counterexamples to the Riemann Hypothesis Re: Proof of most of the World's unsolved Math Problems due to the fact that Natural Numbers are these 0,1,2,3, ...., ...999999
a_plutonium
a_plutonium wrote:
> In this long running thread I will prove these unsolved Math problems:
> (1) Riemann Hypothesis
> (2) Fermat's Last Theorem
> (3) generalized FLT (known as Beals conjecture)
> (4) Goldbach Conjecture
> (5) Infinitude of Twin Primes
> (6) Infinitude of 4-primes, 6-primes, 8-primes,....
> (7) Infinitude of Perfect Numbers
> (8) Infinitude of primes of form (2^n) +1
> (9) the number 1 is the only odd perfect number
>
> I will prove those above listed 1-9 by the key fact that the
> Natural-Numbers are this set:
>
> 0 , 1, 2, 3, 4, 5, 6, 7 , 8 , 9 , 10 , 11, ........., .....9999996,
> ....999997, ......999998, .....9999999
>
> That is new and novel in human history. I discovered that fact around
> 1991. I spent the next decade seeing if the p-adics was this new set. I
> concluded recently it was not. The p-adics and the Surreal numbers are
> mathematical novelties but are not the foundation of Mathematics. The
> foundation has two sets of numbers--- Reals and Natural-Numbers.
Circa 1991-1993 I was under the impression that Natural Numbers were
this set:
0,1,2,3,4,5,6,7,8, 9, 10, 11, .....00000000n, .....000000n+1, ........
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.
Because these are the real and true Natural Numbers, that all
mathematical statements saying "For all Natural Numbers" could not be
proven, or if a proof was given it had a fatal flaw. The Riemann
Hypothesis and Goldbach and Fermat's Last Theorem could never be
proven, nor could the Infinitude of Twin Primes and the thousands of
Erdos problems in Number theory. None could be proven because they were
using 1/10 of the full deck of Natural Numbers, as an analogy to
playing cards can you play bridge or poker with only 5 cards.
I did the below two alleged proofs of Riemann Hypothesis under the old
delusion of Natural Numbers were Finite Integers. This is what I said.
TWO PROOFS OF THE RIEMANN HYPOTHESIS
PROOFS: Two proofs of the Riemann Hypothesis follows as A
and B.
Proof (A) is a geometrical proof. It was proved that the Riemann
Hypothesis is equivalent to the following-- the Moebius function mu of
x, m(x), and adding-up the values of m(x) for all n less than or equal
to N giving M(N). Then M(N) grows no faster than a constant multiple k
of (N^1/2)(N^E) as N goes to infinity (E is arbitrary but greater than
0). Figure1, by setting-up a logarithmic spiral in a rectangle of
whirling squares where the squares are the sequences:
1,1,2,3,5,8,13,21,34,55,89, . . . 2,2,4,6,10,16,26, . . .
3,3,6,9,15,24,39, . . . then every number appears in at least one of
these sequences because every number will start a sequence. Since all
numbers are represented uniquely by prime factors (the unique prime
factorization theorem or called the fundamental theorem of arithmetic)
and The Prime Number Theorem: the distribution of prime numbers is
governed by a logarithmic function, where (An/n)/(1/Ln of n) tends to 1
as n increases, where An denotes the number of primes below the
positive integer n, and where An/n is called the density of the primes
in the first n positive integers. The density of the primes, An/n, is
approximated by 1/(Ln of n), and as n increases, the approximation gets
better. The distribution of prime numbers is governed by a
logarithmic function where these two math concepts-- one of prime
numbers, and the other, logarithms seem unconnected at first
appearance, but in reality they are totally connected. Geometrically,
the logarithmic spiral exhausts every positive integer, see figure 1.
The area of the rectangles containing the logarithmic spiral is always
greater, since the spiral is always inside the rectangles. Thus the
Moebius function k (N^1/2)(N^E) is satisfied since the area of the
logarithmic spiral is less than the rectangle whose area represents the
number N, and whose sides represent its factors. The area of a
logarithmic spiral is represented by A=(r)(e^(Hj)) , and so depending
on where the point of origin for the spiral is taken rsubO determines
k, and depending on the value of H, H determines the E value for N,
when H=0 then the curve is a circle. The logarithmic spiral inside
rectangles of whirling squares implies that for any number N then N^1/2
is the limit of the factors for N, for example, given the number 28,
then 28^1/2 = 5.2915. . and so looking for the factors of 28, it is
useless to try beyond 5 because the factors repeat, 4x7 then repeats as
7x4. But if the Moebius function was false then there must exist a
number M such that M^1/2 is not the limit of the factors for M and the
spiral is outside of the square, which is impossible, hence the Moebius
function is true. Therefore the Riemann Hypothesis is proved. Q.E.D.
My second proof (B) of the Riemann Hypothesis uses a reductio
ad absurdum argument. Euler proved that a formula encoding the
multiplication of primes was equal to the zeta function. Euler's
formula in complex variable form is as follows:
(1/(1-(1/(2^c))))x(1/(1-(1/(3^c))))x(1/(1-(1/(5^c))))x(1/(1-(1/(7^c))))x
(1/(1-(1/(11^c))))x . . . , where c is a complex variable, c=u+iv. The
Riemann zeta function is as follows. Re(c) =
1+(1/(2^c))+(1/(3^c))+(1/(4^c))+. . . , where c is a complex variable,
c=u+iv. Euler's formula involves multiplication of terms and the
Riemann zeta function involves addition of terms of a sequence. Taking
Re(c) > 0, suppose the Riemann Hypothesis is false then there is a 0
such that Re(c)=0 and c not equal 1/2 +iy, which implies there is
another 0 which is not on the 1/2 real line. Which means another real
number other than 1/2 works as an exponent resulting in a zero for the
Riemann zeta function, and a zero in the Euler formula. Thus, Riemann
zeta function subtract Euler formula must equal zero. This implies for
any other real number exponent, either rational or irrational numbers,
such as for example the rational exponents: 1/3,1/4,1/5, . . . (Note:
any other exponent y/x , where y and x are Real numbers and where the
Real number of A^(y/x) such that y not equal 1, immediately transforms
to a number A^y(1/x), so that exponents with a 1 in the numerator
entail all of the Real exponents). To make clear of the above, for
example, 2^2/3 is 4^1/3. So then back to the proof. Then for exponent
1/3 there has to exist a number M not equal 0 where (M+M+M)^1/M =
(MXMXM)^1/M = M. Then for exponent 1/4 there has to exist a number M
not equal 0 where (M+M+M+M)^1/M = (MXMXMXM)^1/M = M, and so on.
Including the infinite number of cases where the x denominator is
irrational are impossible. Only the real number 1/2 works since 2 does
not equal 0, and (2+2)^1/2 = (2X2)^1/2 = 2, and so (2+2)^1/2
- (2X2)^1/2 = 0. In all of Reals and the Complex numbers, 2 is the
only number N which has the encoding ((N+N)^1/N) = ((NxN)^1/N) = N.
Unlike 0, the number 2, its sum equals its product and where the sum
and product is a new number 4. If RH were false, then another number
other than 2 would satisfy a generalized encoding formula ((N+N)^1/N)
= ((NxN)^1/N) = N. False, hence the proof. QED
--------------------------------------------------------------------
I think the above was a valiant attempt. But armed with my new
knowledge that the Natural Numbers are this very larger set that
stretches up to Infinite Integers and ends with the number ....9999999.
This is rather keen and beautiful for we know very much about the start
of the Natural Numbers and the end of the Natural Numbers but we know
very little about that huge center. We do not know how the Natural
Numbers that end in 000s starts into those that end in 1111s. In other
words our minds are very challenged as to see how they go from those
ending in 000s to those ending in 111s or ending later on in 2222s and
later on ending in 33333s. How do they bridge one another? And do we
have Transcendental Natural Numbers such as pi and e in the form of
Natural Numbers or the 5-adic square root of -1. So we have new
challenges in the future.
But the counterexamples of Riemann Hypothesis is any one of these
nonending in 0000s. The numbers ....9999997 or ......13121110987654321
defy the Riemann Hypothesis. The Natural Numbers do not lie or do not
form a straight line as the Riemann Hypothesis implies. The Natural
Numbers have a curve to them as if they spun around in a circle and
come back to the starting point of 0 and then 1, 2, 3. The Riemann
Hypothesis implies the Natural Numbers lie on a 1/2 Real line that is a
straight line. But the Natural Numbers spin around to ....99999. In
10-adics the number ...99999 is considered to be -1 and the number
....999998 is considered to be -2.
So any of these Natural Numbers that does not end in 000s is a
counterexample to the Riemann Hypothesis.
Now I do not know the many theorems in mathematics that relies on a
Riemann Hypothesis being true. But I am confident that if one goes back
and injects Natural Numbers = Infinite Integers that all of those
problems will be settled.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
MuTsun Tsai
> 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.
Before you continue with such nonsense, please multiply your so-called
.....9999 by .....9999 and justify your result.
a_plutonium
The fact that these numbers exist is the most important point about
them. Not that we have a well formed algebra surrounding these
numbers. As future time goes by, we arrive at well formed operations.
This is one of the troubles of modern day mathematics in that we focus
solely on the algebras. We have dressed up the p-adics so much that
they have become simply just another form of Reals.
To multiply ...99999 by ...9999 maybe what Alexander Abian suggested in
1990s in that we transpose them into Reals of 0.9999.... x 0.9999....
which is 1x1 and thus is 1.
To multiply ...333333 by ....000003 gives ......99999999 and wish all
were as easy as this.
But I want to maintain the operations of add, subtract, multiply,
divide as strictly close to that of what is done for "finite integers"
as much as possible. So that we have 99x99 is 9801 and we have 999x999
= 998001 and we have 9999x9999= 99980001 so the picture is coming out
that this .....9999999 x .....9999999 is going to be a number that
looks like this .....00000001 which agrees with Abian.
Proginoskes
a_plutonium wrote:
> MuTsun Tsai wrote:
> > > 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.
But then what is ...9999 + 1? Normal 10-adic addition suggests it is
...000, but then AP has a set of numbers which doesn't pass Peano's
Axioms (0 is the successor of something), so they do not represent the
actual natural numbers after all.
(BTW, all of this was mentioned months ago. I'm sure AP collected the
posts; now he needs to go back and actually read them.)
> > Before you continue with such nonsense, please multiply your so-called
> > .....9999 by .....9999 and justify your result.
>
> The fact that these numbers exist is the most important point about
> them. Not that we have a well formed algebra surrounding these
> numbers. As future time goes by, we arrive at well formed operations.
> This is one of the troubles of modern day mathematics in that we focus
> solely on the algebras. We have dressed up the p-adics so much that
> they have become simply just another form of Reals.
>
> To multiply ...99999 by ...9999 maybe what Alexander Abian suggested in
> 1990s in that we transpose them into Reals of 0.9999.... x 0.9999....
> which is 1x1 and thus is 1.
>
> To multiply ...333333 by ....000003 gives ......99999999 and wish all
> were as easy as this.
>
> But I want to maintain the operations of add, subtract, multiply,
> divide as strictly close to that of what is done for "finite integers"
> as much as possible. So that we have 99x99 is 9801 and we have 999x999
> = 998001 and we have 9999x9999= 99980001 so the picture is coming out
> that this .....9999999 x .....9999999 is going to be a number that
> looks like this .....00000001 which agrees with Abian.
So is AP using the standard + and * of 10-adics, or not? Who knows?
Besides, "rigor" isn't in AP's dictionary, which is why he will never
get any results published; his proofs, when right, just aren't up to
the standards that everyone else has to follow.
--- Christopher Heckman
a_plutonium
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.
Ken Quirici
> MuTsun Tsai wrote:
> > > 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.
> >
> > Before you continue with such nonsense, please multiply your so-called
> > .....9999 by .....9999 and justify your result.
>
> The fact that these numbers exist is the most important point about
> them. Not that we have a well formed algebra surrounding these
> numbers. As future time goes by, we arrive at well formed operations.
> This is one of the troubles of modern day mathematics in that we focus
> solely on the algebras. We have dressed up the p-adics so much that
> they have become simply just another form of Reals.
>
> To multiply ...99999 by ...9999 maybe what Alexander Abian suggested in
> 1990s in that we transpose them into Reals of 0.9999.... x 0.9999....
> which is 1x1 and thus is 1.
>
I believe this result is '1' is correct, in the limit, just as AP's
example.
I tried a long by-hand multiplication for a few digits
at the far right:
999999999999*
999999999999
---------------------
999999999991
99999999991
9999999991
999999991
99999991
9999991
..........
----------------------
0000000000001
Note that in each column from the right, the 'carry' from the
addition just to the right is exactly as many 9's as the column
contains except the bottom one, and the carry zeroes these 9's,
and the bottom '9' is zeroed by the 1 under it.
so the first column in the sum from the right is 1
the second column from the right is 9 + 1 = 0
the third column from the right is 9 + 9 + 1 + carry 1 - so the carry
zeroes the one 9 that isn't zeroed by the bottom 1
the fourth column from the right is 9 + 9 + 9 + 1 + carry 2 - so the
carry zeroes the two 9's that aren't zerod by the bottom 1
etc.
I'm not using the techique, however, that AP suggests, but doing
what seems like the obvious multiplication algorithm. But it
may not be the RIGHT one.
Ken
Proginoskes
This is the standard way to multiply p-adics. Since the "usual +" and
"usual *" form a ring, you know that
(-x)*(-y) = x*y, which simplifies the calculation:
...999 * ...999 = (- ...0001 * -...0001) = ...0001 * ...0001 = ...0001.
AP either doesn't know or doesn't care how he wants to define + and *.
But calculations like the above are a lot easier if you can use the
ring structure.
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> [...]
> TWO PROOFS OF THE RIEMANN HYPOTHESIS
> My second proof (B) of the Riemann Hypothesis uses a reductio
> ad absurdum argument. Euler proved that a formula encoding the
> multiplication of primes was equal to the zeta function. Euler's
> formula in complex variable form is as follows:
> (1/(1-(1/(2^c))))x(1/(1-(1/(3^c))))x(1/(1-(1/(5^c))))x(1/(1-(1/(7^c))))x
>
> (1/(1-(1/(11^c))))x . . . , where c is a complex variable, c=u+iv. The
> Riemann zeta function is as follows. Re(c) =
> 1+(1/(2^c))+(1/(3^c))+(1/(4^c))+. . . , where c is a complex variable,
Actually, this formula for zeta(c) (not Re(c)) is only valid if u > 1.
--- Christopher Heckman
a_plutonium
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.
Proginoskes
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.
a_plutonium
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.
Proginoskes
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
hagman
a_plutonium schrieb:
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)
hagman
Proginoskes schrieb:
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).
a_plutonium
Proginoskes wrote:
> a_plutonium wrote:
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
a_plutonium
a_plutonium wrote:
> 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 I do not know if anyone has proven formally that when you take an
Irrational number and change a few digits that the new number is still
irrational. Whether it is a Irrational Real or an Irrational Infinite
Integer. So that if someone takes square root of 2 = 1.41421.... and
say lops off the 1 leaving 0.41421..... or lops off the 1.41 leaving
0.421.... whether these are still irrational or whether someone adjoins
extra digits say like adjoining 911 to 1.41421.... as in this
0.911141421....... whether that still remains irrational. To me the
answer is obviously yes that the alteration of a finite string of the
irrational number does not affect its overall characteristic of being
irrational. And I think someone has already proven this idea.
Aluminium Holocene Holodeck Zoroaster
AP is really a charlatan. I mean,
it's in the very beginning of every book on p-adics
-- that part that I was able to comprhend, thus far --
that the arhcimedean-valued integers a infinity-adics;
maybe, taht's where he got the idea.
he seems to believe that supplying "potential areas
of research" in a continually updated list, constitutes a theory
of nothing, whereas it's probably just his inability
to express himself in English (or any written format,
whatsoever ... F-999?).
I wouldn't say that this applies to his cousin,
Governeurateur Terminator, since
that guy came to teh USA at age nineteen --
I think, his accent is entirely cultivated, and
that his main problem is his financial sponsors
(Rothschild, George Schultz, Warren Allyoucaneat Buffet,
junkbondtraders etc ad vomitorium).
> This is the standard way to multiply p-adics. Since the "usual +" and
> "usual *" form a ring, you know that
> (-x)*(-y) = x*y, which simplifies the calculation:
>
> ...999 * ...999 = (- ...0001 * -...0001) = ...0001 * ...0001 = ...0001.
thus:
how is that?
> There are cases with complex a,b such that
> sqrt(a/b) not equal to sqrt(a) / sqrt(b) ...
thus:
you have to use tripolar coordinates, though. I've always
called it tetrahedronometry, although, a la the study of trigona,
it can also be called tetragonometry. an alternative
that I've espoused, was inspired by Bucky Fuller,
in a _Posthumous_ publication written by his adjuvant,
when he finally started to attend to teh duals
of the trigonated polyhedra; he used the word,
"polyvertexion," but I prefer polyasteron.
that also dystinguished it from Peter Schoute's polygonometry,
which is just an old, "n-D" thing,
which Coxeter told me of, when I called him up at his home
in Toronto -- correcting me for asking if he knew
of tetrahedronometry!
> > Basically my question here is: what's 4-gonometry.
>
> May be the tetrahedronometry.It should include dihedrals and trihedral
> "solid" angles.Unfortunately, this 3D counterpart did not develop so
> well or used so much like plane trigonometry.It could have involved
> quaternions.
thus:
unfortunately, the neocons & jihadists are sliding us
into Sudan, a dried-up quagmire (British sand,
viz Sandhurst, EMI et al ad vomitorium Seargent Peppers).
also note that China sent in about 10K workers
for the pipeline consortium with Canada, and
it has a larger--percentagewise?-- number of Muslims
(not 100%, though, as in Darfur).
