Discussion:
How can we obtain the real irrational root, (x^5 - x - 1 = 0)
(too old to reply)
bassam king karzeddin
2017-12-06 09:20:33 UTC
Permalink
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form

This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure

However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!

Regards
Bassam King Karzeddin
Dec. 6th, 2017
bassam king karzeddin
2017-12-06 14:41:17 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
https://www.quora.com/How-can-we-obtain-the-real-irrational-root-for-this-polynomial-x-5-x-1-0?__filter__&__nsrc__=2&__sncid__=967814154

Luckily, the question is still running up till now, despite someone re-edited my question as usual, but I preserved it in the first comment of mine for this purpose only since Quora doesn't provide me with revert option button like any other member, but it doesn't any matter for me since I want to avoid blocking me again in order to save those innocent students, for sure

However, within few hours, around 3500 views only with 14 followers wanting answers, but strangely hear only 6 views, which implies mainly that same Trolls who are resisting the climate change are only reading here as if this site is almost blocked due to their devilish actions, wonder!, but anyway it has an advantage that almost nobody would be able to (dirt, modify, delete, ...etc) your content, except reporting as abuse, but still existing

Some few common typo answers where provided, but aren't useful, where I commented them and I don't know if Quora they hide my comments

The problem is that the correct answer was already published by me, where no roots at all exists, even this might sound crazy for any new reader since the since the size of the tragedy is really so huge fictions in mathematics to believe from the first look, but slowly slowly people would understand everything in details

BKK
IV
2017-12-06 18:57:08 UTC
Permalink
Post by bassam king karzeddin
the correct answer was already published by me
Could you please give here to us the source of your mathematical scientific
publication?
konyberg
2017-12-06 20:25:25 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
https://www.quora.com/How-can-we-obtain-the-real-irrational-root-for-this-polynomial-x-5-x-1-0?__filter__&__nsrc__=2&__sncid__=967814154
Luckily, the question is still running up till now, despite someone re-edited my question as usual, but I preserved it in the first comment of mine for this purpose only since Quora doesn't provide me with revert option button like any other member, but it doesn't any matter for me since I want to avoid blocking me again in order to save those innocent students, for sure
However, within few hours, around 3500 views only with 14 followers wanting answers, but strangely hear only 6 views, which implies mainly that same Trolls who are resisting the climate change are only reading here as if this site is almost blocked due to their devilish actions, wonder!, but anyway it has an advantage that almost nobody would be able to (dirt, modify, delete, ...etc) your content, except reporting as abuse, but still existing
Some few common typo answers where provided, but aren't useful, where I commented them and I don't know if Quora they hide my comments
The problem is that the correct answer was already published by me, where no roots at all exists, even this might sound crazy for any new reader since the since the size of the tragedy is really so huge fictions in mathematics to believe from the first look, but slowly slowly people would understand everything in details
BKK
Well, there is a real irrational root: x = 1.1673...
The root may not have a closed form (what do you mean by real?), but not every number do.
Oh, I forgot; you don't believe in such numbers (that's your problem).
When graphing the function f(x) = x^5 - x - 1, we see it cutting the x-axis. That is the real root. Consider the x-axis as a number line. The equation also has four complex solutions.
KON
Pubkeybreaker
2017-12-06 20:47:52 UTC
Permalink
<snip>
Post by konyberg
Well, there is a real irrational root: x = 1.1673...
The root may not have a closed form (what do you mean by real?), but not every number do.
But it does have a closed form. One can represent it in terms of
either Elliptic functions or Theta functions. It does not have an
elementary closed form.
Post by konyberg
Oh, I forgot; you don't believe in such numbers (that's your problem).
When graphing the function f(x) = x^5 - x - 1, we see it cutting the x-axis. That is the real root. Consider the x-axis as a number line. The equation also has four complex solutions.
KON
konyberg
2017-12-06 21:39:19 UTC
Permalink
Post by Pubkeybreaker
<snip>
Post by konyberg
Well, there is a real irrational root: x = 1.1673...
The root may not have a closed form (what do you mean by real?), but not every number do.
But it does have a closed form. One can represent it in terms of
either Elliptic functions or Theta functions. It does not have an
elementary closed form.
Post by konyberg
Oh, I forgot; you don't believe in such numbers (that's your problem).
When graphing the function f(x) = x^5 - x - 1, we see it cutting the x-axis. That is the real root. Consider the x-axis as a number line. The equation also has four complex solutions.
KON
Correct.
KON
bassam king karzeddin
2017-12-07 08:07:23 UTC
Permalink
Post by konyberg
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
https://www.quora.com/How-can-we-obtain-the-real-irrational-root-for-this-polynomial-x-5-x-1-0?__filter__&__nsrc__=2&__sncid__=967814154
Luckily, the question is still running up till now, despite someone re-edited my question as usual, but I preserved it in the first comment of mine for this purpose only since Quora doesn't provide me with revert option button like any other member, but it doesn't any matter for me since I want to avoid blocking me again in order to save those innocent students, for sure
However, within few hours, around 3500 views only with 14 followers wanting answers, but strangely hear only 6 views, which implies mainly that same Trolls who are resisting the climate change are only reading here as if this site is almost blocked due to their devilish actions, wonder!, but anyway it has an advantage that almost nobody would be able to (dirt, modify, delete, ...etc) your content, except reporting as abuse, but still existing
Some few common typo answers where provided, but aren't useful, where I commented them and I don't know if Quora they hide my comments
The problem is that the correct answer was already published by me, where no roots at all exists, even this might sound crazy for any new reader since the since the size of the tragedy is really so huge fictions in mathematics to believe from the first look, but slowly slowly people would understand everything in details
BKK
Well, there is a real irrational root: x = 1.1673...
The root may not have a closed form (what do you mean by real?), but not every number do.
Oh, I forgot; you don't believe in such numbers (that's your problem).
When graphing the function f(x) = x^5 - x - 1, we see it cutting the x-axis. That is the real root. Consider the x-axis as a number line. The equation also has four complex solutions.
KON
Here is the full tragedy that indeed it is too hard to escape from being captured with it, which had been well explained to the mathematicians the size of many fictional stories, one after one till it became full mind covering

I had already explained the devilish purposes of making the negatives, then the invented imaginary, then the artificial continuity, then...finally where the XYZ negative positive coordinates ... till it became almost impossible to recover to normally sensible mathematics that was few thousands of years back, for sure

So, the mathematical illness symptoms are really so many and so huge to unbelievable limit

But only with Number theory and Diophantine Eqns. all that crap invented mathematics would be so easily refuted under every one own eye for sure

And who is on earth that can beat the numbers? wonder!

Just wait till I get some free time, and my answer would be in advance, no real root, or even imaginary roots exist for such a fabricated polynomials, sure, EVEN though I had alrady tought you about it, sure
BKK
bassam king karzeddin
2017-12-07 09:00:28 UTC
Permalink
Post by bassam king karzeddin
Post by konyberg
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
https://www.quora.com/How-can-we-obtain-the-real-irrational-root-for-this-polynomial-x-5-x-1-0?__filter__&__nsrc__=2&__sncid__=967814154
Luckily, the question is still running up till now, despite someone re-edited my question as usual, but I preserved it in the first comment of mine for this purpose only since Quora doesn't provide me with revert option button like any other member, but it doesn't any matter for me since I want to avoid blocking me again in order to save those innocent students, for sure
However, within few hours, around 3500 views only with 14 followers wanting answers, but strangely hear only 6 views, which implies mainly that same Trolls who are resisting the climate change are only reading here as if this site is almost blocked due to their devilish actions, wonder!, but anyway it has an advantage that almost nobody would be able to (dirt, modify, delete, ...etc) your content, except reporting as abuse, but still existing
Some few common typo answers where provided, but aren't useful, where I commented them and I don't know if Quora they hide my comments
The problem is that the correct answer was already published by me, where no roots at all exists, even this might sound crazy for any new reader since the since the size of the tragedy is really so huge fictions in mathematics to believe from the first look, but slowly slowly people would understand everything in details
BKK
Well, there is a real irrational root: x = 1.1673...
The root may not have a closed form (what do you mean by real?), but not every number do.
Oh, I forgot; you don't believe in such numbers (that's your problem).
When graphing the function f(x) = x^5 - x - 1, we see it cutting the x-axis. That is the real root. Consider the x-axis as a number line. The equation also has four complex solutions.
KON
Here is the full tragedy that indeed it is too hard to escape from being captured with it, which had been well explained to the mathematicians the size of many fictional stories, one after one till it became full mind covering
I had already explained the devilish purposes of making the negatives, then the invented imaginary, then the artificial continuity, then...finally where the XYZ negative positive coordinates ... till it became almost impossible to recover to normally sensible mathematics that was few thousands of years back, for sure
So, the mathematical illness symptoms are really so many and so huge to unbelievable limit
But only with Number theory and Diophantine Eqns. all that crap invented mathematics would be so easily refuted under every one own eye for sure
And who is on earth that can beat the numbers? wonder!
Just wait till I get some free time, and my answer would be in advance, no real root or even imaginary roots exist for such a fabricated polynomials, sure, EVEN though I had already taught you about it, sure
BKK
BIG HINT: see my last post here only

https://groups.google.com/forum/#!topic/sci.math/pW5JkeZspFQ
BKK
Zelos Malum
2017-12-07 09:36:08 UTC
Permalink
What are you talking about? They are giving you answers there.
bassam king karzeddin
2017-12-07 12:38:00 UTC
Permalink
Post by Zelos Malum
What are you talking about? They are giving you answers there.
Even the Quintic Equation you want to put your nose in, wonder!
please this not for illiterate people, so keep silent and learn freely and silently, and watch the total huge collapse by only a single Diophantine Equation (COUNTER EXAMPLE)

Since you had never realized that my last post only had already and absolutely killed all those nonsense answers and beyond any little doubt, also people had immediately realized this obvious fact, but none of them would ever dare to confess and tell you the simplest truth loudly, since we know the reasons, and by the way, it truly doesn't NEED any LITTLE PEER-REVIEW, so was PUBLISHED publically in such a dirty places, thus not deserving any reputable Journal or any reputable University to be honoured by its absolute truthness

And truly more and again, a skilled football player boy is much more worth than all the alleged top-most genius mathematicians on earth, including most of their BIG historical figures, for sure

And the wide audience of football would immediately recognize their little masters, but mathematicians would never realize their BIGGEST masters, but on the contrary, they wish to kill them badly and immediately for sure

