus about now, rather than teaching others
that Earth is round. How I empathize with Galileo about now, who
actually did try to teach others that Earth is moving in space at
enormous speed, but why bother.

Proginoskes

unread,
Dec 25, 2005, 12:27:24 AM12/25/05
to

It's good that you've come across the idea of Sequential Ordering and
Countable sets. Because now Cantor's diagonalization theorem proves
that the Infinite Integers aren't countable, so they can't be the
natural numbers.

Also, I give an explicit example of a set of numbers satisfying #1-#4
but not #5.

a_plu...@hotmail.com wrote:
> These are the Peano Axioms as given by Wikipedia:
> 1 There is a natural number 0.
> 2 Every natural number a has a successor, denoted by S(a) or x'.
> 3 There is no natural number whose successor is 0.
> 4 Distinct natural numbers have distinct successors: a = b if and
> only if S(a) = S(b).
> 5 If a property is possessed by 0 and also by the successor of
> every natural number which possesses it, then it is possessed by all
> natural numbers. (This axiom, also known as axiom of induction, ensures
> that the proof technique of mathematical induction is valid.)
>
> It is time to clean up these Axioms.
>

> [A.P. accepts #1, #2, and #3 as-is. His terminology of "endlessly adding
> 1" is equivalent to the statement that the successor function is onto.]


>
> Now we come to a crucial question, if we can build the Adics or the
> Infinite Integers with the endless adding of 1, then it makes
> commonsense that by starting with 0 and 1 that all the other numbers
> created are in a Sequential Ordering. I mean, given only 0 and 1 and
> then endlessly adding 1 to build every Natural Number must be a
> sequential building and thus have internal order to be Counted.

... But the Infinite Integers cannot be sequentially ordered. Cantor's
diagonalization proof again.

> 4. The number 0 has no predecessor, or its denial, that 0 can have a
> predecessor is of no conflict with the other three previous axioms.
>
> Now it is Peano's #4 that I suspect is internally contradictory to the
> Successor axiom.

It isn't contradictory in the mathematical sense; the mathematical
sense of "contradictory" means "causes a contradiction with". So if you
were using normal mathematical terminology, you'd be saying that a
logical contradiction can be reached by applying these axioms.

I suspect that in AP-speak, "contradictory" means you don't like it.

The problem is that without it, you can't talk about THE predecessor of
a number; you can only talk about A predecessor.

> As I said above, if you invent a system where you
> endless add 1 then there must be an order as you crank out each new
> successor number, and that new natural numbers are not cranked out
> simultaneously en masse, but cranked out individually and thus be
> Sequentally Ordered and thus able to be Counted. That means the Adics
> are countable and the Infinite Integers are countable. So #4 is at odds
> with the Successor postulate.

No, it isn't. The thing you fail to see is that the Infinite Integers
can't be sequentially ordered. Did you even LOOK at Cantor's argument?

> Finally #5, and it appears to me as an oddball for it looks more like a
> characteristic of the Natural Numbers for which it was unnecessary to
> devote an axiom to it.

#5 is fundamentally saying that every number is the sucessor of some
other number. If you eliminate it, then the Infinite Integers are
allowed. (You can use that weird ordering you came up with.)

It's also used to prove results like m + n = n + m, for all natural
numbers m and n. This means nothing to you, of course.

> That the Successor function or series should be
> able to prove Induction as a theorem and unnecessary to make it a
> full-fledged axiom itself.

Well, the Parallel Postulate looks like it should follow from Euclid's
other axioms, but it doesn't. And you get other geometries (Lobachevsky
and Riemann) by altering it.

> I suspect in the history there was offered some set that seemed to obey

> all the postulates except the Induction one [...]

In fact, it's not too difficult to create one: Think of the nonnegative
rational (or real) numbers, and let the successor of R be R+1 (normal
addition). Then axioms #1-#4 are satisfied, but #5 isn't. And there are
statements about the natural numbers which aren't true for the rational
(or real) numbers.

> Now the important issue is that if you can create all the Adics or all
> the Infinite Integers with simply one axiom-- Successor of endless
> adding of 1, that this creation must have internal ordering that allows
> you to Count, one by one each number and thus a Countable set.

And since the Infinite Integers, as a set, are uncountable, the
Infinite Integers can't be the natural numbers.

> Another thing that is very messy about the Peano Axioms. It creates all
> of the Adics and the Infinite Integers in step #2 of endless adding of
> 1, but by step #4 where it tosses out all those numbers which lack a
> unique predecessor or successor is a very messy situation. Because here
> the Peano axioms create let us say the Adics or Infinite Integers and
> then by step #4 is going to try to weed out or toss out all of those
> adics or Infinite Integers that seem to disobey the uniqueness. A clean
> wholesome Axiomatics would not be creating more numbers than wanted and
> then inject a axiom that acts as a removal axiom.

This will happen whenever you have more than one condition: There will
be more things that satisfy condition #1 than satisfy condition #1 and
condition #2. (This is basic set theory; A is at least as large as A
intersected B.) So to do what you want, you would have to combine
things into one condition, which doesn't look feasible. And it would be
confusing at that.

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 12:29:54 AM12/25/05
to

a_plu...@hotmail.com wrote:
> It is best to do these major posts as brief as possible so the theme
> sinks in.
>
> The Peano axioms are flawed

In your opinion; you have no proof of this.