>from the beginning of 2002 to the end of 2005, the number of armed conflicts being
waged around the world shrank 15% from 66 to 56. By far the
greatest decline was in
sub-Saharan Africa.
thus:
I'm sure that I read about the quantized galactic redshifts
in a mainstream journal ...
unless it was the LaRouchiac one!
http://www.21stcenturysciencetech.com/
thus quoth:
http://chephip.free.fr/pbg_en/sol143.html
thus quoth:
"The Universe has as many different centers as there are living
beings in it." - Alexander Solzhenitsyn
thus:
Harry Potter-affiliated stuff,
in general -- like both Iraq wars).
this just in:
yesterday's (Tues,. Nov.15) *UCLA Daily Bruin* finally noted that
darfur is entirely Muslim, though downplaying it AMAP.
thus:
Dick Cheeny, Don Rumsfeld and Osama bin Latin form a mission
to Darfur, to prevent a war instead of to start one:
if Darfur is "100% Muslim," then
what's really going on, there?
is it just aother British Quag for USA soldiers to get bogged
into, with Iran, Iraq, Afghanistan et al ad vomitorium,
under auspices of the UN and NATO?
why won't the Bruin publish the fact of Islam on the ground,
therein?
thus:
Why doesn't the [UCLA Daily] Bruin report that
Darfur's populace is "100%" Muslim,
according to the DAC's sponsor,
Terry Saunders?...
"99%" was the figure given
by Brian Steidle, when I finally found
him at the Hammer, after everyone else
had left (he, his friend & I were the
very last to leave!)...
What could it possibly mean?
--The Other Side (if it exists)
Aluminium Holocene Holodeck Zoroaster
it has been proven, that adding or subbtracting a rational
from an irrational (secondroot of 2, minus 1 e.g.) is generally
irrational. in other words, if you don't affect the digital endings
of beginnings, to use Munk's terminology. however,
what if you just change, delete or add digits
in the middle of the sequence;
could it be made rational, or just transcendental (or
possibly surreal) ??
> Now I do not know if anyone has proven formally that when you take an
> Irrational number and change a few digits that the new number is still
> irrational. Whether it is a Irrational Real or an Irrational Infinite
> Integer. So that if someone takes square root of 2 = 1.41421.... and
> say lops off the 1 leaving 0.41421..... or lops off the 1.41 leaving
> 0.421.... whether these are still irrational or whether someone adjoins
> extra digits say like adjoining 911 to 1.41421.... as in this
> 0.911141421....... whether that still remains irrational. To me the
> answer is obviously yes that the alteration of a finite string of the
> irrational number does not affect its overall characteristic of being
> irrational. And I think someone has already proven this idea.
also from your reference:
The first papers in which this startling new evidence was presented
were not warmly embraced by the astronomical community. Indeed, an
article in the Astrophysical Journal carried a rare note from the
editor pointing out that the referees "neither could find obvious
errors with the analysis nor felt that they could enthusiastically
endorse publication." Recognizing the far-reaching cosmological
implications of the single-galaxy results, and undaunted by criticism
from those still favoring the conventional view, the analysis was
extended to pairs of galaxies.
Two galaxies physically associated with one another offer the ideal
test for redshift quantization; they represent the simplist possible
system. According to conventional dynamics, the two objects are in
orbital motion about each other. Therefore, any difference in redshift
between the galaxies in a pair should merely reflect the difference in
their orbital velocities along the same line of sight. If we observe
many pairs covering a wide range of viewing angles and orbital
geometries, the expected distribution of redshift differences should be
a smooth curve. In other words, if redshift is solely a Doppler effect,
then the differences between the measured values for members of pairs
should show no jumps.
But....
> --- quoting from http://www.cs.unc.edu/~plaisted/ce/redshift.html
thus:
could you, please,
attach an edited version of this thread
to all of JSH's future topics,
to be applied automatically after midnight,
you local time?
> > > In the algebraics one can have
> > >non-unit integers arbitrarily close to 1 (a^(1/n) for any integer a and
> > >any positive integer n),
>
>
> I suspect that what you are really groping for is the degree of the
> algebraic integers. Given an algebraic integer (or in fact, an
> algebraic number) a, we define its "degree over Q" to be the degree of
> the extension Q(a)/Q, which is equivalent to the degree of the monic
> irreducible polynomial for a over Q.
thus:
any moron could come-up with the Fibonacci series;
stupid people, however, is another story. (of course,
the internal evidence that Fermat made no known mistakes,
sort-of leads one to believe that
it wasn't his last theorem, at all .-)
> You might as well say that people didn't understand the logic behind
> Fermat's Last Theorem - or the Russian guy (I don't know his name) who has
thus:
you should see if you can find a copy
of Charles B. Officer's book about the unmitigated hypothesis
that all craters are due to impacts, which is really
a reflection of the hegemony of the Alvarez et al pseudo-
finding about the K-T boundary. mostly remaindered, and
he gets quite worked-up in his prose, but, hey. (also,
find the *first* edition of _The Moon_,
by that former British Royal Astronomer, and compare it
with his second edition, on this subject --
he makes a little joke out of it....
beautifully said, though
about our non-programme d'espace;
soon, a thousand points of light will be on the moon,
all representing other countries. I mean,
why do you think that *yehT* whacked KFJ?
> http://en.wikipedia.org/wiki/Sudbury_Basin
> We do know for sure the effects of a major
> meteor splat on the moon from our experience
> in the Sudbury Basin. What the heck is a
> geological study of Copernicus to yield ?
> Awsome is an understatement, that's
> "hands-on" and available right now....
> "insane"?, well perhaps I would argue it's insane to
> bypass lunar geology ((Selinology?)) that has been
> collecting data for billions of years and is there now,
> rather than chasing about based on speculations
> about Big Bangs and Black Holes that are semi-
> religious artifacts of some mathematical calculations.
thus:
I'm just going to say, I guess that
AP is really a charlatan. I mean,
it's in the very beginning of every book on p-adics
-- that part that I was able to comprehend, thus far --
that the arhcimedean-valued integers a infinity-adics;
maybe, taht's where he got the idea....
Dik T. Winter
...
> 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/
a_plutonium
Dik T. Winter wrote:
> In article <1167848912.6...@k21g2000cwa.googlegroups.com> "a_plutonium" <a_plu...@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.
Proginoskes
I didn't ask for a proof. I asked for definitions of + and *. You can't
have primes without * and +.
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> [...]
> Now I do not know if anyone has proven formally that when you take an
> Irrational number and change a few digits that the new number is still
> irrational.
Yes. It is an exercise in undergraduate Number Theory; a direct
consequence of
THEOREM. A real number R is rational iff its base b expansion is
eventually periodic.
(And why do you insert the adjective: "formally"? Aren't all of your
proofs formal?)
--- Christopher Heckman
Proginoskes
Dik T. Winter wrote:
> In article <1167848912.6...@k21g2000cwa.googlegroups.com> "a_plutonium" <a_plu...@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.
I wouldn't; AP hasn't stated how two Infinite Integers are multiplied.
> > 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.
My thinking was about to head in that direction; thanks, Dik, for the
info.
Of course, next AP will claim he's not working with the 10-adics. Which
means only AP and God know how he's defining + and *.
--- Christopher Heckman
carlh...@gmail.com
> Dik T. Winter wrote:
> > In article <1167848912.6...@k21g2000cwa.googlegroups.com> "a_plutonium" <a_plu...@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.
Same procedure, then, if your Infinite Integer ends in 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.
The "lopping off 2's (and 5's)" doesn't help this proof any, since for
every Infinite Integer N, there are an infinite number of irrational
numbers between 0 and 1 that get mapped to N, since "infinity divided
by infinity" can be finite.
For instance, if N = ...2951413, then the following irrational numbers
end up at N:
0.3141592...
0.23141592...
0.223141592...
0.2223141592...
etc.
> > > 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. [...]
Actually, they do, using something called the Division Theorem.
> 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. [...]
Well, then you have to state exactly how addition and multiplications
are defined, and prove your claim that 5 x 11 = 55 really is true.
I'll get the ball rolling for you:
DEFINITION. Let A = ...(a[3])(a[2])(a[1])(a[0]), B =
...(b[3])(b[2])(b[1])(b[0]),
S = ...(s[3])(s[2])(s[1])(s[0]), and , P = ...(p[3])(p[2])(p[1])(p[0])
be infinite integers such that
A + B = S and A * B = P. Then
s[i] = ___ and
p[i] = ___.
All you have to do is to fill in the blanks.
> [...] Again, I have defined multiplication and division
No you haven't. For instance, AP hasn't defined ...123123123 *
...2352352352357.
> 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.
It depends on the context. It is true, for instance, in the p-adics
(where p is not 7), and in the set of rational numbers.
--- Christopher Heckman
Gottfried Helms
> Dik T. Winter wrote:
>> In article <1167848912.6...@k21g2000cwa.googlegroups.com> "a_plutonium" <a_plu...@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.
>
> I wouldn't; AP hasn't stated how two Infinite Integers are multiplied.
Some years ago I discussed that matter a bit; for
instance defining multiplication and the like.
See
http://go.helms-net.de/math/adics/adic2.htm
http://go.helms-net.de/math/adics/adic3.htm
http://go.helms-net.de/math/adics/adic4.htm
but finally I didn't find any special use for it
Also I went in the discussion relatively new to
number-theory, so my treatises are a bit ...<sigh>
but: fwiw -
Gottfried Helms
hagman
a_plutonium schrieb:
It looks like the first of these is not a prime but a cube:
...39608071 * ...39608071 = ...88341041
and
...39608071 * ...88341041 = ...24141911
Are there infinitely AP-primes that are cubes of other numbers?
Also,
...3732653124141911
is a unit:
...41911 * ..45191 = ...00001
Dik T. Winter
> 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.
a_plutonium
Gottfried Helms wrote:
>
> Some years ago I discussed that matter a bit; for
> instance defining multiplication and the like.
>
Well actually Alexander Abian did the same and called them "Plutonium
Numbers". He did a bit of transposing of the Infinite Integers to Reals
of a back and forth transposition.
Dik Winter and most others look at the Infinite Integers and say they
are identical to the p-adics which is not true.
I spent over a decade entertaining the idea that Infinite Integers
could or would be the p-adics. But now I see they are utterly different
from the p-adics. How so? Because they have no Algebras or what little
Algebra they have, is so primitive that those "lovers of Algebra" have
to scatter to the hills. Perhaps that is why I discovered Infinite
Integers and noone else, because when Henkel discovered p-adics he too
was chained to the romance of algebras and this insane fascination to
cloak and dress everything with an Algebra, allowed Infinite Integers
to remain hidden and buried until I came along and saw them.
Infinite Integers are not p-adics because these Infinite Integers have
little to no Algebra and are so primitive in Algebras that to discuss
them from an Algebraic point of view is missing the entire point. The
reason Infinite Integers have little to no algebra is because they are
Circular or form a circle or spiral of at least 2 spirals. The base 2
of just digits 0 and 1 have numbers ending in either 0000s or 11111s
and thus two spirals are formed.
The base 10 would have 10 spirals of those ending in 0000s, 1111s,
2222s, etc, 9999s
Where a spiral between the 111s and 222s or the 888s and 999s.
The successor of ....999999 is 0
So because the geometry of these Infinite Integers is a circle or
spiral we end up with strange things as ....99999 x ....99999 =
.....000001
Imagine for a moment that the Reals from 0 to 1 were mapped onto a
circle and where 0 and 1 were omitted. And so you add say 0.9 + 0.8 and
you get 1.7 but there is no 1.7 since all the numbers of Reals that
exists on this circle are between 0 and 1. So that is why when Dik
multiplys ...9999999911 = 89 * ...9999999999 he is circumnavigating
from the number ...999999 times ....0000089 all the around that circle
and back to the number ....99999911
So if all the Reals that existed were those between 0 and 1 and you
added 0.9 with 0.8 then you would end up with 0.7 because the Reals in
this limited existence would be a circle and not a straight line to
infinity.
In this way, you can have all the Algebras that algebra lovers love to
have on straight lines, but the moment that you have all the numbers
that exist on a globe or a circle geometry then you lose Algebra
because addition and multiplication are no longer tenable on a circular
geometry. So Algebra is confined to straight line flat geometry which
the Reals are. And which the p-adics are because they have been chained
to Algebra.
a_plutonium
Dik T. Winter wrote:
> In article <1167883498.6...@51g2000cwl.googlegroups.com> "a_plutonium" <a_plu...@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.
Aluminium Holocene Holodeck Zoroaster
of integral bases (bigger than one), as explained
by Ore in his _Number Theory and Its History_
(originally by Simon Stevin in the 15thcce). so,
you make the bizarre extrapolation to ...2222. etc.
in base ten -- not even EEEscultura did that! even so,
it is at least a vaguely comprehensible line of thought,
unlike your usual stuff. (I keep-on saying,
those ESL classes are "FREE," dude; or,
just go thespian, bisexual roles in Shakespeare
til you can comprehend Hamlet.)
when you can configure Munk's paper on congruence surds,
then you'll be able to publish. til then,
just continue to pile-on the repeatative strange injury!
> Circular or form a circle or spiral of at least 2 spirals. The base 2
> of just digits 0 and 1 have numbers ending in either 0000s or 11111s
> and thus two spirals are formed.
thus quoth:
Two galaxies physically associated with one another offer the ideal
test
for redshift quantization; they represent the simplist possible system.
According to conventional dynamics, the two objects are in orbital
motion
about each other. Therefore, any difference in redshift between the
galaxies in a pair should merely reflect the difference in their
orbital
velocities along the same line of sight. If we observe many pairs
covering
a wide range of viewing angles and orbital geometries, the expected
distribution of redshift differences should be a smooth curve. In other
words, if redshift is solely a Doppler effect, then the differences
between
the measured values for members of pairs should show no jumps.
But....
http://www.cs.unc.edu/~plaisted/ce/redshift.html
> > > In the algebraics one can have
> > >non-unit integers arbitrarily close to 1 (a^(1/n) for any integer a and
> > >any positive integer n),
thus:
any moron could come-up with the Fibonacci series;
stupid people, however, is another story. (of course,
the internal evidence that Fermat made no known mistakes,
sort-of leads one to believe that
it wasn't his last theorem, at all .-)
thus:
you should see if you can find a copy
of Charles B. Officer's book about the unmitigated hypothesis
that all craters are due to impacts, which is really
a reflection of the hegemony of the Alvarez et al pseudo-
finding about the K-T boundary. mostly remaindered, and
he gets quite worked-up in his prose, but, hey. (also,
find the *first* edition of _The Moon_,
by that former British Royal Astronomer, and compare it
with his second edition, on this subject --
he makes a little joke out of it....
beautifully said, though
about our non-programme d'espace;
soon, a thousand points of light will be on the moon,
all representing other countries. I mean,
why do you think that *yehT* whacked KFJ?
> http://en.wikipedia.org/wiki/Sudbury_Basin
> "insane"?, well perhaps I would argue it's insane to
> bypass lunar geology ((Selinology?)) that has been
> collecting data for billions of years and is there now,
> rather than chasing about based on speculations
> about Big Bangs and Black Holes that are semi-
> religious artifacts of some mathematical calculations.
thus:
I'm just going to say, I guess that
AP is really a charlatan. I mean,
it's in the very beginning of every book on p-adics
-- that part that I was able to comprehend, thus far --
that the arhcimedean-valued integers a infinity-adics;
maybe, taht's where he got the idea....
I wouldn't say that this applies to his cousin,
Governeurateur Terminator, since
that guy came to teh USA at age nineteen --
I think, his accent is entirely cultivated, and
that his main problem is his financial sponsors
(Rothschild, George Schultz, Warren Allyoucaneat Buffet,
junkbondtraders etc ad vomitorium).
thus:
Proginoskes
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
David Bernier
> 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)?
I've been wondering if there might be a natural way to
get a direct limit for the rings with unity
R_n = Z/(10^n Z), n=1,2,3 ... [n in the usual positive integers].
I had a look at the Wikipedia article on direct limits at
< http://en.wikipedia.org/wiki/Direct_limit >
The second example there, which begins with
"Let p be a prime number.", might be relevant.
Since ...9999999999 + ...0000000001 ought to be
...0000000000, a sensisble order might not exist...
David Bernier
Proginoskes
In normal p-Adics, you have a distance d defined by
d(A,B) = p^-(min {i: A[i] is not B[i]}),
where A = ...(A[2])(A[1])(A[0]) and B = ...(B[2])(B[1])(B[0]). Thus A
and B being close means
A[0] = B[0], A[1] = B[1], ..., A[k] = B[k] for some integer k. This is
standard practice.
Only AP and God know what AP means by his Infinite Integers.
> The second example there, which begins with
> "Let p be a prime number.", might be relevant.
>
> Since ...9999999999 + ...0000000001 ought to be
> ...0000000000, a sensisble order might not exist...
No, it won't. In an ordered ring, 0 < 1, which means N < N+1 for all N.
If you "add enough 1s", you'll find out that N < N+1 < N+2 < ... < N-1
< N. So saying "a is less than b" is meaningless, because it's always
true (and always false, for that matter).
But AP's reaction to that would be something like: "You're thinking
about old algebra with old ring axioms. Also old=wrong, so this
argument doesn't apply to Infinite Integers" (in a nutshell).
--- Christopher Heckman
a_plutonium
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.
Proginoskes
a_plutonium wrote:
> 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.
Presented, with examples. S is the successor function which is included
in the Peano Axioms. (BTW, this will be in any book on Peano
Arithmetic.)
