Discussion:
The legend of the real existence of the real Transcendental numbers?
(too old to reply)
bassam king karzeddin
2017-07-11 09:39:46 UTC
Permalink
What is the probability of picking up a random length such that it represents a transcendental number relative to any arbitrary chosen constructible unit length?

Big Hint: the probability exactly equals to zero, sure

Regards

Bassam King Karzeddin
7 July 2017
bassam king karzeddin
2017-07-11 12:12:02 UTC
Permalink
See also AP recent thread here: https://groups.google.com/forum/#!topic/sci.math/CDAML9f0KP0

Let me think of the simplest answer for this type of questions
Naturally, we can’t choose or assume an arbitrary unit such that it represents another transcendental number, otherwise, we must return to the same point of the question, since this becomes as an assumption in mind only and never related to any reality that can be demonstrated by Geometrical representation

But from the basic definition of the transcendental numbers we see that those real transcendental numbers are the roots of polynomials that are certainly not existing polynomials, therefore the probability of picking up a random length such that it represents a transcendental number relative to any arbitrarily chosen unit length exactly equals to zero

Thus, the transcendental numbers generally are not any real existing numbers, but since we do need them for our own needs like (Pi) for APPROXIMATING the circle area for example, then we can simply do present them in many APPROXIMATED constructible form numbers (generally as rational numbers) depending on our required degree of accuracy for mainly our own practical problem that is not necessarily related to mathematics, and so, unfortunately, there are no other choices in mathematics except in mind only as a mere notation (which is more than meaningless)

However, there are many other ways to demonstrate this simple and too obvious fact from the elementary principles of basic mathematics for sure
********************Adding***********************************
Since also with all human knowledge, we couldn’t construct EXACTLY even one real transcendental number as π for example, (but approximately yes by so many known methods (as paper folding, Origami, marked rulers, carpenter square, Cauchy sequences, famous cuts, Newton’s Approximations, …etc)
Then the question arises, how can we construct exactly a number so randomly, if we really know it is even impossible construction with all our tools and knowledge in advance?
So, the probability must be equals to zero, sure

Regards
Bassam King Karzeddin
JULY 7, 2017
konyberg
2017-07-11 13:19:46 UTC
Permalink
Post by bassam king karzeddin
See also AP recent thread here: https://groups.google.com/forum/#!topic/sci.math/CDAML9f0KP0
Let me think of the simplest answer for this type of questions
Naturally, we can’t choose or assume an arbitrary unit such that it represents another transcendental number, otherwise, we must return to the same point of the question, since this becomes as an assumption in mind only and never related to any reality that can be demonstrated by Geometrical representation
But from the basic definition of the transcendental numbers we see that those real transcendental numbers are the roots of polynomials that are certainly not existing polynomials, therefore the probability of picking up a random length such that it represents a transcendental number relative to any arbitrarily chosen unit length exactly equals to zero
Thus, the transcendental numbers generally are not any real existing numbers, but since we do need them for our own needs like (Pi) for APPROXIMATING the circle area for example, then we can simply do present them in many APPROXIMATED constructible form numbers (generally as rational numbers) depending on our required degree of accuracy for mainly our own practical problem that is not necessarily related to mathematics, and so, unfortunately, there are no other choices in mathematics except in mind only as a mere notation (which is more than meaningless)
However, there are many other ways to demonstrate this simple and too obvious fact from the elementary principles of basic mathematics for sure
********************Adding***********************************
Since also with all human knowledge, we couldn’t construct EXACTLY even one real transcendental number as π for example, (but approximately yes by so many known methods (as paper folding, Origami, marked rulers, carpenter square, Cauchy sequences, famous cuts, Newton’s Approximations, …etc)
Then the question arises, how can we construct exactly a number so randomly, if we really know it is even impossible construction with all our tools and knowledge in advance?
So, the probability must be equals to zero, sure
Regards
Bassam King Karzeddin
JULY 7, 2017
If you chose a radius of 0.5 in your compass. Then draw a circle. Isn't PI then constructed?

KON
bassam king karzeddin
2017-07-11 15:47:46 UTC
Permalink
Post by konyberg
Post by bassam king karzeddin
See also AP recent thread here: https://groups.google.com/forum/#!topic/sci.math/CDAML9f0KP0
Let me think of the simplest answer for this type of questions
Naturally, we can’t choose or assume an arbitrary unit such that it represents another transcendental number, otherwise, we must return to the same point of the question, since this becomes as an assumption in mind only and never related to any reality that can be demonstrated by Geometrical representation
But from the basic definition of the transcendental numbers we see that those real transcendental numbers are the roots of polynomials that are certainly not existing polynomials, therefore the probability of picking up a random length such that it represents a transcendental number relative to any arbitrarily chosen unit length exactly equals to zero
Thus, the transcendental numbers generally are not any real existing numbers, but since we do need them for our own needs like (Pi) for APPROXIMATING the circle area for example, then we can simply do present them in many APPROXIMATED constructible form numbers (generally as rational numbers) depending on our required degree of accuracy for mainly our own practical problem that is not necessarily related to mathematics, and so, unfortunately, there are no other choices in mathematics except in mind only as a mere notation (which is more than meaningless)
However, there are many other ways to demonstrate this simple and too obvious fact from the elementary principles of basic mathematics for sure
********************Adding***********************************
Since also with all human knowledge, we couldn’t construct EXACTLY even one real transcendental number as π for example, (but approximately yes by so many known methods (as paper folding, Origami, marked rulers, carpenter square, Cauchy sequences, famous cuts, Newton’s Approximations, …etc)
Then the question arises, how can we construct exactly a number so randomly, if we really know it is even impossible construction with all our tools and knowledge in advance?
So, the probability must be equals to zero, sure
Regards
Bassam King Karzeddin
JULY 7, 2017
If you chose a radius of 0.5 in your compass. Then draw a circle. Isn't PI then constructed?
Yes, sure the (Pi) you can see or draw is definitely must represent a constructible number, once you measure it, even though we can't calculate it exactly but can always be approximated, don't we really?

But the absolute (Pi) measures exists only in the perfect circle, but so, unfortunately, the perfect circle doesn't exist in reality, even you draw your circle to contain all the universe, thus leaving your (Pi) only in mind and not, in reality, so your (Pi) is definitely not any existing number, just like any real algebraic number that is not constructible, sure

BKK
Post by konyberg
KON
Markus Klyver
2017-07-11 18:42:44 UTC
Permalink
Post by bassam king karzeddin
Post by konyberg
Post by bassam king karzeddin
See also AP recent thread here: https://groups.google.com/forum/#!topic/sci.math/CDAML9f0KP0
Let me think of the simplest answer for this type of questions
Naturally, we can’t choose or assume an arbitrary unit such that it represents another transcendental number, otherwise, we must return to the same point of the question, since this becomes as an assumption in mind only and never related to any reality that can be demonstrated by Geometrical representation
But from the basic definition of the transcendental numbers we see that those real transcendental numbers are the roots of polynomials that are certainly not existing polynomials, therefore the probability of picking up a random length such that it represents a transcendental number relative to any arbitrarily chosen unit length exactly equals to zero
Thus, the transcendental numbers generally are not any real existing numbers, but since we do need them for our own needs like (Pi) for APPROXIMATING the circle area for example, then we can simply do present them in many APPROXIMATED constructible form numbers (generally as rational numbers) depending on our required degree of accuracy for mainly our own practical problem that is not necessarily related to mathematics, and so, unfortunately, there are no other choices in mathematics except in mind only as a mere notation (which is more than meaningless)
However, there are many other ways to demonstrate this simple and too obvious fact from the elementary principles of basic mathematics for sure
********************Adding***********************************
Since also with all human knowledge, we couldn’t construct EXACTLY even one real transcendental number as π for example, (but approximately yes by so many known methods (as paper folding, Origami, marked rulers, carpenter square, Cauchy sequences, famous cuts, Newton’s Approximations, …etc)
Then the question arises, how can we construct exactly a number so randomly, if we really know it is even impossible construction with all our tools and knowledge in advance?
So, the probability must be equals to zero, sure
Regards
Bassam King Karzeddin
JULY 7, 2017
If you chose a radius of 0.5 in your compass. Then draw a circle. Isn't PI then constructed?
Yes, sure the (Pi) you can see or draw is definitely must represent a constructible number, once you measure it, even though we can't calculate it exactly but can always be approximated, don't we really?
But the absolute (Pi) measures exists only in the perfect circle, but so, unfortunately, the perfect circle doesn't exist in reality, even you draw your circle to contain all the universe, thus leaving your (Pi) only in mind and not, in reality, so your (Pi) is definitely not any existing number, just like any real algebraic number that is not constructible, sure
BKK
Post by konyberg
KON
No perfect lines or points exist in reality either, so ALL LINES AND POINTS MATHEMATICS ARE FAKERY, FOR SURE. *REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE* *hurr durr*
bassam king karzeddin
2017-07-12 07:37:47 UTC
Permalink
Post by Markus Klyver
Post by bassam king karzeddin
Post by konyberg
Post by bassam king karzeddin
See also AP recent thread here: https://groups.google.com/forum/#!topic/sci.math/CDAML9f0KP0
Let me think of the simplest answer for this type of questions
Naturally, we can’t choose or assume an arbitrary unit such that it represents another transcendental number, otherwise, we must return to the same point of the question, since this becomes as an assumption in mind only and never related to any reality that can be demonstrated by Geometrical representation
But from the basic definition of the transcendental numbers we see that those real transcendental numbers are the roots of polynomials that are certainly not existing polynomials, therefore the probability of picking up a random length such that it represents a transcendental number relative to any arbitrarily chosen unit length exactly equals to zero
Thus, the transcendental numbers generally are not any real existing numbers, but since we do need them for our own needs like (Pi) for APPROXIMATING the circle area for example, then we can simply do present them in many APPROXIMATED constructible form numbers (generally as rational numbers) depending on our required degree of accuracy for mainly our own practical problem that is not necessarily related to mathematics, and so, unfortunately, there are no other choices in mathematics except in mind only as a mere notation (which is more than meaningless)
However, there are many other ways to demonstrate this simple and too obvious fact from the elementary principles of basic mathematics for sure
********************Adding***********************************
Since also with all human knowledge, we couldn’t construct EXACTLY even one real transcendental number as π for example, (but approximately yes by so many known methods (as paper folding, Origami, marked rulers, carpenter square, Cauchy sequences, famous cuts, Newton’s Approximations, …etc)
Then the question arises, how can we construct exactly a number so randomly, if we really know it is even impossible construction with all our tools and knowledge in advance?
So, the probability must be equals to zero, sure
Regards
Bassam King Karzeddin
JULY 7, 2017
If you chose a radius of 0.5 in your compass. Then draw a circle. Isn't PI then constructed?
Yes, sure the (Pi) you can see or draw is definitely must represent a constructible number, once you measure it, even though we can't calculate it exactly but can always be approximated, don't we really?
But the absolute (Pi) measures exists only in the perfect circle, but so, unfortunately, the perfect circle doesn't exist in reality, even you draw your circle to contain all the universe, thus leaving your (Pi) only in mind and not, in reality, so your (Pi) is definitely not any existing number, just like any real algebraic number that is not constructible, sure
BKK
Post by konyberg
KON
No perfect lines or points exist in reality either, so ALL LINES AND POINTS MATHEMATICS ARE FAKERY, FOR SURE. *REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE* *hurr durr*
I don't know what is the relation, isn't so strange that when someone proves something really impossible, then others as you here want to collapse everything around just to invalidate the so obvious proof, as if this is the only proof I gave, or as if there aren't many other proofs as well. wonder!