Bassam King Karzeddin
Dec. 7th, 2017
Python
2017-12-07 12:52:15 UTC
Permalink
Post by bassam king karzeddin
Post by Zelos Malum
What are you talking about? They are giving you answers there.
Even the Quintic Equation you want to put your nose in, wonder!
please this not for illiterate people, so keep silent and learn freely and silently, and watch the total huge collapse by only a single Diophantine Equation (COUNTER EXAMPLE)
Since you had never realized that my last post only had already and absolutely killed all those nonsense answers and beyond any little doubt, also people had immediately realized this obvious fact, but none of them would ever dare to confess and tell you the simplest truth loudly, since we know the reasons, and by the way, it truly doesn't NEED any LITTLE PEER-REVIEW, so was PUBLISHED publically in such a dirty places, thus not deserving any reputable Journal or any reputable University to be honoured by its absolute truthness
And truly more and again, a skilled football player boy is much more worth than all the alleged top-most genius mathematicians on earth, including most of their BIG historical figures, for sure
And the wide audience of football would immediately recognize their little masters, but mathematicians would never realize their BIGGEST masters, but on the contrary, they wish to kill them badly and immediately for sure
Bassam King Karzeddin
Dec. 7th, 2017
Mr King-of-IDIOTS, you are an utter imbecile and mentally ill.
Serg io
2017-12-06 20:18:56 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;

https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0

x≈1.1673039782614186843




Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
bassam king karzeddin
2017-12-07 07:47:31 UTC
Permalink
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
@Serg io, please note that you are providing a rational solution, but mathematicians believe that the real root must be irrational number, hence your provided solution isn't at all a solution, so when shall your solution becomes real irrational number then show it to us

BKK
Serg io
2017-12-11 16:51:27 UTC
Permalink
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
@Serg io, please note that you are providing a rational solution, but mathematicians believe that the real root must be irrational number, hence your provided solution isn't at all a solution, so when shall your solution becomes real irrational number then show it to us
BKK
I am real and rational, you are irrational but are you also imaginary ?
bassam king karzeddin
2017-12-12 06:29:11 UTC
Permalink
Post by Serg io
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
@Serg io, please note that you are providing a rational solution, but mathematicians believe that the real root must be irrational number, hence your provided solution isn't at all a solution, so when shall your solution becomes real irrational number then show it to us
BKK
I am real and rational, you are irrational but are you also imaginary ?
So, you couldn't answer correctly, then you fail and jump into personal abusing, confining to the absolute fact that people generally are the same inhabitants of those old centuries, for sure
BKK
Dan Christensen
2017-12-12 14:11:29 UTC
Permalink
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
@Serg io, please note that you are providing a rational solution, but mathematicians believe that the real root must be irrational number, hence your provided solution isn't at all a solution, so when shall your solution becomes real irrational number then show it to us
BKK
I am real and rational, you are irrational but are you also imaginary ?
So, you couldn't answer correctly, then you fail and jump into personal abusing, confining to the absolute fact that people generally are the same inhabitants of those old centuries, for sure
HA, HA! It is YOU who wants to turn back the clock two thousand years, Crank Boy!

BTW, have any "artificial beings" been in contact with you about your goofy "system" yet? You wrote:

“Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now, for sure.”
-- Dec. 6, 2017

We really have to wonder about you, Crank Boy!


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Serg io
2017-12-12 15:12:56 UTC
Permalink
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
@Serg io, please note that you are providing a rational solution, but mathematicians believe that the real root must be irrational number, hence your provided solution isn't at all a solution, so when shall your solution becomes real irrational number then show it to us
BKK
I am real and rational, you are irrational but are you also imaginary ?
So, you couldn't answer correctly, then you fail and jump into personal abusing, confining to the absolute fact that people generally are the same inhabitants of those old centuries, for sure
BKK
I determine what answers are correct, not you, as you are both
irrational and imaginary. a ghost with ADHD.
bassam king karzeddin
2017-12-13 13:35:38 UTC
Permalink
Post by Serg io
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
@Serg io, please note that you are providing a rational solution, but mathematicians believe that the real root must be irrational number, hence your provided solution isn't at all a solution, so when shall your solution becomes real irrational number then show it to us
BKK
I am real and rational, you are irrational but are you also imaginary ?
So, you couldn't answer correctly, then you fail and jump into personal abusing, confining to the absolute fact that people generally are the same inhabitants of those old centuries, for sure
BKK
I determine what answers are correct, not you, as you are both
irrational and imaginary. a ghost with ADHD.
It seems that the meaning of irrational number solution for our current modern mathematicians minds and also those old mathematicians are still not so clear for them (as if in their back mind say more than 10 or 20 digits of accurate digits), never imagining or truly realizing that even a rational solution in decimal notations and in any number system can fill say only seven trillion galaxy sizes where everyone (mm) cube can store say only one trillion digits, wonder about their so silly and naive understanding what truly an irrational root of a polonomial would ultimatly mean?

And still, they would never understand what does a mind illusion mean exactly? for sure

BKK
bassam king karzeddin
2017-12-16 07:08:32 UTC
Permalink
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
@Serg io, please note that you are providing a rational solution, but mathematicians believe that the real root must be irrational number, hence your provided solution isn't at all a solution, so when shall your solution becomes real irrational number then show it to us
BKK
I am real and rational, you are irrational but are you also imaginary ?
So, you couldn't answer correctly, then you fail and jump into personal abusing, confining to the absolute fact that people generally are the same inhabitants of those old centuries, for sure
BKK
I determine what answers are correct, not you, as you are both
irrational and imaginary. a ghost with ADHD.
It seems that the meaning of irrational number solution for our current modern mathematicians minds and also those old mathematicians are still not so clear for them (as if in their back mind say more than 10 or 20 digits of accurate digits), never imagining or truly realizing that even a rational solution in decimal notations and in any number system can fill say only seven trillion galaxy sizes where everyone (mm) cube can store say only one trillion digits, wonder about their so silly and naive understanding what truly an irrational root of a polonomial would ultimatly mean?
And still, they would never understand what does a mind illusion mean exactly? for sure
BKK
Silence is so wonderful reaction before the absolute truth.
But it is very well understood that the absolute truths are very bitter taste almost for all dogmatic mythmakers, but it doesn't any matter for sure
BKK
Chris M. Thomasson
2018-11-09 06:36:49 UTC
Permalink
Post by Serg io
Post by bassam king karzeddin
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
@Serg io, please note that you are providing a rational solution, but mathematicians believe that the real root must be irrational number, hence your provided solution isn't at all a solution, so when shall your solution becomes real irrational number then show it to us
BKK
I am real and rational, you are irrational but are you also imaginary ?
Only if one of his plots has a non-zero imaginary part.
bassam king karzeddin
2018-01-04 11:21:17 UTC
Permalink
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
Bassam King Karzeddin
Post by bassam king karzeddin
Dec. 6th, 2017
Serg io can't recognize yet the rational approximation solution from exact solution, and he would never understand the difference, for sure
BKK
d***@gmail.com
2018-01-24 23:35:24 UTC
Permalink
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
mathhandbook gives
http://server.drhuang.com/input/?guess=solve%28x%5E5+-+x+-+1+%3D+0%29&inp=x%5E5+-+x+-+1+%3D+0

reference math handbook
www.mathHandbook.com
bassam king karzeddin
2018-01-25 06:42:17 UTC
Permalink
Post by d***@gmail.com
Post by Serg io
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
simple;
https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
x≈1.1673039782614186843
mathhandbook gives
http://server.drhuang.com/input/?guess=solve%28x%5E5+-+x+-+1+%3D+0%29&inp=x%5E5+-+x+-+1+%3D+0
reference math handbook
www.mathHandbook.com
Truly, no one seems to understand the real problems in mathematics, and I don't know why do many sites produce so shamelessly and only a rational number as an approximation to whatever degree of accuracy they might think of and contrary to the common old believes that states the real root must be real irrational algebraic number, so when shall they confess openly and so loudly that no such real root exists in any imaginable physical existing reality? wonder!
And the real root they do believe in is truly and only existing in their tiny minds as a phobia or ghost non-existing number for sure

Isn't your presented real APPROXIMATED root is (11673039782614186843/10^19), which is forever a RATIONAL number, and regardless of whatever your degree of accuracy that you can obtain? wonder!

So, more and more accuracy would take you ultimately to a ratio of two non-existing integers that are absolutely impossible to obtain

In short, the convergence principle concept is absolutely flawed concept, as proved earlier by me quite many times

And once you comprehend that no such real irrational root actually exists, then it becomes much easier for a keen researchers to uncover the obvious fact of those fictional fabricated other complex roots with imaginary parts, that we had already exposed separately in many posts, and proved so rigorously their pure fictionality and invalidities since all the other alleged complex roots for a fabricated polynomial (x^5 - x - 1 = 0), are basically associated strictly with that fictional and non-existing alleged real root, for sure

And remember one basic elementary and so important point that using the notation (≈) instead of (=) would never be a valid mathematical substitutions nor any meaningful mathematics, but certainly good enough for our little practical carpentry works

But mathematics was never any carpentry nor competing with the carpenter talent works

And refusing deliberately to see the so rigorous proven and PUBLISHED facts by All the professional mathematicians is called absolute stubbornness and unimaginable stupidity that would never last anymore, for sure

Regards
Bassam King Karzeddin
Jan. 25th, 2018
4***@gmail.com
2017-12-07 12:47:26 UTC
Permalink
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form assume there is not an irrational one
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
bassam king karzeddin
2017-12-07 08:39:15 UTC
Permalink
On Wednesday, December 6, 2017 at 3:25:38 PM UTC-5,
onsdag 6. desember 2017 15.41.32 UTC+1 skrev bassam
<snip>
Well, there is a real irrational root: x =
1.1673...
The root may not have a closed form (what do you
mean by real?), but not every number do.
But it does have a closed form. One can represent it
in terms of
either Elliptic functions or Theta functions. It
does not have an
elementary closed form.
Oh, I forgot; you don't believe in such numbers
(that's your problem).
When graphing the function f(x) = x^5 - x - 1, we
see it cutting the x-axis. That is the real root.
Consider the x-axis as a number line. The equation
also has four complex solutions.
KON
Theta functions or generally trig.functions were based on any random angle variable, where most of the angles mathematicians believe in are purely fictional and of impossible existence angels that were as a product of human mind hallucinations or brain fart angels of the mathematicians long ago, and the existing angels are only those rare angles in triangles that have at least two constructible sides, where the third side must be constructible as a result

However, we have named some sets of those impossible existing angels with PUBLISHED proofs here and elsewhere and fantastically some of those fictional angels are integer degrees angles say (n) degree, where (n) isn't divisible by 3, Where (PI = 180 DEGREES)

So, those trig.functions produce nothingness at the core unless we deal with them as constructible angels by computers and not the same way we understand them conventionally, so the contradictions are only in mathematicians minds for sure

BKK
Dan Christensen
2017-12-07 13:22:38 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0

There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)

It must be possible to obtain the result to ANY number of significant digits.

Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
bassam king karzeddin
2017-12-07 15:43:46 UTC
Permalink
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
See the responses to my so peculiar claim (in the history of mathematics) coming immediately from fictional well-known hired Trolls as Dan and Python and...etc, but the people who immediately understand it keeps so long silence, wonder!

Does this issue truly need more than 10 minutes to fully understand even by a clever school student or a layperson? wonder!

Why all that hypocrisy in mathematics? wonder!
where are your alleged honesty and morality mathematicians? (in fictional and fabricated history books only most likely) wonder!

What type of people are you professional mathematicians? (cheaters most likely) wonder!

Do you think that your silence would resolve this most important issue ever made in the history of mathematics? wonder!

Did you understand truly why did the KING promise you here and more than ten years back to make a final third world war against your huge ignorance and cowardliness? wonder!

And how are those many fictional characters as (Dan, Python, Jim burse, Me, Zelos, J4, Burgan, Serg io, KON, ...etc) can be cured completely of their brain fart phobias? (despite taking hundreds of lectures freely) wonder!

Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!

And do you think truly whatever sequence of digits you bring with you would beat a STRONG COUNTER-EXAMPLE of non-solvable Diophantine Equation based on non-existing integer solution? wonder!

Or did you truly believe that your absolute stupidity with fabricated artificial continuity based on distances as Delta or Epsilon Would save your short neck from the sharp sword of the KING with the Exact meaning of Perfection of Diophantine Equations

Or maybe you wrongly thought that your fake non-existing Paradise called INFINITY would also save you from such miserable situations? wonder!

But also that imaginary fabrication of numbers as opposed to each other (+/-) wouldn't help you for sure

You have lost the FINAL war against the KING for sure

And the Queen will be finally freed and would soon curse and shame you perpetually and forever, for all your ignorance and be insisting to die with it stubbornly for sure

Now, many school students had grasped the easy trick, and consequently can make so many counterexamples to all your unnecessary buissness inherited foolishness and so naive understanding for sure

Did you truly understand what a counterexample with INTEGERS can make? wonder!

But miracles for sure


Bassam King Karzeddin
Dec. 7th, 217
Dan Christensen
2017-12-07 16:10:01 UTC
Permalink
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means, Crank Boy? I'm sure your super intelligent "future artificial beings" will grasp it quickly enough if you cannot.

BKK actually wrote:

"Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now, for sure." [OMG! What a psycho!]
--Dec. 6, 2017


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
FredJeffries
2017-12-08 16:33:49 UTC
Permalink
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits.
Really?

Please obtain 12,543,678,443,345,340 significant digits.

Oh, yes, show your work.
bassam king karzeddin
2017-12-09 12:02:59 UTC
Permalink
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure

And definitely, you and the vast majority of mainstream mathematicians would never understand what does the word solution exactly mean in perfect mathematics, for 100% sure

Or what are the fabricated polynomials even mean, sure

BKK



Crank Boy? I'm sure your super intelligent "future artificial beings" will grasp it quickly enough if you cannot.
Post by Dan Christensen
"Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now, for sure." [OMG! What a psycho!]
--Dec. 6, 2017
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Dan Christensen
2017-12-09 22:15:16 UTC
Permalink
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try, Crank Boy. That should keep you busy for a couple of centuries before your super intelligent "future artificial beings" take over. (Hee, hee!)

BKK actually wrote here:

"Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now."
-- Dec. 6, 2017


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
bassam king karzeddin
2017-12-10 07:04:19 UTC
Permalink
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try, Crank Boy. That should keep you busy for a couple of centuries before your super intelligent "future artificial beings" take over. (Hee, hee!)
"Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now."
-- Dec. 6, 2017
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
No, I'm not thinking at all of making calculus, since I knew this actually an engineering problem where simply perfect mathematics may not be so helpful in this regard, exactly like approximating the area of a circle that we need in OUR daily life problems, which is an easy task once you consider a circle as a regular constructible polygon with many sides
But the perfect circle doesn't exist in any imaginable reality, sure
BKK
Dan Christensen
2017-12-10 15:51:47 UTC
Permalink
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try, Crank Boy. That should keep you busy for a couple of centuries before your super intelligent "future artificial beings" take over. (Hee, hee!)
"Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now."
-- Dec. 6, 2017
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
No, I'm not thinking at all of making calculus, since I knew this actually an engineering problem where simply perfect mathematics may not be so helpful in this regard, exactly like approximating the area of a circle that we need in OUR daily life problems, which is an easy task once you consider a circle as a regular constructible polygon with many sides
But the perfect circle doesn't exist in any imaginable reality, sure
Both analytic and numerical solutions are required at times in science and engineering. And the theoretical basis for numerical analysis is, of course, calculus. You have to know what you are approximating.

So, no calculus, approximations. Maybe your super intelligent "future artificial beings" will be able to explain it to you. I wouldn't count on it,
though. (Hee, hee!)


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Dan Christensen
2017-12-10 22:05:27 UTC
Permalink
Post by bassam king karzeddin
Post by Dan Christensen
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try, Crank Boy. That should keep you busy for a couple of centuries before your super intelligent "future artificial beings" take over. (Hee, hee!)
"Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now."
-- Dec. 6, 2017
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
No, I'm not thinking at all of making calculus, since I knew this actually an engineering problem where simply perfect mathematics may not be so helpful in this regard, exactly like approximating the area of a circle that we need in OUR daily life problems, which is an easy task once you consider a circle as a regular constructible polygon with many sides
But the perfect circle doesn't exist in any imaginable reality, sure
Both analytic and numerical solutions are required at times in science and engineering. And the theoretical basis for numerical analysis is, of course, calculus. You have to know what you are approximating.

So, no calculus, NO approximations. Maybe your super intelligent "future artificial beings" will be able to explain it to you. I wouldn't count on it,
though. (Hee, hee!)


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
FredJeffries
2017-12-10 15:18:12 UTC
Permalink
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Dan Christensen
2017-12-10 15:59:19 UTC
Permalink
Post by FredJeffries
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Fred seems to think there is a theoretical upper limit on the number of significant digits.

It wouldn't happen to be Archie Pu's infinity border line at 10^666, eh, Fred?


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
FredJeffries
2017-12-10 16:17:22 UTC
Permalink
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Fred seems to think there is a theoretical upper limit on the number of significant digits.
It wouldn't happen to be Archie Pu's infinity border line at 10^666, eh, Fred?
Typical evasive reaction of a bully when backed into a corner.

YOU made the idiotic claim that "the solution can be obtained to ANY number of significant digits."

So obtain 12,543,678,443,345,340 significant digits.
Dan Christensen
2017-12-10 16:32:57 UTC
Permalink
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Fred seems to think there is a theoretical upper limit on the number of significant digits.
It wouldn't happen to be Archie Pu's infinity border line at 10^666, eh, Fred?
Typical evasive reaction of a bully when backed into a corner.
So, do you think there is an theoretical upper limit on the number of significant digits or not? You would rather not say? You would rather play the silly bugger? Oh, well...


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Post by FredJeffries
YOU made the idiotic claim that "the solution can be obtained to ANY number of significant digits."
So obtain 12,543,678,443,345,340 significant digits.
FredJeffries
2017-12-10 17:32:49 UTC
Permalink
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Fred seems to think there is a theoretical upper limit on the number of significant digits.
It wouldn't happen to be Archie Pu's infinity border line at 10^666, eh, Fred?
Typical evasive reaction of a bully when backed into a corner.
So, do you think there is an theoretical upper limit on the number of significant digits or not? You would rather not say? You would rather play the silly bugger? Oh, well...
More cowardly evasion by the bully.

It is IRRELEVANT what I THINK. You made the idiotic claim. This is sci.math. You have to PROVE your claims.

Obtain 12,543,678,443,345,340 significant digits.
Dan Christensen
2017-12-10 22:17:58 UTC
Permalink
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Fred seems to think there is a theoretical upper limit on the number of significant digits.
It wouldn't happen to be Archie Pu's infinity border line at 10^666, eh, Fred?
Typical evasive reaction of a bully when backed into a corner.
So, do you think there is an theoretical upper limit on the number of significant digits or not? You would rather not say? You would rather play the silly bugger? Oh, well...
More cowardly evasion by the bully.
It is IRRELEVANT what I THINK.
So, you would rather NOT clarify your position. That's... OK. I guess.


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
FredJeffries
2017-12-11 02:00:38 UTC
Permalink
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Fred seems to think there is a theoretical upper limit on the number of significant digits.
It wouldn't happen to be Archie Pu's infinity border line at 10^666, eh, Fred?
Typical evasive reaction of a bully when backed into a corner.
So, do you think there is an theoretical upper limit on the number of significant digits or not? You would rather not say? You would rather play the silly bugger? Oh, well...
More cowardly evasion by the bully.
It is IRRELEVANT what I THINK.
So, you would rather NOT clarify your position. That's... OK. I guess.
I will gladly "clarify my position".

My "position" has nothing to do with a "theoretical upper limit" nor with mathematics at all. (I have no idea what you mean by "theoretical upper limit" or what relevance it has to do with your idiotic claim, since you have never given a definition for "theoretical upper limit", unsurprisingly since you have never given any definition for anything at all and have demonstrated that you don't know what a definition in mathematics is).

I freely admit my mathematical ignorance and illiteracy, so it is pointless of you to try to demonstrate THAT.

My "position" is that you will never "obtain" 12,543,678,443,345,340 significant digits for the solution to x^5 - x - 1 = 0 despite your claim that you can "obtain the result to ANY number of significant digits" [your emphasis].

You will not "obtain" the digits, not for any "theoretical" reason, but because you are a mere bullying coward and will neither begin any attempt, nor admit that it is impossible, nor even waffle about what you meant by "obtain".

You have demonstrated that point to my satisfaction, so I shall withdraw from this discussion and allow you and the other bullies to have the last words.
Dan Christensen
2017-12-11 04:37:23 UTC
Permalink
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Fred seems to think there is a theoretical upper limit on the number of significant digits.
It wouldn't happen to be Archie Pu's infinity border line at 10^666, eh, Fred?
Typical evasive reaction of a bully when backed into a corner.
So, do you think there is an theoretical upper limit on the number of significant digits or not? You would rather not say? You would rather play the silly bugger? Oh, well...
More cowardly evasion by the bully.
It is IRRELEVANT what I THINK.
So, you would rather NOT clarify your position. That's... OK. I guess.
I will gladly "clarify my position".
At last!
Post by FredJeffries
My "position" has nothing to do with a "theoretical upper limit" nor with mathematics at all. (I have no idea what you mean by "theoretical upper limit" or what relevance it has...
I see. If you are interested, you may find these articles useful as starting point to understanding the issue here:

https://en.wikipedia.org/wiki/Real_number

https://en.wikipedia.org/wiki/Significant_figures

https://en.wikipedia.org/wiki/Mathematical_proof


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
bassam king karzeddin
2017-12-11 06:33:29 UTC
Permalink
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by FredJeffries
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
Post by Dan Christensen
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
See the graph at https://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There you can get x approx. = 1.167303978261418684256045899854842180721 (40 significant digits)
It must be possible to obtain the result to ANY number of significant digits.
Do you really think your "future artificial beings" will be satisfied with your claim that x^5 - x - 1 = 0 has no solution?
Dan is bringing with him a rational number with only 40 sequence digit for an irrational number that supposed to be with endless digits, wonder!
Again, the solution can be obtained to ANY number of significant digits. Do you understand what that means,
It means nothing at all as long as those rational (n/m) number sequence of alleged sequence digiits are not any exact solution to this insolvable Diaphontine Eqn. (n^5 = m^5 + nm^4), for sure
If you think you can do calculus without irrational numbers, pi and 40 degree angles, do give it a try
And if YOU think "the solution can be obtained to ANY number of significant digits," do obtain 12,543,678,443,345,340 significant digits.
Fred seems to think there is a theoretical upper limit on the number of significant digits.
It wouldn't happen to be Archie Pu's infinity border line at 10^666, eh, Fred?
Typical evasive reaction of a bully when backed into a corner.
So, do you think there is an theoretical upper limit on the number of significant digits or not? You would rather not say? You would rather play the silly bugger? Oh, well...
More cowardly evasion by the bully.
It is IRRELEVANT what I THINK.
So, you would rather NOT clarify your position. That's... OK. I guess.
I will gladly "clarify my position".
At last!
Post by FredJeffries
My "position" has nothing to do with a "theoretical upper limit" nor with mathematics at all. (I have no idea what you mean by "theoretical upper limit" or what relevance it has...
https://en.wikipedia.org/wiki/Real_number
https://en.wikipedia.org/wiki/Significant_figures
https://en.wikipedia.org/wiki/Mathematical_proof
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Can you (Dan) summarize what are those many tonnes of references say about what is the real number? despite they seem to be modified up to very recent dates

To make it much easier task for you

I summarized only in one word, which is "real number is only constructible"

Or so simply, real numbers are distinct distances relative to a unit distance

BKK
Dan Christensen
2017-12-11 14:08:00 UTC
Permalink
Post by bassam king karzeddin
Post by Dan Christensen
https://en.wikipedia.org/wiki/Real_number
https://en.wikipedia.org/wiki/Significant_figures
https://en.wikipedia.org/wiki/Mathematical_proof
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Can you (Dan) summarize what are those many tonnes of references say about what is the real number?
There 3 equivalent approaches. Real numbers are either:

1. Dedekind cuts
2. Cauchy sequences
3. Defined by axioms for a complete ordered field

See https://en.wikipedia.org/wiki/Real_number

Unlike your goofy system, either of these constructs can serve as the basis for most if not all modern science and technology. Must be frustrating as hell for you and your fellow cranks and trolls here.


