is endless adding of 1 equivalent to Dedekind Cut method? Re: counterexamples to the Riemann Hypothesis


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Topic: Science > Physics
User: "a_plutonium"
Date: 02 Jan 2007 01:37:41 PM
Object: is endless adding of 1 equivalent to Dedekind Cut method? Re: counterexamples to the Riemann Hypothesis
a_plutonium wrote:
(snipped)


Numbers which we call finite and call the Finite Integers. But when you
do a little thinking about a process of endlessly adding one, can those
numbers remain tame and lame and self behaved. If you are given 1 with
a process of endlessly adding 1, does that not force every place value
to be filled up with digits and that it spills out and spills over from
....0000000n+1 into something like .....111111111113 and with an
infinite adding of 1 does that eventually fill up into something like
....22222228 and then more adding of 1 goes into .....333333332 and on
and on until you get to the last ten Natural Numbers that can exist,
namely ......9999990, ....9999991, ....9999992, ....9999993,
.....99999994, .....999995, .....9999996, .....9999997, .....99999998,
.....99999999

You see the flaw in Peano Axioms is that when you have a process of
endlessly adding 1, the Natural Numbers do not stop with these
.....000000n and ....00000n+1 numbers but automatically keep going
until they end up with ....9999999.

Now I do not know if in my lifetime I am able to easily explain how it
is that by endlessly adding of 1 that you can bridge the gap of going
from say .....0000000n into that of .....111111111n. Or going from that
of say .....888888n into that of ....9999999n
What we commonly know of as the Finite Integers which I will sometimes
call the Finite Mirage Integers since most people think they stop
before ending up as ....999999, can easily be seen as "living in the
Reals".
Take a moment and look at all the Reals between 0 and 1. You have
....5555 in there as 0.5555... only flipped around. You have ....999999
in there only flipped over as 0.9999.....
So the question, is, is there a similar process of endless adding of 1
for the creation of the Reals? To create the Integers we apply the
Peano axiom of endless adding of 1 to generate all the Integers but we
see those very same numbers flipped around and embedded in the Reals
from 0 to 1. And the Reals were created from Dedekind Cut even though
the Cut does not produce the Real for it seems to have been already
made and the Dedekind Cut is only a process of acknowledgement that a
specific Real exists.
So is the true process of creation of the Reals a transfer of the very
same process that creates the Integers--- endlessly adding 1. And in
the case of the Reals from 0 to 1 is an endless dividing of 1. So to
create the Integers we endlessly add 1 process and to create the Reals
we endlessly divide into 1 and where to create the Real 2 is that of
1/0.5 and to create the Real 3 is that of 1/(1/3)
I think the answer to most of my questions above is a resounding yes.
And that the Axiom system that created the Integers even the Finite
Mirage Integers also creates the Reals in a process of endless adding
by 1 or an endless dividing by 1. So the Peano Axiom System is thrown
out the window almost completely.
And it is time for us to recognize that the Dedekind Cut Method is only
a "acknowledgement exercise" and that the Reals need to have a Axiom
System for themselves independent of the Integers. That the old way of
looking at Integers and Reals was as if they were built out of one
stock-- Natural Numbers. The new way is that Reals have an independence
of Natural Numbers and where both have their own axiomatics to create
them.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.

User: "David R Tribble"

Title: Re: is endless adding of 1 equivalent to Dedekind Cut method? Re: counterexamples to the Riemann Hypothesis 06 Jan 2007 12:21:32 PM
Archimedes Plutonium wrote:

Now I do not know if in my lifetime I am able to easily explain how it
is that by endlessly adding of 1 that you can bridge the gap of going
from say .....0000000n into that of .....111111111n. Or going from that
of say .....888888n into that of ....9999999n

What we commonly know of as the Finite Integers which I will sometimes
call the Finite Mirage Integers since most people think they stop
before ending up as ....999999, can easily be seen as "living in the
Reals".

Take a moment and look at all the Reals between 0 and 1. You have
...5555 in there as 0.5555... only flipped around. You have ....999999
in there only flipped over as 0.9999.....

Huh. You say you can sequentially denumerate the finite and infinite
naturals, going from 0 to 99999 to ...999. Then does that mean that
by simply flipping the digits around like you said, that you can
denumerate all of the reals in [0,1)?
This would be a big deal, since it would be a well-ordering of the
reals. Perhaps you could demonstrate how that would work?
For instance, we could start with your example of n = ...555
being flipped to get x = 0.555... (5/9). So then what would the
successor of n, n+1, be so that we could flip it to get the next
real after x in your ordering?
.
User: "Tony Orlow"

Title: Re: is endless adding of 1 equivalent to Dedekind Cut method? Re:counterexamples to the Riemann Hypothesis 08 Jan 2007 09:23:24 AM
David R Tribble wrote:

Archimedes Plutonium wrote:

Now I do not know if in my lifetime I am able to easily explain how it
is that by endlessly adding of 1 that you can bridge the gap of going
from say .....0000000n into that of .....111111111n. Or going from that
of say .....888888n into that of ....9999999n

What we commonly know of as the Finite Integers which I will sometimes
call the Finite Mirage Integers since most people think they stop
before ending up as ....999999, can easily be seen as "living in the
Reals".

Take a moment and look at all the Reals between 0 and 1. You have
...5555 in there as 0.5555... only flipped around. You have ....999999
in there only flipped over as 0.9999.....


Huh. You say you can sequentially denumerate the finite and infinite
naturals, going from 0 to 99999 to ...999. Then does that mean that
by simply flipping the digits around like you said, that you can
denumerate all of the reals in [0,1)?

This would be a big deal, since it would be a well-ordering of the
reals. Perhaps you could demonstrate how that would work?

For instance, we could start with your example of n = ...555
being flipped to get x = 0.555... (5/9). So then what would the
successor of n, n+1, be so that we could flip it to get the next
real after x in your ordering?

.....55556, making succ(0.555...) equal to 0.6555.... Does this create a
well ordering of the reals? No, you still have an infinite set of
predecessors for any infinite string, and infinite descending sequence.
Oh well.
Tony
.


User: "a_plutonium"

Title: first counterexample that falsifies the Riemann Hypothesis is ....111111111 Re: counterexamples to the Riemann Hypothesis 02 Jan 2007 09:30:24 PM
Now the reason some of these unproven Math conjectures were unbearably
more difficult to find a proof and where some seemed to have so called
"close calls" where just some more artillery was needed to eke out a
proof. For example, many are coming what they think are close to a
proof of Riemann Hypothesis using the old mirage integers of finite
integers, yet noone is coming at all close to say a Goldbach or a Twin
Primes.
And the answer is because the true Natural Numbers are the set
0,1,2,3,...., ....999998, ....99999 and they curve in Space. They do
not form a straight line that the Riemann Hypothesis hints of as on the
1/2 Real line. Because the Natural Numbers have a natural geometry and
curve in space is the reason that some problems are more difficult,
because the conjecture may touch directly on the curvature of the
Natural Numbers whereas in other conjectures the curvature is not so
directly involved. The Riemann Hypothesis is based on the consideration
that the Natural Numbers form a straight line and so the proof is
tolerant of a small and gradual curvature. But by the time the Natural
Numbers wheel around in a gigantic circle to the ....999999s string and
ending in ....99999 which is only one unit away from the starting point
of the number 0.
So the curvature of the Natural Numbers, although ever so slight and
small of a curvature, where we visualize a gigantic circle and every
number starting with 0 then 1 then 2 then 3 are points on this gigantic
circle.
And so the first real big problem and counterexample for the Riemann
Hypothesis is the number ...1111111. It is larger than any number of
the 0000s string where our number 1 is ....000001 and our number 2 is
.....000002. So that as the mathematicians of the 19th century and 20th
century tried to prove the Riemann Hypothesis only got a microscopic
distance on the Natural Numbers and thus could never really see that
they curved and thus the Hypothesis was false.
But a number like ....111111 falsifies the Riemann Hypothesis
And for conjectures like Twin Primes infinitude or Goldbach, the
curvature of the Natural Numbers would not be touched upon and made
those two conjectures abomidably more difficult to ever get any handle
on.
The proof of Infinitude of Twin Primes needs only to see that
....9999913 and .....999911 are twin primes at infinity and thus there
are an infinite number of twin primes between 3,5 and ....999911,
.....999913.
Now as for Goldbach, I will give more details but ask the question of
counterexamples in Infinite Integers. Would not the Real number "e"
when transposed as an infinite integer of ....172, be a Natural Number
which has no p + p' to satisfy Goldbach and the number of square-root
(-1) of the 5-adics when transposed as an Infinite Integer also be a
counterexample. And then the idempotents of the 10-adics offer us
counterexamples. Or how about this great-grand-daddy of even numbers of
......161412108642. So the Goldbach Conjecture is false. But details
when I reach the Goldbach Conjecture in this series of lectures.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.
User: "hagman"

Title: Re: first counterexample that falsifies the Riemann Hypothesis is ....111111111 Re: counterexamples to the Riemann Hypothesis 03 Jan 2007 04:31:15 AM
a_plutonium schrieb:

Now the reason some of these unproven Math conjectures were unbearably
more difficult to find a proof and where some seemed to have so called
"close calls" where just some more artillery was needed to eke out a
proof. For example, many are coming what they think are close to a
proof of Riemann Hypothesis using the old mirage integers of finite
integers, yet noone is coming at all close to say a Goldbach or a Twin
Primes.