See here, I had just asked a little question about why the principle of simple common sense not accepted in mathematics, it got around (200 000) views so far, with almost 40 non-common sense answers too, even though I didn't provide an answer to my own question:

Link: https://www.quora.com/Why-is-the-principle-of-simple-common-sense-not-accepted-in-mathematics

So, no perfect lines ...etc, then did you really hear about something called "distance" in mathematics? wonder!

BKK
bassam king karzeddin
2017-07-12 10:52:18 UTC
Permalink
Post by bassam king karzeddin
Post by Markus Klyver
Post by bassam king karzeddin
Post by konyberg
Post by bassam king karzeddin
See also AP recent thread here: https://groups.google.com/forum/#!topic/sci.math/CDAML9f0KP0
Let me think of the simplest answer for this type of questions
Naturally, we can’t choose or assume an arbitrary unit such that it represents another transcendental number, otherwise, we must return to the same point of the question, since this becomes as an assumption in mind only and never related to any reality that can be demonstrated by Geometrical representation
But from the basic definition of the transcendental numbers we see that those real transcendental numbers are the roots of polynomials that are certainly not existing polynomials, therefore the probability of picking up a random length such that it represents a transcendental number relative to any arbitrarily chosen unit length exactly equals to zero
Thus, the transcendental numbers generally are not any real existing numbers, but since we do need them for our own needs like (Pi) for APPROXIMATING the circle area for example, then we can simply do present them in many APPROXIMATED constructible form numbers (generally as rational numbers) depending on our required degree of accuracy for mainly our own practical problem that is not necessarily related to mathematics, and so, unfortunately, there are no other choices in mathematics except in mind only as a mere notation (which is more than meaningless)
However, there are many other ways to demonstrate this simple and too obvious fact from the elementary principles of basic mathematics for sure
********************Adding***********************************
Since also with all human knowledge, we couldn’t construct EXACTLY even one real transcendental number as π for example, (but approximately yes by so many known methods (as paper folding, Origami, marked rulers, carpenter square, Cauchy sequences, famous cuts, Newton’s Approximations, …etc)
Then the question arises, how can we construct exactly a number so randomly, if we really know it is even impossible construction with all our tools and knowledge in advance?
So, the probability must be equals to zero, sure
Regards
Bassam King Karzeddin
JULY 7, 2017
If you chose a radius of 0.5 in your compass. Then draw a circle. Isn't PI then constructed?
Yes, sure the (Pi) you can see or draw is definitely must represent a constructible number, once you measure it, even though we can't calculate it exactly but can always be approximated, don't we really?
But the absolute (Pi) measures exists only in the perfect circle, but so, unfortunately, the perfect circle doesn't exist in reality, even you draw your circle to contain all the universe, thus leaving your (Pi) only in mind and not, in reality, so your (Pi) is definitely not any existing number, just like any real algebraic number that is not constructible, sure
BKK
Post by konyberg
KON
No perfect lines or points exist in reality either, so ALL LINES AND POINTS MATHEMATICS ARE FAKERY, FOR SURE. *REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE* *hurr durr*
I don't know what is the relation, isn't so strange that when someone proves something really impossible, then others as you here want to collapse everything around just to invalidate the so obvious proof, as if this is the only proof I gave, or as if there aren't many other proofs as well. wonder!
Link: https://www.quora.com/Why-is-the-principle-of-simple-common-sense-not-accepted-in-mathematics
So, no perfect lines ...etc, then did you really hear about something called "distance" in mathematics? wonder!
BKK
Or maybe you (Markus) think that there isn't any perfect distance too? wonder!, OR there isn't anything at all! may be

BKK
Me
2017-07-16 13:40:33 UTC
Permalink
Post by bassam king karzeddin
Or maybe you (Markus) think that there isn't any perfect distance too?
wonder!, OR there isn't anything at all!
Right, there are't ANY *mathematical objects* in "reality". At least not as physical objects "located" in space-time.
Markus Klyver
2017-07-13 00:13:15 UTC
Permalink
Post by bassam king karzeddin
Post by Markus Klyver
Post by bassam king karzeddin
Post by konyberg
Post by bassam king karzeddin
See also AP recent thread here: https://groups.google.com/forum/#!topic/sci.math/CDAML9f0KP0
Let me think of the simplest answer for this type of questions
Naturally, we can’t choose or assume an arbitrary unit such that it represents another transcendental number, otherwise, we must return to the same point of the question, since this becomes as an assumption in mind only and never related to any reality that can be demonstrated by Geometrical representation
But from the basic definition of the transcendental numbers we see that those real transcendental numbers are the roots of polynomials that are certainly not existing polynomials, therefore the probability of picking up a random length such that it represents a transcendental number relative to any arbitrarily chosen unit length exactly equals to zero
Thus, the transcendental numbers generally are not any real existing numbers, but since we do need them for our own needs like (Pi) for APPROXIMATING the circle area for example, then we can simply do present them in many APPROXIMATED constructible form numbers (generally as rational numbers) depending on our required degree of accuracy for mainly our own practical problem that is not necessarily related to mathematics, and so, unfortunately, there are no other choices in mathematics except in mind only as a mere notation (which is more than meaningless)
However, there are many other ways to demonstrate this simple and too obvious fact from the elementary principles of basic mathematics for sure
********************Adding***********************************
Since also with all human knowledge, we couldn’t construct EXACTLY even one real transcendental number as π for example, (but approximately yes by so many known methods (as paper folding, Origami, marked rulers, carpenter square, Cauchy sequences, famous cuts, Newton’s Approximations, …etc)
Then the question arises, how can we construct exactly a number so randomly, if we really know it is even impossible construction with all our tools and knowledge in advance?
So, the probability must be equals to zero, sure
Regards
Bassam King Karzeddin
JULY 7, 2017
If you chose a radius of 0.5 in your compass. Then draw a circle. Isn't PI then constructed?
Yes, sure the (Pi) you can see or draw is definitely must represent a constructible number, once you measure it, even though we can't calculate it exactly but can always be approximated, don't we really?
But the absolute (Pi) measures exists only in the perfect circle, but so, unfortunately, the perfect circle doesn't exist in reality, even you draw your circle to contain all the universe, thus leaving your (Pi) only in mind and not, in reality, so your (Pi) is definitely not any existing number, just like any real algebraic number that is not constructible, sure
BKK
Post by konyberg
KON
No perfect lines or points exist in reality either, so ALL LINES AND POINTS MATHEMATICS ARE FAKERY, FOR SURE. *REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE* *hurr durr*
I don't know what is the relation, isn't so strange that when someone proves something really impossible, then others as you here want to collapse everything around just to invalidate the so obvious proof, as if this is the only proof I gave, or as if there aren't many other proofs as well. wonder!
Link: https://www.quora.com/Why-is-the-principle-of-simple-common-sense-not-accepted-in-mathematics
So, no perfect lines ...etc, then did you really hear about something called "distance" in mathematics? wonder!
BKK
Your "critique" of transcendental numbers like π seem to go as follow: "π does not exist in the real world, hence π does not exist at all and is all fakery and bogus."

Well, no perfect circles exist in the real world either, and nor do perfect points, lines and planes. So all geometry is, by your own logic, ALL FAKERY AND A BIG JOKE.
Me
2017-07-16 13:38:18 UTC
Permalink
Post by bassam king karzeddin
But the absolute (Pi) measures exists only in the perfect circle,
"only" is to strong a word here. Still...
Post by bassam king karzeddin
unfortunately, the perfect circle doesn't exist in reality, even you draw
your circle to contain all the universe,
Assuming this is true....
Post by bassam king karzeddin
thus leaving your (Pi) only in mind and not, in reality,
*No* number exists in "reality", man. There may, say, be two apples on a desk, but you won't find THE NUMBER 2 on it.
Post by bassam king karzeddin
so your (Pi) is definitely not any existing number,
Indeed, (in this -physical- sense) just like ALL numbers.
Post by bassam king karzeddin
just like any real algebraic number that is not constructible, sure
Of course. :-)

Just like natural numbers, negative numbers (or integers), rational numbers, etc. etc.
bassam king karzeddin
2017-07-16 15:44:57 UTC
Permalink
Post by Me
Post by bassam king karzeddin
But the absolute (Pi) measures exists only in the perfect circle,
"only" is to strong a word here. Still...
Post by bassam king karzeddin
unfortunately, the perfect circle doesn't exist in reality, even you draw
your circle to contain all the universe,
Assuming this is true....
Post by bassam king karzeddin
thus leaving your (Pi) only in mind and not, in reality,
*No* number exists in "reality", man. There may, say, be two apples on a desk, but you won't find THE NUMBER 2 on it.
Here I am talking about existence of immaterial concepts, that are out of time or space concept, even though it is concerned about property of mere space that represents absolute nothingness

To illustrate it further, consider the concept of Pythagorean fact!

Does this mere fact exist?
Does this mere fact die with time?
Did this fact have any start?
Isn't this mere fact a real existing fact, whether we know it or don't know it?
Isn't this mere fact a real existing fact even though monkeys do not comprehend it yet? wonder!
So, it is a mere fact in numbers, in algebra, in space too which is the reality we are inside unless you are the only one who is dreaming alone?