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
bassam king karzeddin
2017-12-10 17:06:21 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Look closely at this non-solvable Diaphontine Eqn. (n^5 = m^5 + nm^4), and divide this Eqn. by m^5, then you get easily a rational non-solvable polonomial of this form ((n/m)^5 - (n/m) - 1 = 0), now let (x = (n/m)) and substitute, you get the following non-solvable rational polonomial (x^5 - x - 1 = 0)

And why do I claim this polynomial is so simply rational because any solution provided would be so simply rational number no matter how many digits you are able to provide, and the form of your alleged rational solution (n/m) would be as

this:[N(k)/10^k], where (n = N(k)) is integer with (k + 1) digits in 10base numer system, and (m = 10^k)

Now it is quite too simple to check any alleged root, not of this inevitable form,

But, since mathematicians claim this root must be with an infinite sequence of digits, then you require defining the integer N(K) with an infinite sequence of digits where this is first: impossible achievement task, second: impossible acceptance of existence from the holy grail principles of mathematics, third: an impossible solution to our original non-solvable Diophantine Eqn.

So, after learning all those so easy tricks, can't you simply comprehend that numbers with endless terms or endless missing digits are impossible existence? wonder!

Would you still argue stubbornly against a non-solvable Diaphontine Eqn.? wonder! (Where you simply claim so but without your awareness, sure)

Then why not you claim there is an integer solution to Fermat's Last Theorem? wonder! since they are the same core principles of number theory

But I do appreciate that absolute fact are generally so painful for so many beginners with high skills of calculations since it reveals the absolute truth before a layperson's eyes, so admit the truth by saying no irrational solution doesn't exist nor we can construct it exactly because simply it is not there but only as a well-established illusion in your minds, for sure

Now, one might ask, what about the other four roots, then I must point him immediately to read my many articles already were PUBLISHED about those many similar fictional stories that were planted into your heads, where the range of fictionality is so HUGE and so unbelievable and indeed, it needs few thousands of many independent thinkers to uncover the absolute and physiological truths behind them, sure

In short, no roots at all for this polynomial (x^5 - x - 1 = 0), and forget about the graph OR the curve that must cross the (X-axis) at some point, or (+/-) concepts, or imaginary or complex roots, or the artificial continuity based on Epsilon, Delta DISTANCES, or infinity concept itself, ....

Folks of sci.math are more lucky to have all those mere facts before them since this truly a very rare event as true history and never one man show fabricated history of mathematics we were used to

So, don't stay behind and let those astray old mathematicians guide you anymore, just go to that well-guarded site as SE, and invade it till they surrender to the absolute facts

I do have still many surprising facts and true theorems that never require any definition, decision, or even require our own existence

And this revolution of human mind, wouldn't necessarily stay at the mathematical house, but it will soon expand to all other branches of sciences and all other walks of life against salvation to the pure ignorant

Regards
Bassam King Karzeddin
Dec. 10th, 2017
Python
2017-12-10 17:23:16 UTC
Permalink
[utter delusional bullshit]
https://www.whatclinic.com/psychiatrists/saudi-arabia
bassam king karzeddin
2017-12-10 17:50:02 UTC
Permalink
Post by Python
[utter delusional bullshit]
https://www.whatclinic.com/psychiatrists/saudi-arabia
A true coward with bad intention is generally hiding under many fictional names, where at the same time have no valid argument to express, sure

Why are so shameful of your own name? wonder!

BKK
Dan Christensen
2017-12-10 22:29:34 UTC
Permalink
Post by bassam king karzeddin
Post by Python
[utter delusional bullshit]
https://www.whatclinic.com/psychiatrists/saudi-arabia
A true coward with bad intention is generally hiding under many fictional names, where at the same time have no valid argument to express, sure
Why are so shameful of your own name? wonder!
It's just common sense. With psychos like BKK and JG lurking around here, it would be unwise to use one's real name.


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
bassam king karzeddin
2017-12-16 07:48:39 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
And the big scandal of FTG that the other four complex roots are directly relevant to that fictional non-existing root we have demonstrated so rigorously for sure
And if you need many more JOKES about that decided or invented or fabricated imaginary unit (i) where (i^2 = - 1), (but was never any true discovery), just ask me to learn many more tricks about the absolute meaningless of this another Paradise for mythmakers

So, you have (i^8 = 1), No one on earth can even deny this fabricated fact in mathematics, now factor the terms and enjoy it forever, you get this:

(i - 1)(i + 1)(i^2 + 1)(i^4 + 1) = 0, Oops, we have many more solutions to (i) itself, (i = 1), (i = - 1), (i^2 = -1), (i^4 = -1)

But wait, you have to be so smart to pick up only one solution such that it mustn't contradict what had been decided to you as a solution, as (i^2 = -1), so the mathematics isn't a matter of choice nor a matter of any coherent logic for sure, and it is so easy for you now to factor say (i^{2n} +/- 1 = 0)!
Good luck with this another fictional Paradise for sure
BKK
bassam king karzeddin
2018-01-03 13:45:32 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
And the big scandal of FTG that the other four complex roots are directly relevant to that fictional non-existing root we have demonstrated so rigorously for sure
And if you need many more JOKES about that decided or invented or fabricated imaginary unit (i) where (i^2 = - 1), (but was never any true discovery), just ask me to learn many more tricks about the absolute meaningless of this another Paradise for mythmakers
(i - 1)(i + 1)(i^2 + 1)(i^4 + 1) = 0, Oops, we have many more solutions to (i) itself, (i = 1), (i = - 1), (i^2 = -1), (i^4 = -1)
But wait, you have to be so smart to pick up only one solution such that it mustn't contradict what had been decided to you as a solution, as (i^2 = -1), so the mathematics isn't a matter of choice nor a matter of any coherent logic for sure, and it is so easy for you now to factor say (i^{2n} +/- 1 = 0)!
Good luck with this another fictional Paradise for sure
BKK
So, what was the easy lesson that every student must learn compulsory in order to avoid being swimming endlessly in pure fictional stories and mind endless hallucinations and thinking that was truly a realistic science, whereas the fact it was just a silly mind games that immediately collapse upon itself once you subject it to reality check lists

Observe, how so easy it becomes when you DECIDE a solution for few following basic Non-Solvable Diophantine Equations of the following forms

1) (N + 1 = 0), OK, LET there be a negative integers, and the problem was completely solved creating immediately huge volumes of baseless mathematics

2) (N^2 + 1 = 0), OK, LET there be an imaginary unit numbers (i = sqrt(- 1)), and the problem was completely solved creating immediately huge volumes of baseless mathematics

Ignoring the obvious fact that square root operation was basically originated from a constructible Pythagorean Triangle and were not with double roots

3) (N^3 - 2 = 0), OK, LET there be a anairthemetical cube root of two, and the problem was completely solved creating immediately huge volumes of baseless
mathematics

So, here in this historical particular HISTORICAL scandal in the history of mathematics, that the old ancient Greek had proven simply in half a page or so that no rational cube root of two can possibly exist, but later after few thousands years the mythmakers or the alleged top most historical genius mathematicians created such an alleged root (but again in rational numbers only), where there isn't any other way and it is impossible to deny it, for sure

So, did you learn yet why a human decisions are never any true discovery but mind games that lead to entertainments (particularly for few abnormal people)? for sure

Regads
Bassam King Karzeddin
Jan. 3ed, 2018
bassam king karzeddin
2018-01-06 08:56:12 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
IN SHORT
A POLYNOMIAL (X^5 - X - 1 =/= 0), for whatever magnitude you may obtain for (x), no matter however long sequence it is and in any number system you may adopt, and anyone can so simply checkup this self spoken obvious reality, For sure
Hence, your inherited mathematics is forged with endless numerical counter examples, FOR SURE
So, go and correct it NOW
BKK
bassam king karzeddin
2018-01-13 13:49:34 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
IN SHORT
A POLYNOMIAL (X^5 - X - 1 =/= 0), for whatever magnitude you may obtain for (x), no matter however long sequence it is and in any number system you may adopt, and anyone can so simply checkup this self spoken obvious reality, For sure
Hence, your inherited mathematics is forged with endless numerical counter examples, FOR SURE
So, go and correct it NOW
BKK
So, the alleged polynomial (x^5 - x - 1 = 0), that was fabricated from insolvable Diophantine Equation (n^5 - nm^4 - m^5 = 0), (by dividing all terms by m^5), has no real existing root, nor any other alleged and fabricated complex root in true mathematics, For sure

So, why don't you start correcting the mathematics now? wonder!

Why can't you realize the simple fictions that had been well-established into your skulls, after seeing many simple irrefutable NUMERICAL counterexamples? wonder!

Are the alleged top most professional genius mathematicians so stupid upto this limit? wonder!

Why don't you arrange a world conference that is opened for a laypersons to witness the absolute inherited stupidity of the world mathematicians? wonder!

How many so rigorous proofs that were PUBLISHED by me and few others about such incomprehensible absolute stubbornness and unlimited stupidity? wonder!

Now, can't anyone create so many easy counter examples for many alleged insolvable polynomials? wonder!

So, what are the areas of research that must stop immediately in mathematics?

And when shall you save the human minds from your absolute brain fart hallucinations? wonder!