Addition:
(1) m + 0 = m,
(2) m + S(n) = S(m+n).
(Example: 2 + 2 = S(S(0)) + S(S(0)) = S(S(S(0)) + S(0)) = S(S(S(S(0)))
+ 0)
= S(S(S(S(0)))) = 4.)
Multiplication:
(1) m * 0 = 0,
(2) m * S(n) = (m * n) + m.
(Example: 1 * 2 = S(0) * S(S(0)) = (S(0) * S(0)) + S(0)
= [S(0) * 0 + S(0)] + S(0) = [0 + S(0)] + S(0) = S(0 + 0) + S(0) =
S(0) + S(0)
= S(S(0) + 0) = S(S(0)) = 2.)
Subtraction:
(*) a - b = c iff a = b + c.
Division:
(*) a / b = c iff a = b * c.
(Note that in Peano _Arithmetic_, these operations are defined for
bigger sets of numbers, namely the set of all integers, all rationals,
and all reals.)
> Do you have to show how to mechanically add two numbers like 99998 + 77359?
In order to make + and * well defined, and to make sure they have the
needed properties, yes. Because no one else knows how to do it, yes.
If Infinite Integers are going to replace the Natural Numbers, then in
the future, school children will have to learn about them and about how
to calculate with them. How will this be taught, then?
> Do you have to show how to mechanically divide?
No, but I've given enough information above for you to calculate any of
the four basic functions, as well as two examples. This is MUCH more
than you have provided.
> 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?
No. Since there is no number n such that n + 3 = 2, 2 - 3 is
undefined.
> 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.
No. (See above.)
> [...]
> > > 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.
That is because the one-sided infinite expression is being approached
from the FINITE side; thus 0.125... and 0.123... are close because a
finite number of digits are the same, and we don't care about the ones
later on. In your ordering of the Infinite Integers, ...11103 is close
to ...11195; but you need an infinite number of digits to agree for
this to happen.
That is why the 10-adics are traditionally "ordered" in the following
way:
{those ending in 00} {those ending in 10} ... {those ending in 90}
{those ending in 01} {those ending in 11} ... {those ending in 91}
{those ending in 02} ... {those ending in 99}
i.e., all the 10-adics which end in 0 are grouped together, all the
ones ending in 1 are grouped together, etc. Of course the 10-adics are
not the Infinite Integers.
> 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.
Not necessarily; just extended. Mathematicians didn't throw out the
integers because there were real numbers, did they?
> > (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
So the Infinite Integers ARE the 10-adics. (This is how 10-adic
multiplication is done.)
> > (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.
Not according to what you said about about 2222x3333; that was a base
10 calculation!
If we do it in base 6, we find that
2 x 3 = 10
22 x 33 = 1210
222 x 333 = 123210
2222 x 3333 = 12343210,
so ...222 * ...333 = ...3210. Is ...3210 supposed to equal ...926?
> Whatever you find true for these Infinite Integers is true in all bases of them.
So you're saying that the Infinite Integers ...0001001 and ...000121
are the same, since they're both ...00011 * ...00011?
> 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.
Again, this suggests that Infinite Integers are not "base independent".
> 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
Not if you use the Peano Axioms; the Peano Axioms can be used to prove
that the digits in all natural numbers are eventually 0.
> > > 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.
AP missed the point: If we can't tell when M is larger than N, we don't
know whether N can be subtracted from M or not.
> > (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.
A real mathematician would look for them.
> 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.
Certainly, you can't order such a thing; so you can't talk about one
number being larger than another.
> 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.
IOW, they will not be mathematical objects.
> > 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.
> >
> >>
> 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.
But the arithmetic you're using in this post indicate you ARE using
them again. So AP has gone from "Infinite Integers = 10-adics" to
"Infinite Integers = All-Adics" to "Infinite Integers = Something Else"
to "Infinite Integers = 10-adics". And just in the past 2 years.
--- Christopher Heckman
a_plutonium
(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.
Proginoskes
Tonight, AP took a hit, and refused to pass it around ...
a_plutonium wrote:
> [...]
> 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.
That's like asking: If the sky was green, how would we know where to
stop mowing?
BTW, The Natural Numbers are NOT the Infinite Integers. This can be
proven.
> 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.
??? So there are only two transcendental numbers, except for the rest?
What about e^2? Or pi^2? Or e or pi to any power? Those can be proven
to be transcendental, too.
What about Liousville's Number? (sum(1/10^n!)), which was the first
number to be proven transcendental?
> 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?
The Infinite Integers are the 10-adics. AP posted earlier tonight a
definition of + and * in his Infinite Integers, and they are the same
as + and * in the 10-adics.
So Atom Totality implies that pi = 22/7, which makes 22/7
transcendental? Nice going, AP.
> So physics points to the likelihood
> that there exists only two transcendental numbers.
Which makes "physics" wrong.
It's posts like this that make people think AP isn't playing with a
full deck of cards.
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> [...] these Infinite
> Integers form a circular or spherical geometry and bend back around to
And that is why the Infinite Integers are not the Natural Numbers. This
fact clearly violates the Peano Axiom that states
For all m, S(m) is not 0.
(See http://mathworld.wolfram.com/PeanosAxioms.html ):
1. Zero is a number.
2. If a is a number, the successor of a is a number.
3. zero is not the successor of a number.
4. Two numbers of which the successors are equal are themselves equal.
5. (induction axiom.) If a set S of numbers contains zero and also the
successor of every number in S, then every number is in S.
Even if this "circular" statement is revoked, it still can be proven
that the Infinite Integers are not the Natural Numbers.
Proof:
(1) Suppose that the Infinite Integers model the Natural Numbers.
(Reducio ad Absurdum assumption)
(2) Let S be the set of all Infinite Integers whose digits are
eventually 0.
(3) Clearly 0 is in S. (definition of S)
(4) Now suppose N is in S. (direct proof, assumption)
(5) By the definition of S, N[i] = 0 for i >= k, for some k.
(6) The successor of N has the property that (succ N)[i] = 0 for all i
>= k+1; succ N might require one more nonzero digit than N, but not any more. (property of +)
(7) Then the successor of N is in S. (definition of S)
(8) Thus: If N is in S, then the successor of N is in S. (direct proof;
discharge (4)-(7).)
(9) By Peano Axiom #5, this set S is the set of all Natural Numbers.
(10) By assumption, the set of all Natural Numbers is the set of all
Infinite Integers.(1)
(11) Therefore every Infinite Integer is in S. (substitution, (9) and
(10).)
(12) ...111 is an Infinite Integer. (Fact)
(13) Therefore ...1111 is in S. (Substitution, (11) and (12).)
(14) ...111 does not have any 0's in it. (Fact)
(15) ...1111 is not in S (since the digits are not eventually 0;
definition of S (2), (14))
(16) (13) and (15) form a contradiction.
(17) Hence the Infinite Integers do not model the Natural Numbers.
(indirect proof; discharge (1)-(16).)
QED.
--- Christopher Heckman
a_plutonium
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
a_plutonium
Proginoskes wrote:
> a_plutonium wrote:
(snipped)
Well the above is an example in science where everyone knows add and
multiply and only *incidentally* is a formal definition given for which
that formal definition is full of holes and leaks. Heckman pretends as
though the mechanics of add, multiply, divide, subtract for the Natural
Numbers is incidental and that the above formal definition is primary
foundational. He is wrong. The mechanics of add, multiply, divide,
subtract of the Natural Numbers in mathematics is primary and
foundational, and the formal definition as given above are no more than
a artifice label. Much like the ingredients on a box with a few
instructions. Humanity has added, multiplied, subtracted and divided
for thousands of years before anyone gave formal definition, which
alone tells us that the formal definition is a superficial label or
superficial instruction.
But the above was needed in this discussion for all the jerks who claim
I do not define add, multiply, subtract and divide for Infinite
Integers. I do and the definitions are the same as above. The mechanics
of actually adding or multiplying or dividing or subtracting Infinite
Integers such as ......333333 x .....2222222 x .....9999999 +
......88888 is the same as if one confronted 3 x 2 x 9 + 8
Why are the definitions the same? Because Infinite Integers are the
same as the Natural Numbers because every axiom system that has an
endless adding of 1 is a set of Infinite Integers.
a_plutonium
Proginoskes wrote:
> a_plutonium wrote:
(snipped)
> > Integers form a circular or spherical geometry and bend back around to
> > the starting point of 0. [...]
>
> And that is why the Infinite Integers are not the Natural Numbers. This
> fact clearly violates the Peano Axiom that states
>
> For all m, S(m) is not 0.
>
> (See http://mathworld.wolfram.com/PeanosAxioms.html ):
>
> 1. Zero is a number.
> 2. If a is a number, the successor of a is a number.
> 3. zero is not the successor of a number.
> 4. Two numbers of which the successors are equal are themselves equal.
> 5. (induction axiom.) If a set S of numbers contains zero and also the
> successor of every number in S, then every number is in S.
>
> Even if this "circular" statement is revoked, it still can be proven
> that the Infinite Integers are not the Natural Numbers.
Well I wanted to save the discussion of revamping the Peano Axioms for
later but since you bring it up, I may as well touch on the issues.
By the way, in a previous post of yours you discuss how the Peano
Axioms define addition and multiplication in terms of the Successor
function S(0) but a better way of defining addition and multiplication
is in terms of geometry where a unit square is defined of a square 1 by
1 and so the addition such as 2 +3 is 2 unit squares added 3 unit
squares is 5 unit squares and where multiplication is geometrical also
of 2 unit squares x 3 unit squares equals 6 unit squares. Which points
out one of many flaws of the Peano axioms in that it creates 0 but it
never creates a nonvariable successor. As I pointed out some years
back, that the current Peano Axioms needs to say in the same sentence
where they create 0 that they create 1 along with 0 as a measuring rod.
Because the current Peano Axioms can have a variable rod of 2 to 3 as 1
unit apart in distance but where say 77793 and 77794 are 1.14 distance
apart. By creating 0 and 1 together to mark out what the successor
function is, you prevent this gap and hole of logic of a "variable
successor". But that is only a tiny flaw of the Peano Axioms.
If Peano had been a superior Logician, he would have realized that the
creation of the Reals was over one main feature-- continuity between
numbers. So that between any two numbers there are no holes but filled
with an infinite supply of numbers. Now, seeing that continuity was the
main feature of Reals, Peano would have then asked what is the main
feature of Natural Numbers in order to construct a axiomatics on
Natural Numbers. And the main feature of Natural Numbers was endlessly
adding 1, a quantization of space out to infinity. The Reals were not
concerned with what is infinity, but that was the concern of Natural
Numbers. So if Peano had sat down to pen what the axioms of Natural
Numbers should be with its main feature of answering -- what is
infinity for mathematics using endless adding of 1.
The corrected Peano Axioms would look like this:
(1) There are two numbers, call them 0 and 1 and they are 1 unit
distance apart
(2) There is a successor function using 1 unit distance that produces
all the other Natural Numbers with this endless adding of 1
Nothing more is needed since these two axioms yield the set
0,1,2,3,4,,.....,, ....99999
The important conclusions derived from these axioms is that the above
set of numbers forms a spherical Riemannian geometry.
The Axioms for the Reals would look like this:
(1) There are two numbers call them 0 and 1 and they are 1 unit
distance apart
(2) There are an infinite amount of numbers between 0 and 1
(3) There are operators of add, subtract, multiply, divide, square root
The old mathematics thought that it could get away with only
axiomatizing the Natural Numbers and which those Peano axioms were mere
extended to yield the Rationals and then Reals.
The new mathematics realizes that their is no bridging of the axioms of
Natural Numbers to that of Reals and that these two number systems have
their own independent axiomatics. This is because the Reals also form a
innate geometry of Euclidean geometry whereas the Natural Numbers forms
a Riemannian geometry and those two geometries are not extensions of
one another but require separate axiomatics.
a_plutonium
Proginoskes wrote:
>
> What about e^2? Or pi^2? Or e or pi to any power? Those can be proven
> to be transcendental, too.
>
> What about Liousville's Number? (sum(1/10^n!)), which was the first
> number to be proven transcendental?
>
In that post yesterday I was just exploring with pi and e, and today I
corrected it by saying there exists only 3 Root-transcendentals. There
is a difference between Root Transcendental and merely transcendental,
much like there exists only 3 space vectors but an infinite scalar
product of those 3 vectors. So I am using Root Transcendental as a
vector.
I remember Liousvill numbers from History of Mathematics. I would have
two big reservations about that number as a transcendental. First I
would question the strength or weakness of the proof itself. I would
suspect it is only a "reductio ad absurdum" argument. And in
mathematics, for a proof to really be a proof would have to have both a
direct and indirect method. So I suspect the Liousville Number has no
direct method proof.
Secondly, if it turns out that the Liousville Number does have a direct
and indirect method proof and that the number is truly a transcendental
number, then the question arises whether this number is some scalar
multiple of either pi, e, or (i), which I claim are the only 3 Root
Transcendental Numbers.
So there is a big chore of work for Chris Heckman to do.
Aluminium Holocene Holodeck Zoroaster
that Leibniz didn't. since you refuse to acknowledge that
the infinite-adics are the only ones that are "archimedean,"
you'll have to find another namesake, like Bozonium.
> > And that is why the Infinite Integers are not the Natural Numbers. This
> The old mathematics thought that it could get away with only
> axiomatizing the Natural Numbers and which those Peano axioms were mere
> extended to yield the Rationals and then Reals.
thus:
I do declare. note that Lyn was released by the Clinton parole board,
which you might infer from perusing _The Unauthorized Biography
of [the Original] George_, Copyr.'92 by EIR --
there's a new edition, with a new chapter, but
you can get the whole 600+pp. on Tarpley's site, I think.
>political activities from behind bars until his release in 1994 on
>parole. His imprisonment was protested by public figures from around
>the world. Former U.S. Attorney General Ramsey Clark charged that his
>case "involves a broader range of deliberate and systematic misconduct
>and abuse of power over a longer period of time in an effort to
>destroy a political movement and leader, than any other federal
>prosecution in my time or to my knowledge." [13] In an interview with
>LaRouche's Executive Intelligence Review, the late former U.S. Senator
>and Democratic presidential aspirant Eugene McCarthy called LaRouche
>"a man who has brought Plato and Schiller back into politics ? and was
>sent to jail for it." [14]Quoting "Mr. Tetrahedronometry"
thus:
also, the local library used to have an old _The Universe_
from Rand McNally, which went into the dichotomy
of volcanic and/or impact craters. now, clearly,
the patterns on Moon are very far from '"random,"
what amounts to a basic scheme of a dead system
of plate tectonics (or undead, since
there is still noticable siesmogrammetry, and
telescpic sightings of magma).
one supposed that one needs quite a bit of vulcanism,
to have a "glass house effect," at all;
isn'tthat logical, Mr.Spock?... anyway,
Moore's first edition was, I think, before the landings.
> NB, that is Patrick Moore, _The Moon_,
> first and later editions. he actually went around Earth
> to various volcanic craters for comparison
thus:
and then there's "the value of the Bible in pi,"
the skipcodes of the bible code nazis meet Carl Sagan;
bring *all* of your Ouija Boards!
in case, you were wondering,
who the real Dr. Strangelove was.
> > And he made a molten sea, ten cubits from the one brim to the other: it
> > was round all about, and his height was five cubits: and a line of thirty
> > cubits did compass it about. (I Kings 7, 23)
> http://www.uwgb.edu/DutchS/pseudosc/pibible.htm
thus:
I mean, I think that was where I picked-up
the (occaisonal) usage, "polygona" for polyhedra,
tetragona being plural.
> also, Wenninger (sp.?) in his paper models books,
> uses the "-gon" suffix for polyhedra, somwhere.
>
> > "polyvertexion," but I prefer polyasteron.
thus:
well, twenty-year-old "pending" legislation; good one!
that ain't nothin', compared to what they did
to our companies, to get Lyn through the "rocket docket"
in the Beltway.... well, it'd already been done
before the first trial, which was a mistrial against the goment.
the ch.13s were later declared a grossly fraudulent proceeding
by the original bankruptcy court judge, anyway.
>of "frivolous paperwork."
>_The Big Investment Lie_,
>>http://query.nytimes.com/gst/fullpage.html?res=9A0DEFD8163EF93AA15753C1A960948260
>
>>> not to mention the shape; remember Trinitron?
thus:
I fail to see what desire has to do with it;
did you parse my sentence?
that's Occidental Petroleum, the company
that made Al Gore Jr & Sr what they are today ...
well, they didn't kill the first Senator,
that I know of!
> > the gasoline additive that was tootally unnecessary,
> > as proven by Occidental losing their suit against California.
>
> Ah, so your desire is to have a return to Lead additives in gasoline.
thus:
you should see if you can find a copy
of Charles B. Officer's book about the unmitigated hypothesis
that all craters are due to impacts, which is really
a reflection of the hegemony of the Alvarez et al pseudo-
finding about the K-T boundary. mostly remaindered, and
he gets quite worked-up in his prose, but, hey. (also,
find the *first* edition of _The Moon_,
by Patrick Moore, and compare it
with his second edition, on this subject --
he makes a little joke out of it....
beautifully said, though
about our non-programme d'espace;
soon, a thousand points of light will be on the moon,
all representing other countries. I mean,
why do you think that *yehT* whacked KFJ?