So, constructible numbers only exist the same way, where no one else can construct them except their real creator (which is the defined unity) the only one can give them existence once the one (the unity) exist by definition

So, one by definition is the real God of any existing number, and if we can define the absolute true unity as one, then this one can only construct anything from existing numbers to numbers in the reality, sure!
Post by Me
Post by bassam king karzeddin
so your (Pi) is definitely not any existing number,
Indeed, (in this -physical- sense) just like ALL numbers.
But so, unfortunately, (Pi) was a human Imaginary fetish that they worshipped
Post by Me
Post by bassam king karzeddin
just like any real algebraic number that is not constructible, sure
Of course. :-)
Just like natural numbers, negative numbers (or integers), rational numbers, etc. etc.
No, things you describe aren't alike for sure

BKK
b***@gmail.com
2017-07-16 18:32:29 UTC
Permalink
Ha Ha BKK going full circle. When I want to see absolute
nothingness, I look at BKK, no clue about math.

Like John Gabriel bird brain with his calcluelss, there
is BKK with his Pythagoschizo brabble.
Post by bassam king karzeddin
Post by Me
Post by bassam king karzeddin
But the absolute (Pi) measures exists only in the perfect circle,
"only" is to strong a word here. Still...
Post by bassam king karzeddin
unfortunately, the perfect circle doesn't exist in reality, even you draw
your circle to contain all the universe,
Assuming this is true....
Post by bassam king karzeddin
thus leaving your (Pi) only in mind and not, in reality,
*No* number exists in "reality", man. There may, say, be two apples on a desk, but you won't find THE NUMBER 2 on it.
Here I am talking about existence of immaterial concepts, that are out of time or space concept, even though it is concerned about property of mere space that represents absolute nothingness
To illustrate it further, consider the concept of Pythagorean fact!
Does this mere fact exist?
Does this mere fact die with time?
Did this fact have any start?
Isn't this mere fact a real existing fact, whether we know it or don't know it?
Isn't this mere fact a real existing fact even though monkeys do not comprehend it yet? wonder!
So, it is a mere fact in numbers, in algebra, in space too which is the reality we are inside unless you are the only one who is dreaming alone?
So, constructible numbers only exist the same way, where no one else can construct them except their real creator (which is the defined unity) the only one can give them existence once the one (the unity) exist by definition
So, one by definition is the real God of any existing number, and if we can define the absolute true unity as one, then this one can only construct anything from existing numbers to numbers in the reality, sure!
Post by Me
Post by bassam king karzeddin
so your (Pi) is definitely not any existing number,
Indeed, (in this -physical- sense) just like ALL numbers.
But so, unfortunately, (Pi) was a human Imaginary fetish that they worshipped
Post by Me
Post by bassam king karzeddin
just like any real algebraic number that is not constructible, sure
Of course. :-)
Just like natural numbers, negative numbers (or integers), rational numbers, etc. etc.
No, things you describe aren't alike for sure
BKK
b***@gmail.com
2017-07-16 18:55:26 UTC
Permalink
Proablem is that the old phylosophers were writing so much
incomprehensible nonsense, that couldn't stand the time,
some notions of infinity were only fixed by Gallileo, so
that it is very likely for any crank choosing some old theory,

that it will be anyway full of inconsistencies. You also
find new phylosophers writing a lot of incomprehensible
nonsense, which only makes sense as Ethics, Moral, Religion,
etc... Take for example teleological advises in the form:

Krishnamurti: Do you know what it means?
Needleman: Association with the wise.
Krishnamurti: No, with good people.
Needleman: With good people, Ah!
Krishnamurti: Being good you are wise. Not, being wise you are good.
Needleman: I understand that.
Krishnamurti: Because you are good, you are wise.[2]
https://en.wikipedia.org/wiki/Satsang

Has probably zero mathematical content, except you
might make some social/economic model out of it and
speculate what the benefinit could be. Like what is
the effective rational benefit of being/having vegetarians

for the individual/the society? (*) Now we have a
BKK-ish contructible number veganism. Ha Ha!

BTW: Pythagoras was preaching Vegetarism:
"Bei so reichlichem Gut, das die Erde, die beste der Mütter,
Zeuget, behagt dir nichts als traurige Stücke zu kauen
Mit unseligem Zahn und zu tun nach Art der Zyklopen?"
http://gutenberg.spiegel.de/buch/metamorphosen-4723/66

Was Pythagoras a Mathematician or Cosmologist?
https://plato.stanford.edu/entries/pythagoras/#WasPytMatCos
Post by b***@gmail.com
Ha Ha BKK going full circle. When I want to see absolute
nothingness, I look at BKK, no clue about math.
Like John Gabriel bird brain with his calcluelss, there
is BKK with his Pythagoschizo brabble.
Post by bassam king karzeddin
Post by Me
Post by bassam king karzeddin
But the absolute (Pi) measures exists only in the perfect circle,
"only" is to strong a word here. Still...
Post by bassam king karzeddin
unfortunately, the perfect circle doesn't exist in reality, even you draw
your circle to contain all the universe,
Assuming this is true....
Post by bassam king karzeddin
thus leaving your (Pi) only in mind and not, in reality,
*No* number exists in "reality", man. There may, say, be two apples on a desk, but you won't find THE NUMBER 2 on it.
Here I am talking about existence of immaterial concepts, that are out of time or space concept, even though it is concerned about property of mere space that represents absolute nothingness
To illustrate it further, consider the concept of Pythagorean fact!
Does this mere fact exist?
Does this mere fact die with time?
Did this fact have any start?
Isn't this mere fact a real existing fact, whether we know it or don't know it?
Isn't this mere fact a real existing fact even though monkeys do not comprehend it yet? wonder!
So, it is a mere fact in numbers, in algebra, in space too which is the reality we are inside unless you are the only one who is dreaming alone?
So, constructible numbers only exist the same way, where no one else can construct them except their real creator (which is the defined unity) the only one can give them existence once the one (the unity) exist by definition
So, one by definition is the real God of any existing number, and if we can define the absolute true unity as one, then this one can only construct anything from existing numbers to numbers in the reality, sure!
Post by Me
Post by bassam king karzeddin
so your (Pi) is definitely not any existing number,
Indeed, (in this -physical- sense) just like ALL numbers.
But so, unfortunately, (Pi) was a human Imaginary fetish that they worshipped
Post by Me
Post by bassam king karzeddin
just like any real algebraic number that is not constructible, sure
Of course. :-)
Just like natural numbers, negative numbers (or integers), rational numbers, etc. etc.
No, things you describe aren't alike for sure
BKK
Ross A. Finlayson
2017-07-16 19:01:53 UTC
Permalink
Post by b***@gmail.com
Proablem is that the old phylosophers were writing so much
incomprehensible nonsense, that couldn't stand the time,
some notions of infinity were only fixed by Gallileo, so
that it is very likely for any crank choosing some old theory,
that it will be anyway full of inconsistencies. You also
find new phylosophers writing a lot of incomprehensible
nonsense, which only makes sense as Ethics, Moral, Religion,
Krishnamurti: Do you know what it means?
Needleman: Association with the wise.
Krishnamurti: No, with good people.
Needleman: With good people, Ah!
Krishnamurti: Being good you are wise. Not, being wise you are good.
Needleman: I understand that.
Krishnamurti: Because you are good, you are wise.[2]
https://en.wikipedia.org/wiki/Satsang
Has probably zero mathematical content, except you
might make some social/economic model out of it and
speculate what the benefinit could be. Like what is
the effective rational benefit of being/having vegetarians
for the individual/the society? (*) Now we have a
BKK-ish contructible number veganism. Ha Ha!
"Bei so reichlichem Gut, das die Erde, die beste der Mütter,
Zeuget, behagt dir nichts als traurige Stücke zu kauen
Mit unseligem Zahn und zu tun nach Art der Zyklopen?"
http://gutenberg.spiegel.de/buch/metamorphosen-4723/66
Was Pythagoras a Mathematician or Cosmologist?
https://plato.stanford.edu/entries/pythagoras/#WasPytMatCos
Post by b***@gmail.com
Ha Ha BKK going full circle. When I want to see absolute
nothingness, I look at BKK, no clue about math.
Like John Gabriel bird brain with his calcluelss, there
is BKK with his Pythagoschizo brabble.
Post by bassam king karzeddin
Post by Me
Post by bassam king karzeddin
But the absolute (Pi) measures exists only in the perfect circle,
"only" is to strong a word here. Still...
Post by bassam king karzeddin
unfortunately, the perfect circle doesn't exist in reality, even you draw
your circle to contain all the universe,
Assuming this is true....
Post by bassam king karzeddin
thus leaving your (Pi) only in mind and not, in reality,
*No* number exists in "reality", man. There may, say, be two apples on a desk, but you won't find THE NUMBER 2 on it.
Here I am talking about existence of immaterial concepts, that are out of time or space concept, even though it is concerned about property of mere space that represents absolute nothingness
To illustrate it further, consider the concept of Pythagorean fact!
Does this mere fact exist?
Does this mere fact die with time?
Did this fact have any start?
Isn't this mere fact a real existing fact, whether we know it or don't know it?
Isn't this mere fact a real existing fact even though monkeys do not comprehend it yet? wonder!
So, it is a mere fact in numbers, in algebra, in space too which is the reality we are inside unless you are the only one who is dreaming alone?
So, constructible numbers only exist the same way, where no one else can construct them except their real creator (which is the defined unity) the only one can give them existence once the one (the unity) exist by definition
So, one by definition is the real God of any existing number, and if we can define the absolute true unity as one, then this one can only construct anything from existing numbers to numbers in the reality, sure!
Post by Me
Post by bassam king karzeddin
so your (Pi) is definitely not any existing number,
Indeed, (in this -physical- sense) just like ALL numbers.
But so, unfortunately, (Pi) was a human Imaginary fetish that they worshipped
Post by Me
Post by bassam king karzeddin
just like any real algebraic number that is not constructible, sure
Of course. :-)
Just like natural numbers, negative numbers (or integers), rational numbers, etc. etc.
No, things you describe aren't alike for sure
BKK
j4n bur53
2017-07-16 19:03:57 UTC
Permalink
Or the beneinfinite as well...
https://en.wikipedia.org/wiki/Welfare
Post by b***@gmail.com
speculate what the benefinit could be. Like what is
Me
2017-07-17 12:42:43 UTC
Permalink
Post by bassam king karzeddin
Post by Me
Post by bassam king karzeddin
thus leaving your (Pi) only in mind and not, in reality,
*No* number exists in "reality", man. There may, say, be two apples on a
desk, but you won't find THE NUMBER 2 on it.
Here I am talking about existence of immaterial concepts, that are out of
time or space [...]
Right! Of course the number Pi is such a "concept", just like any other number.
bassam king karzeddin
2017-07-17 15:33:55 UTC
Permalink
Post by Me
Post by bassam king karzeddin
Post by Me
Post by bassam king karzeddin
thus leaving your (Pi) only in mind and not, in reality,
*No* number exists in "reality", man. There may, say, be two apples on a
desk, but you won't find THE NUMBER 2 on it.
Here I am talking about existence of immaterial concepts, that are out of
time or space [...]
Right! Of course the number Pi is such a "concept", just like any other number.
(Pi) is indeed like any other fiction number, but not like any real constructible number, sure