> and I will correct them and clean them up.
> But the major theme is embodied in one axiom-- the Successor axiom of
> endless adding of 1. Let me call it a Endless Adding of 1 Machine.
>
> If a axiom system has such a machine which the Natural Numbers do and
> the Adics also have this very same machine, then the final product or
> outcome of such a machine is an infinite set that is countable. The
> Adics are countable,

No, they're not. Cantor.

> the Infinite Integers are countable

No, they're not. Cantor.

> and in fact
> the two are the same set only one is base dependent and the other is
> base independent.
>
> The Reals are countable.

No, they're not.

> I will show you and pinpoint where the Cantor
> diagonal argument is a utter hoax and a fake.

I doubt it. No one else has been able to do it.

> But here I must dwell on this major theme.
>
> Theme: If you construct numbers and you use Successor Axiom to build
> those numbers then those numbers are Countable.

True.

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 12:44:26 AM12/25/05
to

You've mistaken how Cantor diagonalization works. And since ...9990 is
a different infinite integer from ...0001, you don't have the problem
that 0.999.. = 1.000... that you mentioned in your P.S. And you don't
know how negative numbers are generated. (You should pick up a book on
the Peano Axioms to see how they're used to generate all the complex
numbers. There's a lot which I haven't explained in this thread.)

a_plu...@hotmail.com wrote:
> I need to keep each one of these as brief as possible so the main
> message shines through.
>
> The only reason people of the past believed there are different
> cardinalities of infinite sets is because of the Cantor diagonal
> argument which comes about from a flawed set of axioms of Peano.

(1) The Peano Axioms have not been proved to be flawed; this was shown
by Gerhard Gentzen in 1936; (2) Cantor's argument only depends on set
theory, not the Peano Axioms.

> Infinity is endlessness and there is only one kind of infinity. So let
> me straighten out the diagonal argument.
>
> Cantor claims that the Reals are of a different infinity from counting
> numbers. He sets up any given number of Reals, say 2 Reals, say a 1,000
> Reals and he draws a diagonal line down those list of Reals and he
> changes one digit in each of those listed Reals (purpose of the
> diagonal).

That's not how Cantor's argument works; you've fallen into one of the
more common traps. Cantor says: You claim that the real numbers are
countable. So I say: give me a list which you claim has all of the real
numbers on it. Then I will show you a real number which is not on your
list. So your original claim, that the real numbers are countable, is
false.

You get once chance here. You can also talk with "God" before preparing
your list, to make sure that all of the real numbers are indeed on that
list.

> The trouble here is that the Peano Axioms are flawed and not true to
> what they are because they miss the negative counting numbers.

Throwing in the negative numbers doubles the size of a set. But if you
double the size of a set, it's still the same cardinality. This is
because you can match up the nonnegative integers with the integers in
a 1-1 fashion:

0 with 0
1 with -1
2 with 1
3 with -2
4 with 2
5 with -3
etc.

> The Successor Axiom of Peano creates the negative numbers

There are no negative numbers in the set of natural numbers; otherwise
the condition that S(x) be non-zero is contradicted. The negative
numbers are created differently:

Given S (the set of natural numbers or "nonnegative" rational numbers,
or "nonnegative" real numbers), consider all ordered pairs (x,y), and
say that (x,y) and (z,w) are equivalent if
x + w = y + z. (For instance, (1,0) = (2,1) = (3,2) = ...) Now pick a
set of ordered pairs, never choosing two which are equivalent. Now you
have negative numbers.

> P.S. also in the above I did not mention that the Reals have the
> problem in Cantor Diagonal of not able to eliminate all of those Reals
> that are repeated such as 1.999.... is 2.0000.... and so you have
> repeated Reals in Cantor's diagonal but no repeated counting numbers.

This isn't a problem, because when a new digit is selected for the nth
position, you can always avoid 9 and 0 as well.

This isn't a problem with the Infinite Integers, though, since ...99990
is different from ...00001.

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 12:52:38 AM12/25/05
to

Your axiom system generates sets of natural numbers and throws out some
sets as well. For instance {0, 1, 5}. And there's more to the Peano
Axioms than I've posted in this thread, so you shouldn't be saying
"Peano Axioms don't define X" without checking it first. Get a book.

a_plu...@hotmail.com wrote:
> a_plu...@hotmail.com wrote:
> > These are the Peano Axioms as given by Wikipedia:
> > 1 There is a natural number 0.
> > 2 Every natural number a has a successor, denoted by S(a) or x'.
> > 3 There is no natural number whose successor is 0.
> > 4 Distinct natural numbers have distinct successors: a = b if and
> > only if S(a) = S(b).
> > 5 If a property is possessed by 0 and also by the successor of
> > every natural number which possesses it, then it is possessed by all
> > natural numbers. (This axiom, also known as axiom of induction, ensures
> > that the proof technique of mathematical induction is valid.)
> >
>
> These are what the Peano Axioms should be and they should only be these
> four.
> >
> > 1. There is a natural number 0.
> > 2. There is a natural number 1.

This is unnecessary, given the successor function.

> > 3. There is a successor series of endless adding of 1

But how do you guarantee that 1 + 1 + 1 + 1 is not 0? And it also
allows the set
{0, 1/3, 2/3, 3/3, 4/3, ...}, where "1" is 1/3.

> 4. A number is an infinite series and the Counting Numbers are
> infinite leftward of all possible digit arrangements.

#4 contradicts itself. Cantor.

> Why only these four? Because these four create the Adics and the
> Infinite Integers.

