pi twiddling and the 100 to 10^4 B matrices of Circumferencing #1324 Correcting Math
Archimedes Plutonium
start this thread anew.
Earlier today I posted the below:
When Enrico did Squaring the Circle his program found the first
breakdown in 100 B
as 28.87 radius but here we are Circumferencing the Perimeter and
twiddling on pi as thus:
side of square = 28.87
perimeter = 28.87 x4= 115.48
115.48/3.14 = 36.77 = diameter
start pi twiddling 115.48/36.77 = 3.140603
max pi twiddle is 3.1407 and min is 3.1406
3.1407 x36.77= 115.48
match
So, obviously 28.87 is not the first breakdown in Circumferencing in
100 B.
Enrico, where is the first breakdowns in 100 B, 1000 B and 10,000 B?
I am comfortable with that set up and pi twiddling. I like pi
twiddling more than
the earlier work that twiddled with the side-of-square in Squaring the
Circle.
I like pi twiddling better because it is superior for it allows a
uniformity for Perimeterization
and for Squaring the Circle, whereas twiddling with side or radius
could not all be compared.
With twiddling on pi, we can compare Squaring the Circle or Circling
the Square, or Perimeterizing the Circumference or Circumferencing the
Perimeter. All four are different, yet
all four comparable because the only twiddling was a uniform twiddling
on pi.
But the main advantage of pi twiddling is that it allows us to combine
theory with the actual facts and data and the reason for border
breakdown
of 10^603.
Multiplication can span across two zeroes in a row to make a match of
digits on either side of the two zeroes, but when you have three
zeroes in
a row of pi as 10^-603, multiplication has a density of greater
breakdowns
to span those three zeroes in order to match the digits on either
side.
Think of multiplication and twiddling as bridges to match numbers. But
when the multiplier of pi has three zeroes in a row, the density of
mismatches increases drastically.
Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
Archimedes Plutonium
B matrices,
only he used a 1/10B twiddle whereas I want a pi twiddle.
So I checked to see if his mismatches followed mine:
Enrico wrote:
> Input Diameter, get Pi*D, 4*S
> B=100 1/10B Rule 1st 10 Mismatches
> 41742.80
side of square = 28.87
perimeter = 28.87 x4= 115.48
115.48/3.14 = 36.77 = diameter
start pi twiddling 115.48/36.77 = 3.140603
max pi twiddle is 3.1407 and min is 3.1406
3.1407 x36.77= 115.48
match
side of square = 41742.80
perimeter = 166971.20
166971.20/3.14 = 53175.54 = diameter
start pi twiddling 166971.20/53175.54 = 3.140000082
max pi twiddle is 3.1401 and min pi twiddle is 3.1400
3.1401 x 53175.54 = 166976.51 mismatch
3.1400 x 53175.54 = 166971.19 mismatch
So here mine agrees with Enrico's.
However, I am not sure whether 41742.80 is the first side length
that has a breakdown in the 100 B matrix.
> 41742.96
> 41743.12
> 41743.28
> 41744.87
> 41745.03
> 41745.19
> 41745.35
> 41746.94
> 41747.10
> Input Diameter, get Pi*D, 4*S
> B=1000 1/10B Rule 1st 10 Mismatches
> 20864.927
side of square = 20864.927
perimeter = 83459.708
83459.708/3.141 = 26571.062 = diameter
start pi twiddling 83459.708/26571.062 = 3.141000084
max pi twiddle is 3.141001 and min pi twiddle is 3.141000
3.141001 x 26571.062 = 83459.732 mismatch
3.141000 x 26571.062 = 83459.705 mismatch
Here again, mine agrees with Enrico's, only not sure whether 20864.927
is the first mismatch in the 1000 B matrix. Not sure since Enrico used
a 1/10B twiddle whereas I used a pi twiddle.
> 20865.086
> 20865.325
> 20865.484
> 20865.723
> 20865.882
> 20866.280
> 20866.678
> 20866.837
> 20867.076
One good news of Enrico's above report on 100 B and 1000 B is that the
number for first mismatch has a huge decreasing trendline. Whereas in
Squaring the Circle,the trendline was stuck at 28% of the B number.
Here
we have 41742.80 in 100 B and 20864.927 in 1000 B for a decrease of
1/2
in numeric value. But that could spell problems if that decrease
continues,
because I expect the critical perimeter of equal to or less than B to
occur
at 10^603. And the present rate of decline would reach a critical
perimeter far too soon. I suspect the culprit to be the 1/10B
twiddle.
Archimedes Plutonium
Now what I am trying to do is get a sense and feel for what the three
digits in a row at 10^-603 in pi does to
Circumferencing the Perimeter.
side of square = 20864.927
perimeter = 83459.708
83459.708/3.141 = 26571.062 = diameter
start pi twiddling 83459.708/26571.062 = 3.141000084
max pi twiddle is 3.141001 and min pi twiddle is 3.141000
3.141001 x 26571.062 = 83459.732 mismatch
3.141000 x 26571.062 = 83459.705 mismatch
pretend pi is 3.000 to see if we can lower the numerical value of the
side
side of square = 864.927
perimeter = 4side = 3459.708
3459.708/3.000 = 1153.236 = diameter
start pi twiddling 3459.708/1153.236 = 3.0000000
max pi twiddle is 3.000001 and min pi twiddle is 3.000000
3.000001 x 1153.236 = 3459.709 mismatch
3.000000 x 1153.236 = 3459.708 match
Again, pretending that pi is 3.000
side of square = 64.927
perimeter = 4side = 259.708
259.708/3.000 = 86.569 = diameter
start pi twiddling 259.708/86.569 = 3.000011
max pi twiddle is 3.000012 and min pi twiddle is 3.000011
3.000012 x 86.569 = 259.708 match
3.000011 x 86.569 = 259.707 mismatch
> 20865.086
> 20865.325
> 20865.484
> 20865.723
> 20865.882
More pretending pi is 3.000
side of square = 865.882
perimeter = 4side = 3463.528
3463.528/3.000 = 1154.509 = diameter
start pi twiddling 3463.528/1154.509 = 3.000000
max pi twiddle is 3.000001 and min pi twiddle is 3.000000
3.000001 x 1154.509 = 3463.528 match
3.000000 x 1154.509 = 3463.527 mismatch
> 20866.280
> 20866.678
> 20866.837
> 20867.076
More pretending pi is 3.000
side of square = 867.076
perimeter = 4side = 3468.304
3468.304/3.000 = 1156.101 = diameter
start pi twiddling 3468.304/1156.101 = 3.000000
max pi twiddle is 3.000001 and min pi twiddle is 3.000000
3.000001 x 1156.101 = 3468.304 match
3.000000 x 1156.101 = 3468.303 mismatch
side of square = 67.076
perimeter = 4side = 268.304
268.304/3.000 = 89.434 = diameter
start pi twiddling 268.304/89.434 = 3.000022
max pi twiddle is 3.000023 and min pi twiddle is 3.000022
3.000023 x 89.434 = 268.304 match
3.000022 x 89.434 = 268.303 mismatch
Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
Now I notice there is no gap between the max and min such as the last
example:
3.000023 x 89.434 = 268.304 match
3.000022 x 89.434 = 268.303 mismatch
So what I am expecting is that the large quantity of digits used in
10^603 of its 603 digits rightwards of
the decimal point and say 601 digits leftward of the decimal point
that the three zeroes in pi cause too much
strain and stress of the division and multiplication that the mismatch
of a ** critical side ** takes place for the
first time in this 10^603 B matrix.
What is fortunate about Circumferencing the Perimeter that was
unavailable with Squaring the Circle is that we can
penetrate this 10^603 B matrix by Enrico's programs.
Hopefully Enrico will post, with pi twiddles the first breakdowns in
100, 10^3, 10^4, and perhaps 10^5, and 10^6 and
10^7 and 10^8 to see for an overall trendline. And then ask Enrico to
try to see if a "critical-side" where the perimeter
is equal or smaller than the B-number has a mismatch in 10^603.
Enrico
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies
>
> Now I notice there is no gap between the max and min such as the last
> example:
> 3.000023 x 89.434 = 268.304 match
> 3.000022 x 89.434 = 268.303 mismatch
>
> So what I am expecting is that the large quantity of digits used in
> 10^603 of its 603 digits rightwards of
> the decimal point and say 601 digits leftward of the decimal point
> that the three zeroes in pi cause too much
> strain and stress of the division and multiplication that the mismatch
> of a ** critical side ** takes place for the
> first time in this 10^603 B matrix.
>
> What is fortunate about Circumferencing the Perimeter that was
> unavailable with Squaring the Circle is that we can
> penetrate this 10^603 B matrix by Enrico's programs.
>
> Hopefully Enrico will post, with pi twiddles the first breakdowns in
> 100, 10^3, 10^4, and perhaps 10^5, and 10^6 and
> 10^7 and 10^8 to see for an overall trendline. And then ask Enrico to
> try to see if a "critical-side" where the perimeter
> is equal or smaller than the B-number has a mismatch in 10^603.
>
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies
===========================================================
Earlier, you posted:
But the Twiddle Rule is going to be allowed a maximum of 1/(B^2)
place
value with
the insertion of any digits. A breakdown is thus defined as that in
which no digits up to
1/(B^2) can match the square area. I guess this is what would be a
nightmare for Enrico
to computer program.
This seems to say that you now are using multiple digits of Pi
as variables, each of which can be from 0 to 9.
For B=100 there would be
Pi = 3.1400, 3.1401, 3.1402, ... 3.1499
For B=1000 there would be
P=3.141000, 3.141001, 3.141002, ... 3.141999
etc.
Is this what you mean?
Meanwhile, here's a couple more first 10 breakdowns under the
the original rules:
Input Diameter, get Pi*D, 4*S
B=10000 1/10B Rule 1st 10 Mismatches
5215.5062
5215.5460
5215.5669
5215.5858
5215.6067
5215.6465
5215.9041
5215.9439
5215.9648
5215.9837
Input Diameter, get Pi*D, 4*S
B=100000 1/10B Rule 1st 10 Mismatches
651.91154
651.95133
652.05080
652.09059
652.19006
652.22985
652.26964
652.36911
652.40890
652.50837
Notice that the intervals between breakdowns are getting larger.
With the last one, I had to use a test range of 10000 just to
be reasonably certain of picking up the first breakdown, plus
a final check range of 100000 to confirm it. This is hitting
the practical limit for any exhaustive search strategy.
Enrico
Archimedes Plutonium
>
> ===============
> Earlier, you posted:
>
> But the Twiddle Rule is going to be allowed a maximum of 1/(B^2)
> place
> value with
> the insertion of any digits. A breakdown is thus defined as that in
> which no digits up to
> 1/(B^2) can match the square area. I guess this is what would be a
> nightmare for Enrico
> to computer program.
>
> This seems to say that you now are using multiple digits of Pi
> as variables, each of which can be from 0 to 9.
>
Please let me explain, Enrico. I have a nasty habit of over-curiousity
and excitement
of what is ahead that I jump further than I should jump. I should slow
down and refine
and get it consolidated as to how to twiddle in Circumferencing,
before I ask you to
do anything. Because of my poor refining and asking you to do things
with the computer
before I even know "what I want" leads to these sad situations were I
have to backtrack
and where you have wasted time on programming. All because of my
curiousity and jumping ahead
when I should have refined what I wanted in the first place. I am
sorry for that.
I finally have pi-twiddling down pat. I prefer pi twiddling because I
can use it if I want to
generalize and go to Perimeterization rather than Circumferencing, or
go to Squaring the Circle or Circling the Square. All four of these
processes are different, but pi twiddling can remain the same in all
four processes. But most important advantage is that pi twiddling
reveals why a breakdown of a "critical perimeter" must occur in
10^603,if I am guessing correctly.
In your below you report that 1/10B twiddle delivers the side
651.91154 as a mismatch, but
in pi twiddling, it is not a breakdown but a match. In pi twiddling we
have a max and min
of twiddle and for 10^5 B matrix we have pi as 3.14159. We use the
root of that pi in the twiddle, so we use 3.14159 and now we have to
go to 1/(B^2) on pi so that we have 3.14159abcde where we have to fill
in for the abcde. Please look below at the example shown
100000 B matrix with pi = 3.14159
side of square = 651.91154
perimeter = 4xside = 2607.64616
2607.64616/3.14159 = 830.04025 = diameter
start pi twiddling 2607.64616/830.04025 = 3.141590013255
max pi twiddle is 3.1415900133 and min pi twiddle is 3.1415900132
3.1415900133 x 830.04025 = 2607.64616 match
no need to go any further
> For B=100 there would be
> Pi = 3.1400, 3.1401, 3.1402, ... 3.1499
>
Enrico, in pi-twiddling there is a step in which you divide the
perimeter by the
diameter to obtain a pi for the pi twiddling. Please look at that step
in the
above example. That delivers the "root pi" setting up the next step in
which we
have the maximum and minimum pi twiddle
> For B=1000 there would be
> P=3.141000, 3.141001, 3.141002, ... 3.141999
>
> etc.
>
> Is this what you mean?
>
Please look at the example for it is very clear that the pi twiddle
goes out as far as
1/(B^2) and uses the root-pi and has a minimum and a maximum. Pi
twiddling is very similar
to what you were doing before in Squaring the Circle when you explored
the 10^308 B matrix
and was going out to 616 digits to add a "1" in the 616th digit.
Instead of twiddling on the side of the square in Squaring the Circle,
here we twiddle on pi itself.
The beauty of pi twiddling is that when a breakdown occurs and suppose
it occurs for 10^603 of a **critical perimeter** implies that "no pi"
exists that has a finite Perimeter to match with a finite circle
circumference and that pi breaks-down.
Please, Enrico, can you re program using a pi twiddle for the 100 out
to the 10^5 or even
10^8 if possible? And again, I apologize for sending you out on wild
chases when I do not even have what it is that I want. My curiousity
and excitement gets me into this trouble.
I desperately need your computer results, for without them I am alone
with just theory
and nothing to guide that theory. Your results tell me if I am on a
correct path or not
and forces me to adjust. So please, please can you set up your program
to do pi twiddling
on Circumferencing. The example above is a perfect model. Let me
repeat that example again:
100000 B matrix with pi = 3.14159
side of square = 651.91154
perimeter = 4xside = 2607.64616
2607.64616/3.14159 = 830.04025 = diameter
start pi twiddling 2607.64616/830.04025 = 3.141590013255
max pi twiddle is 3.1415900133 and min pi twiddle is 3.1415900132
3.1415900133 x 830.04025 = 2607.64616 match no need to go any further
Archimedes Plutonium
Enrico
> whole entire Universe is just one big atom
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
==============================================================
>
> 3.1415900133 x 830.04025 = 2607.64616 match
no need to go any further
>
This is what's driving me crazy - How am I supposed to program
that statement? How do I program it to work for any B matrix from
100 to 10^603?
Calculations, comparisons and (IF-Then, Else) statements needed.
As long as you refuse to learn any programming, stuff like this
is going to keep coming up. I'm not going to keep trying to
reverse-engineer your examples to figure out what you're thinking.
Enrico
Archimedes Plutonium
> On Dec 23, 2:14 am, Archimedes Plutonium
>
>
>
> > 3.1415900133 x 830.04025 = 2607.64616 match
no need to go any further
>
> This is what's driving me crazy - How am I supposed to program
> that statement? How do I program it to work for any B matrix from
> 100 to 10^603?
> Calculations, comparisons and (IF-Then, Else) statements needed.
>
> As long as you refuse to learn any programming, stuff like this
> is going to keep coming up. I'm not going to keep trying to
> reverse-engineer your examples to figure out what you're thinking.
>
> Enrico
Enrico, please look at this example again and here is what to
PROGRAMM--
100000 B matrix with pi = 3.14159
side of square = 651.91154
perimeter = 4 x side = 2607.64616
2607.64616/3.14159 = 830.04025 = diameter
Up to there, it is straightforward with nothing new.
Now, program this step:
truncated perimeter divided by truncated diameter or , P/D
2607.64616/830.04025 = 3.141590013255 and let me call this the pi-
root
Now Programm a 1/(B^2) twiddle on that pi-root formed so it would be
3.1415900133 instead of 3.1415900132
Forget about max and min
Now simply check to see if 3.1415900133 x 830.04025 produces
2607.64616 for a match or mismatch
And that is all there is.
So there are three new things to program:
(1) divide the truncated perimeter by the truncated diameter for a "pi
root"
(2) add the 1/(B^2) twiddle to the pi-root
(3) see if the diameter x pi twiddle yields the perimeter for a match
or mismatch
Please continue, Enrico, for your help is vital.
Archimedes Plutonium
Archimedes Plutonium
>
> Input Diameter, get Pi*D, 4*S
> B=100000 1/10B Rule 1st 10 Mismatches
> 651.91154
> 651.95133
> 652.05080
> 652.09059
> 652.19006
> 652.22985
> 652.26964
> 652.36911
> 652.40890
> 652.50837
>
> Notice that the intervals between breakdowns are getting larger.
> With the last one, I had to use a test range of 10000 just to
> be reasonably certain of picking up the first breakdown, plus
> a final check range of 100000 to confirm it. This is hitting
> the practical limit for any exhaustive search strategy.
>
> Enrico
Maybe, Enrico, I should not ask you to change anything but keep on
going with the
1/10B Rule. Because it appears that I can factor out where that 1/10B
Rule intersects
with the 1/(B^2) Rule with pi-twiddling that I seek. So for instance,
your above
10^5 B matrix with 651.91154 breakdown when re-applied into the 100 B
matrix with
651.91 yields a breakdown as witnessed:
100 B matrix with pi = 3.14
side of square = 651.91
perimeter = 4xside = 2607.64
2607.64/3.14 =
830.45 = diameter
start start pi twiddling 2607.64/830.45 = 3.14003
pi twiddle = 3.1401
3.1401 x 830.45 = 2607.69 mismatch
3.1400 x 830.45 = 2607.61 mismatch
Now Enrico reported the first breakdown in 100 B matrix doing the
1/10B Rule was:
> Input Diameter, get Pi*D, 4*S
> B=100 1/10B Rule 1st 10 Mismatches
> 41742.80
And apparently doing the 1/(B^2) Rule with pi twiddling the first
breakdown must
be either 651.91 or smaller.
Now Enrico did find the first breakdown in Squaring the Circle with
the 1/(B^2) Rule
to be that of 28.87 in the 100 B matrix.
So comparing 651.91 with 28.87, both for the 100 B matrix but where
one is perimeter length
and the other is area, I would say the 1st breakdown in 100 B matrix
with 1/(B^2) in pi-twiddling is going to be even smaller than 651.91.
Please Enrico, why give up on this search adventure, just when it is
getting to be highly
interesting. You have alot of time invested already in this adventure,
and would be awfully
silly to drop it just when it has become very interesting and likely
to pay a handsome return.
So please, reprogram for pi twiddling and using 1/(B^2). Your foray
with 10^308 was
all of 1/(B^2), which has not changed, except that the twiddle is on
pi and not the area
of the square.
Please stay with this adventure for the Circumference and that is the
last programming I will ever ask of you. Please program for
Circumferencing of pi twiddle using 1/(B^2)
Archimedes Plutonium
Archimedes Plutonium
Enrico wrote with a 1/10B twiddle:
> Input Diameter, get Pi*D, 4*S
> 5215.5062
Let me see if 215.50 is a breakdown in 100 B matrix
First reviewing the breakdown of 651.91 breakdown:
100 B matrix with pi = 3.14
side of square = 651.91
perimeter = 4xside = 2607.64
2607.64/3.14 = 830.45 = diameter
start pi twiddling 2607.64/830.45 = 3.14003
pi twiddle = 3.1401
3.1401 x 830.45 = 2607.69 mismatch
3.1400 x 830.45 = 2607.61 mismatch
----------------
Now let me see if 215.50 has a breakdown
100 B matrix with pi = 3.14
side of square = 215.50
perimeter = 4xside = 862.00
862.00/3.14 = 274.52 = diameter
start pi twiddling 862.00/274.52 = 3.14002
pi twiddle = 3.1401
3.1401 x 274.52 = 862.02 mismatch
In the 100 B matrix of Squaring the Circle, using 1/(B^2) Rule the
first mismatch was 28%
of the B number at 28.87 and then the 28% persisted for higher
matrices in Squaring the Circle.
But here and now, I am Circumferencing the Perimeter because it is
more simple and its One
Dimension is in keeping with location of the Border between Finite and
Infinity.
I have not yet located the first breakdown of Circumferencing, but the
above breakdown
of side 215.50 tells me it is at least that small or smaller, and
probably smaller.
I do not expect the first breakdown to be near 28.87. So it must be
somewhere between
28.87 and 215.50 in 100 B matrix.
The curious question is whether a pattern develops of a uniform
percentage for Circumferencing.
Here was the 28% pattern in Squaring the Circle, starting with 100 B
and finishing with
10^8 B matrix:
28.87
282.784
2821.5778
28210.05805
282095.632126
2820950.1143194
28209498.00410956
I do not expect such a small and steady decline in first breakdown
with Circumferencing.
What Enrico has found so far, although using a 1/10B Rule, is a 1/2
rate of numeric value
decline. I do not expect such a drastic falling decline but a decline
larger than the
Squaring the Circle of its almost imperceptable decline.
And it may well be that the decline is a chaotic decline, or a
logarithmic decline in
Circumferencing.
What I am banking on, is that the decline in Circumferencing at the
10^306 B matrix
where pi has two zeroes in a row, that we can use the first breakdown
in the 10^306 B matrix and it translates into a breakdown in the
10^308 B matrix due to the fact that the two zeroes in a row help to
insure that premature-breakdown. But I suspect this premature
breakdown is not enough to get the perimeter below the numeric value
of the B number itself.
Critical-perimeter does not happen until pi has three zeroes in a row
at the 10^603 B matrix. Here, I suspect the first breakdown in 10^600
B matrix is a perimeter which translates as a first
breakdown in the 10^601 and 10^602, and finally 10^603 B matrix and
thus, happy days, the
first breakdown of a perimeter that is smaller than the B number
itself, and thus, the border between Finite and Infinity.
This is why pi-twiddling is far superior to twiddling with the square
or some feature of the
circle other than pi itself. By twiddling with pi, we instantly see
the mechanism of how the first breakdown of a perimeter smaller than
the B number itself takes place.
Archimedes Plutonium
breakdown in
100 B matrix for Circumferencing, other than the 215.50. I had no luck
and 215.50
maybe the smallest and first breakdown in 100 B matrix.
Now I used Enrico's previous post:
> Input Diameter, get Pi*D, 4*S
> B=10000 1/10B Rule 1st 10 Mismatches
> 5215.5062
> 5215.5460
> 5215.5669
> 5215.5858
> 5215.6067
> 5215.6465
> 5215.9041
> 5215.9439
> 5215.9648
> 5215.9837
And the reason, I feel, that the 215.50 works is because with his
1/10B twiddle versus
my 1/(B^2) pi-twiddle is a factor of 100 x 100 = 10000 so that Enrico
is in the B=10000
while I am in the B=100
Now this maybe a godsend in a way, because it appears to be accurate,
so that I could
use this trick of the trade for 10^300 B matrix to predict what string
of digits would be the
first breakdown in 10^600 B matrix.
So from looking at Enrico's above, I extracted the 215.50 digit string
to see if it broke-down
in 100 B matrix and it did so:
>
> Now let me see if 215.50 has a breakdown
>
> 100 B matrix with pi = 3.14
> side of square = 215.50
>
perimeter = 4xside = 862.00
>
862.00/3.14 = 274.52 = diameter
> start pi twiddling 862.00/274.52 = 3.14002
> pi twiddle = 3.1401
>
3.1401 x 274.52 = 862.02 mismatch
>
But now, another question, interesting question which I have no clue
of an answer as
to why this happens. But if you noticed in the Squaring of the Circle,
that whenever a
breakdown occurred, there seemed to be a plethora of neighboring
breakdowns. Notice
that when Enrico got his first breakdown of 5215.5062 he got a
plethora of immediate
neighbors which also had a breakdown. So why does that happen? I
really do not know
as of yet, but I must stay focused on hunting down the borderline
ultimate goal, than to
worry about side issues.
But I did take a few minutes to see if this "immediate neighbors with
breakdowns" follows.
> 5215.5062
> 5215.5460
> 5215.5669
> 5215.5858
So I needed to check 215.50 and it broke-down then check 215.54 then
215.56 then
215.58.
----------------
100 B matrix with pi = 3.14
side of square = 215.54
perimeter = 4xside = 862.16
862.16/3.14 = 274.57 = diameter
start pi twiddling 862.16/274.57 = 3.140037
pi twiddle = 3.1401
3.1401 x 274.57 = 862.17 mismatch
-------------------
100 B matrix with pi = 3.14
side of square = 215.56
perimeter = 4xside = 862.24
862.24/3.14 = 274.59 = diameter
start pi twiddling 862.24/274.59 = 3.140099
pi twiddle = 3.1401
3.1401 x 274.59 = 862.24 match
-------------------
100 B matrix with pi = 3.14
side of square = 215.58
perimeter = 4xside = 862.32
862.32/3.14 = 274.62 = diameter
start pi twiddling 862.32/274.62 = 3.140048
pi twiddle = 3.1401
3.1401 x 274.62 = 862.33 mismatch
So of these four checked "sides" three of them had a breakdown.
Why do the breakdowns come in these clusters? I do not know, but do
not have the time
to find out as it is more important to track down the borderline
between Finite and Infinity
and then perhaps come back to answer that mysterious question. Maybe
it is something
like the primes with its twin prime clusters
Archimedes Plutonium
> B=1000 1/10B Rule 1st 10 Mismatches
> 20864.927
> 20865.086
> 20865.325
> 20865.484
> 20865.723
> 20865.882
> 20866.280
> 20866.678
> 20866.837
> 20867.076
And sometime ago, we found that the first mismatches in 100 B to 10^8
B in Squaring the Circle followed this pattern:
28.87
282.784
2821.5778
28210.05805
282095.632126
2820950.1143194
28209498.00410956
Now, from what I can gather the first mismatch in 100 B matrix with
Circumferencing using 1/(B^2) on pi-twiddling yields 215.50 as the
first breakdown.
Now let me see if I can build a similar pattern with 215.50 as what
was found
to be true for 28.87 in Squaring the Circle pattern.
So I have
215.50
2152.050
So the above suggests that the first breakdown in the 1000 B matrix
for Circumferencing
is going to be 2152.050
Let me just quickly see if it is indeed a breakdown.
