Discussion:
The non existence of p'th root of any prime number, for (p>2) prime
bassam king karzeddin
2017-02-19 11:30:46 UTC
Why does the trustiness of Fermat's last theorem implies directly the non existence of the real positive arithmetical p'th root of any prime number
($\sqrt[p]{q}$)?

Where (p) is odd prime number, and (q) is prime number

It is an easy task for school students NOW!

Regards
Bassam King Karzeddin
19/02/17
a***@gmail.com
2017-02-19 18:19:43 UTC
wTf is that supposed to mean?
Post by bassam king karzeddin
($\sqrt[p]{q}$)?
Where (p) is odd prime number, and (q) is prime number
a***@gmail.com
2017-02-20 05:02:38 UTC
the geometrical mean of two primes, their product,
could be of some interest ... so, go for it
Post by bassam king karzeddin
($\sqrt[p]{q}$)?
Where (p) is odd prime number, and (q) is prime number
a***@gmail.com
2017-02-20 05:24:40 UTC
of course, the geometric mean of a twin prime is an even number;
what is the thirdr00t of a hundred and five?
Post by bassam king karzeddin
($\sqrt[p]{q}$)?
a***@gmail.com
2017-02-20 05:27:07 UTC
hah, 00ps secondr00t(143) is not integral.
Post by a***@gmail.com
of course, the geometric mean of a twin prime is an even number;
what is the thirdr00t of a hundred and five?
Post by bassam king karzeddin
($\sqrt[p]{q}$)?
a***@gmail.com
2017-02-20 08:51:12 UTC
I was trying to find if there were pairs (6n -\+ 1) that were both not prime,
but I realized that that is about the same as (6n -\+ 5),
and found a couple of pairs that were both nonprimal, so
bassam king karzeddin
2017-02-20 17:32:25 UTC
Post by a***@gmail.com
of course, the geometric mean of a twin prime is an even number;
what is the thirdr00t of a hundred and five?
Post by bassam king karzeddin
($\sqrt[p]{q}$)?
Post by a***@gmail.com
what is the thirdr00t of a hundred and five?
It can be written in mathematics as this

$\sqrt[3]{105} = \sqrt[3]{3}*\sqrt[3]{5}*\\sqrt[3]{7}$, where

$\sqrt[3]{3} \neq 1.44224957...$
$\sqrt[3]{5} \neq 1.7099759466...$
$\sqrt[3]{7} \neq 1.9129311827...$
$\sqrt[3]{105} \neq 4.7176939803...$

But mathematics replace the non equality notation (\neq) with (=), but at their fake paradise (infinity), (so innocent mistake they made!)

And by the way, there is not any other way to represent them geometrically on a straight line exactly, (I assume you know this impossibility)

But do they really finish at the paradise of mythematickers,

Of course not at all, since the operation is endless (by definition)

From this point of view, anyone can invent numbers so easily, just assumed it in mind, that is all the trick, who cares

So, unlike the Greek, when truly found new numbers (that are not at all rationals), and could exactly locate it even without measurements in rationals, that was indeed the truly number revolution, since backed with rigorous proof by the greatest theorem

But here, we did everything (APPROXIMATELY) just by fool guessing and for business purpose ONLY

Did you see the so huge difference?

So, my concern here is only the ODD prime root of a prime number (which is a fake non existing number) except in the finest minds (supposedly)! wonder

Regards
Bassam King Karzeddin
20/02/17
b***@gmail.com
2017-02-20 17:56:05 UTC
Nope the below is not correct.
Correct are the following two statements:

Let D_n be the decimal representation of 105^(1/3) up
to n digits. Let D be the limes lim_n->oo D_n. We then have:

105^(1/3) =\= D_n for each n

105^(1/3) = D

Be careful with the use of the ellipses. It means
limes, so you can easily make wrong math statements,

and run into contradictions. For example 105^(1/3)
=/= D would imply:

0 =/= 0

Which is nonsense.
105^(1/3) =\= 4.7176939803...
bassam king karzeddin
2017-02-21 08:58:28 UTC
Post by b***@gmail.com
Nope the below is not correct.
Let D_n be the decimal representation of 105^(1/3) up
105^(1/3) =\= D_n for each n
105^(1/3) = D
Be careful with the use of the ellipses. It means
limes, so you can easily make wrong math statements,
and run into contradictions. For example 105^(1/3)
0 =/= 0
Which is nonsense.
105^(1/3) =\= 4.7176939803...
Why people here ignore deliberately what had been taught and proved so easily earlier in sci.math and elsewhere?

Especially you insect brain brusegan!, who had great chance to understand the basic MISTAKE on those fiction numbers while discussion with me

Did not I PUBLISH the following obvious self proved inequality in the whole non zero integer numbers?

Or do you need permit ion on its absolute validity from your famous corrupted Journals and Universities?

Applying your own notation for ( D_n) in That famous INEQUALITY I posted earlier:

Iff: (D_n)^3 < 105*(10)^{3n} <( D_(n+1))^3, where then, (D_n) represents the rational decimal approximation of the cube root of (105), ($\sqrt[3]{105}$), with (n) number of accurate digits after the decimal notation (in 10base number system)

Then there is a positive integer (k(n)), where we can write this INEQUALITY as this:

(D_n)^3 + k(n) = 105*(10)^{3n}, where k(n) is increasing indefinitely when (n) increase

But what a professional mathematician do exactly, is neglecting (k(n)) completely and equating it with zero, where then it becomes so easy approximation for them to call it as real irrational number as this:

(D_n)^3 = 105*(10)^{3n}, and hence, $(D_n)/10^n = \sqrt[3]{105}$, (happy end at the fool’s paradise - infinity) with (REAL CHEATING), whereas the obvious fact that:

(D_n)^3 =/= 105*(10)^{3n}, and thus no real cube root exists for (105), and for sure, but there is another easy proof from Fermat’s last theorem too.

Note this is not a claim just because those described numbers are indeed impossible construction, but because they are not existing at all to be constructed (very easy logic - brains)

Many proofs can be so easily made to this famous illegal numbers, but this would certainly ruin the ambitious dreams of the professional mathematicians of continuously making many of alike fake numbers.

And certainly they would ignore completely what had been taught here, since it is not published in a reputable Journal, or from a very well known mathematicians from so famous universities

The work is already PUBLISHED, whether they like it or dislike it, and the professionals certainly would find it interesting (but not in this century)

So, this is only to document and paint with shame the selfishness, the stupidity and the ill behaviours of the mainstream professional mathematicians who care a lot about their science (by ignoring the facts)

May be they wanted to do it themselves and away from a mature who is teaching them tirelessly

Regards
Bassam King Karzeddin
21/02/17
b***@gmail.com
2017-02-21 10:04:36 UTC
How many times did I tell you, it is not "fiction
numbers", its it Unicorn numbers. When do you get it?

Am Dienstag, 21. Februar 2017 09:58:41 UTC+1 schrieb bassam king karzeddin:
bassam king karzeddin
2017-02-22 08:07:15 UTC
Post by b***@gmail.com
How many times did I tell you, it is not "fiction
numbers", its it Unicorn numbers. When do you get it?
So, still I would not follow your description (Unicorn numbers), since (in other thread you call (sqrt(2)) as Unicorn number, where I do not

So, your interpretation is completely different from mine and regardless who is right

Did you personally or anyone else introduce those numbers and WHEN? with EXACT DATES?
BK
bassam king karzeddin
2017-02-20 12:58:09 UTC
Post by a***@gmail.com
wTf is that supposed to mean?
Post by bassam king karzeddin
($\sqrt[p]{q}$)?
Where (p) is odd prime number, and (q) is prime number
($\sqrt[p]{q}$) is the real positive (supposed!) arithmetical p'th root of the prime number (q), where actually non exists (including all those alleged complex roots with imaginary terms)

So, imagine how fictional is your mathematics, (so unbelievable), and strangely it is still working so smoothly in the supposed finest heads!

I really can not believe it, nor wanting to believe it, but alas it is a perpetual fact indeed!

I had tried all the means to rescue it, but so unfortunately there is not any way (for sure)

But, I finally realized that facts are more better even they are so bitter

And who can stand before the absolute facts then? wonder!

Regards
Bassam King Karzeddin
20/02/17
t***@gmail.com
2017-02-20 17:00:42 UTC
it l00ks like the secondr00t of p times q, but
you have not bothered to use even that, so ...
why should 2 = p, be a pth r00t?
Post by bassam king karzeddin
Post by a***@gmail.com
wTf is that supposed to mean?
($\sqrt[p]{q}$) is the real positive (supposed!) arithmetical p'th root of the prime number (q), where actually non exists (including all those alleged complex roots with imaginary terms)
a***@gmail.com
2017-02-21 20:47:32 UTC
apparently, bssm uses skwarer00t3 to indicate thirdr00t,
which is rather sylli, but it seems to be the format
of some kind of symbolware that he is using, or
perhaps he programmed it
Post by t***@gmail.com
it l00ks like the secondr00t of p times q, but
you have not bothered to use even that, so ...
why should 2 = p, be a pth r00t?
Post by a***@gmail.com
wTf is that supposed to mean?
bassam king karzeddin
2017-02-22 07:31:34 UTC
Post by a***@gmail.com
apparently, bssm uses skwarer00t3 to indicate thirdr00t,
which is rather sylli, but it seems to be the format
of some kind of symbolware that he is using, or
perhaps he programmed it
Post by t***@gmail.com
it l00ks like the secondr00t of p times q, but
you have not bothered to use even that, so ...
why should 2 = p, be a pth r00t?
Post by a***@gmail.com
wTf is that supposed to mean?
I never used the square root operation, I used the CUBE ROOT, or more generally the arithmetical

(P'th) ROOT OF Q, where (P) is ODD PRIME NUMBER, and (Q) is prime number

Most likely you do not understand my notations, then how can you go further I wonder!

Then I CLAIM WITH PROOFs, THIS (P'TH) ROOT DOES NOT EXIST, SINCE IT IS AN INVENTED NUMBER ONLY, (NOT ANY REAL NUMBER),

I also suggested to call them UNREAL NUMBERS just to save the mathematicians from the biggest scandals ever made in the history of mankind

OR FAKE numbers, or non existing numbers, or invented (not discovered) numbers,

Or PARADISE numbers, (since they were cooked up at the smelling kitchen of the paradise of mathematicians called (infinity)

For get about that mad insect brain brusegan calling them unicorn, since I caught him GUILTY when he describes (sqrt(2)) as unicorn number

Since, almost everybody knows about the well established numbers in mathematics up to this date

Gohn Gabriel had made tremendous efforts by recognising most of the unreal numbers, where I also and independently recognised their illusionary existence since many years (refer to my old posts in this regards)

After all, who would believe in a number that starts on earth and never ending even at the FOOL'S paradise of the BIG STUPIDS called (INFINITY),

Not even a layperson would believe this OBVIOUS FICTION STORY, but a MORON or a CRANK or a BIG STUPID or a PROFESSIONAL MODERN MATHEMATICIAN would certainly considerS them as (REAL NUMBERS), WONDER!

NO NUMBER EXISTS WITH ENDLESS TERMS MORONS, and (FOR SURE)

Regards
Bassam King Karzeddin
22/02/17
a***@gmail.com
2017-02-22 08:39:15 UTC
I think that it is just your English, but
it is just as likely a lack in your birthlanguage, and
you should probably attend to that, firstly.

of course, this is a commonplace on usenet etc.,
for reasons that are a)
various or\and b)
obvious!
Post by bassam king karzeddin
Not even a layperson would believe this OBVIOUS FICTION STORY, but a MORON or a CRANK or a BIG STUPID or a PROFESSIONAL MODERN MATHEMATICIAN would certainly considerS them as (REAL NUMBERS), WONDER!
NO NUMBER EXISTS WITH ENDLESS TERMS MORONS, and (FOR SURE)
Regards
Bassam King Karzeddin
22/02/17
bassam king karzeddin
2017-02-22 08:49:00 UTC
Post by a***@gmail.com
I think that it is just your English, but
it is just as likely a lack in your birthlanguage, and
you should probably attend to that, firstly.
of course, this is a commonplace on usenet etc.,
for reasons that are a)
various or\and b)
obvious!
Post by bassam king karzeddin
Not even a layperson would believe this OBVIOUS FICTION STORY, but a MORON or a CRANK or a BIG STUPID or a PROFESSIONAL MODERN MATHEMATICIAN would certainly considerS them as (REAL NUMBERS), WONDER!
NO NUMBER EXISTS WITH ENDLESS TERMS MORONS, and (FOR SURE)
Regards
Bassam King Karzeddin
22/02/17
Why not we avoid completely the mathematics here, and start debating on language issues?
It is also the hobby of some mathematicians here (I forgot their names)

BK
Mutt Buncher
2017-02-20 00:11:26 UTC
Shut up idiot.
a***@gmail.com
2017-02-22 19:49:25 UTC
irrationals are treated arithmetically via continued fractions
-- and it is hard to do that
bassam king karzeddin
2017-02-22 20:13:07 UTC
Post by a***@gmail.com
irrationals are treated arithmetically via continued fractions
-- and it is hard to do that
But continued fractions would certainly yields the same false result, by all means of approximations

It would never realize the true NEW story of their non existence

Otherwise, prove only the existence of cube root of two, $\sqrt[3]{2}$

**Note here I do not ask to construct it exactly, because this was proved impossible construction thousands of years ago, and all the new alleged modern claims of constructions are actually clear cheating, (for sure)

BK

BK
a***@gmail.com
2017-02-23 02:26:47 UTC
daft; much more important is the secondr00t of three, or,
rather, the ssecondr00t of a third as the edge of the hexahedron;;
it is certainly just the pythagorean theorem,
or at least three of them
Post by bassam king karzeddin
**Note here I do not ask to construct it exactly, because this was proved impossible construction thousands of years ago, and all the new alleged modern claims of constructions are actually clear cheating, (for sure)
BK
BK
bassam king karzeddin
2017-02-23 08:08:05 UTC
Post by a***@gmail.com
daft; much more important is the secondr00t of three, or,
rather, the ssecondr00t of a third as the edge of the hexahedron;;
it is certainly just the pythagorean theorem,
or at least three of them
Post by bassam king karzeddin
**Note here I do not ask to construct it exactly, because this was proved impossible construction thousands of years ago, and all the new alleged modern claims of constructions are actually clear cheating, (for sure)
BK
BK
When shall you realize that $\sqrt[3]{2}$ is a notation for cubic root of two? whereas you insist to understand it as the second square root of three.

Then certainly you would go so astray in your understanding of the problem!