Wake up mythmakers up to the truth that can't be hidden by spiders threads anymore, For sure

Regards
Bassam King Karzeddin
Jan. 13th, 2018
bassam king karzeddin
2018-01-15 14:09:17 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
SO, Let us see a decent professional mathematician (if at all existing), who would ever dare again (at least in sci.math Google groups And after witnessing my REFUTATION proofs) claiming any existing root for such a fabricated and forged polynomial
(x^5 - x - 1 = 0) from a non-solvable Diophantine Eqn.of this form:
(n^5 - nm^4 - m^5 = 0), Sure
But avoid those many fictional coword characters, who usually add nothing as always as forever, Sure
BKK
Dan Christensen
2018-01-15 15:05:03 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
SO, Let us see a decent professional mathematician (if at all existing), who would ever dare again (at least in sci.math Google groups And after witnessing my REFUTATION proofs) claiming any existing root for such a fabricated and forged polynomial
(n^5 - nm^4 - m^5 = 0), Sure
But avoid those many fictional coword characters, who usually add nothing as always as forever, Sure
BKK
It would seem that you can obtain a root of this polynomial accurate to any number of decimal places. See the graph at http://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0

There, you can obtain
x≈1.1673039782614186842560458998548421807205603715254890391400824492756519034295271

More decimal places are available at the link by clicking on More Digits.

Maybe you should have another look that "proof" of yours, BKK.


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
bassam king karzeddin
2018-01-16 09:14:16 UTC
Permalink
Post by Dan Christensen
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
SO, Let us see a decent professional mathematician (if at all existing), who would ever dare again (at least in sci.math Google groups And after witnessing my REFUTATION proofs) claiming any existing root for such a fabricated and forged polynomial
(n^5 - nm^4 - m^5 = 0), Sure
But avoid those many fictional coword characters, who usually add nothing as always as forever, Sure
BKK
It would seem that you can obtain a root of this polynomial accurate to any number of decimal places. See the graph at http://www.wolframalpha.com/input/?i=x%5E5+-+x+-+1+%3D+0
There, you can obtain
x≈1.1673039782614186842560458998548421807205603715254890391400824492756519034295271
More decimal places are available at the link by clicking on More Digits.
Maybe you should have another look that "proof" of yours, BKK.
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
No solution exist for our Diophantine Eqn. (n^5 - nm^4 - m^5 = 0), which is the origin of your fabricated polynomial (x^5 - x - 1 = 0), where your alleged solution (x = n/m) doesn't exist, but its non-existing ghost number (n/m) that seems a solution for very poor minds as you, and that is why your alleged solution would forever appear rational number in this form [A(n)/10^n], where n is positive integer and A(n) is positive integer but with (n + 1) sequence digits (say in 10base number system here), no matter if you fill up the galaxy with your obtained digits

But the true irrational solution would require your n and A(n) to be integers with infinite sequence of digits, where this is not permissible in elementary fundamental mathematics beside being impossible task since no infinity exists at all, and also impossible ratio

And that is why you keep running aimlessly and forever after an illusion only in your mind, hence no real root exists, nor any other roots exist also because they are associated with that fictional root, for sure

It is so unbelievable that almost all the alleged top most genius mathematicians or a laypersons as well can't realize this very easy and very dirty trick that had blocked human minds shamelessly up to this date, wonder!

But certainly so many coward professional mathematicians had realized this simplest facts long ago and immediately once stated and well explained even to school kids, where this doesn't require any peer-review nor any advanced mathematics but only requires a very little drop of humanity first beside a very little drop of honosity or nobility second and a much less drop of intelligence last to realize this big scandal in mathematics, For more than sure

Get it you real morons beyond imaginations, it is toooooo.... Easyyyy....

After all, who can beat a Diophantine Equations? wonder!

BKK
e***@gmail.com
2018-01-15 15:00:27 UTC
Permalink
Please read and do you have computer resources to check?

PROOF of twin primes (p,q) then composite integers 3(p) and 3(q) with 3(q) - 3(p)=6, then between those two distinct composite integers, there is always at least one more prime, this is true if and only if there are infinitely many twin primes. Q.E.D. - Martin Michael Musatov, Jan 15 2018 (OEIS)
REFERENCES
P. Ribenboim, The New Book
bassam king karzeddin
2018-01-17 08:58:20 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
And when shall you understand the most basic thing in the so called very elementary mathematics? wonder!

BKK
Zelos Malum
2018-01-17 13:12:04 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
And when shall you understand the most basic thing in the so called very elementary mathematics? wonder!
BKK
We understand it, you do not. You cannot even cite definitions.
bassam king karzeddin
2018-01-20 18:58:16 UTC
Permalink
Post by Zelos Malum
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
And when shall you understand the most basic thing in the so called very elementary mathematics? wonder!
BKK
We understand it, you do not. You cannot even cite definitions.
You understand things as they are, exactly the way they taught you that, no wonder!
So, you are a obedient bileaver, for sure
BKK
Zelos Malum
2018-01-25 11:21:10 UTC
Permalink
Post by bassam king karzeddin
You understand things as they are, exactly the way they taught you that, no wonder!
So, you are a obedient bileaver, for sure
I have studied things, that is better than you that have not.

I understand it and seek to understand it carefully, unlike you.
bassam king karzeddin
2018-01-25 13:51:39 UTC
Permalink
Post by Zelos Malum
Post by bassam king karzeddin
You understand things as they are, exactly the way they taught you that, no wonder!
So, you are a obedient bileaver, for sure
I have studied things, that is better than you that have not.
I understand it and seek to understand it carefully, unlike you.
You have understood nothing I swear

You are a big loser as your alleged top most genius historian and current mathematicians, for sure

BKK
Zelos Malum
2018-03-26 07:01:24 UTC
Permalink
Projection much? Are you jealous that I know much more than you will ever do? Despite being MUCH younger?
bassam king karzeddin
2018-01-24 18:02:34 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
So, let us see any expert professional who would still openly dare (with true identity name and proven record of performance) OR claim that such a polynomial (x^5 - x - 1 = 0), have any exact and truly existing alleged root, wither real or complex it doesn't matter for sure

**Note: that my refutation of any existence to the real alleged root is also refutation to all other four alleged complex roots as well**

SURE
BKK
bassam king karzeddin
2018-01-29 14:26:04 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Summary, no one so far could legalize mathematically any existence of any roots of such a fabricated polynomial (x^5 - x - 1 = 0), from originally inolvidable Diophantine Eqn. (n^5 - nm^4 - m^d = 0), where (x = n/m) are non-zero integers.

Hence no true existence of this famous polynomial in mathematics, for sure

Pease, keep it for the record and truly for historical events in mathematics

BKK
konyberg
2018-01-29 21:28:36 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Summary, no one so far could legalize mathematically any existence of any roots of such a fabricated polynomial (x^5 - x - 1 = 0), from originally inolvidable Diophantine Eqn. (n^5 - nm^4 - m^d = 0), where (x = n/m) are non-zero integers.
Hence no true existence of this famous polynomial in mathematics, for sure
Pease, keep it for the record and truly for historical events in mathematics
BKK
The exact real root of x^5 - x - 1 = 0 given in Bring radicals equals

x = (4F3)(-1/20,3/20,7/20,11/20; 1/4,2/4,3/4; 3125/256)
+ (1/4)(4F3)(4/20,8/20,12/20,16/20; 2/4,3/4,5/4; 3125/256)
- (5/32)(4F3)(9/20,13/20,17/20,21/20; 3/4,5/4,3/2; 3125/256)
+ (5/32)(4F3)(14/20,18/20,22/20,26/20; 5/4,6/4,7/4; 3125/256)

KON
bassam king karzeddin
2018-01-30 06:39:23 UTC
Permalink
Post by konyberg
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Summary, no one so far could legalize mathematically any existence of any roots of such a fabricated polynomial (x^5 - x - 1 = 0), from originally inolvidable Diophantine Eqn. (n^5 - nm^4 - m^d = 0), where (x = n/m) are non-zero integers.
Hence no true existence of this famous polynomial in mathematics, for sure
Pease, keep it for the record and truly for historical events in mathematics
BKK
The exact real root of x^5 - x - 1 = 0 given in Bring radicals equals
x = (4F3)(-1/20,3/20,7/20,11/20; 1/4,2/4,3/4; 3125/256)
+ (1/4)(4F3)(4/20,8/20,12/20,16/20; 2/4,3/4,5/4; 3125/256)
- (5/32)(4F3)(9/20,13/20,17/20,21/20; 3/4,5/4,3/2; 3125/256)
+ (5/32)(4F3)(14/20,18/20,22/20,26/20; 5/4,6/4,7/4; 3125/256)
KON
This is definitely non-exact root, since F function is taylor series summation again to Infinity, where actually no Infinity exists, and it would ultimately amount to a fraction which is absolutely a rational number again and again and again ..., hence no radical roots for (100%) sure.

The so common mistake or old belief among the professional mathematicians is that they thought of a real root as a point on the extension of a real number line, where this is actually proved false by only very elementary number theory in just few seconds, beside many advanced methods discussed by many others (JG, WM, AP, ...), that are so available in front pages here around you at mainly sci.math Google, for sure

BKK
bassam king karzeddin
2018-01-31 10:29:08 UTC
Permalink
Post by konyberg
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Summary, no one so far could legalize mathematically any existence of any roots of such a fabricated polynomial (x^5 - x - 1 = 0), from originally inolvidable Diophantine Eqn. (n^5 - nm^4 - m^d = 0), where (x = n/m) are non-zero integers.
Hence no true existence of this famous polynomial in mathematics, for sure
Pease, keep it for the record and truly for historical events in mathematics
BKK
The exact real root of x^5 - x - 1 = 0 given in Bring radicals equals
x = (4F3)(-1/20,3/20,7/20,11/20; 1/4,2/4,3/4; 3125/256)
+ (1/4)(4F3)(4/20,8/20,12/20,16/20; 2/4,3/4,5/4; 3125/256)
- (5/32)(4F3)(9/20,13/20,17/20,21/20; 3/4,5/4,3/2; 3125/256)
+ (5/32)(4F3)(14/20,18/20,22/20,26/20; 5/4,6/4,7/4; 3125/256)
KON
Isn't truly my general formula for a real roots for a polynomials of general form also in terms of your current modern mathematics notations and definitions? wonder!

(x^n + x^m = 1)

https://mathoverflow.net/questions/208169/quintic-equation

that had been posted to SE (in below link) even far better results than those "Bring Radicals"

And guess what was the reward by those many mentally and very ill incompetent professionals mathematicians at that unnamed site? wonder!