They also create the integers modulo 2, modulo 3, etc. And you're not
even saying that 0 and 1 are distinct, so you're also allowing the
"natural numbers" to be {0}.

Nope. Try again.

> Peano Axioms have too many flaws. He forgets to create "1".

No; he has a name for "1": S(0). He has a name for 2: S(S(0)). And so
on.

> He has two
> axioms that build his system of numbers and the other three are a
> removal technique on his two axioms of construction. This is not
> logical to devise a system where you construct on one side and tear
> down and throw away on the other side.

Well, _your_ system does the same thing. For instance, the set {0,1,5}
passes the first two conditions but not the third.

> According to Peano, negative counting numbers are simply defined, not
> constructed, and this is a major flaw.

No; there is a construction for them; you just don't know it. (I can't
tell you _everything_ about the Peano Axioms in this thread. Go buy a
book. It's not as hard a subject as the p-adics.)

> According to Peano, the
> Rationals and then Reals are also defined and not constructed which
> increases the flaws.

Ditto. (In fact, I showed how the rational numbers are defined in Peano
Arithmetic. You just didn't read it.)

> But the overall major flaw of Peano is to build a system of numbers for
> which he has a construction process by two axioms and then the other
> three axioms are removal and tossing out of constructions. Mathematics
> of such importance as the Counting Numbers would not have removal and
> deciding axioms but be purely construction axioms.

Your axiom system fails for the same reason, but you think it's okay.

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 1:03:22 AM12/25/05
to
> [...]

This is not the approach to take with A.P. Whenever he sees an
equation, he skips over the rest of the post without reading it. Ditto
if you have a rigorous proof.

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 1:07:18 AM12/25/05
to

The Ghost In The Machine wrote:
> In sci.logic, a_plu...@hotmail.com
> <a_plu...@hotmail.com>
> wrote
> on 24 Dec 2005 00:30:27 -0800
> <1135413027....@g49g2000cwa.googlegroups.com>:
> > These are the Peano Axioms as given by Wikipedia:
> > 1 There is a natural number 0.
> > 2 Every natural number a has a successor, denoted by S(a) or x'.
> > 3 There is no natural number whose successor is 0.
> > 4 Distinct natural numbers have distinct successors: a = b if and
> > only if S(a) = S(b).
> > 5 If a property is possessed by 0 and also by the successor of
> > every natural number which possesses it, then it is possessed by all
> > natural numbers. (This axiom, also known as axiom of induction, ensures
> > that the proof technique of mathematical induction is valid.)
> > [...]

> > Finally #5, and it appears to me as an oddball for it looks more like a
> > characteristic of the Natural Numbers for which it was unnecessary to
> > devote an axiom to it. That the Successor function or series should be
> > able to prove Induction as a theorem and unnecessary to make it a
> > full-fledged axiom itself.
>
> #5 is a funny axiom in any event. It's almost a meta-axiom, as,
> rather than defining an axiom about a number, it defines an axiom
> about *predicates* involving a number -- or, in some formulations,
> an axiom about sets of numbers. [...]

The most important part of #5 is that the successor function is "almost
onto"; that any natural number except 0 is the successor of some
natural number. (Pause) Although I guess the more general wording also
keeps you from having the following situation:

S(0) = 1, S(1) = 2, S(2) = 3, ...
S(1/2) = 3/2, S(3/2) = 1/2,

and letting the "set of natural numbers" be {0,1,2,3,...} union
{1/2,3/2}.

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 1:09:22 AM12/25/05
to

ken.q...@excite.com wrote:
> [...]

> A separate point - you say somewhere in your chain of notes that
> the infinite adics are countable. My understanding is that they
> are uncountable, and this point has been made before.

Yes; Cantor's diagonalization argument works, and even better than it
does for real numbers, since ...9990 is different from ...0001.

Now it looks like AP has turned into a "Cantor crank".

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 1:30:26 AM12/25/05
to
> adding. [...]

This is necessary because you've gotten rid of the successor function.
But how do you keep 1 from equalling 0?

> > a. 0+0 = 0.
> > b. a+b' = (a+b)'.
> > c. a'+b = (a+b)'.
> >
> > therefore one can prove that a+1 = a+0' = (a+0)' = a', and
> > 1+a = 0'+a = (0+a)' = a'.
> >
> > Of course one might define the two numbers in a single axiom:
> >
>
> It is probably a good idea to define the two numbers into one single
> postulate because they are in a sense paired--- nothing and something.

"Nothing" and "something" are antonyms; they're complete opposites. 0
and 1 don't quite have the same relationship, because of the existence
of 2.

> > 1. There are natural numbers 0 and 1, with 1 = 0'
> >
> > though that might be construed as cheating. :-) But from the looks
> > of it thus far you're attacking a non-problem.
>
> No, I am revealing the gaps of logic. Without stating the existence of
> 1, we cannot make a Successor Axiom which uses 1. We know 0 exists but
> we have no clue that 1 exists and yet it is assumed in the Successor

> axiom. So that is a flaw of the Peano axioms and must be fixed. [...]

1 is only a problem with your system of axioms; it's not in Peano's,
where there's a successor function with certain properties, and he can
simply define 1 to be S(0).

> Well I am wondering about the set of negative integers for they would
> obey all of the Peano axioms except for 0 has no predecessor. So it
> seems like a artifical contrivance imposed by human mind rather than
> something intrinsic and innate in the structure of mathematics.