> http://en.wikipedia.org/wiki/Sudbury_Basin
thus:
that guy came to teh USA at age nineteen --
I think, his accent is entirely cultivated, and
that his main problem is his financial sponsors
(Rothschild, George Schultz, Warren Allyoucaneat Buffet,
junkbondtraders etc ad vomitorium).
thus:
you have to use tripolar coordinates, though. I've always
called it tetrahedronometry, although, a la the study of trigona,
it can also be called tetragonometry. an alternative
that I've espoused, was inspired by Bucky Fuller,
in a _Posthumous_ publication written by his adjuvant,
when he finally started to attend to teh duals
of the trigonated polyhedra; he used the word,
"polyvertexion," but I prefer polyasteron....
that also dystinguished it from Peter Schoute's polygonometry,
which is just an old, "n-D" thing,
which Coxeter told me of, when I called him up at his home
in Toronto -- correcting me for asking if he knew
of tetrahedronometry!
> > Basically my question here is: what's 4-gonometry.
> May be the tetrahedronometry.It should include dihedrals and trihedral
thus:
unfortunately, the neocons & jihadists are sliding us
into Sudan, a dried-up quagmire (British sand,
viz Sandhurst, EMI et al ad vomitorium Seargent Peppers)....
thus quoth:
http://chephip.free.fr/pbg_en/sol143.html
--The Other Side (if it exists ... nah !-)
a_plutonium
using the reductio ad absurdum over the assumption of a finite string
of algebraic digits. Need I say more. That these are not what I
consider mathematical proofs but something on the form of a indication.
In mathematics, when something is true can be proven by a Direct Method
Proof. If something in mathematics has only a Indirect Method, does not
count as a proof.
So I can sense where Hermite and all the others who followed on this
method is that thiers is not a proof but a argument.
a_plutonium
a_plutonium wrote:
(snipped)
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.
Eric Gisse
a_plutonium wrote:
[snip archie poo]
Explore your way to a fucking complex analysis textbook.
Proginoskes
a_plutonium wrote:
> 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.
Let's see: By that logic, negative numbers are not positive integers,
and fractions of the form 1/(n^2+3) are not positive integers.
Therefore, every negative number is of the form 1/(n^2+3).
>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).
i is not transcendental; you should know this.
i is a root of the polynomial x^2 + 1, a polynomial all of whose
coefficients are integers.
--- Christopher Heckman
Proginoskes
> multiply and only *incidentally* is a formal definition given for which
> that formal definition is full of holes and leaks. Heckman pretends as
> though the mechanics of add, multiply, divide, subtract for the Natural
> Numbers is incidental and that the above formal definition is primary
> foundational. He is wrong. The mechanics of add, multiply, divide,
> subtract of the Natural Numbers in mathematics is primary and
> foundational, and the formal definition as given above are no more than
> a artifice label.
Get me a shovel.
I assume AP is saying that addition and the other operations are
assumed to satisfy certain conditions, and that this definition is
forced. Anyone who looks at a book on Peano Arithmetic will know that
after these definitions are given, the usual properties of these
operations are proved. For instance, a+b = b+a, a+(b+c) = (a+b)+c, etc.
> Much like the ingredients on a box with a few
> instructions. Humanity has added, multiplied, subtracted and divided
> for thousands of years before anyone gave formal definition, which
> alone tells us that the formal definition is a superficial label or
> superficial instruction.
Similarly, humanity has existed for thousands of years without Atom
Totality Theory, which alone tells us that Atom Totality is
superficial.
Humanity has existed just as long without Infinite Integers, which
alone tells us that Infinite Integers are superficial.
> But the above was needed in this discussion for all the jerks who claim
> I do not define add, multiply, subtract and divide for Infinite
> Integers. I do and the definitions are the same as above. The mechanics
> of actually adding or multiplying or dividing or subtracting Infinite
> Integers such as ......333333 x .....2222222 x .....9999999 +
> ......88888 is the same as if one confronted 3 x 2 x 9 + 8
When AP's arm was finally twisted enough, he gave a definition of
addition and multiplication, which turns out to be the usual 10-adic
arithmetic, despite what he wants.
> Why are the definitions the same? Because Infinite Integers are the
> same as the Natural Numbers because every axiom system that has an
> endless adding of 1 is a set of Infinite Integers.
The Infinite Integers are NOT the Natural Numbers. I posted an
air-tight proof of this.
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> Proginoskes wrote:
> > a_plutonium wrote:
> (snipped)
> > > [...] these Infinite
> > > Integers form a circular or spherical geometry and bend back around to
> > > the starting point of 0. [...]
> >
> > And that is why the Infinite Integers are not the Natural Numbers. This
> > fact clearly violates the Peano Axiom that states
> >
> > For all m, S(m) is not 0.
> >
> > (See http://mathworld.wolfram.com/PeanosAxioms.html ):
> >
> > 1. Zero is a number.
> > 2. If a is a number, the successor of a is a number.
> > 3. zero is not the successor of a number.
> > 4. Two numbers of which the successors are equal are themselves equal.
> > 5. (induction axiom.) If a set S of numbers contains zero and also the
> > successor of every number in S, then every number is in S.
> >
> > Even if this "circular" statement is revoked, it still can be proven
> > that the Infinite Integers are not the Natural Numbers.
>
>
> Well I wanted to save the discussion of revamping the Peano Axioms for
> later but since you bring it up, I may as well touch on the issues.
I didn't say anything about revamping Peano's Axioms. The remark about
"if the 'circular' statement [being] revoked' refers to the fact that
the successor of ...999 is 0.
> By the way, in a previous post of yours you discuss how the Peano
> Axioms define addition and multiplication in terms of the Successor
> function S(0) but a better way of defining addition and multiplication
> is in terms of geometry where a unit square is defined of a square 1 by
> 1 and so the addition such as 2 +3 is 2 unit squares added 3 unit
> squares is 5 unit squares and where multiplication is geometrical also
> of 2 unit squares x 3 unit squares equals 6 unit squares.
No, it isn't. The Peano definitions don't require geometry. AP's
definitions require the Peano Axioms and geometry, which is more
"baggage". (This is why a single function is included in the Peano
Axioms, as opposed to an addition operation.)
> Which points
> out one of many flaws of the Peano axioms in that it creates 0 but it
> never creates a nonvariable successor.
Shovel time. The successors S(0), S(S(0)), S(S(S(0))) are all fixed
entities.
> As I pointed out some years
> back, that the current Peano Axioms needs to say in the same sentence
> where they create 0 that they create 1 along with 0 as a measuring rod.
> Because the current Peano Axioms can have a variable rod of 2 to 3 as 1
> unit apart in distance but where say 77793 and 77794 are 1.14 distance
> apart.
No, they don't. No mention is made of the "distance" between N and the
successor of N.
Have you even READ a book about the Peano Axioms?
> By creating 0 and 1 together to mark out what the successor
> function is, you prevent this gap and hole of logic of a "variable
> successor". But that is only a tiny flaw of the Peano Axioms.
>
> If Peano had been a superior Logician, he would have realized that the
> creation of the Reals was over one main feature-- continuity between
> numbers. So that between any two numbers there are no holes but filled
> with an infinite supply of numbers.
But between any two numbers, there are already an infinite number of
rational numbers.
> Now, seeing that continuity was the
> main feature of Reals, Peano would have then asked what is the main
> feature of Natural Numbers in order to construct a axiomatics on
> Natural Numbers. And the main feature of Natural Numbers was endlessly
> adding 1,
Which is the same thing as the successor function.
When you're talking about starting off with addition, there's
considerably more baggage; you need to define what a+b is for every
PAIR of Natural Numbers. For the successor function, you only need to
define S(a) for all SINGLE Natural Numbers. Furthermore, there are only
three properties that S needs to have; addition would require
significantly more, to guarantee that it represents the Natural Numbers
(well defined, commutivity, associativity, neutral element).
> a quantization of space out to infinity. The Reals were not
> concerned with what is infinity, but that was the concern of Natural
> Numbers. So if Peano had sat down to pen what the axioms of Natural
> Numbers should be with its main feature of answering -- what is
> infinity for mathematics using endless adding of 1.
>
> The corrected Peano Axioms would look like this:
> (1) There are two numbers, call them 0 and 1
1 is only a symbol to stand for S(0). Similarly, 3 is a symbol which
stands for S(S(S(0))).
> and they are 1 unit distance apart
What is this "distance" thing that AP speaks of? More baggage.
> (2) There is a successor function using 1 unit distance that produces
> all the other Natural Numbers with this endless adding of 1
This is just a combination of the 2nd and 5th axioms of Peano.
> Nothing more is needed since these two axioms yield the set
> 0,1,2,3,4,,.....,, ....99999
Wrong. Let S = {0, 1, 2}, and have addition defined by:
0 + 0 = 0; 0 + 1 = 1; 0 + 2 = 2;
1 + 0 = 1; 1 + 1 = 2; 1 + 2 = 0;
2 + 0 = 2; 2 + 1 = 0; 2 + 2 = 1.
Then AP's axioms are both true, but clearly S wouldn't be considered to
be the set of Natural Numbers.
(This makes me wonder whether AP even gave a second thought to seeing
whether his proposed axioms actually work. A true mathematician would.)
> The important conclusions derived from these axioms is that the above
> set of numbers forms a spherical Riemannian geometry.
>
> The Axioms for the Reals would look like this:
> (1) There are two numbers call them 0 and 1 and they are 1 unit
> distance apart
> (2) There are an infinite amount of numbers between 0 and 1
> (3) There are operators of add, subtract, multiply, divide, square root
Again, the Axioms for the Reals don't produce the set of real numbers
(up to isomorphism). You can let S = {0, 1, 1/2, 1/3, 1/4, 1/5, ...},
which satisfies (1) and (2), and use arbitrary definitions to satisfy
(3). For instance, if x and y are in S, define
x + y = 0
x * y = 0
x - y = 0
x / y = 0
sqrt(x) = 0
And again, this easy counterexamples suggests that AP doesn't think
about what he posts.
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> Proginoskes wrote:
>
> >
> > What about e^2? Or pi^2? Or e or pi to any power? Those can be proven
> > to be transcendental, too.
> >
> > What about Liousville's Number? (sum(1/10^n!)), which was the first
> > number to be proven transcendental?
> >
>
> In that post yesterday I was just exploring with pi and e, and today I
> corrected it by saying there exists only 3 Root-transcendentals. There
> is a difference between Root Transcendental and merely transcendental,
> much like there exists only 3 space vectors but an infinite scalar
> product of those 3 vectors. So I am using Root Transcendental as a
> vector.
>
> I remember Liousvill numbers from History of Mathematics. I would have
> two big reservations about that number as a transcendental. First I
> would question the strength or weakness of the proof itself. I would
> suspect it is only a "reductio ad absurdum" argument. And in
> mathematics, for a proof to really be a proof would have to have both a
> direct and indirect method. So I suspect the Liousville Number has no
> direct method proof.
And AP would be wrong. EVERY indirect proof can be turned into a direct
proof, and vice versa. I posted demonstrations, which AP has
conveniently ignored (and refused to include in his response to other
issues in that post).
> Secondly, if it turns out that the Liousville Number does have a direct
> and indirect method proof and that the number is truly a transcendental
> number, then the question arises whether this number is some scalar
> multiple of either pi, e, or (i), which I claim are the only 3 Root
> Transcendental Numbers.
>
> So there is a big chore of work for Chris Heckman to do.
No, AP is committing a Logical Fallacy here called "Shifting the Burden
of Proof". He claims that there are only two (or three) transcendental
numbers. I challenge this, and AP has the burden of proving his claim,
not challenging me to prove something.
Besides, no one except AP and God know what AP means by a "Root
Transcendental".
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> I looked it up and Hermite did a proof of the transcendentalness of e
> using the reductio ad absurdum over the assumption of a finite string
> of algebraic digits. Need I say more. That these are not what I
> consider mathematical proofs but something on the form of a indication.
Saying something is true doesn't make it true. I can say "Archimedes
Plutonium is a 2-year old girl living in Antarctica", but that doesn't
mean it's true.
> In mathematics, when something is true can be proven by a Direct Method
> Proof. If something in mathematics has only a Indirect Method, does not
> count as a proof.
But any direct proof can be converted into an indirect proof, and
vice-versa. Since AP evidently didn't read this before, I'll post it
again:
Suppose you have a direct proof which proceeds as follows: (Think of
this as a template.)
(1) Assume P
(2)-(n) (various statements logically following from P, no discharging)
(n+1) Deduce Q
(n+2) Discharge assumption [that means you cannot use statements
(1)-(n+1) in the rest of the proof]; you have "P implies Q" now.
The same proof can be written as:
(1) Assume "P implies Q" is false; deduce that P is true and Q is false
from this assumption.
(2)-(n) (the same statements (2)-(n) above)
(n+1) Deduce Q is true.
(n+2) But now you have Q is false (by assumption) and Q is true
(deduced from (n+1)).
(n+3) Contradiction; discharge the assumption that "P implies Q" is
false; deduce that "P implies Q" is true.
Similarly, an indirect (contradiction) proof can be turned into a
direct proof. Suppose your indirect proof proceeds as follows:
(1) Assume P is true.
(2)-(n) (various statements logically following from P, no discharging)
(n+1) Deduce a contradiction: Namely some statement Q is true and false
at the same time.
(n+2) Contradiction; discharge the assumption that P is true; hence P
is false.
Then you can construct the following direct proof:
(1) Assume P is true.
(2)-(n) (the same statements (2)-(n) above)
(n+1) Deduce a contradiction: Q is true and false at the same time.
(n+2) Discharge the assumption that P is true; now you have a statement
of the form "P implies [Q is true and Q is false]".
(n+3) The statement "Not [Q is true and Q is false]" is true (no matter
what Q is)
(n+4) Use Modus Tolens on (n+2) and (n+3), deduce "Not P" is true; that
is, P is false.
> So I can sense where Hermite and all the others who followed on this
> method is that thiers is not a proof but a argument.
It's more of a proof than AP has ever done.
--- Christopher Heckman
Proginoskes
There is no proof of this suggestion.
> 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.
Well, you can start with i: Find an Infinite Integer N such that N*N =
-1 = ...999.
--- Christopher Heckman
a_plutonium
>
> The Infinite Integers are NOT the Natural Numbers. I posted an
> air-tight proof of this.
>
> --- Christopher Heckman
>From the man who cannot even prove Euclid Infinitude of Primes
(indirect method) is quite a boasting.
So how does your alleged proof answer this question:
Given an infinite set of boxes lined up in a row. Something like this:
______________
| | | | | | | | | | | | | | |
There are an infinity of these boxes and in each box starting from the
first on on the right you pour in a 1 unit of water and which each box
can hold 9 units of water.
So you start a Peano Machine of a Successor function that pumps water
into the first box and which the overflow goes into the next box and
when it is full of 9 units of water the overflow goes to the next box
to the left filling it with 9 units of water and on down the line of
these infinite boxes.
So tell us how and why there are any boxes to the left that is not
filled with water?
And how your alleged proof overcomes that.
a_plutonium
Proginoskes wrote:
>
> Wrong. Let S = {0, 1, 2}, and have addition defined by:
> 0 + 0 = 0; 0 + 1 = 1; 0 + 2 = 2;
> 1 + 0 = 1; 1 + 1 = 2; 1 + 2 = 0;
> 2 + 0 = 2; 2 + 1 = 0; 2 + 2 = 1.
>
You are wrong since you did not apply the endless adding of 1. Why stop
with 2.
> Then AP's axioms are both true, but clearly S wouldn't be considered to
> be the set of Natural Numbers.
>
> (This makes me wonder whether AP even gave a second thought to seeing
> whether his proposed axioms actually work. A true mathematician would.)
>
> > The important conclusions derived from these axioms is that the above
> > set of numbers forms a spherical Riemannian geometry.
> >
> > The Axioms for the Reals would look like this:
> > (1) There are two numbers call them 0 and 1 and they are 1 unit
> > distance apart
> > (2) There are an infinite amount of numbers between 0 and 1
> > (3) There are operators of add, subtract, multiply, divide, square root
>
> Again, the Axioms for the Reals don't produce the set of real numbers
> (up to isomorphism). You can let S = {0, 1, 1/2, 1/3, 1/4, 1/5, ...},
> which satisfies (1) and (2), and use arbitrary definitions to satisfy
> (3). For instance, if x and y are in S, define
>
> x + y = 0
> x * y = 0
> x - y = 0
> x / y = 0
> sqrt(x) = 0
>
> And again, this easy counterexamples suggests that AP doesn't think
> about what he posts.
The specification of the operators eliminates your fictional
counterexample
a_plutonium
transcendental numbers and why an Infinite Integer exists to solve that
equation for i.
Notice that the equation is for e^(i*pi)= -1 where I replace -1 with
....999999
But the equation could be for -2 where we adjust i to compensate for a
-2 instead of a -1
or we could substitute any n for -1 and make a compensation for the i.
So what this tells us is that any Natural Number is expressible in
terms of e, pi, and i. Like the analogy I made before that these three
numbers are the unit vectors of Natural Numbers and which all the other
Natural Numbers can be expressed.
Proginoskes
a_plutonium wrote:
> >
> > The Infinite Integers are NOT the Natural Numbers. I posted an
> > air-tight proof of this.
>
Only AP seems to think this, though.
> (indirect method) is quite a boasting.
>
> So how does your alleged proof answer this question:
>
> Given an infinite set of boxes lined up in a row. Something like this:
> ______________
> | | | | | | | | | | | | | | |
>
> There are an infinity of these boxes and in each box starting from the
> first on on the right you pour in a 1 unit of water and which each box
> can hold 9 units of water.
>
> So you start a Peano Machine of a Successor function that pumps water
> into the first box and which the overflow goes into the next box and
> when it is full of 9 units of water the overflow goes to the next box
> to the left filling it with 9 units of water and on down the line of
> these infinite boxes.
Okay; let's suppose that the machine takes 1 minute to pour 1 unit of
water in the box.
> So tell us how and why there are any boxes to the left that is not
> filled with water?
> And how your alleged proof overcomes that.
It's a proof by induction.