And the answer is because the true Natural Numbers are the set
0,1,2,3,...., ....999998, ....99999 and they curve in Space. They do
not form a straight line that the Riemann Hypothesis hints of as on the
1/2 Real line. Because the Natural Numbers have a natural geometry and
curve in space is the reason that some problems are more difficult,
because the conjecture may touch directly on the curvature of the
Natural Numbers whereas in other conjectures the curvature is not so
directly involved. The Riemann Hypothesis is based on the consideration
that the Natural Numbers form a straight line and so the proof is
tolerant of a small and gradual curvature. But by the time the Natural
Numbers wheel around in a gigantic circle to the ....999999s string and
ending in ....99999 which is only one unit away from the starting point
of the number 0.

So the curvature of the Natural Numbers, although ever so slight and
small of a curvature, where we visualize a gigantic circle and every
number starting with 0 then 1 then 2 then 3 are points on this gigantic
circle.

And so the first real big problem and counterexample for the Riemann
Hypothesis is the number ...1111111. It is larger than any number of
the 0000s string where our number 1 is ....000001 and our number 2 is
....000002. So that as the mathematicians of the 19th century and 20th
century tried to prove the Riemann Hypothesis only got a microscopic
distance on the Natural Numbers and thus could never really see that
they curved and thus the Hypothesis was false.

But a number like ....111111 falsifies the Riemann Hypothesis

And for conjectures like Twin Primes infinitude or Goldbach, the
curvature of the Natural Numbers would not be touched upon and made
those two conjectures abomidably more difficult to ever get any handle
on.

The proof of Infinitude of Twin Primes needs only to see that
...9999913 and .....999911 are twin primes at infinity and thus there
are an infinite number of twin primes between 3,5 and ....999911,
....999913.

Call to AP-numbers twin-slime if their difference is 2 and the last
digits are odd and all but the last two digits are identical.
The proof of Infinitude of Twin Slimes needs only to see that
...9999913 and .....999911 are twin slimes at infinity and thus there
are an infinite number of twin slimes between 3,5 and ....999911,
....999913.
Unfortunately, however, there are only finitely many twin slimes (about
500)
.

User: "Proginoskes"

Title: Re: first counterexample that falsifies the Riemann Hypothesis is ....111111111 Re: counterexamples to the Riemann Hypothesis 02 Jan 2007 09:41:18 PM
a_plutonium wrote:

Now the reason some of these unproven Math conjectures were unbearably
more difficult to find a proof and where some seemed to have so called
"close calls" where just some more artillery was needed to eke out a
proof. For example, many are coming what they think are close to a
proof of Riemann Hypothesis using the old mirage integers of finite
integers, yet noone is coming at all close to say a Goldbach or a Twin
Primes.

And the answer is because the true Natural Numbers are the set
0,1,2,3,...., ....999998, ....99999 and they curve in Space.

But the Infinite Integers was not the context of the original
conjectures, and as someone once said, "It is insanity to think that by
ignoring key essential elements of the domain of proving elements that
you are still doing mathematics." *

The proof of Infinitude of Twin Primes needs only to see that
...9999913 and .....999911 are twin primes at infinity and thus there
are an infinite number of twin primes between 3,5 and ....999911,
....999913.

This doesn't follow. (1) There are an infinite number of reals between
0 and 10, but there are only finitely many INTEGERS between 0 and 10.

Now as for Goldbach, I will give more details but ask the question of
counterexamples in Infinite Integers. Would not the Real number "e"
when transposed as an infinite integer of ....172, [...]

If I was going to assign e to an Infinite Integer (a la the 10-adics),
I would first want to calculate the limit of (1 + 1/n)^n as n
approaches infinity, if I ended up with a particular Infinite Integer.
--- Christopher Heckman
* Archimedes Plutonium, "why the 4 Color Mapping is a fake and a dude
[sic]; the domain space of a proof cannot be ignored", sci.math,
sci.logic, Fri, Sep 30, 2005, 10:30 am.
.
User: "a_plutonium"

Title: Re: first counterexample that falsifies the Riemann Hypothesis is ....111111111 Re: counterexamples to the Riemann Hypothesis 03 Jan 2007 01:47:07 AM
Proginoskes wrote:

a_plutonium wrote:

(snipped)


The proof of Infinitude of Twin Primes needs only to see that
...9999913 and .....999911 are twin primes at infinity and thus there
are an infinite number of twin primes between 3,5 and ....999911,
....999913.


This doesn't follow. (1) There are an infinite number of reals between
0 and 10, but there are only finitely many INTEGERS between 0 and 10.

It follows because if you can get one representative at the end portion
of the Infinite Integers it is simple to build or custom craft an
infinitude of that class of number. So I have Twin Primes of 3,5 at the
beginning of Infinite Integers and a pair at the end of .....999911 and
.....9999913. So I assert there is an infinitude of twin primes in
between those extreme ends, or let me say in the middle.
So I construct an infinite set of Twin Primes. I can choose the prime
pattern of the number e in Reals which if memory serves is 2.71828....
and transpose it as an Infinite Integer .....828172. Now I build,
custom build an infinite set of twin-primes.
......82817299911 paired to .....82817299913
now for the next pair I lop off the 2
.......8281799911 paired to .....8281799913
the next pair I lop off the 7
ad infinitum
Or if I like a different infinite prime such as this number which I
call as the grand-daddy of prime numbers ..........31292319171311753
where I delete 2. Now is this infinite integer always prime as I lop
off a digit? I think so.
So in this manner of proof method, I see if a Twin Prime is at the
beginning and if a Twin Prime is at the end in the ....99999 series. If
so, I can construct an infinite class of that number species.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.
User: "Proginoskes"

Title: Re: first counterexample that falsifies the Riemann Hypothesis is ....111111111 Re: counterexamples to the Riemann Hypothesis 03 Jan 2007 02:25:45 AM
a_plutonium wrote:

Proginoskes wrote:

a_plutonium wrote:

(snipped)


The proof of Infinitude of Twin Primes needs only to see that
...9999913 and .....999911 are twin primes at infinity and thus there
are an infinite number of twin primes between 3,5 and ....999911,
....999913.


This doesn't follow. (1) There are an infinite number of reals between
0 and 10, but there are only finitely many INTEGERS between 0 and 10.


It follows because if you can get one representative at the end portion
of the Infinite Integers it is simple to build or custom craft an
infinitude of that class of number. So I have Twin Primes of 3,5 at the
beginning of Infinite Integers and a pair at the end of .....999911 and
....9999913. So I assert there is an infinitude of twin primes in
between those extreme ends, or let me say in the middle.

In short: You have symmetry between the ends. But that is not
sufficient for an infinite number of objects.
My example (1) above shows this as well; if N is an integer between 0
and 10, then 10-N is also an integer between 0 and 10, but there are
only finitely many integers between 0 and 10.

So I construct an infinite set of Twin Primes. I can choose the prime
pattern of the number e in Reals which if memory serves is 2.71828....

You can't just willy-nilly choose which numbers end up being primes,
and which end up being composite; in that case, you have primes in name
only, not real primes. Primes are primes because they satisfy the
definition of a prime, which is:
If N = A*B, then A or B is a unit (i.e., 1/A or 1/B exists).
Otherwise it's like a con artist who sells you an immortality potion
which turns out to be just tap water; it wasn't (from your point of
view) what he said it was.

and transpose it as an Infinite Integer .....828172. Now I build,
custom build an infinite set of twin-primes.

.....82817299911 paired to .....82817299913
now for the next pair I lop off the 2
......8281799911 paired to .....8281799913
[...]

I will not pay for your snake oil until you show it does what you claim
it does.
--- Christopher Heckman
.
User: "hagman"

Title: Re: first counterexample that falsifies the Riemann Hypothesis is ....111111111 Re: counterexamples to the Riemann Hypothesis 03 Jan 2007 04:40:02 AM
Proginoskes schrieb:

a_plutonium wrote:

Proginoskes wrote:

a_plutonium wrote:

(snipped)


The proof of Infinitude of Twin Primes needs only to see that
...9999913 and .....999911 are twin primes at infinity and thus there
are an infinite number of twin primes between 3,5 and ....999911,
....999913.


