
Re: A very simple question to secondary school students?
Posted:
Nov 11, 2017 7:26 AM


On Wednesday, November 8, 2017 at 11:12:45 AM UTC+3, bassam king karzeddin wrote: > 1) How can you approximate the arithmetical cube root of say (10) without using the decimal notation, (.), in any number system, say simply 10base number system? > > 2) What ultimately that you must discover about the arithmetical exact cube root of (10) but again without using the decimal notation? > > Regards > Bassam King Karzeddin > Nov. 08th, 2017
So, unfortunately, it seems that school students never or rarely read here, but never mind to prove this again and again for any future interested (but clever) students
Strictly as per the requirement and respect for the OP question above mentioned
We already know how the current modern mathematics can do (in say 10base number system that is not any different from any other number system), where also the ancient mathematician could do approximately in simple fractions as (rational numbers)
so, the arithmetical cube root of say (10), denoted by 10^{1/3} without using the decimal notation, (.), in any number system, say simply 10base number system?
First approximation is clearly 2, thus, 10^{1/3} =/= 2 Second approximation is (21/10), thus, 10^{1/3} =/= 21/10 Third approximtion is (215/100), thus, 10^{1/3} =/= 43/20 Forth approximation is (2154/1000), thus, 10^{1/3} =/= 1077/500 Fifth approximation is (21544/10000), thus, 10^{1/3} =/= 2693/1250 Sixth approximation is (215443/100000), thus, 10^{1/3} =/= 215443/100000 Seventh approximation is (2154434/1000000), thus, 10^{1/3} =/= 1077217/500000 Eight approximation is (21544346/10^7), thus, 10^{1/3} =/= 10772173/5*10^6 Nineth approximation is (215443469/10^8), thus, 10^{1/3} =/= 215443469/10^8 Tenth approximation is (215443469/10^9), thus, 10^{1/3} =/= 215443469/10^9 Eleventh approximation is (215443469/10^10),thus,10^{1/3} =/= 215443469/10^10
Note that inequality is always there because approximation is always in rational numbers that can never equate our irrational number, and generally, it can be always approximated or expressed symbolically as a rational number of this simple and direct form as [N(m)/10^m], where m is nonnegative integer, and N(m) is positive integer with (m) digits, therefore the m'th approximation would be the ratio of +ve integer with m digits to an integer equals to
10^{m  1), where strictly, 10^{1/3} =/= N(m)/10^{m  1},
No matter if your integer N(m) can fill say only seven trillion galaxies size, where every trillion of digits can be stored only say in one (mm) cube
Since mathematics require both integers tending to infinity, where then this ratio becomes a ratio of two nonexisting integers (since this is obviously impossible and forever and for sure), which implies strictly that our 10^{1/3} is purely a fiction and nonexisting number, reminding also, it is impossible to construct it or describe its exact existence in geometry, hence a human brain fart number for sure
So, any clever student wouldn't let that simple decimal notation deceive him anymore as many alleged topmost genius mathematicians were badly deceived up to our dates so unfortunately
Nor they must be deceived any more about many alleged big theorems that legalize such real numbers especially after seeing many simple and rigorous proofs
But it must be understood the approximation purpose of our practical needs, for instance, to make a water tank of cube shape and twometer cube volume
Same thing you may so easily deduce about any alleged real number with infinite sequence of digits or terms after that decimal notation, and regardless whether you know or don't know the pattern of the sequence or repeated digits
Regards BKK

