The non existence of p'th root of any prime number, for (p>2) prime

782 views
Skip to first unread message

bassam king karzeddin

unread,
Feb 19, 2017, 12:30:04 PM2/19/17
to
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

abu.ku...@gmail.com

unread,
Feb 19, 2017, 1:19:52 PM2/19/17
to
wTf is that supposed to mean?

Mutt Buncher

unread,
Feb 19, 2017, 7:12:26 PM2/19/17
to
Shut up idiot.

abu.ku...@gmail.com

unread,
Feb 20, 2017, 12:02:52 AM2/20/17
to
the geometrical mean of two primes, their product,
could be of some interest ... so, go for it

abu.ku...@gmail.com

unread,
Feb 20, 2017, 12:24:44 AM2/20/17
to
of course, the geometric mean of a twin prime is an even number;
what is the thirdr00t of a hundred and five?

abu.ku...@gmail.com

unread,
Feb 20, 2017, 12:27:23 AM2/20/17
to
hah, 00ps secondr00t(143) is not integral.

abu.ku...@gmail.com

unread,
Feb 20, 2017, 3:51:20 AM2/20/17
to
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

unread,
Feb 20, 2017, 7:58:17 AM2/20/17
to
($\sqrt[p]{q}$) 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

thugst...@gmail.com

unread,
Feb 20, 2017, 12:00:48 PM2/20/17
to
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?

> > wTf is that supposed to mean?

bassam king karzeddin

unread,
Feb 20, 2017, 12:32:30 PM2/20/17
to
So, you asked for it

> 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

burs...@gmail.com

unread,
Feb 20, 2017, 12:56:10 PM2/20/17
to
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.

Am Montag, 20. Februar 2017 18:32:30 UTC+1 schrieb bassam king karzeddin:
> 105^(1/3) =\= 4.7176939803...

bassam king karzeddin

unread,
Feb 21, 2017, 3:58:41 AM2/21/17
to
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

burs...@gmail.com

unread,
Feb 21, 2017, 5:04:42 AM2/21/17
to
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

unread,
Feb 21, 2017, 6:20:05 AM2/21/17
to
> 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.
>
> Am Montag, 20. Februar 2017 18:32:30 UTC+1 schrieb
> bassam king karzeddin:
> > 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.


burs...@gmail.com

unread,
Feb 21, 2017, 7:23:04 AM2/21/17
to
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?

burs...@gmail.com

unread,
Feb 21, 2017, 7:28:36 AM2/21/17
to
Am Dienstag, 21. Februar 2017 13:23:04 UTC+1 schrieb burs...@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.

robersi

unread,
Feb 21, 2017, 8:50:04 AM2/21/17
to
> 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.

abu.ku...@gmail.com

unread,
Feb 21, 2017, 3:47:36 PM2/21/17
to
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

bassam king karzeddin

unread,
Feb 22, 2017, 2:31:40 AM2/22/17
to
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

bassam king karzeddin

unread,
Feb 22, 2017, 2:59:41 AM2/22/17
to
It is good that you recognized it so easily!, but did the top professional mathematicians recognize this also? I wonder!

BK

bassam king karzeddin

unread,
Feb 22, 2017, 3:07:24 AM2/22/17
to
On Tuesday, February 21, 2017 at 1:04:42 PM UTC+3, burs...@gmail.com wrote:
> 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?
>
> Am Dienstag, 21. Februar 2017 09:58:41 UTC+1 schrieb bassam king karzeddin:

BK

bassam king karzeddin

unread,
Feb 22, 2017, 3:37:43 AM2/22/17
to
On Tuesday, February 21, 2017 at 3:23:04 PM UTC+3, burs...@gmail.com wrote:
> 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,

> 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

> What we have is simply:
>
> D_n = floor((105)^(1/3)*10^n)/10^n.

FLOOR!, so great it might mean (=), why not BIG MORON?
>
> From this follows immediately:
>
> 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
>
> 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?

