On Sunday, April 23, 2017 at 1:20:05 PM UTC-5, bassam
for
Post by bassam king karzeddinPost by bassam king karzeddina given Polynomial of (nth) degree, and becomes
easily clearer if the given polynomial of odd
degree,
first by
or
form,
Post by bassam king karzeddinPost by bassam king karzeddinfor a solution as (n/m), for some nonzero integer
(n),
And if your solution requires both your integers
(n
digits,
constructible
does
Post by bassam king karzeddinPost by bassam king karzeddinnot exist for that polynomial or that same
Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
It seems that I have to make all the work ready for
everyone professional independently and step by step
in order to make the idea clearer, despite providing
you all the scattered pieces essential in my posts to
complete this easy work for sure
Post by bassam king karzeddinso let us consider polynomials with rational
coefficients as the first step, and jumping on the
first and second degree polynomials, and choosing the
third degree polynomial (reduced form) to explain
this simple idea
Post by bassam king karzeddin(x^3 + ax + b = 0)
In any case a real (x) can be obtained as a
solution, and (x) is ultimately expressed as a
fraction in decimal representation (i.e rational
number), to some finite sequence of digits as a
rational solution or approximate solution once (x)
considered irrational number
Post by bassam king karzeddinSo, let (x = n/m), and (a = a_1/a_2), and (b =
b_1/b_2), where all the terms (n, m, a_1, a_2, b_1,
b_2) are integers, and (m*(a_2)*(b_2) =/= 0)
Post by bassam king karzeddinThen we can simply get the following Diophantine
equation by direct substitution
Post by bassam king karzeddin(a_2)*(b_2)*n^3 + (a_1)*(b_2)*n*m^2 +
(a_2)*(b_1)*m^3 = 0
Post by bassam king karzeddinSo, when does this integer equation have solution?
Oops, my time is out now, to be completed later
Any way it is easy beyond limits, also extending
this simplest idea is not any harder problem, but the
point is that all polynomials with rational
coefficients are actually and only Diophantine
equations for sure
Post by bassam king karzeddinRegards
Bassam King Karzeddin
23th, April, 2017
I'm obviously missing your point, but wouldn't the
rational root theorem allow you to avoid these lovely
Diophantine equations
Don
The rational root theorem is checking if a rational number is the solution expressed in simple form as (n/m), for some finite integers
But when there is not any real solution the alleged solution of cubic equation formula for the real solution pretends a fake solution again in the same simple form above (n/m), but in this case both integers are with infinite sequence of digits, where simply they take few digits as a solution to convince the poor student that was really a real solution
But the simplest obvious fact that such a solution can never exists, being unreal and non existing for sure, since such division of two integers where each consists of infinite sequence of digits is not permissible in the holy grail principle of mathematics nor defined plus also impossible task or achievement for sure
This is actually the true meaning of non solvability of some Diophantine equation
Then there is not any shame if the currents mathematics confesses the truth and simply say the obvious fact, that solution generally does not exist, but only approximation solution instead of committing three obvious crimes against human minds (this is the whole point)
Crime 1) permitting the division operation of two undefined integers where each of them must consists of unknown infinite sequence of digits
Crime 2) Convincing the student that solution exists at infinity, which is not true absolutely
Crime 3) Obtaining the solution again in rational or generally constructible form ( however there is no other way except meaningless notations in mind only) and convincing the innocents students that is irrational solution
We must realize also that the ONLY proved root operation is the SQUARE root operation by the Pythagoras theorem, whereas other higher root operations (not purely even) were never proved but wrongly concluded, check it from history please
One day (before more than two thousands years), it was a very big discovery and a real revolution in the mathematical science for irrational numbers by the Pythagoreans, as Sqrt(2), but we never heard of alike discovery for 2^{1/3}, or higher valid proved root operation and purely even
Regards
Bassam King Karzeddin
24th, April, 2017