This is an example of why you have to work with the axioms before you
pass judgment on them. Peano Arithmetic (the consequences of Peano's
Axioms) include negative numbers, as well as rational numbers (and I
showed you how to create them elsewhere in this thread), real numbers,
complex numbers. You simply don't know it, so you're arguing that no
one else knows how to do it, either.

> No, the logic is simple and straightforward. The Adics are built from
> the same Successor axiom as the Naturals of endless adding of 1, so the
> logic then goes that if endless adding of 1 builds the Natural Numbers
> and also build the Adics, call it the Endless adding of 1 Machine. So
> if this same machine builds Adics and builds the Counting Numbers then
> the Adics must be countable. Even a layman can understand

The problem here, of course, is that layman's intuition turns out to be
WRONG when dealing with infinite sets. A very good example of this is:
When someone proved to Cantor that the set of points on a line and the
set of points have the same cardinality, he said, "I see it but I don't
believe it." That is, he saw the proof was valid, but the result wasn't
intuitive. And if this can happen to Cantor, it can happen to a layman.

This is why mathematics relies on precise arguments, as opposed to
intuition.

> that if you
> add 1 to a number to get a new number and to get all the numbers of
> that set then:
>
> endless adding of 1 Machine is equal to the ability to order and count
> those numbers.

What if addition of 1 turns out to have the following properties?

0 + 1 = 1
1 + 1 = 2
2 + 1 = 3 (etc.)
X + 1 = Y
Y + 1 = Z
Z + 1 = X,

where your set of numbers is the symbols {X,Y,Z,0,1,2,3,...} ? Yes, all
numbers (except 0) can be obtained by adding 1 to a number, but if you
start looking at

0
0 + 1
0 + 1 + 1
0 + 1 + 1 + 1
0 + 1 + 1 + 1 + 1,
etc.,

you will NOT get X at any point, or Y, or Z. (And don't say that this
won't happen, because it is allowed by your axioms.)

> Since Adics are 1-1 correspondence with Reals then Reals are also
> countable.

I've discussed this elsewhere.

> > > Finally #5, and it appears to me as an oddball for it looks more like a
> > > characteristic of the Natural Numbers for which it was unnecessary to
> > > devote an axiom to it. That the Successor function or series should be
> > > able to prove Induction as a theorem and unnecessary to make it a
> > > full-fledged axiom itself.
> >
> > #5 is a funny axiom in any event. It's almost a meta-axiom, as,
> > rather than defining an axiom about a number, it defines an axiom
> > about *predicates* involving a number -- or, in some formulations,
> > an axiom about sets of numbers.
>
> I really believe it is a summary of the axioms of successor and
> existence of 0 (and the missing existence axiom of 1). It should not be
> an axiom but proven as a theorem.

No; it's actually got more meaning. In fact, it throws out the
possiblity I have above (with {X,Y,Z, 0,1,2,3,...} It actually says two
things:

(1) Every natural number, except 0, is the successor of some natural
number.
(2) Every natural number can be expressed as 0 plus a finite number of
1s.

And if it's omitted, then my {X,Y,Z,0,1,2,3,...} example is allowed as
the "natural numbers".

> It would be interesting to revisit the history of Peano when he was
> writing these things and to see what prompted him to think that Math
> Induction was needed.

To prove things like
(1 + 1 + 1 + 1 + ... + 1) + (1 + 1) = (1 + 1) + (1 + 1 + 1 + ... + 1)
(a googol times) (a google times)

You're not allowed to use associativity or commutativity of addition.
(Any math student knows about things like this, btw.)

> I do not see your operation as succession; but as a correspondence
> mapping and not a Succession function. It is easy to fool oneself in
> thinking that a dreamed up function is a Succession function.

That's why the Peano Axioms are there; if the successor function passes
all the tests, it's a real successor function, and what we're dealing
with are the natural numbers (in some form).

If I hand you a round red fruit and ask if it's a banana, how long does
it take for you to answer?

> One must
> keep in mind that a Succession function on par with endless adding of 1
> goes to infinity and so one can be fooled into thinking that some
> "cycling type function" such as a trig function is a succession when it
> is not.

And that's another reason for the "oddball" (Axiom #5).

> If it were a true Succession, you would be able to give the
> first three numbers and the reader would then be able to generate every
> number thereafter without ever recourse to your prescription.

No, if you give the first ONE number, and the successor function, the
reader can generate every number thereafter. (Technically, as many as
he/she wants.)

And note that your arrangement

0, ...000, 1, ...111, 2, ...222, 3, ...

doesn't pass this test, it's not a real successor function, either.

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 1:44:42 AM12/25/05
to

a_plu...@hotmail.com wrote:
> Ken Quirici writes:
>
> The point however is that Cantor's argument is not I believe intended
> to construct the reals, one -at-a-time (which could only result, as I
> said above, in countably many). It is rather to show that the
> assumption that the reals are countable leads to a contradiction,
> and hence must be false.
>
> A separate point - you say somewhere in your chain of notes that
> the infinite adics are countable. My understanding is that they
> are uncountable, and this point has been made before.
>
> A.P. writes:
> [...]

> The commonsense argument is that to construct the Natural Numbers, Ken,
> you use what I call the Endless adding of 1 Machine. As you add one you
> also count that number. The Adics are built from only this Machine,
> hence the Adics are Countable [...]

But common sense doesn't work with arguments involving infinite sets.
Let's see what happens when you really do this repeated addition of 1,
and do a reducto ad absurdum argument.