>
> 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)
>
> Am Dienstag, 21. Februar 2017 12:20:05 UTC+1 schrieb 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

abu.ku...@gmail.com

unread,
Feb 22, 2017, 3:39:19 AM2/22/17
to
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!

bassam king karzeddin

unread,
Feb 22, 2017, 3:49:06 AM2/22/17
to
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

burs...@gmail.com

unread,
Feb 22, 2017, 3:55:03 AM2/22/17
to
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!

burs...@gmail.com

unread,
Feb 22, 2017, 4:02:24 AM2/22/17
to
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
https://www.youtube.com/watch?v=QpDDWBRHNRM

bassam king karzeddin

unread,
Feb 22, 2017, 4:41:12 AM2/22/17
to
On Wednesday, February 22, 2017 at 11:55:03 AM UTC+3, burs...@gmail.com wrote:
> 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!

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

burs...@gmail.com

unread,
Feb 22, 2017, 10:02:44 AM2/22/17
to
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:

burs...@gmail.com

unread,
Feb 22, 2017, 10:09:36 AM2/22/17
to
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.

burs...@gmail.com

unread,
Feb 22, 2017, 10:17:12 AM2/22/17
to
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.

bassam king karzeddin

unread,
Feb 22, 2017, 11:10:22 AM2/22/17
to
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


burs...@gmail.com

unread,
Feb 22, 2017, 1:23:00 PM2/22/17
to
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:

burs...@gmail.com

unread,
Feb 22, 2017, 1:31:42 PM2/22/17
to
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.

bassam king karzeddin

unread,
Feb 22, 2017, 1:49:10 PM2/22/17
to
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

burs...@gmail.com

unread,
Feb 22, 2017, 2:02:23 PM2/22/17
to
You must be insane, I wrote K(n) > 0, so I cannot
set K(n) = 0. I wrote:

Am Mittwoch, 22. Februar 2017 19:31:42 UTC+1 schrieb burs...@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?

burs...@gmail.com

unread,
Feb 22, 2017, 2:09:07 PM2/22/17
to
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

abu.ku...@gmail.com

unread,
Feb 22, 2017, 2:49:31 PM2/22/17
to
irrationals are treated arithmetically via continued fractions
-- and it is hard to do that

bassam king karzeddin

unread,
Feb 22, 2017, 3:13:13 PM2/22/17
to
On Wednesday, February 22, 2017 at 10:49:31 PM UTC+3, abu.ku...@gmail.com wrote:
> 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

bassam king karzeddin

unread,
Feb 22, 2017, 3:18:59 PM2/22/17
to
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

burs...@gmail.com

unread,
Feb 22, 2017, 5:08:23 PM2/22/17
to
By juggling you mean Gabrieloconfusion and Johno
stupidics, the new religion of all cranks?

abu.ku...@gmail.com

unread,
Feb 22, 2017, 9:26:58 PM2/22/17
to
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

bassam king karzeddin

unread,
Feb 23, 2017, 3:08:18 AM2/23/17
to
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

bassam king karzeddin

unread,
Feb 23, 2017, 3:25:08 AM2/23/17
to
On Thursday, February 23, 2017 at 1:08:23 AM UTC+3, burs...@gmail.com wrote:
> 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

burs...@gmail.com

unread,
Feb 23, 2017, 5:14:01 AM2/23/17
to
You must be insane for sure, I wrote K(n) > 0, so
I cannot set K(n) = 0. I wrote:

Am Mittwoch, 22. Februar 2017 19:31:42 UTC+1 schrieb burs...@gmail.com:
> K(n) = (105 - D_n^3)*10^(3*n) > 0

Best Regards for sure.

abu.ku...@gmail.com

unread,
Feb 23, 2017, 2:02:25 PM2/23/17
to
I did notice, a few days ago, that it was just your sylliware
that made that error about secondr00ting -- and,
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.

bassam king karzeddin

unread,
Feb 25, 2017, 2:43:44 AM2/25/17
to
On Thursday, February 23, 2017 at 1:14:01 PM UTC+3, burs...@gmail.com wrote:
> You must be insane for sure, I wrote K(n) > 0, so
> I cannot set K(n) = 0. I wrote:
>
> Am Mittwoch, 22. Februar 2017 19:31:42 UTC+1 schrieb burs...@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!