1000 B matrix with pi = 3.141
side of square = 2152.050
perimeter = 4xside = 8608.200
8608.200/3.141 = 2740.592 = diameter
start pi twiddling 8608.200/2740.592 = 3.141000193
pi twiddle = 3.141001
3.141001 x 2740.592 = 8608.202 mismatch
3.141000 x 2740.592 = 8608.199 mismatch
Well, yes indeed it is a breakdown. Whether it is the very first
breakdown
in 1000 B matrix for Circumferencing is questionable. But all of a
sudden
B matrices have become highly predictable.
And let me note about the anatomy of the breakdown. The source of the
breakdown is a
string of zeroes in the twiddle zone. So for 1000 B, if the pi-twiddle
is 3.141 000
with those three in a row zeroes and then you make the twiddle of
3.141001 is usually
going to cause a breakdown. If the pi-twiddle had been 3.141000999
then it would not
be a mismatch. But if it was 3.141000211 then it usually is a
mismatch.
So to cause a mismatch requires a string of zeroes in a row in the
twiddle-zone. And somehow
the multiplication and division involved with a specific string of
digits brings about that
string of zeroes and thus answering why a pattern of 215.50 is
repeating in larger B matrices.
Now the above is all good news for the hunting down of the borderline
between Finite and Infinity. So that if there is this pattern of
215------- then at 10^603 with pi having itself three zeroes in a row
and the 215---- causing more zeroes in a row, that I suspect the
combination of the zeroes inherent in pi at 10^603 with the zeroes
forced by the 215---- that
the breakdown is a perimeter that is smaller than the B number of
10^603 itself.
The reason none of this was achievable in Squaring the Circle, is
because that is a two dimensional geometry and finite versus infinite
is a one dimensional geometry such as Circumferencing the Perimeter.
Archimedes Plutonium
Alright pretty news. I have found a method-of-proof that infinity
starts at 10^603. That means the actual computer programming is the
proof itself and the method-of-proof is the theory behind it. So the
computer programming is the verification that the theory is true. I
will leave the verification to anyone who sets up the computer
programming and there can be lots of folks who verify this proof.
Statement: The border between Finite versus Infinity is 10^603, which
means that mathematics as a reliable and trustworthy and the science
of precision stops at 10^603 of being reliable and trustworthy, since
the Algebra of Numbers stops at 10^603 of giving
a match of a finite perimeter with a finite circumference. After
10^603, mathematics is no
longer the science of precision because it no longer can reliably tell
you if a circle circumference is matched with a square perimeter.
Mathematics beyond 10^603 is poking in the dark and surrounded by
imprecision. Aristotelian Logic of either true or false with
nothing in the middle no longer holds up beyond 10^603 where Quantum
Duality Logic takes over completely. Mathematics is the science of
precision and that precision ends at 10^603 of
its Algebra.
Method-of-Proof: The method-of-proof uses Circumferencing of the
Perimeter and does a
pi-twiddling. This method parallels the historical Squaring of the
Circle which dates
back to Ancient Greek times. The problem with present day current
mathematics is that everyone expects math to be the science of
precision, yet they neglect to precisely define Finite versus Infinity
and have everyone assuming their opinion of what finite means and what
infinite means. Current math has finite and infinity as opinions, not
as precise definitions. Current math has teachers assuming what finite
and infinite means and has their students making the same opinionated
assumption. Assumption as a definition is not mathematics at all, and
somewhere in the fork of the road of the history of mathematics, the
definition that is precise of Finite versus Infinity could have been
taken the correct fork
in the road rather than the fork that leads to fake math of the
future, ie, present day math.
The fork in the road of math history was where it is questioned about
finite versus infinity that the two concepts run into each other or
move along one another or transition from finite into infinity.
Somehow in the long road of the history of mathematics, the concept
that finite has a border with infinity, seemed to have never come up
and thus the fake fork in the road was taken
and leaving modern day mathematics is such miserable and terrible
state of condition. Our modern day mathematics is in the sick-ward.
With the history of mathematics and the Squaring of the Circle, it
should have been asked for the border or borderline between Finite and
Infinity.
So that if any mathematician was asked what does it mean to be Finite
versus Infinite. All that he or she had to say was the border between
the two concepts, because they transition between one and the other.
To find that border, that is natural intrinsic and embedded inside of
mathematics means simply to
find where Circumferencing of the Circle no longer has a finite square
perimeter to match.
What this method of proof is going to show is that we can move along
and transition along in finding a finite Circumference that matches a
finite perimeter of a square until we reach 10^603 where pi has three
zero digits in a row
and which then forces or causes the first breakdown in that 10^603 B
matrix to be the same perimeter breakdown that was found in the
10^600 B matrix. Because we can use the 10^600 breakdown perimeter, it
is far smaller than the
number 10^603 itself and thus we have found a perimeter of a square
that has no matching circle circumference and there can never be a
matching circumference because of the inherent
intrinsic mathematics itself. Mathematics as precision stops at 10^603
and Quantum Physics takes over.
Yesterday I posted that 2152.050 is a breakdown
in the 1000 B matrix but today I tried 2152.550 to see if it is a
breakdown, and although the other is a bit smaller of a side, this
number
of 2152.550 is more suited for the argument of method of proof.
Method of Proof Argument:
In Squaring the Circle we had this pattern of first breakdown of the B
matrices starting with the 100 B and showing the 10^8 B:
28.87
282.784
2821.5778
28210.05805
282095.632126
2820950.1143194
28209498.00410956
215.50
2152.050
Let me try 2152.550 as more suitable
side of square = 2152.550
perimeter = 4xside = 8610.200
8610.200/3.141 = 2741.228 = diameter
start pi twiddling 8610.200/2741.228 = 3.1410010
pi twiddle = 3.141002
3.141002 x 2741.228 = 8610.202 mismatch
3.141001 x 2741.228 = 8610.199 mismatch
Trouble with squaring the circle is that it is 2dimensional when
finite versus infinite is 1dimensional, and that is why squaring the
circle is not amenable to the proof argument but circumferencing the
perimeter is amenable.
Alright, so how does it actually work?
It works because at 10^603,with the three zero
digits in a row in pi causes a first breakdown
of circumferencing by using the same first breakdown we found in
10^600 B matrix. In other words, we can find a breakdown number in
10^600 and transfer that number intact into the 10^603 B matrix and
due to the action of those three in a row zero digits, forces that
smaller side-perimeter of 10^600 to have a breakdown in 10^603.
Now tonight I hope to do some experimenting to show that I can force a
breakdown in 1000 B matrix by pretending pi is 3.100 and not 3.141
and where a breakdown of 215.50 in 100 B matrix
works in the 1000 B matrix. And so by Mathematical Induction and a bit
more case examples we can thence say that 10^603 is indeed the border
between Finite versus Infinity.
Archimedes Plutonium
Usually I snip to save space but these two posts are historical, and
this post follows from the above.
I went ahead to see if the theory is sound; by that I mean, to see if
a breakdown in a prior B matrix
can be a breakdown in a higher order B matrix if pi has zeroes as the
last digits.
So here I take the breakdown of side 2152.550 in the 10^3 B matrix and
pretend that pi is a pretend-pi
of 3.1400 rather than 3.1415 which it really is. I want to see if the
zeroes in a row in pi allow the breakdown of
a previous B matrix to also be a breakdown in that higher order B
matrix. This is important for it lowers the breakdown number for the
perimeter and allowing the perimeter to be smaller in size than the B
number itself, and thus, happy-days, we have the border between Finite
and Infinity.
215.50 as first breakdown in 100 B matrix
2152.550 as near to the first breakdown in 1000 B matrix
pretend for 10^4 B matrix
Pretending 10000 B matrix with pi as pretend pi = 3.1400
side of square = 2152.5500
perimeter = 4xside = 8610.2000
8610.2000/3.1400 = 2742.1019 = diameter
start pi twiddling 8610.2000/2742.1019 = 3.140000 012
pi twiddle = 3.14000002
3.14000002 x 2742.1019 = 8610.2002 mismatch
3.14000001 x 2742.1019 = 8610.1999 mismatch
This is truly a pretty and beautiful gift for the holidays for me
anyway. I have finally solved the theory portion
and now, all that remains is the world's supercomputers to verify that
either 10^308 B matrix is a breakdown of a perimeter smaller than
10^308 due to the fact that pi has two zeroes in a row at 10^-308, or
that 10^603 is the
border between Finite and Infinity due to the fact that pi has three
zeroes in a row at 10^-603.
So all I need is the supercomputers of the world to verify that
Circumferencing of the Perimeter fails mathematics
at 10^603 ( I believe it is 10^603, but science has a way of
surprizing us, for example the human genome was expected to have twice
as many genes as what finally was found).
So can we have some supercomputers of the world train their programs
to finding the exact 1st breakdown of each B matrix all the way up to
10^603?? This answer sheet by the computer will have at least 603
numbers starting with the
100 B matrix showing what number is the first breakdown in that
specific B matrix using a pi twiddle of 1/(B^2).
I can do a Mathematical Induction but that is not as good as actual
verification by computers.
So, here in mathematics, for the first time we have computers doing
the proof of the most important piece of mathematics ever since the
Pythagorean theorem started the mathematics adventure. And the
computer is essential in this proof, since it verifies and validates.
There is no escaping the conclusions. The conclusion that every finite
circle and square have an end at 10^603 because one cannot match the
other.
And another feature that I have not spent enough time on. Is that
Algebra has a universe of validity, which ends at
10^603. That numbers in mathematics follow Aristotelian Logic only up
to 10^603 and that is the end of mathematics
as we know it. From 10^603 and beyond is untrustworthy mathematics
because it is ruled by Quantum duality Logic and
our answers or definitions or concepts breakdown once we cross this
borderline at 10^603.
Archimedes Plutonium
(most snipped to save space)
>
> This is truly a pretty and beautiful gift for the holidays for me
> anyway. I have finally solved the theory portion
> and now, all that remains is the world's supercomputers to verify that
> either 10^308 B matrix is a breakdown of a perimeter smaller than
> 10^308 due to the fact that pi has two zeroes in a row at 10^-308, or
> that 10^603 is the
> border between Finite and Infinity due to the fact that pi has three
> zeroes in a row at 10^-603.
>
> So all I need is the supercomputers of the world to verify that
> Circumferencing of the Perimeter fails mathematics
> at 10^603 ( I believe it is 10^603, but science has a way of
> surprizing us, for example the human genome was expected to have twice
> as many genes as what finally was found).
>
>
As I mentioned previously that this is the first time in the history
of mathematics where a computer
reply back is going to be better than a mathematical proof, for it not
only tells more about the subject
under scrutiny but even validates any proof on the subject. What I am
trying to say is that if some
mathematician was able to concoct a proof that 10^603 has its first
breakdown of a perimeter smaller than
the B number and that Infinity begins at 10^603, that is not as good
of a math result as having the
computer list the breakdowns and showing why the breakdowns occurred.
Here I can offer an analogy, that a math
proof on this subject would be like sitting in a classroom on the
theory of the auto engine whereas the actual
building of the auto engine is like a computer run through of the B
matrices.
Now I think I have some sort of lead into why 10^308 B matrix will not
have a breakdown of a **critical perimeter**
which is a perimeter smaller than the B number itself. Not only does
10^308 have just two zeroes in a row in pi, whereas 10^603 has three
zeroes in a row, but that 10^308 is a even numbered exponent whereas
10^603 has an odd numbered exponent. It is seen that in Squaring the
Circle, the even and odd numbered exponents follow some different
track and I suspect the even and odd exponents contributes to what B
matrix has its first breakdown of a critical-perimeter. As of yet, I
fail to see the mechanism of why even or odd contributes, but it
likely has some contributing factor as to why 10^308 fails to have
that important critical perimeter while 10^603 does have the critical
perimeter. And here again, if the supercomputers would run through the
matrices, we can thence find out whether or not even or odd exponents
contributes.
Archimedes Plutonium
between Finite and Infinity is discovered
as 10^603 (waiting for verification).
Of course, for the first time in history of science we have a
precision definition of what it means to be Finite versus Infinite.
We just state the number 10^603 as the border. Anything under it is
Finite, and over is Infinite. This number 10^603 also
defines the boundary of mathematics, because numbers over 10^603 and
their operations are no longer trustworthy or
mathematical.
So here we define the boundary of Algebra and its operators as that of
10^-603 to that of 10^603. Picture those numbers
as a vast Matrix or Array, and append as an imaginary number the
number 0. Now we can picture a 10^603 matrix having all the numbers
from 10^-603 to that of 10^603 as a huge matrix and if we chose any
number out of that matrix as the side of the square, we can find a
diameter such that the perimeter equals the circumference. Of course,
zero is an imaginary number in this new world of Algebra. Now we can
admit negative numbers by fitting each of the positives with a
negative sign, except zero of course.
So Algebra becomes a altogether different subject from what 20th
century and up to 2010 saw as Algebra. Ever since Galois theory was
started in the 19th century, Algebra has become this huge snake with
painted legs on it. Most past centuries of mathematics have gone to
extremes by over-hyping a subject of mathematics. The 20th century
overhyped
Algebra of Galois theory.
How much of Algebra is lost when it is recognized that Algebra is
restricted to a 10^603 Matrix and where numbers
beyond 10^603 is not even legitimate mathematics because the numbers
and the operators are no longer trustworthy or
reliable? Does math even retain the concepts of Galois Field, Ring,
Group? A century of overhype of Algebra is really a
sort of Dark Ages of mathematics, until they regain their senses and
pull back and put Algebra into its proper perspectives.
But let me also touch on these famous conjectures in light of a
precision definition of Finite versus Infinite:
In historical order of approximate dates:
1) Kepler Packing conjecture 1615
2) Fermat's Last Theorem 1640
3) Goldbach conjecture 1742
4) Riemann Hypothesis 1857
So we have a precision definition of Finite versus Infinite. And then
we recognize that mathematicians
can no longer make spurious conjectures about infinity such as a
Kepler Packing of "infinite space" when
that space needs to be precision defined as the border of 10^603. Or
the spurious Fermat's Last Theorem
that says no integer triples can solve a^n + b^n = c^n , since the
conjecture is bounded by 10^603 as
being mathematical. Or in Goldbach's conjecture we can no longer let
the mind roam beyond 10^603 looking
for even numbers with a duet of primes adding up to equal that even
number. Or in the Riemann Hypothesis
we can no longer think that a Zeta function beyond 10^603 exists or is
a part of mathematics but probably in a
shadowy world of nonmath.
In Old Math, where no-one bothers to precision define Finite versus
Infinite and where those concepts are left as
opinions, then you have a morass of unsolvable conjectures.
But once you precision define what it means to be Finite and what it
means to be Infinite, then those four conjectures
look to be easy to prove or dismissed as false. I shall dwell on each
of them in future posts.
But let me end this post by clarification of Infinity. What does
Infinity mean when infinity starts at 10^603. This is a new idea that
is going to be hard to tackle by most everyone, especially the older
folks who have been indoctrinated by old
school math.
What does it mean for a set to be infinite when infinity starts at
10^603? Does it mean that if you have to have 10^603 or more numbers
that have that characteristic? Does it mean that we have to list
10^603 primes in order to say that the primes are infinite? Or does it
mean that we have a large number set of primes going up to 10^603 and
across that borderline are sitting some more primes, which would be a
set of primes that is smaller than a cardinality of 10^603 but because
there are primes across the borderline of finite with infinity that we
can say the primes are infinite?
What I like to think is the best answer of what is a infinite set, is
that infinity as a concept cannot be applied to set cardinality. If
the border of Finite to Infinity is 10^603, to me anyway, means that
mathematics ends at 10^603 and where
Quantum Logic Physics takes over at the borderline. So to ask whether
the Primes are Infinite means that they exist beyond 10^603 but
mathematics cannot retrieve them. A case example is in AP-adics of a
number such as
99999.....99997? Is it prime or composite? Here, mathematics can never
tell. Can never even tell if the number exists
because mathematics stopped at 10^603 and has entered a shadowy logic
of Quantum Physics.
So to me, all of mathematics ends at 10^603 and to ask if a set is
infinite and what is infinite cardinality is no longer
doing mathematics but rather opinion.
So what if the above conjectures were reworded as such:
1) revised Kepler Packing conjecture 2010: given a border of 10^603, a
pure hexagonal closed packing is not the
maximum packing but requires a mix of packing to achieve a maximum
density of equal spheres.
2) revised Fermat's Last Theorem 2010: since mathematics is bounded by
10^603, that there are no integral
solutions to a^n +b^n = c^n, and if there were, then there would be a
matching of circumference with perimeter
far beyond 10^603. In other words, the proof that 10^603 is a
breakdown of Circumferencing is a proof that can
be used to prove that FLT has no solutions up to 10^603. And here
again, it is all due to pi having three zeroes
in a row in 10^-603 place value. Who would have ever thought that the
three zeroes in a row in pi is the ultimate
proof of FLT? I certainly would not have, before 2010.
3) revised Goldbach conjecture 2010: GC is true for all even integers
up to 10^603. Here we can just simply have
the computers grind out all the even integers with their prime added
duet.
4) revised Riemann Hypothesis 2010: This one is a bit more ticklish as
to whether it is true like FLT for all the numbers
up to 10^603 or whether it is patently false such as Kepler's Packing
with the 10^603 border. Does RH breakdown before 10^603 with
counterexamples or does RH breakdown exactly at 10^603 due to those
three zero digits in pi? Or does
RH breakdown after it crosses over into infinite territory of beyond
10^603? My gut intuition is that pi is so connected
to the Zeta Function that RH breaksdown exactly at 10^603 where the
border is. So that would mean that RH is true
under the boundary of 10^603.
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
Amazing what a nice hot bathe does to stimulate math thinking. I
believe I have the outline of a simple
and easy proof that Fermat's Last Theorem is true up until 10^603 but
not true thereafter.
I am guessing that no-one in mathematics ever tried to prove FLT by
restricting its boundary of validity, but
that all those "attempters of the proof" jumped into the fight with
infinity as unlimited boundary. But maybe
if someone said, what happens to FLT if we impose a boundary of say
10^500, can we thence have any easy proof that
FLT is true for all integers up until 10^500? And with that small
success, impose a higher boundary of 10^1000. But I am guessing no-one
had that sort of smarts, and of course 10^500 and 10^1000 are ad hoc.
But the boundary where
Finite ends and Infinity starts because that is where circle
circumference cannot be matched by a square perimeter
due to pi having three zero digits in a row at 10^-603 place value, is
not ad hoc but natural ingrained intrinsic to the heart of
mathematics.
So in my hot bathe, I thought of a concept of (a) encapsulating.
Encapsulating is Perimeter + Perimeter = Circumference where we have
1^3 + 2^3 = c^3 for FLT when n=3. And so if FLT is false there can
exist a c that
satisfies that where 1 + 8 = c^3. In encapsulating I forget about the
fact that 1 is a cube of 1 and 8 is a cube
of 2 and ask if there is a a^3 + b^3 = c^3. And what the Encapsulating
concept eases, is the need to worry about exponents. It eliminates
exponent 4 then 5 etc etc. Encapsulating lets us worry only about
exponent 3 in FLT, because
all the other exponents fall under the same argument.
I thought of a second concept that I need in this proof of FLT with
its border at 10^603. I call it Integerization and the idea is simple
in that in FLT all rationals are integerizable. And the rational that
I want to integerize
are the digits of pi. So that pi as 3.14 becomes 314 and pi as 3.14159
becomes 314159. So in integerization in this
proof argument we eliminate the decimal point in pi.
Now I need in this proof the AP theorem that Circumferencing the
Perimeter has its breakdown in 10^603 B matrix where pi has those
three zeroes in a row and thus 10^603 is the border between
mathematics and nonmath, or as some would say the border between
Finite versus Infinite.
Proof of Fermat's Last Theorem is true only out to 10^603: We need
only consider exponent 3 since all the other exponents follow. Let us
consider first the two integers of 1 and 2 as a and b respectively in
a^3 + b^3 = c^3. Then
consider 1 and 3 and pair up every integer out to 10^603.
We will show that no c^3 can exist with boundary 10^603. Encapsulate
and then Integerize pi out to 10^603.
We know from the AP theorem that all finite square perimeters are
matched by finite circle circumferences. We transform the a^3 + b^3 =
c^3 into square perimeter + square perimeter = circle circumference.
We have
encapsulated the a^3 and b^3 as square perimeters and the c^3 as a
circle circumference.
We have made pi an integer so that no loss to the argument.
Now we ask, is it possible that there exists a circle circumference
that can match the square perimeter + square perimeter?
And the elegance of this proof is that the answer is no. But barely a
no.
If you take pi and integerize it out to 10^-603, that all along the
string of pi digits, nowhere is it the cube root
except at 10^-603.
Because pi has those three zero digits in a row, if you take the cube
root of pi, you may or may not have a even
cube root. The number 1000 has the even cube root of 10.
The fact that nowhere along the string of pi is pi evenly cube
rootable, evenly 4 exponent rootable, evenly 5 exponent rootable,
means that FLT is true up until 10^603. But then, unlike Wiles, with
his fake proof of a fake
asserted Fermat's Last Theorem, if you continue with pi far enough,
there is a moment in the digits of pi where
FLT integerized will have a large number solutions of a, b, c for
exponent 3. It maybe the case that integerized pi at 10^-603 is
even cube rootable. I need someone with a computer to check to see if
pi integerized is evenly cube rootable at 10^-603 with its three zero
digits in a row.
Before in the 1990s when I was working on FLT, I claimed that FLT is
false because of the p-adic counterexamples, but
here, with more wisdom, I can now see where Wiles's FLT falls apart
without having to use p-adic theory.
FLT is only true when given its boundary, restricting its domain of
numbers. Fermat in the 1600s and Wiles in the
1900s, saw this conjecture without ever precision defining what is
meant by Finite versus Infinite and without ever
realizing there is a border between Finite and Infinite and that this
borderline has to be incorporated into the Fermat's Last Theorem
Conjecture.
When Pierre Fermat or Andrew Wiles never precision define infinity
from finite, then there never is a proof of that
conjecture. But once they understand that a definition of infinity
needs to be inserted into the conjecture, then a simple and easy proof
is found for FLT, not some 100 page unconvincing hornswaggle.
Archimedes Plutonium
that are legitimate
for mathematics, the 10^603 with its 10^-603 inverses and where 0 is
appended imaginary
number and where negative numbers are those positives with a negative
sign attached.
So mathematics, the New Mathematics precisely defines what it means to
be Finite versus
Infinite. And in the New Mathematics, every math conjecture is no
longer unsolved but
solvable by tomorrow since we can train every supercomputer to run
through all the
numbers in the Algebra Array.
Say some joker over there blurts out the P NP problem (that silly
computer engineering problem that is not
mathematics). But say they blurt out that silly question of the NP
conjecture. Then the
Array of Algebra starts to whine through the supercomputers and by
tomorrow morning the supercomputers
put to rest, forever that silly NP conjecture.
So say joker over yonder feels doubtful of the proof that Fermat's
Last Theorem is truthful about exponent
3. Well, the supercomputer is trained to go through all numbers of the
Algebra Array up to 10^603 and including
10^603 and by next morning it shows the doubter that there are no
solutions and thus true.
But now some one pops there head up and asks a proof of the quintic
that there are no solutions.
Now I bring up this topic because, this is the moment in history of
mathematics, that I suspect should have
been the moment where the TRUE MATH PATH should have gone down in that
fork of the road, instead of the
fake path that led towards the fake present day mathematics community
that is drowning and suffocating in
their delusions of infinity.
What does Squaring the Circle have in common with the Quintic quest?
What is common to both is Algebra of numbers.
What numbers exist and do not exist for mathematics.
In the quest of the Quintic, someone should have been bright enough or
sharp enough to recognize that some numbers do not exist in
mathematics such as square root of -1 or the numbers smaller than
10^-603 yet larger than 0. Or numbers
that reside beyond 10^603. By the time of the Quintic Quest, there
should have been a mathematician or two who
recognized that Algebra is limited and that the Quintic is a border
between valid Algebra and cooked up imaginary Algebra. The reason that
there are no solutions to the Quintic is because mathematics has
natural and intrinsic borderlines and by the time of the Quintic
quest, started in the Middle Ages, Tartaglia comes to mind, is the
time in
which the history of mathematics should have produced one or two
mathematicians that pressed for a boundary or
limits of validity on mathematics, the true mathematics, rather than
those that would make Finite and Infinity opinions.
Is there no solutions for the Quintic? And do we really need Galois
Groups, Rings and Fields for the Quintic? Assuredly, when we run the
Quintic through the Array of Algebra of its colossal 10^603 B
matrices, there is no
solution to the Quintic. So that B matrices not only finds a border
between Finite and Infinity at 10^603 for Circumferencing, but that B
matrices easily finds a proof of Fermat's Last Theorem and that the
Quintic has
no solutions.
In other words, for all those laypersons reading this. When
mathematics is on its true and correct path, that any conjecture given
to the professional mathematician, he/she is able to answer that
conjecture over night with the
supercomputer running through the entire Array of Algebra numbers.
When mathematics is true, then it no longer has
any conjecture that is unsolved for a long period of time, but is
solveable overnight. When mathematics is fake
math, a huge log jam of unsolved problems accumulates. Not because
those problems have merit, but only because those problems are using
imprecise-definitions such as finite versus infinite.
Archimedes Plutonium
its first even square rootable? For example 314 has square root
17.72.. then 3141 is 56.04.. so where is the
integerized pi have its first even square rootable? Is it at 10^-308
place value where it
has two zero digits in a row?
Then, where is integerized pi first even cube-rootable? Is it at
10^-603 place value where it has three zero digits in a row?
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
> With respect to Fermat's Last Theorem, where is integerized pi have
> its first even square rootable? For example 314 has square root
> 17.72.. then 3141 is 56.04.. so where is the
> integerized pi have its first even square rootable? Is it at 10^-308
> place value where it
> has two zero digits in a row?
>
> Then, where is integerized pi first even cube-rootable? Is it at
> 10^-603 place value where it has three zero digits in a row?
>
If memory is correct, although this was a long time ago, where I read
some
near misses to the Fermat's Last Theorem. I seem to recall the FLT
missed by
just one. If I recall it was in exponent 3.
So what is the closest miss of FLT to date?
Enrico
<plutonium.archime...@gmail.com> wrote:
> On Dec 27, 10:07 pm, Archimedes Plutonium
>
> <plutonium.archime...@gmail.com> wrote:
> > With respect to Fermat's Last Theorem, where is integerized pi have
> > its first even square rootable? For example 314 has square root
> > 17.72.. then 3141 is 56.04.. so where is the
> > integerized pi have its first even square rootable? Is it at 10^-308
> > place value where it
> > has two zero digits in a row?
>
> > Then, where is integerized pi first even cube-rootable? Is it at
> > 10^-603 place value where it has three zero digits in a row?