0: ....0000 ....000x
1: ....0001 ....00x1
2: ....0002 ....0x02
3: ....0003 ....x003
...

On the left is a natural number n, and immediately to the right is the
Infinite Integer which is the sum of n 1's (where a sum of zero 1's is
defined to be 0, as usual). I'll explain what the x's mean in a moment.

Now, if the Infinite Integers (I.I.'s) are really the same cardinality
as the natural numbers, for every natural number there's an I.I., and
vice versa. And every I.I. is on the right-hand side somewhere.

Now let's look at the right-most digit of the I.I. next to 0; it's a 0.
The second right-most digit of the I.I. next to 1 is the one
immediately to the left of the 1, and it turns out to be 0.
The third right-most digit of the I.I. next to 2 is 0 as well, and so
on.

In fact, the (n+1)th right-most digit of the nth I.I. is 0, for every
natural number n. (These are the digits in positions marked by the x's;
so every x represents a 0.) ***

Now, let's consider the I.I. ...1111 (all 1's). It's an Infinite
Integer, so it should be on the right-hand side somewhere. So suppose
that it's in the kth position.

Since all the digits of ...1111 are 1, the (k+1)th right-most digit of
...1111 is 1. But, according to ***, the (k+1)th right-most digit of
the kth Infinite Integer is 0. This leads us to the conclusion that 0 =
1.

This is clearly a contradiction, so our last assumption was false;
namely that every Infinite Integer is on the list.

Now, this means we can't get ...1111 by endlessly adding 1 every time,
where "endlessly" means "once for every natural number". But since you
_can_ get ...1111 by adding 1 repeatedly, this means we have a
different degree of "endlessness" when dealing with the Infinite
Integers. That means the Infinite Integers are a bigger set than the
natural numbers.

--- Christopher Heckman

Proginoskes

unread,
Dec 25, 2005, 1:46:04 AM12/25/05
to

a_plu...@hotmail.com wrote:
> I am going to give a list of logical flaws of most every practicing
> mathematician today. Flaws that they believe in and that are fakeries.
> Flaws of the Peano axioms and the flaw of most of Cantor's work. I will
> just list them below:

I am going to give a list of logical flaws of A.P. Flaws that he
believes in and that are fakeries. Flaws of his new axiom system and
the flaw of how he does not understand Cantor's work. I will just list
them below:

(1) The Peano Axioms are flawed. No one has shown this in a rigorous
sense.

(2) His suggested axiom system is not "flawed" like the Peano Axioms
are.

> > 1. There is a natural number 0.
> > 2. There is a natural number 1.

> > 3. There is a successor series of endless adding of 1


> > 4. A number is an infinite series and the Counting Numbers are
> > infinite leftward of all possible digit arrangements.

Axioms #1 and #2 create lots of sets which could be natural numbers,
but #3 gets rid of some of them. (One example is {0,1,5}.)

(3) #4 is consistent.

(4) Cantor's diagonalization argument doesn't apply to the Infinite
Integers/The Infinite Integers can be sequentially ordered. Cantor's
argumetn works, and a lot nicer than it does for real numbers, since


...9990 is different from ...0001.

(5) In Peano Arithmetic, there are only nonnegative integers. I gave a
construction of the rational numbers earlier, I just posted a
construction of negative numbers.

(6) He doesn't need to check out a book on Peano Axioms to understand
how they work.

(7) "Endless adding of 1s" is the same as Peano's successor function.
(1) No, since nothing in AP's axioms prevents us from having 1 + 1 = 0,
or even 1 = 0. (2) No, since the size of the natural numbers is
different from the size of the Infinite Integers.

(8) He's found a problem with Cantor's argument. No, he's just fallen
into a common trap: That the list of numbers which he gives to Cantor
can be modified afterwards.

(9) Proofs don't have to be rigorous. (Maybe he should create a group
called sci.illogic.)

(10) Peano didn't create "1". Yes, he did; he calls it S(0).

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 25, 2005, 1:46:08 AM12/25/05
to

a_plu...@hotmail.com wrote:

> > > New Array:
> > > 0, ....000000, 1, ....11111, 2, .....22222, 3, .....333333, 4,
> > > ....44444, 5, ....55555, 6, ....66666, 7, ....777777, 8, .....88888, 9,
> > > ....99999, 10, ....10101010, 11, .....111111, 12 ad infinitum
> > >
> > > That is the New Array and I went to 11 so as to show the ....11111
> > > repetition with that of ....1111 for 11 and for 1.
> >
> > This doesn't work either; and #4 is the property that fails. This is
> > because 1 and 11 have the same successor, and 1 and 11 are different.
> >
>
> No, I overcome #4 by pairings. You have to stipulate a pair of
> predecessors.
> So to put the Infinite Integers into a sequental ordering so they can
> be counted, apply the trick of pairing. With the Windings the
> successors cannot be unique but by pairing the Successors can be
> unique.
>
> But another major problem crops up. With Windings I can get all
> possible digit arrangements and especially the irrational Infinite
> Integers such as digitized e and digitized pi and digitized square root
> 2. But can I get the irrational Infinite Integers by this New-Array?
> Does it only produce rational Infinite Integers. If it cannot produce
> the irrational Infinite Integers my work is not finished.
>
> Can I say the New Array yields the irrational Infinite Integers in that
> they converge at infinity?