BKK
bassam king karzeddin
2017-07-12 12:02:18 UTC
Permalink
Post by konyberg
Post by bassam king karzeddin
See also AP recent thread here: https://groups.google.com/forum/#!topic/sci.math/CDAML9f0KP0
Let me think of the simplest answer for this type of questions
Naturally, we can’t choose or assume an arbitrary unit such that it represents another transcendental number, otherwise, we must return to the same point of the question, since this becomes as an assumption in mind only and never related to any reality that can be demonstrated by Geometrical representation
But from the basic definition of the transcendental numbers we see that those real transcendental numbers are the roots of polynomials that are certainly not existing polynomials, therefore the probability of picking up a random length such that it represents a transcendental number relative to any arbitrarily chosen unit length exactly equals to zero
Thus, the transcendental numbers generally are not any real existing numbers, but since we do need them for our own needs like (Pi) for APPROXIMATING the circle area for example, then we can simply do present them in many APPROXIMATED constructible form numbers (generally as rational numbers) depending on our required degree of accuracy for mainly our own practical problem that is not necessarily related to mathematics, and so, unfortunately, there are no other choices in mathematics except in mind only as a mere notation (which is more than meaningless)
However, there are many other ways to demonstrate this simple and too obvious fact from the elementary principles of basic mathematics for sure
********************Adding***********************************
Since also with all human knowledge, we couldn’t construct EXACTLY even one real transcendental number as π for example, (but approximately yes by so many known methods (as paper folding, Origami, marked rulers, carpenter square, Cauchy sequences, famous cuts, Newton’s Approximations, …etc)
Then the question arises, how can we construct exactly a number so randomly, if we really know it is even impossible construction with all our tools and knowledge in advance?
So, the probability must be equals to zero, sure
Regards
Bassam King Karzeddin
JULY 7, 2017
If you chose a radius of 0.5 in your compass. Then draw a circle. Isn't PI then constructed?
KON
What do I mean exactly by earlier reply is that the circle you can observe before your eyes is actually not a perfect circle since this circle doesn't exist for sure

But you see actually a regular constructible polygon with many sides, where the ability of visibility for human eyes is limited and can't distinguish this fact in mind thinking that this is a perfect circle

Doesn't (Pi) increase absolutely in mathematics? wonder

Or do you think truly that there is any constructible number that is less than (Pi)?

And do not let that misleading concept of convergence mislead you since it is almost the same meaning as divergence concept but in opposite magnitude infinitism direction, (never and forever vanishing), unless WE DECIDE to make it vanishing deliberately for our own irrelevant little purposes

Thus, if we can distinguish our ability of visibility then the problem becomes so easy where you have to define the number of sides of that regular constructible polygon you see by your eyes and then calculate the side length and multiply by the number of sides where you feel very close from absolute value of (Pi), exactly the same way they do to APPROXIMATE (pi)

In short, the (Pi) you see is a constructible number always and forever

But the (Pi) you seek in mind isn't anywhere in reality, but only in human MIND as very famous example of the exact definition of an illusion for sure

Read here "the mathematics had been broken by the death of (Pi)"

However, It doesn't any matter if you don't like it

https://groups.google.com/forum/#!msg/sci.math/fWHaxeAPats/I11g_IDkAAAJ

BKK
j4n bur53
2017-07-16 18:58:08 UTC
Permalink
The problem is, that pythagoras is probably the
only legend here, and not the real numbers.
Post by b***@gmail.com
Was Pythagoras a Mathematician or Cosmologist?
https://plato.stanford.edu/entries/pythagoras/#WasPytMatCos
bassam king karzeddin
2017-07-17 16:43:48 UTC
Permalink
Post by j4n bur53
The problem is, that pythagoras is probably the
only legend here, and not the real numbers.
So, you pronounced it correctly this time

The problem is only Pythagoras, and do you really comprehend what does it mean when the Pythagorean theorem is the issue? wonder!

And do you think so foolishly that the Pythagorean theorem needs any of your meaningless definitions to be absolutely true and valid? wonder again!

Or do you think foolishly that anyone can beat a real true theorem? No wonder here!

Same applies to any absolute facts without at all your dying meaningless definitions for sure
And did you understand how did the top legend mathematicians (sleeping peacefully in the history) betray the Pythagoreans and made the Queen for sale?

Most likely you didn't even read my old posts in those regards

So, wait and see if you are lucky enough to witness how the Queen would be liberated from all those humans Devilish desires, where those would be judged again with the Equal sign operations,(===...), that they had neglected purposely to confess it exactly, where then no escape from a balance (in any place or in any century for sure)

BKK
Post by j4n bur53
Post by b***@gmail.com
Was Pythagoras a Mathematician or Cosmologist?
https://plato.stanford.edu/entries/pythagoras/#WasPytMatCos
b***@gmail.com
2017-07-11 13:12:15 UTC
Permalink
Well to begin with this is complete nonsense. The
finite decimal representations, i.e. those representations
that stop somewhere, have the form:

d/10^n

So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.

So their union gives the reals:

I union Q = R

And they are disjoint:

I intersect Q = {}

So they form a partition of the real numbers into
two kind of numbers. Which can be also written
as follows:

R \ Q = I

Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
of irrational numbers I. We can write:

R \ Q \ AI = T

Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
numbers (Q), we get:

R \ A = T

Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
a prefix belongs to a transcendental number is:

The probability is 0

Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,

by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
then have:

P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)

= 1/2n

So that we have for lim n->oo Rn = R, finally:

P(T |
https://math.stackexchange.com/q/37121/4414
Post by bassam king karzeddin
What is the probability of picking up a random length such that it represents a transcendental number relative to any arbitrary chosen constructible unit length?
Big Hint: the probability exactly equals to zero, sure
Regards
Bassam King Karzeddin
7 July 2017
b***@gmail.com
2017-07-11 13:20:24 UTC
Permalink
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
that stop somewhere, have the form:

d/10^n

So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.

So their union gives the reals:

I union Q = R

And they are disjoint:

I intersect Q = {}

So they form a partition of the real numbers into
two kind of numbers. Which can be also written
as follows:

R \ Q = I

Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
of irrational numbers I. We can write:

R \ Q \ AI = T

Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
numbers (Q), we get:

R \ A = T

Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
a prefix belongs to a transcendental number is:

The probability is 0

Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,

by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
then have:

P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)

= 1/2n

So that we have for lim n->oo Rn = R, finally:

P(T | C) = lim n->oo 1/2n = 0

The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
same result, on other ground that P(B n C) = 0 already:

P(A | C) = 0

So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different

probabilities.

https://math.stackexchange.com/q/37121/4414

https://math.stackexchange.com/q/2070246/4414
j4n bur53
2017-07-11 13:24:31 UTC
Permalink
I dont know whether the below is allowed in
probability theory, but you get the idea.
Markus Klyver
2017-07-11 15:19:34 UTC
Permalink
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
https://math.stackexchange.com/q/2070246/4414
What? The probability that a number chosen at uniformly in [a, b], with a < b, is transcendental is exactly equal to 1. The measure of [a, b] is b-a and the measure of the set of transcendentals in [a, b] is also b-a.
b***@gmail.com
2017-07-11 16:08:55 UTC
Permalink
Yes thats also correct. But lets say I give you
an integer number k, how much is the probability
for a real number x, that floor(x)=k ?

zero, right?
Post by Markus Klyver
What? The probability that a number chosen at uniformly in [a, b], with a < b, is transcendental is exactly equal to 1. The measure of [a, b] is b-a and the measure of the set of transcendentals in [a, b] is also b-a.
b***@gmail.com
2017-07-11 16:28:34 UTC
Permalink
I am assuming choosing uniformly from R, is something
that exists, and that can be modelled as choosing uniformly
from Rn = [n,n), and then taking the limes n->oo Rn = R.

But I am not sure whether this is possible. Otherwise
we could use this function to randomly generate a
value in R+:

f(x) = x/(1-x)

If the random variable X is uniformly distributed on [0,1),
then Y = f(X) is what on R+? How much is the probability
floor(Y) = k? Depends on k now?

P(k =< f(X) < k+1) = ?

Lets see P(f(X) < c) = P(X < f^(-1)(c)) = c/(1+c), so
that we now have

P(k =< f(X) < k+1) = P(f(X) < k+1) - P(f(X) < k)

= (k+1)/(k+2)-k/(k+1)

= 1/(k^2 + 3 k + 2)

So I guess BKKs problem is underspecified.
What is the used distribution probability?
Post by b***@gmail.com
Yes thats also correct. But lets say I give you
an integer number k, how much is the probability
for a real number x, that floor(x)=k ?
zero, right?
Post by Markus Klyver
What? The probability that a number chosen at uniformly in [a, b], with a < b, is transcendental is exactly equal to 1. The measure of [a, b] is b-a and the measure of the set of transcendentals in [a, b] is also b-a.
bassam king karzeddin
2017-07-11 17:12:43 UTC
Permalink
Post by b***@gmail.com
I am assuming choosing uniformly from R, is something
that exists, and that can be modelled as choosing uniformly
from Rn = [n,n), and then taking the limes n->oo Rn = R.
But I am not sure whether this is possible. Otherwise
we could use this function to randomly generate a
f(x) = x/(1-x)
If the random variable X is uniformly distributed on [0,1),
then Y = f(X) is what on R+? How much is the probability
floor(Y) = k? Depends on k now?
P(k =< f(X) < k+1) = ?
Lets see P(f(X) < c) = P(X < f^(-1)(c)) = c/(1+c), so
that we now have
P(k =< f(X) < k+1) = P(f(X) < k+1) - P(f(X) < k)
= (k+1)/(k+2)-k/(k+1)
= 1/(k^2 + 3 k + 2)
So I guess BKKs problem is underspecified.
What is the used distribution probability?
Post by b***@gmail.com
Yes thats also correct. But lets say I give you
an integer number k, how much is the probability
for a real number x, that floor(x)=k ?
zero, right?
Post by Markus Klyver
What? The probability that a number chosen at uniformly in [a, b], with a < b, is transcendental is exactly equal to 1. The measure of [a, b] is b-a and the measure of the set of transcendentals in [a, b] is also b-a.
Before I go into this, let us REMEMBER the so easy basic and so elementary definition of a real transcendental number, where it states that a real transcendental must be a root of a polynomial with infinite terms, doesn't it? wonder!