>
> If memory is correct, although this was a long time ago, where I read
> some
> near misses to the Fermat's Last Theorem. I seem to recall the FLT
> missed by
> just one. If I recall it was in exponent 3.
>
> So what is the closest miss of FLT to date?
>
>
>
>
> > whole entire Universe is just one big atom
>
> - Show quoted text -
==============================================================
>
> If memory is correct, although this was a long time ago, where I read
> some
> near misses to the Fermat's Last Theorem. I seem to recall the FLT
> missed by
> just one. If I recall it was in exponent 3.
>
> So what is the closest miss of FLT to date?
>
> However, there are some impressive 'near misses'.
> Consider the cubes of 64 and 94:
> 64^3 + 94^3 - 103^3 = 1
> Can anyone come up with even more impressive 'near misses'?
There was reportedly one in one of the Simpsons Halloween episodes.
Google google...
1782^12 + 1841^12 = 1922^12
Well, my calculator says it's correct...
-Arthur,
an ingenious parity
Archimedes Plutonium
>
> > However, there are some impressive 'near misses'.
> > Consider the cubes of 64 and 94:
> > 64^3 + 94^3 - 103^3 = 1
> > Can anyone come up with even more impressive 'near misses'?
>
> There was reportedly one in one of the Simpsons Halloween episodes.
> Google google...
>
> 1782^12 + 1841^12 = 1922^12
>
(snipped)
Thanks Enrico. And thanks to http://www.alpertron.com.ar/BIGCALC.HTM
Alpern's big calculator that we can verify.
I get 262144 + 830584 = 1092728 and where 103^3 = 1092727
Now there is nothing within mathematics that would logical permit one
to say that if there is a a^3 + b^3 = c^3
that misses by one above and by one below, such as 103 misses by one
below. There is no logical impellment
that if we go to infinity that we will run into a a,b,c that actually
hits dead on as equal. This sort of reminds me
of the "no odd perfect numbers conjecture." Wherein in odd perfect
numbers is bounded above and bounded below
but we seem to not find any dead center hits.
What I have said above is nothing new. Except what I say now, is very
new and special. When Fermat proposed his
FLT and when Wiles allegedly thought he had a proof and everyone in
between Fermat and Wiles, all used a notion
of infinity and finiteness. FLT was never explored with a full
precision definition of what infinity means.
So to say that FLT is true, means that the set {1, 8, 27, 64,
125, . .} for exponent 3 has no triplets wherein A + B = C.
And where 64^3 , 94^3 and 103^3 are members of that set.
The trouble that Fermat and Wiles and everyone in between failed to
recognize is that they never defined "infinity" with
precision. The failed to recognize that Finiteness passes, or goes, or
transitions into infinity and thus a borderline
exists between Finite and Infinity. So that, as we recognize that
10^603 is the last number where every finite square perimeter is meet
with a equal circle circumference. So that if we revised Fermat's Last
Theorem to say: For a^n + b^n
= c^n where n is > 2 and where a boundary of 10^603 is the end of
math, then FLT is true.
Because we have an example of where a near miss occurs and missing by
only 1, means to me that we can build
a number that is a counterexample of FLT for exponent 3.
The cube root of 64 plus the cube root of 94 equals a number call it
C. Since in Fermat's time up to Wiles and including
everyone in between, and since they never nailed down a precision
definition of Infinity, that we can go ahead and
engineer this counterexample of FLT for exponent 3. And by doing so,
we have proven that the original FLT, blemished
for its ill-notion of what infinity is, that the original FLT is false
and that it needs a borderline Wall, so to speak to make
sure that there are no counterexamples.
Now if I were doing FLT and had the use of Hensel p-adics, the
counterexamples are immediately there. But many in mathematics would
clamor to say that p-adics are not integers. And even if I used my AP-
adics, infinite integers invented by me to get around base dependency,
many in mathematics would clamor against AP-adics.
But I think I can fetch counterexamples of FLT in exponent 3, perhaps
4 would be easier, straight out of the integers
that Fermat and Wiles are accustomed with and that no-one could clamor
against. I may not be able to write those triples
out or cube them because they are so huge.
But you see, everyone in mathematics from Fermat to Wiles, were never
really accustomed to huge large numbers. The
numbers Fermat and Wiles are used to are these tiny numbers like
103^3. The numbers I speak of are numbers such as
where pi has 4 zeroes in a row or has 9 zeroes in a row. In the the pi
website where it allows you to search pi, where is pi digits that has
say 9 zeroes in a row.
Let me give you an example: we know 10^3 = 1000 and that 5^3 = 125 so
we have to have a cube of a number to
fill in for 875 worth to make equal. Now 9 x 9 x 9 = 729. So it looks
bad for thinking that FLT could have a counterexample, except for the
fact that Fermat to Wiles never had a precision definition of
infinity, and so I can
thence go to 100^3 and 50^3 and then narrow that gap, and go to
infinity until they are equal.
So in other words, modern mathematics with a ill-defined infinity
should have seen that FLT has boatloads of counterexamples as we go
unlimited and unfettered to infinity.
But if we understood from the start that we had to precision define
Infinity to mean where perimeter and circumference
no longer match because of intrinsic pi of mathematics and that
infinity starts at 10^603, then there is hope that Fermat's Last
Theorem truly has no solutions when 10^603 is the border.
Archimedes Plutonium
Archimedes Plutonium
hard work
to program a computer.
Enrico found the 1st mismatches in Squaring the Circle with a twiddle
of 1/(B^2)
This is a accurate table, as far as I can discern:
28.87
282.784
2821.5778
28210.05805
282095.632126
2820950.1143194
28209498.00410956
It uses truncations and is for the B matrices of 100 to that of 10^8.
Those numbers
above are the first breakdowns of radii where a finite circle area
cannot be matched
by a finite square area. The trouble is that none of those radii are
critical-radii. A critical radius is one in which the area is smaller
than the B number itself.
The reason I stopped looking in Squaring the Circle or its counterpart
of Circling the Square
is that it is 2 dimensional and the border of finite with infinity is
1 dimensional. Infinity is a 1 dimensional concept, so I had to
abandon area and go for length in perimeter and
circumference.
In Circumferencing the Perimeter we start with a square-side and find
its perimeter and then
we pi-twiddle and see if we can match it with a circumference. We have
truncations also.
215.50
2152.050
Here is an example of on square side in the Circumferencing the
Perimeter:
Let me try 2152.550 as more suitable
side of square = 2152.550
perimeter = 4xside = 8610.200
8610.200/3.141 = 2741.228 = diameter
start pi twiddling 1/(B^2)twiddle 8610.200/2741.228 = 3.1410010
pi twiddle = 3.141002
3.141002 x 2741.228 = 8610.202 mismatch
3.141001 x 2741.228 = 8610.199 mismatch
Can someone set up a computer program that incorporates the above
example and find that
accurate table of first breakdowns in B matrices from 100 to 10^8?
Is 215.50 the first square-side to have a breakdown in the 100 B
matrix? Please fill in the
first breakdowns all the way out to 10^8.
P.S. Can someone set up a computer program to tell me all of the even
cube roots of integerized pi lie? 314 is integerized pi to three
places and 3141 is integerized pi to
four places and neither are evenly cube rootable. I want to know along
the integerized pi
string of digits, where it is evenly cube rootable. I suspect that pi
is cube rootable only
at the 603rd digits where pi has those three zeroes in a row. The
reason I want this information is because pi could tell us where the
first counterexamples to FLT exist when
we have infinity as imprecisely defined. In other words, pi could lead
us to how large of a
large number we have to go to in order to engineer the counterexample.
Or, pi may tell us that 10^603 is the counterexample of FLT for
exponent 3.
plutonium....@gmail.com
or how it relates to finiteness, then here is a outline of a technique
to find the counterexamples that disprove a Fermat's Last Theorem
that is emanating from a ill-defined infinity.
We see already that 64^3 + 93^3 = 103^3 add 1
Now I should use that pattern of 64 with 93 with 103 and extend it to
infinity to find that counterexample, but I am rather lazy in using 64, 93 and
103 so I will use the more convenient 50, 90 and 100 in exponent 3.
First we start with 10^3 = 1000
and half of 10 is 5, so 5^3 = 125 and now all we need is c^3 to
make up for 875, but the nearest we can come is 9^3 = 729 for a overall
deficit of about 14.6%
So we skip to the next level of 100^3 and 50^3 and 95^3 which leaves
us with a deficit not so severe as with the previous level, since the deficit
is now only 1.76% rather than the 14.6%. So we are making progress and improvement.
So we jump to the next level of 1000^3 with 500^3
And we can also do a bit of surplus rather than deficits, by running say
100^3 with 55^3 and 95^3 with each higher level so that we bound the
convergence to a counterexample for Fermat's Last Theorem.
Now the final answer is a triplet that is a counterexample to FLT, and the reason it works and is sure to work is that in FLT as understood by Fermat and Wiles and everyone in between, is that they never understood that they were working with a imprecise meaning of infinity. So that if a triplet of 100, 50, 95 is not a FLT triplet because it still has a deficit or a surplus and not equal, well, with an ill-defined infinity, we just keep going until we
do have that equality.
So the proof of FLT as Pierre Fermat understood mathematics and as Wiles
understood mathematics with use of a mushy ill defined finite and infinity, there are always solutions to a^n + b^n = c^n provided you have the patience in going out far enough to obtain that solution. But Fermat to Wiles with everyone in between, never really had the patience to fetch one of these huge triplet numbers that is an equality, but they exist.
The Wiles alledged proof is nothing but a mere hornswaggle, much like the nonsense one sees from a crank of mathematics, since the concepts of a precision infinity or finite was unknown to Andrew Wiles. And to plaster over Wiles self confusion, he supposes the nonexistence of some elliptic curves. Much like a modern day nutter supposing that "what if Sept 11, 2001 had never happened would the Great Recession of 2008 never happened." Now Andrew Wiles thinks his work on FLT is good math but anyone with a acumen of Logic, knows that Wiles is just a hot air buffoon of mathematics. And the people who have accepted Wiles's contraptions, just simply do not know mathematics at all.
I once made a math survey on the East Coast of the USA asking college University Math professors their definition of Finite versus Infinity.
It is a shame I did not do the survey at Princeton University. I did get a few inputs from Harvard University when there in the 1990s.
What I got in the survey is that most mathematic professors at colleges and Universities believe that Finite versus Infinite is distinguished by finite ending in a zeroes digits to the leftwards, so that the number 22 is finite because it is ....0000022. And it did not matter to these professors whether a number like pi which is 3.14..... is nonrepeating in its rightwards digit string so long as its leftward string is ending in 00s digits so that pi is a finite number.
So that is the modern day convoluted understanding of what it means to be
finite versus infinite.
Certainly, in Wiles's hornswaggle of a contraption of a proof of FLT, never once does Wiles make clear that he is or is not following that definition of infinity or finiteness.
What does the existence or nonexistence of elliptic curves have anything to do with whether there is a triple in the set {1, 8, 27, 64, . .} that have A + B= C? Nothing, and especially when those elliptic curves are reductio ad absurdum. Suppose 9/11/2001 never happened, what connection would it have to whether or not the Great Recession of 2008 happened? This is a phony history and economics question, but worse is a phony mathematics reductio ad absurdum on Fermat's Last Theorem.
I propose a new survey on all mathematics professors of colleges and universities. Two surveys in fact:
Survey One:
ask the math professor what his/her definitions are for Finite versus Infinite
Survey Two:
ask the math professor these true false questions:
1) Does finite run or go or transition into infinite? true or false
2) Is there a border or borderline between finite and infinity? true or false
Now if you are in a math class, do not ask your professor, for he/she likely will give you a bad grade, no matter how well you honestly do.
Math professors hate to admit that they were fooled for most of their lives on an issue of finiteness and infinity and would rather ignore the fact they are ignorant of finite and infinity.
Archimedes Plutonium
<plutonium.archime...@gmail.com>
wrote:
(snipped to save space)
So are those accurate?
215.50 first breakdown in 100 B matrix
2152.050 first breakdown in 1000 B matrix of Circumferencing the
Perimeter
And can someone fill in the table out to 10^8 B matrix?
adamk
Ap: you are a legend in your own mind...and
nowhere else.
Loser.
Archimedes Plutonium
Adam, I need more substantive information of these surveys of math
professors at
Penn State than that of "bananas and losers".
FIRST SURVEY:
1) Define what is meant by a Finite versus Infinity and how to
recognize one from the other, such as what is a Finite Number versus
an Infinite Number?
If you look at any encycopedia of mathematics and even Wikipedia fail
to distinguish between a Finite number and an Infinite number, yet
these two concepts are used every time a mathematics proof involving
numbers is undertaken.
SECOND SURVEY:
Is a true false questionnaire on the topic of Finite versus Infinity
to be given to all mathematics
professors and editors of
mathematics.
1) Does Finite run or go or transition or move from finiteness into
the infinite?
2) Is there a border or borderline between Finiteness and Infinity?
3) Is there a largest Finite number and thus, a smallest infinite
number?
adamk
without _ever_ having provided any proof at all
-- and I mean proof, and not just unfounded
opinion -- that there is anything wrong to start with.
That is incredible indeed.
Archimedes Plutonium
215.50
2152.050
21521.5000
Circumferencing the
Perimeter:
for B matrix 10^4
side of square = 21521.5000
perimeter = 4xside = 86086.0000
86086.0000/3.1415 = 27402.8330 = diameter
start pi twiddling 1/(B^2)twiddle 86086.0000/27402.8330 = 3.141500004
pi twiddle = 3.14150001
3.14150001 x 27402.8330 = 86086.0001 mismatch
3.14150000 x 27402.8330 = 86085.9998 mismatch
Apparently the digit pattern arrangement of 215---- leads to
breakdowns in Circumferencing, but whether they are the
first breakdowns for those B matrices is not yet established.
Archimedes Plutonium
I think I may have spotted where the Riemann Hypothesis has its first
breakdown in
counterexamples of nontrivial zeroes on the 1/2 Real strip. The
question being, as in
FLT, whether the breakdown is before 10^603 or long after the 10^603
as the border between
Finite and Infinity.
Archimedes Plutonium
Reading from the book Prime Obsession by John Derbyshire on page 104.
What I am trying to resolve is whether the Euler encoding of the
multiplication of primes is really and truly equal to that of the zeta
function of the addition series.
Here is what we do know for sure, is that pi is a series involving
primes and this series allows us to
fetch three zeroes in a row in the 10^-603 place value.
But that a multiplication of primes as the Euler Encoding would not
possibly be able to fetch 3 zero digits in a row
in multiplication. So that the Euler multiplication encoding is not
equal to the Riemann Zeta Function and that all this huff puff
hallyballoo about the primes being as perfectly spaced as possible was
just a fanciful wish.
What I have to do here, is see if I can analyze where RH has its first
counterexample, whether it is below the 10^603 mark or smack dab on
the mark of 10^603 where Circumferencing the Perimeter breaks down
also, or whether RH breaksdown some distance beyond the 10^603 mark.
Archimedes Plutonium
integerized pi and find each spot up to about
its 700 digit of where integerized pi is evenly cube rootable. For
example 314 is pi integerized to 3 places and that number is not
evenly cube rootable. So is integerized pi only cube rootable at the
spot where pi has those three
zero digits in a row? That would be a valuable clue as to where the
first counterexamples of FLT lie.
Now here is another method of finding the first counterexamples to
FLT:
Thank goodness for this big calculator website which furnishes cube
roots:
http://www.alpertron.com.ar/BIGCALC.HTM
Cube root of 9 is:
2.080 083 823 051 904 114 530 056 824 357 885 386 337 805 340 373 262
109 697 591 080 200 106 311 397 268 773 606 056 636 790 757 486 728
671 592 086 574 520 538 907 806 551 432 406 435 155 956 414 938 597
044 743 429 752 216 294 112 964 022 766 857 669 747 769 143 933 688
804 813 536 021 599 545 811 696 696 536 066 894 994 770 998 260 332
103 790 432 402 956 475 010 988 207 691 299 963 138 219 077 694 067
605 530 360 769 279 867 841 312 933 591 633 086 534 769 408 692 721
478 724 093 497 104 836 551 394 365 820 070 260 491 731 141 474 317
009 878 047 273 127 944 297 002 343 787 245 304 869 234 558 012 240
906 823 440 092 285 995 194 896 338 315 133 702 469 3
Now we know that FLT for exp3 asks whether in this set {1, 8, 27,
64..} there is a A + B that equals C.
So let me just take the first two of 1 and 8 and play a trick of 1 + 8
= 9 and although 9 is not in that
set, let me expand those triplets to where they are integers and where
they do falsify FLT.
We know the cube root of 1 is 1.000.. and 8 is 2.0000... with never a
worry about a zero digit rightwards.
But let us examine the cube root of 9 and ask where in that expansion
do we run into three zero digits in a row? In the above, I see several
spots where there are two zero digits in a row.
So this method works by integerizing in steps until we reach where
cube root of 9 has three zero digits in a row.
The first step is
1^3 + 2^3 compare 2^3
second step is
10^3 + 20^3 compare 20^3
third step is
100^3 + 200^3 compare 208^3
As you can see, I am borrowing the digit arrangement of the cube root
of 9 and walking out on that arrangement
as integerized until I reach a moment where cube root of 9 has three
zero digits in a row. At that spot, I may
find where the integerized steps lead to an equality. If not an
equality, enough of a leeway to see where I need
to make adjustments to reach an equality.
Now I suspect that the spot where cube root of 9 has three zero digits
in a row is not that far away, at least not
as far away as 603 digits. And then I would ask where cube root of 9
has six zero digits in a row.
The point that must be hammered home, is that Fermat up to Wiles and
everyone in between, thought of infinity
as a concept that needs no precision definition. And thus, their
infinity allows for solutions to all exponents
in FLT, provided if you are willing to go out in distance far enough
to fetch those counterexamples.
When a mathematician cares to precision define Finite versus Infinity
by stating the border between the two such as
10^603, then FLT only has solutions to exponent 2, and none beyond
exponent 2.
David R Tribble
> Reading from the book Prime Obsession by John Derbyshire on page 104.
Excellent book.
> What I am trying to resolve is whether the Euler encoding of the
> multiplication of primes is really and truly equal to that of the zeta
> function of the addition series.
>
> Here is what we do know for sure, is that pi is a series involving
> primes and this series allows us to
> fetch three zeroes in a row in the 10^-603 place value.
>
> But that a multiplication of primes as the Euler Encoding would not
> possibly be able to fetch 3 zero digits in a row
> in multiplication. So that the Euler multiplication encoding is not
> equal to the Riemann Zeta Function and that all this huff puff
> hallyballoo about the primes being as perfectly spaced as possible was
> just a fanciful wish.
1. So exactly which step in the derivation (shown in the book)
is wrong?
2. If the primes are not spaced according to the PNT, how are
they spaced?
David R Tribble
> The point that must be hammered home, is that Fermat up to Wiles and
> everyone in between, thought of infinity
> as a concept that needs no precision definition. And thus, their
> infinity allows for solutions to all exponents
> in FLT, provided if you are willing to go out in distance far enough
> to fetch those counterexamples.
So are you saying that the Modularity Theorem (a.k.a. the
Taniyama-Shimura-Weil Conjecture) is false?
adamk
Putonium has been claiming for years now that he is
correcting mathematics (what would we do if our hero
was not here to correct us?), and has not yet shown
_anything_ to be wrong.
Putonium accusses mathematicians of lying, yet
has never produced any proof .
Putonium talks and talks: putonium is a big talker
who cannot support anything he says. Because he lives
in a bubble.
And, Putonium: I am not interested in any advice you
may have for me, since I only consider advice from those
I respect. And I feel nothing but contempt for you,
libelous liar . So spare me your stupid (what else,
coming from you?) advice, and go twiddle with
Enrico's banana.
Loser.
adamk
> of infinite
> or how it relates to finiteness, then here is a
> outline of a technique
> to find the counterexamples that disprove a Fermat's
> Last Theorem
> that is emanating from a ill-defined infinity.
Of course, you exclude the possibility that _you_
do not understand it.
Because you have the charming habit of making
claims without any evidence.
>
> We see already that 64^3 + 93^3 = 103^3 add 1
And to you, that is a counterexample, right?
You keep claiming, and you _never_ produce anything
come up with the damn counterexample for once!!!!!!
>
> Now I should use that pattern of 64 with 93 with 103
> and extend it to
> infinity to find that counterexample, but I am rather
> lazy in using 64, 93 and
> 103 so I will use the more convenient 50, 90 and 100
> in exponent 3.
No one really cares what you should or not do.
DO YOU HAVE A COUNTEREXAMPLE OR NOT????????
As usual, the answer is no. You are making things
up.
> <snip> garbage, fantasy and self-delusion.
>
>
> So we jump to the next level of 1000^3 with 500^3
>
> And we can also do a bit of surplus rather than
> deficits, by running say
> 100^3 with 55^3 and 95^3 with each higher level so
> that we bound the
> convergence to a counterexample for Fermat's Last
> Theorem.
So you do have a triplet, or not?
>
> Now the final answer is a triplet that is a
> counterexample to FLT, and the reason it works and is
> sure to work is that in FLT as understood by Fermat
> and Wiles and everyone in between, is that they never
> understood that they were working with a imprecise
> meaning of infinity.
A imprecise a infinity? A infinity is only a imprecise
to those who do not a understand it.
So that if a triplet of 100, 50,
> 95 is not a FLT triplet because it still has a
> deficit or a surplus and not equal, well, with an
> ill-defined infinity, we just keep going until we
> do have that equality.
So , it comes down to this: you DO NOT have a
counterexample.
But you still claim that FLT is wrong.
Please explain how this follows. You claim
a result is wrong without showing a counterexample.
>
> So the proof of FLT as Pierre Fermat understood
> mathematics and as Wiles
> understood mathematics with use of a mushy ill
> defined finite and infinity, there are always
> solutions to a^n + b^n = c^n provided you have the
> patience in going out far enough to obtain that
> solution. But Fermat to Wiles with everyone in
> between, never really had the patience to fetch one
> of these huge triplet numbers that is an equality,
> but they exist.
Really? So why don't you then show us one such
triplet?
>
> The Wiles alledged proof is nothing but a mere
> hornswaggle, much like the nonsense one sees from a
> crank of mathematics, since the concepts of a
> precision infinity or finite was unknown to Andrew
> Wiles.
And the concept of reality is unknown to you:
you keep making claims without any evidence.
PRODUCE A COUNTEREXAMPLE, OR SHUT UP.
And to plaster over Wiles self confusion, he
> supposes the nonexistence of some elliptic curves.
> Much like a modern day nutter supposing that "what if
> Sept 11, 2001 had never happened would the Great
> Recession of 2008 never happened."
Or, even more fantastic: what if Zirconium was
not a self-deluded moron?
Now Andrew Wiles
> thinks his work on FLT is good math but anyone with a
> acumen of Logic, knows that Wiles is just a hot air
> buffoon of mathematics.
But you, Zirconium , you do have a acumen, a right?
I will e-mail this comment to Wiles, where you
call him a buffoon.
And the people who have
> accepted Wiles's contraptions, just simply do not
> know mathematics at all.
And you know absolutely nothing, imbecile.
>
> I once made a math survey on the East Coast of the
> USA asking college University Math professors their
> definition of Finite versus Infinity.
>
> It is a shame I did not do the survey at Princeton
> University. I did get a few inputs from Harvard
> University when there in the 1990s.
>
> What I got in the survey is that most mathematic
> professors
From an informal, non-randomized experiment done
by a primate like you, you reached a conclusion
about all math proffessors?
So statistics is yet
another area where you excel at being an imbecile!
at colleges and Universities believe that
> Finite versus Infinite is distinguished by finite
> ending in a zeroes digits to the leftwards, so that
> the number 22 is finite because it is ....0000022.
> And it did not matter to these professors whether a
> number like pi which is 3.14..... is nonrepeating in
> its rightwards digit string so long as its leftward
> string is ending in 00s digits so that pi is a finite
> number.
Which you actually made up. Or did they actually say that? Show us the data. Er, sorry, I forgot, you have
a desdain for evidence; you just make things up.
>
> So that is the modern day convoluted understanding of
> what it means to be
> finite versus infinite.
Yes, from a poorly-designed sample, which you cannot
show us, you reached a conclusion about the overall
modern understanding. Idiot.
>
> Certainly, in Wiles's hornswaggle of a contraption of
> a proof of FLT, never once does Wiles make clear that
> he is or is not following that definition of infinity
> or finiteness.
And you never show any counterexample, even after
repeatedly claiming that FLT is wrong. Because you
are making it up.
>
> What does the existence or nonexistence of elliptic
> curves have anything to do with whether there is a
> triple in the set {1, 8, 27, 64, . .} that have A +
> B= C? Nothing, and especially when those elliptic
> curves are reductio ad absurdum. Suppose 9/11/2001
> never happened,
Assume Zirconium is not a moron.
what connection would it have to
> whether or not the Great Recession of 2008 happened?
> This is a phony history and economics question, but
> worse is a phony mathematics reductio ad absurdum on
> Fermat's Last Theorem.
>
> I propose a new survey on all mathematics professors
> of colleges and universities. Two surveys in fact:
>
> Survey One:
>
> ask the math professor what his/her definitions are
> for Finite versus Infinite
>
>
> Survey Two:
>
> ask the math professor these true false questions:
>
> 1) Does finite run or go or transition into infinite?
> true or false
>
> 2) Is there a border or borderline between finite and
> infinity? true or false
>
> Now if you are in a math class, do not ask your
> professor, for he/she likely will give you a bad
> grade, no matter how well you honestly do.
I don't take advice from buffoons like you. Do
your own fucking survey. And then fudge the results,
so you can keep making things up.
>
> Math professors hate to admit that they were fooled
> for most of their lives on an issue of finiteness and
> infinity and would rather ignore the fact they are
> ignorant of finite and infinity.
And they have to admit it because you said so,
without counterexamples, right?
>
> Archimedes Plutonium
> http://www.iw.net/~a_plutonium/
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies
And people are made of marshmallows.
Loser.
David Bernier
The Riemann-von Mangoldt formula for psi(x) is a truly marvelous thing.
First, the definition of the von Mangoldt function Lambda(n),
for an integer n >=1:
If n has no prime factors, Lambda(n) = 0.
If n has two or more distinct prime factors, Lambda(n) = 0.
The remaining case is where n has just one prime factor, p.
Then n could be p, p^2, p^3, ...
Then Lambda(p) = Lambda(p^2) = Lambda(p^3) = ... = log(p) [base 'e' ].
< http://en.wikipedia.org/wiki/Von_Mangoldt_function > .
psi(x) := sum_{1 <= n <= x} Lambda(n) .
[ This is the summatory von Mangoldt function, or Chebyshev function,
according to Wikipedia].
