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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

($\sqrt[p]{q}$)?

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

Feb 19, 2017, 1:19:52 PM2/19/17

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wTf is that supposed to mean?

Feb 19, 2017, 7:12:26 PM2/19/17

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Shut up idiot.

Feb 20, 2017, 12:02:52 AM2/20/17

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the geometrical mean of two primes, their product,

could be of some interest ... so, go for it

could be of some interest ... so, go for it

Feb 20, 2017, 12:24:44 AM2/20/17

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of course, the geometric mean of a twin prime is an even number;

what is the thirdr00t of a hundred and five?

what is the thirdr00t of a hundred and five?

Feb 20, 2017, 12:27:23 AM2/20/17

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hah, 00ps secondr00t(143) is not integral.

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

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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

but I realized that that is about the same as (6n -\+ 5),

and found a couple of pairs that were both nonprimal, so

Feb 20, 2017, 7:58:17 AM2/20/17

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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

Feb 20, 2017, 12:00:48 PM2/20/17

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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?

you have not bothered to use even that, so ...

why should 2 = p, be a pth r00t?

> > wTf is that supposed to mean?

Feb 20, 2017, 12:32:30 PM2/20/17

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> what is the thirdr00t of a hundred and five?

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

Feb 20, 2017, 12:56:10 PM2/20/17

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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...

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...

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

to

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

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

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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:

numbers", its it Unicorn numbers. When do you get it?

Am Dienstag, 21. Februar 2017 09:58:41 UTC+1 schrieb bassam king karzeddin:

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
> 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.

>

> 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
Why people here ignore deliberately what had been taught and proved so easily earlier in sci.math and elsewhere?

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?

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.

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?

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?

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
> 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.

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.

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
> implies directly the non existence of the real

> ($\sqrt[p]{q}$)?

>

>

> 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 an easy task for school students NOW!

>

> Regards

> Bassam King Karzeddin

> 19/02/17

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

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

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

to

(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

Feb 22, 2017, 2:59:41 AM2/22/17

to

BK

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
> How many times did I tell you, it is not "fiction

> numbers", its it Unicorn numbers. When do you get it?

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
> Am Dienstag, 21. Februar 2017 09:58:41 UTC+1 schrieb bassam king karzeddin:

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,
> Nope,

>

> a positive decimal representation approaches a real

> from below, so you don't have an interval enclosing

> the irrational number, thats B.S.

>

> 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.

>

> 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
> From this follows immediately:

>

> D_n =< D_n+1 for each n

>

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

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)?
> 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.

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?

>

> 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
> 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)

Bassam King Karzeddin

22/02/17

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!

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!

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

to

It is also the hobby of some mathematicians here (I forgot their names)

BK

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!

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!

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

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

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,
> 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!

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

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:

here is what I wrote:

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?

Am Mittwoch, 22. Februar 2017 10:41:12 UTC+1 schrieb bassam king karzeddin:

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.

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

without the division /10^n, you cannot take the limes.

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.

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.

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

to

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

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:

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

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:

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.

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.

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

to

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

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?

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

John Gabriel birdbrains clone?

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

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)

K(n) = 105*10^(3*n) - D'_n^3

Feb 22, 2017, 2:49:31 PM2/22/17

to

irrationals are treated arithmetically via continued fractions

-- and it is hard to do that

-- and it is hard to do that

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
> irrationals are treated arithmetically via continued fractions

> -- and it is hard to do that

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

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

to

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

Feb 22, 2017, 5:08:23 PM2/22/17

to

By juggling you mean Gabrieloconfusion and Johno

stupidics, the new religion of all cranks?

stupidics, the new religion of all cranks?

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

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

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

to

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

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!
> By juggling you mean Gabrieloconfusion and Johno

> stupidics, the new religion of all cranks?

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,

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

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

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

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.

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.

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),
> 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 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

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.

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.

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!

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!

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)
> 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,

This is absolutely correct

>

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

Yes
> D_n =/= D for each n (because D is irrational)

>

> Or in summary:

>

> D_n < D for each n

Bravo
> 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.

No, the Queen says it is not up to any body to allow its existence,
> 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 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

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)

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)

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!

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!

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.

c = 2^(1/3)

Is a solution.

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)
> 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

this is also magic
> A =< B <= A < B

>

> A =/= B <= A < B

I do not know who sets this nonsense logic
> 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

Maths works fine with axioms of morons, to generate their prior decisions only
> 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

I swear you do not have any clue how all the true mathematics is living peacefully in complete perpetual symphony
> 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.

But the challenge for you and everybody else on this planet, is to show (c) in R, exactly, when (a, b) are nonzero integers!
> Fermats

>

> a^3 + b^3 = c^3

>

> has solution a,b,c<>0 in R.

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

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

>

> 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,

Bassam King Karzeddin

27/02/17

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

to

So, this is really the bad news for you, and for sure

BK

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:

the same logic.

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

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
>> > 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

the same logic.

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

to

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

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

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

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

to

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

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

> > 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

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.

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.

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.

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.

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