In the beginning, the box to the far right is empty, and so are all the
boxes to the left.
One minute later, the box to its left is empty, and so are all the
boxes to the left. You have at most 1 box with liquid in it.
One minute later, the box to ITS left is empty, and so are all the
boxes to the left. You have at most 2 boxes with liquid in it.
At any point in time, there will only be a finite number of boxes which
can contain liquid.
You simply never even get to a point where all of the boxes are full;
there's not enough time.
No matter how much time you have, you will never fill the boxes.
--- Christopher Heckman
Proginoskes
AP doesn't seem to realize that Peano's Axioms have the property that
any set of "numbers" and any "successor function" with certain
properties will lead to one particular set. I've deduced this based on
the fact that he doesn't know what his axiom systems allow.
a_plutonium wrote:
> Proginoskes wrote:
>
> >
> > Wrong. Let S = {0, 1, 2}, and have addition defined by:
> > 0 + 0 = 0; 0 + 1 = 1; 0 + 2 = 2;
> > 1 + 0 = 1; 1 + 1 = 2; 1 + 2 = 0;
> > 2 + 0 = 2; 2 + 1 = 0; 2 + 2 = 1.
> >
>
> You are wrong since you did not apply the endless adding of 1. Why stop
> with 2.
You are wrong because you omitted the important part of the post: The
so-called axioms which characterize the Natural Numbers:
> (1) There are two numbers, call them 0 and 1 and they are 1 unit distance apart
> (2) There is a successor function using 1 unit distance that produces
> all the other Natural Numbers with this endless adding of 1
Okay, let's say 1 = 0 + 1,
2 = 1 + 1,
3 = 2 + 1,
4 = 3 + 1,
5 = 4 + 1,
etc.
I can keep on doing this forever. However, there's nothing your axioms
to keep 3 from equalling 6.
The axioms are satisfied: (1) I have numbers called 0 and 1, which I
can put at (0,0) and (1,0) in the plane. In fact, I can do more: I have
another called 2, which can be put at (0.5, sqrt(3)/2), another called
3, which I can put at (0,0), 4, which I can put at (1,0), etc. So each
number N is at distance 1 from N+1.
Also, endless adding of 1 _does_ produce all of the Natural Numbers.
> > Then AP's axioms are both true, but clearly S wouldn't be considered to
> > be the set of Natural Numbers.
> >
> > (This makes me wonder whether AP even gave a second thought to seeing
> > whether his proposed axioms actually work. A true mathematician would.)
> >
> > > The important conclusions derived from these axioms is that the above
> > > set of numbers forms a spherical Riemannian geometry.
> > >
> > > The Axioms for the Reals would look like this:
> > > (1) There are two numbers call them 0 and 1 and they are 1 unit
> > > distance apart
> > > (2) There are an infinite amount of numbers between 0 and 1
> > > (3) There are operators of add, subtract, multiply, divide, square root
> >
> > Again, the Axioms for the Reals don't produce the set of real numbers
> > (up to isomorphism). You can let S = {0, 1, 1/2, 1/3, 1/4, 1/5, ...},
> > which satisfies (1) and (2), and use arbitrary definitions to satisfy
> > (3). For instance, if x and y are in S, define
> >
> > x + y = 0
> > x * y = 0
> > x - y = 0
> > x / y = 0
> > sqrt(x) = 0
> >
> > And again, this easy counterexamples suggests that AP doesn't think
> > about what he posts.
>
> The specification of the operators eliminates your fictional
> counterexample
There is no "specification of the operators" in the axioms. The axioms
are:
> (1) There are two numbers call them 0 and 1 and they are 1 unit distance apart
> (2) There are an infinite amount of numbers between 0 and 1
> (3) There are operators of add, subtract, multiply, divide, square root
And they are all true.
--- Christopher Heckman
Proginoskes
Why get so complicated about things? I'll clean up your argument a bit:
Let's let
A = ...555, and B = ...444.
"Now here is an argument as to why A and B are the only two
transcendental numbers and why an Infinite Integer exists to solve that
equation for B.
Notice that the equation is for A+B = -1 where I replace -1 with
....999999
But the equation could be for -2 where we adjust B to compensate for a
-2 instead of a -1
or we could substitute any n for -1 and make a compensation for the i.
So what this tells us is that any Natural Number is expressible in
terms of A and B. Like the analogy I made before that these two
numbers are the unit vectors of Natural Numbers and which all the other
Natural Numbers can be expressed."
a_plutonium wrote:
> Now here is an argument
As AP has pointed out, an argument is not a proof.
> as to why e, pi and i are the only three
> transcendental numbers and why an Infinite Integer exists to solve that
> equation for i.
>
> Notice that the equation is for e^(i*pi)= -1 where I replace -1 with
> ....999999
>
> But the equation could be for -2 where we adjust i to compensate for a
> -2 instead of a -1
But that would give you a different number. What AP is proposing, in a
nutshell, is something like:
We want x to satisfy the equation 2 * x = 1, as well as the equation 2
* x = 3.
Can't be done. The value of 'i' isn't allowed to be changed.
> or we could substitute any n for -1 and make a compensation for the i.
>
> So what this tells us is that any Natural Number is expressible in
> terms of e, pi, and i.
If you're allowed to change the value of i.
> Like the analogy I made before that these three
> numbers are the unit vectors of Natural Numbers and which all the other
> Natural Numbers can be expressed.
Here's a hint for AP: If you're going to use a term which has a meaning
(like "transcendental"), either stick to that meaning, or say out in
front that you're changing the meaning to something else.
--- Christopher Heckman
P.S. For some reason, AP's recent posts remind me of a fifth grader I
once saw who was filling glasses partially full with water, pouring in
food coloring, then pouring the water from one glass into another.
After a few minutes of this, I asked him, "What are you doing?" He
replied: "Chemistry."
hagman
Proginoskes schrieb:
> Besides, no one except AP and God know what AP means by a "Root
> Transcendental".
It just occured to me that God is a transcendental concept as well, and
is "Root Transcendental" in the sense of prima causa.
On the other end, AP is transcendental in the sense that his claims "go
beyond" reason.
So is there any relation between (e,pi) and (God,AP) in Atom Totality
Theory?
hagman
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?)
a_plutonium
Proginoskes wrote:
> a_plutonium wrote:
> > >
> > > The Infinite Integers are NOT the Natural Numbers. I posted an
> > > air-tight proof of this.
> >
> > From the man who cannot even prove Euclid Infinitude of Primes
>
> Only AP seems to think this, though.
>
Only Dik Winter agreed that both AP and Chris Heckman had a valid
Infinitude of Primes Euclid Indirect Method, but that is not saying
much because Dik rendered an invalid proof attempt himself when he
included (5b) and (5c) and admitting that he has taught a wrong proof
for 10 years, and will probably continue to teach invalid arguments. So
we have two people who cannot give a valid Euclid Infinitude of
Primes-- Dik Winter and Chris Heckman. Pity all the future students
that receive a muddleheaded Infinitude of Primes from them.
> > (indirect method) is quite a boasting.
> >
> > So how does your alleged proof answer this question:
> >
> > Given an infinite set of boxes lined up in a row. Something like this:
> > ______________
> > | | | | | | | | | | | | | | |
> >
> > There are an infinity of these boxes and in each box starting from the
> > first on on the right you pour in a 1 unit of water and which each box
> > can hold 9 units of water.
> >
> > So you start a Peano Machine of a Successor function that pumps water
> > into the first box and which the overflow goes into the next box and
> > when it is full of 9 units of water the overflow goes to the next box
> > to the left filling it with 9 units of water and on down the line of
> > these infinite boxes.
>
> Okay; let's suppose that the machine takes 1 minute to pour 1 unit of
> water in the box.
Where is the element of "time" within the Peano axioms?
>(See http://mathworld.wolfram.com/PeanosAxioms.html ):
> 1. Zero is a number.
> 2. If a is a number, the successor of a is a number.
> 3. zero is not the successor of a number.
> 4. Two numbers of which the successors are equal are themselves equal.
> 5. (induction axiom.) If a set S of numbers contains zero and also the
> successor of every number in S, then every number is in S.
There is no time constraint in mathematics. We instantly know the
Primes in Euclid proof are infinite and there was no time constraint in
proving infinitude of primes. There is no time factor in any of
mathematics as regards to proof.
What there is above in those Peano Axioms is this Successor Machine
that is the endless flow of water that fills every one of those boxes
and instantly produces ....999999999
>
> > So tell us how and why there are any boxes to the left that is not
> > filled with water?
> > And how your alleged proof overcomes that.
>
> It's a proof by induction.
>
> In the beginning, the box to the far right is empty, and so are all the
> boxes to the left.
>
> One minute later, the box to its left is empty, and so are all the
> boxes to the left. You have at most 1 box with liquid in it.
>
> One minute later, the box to ITS left is empty, and so are all the
> boxes to the left. You have at most 2 boxes with liquid in it.
>
> At any point in time, there will only be a finite number of boxes which
> can contain liquid.
>
> You simply never even get to a point where all of the boxes are full;
> there's not enough time.
>
> No matter how much time you have, you will never fill the boxes.
>
> --- Christopher Heckman
What a silly argument. And you are a college professor teaching
mathematics?
There is no *time factor* in the axioms of Peano.
There is a Successor axiom of endless adding of 1 and so the Peano
Axioms when written and before the ink is dry, the number .....9999999
is produced.
That pictorial is a proof that Natural-Numbers are the Infinite
Integers and that pictorial displays to us that the Peano Axioms are
self-contradictory. When Peano said that 0 is not the successor of any
number-- he immediately made his axioms contradictory because
....999999 precedes 0. Whether his axiom of Math Induction is allowable
is up for debate. It may be that Math Induction is a technique to be
used some of the time but not the status of an axiom of mathematics. I
sense that Successor function does not allow for such a nice counting
of Natural Numbers for which Math Induction is a counting process. How
does one transition from those numbers ending with a 000s string to
those ending with a 1111s string. So I doubt Math Induction is
consistent with Successor, but that they are contradictory.
a_plutonium
a_plutonium wrote:
> > >
> > > Given an infinite set of boxes lined up in a row. Something like this:
> > > ______________
> > > | | | | | | | | | | | | | | |
> > >
> > > There are an infinity of these boxes and in each box starting from the
> > > first on on the right you pour in a 1 unit of water and which each box
> > > can hold 9 units of water.
> > >
> > > So you start a Peano Machine of a Successor function that pumps water
> > > into the first box and which the overflow goes into the next box and
> > > when it is full of 9 units of water the overflow goes to the next box
> > > to the left filling it with 9 units of water and on down the line of
> > > these infinite boxes.
Now this pictorial of endless adding of 1 as water flowing into boxes
and when full filling up the next box to the left ad infinitum may be
instructive in another way. Maybe this pictorial can somehow pinpoint
how the Natural Numbers transition from say a ....00000n to that of a
....11111n
Notice that in the pictorial that the number 10 is never really visited
whereas the first box goes from 0 to 1 to 2 to 3 to 4 to 5 to 6 to 7 to
8 to 9 and then as it begins to overflow into the next box on the left
we have 19, then 29 then 39.
So what does this tell us? Can we construct a water system of boxes
that starts with 0 then goes to 1 then to 2 and never skips any number
and goes out to ....9999999?
Part of the problem is place value carry over. But I would think we can
construct a watering system that is a continuous flow of water that
represents going from 0 to 1 to 2 to 3 and on out to ...999999
Or maybe this is impossible. And that the moment one asserts a function
of infinite adding of 1 creates numbers from 0 to ...99999 but does not
allow anyone to count them completely
David R Tribble
> 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?
David R Tribble
> So what does this tell us? Can we construct a water system of boxes
> that starts with 0 then goes to 1 then to 2 and never skips any number
> and goes out to ....9999999?
How could you? If every time we take a finite number and add 1
to it to get the next finite number, how can you ever get an infinite
number? Is there some finite natural number that is so large
that adding 1 to it somehow breaks the rule and gives you an
infinite number? What is this magical "almost infinite" finite
number?
a_plutonium
Okay, I am justified in Infinite Integers in calling e and pi as e =
....828172 and pi = .....562951413, because the diameter of the
Natural Numbers since the endpoint is ...9999999 is that of a number
approx .....3333333333. Keep in mind that the Natural Numbers forms a
circle and distances on a circle allow for a variable pi (check any
Riemannian surface and pi varies on that surface.)
So I am allowed those numbers for e and pi
So now in the equation we have
e^(pi*i) = ....99999 and substituting for e as ...828172 we have
....828172 ^(pi*i) =....999999
Now since pi is variable on Riemannian surfaces we can say that pi is
of value 3 for the Natural Numbers and thus giving us this:
......828172 ^(3*i) =....999999
So what does that make or force as a value for (i)? It forces the value
of i to be approx that of idempotent of.....9999994 leaving 3*i as
equal to 2
thus we end up with .........828172 ^2 =....999999 but that is only an
approximation and not equal. The point is that we are getting closer
and closer to the value of i
>
> 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.
>
I can do it in Reals also. But I ignore the problems of defining
negative-Reals as Doubly Infinites. I dispose of all the negative-Reals
and force the Reals into Doubly Infinites. This not only eliminates
negative Reals but changes the geometry of the Reals from that of flat
plane Euclidean to that of a Riemannian surface.
Here I use the value of pi and e as Reals. So I start with the
equation:
e^(pi*i) = ....99999
And substituting for e and pi gives:
2.71.....^(3.14....*i) = ....99999
Now can I construct a meaningful value for (i)? I think so only it will
be another approximation and it will have to end up with eliminating
everything rightwards of the decimal-point for e and somehow making the
2 in e as a ....9999999. So I think this can be accomplished because of
an infinite-exponentiation. An infinite exponent yields an inifinite
string of digits and hopefully ....999999
So I want an (i) that when multiplied by pi yields an infinite string.
And using this infinite string will then eliminate the 0.71... portion
and transform the 2 of e into ....999999
Now let us get a rough and crude estimate of the value of this (i) by
thinking of pi as 3 and e as 2. Then the value of (i) is this number
approximately
infinity 1000000.......0000000.
Now the number .....9999999 can be rewritten as
infinity99999.....999999 Where the term "infinity indicates that the
terminal digit is 9 for ....999999 and where the terminal digit is 1
for 10000.....0000000
So that we finally have for our equation
e^(pi*i) = ....99999
becomes that of 2^ (3 * 10000.....000000) = .....99999999
Thus we have 3 *10000.....00000 which is equal to 300000.....000000
What can we think of a number as 3000000.....000000?
Well, it is like a "cube" and sort of like 2^3 which equals 8. Now 8 is
not 9 and the number 2^3000....00000 is ....88888888 but I am getting
closer to ....999999
So I am getting closer to the value of (i) in both Natural Numbers =
Infinite Integers and in Reals.
And the above number 100000.....00000 gives us new light, new insight
and new meaning of the old concept of imaginary-number in Reals. The
imaginary number of square root of -1 has meaning and value in p-adics.
But it also has meaning and value in Natural Numbers and Reals for
imaginary numbers are these type of numbers:
100000.....0000000 where .....999999 (Natural-Numbers) is the last
number in existence.
a_plutonium
You do not understand the problem and what I wrote, or the question I
asked.
The water box pictorial proves that the Peano Axioms gives us
immediately and directly the number ...99999. Endless adding of 1
yields instantly the Natural Number ....99999.
So the mystery is between that of 0,1,2 and ....99999. We already have
an infinite-integer, so how did we end up with it?
The trouble is explaining (my question) of how that endless adding of 1
transitions from say ....00000n to that of ....11111n or how that
endless adding of 1 transitions from ....11111n to that of .....22222n.
So the Peano Axioms gives us already ....999999, and the question and
mystery is how it smoothly navigates in the middle between 0, 1 and
.....9999999.
David has committed fallacy of finite tackling infinite.
hagman
a_plutonium schrieb:
> Proginoskes wrote:
> > a_plutonium wrote:
> > > (indirect method) is quite a boasting.
> > >
> > > So how does your alleged proof answer this question:
> > >
> > > Given an infinite set of boxes lined up in a row. Something like this:
> > > ______________
> > > | | | | | | | | | | | | | | |
> > >
> > > There are an infinity of these boxes and in each box starting from the
> > > first on on the right you pour in a 1 unit of water and which each box
> > > can hold 9 units of water.
> > >
> > > So you start a Peano Machine of a Successor function that pumps water
> > > into the first box and which the overflow goes into the next box and
> > > when it is full of 9 units of water the overflow goes to the next box
> > > to the left filling it with 9 units of water and on down the line of
> > > these infinite boxes.
> >
> > Okay; let's suppose that the machine takes 1 minute to pour 1 unit of
> > water in the box.
>
> Where is the element of "time" within the Peano axioms?
Hello???? Note who came up with that BS!
Where's the element of "water" in Peano axioms?
hagman
a_plutonium schrieb:
> >(See http://mathworld.wolfram.com/PeanosAxioms.html ):
> > 1. Zero is a number.
> > 2. If a is a number, the successor of a is a number.
> > 3. zero is not the successor of a number.
> > 4. Two numbers of which the successors are equal are themselves equal.
> > 5. (induction axiom.) If a set S of numbers contains zero and also the
> > successor of every number in S, then every number is in S.
>
> There is no time constraint in mathematics. We instantly know the
> Primes in Euclid proof are infinite and there was no time constraint in
> proving infinitude of primes. There is no time factor in any of
> mathematics as regards to proof.
>
> What there is above in those Peano Axioms is this Successor Machine
> that is the endless flow of water that fills every one of those boxes
> and instantly produces ....999999999
Let S be the set of all numbers with the property that at most finitely
many digits are non-zero.
(For example ...0000097134561050678346215061 is in this set).