I also explained it in language before putting it in mathematical computer notations

Please learn the notations first, otherwise you would miss the whole issue

BK
a***@gmail.com
2017-02-23 19:02:11 UTC
I did notice, a few days ago, that it was just your sylliware
might as well, since it is totally mainstream,
to associate the regular tetragon with secondr00ting.

all that I am saying is that it is moreimportant
to understand the secondr00t of three, than the thirdr00t of 2,
because it is an important aspect of mensuration
(of the hexahedron and its dual, the octahedron).
When shall you realize that $\sqrt[3]{2}$ is a notation for cubic root of two? whereas you insist to understand it as the second secondr00t of three.
Then certainly you would go so astray in your understanding of the problem!
I also explained it in language before putting it in mathematical computer notations
Please learn the notations first, otherwise you would miss the whole issue
BK
a***@gmail.com
2017-02-25 11:58:33 UTC
rather to use the diameter of the hexahedron as unit,
whence the facet's diagonal is r00t(2/3) and
the edge is r00t(1/3); thence,
the unit-diameter octahedron has edges of r00t(1/2);
for the tetrahedron, the diameter is the edgelength, already!
all that I am saying is that it is more important
to understand the secondr00t of three, than the thirdr00t of 2,
because it is an important aspect of mensuration
(of the hexahedron and its dual, the octahedron).
bassam king karzeddin
2017-02-21 08:34:16 UTC
Post by b***@gmail.com
Nope the below is not correct.
Let D_n be the decimal representation of 105^(1/3) up
to n digits. Let D be the limes lim_n->oo D_n. We
105^(1/3) =\= D_n for each n
105^(1/3) = D
Be careful with the use of the ellipses. It means
limes, so you can easily make wrong math statements,
and run into contradictions. For example 105^(1/3)
0 =/= 0
Which is nonsense.
Am Montag, 20. Februar 2017 18:32:30 UTC+1 schrieb
105^(1/3) =\= 4.7176939803...
Why people here ignore deliberately what had been taught and proved so easily earlier in sci.math and elsewhere?

Especially you insect brain brusegan!, who had great chance to understand the basic flow on those fiction numbers while discussion with me

Did not I PUBLISH the following obvious self proved inequality in the whole non zero integer numbers?

Or do you need permit ion on its absolute validity from your famous corrupted Journals and Universities?

Applying your own notation for ( D_n) in The famous INEQUALITY:

Iff: (D_n)^3 < 105*(10)^{3n} <( D_(n+1))^3, where then, (D_n) represents the rational decimal of the cube root of (105), ($\sqrt[3]{105}$), with (n) number of accurate digits after the decimal notation (in 10base number system)

Then there is a positive integer (k(n)), where we can write this INEQUALITY as this:

(D_n)^3 + k(n) = 105*(10)^{3n}, where k(n) is increasing indefinitely when (n) increase

But what a professional mathematician do exactly, is neglecting (k(n)) completely and equating it with zero, where then it becomes so easy approximation for them to call it as real irrational number as this:

(D_n)^3 = 105*(10)^{3n}, and hence, $(D_n)/10^n = \sqrt[3]{105}$, (happy end at the fool’s paradise - infinity) with (REAL CHEATING), whereas the obvious fact that:

(D_n)^3 =/= 105*(10)^{3n}, and thus no real cube root exists for (105), and for sure, but there is another easy proof from Fermat’s last theorem too.

Regards
Bassam King Karzeddin
21/02/17
b***@gmail.com
2017-02-21 12:22:56 UTC
Nope,

a positive decimal representation approaches a real
from below, so you don't have an interval enclosing
the irrational number, thats B.S.

What we have is simply:

D_n = floor((105)^(1/3)*10^n)/10^n.

From this follows immediately:

D_n =< D_n+1 for each n

D_n =< (105)^(1/3) for each n

We only have D_n =\= (105)^(1/3) for each n, since
(105)^(1/3) is an irrational number, whereas D_n
for each n is a rational number.

Do you have any clue what a decimal representation is?
Or are you just even more stupid and clueless than
bird brain John Gabriel birdbrains?
Post by bassam king karzeddin
Iff: (D_n)^3 < 105*(10)^{3n} <( D_(n+1))^3, where then, (D_n) represents the rational decimal of the cube root of (105), ($\sqrt[3]{105}$), with (n) number of accurate digits after the decimal notation (in 10base number system)
b***@gmail.com
2017-02-21 12:28:31 UTC
Post by b***@gmail.com
We only have D_n =\= (105)^(1/3) for each n, since
(105)^(1/3) is an irrational number, whereas D_n
for each n is a rational number.
Well we could also have D_n =\= r, for a rational r, when
r doesn't have an exact finite representation. For example
0.333... = 1/3, here D_n =\= r, although r rational.

But when r is irrational, we know for sure D_n =\= r,
irrespective of the integer basis. If we consider non-
integer basis, what Erdös et al. did, things are different.
bassam king karzeddin
2017-02-22 08:37:37 UTC
Post by b***@gmail.com
Nope,
a positive decimal representation approaches a real
from below, so you don't have an interval enclosing
the irrational number, thats B.S.
This is good point you said,
Post by b***@gmail.com
APPROACHES FROM BELOW
what is that supposed to mean MORON?

Of course you mean it would substitute the equality sign (=), but never for sure, since every well established notation means something unique
Post by b***@gmail.com
D_n = floor((105)^(1/3)*10^n)/10^n.
FLOOR!, so great it might mean (=), why not BIG MORON?
Post by b***@gmail.com
D_n =< D_n+1 for each n
D_n =< (105)^(1/3) for each n
and where is k(n) gone BIG STUPID, did not you see it growing indefinitely as (n) increases indefinitely

You did not even bothered (as usual) to include the context, just to hide the issue, as if anyone reads your nonsense defence is so stupid as you here

What immediately follows is your infinite stupidity, by IGNORING and NEGLECTING an essential term and foolishly considering it zero (k(n) = 0), which is obviously a clear cheating that a layperson can recognize
Post by b***@gmail.com
We only have D_n =\= (105)^(1/3) for each n, since
(105)^(1/3) is an irrational number, whereas D_n
for each n is a rational number.
And when shall your D_n TURNS irrational number (MAGICALLY)?

OK, I would provide you with a memory to store your digits, where you can store every trillion digitS only in one (mm) cube, the I would offer you a trillion galaxy size to store your endless digits, and once you consume them (you run out of memory) then I would offer you many more, but note that after consuming your memory then your number is only RATIONAL, IS NOT IT?

It is for every RATIONAL number MORON, when shall you get it big stupid?
Post by b***@gmail.com
Do you have any clue what a decimal representation is?
Or are you just even more stupid and clueless than
bird brain John Gabriel birdbrains?
Maybe you have still all the clues to all the problems, and frankly you do not have a section of John Gabriel intelligence yet, but the problem that you do not know (for sure)
Post by b***@gmail.com
Post by bassam king karzeddin
Iff: (D_n)^3 < 105*(10)^{3n} <( D_(n+1))^3, where then, (D_n) represents the rational decimal of the cube root of (105), ($\sqrt[3]{105}$), with (n) number of accurate digits after the decimal notation (in 10base number system)
Enjoy your infinite stupidity MORONS, WITH endless fake numbers

Bassam King Karzeddin
22/02/17
b***@gmail.com
2017-02-22 08:54:58 UTC
Yes the english is even worse than mine, or his writing
is full of logical errors. I guess he means "your D_n turn
irrational" and not "your D_n turns irrational".

Since D_n is an indexed object, its basically the sequence
D_1, D_2, ... . So they are many, plural and hence I would
prefer the verb form "turn" and not "turns".

Answer is simple never, the statement is:

D_n =\= 105^(1/3) for each n

Has the the "for each n" qualification. So its always the
case that D_n =\= 105^(1/3) for any n. So we have
D_1 =\= 105^(1/3), D_2 =\= 105^(1/3), ...

I said this already. I also defined the D as lim n->oo
D_n, D is not some "last D_n", its a new object. We have
D = 105^(1/3). Learn some math man!
Post by bassam king karzeddin
And when shall your D_n TURNS irrational number (MAGICALLY)?
b***@gmail.com
2017-02-22 09:02:19 UTC
Also grammatically not only semantically, "turn"
is better than "turns", since the verb was preceeded

by a modal verb, "shall". So the verb needs to
be in indefinite form anyway.

Future simple tense - will and shall

Post by b***@gmail.com
Since D_n is an indexed object, its basically the sequence
D_1, D_2, ... . So they are many, plural and hence I would
prefer the verb form "turn" and not "turns".
bassam king karzeddin
2017-02-22 09:41:02 UTC
Post by b***@gmail.com
Yes the english is even worse than mine, or his writing
is full of logical errors. I guess he means "your D_n turn
irrational" and not "your D_n turns irrational".
Since D_n is an indexed object, its basically the sequence
D_1, D_2, ... . So they are many, plural and hence I would
prefer the verb form "turn" and not "turns".
D_n =\= 105^(1/3) for each n
Has the the "for each n" qualification. So its always the
case that D_n =\= 105^(1/3) for any n. So we have
D_1 =\= 105^(1/3), D_2 =\= 105^(1/3), ...
I said this already. I also defined the D as lim n->oo
D_n, D is not some "last D_n", its a new object. We have
D = 105^(1/3). Learn some math man!
And finally you admit that D = 105^(1/3), after so much explanations, and of course never any possible decimal representations,

But, note that your notation are only symbol in mind, where also it is impossible to represent it exactly on a straight line (as the case of sqrt(2)), (Here I assume you comprehend what do I mean exactly)

And if you still insists that it is at your paradise (infinity):

and its representation is as you describe (D_n/10^n), as (n-->infinity)

Where here in our example $\sqrt[3]{105} = D_n/10^n$, where (D_n) is integer with (n+1) sequence of digits,

So it is a ratio of two integers where each integer with infinite sequence of digits, right?

But unfortunately this is not allowed in the holy grail principles of your mathematics, is not it?

Then your number is a meaningless number, got it?

So, keep it symbolically assumed in mind only, and never call it a real number anymore

And you would never understand what do I mean, (for sure)

Because, the whole issue it was illegally well established or decided (that is the whole point)

But fiction is fiction, no matter if it stays for thousands years

BK
Post by b***@gmail.com
Post by bassam king karzeddin
And when shall your D_n TURNS irrational number (MAGICALLY)?
b***@gmail.com
2017-02-22 15:02:28 UTC
I never wrote $\sqrt[3]{105} = D_n/10^n$,
here is what I wrote:

D_n = floor((105)^(1/3)*10^n)/10^n.

D_n =\= 105^(1/3) for each n

And here is what you get for n=1..9:

n D_n
1 4.7
2 4.71
3 4.717
4 4.7176
5 4.71769
6 4.717693
7 4.7176939
8 4.71769398
9 4.717693980

You seem to have some serious reading disability.
Maybe see a doctor and get some glasses.

Or is it just your stupidity, same stupidity as
with bird brain John Gabriel birdbrains?

Am Mittwoch, 22. Februar 2017 10:41:12 UTC+1 schrieb bassam king karzeddin:
b***@gmail.com
2017-02-22 15:09:30 UTC
If you use D'_n as follows:

D'_n = floor((105)^(1/3)*10^n)

without the division /10^n, you cannot take the limes.
Post by b***@gmail.com
I never wrote $\sqrt[3]{105} = D_n/10^n$,
D_n = floor((105)^(1/3)*10^n)/10^n.
D_n =\= 105^(1/3) for each n
n D_n
1 4.7
2 4.71
3 4.717
4 4.7176
5 4.71769
6 4.717693
7 4.7176939
8 4.71769398
9 4.717693980
You seem to have some serious reading disability.
Maybe see a doctor and get some glasses.
Or is it just your stupidity, same stupidity as
with bird brain John Gabriel birdbrains?
b***@gmail.com
2017-02-22 15:17:07 UTC
Last remark, of course you get nevertheless:

D'_n/10^n =/= 105^(1/3) for each n

Note the floor() in the definition of D'_n. On
the other hand if you omit the floor, you get
something completely different:

D"_n = 105^(1/3)*10^n

And then you have of course:

D"_n/10^n = 105^(1/3) for each n

Hope this helps.
Post by b***@gmail.com
D'_n = floor((105)^(1/3)*10^n)
without the division /10^n, you cannot take the limes.
Post by b***@gmail.com
I never wrote $\sqrt[3]{105} = D_n/10^n$,
D_n = floor((105)^(1/3)*10^n)/10^n.
D_n =\= 105^(1/3) for each n
n D_n
1 4.7
2 4.71
3 4.717
4 4.7176
5 4.71769
6 4.717693
7 4.7176939
8 4.71769398
9 4.717693980
You seem to have some serious reading disability.
Maybe see a doctor and get some glasses.
Or is it just your stupidity, same stupidity as
with bird brain John Gabriel birdbrains?
bassam king karzeddin
2017-02-22 16:10:16 UTC
Post by b***@gmail.com
D'_n/10^n =/= 105^(1/3) for each n
Note the floor() in the definition of D'_n. On
the other hand if you omit the floor, you get
D"_n = 105^(1/3)*10^n
D"_n/10^n = 105^(1/3) for each n
Hope this helps.
Post by b***@gmail.com
D'_n = floor((105)^(1/3)*10^n)
without the division /10^n, you cannot take the limes.
Post by b***@gmail.com
I never wrote $\sqrt[3]{105} = D_n/10^n$,
D_n = floor((105)^(1/3)*10^n)/10^n.
D_n =\= 105^(1/3) for each n
n D_n
1 4.7
2 4.71
3 4.717
4 4.7176
5 4.71769
6 4.717693
7 4.7176939
8 4.71769398
9 4.717693980
You seem to have some serious reading disability.
Maybe see a doctor and get some glasses.
Or is it just your stupidity, same stupidity as
with bird brain John Gabriel birdbrains?
You repeat what everyone had been ALREADY taught exactly, you can not or do not like therefore see anything that upside down the whole obvious fallacy, wonder!

Where did you hide k(n)? Why do you ignore it moron? is it indeed equals to zero?, or the ploughing is becoming an inherited art with you as many?

DID YOU NOTICE HOW INDEFINITELY K(n) goes as (n) increases indefinitely?

Do you want me to make it with numerical examples?

And who can stand still against INTEGER examples?

But still you would so simply refuse to comprehend, since comprehension the obvious truth is so painful for the expert professionals (for sure)

because what else can you do then?, nothing I suppose, but continuous denial

BK
b***@gmail.com
2017-02-22 18:22:55 UTC
Plesee define k(n) more closely. It cannot be
rational or integer, if it has to do with:

D_n - 105^(1/3)

You cannot define k(n) as a rational or integer
since D_n is rational (in my definition)
and 105^(1/3) is irrational (you say so and
I sayso), so as a result there is no

k(n), which would be rational or integer,
if it were so, this would disprove that
103^(1/3) is irrational. On the other hand
you could try:

D_n^3 - 105

And show that this is negative. But I already
stated the following two because of floor and
irrationality:

D_n =< 105^(1/3) for each n

D_n =\= 105^(1/3) for each n

Or together:

D_n < 105^(1/3) for each n

Which implies (raising them power 3 on both sides):

D_n^3 < 105

Which implies (subtracting 105 on both sides):

D_n^3 - 105 < 0

So you wouldn't tell me anything new. Basically it
deepens the impression that you are damned stupid,
similar to bird brain John Gabriel bird brains.