There is an expression for psi(x) in terms of the non-trivial zeta zeros
that is referred to as the "Riemann-von Mangoldt explicit formula"
or "von Mangoldt explicit formula". You can find it here,
following the text "Von Mangoldt's formula for psi(x):"
< http://web.viu.ca/pughg/Psi/ > .
There is also an applet there. As more zeta zeros are included,
the graph changes less and less. I think there are jump
discontinuities in psi(x) at primes and prime powers,
and I'm not sure how the "explicit formula" behaves point-wise
at the jump discontinuities.
David Bernier
> Here is what we do know for sure, is that pi is a series involving
> primes and this series allows us to
> fetch three zeroes in a row in the 10^-603 place value.
>
> But that a multiplication of primes as the Euler Encoding would not
> possibly be able to fetch 3 zero digits in a row
> in multiplication. So that the Euler multiplication encoding is not
> equal to the Riemann Zeta Function and that all this huff puff
> hallyballoo about the primes being as perfectly spaced as possible was
> just a fanciful wish.
>
> What I have to do here, is see if I can analyze where RH has its first
> counterexample, whether it is below the 10^603 mark or smack dab on
> the mark of 10^603 where Circumferencing the Perimeter breaks down
> also, or whether RH breaksdown some distance beyond the 10^603 mark.
>
>
> Archimedes Plutonium
> http://www.iw.net/~a_plutonium/
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies
>
--
$ gpg --fingerprint davi...@videotron.ca
pub 2048D/653721FF 2010-09-16
Key fingerprint = D85C 4B36 AF9D 6838 CC64 20DF CF37 7BEF 6537 21FF
uid David Bernier (Biggy) <davi...@videotron.ca>
Archimedes Plutonium
As to 1) I am not sure yet whether anything in the derivation is
wrong.
It maybe the interpretation at the point of infinity, such as the math
community with ill-defined infinity says that 1.9999.. is 2.00.. so
the interpretation
at the end begets the zeroes in pi.
What I do know is that pi is this series:
pi = 4 - 4/3 + 4/5 -4/7 + . .
And the Zeta function is
Z = 1 + 1/2^s + 1/3^s + 1/4^s + ..
The Euler Zeta function is E = 1/1-(1/2^s)x 1/1-(1/3^s) x . .
I have never seen a pi of multiplication of anything resembling
primes.
My hunch is that we know that you can get zeroes digits in pi by
adding or subtracting and especially
the three zeroes in a row at the 10^-603 place value.
So my hunch is that since Euler's Zeta is multiplication involving
primes then how does that multiplication
deliver those three zeroes in a row of pi at the 10^-603 place value.
Keeping in mind that multiplication using
primes can only seem to deliver a zero digit when 5 x another prime
takes place but how do you get three zeroes in a row with some
arrangement of the 5 digit?
So, my hunch is that something is amiss at this juncture of Euler's
Zeta of multiplication retrieving three zeroes in a row for pi whereas
the Zeta function of add and subtract can deliver those three zeroes.
2) as for PNT, think of snapshots of primes for the 100 matrix then
the 1000 matrix then the 10000 matrix. In the
100 matrix there are x/Ln(x) or 25 primes and then for 1000 there are
roughly 1000/6 = 166 etc etc. Now if Old Math
insists on never a precision definition of infinite we can look at a
snapshot of AP-adics of the what I called the
last snapshot of Infinite Integers listing them backwards as
99999....9999 then 9999....99998 then 9999....99997 etc etc. To make a
long story short, if we look at these numbers as a snapshot with Old
Math of PNT and a ill-defined-infinity, that PNT would require a rough
amount of primes in that snapshot, but according to the late Dik
Winter in several of his posts where no matter how large that snapshot
is, they are all composites with no primes. And according to LWalk who
participated in that debate reckoned on whether the 9999....9999
series was even or odd and that LWalk could then tag 9999....9997 as a
prime. So suppose LWalk was correct in that 9999....9997 was prime and
having thus cemented the fact that the digits were odd (do not
remember whether they had to be odd for 999...9997 to be prime) and
working backwards thence we can thus determine whether 9999...99991
was prime or composite. The upshot is that in doing so, we see that
the Old Math of (a) ill defined infinity coupled with (b) PNT and
coupled with (c)
AP-adics that we cannot have the PNT as a true theorem of mathematics.
That as the Finite numbers of mathematics ends at 10^603, so do the
theorems of mathematics hold true for only numbers from 0 to 10^603
and once that border is crossed, all of Old Math theorems are in
jeopardy of no longer being true.
P.S. As for RH and the Euler multiplication encoding, I am only
playing on a hunch. That I fail to see the symmetry of pi as a series
but not a multiplication and how to retrieve those three zeroes in a
row at 10^-603 or when pi has
alot of zeroes in a row, how to retrieve them via multiplication.
If that hunch pans out as true, then the RH instantly loses alot of
its status, because no longer is there much interest because the
"primes" no longer are a center of attention in RH. And the Zeta
function becomes just another
ho hum function.
Archimedes Plutonium
I actually do not know what that theorem or that conjecture is about.
But I do know that the
authors of those two pieces of mathematics had no clue of what
"infinity and finite" were,
and I am certain that if those two pieces of mathematics involves a
question of finite and
infinity, such as FLT or RH involve infinity, that those two pieces of
math were not mathematics
at all but just some floating around opinions.
Can the Modularity piece accommodate the idea of a borderline at
10^603 where beyond is nonmath?
Can the TSW Conjecture accommodate a borderline of math at 10^603? I
tend to think a precision definition
of finite versus infinity sinks both those pieces of Old Math, not
even knowing what they say.
When mathematics lacks a precision definition of finite versus
infinite, then alot of fake structures
are built up which come tumbling down once the definitions are
precision. This is what happened to
the Ptolemy epicycles
Archimedes Plutonium
Thanks for that information David. Maybe I am wrong, but I do not see
that as
applying to using pi digits to resolve RH. I do not think it helps my
search, because
if I recall the latest nontrivial zeros of RH are only out to about
10^25, and my
memory is bad on that. I do recall someone saying that it is
impossible to get
the zeta function out to 10^300 with our technology and no chances of
some future
breakthrough.
So what I need to find are equivalent statements to the RH such as the
Moebius function,
to see if RH breaks down at when pi has two zeroes in a row at 10^-308
and then when
pi has three zeroes in a row at 10^-603.
I am confident that the connection of the digits of pi to the Riemann
Hypothesis is the key to unlocking whether RH is true or false and
where it breaks down.
Maybe the Moebius Function is the appropriate channel that can deal
with the question of whether those three zero digits in a row of pi at
10^603 also is the moment that RH has counterexamples. Or, maybe it
breaks down earlier when pi has two digits of zeroes in a row at
10^308.
David Bernier
Several equivalents to RH are given here:
< http://aimath.org/pl/rhequivalences > .
David Bernier
> I am confident that the connection of the digits of pi to the Riemann
> Hypothesis is the key to unlocking whether RH is true or false and
> where it breaks down.
> Maybe the Moebius Function is the appropriate channel that can deal
> with the question of whether those three zero digits in a row of pi at
> 10^603 also is the moment that RH has counterexamples. Or, maybe it
> breaks down earlier when pi has two digits of zeroes in a row at
> 10^308.
>
> Archimedes Plutonium
> http://www.iw.net/~a_plutonium/
> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud are galaxies
Archimedes Plutonium
(snipped)
>
> Several equivalents to RH are given here:
>
> <http://aimath.org/pl/rhequivalences> .
>
> David Bernier
David, I was just about to make the post below when yours came up
so let me ask you a question nagging on my mind.
When we apply the border of doing math at 10^603, that would change
all
the convergences in Series, however slight or tiny of a change, still
a change
and it would also mean that the Zeta function is not equal to the
Euler encoding.
Would such a application totally alter the Riemann Hypothesis, or
would it be a
cosmetic small change? Would the nontrivial zeroes still appear much
the same as before?
Through the years, I seemed to have coddled the Riemann Hypothesis RH.
Perhaps because of its
complications and here even with a borderline between finite and
infinity, I have not
looked at RH objectively. I guess habits reside more over what we do
than does reason or logic.
I have never really put the borderline of mathematics at 10^603
foremost when looking at
RH, as if I was expecting to get RH to reproduce the borderline.
And funny how my worst mistake in math for 2010 comes near the last
day of 2010.
This morning (yesterday) I started with a position that the Euler
multiplication encoding was not the same as the zeta function. Here I
am still on habit and not reasoning.
Mathematics of Finite versus Infinity should have been made clear at
the very beginning of mathematics, but we could not expect the Ancient
Greeks to have Circumferenced the Perimeter and found this border of
10^603. What we can expect is that we find this border by 2010, and
then, instead of assuming RH was mostly correct we should have started
over completely with RH as if we were in the 1800s when Riemann was
actually doing RH.
Proof of Goldbach Conjecture: we start with the border of finite and
infinity and so mathematics stops at 10^603. So to prove Goldbach we
still have to go through all the even
numbers from 4 to 10^603. And the easiest proof is to simply train all
the supercomputers to
list all the even numbers with their two primes that satisfy. Sheer
computer brute force
outputs is the proof. There maybe a surprize but I doubt it.
Proof of the geometry problems outstanding such as Kepler Packing and
Poincare Conjecture:
Kepler Packing starts with that borderline of 10^603 and means that
the proof involved is for the most part, or bulk of the packing is
hexagonal closed packing, except for the border region of its last
rows where a hybrid technique I called oblong packing serves as the
maximum dense packing. Poincare Conjecture is patently false since a
borderline of 10^603 means an inverse of 10^-603 and so there exists
no points or numbers between 0 and 10^-603.
So the fabric of geometry is full of tiny holes quantized by 10^-603.
So Poincare Conjecture
is totally false. Whereas in the Kepler Packing, the bulk of it is
hexagonal closed packing except a hybrid at the last row at the
border.
Proof of Fermat's Last Theorem: And maybe FLT is what has caught me on
a snag with RH, because both speak of a counterexample. So why not run
FLT through a computer program as the proof of Goldbach is done? Well,
FLT is more involved than Goldbach for computer time. FLT is likely to
be true given that imposition of the border at 10^603. Maybe an easy
proof will be
to run FLT for exponent 3 and 4, all the way to 10^603 and if no
counterexample is found, that there is some math reasoning that the
other exponents cannot have a counterexample either. And of course
since 10^603 is the end of mathematics, that the exponent in FLT also
stops making sense with a number far smaller than 603. So perhaps the
easiest proof of FLT
is raw computer work of exp 3 and 4 out to 10^603.
Proof of Riemann Hypothesis: So far I have treated this conjecture
with kid's gloves, and maybe I should just get right down rough and
ready in all out fight. Apply 10^603, which heretofore I have been
reluctant. Then that means the Zeta function ends with a 1/10^603
and then the Euler encoding of multiplication would require the last
prime in the vicinity
of 10^603. So the Zeta function is not equal to the Euler encoding. Is
that enough to say RH is false? In this new stark reality
the entire subject of Series in mathematics must be revised. Series
that converge, have to be
reevaluated because infinity is 10^603. In this light, we ask the
reverse question of what
Riemann asked whether all the nontrivial zeroes lie on the 1/2 Real
critical strip, and here we would ask whether any nontrivial zeroes
lies on the critical strip. So does the proof of RH then rely on
computer output of all the numbers to 10^603? Maybe the revision is
minor such as what was seen in Kepler Packing that most of it was true
with a small modification at the border.
Archimedes Plutonium
>
> <http://aimath.org/pl/rhequivalences> .
>
> David Bernier
>
David, at the bottom of that website talks about the Hardy Littlewood
equivalency to a Series of << x(1/4) with the next one being the Carey
Series. Tell me David, what happens with those Series Equivalencies if
we injected the end of math at 10^603 into those Series. Would that
falsify RH or support RH?
David Bernier
I'll reply according to standard math., which is the kind I
understand. Euler discovered the Euler product identity,
which holds for zeta(s) when s is real and > 1.
zeta(s) can be extended to complex s with Re(s) > 1
without convergence problems. Re(s) > 1 means that
s = x + i*y, x, y real, and x>1.
zeta(1) is divergent. It's possible to extend zeta
complex-differentiably to all the complex numbers except 1,
or C \ {1}. So there's just one zeta function that extends
Euler's zeta while maintaining analyticity, i.e.
continuous complex-differentiability. Riemann showed the
way in how zeta zeros in the critical strip
0 <= Re(s) <= 1 relate to the distribution of primes,
the prime counting function pi(x). This is
advanced material, but known for 100 years or more.
For example, I think there's an explicit formula for
pi(x), but I don't know when it was proved.
According to you, there is a border...
But then what do you mean by Infinity ?
David Bernier
Archimedes Plutonium
> Archimedes Plutonium wrote:
(snipped)
>
> I'll reply according to standard math., which is the kind I
> understand. Euler discovered the Euler product identity,
> which holds for zeta(s) when s is real and > 1.
> zeta(s) can be extended to complex s with Re(s) > 1
> without convergence problems. Re(s) > 1 means that
> s = x + i*y, x, y real, and x>1.
>
> zeta(1) is divergent. It's possible to extend zeta
> complex-differentiably to all the complex numbers except 1,
> or C \ {1}. So there's just one zeta function that extends
> Euler's zeta while maintaining analyticity, i.e.
> continuous complex-differentiability. Riemann showed the
> way in how zeta zeros in the critical strip
> 0 <= Re(s) <= 1 relate to the distribution of primes,
> the prime counting function pi(x). This is
> advanced material, but known for 100 years or more.
> For example, I think there's an explicit formula for
> pi(x), but I don't know when it was proved.
>
> According to you, there is a border...
>
> But then what do you mean by Infinity ?
>
> David Bernier
Thanks David, for what you said above, correct me if wrong. Is that
the Riemann Hypothesis
is crucially dependent on the idea that Euler's Zeta encoding is equal
to the Riemann Zeta function.
If those two encodings of Euler's primes in multiplication and
Riemann's zeta in addition of terms are
not equal then there is no Riemann Hypothesis and the entire
conjecture has no underpinning.
So with Mathematics ending at 10^603 there is no Riemann Hypothesis
because the Euler primes
and the zeta function are not equal to each other.
With mathematics ending at 10^603, Goldbach conjecture is proven true
with a computer running through
all even numbers from 4 to 10^603. Kepler Packing is proven true with
a modification of the last rows of
spheres in a oblong packing and the bulk packing of hexagonal closed
packing. Poincare Conjecture is proven
to be false because there are tiny holes everywhere in geometry of
smaller than 10^-603.
FLT is probably proven true with that 10^603 border where a computer
can easily run through all the cases for
exponent 3 and 4 and by numerical analysis prove that no pythagorean
triple is possible for higher exponents. However, the Wiles's sham
proof is sham because a ill-defined infinity does yield boatloads of
solutions.
So it is a surprize to me David, that the Riemann Hypothesis did not
behave as FLT, Goldbach or Kepler with modifications. RH ends up
behaving like that of Poincare Conjecture as purely false and where no
modification is ever
going to revive it.
David, please answer another question. Please write a page on what
mathematics has for this so called P NP problem.
So far I have seen only a P NP that talks of computer engineering with
time and a recursion question. Can you explain
this problem, David, in purely a mathematics setting without any
reference to computers and engineering. I feel it is
a problem that knowing mathematics ends at 10^603 then P NP ends at
10^603. Perhaps the easiest of all conjectures to disprove.
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
> On Dec 31, 3:56 am, David Bernier <david...@videotron.ca> wrote:
>
>
>
> > Archimedes Plutonium wrote:
> (snipped)
>
> > I'll reply according to standard math., which is the kind I
> > understand. Euler discovered the Euler product identity,
> > which holds for zeta(s) when s is real and > 1.
> > zeta(s) can be extended to complex s with Re(s) > 1
> > without convergence problems. Re(s) > 1 means that
> > s = x + i*y, x, y real, and x>1.
>
> > zeta(1) is divergent. It's possible to extend zeta
> > complex-differentiably to all the complex numbers except 1,
> > or C \ {1}. So there's just one zeta function that extends
> > Euler's zeta while maintaining analyticity, i.e.
> > continuous complex-differentiability. Riemann showed the
> > way in how zeta zeros in the critical strip
> > 0 <= Re(s) <= 1 relate to the distribution of primes,
> > the prime counting function pi(x). This is
> > advanced material, but known for 100 years or more.
> > For example, I think there's an explicit formula for
> > pi(x), but I don't know when it was proved.
>
> > According to you, there is a border...
>
> > But then what do you mean by Infinity ?
>
> > David Bernier
>
I should address David's question in detail rather than to glide over
it.
And it is not that David thinks he is doing the correct math, but a
failure
to see that David's math is imprecision math, hence no math.
In the Riemann Zeta function we have Z(1) = 1 + 1/2 + 1/3 + . .
And most in Old Math recognize that as the harmonic series, but their
next thought is a travesty of
math. Their next thought is that using an imprecise definition of
finite versus infinity, they come
to a mistaken conclusion that the series diverges to infinity, yet
they never defined infinity with any
sort of precision and that is the boat that David is stuck in also.
So now, what if the Old Math community and David decided to define
infinity with precision.
They would then see that finite transitions into infinity meaning
there must be a border
between them. David and the Old Math community would then look at
where in mathematics a
natural intrinsic border exists and the answer is that in
Circumferencing the Perimeter the
Algebra of mathematics ends at producing a match of a finite square
perimeter with a finite
circle circumference at 10^603. So that David and his Old Math
companions then have a
borderline between finite and infinity and thus defined infinity with
precision.
Now David and Old Math community comes back to the question of the
Riemann Zeta Function
and harmonic series of 1 + 1/2 + 1/3 + . .
When you never have a precision definition of infinite, you are easily
fooled into thinking that
this series diverges but when you recognize that the last term in the
sequence is 1/10^603 then
you instantly recognize it converges to a finite number. What is the
convergence anyway, David?
Looks like it is going to be somewhere around what? 1 + .5 + .33 + .25
+ .20 + .16 + .14 + ..
What is that convergence when Infinity is 10^603, David? Is it a
convergence of about 4. or 5. something?
So now let us look at the Euler multiplication zeta with its primes.
Looking at it term by term we realize
that the last prime to plug into Euler is the prime before 10^603 and
as we plug all those primes from 2,3,5
to the last prime, do we ever have a situation where the addition Zeta
ever equals the multiplication Zeta?
The answer is no.
So the difference between the Old Math and its staleness is that the
Old Math hobbles along without ever
defining Finite versus Infinity and thus has bogus claims of Series
and bogus conjectures such as the
Riemann Hypothesis.
Another area of mathematics, for those handicapped with algebra is the
geometry area where this series nonsense
crops up due to imprecise finite versus infinity. In Geometry with the
surface area and volume of the pseudosphere
where the ill-defined-infinity is claimed, then the pseudosphere
although it extends its arms to infinity has
this finite volume and finite surface area. But now apply precision
definition of infinity as 10^603 and we instantly
recognize that the pseudosphere volume and surface area must be finite
and that the formulas for volume and area of
pseudosphere must incorporate 10^603.
Now the Old Math community thinks the Euler Zeta is equal to the
Riemann Zeta, but the Old Math never defined
infinity and so they were never asked to show where the terms of the
Euler Zeta ever equaled the terms in the
Riemann Zeta. In fact, when going along term by term in the two zetas,
they are never equal. So here we have a famous conjecture that was
built from notions-of-infinity and not well defined infinity, and no
wonder never a proof,
because the conjecture is utterly false at the starting block.
Archimedes Plutonium
In New and True Math the harmonic series 1+ 1/2 + 1/3 + . . does not
diverge but converges
since infinity is the border of 10^603 and thus the last term of the
series if 1/10^603
So what is the sum of that Series? I just made a silly guess of about
4 to 5, but it is probably
a lot higher than 5.
Now this also raises the interesting question of the smallest
divergent Series and that would
have to be 1 + 1 + . . all the way out to 10^603 terms and the sum is
10^603 and that is the
smallest divergent Series in New True Math. And that is the way it
really should be, should it
not that the endless adding of 1 should reach infinity, and that the
Old Math harmonic Series
of 1 + 1/2 + 1/3+ . . should have alerted a smart person that much of
mathematics was misguided
and had a deep internal flaw. That flaw being an imprecise definition
of Finite versus Infinity.
Now tonight I want to extend the first breakdown in Circumferencing
the Perimeter to the 10^5
B matrix:
215.50 for 10^2 B matrix
2152.050 for 10^3 B matrix
21521.5000 for 10^4 B matrix
215215.21500 for 10^5 B matrix
Circumferencing the
Perimeter:
for B matrix 10^5 where pi is
3.14159
side of square = 215215.21500
perimeter = 4xside = 860860.86000
860860.86000/3.14159 = 274020.75382 = diameter
start pi twiddling 1/(B^2)twiddle 860860.86000/274020.75382 =
3.141590000024
pi twiddle = 3.1415900001
3.1415900001 x 274020.75382 = 860860.86002 mismatch
3.1415900000 x 274020.75382 = 860860.85999 mismatch
Caveat: the above are breakdowns but I have not proven they are the
very first breakdowns in
each of those B matrices.
Now I can show in my next post how the B matrix at 10^603 is going to
cause a breakdown
in a ** Critical Perimeter** which is a perimeter that is smaller than
the B number itself
and why the border between Finite and Infinity is at 10^603.
David Bernier
The Euler product formula says that two things are equal, A = B,
where A is a sum over all positive integers 'n',
and B is an infinite product over all primes 'p' :
I think I prefer the "Another proof" at Wikipedia:
<
http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function
> .
The Euler product formula is proved, for Re(s) > 1.
The Riemann Hypothesis says something about where the zeros
of the Riemann zeta function can be found. The Riemann zeta
function extends the Euler zeta function.
> If those two encodings of Euler's primes in multiplication and
> Riemann's zeta in addition of terms are
> not equal then there is no Riemann Hypothesis and the entire
> conjecture has no underpinning.
But the Euler product is accepted as true for Re(s) > 1.
> So with Mathematics ending at 10^603 there is no Riemann Hypothesis
> because the Euler primes
> and the zeta function are not equal to each other.
"natural numbers end at 10^603" isn't compatible with
standard math. For me to understand you, it might be
possible if you gave your axioms, just like there are axioms
in Euclidean geometry. I would need a list of unquestioned
assumptions ...
David Bernier
> With mathematics ending at 10^603, Goldbach conjecture is proven true
> with a computer running through
> all even numbers from 4 to 10^603. Kepler Packing is proven true with
> a modification of the last rows of
> spheres in a oblong packing and the bulk packing of hexagonal closed
> packing. Poincare Conjecture is proven
> to be false because there are tiny holes everywhere in geometry of
> smaller than 10^-603.
>
> FLT is probably proven true with that 10^603 border where a computer
> can easily run through all the cases for
> exponent 3 and 4 and by numerical analysis prove that no pythagorean
> triple is possible for higher exponents. However, the Wiles's sham
> proof is sham because a ill-defined infinity does yield boatloads of
> solutions.
>
[...]
David R Tribble
> So with Mathematics ending at 10^603 there is no Riemann Hypothesis
> because the Euler primes
> and the zeta function are not equal to each other.
Since all primes are finite by definition, and if the largest prime
is less than 10^603, this would present a problem with Euclid's
Proof of the infinitude of primes.
Specifically, taking the product of all the primes less than 10^603
and adding 1 would produce an infinite integer, and thus not a prime.
David R Tribble
> FLT is probably proven true with that 10^603 border where a computer
> can easily run through all the cases for
> exponent 3 and 4 and by numerical analysis prove that no pythagorean
> triple is possible for higher exponents.
Easily? Hardly.
Assuming that a computer could examine 1 trillion (10^12)
cases per second, it would take 10^603 / 10^12 seconds to
check for all possible cases less than 10^603.
That's 10^591 seconds, or about 3.168 x 10^583 years.
Jesse F. Hughes
So what?
With the recent advances in computing power, I'll bet we can *double*
the number of cases per second within a year or two! That'll crack that
nut!
--
Jesse F. Hughes
"Our enemies are innovative and resourceful, and so are we. They never
stop thinking about new ways to harm our country and our people, and
neither do we."-- George W. Bush
spudnik
could be a BIG problem.
> neither do we."-- George W. BusH
--GMMXI, number of entities using my account w/o auth.
htp://wlym.com
Archimedes Plutonium
To be infinite in physics means the end of physics such as "no more
strong nuclear force"
or a perfect vaccuum or reaching absolute zero Kelvin or "no more
Coulomb force."
For mathematics it is the e picture, that mathematics as a subject
ends with 10^603,
not because we can continue searching for primes but because we lose
precision in what we are doing
since we lose Aristotelian Logic and is doing Duality Logic. We no
longer can be sure that
9999....9997 is prime or composite.
So what does a 22nd century mathematician ask about how many primes
there are? He/she can do Euclid's
proof and fetch a set that has a cardinality of 10^603 of only primes.
But does it matter? Because more
than half of that set was numbers fetched from "across the border" as
untrustworthy of actually being prime.
What you fail to understand in this thread, Tribble, is that infinity
is not a quantity, but like physics, infinity is a border or boundary.
So in physics we merely ask "what if there is no strong nuclear force"
then we have infinity. In mathematics when we ask, are the primes
infinite, is asking do they continue in a pattern up to and including
10^603 and that some are across that border.
I do not expect you to understand any of this, after living a life
under a "deluded math with never defining what Finite even means."
Archimedes Plutonium
Enrico easily set up a computer to check B matrices for 10^308. And
FLT for exp3 is
far easier.
Let me show you how.
Set the computer program to fetch the set of this:
{1, 8, 27, 64, . .}
That set is every natural from 1 to cube root of 10^603. Now the
computer has an easy time
of fetching that set. Perhaps taking 1 day to do it. Now you program
into the computer to
look for where A +B equals another member of that set. In total,
perhaps 2 days of computer time.
Do the same for exp4.
Figure out where exp conflicts with 10^603 so that the members of the
set are too sparse to be able
to have a counterexample, for unlike Fermat and Wiles they had to go
far beyond exponent 603 whereas
I have to only go to say exponent 20 because of the sparsity of
numbers in that set.
So all told a computer proof of FLT would take perhaps a week at most.
And then, someone can rig the computer to find where the first
counterexamples are for Wiles's sham proof.
And also, let me note that the computer proof of Goldbach may take
longer than the computer proof of
FLT because of the larger set to worry about in Goldbach since you
have all those primes to contend with.
Archimedes Plutonium
(snipped)
>
> > Thanks David, for what you said above, correct me if wrong. Is that
> > the Riemann Hypothesis
> > is crucially dependent on the idea that Euler's Zeta encoding is equal
> > to the Riemann Zeta function.