This doesn't follow. (1) There are an infinite number of reals between
0 and 10, but there are only finitely many INTEGERS between 0 and 10.


It follows because if you can get one representative at the end portion
of the Infinite Integers it is simple to build or custom craft an
infinitude of that class of number. So I have Twin Primes of 3,5 at the
beginning of Infinite Integers and a pair at the end of .....999911 and
....9999913. So I assert there is an infinitude of twin primes in
between those extreme ends, or let me say in the middle.


In short: You have symmetry between the ends. But that is not
sufficient for an infinite number of objects.

My example (1) above shows this as well; if N is an integer between 0
and 10, then 10-N is also an integer between 0 and 10, but there are
only finitely many integers between 0 and 10.

So I construct an infinite set of Twin Primes. I can choose the prime
pattern of the number e in Reals which if memory serves is 2.71828....


You can't just willy-nilly choose which numbers end up being primes,
and which end up being composite; in that case, you have primes in name
only, not real primes. Primes are primes because they satisfy the
definition of a prime, which is:

If N = A*B, then A or B is a unit (i.e., 1/A or 1/B exists).

You forgot the additional condition that a prime must not be a unit!
And as
3 x ...3333333333 = ...99999999
and ...999999999 x ..999999999 = 1
we have that 3 is a unit.
In fact, there are units galore in AP-numbers (hence only few primes,
not to mention twin primes).
.

User: "a_plutonium"

Title: irrational-infinite-integers forms a set of infinite primes Re: counterexamples to the Riemann Hypothesis 03 Jan 2007 12:28:32 PM
Proginoskes wrote:

a_plutonium wrote:

It follows because if you can get one representative at the end portion
of the Infinite Integers it is simple to build or custom craft an
infinitude of that class of number. So I have Twin Primes of 3,5 at the
beginning of Infinite Integers and a pair at the end of .....999911 and
....9999913. So I assert there is an infinitude of twin primes in
between those extreme ends, or let me say in the middle.


In short: You have symmetry between the ends. But that is not
sufficient for an infinite number of objects.

My example (1) above shows this as well; if N is an integer between 0
and 10, then 10-N is also an integer between 0 and 10, but there are
only finitely many integers between 0 and 10.

So I construct an infinite set of Twin Primes. I can choose the prime
pattern of the number e in Reals which if memory serves is 2.71828....


You can't just willy-nilly choose which numbers end up being primes,
and which end up being composite; in that case, you have primes in name
only, not real primes. Primes are primes because they satisfy the
definition of a prime, which is:

If N = A*B, then A or B is a unit (i.e., 1/A or 1/B exists).

Otherwise it's like a con artist who sells you an immortality potion
which turns out to be just tap water; it wasn't (from your point of
view) what he said it was.

and transpose it as an Infinite Integer .....828172. Now I build,
custom build an infinite set of twin-primes.

.....82817299911 paired to .....82817299913
now for the next pair I lop off the 2
......8281799911 paired to .....8281799913
[...]


I will not pay for your snake oil until you show it does what you claim
it does.

--- Christopher Heckman

You want proof, okay, proof is what you get.
In my construction of an infinite set of Twin Primes I used the
transcendental number in Reals of e. Transcendental Numbers are
Irrational Numbers and since they are irrational there is never a
repeating pattern of digits. Because there is no pattern of digits
renders them prime. Why? Because they are not divisible. Having no
pattern of digits renders them nondivisible and hence prime. Irrational
is the same as prime for integers of these infinite integers.
Now here is a list of some Irrational Infinite Integers derived from
the square root of 2,3,5
.......3732653124141
......88657080502371
......7797606322
Now since irrational numbers are an infinite set between 0 and 1 in
Reals then there is an infinite set of Irrational-Infinite-Integers.
So that if I see Twin Primes at the beginning of 3,5 or 5,7 and I see
Twin Primes at the end of the Natural-Numbers of .....9999999911 and
........99999913 then I will have proved there is an infinite set of
Twin Primes in between those two extreme ends because I can manually
construct an infinite set of Twin Primes. I simply take the set of
square roots and transpose them into Irrational Infinite Integers and I
augment the end piece as both 911 and 913 onto the last three digits of
that set of those Irrational Infinite Integers.
.......3732653124141 yields the twin primes of ......3732653124141911
and
.......3732653124141913
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.
User: "Dik T. Winter"

Title: Re: irrational-infinite-integers forms a set of infinite primes Re: counterexamples to the Riemann Hypothesis 03 Jan 2007 05:33:08 PM
In article <1167848912.658364.259440@k21g2000cwa.googlegroups.com> "a_plutonium" <a_plutonium@hotmail.com> writes:
....

In my construction of an infinite set of Twin Primes I used the
transcendental number in Reals of e. Transcendental Numbers are
Irrational Numbers and since they are irrational there is never a
repeating pattern of digits. Because there is no pattern of digits
renders them prime. Why? Because they are not divisible. Having no
pattern of digits renders them nondivisible and hence prime. Irrational
is the same as prime for integers of these infinite integers.

....

.....7797606322

I would state that this number is divisible by 2.

Now since irrational numbers are an infinite set between 0 and 1 in
Reals then there is an infinite set of Irrational-Infinite-Integers.

Not so quick.

So that if I see Twin Primes at the beginning of 3,5 or 5,7 and I see
Twin Primes at the end of the Natural-Numbers of .....9999999911 and
.......99999913 then I will have proved there is an infinite set of

According to your own definition of multiplication of those numbers,
...9999999911 = 89 * ...9999999999
and
...9999999913 = 87 * ...9999999999
In what way are they prime?

......3732653124141 yields the twin primes of ......3732653124141911
and
......3732653124141913

and ......3732653124141909
and ......3732653124141907
so actually a quadruple prime?
Note, moreover, that in your infinite integers 3 is *not* prime,
it is 7 * ...1428571428571429.
And it actually is easy to prove that in the 10-adics (which are your
infinite integers) there are only two primes: 2 and 5.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.
User: "a_plutonium"

Title: Re: irrational-infinite-integers forms a set of infinite primes Re: counterexamples to the Riemann Hypothesis 03 Jan 2007 10:04:58 PM
Dik T. Winter wrote:

In article <1167848912.658364.259440@k21g2000cwa.googlegroups.com> "a_plutonium" <a_plutonium@hotmail.com> writes:
...

In my construction of an infinite set of Twin Primes I used the
transcendental number in Reals of e. Transcendental Numbers are
Irrational Numbers and since they are irrational there is never a
repeating pattern of digits. Because there is no pattern of digits
renders them prime. Why? Because they are not divisible. Having no
pattern of digits renders them nondivisible and hence prime. Irrational
is the same as prime for integers of these infinite integers.

...

.....7797606322


I would state that this number is divisible by 2.

Excuse my typing error of haste. I should have mentioned a lopping off
of the two 2s in the transpose of sqrt 5.

Now since irrational numbers are an infinite set between 0 and 1 in
Reals then there is an infinite set of Irrational-Infinite-Integers.


Not so quick.

So that if I see Twin Primes at the beginning of 3,5 or 5,7 and I see
Twin Primes at the end of the Natural-Numbers of .....9999999911 and
.......99999913 then I will have proved there is an infinite set of


According to your own definition of multiplication of those numbers,
...9999999911 = 89 * ...9999999999
and
...9999999913 = 87 * ...9999999999
In what way are they prime?

In the 1990s decade I went along with how p-adics looks at Infinite
Integers. Trouble with p-adics is that it is base dependent, and
imagine if Reals or Counting Numbers were base dependent, how confused
everyone would be.
Algebras are great on the Reals, but not on Natural Numbers. So
although you would think that .....9999999911 is divisible by 89, it is
not. Maybe your p-adics tells you ....9999911 is divisible by 89, the
old definition of division of old Natural Numbers starts to come out
with a number like this 112...
So there is a disconnect here of definition of multiplication and
division. They do not link and join together. Maybe in your Algebra of
Reals there is some flaw when transfering Algebras to Infinite
Integers.
As I said earlier, I am keeping the definition of add, multiply,
divide, subtract as close to what it was for the Counting Numbers. So
that when I multiply 5 x 11 = 55 and then 55/5 = 11 the numbers all
agree. So neither ....999999911 and .....99999913 are evenly divisible
by 89, or 87 or ....999999.
As I said, these Infinite Integers are not Adics. My least worry is
what Algebra the Infinite Integers do possess.

