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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

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 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 21, 2017 3:34 AM

> 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.

Regards
Bassam King Karzeddin
21/02/17

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