>
> The Euler product formula says that two things are equal, A = B,
> where A is a sum over all positive integers 'n',
> and B is an infinite product over all primes 'p' :
Make that more clear David, term by term. So the first term in A is 1
with exponent 2, and the first term in B is
1/(1/4) or 4. So, term by term David or staggered terms, when is A
ever equal to B. Or did you just learn the Zeta function by
memorization that everyone says they are equal and so you never
questioned it term by term.
The question I was trying to pry out of you is whether you ever saw a
term at which A ever really did equal a sequence of terms on B.
So apparently, the equality in Old Math on the Zeta function is the
magical belief that "at Infinity" they are equal, but
in reality, they are never equal. Especially since mathematics ends at
10^603, there is never a sequence of terms in which A equals B.
David, you are supposed to have thought about the Zeta function and
the Riemann Hypothesis for years and years,
and yet you seem to not even know the first thing about this Riemann
Hypothesis at all. You believe they are equal
but can not even point to where the terms are equal.
>
> I think I prefer the "Another proof" at Wikipedia:
> <http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_t...
> > .
>
> The Euler product formula is proved, for Re(s) > 1.
> The Riemann Hypothesis says something about where the zeros
> of the Riemann zeta function can be found. The Riemann zeta
> function extends the Euler zeta function.
>
Again, show me at least what terms of A ever equal terms in B. Not
even the first term of
A and first term of B equal each other. The only place, then, that
they equal is some mythical
fairytale land called infinity that they ever equal, but you and every
mathematician since Riemann
have ever defined "infinity."
> > If those two encodings of Euler's primes in multiplication and
> > Riemann's zeta in addition of terms are
> > not equal then there is no Riemann Hypothesis and the entire
> > conjecture has no underpinning.
>
> But the Euler product is accepted as true for Re(s) > 1.
>
Math and science must be better than "accepted". The Euler Product is
a result of Old Math
that never sat down and precision defined "Finite versus Infinity".
David, take a second out of your time and look at this piece of Old
Math:
Harmonic Series = 1 + 1/2 + 1/3 + 1/4 + ..
You and Old Math when asked what that equals, will say it equals
something you have never defined with
mathematical precision. You will say it is equal to infinity or is
divergent.
Now, if instead you say that Finite transitions into Infinity and such
must have a border call it B. You can admit
David that you have always believed that Finite passes into the
Infinite, and it is okay for you to admit that
since every math professor would admit that.
Now, reexamine the Harmonic Series for if infinity is B then the last
term is 1/B which means the Harmonic
Series never reaches infinity but is always a finite number.
Because current day math that you so deeply believe in David has the
Harmonic Series as equal to infinity
is a barometer or thermometer of how diseased and cancered current
math is. Because to a logical person,
the *** smallest series*** that diverges should be 1 + 1 + 1 + . . for
it equals B itself and thus is infinite.
Because, David, you accept current math that believes 1 + 1/2 + 1/3
+ . . diverges and equal infinity is an
indication of how way out in left field you understand true
mathematics, because, you like most every
current math professor refuses to admit that you have a lousy
definition of both finite and infinite and continues
to make these silly mistakes.
> > So with Mathematics ending at 10^603 there is no Riemann Hypothesis
> > because the Euler primes
> > and the zeta function are not equal to each other.
>
> "natural numbers end at 10^603" isn't compatible with
> standard math. For me to understand you, it might be
> possible if you gave your axioms, just like there are axioms
> in Euclidean geometry. I would need a list of unquestioned
> assumptions ...
>
> David Bernier
What you call "standard math" is math done by people who are too lazy
to look at the fact that they
have no precision definition of "finite versus infinite". And as long
as you never look at that definition,
for you do not even have a definition of a finite number, and as long
as you keep that up, you are not
doing mathematics but your own blithering opinion. You are not doing
science, nor math, but is doing
some David artwork.
Standard math to you seems to mean what you expect a math professor to
agree with you, but never what is
true mathematics.
Define "finite" as what you think Standard Math would like you to say
David and then define "infinite". You can't
and you probably will go running away and hide, and hide behind the
apron of some math professor.
The thing that amazes me about your posts David, is that you can study
and learn mathematics and get good
grades and can understand when mathematicians define circle or
rectangle or define matrix or determinant. But when
I ask you to define Finite number versus Infinite number, you run
squirrely all over the place and end up with
"standard math". Why not be honest and simply say that standard-math
fails to define with precision the concept of
Finite versus Infinity. Be honest for a change. I know I have been
hard on you in this post, but sometimes a slap in the face is the only
way of drawing a person in a coma out of his comatose. Admit that you
never defined with precision
finite versus infinite and that your understanding of math is poor
because of that lack of definition.
adamk
> > AP wants to twiddle enrico's banana.
> >
> > Ap: you are a legend in your own mind...and
> > nowhere else.
> >
> > Loser.
>
> Adam, I need more substantive information of these
> surveys of math
> professors at
> Penn State than that of "bananas and losers".
Good for you!. Why the hell should I care? Do your
own surveys. You insult me, everyone, including me
and then you expect me to do work for you --for which
you will surely take full credit?
>
> FIRST SURVEY:
> 1) Define what is meant by a Finite versus Infinity
> and how to
> recognize one from the other, such as what is a
> Finite Number versus
> an Infinite Number?
Wow, you still don't get it. Try imagining how
stupid you are, and then you will get an idea of the
meaning of infinity.
> If you look at any encycopedia of mathematics and
> even Wikipedia
Even Wikipedia!
Do tell, liar: how many encyclopedias have you looked
at?
fail
> to distinguish between a Finite number and an
> Infinite number, yet
> these two concepts are used every time a mathematics
> proof involving
> numbers is undertaken.
False. At best this is used _implicitly_, most
of the time, idiot. Only if you are working on foundations will you make explicit use.
> SECOND SURVEY:
> Is a true false questionnaire on the topic of Finite
> versus Infinity
> to be given to all mathematics
professors and
> editors of
> mathematics.
> 1) Does Finite run or go or transition or move from
> finiteness into
> the infinite?
> 2) Is there a border or borderline between Finiteness
> and Infinity?
> 3) Is there a largest Finite number and thus, a
> smallest infinite
> number?
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
>
> Now tonight I want to extend the first breakdown in Circumferencing
> the Perimeter to the 10^5
> B matrix:
>
> 215.50 for 10^2 B matrix
> 2152.050 for 10^3 B matrix
> 21521.5000 for 10^4 B matrix
> 215215.21500 for 10^5 B matrix
>
> Circumferencing the Perimeter: for B matrix 10^5 where pi is
> 3.14159
> side of square = 215215.21500
> perimeter = 4xside = 860860.86000
> 860860.86000/3.14159 = 274020.75382 = diameter
> start pi twiddling 1/(B^2)twiddle 860860.86000/274020.75382 =
> 3.141590000024
> pi twiddle = 3.1415900001
> 3.1415900001 x 274020.75382 = 860860.86002 mismatch
> 3.1415900000 x 274020.75382 = 860860.85999 mismatch
>
Now I spent some time tonight trying to duplicate a pretend-pi
breakdown
such as this post in December of 2010:
215.50 as first breakdown in 100 B matrix
2152.550 as near to the first breakdown in 1000 B matrix
pretend for 10^4 B matrix
Pretending 10000 B matrix with pi as
pretend pi = 3.1400
side of square = 2152.5500
perimeter = 4xside = 8610.2000
8610.2000/3.1400 = 2742.1019 = diameter
start pi twiddling 8610.2000/2742.1019 = 3.140000 012
pi twiddle = 3.14000002
3.14000002 x 2742.1019 = 8610.2002 mismatch
3.14000001 x 2742.1019 = 8610.1999 mismatch
I spent some time tonight trying to achieve a breakdown of the 10^5 B
matrix breakdown:
> 215215.21500 for 10^5 B matrix
By using a pretend-pi of 3.141000 in the 10^6 B matrix and the number
215215.215000. Why am I doing this?
Because, by pretending pi has three zero digits in a row as 3.141000
rather than 3.141592 and achieving a breakdown of 215215.215000 would
be a critical diameter and thus a critical perimeter as smaller than
the B number 10^6.
Seems I could not duplicate that earlier pretend-pi where I only had
two zeroes in a row as shown above.
The important news of this post, is that the zeroes in a row in pi
digits at 10^603 causes a breakdown in matching
of the Circumference to the Perimeter and is smaller of a perimeter
than 10^603. That implies that not all finite
perimeters are matched by a finite circumference and thus, we must
admit the end of finiteness and the border to infinity.
Now maybe I have to fiddle around with those digits of 215215.21500 of
the 10^5 B matrix with a pretend-pi to achieve that critical perimeter
breakdown. But since it worked before, it is not pressing
demonstration.
And hopefully someone will program a computer to see if the 10^308 has
a critical perimeter breakdown or whether the first such incident
happens only in the 10^603 B matrix.
Archimedes Plutonium
not precision define finite versus infinite
would harbor a counterexample in only exponent 3. Here is the
reasoning:
These are solution sets for exponents in FLT and to gain a
counterexample one must show that a
A,B,C in a solution set has A+B = C
exp3
{1, 8, 27, 64, 125, 216, 343, 512, 729, . .}
exp4
{1, 16, 81, 256, 625,. .}
exp5
{1, 32, 243, 1024,. . }
Now if you look at the solution sets of exp4 and greater that a
particular A picked out has its successor
number more than twice as large. That means it is impossible to have a
A+B = C. That trick thus eliminates all
exponents of 4 or larger. Leaving only exponent 3 as a solution set
that can harbor a counterexample to FLT.
That means to prove FLT by computers out to 10^603 is easily a day
task by a supercomputer going through just the exp3 solution set.
Now we already know that exp3 in FLT misses a counterexample by a mere
1, witness:
64^3 + 94^3 = 103^3 add 1
Do we have a triple that overshots by 1? That would not comprise a
proof that a counterexample exists but would
be supporting evidence of the likelihood a counterexample exists.
Maybe we should wait for the supercomputer to run through the solution
set of exp3 out to 10^603 which would have
numbers with 201 digits in them. There maybe a counterexample existing
in that exp3 before we reach 10^603. If not
we can examine all the near misses for clues as to a pattern that
guides us in locating the first counterexample beyond 10^603.
Now I do not recall if Wiles's proof dealt exclusively with exp3,
having shucked the other exponents because of the
impossibility as stated above. Anyone recall?
Archimedes Plutonium
I looked it up in a math history book and apparently no-one before saw
that simple trick because
Euler showed it for n=3, Fermat for n=4, Legendre and Dirichlet for
n=5, Lame for n=7, Kummer for
all primes less than 100 with exceptions. So I wonder what cases that
Wiles alleged proof covered? Was it all cases?
Apparently it must have been because it is alleged that Euler did it
for n=3.
And apparently none of these famous mathematicians ever wrote out what
the solution set was, as I did above,
for if you did, you could not escape noticing that for n greater than
3 , any particular B is over twice as large
as its predecessor, thus, by logic no A+B could equal a C.
So I guess that the Wiles alleged covered the entire gamut of
exponents.
Now I wonder what Euler did to convince himself that exp 3 had no
solutions? Here I am thinking that he made some
sort of mistake.
Archimedes Plutonium
Sorry, my last two posts were mistakes, for I did not go out far
enough to see a pattern
where the successor becomes smaller relative to the predecessor.
exp4
{1, 16, 81, 256, 625, 1296, 2401, 4096,. .}
Anyone know how Euler came up with no solutions for n=3?
Archimedes Plutonium
Alright, so I made a mistake, but you have to be willing to see if
some fruit lies
in that mistake. So I quickly come to a question. We know FLT wants no
solutions
for A+B = C but can we have the situation in a even exponent of where A
+A = C?
And another question comes to mind, is each solution set have a sort
of built in convergence
point where the closest approach to A+B= C occurs only once and from
there on out, a
divergence in A+B= C. For instance in the exp3, is the closest
convergence point at
64^3 + 94^3 = 103^3 add 1 and it never happens again in exp3?
So here again, we need the supercomputer to go through all exponents
where no number exceeds
10^603. We need to see patterns, which only raw computer power can
hand us reams of near misses.
This has never been done before for FLT is to examine a large list of
near misses. And before
only theory was applied. It is like a physicist who never sees data
from experiments and only
dreams up theories.
Archimedes Plutonium
And I had
no sleep tonight for this thing kept my awake and I found the
solution, the inner
tickings of the Fermat equation a^n + b^n = c^n. The inner tickings of
that entire equation
both when exp2 having solutions and when exp 3 and higher do have
solutions only they come
rarely. I have found a way of extracting the counterexamples in the
case of exp3, without blindly having computers find the first set of
counterexamples.
It is late for me already and so let me just outline the proof, why
FLT works for 2 and rarely for higher exponents. And also, I still
would need a computer to prove there are no
solutions for n = or greater than 3 until after 10^603.
So let me pan out the basics of the proof and the easy acquisition of
the first counterexamples.
First let me profusely thank this website, for without it none was
possible.
http://www.alpertron.com.ar/BIGCALC.HTM
Back in December,last month I posted about searching for a
counterexample in exp3 by starting with:
1^3 + 2^3 = 9 take cube root of 9
2.080 083 823 051 904 114 530 056 824 357 885 386 337 805 340 373 262
109 697 591 080 200 106 311 397 268 773 606 056 636 790 757 486 728
671 592 086 574 520 538 907 806 551 432 406 435 155 956 414 938 597
044 743 429 752 216 294 112 964 022 766 857 669 747 769 143 933 688
804 813 536 021 599 545 811 696 696 536 066 894 994 770 998 260 332
103 790 432 402 956 475 010 988 207 691 299 963 138 219 077 694 067
605 530 360 769 279 867 841 312 933 591 633 086 534 769 408 692 721
478 724 093 497 104 836 551 394 365 820 070 260 491 731 141 474 317
009 878 047 273 127 944 297 002 343 787 245 304 869 234 558 012 240
906 823 440 092 285 995 194 896 338 315 133 702 469 3
Then, then to go with
1^3 + 208^3 = 208^3
and if it does not work go down that string
for 2.080 = 8.998 912
for 2.080 083 823 051 = 8.999 999 999 988 264 370 867 656 846 408
001 651
What I am looking for is when the cube root, cubed is multiplied back
to have
a end string looking like this 9.00000000000..01 So far the cube root
of 9 is looking
like this 8.9999999.. but sometime in its long march, it does surface
as this
9.0000000..01, but never as this 9.0000000..000
Now I do not want to play this on 2^3 and the cube root of 9 because
obviously I need
the 1^3 to play this where I can utilize the 1^3 to retrieve that end
lonely digit of 1.
So I want a different triple in a^3 + b^3 = c^3 and I think the one I
need is
something like this 1^3 + 2^3 => 7.999999...
Now with infinity ill defined that number 7.9999... is what Old Math
thought was 8.0000
and 2^3 is 8.
So now I have things set up, ready to go. I take the computer and find
the cube root of
7.99999... to as far out as it will go with accuracy and no rounding
off. If it takes me out
to 700 digits,so much the better
Then I go along that digit string until I find a moment in which it
delivers to me
not 7.99999999.... or less but where it multiplies out as
8.000000....01
The instant it does that I have both sides of the FLT equation with
equality for I have
on the one side 1 + 8000...00 and on the other side I have 8.000...01,
and thus equality.
How far out I have to go before those 9s turn into 0s with a 1 at the
end is questionable.
For exp3, I suspect that somewhere beyond 603 digits is the first
counterexample. For exp4
I suspect it takes even longer.
Now if this really is the mechanism, or inner tickings of FLT, it
should also explain the ease of Pythagorean triples in exp2. So let me
explore that.
for exp2
solution set {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, . . }
And immediately we see 9+16=25 and 36+64 = 100
1^2 + 2^2 = 1.99999^2 take square root of 1.99999.... to as many
digits
as the computer allows. We are playing on the 99s digits.
Now we run through that square root and wait for the moment that it
delivers
a 4.000000...01 string. Suppose it is at the 10^40 place value. Then
we
have the integers of 1 + 40000..00 = 40000..01
The only thing different from exp2 and exp3 is that both have
counterexamples,
only that exp2 has boatloads and easy to find whereas exp3 may have 1
counterexample
before reaching 10^700.
The trick is that in Old Math where infinity is ill-defined and where
it is thought
that 1.99999.... is the same as 2.0000..... but where the New Math
knows better,that
they are not the same and that at a large number, the exponent
switches from 99s to that
of 0001 with 1 digit stuck lonely at the end.
Now there maybe a fatal flaw in my argument, but I am too tired now to
look. What I mean is that I did this searching many years ago looking
for something else where I noticed that taking the cube of a number
which I had fetched the cube root that the answer alternated back and
forth between z.9999999....999 and z.0000000....0001 So if my memory
of this experience is false, that I have a bad memory and that it
never fluctuated between 999s and 001, then my above pronounced proof
is flawed. But if my memory is correct that as we take a cube root and
then cube the string along its digits that the answer fluctuates
between 99999s and 00001,
then that is the ultimate proof of FLT for it explains the intricacies
of why it has no solutions, why it has solutions and when it has
solutions. It is a dance between 999s and 0000s is the explanation.
And in Old Math, 999s were equal to 000s at an ill-defined infinity.
In new true math 999s are distinct from 000s since the border of
infinity is 10^603.
spudnik
to see that it is the same as "the real number, 2,"
which is technically defined as 2.0000... ... hey;
didn't Saturday Night Live have a regular skit
with that oine guy, "Incredibly Bad Math?"
* carry the (natural) ones!
David R Tribble
>> FLT is probably proven true with that 10^603 border where a computer
>> can easily run through all the cases for
>> exponent 3 and 4 and by numerical analysis prove that no pythagorean
>> triple is possible for higher exponents.
>
David R Tribble wrote:
>> Easily? Hardly.
>> Assuming that a computer could examine 1 trillion (10^12)
>> cases per second, it would take 10^603 / 10^12 seconds to
>> check for all possible cases less than 10^603.
>> That's 10^591 seconds, or about 3.168 x 10^583 years.
>
Archimedes Plutonium wrote:
> Enrico easily set up a computer to check B matrices for 10^308. And
> FLT for exp3 is
> far easier.
>
> Let me show you how.
>
> Set the computer program to fetch the set of this:
> {1, 8, 27, 64, . .}
>
> That set is every natural from 1 to cube root of 10^603. Now the
> computer has an easy time
> of fetching that set. Perhaps taking 1 day to do it.
Your calculator appears to be broken.
The set contains cubes for every natural from 1 to the
cube root of 10^603, which is 10^201. Assuming that the
computer can generate a trillion (10^12) cubes per second,
this will take 1.157 x 10^184 days, which is about
3.169 x 10^181 years.
Just to generate the set of cubes.
Of course, you couldn't actually store the entire set at
once, since the Earth contains only about 10^50 atoms.
> Now you program into the computer to
> look for where A +B equals another member of that set. In total,
> perhaps 2 days of computer time.
Searching with each pair of numbers within the set is an
O(n^2) operation, which will take (10^201)^2, or 10^402,
operations. At a trillion pairs per second, that's 10^390
seconds, or about 3.168 x 10^382 years.
Assuming we only have to search half of the pairs to find
a match, that would likewise cut the search time in half.
> So all told a computer proof of FLT would take perhaps a week at most.
> And then, someone can rig the computer to find where the first
> counterexamples are for Wiles's sham proof.
We all wait anxiously for your results.
But not for too many centuries.
Archimedes Plutonium
Don't bother me and this thread again over computer time. If you
applied any of your above
to Enrico going through the spread sheet of B matrix, Tribble would
calculate that Enrico
would not have returned for two centuries with any answers. Whereas in
reality, Enrico
discovered the B matrices of Squaring the Circle of its first
breakdowns out to 10^9 in a
matter of two days.
So you are making spurious and erroneous assumptions such as the
computer taking lunch breaks.
So don't bother me with you pessimistic nonsense of assumptions.
Go and straighten out those redneck bigots of editors of Glenfart and
DMack on the Wikipedia Usenent celebrities.
notworth saving
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
> Thank goodness for my mistakes, because it only spurs me on forward.
> And I had
> no sleep tonight for this thing kept my awake and I found the
> solution, the inner
> tickings of the Fermat equation a^n + b^n = c^n. The inner tickings of
> that entire equation
> both when exp2 having solutions and when exp 3 and higher do have
> solutions only they come
> rarely. I have found a way of extracting the counterexamples in the
> case of exp3, without blindly having computers find the first set of
> counterexamples.
>
> It is late for me already and so let me just outline the proof, why
> FLT works for 2 and rarely for higher exponents. And also, I still
> would need a computer to prove there are no
> solutions for n = or greater than 3 until after 10^603.
>
> So let me pan out the basics of the proof and the easy acquisition of
> the first counterexamples.
>
> First let me profusely thank this website, for without it none was
Let me start with exponent 2, for which the other exponents apply as
well. And although
we already know exponent 2 has solutions such as 3,4,5, the solution
to exponent 2 that I am
going to retrieve will look totally unfamilar to what everyone else
has been used to in a
Pythagorean triple.
The basic idea is a play on numbers of say 10.000000..... which Old
Math never defined infinity and thought that
9.999999...... was equal but in New Math where the border between
finite and infinity is 10^603, that the numbers
10.0000.... and 99.999999..... are two different numbers. So let me
show you how to retrieve a Pythagorean triple
that looks like this in exponent 2, and higher exponents follow the
same pattern of retrieval:
1^2 + 10.000...^2 = 9.99999....^2
Now everyone in Old Math will look at that in amazement and wonder how
in the world do I retrieve a P-triple out of that mess. And the answer
is simple that as I take the square root of 100 and then of 99.9999
out to some decimal points of a long way out and then I square that
square root. I reach a point in the squaring of the square root
where the digits retrieved are not that of 99.99999.... but rather
where the digits retrieved are that of 100.0000..01 with a lonely one
digit sticking at the end. It may take 600 digits to find where this
happens in
exponent 2 and it may take thousands of digits to find where it
happens in exponent 3 but every exponent in Fermat's Last Theorem has
a solution set, not just exponent 2, but every exponent only it is a
huge number of digits.
Now here I worked with 1000, and 999.9999.... and I see them diverging
in square root at about where I spaced them
differently. So what I aim to do is to go out beyond that divergence
and square that arrangement until I receive
a number that looks like this 1000.0000000..01 and thus my job is done
because the 1^2 picks up that lonely 1 digit
sticking on the end.
sqrt1000
31.622 776 601 683 793 319 988 935 444 327 185 337 195 551 393 252 168
268 575 048 527 925 944 386 392 382 213
sqrt 999.99999 to sixty digits
31.622 776 601 683 793 319 988 935 444 327 185 337 195 551 393 252 168
268 575 048 512 114 556 085 550 485 553 448 013 361 629 410 283
275 697 144 902 316 417 197 964 292
Archimedes Plutonium
Here I took the cube root of 999.99999 to thirty digits:
9.999 999 999 999 999 999 999 999 999 999 996 666 666 666 666 666 666
666 666 666 666 665 555 555 555 555 555 555 555 555 555 555 554 938
271 604 938 271 604 938 271 604 938 271 193 415 637 860 082 304 526
748 971 193 415 336 076 817 558 299 039 780 521 262 002 743 249 504
648 681 603 414 113 702 179 545 800 754 966 214 499 822 181 578 011
481 989 533 957 391 319 073 989 398 635 031 922 640 518 756 205 612
037 209 524 042 493 271 438 723 403 992 373 774 532 095 022 986 768
260 467 401 867 612 634 188 792 219 504 942 083 741 896 836 850 051
So a computer can take it out much further, and what I seek is this
happenstance:
1^3 + 10^3 = 9.9999...^3
Now it is difficult to see how that is an equality. But what is
required is that as I go along that string of digits of the cube root
of 999.99999.... is a moment in which that cubing of the cube root
produces a number that looks like
this 1000.000000..01
The likely spot such a number is produced is where the cube root
digits have a row of zeroes with a 1 digit ending such as 01 or 001 or
0001.
Now the reason that every exponent of FLT has solutions is because
every exponent can go out a huge distance and retrieve these numbers
like 100000000....0001 which the 1^3 will equilibrate with that lonely
1 digit sticking at the
end. No matter what exponent, a number that looks like 100000.....0001
exists, only it would have a huge number of digits.
adamk
Not so, you fucking baboon. The series grows without bound, in a precise way, you worthless fucking idiot. Is that too hard for you to understand?
> is a barometer or thermometer of how diseased and
> cancered current
> math is.
True. Hundreds of real-world and otherwise, applications
without any major problem. Buildings stand, bridges stand, electornics work, etc. And no major contradiction is found within mathematics.
But to a fucking idiot like you, that is a failure.
Because to a logical person,
i.e., not you.
> the *** smallest series*** that diverges
Define your terms, fuckwit. Do you get, idiot, that if you use your own terminology others have never seen, that no one can understand what you say?
should be 1
> + 1 + 1 + . . for
> it equals B itself and thus is infinite.
>
> Because, David, you accept current math that believes
> 1 + 1/2 + 1/3
> + . . diverges and equal infinity
False. Get your facts straight, for once.
is an
> indication of how way out in left field you
> understand true
> mathematics,
another undefined term, idiot.
because, you like most every
> current math professor refuses to admit that you have
> a lousy
> definition of both finite and infinite and continues
> to make these silly mistakes.
You have not pointed out yet a _single_ mistake, other
that someone disagreeing with you.
>
>
> > > So with Mathematics ending at 10^603
Which you have neither defined nor proved.
there is no
> Riemann Hypothesis
> > > because the Euler primes
> > > and the zeta function are not equal to each
> other.
> >
> > "natural numbers end at 10^603" isn't compatible
> with
> > standard math. For me to understand you, it might
> be
> > possible if you gave your axioms, just like there
> are axioms
> > in Euclidean geometry. I would need a list of
> unquestioned
> > assumptions ...
> >
> > David Bernier
>
> What you call "standard math" is math done by people
> who are too lazy
> to look at the fact that they
> have no precision definition of "finite versus
> infinite".
Meaning _you_ do not understand the difference.
And as long
> as you never look at that definition,
> for you do not even have a definition of a finite
> number,
More lies.
and as long
> as you keep that up, you are not
> doing mathematics but your own blithering opinion.
> You are not doing
> science, nor math, but is doing
> some David artwork.
>
> Standard math to you seems to mean what you expect a
> math professor to
> agree with you, but never what is
> true mathematics.
>
> Define "finite" as what you think Standard Math would
> like you to say
Get of your fat ass and pick a book, loser.
> David and then define "infinite". You can't
> and you probably will go running away and hide, and
> hide behind the
> apron of some math professor.
Yes. Very likely. And the bubble of self-delusion
you have created to hide your life of constant failures.