......3732653124141 yields the twin primes of ......3732653124141911
and
......3732653124141913


and ......3732653124141909
and ......3732653124141907
so actually a quadruple prime?

Yes, there are going to be very strange numbers in Infinite Integers.
For example, there are even irrationals and odd irrationals. There are
even transcendental and odd transcendental. No wonder Goldbach
Conjecture could never be proven when you have even irrationals and
even transcendental natural numbers to work with.


Note, moreover, that in your infinite integers 3 is *not* prime,
it is 7 * ...1428571428571429.

This is what I mean about saying that the p-adics are just a different
name for Reals in that the Algebras are making smoke and mirrors.
Again, I have defined multiplication and division and ....000003 is not
divisible by ...00007 nor by ...1428571428571429. So that none of what
Dik says above has any meaning.
I dare Dik to say to someone in elementary Number theory that 3 is
evenly divisible by 7.
Give up the game, Dik, these are Infinite Integers, not some algebra
game.


And it actually is easy to prove that in the 10-adics (which are your
infinite integers) there are only two primes: 2 and 5.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

Infinite Integers are numbers that exist just as 1,2,3,4 exists and
they are not governed by some trite game of Algebras where you play
around with p-adics, or another trite game of surreal numbers or some
other trite game.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.
User: "Dik T. Winter"

Title: Re: irrational-infinite-integers forms a set of infinite primes Re: counterexamples to the Riemann Hypothesis 04 Jan 2007 07:53:50 AM
In article <1167883498.620452.129920@51g2000cwl.googlegroups.com> "a_plutonium" <a_plutonium@hotmail.com> writes:

Dik T. Winter wrote:

....

...

.....7797606322


I would state that this number is divisible by 2.


Excuse my typing error of haste. I should have mentioned a lopping off
of the two 2s in the transpose of sqrt 5.

Ah. And why would you lop them off? If the first digit of the irrational
is a 5, would you also lop it off?

According to your own definition of multiplication of those numbers,
...9999999911 = 89 * ...9999999999
and
...9999999913 = 87 * ...9999999999
In what way are they prime?

....

Algebras are great on the Reals, but not on Natural Numbers. So
although you would think that .....9999999911 is divisible by 89, it is
not. Maybe your p-adics tells you ....9999911 is divisible by 89, the
old definition of division of old Natural Numbers starts to come out
with a number like this 112...

So there is a disconnect here of definition of multiplication and
division. They do not link and join together. Maybe in your Algebra of
Reals there is some flaw when transfering Algebras to Infinite
Integers.

As I said earlier, I am keeping the definition of add, multiply,
divide, subtract as close to what it was for the Counting Numbers.

Pray define your addition, subtraction, multiplication and division so
that we can check your results.

So
that when I multiply 5 x 11 = 55 and then 55/5 = 11 the numbers all
agree. So neither ....999999911 and .....99999913 are evenly divisible
by 89, or 87 or ....999999.

This is nonsense. Als when I calculate 89 * ...999999999 = ...99999911
they also agree, at least using the only attempt of you to define
multiplication.

As I said, these Infinite Integers are not Adics. My least worry is
what Algebra the Infinite Integers do possess.

In that case *define* what algegra these infinite integers do possess.

......3732653124141 yields the twin primes of ......3732653124141911
and
......3732653124141913


and ......3732653124141909
and ......3732653124141907
so actually a quadruple prime?


Yes, there are going to be very strange numbers in Infinite Integers.
For example, there are even irrationals and odd irrationals. There are
even transcendental and odd transcendental.

Odd. How do you *define* all this?

No wonder Goldbach
Conjecture could never be proven when you have even irrationals and
even transcendental natural numbers to work with.

No wonder? I thought that Goldbach was about the integers, not about
the infinite integers.

Note, moreover, that in your infinite integers 3 is *not* prime,
it is 7 * ...1428571428571429.


This is what I mean about saying that the p-adics are just a different
name for Reals in that the Algebras are making smoke and mirrors.

Again, I have defined multiplication and division and ....000003 is not
divisible by ...00007 nor by ...1428571428571429. So that none of what
Dik says above has any meaning.

Pray *give* your definitions of multiplication and division so that we
can verify your statements.

I dare Dik to say to someone in elementary Number theory that 3 is
evenly divisible by 7.

Well, I have done quite a bit in number theory, so I can state it to
myself. And I would ask: "what is the domain of discourse?" And I
would answer: "the 10-adics." And I would reply: "yes, true."
And yes, in all the n-adics 3 is divisible by 7 unless n is divisible
by 3 or by 7. And in Z_23, 3 is even the square of 7. And in the
algebraic integers 3 is not prime although it is not divisible by 7.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.
User: "a_plutonium"

Title: Re: irrational-infinite-integers forms a set of infinite primes Re: counterexamples to the Riemann Hypothesis 04 Jan 2007 12:21:39 PM
Dik T. Winter wrote:

In article <1167883498.620452.129920@51g2000cwl.googlegroups.com> "a_plutonium" <a_plutonium@hotmail.com> writes:

Dik T. Winter wrote:

...

...

.....7797606322


I would state that this number is divisible by 2.


Excuse my typing error of haste. I should have mentioned a lopping off
of the two 2s in the transpose of sqrt 5.


Ah. And why would you lop them off? If the first digit of the irrational
is a 5, would you also lop it off?

Yes, as you do not know that such a string is divisible by 5.
(snipped)


As I said earlier, I am keeping the definition of add, multiply,
divide, subtract as close to what it was for the Counting Numbers.


Pray define your addition, subtraction, multiplication and division so
that we can check your results.

INFINITE INTEGERS operations defined:
addition-- define addition the same as any addition of Counting Numbers
where we do the same for infinite strings
multiplication-- define multiplication the same as any multiplication
of Counting Numbers where we start the multiplication between the first
digits of both numbers to be multiplied.
division-- define as in Counting Numbers but if one number is larger
than the other then we cannot divide larger numbers into smaller ones.
subtraction-- same as with Counting Numbers only you can only subtract
a smaller number from a larger number
Square-root-- same as in Counting Numbers and here is a website that
outlines the mechanics
--- quoting from website
http://math.arizona.edu/~kerl/doc/square-root.html
46656
First, divide the number to be square-rooted into pairs of digits,
starting at the decimal point. That is, no digit pair should straddle a
decimal point. (For example, split 1225 into "12 25" rather than "1 22
5"; 6.5536 into "6. 55 36" rather than"6.5 53 6".)
Then you can put some lines over each digit pair, and a bar to the
left, somewhat as in long division.
+--- ---- ----
| 4 66 56
Find the largest number whose square is less than or equal to the
leading digit pair. In this case, the leading digit pair is 4; the
largest number whose square is less than or equal to 4 is 2.
Put that number on the left side, *and* above the first digit pair.
2
+--- ---- ----
2 | 4 66 56
--- end quoting from website
http://math.arizona.edu/~kerl/doc/square-root.html

So
that when I multiply 5 x 11 = 55 and then 55/5 = 11 the numbers all
agree. So neither ....999999911 and .....99999913 are evenly divisible
by 89, or 87 or ....999999.


This is nonsense. Als when I calculate 89 * ...999999999 = ...99999911
they also agree, at least using the only attempt of you to define
multiplication.

As I said, these Infinite Integers are not Adics. My least worry is
what Algebra the Infinite Integers do possess.


In that case *define* what algegra these infinite integers do possess.

No, what is nonsense is that you impose Algebra on numbers for which
there is no Algebra. The numbers exist, but you, in a silly manner
wants to impose something that is not possible. So whether you remain
to be and act stubborn and idiotic is up to you Dik to remain stubborn
and idiotic.
The thing you do not see or understand Dik, is that these Infinite
Integers form a circular or spherical geometry and bend back around to
the starting point of 0. You cannot impose a Algebra on the coordinates
of a globe in geography or geometry and expect that Algebra to be the
same as the Algebra on a flat surface. So stop acting ridiculously
stupid about 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
.
User: "Proginoskes"

Title: Re: irrational-infinite-integers forms a set of infinite primes Re: counterexamples to the Riemann Hypothesis 04 Jan 2007 07:41:21 PM
a_plutonium wrote:

Dik T. Winter wrote:
[...]

Pray define your addition, subtraction, multiplication and division so
that we can check your results.


INFINITE INTEGERS operations defined:
addition-- define addition the same as any addition of Counting Numbers
where we do the same for infinite strings

(1) Since the Infinite Integers are created by "infinite adding of 1",
and 1 = ...0001, then can you guarantee that N+...0001 is the Infinite
Integer right after N?
(2) Fill in the blank:
....(a[3])(a[2])(a[1])(a[0]) + ...(b[3])(b[2])(b[1])(b[0]) =
______________,
where a[i] and b[i] are digits between 0 and 9 for all i.

multiplication-- define multiplication the same as any multiplication
of Counting Numbers where we start the multiplication between the first
digits of both numbers to be multiplied.