So, please provide us with that polynomial that has at least one real root in order to believe in you and in the well-established mathematics for sure!

Hint, ask AP since he has a good idea about the subject, sure

BKK
b***@gmail.com
2017-07-11 17:50:59 UTC
Permalink
No, a transcendental number must not necessarely
be the root of an "infinite polynomial", what
ever this should mean.

Where did you get that definition from, such a
definition is nowhere implied? The definition is
only that it is not the root of a finite polynomial.

But for example e = e^1, can be viewed as the
value of an "infinite polynomial", i.e. a series:

f(x) = 1 + x + 1/2 x^2 + 1/6 x^3 + ...

You should know:

f(1) = e

And every decimal expansion:

r = d0.d1 d2 d3 ....

Leads to a series:

g(x) = d0 + d1*x + d2*x^2 + ...

With the property:

r = g(1/10)

For your claim to say r, is also the root of some
series h(x), i.e. we will have:

h(r) = 0

You would for example show that such a series
reversal exists, i.e. going from g to h, as was
for example done Morse and Feshbach (1953):

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

We would then have:

g^(-1)(r) - 1/10 = 0

Bye
Post by bassam king karzeddin
Post by b***@gmail.com
I am assuming choosing uniformly from R, is something
that exists, and that can be modelled as choosing uniformly
from Rn = [n,n), and then taking the limes n->oo Rn = R.
But I am not sure whether this is possible. Otherwise
we could use this function to randomly generate a
f(x) = x/(1-x)
If the random variable X is uniformly distributed on [0,1),
then Y = f(X) is what on R+? How much is the probability
floor(Y) = k? Depends on k now?
P(k =< f(X) < k+1) = ?
Lets see P(f(X) < c) = P(X < f^(-1)(c)) = c/(1+c), so
that we now have
P(k =< f(X) < k+1) = P(f(X) < k+1) - P(f(X) < k)
= (k+1)/(k+2)-k/(k+1)
= 1/(k^2 + 3 k + 2)
So I guess BKKs problem is underspecified.
What is the used distribution probability?
Post by b***@gmail.com
Yes thats also correct. But lets say I give you
an integer number k, how much is the probability
for a real number x, that floor(x)=k ?
zero, right?
Post by Markus Klyver
What? The probability that a number chosen at uniformly in [a, b], with a < b, is transcendental is exactly equal to 1. The measure of [a, b] is b-a and the measure of the set of transcendentals in [a, b] is also b-a.
Before I go into this, let us REMEMBER the so easy basic and so elementary definition of a real transcendental number, where it states that a real transcendental must be a root of a polynomial with infinite terms, doesn't it? wonder!
So, please provide us with that polynomial that has at least one real root in order to believe in you and in the well-established mathematics for sure!
Hint, ask AP since he has a good idea about the subject, sure
BKK
b***@gmail.com
2017-07-11 17:53:07 UTC
Permalink
For transcendental r, we know g^(-1)(_) will not
be a finite polynomial, since otherwise we would have
a contradiction, and r would be algebraic.
Post by b***@gmail.com
g^(-1)(r) - 1/10 = 0
b***@gmail.com
2017-07-11 18:04:19 UTC
Permalink
But main problem is convergence of the reverse
series. How do you proof that g^(-1)(r) converges?

g^(-1)(_) is only a formal power series. The original
g(_) will converge, since the digits dj are form
the value set {0,..,9}, so you find convergence

but g^(-1)(_) what do we know about its convergence?
And if the convergence fails, what about starting
with a different g(_), to completely answer the question.
Post by b***@gmail.com
For transcendental r, we know g^(-1)(_) will not
be a finite polynomial, since otherwise we would have
a contradiction, and r would be algebraic.
Post by b***@gmail.com
g^(-1)(r) - 1/10 = 0
Markus Klyver
2017-07-11 18:44:24 UTC
Permalink
Post by bassam king karzeddin
Post by b***@gmail.com
I am assuming choosing uniformly from R, is something
that exists, and that can be modelled as choosing uniformly
from Rn = [n,n), and then taking the limes n->oo Rn = R.
But I am not sure whether this is possible. Otherwise
we could use this function to randomly generate a
f(x) = x/(1-x)
If the random variable X is uniformly distributed on [0,1),
then Y = f(X) is what on R+? How much is the probability
floor(Y) = k? Depends on k now?
P(k =< f(X) < k+1) = ?
Lets see P(f(X) < c) = P(X < f^(-1)(c)) = c/(1+c), so
that we now have
P(k =< f(X) < k+1) = P(f(X) < k+1) - P(f(X) < k)
= (k+1)/(k+2)-k/(k+1)
= 1/(k^2 + 3 k + 2)
So I guess BKKs problem is underspecified.
What is the used distribution probability?
Post by b***@gmail.com
Yes thats also correct. But lets say I give you
an integer number k, how much is the probability
for a real number x, that floor(x)=k ?
zero, right?
Post by Markus Klyver
What? The probability that a number chosen at uniformly in [a, b], with a < b, is transcendental is exactly equal to 1. The measure of [a, b] is b-a and the measure of the set of transcendentals in [a, b] is also b-a.
Before I go into this, let us REMEMBER the so easy basic and so elementary definition of a real transcendental number, where it states that a real transcendental must be a root of a polynomial with infinite terms, doesn't it? wonder!
So, please provide us with that polynomial that has at least one real root in order to believe in you and in the well-established mathematics for sure!
Hint, ask AP since he has a good idea about the subject, sure
BKK
???

p(x) = x has one real root.
Markus Klyver
2017-07-11 18:47:47 UTC
Permalink
Post by b***@gmail.com
Yes thats also correct. But lets say I give you
an integer number k, how much is the probability
for a real number x, that floor(x)=k ?
zero, right?
Post by Markus Klyver
What? The probability that a number chosen at uniformly in [a, b], with a < b, is transcendental is exactly equal to 1. The measure of [a, b] is b-a and the measure of the set of transcendentals in [a, b] is also b-a.
That depends on the distribution. A uniform distribution over ℝ does not make much sense to me. What would its PDF/CDF be? The uniform distribution is only defined for intervals [a, b].
bassam king karzeddin
2017-07-11 16:15:21 UTC
Permalink
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So, tell me, please what is the suitable value for an integer (n), to make your alleged real transcendental or real algebraic number or any infinite decimal representation of any constructible number as really a real number? wonder

Of course, whatever finite integer (n) you choose, your number is always a rational number, and you have no other choice, don't you really?

But the problem with mathematicians that they claim funnily (n) must be an integer with infinite sequence of digits where at the same time (d) also must be with infinite sequence of digits, where also they don't accept those integers with infinite sequence of digits, and even if they were to accept them, they would not save the situation and make those ratios (d/10^n) as existing real numbers, for sure

[snip all this ignorance]

Numbers are only constructible

BKK
bassam king karzeddin
2017-07-13 15:15:43 UTC
Permalink
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)

This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers

Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic

Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!

Or maybe you don't comprehend any of my questions/or answers, wonder!

But most likely you don't understand at all for sure

BKK
b***@gmail.com
2017-07-13 15:37:10 UTC
Permalink
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
b***@gmail.com
2017-07-13 15:45:26 UTC
Permalink
On stackexchange you find nice stuff like:

Get a (two-dimensional) dog and a very long (one-dimensional) leash. Send your dog out exploring, letting the leash play out. When the dog returns, try to pull in the leash. (Meaning, you try to reel in the loop with you and the dog staying put.) On a sphere, the leash can always be pulled in; on a torus, sometimes it can't be.

Which wont make it probably into wikipedia.

https://math.stackexchange.com/a/854394/4414
Post by b***@gmail.com
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
bassam king karzeddin
2017-07-13 16:21:13 UTC
Permalink
Post by b***@gmail.com
Get a (two-dimensional) dog and a very long (one-dimensional) leash. Send your dog out exploring, letting the leash play out. When the dog returns, try to pull in the leash. (Meaning, you try to reel in the loop with you and the dog staying put.) On a sphere, the leash can always be pulled in; on a torus, sometimes it can't be.
Which wont make it probably into wikipedia.
https://math.stackexchange.com/a/854394/4414
Post by b***@gmail.com
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
I will tell you much easier one, that is certainly easily proven, if you have a dog, then mark equally two points arbitrarily (one on the dog tail, and one on the dog ass), same do for yourself but since you don't have a tail, let one point be marked on your ass, and the other on your face, then my theorems says, whenever, and wherever you and your dog are, then all the marked points must lie EXACTLY on a surface of a sphere, no matter even if you send your dog anywhere in space, for more than sure I swear

BKK
b***@gmail.com
2017-07-13 16:49:50 UTC
Permalink
The algebraic numbers are like living on a sphere,
you can always find a polynomial p(x) in Z[x], so
that p(r)=0 (you can reel in the leash).

For transcendental numbers this is not possible.
You always get stuck, take any polynomial p(x) in
Z[X], you will have p(r)<>0.

if you have two transcendental numbers r1 and r2,
we might indeed have p(r1)-p(r2)=0 for some polynomial,
call such transcendental numbers friends.

The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.

Proof: Left as an exercise.