>
> The thing that amazes me about your posts David, is
> that you can study
> and learn mathematics and get good
> grades and can understand when mathematicians define
> circle or
> rectangle or define matrix or determinant. But when
> I ask you to define Finite number versus Infinite
> number, you run
> squirrely all over the place and end up with
> "standard math". Why not be honest and simply say
> that standard-math
> fails to define with precision the concept of
> Finite versus Infinity.
Liar.
> Be honest for a change.
Ditto. And stop being such a fucking idiot too, while
you're at it.
I
> know I have been
> hard on you in this post,
You're hard on everyone but yourself. You give yourself plenty of leeway to lie and cheat. And you try to exploit people by having them do heavy work for you without
any pay.
but sometimes a slap in the
> face is the only
> way of drawing a person in a coma out of his
> comatose.
give yourself one, then. Pound yourself into a rock
a few times. I bet you'll make more sense afterwards.
Admit that you
> never defined with precision
> finite versus infinite and that your understanding of
> math is poor
> because of that lack of definition.
Get of your fat ass, pick up a math book and read it,
you fucktwat.
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
> On Jan 1, 10:05 am, David Bernier <david...@videotron.ca> wrote:
(snipped)
>
> > > because the Euler primes
> > > and the zeta function are not equal to each other.
>
> > "natural numbers end at 10^603" isn't compatible with
> > standard math. For me to understand you, it might be
> > possible if you gave your axioms, just like there are axioms
> > in Euclidean geometry. I would need a list of unquestioned
> > assumptions ...
>
> > David Bernier
Let me say some more as to why I had to read the riot act against
David.
Mathematics, for the most part is precision, and not about some axiom
system that is given to you David. Because if you cannot recognize a
lousy definition
in your own standard-math axioms, it makes no difference if I offered
you a alternate
set of axioms.
Why is it that you, David cannot recognize when you have a lousy
definition or no definition
at all for Finite versus Infinity. And it is not a matter that
Standard Math has a good definition
of Finite versus Infinite but simply has none at all.
Now maybe the fault with modern math is that no-one in teaching of
math ever spends time on
scrutiny and examination of mathematics definitions. There has never
been a book written about subject
and so it is probably never even mentioned in a 6 year math degree
program. Having precision definitions in
mathematics is critical. And yet on the Internet of sci.math and in
particular here with David Bernier
who cannot recognize that he has no definition of Finite number versus
infinite number but only a hazy
shadowy notion. So it is not a matter, David of me offering you a
alternate set of axioms, for you cannot even
recognize a lousy definition from a good definition in your so called
Standard Axioms.
Now maybe a remedy for this horrible shortcoming of mathematics
graduates is to write a book specifically on
what constitutes a precision definition in mathematics and examine
whether current math is loaded with lousy
definitions. Gauss, we can all admit was a superlative mathematician
and yet he said that whenever there is a
breakdown in math it usually occurs because person A and person B have
a mixed understanding of a definition. Yet
the problem is that no-one in math today heeds the advice of Gauss,
except AP on Finite versus Infinite.
Now getting on with math.
I was trying to pry out of David, whether he can see were a specific
term of the Euler zeta ever actually equals a
sequence of terms in the Riemann zeta? We can compare term for term or
staggered terms, and ask, is it ever the case that the Euler zeta
equals the Riemann zeta? As far as I can see, it is never the case
that a sequence of the terms sets those two zetas equal. So if never a
term by term equality in comparison, then how in the world does the
two zetas ever really be equal to one another? And here is a possible
connection with the solving of the FLT. For we see that to find those
counterexamples in each exponent of a^n + b^n = c^n we have to coax
along the idea that when you ill-define infinity you mistakenly think
that 1.9999..... is the same or equal to 2.00000..... But when you
precision
define infinity, those two numbers are always distinctly different.
So, if you follow my drift, can you sense that when Euler thought that
he had proven the multiplication zeta was the same as the addition
Riemann zeta, that they really were not equal because the Euler Zeta
ends up being one of those
1.999999..... and the Riemann zeta ends up being 2.00000.... and
distinctly different.
So here maybe a connection to Riemann Hypothesis and Fermat's Last
Theorem and that both were never provable so long as both never
precision defined finite versus infinite. And once a precision
definition is applied, then the Euler zeta does not equal the Riemann
Zeta, and so all that hallyballoo about primes in perfect order was
just noise in the
wind.
P.S. But really, a large proportion of being able to do mathematics
and good math, depends on a person able to see
when they have a lousy definition. So why never anyone spending time
on writing a textbook on what it takes to have
a good definition in mathematics? Probably because no-one until 1990s
had a clear enough logical mind in mathematics to start such a
project. By 1993 no-one in the history of mathematics had a valid
Euclid Infinitude of Primes proof
which easily gives a Infinitude of Twin Primes proof. This tells us
how primitive was the logical abilities of the
mathematics community up till 1993. Could not even do a valid Euclid
IP proof Indirect and could not even recognize
they needed a precision definition of Finite versus Infinite.
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
>
> So a computer can take it out much further, and what I seek is this
> happenstance:
>
> 1^3 + 10^3 = 9.9999...^3
>
> Now it is difficult to see how that is an equality. But what is
> required is that as I go along that string of digits of the cube root
> of 999.99999.... is a moment in which that cubing of the cube root
> produces a number that looks like
> this 1000.000000..01
>
> The likely spot such a number is produced is where the cube root
> digits have a row of zeroes with a 1 digit ending such as 01 or 001 or
> 0001.
>
> Now the reason that every exponent of FLT has solutions is because
> every exponent can go out a huge distance and retrieve these numbers
> like 100000000....0001 which the 1^3 will equilibrate with that lonely
> 1 digit sticking at the
> end. No matter what exponent, a number that looks like 100000.....0001
> exists, only it would have a huge number of digits.
>
So let me give you a flavor of this procedure that extracts
counterexamples to
FLT in all exponents.
For exp3 and using the equation 1^3 + 10^3 = 9.999....^3
where our task is to cube root 999.999999.... then go along that
string of
digits and craft or twiddle with the digits until you produce an
arrangement that
when cubed delivers a 1000.0000000..01 string
Now witness here I have cube root of 999.9999 out a little ways
31.622 776 6
And now what I do is make a twiddle on that ending 6 digit and boost
it to be 7
and then I cube this number 31.622 776 7
and get this number
1 000.000 006 218 062 89
Now I have a far ways of going before I find a number when cubed
delivers 1000.0000..01
But I just wanted to show you how the program of fetching
counterexamples in all exponents of
FLT works.
And the counterexamples are not going to be small and lovely like
6,8,10 is in exp2 of FLT
but are going to be likely having thousands of digits.
Now the nagging question that some will have as to how is this
possible when Euler proved no counterexamples in n=3, Fermat for n=4,
Legendre and Dirichlet for
n=5, Lame for n=7, Kummer for
all primes less than 100 with exceptions.
So how is that possible that AP finds counterexamples to all exponents
of FLT and yet those
illustrious mathematicians of the past gave a so called proof against
such a situation?
The answer is that Euler to Fermat to Legendre and Dirichlet and Lame
and Kummer never defined what mathematics means for Finite versus
Infinity. The fatal flaw in all those proofs is a imprecise infinity.
And to Euler to Kummer all thought that 2.0000 is the same as 1.999...
There alleged proofs all fail because they have assumed 2.00.... is
the same as
1.9999....
Archimedes Plutonium
Sorry, I somehow comingled squaring of 31 with cubing and 1000 instead
of 100
But the demonstration of the procedure is valid even if I made an
error of
square and cube.
The procedure is valid in extracting a counterexample to any given
exponent of
FLT
> Archimedes Plutoniumhttp://www.iw.net/~a_plutonium/
Archimedes Plutonium
FLT for
every exponent but let me focus on exp3.
The trick is that we play on 1000.0000.... is not the same as
999.9999.... since
infinity begins at 10^603.
So every FLT exponent has a solution triple and the fastest and
easiest way of accessing
what that triple solution set is for a exponent such as 3, is to use
the general equation of this:
1^3 + 10^3 = 9.9999..^3 with a twiddle
Now with a twiddle means that we fetch the cube root of 999.99999..
out to some designated
place value and then we search and pick and hunt through that digit
arrangement where we actively make a twiddle change of one digit or
several digits until we are able to have a digit string that when
cubed delivers this end product 1000.00000..01
That product may have hundreds or thousands of zeroes and the number
which begot that
product has hundreds or thousands of digits. And then we eliminate the
decimal point and make the rational number into an integer and we do
the same for the 10^3 compensating it
to be a integer on par with the twiddle product. That leaves the end
stage of extracting the
counterexample looking like this:
1 + 10000..00 = 10000..01
There maybe some hardships such as compensating the 10^3 to have on-
par to the integerized twiddle product. If there are unequilibrated
number of digits in this stage,
we have to drop that twiddle product and go out much further.
Now I could have started or used something like this 1^3 + 2^3 =
1.999....^3 looking for
a final product like this 200000..01 But I chose the 10 number since
it is easy and can be applied in all exponents.
So what I am offering here is a technique in which to manufacture a
counterexample to every
exponent in Fermat's Last Theorem.
And the reason FLT is false for every exponent is because we can
manufacture these numbers like 1000..01, unless you proscribe that
Infinity starts at 10^603, and in that case, FLT is a true theorem.
But the reason that counterexamples exist
in every exponent is because numbers that look like this 100000..01 or
like this 20000..01
and are composite and roots are infinite. We do not recognize that
with numbers like 101 or
1001 or 500001 but when you have unlimited supplies of numbers that
end with a lonely
1 digit, that there is no exponent of FLT that can escape being solved
by a P-triple, provided
you are willing to go out horrendously far enough with digits.
Now this number 31.622 776 6 is the first few digits of the square
root of 999.999...
And if I did a twiddle on that last 6 digit to get this 31.622 776 7
And then I squared it getting this:
1 000.000 006 218 062 89
Now it is far from being 1000.000..01 but it is a progress in the
right direction.
Remember we have hundreds and thousands of digits to play with so that
if I can
produce with ease, above then it is just a matter of time to fetch out
one of these huge
numbers so that we have the first counterexample of FLT for exp3 that
looks like this:
1 + 100000..00 = 100000..01
Archimedes Plutonium
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
> numbers so that we have the first counterexample of FLT for exp3 that
> looks like this:
>
> 1 + 100000..00 = 100000..01
>
Looks like it is close to the time to wrap up this book, set it aside
for a later date
and pick back up with a new edition. This book is almost 1400 pages
but with mistakes made
in a revised edition, it should be about 1/2 as long.
The earlier mistake was to think that I could easily show exp4 and
larger could not have a
P-triple because the predecessor of a number was less than 1/2 as
large such as in exp4 81 is
less than 1/2 of 256 so no way can you have a A+B = C, until I
realized that as you go out far enough the predecessors are larger
than 1/2 of the successor, as noted by 2401 then 4096
in exp4. So that was a silly mistake by me. But what does that pattern
develop into when
we go out really far? One would think there is always some "distance
of spacing that cannot
be broached".
These are solution sets for exponents in FLT and to gain a
counterexample one must show that a
A,B,C in a solution set has A+B =
C
exp3
{1, 8, 27, 64, 125, 216, 343, 512, 729, . .}
exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096,. .}
exp5
{1, 32, 243, 1024,. . }
Which leads to my second big mistake in thinking that I can fetch
numbers
in an exponent separated by 1 for
1 + 100000..00 = 100000..01
So for exponent 3, what is the minimum untouchable distance separating
apart a
predecessor with a successor? Would it not be a distance of 7 as with
1 to 8?
So that no two numbers in that solution space of exp3 can be closer to
each
other than with 7 distance. And what is the nature of that distance of
separation
as the numbers get larger? Does the separation distance have some sort
of
equilibrium stable state of distance apart?
So my second mistake is thinking that a B and C in A+B = C could be
separated by
a distance of 1 as the numbers got huge.
But all is not lost, for a mistake leads to perhaps a insight
elsewhere. I was talking
of this technique with the numbers
64^3 + 94^3 = 103^3 add 1
before I went over to 1^3 + 1000..00^3 = 1000..01^3
So let me return to the above near miss and apply this technique:
640000..00^3 + 940000..00^3 = 1030000..01^3
Here I have no separation by 1 in the exp3 solution set.
Here I take the cube root of 103^3 and go along that digit string
until I
am able to drag out a 1030000..01 number.
But then I have to recalibrate the 64 and 94 numbers so that they are
at
the same level as the new 103 number.
There is one thing nasty about these math conjectures, in that they
seem to
always have a way of slipping away.
I may call an end to this book soon, rather than make another big
mistake, for
it indicates I am too tired and need a rest from math.
If I leave soon, I will have my main goal accomplished-- infinity is
at 10^603
by Circumferencing the Perimeter. But I will not have verified that
claim. Only the
circumstantial evidence of the 215215.21500 in 10^5 B matrix
215.50 for 10^2 B matrix
2152.050 for 10^3 B matrix
21521.5000 for 10^4 B matrix
215215.21500 for 10^5 B matrix
I have not verified that a critical perimeter has its first breakdown
at 10^603 B matrix
with its three zero digits in a row in pi. The breakdown may occur
earlier in 10^308 with
its two zero digits in a row in pi.
But much work was done to penetrate Riemann Hypothesis and FLT. I am
convinced RH is totally
false with nothing to reconfigure to make it true as reconfiguring of
Kepler Packing.
And surprizingly, it seems as though the toughest one of these old
conjectures that hangs on
the worst is FLT.
With Goldbach and the 10^603 border is straightforward of a countdown
from even numbers
at 10^603. A genuine computer proof for the first time. I expect no
glitches with Goldbach.
With RH, and 10^603, there is no way that Euler zeta is equal to
Riemann zeta and that takes
all the air out of the balloon.
With Poincare conjecture and no numbers between 0 and 10^-603 means
holes in geometry all
over the place.
With FLT, again, the question is with 10^603 are there or are there no
P-triples for exp3
and higher before 10^603 is reached? If not FLT is true. And then the
question is whether
the Wiles FLT has any merit, if a counterexample is found after
10^603.
Knowing that 64^3 + 94^3 = 103^3 add 1 is a near miss to FLT, and
wondering if that technique
can expand that out to see if a equality is reached, is promising.
Whether the 64 and 94 are
flexible enough once a twiddle of 1 in the 103 root is instilled. If
there is an equality, it probably is going to be in the thousands of
digits for each of the 64 root and 94 root and
103 root. It is not going to look pretty. And if such a triple is
found could well take up an entire book in just stating the numbers.
David R Tribble
>> Enrico easily set up a computer to check B matrices for 10^308. And
>> FLT for exp3 is
>> far easier.
>>
>> Let me show you how.
>> Set the computer program to fetch the set of this:
>> {1, 8, 27, 64, . .}
>>
>> That set is every natural from 1 to cube root of 10^603. Now the
>> computer has an easy time
>> of fetching that set. Perhaps taking 1 day to do it.
>
David R Tribble wrote:
>> Your calculator appears to be broken.
>> The set contains cubes for every natural from 1 to the
>> cube root of 10^603, which is 10^201. Assuming that the
>> computer can generate a trillion (10^12) cubes per second,
>> this will take 1.157 x 10^184 days, which is about
>> 3.169 x 10^181 years.
>> Just to generate the set of cubes.
>
Archimedes Plutonium wrote:
> Don't bother me and this thread again over computer time.
Of course. Why be bothered by petty little facts?
> If you applied any of your above
> to Enrico going through the spread sheet of B matrix, Tribble would
> calculate that Enrico
> would not have returned for two centuries with any answers. Whereas in
> reality, Enrico
> discovered the B matrices of Squaring the Circle of its first
> breakdowns out to 10^9 in a
> matter of two days.
Because he did not do it using the linear searching method you
describe. He couldn't have, because it is physically impossible.
> So you are making spurious and erroneous assumptions such as the
> computer taking lunch breaks.
Again, your calculator must be broken. Perhaps you can
pinpoint exactly where my calculations are incorrect.
> So don't bother me with you pessimistic nonsense of assumptions.
Or easily verified facts. Got it.
Archimedes Plutonium
Okay I concede, that the entire package of 10^603 is out of reach for
computer checking.
However, by numerical analysis, we can knock out enough of that
package that will
leave the rest of the package individually verifiable by computer that
should not take
more than say a year time in all.
What I mean is that we go hunting down a near miss of
64^3 + 94^3 = 103^3
Take the cube root of each one of those and crawl down the root string
of digits by a computer
to see if a twiddle on the 103 root can be boosted by one more
increment so that we have equality.
And, you must concede also, for about a year ago, it was you always
complaining that 10^500 did
not feel like infinity. Well, looks as though you now are feeling
infinity in a big way, eh.
AP
David R Tribble
> Go and straighten out those redneck bigots of editors of Glenfart and
> DMack on the Wikipedia Usenent celebrities.
Why not, instead, you learn the proper rules of Wikipedia editing,
text formatting, and etiquette? A good start would be to stop
insulting people.
After all this time, it seems that you still don't understand how
Wikipedia works. It is not a forum for your personal beliefs and
theories, nor is it a place for personal attacks in any form.
David R Tribble
> Mathematics, for the most part is precision, and not about some axiom
Well, there's your problem.
Archimedes Plutonium
Wikipedia has become a place where loser editors can behave like
little hitler despots
over people who never want to be in that rag encyclopedia
Wikipedia does not even allow for updates when those lousy tin badge
despots run rampant
Most every article in Wikipedia is lorded over by less than average
minds on that subject
matter. The model of an encyclopedia is when Maxwell edited
Encyclopedia Britannica. Wikipedia is closer to an encyclopedia edited
by prison inmates than edited by objective
people of science.
The reason I attack the editors of Wikipedia is that every time I want
to make a change
no matter how small of a change, I get nothing but a gang of creeps
like DMack Glenfar,
Dingley and a slew more opposed to every tiny little bit of change.
And you want me to
act like an angel when they act like devils.
I rather you pull the article completely so I never see those wretched
editors
Archimedes Plutonium
Well, my two recent big mistakes on FLT does pay at least one dividend
check.
As I was doing FLT, an idea crossed my mind about a technique of
double Math Induction.
I do not know if anyone ever tried or thought of combining Math
Induction, sort of reminds me of double-integral or multivariable
calculus.
PROOF OF GOLDBACH that every even integer from 4 onwards is the sum of
two primes.
PROOF: This much we know for certain that every even integer N from 4
onwards is the sum of
N/2 primes. Proof: we simply add 2s. Now we know for certain also,
that Goldbach is true out
to what our most recent computer verified that it is true to 10^25,
which I am sure Tribble
can verify. Now this proof is Indirect, so those squeamish about
reductio ad absurdum best not read any further. Suppose Goldbach is
false, then somewhere out there beyond 10^25 lies a
even number P which is not the sum of two primes. But, however, it
must be the sum of three primes? How is that true? Because its
predecessor P-2, even number was the sum of two primes and so we
include the prime 2 which proves this number is the sum of three
primes. But now we run into a contradiction by Mathematical Induction
of Fermat's Infinite Descent and using P-2, P, and P+2. We have the
situation that all the even up to and including P-2 were Goldbach
compliant and where P is not Goldbach compliant, but that all the even
numbers from 4 to
P were the sum of 3 primes. So the situation is that P is 3 prime
summable but not 2 prime summable. Is that possible? Well, it involves
the question of P+2 that we run into this contradiction. If P+2 is 2
prime summable then P must also be 2 prime summable. Hence Goldbach.
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
correction there, I should say all the even numbers from 6 to P
> P were the sum of 3 primes. So the situation is that P is 3 prime
> summable but not 2 prime summable. Is that possible? Well, it involves
> the question of P+2 that we run into this contradiction. If P+2 is 2
> prime summable then P must also be 2 prime summable. Hence Goldbach.
>
I see I am going to have to explain and clarify that contradiction a
lot
more than written above. I see the contradiction in my mind's eye, but
not able to make it clear for others to understand the contradiction.
Basically the contradiction is that from 4 to P-2 they are all two-
prime summable
and from 6 to P they are all three-prime summable. But now we examine P
+2
and it cannot be two prime summable nor can it be three prime summable
by
Fermat Infinite Descent. Hence P was two prime summable.
If Goldbach was a false conjecture, then it causes one even number P
and then the
successor even numbers to generate a big hole that destroys both
2prime summable and
3prime summable. If Goldbach is false, it degenerates all the
successor even numbers.
And ultimately the reason that Goldbach has to be true is because if a
hole is opened up
by P then P+2 then P+4 etc etc contradicts the proof that every even
Number K is K/2 prime
summable. So here I have a Mathematical Induction of successors rather
than Fermat's backward infinite descent.
David R Tribble
You are saying that for a given P-2 = A + B (where P-2 is even, and
A and B are prime), that P-2+2 = P = A + B + 2 is the sum of three
primes: A, B, and 2.
But what if A+2 or B+2 is also a prime? Then P is not the sum of
three primes, but the sum of only two primes.
Simple example: P = 14
Then you have
P-2 = 12 = 5 + 7
P = 14 = 2 + 11 + 2
P+2 = 16 = 5 + 11
But
P = 2 + 13
So P=14 is not the sum of three primes.
Archimedes Plutonium
PROOF OF GOLDBACH that every even integer from 4 onwards is the sum of
two primes.
PROOF: This much we know for certain that every even integer N from 4
onwards is the sum of
N/2 primes. Proof: we simply add 2s. Now we know for certain also,
that Goldbach is true out
to what our most recent computer verified that it is true to 10^25,
which I am sure Tribble
Suppose Goldbach is
false, then somewhere out there beyond 10^25 lies a
even number P which is not the sum of two primes. But, however, it
predecessor P-2, even number was the sum of two primes and so we
include the prime 2 which proves this number is the sum of three
primes. But now we run into a contradiction by Mathematical Induction
of Fermat's Infinite Descent and using P-2, P, and P+2. We have the
situation that all the even up to and including P-2 were Goldbach
compliant and where P is not Goldbach compliant, but that all the even
P is 3 prime
summable but not 2 prime summable. Is that possible? Well, it involves
the question of P+2 that we run into this contradiction. If P+2 is 2
prime summable then P must also be 2 prime summable. Hence Goldbach.
I am striving for clarity on this. The contradiction occurs at the
site
of the even number not being 2prime summable and its immediate
neighbors.
Let me throw in a example to guide the explanation and proof:
Suppose 8 was the last 2summable prime as 5+3 and that 10 is only
3summable as
5+3+2. Now let me include an additional new concept of the total
summations-matrix of
primes of even numbers up to a given L even number. So for 4 we have
2+2. For
6 we have 3+3 and 2+2+2 and 2+2 for this matrix on 6
2+2
2+2+2
3+3
For 8 we have this matrix (includes all previous matrices)
2+2
2+2+2
2+2+2+2
3+3
3+3+2
5+3
In other words the matrix includes all the Goldbach-related additions
of primes up to that given
even number L. This matrix is important because as we "suppose
Goldbach is false with P"
will deny the existence of Goldbach solutions previously begot.
So now the example of 10 not being 2 prime summable but 3 prime
summable as 5+3+2. This means we cannot have 5+5 nor 7+3 in that
matrix. And now we go to number 12 or P+2 and ask if it
is 3prime summable? Since we are not allowed 5+5 nor 7+3, 12 is not 3
prime summable.
David R Tribble
> And, you must concede also, for about a year ago, it was you always
> complaining that 10^500 did
> not feel like infinity. Well, looks as though you now are feeling
> infinity in a big way, eh.
10^500 is not infinity (not non-finite). Neither is 10^603,
for exactly the same reasons. These are fairly small finite
numbers, after all. You're not even close.
Archimedes Plutonium
No David, I am not adding the 2 to any of the previous Goldbach duet
of primes
12 is 2prime summable meaning A + B as 5 and 7, and so 14 is 3prime
summable as A+B+C with
5+7+2
The gist of the proof is that If Goldbach is false means a even number
P is not 2prime summable but is 3prime summable borrowing the two
primes of P-2 and including the prime
2 to make a trio of primes A,B,C. And now looking at P+2 and questions
of its prime sums
contradicts the idea that P was not Goldbach summable.
I still do not have the clarity I want.
Enrico
>
> - Show quoted text -
=======================================================
>
> I still do not have the clarity I want.
>
Try defining your terms and procedures starting with
unambiguous basic items. Then use the newly defined
items to build the mathematical structure you need
for your proof.
I run into this when I start writing a program -
invariably, I run into items that turn out to be
different from what I thought when I try to write
instructions for their use.
Enrico
Archimedes Plutonium
Thanks for the advice. I found the clarity now. Let me see if I can
convince you of this
Proof of Goldbach.
The piece I had missing is the use of the truth that every even number
S is S/2 prime summable.
For example, 4 is 2+2 and 6 is 2+2+2 and 8 is 2+2+2+2 and 10 is
2+2+2+2+2
Now in Goldbach conjecture we like to reduce that summability of
primes to just two primes A+B
and not S/2 primes.
So now, using S/2, which I did not use before, what I get is that P is
violation of A+B, but it is not
in violation of A+B+C having borrowed the two primes of P-2 and using
a 2 for C.
Fine so far. For we have that P-2 is 2prime summable and 3prime
summable. We also have that
P is not 2prime summable but is 3prime summable.
Now we look towards P+2, and it may or may not be 2prime summable. And
we do not care if it
is or is not 2prime summable, but because P is not 2prime summable
then P+2 is not 3prime summable.
Therein lies the contradiction, that if P is not 2prime summable yet
3prime summable which implies
that P+2 is not 3prime summable, contradiction and that P must be
2prime summable.
Let me show this with example of that Goldbach matrix I was starting
in the last post and let me pretend that
10 is the last Goldbach and that 12 is a Goldbach failure.
4 = 2+2
6= 2+2+2 for 3prime summable, and splitting apart one of the 2s of
2+2+2 and adding it to the remaining
two 2s gives 3+3 for a 2prime summable
8 = 2+2+2+2 for 4prime summable which is far larger than what we want,
so we reduce that by removal of
one of the 2s splitting it up into 1 and 1 and adding it to two of the
other 2s like this, 3+3+2 and now 8 has
been turned into 3prime summable. Now we do a final matrix maneuver
and see if we can make 8 2prime summable instead of its 3prime
summable. We first see if we can split up the remaining 2 but that
only gives us 4+4
but we see that adding the last 2 to that of 3 gives us 3+5 for 2prime
summability.
10= 2+2+2+2+2 which is 5prime summable and turns into 3+3+2+2 for
4prime summability which turns into
3+7 or also 5+5 for 2prime summability. And NOTE, that this is
supposed as the last Goldbach behaving number and
that 12 is not Goldbach compliant.
