*Post by bassam king karzeddin*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 non-negative 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 non-existing integers (since this is obviously impossible and forever and for sure), which implies strictly that our 10^{1/3} is purely a fiction and non-existing 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 top-most 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 two-meter 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