Now look at the set of befriended equivalence classes,
i.e. { [r] | r in R }. How many equivalence classes form
the algebraic numbers, and how many equivalence classes

form the transcendental numbers?
Post by b***@gmail.com
Get a (two-dimensional) dog and a very long (one-dimensional) leash.
Send your dog out exploring, letting the leash play out. When the
dog returns, try to pull in the leash. (Meaning, you try to reel in
the loop with you and the dog staying put.) On a sphere, the leash
can always be pulled in; on a torus, sometimes it can't be.
Post by b***@gmail.com
Which wont make it probably into wikipedia.
https://math.stackexchange.com/a/854394/4414
Post by b***@gmail.com
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
b***@gmail.com
2017-07-13 17:05:09 UTC
Permalink
Actually I have no such proof. Just guessed.
p1,p2,p3 non-constant integer polynomials.
Counter example or proof?
Post by b***@gmail.com
The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.
Proof: Left as an exercise.
b***@gmail.com
2017-07-13 22:09:40 UTC
Permalink
Two algebraic numbers are trivially befriended, since
by definition for an algerbaic number there is a polynomial,
where the algebraic number is a root of.

So if we have:

p_a(a) = 0

p_b(b) = 0

We can take the polynomial:

p(x) = p_a(x)*p_b(x)

Then:

p(a) - p(b) = 0

But it seems also that algebraic and transcendental
numbers cannot be befriended. Whats the full
proof here?

Transcendental numbers can be also befriended, take
for example

a = π and

b = -π-1

They are befriended via the polynomial

p(x) = x^2 + x.

Just try it by yourself:

p(a) - p(b) =

p(π) - p(-π-1) =

π^2+π - (-π-1)^2-(-π-1) =

π^2 + π - π^2 - 2*π -1 + π + 1 =

0

The transitivity needs to be proved also for transcendental
numbers. If we have transitivity, than we have equivalence
classes among real numbers.

Whereby the algebraic numbers oollapse into a single
equivalence class. Next step, elaborate whether the
befriended equivalence classes of

transcendental numbers are uncountable.

https://math.stackexchange.com/q/2357627/4414
Post by b***@gmail.com
Actually I have no such proof. Just guessed.
p1,p2,p3 non-constant integer polynomials.
Counter example or proof?
Post by b***@gmail.com
The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.
Proof: Left as an exercise.
bassam king karzeddin
2017-07-15 08:23:23 UTC
Permalink
Post by b***@gmail.com
Two algebraic numbers are trivially befriended, since
by definition for an algerbaic number there is a polynomial,
where the algebraic number is a root of.
p_a(a) = 0
p_b(b) = 0
p(x) = p_a(x)*p_b(x)
p(a) - p(b) = 0
But it seems also that algebraic and transcendental
numbers cannot be befriended. Whats the full
proof here?
Transcendental numbers can be also befriended, take
for example
a = π and
b = -π-1
They are befriended via the polynomial
p(x) = x^2 + x.
p(a) - p(b) =
p(π) - p(-π-1) =
π^2+π - (-π-1)^2-(-π-1) =
π^2 + π - π^2 - 2*π -1 + π + 1 =
0
The transitivity needs to be proved also for transcendental
numbers. If we have transitivity, than we have equivalence
classes among real numbers.
Whereby the algebraic numbers oollapse into a single
equivalence class. Next step, elaborate whether the
befriended equivalence classes of
transcendental numbers are uncountable.
https://math.stackexchange.com/q/2357627/4414
You keep buttering things to make it as poised food for others

And when you make it separately at that well-moderated site (MSE), it is ethical to mention original references of discussion here, or simply you can tell them that BKK claims (with his alleged proofs) that real transcendental numbers are non-existing numbers, but I know those vast majority of morons at MSE can't tolerate any new claim for sure, thus they would immediately delete the issue as if they really can protect those numbers from not being collapsed forever, wonder!
Post by b***@gmail.com
Post by b***@gmail.com
Actually I have no such proof. Just guessed.
p1,p2,p3 non-constant integer polynomials.
Counter example or proof?
Post by b***@gmail.com
The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.
Proof: Left as an exercise.
The very simple exercise is left for many and not necessarily very clever schools students to disprove the fiction of those trans. numbers also in many other ways for sure

BKK
bassam king karzeddin
2017-07-13 17:32:49 UTC
Permalink
Post by b***@gmail.com
The algebraic numbers are like living on a sphere,
you can always find a polynomial p(x) in Z[x], so
that p(r)=0 (you can reel in the leash).
For transcendental numbers this is not possible.
You always get stuck, take any polynomial p(x) in
Z[X], you will have p(r)<>0.
if you have two transcendental numbers r1 and r2,
we might indeed have p(r1)-p(r2)=0 for some polynomial,
call such transcendental numbers friends.
The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.
Proof: Left as an exercise.
Now look at the set of befriended equivalence classes,
i.e. { [r] | r in R }. How many equivalence classes form
the algebraic numbers, and how many equivalence classes
form the transcendental numbers?
Post by b***@gmail.com
Get a (two-dimensional) dog and a very long (one-dimensional) leash.
Send your dog out exploring, letting the leash play out. When the
dog returns, try to pull in the leash. (Meaning, you try to reel in
the loop with you and the dog staying put.) On a sphere, the leash
can always be pulled in; on a torus, sometimes it can't be.
Post by b***@gmail.com
Which wont make it probably into wikipedia.
https://math.stackexchange.com/a/854394/4414
Post by b***@gmail.com
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
So, before going into those alleged real algebraic numbers (that are not constructible numbers), let us see exactly only one of them in reality since they were illegally called real numbers

But, stop here, nobody would make impossibilities except in legendary stories

What do I mean by reality, is simply the common sense anyone must have, especially the mathematicians

Why should I need more classification of real numbers if they are called real but of impossible existence in the real reality, especially that we can measure any length to an unbelievable degree of accuracy with only Constructible numbers or even with rationales alone, wonder!

Is it only for entertainment of some abnormal people who keep producing more nonsense, so let them entertain alone and away from societies they exploit badly and harm them too for mere obvious so silly mind games that are completely useless since they were based on deliberate planted artificial fallacies for sure

This is an example of very simple COMMON SENSE reality even with irrational numbers
Sqrt(10) is the diagonal of a rectangle of sides (1, 3) Units, (FINISHED)
So, where is the reality of those other named real irrational numbers but algebraic? wonder!

Yes, in mind, what a real more scandal than this fiction story in the history of mathematics than this, yes, many more for sure

But, the human mind usually can contain much more fictions if we keep silent

So, what can you very poor mind do to save your own sons and daughters, from this real danger of obvious ignorance that is also expanding endlessly

Butter keep silent and let it go, lazy

BKK
b***@gmail.com
2017-07-13 18:10:08 UTC
Permalink
Whats this butter thing? You shouldn't use non-existing
words in english. The devil will grab you by the nuts.

http://www.duden.de/rechtschreibung/Butter
Post by bassam king karzeddin
Butter keep silent and let it go, lazy
Markus Klyver
2017-07-13 18:26:29 UTC
Permalink
Post by bassam king karzeddin
Post by b***@gmail.com
The algebraic numbers are like living on a sphere,
you can always find a polynomial p(x) in Z[x], so
that p(r)=0 (you can reel in the leash).
For transcendental numbers this is not possible.
You always get stuck, take any polynomial p(x) in
Z[X], you will have p(r)<>0.
if you have two transcendental numbers r1 and r2,
we might indeed have p(r1)-p(r2)=0 for some polynomial,
call such transcendental numbers friends.
The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.
Proof: Left as an exercise.
Now look at the set of befriended equivalence classes,
i.e. { [r] | r in R }. How many equivalence classes form
the algebraic numbers, and how many equivalence classes
form the transcendental numbers?
Post by b***@gmail.com
Get a (two-dimensional) dog and a very long (one-dimensional) leash.
Send your dog out exploring, letting the leash play out. When the
dog returns, try to pull in the leash. (Meaning, you try to reel in
the loop with you and the dog staying put.) On a sphere, the leash
can always be pulled in; on a torus, sometimes it can't be.
Post by b***@gmail.com
Which wont make it probably into wikipedia.
https://math.stackexchange.com/a/854394/4414
Post by b***@gmail.com
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
So, before going into those alleged real algebraic numbers (that are not constructible numbers), let us see exactly only one of them in reality since they were illegally called real numbers
But, stop here, nobody would make impossibilities except in legendary stories
What do I mean by reality, is simply the common sense anyone must have, especially the mathematicians
Why should I need more classification of real numbers if they are called real but of impossible existence in the real reality, especially that we can measure any length to an unbelievable degree of accuracy with only Constructible numbers or even with rationales alone, wonder!
Is it only for entertainment of some abnormal people who keep producing more nonsense, so let them entertain alone and away from societies they exploit badly and harm them too for mere obvious so silly mind games that are completely useless since they were based on deliberate planted artificial fallacies for sure
This is an example of very simple COMMON SENSE reality even with irrational numbers
Sqrt(10) is the diagonal of a rectangle of sides (1, 3) Units, (FINISHED)
So, where is the reality of those other named real irrational numbers but algebraic? wonder!
Yes, in mind, what a real more scandal than this fiction story in the history of mathematics than this, yes, many more for sure
But, the human mind usually can contain much more fictions if we keep silent
So, what can you very poor mind do to save your own sons and daughters, from this real danger of obvious ignorance that is also expanding endlessly
Butter keep silent and let it go, lazy
BKK
Are you happier with the name "complete numbers"? I mean, you can call real numbers whatever you want. It's the same thing.
bassam king karzeddin
2017-07-13 18:45:04 UTC
Permalink
Post by Markus Klyver
Post by bassam king karzeddin
Post by b***@gmail.com
The algebraic numbers are like living on a sphere,
you can always find a polynomial p(x) in Z[x], so
that p(r)=0 (you can reel in the leash).
For transcendental numbers this is not possible.
You always get stuck, take any polynomial p(x) in
Z[X], you will have p(r)<>0.
if you have two transcendental numbers r1 and r2,
we might indeed have p(r1)-p(r2)=0 for some polynomial,
call such transcendental numbers friends.
The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.
Proof: Left as an exercise.
Now look at the set of befriended equivalence classes,
i.e. { [r] | r in R }. How many equivalence classes form
the algebraic numbers, and how many equivalence classes
form the transcendental numbers?
Post by b***@gmail.com
Get a (two-dimensional) dog and a very long (one-dimensional) leash.
Send your dog out exploring, letting the leash play out. When the
dog returns, try to pull in the leash. (Meaning, you try to reel in
the loop with you and the dog staying put.) On a sphere, the leash
can always be pulled in; on a torus, sometimes it can't be.
Post by b***@gmail.com
Which wont make it probably into wikipedia.
https://math.stackexchange.com/a/854394/4414
Post by b***@gmail.com
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
So, before going into those alleged real algebraic numbers (that are not constructible numbers), let us see exactly only one of them in reality since they were illegally called real numbers
But, stop here, nobody would make impossibilities except in legendary stories
What do I mean by reality, is simply the common sense anyone must have, especially the mathematicians
Why should I need more classification of real numbers if they are called real but of impossible existence in the real reality, especially that we can measure any length to an unbelievable degree of accuracy with only Constructible numbers or even with rationales alone, wonder!
Is it only for entertainment of some abnormal people who keep producing more nonsense, so let them entertain alone and away from societies they exploit badly and harm them too for mere obvious so silly mind games that are completely useless since they were based on deliberate planted artificial fallacies for sure
This is an example of very simple COMMON SENSE reality even with irrational numbers
Sqrt(10) is the diagonal of a rectangle of sides (1, 3) Units, (FINISHED)
So, where is the reality of those other named real irrational numbers but algebraic? wonder!
Yes, in mind, what a real more scandal than this fiction story in the history of mathematics than this, yes, many more for sure
But, the human mind usually can contain much more fictions if we keep silent
So, what can you very poor mind do to save your own sons and daughters, from this real danger of obvious ignorance that is also expanding endlessly
Butter keep silent and let it go, lazy
BKK
Are you happier with the name "complete numbers"? I mean, you can call real numbers whatever you want. It's the same thing.
It was already called, the constructible number, before thousands of years by the Pythagorean's, but what did I add is the general definition provided in my post, What is the real number? however, I may improve it further if I have time

