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

Replies: 71   Last Post: Dec 16, 2017 8:46 AM

 Messages: [ Previous | Next ]
 bassam king karzeddin Posts: 1,268 Registered: 12/8/04
Re: The non existence of p'th root of any prime number, for (p>2)
prime

Posted: Feb 27, 2017 1:10 PM

> > 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

Let us see how simple it is, and for simplicity, I would like to repeat it in decimal 10base number system, and once you get it, please teach and educate your teachers about it before you would certainly get brain washed and become as ignorant and a victim of old corrupted fake unnecessary mathematics

ONE Example of Old claim of mine, the famous number representing the arithmetical cube root of two does not exist, denoted by (2^(1/3)), or ($\sqrt[3]{2}$)

The proof:

I had introduced this general self proved Diophantine equation

IFF: (D(n))^p + K(n) = q*s^{n*p} < (D(n) + 1)^p

where (p, q) are prime numbers, (n) is non negative integer, (s > 1) is positive integer, then this Diophantine equation implies that D(n) is positive integer that must represent the approximation of the arithmetical p th root of q, with (n) number of accurate digits in (s) base number system, denoted by (q^(1/p)), or ($\sqrt[p]{q}$), K(n) is therefore positive integer

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

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

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

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

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

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

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

??? ??????.. ????????? ???????.

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

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

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

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

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

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

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

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

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

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

Regards
Bassam King Karzeddin
27/02/17

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