Discussion:
Why actually Polynomials are basically Diophantine equations?
(too old to reply)
bassam king karzeddin
2017-04-22 11:49:12 UTC
Permalink
Raw Message
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),

And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation

Regards
Bassam King Karzeddin
22th, April, 2017
bassam king karzeddin
2017-04-23 18:16:29 UTC
Permalink
Raw Message
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for
a given Polynomial of (nth) degree, and becomes
easily clearer if the given polynomial of odd degree,
where then it is more than trivial case even first by
trial and error to substitute a rational number or
generally a constructible number in its simple form,
for a solution as (n/m), for some nonzero integer
(n),
And if your solution requires both your integers (n
n and m) to be each with infinite sequence of digits,
then your solution actually either a constructible
number or an approximate solution that actually does
not 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

so 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

(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

So, 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)

Then we can simply get the following Diophantine equation by direct substitution

(a_2)*(b_2)*n^3 + (a_1)*(b_2)*n*m^2 + (a_2)*(b_1)*m^3 = 0

So, 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

Regards
Bassam King Karzeddin
23th, April, 2017
Don Redmond
2017-04-24 00:14:58 UTC
Permalink
Raw Message
Post by bassam king karzeddin
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for
a given Polynomial of (nth) degree, and becomes
easily clearer if the given polynomial of odd degree,
where then it is more than trivial case even first by
trial and error to substitute a rational number or
generally a constructible number in its simple form,
for a solution as (n/m), for some nonzero integer
(n),
And if your solution requires both your integers (n
n and m) to be each with infinite sequence of digits,
then your solution actually either a constructible
number or an approximate solution that actually does
not 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
so 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
(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
So, 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)
Then we can simply get the following Diophantine equation by direct substitution
(a_2)*(b_2)*n^3 + (a_1)*(b_2)*n*m^2 + (a_2)*(b_1)*m^3 = 0
So, 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
Regards
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
b***@gmail.com
2017-04-24 01:06:43 UTC
Permalink
Raw Message
Probably there is no point, except a confused school skipper
BKK. Lets look at the cuboid, in the reals it has obviously
solutions. We can use the following problem formulation, where
a,b,c are the sides of the cuboid:

a^2 + b^2 = p^2

a^2 + c^2 = q^2

b^2 + c^2 = r^2

a^2 + r^2 = s^2

A cuboid is just a brick, and it obviously has all the
four diagonals (in the reals). The above are polynomial
equations. Now if we ask for integer solution, the problem is
usually called a diophantine problem. Typically its

not only equations but also inequations, in the present
case we would have a,b,c>0 as well. And as of now, there
seems no definitive answer yet to the cuboid problem itself.
So this example Diophantine equation seems to be an open problem.

"In mathematics, a Diophantine equation is a polynomial
equation, usually in two or more unknowns, such that
only the integer solutions are sought or studied"
https://en.wikipedia.org/wiki/Diophantine_equation

"A perfect cuboid (also called a perfect box) is an
Euler brick whose space diagonal also has integer length.
As of May 2015, no example of a perfect cuboid had been
found and no one has proven that none exist."
https://en.wikipedia.org/wiki/Euler_brick#Perfect_cuboid
Post by Don Redmond
Post by bassam king karzeddin
Regards
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
b***@gmail.com
2017-04-24 01:31:30 UTC
Permalink
Raw Message
Now consider the variation asking for a rational cuboid,
with rational diagonals. If we use pairs (p,q) to model
a rational p/q we can indeed get a new Diophantine equation.

But obviously a rational cuboid would also imply an integer
cuboid, just take the least common multiple z of the
denominators, and then measure everything in 1/z.