And by this definition, the fundamental theorem of algebra would soon be replaced, and many ancient unsolved puzzles must be cleared so easily

BKK
Markus Klyver
2017-07-13 23:56:20 UTC
Permalink
Post by bassam king karzeddin
Post by Markus Klyver
Post by bassam king karzeddin
Post by b***@gmail.com
The algebraic numbers are like living on a sphere,
you can always find a polynomial p(x) in Z[x], so
that p(r)=0 (you can reel in the leash).
For transcendental numbers this is not possible.
You always get stuck, take any polynomial p(x) in
Z[X], you will have p(r)<>0.
if you have two transcendental numbers r1 and r2,
we might indeed have p(r1)-p(r2)=0 for some polynomial,
call such transcendental numbers friends.
The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.
Proof: Left as an exercise.
Now look at the set of befriended equivalence classes,
i.e. { [r] | r in R }. How many equivalence classes form
the algebraic numbers, and how many equivalence classes
form the transcendental numbers?
Post by b***@gmail.com
Get a (two-dimensional) dog and a very long (one-dimensional) leash.
Send your dog out exploring, letting the leash play out. When the
dog returns, try to pull in the leash. (Meaning, you try to reel in
the loop with you and the dog staying put.) On a sphere, the leash
can always be pulled in; on a torus, sometimes it can't be.
Post by b***@gmail.com
Which wont make it probably into wikipedia.
https://math.stackexchange.com/a/854394/4414
Post by b***@gmail.com
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
So, before going into those alleged real algebraic numbers (that are not constructible numbers), let us see exactly only one of them in reality since they were illegally called real numbers
But, stop here, nobody would make impossibilities except in legendary stories
What do I mean by reality, is simply the common sense anyone must have, especially the mathematicians
Why should I need more classification of real numbers if they are called real but of impossible existence in the real reality, especially that we can measure any length to an unbelievable degree of accuracy with only Constructible numbers or even with rationales alone, wonder!
Is it only for entertainment of some abnormal people who keep producing more nonsense, so let them entertain alone and away from societies they exploit badly and harm them too for mere obvious so silly mind games that are completely useless since they were based on deliberate planted artificial fallacies for sure
This is an example of very simple COMMON SENSE reality even with irrational numbers
Sqrt(10) is the diagonal of a rectangle of sides (1, 3) Units, (FINISHED)
So, where is the reality of those other named real irrational numbers but algebraic? wonder!
Yes, in mind, what a real more scandal than this fiction story in the history of mathematics than this, yes, many more for sure
But, the human mind usually can contain much more fictions if we keep silent
So, what can you very poor mind do to save your own sons and daughters, from this real danger of obvious ignorance that is also expanding endlessly
Butter keep silent and let it go, lazy
BKK
Are you happier with the name "complete numbers"? I mean, you can call real numbers whatever you want. It's the same thing.
It was already called, the constructible number, before thousands of years by the Pythagorean's, but what did I add is the general definition provided in my post, What is the real number? however, I may improve it further if I have time
And by this definition, the fundamental theorem of algebra would soon be replaced, and many ancient unsolved puzzles must be cleared so easily
BKK
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
b***@gmail.com
2017-07-14 00:03:52 UTC
Permalink
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
Markus Klyver
2017-07-14 00:14:07 UTC
Permalink
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
https://en.wikipedia.org/wiki/Transcendental_number#Properties
bassam king karzeddin
2017-07-15 08:08:10 UTC
Permalink
Post by Markus Klyver
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
https://en.wikipedia.org/wiki/Transcendental_number#Properties
Oops, this is very recently updated (6th July 2017), but still, even though we had been writing here about them for many years, anyway, they haven't got the right idea yet, and all their contents amounts to fiction knowledge so far

But watch out soon in the future, they would make much more updates till they know exactly that they were completely drawn into mere fictions that had been proved to school students for sure

BKK
Markus Klyver
2017-07-14 00:15:03 UTC
Permalink
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
Also see https://en.wikipedia.org/wiki/Transcendental_number_theory
b***@gmail.com
2017-07-14 01:10:25 UTC
Permalink
I mean something funny you dont find on wiki.
https://groups.google.com/d/msg/sci.math/0jeLoUPAWeQ/VWTCcQvyBgAJ
Also see https://en.wikipedia.org/wiki/...
Ross A. Finlayson
2017-07-14 02:13:27 UTC
Permalink
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
The title is of a retro-finitist crankety-troll,
"transcendental numbers" are all the non-algebraic
real numbers, which exist.

And retro-finitist crankety trolls can get bent.

"Retro-finitism" is because mathematics already
has infinity and trying to take it back (instead
of, say, leaving it back) has the retro-finitism
is an ignorance (vis-a-vis foundations besides
models of numbers).

Transcendental numbers are where the algebraic
numbers don't yet have properties of being
continuous, as a body.

So, they exist. The real numbers are continuous.

They are: continuous, the real numbers.



There's Mahler on transcendentals with a
descriptive approach, there are also ways
to describe the transcendentals in terms
of say, completions after exhausting the
algebraics (or, for example, rationals,
rational numbers).

This points to the "signal continuity".
b***@gmail.com
2017-07-14 03:15:39 UTC
Permalink
I don't know whether its is somewhere in the works
of Kurt Mahler, that a ~ equivalence relation between
real numbers a,b such that:

The equivalence classe of [1] = { a ~ 1 | a in } equals
the algebraic numbers. Maybe I have picked it up from
somewhere else.

Dont remember. So two real numbers a,b are friends,
if there exists a non-constant integer polynomial p
such that:

a ~ b iff exists p in Z[x] (deg(p)>0 & p(a)-p(b) = 0)

Is this relation transitive, so that we can form
equivalence classes? It is surely symmetric,
and reflexive.

For symmetric, if a ~ b then there was a p in Z[x],
with p(a)-p(b)=0. But then we also have p(b)-p(a)=0.
So we also have b ~ a.

For reflexiv, we need to show a ~ a. Which happens
for any polynomial, trivially p(a)-p(a)=0. What
remains to show is:

a ~ b & b ~ c => a ~ c

Anybody up to a proof or disproof?
Post by Ross A. Finlayson
There's Mahler on transcendentals with a
descriptive approach, there are also ways
b***@gmail.com
2017-07-14 03:22:59 UTC
Permalink
We can already show this without transitivity,
namely if an algebraic number a, has minimal
polynomial A(x):

A(a) = 0

When we can take this polynomial:

p(x) = (x-1)*A(x)

And we see:

p(1)-p(a) = (1-1)*A(1)-(a-1)*A(a)

= 0*A(1)-(a-1)*0

= 0

Hence 1 ~ a.
Post by b***@gmail.com
The equivalence classe of [1] = { a ~ 1 | a in } equals
the algebraic numbers. Maybe I have picked it up from
somewhere else.
b***@gmail.com
2017-07-14 15:03:44 UTC
Permalink
Now I think quantifier elimination would do the
job partially. We can easily view the expression:

p(x) - p(y)

As a bivariate polynomial:

q(x,y)

Now its easy to see that we will have:

q1(a,b) * q2(b,c) = 0

If we find new bivariate polynomial such that:

q3(x,z) = 0 iff exists y (q1(x,y) * q2(y,z) = 0)

We are almost done. We would then need to further
find a polynomial p3 such that:

q3(x,z) = p3(x)-p3(z)

But somehow, playing around today I had a lot of
set backs. Except for a small little new idea.
The interesting case is anyway:

a <> b, b <> c, a <> c

So instead of looking at p(x)-p(y), we could
also look at the following:

p(x) - p(y)
----------- = 0
x - y

Now p(x)-p(y) is divided by x-y, the result is
a new binvariate polynomial, which has a symmetry,
switching the variables doesn't change the sign anymore:

p(x) - p(y)
----------- = r(x,y) = r(y,x)
x - y

Now I am stuck, except for a further idea:

1 1 1 1
----- - ----- = (y - z) ----- -----
x - y x - z x - y x - z

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

But still stuck. It could be also that the
transitivity doesn't hold.

