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

Replies: 46   Last Post: Oct 12, 2017 1:41 AM

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 bursejan@gmail.com Posts: 4,592 Registered: 9/25/16
Re: The non existence of p'th root of any prime number, for (p>2) prime
Posted: Feb 21, 2017 7:22 AM

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?

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)