Here are a few questions for you:
Q1: Is 0 an element of S?
Q2: Is there any number n such that n is in S but the successor of n is
not in S?
Q3: Is ...99999999999 in S?
Q4: Do the Peano Axioms instantly produce ...9999999999?
hagman
a_plutonium schrieb:
I've never noticed that.
Aha.
What is this number times 10?
>
> Now the number .....9999999 can be rewritten as
> infinity99999.....999999 Where the term "infinity indicates that the
> terminal digit is 9 for ....999999 and where the terminal digit is 1
> for 10000.....0000000
So, do "...9999999" and "10000....00000" have the same number of
digits?
More specifically, does the former have a "9" at the place where the
latter has a "1"?
Since the former is -1 and the latter an approximation to i (let's call
this approximation j for the moment), don't your arithmetic rules imply
that (j-1)*10+9=-1?
10000...00000 -1 = 09999...99999
09999...99999 * 10 = 99999...99990
99999...99990 + 9 = 99999...99999
Do you believe that (j-1)*10+9=-1 implies j=0?
Why not?
> becomes that of 2^ (3 * 10000.....000000) = .....99999999
Is that a calculation? If so, according to which rules?
Is 3*10000...00000 the smallest non-zero number n such that 2^n is odd?
> And the above number 100000.....00000 gives us new light, new insight
> and new meaning of the old concept of imaginary-number in Reals. The
> imaginary number of square root of -1 has meaning and value in p-adics.
Could you repeat the rules of multiplication so that we can verify that
i*i=-1 holds for that number?
Dik T. Winter
...
> Only Dik Winter agreed that both AP and Chris Heckman had a valid
> Infinitude of Primes Euclid Indirect Method, but that is not saying
> much because Dik rendered an invalid proof attempt himself when he
> included (5b) and (5c) and admitting that he has taught a wrong proof
> for 10 years, and will probably continue to teach invalid arguments.
Where did I admit any somesuch? I think it would be pretty stupid to
admit that because I never even have taught that subject. And I also
ever have admitted it was wrong.
> So
> we have two people who cannot give a valid Euclid Infinitude of
> Primes-- Dik Winter and Chris Heckman. Pity all the future students
> that receive a muddleheaded Infinitude of Primes from them.
I do not think there will be much more than 0 that receive it from me.
However, if I remember your biography right, you have been teaching
mathematics. Was it not in Australia? Did you ever teach that theorem
there? And if so, what method did you use?
> There is no *time factor* in the axioms of Peano.
> There is a Successor axiom of endless adding of 1 and so the Peano
> Axioms when written and before the ink is dry, the number .....9999999
> is produced.
I think not. At exactly what step produce the Peano axioms a "number"
where the representation does not have only 0's to the left of some
finite position?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
> Now here is an argument as to why e, pi and i are the only three
> transcendental numbers and why an Infinite Integer exists to solve that
> equation for i.
How do you define transcendental? According to the common definition 'i'
is not transcendental but algebraic.
Dik T. Winter
> But the equation could be for -2 where we adjust B to compensate for a
> -2 instead of a -1
> or we could substitute any n for -1 and make a compensation for the i.
Ah, your compensations are at least trivial.
> a_plutonium wrote:
...
> > as to why e, pi and i are the only three
> > transcendental numbers and why an Infinite Integer exists to solve that
> > equation for i.
...
> > But the equation could be for -2 where we adjust i to compensate for a
> > -2 instead of a -1
>
> But that would give you a different number. What AP is proposing, in a
> nutshell, is something like:
But here the compensation is less simple. We need to adjust i to
'i + ln(2)/pi'.
> Can't be done. The value of 'i' isn't allowed to be changed.
Of course.
> > So what this tells us is that any Natural Number is expressible in
> > terms of e, pi, and i.
>
> If you're allowed to change the value of i.
And if you are not allowed to you need ln(n) for every n.
> Here's a hint for AP: If you're going to use a term which has a meaning
> (like "transcendental"), either stick to that meaning, or say out in
> front that you're changing the meaning to something else.
His meaning of "transcendental" is already non-standard.
Proginoskes
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.
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
Proginoskes
a_plutonium wrote:
> Proginoskes wrote:
> > a_plutonium wrote:
> > > >
> > > > The Infinite Integers are NOT the Natural Numbers. I posted an
> > > > air-tight proof of this.
> > >
> > > From the man who cannot even prove Euclid Infinitude of Primes
> >
> > Only AP seems to think this, though.
> >
>
> Only Dik Winter agreed that both AP and Chris Heckman had a valid
> Infinitude of Primes Euclid Indirect Method, but that is not saying
> much because Dik rendered an invalid proof attempt himself when he
> included (5b) and (5c) and admitting that he has taught a wrong proof
> for 10 years, and will probably continue to teach invalid arguments.
I must have missed that.
> So we have two people who cannot give a valid Euclid Infinitude of
> Primes-- Dik Winter and Chris Heckman. Pity all the future students
> that receive a muddleheaded Infinitude of Primes from them.
>
>
> > > (indirect method) is quite a boasting.
> > >
> > > So how does your alleged proof answer this question:
> > >
> > > Given an infinite set of boxes lined up in a row. Something like this:
> > > ______________
> > > | | | | | | | | | | | | | | |
> > >
> > > There are an infinity of these boxes and in each box starting from the
> > > first on on the right you pour in a 1 unit of water and which each box
> > > can hold 9 units of water.
> > >
> > > So you start a Peano Machine of a Successor function that pumps water
> > > into the first box and which the overflow goes into the next box and
> > > when it is full of 9 units of water the overflow goes to the next box
> > > to the left filling it with 9 units of water and on down the line of
> > > these infinite boxes.
> >
> > Okay; let's suppose that the machine takes 1 minute to pour 1 unit of
> > water in the box.
>
> Where is the element of "time" within the Peano axioms?
(1) Because you've set up an experiment which is done in space-time, I
can use this concept.
(2) You can only add one 1 at a time; 0 + 1 = 1, not ...999.
> >(See http://mathworld.wolfram.com/PeanosAxioms.html ):
> > 1. Zero is a number.
> > 2. If a is a number, the successor of a is a number.
> > 3. zero is not the successor of a number.
> > 4. Two numbers of which the successors are equal are themselves equal.
> > 5. (induction axiom.) If a set S of numbers contains zero and also the
> > successor of every number in S, then every number is in S.
>
> There is no time constraint in mathematics.
There is no space constraint in mathematics, either, but that doesn't
seem to bother YOU. ("0 and 1 must be at a distance 1 from each
other.")
> We instantly know the
> Primes in Euclid proof are infinite and there was no time constraint in
> proving infinitude of primes. There is no time factor in any of
> mathematics as regards to proof.
>
> What there is above in those Peano Axioms is this Successor Machine
> that is the endless flow of water that fills every one of those boxes
> and instantly produces ....999999999
So AP would have us believe that 0 + 1 = ...999 = -1.
> > > So tell us how and why there are any boxes to the left that is not
> > > filled with water?
> > > And how your alleged proof overcomes that.
> >
> > It's a proof by induction.
> >
> > In the beginning, the box to the far right is empty, and so are all the
> > boxes to the left.
> >
> > One minute later, the box to its left is empty, and so are all the
> > boxes to the left. You have at most 1 box with liquid in it.
> >
> > One minute later, the box to ITS left is empty, and so are all the
> > boxes to the left. You have at most 2 boxes with liquid in it.
> >
> > At any point in time, there will only be a finite number of boxes which
> > can contain liquid.
> >
> > You simply never even get to a point where all of the boxes are full;
> > there's not enough time.
> >
> > No matter how much time you have, you will never fill the boxes.
> >
> >
>
> What a silly argument. And you are a college professor teaching
> mathematics?
Yes. Are YOU a college professor teaching mathematics?
> There is no *time factor* in the axioms of Peano.
> There is a Successor axiom of endless adding of 1 and so the Peano
> Axioms when written and before the ink is dry, the number .....9999999
> is produced.
>
> That pictorial is a proof that Natural-Numbers are the Infinite
> Integers and that pictorial displays to us that the Peano Axioms are
> self-contradictory. When Peano said that 0 is not the successor of any
> number-- he immediately made his axioms contradictory
No, he didn't. (You don't have a good grasp on infinite sets, do you?)
> because
> ....999999 precedes 0. Whether his axiom of Math Induction is allowable
> is up for debate. It may be that Math Induction is a technique to be
> used some of the time but not the status of an axiom of mathematics. I
> sense that Successor function does not allow for such a nice counting
> of Natural Numbers for which Math Induction is a counting process. How
> does one transition from those numbers ending with a 000s string to
> those ending with a 1111s string.
One doesn't, because ...111 is not a Natural Number.
BTW, if there is something wrong with the result that the Infinite
Integers are not the Natural Numbers, then there must be an error in
the proof. Find it.
> So I doubt Math Induction is
> consistent with Successor, but that they are contradictory.
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> a_plutonium wrote:
>
> > > >
> > > > Given an infinite set of boxes lined up in a row. Something like this:
> > > > ______________
> > > > | | | | | | | | | | | | | | |
> > > >
> > > > There are an infinity of these boxes and in each box starting from the
> > > > first on on the right you pour in a 1 unit of water and which each box
> > > > can hold 9 units of water.
> > > >
> > > > So you start a Peano Machine of a Successor function that pumps water
> > > > into the first box and which the overflow goes into the next box and
> > > > when it is full of 9 units of water the overflow goes to the next box
> > > > to the left filling it with 9 units of water and on down the line of
> > > > these infinite boxes.
>
> Now this pictorial of endless adding of 1 as water flowing into boxes
... which itself requires a time element ...
> and when full filling up the next box to the left ad infinitum may be
> instructive in another way. Maybe this pictorial can somehow pinpoint
> how the Natural Numbers transition from say a ....00000n to that of a
> ....11111n
According to another of AP's post, this doesn't happen; the boxes go
from ...000 to ...999 instantly.
> Notice that in the pictorial that the number 10 is never really visited
> whereas the first box goes from 0 to 1 to 2 to 3 to 4 to 5 to 6 to 7 to
> 8 to 9 and then as it begins to overflow into the next box on the left
> we have 19, then 29 then 39.
So ...0009 + 1 is really ...00019? That ...00020 isn't an Infinite
Integer at all?
> So what does this tell us?
That AP is on drugs, stupid, crazy, or clueless?
--- Christopher Heckman
Proginoskes
Don't go down that road. It gives AP conniption fits.
The only sane possibility is to accept the fact that the Infinite
Integers do not satisfy Peano's Axioms, and therefore are not Natural
Numbers.
--- Christopher Heckman
Proginoskes
In today's post, AP does not define A^B for arbitrary Infinite
Integers, and follows the non-consequences of this non-definition to
conclude that (-1) + 1 = Infinity.
(1) Where did we get geometry from?
(2) How do you know the Infinite Integers are laid out in a circle? Why
not a figure 8, or a parabola?
> Keep in mind that the Natural Numbers forms a
> circle and distances on a circle allow for a variable pi (check any
> Riemannian surface and pi varies on that surface.)
>
> So I am allowed those numbers for e and pi
Why is e = ...828172? What is the defintion of e? Can you even take
limits in the Infinite Integers? (If the answer to the last question is
no, then e cannot be defined.)
> So now in the equation we have
>
> e^(pi*i) = ....99999 and substituting for e as ...828172 we have
> ....828172 ^(pi*i) =....999999
>
> Now since pi is variable on Riemannian surfaces we can say that pi is
> of value 3 for the Natural Numbers and thus giving us this:
>
> ......828172 ^(3*i) =....999999
>
> So what does that make or force as a value for (i)? It forces the value
> of i to be approx that of idempotent of.....9999994 leaving 3*i as
> equal to 2
You never defined A^B where A and B are Infinite Integers, though.
> thus we end up with .........828172 ^2 =....999999 but that is only an
> approximation and not equal. The point is that we are getting closer
> and closer to the value of i
There is no i, because there are no Infinite Integers with the defining
property of i. (IOW, N^2+1 is never 0, for any Infinite Integer.)
> > 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.
> >
>
> I can do it in Reals also. But I ignore the problems of defining
> negative-Reals as Doubly Infinites. I dispose of all the negative-Reals
> and force the Reals into Doubly Infinites. This not only eliminates
> negative Reals
Watch out, AP, the IRS will come after you for this!
> but changes the geometry of the Reals from that of flat
> plane Euclidean to that of a Riemannian surface.
>
> Here I use the value of pi and e as Reals. So I start with the
> equation:
>
> e^(pi*i) = ....99999
>
> And substituting for e and pi gives:
>
> 2.71.....^(3.14....*i) = ....99999
>
> Now can I construct a meaningful value for (i)? I think so only it will
> be another approximation and it will have to end up with eliminating
> everything rightwards of the decimal-point for e and somehow making the
> 2 in e as a ....9999999. So I think this can be accomplished because of
> an infinite-exponentiation. An infinite exponent yields an inifinite
> string of digits and hopefully ....999999
There is no "hopefully" in real mathematics.
Furthermore, A^B has never been defined, where A and B are Infinite
Integers.
AP must be smoking something really good here.
> So I want an (i) that when multiplied by pi yields an infinite string.
> And using this infinite string will then eliminate the 0.71... portion
> and transform the 2 of e into ....999999
>
> Now let us get a rough and crude estimate of the value of this (i) by
> thinking of pi as 3 and e as 2. Then the value of (i) is this number
> approximately
>
> infinity 1000000.......0000000.
Of course, Infinity is not an Infinite Integer, because it is not of
that form.
> Now the number .....9999999 can be rewritten as
> infinity99999.....999999 Where the term "infinity indicates that the
> terminal digit is 9 for ....999999 and where the terminal digit is 1
> for 10000.....0000000
>
> So that we finally have for our equation
>
> e^(pi*i) = ....99999
>
> becomes that of 2^ (3 * 10000.....000000) = .....99999999
>
> Thus we have 3 *10000.....00000 which is equal to 300000.....000000
>
> What can we think of a number as 3000000.....000000?
>
> Well, it is like a "cube" and sort of like 2^3 which equals 8. Now 8 is
> not 9 and the number 2^3000....00000 is ....88888888 but I am getting
> closer to ....999999
>
> So I am getting closer to the value of (i) in both Natural Numbers =
> Infinite Integers and in Reals.
>
> And the above number 100000.....00000 gives us new light, new insight
> and new meaning of the old concept of imaginary-number in Reals. The
> imaginary number of square root of -1 has meaning and value in p-adics.
> But it also has meaning and value in Natural Numbers and Reals for
> imaginary numbers are these type of numbers:
>
> 100000.....0000000 where .....999999 (Natural-Numbers) is the last
> number in existence.
So AP is saying (-1) + 1 = Infinity? Far out, man!
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> David R Tribble wrote:
> > Archimedes Plutonium wrote:
> > > So what does this tell us? Can we construct a water system of boxes
> > > that starts with 0 then goes to 1 then to 2 and never skips any number
> > > and goes out to ....9999999?
> >
> > How could you? If every time we take a finite number and add 1
> > to it to get the next finite number, how can you ever get an infinite
> > number? Is there some finite natural number that is so large
> > that adding 1 to it somehow breaks the rule and gives you an
> > infinite number? What is this magical "almost infinite" finite
> > number?
>
> You do not understand the problem and what I wrote, or the question I
> asked.
>
> The water box pictorial proves that the Peano Axioms gives us
> immediately and directly the number ...99999. Endless adding of 1
> yields instantly the Natural Number ....99999.
In other words, 0 + Infinity = ...999 = -1.
> So the mystery is between that of 0,1,2 and ....99999. We already have
> an infinite-integer, so how did we end up with it?
>
> The trouble is explaining (my question) of how that endless adding of 1
> transitions from say ....00000n to that of ....11111n or how that
> endless adding of 1 transitions from ....11111n to that of .....22222n.
It's no mystery, because it doesn't happen. I proved this, and AP can't
find any mistakes in that proof.
> So the Peano Axioms gives us already ....999999, and the question and
> mystery is how it smoothly navigates in the middle between 0, 1 and
> .....9999999.
>
> David has committed fallacy of finite tackling infinite.
He hasn't committed any fallacies, since he didn't claim to have proven
anything; he just asked questions.
--- Christopher Heckman
Proginoskes
This is an outline of my proof that the Infinite Integers are not the
Natural Numbers, according to Peano's Axioms. AP evidently can't find
anything wrong with it and has resorted to ad hominem attacks and Straw
Man arguments (the water-pouring machine), both of which are logical
fallacies.
--- Christopher Heckman
Proginoskes
No one other than AP has.
Infinity is not a number. It isn't even an "Infinite Integer", because
it's not of the right form.
> > Now the number .....9999999 can be rewritten as
> > infinity99999.....999999 Where the term "infinity indicates that the
> > terminal digit is 9 for ....999999 and where the terminal digit is 1
> > for 10000.....0000000
>
> So, do "...9999999" and "10000....00000" have the same number of
> digits?
> More specifically, does the former have a "9" at the place where the
> latter has a "1"?
> Since the former is -1 and the latter an approximation to i (let's call
> this approximation j for the moment), don't your arithmetic rules imply
> that (j-1)*10+9=-1?
> 10000...00000 -1 = 09999...99999
> 09999...99999 * 10 = 99999...99990
> 99999...99990 + 9 = 99999...99999
> Do you believe that (j-1)*10+9=-1 implies j=0?
> Why not?
>
> > becomes that of 2^ (3 * 10000.....000000) = .....99999999
>
> Is that a calculation? If so, according to which rules?
> Is 3*10000...00000 the smallest non-zero number n such that 2^n is odd?