They had deleted almost all my posts in this regard, and gave me only six downvotes so far for such an irrefutable formula in their current mathematics

But, how strangely this is still made visible (despite being closed) mainly due to a useless professional answer by a known one at their narrow society, beside making or publishing my formula in my same question

I don't know yet, if it is still visible by anyone except me

So, substitute x = (1/y, y =/= 0), and our polynomial in question becomes

as (y^5 + y^4 = 1), then, solve immediately and so easily in my formula, for (n = 5), (m = 4)


BKK
bassam king karzeddin
2018-02-01 13:42:37 UTC
Permalink
Post by bassam king karzeddin
Post by konyberg
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Summary, no one so far could legalize mathematically any existence of any roots of such a fabricated polynomial (x^5 - x - 1 = 0), from originally inolvidable Diophantine Eqn. (n^5 - nm^4 - m^d = 0), where (x = n/m) are non-zero integers.
Hence no true existence of this famous polynomial in mathematics, for sure
Pease, keep it for the record and truly for historical events in mathematics
BKK
The exact real root of x^5 - x - 1 = 0 given in Bring radicals equals
x = (4F3)(-1/20,3/20,7/20,11/20; 1/4,2/4,3/4; 3125/256)
+ (1/4)(4F3)(4/20,8/20,12/20,16/20; 2/4,3/4,5/4; 3125/256)
- (5/32)(4F3)(9/20,13/20,17/20,21/20; 3/4,5/4,3/2; 3125/256)
+ (5/32)(4F3)(14/20,18/20,22/20,26/20; 5/4,6/4,7/4; 3125/256)
KON
Isn't truly my general formula for a real roots for a polynomials of general form also in terms of your current modern mathematics notations and definitions? wonder!
(x^n + x^m = 1)
https://mathoverflow.net/questions/208169/quintic-equation
that had been posted to SE (in below link) even far better results than those "Bring Radicals"
And guess what was the reward by those many mentally and very ill incompetent professionals mathematicians at that unnamed site? wonder!
They had deleted almost all my posts in this regard, and gave me only six downvotes so far for such an irrefutable formula in their current mathematics
But, how strangely this is still made visible (despite being closed) mainly due to a useless professional answer by a known one at their narrow society, beside making or publishing my formula in my same question
I don't know yet, if it is still visible by anyone except me
So, substitute x = (1/y, y =/= 0), and our polynomial in question becomes
as (y^5 + y^4 = 1), then, solve immediately and so easily in my formula, for (n = 5), (m = 4)
BKK
So, who could ever play the game of Infinity better than me? wonder!

Thought I confess the full denial of its mere existence, but that old formula started in 1986, was obtained in 1990, when I was so ignorant or so brainwashed as you are exactly now, for sure

But frankly, that result with so many downvotes from SE professional morons, could not hide it yet and not for its or my own sake, despite deleting many more other interesting results of mine, they truly want to bury or hide a very BIG secret that was annoying them so desperately I swear

In short, they wanted to avoid eventual scandals that are so inevitable event in their damn shameful history as well

But the KING, would never let it go so easily and so unbanished, for sure

BKK
bassam king karzeddin
2018-02-10 13:37:40 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
Post by konyberg
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Summary, no one so far could legalize mathematically any existence of any roots of such a fabricated polynomial (x^5 - x - 1 = 0), from originally inolvidable Diophantine Eqn. (n^5 - nm^4 - m^d = 0), where (x = n/m) are non-zero integers.
Hence no true existence of this famous polynomial in mathematics, for sure
Pease, keep it for the record and truly for historical events in mathematics
BKK
The exact real root of x^5 - x - 1 = 0 given in Bring radicals equals
x = (4F3)(-1/20,3/20,7/20,11/20; 1/4,2/4,3/4; 3125/256)
+ (1/4)(4F3)(4/20,8/20,12/20,16/20; 2/4,3/4,5/4; 3125/256)
- (5/32)(4F3)(9/20,13/20,17/20,21/20; 3/4,5/4,3/2; 3125/256)
+ (5/32)(4F3)(14/20,18/20,22/20,26/20; 5/4,6/4,7/4; 3125/256)
KON
Isn't truly my general formula for a real roots for a polynomials of general form also in terms of your current modern mathematics notations and definitions? wonder!
(x^n + x^m = 1)
https://mathoverflow.net/questions/208169/quintic-equation
that had been posted to SE (in below link) even far better results than those "Bring Radicals"
And guess what was the reward by those many mentally and very ill incompetent professionals mathematicians at that unnamed site? wonder!
They had deleted almost all my posts in this regard, and gave me only six downvotes so far for such an irrefutable formula in their current mathematics
But, how strangely this is still made visible (despite being closed) mainly due to a useless professional answer by a known one at their narrow society, beside making or publishing my formula in my same question
I don't know yet, if it is still visible by anyone except me
So, substitute x = (1/y, y =/= 0), and our polynomial in question becomes
as (y^5 + y^4 = 1), then, solve immediately and so easily in my formula, for (n = 5), (m = 4)
BKK
So, who could ever play the game of Infinity better than me? wonder!
Thought I confess the full denial of its mere existence, but that old formula started in 1986, was obtained in 1990, when I was so ignorant or so brainwashed as you are exactly now, for sure
But frankly, that result with so many downvotes from SE professional morons, could not hide it yet and not for its or my own sake, despite deleting many more other interesting results of mine, they truly want to bury or hide a very BIG secret that was annoying them so desperately I swear
In short, they wanted to avoid eventual scandals that are so inevitable event in their damn shameful history as well
But the KING, would never let it go so easily and so unbanished, for sure
BKK
So, it is worth to repeat the true reason of the impossibility of solving the general quintic or general higher degree equations by radicals, that seems actually and not necessarily from (Galois, Able, Raffaini) proofs, but so simply from the forged polynomial of original unsolvable Diophantine Equations, for sure

And so many further reasons of illegal Fundamental Theorem of Algebra as well, wonder!

Where everything was well-demonstrated by so elementary means in my posts, sure
BKK
bassam king karzeddin
2018-02-11 14:15:03 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Here in this thread, we had already taught you a very good lesson that you had never heard of before

It was only one of many great secrets about the unsolvability of the general polynomials of degree higher than four by radicals, where the secret was because they were fabricated (non-existing polynomials) from originally non-solvable Diophantine Equations, for sure

And all the complexity or forged fabricated stories added to the issue were simply a very devilish acts by the true human devils on earth for their own hidden and very narrow purposes for sure

And one of those so many nobilish or fabricated and so unbelievable historical stories in those old centuries, tells that a little boy called (Galois under twenty years old) wrote a handwritten script that proved the impossibility of solutions by radicals just imagine the night before his death (in a quarrel for a girl), where people felt so sad and so sympathy for his alleged super intellectuality and hence easily or blindly adopted the reasoning that was actually well-cooked up slowly by a few dogmatic profissional mathematicians in power, ten years after his death! wonder!

But someone might ask me that how do you come up with this freak conclusion where no apparent evidence in hand? wonder!

Then I would tell him frankly to read the very BIG evidence before your big eyes that is also published in this thread about unsolvable Diophantine Equation (n^5 = nm^4 + m^5), and to read more carefully about the absolute falsifiability of the so naive decision of making those called imaginary or generally complex numbers that was disproved completely in my posts and in so many ways that was already PUBLISHED for you

But it seems that the older devils didn't want to uncover the earlier crime of solving illegally the unsolvable Diophantine Equation (a^3 = 2b^3) in rational numbers and so shamefully claiming it as irrational at fictional infinity, by adding meaning less notations as three dots (...), where actually no infinity at all exists since natural integers are basically endless and with no existing set at all that contain them , where this fact was proved impossible in quarter a page by the old Greeks thinkers

And the second obvious reason is this, if old ancient mathematicians were so nobel up to this legendary state even with almost no facilities of communications or vast education as today (except by posts), then how would their grandchildren mathematicians of today that have almost all the facilities for almost everything had become so immoral, so selfish, so stubborn and foolish, so deniers and so egostics as well, and up to this limit of no nobility or honosity at all, wonder!

In short, the current truly description of today's mathematicians reflects obvious counter facts about their ancient mathematicians, for sure

Regards
Bassam King Karzeddin
Feb. 11th, 2018
bassam king karzeddin
2018-02-13 14:02:33 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Here in this thread, we had already taught you a very good lesson that you had never heard of before
It was only one of many great secrets about the unsolvability of the general polynomials of degree higher than four by radicals, where the secret was because they were fabricated (non-existing polynomials) from originally non-solvable Diophantine Equations, for sure
And all the complexity or forged fabricated stories added to the issue were simply a very devilish acts by the true human devils on earth for their own hidden and very narrow purposes for sure
And one of those so many nobilish or fabricated and so unbelievable historical stories in those old centuries, tells that a little boy called (Galois under twenty years old) wrote a handwritten script that proved the impossibility of solutions by radicals just imagine the night before his death (in a quarrel for a girl), where people felt so sad and so sympathy for his alleged super intellectuality and hence easily or blindly adopted the reasoning that was actually well-cooked up slowly by a few dogmatic profissional mathematicians in power, ten years after his death! wonder!
But someone might ask me that how do you come up with this freak conclusion where no apparent evidence in hand? wonder!
Then I would tell him frankly to read the very BIG evidence before your big eyes that is also published in this thread about unsolvable Diophantine Equation (n^5 = nm^4 + m^5), and to read more carefully about the absolute falsifiability of the so naive decision of making those called imaginary or generally complex numbers that was disproved completely in my posts and in so many ways that was already PUBLISHED for you
But it seems that the older devils didn't want to uncover the earlier crime of solving illegally the unsolvable Diophantine Equation (a^3 = 2b^3) in rational numbers and so shamefully claiming it as irrational at fictional infinity, by adding meaning less notations as three dots (...), where actually no infinity at all exists since natural integers are basically endless and with no existing set at all that contain them , where this fact was proved impossible in quarter a page by the old Greeks thinkers
And the second obvious reason is this, if old ancient mathematicians were so nobel up to this legendary state even with almost no facilities of communications or vast education as today (except by posts), then how would their grandchildren mathematicians of today that have almost all the facilities for almost everything had become so immoral, so selfish, so stubborn and foolish, so deniers and so egostics as well, and up to this limit of no nobility or honosity at all, wonder!
In short, the current truly description of today's mathematicians reflects obvious counter facts about their ancient mathematicians, for sure
Regards
Bassam King Karzeddin
Feb. 11th, 2018
bassam king karzeddin
2018-02-13 16:50:09 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
What a tragic and so easy secret that was REVEALED by the KING behind you general quintic equation and higher degrees that can't be generally solved by radicals? wonder!

But what about all the other fabricated and so tragic and very interesting stories, especially that one which was written by a genius boy under 20 years old called Galois, just the night before he died in a quarrel for a girl? wonder!

Actually, in that last night he must had been too busy with many other troubles that girl caused him death only in the next day, but the legendary story says that many of his school boys friends asked him to write it immediately before you die, where he stayed late in the night till morning writing his greatest theorems in mathematics, where also he didn't forget to submit it immediately the next day just few hours before his tragic death

Where the dogmatic masters mathematicians those days took only ten years to understand those hand written script by that so unlucky boy, and finally and suddenly discovered the real treasure, that is above your nutty empty skull box even these days, for sure

However, many alike stories are also there, and very similar to this alleged tragedy, were death was mainly the tragic consequences of those very poor and so unlucky boys, that were behind your greatest ever theorems in mathematics

So, what would you expect from such an obvious forged and so fictional stories to produce you? wonder!

Only fictions and mostly more devilish human brain fart hallucination, for sure

Wake up, up to reality you stubborn and blind so obedient believers professionalism mathematicians

What a history of mathematics is this? wonder!