Okay I paired the Infinite Integers with the Counting Numbers, but let
us try a new trick. Let us pair the Infinite Integers with the Reals.
The Positive Reals are ordered. I drop the Counting Numbers in the
pairing and replace with Reals as such:

0, ....000000, 1.00..., ....001, 2.71...., .....172, 3.333....,
.....33333,
3.14159....., ......951413 , 3.333...., .....33333, ad infinitum

What I get out of this is all the rational and irrational Infinite
Integers. They are all in the list. Now I delete each and every Real
and there is no longer any pairing. What remains is a Ordered list as
per the Reals. But is it a Sequental Ordering by unit 1? Obviously not.
And are there repeats? Obviously yes for example ....33333 could be
3.33... or 33.333....

So why do this? I do it to see how I can get the Infinite Integers
complete list of irrational and rational Infinite Integers and preserve
Order, even though it is not unitary Sequental Order. Is it all
possible digit arrangements? That is a deep question for it begs to ask
whether the Reals are all possible digit arrangements.

P.S. I could make it easy on myself and just say the countable Infinite
Integers go from ...0000, .....000001 all the way to .....9999999 in
succession by adding 1 and that every possible digit arrangement is
trespassed in the infinite adding of 1 starting with ....00000. But I
would like to know the bridges of succession from ....00000xyz to a
number such as .....1111111 or .....2222222.

So the Reals as a pairing tells me that ....11111 is between
......00001 and ....0000012.

Perhaps what I am hoping deep in side me is that if I talk this thing
to death, that the answer will reveal itself. A situation in which
constant talking about it sifts out the answer. The old saying of "beat
the problem to death".

Proginoskes

unread,
Dec 25, 2005, 1:55:59 AM12/25/05
to

a_plu...@hotmail.com wrote:
> [...]

> Perhaps what I am hoping deep in side me is that if I talk this thing
> to death, that the answer will reveal itself. A situation in which
> constant talking about it sifts out the answer. The old saying of "beat
> the problem to death".

True. But you don't have to post everything you say.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 25, 2005, 2:18:08 AM12/25/05
to

A.P. writes:

Time out Chris. Your posts have become too acrimonious. When you
started these dialogues about math some many months ago another poster
chimed in and asked me to admit mistakes if mistakes are revealed. I
did that. I did it with Goldbach conjecture and some others and I did
it recently with Windings.

But apparently you cannot do it when you make mistakes. Your posts are
degenerating into nothing but contraryness. Contrary to every thing I
post and I have no need or interest in reading a poster who takes 180
degree position to everything I say. There was a chap in 1993 in
sci.physics who posted everything contrary to whatever a poster said
and there was a chap from Princeton who engaged in my threads and who
also posted every contrary statement to what I said. People who are
contrary quickly become depressing. And if there is any kernel of value
in their posts it is not worth it to have to read 99% depressing
contrariness.

The above post is a clear example. You should be giving me credit for
pointing out the gap in Peano's Axioms for it has missing the postulate
where it says both 0 and 1 exists.

Now why you want to deny me that credit is perhaps because your posts
have ceased being a discussion about mathematics and has become for you
a ad hominem whipping post.

When I do make a significant contribution to mathematics, dispense with
your acrimony and contrariness and acknowledge my good deed.

Peano cannot define 1 as S(0) because 1 is needed in the S itself. So
when Peano says 0 exists, he must simultaneously say 1 exists and then
he can push onwards to the S postulate.

So, do the gentlemen thing and give me the credit where credit is due.

Virgil

unread,
Dec 25, 2005, 2:43:50 AM12/25/05
to
In article <1135489466.0...@g49g2000cwa.googlegroups.com>,
"Proginoskes" <CCHe...@gmail.com> wrote:

AP also ignores the Cantor First proof, which does not rely on any
representations of real numbers at all, but is entirely based on the
least upper bound, greatest lower bound property of the reals, as is
proved in both the Dedekind and Cauchy models of the reals.

Virgil

unread,
Dec 25, 2005, 2:47:38 AM12/25/05
to
In article <1135490962....@f14g2000cwb.googlegroups.com>,
"Proginoskes" <CCHe...@gmail.com> wrote:

AP has long been a crank of the crankiest sort, though he does switch
subjects from time to time.

Proginoskes

unread,
Dec 25, 2005, 3:10:03 AM12/25/05
to

When someone repeatedly ignores what I've said in previous posts, yes,
the kid gloves come off.

> When you
> started these dialogues about math some many months ago another poster
> chimed in and asked me to admit mistakes if mistakes are revealed. I
> did that. I did it with Goldbach conjecture and some others and I did
> it recently with Windings.
>
> But apparently you cannot do it when you make mistakes.

But I haven't made any. Can you provide any examples of (what you think
are) mistakes?

> Your posts are degenerating into nothing but contraryness. [...]

Well, in pointing out that you've got some things wrong, I must (ipso
facto) take a contrary position. Especially when you say things like
"Cantor's diagonalization proof is wrong", which is clearly false.

> The above post is a clear example. You should be giving me credit for
> pointing out the gap in Peano's Axioms for it has missing the postulate
> where it says both 0 and 1 exists.

I worked extensively with the Peano Axioms during a mentoring program
when I was in high school. I went through all the constructions of
negative numbers, rational numbers, real numbers, complex numbers,
filling in all of the gaps, including in things where I proved that
addition is commutative and associative (which is not obvious from the
definition), things that I'm not expecting that you do. But I know
whereof I speak because I've been there.