Am Mittwoch, 22. Februar 2017 17:10:22 UTC+1 schrieb bassam king karzeddin:
b***@gmail.com
2017-02-22 18:31:37 UTC
There are now two ways to define a k(n) resp K(n), which
we would know negative non-zero, or if you change sides
as follows positive non-zero:

This one would be rational:

k(n) = 105 - D_n^3 > 0

This one would be integer:

K(n) = (105 - D_n^3)*10^(3*n) > 0

Please note the factor 10^(3*n), you cannot make it integer
only with a factor 10^n, but you know that already. For
the limes we can use the rational k(n), which approaches zero.

The K(n) cannot be used for the limes unmodified as is.
Post by b***@gmail.com
D_n^3 - 105 < 0
So you wouldn't tell me anything new. Basically it
deepens the impression that you are damned stupid,
similar to bird brain John Gabriel bird brains.
bassam king karzeddin
2017-02-22 18:49:04 UTC
Post by b***@gmail.com
Plesee define k(n) more closely. It cannot be
D_n - 105^(1/3)
You cannot define k(n) as a rational or integer
since D_n is rational (in my definition)
and 105^(1/3) is irrational (you say so and
I sayso), so as a result there is no
k(n), which would be rational or integer,
if it were so, this would disprove that
103^(1/3) is irrational. On the other hand
D_n^3 - 105
And show that this is negative. But I already
stated the following two because of floor and
D_n =< 105^(1/3) for each n
D_n =\= 105^(1/3) for each n
D_n < 105^(1/3) for each n
D_n^3 < 105
D_n^3 - 105 < 0
So you wouldn't tell me anything new. Basically it
deepens the impression that you are damned stupid,
similar to bird brain John Gabriel bird brains.
Let us stick to our given example here about the cube root of (105), $\sqrt[3]{105}$, so

K(n) = 105*(10)^{3n} - (D_n)^3,

where (D_(n+1))^3 > 105*(10)^{3n}, (n) is positive integer representing the number of accurate digits after the decimal notation in 10base number system, and D_n, K(n) are positive integers

Can you simply show how K(n) goes indefinitely large as (n) increase indefinitely