This is because the equations are homogenous in power 2.
Post by b***@gmail.com
Probably there is no point, except a confused school skipper
BKK. Lets look at the cuboid, in the reals it has obviously
solutions. We can use the following problem formulation, where
a^2 + b^2 = p^2
a^2 + c^2 = q^2
b^2 + c^2 = r^2
a^2 + r^2 = s^2
A cuboid is just a brick, and it obviously has all the
four diagonals (in the reals). The above are polynomial
equations. Now if we ask for integer solution, the problem is
usually called a diophantine problem. Typically its
not only equations but also inequations, in the present
case we would have a,b,c>0 as well. And as of now, there
seems no definitive answer yet to the cuboid problem itself.
So this example Diophantine equation seems to be an open problem.
"In mathematics, a Diophantine equation is a polynomial
equation, usually in two or more unknowns, such that
only the integer solutions are sought or studied"
https://en.wikipedia.org/wiki/Diophantine_equation
"A perfect cuboid (also called a perfect box) is an
Euler brick whose space diagonal also has integer length.
As of May 2015, no example of a perfect cuboid had been
found and no one has proven that none exist."
https://en.wikipedia.org/wiki/Euler_brick#Perfect_cuboid
Post by Don Redmond
Post by bassam king karzeddin
Regards
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
bassam king karzeddin
2017-04-24 08:58:51 UTC
Permalink
Raw Message
Post by Don Redmond
Post by bassam king karzeddin
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for
a given Polynomial of (nth) degree, and becomes
easily clearer if the given polynomial of odd degree,
where then it is more than trivial case even first by
trial and error to substitute a rational number or
generally a constructible number in its simple form,
for a solution as (n/m), for some nonzero integer
(n),
And if your solution requires both your integers (n
n and m) to be each with infinite sequence of digits,
then your solution actually either a constructible
number or an approximate solution that actually does
not 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
so 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
(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
So, 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)
Then we can simply get the following Diophantine equation by direct substitution
(a_2)*(b_2)*n^3 + (a_1)*(b_2)*n*m^2 + (a_2)*(b_1)*m^3 = 0
So, 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
Regards
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 STORY for 2^{1/3}, or higher root operations (not purely even integer root operation), wonder!

Regards
BK
b***@gmail.com
2017-04-24 09:11:27 UTC
Permalink
Raw Message
The cuboid is a nice example, where in the reals sqrt(3)
exists, but as Diophantine equation, sqrt(3) doesn't exists.
Just take a cuboid a=1,b=1,c=1 the space diagonal will be
sqrt(3), but its not an integer and also not a rational.

The diophanitine equation is:

a^2 + b^2 = p^2

a^2 + c^2 = q^2

b^2 + c^2 = r^2

a^2 + r^2 = s^2

So diaphantine equations don't model the reality. How
can there be ideal bricks (cuboid), whichs are excluded by
diophantine equations, but which exist in the real world?
(well the mathematical bick is an idealization, neverthelss)

So there is no story linking diophantine to reals. Diophantines
are about integer, but not about reals. You cannot transfer
any results from diophantine to reals. Do you think a a=1,
b=1, c=1 cuboid or any other cuboid is unreal?

Here is a picture of some bricks:
Loading Image...
Post by bassam king karzeddin
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 STORY for 2^{1/3}, or higher root operations (not purely even integer root operation), wonder!
Regards
BK
bassam king karzeddin
2017-05-08 18:03:41 UTC
Permalink
Raw Message
Post by bassam king karzeddin
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for
a given Polynomial of (nth) degree, and becomes
easily clearer if the given polynomial of odd degree,
where then it is more than trivial case even first by
trial and error to substitute a rational number or
generally a constructible number in its simple form,
for a solution as (n/m), for some nonzero integer
(n),
And if your solution requires both your integers (n
n and m) to be each with infinite sequence of digits,
then your solution actually either a constructible
number or an approximate solution that actually does
not 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
so 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
(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
So, 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)
Then we can simply get the following Diophantine equation by direct substitution
(a_2)*(b_2)*n^3 + (a_1)*(b_2)*n*m^2 + (a_2)*(b_1)*m^3 = 0
So, 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
Regards
Bassam King Karzeddin
23th, April, 2017
this is to say that if the solution isn't rational, then must be irrational (according to your ruined mathematics), and hopelessly and so madly they generally bring that irrational root again in rational form (but with many finite numbers of digits, to convince the so innocent like you that is irrational, and so, unfortunately, the do not have any other choice for sure, and that irrational naturally would remain poisoned forever in their heads as a meaningless notations (not even physically or geometrically represented exactly)

And the poor mind would shake his head happily with his new finding and so peculiar ability to understand things that normal people can not, and funny believing himself absolutely and strictly beyond any little doubt, a real tragedy with those top mathematicians for sure

What is this world of mad distinguished people is this really, wonder!