>
> > And the above number 100000.....00000 gives us new light, new insight
> > and new meaning of the old concept of imaginary-number in Reals. The
> > imaginary number of square root of -1 has meaning and value in p-adics.
>
> Could you repeat the rules of multiplication so that we can verify that
> i*i=-1 holds for that number?
AP doesn't worry about details like consistency.
--- Christopher Heckman
Proginoskes
Dik T. Winter wrote:
> In article <1168104847.7...@11g2000cwr.googlegroups.com> "a_plutonium" <a_plu...@hotmail.com> writes:
> ...
> > Only Dik Winter agreed that both AP and Chris Heckman had a valid
> > Infinitude of Primes Euclid Indirect Method, but that is not saying
> > much because Dik rendered an invalid proof attempt himself when he
> > included (5b) and (5c) and admitting that he has taught a wrong proof
> > for 10 years, and will probably continue to teach invalid arguments.
>
> Where did I admit any somesuch? I think it would be pretty stupid to
> admit that because I never even have taught that subject. And I also
> ever have admitted it was wrong.
Good. I was wondering about that.
I guess in some parallel universe (which really DOES look like a
plutonium atom), you did apologize.
So AP is a liar as well.
> > So
> > we have two people who cannot give a valid Euclid Infinitude of
> > Primes-- Dik Winter and Chris Heckman. Pity all the future students
> > that receive a muddleheaded Infinitude of Primes from them.
>
> I do not think there will be much more than 0 that receive it from me.
> However, if I remember your biography right, you have been teaching
> mathematics. Was it not in Australia? Did you ever teach that theorem
> there? And if so, what method did you use?
>
> > There is no *time factor* in the axioms of Peano.
> > There is a Successor axiom of endless adding of 1 and so the Peano
> > Axioms when written and before the ink is dry, the number .....9999999
> > is produced.
>
> I think not. At exactly what step produce the Peano axioms a "number"
> where the representation does not have only 0's to the left of some
> finite position?
Great. Now YOU'LL give him a conniption fit, too.
(This is of course the basic idea of my proof that the Infinite
Integers are not the Natural Numbers. AP can't find anything wrong with
this proof, and has resorted to ad hominem attacks (which evidently
include distorting the truth) and a Straw Man argument. These are both
logical fallacies and therefore unacceptable counterarguments.)
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> Proginoskes wrote:
> > a_plutonium wrote:
> > (This makes me wonder whether AP even gave a second thought to seeing
> > whether his proposed axioms actually work. A true mathematician would.)
> >
> > > The important conclusions derived from these axioms is that the above
> > > set of numbers forms a spherical Riemannian geometry.
> > >
> > > The Axioms for the Reals would look like this:
> > > (1) There are two numbers call them 0 and 1 and they are 1 unit
> > > distance apart
> > > (2) There are an infinite amount of numbers between 0 and 1
> > > (3) There are operators of add, subtract, multiply, divide, square root
> >
> > Again, the Axioms for the Reals don't produce the set of real numbers
> > (up to isomorphism). You can let S = {0, 1, 1/2, 1/3, 1/4, 1/5, ...},
> > which satisfies (1) and (2), and use arbitrary definitions to satisfy
> > (3). For instance, if x and y are in S, define
> >
> > x + y = 0
> > x * y = 0
> > x - y = 0
> > x / y = 0
> > sqrt(x) = 0
> >
> > And again, this easy counterexamples suggests that AP doesn't think
> > about what he posts.
>
> The specification of the operators eliminates your fictional
> counterexample
Okay; the set of algebraic numbers also satisfies all of the axioms,
including the definitions of the arithmetic operations.
--- Christopher Heckman
Proginoskes
Proginoskes wrote:
> [...]
> (See http://mathworld.wolfram.com/PeanosAxioms.html ):
>
> 1. Zero is a number.
> 2. If a is a number, the successor of a is a number.
> 3. zero is not the successor of a number.
> 4. Two numbers of which the successors are equal are themselves equal.
> 5. (induction axiom.) If a set S of numbers contains zero and also the
> successor of every number in S, then every number is in S.
> Proof:
> (1) Suppose that the Infinite Integers model the Natural Numbers.
> (Reducio ad Absurdum assumption)
> (2) Let S be the set of all Infinite Integers whose digits are
> eventually 0.
> (3) Clearly 0 is in S. (definition of S)
> (4) Now suppose N is in S. (direct proof, assumption)
> (5) By the definition of S, N[i] = 0 for i >= k, for some k.
> (6) The successor of N has the property that (succ N)[i] = 0 for all i
> >= k+1; succ N might require one more nonzero digit than N, but not any more. (property of +)
> (7) Then the successor of N is in S. (definition of S)
> (8) Thus: If N is in S, then the successor of N is in S. (direct proof;
> discharge (4)-(7).)
> (9) By Peano Axiom #5, this set S is the set of all Natural Numbers.
> (10) By assumption, the set of all Natural Numbers is the set of all
> Infinite Integers.(1)
> (11) Therefore every Infinite Integer is in S. (substitution, (9) and
> (10).)
> (12) ...111 is an Infinite Integer. (Fact)
> (13) Therefore ...1111 is in S. (Substitution, (11) and (12).)
> (14) ...111 does not have any 0's in it. (Fact)
> (15) ...1111 is not in S (since the digits are not eventually 0;
> definition of S (2), (14))
> (16) (13) and (15) form a contradiction.
> (17) Hence the Infinite Integers do not model the Natural Numbers.
> (indirect proof; discharge (1)-(16).)
>
> QED.
UPDATE.
Since I posted this proof, AP has sent out ad hominem attacks, lied
(Dik Winter never did admit his proof was wrong, and had not taught a
proof-based course, as AP claimed), and built a Straw Man (a
water-pouring machine, which requires water, space, and gravity, none
of which are a part of mathematics*), all of which are logical
fallacies. One thing he has not done is to point out where he thinks
there are any mistakes in this proof, and my theory is that he can't.
--- Christopher Heckman
* You know, I think I can anticipate another logical fallacy coming up
soon: AP will claim that "Physics is King", and that mathematics is
only a subordinate. This is, of course, a non sequitur.
a_plutonium
hagman wrote:
> >
> > Okay, I am justified in Infinite Integers in calling e and pi as e =
> > ....828172 and pi = .....562951413, because the diameter of the
> > Natural Numbers since the endpoint is ...9999999 is that of a number
> > approx .....3333333333. Keep in mind that the Natural Numbers forms a
> > circle and distances on a circle allow for a variable pi (check any
> > Riemannian surface and pi varies on that surface.)
>
> I've never noticed that.
That tells me you are not a mathematician. Every mathematician worth
his weight in salt knows that a circle drawn on a globe has a variable
pi depending on the size of the circle because of the curvature of the
globe.
The number ....9999 times 10 is .....999990
>
> >
> > Now the number .....9999999 can be rewritten as
> > infinity99999.....999999 Where the term "infinity indicates that the
> > terminal digit is 9 for ....999999 and where the terminal digit is 1
> > for 10000.....0000000
>
> So, do "...9999999" and "10000....00000" have the same number of
> digits?
Yes, and the subtraction since ....99999 is the larger is this new
number
899999......9999999
> More specifically, does the former have a "9" at the place where the
> latter has a "1"?
What a silly question, since the number ....999999 has only 9 as
digits.
> Since the former is -1 and the latter an approximation to i (let's call
> this approximation j for the moment), don't your arithmetic rules imply
> that (j-1)*10+9=-1?
> 10000...00000 -1 = 09999...99999
> 09999...99999 * 10 = 99999...99990
> 99999...99990 + 9 = 99999...99999
> Do you believe that (j-1)*10+9=-1 implies j=0?
> Why not?
>
.....99999 is not -1. It can be replaced for -1, just like an actor is
not Alexander the Great but an actor pretending to be such.
The numbers in mathematics of i, j, k are about to be replaced by these
Infinite Integers where i could be the number 10000.....00000. What is
j and k, I do not yet know.
> > becomes that of 2^ (3 * 10000.....000000) = .....99999999
>
> Is that a calculation? If so, according to which rules?
> Is 3*10000...00000 the smallest non-zero number n such that 2^n is odd?
>
I was estimating what numbers would deliver ....9999999 for e, pi, i
Obviously a cube of 2 is only 8, and I need a 9 and the number
30000......00000 as the superexponent makes that into ....88888888, but
I needed ....9999999
But that is a good first approximation.
> > And the above number 100000.....00000 gives us new light, new insight
> > and new meaning of the old concept of imaginary-number in Reals. The
> > imaginary number of square root of -1 has meaning and value in p-adics.
>
> Could you repeat the rules of multiplication so that we can verify that
> i*i=-1 holds for that number?
Well we know that .....999999 x .....999999 = ....000001
and we know that 10000......000000 x 10000....000000 = 100000....00000.
Now this new number has one more digit than does ....999999 and
....99999 is the largest digit possible, so that if we subtract a
larger number of 10000.....00000 from the smaller ....99999 we end up
with a -1
I will have to look up and see what j and k were, but I have a
suspicion that j and k were just spurious concoctions of no real value
or meaning.
a_plutonium
I am using the old terms for now, but they will have to change
drastically when the dust begins to settle on Infinite Integers. Old
mathematics had artifices such as negative numbers, such as i,j,k with
no numeric value. And the old term of "transcendental" as that being
not algebraic was a rather shoddy definition of what is really going
on. To define something which it is "not" is a poor definition. We want
definitions of what something "is" rather than what it is "not".
So I am keeping and using the old terms even though they are going to
be changed.
For instance, in Infinite Integers you already noted that there is a
quadrapole of primes such as ....82817213 and ....82817211 and
....82817209 and 82817207 and easily proven that there are an infinite
set of these quadrapole primes so here is a new term already in
quadrapole.
But we have the new terms of these:
1) prime irrationals an even irrationals and odd irrationals
2) prime transcendental and even transcendental and odd transcendental
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. And since most numbers are fixed numbers 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.
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.
Proginoskes
a_plutonium wrote:
> hagman wrote:
> [...]
> > > Now the number .....9999999 can be rewritten as
> > > infinity99999.....999999 Where the term "infinity indicates that the
> > > terminal digit is 9 for ....999999 and where the terminal digit is 1
> > > for 10000.....0000000
> >
> > So, do "...9999999" and "10000....00000" have the same number of
> > digits?
>
> Yes, and the subtraction since ....99999 is the larger is this new
> number
> 899999......9999999
So ...9999 is greater than 1...000? That would make -1 greater than
Infinity.
> > More specifically, does the former have a "9" at the place where the
> > latter has a "1"?
>
> What a silly question, since the number ....999999 has only 9 as
> digits.
Actually, it isn't, since the 1 in 1...000 isn't in the nth digit, for
any positive (finite) integer n.
The way that I make sense of all this place-values is the following: A
normal Infinite Integer can be represented as a function from N (the
Natural Numbers) to {0, 1, 2, ..., 9}.
For instance, the Infinite Integer ...2345 is represented by the
function f(n) where
f(0) = 5, f(1) = 4, f(2) = 3, f(3) = 2, etc.
With an Infinite Integer expressed in terms of Natural Numbers, it
becomes (reasonably) easy to define what the Infinite Integers f+g,
f-g, f*g are, given Infinite Integers f and g, since all of these
operations proceed "right to left" (least significant digit to the more
significant digits). (This is assuming, of course, that you have enough
time to do an infinite number of calculations. 8-))
Division and taking square roots can get messy. If you are using the
traditional algorithm to calculate the square root of ...888, you have
to break the 8's into blocks of two (from the right). The algorithm
gets stuck at this point, because (1) there is no "most significant
digit", and (2) Even if there was, you don't know whether you should
start off with the first group being 8 or 88.
For 1...000, the idea is to take one of these functions f and add a new
element (w: "omega" is a good choice) to the domain of f. So a
"w-Infinite Integer" (for lack of a good name) can be represented by a
function from {w, 0, 1, 2, ...} to {0, 1, ..., 9}. 1...000 in this
context would be represented by the function f satisfying the following
conditions:
(1) f(n) = 0, for any Natural Number n
(2) f(w) = 1.
Infinite Integers can be expressed as a w-Infinite Integer whose
"first" (wth) digit is 0.
The largest w-Infinite Integer would be 9...999.
To add two w-Infinite Integers, you would add the "Infinite Integer"
part, and the extra digits separately. Then, if there is some k such
that you are carrying when calculating all places to the left of the
k'th, then you add 1...000:
2...777 + 3...555 := 5...3332 + 1...000 = 6...3332.
Of course, you could include more digits to the left of the wth,
creating "(w+1)-Infinite Integers", "2w-Infinite Integers", even
"w^2-Infinite Integers". And that's for a start. 8-)
In short: Infinite Integers are weird things already, and w-Infinite
Integers even more so.
> [...]
> .....99999 is not -1. It can be replaced for -1, just like an actor is
> not Alexander the Great but an actor pretending to be such.
Well, Archimedes Plutonium didn't post this last reply, but rather a
human being using the name of Archimedes Plutonium.
> The numbers in mathematics of i, j, k are about to be replaced by these
> Infinite Integers where i could be the number 10000.....00000. What is
> j and k, I do not yet know.
Beats me, as well. In quaternions, i, j, and k are all distinct "square
roots of -1" with the properties that
ij = k and ji = -k.
> > > becomes that of 2^ (3 * 10000.....000000) = .....99999999
> >
> > Is that a calculation? If so, according to which rules?
> > Is 3*10000...00000 the smallest non-zero number n such that 2^n is odd?
> >
>
> I was estimating what numbers would deliver ....9999999 for e, pi, i
> Obviously a cube of 2 is only 8, and I need a 9 and the number
> 30000......00000 as the superexponent makes that into ....88888888, but
> I needed ....9999999
>
> But that is a good first approximation.
What you need first is a definition of what A^B is for Infinite
Integers A,B. About the only things that would be required are that A^B
is the usual exponentiation when the digits of A and B are eventually
0, and the usual exponent laws.
> > > And the above number 100000.....00000 gives us new light, new insight
> > > and new meaning of the old concept of imaginary-number in Reals. The
> > > imaginary number of square root of -1 has meaning and value in p-adics.
> >
> > Could you repeat the rules of multiplication so that we can verify that
> > i*i=-1 holds for that number?
>
>
> Well we know that .....999999 x .....999999 = ....000001
>
> and we know that 10000......000000 x 10000....000000 = 100000....00000.
We don't know that; only AP knows that.
Someone posted a proof that there is no Infinite Integer which can
equal i (since n^2 cannot end in 99, for all Natural Numbers n), but
that's only in base 10. If we were working with base 5, then there
_would_ be a number whose square is -1. (For a proof, look up Hensel's
Lemma.)
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> Dik T. Winter wrote:
> > In article <1168072017.3...@38g2000cwa.googlegroups.com> "a_plutonium" <a_plu...@hotmail.com> writes:
> > > Now here is an argument as to why e, pi and i are the only three
> > > transcendental numbers and why an Infinite Integer exists to solve that
> > > equation for i.
> >
> > How do you define transcendental? According to the common definition 'i'
> > is not transcendental but algebraic.
>
> drastically when the dust begins to settle on Infinite Integers. Old
> mathematics had artifices such as negative numbers, such as i,j,k with
> no numeric value. And the old term of "transcendental" as that being
> not algebraic was a rather shoddy definition of what is really going
> on. To define something which it is "not" is a poor definition. We want
> definitions of what something "is" rather than what it is "not".
Infinite sets can be defined differently from "not finite" in the
following way:
DEFINITION. A set S is infinite iff there is a bijection f from S to
some proper subset T of S.
For instance, if S = {0, 1, 2, 3, ...}, S is infinite; the proof of
this is to take the proper subset
T = {1, 2, 3, 4, ...} and set up the bijection f from S to T by the
formula
f(n) = n+1.
> [...]
> 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
... except that most numbers are not computable. So the analogy breaks
down.
> 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
a_plutonium
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
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.
a_plutonium
Proginoskes wrote:
> a_plutonium wrote:
>
> Division and taking square roots can get messy. If you are using the
> traditional algorithm to calculate the square root of ...888, you have
> to break the 8's into blocks of two (from the right). The algorithm
> gets stuck at this point, because (1) there is no "most significant
> digit", and (2) Even if there was, you don't know whether you should
> start off with the first group being 8 or 88.
There are many algorithms and one of my favorite is the one where you
give an upper and lower bound and sandwich the desired square root
inbetween. For example the square root of 10 starts out with knowing it
is between 3 and 4. So we have the first number of 3 and now we
sandwich in between to get the next digit of 3.1 since 3.2 overshots
10.
So some difficult Infinite Integers are much easier to solve for square
root using this algorithm.
>
> For 1...000, the idea is to take one of these functions f and add a new
> element (w: "omega" is a good choice) to the domain of f. So a
> "w-Infinite Integer" (for lack of a good name) can be represented by a
> function from {w, 0, 1, 2, ...} to {0, 1, ..., 9}. 1...000 in this
> context would be represented by the function f satisfying the following
> conditions:
>
That is somewhat cute and gives the omega idea some validity, but the
omega idea was from the old fallacious period of Cantor who led the
world of mathematics into falsehoods. He should have known and the
successor generations of mathematicians should have known that
"infinity" comes in only one type. And there is an easy proof of this
claim because if you make the world with a hierarchy of infinities you
make the world with various levels of and various types of the number
0. And when Cantor did his set theory nonsense that the Reals have a
different class of infinity from the Natural Numbers, if he had his
wits about him, he would have realized that if Reals were a different
infinity from Natural Numbers then the world has two different numbers
for 0 which are totally different from one another. The world does not
have that. Emptiness, or nothing, or zero comes in only one type, one
class, one concept, likewise the inverse of nothing is infinity. So if
0 is of one type then infinity is of one type.