Bassam King Karzeddin

Feb. 13th, 2018
bassam king karzeddin
2018-02-17 07:59:08 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
What a tragic and so easy secret that was REVEALED by the KING behind you general quintic equation and higher degrees that can't be generally solved by radicals? wonder!
But what about all the other fabricated and so tragic and very interesting stories, especially that one which was written by a genius boy under 20 years old called Galois, just the night before he died in a quarrel for a girl? wonder!
Actually, in that last night he must had been too busy with many other troubles that girl caused him death only in the next day, but the legendary story says that many of his school boys friends asked him to write it immediately before you die, where he stayed late in the night till morning writing his greatest theorems in mathematics, where also he didn't forget to submit it immediately the next day just few hours before his tragic death
Where the dogmatic masters mathematicians those days took only ten years to understand those hand written script by that so unlucky boy, and finally and suddenly discovered the real treasure, that is above your nutty empty skull box even these days, for sure
However, many alike stories are also there, and very similar to this alleged tragedy, were death was mainly the tragic consequences of those very poor and so unlucky boys, that were behind your greatest ever theorems in mathematics
So, what would you expect from such an obvious forged and so fictional stories to produce you? wonder!
Only fictions and mostly more devilish human brain fart hallucination, for sure
Wake up, up to reality you stubborn and blind so obedient believers professionalism mathematicians
What a history of mathematics is this? wonder!
Bassam King Karzeddin
Feb. 13th, 2018
So, only the truth that must be published, where fictions must be recognized from the first glance, for sure

BKK
bassam king karzeddin
2018-02-20 15:16:09 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
What a tragic and so easy secret that was REVEALED by the KING behind you general quintic equation and higher degrees that can't be generally solved by radicals? wonder!
But what about all the other fabricated and so tragic and very interesting stories, especially that one which was written by a genius boy under 20 years old called Galois, just the night before he died in a quarrel for a girl? wonder!
Actually, in that last night he must had been too busy with many other troubles that girl caused him death only in the next day, but the legendary story says that many of his school boys friends asked him to write it immediately before you die, where he stayed late in the night till morning writing his greatest theorems in mathematics, where also he didn't forget to submit it immediately the next day just few hours before his tragic death
Where the dogmatic masters mathematicians those days took only ten years to understand those hand written script by that so unlucky boy, and finally and suddenly discovered the real treasure, that is above your nutty empty skull box even these days, for sure
However, many alike stories are also there, and very similar to this alleged tragedy, were death was mainly the tragic consequences of those very poor and so unlucky boys, that were behind your greatest ever theorems in mathematics
So, what would you expect from such an obvious forged and so fictional stories to produce you? wonder!
Only fictions and mostly more devilish human brain fart hallucination, for sure
Wake up, up to reality you stubborn and blind so obedient believers professionalism mathematicians
What a history of mathematics is this? wonder!
Bassam King Karzeddin
Feb. 13th, 2018
***
bassam king karzeddin
2018-02-18 09:37:56 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
So, how many of you truly or secretly had realized completely and so easily the real true and so elementary reasons of the impossibility of solving the Quinitic Eqns. and higher degree polynomials generally with radicals mainly from my posts? wonder!

But, dare you pronounce it loudly and so bravely? for sure

Since basically you know that you are too... cow... to say it frankly, for sure

BKK
bassam king karzeddin
2018-02-20 13:33:42 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
So, how many of you truly or secretly had realized completely and so easily the real true and so elementary reasons of the impossibility of solving the Quinitic Eqns. and higher degree polynomials generally with radicals mainly from my posts? wonder!
But, dare you pronounce it loudly and so bravely? for sure
Since basically you know that you are too... cow... to say it frankly, for sure
BKK
It seems good enough now that almost everybody (who read my relevant recent posts) do understand quite well the actual true reason for the impossibility of solving polynomials of degrees higher than fourth generally by radicals

Which is too short to comprehend fully, but do you truly understand the old published reasons in many thousands documented sources as well? wonder!

And if not, or too... difficult to truly comprehend, then why? wonder!

So, go now and dig under those wrong published sources and point out their actual fallacy before the KING does that for you, for sure

BKK
bassam king karzeddin
2018-02-21 07:48:47 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
Even in the current mathematics, there is so much history with so many (tons of books, papers, articles, lectures. courses, ... etc) about tiny polynomials solutions as (x^2 + x = 1), or more about (x^3 + x = 1), or (x^4 + x = 1), or
(x^5 + x = 1),

but when it comes to polynomials of more general trinomil forms as this for instance
(x^n + x^m = 1), where (n > m) are positive integers

Then it becomes so boring, or so useless or maybe so trivial or so intolerable or most likely big secrets that must not be taught to them so cheap sheep public mathematicians hence must be buried immediately no matter if the (one line formula) is so simply self-proved identity and too easy to implement and understand

See the so shameful acts and rewords of your likes at that moronS site SE

https://mathoverflow.net/questions/208169/quintic-equation

And if it is still visible with so many downvotes that is because it was provided in the question itself (as smuggling) where another known professional answer was provided with many upvotes, since they had deleted many answers of mine with much more details about the same formula

Can't you realize the so obvious shame yet? wonder!

But truly, this formula and more discovered in (1986) of mine is perpetual, despite all the biggest shameful crimes about it committed by the so tiny inferior professional mathematicians, for sure

Bassam King Karzeddin
Feb. 21st, 2018
bassam king karzeddin
2018-03-03 18:18:33 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
So, what is the exact real root in the mathematics of a polynomial :

(x^9 = x^3 + 1)? wonder!

Don't be so shy full to say it? no matter how long it is

And forget about other 8 roots

BKK
konyberg
2018-03-03 19:38:47 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
(x^9 = x^3 + 1)? wonder!
Don't be so shy full to say it? no matter how long it is
And forget about other 8 roots
BKK
x = 2^(-1/9)*3^(-2/9)*((9-69^(1/2))^(1/3)+(9+69^(1/2))^(1/3))^(1/3)

KON
bassam king karzeddin
2018-03-04 07:58:08 UTC
Permalink
Post by konyberg
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
(x^9 = x^3 + 1)? wonder!
Don't be so shy full to say it? no matter how long it is
And forget about other 8 roots
BKK
x = 2^(-1/9)*3^(-2/9)*((9-69^(1/2))^(1/3)+(9+69^(1/2))^(1/3))^(1/3)
KON
@ KON

This is a known cubic formula solution notation based on the invalid cubic root operations, comes down to a mere notation in mind only that is impossible to construct exactly, hence not any real number nor any real existing solution

And it's endless decimal or rational representation can never exist because it would constitute a counterexample to Fermat's last theorem

So, can you figure out this dilemma or a clear contradiction to the real solution of cubic formula? wonder!

However, Diophantine Eqns. would reveal more important facts about the pure juggling and total fictionality of another alleged complex roots, system coordination and so many other things that had been wrongly rooted in mathematics as a real knowledge, where the facts they were the only juggling with true mathematics by deceiving the human minds so mercilessly

This would still take a very long time to remove many covers before discovering the full truth that had been lost and scattered to unbelievable limits of devilish acts, wonder!

But at the end, the integer equations acquire the absolute facts that are much stronger than any sweet fallacies, for sure

In short, only square root operation which was proven rigorously in mathematics (based on the Pythagorean theorem), whereas another non-square root operation was a mere imitation and pure fabrication based on intuitive wrong conclusions, sure

The numerical counterexample is only here before anyone's eyes if he likes to see the better facts, sure

Regards
Bassam King Karzeddin
March, 4th, 2018
Bill
2018-03-04 05:27:14 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
(x^9 = x^3 + 1)? wonder!
Note that if you let u=x^3, then it reduces to
x^3 = x + 1,
and this can be solved by "Cardano's method" (see wikipedia.org, for
instance, for further details). Take the 3rd root, and you are done.
Easy-peazy?
Post by bassam king karzeddin
Don't be so shy full to say it? no matter how long it is
And forget about other 8 roots
BKK
bassam king karzeddin
2018-03-24 09:51:05 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
In other so simple words, and if you are truly keen to understand how so easy to make many fictions and mainly in the solid foundations of mathematics, so what you have to do is invent some problem that you know for sure it doesn't have any real solution, then you have to establish some fictional paradises for the hopeless poor minds that make them think utterly that there is indeed o near solution to endless operations, such most two famous paradises are:

1) Infinity, where this damn ill concept is actually working to limit the natural numbers (as if there is the largest integer) or as if the natural numbers do ever get completely exhausted wonder!

Whereas the simplest facts about natural numbers being as a continuous chain of endless successive integers, where this only invalidates strictly the whole fictional concept of Infinity or any typeS of a like alleged Infinities

You must notice that the word "endless" is much stronger than Infinity since the later is a purely a fake try to limit the natural numbers despite being not any number nor anything else in mathematics (from its own basic definition), hence a pure fiction in minds and a business useless tool to produce huge volumes of fictional mathematics that ruins physics to its roots

So, what you have when you are calculating your endless series are ONLY integers (and forever without any stop)

2) The other paradise of imaginary numbers that had been DECIDED utterly in mathematics where you must need an instruction catalogue (with some period guarantee) on how to use it (such that it MUSTN'T contradict what had been decided blindly and so FOOLISHLY)

But truly all that (two paradises) fail strictly and beyond any little doubt with only this solid NUMERICAL COUNTER EXAMPLE OF QUINTIC polynomials and many more (x^5 - x - 1 = 0)

See here in the below link (before they delete it), for more understanding how to make five fictional roots from nothing to so innocent people as YOU (for sure)

Link: https://www.quora.com/What-are-the-ways-to-understand-the-proof-that-there-is-no-formula-for-expressing-the-roots-of-the-general-quintic-equation-via-radicals/answer/Bassam-Karzeddin-1

Regards
Bassam King Karzeddin
March 24th, 2018
Zelos Malum
2018-04-06 10:44:15 UTC
Permalink
https://www.reddit.com/r/badmathematics/comments/89u8of/a_grabbag_of_fallacies_professional/

Enjoy Bassam
bassam king karzeddin
2018-04-07 19:37:36 UTC
Permalink
Post by Zelos Malum
https://www.reddit.com/r/badmathematics/comments/89u8of/a_grabbag_of_fallacies_professional/
Enjoy Bassam
So, Zelos, you may be taking my topics to "bad section" mathematics, trying to find allies there, wonder!

But, you saw yourself there, how everybody was silenced once I responded, where they deleted your some bad usual comments most likely,

So, dare you to take it there to bad sections because I can so easily make so much fun with people of your attitudes and negligible ability you Dwarf, for sure
BK
Zelos Malum
2018-04-09 05:34:48 UTC
Permalink
Post by bassam king karzeddin
Post by Zelos Malum
https://www.reddit.com/r/badmathematics/comments/89u8of/a_grabbag_of_fallacies_professional/
Enjoy Bassam
So, Zelos, you may be taking my topics to "bad section" mathematics, trying to find allies there, wonder!
But, you saw yourself there, how everybody was silenced once I responded, where they deleted your some bad usual comments most likely,
So, dare you to take it there to bad sections because I can so easily make so much fun with people of your attitudes and negligible ability you Dwarf, for sure
BK
They weren't silenced you moron, they didn't delete it, I did. You are so fucking stupid :)
Mike Hart
2018-03-26 20:26:35 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
The polynomial P(x)=x^5-x-1 is a continuous function, as is every polynomial.

P(1)=-1<0

P(2)=29>0

By Bolzano's Theorem the is a value k between 1 and 2 where P(k)=0

So it has, at least, one real solution between 1 and 2. Period.