But, if you blindly and stubbornly set (K(n) = 0), then you would get what everybody might think that is true $\sqrt[3]{105}, which is not at all BK b***@gmail.com 2017-02-22 19:02:17 UTC Permalink You must be insane, I wrote K(n) > 0, so I cannot Post by b***@gmail.com K(n) = (105 - D_n^3)*10^(3*n) > 0 Could you be anymore stupid? Are you a bird brain John Gabriel birdbrains clone? Post by b***@gmail.com But, if you blindly and stubbornly set (K(n) = 0) b***@gmail.com 2017-02-22 19:08:52 UTC Permalink BTW, I am strictly using my definition of D_n, since I introduced it and you said you adopted it. When you adopt something, you should change the meaning, thats fucking retard: D_n = floor((105)^(1/3)*10^n)/10^n. On the other hand you might now have, since you retard moron fool changed the meaning, the following: D'_n = floor((105)^(1/3)*10^n). But K(n) will be the same either way: K(n) = (105 - D_n^3)*10^(3*n) Or the other way: K(n) = 105*10^(3*n) - D'_n^3 Post by b***@gmail.com You must be insane, I wrote K(n) > 0, so I cannot Post by b***@gmail.com K(n) = (105 - D_n^3)*10^(3*n) > 0 Could you be anymore stupid? Are you a bird brain John Gabriel birdbrains clone? bassam king karzeddin 2017-02-22 20:18:45 UTC Permalink Post by b***@gmail.com BTW, I am strictly using my definition of D_n, since I introduced it and you said you adopted it. When you adopt something, you should change the D_n = floor((105)^(1/3)*10^n)/10^n. On the other hand you might now have, since you retard moron fool changed the meaning, D'_n = floor((105)^(1/3)*10^n). K(n) = (105 - D_n^3)*10^(3*n) K(n) = 105*10^(3*n) - D'_n^3 Post by b***@gmail.com You must be insane, I wrote K(n) > 0, so I cannot Post by b***@gmail.com K(n) = (105 - D_n^3)*10^(3*n) > 0 Could you be anymore stupid? Are you a bird brain John Gabriel birdbrains clone? You failed completely to understand proofs even with INTEGERS and I wonder who can resist a proof by INTEGER NUMBERS, I do not have enough time now to refute it only by INTEGERS Read my older post regarding this issue and learn how INTEGERS can defeat all the jugglers BK b***@gmail.com 2017-02-22 22:08:18 UTC Permalink By juggling you mean Gabrieloconfusion and Johno stupidics, the new religion of all cranks? Post by bassam king karzeddin Read my older post regarding this issue and learn how INTEGERS can defeat all the jugglers bassam king karzeddin 2017-02-23 08:24:55 UTC Permalink Post by b***@gmail.com By juggling you mean Gabrieloconfusion and Johno stupidics, the new religion of all cranks? I see how do you spend 48 hours a day for protecting the nonsense mathematics! So, the question, how much they pay you for that dirty work? How can you neglect K(n) to justify your results by hook or rock? Here is the INTEGER equation again K(n) = 105*(10)^{3n} - (D_n)^3, By considering (K(n) = 0), then yes your result is so simply justified and to any degree of accuracy But, can not even a school student notice this obvious juggling and big cheating Are you unable to show K(n) even for small (n) < 50, Had you noticed how shameful the grand famous mathematicians passed those obvious fiction numbers into your tiny skulls Or is it too embarrassing to talk about it? Can not you do a little trick and save the situation, but really it is too hard to beat the numbers so, show it in integers moron NUMBERS ARE THEMSELVES TALKING THE TRUTH, AND NOBODY CAN HIDE A TRUTH BY A SPIDER THREADS ANY MORE, AND FOR SURE BK Post by b***@gmail.com Post by bassam king karzeddin Read my older post regarding this issue and learn how INTEGERS can defeat all the jugglers b***@gmail.com 2017-02-23 10:13:47 UTC Permalink You must be insane for sure, I wrote K(n) > 0, so Post by b***@gmail.com K(n) = (105 - D_n^3)*10^(3*n) > 0 Best Regards for sure. Post by b***@gmail.com Post by b***@gmail.com By juggling you mean Gabrieloconfusion and Johno stupidics, the new religion of all cranks? I see how do you spend 48 hours a day for protecting the nonsense mathematics! So, the question, how much they pay you for that dirty work? How can you neglect K(n) to justify your results by hook or rock? Here is the INTEGER equation again K(n) = 105*(10)^{3n} - (D_n)^3, By considering (K(n) = 0), then yes your result is so simply justified and to any degree of accuracy But, can not even a school student notice this obvious juggling and big cheating Are you unable to show K(n) even for small (n) < 50, Had you noticed how shameful the grand famous mathematicians passed those obvious fiction numbers into your tiny skulls Or is it too embarrassing to talk about it? Can not you do a little trick and save the situation, but really it is too hard to beat the numbers so, show it in integers moron NUMBERS ARE THEMSELVES TALKING THE TRUTH, AND NOBODY CAN HIDE A TRUTH BY A SPIDER THREADS ANY MORE, AND FOR SURE BK Post by b***@gmail.com Post by bassam king karzeddin Read my older post regarding this issue and learn how INTEGERS can defeat all the jugglers bassam king karzeddin 2017-02-25 07:43:38 UTC Permalink Post by b***@gmail.com You must be insane for sure, I wrote K(n) > 0, so Post by b***@gmail.com K(n) = (105 - D_n^3)*10^(3*n) > 0 And if K(n) =/= 0, then D is greater than any possible decimal representation, thus No real cube root exists for (105), and in a more general result, "No p'th root exists for any prime number (q), where (p) is odd prime number" And still you would not like to get it, because it is so simply a very solid contradiction to what had been adopted in mathematics as real numbers and for many centuries! Post by b***@gmail.com Best Regards for sure. Bassam King Karzeddin 25/02/17 b***@gmail.com 2017-02-25 11:43:28 UTC Permalink You have the same problem as with WM schema: 1 1 2 1 2 3 ... WM says the "set limit" is |N, but none of the rows is |N. Same here: D_n =< D for each n (because D_n decimal rep) D_n =/= D for each n (because D is irrational) Or in summary: D_n < D for each n So the limes D is not one of the D_n. Whether it exists or not, depends whether your theory allow to exist it or not. You can also say it doesn't exists. Its up to you what you define Unicorn numbers to be. Post by bassam king karzeddin And if K(n) =/= 0, then D is greater than any possible decimal representation, thus No real cube root exists for (105), bassam king karzeddin 2017-02-26 16:16:52 UTC Permalink Post by b***@gmail.com 1 1 2 1 2 3 ... WM says the "set limit" is |N, but none of the rows D_n =< D for each n (because D_n decimal rep) This is big mistake, No equality (for sure) D_n < D, This is absolutely correct Post by b***@gmail.com D_n =/= D for each n (because D is irrational) Yes Post by b***@gmail.com D_n < D for each n Bravo Post by b***@gmail.com So the limes D is not one of the D_n. Whether it exists or not, depends whether your theory allow to exist it or not. You can also say it doesn't exists. Its up to you what you define Unicorn numbers to be. No, the Queen says it is not up to any body to allow its existence, No choice answer must be allowed in mathematics, but the only unique truth answer which must be obeyed blindly to avoid any fiction this is the whole point that maythematickers keep fabricating for their endless myth production It is so simple and clear shining fact and beyond any little doubt Hence, no such numbers exist but approximation to something assumed existing in mind, but physically impossible And this is the true deep meaning of non existence of a solution to some famous Diophantine equation, where generally comprehended by mathematicians BK Post by b***@gmail.com Post by bassam king karzeddin And if K(n) =/= 0, then D is greater than any possible decimal representation, thus No real cube root exists for (105), b***@gmail.com 2017-02-26 17:56:13 UTC Permalink In case you don't know how conjunction (&) works in logic/math, we have of course: A =< B & A =/= B <=> A < B A =< B <= A < B A =/= B <= A < B If this is news for you, then I have bad news for you, you don't have a clue about math. Further msth is neither Radio Ga Ga nor Radio Erivan, math works of course with axioms and different theories, you don't have D in Q, otherwise your reference to Diophantine equation doesn't make any sense. Since in R there is no such thing as Diophantine equation, for example Fermats a^3 + b^3 = c^3 has solution a,b,c<>0 in R. Again if this is news for you, then I have bad news for you, you don't have a clue about math. Diophantine equations require integers from Z, and via p/q in Q for p,q in Z, you can also look at rational problems. https://en.wikipedia.org/wiki/Diophantine_equation (Here is a receipt how you can turn a +, * equation over Q, into a +, * equation over Z: Step 1: For each Q variable x, introduce two xn,xd variables in Z and an additional equation xd <> 0. Step 2: For each combination x*y or x+y in Q, now expressed from pairs (xn,xd) and (yn,yd), express this as pairs: (xn,xd) + (yn,yd) = (xn*yd+yn*xd,xd*yd) (xn,xd) * (yn,yd) = (xn*yn,xd*yd) Step 3: For each equation x = y in Q, now expressed from pairs (xn,xd) and (yn,yd), express this as equations with new variables m and n: xn * m = yn * n xd * m = yd * n n <> 0 m <> 0 End of proceedure) Post by bassam king karzeddin No, the Queen says it is not up to any body to allow its existence, No choice answer must be allowed in mathematics, b***@gmail.com 2017-02-26 18:01:50 UTC Permalink Take for example a=1 and b=1, then c = 2^(1/3) Is a solution. Post by b***@gmail.com doesn't make any sense. Since in R there is no such thing as Diophantine equation, for example Fermats a^3 + b^3 = c^3 has solution a,b,c<>0 in R. bassam king karzeddin 2017-02-27 15:20:23 UTC Permalink Post by b***@gmail.com Take for example a=1 and b=1, then c = 2^(1/3) Is a solution. Post by b***@gmail.com doesn't make any sense. Since in R there is no such thing as Diophantine equation, for example Fermats a^3 + b^3 = c^3 has solution a,b,c<>0 in R. But R is only in mind, and mind is so absent, thus no mind exists, hence no solution even in R for c, (when a, b are non zero integers) So, this is really the bad news for you, and for sure BK bassam king karzeddin 2017-02-27 15:16:02 UTC Permalink Post by b***@gmail.com In case you don't know how conjunction (&) A =< B & A =/= B <=> A < B This is certainly not any logic, if (A =/= B), then either (A > B),or(A < B) Post by b***@gmail.com A =< B <= A < B this is also magic Post by b***@gmail.com A =/= B <= A < B I do not know who sets this nonsense logic Post by b***@gmail.com If this is news for you, then I have bad news for you, you don't have a clue about math. Further msth is neither Radio Ga Ga nor Radio Erivan, math works of course with axioms Maths works fine with axioms of morons, to generate their prior decisions only Post by b***@gmail.com and different theories, you don't have D in Q, otherwise your reference to Diophantine equation doesn't make any sense. Since in R there is no such thing as Diophantine equation, for example I swear you do not have any clue how all the true mathematics is living peacefully in complete perpetual symphony Post by b***@gmail.com Fermats a^3 + b^3 = c^3 has solution a,b,c<>0 in R. But the challenge for you and everybody else on this planet, is to show (c) in R, exactly, when (a, b) are nonzero integers! Did not I publish this challenge some days ago? What do I mean exactly, refuting your repeated well established mythematics about the fiction stories of real numbers "It is impossible to make a cube with positive real number side, equivalent to the sum or a difference of two other non zero integer sides cubes" And please note that, you must not lose the sense of exactness while dealing with Diophantine equations, denoted by (=) sign, This impossibility is actually not only mathematical, but also physical, but not in the scope of physics or mathematics yet, Post by b***@gmail.com Again if this is news for you, then I have bad news for you, you don't have a clue about math. Diophantine equations require integers from Z, and via p/q in Q for p,q in Z, you can also look at rational problems. https://en.wikipedia.org/wiki/Diophantine_equation The worst news not only for you, but also for physics and mathematics too, is that your alleged R, or even C, can not solve these puzzles (I do create), for sure Post by b***@gmail.com (Here is a receipt how you can turn a +, * equation Step 1: For each Q variable x, introduce two xn,xd variables in Z and an additional equation xd <> 0. Step 2: For each combination x*y or x+y in Q, now expressed from pairs (xn,xd) and (yn,yd), express this (xn,xd) + (yn,yd) = (xn*yd+yn*xd,xd*yd) (xn,xd) * (yn,yd) = (xn*yn,xd*yd) Step 3: For each equation x = y in Q, now expressed from pairs (xn,xd) and (yn,yd), express this xn * m = yn * n xd * m = yd * n n <> 0 m <> 0 And the Queen says nothing would defeat the obvious so sensible truth Post by b***@gmail.com End of proceedure) Post by bassam king karzeddin No, the Queen says it is not up to any body to allow its existence, No choice answer must be allowed in mathematics, Regards Bassam King Karzeddin 27/02/17 j4n bur53 2017-02-27 16:01:29 UTC Permalink No, the folllowing is of course true: A =< B & A =/= B <=> A < B Lets try some values for A and B: A | B | A =< B | A =\= B | A =< B & A =\= B | A < B --+---+--------+---------+------------------+------- 1 | 3 | Yes | Yes | Yes | Yes 2 | 2 | Yes | No | No | No 3 | 1 | No | Yes | No | No You see the last to column give the same truth values, hence we have A =< B & A =/= B <=> A < B Or do you have a counter example? Post by bassam king karzeddin Post by b***@gmail.com In case you don't know how conjunction (&) A =< B & A =/= B <=> A < B This is certainly not any logic, if (A =/= B), then either (A > B),or(A < B) Post by b***@gmail.com A =< B <= A < B this is also magic Post by b***@gmail.com A =/= B <= A < B I am waiting with any challenges until we talk about the same logic. Post by bassam king karzeddin But the challenge for you and everybody else on this planet bassam king karzeddin 2017-02-27 18:21:23 UTC Permalink Post by b***@gmail.com A =< B & A =/= B <=> A < B A | B | A =< B | A =\= B | A =< B & A =\= B | A < B --+---+--------+---------+------------------+------- 1 | 3 | Yes | Yes | Yes | Yes 2 | 2 | Yes | No | No | No 3 | 1 | No | Yes | No | No You see the last to column give the same truth values, hence we have A =< B & A =/= B <=> A < B Or do you have a counter example? Post by bassam king karzeddin Post by b***@gmail.com In case you don't know how conjunction (&) A =< B & A =/= B <=> A < B This is certainly not any logic, if (A =/= B), then either (A > B),or(A < B) Post by b***@gmail.com A =< B <= A < B this is also magic Post by b***@gmail.com A =/= B <= A < B I am waiting with any challenges until we talk about the same logic. Post by bassam king karzeddin But the challenge for you and everybody else on this planet I have plenty of counter examples with INTEGERS only Let us see how simple it is, and for simplicity, I would like to repeat it in decimal 10base number system, and once you get it, please teach and educate your teachers about it before you would certainly get brain washed and become as ignorant and a victim of old corrupted fake unnecessary mathematics ONE Example of Old claim of mine, the famous number representing the arithmetical cube root of two does not exist, denoted by (2^(1/3)), or ($\sqrt[3]{2}$) The proof: I had introduced this general self proved Diophantine equation IFF: (D(n))^p + K(n) = q*s^{n*p} < (D(n) + 1)^p where (p, q) are prime numbers, (n) is non negative integer, (s > 1) is positive integer, then this Diophantine equation implies that D(n) is positive integer that must represent the approximation of the arithmetical p’th root of q, with (n) number of accurate digits in (s) base number system, denoted by (q^(1/p)), or ($\sqrt[p]{q}$), K(n) is therefore positive integer In our chosen case here for (2^(1/3)), we have (s = 10), p = 3, q = 2, so our Diophantine equation : (D(n))^3 + K(n) = 2*(10)^{3n} < (D(n) + 1)^3 Now, we assume varying integer for (n = 0, 1, 2, 3, …, n), and we then, can easily observe how indefinitely K(n) goes as we increase (n) For (n = 0), D(0) = 1, K(0) = 1, showing the first approximation with zero digit after the decimal notation For (n = 1), D(1) = 12, K(1) = 272, showing the second approximation with one digit after the decimal notation For (n = 2), D(2) = 125, k(2) = 46875, showing the THIRD approximation with TWO digits after the decimal notation For (n = 3), D(3) = 1259, K(3) = 4383021, showing the FOURTH approximation with THREE digits after the decimal notation For (n = 4), D(4) = 12599, K(n) = 100242201, showing the fifth approximation with four digits after the decimal notation ……… ……………….. ……………………… …………………. For (n = 10), D(10) = 12599210498, K(n) = 451805284631045974008, showing the eleventh approximation with 10 digits after the decimal notation Now, can you imagine how large the existing term integer K(n) would be for large (n)? Or do you think that K(n) would become zero at infinity?, of course never, it would grow indefinitely with (n) to the alleged infinity Numbers are themselves talking the truth, and who is on earth that can face numbers with perpetual denial, even the proofers of the most famous theorems could not get it yet, WONDER! “NO NUMBER EXISTS WITH ENDLESS TERMS” This is really the true absolute meaning of the truthiness of Fermat’s last theorem! But what do the top professional mathematics do exactly, they simply consider (K(n) = zero), in order to justify the ILLEGAL real number (2^(1/3)), so they set K(n) = 0, then obviously our exact Diophantine equation is foraged to becomes as this (D(n))^3 = 2*(10)^{3n}, take cube root of both sides then you get D(n)/10^n = 2^(1/3), when (n) tends to infinity, where even this expression is not permissible in the holy grail principles of mathematics , since you would have a ratio of two integers where each of them is with infinite sequence of digits So, happy real fake numbers at the fake paradise of all the top mathematicians on earth It is also understood that those roots for (p > 2), are proved rigorously as impossible constructions in order to justify them or accept them as real existing numbers, and all the alleged claimed constructions are only mere cheating (if you investigate carefully into them) But it is well understood for practical needs as that if we want to make a cube that contains two unit cubes for instance, then this is not at all difficult even for a carpenter, we can use those approximation as much as we do like, (this is actually not mathematics works, but application works) So, the hope is held now on the clever school students mainly, whom they would understand those fiction numbers so easily, and certainly they would remove this long standing shame upon themselves and upon mathematics Regards Bassam King Karzeddin 27/02/17 b***@gmail.com 2017-02-27 18:26:19 UTC Permalink You know the meaning of the phrase "counter example"? Its a set of values, that makes a proposed conjecture FALSE. So far we have as a proposed conjecture by me (and you), D_n < D. And of course the below logical law. If you show me that D_n < D, then this is not a "counter example", this is only in support of my claim. Try again please? What do you exactly refute with a "counter example"? a) D_n < D b) A =< B & A =/= B <=> A < B c) Your grand mothers dentals Post by b***@gmail.com A =< B & A =/= B <=> A < B Or do you have a counter example? bassam king karzeddin 2017-02-27 18:57:44 UTC Permalink Post by b***@gmail.com You know the meaning of the phrase "counter example"? Its a set of values, that makes a proposed conjecture FALSE. So far we have as a proposed conjecture by me (and you), D_n < D. And of course the below logical law. If you show me that D_n < D, then this is not a "counter example", this is only in support of my claim. Try again please? What do you exactly refute with a "counter example"? a) D_n < D b) A =< B & A =/= B <=> A < B c) Your grand mothers dentals Post by b***@gmail.com A =< B & A =/= B <=> A < B Or do you have a counter example? I simply proved the nonexistence of 2^(1/3) as a real number, not because it is indeed proved rigorously as impossible construction, but because it is basically does not exist to be constructed (so simple reasoning with the proof provided) your D finally at infinity, thus it is a ratio of two integers where each integer consists of infinite sequence of digits, which DOES NOT EXIST, and the absurd logic you follow is meaning less, since it states (A =< B), this is double meaning which makes the confusion So, D is greater than any possible digital representation (absolutely), and if you list your endless representation with integer sequence, then the largest integer would be your D, but unfortunately, No largest integer exists (this is accepted by mathematicians), thus No D exists (absolutely) IT is not a big magic, mathematics dose not accept integers with endless digits generally BK b***@gmail.com 2017-02-27 19:25:01 UTC Permalink I never defined D as such, where do you read this? The sequence D_n goes on and on, never ends, there is no finally. And D is the least upper bound, a new object. I wrote this already, namely: D_n =< D (upper bound) D_n =/= D (new object) Or together: D_n < D Since D is a new object its not one of the D_n, and there is also no final D_n, since the sequence D_1, D_2, ... goes on and on. And always D_n =/= D. Please don't put words in my mouth I never said. Could you be more careful when posting. Dont post nonsense. The following I never said: - I never said there is a final D_oo among the D_n. - Consequently I never said D_oo = D, how should I say that? I cannot say that if I don't define a final D_oo. - The notation lim n->oo D_n = D, doesn't mean D_oo = D. The notation means the following: a) D_n < D, and b) for each D' (D_n < D' implies D =< D'). The notation does say there is a final D_oo, where did you get that from? You are old enough to get the correct definition from internet of lim n->oo for a sequence. Post by bassam king karzeddin your D finally at infinity, thus it is a ratio of two integers where each integer consists of infinite sequence of digits, which DOES NOT EXIST, b***@gmail.com 2017-02-27 19:30:25 UTC Permalink Lets make the n quantifier correct: a) for each n (D_n < D), and b) for each D' (for each n (D_n < D') implies D =< D'). Post by b***@gmail.com - The notation lim n->oo D_n = D, doesn't mean D_oo = D. a) D_n < D, and b) for each D' (D_n < D' implies D =< D'). The notation doesn't say there is a final D_oo, where did you get that from? You are old enough to get the correct definition from internet of lim n->oo for a sequence. bassam king karzeddin 2017-02-27 20:15:38 UTC Permalink Post by b***@gmail.com I never defined D as such, where do you read this? The sequence D_n goes on and on, never ends, there is no finally. And D is the least upper bound, a new but the whole issue that there is no least upper bound, as if you are trying to define the greatest number that is less than say one, of course does not exist Post by b***@gmail.com D_n =< D (upper bound) Please realize that your assumed D, is mainly conceptual (in mind only), not attainable (impossible), since endless, for sure, thus an obvious illusion, see the definition of an illusion please Post by b***@gmail.com D_n =/= D (new object) Thus D = NOTHINGNESS Post by b***@gmail.com D_n < D D_n is only rational, but D is nothingness, then they are not comparable Post by b***@gmail.com Since D is a new object its not one of the D_n, and there is also no final D_n, since the sequence D_1, D_2, ... goes on and on. And always D_n =/= D. So, this is actually the definition of an illusion (for sure) Post by b***@gmail.com Please don't put