So this Omega stuff is just more phony baloney.
>
> > The numbers in mathematics of i, j, k are about to be replaced by these
> > Infinite Integers where i could be the number 10000.....00000. What is
> > j and k, I do not yet know.
>
> Beats me, as well. In quaternions, i, j, and k are all distinct "square
> roots of -1" with the properties that
> ij = k and ji = -k.
>
Okay, I think I know today what i, j, and k are as mathematical
numbers. They are what this new number of 100000.....000000 represents
when .....999999 is the last and largest Infinite Integer. We need the
negative sign. Just like in Reals the smallest Real if you could not
have negative-Reals is the number 0. The largest Infinite Integer is
....99999 and the smallest Infinite Integer is 0. So, now when you
reach ....999999 by endlessly adding 1, can you go any further without
going back to 0 and then 1 and on up to ....999999 again? In Reals we
went past 0 by throwing up a negative-sign on the positive Reals and
thus creating an infinite supply of new Reals in the opposite direction
of the positive Reals.
Now one must keep in mind that the Reals form an intrinsic geometry and
that is flat plane Euclidean Geometry. When you restrict numbers as
infinite rightwards with a finite portion leftwards, in that
restriction, you create or give birth to a geometry that is intrinsic
to those numbers. Likewise when you say the numbers are infinite
leftward strings you automatically create and give birth to a geometry
that accommodates those numbers and in the case of the Natural Numbers
as 0,1,2,3,....., .....999999 the geometry that is intrinsically formes
is that of a Riemannian surface, such as the surface of a globe or
sphere or 3-d ellipsoid etc.
So, now, like in Reals would be only half the set of all Reals if we
ignored negative Reals. Likewise for Natural Numbers, for they go from
0,1,2,3,4,....., ....9999999 but that is only a transit on a
great-circle or line of longitude on a sphere. When one travels from
the North Pole and gets to the South Pole and continues on that same
line of longitude and returns to the North Pole, well, he can then pick
another line of longitude that takes him through different cities on a
different line of longitude.
So, as the positive-Reals creates negative-Reals, the positive Natural
Numbers of 0,1,2,3,4,...., .....999999 creates Negative-Natural-Numbers
of 0,-1,-2,-3,-4,(-...99999)
Now we can assign numerical value to i, j, and k where i is perhaps
(10000.....00000)
And where either j or k are either (-10000....00000) or
(-......9999999) depending on which would satisfy the algebra of ij =
k and ji = -k.
By the way, physics needs this i, j, and k and when physics needs
something then I am more obliged to heed that demand. Since all of
mathematics is a secondary offshoot of physics.
Dik T. Winter
> Dik T. Winter wrote:
...
> > I think not. At exactly what step produce the Peano axioms a "number"
> > where the representation does not have only 0's to the left of some
> > finite position?
>
> Great. Now YOU'LL give him a conniption fit, too.
Oh, I have seen that already many times. It did already occur when his
name still was Ludwig Plutoniom, long, long ago.
Dik T. Winter
> hagman wrote:
> > > Okay, I am justified in Infinite Integers in calling e and pi as e =
> > > ....828172 and pi = .....562951413, because the diameter of the
> > > Natural Numbers since the endpoint is ...9999999 is that of a number
> > > approx .....3333333333. Keep in mind that the Natural Numbers forms a
> > > circle and distances on a circle allow for a variable pi (check any
> > > Riemannian surface and pi varies on that surface.)
> >
> > I've never noticed that.
>
> That tells me you are not a mathematician. Every mathematician worth
> his weight in salt knows that a circle drawn on a globe has a variable
> pi depending on the size of the circle because of the curvature of the
> globe.
Every mathematician worth his weight in salt knows that on a globe the
circumference of a circle is *not* equal to 2.pi.r, when you measure
r along the globe. And that the area is *not* equal to pi.r^2 when
you want to measure the area across the globe surface. Pi is a number
that has been defined as the ratio between the circumference of a circle
and its diameter *on a flat surface*.
Dik T. Winter
> Dik T. Winter wrote:
> > In article <1168072017.3...@38g2000cwa.googlegroups.com> "a_plutonium" <a_plu...@hotmail.com> writes:
> > > Now here is an argument as to why e, pi and i are the only three
> > > transcendental numbers and why an Infinite Integer exists to solve that
> > > equation for i.
> >
> > How do you define transcendental? According to the common definition 'i'
> > is not transcendental but algebraic.
>
> drastically when the dust begins to settle on Infinite Integers.
You do use old terms with new meanings. How do you define 'transcendental'?
> Old
> mathematics had artifices such as negative numbers, such as i,j,k with
> no numeric value.
What do you mean with the latter?
> And the old term of "transcendental" as that being
> not algebraic was a rather shoddy definition of what is really going
> on. To define something which it is "not" is a poor definition. We want
> definitions of what something "is" rather than what it is "not".
In that case you should not use the terms "transcendental" or "irrational",
because both are defined in terms of something they are not. Also you
should not use the term "infinite" because that is only defined in terms
of something it is not.
> So I am keeping and using the old terms even though they are going to
> be changed.
You are using them, but with a different meaning.
> For instance, in Infinite Integers you already noted that there is a
> quadrapole of primes such as ....82817213 and ....82817211 and
> ....82817209 and 82817207 and easily proven that there are an infinite
> set of these quadrapole primes so here is a new term already in
> quadrapole.
But, how do you *prove* they all are prime? Using your arithmetic for
multiplication, neither is prime (if we assume that multiplication is
the inverse of division). Now suppose that ...82817207 is not divisible
by 3. Using normal arithmetic means that if you divide it by 3 you
get a remainder of 1 or 2. Using the first you would get that
...82817209 is divisible by 3, using the second you get that ...82817211
is divisible by 3. So to verify your proof we need to know how to
perform division. So, what is the result of dividing those number
by 3?
> But we have the new terms of these:
> 1) prime irrationals an even irrationals and odd irrationals
> 2) prime transcendental and even transcendental and odd transcendental
But no definitions.
> 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. And since most numbers are fixed numbers 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.
A strange definition of the word "number". A number is fixed in common
terminology.
> 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.
Eh? For e and pi we can fetch any digit we want. Why do you think that
is not possible?
> Same thing for e, only it is approx 19 subshells
> occupied inside of approx 7 shells.
e is (by definition) sum{n = 0 .. oo} 1/n!. You can get any digit you
want.
> So I think i is algebraic and I just loosely group i with e and pi.
Interesting. What are the digits of i? You are using (behind the
screen) 10-adics. And in the 10-adics the square root of -1 does not
exist.
> 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.
Yeah. How do I derive sqrt(2) using these numbers?
> 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.
In that case, use the old terms with the old definitions.
Dik T. Winter
> a_plutonium wrote:
...
> > I am using the old terms for now, but they will have to change
> > drastically when the dust begins to settle on Infinite Integers. Old
> > mathematics had artifices such as negative numbers, such as i,j,k with
> > no numeric value. And the old term of "transcendental" as that being
> > not algebraic was a rather shoddy definition of what is really going
> > on. To define something which it is "not" is a poor definition. We want
> > definitions of what something "is" rather than what it is "not".
>
> Infinite sets can be defined differently from "not finite" in the
> following way:
>
> DEFINITION. A set S is infinite iff there is a bijection f from S to
> some proper subset T of S.
That is Dedekind-infinite in standard terminology. The two concepts are
the same only under a weak form of AC. Without it there are infinite sets
that are Dedekind-finite.
> > 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
I do not think so. I never heard about the concept of "growing" number.
a_plutonium
Dik T. Winter wrote:
>
> Every mathematician worth his weight in salt knows that on a globe the
> circumference of a circle is *not* equal to 2.pi.r, when you measure
> r along the globe. And that the area is *not* equal to pi.r^2 when
> you want to measure the area across the globe surface. Pi is a number
> that has been defined as the ratio between the circumference of a circle
> and its diameter *on a flat surface*.
Dik can you give us the range of values of pi for all circles drawn on
a sphere?
And can you give us the range of values of pi for all circles drawn on
a 3-d ellipse-- the ellipsoid?
There is a curious question arising in my mind about now. That a sphere
has two poles for all great-circles except for the equator. What about
all other smooth shaped Riemannian objects like a ellipsoid. Can we say
that all smooth Riemannian objects have two poles? Obviously an
ellipsoid would not have an equator. Any theorems pertaining to poles?
a_plutonium
Dik T. Winter wrote:
(snipped)
>
> > For instance, in Infinite Integers you already noted that there is a
> > quadrapole of primes such as ....82817213 and ....82817211 and
> > ....82817209 and 82817207 and easily proven that there are an infinite
> > set of these quadrapole primes so here is a new term already in
> > quadrapole.
>
> But, how do you *prove* they all are prime? Using your arithmetic for
> multiplication, neither is prime (if we assume that multiplication is
> the inverse of division). Now suppose that ...82817207 is not divisible
> by 3. Using normal arithmetic means that if you divide it by 3 you
> get a remainder of 1 or 2. Using the first you would get that
> ...82817209 is divisible by 3, using the second you get that ...82817211
> is divisible by 3. So to verify your proof we need to know how to
> perform division. So, what is the result of dividing those number
> by 3?
>
I am happier proving easy things than hard ones. Irrationals by
definition do not have a ratio (other than 1). Since they can not be
put into a ratio means they are not evenly divisible. Sometimes I type
too fast and make typing errors like the time I said ....828172 was a
prime when obviously it is divisible by 2. That number is the transpose
of e into Infinite Integers. But I can slip in front of the "2" that of
7, 9, 11,13 and know that it is still irrational and thus not having a
"ratio" and thus not evenly divisible and finally it is prime.
With irrationals like e, we worry in division as to its end digits and
not that it starts with 2, but in transposing e to Infinite Integers
the 2 becomes problematic since it now ends the string and obviously
divisible by 2 and obviously able to put into a "ratio" of 2/z and thus
not prime. So on some of the Irrationals when transposed will have to
delete the "2" or "0" or "5" or "6" or "8".
As to why "9" serves because many prime numbers end in 9 digit like 19.
And so if ....82817 is irrational then ....82817209 is still
irrational.
> Eh? For e and pi we can fetch any digit we want. Why do you think that
> is not possible?
I was explaining why pi and e are transcendental. It is because they
are not fixed, but growing. They are dependent on "time itself". They
are not fully formed.
>
> > Same thing for e, only it is approx 19 subshells
> > occupied inside of approx 7 shells.
>
> e is (by definition) sum{n = 0 .. oo} 1/n!. You can get any digit you
> want.
>
Since e and pi are growing you cannot get all the digits at once.
Algebraic numbers are numbers where all the digits pre-exist and you
can get all of them, but not with pi or e.
So I was giving a deeper meaning what Transcendental means.
> > So I think i is algebraic and I just loosely group i with e and pi.
>
> Interesting. What are the digits of i? You are using (behind the
> screen) 10-adics. And in the 10-adics the square root of -1 does not
> exist.
The number i is algebraic and it is this value 100000.....000000. At
least that is what I thought it was. But after spending some time with
the numbers (-100000.....00000) and (-....999999) that maybe i or j or
k are one of them.
>
> > 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.
>
> Yeah. How do I derive sqrt(2) using these numbers?
Simple question. How do you derive sqrt 2, well the first counting
number that has a square root is 4 which is 2. Get with the program,
Dik.
>
> > 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.
>
> In that case, use the old terms with the old definitions.
Okay, i is algebraic.
Proginoskes
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
Proginoskes
a_plutonium wrote:
> Proginoskes wrote:
> > a_plutonium wrote:
>
> >
> > Division and taking square roots can get messy. If you are using the
> > traditional algorithm to calculate the square root of ...888, you have
> > to break the 8's into blocks of two (from the right). The algorithm
> > gets stuck at this point, because (1) there is no "most significant
> > digit", and (2) Even if there was, you don't know whether you should
> > start off with the first group being 8 or 88.
>
> There are many algorithms and one of my favorite is the one where you
> give an upper and lower bound and sandwich the desired square root
> inbetween. For example the square root of 10 starts out with knowing it
> is between 3 and 4. So we have the first number of 3 and now we
> sandwich in between to get the next digit of 3.1 since 3.2 overshots
> 10.
>
> So some difficult Infinite Integers are much easier to solve for square
> root using this algorithm.
But there are Infintie Integers which you can't even get a starting
bound, such as ...888. The algorithm is useless in that case.
> > For 1...000, the idea is to take one of these functions f and add a new
> > element (w: "omega" is a good choice) to the domain of f. So a
> > "w-Infinite Integer" (for lack of a good name) can be represented by a
> > function from {w, 0, 1, 2, ...} to {0, 1, ..., 9}. 1...000 in this
> > context would be represented by the function f satisfying the following
> > conditions:
> >
>
> That is somewhat cute and gives the omega idea some validity, but the
> omega idea was from the old fallacious period of Cantor who led the
> world of mathematics into falsehoods.
> [...]
> So this Omega stuff is just more phony baloney.
The omega idea was my way of making sense of the notation 1...000 and
was intended for people who are trying to get a grasp on the idea. In
short, it wasn't meant for AP's eyes.
> [...] The largest Infinite Integer is
> ....99999 and the smallest Infinite Integer is 0. So, now when you
> reach ....999999 by endlessly adding 1, can you go any further without
> going back to 0 and then 1 and on up to ....999999 again? [...]
My 2 cents' worth is that it's easy to identify the Infinite Integers
whose digits are eventually 0 with the nonnegative integers, and the
Infinite Integers whose digits are eventually 9 with the negative
integers. The rest of the Infinite Integers would be the "extra cards"
in a deck of cards.
I knew it! I knew that AP would bring up his "Physics is King" motto
again:
On 2007, Jan 6, 7:09 pm, I wrote in the "The Infinite Integers are not
the Natural Numbers" thread at sci.math:
> * You know, I think I can anticipate another logical fallacy coming up
> soon: AP will claim that "Physics is King", and that mathematics is
> only a subordinate. This is, of course, a non sequitur.
--- Christopher Heckman
Proginoskes
Dik T. Winter wrote:
> In article <1168156693.7...@51g2000cwl.googlegroups.com> "Proginoskes" <CCHe...@gmail.com> writes:
> > a_plutonium wrote:
> ...
> > > I am using the old terms for now, but they will have to change
> > > drastically when the dust begins to settle on Infinite Integers. Old
> > > mathematics had artifices such as negative numbers, such as i,j,k with
> > > no numeric value. And the old term of "transcendental" as that being
> > > not algebraic was a rather shoddy definition of what is really going
> > > on. To define something which it is "not" is a poor definition. We want
> > > definitions of what something "is" rather than what it is "not".
> >
> > Infinite sets can be defined differently from "not finite" in the
> > following way:
> >
> > DEFINITION. A set S is infinite iff there is a bijection f from S to
> > some proper subset T of S.
>
> That is Dedekind-infinite in standard terminology. The two concepts are
> the same only under a weak form of AC. Without it there are infinite sets
> that are Dedekind-finite.
The Axiom of Choice seems "obviously true" to me. Yes, I am aware of
the Banach-Tarski paradox, but AC still looks "obviously true". In
practice, I only deal with finite sets anyway. 8-)
> > > 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
>
> I do not think so. I never heard about the concept of "growing" number.
I don't understand what AP is after, either. This is only an educated
guess on my part.
Proginoskes
a_plutonium wrote:
> Dik T. Winter wrote:
>
> >
> > Every mathematician worth his weight in salt knows that on a globe the
> > circumference of a circle is *not* equal to 2.pi.r, when you measure
> > r along the globe. And that the area is *not* equal to pi.r^2 when
> > you want to measure the area across the globe surface. Pi is a number
> > that has been defined as the ratio between the circumference of a circle
> > and its diameter *on a flat surface*.
>
> Dik can you give us the range of values of pi for all circles drawn on
> a sphere?
There are other ways to define distance in the Euclidean plane; for
instance, you could define
dist((x1,y1),(x2,y2)) = [(x1-x2)^p + (y1-y2)^p]^(1/p)
for any p >= 1 (p = 2 gives the usual Euclidean distance). If you
choose one of these "distance functions" and sketch the graph of
dist((x,y),(0,0)) = 1, and then measure the perimeter carefully, you
can wind up with a value of pi anywhere between 2 and 4.
> And can you give us the range of values of pi for all circles drawn on
> a 3-d ellipse-- the ellipsoid?
>
> There is a curious question arising in my mind about now. That a sphere
> has two poles for all great-circles except for the equator. [...]
If this statement is supposed to mean "All great circles except the
equator pass through the poles", then it is false.
--- Christopher Heckman
Proginoskes
a_plutonium wrote:
> Dik T. Winter wrote:
> > > So I think i is algebraic and I just loosely group i with e and pi.
> >
> > Interesting. What are the digits of i? You are using (behind the
> > screen) 10-adics. And in the 10-adics the square root of -1 does not
> > exist.
>
> The number i is algebraic and it is this value 100000.....000000. At
> least that is what I thought it was.
If so, then i isn't an Infinite Integer. Infinite Integers have no
left-most digit, whereas 10000.....000000 does.
> [...]
> > > 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.
> >
> > Yeah. How do I derive sqrt(2) using these numbers?
>
> Simple question. How do you derive sqrt 2, well the first counting
> number that has a square root is 4 which is 2. Get with the program,
> Dik. [...]
This says that the square root of 4 is 2, not that the square root of 2
is 4.
And I believe he was asking: How do you CALCULATE sqrt(2) using these
numbers? And the answer to that is: There is no Infinite Integer which
can be sqrt(2), since all perfect squares end in 0, 1, 4, 5, 6, or 9.
That is, because "2 is not a quadratic residue of 10."
--- Christopher Heckman