Where is all this fuss from?
Python
2018-03-26 20:42:29 UTC
Permalink
Post by Mike Hart
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
The polynomial P(x)=x^5-x-1 is a continuous function, as is every polynomial.
P(1)=-1<0
P(2)=29>0
By Bolzano's Theorem the is a value k between 1 and 2 where P(k)=0
So it has, at least, one real solution between 1 and 2. Period.
Where is all this fuss from?
Bassam King Karzeddin suffers of deep brain damage, he is not the only
one here, you may notice Wolfgang Mueckenheim from Augsburg (aka WM) and
John Gabriel both suffer of similar diseases.
bassam king karzeddin
2018-07-07 18:18:33 UTC
Permalink
Post by Mike Hart
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
The polynomial P(x)=x^5-x-1 is a continuous function, as is every polynomial.
P(1)=-1<0
P(2)=29>0
By Bolzano's Theorem the is a value k between 1 and 2 where P(k)=0
So it has, at least, one real solution between 1 and 2. Period.
Where is all this fuss from?
Is continuity of real numbers truly proven or decided? wonder!

One has to read from the beginning, and so realize that some problems with polynomial aren't solvable even in real numbers


BK
bassam king karzeddin
2018-11-08 17:52:47 UTC
Permalink
Post by bassam king karzeddin
Post by Mike Hart
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
The polynomial P(x)=x^5-x-1 is a continuous function, as is every polynomial.
P(1)=-1<0
P(2)=29>0
By Bolzano's Theorem the is a value k between 1 and 2 where P(k)=0
So it has, at least, one real solution between 1 and 2. Period.
Where is all this fuss from?
Is continuity of real numbers truly proven or decided? wonder!
One has to read from the beginning, and so realize that some problems with polynomial aren't solvable even in real numbers
BK
If the least real number that is greater than 7 doesn't exist in mathematics, nor the largest real number that is less than 7 exists either, then real numbers are discrete numbers, (not continuous), for sure

Then how did they decided such continuity of real numbers which is so illegal and against the simplest ever shown logic above? wonder!

Most likely, they never respect your ability to think loudly, nor they expected you to dare to ask such questions

But, is it STILL possible with today so open world? wonder!

OR maybe it is the necessity of business making that a curve must cross the X-axis in order to comply deliberately with the fabricated or fundamental theorem of algebra? no wonder!

Otherwise, things would be uncovered to be inconsistent in Whatever ZFC, SURE

BKK
Zelos Malum
2018-11-09 06:29:40 UTC
Permalink
Post by bassam king karzeddin
If the least real number that is greater than 7 doesn't exist in mathematics, nor the largest real number that is less than 7 exists either, then real numbers are discrete numbers, (not continuous), for sure
Incorrect, the fact that they are dense does not mean you cannot have continuity.
bassam king karzeddin
2019-04-13 16:57:35 UTC
Permalink
Post by Zelos Malum
Post by bassam king karzeddin
If the least real number that is greater than 7 doesn't exist in mathematics, nor the largest real number that is less than 7 exists either, then real numbers are discrete numbers, (not continuous), for sure
Incorrect, the fact that they are dense does not mean you cannot have continuity.
Dense? how dense they are?

BKK
Python
2019-04-14 00:25:37 UTC
Permalink
BKK is King of Idiots.
bassam king karzeddin
2019-04-15 17:13:20 UTC
Permalink
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
had you ever obtained it? no wonder!
BKK
konyberg
2019-04-15 17:16:51 UTC
Permalink
Post by bassam king karzeddin
Post by bassam king karzeddin
How can one obtain the real irrational root of this Quintic polynomial (x^5 - x - 1 = 0), but not in rational number form
This question had been asked at Quora and seems that immediately was frozen by the professional mathematicians for very big reasons that they never wanted the innocent students to know about it at all, for sure
However, they had never learned yet that obvious and very simple facts can't be hidden by a spider's threads anymore, otherwise why the scientists and engineers had already made it very easy world communications? wonder!
Regards
Bassam King Karzeddin
Dec. 6th, 2017
had you ever obtained it? no wonder!
BKK
I did show you the root. You just wouldn't accept it.
KON
Vinicius Claudino Ferraz
2019-04-15 18:40:06 UTC
Permalink
f_1(t) = 1 + 0 t + 0 t² + 0 t³ + 0 t^4 + 1/120 t^5 + 0 t^6 + 0 t^7 + 0 t^8 + ...
f_2(t) = 0 + 1 t + 0 t² + 0 t³ + 0 t^4 + 1/120 t^5 + 1/720 t^6 + 0 t^7 + 0 t^8 + ...
f_3(t) = 0 + 0 t + 1/2 t² + 0 t³ + 0 t^4 + 0 t^5 + 1/720 t^6 + 1/5040 t^7 + 0 t^8 + ...
f_4(t) = 0 + 0 t + 0 t² + 1/6 t³ + 0 t^4 + 0 t^5 + 0 t^6 + 1/5040 t^7 + 1/40320 t^8 + ...
f_5(t) = 0 + 0 t + 0 t² + 0 t³ + 1/24 t^4 + 0 t^5 + 0 t^6 + 0 t^7 + 1/40320 t^8 + ...

f_i(t) = exp rt
ln f = rt
1/t ln f = r

My delphi is too poor. I will begin by Baskara. Maybe after Sunday I will have a fraction as desired solution. :-)

I was thinking who is right wing, who is left wing?

I like luis inacio LULA da silva, who is arrested.
i
Post by konyberg
I did show you the root. You just wouldn't accept it.
KON
Vinicius Claudino Ferraz
2019-04-15 20:29:29 UTC
Permalink
Hey, look at my problem.

y'' + 5 y' + 6 y = 0
6y = 6c_1 exp -2t + c_2 exp -3t
5y' = -10 c_1 exp -2t - 3 c_2 exp -3t
y'' = 4 c_1 exp -2t + 9 c_2 exp -3t

exp -2t = \sum (-2)^n/n! t^n = f_1 or f_2
exp -3t = \sum (-3)^n/n! t^n = f_2 or f_1

But when I put numbers on this, what happens?

f_1(t) = 1 + 0 t + 6/2 t² + 30/6 t³ + 186/24 t^4 + 1110/120 t^5 + 6666/720 t^6 + 39990/5040 t^7 + 239946/40320 t^8 + ...
f_2(t) = 0 + 1 t + 5/2 t² + 31/6 t³ + 185/24 t^4 + 1111/120 t^5 + 6665/720 t^6 + 39991/5040 t^7 + 239945/40320 t^8 + ...
f_1'(t) = 0 + 6 t + 30/2 t² + 186/6 t³ + 1110/24 t^4 + 6666/120 t^5 + 39990/720 t^6 + 239946/5040 t^7 + 1439670/40320 t^8 + ...
f_2'(t) = 1 + 5 t + 31/2 t² + 185/6 t³ + 1111/24 t^4 + 6665/120 t^5 + 39991/720 t^6 + 239945/5040 t^7 + 1439671/40320 t^8 + ...

That's a problem on series_1(t) = series_2(t)

(series_1 - series_2)(t) = 0, forall t
Post by Vinicius Claudino Ferraz
My delphi is too poor.
Vinicius Claudino Ferraz
2019-04-15 18:44:37 UTC
Permalink
i'm polytically left-wing.
i am almost Cuban and Russian and Chinese :-)

i am a progressist mathematician.
i am a liberalist mathematician. i can think whatever.
i think right wing is obviously the devil like adolfo Hitler.
i also think there are several forces against progress and liberty.

and i don't think mathematics has something to do with the ownership of the means of production.

and i want free will for everybody to get in and get out any country whenever we want.
the least important must be treated as the most important ones.
so give special presidential passports, Vatican passports, for me and my Latin brothers and si
Vinicius Claudino Ferraz
2019-04-15 20:33:31 UTC
Permalink
Hey, take a look at my problem:

y'' + 5 y' + 6 y = 0
6y = 6c_1 exp -2t + 6 c_2 exp -3t
5y' = -10 c_1 exp -2t - 15 c_2 exp -3t
y'' = 4 c_1 exp -2t + 9 c_2 exp -3t

exp -2t = \sum (-2)^n/n! t^n = f_1 or f_2
exp -3t = \sum (-3)^n/n! t^n = f_2 or f_1

But, with numbers:

f_1(t) = 1 + 0 t + 6/2 t² + 30/6 t³ + 186/24 t^4 + 1110/120 t^5 + 6666/720 t^6 + 39990/5040 t^7 + 239946/40320 t^8 + ...
f_2(t) = 0 + 1 t + 5/2 t² + 31/6 t³ + 185/24 t^4 + 1111/120 t^5 + 6665/720 t^6 + 39991/5040 t^7 + 239945/40320 t^8 + ...
f_1'(t) = 0 + 6 t + 30/2 t² + 186/6 t³ + 1110/24 t^4 + 6666/120 t^5 + 39990/720 t^6 + 239946/5040 t^7 + 1439670/40320 t^8 + ...
f_2'(t) = 1 + 5 t + 31/2 t² + 185/6 t³ + 1111/24 t^4 + 6665/120 t^5 + 39991/720 t^6 + 239945/5040 t^7 + 1439671/40320 t^8 + ...

That's a problem on series_1(t) = series_2(t)
(series_1 - series_2)(t) = 0, forall t
Vinicius Claudino Ferraz
2019-04-15 20:42:33 UTC
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Hey, take a look at my problem:

y'' + 5 y' + 6 y = 0
y'' = - 6 y - 5 y'

01,00
-6,-5

6y = 6c_1 exp -2t + 6 c_2 exp -3t
5y' = -10 c_1 exp -2t - 15 c_2 exp -3t
y'' = 4 c_1 exp -2t + 9 c_2 exp -3t

exp -2t = \sum (-2)^n/n! t^n = f_1 or f_2
exp -3t = \sum (-3)^n/n! t^n = f_2 or f_1

z = exp at
z' = a exp at
z'' = a^2 exp at
z = \sum a^n/n! t^n

But, with numbers:

f_1(t) = 1 + 0 t - 6/2 t² + 30/6 t³ - 114/24 t^4 + 390/120 t^5 - 1266/720 t^6 + 3990/5040 t^7 - 12354/40320 t^8 + ...
f_2(t) = 0 + 1 t - 5/2 t² + 19/6 t³ - 65/24 t^4 + 211/120 t^5 - 665/720 t^6 + 2059/5040 t^7 - 6305/40320 t^8 + ...
f_1'(t) = 0 - 6 t + 30/2 t² - 114/6 t³ + 390/24 t^4 - 1266/120 t^5 + 3990/720 t^6 - 12354/5040 t^7 + 37830/40320 t^8 + ...
f_2'(t) = 1 - 5 t + 19/2 t² - 65/6 t³ + 211/24 t^4 - 665/120 t^5 + 2059/720 t^6 - 6305/5040 t^7 + 19171/40320 t^8 + ...

That's a problem on series_1(t) = series_2(t)
(series_1 - series_2)(t) = 0, forall t
Vinicius Claudino Ferraz
2019-04-15 21:10:14 UTC
Permalink
I think we need 4 constants:

α exp (-2t) + β exp (-3t) + γ f_1(t) + λ f_2(t) = 0, forall t, exists (αβγλ).
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