words in my mouth I never said. Could I am not putting words in your mouth, but in your skull Post by b***@gmail.com you be more careful when posting. Dont post nonsense. I always post common sense Post by b***@gmail.com - I never said there is a final D_oo among the D_n. The mathematics you follow says so and so clearly Even if you do not say so, but the meaning is the same Post by b***@gmail.com - Consequently I never said D_oo = D, how should I say that? I cannot say that if I don't define a final D_oo. Nobody can define D at any other cases, how can a non existence be brought to real existence? Post by b***@gmail.com - The notation lim n->oo D_n = D, doesn't mean D_oo = D. Yes, even at that (unattainable state of mind), D_oo =/= D, and D =/= TREE, OR D =/= ELEPHANT OR D =/= Dinosaur, and (for sure) Post by b***@gmail.com a) D_n < D, and b) for each D' (D_n < D' implies D =< D'). Most likely at the paradise of mythematickers only Post by b***@gmail.com The notation does say there is a final D_oo, where did you get that from? You are old enough to get the correct definition from internet of lim n->oo for a sequence. All those integer demonstrations can not convince a professional mathematicians (for sure) And all your argument were just to divert the INTEGER proof, So, refute the provided proof if you or anybody else can, instead of wasting time on what you said, or what I said Post by b***@gmail.com Post by bassam king karzeddin your D finally at infinity, thus it is a ratio of two integers where each integer consists of infinite sequence of digits, which DOES NOT EXIST, It does not matter what do you or anybody else say, to the obvious truthness of fiction numbers adopted in mathematics Also it does not matter going in the details of so alleged too advanced mathematics such as SH*T theory to prove the mathematical number 2^(1/3) Also it does not matter to use many concepts to be finally replaced with the greatest mathematical notation of so simple common sense (=), such notations (approaching fast or approaching fast, limits, Newton's approximation, famous cuts,infinity, intermediate value theorem, ..., etc), all those terms can not substitute the (=) sign, for sure, Diophantine equation are exact, The proofs are actually so many, the simplest and more than sufficient one is associating a number with infinity, where infinity it self is not a number, therefore any number associated with it is definitely not a number And since infinity is not a number by definition, then it is greater than any number (too wonderful definition), infinity Not a number but greater than any number, (toooooo funnnnnnny) Even a layperson would keep laughing upon the top mathematicians, (of course I EXCLUDE YOU), BK b***@gmail.com 2017-02-27 20:30:18 UTC Permalink You seem to have severe problems with the meaning of (=/=) and (=<). Of course also in the rationals, you have D with D_n =/= D. For example take D=5. For the decimal representation D_n of 105^(1/3), we always have D_n =/= D for D=5. So your conclusion D=NOTHINGNESS from D_n =/= D is wrong. What you would need to show is that there is no least upper bound in the rationals. How do you do this correctly? (BTW: I don't know how to do it just now, I have some ideas, but not 100% sure) Post by bassam king karzeddin Post by b***@gmail.com D_n =/= D (new object) Thus D = NOTHINGNESS b***@gmail.com 2017-02-27 20:52:21 UTC Permalink Well the proof is not that difficult: If E is the least upper bound of D_n, then E must have the same decimal representation. Define E_n = floor(E*10^n)/10^n. Now assume E_k != D_k for some k. Since E is an upper bound, it must be E_k > D_k. But then there is room for an F upper bound and below E. Just take the first D_j with j>k where the last digit is not 9, and increment it the last digit. We can assume D_n are normalized so this is possible. Hence E is not anymore least, and therefore a contradiction. The rest with impossible rational least upper bound follows here in that D_n was from an irrational, i.e. the Diophantine Argument. So we know in Q, the limes doesn't exist. The construction of R now works, in that we add all such limeses. (P.S. Wonder how this would work our in non-standard analysis). Post by b***@gmail.com You seem to have severe problems with the meaning of (=/=) and (=<). Of course also in the rationals, you have D with D_n =/= D. For example take D=5. For the decimal representation D_n of 105^(1/3), we always have D_n =/= D for D=5. So your conclusion D=NOTHINGNESS from D_n =/= D is wrong. What you would need to show is that there is no least upper bound in the rationals. How do you do this correctly? (BTW: I don't know how to do it just now, I have some ideas, but not 100% sure) Post by bassam king karzeddin Post by b***@gmail.com D_n =/= D (new object) Thus D = NOTHINGNESS bassam king karzeddin 2017-03-01 14:24:27 UTC Permalink Post by b***@gmail.com Well the proof is not that difficult: If E is the least upper bound of D_n, then E must have the same decimal representation. Define E_n = floor(E*10^n)/10^n. Now assume E_k != D_k for some k. Since E is an upper bound, it must be E_k > D_k. But then there is room for an F upper bound and below E. Just take the first D_j with j>k where the last digit is not 9, and increment it the last digit. We can assume D_n are normalized so this is possible. Hence E is not anymore least, and therefore a contradiction. The rest with impossible rational least upper bound follows here in that D_n was from an irrational, i.e. the Diophantine Argument. So we know in Q, the limes doesn't exist. The construction of R now works, in that we add all such limeses. (P.S. Wonder how this would work our in non-standard analysis). Post by b***@gmail.com You seem to have severe problems with the meaning of (=/=) and (=<). Of course also in the rationals, you have D with D_n =/= D. For example take D=5. For the decimal representation D_n of 105^(1/3), we always have D_n =/= D for D=5. So your conclusion D=NOTHINGNESS from D_n =/= D is wrong. What you would need to show is that there is no least upper bound in the rationals. How do you do this correctly? (BTW: I don't know how to do it just now, I have some ideas, but not 100% sure) Post by bassam king karzeddin Post by b***@gmail.com D_n =/= D (new object) Thus D = NOTHINGNESS And see this well exposed guy by AP, going around himself, avoiding the so obvious proof, trying hopelessly flooring the integers, then approaching the integer equation from the sky next time may be, or imagining limes or limits or so like ill imaginations This is not real analysis where you can so simply conclude whatever you like and call it a theorem, but this is EXACT analysis based on INTEGERS, where simply you can not cheat any more Integers can not be limited or floored or approached or nearly approximated Integers are not also as particles of spherical shape with radius epsilon, that you can land on safely in your dreams And indeed there is a real meaning for unsolvable Diophantine equations by integers, the meaning is most likely behind the physical or mathematical knowledge together No solution exists for a certain Diophantine equation, then obviously there is no any real existing number that can solve it also, (but only approximation, which is not mathematics at all and not much better than trial and error approximation) he would finally arrives at what is the ratio of say (1 :: oo), then he would claims it is zero, but if you ask him what is the ratio of (0 :: oo), then he would tell you the same thing, and also the same thing for the ratio of (n :: oo), not different answer from any professional, and all those from the well current established mythematics as well, WONDER! So, the provided proof here and elsewhere is the real mirror where the plenty of common professionals look into, but see long ears and bull's horns, where then get shocked and trying their best to hide the whole issue by reporting it as an abuse, instead of being a colorful front page news for a great discovery, even though it was not any great discovery but a great scandal made deliberately by Jugglers in the history of mathematics, where those Jugglers are sleeping peacefully in the history Thinking that a judgement day would never come, but the one (by definition) who is the real creator of any created existing number would not keep silent any more, by bringing them back for a final balance as he promised earlier, where then no place to hide, or no century to hide Thus, "the real number is any arbitrary existing length on a straight line" Where as rest of numbers are fabricated as Mad Man Make All of my claims were proved rigorously in my posts And all numbers were made by a Mad Man would be thrown out to the sea for the sake of the fish entertainments Regards Bassam King Karzeddin 01/03/17 John Gabriel 2017-03-01 14:34:22 UTC Permalink Post by bassam king karzeddin Post by b***@gmail.com Well the proof is not that difficult: If E is the least upper bound of D_n, then E must have the same decimal representation. Define E_n = floor(E*10^n)/10^n. Now assume E_k != D_k for some k. Since E is an upper bound, it must be E_k > D_k. But then there is room for an F upper bound and below E. Just take the first D_j with j>k where the last digit is not 9, and increment it the last digit. We can assume D_n are normalized so this is possible. Hence E is not anymore least, and therefore a contradiction. The rest with impossible rational least upper bound follows here in that D_n was from an irrational, i.e. the Diophantine Argument. So we know in Q, the limes doesn't exist. The construction of R now works, in that we add all such limeses. (P.S. Wonder how this would work our in non-standard analysis). Post by b***@gmail.com You seem to have severe problems with the meaning of (=/=) and (=<). Of course also in the rationals, you have D with D_n =/= D. For example take D=5. For the decimal representation D_n of 105^(1/3), we always have D_n =/= D for D=5. So your conclusion D=NOTHINGNESS from D_n =/= D is wrong. What you would need to show is that there is no least upper bound in the rationals. How do you do this correctly? (BTW: I don't know how to do it just now, I have some ideas, but not 100% sure) Post by bassam king karzeddin Post by b***@gmail.com D_n =/= D (new object) Thus D = NOTHINGNESS And see this well exposed guy by AP, going around himself, avoiding the so obvious proof, trying hopelessly flooring the integers, then approaching the integer equation from the sky next time may be, or imagining limes or limits or so like ill imaginations This is not real analysis where you can so simply conclude whatever you like and call it a theorem, but this is EXACT analysis based on INTEGERS, where simply you can not cheat any more Integers can not be limited or floored or approached or nearly approximated Integers are not also as particles of spherical shape with radius epsilon, that you can land on safely in your dreams And indeed there is a real meaning for unsolvable Diophantine equations by integers, the meaning is most likely behind the physical or mathematical knowledge together No solution exists for a certain Diophantine equation, then obviously there is no any real existing number that can solve it also, (but only approximation, which is not mathematics at all and not much better than trial and error approximation) he would finally arrives at what is the ratio of say (1 :: oo), then he would claims it is zero, but if you ask him what is the ratio of (0 :: oo), then he would tell you the same thing, and also the same thing for the ratio of (n :: oo), not different answer from any professional, and all those from the well current established mythematics as well, WONDER! So, the provided proof here and elsewhere is the real mirror where the plenty of common professionals look into, but see long ears and bull's horns, where then get shocked and trying their best to hide the whole issue by reporting it as an abuse, instead of being a colorful front page news for a great discovery, even though it was not any great discovery but a great scandal made deliberately by Jugglers in the history of mathematics, where those Jugglers are sleeping peacefully in the history Thinking that a judgement day would never come, but the one (by definition) who is the real creator of any created existing number would not keep silent any more, by bringing them back for a final balance as he promised earlier, where then no place to hide, or no century to hide Thus, "the real number is any arbitrary existing length on a straight line" Where as rest of numbers are fabricated as Mad Man Make All of my claims were proved rigorously in my posts And all numbers were made by a Mad Man would be thrown out to the sea for the sake of the fish entertainments Regards Bassam King Karzeddin 01/03/17 I've read some of his posts and he seems to have some moments of sanity, but then he writes the most absurd rubbish. Don't think there is a classification for AP in the ICDs. He definitely has a very high emotional IQ. Sadly, I can't say the same for his IQ. I kind of feel sorry for him, but see no purpose in trying to engage him in any way. He can be pretty annoying at times, but hey, one doesn't have to read his posts. bassam king karzeddin 2017-03-01 15:34:02 UTC Permalink Post by John Gabriel Post by bassam king karzeddin Post by b***@gmail.com Well the proof is not that difficult: If E is the least upper bound of D_n, then E must have the same decimal representation. Define E_n = floor(E*10^n)/10^n. Now assume E_k != D_k for some k. Since E is an upper bound, it must be E_k > D_k. But then there is room for an F upper bound and below E. Just take the first D_j with j>k where the last digit is not 9, and increment it the last digit. We can assume D_n are normalized so this is possible. Hence E is not anymore least, and therefore a contradiction. The rest with impossible rational least upper bound follows here in that D_n was from an irrational, i.e. the Diophantine Argument. So we know in Q, the limes doesn't exist. The construction of R now works, in that we add all such limeses. (P.S. Wonder how this would work our in non-standard analysis). Post by b***@gmail.com You seem to have severe problems with the meaning of (=/=) and (=<). Of course also in the rationals, you have D with D_n =/= D. For example take D=5. For the decimal representation D_n of 105^(1/3), we always have D_n =/= D for D=5. So your conclusion D=NOTHINGNESS from D_n =/= D is wrong. What you would need to show is that there is no least upper bound in the rationals. How do you do this correctly? (BTW: I don't know how to do it just now, I have some ideas, but not 100% sure) Post by bassam king karzeddin Post by b***@gmail.com D_n =/= D (new object) Thus D = NOTHINGNESS And see this well exposed guy by AP, going around himself, avoiding the so obvious proof, trying hopelessly flooring the integers, then approaching the integer equation from the sky next time may be, or imagining limes or limits or so like ill imaginations This is not real analysis where you can so simply conclude whatever you like and call it a theorem, but this is EXACT analysis based on INTEGERS, where simply you can not cheat any more Integers can not be limited or floored or approached or nearly approximated Integers are not also as particles of spherical shape with radius epsilon, that you can land on safely in your dreams And indeed there is a real meaning for unsolvable Diophantine equations by integers, the meaning is most likely behind the physical or mathematical knowledge together No solution exists for a certain Diophantine equation, then obviously there is no any real existing number that can solve it also, (but only approximation, which is not mathematics at all and not much better than trial and error approximation) he would finally arrives at what is the ratio of say (1 :: oo), then he would claims it is zero, but if you ask him what is the ratio of (0 :: oo), then he would tell you the same thing, and also the same thing for the ratio of (n :: oo), not different answer from any professional, and all those from the well current established mythematics as well, WONDER! So, the provided proof here and elsewhere is the real mirror where the plenty of common professionals look into, but see long ears and bull's horns, where then get shocked and trying their best to hide the whole issue by reporting it as an abuse, instead of being a colorful front page news for a great discovery, even though it was not any great discovery but a great scandal made deliberately by Jugglers in the history of mathematics, where those Jugglers are sleeping peacefully in the history Thinking that a judgement day would never come, but the one (by definition) who is the real creator of any created existing number would not keep silent any more, by bringing them back for a final balance as he promised earlier, where then no place to hide, or no century to hide Thus, "the real number is any arbitrary existing length on a straight line" Where as rest of numbers are fabricated as Mad Man Make All of my claims were proved rigorously in my posts And all numbers were made by a Mad Man would be thrown out to the sea for the sake of the fish entertainments Regards Bassam King Karzeddin 01/03/17 I've read some of his posts and he seems to have some moments of sanity, but then he writes the most absurd rubbish. Don't think there is a classification for AP in the ICDs. He definitely has a very high emotional IQ. Sadly, I can't say the same for his IQ. I kind of feel sorry for him, but see no purpose in trying to engage him in any way. He can be pretty annoying at times, but hey, one doesn't have to read his posts. I was mainly describing the hopeless trials by (brusegan) to invalidate my proof, by mixing up many issues together to divert the whole issue about very simple facts that are based mainly on common sense and do not require any alleged advanced mathematics to prove or disprove, were also no one can refute it by all the means one has AP has some interesting point about what is finite and what is infinite and regardless of his alleged defined infinity as (10^666), that sounds crazy. Where then no meaning to infinity arise, or no meaning of finite or infinite, because the term (finite) is not well defined, so the infinite as well In practice, our calculations are forever finite, and there is not any chance of being infinite calculation, because this is actually impossible For example, if I offer a memory size that can store every trillion of digits of say (pi) only in one mm cube, then a memory size of a trillion galaxy together would finally be a rational number only, which is a finite or simple fraction, then the whole universe memory size would be also another rational number, so where is that irrational number hiding? Sure, in the tiny skulls of the top professionals minds only! Then the obvious conclusion is how the finite was defined first in order to define the counter meaning And infinity is really meaningless concept, but was milked by the Jugglers so illegally to fabricate theorems and formulas without stating the truth or define concepts correctly but foolishly Otherwise (80%) of current mathematics is actually not mathematics at all, but carpentry works that can be made by trial and error methods even befor B.C BK John Gabriel 2017-03-01 15:43:16 UTC Permalink Post by bassam king karzeddin Otherwise (80%) of current mathematics is actually not mathematics at all, but carpentry works that can be made by trial and error methods even befor B.C Exactly! :-) b***@gmail.com 2017-03-01 15:44:50 UTC Permalink JG and AP made it temporarely into the crank dome. But obviously they are fake cranks, produced 0% math so far. For you BKK the diagnosis is much simpler, you will never make it into the crank dome, its obvious from the beginning that youre a fake crank, a forever finite fake crank, by the holy grail principle of mathematics. (FOR SURE) Am Mittwoch, 1. März 2017 16:34:07 UTC+1 schrieb bassam king karzeddin: bassam king karzeddin 2017-03-01 16:44:44 UTC Permalink Post by b***@gmail.com JG and AP made it temporarely into the crank dome. But obviously they are fake cranks, produced 0% math so far. For you BKK the diagnosis is much simpler, you will never make it into the crank dome, its obvious from the beginning that youre a fake crank, a forever finite fake crank, by the holy grail principle of mathematics. (FOR SURE) So frustrated by avoiding the simplest proof Does this proof need even peer review?, or anyone acceptance! Does this proof need publishing?, it is already published, doesn't it? Can not even you or any other professional write it secretly and more professionally to be accepted by your moron masters? (do not forget to add so many references especially for those whom they would accept to publish it officially) But frankly they would not accept it at any case, and not because it is wrong, but because it is completely right and rigorous, since it would be a historical proof of the biggest ever scandal in the foraged history of mathematics for their proven forgery and infinite stupidity and dishonesty! What does the simple word (impossible) mean in the one of the oldest most famous problems of (impossibility of doubling the cube problem)? And do you remember the simplest proof (half a page only) from the secondary school days? Wonder how did they find it real even after a rigorous counter proof! But, forgery is a perpetual art of your grand mathematicians and your alikes today, and their dreams of adding more fiction numbers would immediately evaporate once they confess the truth! But the clever innocent students would get the correct lisson (for sure), and their lives and time must not be butchered by unnecessary Jugglers any more And, naturally art is a buisness making from scratch for the empty skulls whom they worship fictions only, because the fact is not at all better but it is more bitter and disappointing! And my finite is much bigger than you can imagine (for sure) So strangely you are using my words to describe me again Then, go and disprove me if you have the guts or the ability! WONDER! Bassam King Karzeddin 01/03/17 b***@gmail.com 2017-03-01 16:50:07 UTC Permalink I dosn't matter what you write, you will never make it into the crank dome as a fake crank, even not temporarily. Am Mittwoch, 1. März 2017 17:44:49 UTC+1 schrieb bassam king karzeddin: a***@gmail.com 2017-03-01 20:32:09 UTC Permalink as i have said many times, the pr00f that Brun's constant is irrational, is a pr00f that there are an infinity of twin primes, but of course, it does not "output twin primes" at this time, any more than the pr00f of the infinitude of the primes, hits all of the primes. and, you cannot bother wit Post by bassam king karzeddin Otherwise (80%) of current mathematics is actually not mathematics at all, but carpentry works that can be made by trial and error methods even befor B.C BK a***@gmail.com 2017-03-02 02:00:06 UTC Permalink once you now them to 23, that is just fine Post by a***@gmail.com the pr00f that Brun's constant is irrational, is a pr00f that there are an infinity of twin primes, but of course, it does not "output twin primes" at this time, any more than the pr00f of the infinitude of the primes, hits all of the primes. bassam king karzeddin 2017-03-02 07:02:22 UTC Permalink Post by a***@gmail.com as i have said many times, the pr00f that Brun's constant is irrational, is a pr00f that there are an infinity of twin primes, but of course, it does not "output twin primes" at this time, any more than the pr00f of the infinitude of the primes, hits all of the primes. and, you cannot bother wit Post by