(1) What if there are infinitely many digits? That is, what should
....2222 * ..3333 be? Your definition doesn't cover this case.
(2) Is ...00011 * ...00011 equal to ...000121 (doing operations in base
3 or higher) or ...0001001 (doing operations in base 2)?

division-- define as in Counting Numbers but if one number is larger
than the other then we cannot divide larger numbers into smaller ones.

(1) How do I tell if M is larger than N?
(2) If you've defined addition and multiplication already, and they
satisfy the ring axioms, then you don't need a special definition for
division; you can just use the Division Theorem, which states that for
any nonnegative integers a,b, with b nonzero, that you can write
a = q * b + r, with 0 <= r < b,
in exactly one way; then you would (presumably) define a/b to be q.
Which raises the question ...
(3) What if there is a remainder?

subtraction-- same as with Counting Numbers only you can only subtract
a smaller number from a larger number

(1) What is ...00010 - ...00001 then?
(2) Again, if addition satisfies certain properties, you can define A-B
to be C if C is an infinite integer, and
A = B+C.

Square-root-- same as in Counting Numbers and here is a website that
outlines the mechanics

(1) If you have multiplication defined, you can define square root.

--- quoting from website
http://math.arizona.edu/~kerl/doc/square-root.html

First, divide the number to be square-rooted into pairs of digits,
starting at the decimal point. [...]
Find the largest number whose square is less than or equal to the
leading digit pair. [...]

(1) Infinite Integers have no "leading digit pair", by definition.
(2) Assuming that Infinite Integers can have a leading digit pair, what
is that leading digit pair, if the number you want to take the square
root of is ...12341234? Is it 12 or 34?
(3) In order to use this procedure, you need to know various properties
of + and *; for instance, the distributive property needs to hold. This
needs to be shown.
(4) What if there's a remainder?

As I said, these Infinite Integers are not Adics. My least worry is
what Algebra the Infinite Integers do possess.

If the operations on the Infinite Integers are supposed to be "base
free", then the results of these operations are not well-defined; that
means that A+B might end up being C or C+1 or C+...111. (See my example
above for ...00011 * ...00011.) Then you would have various
contradictions, which means the whole idea is meaningless.

In that case *define* what algebra these infinite integers do possess.


No, what is nonsense is that you impose Algebra on numbers for which
there is no Algebra.

Then a lot of Number Theory must be done from scratch. The Division
Theorem and the square root algorithm above (to name a couple) require
the algebraic structures in order to be used.

The numbers exist, but you, in a silly manner
wants to impose something that is not possible.

Then what AP is doing isn't mathematics.

The thing you do not see or understand Dik, is that these Infinite
Integers form a circular or spherical geometry and bend back around to
the starting point of 0.

(1) Then that means every Infinite Integer M is less than any other
Infinite Integer N.
(2) As a consequence of (1), the definitions of division and
subtraction become meaningless; you _can_ divide 3 by 7 after all.

You cannot impose a Algebra on the coordinates
of a globe in geography or geometry and expect that Algebra to be the
same as the Algebra on a flat surface.

That's certainly true. However, if you don't have any structure --- if
you make definitions in a non-rigorous way --- you can't actually work
with what you have. And in that case you are not doing mathematics.

So stop acting ridiculously stupid about Infinite Integers.

Considering that only one person knows what Infinite Integers are, this
statement is a bit harsh. It's not like he can look them up on
Wikipedia, or anything. Especially when you change the rules.
--- Christopher Heckman
.
User: "a_plutonium"

Title: Natural Numbers produce Riemannian geometry and thus no Algebras exist for them 04 Jan 2007 11:07:52 PM
Proginoskes wrote:

a_plutonium wrote:

Dik T. Winter wrote:
[...]

Pray define your addition, subtraction, multiplication and division so
that we can check your results.


Okay, I am going to put Dik Winter and Chris Heckman on the spot here.
What is the definition of add, subtract, multiply divide for the old
Peano Natural Numbers. Do you have to show how to mechanically add two
numbers like 99998 + 77359? Do you have to show how to mechanically
divide? Or subtract? Do you mechanically have to state in subtraction
that one number has to be larger than the other since there are no
negative Natural Number? Do you have to state that in division, that
there is no fractional number as endresult and you say a remainder of
such and such.
So, Dik and Chris, what is the formal definition of add, subtract,
multiply, divide for Natural Numbers of this set 0,1,2,3,...... I
remember my old book on Peano Axiomatics simply said *definition of
add* *definition of multiply*. So, what exactly is the definition of
add, multiply of the old Natural Numbers.

INFINITE INTEGERS operations defined:
addition-- define addition the same as any addition of Counting Numbers
where we do the same for infinite strings


(1) Since the Infinite Integers are created by "infinite adding of 1",
and 1 = ...0001, then can you guarantee that N+...0001 is the Infinite
Integer right after N?

For the time being I do not anticipate that the endless adding of 1
will be changed for it is the essence of the Natural Numbers. The
essence of the Reals is continousness. The essence of the Natural
Numbers is this quantized unit distance apart of endless adding of 1.
It may perhaps be modified and the reason I say that is because if you
endless add 1 you end up with a curve a circle or globe and so the unit
of 1 distance is on a globe and that is why it is so difficult to
imagine how .....000000s string ends up into a ....111111s string. With
Reals between 0 and 1 we do not question how 0.0000s blend into
0.1111s or then into 0.2222s for in Reals we simply say between any two
Reals is infinitely more. But in integers it is a process of endless
adding 1, with Reals between two Reals we fill up that gap with
infinitely many more.
Probably Math Induction will be thrown out. The number "0" will have a
predeccessor-- .....99999999.

(2) Fill in the blank:
...(a[3])(a[2])(a[1])(a[0]) + ...(b[3])(b[2])(b[1])(b[0]) =
______________,
where a[i] and b[i] are digits between 0 and 9 for all i.

multiplication-- define multiplication the same as any multiplication
of Counting Numbers where we start the multiplication between the first
digits of both numbers to be multiplied.


(1) What if there are infinitely many digits? That is, what should
...2222 * ..3333 be? Your definition doesn't cover this case.

This is simple to do. Mechanically we take 222 x 333 and find out it is
73926. Next we use the calculator for 2222x3333 = 7405926 Next we
calculate 22222x33333 And we begin to see that the first three digits
that emerges is .....926.
We simple do every multiplication as if the numbers were finite strings

(2) Is ...00011 * ...00011 equal to ...000121 (doing operations in base
3 or higher) or ...0001001 (doing operations in base 2)?

We drop bases. These Infinite Integers are base independent. Whatever
you find true for these Infinite Integers is true in all bases of them.
So that in the case of where Dik says
where an Infinite Integer ends with a 5 digit and is thus not a prime,
that number is not a prime no matter what base you represent these
infinite integers. So unlike the P-adics, I only have to work with
these base 10 Infinite Integers.
And good riddance to the p-adics because I want to find truth without
shuffling through bases.
......000011 x ....0000011 is simply the same as the old familar
11x11=121.
The Revolutionary key here is that the old mathematicians and believers
of old Natural Numbers thinks that endless adding of 1 will stop nicely
and that only Natural Numbers can have a ....00000n. What is a shock to
them is that if you endlessly add 1, it does not stop nicely but goes
on to form ....111111 then later ....22222 then much later ....99999999


division-- define as in Counting Numbers but if one number is larger
than the other then we cannot divide larger numbers into smaller ones.


(1) How do I tell if M is larger than N?

Which one of the two you can subtract tells you which is larger in most
cases. In the old Natural Numbers we automatically knew which of the
two numbers was larger and never really made a fuss over it. As well as
in division we never made a fuss about the fact that you can only
divide smaller number into a larger one.
As for irrational-infinite-integers sometimes you cannot tell. But this
is okay because we seldom know what digits there are way out on most
irrational numbers in the Reals.

