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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




Re: The non existence of p'th root of any prime number, for (p>2) prime
Posted:
Feb 21, 2017 3:58 AM


On Monday, February 20, 2017 at 8:56:10 PM UTC+3, burs...@gmail.com wrote: > 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 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