bassam king karzeddin Otherwise (80%) of current mathematics is actually not mathematics at all, but carpentry works that can be made by trial and error methods even befor B.C BK But the largest prime number does not exist (for sure) Also the same for the largest pair of twin primes that never exist And what does this suggest for you? B.K bassam king karzeddin 2017-03-02 08:15:48 UTC Permalink Post by a***@gmail.com as i have said many times, the pr00f that Brun's constant is irrational, is a pr00f that there are an infinity of twin primes, but of course, it does not "output twin primes" at this time, any more than the pr00f of the infinitude of the primes, hits all of the primes. and, you cannot bother wit Post by bassam king karzeddin Otherwise (80%) of current mathematics is actually not mathematics at all, but carpentry works that can be made by trial and error methods even befor B.C BK How can an irrational number as ( Brun's Constant), that is increasing absolutely for ever be a constant then? May be the term constant was not well defined in mathematics!Wonder Bassam King Karzeddin 02/03/17 bassam king karzeddin 2017-03-02 08:35:22 UTC Permalink Post by bassam king karzeddin Post by a***@gmail.com as i have said many times, the pr00f that Brun's constant is irrational, is a pr00f that there are an infinity of twin primes, but of course, it does not "output twin primes" at this time, any more than the pr00f of the infinitude of the primes, hits all of the primes. and, you cannot bother wit Post by bassam king karzeddin Otherwise (80%) of current mathematics is actually not mathematics at all, but carpentry works that can be made by trial and error methods even befor B.C BK How can an irrational number as ( Brun's Constant), that is increasing absolutely for ever be a constant then? May be the term constant was not well defined in mathematics!Wonder But I think the term (constant) was well defined in mathematics but so foolishly BK Post by bassam king karzeddin Bassam King Karzeddin 02/03/17 John Gabriel 2017-03-02 12:49:15 UTC Permalink Post by bassam king karzeddin Post by bassam king karzeddin Post by a***@gmail.com as i have said many times, the pr00f that Brun's constant is irrational, is a pr00f that there are an infinity of twin primes, but of course, it does not "output twin primes" at this time, any more than the pr00f of the infinitude of the primes, hits all of the primes. and, you cannot bother wit Post by bassam king karzeddin Otherwise (80%) of current mathematics is actually not mathematics at all, but carpentry works that can be made by trial and error methods even befor B.C BK How can an irrational number as ( Brun's Constant), that is increasing absolutely for ever be a constant then? May be the term constant was not well defined in mathematics!Wonder But I think the term (constant) was well defined in mathematics but so foolishly BK Post by bassam king karzeddin Bassam King Karzeddin 02/03/17 The term constant implies that a magnitude has a measure, but obviously this is untrue in the case of pi, e and sqrt(2) which have no measure. bassam king karzeddin 2017-03-02 13:15:48 UTC Permalink Post by a***@gmail.com as i have said many times, the pr00f that Brun's constant is irrational, is a pr00f that there are an infinity of twin primes, but of course, it does not "output twin primes" at this time, any more than the pr00f of the infinitude of the primes, hits all of the primes. and, you cannot bother wit Post by bassam king karzeddin Otherwise (80%) of current mathematics is actually not mathematics at all, but carpentry works that can be made by trial and error methods even befor B.C BK Still There is only one chance for you with that Burn's constant If you can construct it exactly, then it is a constant, otherwise not any number, but a fiction number at your unattainable (paradise - infinity) And I wonder who wants to stand on your shoulders! and WHY? BK a***@gmail.com 2017-03-01 20:28:09 UTC Permalink I'm quite sure that no one can read his massive typeset, even though it appears to be in English. Post by John Gabriel trying to engage him in any way. He can be pretty annoying at times, but hey, one doesn't have to read his posts. a***@gmail.com 2017-03-02 02:03:06 UTC Permalink hopefully, this is last episode of this tale: this guy wrote this parody of a.p, but I thought that it was actually a.p. just don't ask him about it! Post by a***@gmail.com I'm quite sure that no one can read his massive typeset, bassam king karzeddin 2017-03-01 14:58:21 UTC Permalink Post by b***@gmail.com Well the proof is not that difficult: If E is the least upper bound of D_n, then E must have the same decimal representation. Define E_n = floor(E*10^n)/10^n. Now assume E_k != D_k for some k. Since E is an upper bound, it must be E_k > D_k. But then there is room for an F upper bound and below E. Just take the first D_j with j>k where the last digit is not 9, and increment it the last digit. We can assume D_n are normalized so this is possible. Hence E is not anymore least, and therefore a contradiction. The rest with impossible rational least upper bound follows here in that D_n was from an irrational, i.e. the Diophantine Argument. So we know in Q, the limes doesn't exist. The construction of R now works, in that we add all such limeses. (P.S. Wonder how this would work our in non-standard analysis). Post by b***@gmail.com You seem to have severe problems with the meaning of (=/=) and (=<). Of course also in the rationals, you have D with D_n =/= D. For example take D=5. For the decimal representation D_n of 105^(1/3), we always have D_n =/= D for D=5. So your conclusion D=NOTHINGNESS from D_n =/= D is wrong. What you would need to show is that there is no least upper bound in the rationals. How do you do this correctly? (BTW: I don't know how to do it just now, I have some ideas, but not 100% sure) Post by bassam king karzeddin Post by b***@gmail.com D_n =/= D (new object) Thus D = NOTHINGNESS And see (brusegan) this well exposed guy by AP, going around himself, avoiding the so obvious proof, trying hopelessly flooring the integers, then approaching the integer equation from the sky next time may be, or imagining limes or limits or so like ill imaginations This is not real analysis where you can so simply conclude whatever you like and call it a theorem, but this is EXACT analysis based on INTEGERS, where simply you can not cheat any more Integers can not be limited or floored or approached or nearly approximated Integers are not also as particles of spherical shape with radius epsilon, that you can land on safely in your dreams And indeed there is a real meaning for unsolvable Diophantine equations by integers, the meaning is most likely behind the physical or mathematical knowledge together No solution exists for a certain Diophantine equation, then obviously there is no any real existing number that can solve it also, (but only approximation, which is not mathematics at all and not much better than trial and error approximation) he would finally arrives at what is the ratio of say (1 :: oo), then he would claims it is zero, but if you ask him what is the ratio of (0 :: oo), then he would tell you the same thing, and also the same thing for the ratio of (n :: oo), not different answer from any professional, and all those from the well current established mythematics as well, WONDER! So, the provided proof here and elsewhere is the real mirror where the plenty of common professionals look into, but see long ears and bull's horns, where then get shocked and trying their best to hide the whole issue by reporting it as an abuse, instead of being a colorful front page news for a great discovery, even though it was not any great discovery but a great scandal made deliberately by Jugglers in the history of mathematics, where those Jugglers are sleeping peacefully in the history Thinking that a judgement day would never come, but the one (by definition) who is the real creator of any created existing number would not keep silent any more, by bringing them back for a final balance as he promised earlier, where then no place to hide, or no century to hide Thus, "the real number is any arbitrary existing length on a straight line that is created by a defined unity" Where as rest of numbers are fabricated as Mad Man Make All of my claims were proved rigorously in my posts And all numbers were made by a Mad Man would be thrown out to the sea for the sake of the fish entertainments Regards Bassam King Karzeddin 01/03/17 Buck Futter 2017-02-28 02:08:08 UTC Permalink Go suck on a homeless winos ball-sack, you boring idiot. "bassam king karzeddin" wrote in message news:cd294039-0177-45b5-b543-***@googlegroups.com... <nothing> bassam king karzeddin 2017-02-28 12:36:22 UTC Permalink Post by Buck Futter Go suck on a homeless winos ball-sack, you boring idiot. <nothing> Note who are the new trolls here, from their first post only Glad that my irrefutable proof, which was based on exact analysis with integers have dragged out those worms from their tiny holes I had already pointed out some other worms from their first post in other hot issues of mine So, students should be aware of those hired meaningless worms, the facts are really much better even if they are more bitter Regards Bassam King Karzeddin 28/02/17 a***@gmail.com 2017-02-28 07:41:01 UTC Permalink you should study continued fractions, but nobody gAf about your sylli notions of number, d00d bassam king karzeddin 2017-02-28 12:41:38 UTC Permalink Post by a***@gmail.com you should study continued fractions, but nobody gAf about your sylli notions of number, d00d And once you are clearly incapable of refuting my proof, then you should go and study something else, other than mathematics And better if you do choose a farm or a factory where then you might be able to show your rare talent BK a***@gmail.com 2017-02-27 19:27:10 UTC Permalink I showed the infinitude of the twin primes, although there is no reason, at all, that there should only be a finite number of them ... so, biG Phi = 1.90...., and one seldome actually needs more than two digits thereof. Post by bassam king karzeddin I simply proved the nonexistence of 2^(1/3) as a real number, not because it is indeed proved rigorously as impossible a***@gmail.com 2017-02-28 18:28:27 UTC Permalink biG Phi is irrational, therefore there must be an infinitude of Twin primeS; deal with it, or bite yourself! Post by a***@gmail.com I showed the infinitude of the twin primes, although there is no reason, at all, that there should only be a finite number of them ... so, biG Phi = 1.90...., and one seldome actually needs more than two digits thereof. Post by bassam king karzeddin I simply proved the nonexistence of 2^(1/3) as a real number, not because it is indeed proved rigorously as impossible bassam king karzeddin 2017-02-28 18:37:28 UTC Permalink Post by a***@gmail.com biG Phi is irrational, therefore there must be an infinitude of Twin primeS; deal with it, or bite yourself! Post by a***@gmail.com I showed the infinitude of the twin primes, although there is no reason, at all, that there should only be a finite number of them ... so, biG Phi = 1.90...., and one seldome actually needs more than two digits thereof. Post by bassam king karzeddin I simply proved the nonexistence of 2^(1/3) as a real number, not because it is indeed proved rigorously as impossible And once you define clearly what is meant exactly by finite, then I shall gladly deal with it, (FOR SURE) Hint: I provided you earlier with some hint to it! BK a***@gmail.com 2017-02-28 20:13:34 UTC Permalink finite is just a truncation of infinite, decimally e.g, but it has no bearing on simple fractions, i.e ...00001.3333... (4/3 in any base at least base_5 .-) Post by bassam king karzeddin Hint: I provided you earlier with some hint to it! BK you are an illiterate clown, at least as regards English ... and probably any other language, but nobody gAf about that, since most of us are momlinguals! bassam king karzeddin 2017-03-01 11:20:55 UTC Permalink Post by a***@gmail.com finite is just a truncation of infinite, decimally e.g, but it has no bearing on simple fractions, i.e ...00001.3333... (4/3 in any base at least base_5 .-) Post by bassam king karzeddin Hint: I provided you earlier with some hint to it! BK you are an illiterate clown, at least as regards English ... and probably any other language, but nobody gAf about that, since most of us are momlinguals! Even your chosen words are not from any dictionary, then say what is your first language, if it is Arabic, then cool, since english is my last language BK bassam king karzeddin 2017-03-01 13:31:05 UTC Permalink Post by bassam king karzeddin Post by a***@gmail.com finite is just a truncation of infinite, decimally e.g, but it has no bearing on simple fractions, i.e ...00001.3333... (4/3 in any base at least base_5 .-) Post by bassam king karzeddin Hint: I provided you earlier with some hint to it! BK you are an illiterate clown, at least as regards English ... and probably any other language, but nobody gAf about that, since most of us are momlinguals! Even your chosen words are not from any dictionary, then say what is your first language, if it is Arabic, then cool, since english is my last language BK And your peculiar notations are more than horrible, but your maths is always incomprehensible I wonder, who are you and what do you want exactly? BK bassam king karzeddin 2017-03-01 17:49:25 UTC Permalink Post by bassam king karzeddin Post by b***@gmail.com A =< B & A =/= B <=> A < B A | B | A =< B | A =\= B | A =< B & A =\= B | A < B --+---+--------+---------+------------------+------- 1 | 3 | Yes | Yes | Yes | Yes 2 | 2 | Yes | No | No | No 3 | 1 | No | Yes | No | No You see the last to column give the same truth values, hence we have A =< B & A =/= B <=> A < B Or do you have a counter example? Post by bassam king karzeddin Post by b***@gmail.com In case you don't know how conjunction (&) A =< B & A =/= B <=> A < B This is certainly not any logic, if (A =/= B), then either (A > B),or(A < B) Post by b***@gmail.com A =< B <= A < B this is also magic Post by b***@gmail.com A =/= B <= A < B I am waiting with any challenges until we talk about the same logic. Post by bassam king karzeddin But the challenge for you and everybody else on this planet I have plenty of counter examples with INTEGERS only Let us see how simple it is, and for simplicity, I would like to repeat it in decimal 10base number system, and once you get it, please teach and educate your teachers about it before you would certainly get brain washed and become as ignorant and a victim of old corrupted fake unnecessary mathematics ONE Example of Old claim of mine, the famous number representing the arithmetical cube root of two does not exist, denoted by (2^(1/3)), or ($\sqrt[3]{2}$) I had introduced this general self proved Diophantine equation IFF: (D(n))^p + K(n) = q*s^{n*p} < (D(n) + 1)^p where (p, q) are prime numbers, (n) is non negative integer, (s > 1) is positive integer, then this Diophantine equation implies that D(n) is positive integer that must represent the approximation of the arithmetical p’th root of q, with (n) number of accurate digits in (s) base number system, denoted by (q^(1/p)), or ($\sqrt[p]{q}$), K(n) is therefore positive integer (D(n))^3 + K(n) = 2*(10)^{3n} < (D(n) + 1)^3 Now, we assume varying integer for (n = 0, 1, 2, 3, …, n), and we then, can easily observe how indefinitely K(n) goes as we increase (n) For (n = 0), D(0) = 1, K(0) = 1, showing the first approximation with zero digit after the decimal notation For (n = 1), D(1) = 12, K(1) = 272, showing the second approximation with one digit after the decimal notation For (n = 2), D(2) = 125, k(2) = 46875, showing the THIRD approximation with TWO digits after the decimal notation For (n = 3), D(3) = 1259, K(3) = 4383021, showing the FOURTH approximation with THREE digits after the decimal notation For (n = 4), D(4) = 12599, K(n) = 100242201, showing the fifth approximation with four digits after the decimal notation ……… ……………….. ……………………… …………………. For (n = 10), D(10) = 12599210498, K(n) = 451805284631045974008, showing the eleventh approximation with 10 digits after the decimal notation Now, can you imagine how large the existing term integer K(n) would be for large (n)? Or do you think that K(n) would become zero at infinity?, of course never, it would grow indefinitely with (n) to the alleged infinity Numbers are themselves talking the truth, and who is on earth that can face numbers with perpetual denial, even the proofers of the most famous theorems could not get it yet, WONDER! “NO NUMBER EXISTS WITH ENDLESS TERMS” This is really the true absolute meaning of the truthiness of Fermat’s last theorem! But what do the top professional mathematics do exactly, they simply consider (K(n) = zero), in order to justify the ILLEGAL real number (2^(1/3)), so they set K(n) = 0, then obviously our exact Diophantine equation is foraged to becomes as this (D(n))^3 = 2*(10)^{3n}, take cube root of both sides then you get D(n)/10^n = 2^(1/3), when (n) tends to infinity, where even this expression is not permissible in the holy grail principles of mathematics , since you would have a ratio of two integers where each of them is with infinite sequence of digits So, happy real fake numbers at the fake paradise of all the top mathematicians on earth It is also understood that those roots for (p > 2), are proved rigorously as impossible constructions in order to justify them or accept them as real existing numbers, and all the alleged claimed constructions are only mere cheating (if you investigate carefully into them) But it is well understood for practical needs as that if we want to make a cube that contains two unit cubes for instance, then this is not at all difficult even for a carpenter, we can use those approximation as much as we do like, (this is actually not mathematics works, but application works) So, the hope is held now on the clever school students mainly, whom they would understand those fiction numbers so easily, and certainly they would remove this long standing shame upon themselves and upon mathematics Regards Bassam King Karzeddin 27/02/17 And once they had succeeded in establishing running this obvious fake root as a real existing number, they were not completely satisfied, since this is actually a cubic equation (x^3 = 2), that must have other two roots, but it was so clear that no other real roots exist for this puzzle, so what to do? and how can we create more games from scratch?, they were engaged in more wonderful game as (fundamental theorem of algebra) of course from a scratch They thought about this silly question! Are there any solution for this algebraic equation (x^2 + 1 = 0), of course it was never any real listed problem, since obviously no real solution Then, someone suggested, let us assume an imaginary solution and the problem miracally and instantly was solved by those invented (but not discovered) numbers called imaginary numbers, (making ultra milking of negative numbers which are only mirror image of those real numbers, but they named them the same as real numbers (since then who was there to object such unlimited foolishness), and of course you know the rest, and the wonderful theorem of algebra was then ready manufactured item by those silly assumptions, therefore the cubic equation (x^3 = 2) acquired it is designed number of roots to match the theory they invented I had well explained the obvious illogical step in my earlier posts regarding fabricating the complex numbers They oddly divided the existing space around us, into positive and negative coordination, where actually this is a real JOKE that remains evident even for a layperson Look yourself at the space around you, consider any reference frame location as (0, 0, 0) coordinate, then you can move along any axis in terms of direction and opposite direction only but in positive sense and never in negative sense All that just to justify the imaginary numbers, where the real space would remain perpetual evidence of this obvious ignorance and forgery for ever And guess how physics was polluted by those meaning less assumptions, (positive / negative, finite / infinite, natural symmetry / abnormal symmetry, space / Euclidean space, ... ) So, speed of light is fixed then, and anti mass or anti anything was established from mathematical mere invented concepts and (so easily) History of mathematics is really history of many Jugglers, especially the alleged nobility they talk about, and the appreciation of any new knowledge It is indeed interesting fiction stories for entertainments of kids mainly My advice for a amateurs , ever trust the nobility of a mathematician's, it is only in fiction stories from their history Regards Bassam King Karzeddin 01/03/17 John Gabriel 2017-03-01 18:18:43 UTC Permalink Post by bassam king karzeddin Post by bassam king karzeddin Post by b***@gmail.com A =< B & A =/= B <=> A < B A | B | A =< B | A =\= B | A =< B & A =\= B | A < B --+---+--------+---------+------------------+------- 1 | 3 | Yes | Yes | Yes | Yes 2 | 2 | Yes | No | No | No 3 | 1 | No | Yes | No | No You see the last to column give the same truth values, hence we have A =< B & A =/= B <=> A < B Or do you have a counter example? Post by bassam king karzeddin Post by b***@gmail.com In case you don't know how conjunction (&) A =< B & A =/= B <=> A < B This is certainly not any logic, if (A =/= B), then either (A > B),or(A < B) Post by b***@gmail.com A =< B <= A < B this is also magic Post by b***@gmail.com A =/= B <= A < B I am waiting with any challenges until we talk about the same logic. Post by bassam king karzeddin But the challenge for you and everybody else on this planet I have plenty of counter examples with INTEGERS only Let us see how simple it is, and for simplicity, I would like to repeat it in decimal 10base number system, and once you get it, please teach and educate your teachers about it before you would certainly get brain washed and become as ignorant and a victim of old corrupted fake unnecessary mathematics ONE Example of Old claim of mine, the famous number representing the arithmetical cube root of two does not exist, denoted by (2^(1/3)), or ($\sqrt[3]{2}$) I had introduced this general self proved Diophantine equation IFF: (D(n))^p + K(n) = q*s^{n*p} < (D(n) + 1)^p where (p, q) are prime numbers, (n) is non negative integer, (s > 1) is positive integer, then this Diophantine equation implies that D(n) is positive integer that must represent the approximation of the arithmetical p’th root of q, with (n) number of accurate digits in (s) base number system, denoted by (q^(1/p)), or ($\sqrt[p]{q}$), K(n) is therefore positive integer (D(n))^3 + K(n) = 2*(10)^{3n} < (D(n) + 1)^3 Now, we assume varying integer for (n = 0, 1, 2, 3, …, n), and we then, can easily observe how indefinitely K(n) goes as we increase (n) For (n = 0), D(0) = 1, K(0) = 1, showing the first approximation with zero digit after the decimal notation For (n = 1), D(1) = 12, K(1) = 272, showing the second approximation with one digit after the decimal notation For (n = 2), D(2) = 125, k(2) = 46875, showing the THIRD approximation with TWO digits after the decimal notation For (n = 3), D(3) = 1259, K(3) = 4383021, showing the FOURTH approximation with THREE digits after the decimal notation For (n = 4), D(4) = 12599, K(n) = 100242201, showing the fifth approximation with four digits after the decimal notation ……… ……………….. ……………………… …………………. For (n = 10), D(10) = 12599210498, K(n) = 451805284631045974008, showing the eleventh approximation with 10 digits after the decimal notation Now, can you imagine how