(2) If you've defined addition and multiplication already, and they
satisfy the ring axioms, then you don't need a special definition for
division; you can just use the Division Theorem, which states that for
any nonnegative integers a,b, with b nonzero, that you can write

Well I do not know with all the changes that must take place whether an
Algebra can exist for the Natural Numbers= Infinite Integers. As I said
so often before that these numbers form a geometry that is curved not a
straight line for Reals. And so, can you have an Algebra existing on a
curved geometry? I think not. Betweenness fails in spherical curvature,
and so Algebra fails.
Here the important question arises: If you build a Number System such
as the Reals for the feature of continuity, no holes, but every point
occupied by a number then can you have only a Euclidean geometry. Now
the next important question is if you build another Number System,
Number System number #2 whose feature is quantized distance spacing of
every one of its members (unit apart). The deep question is that does
this requirement end up being a Spherical geometry (nonEuclidean
geometry (Riemannian geometry)).
So the deep question is -- you want continuity then you have to have
Reals and Algebra. But if you want a quantized Number System (Natural
Numbers) then they carve a geometry which has to be Riemannian geometry
and hence no Algebras are allowed.
So the Natural Numbers are going to end up having no Algebras.

Considering that only one person knows what Infinite Integers are, this
statement is a bit harsh. It's not like he can look them up on
Wikipedia, or anything. Especially when you change the rules.

--- Christopher Heckman

Dik is acting sarcastic and jokingly and not serious. So my jab was
warranted.
I am not changing the rules, I am discovering and learning more and
more as I go on. Sometimes we have to change direction and routes. I
gave the P-adics a decade of time to see if they were the Infinite
Integers, they are not, and so I do not change rules but find out they
are not.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.
User: "a_plutonium"

Title: Only two transcendental numbers exist in mathematics Re: Natural Numbers produce Riemannian geometry and thus no Algebras exist for them 05 Jan 2007 01:15:42 AM
a_plutonium wrote:
(snipped)


Here the important question arises: If you build a Number System such
as the Reals for the feature of continuity, no holes, but every point
occupied by a number then can you have only a Euclidean geometry. Now
the next important question is if you build another Number System,
Number System number #2 whose feature is quantized distance spacing of
every one of its members (unit apart). The deep question is that does
this requirement end up being a Spherical geometry (nonEuclidean
geometry (Riemannian geometry)).

So the deep question is -- you want continuity then you have to have
Reals and Algebra. But if you want a quantized Number System (Natural
Numbers) then they carve a geometry which has to be Riemannian geometry
and hence no Algebras are allowed.

Perhaps maybe the Infinite Integers would shed light on a question that
has bothered me since the early 1990s. If the world has only one type
of infinity, where Cantors arguments are trashcanned because the
Natural Numbers are equinumerous with the Reals. Then does the argument
that there exists an infinite number of transcendental numbers also is
trash. Personally, my intuition said in the 1990s that there are only
two transcendental numbers in existence, pi and e, discounting all the
nxpi or pi+n. Call them Root Transcendental. So does the world have
only two root-transcendental numbers in existence? That is the question
and can Infinite Integers help solve that question?
I do not know. But given the idea that the Natural Numbers as Infinite
Integers creates a Riemannian geometry such as the surface of a sphere
then pi would be at home in that Number System. Would it be a pi where
we lopp off the decimal point such as this Infinite Integer .....951413
? Now we do know the last number of Infinite Integers is ...999999 so
what is the circumference of the Infinite Integers? Of course the
circumference is ...999999 units long.
What would the diameter of the Natural Numbers = Infinite Integers,
thus, be? This is harder.
So let us assume that pi of the Natural Numbers is .....951413.
And now we know the formula of c = pi x diameter or c/pi = diameter
Which gives us ....9999999/ ....951413. Now I am going to have to
analyze this. Is it that infinite-integer-pi is that pi x pi =
......999999?
And then what about e as ....172? Could it be that infinite-integer-e
like pi has this relationship of e x e = .....999999?
So to speak idempotents of ....999999 for Infinite Integers? And the
only two idempotents and thus there exists 2 and only 2 transcendental
numbers in Infinite Integers and in Reals?
This has to be analyzed much more than this brief escapade. But I think
my intuition is correct that the world of Mathematics has only two
transcendental numbers and where I get that conviction is from physics
where there are only two parameters on an Atom Totality that have
importance,, the number of subshells and shells where plutonium has 22
subshells inside 7 shells of which 19 are occupied thus giving rise to
two special numbers of pi and e. So physics points to the likelihood
that there exists only two transcendental numbers.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.
User: "a_plutonium"

Title: Correction: only three Transcendental Roots-- e^(i*pi)=....99999 Re: Only two transcendental numbers exist in mathematics 05 Jan 2007 12:38:22 PM
Okay today I have a clearer picture of this. One of Cantor's fake
conclusions was that there is a larger set of infinity of
transcendental numbers then there was of rational numbers. When the
world of mathematics gets its senses together and has a modicum of
commonsense, it realizes that the world of mathematics has only one
type of infinity. And what that does to the axioms of mathematics is
adjust them so that Natural Numbers = Infinite Integers. In other words
the set that is the Reals is equinumerous to the set that is the
Natural Numbers. The world has one and only one type of Infinity and
not this nonsense of levels of infinity which someone called aleph 1
and aleph 2 and other assorted nonsense.
Infinity is simply the idea of "never ending" and this idea does not
have levels of never ending. There is one type of never ending, and
because mathematicians prior to myself did not have the commonsense to
craft the axioms of mathematics around the idea that the world has one
and only one type of infinity, they got themselves into trouble and
into a tangle of nonsense. More on this when I get to the cleaning up
and revamping of the Peano Axioms. Today I want to discuss the idea
that there are only 3 Root-Transcendentals. What I mean by root is
similar to the idea in mathematics of vector and scalar. Where you have
only 3 unit vectors call them A, B, C and where you have an infinity of
these unit vectors when you multiply or divide or add or subtract with
scalars upon those unit vectors. So you have an infinity resulting from
those unit vectors but only 3 such vectors in existence.
So I am going to argue that the world of mathematics has only 3
Root-Transcendental Numbers. And all other Transcendental Numbers are
some scalar of these Root-Transcendentals.
These 3 Root Transcendentals are pi, e, and i (the square root of -1).
Only here we amplify what the negative numbers are in reality.
The number (-1) is the Natural-Number = Infinite Integers of that of
.....9999999999
Now we look at the old equation of e^(pi*i) = -1
And we know that circumference/diameter = pi for a circle
And we know that the last and final Infinite Integer is ....999999 so
the circumference of all the Natural Numbers is ....99999 units long.
Now if we approx pi to be that of 3 then the diameter of all the
Natural-Numbers is .....33333333
So the old equation of e^(pi*i) = -1 is rewritten to be e^(pi*i) =
.....99999999
Now if we approx the known numbers in this equation e^(pi*i) =
.....99999999
we have e = ....828172 and pi = .....562951413 transposed into
Infinite Integers.
So we have all the numbers except for i. But we know that i is the
square root of -1 which in this case is ....99999.
So in my previous post I spoke of pi and e acting and behaving like
idempotents and this is true because the number i depends on
....9999999. So e, pi, i where i depends on ....99999 are multiplied in
such a way as to yield ....999999 and this is idempotent behaviour.
So, if I crudely start the calculations using 2.7 for e and using 3.14
for pi then what number do I need for i such that 2.7^ (3.14x i) =
.......9999999. The number I need for i in this case would decrease the
value of 3.14 to be slightly larger than 2, keeping in mind that the e
in Infinite Integers is ......828172
Since ....828172 x .....828172 approximates closely to that of
......9999999
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.
User: "hagman"

Title: Re: Correction: only three Transcendental Roots-- e^(i*pi)=....99999 Re: Only two transcendental numbers exist in mathematics 06 Jan 2007 06:11:33 AM
a_plutonium schrieb:

Infinity is simply the idea of "never ending" and this idea does not
have levels of never ending.

If "never ending" has no levels, why should "ending" have levels?

So I am going to argue that the world of mathematics has only 3
Root-Transcendental Numbers. And all other Transcendental Numbers are
some scalar of these Root-Transcendentals.

These 3 Root Transcendentals are pi, e, and i (the square root of -1).
Only here we amplify what the negative numbers are in reality.

Great, so i is one of the few transcendentals that are the root of a
polynomial.

The number (-1) is the Natural-Number = Infinite Integers of that of
....9999999999

Now we look at the old equation of e^(pi*i) = -1

And we know that circumference/diameter = pi for a circle

And we know that the last and final Infinite Integer is ....999999 so
the circumference of all the Natural Numbers is ....99999 units long.

Wouldn't it be exactly one unit longer than this?
A unit triangle with vertices labeled "0", "1", "2" has a circumference
of 3, not 2.

Now if we approx pi to be that of 3 then the diameter of all the
Natural-Numbers is .....33333333

In Indiana, maybe.

So the old equation of e^(pi*i) = -1 is rewritten to be e^(pi*i) =
....99999999

Now if we approx the known numbers in this equation e^(pi*i) =
....99999999
we have e = ....828172 and pi = .....562951413 transposed into
Infinite Integers.