The fact is simply: Peano didn't need a special postulate saying that 1
existed; it is a consequence of the existence of 0 and the existence of
the successor function. He just defined it as S(0), and that's all he
had to do. (It would be worthwhile for you to buy a book on Peano
Arithmetic, where it actually gives the definitions of +, *, negative
numbers, rational numbers, real numbers, complex numbers. In fact, it
should be easier to read than either of your two p-adic books.)

Should you be giving me credit for showing that 2 exists?

> Now why you want to deny me that credit is perhaps because your posts
> have ceased being a discussion about mathematics and has become for you
> a ad hominem whipping post.

Like I said above: When someone claims Cantor's diagonalization proof
is wrong, or when they repeatedly ignore what I've said in previous
posts, yes, the kid gloves come off.

But I've always approached your ideas with an open mind, even when you
haven't bothered to define things precisely. (It's not ideas that make
someone a crank, it's their reaction to criticism of them.)

It's also interesting that you haven't checked out any of my posts in
other threads, or in other newsgroups. As far as I can tell, you only
post to sci.math, sci.logic, and sci.physics, and only about your
theories. You don't help people solve problems; in fact maybe you don't
even check out threads where people do ask problems.

In fact, if you search Google Groups for my posts, where "plutonium" is
not listed as being in the post, you'll see that I've posted to:

"is this proof of the product rule rigorous (any loop holes)?"
"Cross products"
"number of grid network combinations"
"New model for primality test"
"HELP finding domain of this derivative"
"Complex numbers without reference to reals"
"? solving max function"
"I think we don't need the 0 !!!"

in the last week alone. (And it's been a slow week, since Winter
Vacation has started.)

> When I do make a significant contribution to mathematics, dispense with
> your acrimony and contrariness and acknowledge my good deed.

I will. But, sorry, "1 exists" isn't a significant contribution. (I
have commended your notation for p-adics, even in other threads, and
actually still use it.)

> Peano cannot define 1 as S(0) because 1 is needed in the S itself.

If you write S(x) = x + 1, that's true, but this isn't the only way to
define S. (In fact, when addition is defined --- later on ---
x + S(0) is defined to be S(x).)

What Peano is saying (in other words), is that S is some function from
some set to itself (with certain properties), and the following are the
natural numbers:

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

where you keep applying S's ad infinitum. Period. You don't need a "1"
to do that, and you don't need a "+" to do that. In particular, Peano's
Axioms as listed at Wikipedia (under "The axioms") don't need them.

But what you're doing is working on your own axiom system for what
natural numbers should be. That's okay, and an interesting approach,
but you shouldn't take concepts from one axiom system and insist that
they MUST be on the other side as well.

--- Christopher Heckman

a_plu...@hotmail.com

unread,
Dec 25, 2005, 3:58:35 AM12/25/05
to

Proginoskes wrote:

>
> What Peano is saying (in other words), is that S is some function from
> some set to itself (with certain properties), and the following are the
> natural numbers:
>
> 0, S(0), S(S(0)), S(S(S(0))), S(S(S(S(0)))), ...
>
> where you keep applying S's ad infinitum. Period. You don't need a "1"
> to do that, and you don't need a "+" to do that. In particular, Peano's
> Axioms as listed at Wikipedia (under "The axioms") don't need them.
>
> But what you're doing is working on your own axiom system for what
> natural numbers should be. That's okay, and an interesting approach,
> but you shouldn't take concepts from one axiom system and insist that
> they MUST be on the other side as well.
>
> --- Christopher Heckman

Rethink your position, Chris. If you do not postulate that 1 exists
along with 0 existing then S(0) can be 0.5 and the Natural Numbers
would be 0, 1/2, 1, 3/2, 2, ....

Unless you pair the existence of 1 to 0, S is arbitrary and the spacing
between 0 and S(0) is arbitrary.

As the Peano Axioms exist presently, it is assumed that S is a spacing
of 1, but that is why we invent axiomatics to get rid of presumptions
and assumptions. So unless you postulate the existence of 0 and 1, the
Counting Numbers can take on horrible forms such as 0, irrational
number r, r+r, r+r+r, ....

You assume, and so does Peano, that S(0) is 1. Unless you include 1
existing, because S is really 0+1, then the S postulate is half-baked.

I do not care how much time you spent on Peano axioms, you are fallible
as I am fallible. But you seem incapable to giving credit where credit
is due, perhaps you have a streak of arrogance that hinders you. So
unless you can admit your mistake and give me credit, our conversation
is ended.

a_plu...@hotmail.com

unread,
Dec 25, 2005, 4:28:09 AM12/25/05
to
Someone wrote:
Another possibility is the set C1, where C1 is the continuum
half-open interval [0,2*pi) (or, if one prefers, the
unit circle) and the operation of succession is

a' = a+1 if a+1 in C1
a+1-2*pi if a+1 is not in C1

This meets axioms 1-4 but C1 also contains elements such as 0.5,
pi, and e, which cannot be generated.

A.P. writes:

Is that really a Succession function of the Peano Axioms? I don't think
so. Looks to me more of an "assignment than a succession function". To
be a Succession means it has an infinite elements because of the
endless adding of 1, so that eliminates finite-fields whereas say
0,5,10,15,20, .... qualifies as a succession function and it shows why
we have to postulate that 0 and 1 exists.

But the other feature of a Succession function is that it dispenses
with a prescription. Just give us its first say 10 numbers and the
succession will provide the rest of the numbers which set C1 fails.