BK
bassam king karzeddin
2017-04-24 08:48:06 UTC
Permalink
Raw Message
On Sunday, April 23, 2017 at 1:20:05 PM UTC-5, bassam
Post by bassam king karzeddin
Post by bassam king karzeddin
It is not so hard to recognize this simple truth
for
Post by bassam king karzeddin
Post by bassam king karzeddin
a given Polynomial of (nth) degree, and becomes
easily clearer if the given polynomial of odd
degree,
Post by bassam king karzeddin
Post by bassam king karzeddin
where then it is more than trivial case even
first by
Post by bassam king karzeddin
Post by bassam king karzeddin
trial and error to substitute a rational number
or
Post by bassam king karzeddin
Post by bassam king karzeddin
generally a constructible number in its simple
form,
Post by bassam king karzeddin
Post by bassam king karzeddin
for a solution as (n/m), for some nonzero integer
(n),
And if your solution requires both your integers
(n
Post by bassam king karzeddin
Post by bassam king karzeddin
n and m) to be each with infinite sequence of
digits,
Post by bassam king karzeddin
Post by bassam king karzeddin
then your solution actually either a
constructible
Post by bassam king karzeddin
Post by bassam king karzeddin
number or an approximate solution that actually
does
Post by bassam king karzeddin
Post by bassam king karzeddin
not 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 karzeddin
so 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 karzeddin
So, 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 karzeddin
Then 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 karzeddin
So, 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 karzeddin
Regards
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
b***@gmail.com
2017-04-24 11:01:57 UTC
Permalink
Raw Message
But they also discovered sqrt(2), etc... as irrational, can be
also shown by means of diophantine equation. But sqrt(2) is
constructible.

So diophantine equations don't buy you anything to understand
constructible numbers. And subsequetly they are also not
adequate for real numbers.

This should answer the question whats the difference between
polynomials in the reals and diophantie equations.
Post by bassam king karzeddin
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
bassam king karzeddin
2017-04-24 19:01:08 UTC
Permalink
Raw Message
Post by b***@gmail.com
But they also discovered sqrt(2), etc... as irrational, can be
also shown by means of diophantine equation. But sqrt(2) is
constructible.
So diophantine equations don't buy you anything to understand
constructible numbers. And subsequetly they are also not
adequate for real numbers.
This should answer the question whats the difference between
polynomials in the reals and diophantie equations.
Post by bassam king karzeddin
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
Did not I say deliberately, to jump over the linear or Quadratic polynomials moron

Did not I mention (in my third post here) that they do always fabricate a rational solution no matter however long it is at your Fake Paradise (infinity), or generally in a constructible form, where this might be exact or may be an approximation too, wonder

But your canny habit of spoiling rare and hot topics is well understood for sure

Maybe you need more hints, but even though would not work with you, for sure

BK
b***@gmail.com
2017-04-24 19:18:00 UTC
Permalink
Raw Message
You said "It is not so hard to recognize this simple truth
for a given Polynomial of (nth) degree". The cuboid equation
is *quadratic*:

a^2 + b^2 = p^2

a^2 + c^2 = q^2

b^2 + c^2 = r^2

a^2 + r^2 = s^2

http://mathworld.wolfram.com/PerfectCuboid.html

Still diophantine probably says it doesn't exists,
but bricks obviously exist, and arent unreal.
If you take a=1, b=1 and c=1, you get insolvable
diophantine. If you allow rationals you also get

an insolvable equation. But bricks are not unreal:

Here is a picture of some bricks:
http://mailbricks.com/wp-content/uploads/2015/01/bricks.jpg
Post by bassam king karzeddin
Post by b***@gmail.com
But they also discovered sqrt(2), etc... as irrational, can be
also shown by means of diophantine equation. But sqrt(2) is
constructible.
So diophantine equations don't buy you anything to understand
constructible numbers. And subsequetly they are also not
adequate for real numbers.
This should answer the question whats the difference between
polynomials in the reals and diophantie equations.
Post by bassam king karzeddin
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
Did not I say deliberately, to jump over the linear or Quadratic polynomials moron
Did not I mention (in my third post here) that they do always fabricate a rational solution no matter however long it is at your Fake Paradise (infinity), or generally in a constructible form, where this might be exact or may be an approximation too, wonder
But your canny habit of spoiling rare and hot topics is well understood for sure
Maybe you need more hints, but even though would not work with you, for sure
BK
bassam king karzeddin
2018-02-13 15:30:50 UTC
Permalink
Raw Message
Post by b***@gmail.com
You said "It is not so hard to recognize this simple truth
for a given Polynomial of (nth) degree". The cuboid equation
a^2 + b^2 = p^2
a^2 + c^2 = q^2
b^2 + c^2 = r^2
a^2 + r^2 = s^2
http://mathworld.wolfram.com/PerfectCuboid.html
Still diophantine probably says it doesn't exists,
but bricks obviously exist, and arent unreal.
If you take a=1, b=1 and c=1, you get insolvable
diophantine. If you allow rationals you also get
http://mailbricks.com/wp-content/uploads/2015/01/bricks.jpg
Oops, this problem is interesting and bricks do always exist up to your choice in constructible numbers generally, not any big deal

But you can't make them all rationals, for sure

Is that all difficult for you? wonder!