>
> Best Regards for sure.
>

Bassam King Karzeddin
25/02/17

burs...@gmail.com

unread,
Feb 25, 2017, 6:43:33 AM2/25/17
to
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.

abu.ku...@gmail.com

unread,
Feb 25, 2017, 6:58:45 AM2/25/17
to
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!

bassam king karzeddin

unread,
Feb 26, 2017, 11:17:07 AM2/26/17
to
On Saturday, February 25, 2017 at 2:43:33 PM UTC+3, burs...@gmail.com wrote:
> 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)

This is big mistake, No equality (for sure)

D_n < D,

This is absolutely correct
>
> D_n =/= D for each n (because D is irrational)

Yes
>
> Or in summary:
>
> D_n < D for each n

Bravo
>
> 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

burs...@gmail.com

unread,
Feb 26, 2017, 12:56:28 PM2/26/17
to
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)

abu.ku...@gmail.com

unread,
Feb 26, 2017, 12:57:12 PM2/26/17
to
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!

burs...@gmail.com

unread,
Feb 26, 2017, 1:01:58 PM2/26/17
to
Take for example a=1 and b=1, then

c = 2^(1/3)

Is a solution.

bassam king karzeddin

unread,
Feb 27, 2017, 10:16:18 AM2/27/17
to
On Sunday, February 26, 2017 at 8:56:28 PM UTC+3, burs...@gmail.com wrote:
> In case you don't know how conjunction (&)
> works in logic/math, we have of course:
>
> A =< B & A =/= B <=> A < B

This is certainly not any logic, if (A =/= B), then either (A > B),or(A < B)
>
> A =< B <= A < B
this is also magic
>
> A =/= B <= A < B

I do not know who sets this nonsense logic
>
> 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
>
> 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
>
> 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,

> 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

> (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
>

And the Queen says nothing would defeat the obvious so sensible truth


> End of proceedure)
>
> Am Sonntag, 26. Februar 2017 17:17:07 UTC+1 schrieb 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

bassam king karzeddin

unread,
Feb 27, 2017, 10:20:28 AM2/27/17
to
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

j4n bur53

unread,
Feb 27, 2017, 11:01:36 AM2/27/17
to
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?

bassam king karzeddin schrieb:
> On Sunday, February 26, 2017 at 8:56:28 PM UTC+3, burs...@gmail.com wrote:
>> > In case you don't know how conjunction (&)
>> > works in logic/math, we have of course:
>> >
>> > A =< B & A =/= B <=> A < B
> This is certainly not any logic, if (A =/= B), then either (A > B),or(A < B)
>> >
>> > A =< B <= A < B
> this is also magic
>> >
>> > A =/= B <= A < B

I am waiting with any challenges until we talk about
the same logic.

bassam king karzeddin

unread,
Feb 27, 2017, 1:21:29 PM2/27/17
to
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

burs...@gmail.com

unread,
Feb 27, 2017, 1:26:24 PM2/27/17
to
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

bassam king karzeddin

unread,
Feb 27, 2017, 1:57:49 PM2/27/17
to
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

bassam king karzeddin

unread,
Feb 27, 2017, 2:10:06 PM2/27/17
to
> > 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.
> >
> > Am Montag, 20. Februar 2017 18:32:30 UTC+1 schrieb
> > bassam king karzeddin:
> > > 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
> 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
> 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
> 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
> s 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

burs...@gmail.com

unread,
Feb 27, 2017, 2:25:07 PM2/27/17
to
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.

abu.ku...@gmail.com

unread,
Feb 27, 2017, 2:27:24 PM2/27/17
to
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.

burs...@gmail.com

unread,
Feb 27, 2017, 2:30:35 PM2/27/17