large the existing term integer K(n) would be for large (n)? Or do you think that K(n) would become zero at infinity?, of course never, it would grow indefinitely with (n) to the alleged infinity Numbers are themselves talking the truth, and who is on earth that can face numbers with perpetual denial, even the proofers of the most famous theorems could not get it yet, WONDER! “NO NUMBER EXISTS WITH ENDLESS TERMS” This is really the true absolute meaning of the truthiness of Fermat’s last theorem! But what do the top professional mathematics do exactly, they simply consider (K(n) = zero), in order to justify the ILLEGAL real number (2^(1/3)), so they set K(n) = 0, then obviously our exact Diophantine equation is foraged to becomes as this (D(n))^3 = 2*(10)^{3n}, take cube root of both sides then you get D(n)/10^n = 2^(1/3), when (n) tends to infinity, where even this expression is not permissible in the holy grail principles of mathematics , since you would have a ratio of two integers where each of them is with infinite sequence of digits So, happy real fake numbers at the fake paradise of all the top mathematicians on earth It is also understood that those roots for (p > 2), are proved rigorously as impossible constructions in order to justify them or accept them as real existing numbers, and all the alleged claimed constructions are only mere cheating (if you investigate carefully into them) But it is well understood for practical needs as that if we want to make a cube that contains two unit cubes for instance, then this is not at all difficult even for a carpenter, we can use those approximation as much as we do like, (this is actually not mathematics works, but application works) So, the hope is held now on the clever school students mainly, whom they would understand those fiction numbers so easily, and certainly they would remove this long standing shame upon themselves and upon mathematics Regards Bassam King Karzeddin 27/02/17 And once they had succeeded in establishing running this obvious fake root as a real existing number, they were not completely satisfied, since this is actually a cubic equation (x^3 = 2), that must have other two roots, but it was so clear that no other real roots exist for this puzzle, so what to do? and how can we create more games from scratch?, they were engaged in more wonderful game as (fundamental theorem of algebra) of course from a scratch They thought about this silly question! Are there any solution for this algebraic equation (x^2 + 1 = 0), of course it was never any real listed problem, since obviously no real solution Then, someone suggested, let us assume an imaginary solution and the problem miracally and instantly was solved by those invented (but not discovered) numbers called imaginary numbers, (making ultra milking of negative numbers which are only mirror image of those real numbers, but they named them the same as real numbers (since then who was there to object such unlimited foolishness), and of course you know the rest, and the wonderful theorem of algebra was then ready manufactured item by those silly assumptions, therefore the cubic equation (x^3 = 2) acquired it is designed number of roots to match the theory they invented I had well explained the obvious illogical step in my earlier posts regarding fabricating the complex numbers They oddly divided the existing space around us, into positive and negative coordination, where actually this is a real JOKE that remains evident even for a layperson Look yourself at the space around you, consider any reference frame location as (0, 0, 0) coordinate, then you can move along any axis in terms of direction and opposite direction only but in positive sense and never in negative sense All that just to justify the imaginary numbers, where the real space would remain perpetual evidence of this obvious ignorance and forgery for ever And guess how physics was polluted by those meaning less assumptions, (positive / negative, finite / infinite, natural symmetry / abnormal symmetry, space / Euclidean space, ... ) So, speed of light is fixed then, and anti mass or anti anything was established from mathematical mere invented concepts and (so easily) History of mathematics is really history of many Jugglers, especially the alleged nobility they talk about, and the appreciation of any new knowledge It is indeed interesting fiction stories for entertainments of kids mainly My advice for a amateurs , ever trust the nobility of a mathematician's, it is only in fiction stories from their history Regards Bassam King Karzeddin 01/03/17 Congratulations Bassam! You are now officially a fake crank. That is far better than being a crank. Fake crank is a double negative, which means you are NOT a crank. It is written like this in set theory: ~crank = not a crank. Well done! Chuckle. b***@gmail.com 2017-03-01 19:05:54 UTC Permalink Well there is also the stinky tofu fake crank, and the I don't know english fake crank. Am Mittwoch, 1. März 2017 19:18:50 UTC+1 schrieb John Gabriel: bassam king karzeddin 2017-03-01 19:19:05 UTC Permalink Post by John Gabriel Post by bassam king karzeddin Post by bassam king karzeddin Post by b***@gmail.com A =< B & A =/= B <=> A < B A | B | A =< B | A =\= B | A =< B & A =\= B | A < B --+---+--------+---------+------------------+------- 1 | 3 | Yes | Yes | Yes | Yes 2 | 2 | Yes | No | No | No 3 | 1 | No | Yes | No | No You see the last to column give the same truth values, hence we have A =< B & A =/= B <=> A < B Or do you have a counter example? Post by bassam king karzeddin Post by b***@gmail.com In case you don't know how conjunction (&) A =< B & A =/= B <=> A < B This is certainly not any logic, if (A =/= B), then either (A > B),or(A < B) Post by b***@gmail.com A =< B <= A < B this is also magic Post by b***@gmail.com A =/= B <= A < B I am waiting with any challenges until we talk about the same logic. Post by bassam king karzeddin But the challenge for you and everybody else on this planet I have plenty of counter examples with INTEGERS only Let us see how simple it is, and for simplicity, I would like to repeat it in decimal 10base number system, and once you get it, please teach and educate your teachers about it before you would certainly get brain washed and become as ignorant and a victim of old corrupted fake unnecessary mathematics ONE Example of Old claim of mine, the famous number representing the arithmetical cube root of two does not exist, denoted by (2^(1/3)), or ($\sqrt[3]{2}$) I had introduced this general self proved Diophantine equation IFF: (D(n))^p + K(n) = q*s^{n*p} < (D(n) + 1)^p where (p, q) are prime numbers, (n) is non negative integer, (s > 1) is positive integer, then this Diophantine equation implies that D(n) is positive integer that must represent the approximation of the arithmetical p’th root of q, with (n) number of accurate digits in (s) base number system, denoted by (q^(1/p)), or ($\sqrt[p]{q}$), K(n) is therefore positive integer (D(n))^3 + K(n) = 2*(10)^{3n} < (D(n) + 1)^3 Now, we assume varying integer for (n = 0, 1, 2, 3, …, n), and we then, can easily observe how indefinitely K(n) goes as we increase (n) For (n = 0), D(0) = 1, K(0) = 1, showing the first approximation with zero digit after the decimal notation For (n = 1), D(1) = 12, K(1) = 272, showing the second approximation with one digit after the decimal notation For (n = 2), D(2) = 125, k(2) = 46875, showing the THIRD approximation with TWO digits after the decimal notation For (n = 3), D(3) = 1259, K(3) = 4383021, showing the FOURTH approximation with THREE digits after the decimal notation For (n = 4), D(4) = 12599, K(n) = 100242201, showing the fifth approximation with four digits after the decimal notation ……… ……………….. ……………………… …………………. For (n = 10), D(10) = 12599210498, K(n) = 451805284631045974008, showing the eleventh approximation with 10 digits after the decimal notation Now, can you imagine how large the existing term integer K(n) would be for large (n)? Or do you think that K(n) would become zero at infinity?, of course never, it would grow indefinitely with (n) to the alleged infinity Numbers are themselves talking the truth, and who is on earth that can face numbers with perpetual denial, even the proofers of the most famous theorems could not get it yet, WONDER! “NO NUMBER EXISTS WITH ENDLESS TERMS” This is really the true absolute meaning of the truthiness of Fermat’s last theorem! But what do the top professional mathematics do exactly, they simply consider (K(n) = zero), in order to justify the ILLEGAL real number (2^(1/3)), so they set K(n) = 0, then obviously our exact Diophantine equation is foraged to becomes as this (D(n))^3 = 2*(10)^{3n}, take cube root of both sides then you get D(n)/10^n = 2^(1/3), when (n) tends to infinity, where even this expression is not permissible in the holy grail principles of mathematics , since you would have a ratio of two integers where each of them is with infinite sequence of digits So, happy real fake numbers at the fake paradise of all the top mathematicians on earth It is also understood that those roots for (p > 2), are proved rigorously as impossible constructions in order to justify them or accept them as real existing numbers, and all the alleged claimed constructions are only mere cheating (if you investigate carefully into them) But it is well understood for practical needs as that if we want to make a cube that contains two unit cubes for instance, then this is not at all difficult even for a carpenter, we can use those approximation as much as we do like, (this is actually not mathematics works, but application works) So, the hope is held now on the clever school students mainly, whom they would understand those fiction numbers so easily, and certainly they would remove this long standing shame upon themselves and upon mathematics Regards Bassam King Karzeddin 27/02/17 And once they had succeeded in establishing running this obvious fake root as a real existing number, they were not completely satisfied, since this is actually a cubic equation (x^3 = 2), that must have other two roots, but it was so clear that no other real roots exist for this puzzle, so what to do? and how can we create more games from scratch?, they were engaged in more wonderful game as (fundamental theorem of algebra) of course from a scratch They thought about this silly question! Are there any solution for this algebraic equation (x^2 + 1 = 0), of course it was never any real listed problem, since obviously no real solution Then, someone suggested, let us assume an imaginary solution and the problem miracally and instantly was solved by those invented (but not discovered) numbers called imaginary numbers, (making ultra milking of negative numbers which are only mirror image of those real numbers, but they named them the same as real numbers (since then who was there to object such unlimited foolishness), and of course you know the rest, and the wonderful theorem of algebra was then ready manufactured item by those silly assumptions, therefore the cubic equation (x^3 = 2) acquired it is designed number of roots to match the theory they invented I had well explained the obvious illogical step in my earlier posts regarding fabricating the complex numbers They oddly divided the existing space around us, into positive and negative coordination, where actually this is a real JOKE that remains evident even for a layperson Look yourself at the space around you, consider any reference frame location as (0, 0, 0) coordinate, then you can move along any axis in terms of direction and opposite direction only but in positive sense and never in negative sense All that just to justify the imaginary numbers, where the real space would remain perpetual evidence of this obvious ignorance and forgery for ever And guess how physics was polluted by those meaning less assumptions, (positive / negative, finite / infinite, natural symmetry / abnormal symmetry, space / Euclidean space, ... ) So, speed of light is fixed then, and anti mass or anti anything was established from mathematical mere invented concepts and (so easily) History of mathematics is really history of many Jugglers, especially the alleged nobility they talk about, and the appreciation of any new knowledge It is indeed interesting fiction stories for entertainments of kids mainly My advice for a amateurs , ever trust the nobility of a mathematician's, it is only in fiction stories from their history Regards Bassam King Karzeddin 01/03/17 Congratulations Bassam! You are now officially a fake crank. That is far better than being a crank. Fake crank is a double negative, which means you are NOT a crank. It is written like this in set theory: ~crank = not a crank. Well done! Chuckle. I really want to help and rescue the mathematicians to a better world and rules, but it seems the bad traits of humans are accumulated in their genes, The big obstacles, and the task seems impossible without the efforts of engineers technology, also biology must be involved in this ridiculous issue with those most failures category of human being They still can not comprehend their silly role of their existence They still are unaware of how dwarfs are they in a presence of a skilled football player boy only They are indeed so unnoticeable among people, but oddly they think that they still have the finest heads on earth, Wonder? How is driving the whole world into this incomprehensible abnormal state? Is there invisible Devil behind those unbelievable situations, and is this the era of free knowledge or vast ignorance I wonder always Where is the whole world going by those plenty of unnoticeable mathematickers? I also confirm that they do not like to learn anything new and created in a so natural way, but then do not have any choice also with the designed new alleged knowledge, but complete and blind obedience I keep thinking of how to save those vast majority of obvious idiots Many more questions but time is so limited BK b***@gmail.com 2017-03-01 19:27:27 UTC Permalink Something tells me, you could be qualified as a tele- vision preacher. But what concerns math, I have my doubts. What is the result of 2+2? Is it still 4, or something else? Am Mittwoch, 1. März 2017 20:19:11 UTC+1 schrieb bassam king karzeddin: bassam king karzeddin 2017-03-02 08:09:15 UTC Permalink Post by b***@gmail.com Something tells me, you could be qualified as a tele- vision preacher. But what concerns math, I have my doubts. What is the result of 2+2? Is it still 4, or something else? What a meaningless puzzle is this, of course you did not invent it yourself, and you should show its source, but most likely another reputable Devil had introduce it to perpetuate his alleged talent and philosophy for passing something evilish of his fake mathematics into your too tiny skulls, where then other sections of mythematics would be established as a true science Those are not better than mind games that most likely are very harmful to human minds, but frankly chess is a better game! But still you would have more doubts I wonder! Wake up to the absolute truths mathematicians BK b***@gmail.com 2017-03-01 19:41:29 UTC Permalink Advice to them, make yourself noticable: An opportunity for all! Breakthrough Prize in Mathematics https://groups.google.com/d/msg/sci.math/BAzP3a6_oTQ/N1ITY_XICwAJ LoL, laughing my ass off... (I could explain further why) Post by bassam king karzeddin Where is the whole world going by those plenty of unnoticeable mathematickers? bassam king karzeddin 2017-03-02 07:44:40 UTC Permalink Post by b***@gmail.com An opportunity for all! Breakthrough Prize in Mathematics https://groups.google.com/d/msg/sci.math/BAzP3a6_oTQ/N1ITY_XICwAJ LoL, laughing my ass off... (I could explain further why) Post by bassam king karzeddin Where is the whole world going by those plenty of unnoticeable mathematickers? Your link does not show any problem, but a bunch of meaningless talk So dream on prizes, those are not opened for anyone (for sure), Imaginary numbers had been refuted badly by me, did not I publish the irrefutable proof yet ? Wonder Or is it the proofer which a matter of more importance?, then wait some time to see yourself how many claimers or reputable proofer would arise, if not already happening Hilbert space!, I do confirm it is not any real space if it is not Euclidean space, there is not any non Euclidean space (for sure) Did not I publish the refutation of non Euclidean space also? Wonder! The easiest proof that was deleted shamelessly by MSE-Physics, "Any for points not co planer in stationary positions or in random motion lie exactly on a surface of a sphere with constructible radius" (for sure) And you have the whole universe to test it physically and report the results here for public So, where are the other non Euclidean spaces I wonder? non existent (for sure) What funny prizes you are talking about? Wonder If you really want any prize, then go and imitate Wiles for FLT proof But most likely nobody would like to hear you, since you have no reputable position nor in any reputable place (for sure) Good luck Bassam King Karzeddin 02/03/17 bassam king karzeddin 2017-03-02 12:43:13 UTC Permalink Post by bassam king karzeddin Post by b***@gmail.com An opportunity for all! Breakthrough Prize in Mathematics https://groups.google.com/d/msg/sci.math/BAzP3a6_oTQ/N1ITY_XICwAJ LoL, laughing my ass off... (I could explain further why) Post by bassam king karzeddin Where is the whole world going by those plenty of unnoticeable mathematickers? Your link does not show any problem, but a bunch of meaningless talk So dream on prizes, those are not opened for anyone (for sure), Imaginary numbers had been refuted badly by me, did not I publish the irrefutable proof yet ? Wonder Or is it the proofer which a matter of more importance?, then wait some time to see yourself how many claimers or reputable proofer would arise, if not already happening Hilbert space!, I do confirm it is not any real space if it is not Euclidean space, there is not any non Euclidean space (for sure) Did not I publish the refutation of non Euclidean space also? Wonder! The easiest proof that was deleted shamelessly by MSE-Physics, "Any for points (not co planer) in their stationary positions or in their random motions lie exactly on a surface of a sphere with a constructible radius at any moment of time" (proved so easily) And you have the whole universe to test it physically and report the results here for public So, where are the other non Euclidean spaces I wonder? non existent (for sure) But the properties of contained energy within space is something different from the perpetual property of space itself which is the (nothingness itself), but so strangely contains every thing, WONDER! Post by bassam king karzeddin What funny prizes you are talking about? Wonder If you really want any prize, then go and imitate Wiles for FLT proof But most likely nobody would like to hear you, since you have no reputable position nor in any reputable place (for sure) Good luck Bassam King Karzeddin 02/03/17 BK a***@gmail.com 2017-02-26 17:57:07 UTC Permalink one normally need only two places for most applcations; pi is => 31/10, in some base at least four; so, for the ration of the diameter to the area of the sphere, we use 7/22, base at least eight ... or, 10/31, base at least four -- 00ps, repeating my self about rational expressions! Post by bassam king karzeddin Hence, no such numbers exist but approximation to something assumed existing in mind, but physically impossible And this is the true deep meaning of non existence of a solution to some famous Diophantine equation, where generally comprehended by mathematicians BK Post by bassam king karzeddin And if K(n) =/= 0, then D is greater than any possible decimal representation, thus No real cube root exists for (105), robersi 2017-02-21 11:58:18 UTC Permalink Post by bassam king karzeddin Why does the trustiness of Fermat's last theorem implies directly the non existence of the real positive arithmetical p'th root of any prime number ($\sqrt[p]{q}$)? Where (p) is odd prime number, and (q) is prime e number It is an easy task for school students NOW! Regards Bassam King Karzeddin 19/02/17 because it is directly. that's why. bassam king karzeddin 2017-02-22 07:59:31 UTC Permalink Post by robersi Post by bassam king karzeddin Why does the trustiness of Fermat's last theorem implies directly the non existence of the real positive arithmetical p'th root of any prime number ($\sqrt[p]{q}$)? Where (p) is odd prime number, and (q) is prime e number It is an easy task for school students NOW! Regards Bassam King Karzeddin 19/02/17 because it is directly. that's why. It is good that you recognized it so easily!, but did the top professional mathematicians recognize this also? I wonder! BK bassam king karzeddin 2017-02-27 18:10:50 UTC Permalink Post by b***@gmail.com Post by b***@gmail.com Nope the below is not correct. Let D_n be the decimal representation of 105^(1/3) up Post by b***@gmail.com to n digits. Let D be the limes lim_n->oo D_n. We 105^(1/3) =\= D_n for each n 105^(1/3) = D Be careful with the use of the ellipses. It means limes, so you can easily make wrong math statements, Post by b***@gmail.com and run into contradictions. For example 105^(1/3) 0 =/= 0 Which is nonsense. Am Montag, 20. Februar 2017 18:32:30 UTC+1 schrieb 105^(1/3) =\= 4.7176939803... Why people here ignore deliberately what had been taught and proved so easily earlier in sci.math and elsewhere? Especially you insect brain brusegan!, who had great chance to understand the basic flow on those fiction numbers while discussion with me Did not I PUBLISH the following obvious self proved d inequality in the whole non zero integer numbers? Or do you need permit ion on its absolute validity y from your famous corrupted Journals and Universities? Applying your own notation for ( D_n) in The famous Iff: (D_n)^3 < 105*(10)^{3n} <( D_(n+1))^3, where then, (D_n) represents the rational decimal of the cube root of (105), ($\sqrt[3]{105}$), with (n) number of accurate digits after the decimal notation (in 10base number system) Then there is a positive integer (k(n)), where we can (D_n)^3 + k(n) = 105*(10)^{3n}, where k(n) is s increasing indefinitely when (n) increase But what a professional mathematician do exactly, is neglecting (k(n)) completely and equating it with zero, where then it becomes so easy approximation for (D_n)^3 = 105*(10)^{3n}, and hence,$ (D_n)/10^n =
\sqrt[3]{105}$, (happy end at the fool’s paradise - infinity) with (REAL CHEATING), whereas the obvious (D_n)^3 =/= 105*(10)^{3n}, and thus no real cube root exists for (105), and for sure, but there is another easy proof from Fermat’s last theorem too. Regards Bassam King Karzeddin 21/02/17 Let us see how simple it is, and for simplicity, I would like to repeat it in decimal 10base number system, and once you get it, please teach and educate your teachers about it before you would certainly get brain washed and become as ignorant and a victim of old corrupted fake unnecessary mathematics ONE Example of Old claim of mine, the famous number representing the arithmetical cube root of two does not exist, denoted by (2^(1/3)), or ($\sqrt[3]{2}$) The proof: I had introduced this general self proved Diophantine equation IFF: (D(n))^p + K(n) = q*s^{n*p} < (D(n) + 1)^p where (p, q) are prime numbers, (n) is non negative integer, (s > 1) is positive integer, then this Diophantine equation implies that D(n) is positive integer that must represent the approximation of the arithmetical p th root of q, with (n) number of accurate digits in (s) base number system, denoted by (q^(1/p)), or ($\sqrt[p]{q}\$), K(n) is therefore positive integer