Counterexamples in mathematics can be risky because what we think is a
counterexample is very much irrelevant or non-sequitur.

I wish Dik Winter would get back to the Internet for I need to know if
the p-adics obey Math Induction. If my ideas are correct, then the
p-adics do obey Math Induction of the Peano Axioms. The reason I say
this is because Math Induction is purely a summary of the Successor
postulate coupled with 0 and 1 exists which applies to the p-adics and
so the P-adics should obey Math Induction. Which is a rather grotesque
thought when we think about it. P-adics should obey Math Induction
because they are really just a fancy way of rewriting the Rationals and
we know the Rationals are a transformation of the Natural Numbers.

a_plu...@hotmail.com

unread,
Dec 25, 2005, 2:19:41 PM12/25/05
to

a_plu...@hotmail.com wrote:
> Ken Quirici writes:
>
> The point however is that Cantor's argument is not I believe intended
> to construct the reals, one -at-a-time (which could only result, as I
> said above, in countably many). It is rather to show that the
> assumption that the reals are countable leads to a contradiction,
> and hence must be false.
>
> A separate point - you say somewhere in your chain of notes that
> the infinite adics are countable. My understanding is that they
> are uncountable, and this point has been made before.
>

Let me say it differently.

What creates the both the Natural Numbers and the Adics is the Endless
adding of 1 Machine. Create 0 and 1 and then this Machine creates all
the other Natural Numbers and the Adics.

So we run into a question here. Does this machine not only create each
Adic or Natural Number by successive adding of 1, but also Count each
Adic as it is created? Of course it does. And that this Machine of
endless adding of 1 is a Counting Machine also.

Now the Adics are equinumerous with the Reals and since the Adics are
Countable then the Reals are Countable.

Now we drop back to question the Natural Numbers since they are built
from this same Machine that produces them and also counts them. Can the
other axioms of Peano that demand a unique successor weed out the
Natural Numbers so that they can be both infinite and not as
equinumerous as the Adics and the Reals.

My opinion is still held out on that question. The question of whether
Peano's axiom have internal conflict and are self contradictory. I
suspect they are inconsistent and the reason I say that is because any
person with Commonsense knows that Infinity is a process and that this
process is a unique process in that Endlessness is a property that is
unique and that there are not two kinds of endlessness. Endlessness is
only one type. And Nothingness is only one type and that Endlessness
and Nothingness are linked to one another as inverses. So that if
Natural Numbers is a different infinity from Reals, then there must be
two types of Nothingness and thus two different types of the number
Zero (0).

Quakes and flakes like Virgil and Chris Heckman have to answer the
CommonMan argument I laid out above, before they can endorse Cantor
because Cantor was not a mathematician but a entertainment phenomenon
for which modern day mathematicians have fallen into Cantor's bait.

a_plu...@hotmail.com

unread,
Dec 25, 2005, 2:45:06 PM12/25/05
to
And while you are in the mood to apologize to me and admit your errors
Chris, you may as well admit to these facts of the past several months:

(1) I did admit to having a flawed Goldbach proof with the Array
set-up.
(2) Chris needs to congratulate me for giving the proof of no Odd
Perfect Numbers, the world's first proof of this conjecture.
(3) I did admit to my mistake that the 4 Color Mapping is a valid piece
of mathematics, however, there is a confusion because 4 Color Mapping
is not geometry but purely Topology, and that the real Color Mapping is
2 Color Mapping proven by the Jordan Curve theorem. So that is a piece
of messiness in the body of mathematics literature that needs to be
straightened out. That 2-Color Mapping is Geometry and 4-Color Mapping
is Topology. So although I admitted my mistake and that 4 Color Mapping
is true mathematics, it is still needed to clean-up that misconception
that 4-Color Mapping is geometry when it is stuck deep inside of
topology and that is what festered my misconception. So I deserve part
credit for straightening out that misconception.

(4) I did admit that the Windings of Infinite Integers is not
satisfactory and am working on resolving this.

(5) Chris needs to admit that there is a CommonPerson argument that
infinity is a property and not a number and that this Endlessness issue
must question whether Endlessness comes in flavors or varieties or
whether Endlessness is unique. And whether Endlessness is connected to
Nothingness which is represented by the number zero 0. Whether
Endlessness is the inverse of zero or nothingness. Chris seems
oblivious to these ideas. And he needs to sink his head into these
ideas instead of blurting out Cantor like some automoton.

So is Endlessness unique and it comes in only one type, one form and is
it connected like an inverse to the number 0. Because that would
suggest that if Infinity is not unique and comes in different
cardinalities that the number zero comes in various flavors and
varieties and types which are all different from one another.

So while Chris is on a apology mood and apology path and on a path to
admitting his mistakes and errors, add these to his list.

mountain man

unread,
Dec 25, 2005, 6:21:36 PM12/25/05
to
<a_plu...@hotmail.com> wrote in message
news:1135538381....@g44g2000cwa.googlegroups.com...

> My opinion is still held out on that question. The question of whether
> Peano's axiom have internal conflict and are self contradictory. I
> suspect they are inconsistent

Every, any and all sequences of axioms cannot ever model nature.
See Godel, Turing, Chaitin. This has been essentially "proved" by
using the very formalisms upon which and by which mathematics
and logic are themselves constructed.

--
Pete Brown
Falls Creek
OZ
www.mountainman.com.au

It is loading more messages.
0 new messages