If this "pi" is approximately 3, then AP seems to have the idea that 10
is small
and 100 is even smaller (the "theory" looks more 10adic from post to
post).

So we have all the numbers except for i. But we know that i is the
square root of -1 which in this case is ....99999.

Assume i*i = ...9999.
Then the last digit of i must be 3 or 7.
Assume i = .....d3 for some digit d, i.e. i=3+10*d+100*something.
Then i*i = 9 + 10*(6*d) + 100*whatever.
But 6*d is even, i.e. doesn't produce 9 as second digit.
Assume i = .....d7 for some digit d, i.e. i=7+10*d+100*something.
Then i*i = 49 + 10*(14*d) + 100*whatever
= 9 + 10*(4+4*d) + 100*yadayada.
Again, 4+4*d is even, i.e. doesn't produce 9 as second digit.

So, if I crudely start the calculations using 2.7 for e and using 3.14
for pi then what number do I need for i such that 2.7^ (3.14x i) =
......9999999. The number I need for i in this case would decrease the
value of 3.14 to be slightly larger than 2,

IIRC, "larger" has not yet been properly defined?

keeping in mind that the e
in Infinite Integers is ......828172

Since ....828172 x .....828172 approximates closely to that of
.....9999999

But
....828172 * ...828172 = ...861584
(Also, why should something like e^2=-1 hold?)
.
User: "Proginoskes"

Title: Re: Correction: only three Transcendental Roots-- e^(i*pi)=....99999 Re: Only two transcendental numbers exist in mathematics 06 Jan 2007 07:28:52 PM
hagman wrote:

a_plutonium schrieb:

Infinity is simply the idea of "never ending" and this idea does not
have levels of never ending.


If "never ending" has no levels, why should "ending" have levels?

So I am going to argue that the world of mathematics has only 3
Root-Transcendental Numbers. And all other Transcendental Numbers are
some scalar of these Root-Transcendentals.

These 3 Root Transcendentals are pi, e, and i (the square root of -1).
Only here we amplify what the negative numbers are in reality.


Great, so i is one of the few transcendentals that are the root of a
polynomial.

AP isn't using the word "transcendental" in the way that everyone else
does.
"Root" means that these numbers generate all of the Infinite Integers.

The number (-1) is the Natural-Number = Infinite Integers of that of
....9999999999

Now we look at the old equation of e^(pi*i) = -1

And we know that circumference/diameter = pi for a circle

And we know that the last and final Infinite Integer is ....999999 so
the circumference of all the Natural Numbers is ....99999 units long.


Wouldn't it be exactly one unit longer than this?
A unit triangle with vertices labeled "0", "1", "2" has a circumference
of 3, not 2.

Now if we approx pi to be that of 3 then the diameter of all the
Natural-Numbers is .....33333333


In Indiana, maybe.

So the old equation of e^(pi*i) = -1 is rewritten to be e^(pi*i) =
....99999999

Now if we approx the known numbers in this equation e^(pi*i) =
....99999999
we have e = ....828172 and pi = .....562951413 transposed into
Infinite Integers.


If this "pi" is approximately 3, then AP seems to have the idea that 10
is small
and 100 is even smaller (the "theory" looks more 10adic from post to
post).

So we have all the numbers except for i. But we know that i is the
square root of -1 which in this case is ....99999.


Assume i*i = ...9999.
Then the last digit of i must be 3 or 7.
Assume i = .....d3 for some digit d, i.e. i=3+10*d+100*something.
Then i*i = 9 + 10*(6*d) + 100*whatever.
But 6*d is even, i.e. doesn't produce 9 as second digit.
Assume i = .....d7 for some digit d, i.e. i=7+10*d+100*something.
Then i*i = 49 + 10*(14*d) + 100*whatever
= 9 + 10*(4+4*d) + 100*yadayada.
Again, 4+4*d is even, i.e. doesn't produce 9 as second digit.

I'm glad I'm not the one that has to tell AP that there is no square
root of -1 in the Infinite integers ...
--- Christopher Heckman

So, if I crudely start the calculations using 2.7 for e and using 3.14
for pi then what number do I need for i such that 2.7^ (3.14x i) =
......9999999. The number I need for i in this case would decrease the
value of 3.14 to be slightly larger than 2,


IIRC, "larger" has not yet been properly defined?

keeping in mind that the e
in Infinite Integers is ......828172

Since ....828172 x .....828172 approximates closely to that of
.....9999999


But

...828172 * ...828172 = ...861584

(Also, why should something like e^2=-1 hold?)

.


User: "a_plutonium"

Title: solving for i in e^(i*pi)=....99999 for Infinite Integers 05 Jan 2007 11:18:43 PM
a_plutonium wrote:
(snipped)


These 3 Root Transcendentals are pi, e, and i (the square root of -1).
Only here we amplify what the negative numbers are in reality.

The number (-1) is the Natural-Number = Infinite Integers of that of
....9999999999

Now we look at the old equation of e^(pi*i) = -1

And we know that circumference/diameter = pi for a circle

And we know that the last and final Infinite Integer is ....999999 so
the circumference of all the Natural Numbers is ....99999 units long.
Now if we approx pi to be that of 3 then the diameter of all the
Natural-Numbers is .....33333333

So the old equation of e^(pi*i) = -1 is rewritten to be e^(pi*i) =
....99999999

Now if we approx the known numbers in this equation e^(pi*i) =
....99999999
we have e = ....828172 and pi = .....562951413 transposed into
Infinite Integers.
So we have all the numbers except for i. But we know that i is the
square root of -1 which in this case is ....99999.

So in my previous post I spoke of pi and e acting and behaving like
idempotents and this is true because the number i depends on
...9999999. So e, pi, i where i depends on ....99999 are multiplied in
such a way as to yield ....999999 and this is idempotent behaviour.

So, if I crudely start the calculations using 2.7 for e and using 3.14
for pi then what number do I need for i such that 2.7^ (3.14x i) =
......9999999. The number I need for i in this case would decrease the
value of 3.14 to be slightly larger than 2, keeping in mind that the e
in Infinite Integers is ......828172

Since ....828172 x .....828172 approximates closely to that of
.....9999999

Okay, I was thinking that perhaps I can do the same for Reals but it
looks like there is trouble with getting Doubly Infinites of a number
like ....9999999.1544..... and that of 0.1544.... as a negative Real.
The Infinite Integers form a circle or sphere but the Reals form a
straight line, but if I inject ...99999 to the negative Reals could I
get the Reals to also form a circle? And although the Infinite Integers
would have holes or gaps between every number, these Reals form a
circle with numbers filling in all the holes. So Infinite Integers form
a circle with holes between numbers and these Doubly Infinite Reals
form a circle with all the gaps filled by numbers.
That is what I was thinking but it seems to not work.
What I do not like about the Infinite Integers solving e^(pi*i)
=....99999999 is that it uses pi and e as morphed numbers of ....951413
and .....828172
Now if we take e to be 3 and pi as 3, then what number as an Infinite
Integer would i be in order that we have 3^(3*i) = .....999999 Here I
have the trouble of making sense of an infinite exponentiation.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
.
User: "a_plutonium"

Title: reformulating the NP Conjecture as to whether numbers are fixed or growing 07 Jan 2007 12:59:30 PM
Proginoskes wrote:

a_plutonium wrote:

(snipped)

My idea of transcendental number is that it is a number that is growing
and not "fixed" whereas an algebraic number is fixed and thus can be
completely known.


This closely parallels the concept of a "computable number":
http://en.wikipedia.org/wiki/Computable_number

And since most numbers are fixed numbers


Maybe this leads into a "clear statement" of the NP conjecture and
involves what I call fixed numbers as algebraic and not fixed numbers
such as pi and e. Since all numbers derive from the Physical Universe
in action-- the Atom Totality-- and because this Cosmos is growing from
22 subshells inside 7 shells of which only 19 are occupied forces all
mathematics in the cosmos to register by those mathematicians a value
of pi as the irrational number close to 22/7 and the number value for e
as the irrational number close to 19/7. Since these two numbers are
not fixed and growing is the reason they are Transcendental.
So, if the NP conjecture can be stated in terms of how many numbers are
transcendental as compared to how many are algebraic. Well, I have a
proof for that in the works. Because if we define Vector Numbers as
compared to Scalar Numbers that there exists 2 and only 2
Transcendental Numbers which are e and pi. There is an infinitude of
Scalar Numbers such as pi/2 or pi^2. But there is only two existing
root-transcendental numbers.

... except that most numbers are not computable. So the analogy breaks
down.