12 = 2+2+2+2+2+2 for 6prime summable and turns into quickly by busting
up two 2s into 3+3+3+3 for 4prime summability or turns into 3+3+2+2+2
for a 5prime summability and turns into a 5+5+2 for a 3prime
summability
and carries over from the 10= 5+5. But we SUPPOSED that 12 was not
2prime summable so we cannot have
that 2 added to the 5.
Now for the Contradiction Step: 14 = 2+2+2+2+2+2+2 which starts as
7prime summable and quickly can be reduced
by splitting a 2 and making the other 2s as 3s as this 3+3+2+2+2+2 for
6prime summable and we can go to smaller
prime summability but because 10 was supposed as the last Goldbach we
are not allowed to find a 3prime summable
in 14 because 10 was not 2prime summable. Contradiction because
clearly we have 2+5+7 as 3prime summable.
So thanks Enrico for the concern of clarity. It looks as though I did
not need a Mathematical Induction.
The Indirect Proof shows us what the weakpoint is in Goldbach. If
Goldbach is false, then a even number P is
not 2prime summable but is 3prime summable, thence looking at P+2, it
must not be 3prime summable because
P was not 2prime summable, contradiction because P+2 must be 3prime
summable, hence proof.
Also, thanks to David Twiddle for his question helped me get over the
clarity issue.
Archimedes Plutonium
That should have read 12 was not 2prime summable and changed in the
original.
> clearly we have 2+5+7 as 3prime summable.
>
> So thanks Enrico for the concern of clarity. It looks as though I did
> not need a Mathematical Induction.
>
> The Indirect Proof shows us what the weakpoint is in Goldbach. If
> Goldbach is false, then a even number P is
> not 2prime summable but is 3prime summable, thence looking at P+2, it
> must not be 3prime summable because
> P was not 2prime summable, contradiction because P+2 must be 3prime
> summable, hence proof.
>
> Also, thanks to David Twiddle for his question helped me get over the
> clarity issue.
>
Let me stress again what the weakpoint in the Goldbach conjecture is
and why
it must be true. The weakpoint is that if a even number is not the sum
of two
primes then it is at best 3prime summable, but its successor neighbor
is not
3prime summable because it cannot borrow the two primes of its
predecessor.
That is a contradiction to the summability matrix of every even number
S having
S/2prime summability.
Archimedes Plutonium
Alright, I spent the day, considering if I really had a proof and come
to the conclusion I really do.
The worrying point is whether I really have a contradiction after
assume P is nonGoldbachian
implies P+2 is not 3prime summable. I was worried that I had no
contradiction there because I
could not see how P+2 had a prestated condition of being 3prime
summable. So I was worried
that "so what, so what if P+2 is not 3prime summable."
And as the hours passed by and me stewing over whether I had or had
not a contradiction. Then
the lights came on.
If P is assumed not 2prime summable (Goldbachian) and because of that
condition that P+2 cannot have
3prime summable for that at least one of the odd primes is missing.
That leaves P+2 with the odd prime
and the even prime 2. In other words, if you assume P is
nonGoldbachian implies P+2 is not 3prime summable
and that implies P+2 is also nonGoldbachian.
So I do achieve a contradiction. And it is obvious to me now, once the
light came on, that if Goldbach is false
that we have two consequences to consider. (a) That a even number like
P with a Goldbach failure is a hole, an isolated
case. (b) And the other alternative, that if P is a Goldbach failure
it generates a complete failure of all future even numbers such as P+2
then P+4 etc etc.
So the proof offered does work and the contradiction is that a failure
of an even number such as P means that P is still
3prime summable but not 2prime summable. That implies that P+2 is
3prime summable because there are no two odd primes to transfer to P
+2, but only one odd prime and the even prime of 2. Hence P+2 is also
**not 2prime summable**
and therefore every even number larger than P fails Goldbach.
Now I am the author of precision definition of Finite versus Infinite
with the border at 10^603. So as a final cap on the proof of Goldbach,
all I have to show, now is a even number larger than 10^603 that is
2prime summable. And thus Goldbach Conjecture is conquered and we do
not need computers running through every even number to 10^603.
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
When I type fast and do not bother to reread before sending then I
easily
make a mistake of not including a "not" in front of the 3prime
summable.
I changed it on the original with a "sic" sign.
> 3prime summable because there are no two odd primes to transfer to P
> +2, but only one odd prime and the even prime of 2. Hence P+2 is also
> **not 2prime summable**
> and therefore every even number larger than P fails Goldbach.
>
> Now I am the author of precision definition of Finite versus Infinite
> with the border at 10^603. So as a final cap on the proof of Goldbach,
> all I have to show, now is a even number larger than 10^603 that is
> 2prime summable. And thus Goldbach Conjecture is conquered and we do
> not need computers running through every even number to 10^603.
>
Can David Tribble or Enrico please tell me of a even number larger
than
10^603 that is Goldbach compliant? I suppose someone has already shown
a Goldbach compliant even number larger than 10^603. Once I get that
information
I know that all the even numbers from 4 to 10^603 are Goldbach true.
Archimedes Plutonium
yet, as Enrico
asked for. But I am still mulling on the proof. I am convinced it is a
proof, but
I want to point out some features that I have never before seen in a
mathematical
proof and a strange coincidence, even though I do not believe in
coincidences.
I often start a proof thinking I have a Mathematical Induction and
rarely does
it come out to be Mathematical Induction. In this case it was a simple
Reductio
Ad Absurdum but a most strange indirect method.
(1) In the proof P is supposed the last even number that is Goldbach
2prime summable.
And from that assumption P is 3prime summable borrowing from P-2,
which implies that P+2
cannot be 3prime summable which further implies that P+2 is also not
2prime summable.
So a chain reaction was initiated by supposing P was not Goldbachian
(2prime summable)
which causes all even numbers beyond P to be failing in Goldbach. Now
that is the most
odd happenstance I have ever seen in a math proof that if you suppose
something all the
numbers beyond have a breakdown.
(2) So it is Mathematical Induction in some way? That supposing P
nonGoldbachian causes
a chain reaction that all even numbers beyond P are nonGoldbachian?
But also, is it really
a contradiction because we know not whether P is say so far out there
as to be an infinite
even integer? Which brings up another strange and bizarre coincidence.
(3) So do I need the new theory I authored that Infinity starts at
10^603 and thus, the proof of Goldbach needs to peg itself to 10^603
as to whether a large even number of 10^603 is Goldbachian so that the
proof can claim all even numbers from 4 to 10^603 are Goldbachian? Do
you see the coincidence? That I discover a boundary of infinity at
10^603
and then do a proof of Goldbach that requires that boundary in the
proof?
Is it fate that no-one could do a proof of Goldbach unless they had a
boundary between Finite and Infinite?
Archimedes Plutonium
And especially to define terms.
PROOF OF GOLDBACH Conjecture (GC) that all even numbers starting with
4
are the sum of two primes.
Let me start the proof with a Example so that the proof itself can
be readily understood by referring to the example supplied.
EXAMPLE:
In this example I am going to pretend that the last even
integer to be Goldbach compliant is 10 and that 12 is a Goldbach
failure.
10 is the last Goldbach number and that 12 is a Goldbach failure.
In this proof I am going to use a concept that I have never seen
anyone else use in attempting GC. If we are given any even number
starting with 4 we take that even number, call it S, and divide it by
2 to receive S/2, for example 6 is 6/2 = 3. The S/2 is how many 2s
are contained in S. So that 6 has three 2s contained, and 2 is a prime
number. So that 6 is 2+2+2 which I will call 3prime-summable. In the
Goldbach Conjecture, it says that all even numbers are 2prime-
summable,
meaning there is a pair of primes when added together equals the even
number.
So we have the concept of S/2 and given any even number such as R,
then
R is R/2prime-summable since we add the R/2 of the prime 2 together to
equal R.
And we have the concept of prime-summable.
4 = 2+2 and as we see, 4 is only 2prime-summable
6= 2+2+2 for 3prime-summable, and splitting apart one of the 2s of
2+2+2 and adding it to the remaining
two 2s gives 3+3 for a 2prime-summable
8 = 2+2+2+2 for 4prime-summable which is far larger than what we
want,
so we reduce that by removal of
one of the 2s splitting it up into 1 and 1 and adding it to two of
the
other 2s like this, 3+3+2 and now 8 has
been turned into 3prime-summable. Now we do a final matrix maneuver
and see if we can make 8 as 2prime-summable instead of its 3prime
-summable. We first see if we can split up the remaining 2 but that
only gives us 4+4 which are not primes,
but we see that adding the last 2 to that of 3 gives us 3+5 for
2prime-
summability.
10= 2+2+2+2+2 which is 5prime-summable and turns into 3+3+2+2 for
4prime-summability which turns into
3+7 or also 5+5 for 2prime-summability. And NOTE, that this is
supposed as the last Goldbach behaving number and
that 12 is not Goldbach compliant.
12 = 2+2+2+2+2+2 for 6prime-summable and turns into quickly by
busting
up two 2s into 3+3+3+3 for 4prime-summability or alternatively turns
into 3+3+2+2+2
for a 5prime-summability and turns into a 5+5+2 for a 3prime-
summability
and carries over from the 10= 5+5. But we SUPPOSED that 12 was not
2prime-summable so we cannot have
that 2 added to the 5.
Now for the Contradiction Step:
14 = 2+2+2+2+2+2+2 which starts as
7prime-summable and quickly can be reduced
by splitting a 2 and making the other 2s as 3s as this 3+3+2+2+2+2
for
6prime-summable and we can go to smaller
prime summability but because 12 was supposed as not Goldbach
compliant we
are not allowed to find a 3prime-summable
in 14 because 12 had no 2prime-summability to carry or transfer over
into this 14. Contradiction in this example because
clearly we have 2+5+7 as 3prime-summable, and which reduces to 2prime-
summability.
In the proof, by Reductio ad Absurdum, what happens is that
supposition that P is not 2prime-summable, yet it is 3prime-summable
due to the transfer of the P-2 set, and because P is not 2prime-
summable causes
the P+2 to be not 3prime-summable since it had no two odd primes from
P.
P+2 did have one odd prime but the other was a 2 prime. Since P+2
received
only two primes from P, one of them being an odd prime and the other
being
a 2 prime, means that P+2 itself is NonGoldbachian, just as P was
supposed
as NonGoldbachian. Thus a chain reaction has been set into motion
where
assuming P was nonGoldbachian that all the future even numbers beyond
P
are also nonGoldbachian.
To be continued..
Archimedes Plutonium
I have never seen anything like this in math before, where the
contradiction is Mathematical Induction
but requires a border between Finite and Infinity for the
contradiction to take affect. In the proof of
Goldbach, we supposed P was nonGoldbachian which then causes a chain
reaction that all even
numbers larger than P are also nonGoldbachian. But that is no
contradiction unless someone imposes
a barrier or border or boundary on P. So if someone finds that 10^604
an even number itself has
a Goldbach solution, then I can come in with the proof and say that
all even numbers starting with 4
to all even numbers 10^603 are Goldbach compliant.
Funny how the proof of Goldbach comes at a time in history when
someone recognizes that mathematics
is dead unless it defines finite and infinity with precision.
Archimedes Plutonium
(snipped to save some space)
>
> I have never seen anything like this in math before, where the
> contradiction is Mathematical Induction
> but requires a border between Finite and Infinity for the
> contradiction to take affect. In the proof of
> Goldbach, we supposed P was nonGoldbachian which then causes a chain
> reaction that all even
> numbers larger than P are also nonGoldbachian. But that is no
> contradiction unless someone imposes
> a barrier or border or boundary on P. So if someone finds that 10^604
> an even number itself has
> a Goldbach solution, then I can come in with the proof and say that
> all even numbers starting with 4
> to all even numbers 10^603 are Goldbach compliant.
>
> Funny how the proof of Goldbach comes at a time in history when
> someone recognizes that mathematics
> is dead unless it defines finite and infinity with precision.
>
Enrico, I cannot finish this proof unless I have a even number beyond
10^603
that is Goldbach compliant. So is 3 and the number ((10^604)-3)
Goldbach primes?
Enrico
> whole entire Universe is just one big atom
>
> - Show quoted text -
=========================================================
>
> Enrico, I cannot finish this proof unless I have a even number beyond
> 10^603
> that is Goldbach compliant. So is 3 and the number ((10^604)-3)
> Goldbach primes?
>
No.
10^604-3 is divisible by 13
Get a list of big primes numbers and take two of them
that add up to greater than 10^604. I guess that should
give you the even number beyond 10^603 that you need.
Enrico
Archimedes Plutonium
Thanks that will allow me to complete the proof. I should have thought
of that
myself.
--- quoting from
http://primes.utm.edu/top20/page.php?id=39
1
U(81839)
17103
p54
Apr 2001
Fibonacci number
2
U(50833)
10624
CH4
Oct 2005
Fibonacci number
3
U(37511)
7839
x13
Jun 2005
Fibonacci number
4
U(35999)
7523
p54
Jul 2001
Fibonacci number, cyclotomy
5
U(30757)
6428
p54
Jul 2001
Fibonacci number, cyclotomy
6
U(25561)
5342
p54
Jul 2001
Fibonacci number
7
U(14431)
3016
p54
Apr 2001
Fibonacci number
8
U(9677)
2023
c2
Nov 2000
Fibonacci number, ECPP
9
U(9311)
1946
DK
Mar 1995
Fibonacci number
10
U(5387)
1126
WM
Dec 1990
Fibonacci number
--- end of quote ---
Alright, so I take the last two and add them, and is well larger than
10^604.
So that in the proof of Goldbach, we start out, as we should in all
proofs of mathematics by saying
Mathematics is good only up to 10^603 where a finite perimeter is
matched by a finite circumference.
Beyond 10^603 is infinity and a logic that is no longer able to give
trustworthy mathematics.
So in the proof of Goldbach, when we suppose GC is false and that P is
a failure of Goldbach, results
in the fact that all even integers beyond P are also failures of
Goldbach. In Old Math, we have no contradiction
at that moment because P could be an infinite integer since Old Math
never defined finite versus infinity.
But by Circumferencing the Perimeter we find that the first place
where Algebra of Numbers breaksdown is
10^603 and so that is the borderline. So to complete Goldbach proof I
need a contradiction which is begot
because those last two huge Fibonacci primes add up to a number that
is even and larger than 10^603.
So that I know all even numbers from 4 to those Fibonacci-primes-sum
obey Goldbach. That is the contradiction
to Mathematics since mathematics is only good out to 10^603 and thus
all even numbers in good-math
obey Goldbach.
An example of where Goldbach is disobeyed is given by either AP-adics
or by Hensel p-adics.
Now this gives me new encouragement that I can find a proof of FLT,
Fermat's Last Theorem. Suppose false. Then
there is a P-triple in exp3 that obeys a^3 + b^3 = c^3. Suppose we
have a P-triple (a,b,c) that obeys FLT in exp3.
Like Goldbach, what is going to be wrong with that picture? It was
easy to find it for Goldbach, but I am afraid much
tougher for FLT.
But with my recent foray into FLT, and that second mistake of mine if
you recall 1^3 + 1000..00^3 = 1000..01^3
Now my second mistake looks awfully bad for newcomers, but to me, who
has worked through P-adics of Hensel
and my own AP-adics, such bizarre numbers as monsters or beasts of
1000..00^3 versus 1000..01^3 and separated
only by 1 unit is rather commonplace and not bizarre at all.
So this concept in FLT, that of the spacing metric distance of one
neighbor number to another neighbor number. Has
anyone given this distance length spacing a technical name in the
literature?
Here is the solution set of FLT for exponent 3:
exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, . .}
So, as the numbers get larger is the distance spacing always greater
than 8-1=7. So that 7 is the smallest length
of neighbors?
Has mathematics ever researched the spacing length and come up with
any theorems on that distance length? Is it
always decreasing as the numbers get larger. Sort of like as the
numbers get larger the multiplication shrinks the distance spacing
between predecessor and successor. Is this shrinkage ever allow for A
+A = C where A=B?
So it seems to me, a perfectly logical question that if A+A = C where
A=B in FLT for exp3, if there ever is a case of
that occurring when the solution set goes to infinity would be the
proof of FLT itself.
So if I started a Reductio Ad Absurdum on FLT tonight, I would want to
well define this distance spacing, because if I
can get A+A= C rather than looking for A+B=C would prove FLT in a
snap. And it would likely be in a even numbered
exponent than an odd numbered exponent such as exp3.
So the proof of FLT, where the border of finite and infinity is
10^603. Suppose FLT is false implies there exists, not only
a A+B = C in a expN but that there exists a A+A=C wherein we can
always convert that to the FLT proper.
So we suppose that exists, then ask how would that supposition tear up
the fabric of mathematics? In Goldbach the tearing of the fabric is
that you have all even integers as nonGoldbachian once one such
integer arises.
So here, if I may borrow the logic of Goldbach and transfer that logic
over to FLT, the question immediately becomes
Suppose FLT is false and that there exists a A +B= C, which is
immediately transfered in logic to saying there exists
a A+A=C. But if that should exist, does it not immediately imply that
the distance spacing or length of predecessor and
successor terms has every length with the same characteristic of A+A=C
or A+B=C once the first such length is
found?
Now for exponent 2 do we ever have the case of A+A=C?
solution set {1, 4, 9, 16, 25, 36, . . .]
I do not yet know but some of those come awfully close such as 9 +9 =
16+2
So if we go out far enough in exp2 do we end up finding a A+A=C?
Probably yes.
So I think the key to FLT as a Indirect proof method is this idea of
length of spacing between numbers in the solution set. And that if we
suppose FLT to be true for exponent 3 and larger that we end up with a
very huge number which
is A+A=C
But that huge number is far beyond 10^603 that we have thus proven FLT
is true for good-mathematics.
This would be bad and horrible news to Wiles because his ideas never
incorporated a precision definition of finite
and infinite. And that finding a A+A=C is a counterexample to Wiles
and would show his proof is a sham.
Archimedes Plutonium
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
>
> Suppose FLT is false and that there exists a A +B= C, which is
> immediately transfered in logic to saying there exists
> a A+A=C. But if that should exist, does it not immediately imply that
> the distance spacing or length of predecessor and
> successor terms has every length with the same characteristic of A+A=C
> or A+B=C once the first such length is
> found?
I am sure the contradiction for FLT does not parallel the Goldbach and
that
the fabric of mathematics would be torn with something else.
But let me borrow some more from the proof of Goldbach that is helpful
in the
proof of FLT.
>
> Now for exponent 2 do we ever have the case of A+A=C?
>
> solution set {1, 4, 9, 16, 25, 36, . . .]
>
> I do not yet know but some of those come awfully close such as 9 +9 =
> 16+2
>
> So if we go out far enough in exp2 do we end up finding a A+A=C?
> Probably yes.
>
In Goldbach we wanted two primes to equal an even number which we can
represent
as A+B = C. But notice in Goldbach that the primes can be identical
such as 5+5=10
so that we have A+A=C, and for 10 we also have 3+7=10 as A+B=C.
So what would happen if we altered the statement of Goldbach to be a
statement that
excludes identical primes. So we strictly require A+B=C and not allow A
+A=C.
Well, we would eliminate 4 and 6. So we would have to start the
Goldbach at 8.
Now I know of no even number greater than 8 that has only a A+A=C
representative
to satisfy Goldbach. As far as I know all even numbers greater than 8
which have
a A+A=C representative also has a A+B=C representative.
So we can ask the question, where in Goldbach when we start with 8, do
we have an
even number that is only represented by Goldbach primes of A+A=C?
So now let me transfer that concept over to FLT. And ask, whether the
first counterexample of
FLT for exp3 or exp4, will be a counterexample of the type where A
+A=C, or be a counterexample of
the type where A+B=C?
Now looking at FLT for exp2, we first have a P-triple of the type of A
+B=C and not of A+A=C.
But I think the FLT for exp2 is of no guidance for FLT of exp3 and
larger.
My hunch is that the first sighting of a A+A=C solution in exp3 or
higher exp is going to come
first rather than a A+B=C counterexample solution. My hunch is that if
A+A=C solution exists in
FLT, it may come in at say 10^300 and the first solutions of A+B=C may
come in at say 10^700 range.
Archimedes Plutonium
to sci.math
on proving FLT before I render this latest update.
This new latest update borrows from the proof of Goldbach, and
essentially says that
suppose FLT were false for exp3 or exp4, then you have a situation
asking for a a^3+b^3=c^3
ditto for exp 4, but that a cube root of 2 and a quadratic root of 2
enters into the situation.
Suppose FLT is false for exp 3 then you have a A+B=C in the solution
set {1, 8, 27,....}
Yet if you ask for a A+A=C which exists before a A+B=C would require
2(a^3)= c^3.
Which requires a cube root of 2 to be what is known as rational and
not irrational. Likewise
for exp4 and all higher exponents that you require the number 2 to be
rational and not irrational.
--- quoting from my old posts of circa 1993 ---
PROOF OF FERMAT'S LAST THEOREM (FLT) Sorry if some of
the symbols did not copy through in the proof.
For the set of positive integers {1,2,3,4,5, . . . }, a
given positive integral exponent n >2, the equation
a^n + b^n = c^n is impossible to solve for positive integers
a, b, c. I want to give a simplified idea of the statement
of Fermat's Last Theorem. Consider the sets
{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, . . .}
{1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289, . . .}
{1,8,27,64,125,216,343,512,729,1000,1331,1728,2197,2744, . . .}
{1,24,34,44,54,64,74,84,94,104,114,124,134,144,154,164,174, . . .}
{1,25,35,45,55,65,75,85,95,105,115,125,135,145,155,165,175, . . .}
.
.
.
The first set consists of all the positive integers, the second
the squares of all the positive integers, the third the cubes,
the fourth is the fourth power of all the positive integers and
the fifth the fifth power. Now with these sets in mind Fermat's
Last Theorem asks for the second set is the sum of two squares
ever itself a square, i.e., are there two numbers in the set
when added together gives another number contained within the
set. It is easy to see that 9,16 and 25 satisfies Fermat's Last
Theorem. Looking now at the third set, the set of cubes. We
again ask is a cube ever the sum of two cubes? And for the
fourth set, the fifth set, and on out endlessly.
First a discussion before I go into the proof, in order to
make the engine of the proof clear. What I will show is that
Fermat's Last Theorem is related to the fact that for
a^2 + b^2 = c^2 there exists a number such that N+N=NxN=M,
where N not= M. This number N is 2, and M is 4. That 2 is a
building block for the smallest Pythagorean triple for
exponent 2, in order to build the smallest Pythagorean triple
which is (3,4,5). But for exponent 3 there does not exist a
number such that N+N+N=NxNxN=M, where N not= M, for if there
was such a number, then a smallest Pythagorean triple is
immediately constructible for exponent 3 because such a number
is equal under addition and multiplication, likewise exponent 4,
and all larger number exponent. PROOF: A constructive
proof follows from looking for a positive integer N, if it
exists for a given n, which is equal under the operation of
multiplication and addition. For n=2, does there exist an N
such that N+N=NxN=M. Then it will be shown that N and M are
unique. Once given an N and M for a given n, then N and M
will allow the construction of a Pythagorean triple (P-triple)
(a1,b1,c1), all because of the property that N is equal under
the operation of multiplication and addition, and from the
unique property of 1, where 1^n=1. It is shown from the
uniqueness of the properties of N, M, and 1 that (a1,b1,c1)
is the smallest P-triple for that given n. By smallest
P-triple, means for every P-triple which satisfies
an + bn = cn for a given n, (a1,b1,c1> (a2,b2,c2)
(a3,b3,c3), . . . then b1 is the smallest positive
integer of b2,b3,b4, . . . and where a1 less than b1 with
a2 less than b2 ,a3 less than b3, . . . Then for a given n,
if no smallest P-triple is constructible then no P-triple
exists for that given n. It is shown for any n, if an N
exists and because N is equal under multiplication and
addition and 1^n=1, then the general form for the smallest
P-triple, for n=2 is a1=1+N, b1=N+N,c1=1+N+N, for n=3
a1=1+N+N,b1=N+N+N,c1=1+N+N+N and so on. For n=2, a number
N which satisfies N+N=NxN=M is the positive integer 2 and
the result of the operation is 4. The number 2 is unique
for the positive integers (and in all of mathematics for
that matter) under the operation of addition and
multiplication resulting in the same sum and product of 4,
since 1+1=2 and 1x1=1; 3+3=6 and 3x3=9; 4+4=8 and 4x4=16;
5+5=10 and 5x5=25, for it is shown that as these positive
integers get larger then the divergence of multiplication
subtract addition (for 3 it is 9-6=3, for 4 it is 16-8=8)
gets much larger and so using mathematical induction
therefore 2 is unique. And since 2 is unique then the
summation equal to the product is unique, so 4 is unique.
Since M is the unique product and sum of N, then M is the
smallest product and the smallest sum. It is proven that 1
is the identity with respect to multiplication such that
1^n=1. Mathematically stated for n=2, and substituting N
with 2 in (1+N)^n+(N+N)^n=((1+N)+N)^n gives
(1+2)^2+(2+2)^2=(1+2+2)^2 which is (1+2)(1+2)+(2+2)(2+2)=
(1+2+2)(1+2+2) which gives (1+4+4)+4+4+4+4=1+4+4+4+4+4+4.
So from the existence of the unique numbers 2 and 4 which
solves N+N=NxN=M allows the construction of the smallest
positive integer P-triple (a1,b1,c1) for n=2 it is (3,4,5).
Showing for the general case of n, suppose there exists a
smaller P-triple (as,bs,cs) than (a1,b1,c1). If (as,bs,cs)
exists then b1>bs, where as less than bs which implies the
existence of two different positive integers which satisfy
the property of equality under multiplication and addition,
starting with the case n=2 where N=2 that (a1,b1,c1) is the
smallest P-triple for n=2, and applying mathematical
induction it follows for any n with corresponding N. For
n=3, no positive integer exists which has the property
N+N+N=NxNxN=M, since for 1+1+1=3 and 1x1x1=1; 2+2+2=6 and
2x2x2=8; 3+3+3=9 and 3x3x3=27 the divergence of (NxNxN)-(N+N+N)
increases with larger positive integers and by mathematical
induction no positive integer exists which has the property
N+N+N=NxNxN=M. If an N exists which has the property
N+N+N=NxNxN=M, then (1+N+N)^3+(N+N+N)^3=(1+N+N+N)^3 is
possible where (1+N+N)=a1, (N+N+N)=b1, (1+N+N+N)=c1 forms
the P-triple (a1,b1,c1>. Since no N exists for n=3 then no
smallest P-triple is constructible. Thus no P-triple exists
for n=3. For n=3, (5,6,7) come the closest but off by 2,
since (2+2+2) is almost (2x2x2). The same argument for n=3
applies for n=4 and larger. Therefore Fermat's last
theorem is proved. Q.E.D.