And assuming that I show you my simple elementary solution, then wouldn't you especially or anyone else try to teach me again my same solution shamelessly under the sunlight and every one's big eyes? wonder!

But most likely many people had solved it even few thousands years back I guess, but please check its history (bursigan), since you are an encyclopedia source, for sure

BKK
Post by b***@gmail.com
Post by bassam king karzeddin
Post by b***@gmail.com
But they also discovered sqrt(2), etc... as irrational, can be
also shown by means of diophantine equation. But sqrt(2) is
constructible.
So diophantine equations don't buy you anything to understand
constructible numbers. And subsequetly they are also not
adequate for real numbers.
This should answer the question whats the difference between
polynomials in the reals and diophantie equations.
Post by bassam king karzeddin
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
Did not I say deliberately, to jump over the linear or Quadratic polynomials moron
Did not I mention (in my third post here) that they do always fabricate a rational solution no matter however long it is at your Fake Paradise (infinity), or generally in a constructible form, where this might be exact or may be an approximation too, wonder
But your canny habit of spoiling rare and hot topics is well understood for sure
Maybe you need more hints, but even though would not work with you, for sure
BK
bassam king karzeddin
2017-07-30 11:12:27 UTC
Permalink
Raw Message
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
So yes, there isn't any polynomial but basically Diophantine Eqns for sure

BKK
bassam king karzeddin
2017-08-06 12:57:56 UTC
Permalink
Raw Message
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
Yes, every alleged polynomial equation can be reduced to its origin which is only Diophantine Equations for sure

And the rest is SO obvious brain fart of the alleged most genius mathematicians in the history of science and mathematics for sure
BKK
bassam king karzeddin
2017-09-18 08:01:51 UTC
Permalink
Raw Message
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
Sure, no polynomials exist, only Diophantine equation which is their origin, for sure

But business in mathematics was so necessary to show that unneeded alleged rare genius talents (really so funny history of mathematics was)
BKK
b***@gmail.com
2017-09-18 09:08:18 UTC
Permalink
Raw Message
The year 2525, BKK still writing "sure" and "for sure".
Post by bassam king karzeddin
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
Sure, no polynomials exist, only Diophantine equation which is their origin, for sure
But business in mathematics was so necessary to show that unneeded alleged rare genius talents (really so funny history of mathematics was)
BKK
bassam king karzeddin
2018-02-01 08:54:08 UTC
Permalink
Raw Message
Post by b***@gmail.com
The year 2525, BKK still writing "sure" and "for sure".
The year 2525 would arrive with only correct mathematics stated strictly by the KING, For sure
BKK
Post by b***@gmail.com
Post by bassam king karzeddin
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
Sure, no polynomials exist, only Diophantine equation which is their origin, for sure
But business in mathematics was so necessary to show that unneeded alleged rare genius talents (really so funny history of mathematics was)
BKK
bassam king karzeddin
2018-02-12 10:32:25 UTC
Permalink
Raw Message
Post by b***@gmail.com
The year 2525, BKK still writing "sure" and "for sure".
Post by bassam king karzeddin
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
Sure, no polynomials exist, only Diophantine equation which is their origin, for sure
But business in mathematics was so necessary to show that unneeded alleged rare genius talents (really so funny history of mathematics was)
BKK
FOR SURE
bkk
Zelos Malum
2018-02-15 13:51:19 UTC
Permalink
Raw Message
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
what is it with your obsession about Diophantine equations? They are not that special
bassam king karzeddin
2018-02-20 15:21:04 UTC
Permalink
Raw Message
Post by Zelos Malum
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
what is it with your obsession about Diophantine equations? They are not that special
What a moron indeed? wonder!

BKK
bassam king karzeddin
2018-03-03 18:21:03 UTC
Permalink
Raw Message
Post by bassam king karzeddin
It is not so hard to recognize this simple truth for a given Polynomial of (nth) degree, and becomes easily clearer if the given polynomial of odd degree, where then it is more than trivial case even first by trial and error to substitute a rational number or generally a constructible number in its simple form, for a solution as (n/m), for some nonzero integer (n),
And if your solution requires both your integers (n and m) to be each with infinite sequence of digits, then your solution actually either a constructible number or an approximate solution that actually does not exist for that polynomial or that same Diophantine equation
Regards
Bassam King Karzeddin
22th, April, 2017
Loading...