Well if this NP Conjecture can be reassembled to be about
transcendental versus algebraic, then there are 2 and only 2 Root
Transcendental Numbers or 2 and only 2 Vector Transcendental Numbers
and they are pi and e. There is an infinite supply of Transcendental
Scalar Numbers such as e/5 or e^3

there should
be a small finite set of transcendentals such as these three--- pi, e,
and perhaps i. It maybe that there exists just two transcendentals in
all of mathematics of pi and e.


Or maybe only one?

Let me explain what I mean by fixed. The square root of 2 is irrational
but not transcendental and we can fetch any digit out we want. But a
number like pi or e are numbers that exist in the Universe at large for
which the Universe itself has not fully made that number. The Universe
is a spherical shaped object which has a circumference that is
approximated by 22 subshells inside an approximate 7 shells, so that
22/7 is a rough approximation of what the number pi in mathematics will
be, but because these subshells and shells are growing and not fixed,
then this number pi is different from all the other numbers that are
fixed in the Universe. Same thing for e, only it is approx 19 subshells
occupied inside of approx 7 shells.

So I think i is algebraic and I just loosely group i with e and pi.

I think these three numbers form vectors in the equation e^(pi*i) =-1.
And this is important in that all the numbers can be derived in two
manners. One we can do a process of infinitely adding 1 and get
0,1,2,3,....., .....999999 Or we can get all these same numbers by
defining e, pi and i and then using this equation as a process of
generating all the numbers. So I see e, pi, i as vectors for which all
the numbers are generated as a scalar product of these three vectors.
And the same holds true for Reals.

So I will continue to use the old terms simply because nearly all of
present day mathematics is going to have to be changed in a very big
way.


When responding to posts, AP should keep this in mind; that there are
big changes, and he will have to communicate these on a fundamental
level.

For instance, how will computaitons involving Infinite Integers be
taught in elementary school?

--- Christopher Heckman

Well throughout High School if any students are taught physics, it is
generally all Newtonian Physics, so the old Counting Numbers are
reasonable teachable material.
For those sharp High School students will begin to pick up on Quantum
Mechanics and how it differs from Newtonian. Likewise, those few
brilliant High School students who want the real truth behind
mathematics will wonder away from the Peano old math and charlatan
Cantor and Godel and begin learning 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
.
User: "Proginoskes"

Title: Major idea of P and NP given: Was Re: reformulating the NP Conjecture as to whether numbers are fixed or growing 08 Jan 2007 12:33:42 AM
I hope AP reads this; I put a good deal of effort into the explanation,
since it's clear that he doesn't understand what the NP Conjecture is.
a_plutonium wrote:

Proginoskes wrote:

a_plutonium wrote:

(snipped)

My idea of transcendental number is that it is a number that is growing
and not "fixed" whereas an algebraic number is fixed and thus can be
completely known.


This closely parallels the concept of a "computable number":
http://en.wikipedia.org/wiki/Computable_number

And since most numbers are fixed numbers



Maybe this leads into a "clear statement" of the NP conjecture and
involves what I call fixed numbers as algebraic and not fixed numbers
such as pi and e.

Probably not.
Let's start off the P vs. NP question with the following variation,
which takes place on the set of postive integers:
GIVEN: A subset S of the natural numbers, and a natural number n
(expressed in binary, to be specific).
OUTPUT: A true statement of whether n is in S.
Mathematically, this may be a trivial problem to solve. But now suppose
that we want to write a (deterministic; i.e., nonrandom) program to
decide whether n is in S or not.
For some subsets S, this will be a short program. For instance, if S =
{1, 3, 5, 7, 9, ...}, you can check the final bit and return YES if it
is 1, and NO otherwise. If S is the set of prime numbers, some more
work is required; certainly there is a finite bound, though.
P is a set whose elements are sets S which have an algorithm which runs
"quickly", namely in at most a constant time (log n)^k computations,
where k is a constant. (This is known as "polynomial time"; hence P
for Polynomials.) The set of odd numbers is in P; all sets with exactly
3 elements are in P; and the set of primes is in P. There are sets
which are not in P.
NP is the same sort of object as P: a set of subsets of the natural
numbers. However, whether a set S is in NP depends on a different type
of program, called nondeterministic programs. These programs, when
started, will generate a natural number R "at random" which isn't too
big, and allow you to use this number as part of the algorithm. The
"random" number generator has a strange property: If there is a number
R which results in an output of YES, then the "random" number generator
will give you one of those numbers.
Then NP will consist of all sets S such that there is a
nondeterministic program which runs in "polynomial time". (Again, this
is a polynomial with respect to log n; the P stand for polynomial, and
the N stands for Nondeterministic.)
For instance, let S be the set of all composite integers. Then the
following is a nondeterministic program:
(1) Generate R
(2) Return NO if R = 1 or R >= n.
(2) Divide n by R
(3) Return YES if the remainder is 0; otherwise, return NO.
(If n is prime, then this program will return NO in any case; if n is
composite, then the right choice of R will give an answer of YES. The
"randomization" guarantees that, if n is composite, that a nontrivial
factor of n will be chosen.) This program will run in polynomial time,
since division can be done in polynomial time.
Clearly, every set in P is also in NP. (Your nondeterministic program
can just ignore the number R given to you.) The P vs. NP problem is
whether every set in NP is in P, which boils down to whether you can
similate "randomization" in polynomial time.
There are sets S in NP called "NP-complete problems". These sets have
the property: "If S is in P, then P = NP." Many many many NP-complete
problems are known (usually expressed in terms of Graph Theory, where
they have meaningful descriptions).
--- Christopher Heckman

Since all numbers derive from the Physical Universe
in action-- the Atom Totality-- and because this Cosmos is growing from
22 subshells inside 7 shells of which only 19 are occupied forces all
mathematics in the cosmos to register by those mathematicians a value
of pi as the irrational number close to 22/7 and the number value for e
as the irrational number close to 19/7. Since these two numbers are
not fixed and growing is the reason they are Transcendental.

So, if the NP conjecture can be stated in terms of how many numbers are
transcendental as compared to how many are algebraic. Well, I have a
proof for that in the works. Because if we define Vector Numbers as
compared to Scalar Numbers that there exists 2 and only 2
Transcendental Numbers which are e and pi. There is an infinitude of
Scalar Numbers such as pi/2 or pi^2. But there is only two existing
root-transcendental numbers.


... except that most numbers are not computable. So the analogy breaks
down.


Well if this NP Conjecture can be reassembled to be about
transcendental versus algebraic, then there are 2 and only 2 Root
Transcendental Numbers or 2 and only 2 Vector Transcendental Numbers
and they are pi and e. There is an infinite supply of Transcendental
Scalar Numbers such as e/5 or e^3



there should
be a small finite set of transcendentals such as these three--- pi, e,
and perhaps i. It maybe that there exists just two transcendentals in
all of mathematics of pi and e.


Or maybe only one?

Let me explain what I mean by fixed. The square root of 2 is irrational
but not transcendental and we can fetch any digit out we want. But a
number like pi or e are numbers that exist in the Universe at large for
which the Universe itself has not fully made that number. The Universe
is a spherical shaped object which has a circumference that is
approximated by 22 subshells inside an approximate 7 shells, so that
22/7 is a rough approximation of what the number pi in mathematics will
be, but because these subshells and shells are growing and not fixed,
then this number pi is different from all the other numbers that are
fixed in the Universe. Same thing for e, only it is approx 19 subshells
occupied inside of approx 7 shells.

So I think i is algebraic and I just loosely group i with e and pi.

I think these three numbers form vectors in the equation e^(pi*i) =-1.
And this is important in that all the numbers can be derived in two
manners. One we can do a process of infinitely adding 1 and get
0,1,2,3,....., .....999999 Or we can get all these same numbers by
defining e, pi and i and then using this equation as a process of
generating all the numbers. So I see e, pi, i as vectors for which all
the numbers are generated as a scalar product of these three vectors.
And the same holds true for Reals.

So I will continue to use the old terms simply because nearly all of
present day mathematics is going to have to be changed in a very big
way.


When responding to posts, AP should keep this in mind; that there are
big changes, and he will have to communicate these on a fundamental
level.

For instance, how will computaitons involving Infinite Integers be
taught in elementary school?



Well throughout High School if any students are taught physics, it is
generally all Newtonian Physics, so the old Counting Numbers are
reasonable teachable material.

For those sharp High School students will begin to pick up on Quantum
Mechanics and how it differs from Newtonian. Likewise, those few
brilliant High School students who want the real truth behind
mathematics will wonder away from the Peano old math and charlatan
Cantor and Godel and begin learning 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

.
User: "a_plutonium"

Title: the Solution that solves the NP Problem of mathematics 08 Jan 2007 01:50:11 PM