----
From: dre...@jaffna.berkeley.edu (Roland Dreier)
Newsgroups: sci.math
Subject: Re: 1 page proof of FLT; this is what Fermat
had in mind when he wrote "margin to small".
Fermat reasoned that the proof was so simple that
others would easily rediscover it.
Date: 13 Aug 93 13:32:17
Organization: U.C. Berkeley Math. Department.
Lines: 12
Message-ID: (DREIER.93Aug13133...@jaffna.berkeley.edu>
References: (CBpno3....@dartvax.dartmouth.edu>
In article (CBpno3....@dartvax.dartmouth.edu>
Ludwig.Pluton...@dartmouth.edu (Ludwig Plutonium) writes:
>For the case n=3 suppose there exists a P-triple
>which satisfies FLT. Then there has to exist a smallest
>P-triple for n=3. If true implies there exists a number N
>which has the property N+N+N=NxNxN=M.
Not that I expect a coherent answer, but let me ask anyway:
why does this follow? In other words, given positive
integers a,b,c with a^3+b^3=c^3, how do I get a positive
integer N with N+N+N=N*N*N ?
--
Roland "Mr. Excitement" Dreier dre...@math.berkeley.edu
----
Newsgroups: sci.math
From: Ludwig Plutonium
Subject: In answer to Mr. Roland Dreier
Date: Sat, 14 Aug 1993 16:06:53 GMT
Lines: 6
Fermat's Last Theorem is arithmetically equivalent to this:
For exponent 2 a smallest P-triple exists because 2+2=2x2.
For exponent 3 no smallest P-triple can exist because no
number N exists such that N+N+N=NxNxN. If such a number exists,
then a smallest P-triple is immediately constructible for
exponent 3. The same argument for exponent 4,5,6, . . .
----
Newsgroups: sci.math
From: Ludwig Plutonium
Subject: Re: 1 page proof of FLT; this is what Fermat
had in mind when he wrote "margin to small". Fermat
reasoned that the proof was so simple that others would
easily rediscover it.
Date: Sat, 14 Aug 1993 17:57:14 GMT
Lines: 14
Fermat reasoned that the proof was so easy, so simple that
others later would see it and not worth Fermat's time.
Fermat saw the uniqueness of the number 2 the only number
in all of math where its sum equals its product allowing
for the construction of a P-triple when exponent is 2 but
when exponent is 3 or greater there are no numbers where
sum equals product. A 1 page proof too long for Fermat's
margin.
The uniqueness of 2 also proves the Riemann Hypothesis
when the Euler formula--multiplication of terms, call it
E and the zeta function--addition of terms call it Z,
when E subtract Z equals zero only real component 1/2
works otherwise there must exist another number in math
such that n+n+n=nxnxn (for exp1/3) This is just a sketch
of my one paragraph proof of the Riemann Hypothesis anyone
wanting the full proof I will gladly email.
----
From: dre...@durban.berkeley.edu (Roland Dreier)
Newsgroups: sci.math
Subject: Re: 1 page proof of FLT
Date: 18 Aug 93 14:55:02
Organization: U.C. Berkeley Math. Department.
Lines: 42
Message-ID: (DREIER.93Aug18145...@durban.berkeley.edu>
References: (CBxp0H....@dartvax.dartmouth.edu>
(24s7de$...@outage.efi.com>
(CByoqr....@dartvax.dartmouth.edu>
(1993Aug18.173020.10...@Princeton.EDU>
In article (1993Aug18.173020.10...@Princeton.EDU>
kin...@fine.princeton.edu (Kin Chung) writes:
In article (CByoqr....@dartvax.dartmouth.edu>
Ludwig.Pluton...@dartmouth.edu (Ludwig Plutonium) writes:
LP Hardy in Math..Apology said words to the effect that the
LP understanding of any math proof is like pointing out a peak in the
LP fog of a mtn range and you can only point so long and do other
LP helps and hope the other person will see it and say Oh yes now I
LP see it. But you can not exchange eyeballs. Again I repeat the
LP arithmetic equivalent of FLT is that for exp2 there exists a
LP number equal under add & multiply i.e. 2+2=2x2=4. Immediately a
LP smallest P triple is constructible for exp2 i.e. (3,4,5). But no
LP number exists like 2 for exp3 or higher in order to construct P-
LP triples for these higher exp. I am very sorry that I cannot make
it
LP any clearer than that. Time to take a break and reread Hardy Math
LP Apology.
KC You also say that a smallest P-triple is constructible for exp2
KC immediately from the existence of a number N such that
KC N+N=NxN, namely N=2. How do you construct a P-triple given N
KC with this property? Please note that I am not asking how you do
KC it for exp3, but for exp2.
Before I continue, let me say that this post does not in any way
constitute
an endorsement of LvP's "proof"; what I am about to explain does not
extend to exponent 3 in the least. However, things are rather easy for
exponent two. (Not to be critical, but you really could have figured
this
out yourself :-)
So suppose we have an N with 2xN=N+N=NxN. Set a=N+1, b=N+N=NxN.
Then we get
a^2 = (N+1)^2 = N^2+2xN+1 = 2xN^2+1
also
b^2 = (N+N)^2 = 4xN^2.
So
a^2+b^2 = 6xN^2+1.
Now set c=2xN+1. Then
c^2 = (2xN+1)^2 = 4xN^2 + 4xN + 1 = 4xN^2 + 2xN^2
+ 1
= 6xN^2+1.
So magically a^2+b^2=c^2, just as desired! !
If you can figure out how to do that for exponent 3, make yourself
famous.
Roland
--
Roland "Mr. Excitement" Dreier
dre...@math.berkeley.edu
----
****** FLT proved false in Infinite Integers in Adics *****
----
Revolutionary New & Important Math
Naturals Numbers = All-adics = Infinite Integers = Riemannian geometry
World's first proof of Fermat's Last Theorem FLT
by Archimedes Plutonium
It was in 1993, where in sci.math searching for a correct proof of
Fermat's Last Theorem, FLT, I noticed that the definition of what it
means to be "finite" as in a finite number such as a Natural, say for
example 3 is really a infinite integer with endless string of zeroes
to
the
left as such ...00003. Reals are rightward infinite strings and so I
realized
by the end of 1993 that our concept of "finite" Naturals was a kooky
idea and
that they too should have an infinite componentry to them also. It was
in 1993
that I learned that Infinite Integers already existed in a body of
math
knowledge discovered mostly by K. Hensel around 1899, almost a hundred
years ago.
Combining p-adics and n-adics I call them the All-Adics as a
shortened name.
(snipped)
By year 2007 I would refine the correcting of the Peano Axioms since
they are a self-contradictory set of axioms. They are not sound and
they
are inconsistent. And I would offer a proof that the world cannot have
different types of infinity. What is lost is the notion that a set of
infinite numbers
is countable. The only infinity that exists is the infinity of the
Reals and the
Natural Numbers has to match the cardinality of the Reals.
----
Subject: WORLD'S 1st VALID PROOF OF FLT;
Message-ID: (CE320q...@dartvax.dartmouth.edu>
Date: Tue, 28 Sep 1993 21:21:57 GMT
I started the ball rolling by claiming the proof of FLT hinges on
the fact that 2+2=2x2=4 is a building block of all P-triples that
exist.
Then Karl Heuer proved that there exists at least one other number
which is unique which satisfies 4N=N^4=M. Then a quick pursuit of a
counterexample for FLT ensued. This is Karl's amazing production with
probably alot more fruit to come.
>In (278vgj$...@paperboy.osf.org> k...@dme3.osf.org writes:
>>Ludwig.Pluton...@dartmouth.edu (Ludwig Plutonium) writes:
>>>The eventual arithmetic proof of FLT, I am confident, will come
>>>from the counting numbers; P-triples are possible only in exp2
>>>because 2+2=2x2=4.
>>I have my doubts as to the connection between that equation and
>>FLT; however, you may be interested to know that other solutions
>>are possible if you allow those left-infinite decimal strings that
>>we discussed earlier. When k=4, there is a unique nonzero solution
>>to N+N+N+N = N*N*N*N = M. Here is the answer, worked out to 60
>>decimal places. You can check it by doing the arithmetic yourself,
>>right to left.
>> N = . . .8217568575974462578891103859665245689398767183
>> 82655349981184
>> M = . . .2870274303897850315564415438660982757595068735
>> 30621399924736
>>Karl Heuer k...@osf.org
William Schneeberger then produced some simple counterexamples
of FLT. However, they are infinite integers. I will reproduce his
counterexamples in a future posting.
I now see the whole picture very clearly. FLT was outstanding
because there is no proof of FLT in the general case. The general
theorem of FLT has no proof because [infinite] integers are just as
real as finite integers. All attempts at a proof of the general
equation of FLT are doomed to failure. Wiles proof is invalid. Not
only does his method violate the Fourier Theorem. But the
counterexamples clearly show that FLT is false. FLT IS FALSE.
This is the reason FLT was impossible to prove. Wiles alleged
proof is a fake. The counterexamples cannot go away.
[lines deleted]
Subject: Re: What are adics?
Date: 24 Jun 1995 04:14:48 GMT
Message-ID: (3sg3fo$...@dartvax.dartmouth.edu>
[deletions]
> OK, but we still don't know what p-adics or (the AP-generalization)
> adics are.
>From Koblitz, P-ADIC ANALYSIS: a Short Course on Recent Work, 1980,
page 7
Let p be a prime number, fixed once and for all. The "p-adic numbers"
are all expressions of the form
a_mp^m + a_m+1p^m+1 + a_m+2p^m+2 + . . ., where the a_i element
{0,1,2,...,p-1} are digits, and m is any integer. These expressions
form a field (+ and x are defined in the obvious way), which contains
the nonnegative integers n = a_0 + a_1p + ...+a_rp^r ("n written to
the
base p"), and hence contains the field of rational numbers Q.
----end of Koblitz quote -----
That is the definition of p-adics. And here is a look at three
10-adics. They look like Naturals and the third one is the familiar
number 1 only written out in full with all the zeroes leftward.
The expression a^n+b^n=c^n is true for all n, in 10-adics,
given the following values.
a= ...9977392256259918212890625
b= ...0022607743740081787109376
c= ...0000000000000000000000001
You will notice that all p-adics and n-adics (n adics are two or more
primes such as the 10-adics is 2 and 5) are INFINITE strings leftward.
--- end quoting my old posts of 1993 ---
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped to save space)
Let me comment on three concepts developed in 1993 that will follow
into this
updated 2011 proof of FLT, since I had no knowledge that mathematics
has a natural
borderline of Finite into the Infinite at 10^603.
>
> --- quoting from my old posts of circa 1993 ---
>
> PROOF OF FERMAT'S LAST THEOREM (FLT) Sorry if some of
> the symbols did not copy through in the proof.
>
> For the set of positive integers {1,2,3,4,5, . . . }, a
> given positive integral exponent n >2, the equation
> a^n + b^n = c^n is impossible to solve for positive integers
> a, b, c. I want to give a simplified idea of the statement
> of Fermat's Last Theorem. Consider the sets
> {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, . . .}
> {1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289, . . .}
> {1,8,27,64,125,216,343,512,729,1000,1331,1728,2197,2744, . . .}
> {1,24,34,44,54,64,74,84,94,104,114,124,134,144,154,164,174, . . .}
> {1,25,35,45,55,65,75,85,95,105,115,125,135,145,155,165,175, . . .}
>
With my recent successful proof of Goldbach I really need to extend
that solution set sequence because
it fails to recognize that Fermat's Last Theorem must start with exp0
and to start FLT at some
spurious numbered exponent of exp3 is one of the reasons FLT was never
validly proven. This is as important
for FLT as starting Goldbach with the concept of N/2prime-summable in
that 4 and 6 are 2prime-summable but
not 3prime-summable and that 8 is the first even number to be 3prime-
summable.
So in these difficult to prove conjectures, if you start them without
their full origins in theory, you probably
will be skunked in the end with a failure of proof.
So I need to start FLT with this solution set for a^0+b^0 = c^0:
{1, 1, 1, 1, ....}
And include that so as to get the full theory.
So that in the proof of FLT, the proof comes out of the full picture
of exponents with all its solution sets.
A full picture of the Goldbach was essential in garnering the proof.
The above is important as to why exp2 stands out as a oddball compared
to all the other
exponents. And a true and valid proof of FLT must reconcile why exp2
is different. In the
Wiles and Taylor offering, they divorce exp2 as irrelevant while they
focus on elliptic curves
for the nonexistence of special elliptic curves. Essentially their
proof format is to prove the
nonexistence of something, which when a rational person thinks about
it, is like proving the nonexistence
of ghosts and witches. Reductio ad Absurdum proofs cannot be valid for
the existence or nonexistence of
a mathematical entity. Reductio ad Absurdum are only valid by showing
a tear or a destruction of the fabric
of all of mathematics. So the Wiles and Taylor offering should have
been dismissed in the 1990s for the simple
fact of the error of logic that a Reductio ad Absurdum can never be a
valid proof when it conjures up the existence or nonexistence of a
mathematical entity. The Infinitude of Primes proof is reductio ad
absurdum that shows a
tear in the math fabric, not the existence of a Euclid prime.
But it seems as though hardly any of modern day mathematics professors
has a good logical sound mind, for you
can show them they do not understand Reductio ad Absurdum, and they do
not understand they lack a precision definition of what it means to be
finite as to being infinite.
This is why FLT for exp2 is special because of 2 as a building block
of 2^2
> The expression a^n+b^n=c^n is true for all n, in 10-adics,
> given the following values.
> a= ...9977392256259918212890625
> b= ...0022607743740081787109376
> c= ...0000000000000000000000001
>
> You will notice that all p-adics and n-adics (n adics are two or more
> primes such as the 10-adics is 2 and 5) are INFINITE strings leftward.
>
> --- end quoting my old posts of 1993 ---
Alright, I cannot write out this updated proof now but will write it
out over a stretch of time
but here is the outline.
PROOF of FLT: We have the full solution sets. We notice that in the
exp1 solution set that the first FLT
solutions of A+A= C occur first with 1+1=2. Then we have later the A
+B=C such as 1+2=3.
This is important because the A+A=C existence is far more important
than the A+B=C existence and drives
the end conclusion of the proof.
Now in this proof we must show why exp2 is so exceptional compared to
any other exponent other than exp1.
And the reason being is the qualities of the number 2 to be a building
block for A+B=C in exp2, but it is
a distraction for higher exponents.
So the proof in basic outline form is that if exp3 or higher has a
solution of A+B=C, it must have a solution
of A+A=C first. Now some will carp and complain that exp2 does not
follow this rule. But what I am saying is that
if you removed all the P-triples in exp2 that were created due to the
specialness of 2 as generating 4-1, 4 and 4+1
to yield the P-triple 3,4,5, and if you removed all the multiples of
<3,4,5>, tell me, would you have any P-triples
in exp2 if all those crafted from <3,4,5> were removed?
So the trouble with FLT and why it became so horribly difficult is
that everyone was focused on exp2 and how it
behaved that they lost sight of how to gain a proof of FLT overall.
So remove the P-triples of FLT in exp2 that are caused by the
specialness of 2 and 4 with 4-1 and 4+1. We have
no more P-triples in exp2 and then exp2 looks as barren of P-triples
as does any higher exponent. And that the
square root of 2 comes into play in exp2 of FLT where there would not
be any P-triple of either form A+A=C or
the form A+B=C.
So here is the Proof in basic outline:
Exp3 and higher have to deal with a solution for A+A=C and that
involves cube root of 2 or quadratic root of
2 for exp4, etc etc. That implies that the root of 2 is rational and
not irrational hence there is no A+A=C
solution. That implies there is never a A+B=C solution for any exp
save for the specialness of 2 in exp2.
But that does not constitute a mathematical proof because there is no
contradiction, unless you pronounce that
mathematics is having a border of Finite with Infinity at 10^603. That
means we have two choices:
Either exp3 and higher do not have a solution set due to the root of 2
being irrational. Or, we have solutions
in every exp3 and higher as given by
a= ...9977392256259918212890625
b= ...
0022607743740081787109376
c= ...0000000000000000000000001
The expression a^n+b^n=c^n is true for all n given the following
values.
In the 1990s, Wiles and Taylor could argue against those a,b,c as
solutions by saying they were not finite numbers.
But neither Wiles nor Taylor can argue against it now because they are
boxed in. Boxed in because if they say
that FLT has no solutions in exp3 because that would mean cube root of
2 is rational and not irrational, they would
contradict themselves because of the above three solutions.
So clearly, they are boxed in becuase of their lack of a precision
definition of what finite and infinite means.
Taylor and Wiles never define finite and infinite so if they pick cube
root of 2 is irrational because it is an
"infinite string" as the definition of not finite, then they
contradict themselves by repulsing the above solutions
for they are also an "infinite string".
So the real valid proof of FLT is that you need the border to put an
end to the contradiction. The border at 10^603
reconciles the fact that FLT is true for all exponents greater than 2,
given that math ends at 10^603.
So here again, the key element to proving both Goldbach and FLT is
that borderline.
adamk
Gee, what a hypocrite: you _repeatedly_, and _without absis_ put question the integrity of those
who disagree with you; you also call them inept
buffoons --not always to their face--if not worse,
just for disagreeing with you.
Never mind that you come in here to a _Mathematics_
site, and you tell us that you are here to correct us.
What a condescending asshole!. Oh yes, o superior one,
I have been doing Mathematics for years now, but I
am too simple for thy highness. Oh, please, fat-assed
unemployed highness, without a degree, without a single
published paper, with repeated baseless claims: please
enlighten me!
You have been claiming to correct mathematics
indefinitely, without _ever_ clearly showing
nor producing a _single_ counterexample. Do you
believe that puts you in a position of condescending
and correcting others? It just makes you look like
a hypocrite, or just a lunatic.
And in return, you expect respect.
Archimedes Plutonium
Outline of proof of Fermat's Last Theorem which is extendable to
Beal's Conjecture also and will
do Beal's separately with time permitting.
Outline of Proof:
Let me broadly outline the proof. We have to take all of the solution
sets into account starting with exp0
and then only working our way to the exp3 and higher. The grave
problem in proving FLT is that you need
a precision definition of Finite versus Infinite with that borderline,
for without it the contradiction makes little
sense. And the other problem was that the Pythagorean triples found in
exp2 have become more of a distraction
in proving FLT rather than any sort of help. If exp2 had never had P-
triples, I bet Fermat himself would have found
the "trick of the proof" that is the mechanism that burnishes the
proof, specifically that the root of 2 is irrational always
versus rational (with ill defined infinity). That mechanism root of 2
is irrational coupled with the idea that
a P-triple is a A+B = C in a solution-set, but we can also expect a A
+A=C in the solution-set. And we can expect a
A+A= C before we have a A+B=C. Now here I borrowed from the Goldbach
proof. Imagine for a moment that someone requested you to only allow
distinct primes in Goldbach so that 5+5=10 is not Goldbachian but
rather 3+7=10 is
Goldbachian so that A+A=C is not allowed but A+B=C is allowed. Imagine
if that requirement had been set down. Then Goldbach Conjecture would
have to be restated
and where the starting even number is no longer 4 and larger but
starts with 8. Same thing with FLT, in that we are all
accustomed to looking for a assume that A+B=C is true, when we should
be placing the Reductio Ad Absurdum of
FLT by saying Suppose there is a A+A=C in any solution set from exp3
and larger.
So you can anticipate how the proof is going to go. Assume there
exists a A+A=C in exp3 and higher. Implies that
you have 2(a^3) = c^3. This implies that you have the cube root of 2
and that you have a rational formed from the cube root of 2 and thus a
integer from the cube root of 2 providing a integer of c. This is
impossible. Same thing for the higher
exponents in transforming the root of 2 as irrational to be a integer.
But then the questions of the Proof of FLT will swirl around the fact
that FLT has P-triples in exp2 yet no A+A=C in exp2. This is not going
to be a fanciful show here with exp2. The idea is that the P-triples
in exp2 arise not from why
there are no P-triples in higher exponents but arises out of a
pecularity of the number 2 as it is equal under addition and
multiplication as thus 2+2 = 2x2 and no other number save 0 does that.
So the P-triples in exp2 are not legitimate P-triples but arise only
out of that pecular feature of the unique number 2. And to bear
witness to this illegitimate P-triples in exp2 is the fact that the
only P-triples in exp2 all arise from <3,4,5> and taking multiples of
that P-triple. There are no
other P-triples in exp2 except for those produced initially from
<3,4,5>. At least I hope so, or else this is going to be my
third big mistake this week.
So if we ignored all the P-triples in exp2 because they arise from the
peculiar strange behaviour of the number 2, if we ignore all those P-
triples in exp2, then exp2 is the same as exp3 in that the square root
of 2 for 2(a^2) = c^2, makes a irrational number of the square root of
2 furnishing a integer equal to c^2, and clearly impossible.
So the reason that FLT has no P-Triples in exp3 and higher is that A
+A=C forces a irrational number of the root of 2
to be turned into a actual integer.
Solution sets:
exp0 =
{1,1,1,1,...}
exp1 =
{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, . . .}
exp2 =
{1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289, . . .}
exp3 =
{1,8,27,64,125,216,343,512,729,1000,1331,1728,2197,2744, . . .}
exp4 =
{1,16, 81,256, 625, 1296, 2401,. . .}
From this post of 1993 where I was discussing the unique feature of
the number 2 as a building-block
of forming P-triples. But if we ignore those P-triples and focus on P-
triples in exp2 not formed from
<3,4,5> there are none, and the reason is the same as in exp3, because
having a 2(a^2) = c^2 means
the square root of 2 is turned into a integer when it is irrational.
(snipped)
Now I come to the end stage of the proof. It is Reductio ad Absurdum
and requires a contradiction.
The contradiction is obvious that if you suppose FLT false then there
exists a A+B=C, and implies
there must exist a A+A= C which is impossible since it forces a
irrational number of the root of 2
to be a rational number.
But the proof is not over with since FLT with ill-defined infinity
does have P-triples in all exponents
higher than 2 and even exp2 with ignoring the <3,4,5> based P-triple
has these solutions:
>
> > The expression a^n+b^n=c^n is true for all n,
> > given the following values.
> > a= ...9977392256259918212890625
> > b= ...0022607743740081787109376
> > c= ...0000000000000000000000001
>
>
> PROOF of FLT: We have the full solution sets. We notice that in the
> exp1 solution set that the first FLT
> solutions of A+A= C occur first with 1+1=2. Then we have later the A
> +B=C such as 1+2=3.
>
When I write the full proof it will have to start with showing how A
+A=C precedes
A+B=C
>
> Now in this proof we must show why exp2 is so exceptional compared to
> any other exponent other than exp1.
> And the reason being is the qualities of the number 2 to be a building
> block for A+B=C in exp2, but it is
> a distraction for higher exponents.
>
> So the proof in basic outline form is that if exp3 or higher has a
> solution of A+B=C, it must have a solution
> of A+A=C first. Now some will carp and complain that exp2 does not
> follow this rule. But what I am saying is that
> if you removed all the P-triples in exp2 that were created due to the
> specialness of 2 as generating 4-1, 4 and 4+1
> to yield the P-triple 3,4,5, and if you removed all the multiples of
> <3,4,5>, tell me, would you have any P-triples
> in exp2 if all those crafted from <3,4,5> were removed?
>
At least I am banking on that to be true, or else I have made that
third
big mistake this week.
> So the trouble with FLT and why it became so horribly difficult is
> that everyone was focused on exp2 and how it
> behaved that they lost sight of how to gain a proof of FLT overall.
>
So here is the contradiction at the end of FLT proof:
>
> Either exp3 and higher do not have a solution set due to the root of 2
> being irrational. Or, we have solutions
> in every exp3 and higher as given by
>
> a= ...9977392256259918212890625
> b= ...0022607743740081787109376
> c= ...0000000000000000000000001
>
> The expression a^n+b^n=c^n is true for all n given the following
> values.
So what is the problem here at the end? The problem is that, although
we have a contradiction,
we have a vicious circle of contradiction. FLT is true because of root
of 2 but FLT is false
because there is always a solution P-triple as shown above.
The way to navigate out of this circle-of-contradiction is to realize
that never a clear precise
definition of Finite versus Infinite was given since Fermat to Wiles
and everyone in between.
So when we define Finite versus Infinite as the border of 10^603 since
that is the last place that
a finite perimeter is matched with a finite circumference. Then the
FLT proof is finally laid to rest
as true for all integers from 0 to 10^603. Beyond 10^603, FLT is iffy
since Aristotelian Logic stops
at the border and where quantum-duality logic takes over. Mathematics
as a subject of precision,
itself stops at that border, and beyond is where Physics takes over.
Mathematics is governed by Aristotelian Logic, the engine of
mathematics, and it allows precision
definitions. So once we reach the border between Finite and Infinity
we cease doing mathematics.
So when we say in mathematics "this is infinite" what we mean is that
it goes up to 10^603 and
passes over that border and that there are 10^603 such items. The
primes in math are infinite
because they go up to 10^603 and cross over into that infinity
territory and that there are 10^603
primes that can be formed into a set.
Archimedes Plutonium
<plutonium.archime...@gmail.com> wrote:
(snipped)
>
> > So the proof in basic outline form is that if exp3 or higher has a
> > solution of A+B=C, it must have a solution
> > of A+A=C first. Now some will carp and complain that exp2 does not
> > follow this rule. But what I am saying is that
> > if you removed all the P-triples in exp2 that were created due to the
> > specialness of 2 as generating 4-1, 4 and 4+1
> > to yield the P-triple 3,4,5, and if you removed all the multiples of
> > <3,4,5>, tell me, would you have any P-triples
> > in exp2 if all those crafted from <3,4,5> were removed?
>
> At least I am banking on that to be true, or else I have made that
> third
> big mistake this week.
>
I should look up these things before I speak about them.
From Wikipedia it says:
There are 16 primitive Pythagorean triples with c ≤ 100:
( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17)
( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85)
(16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65)
(36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)
So I have to retrench that part of the proof. I need to reconcile
why exp2 has P-triples yet no A+A=C and thus 2(a^2) = c^2
But there is one interesting feature that I noticed that reminds me
of the 2nd mistake I made in a week-- 1^3 + 1000..00^3 = 1000..01^3
Seeing those (11, 60, 61) that a P-triple in exp3 of similar nature
is not farfetched, especially with p-adics and AP-adics and even a
large number with thousands of digits.
So this portion of the proof needs explanation. It is not a fatal flaw
but it needs some tie in to the premiss that A+A=C needs to come
before
A+B=C.
This is why FLT is such a hard problem for it seems to always bounce
back
and bite its authors.