In our chosen case here for (2^(1/3)), we have (s = 10), p = 3, q = 2, so our Diophantine equation :

(D(n))^3 + K(n) = 2*(10)^{3n} < (D(n) + 1)^3

Now, we assume varying integer for (n = 0, 1, 2, 3, …, n), and we then, can easily observe how indefinitely K(n) goes as we increase (n)

For (n = 0), D(0) = 1, K(0) = 1, showing the first approximation with zero digit after the decimal notation

For (n = 1), D(1) = 12, K(1) = 272, showing the second approximation with one digit after the decimal notation
For (n = 2), D(2) = 125, k(2) = 46875, showing the THIRD approximation with TWO digits after the decimal notation

For (n = 3), D(3) = 1259, K(3) = 4383021, showing the FOURTH approximation with THREE digits after the decimal notation

For (n = 4), D(4) = 12599, K(n) = 100242201, showing the fifth approximation with four digits after the decimal notation

……… ……………….. ……………………… ………………….

For (n = 10), D(10) = 12599210498, K(n) = 451805284631045974008, showing the eleventh approximation with 10 digits after the decimal notation

Now, can you imagine how large the existing term integer K(n) would be for large (n)?
Or do you think that K(n) would become zero at infinity?, of course never, it would grow indefinitely with (n) to the alleged infinity

Numbers are themselves talking the truth, and who is on earth that can face numbers with perpetual denial, even the proofers of the most famous theorems could not get it yet, WONDER!
“NO NUMBER EXISTS WITH ENDLESS TERMS”

This is really the true absolute meaning of the truthiness of Fermat’s last theorem!

But what do the top professional mathematics do exactly, they simply consider (K(n) = zero), in order to justify the ILLEGAL real number (2^(1/3)), so they set K(n) = 0, then obviously our exact Diophantine equation is foraged to becomes as this

(D(n))^3 = 2*(10)^{3n}, take cube root of both sides then you get

D(n)/10^n = 2^(1/3), when (n) tends to infinity, where even this expression is not permissible in the holy grail principles of mathematics , since you would have a ratio of two integers where each of them is with infinite sequence of digits

So, happy real fake numbers at the fake paradise of all the top mathematicians on earth

It is also understood that those roots for (p > 2), are proved rigorously as impossible constructions in order to justify them or accept them as real existing numbers, and all the alleged claimed constructions are only mere cheating (if you investigate carefully into them)

But it is well understood for practical needs as that if we want to make a cube that contains two unit cubes for instance, then this is not at all difficult even for a carpenter, we can use those approximation as much as we do like, (this is actually not mathematics works, but application works)

Regards
Bassam King Karzeddin
27/02/17