Working with variables, i.e. indeterminates algebraically,
has the advantage, that identities will also hold

for real values. If we avoid using A(r)=0, i.e.
some of the criteria for algebraic numbers, we get results

for transcendental numbers as well.
Post by b***@gmail.com
We can already show this without transitivity,
namely if an algebraic number a, has minimal
A(a) = 0
p(x) = (x-1)*A(x)
p(1)-p(a) = (1-1)*A(1)-(a-1)*A(a)
= 0*A(1)-(a-1)*0
= 0
Hence 1 ~ a.
Post by b***@gmail.com
The equivalence classe of [1] = { a ~ 1 | a in } equals
the algebraic numbers. Maybe I have picked it up from
somewhere else.
b***@gmail.com
2017-07-14 15:06:18 UTC
Permalink
This could have been a nice trajectory for John
Gabriels differential calculus, but the bird brain

started posting there are no real numbers, 0.999... <> 1,
Peano/Cantor/etc is stupid. Everybody knows the story.
Post by b***@gmail.com
Working with variables, i.e. indeterminates algebraically,
has the advantage, that identities will also hold
for real values. If we avoid using A(r)=0, i.e.
some of the criteria for algebraic numbers, we get results
for transcendental numbers as well.
Ross A. Finlayson
2017-07-15 00:45:23 UTC
Permalink
Post by b***@gmail.com
Now I think quantifier elimination would do the
p(x) - p(y)
q(x,y)
q1(a,b) * q2(b,c) = 0
q3(x,z) = 0 iff exists y (q1(x,y) * q2(y,z) = 0)
We are almost done. We would then need to further
q3(x,z) = p3(x)-p3(z)
But somehow, playing around today I had a lot of
set backs. Except for a small little new idea.
a <> b, b <> c, a <> c
So instead of looking at p(x)-p(y), we could
p(x) - p(y)
----------- = 0
x - y
Now p(x)-p(y) is divided by x-y, the result is
a new binvariate polynomial, which has a symmetry,
p(x) - p(y)
----------- = r(x,y) = r(y,x)
x - y
1 1 1 1
----- - ----- = (y - z) ----- -----
x - y x - z x - y x - z
https://en.wikipedia.org/wiki/Resolvent_formalism
But still stuck. It could be also that the
transitivity doesn't hold.
Working with variables, i.e. indeterminates algebraically,
has the advantage, that identities will also hold
for real values. If we avoid using A(r)=0, i.e.
some of the criteria for algebraic numbers, we get results
for transcendental numbers as well.
Post by b***@gmail.com
We can already show this without transitivity,
namely if an algebraic number a, has minimal
A(a) = 0
p(x) = (x-1)*A(x)
p(1)-p(a) = (1-1)*A(1)-(a-1)*A(a)
= 0*A(1)-(a-1)*0
= 0
Hence 1 ~ a.
Post by b***@gmail.com
The equivalence classe of [1] = { a ~ 1 | a in } equals
the algebraic numbers. Maybe I have picked it up from
somewhere else.
Into the Diophantine of integer polynomials like that,
it seems there's always a polynomial that the difference
of roots of other polynomials is the root of it.

This seems to be via just plugging in the values.

Where '~' is "being algebraic", there's also the pair-wise,
for pretty directly establishing transitivity.
b***@gmail.com
2017-07-15 09:40:22 UTC
Permalink
For a ~ b and b ~ c, it not necesseraly follows a ~ c,
at least I did non yet see a proof.
You know the relationship is called friend, such relation-
ships are not necessarily transitivive.

You might be befriended with Tillerson, and he might
be befriended with Trump, which doesn't imply you
are also befriended with Trump. A valid proof must
show that there exists a p3 in Z[x],

the proof could be constructive, giving an algorithm
to find p3 from p1 and p2, or less constructive
only showing the existence of p3.
Post by Ross A. Finlayson
Where '~' is "being algebraic", there's also the pair-wise,
for pretty directly establishing transitivity.
Markus Klyver
2017-07-15 15:06:05 UTC
Permalink
Post by b***@gmail.com
For a ~ b and b ~ c, it not necesseraly follows a ~ c,
at least I did non yet see a proof.
You know the relationship is called friend, such relation-
ships are not necessarily transitivive.
You might be befriended with Tillerson, and he might
be befriended with Trump, which doesn't imply you
are also befriended with Trump. A valid proof must
show that there exists a p3 in Z[x],
the proof could be constructive, giving an algorithm
to find p3 from p1 and p2, or less constructive
only showing the existence of p3.
Post by Ross A. Finlayson
Where '~' is "being algebraic", there's also the pair-wise,
for pretty directly establishing transitivity.
Transitivity necessarily follows if ~ is an equivalence relation.
b***@gmail.com
2017-07-15 16:48:25 UTC
Permalink
Well for the "friends" relationship, so far we only know
symmetry, reflexivity. Transitvity is still open.

If these 3 properties come together, then only we know
that ~ is a equivalence relation. So far its only a sign

denoting a relation, not yet necessarely an equivalence relation.
Can you prove transitivity of the ~ relation?
Post by Markus Klyver
Transitivity necessarily follows if ~ is an equivalence relation.
b***@gmail.com
2017-07-15 16:54:19 UTC
Permalink
Or disproof? Maybe just construct 3 transcendent numbers,
a, b, c with the following propety:

"a ~ b" and "b ~ c" and "not (a ~ c)"

We already know that "not (t ~ 1/t)" for transcendental
numbers. But I dont see some transcendental number s

such that:

"t ~ s" and "s ~ 1/t"

I also don't know whether the counter example should
be search inside "not (t ~ 1/t)".

Maybe some special functions can be used. They have sometimes
many identities that are algebraically, maybe they

amount to the friends relationship sometimes.
Post by b***@gmail.com
Well for the "friends" relationship, so far we only know
symmetry, reflexivity. Transitvity is still open.
If these 3 properties come together, then only we know
that ~ is a equivalence relation. So far its only a sign
denoting a relation, not yet necessarely an equivalence relation.
Can you prove transitivity of the ~ relation?
Post by Markus Klyver
Transitivity necessarily follows if ~ is an equivalence relation.
bassam king karzeddin
2017-07-16 11:46:51 UTC
Permalink
Post by b***@gmail.com
For a ~ b and b ~ c, it not necesseraly follows a ~ c,
at least I did non yet see a proof.
You know the relationship is called friend, such relation-
ships are not necessarily transitivive.
You might be befriended with Tillerson, and he might
be befriended with Trump, which doesn't imply you
are also befriended with Trump. A valid proof must
show that there exists a p3 in Z[x],
the proof could be constructive, giving an algorithm
to find p3 from p1 and p2, or less constructive
only showing the existence of p3.
Post by Ross A. Finlayson
Where '~' is "being algebraic", there's also the pair-wise,
for pretty directly establishing transitivity.
Exactly like a real monkey, jumping from a tree to another taller tree in a jungle, running away from answering my simplest sensible and so intuitive question despite providing him with so many obvious answers as well

BKK
b***@gmail.com
2017-07-15 09:49:25 UTC
Permalink
For transcendental numbers, you cannot plug them
in, for example if you take a bivariate polynomial

p(x,y) in Z[x,y]

If you plug in a real value "a", first you get a
a polynomial over R[x]:

p(x,a) in R[x]

Only if a is algebraic, then you can find a polynomial
p_a such that:

p_a(x) in Z[x]

p_a(x) = p(x,a)

This answer might show a way, its not extremly constructive:
https://math.stackexchange.com/a/1432275/4414

Now if a is transcendental, we dont have this method available.
Post by Ross A. Finlayson
Into the Diophantine of integer polynomials like that,
it seems there's always a polynomial that the difference
of roots of other polynomials is the root of it.
This seems to be via just plugging in the values.
b***@gmail.com
2017-07-15 09:54:20 UTC
Permalink
On a similar line of though, it is probably possible
to show if a is algebraic and b is transcendental,
the they are not befrieded:

a in A and b in T implies ~ (a ~ b)
Post by b***@gmail.com
But it seems also that algebraic and transcendental
numbers cannot be befriended. Whats the full
proof here?
https://groups.google.com/d/msg/sci.math/0jeLoUPAWeQ/RKQ04nUHBwAJ
Post by b***@gmail.com
For transcendental numbers, you cannot plug them
in, for example if you take a bivariate polynomial
p(x,y) in Z[x,y]
If you plug in a real value "a", first you get a
p(x,a) in R[x]
Only if a is algebraic, then you can find a polynomial
p_a(x) in Z[x]
p_a(x) = p(x,a)
https://math.stackexchange.com/a/1432275/4414
Now if a is transcendental, we dont have this method available.
bassam king karzeddin
2017-07-16 12:45:32 UTC
Permalink
Post by b***@gmail.com
On a similar line of though, it is probably possible
to show if a is algebraic and b is transcendental,
a in A and b in T implies ~ (a ~ b)
Post by b***@gmail.com
But it seems also that algebraic and transcendental
numbers cannot be befriended. Whats the full
proof here?
https://groups.google.com/d/msg/sci.math/0jeLoUPAWeQ/RKQ04nUHBwAJ
probably possible, befriend numbers, it seems that, implies that, ...etc

What nonsensible terms you might use also to justify fiction numbers in mind? wonder!

And before you show your confusions to everyone around, make sure of what you write, sure

Are there brother numbers in mathematics too? wonder!

BKK
Post by b***@gmail.com
Post by b***@gmail.com
For transcendental numbers, you cannot plug them
in, for example if you take a bivariate polynomial
p(x,y) in Z[x,y]
If you plug in a real value "a", first you get a
p(x,a) in R[x]
Only if a is algebraic, then you can find a polynomial
p_a(x) in Z[x]
p_a(x) = p(x,a)
https://math.stackexchange.com/a/1432275/4414
Now if a is transcendental, we dont have this method available.
bassam king karzeddin
2017-07-16 11:49:26 UTC
Permalink
Post by b***@gmail.com
For transcendental numbers, you cannot plug them
in, for example if you take a bivariate polynomial
p(x,y) in Z[x,y]
If you plug in a real value "a", first you get a
p(x,a) in R[x]
Only if a is algebraic, then you can find a polynomial
p_a(x) in Z[x]
p_a(x) = p(x,a)
https://math.stackexchange.com/a/1432275/4414
Now if a is transcendental, we dont have this method available.
And you would never have any method available because there isn't any method for sure

BKK
Post by b***@gmail.com
Post by Ross A. Finlayson
Into the Diophantine of integer polynomials like that,
it seems there's always a polynomial that the difference
of roots of other polynomials is the root of it.
This seems to be via just plugging in the values.
b***@gmail.com
2017-07-16 12:51:31 UTC
Permalink
Can you prove this? You know Able as able (pun) to prove that
some equations don't have solutions in radicals.

Its not exactly the same question, but you seem to be expert
in constructible numbers, so can you prove or disprove the

transitivity of befriended numbers? (P.S. its already trivially
transitive for algebraic numbers, what about transcendental numbers)
Post by bassam king karzeddin
And you would never have any method available because there isn't any method for sure
b***@gmail.com
2017-07-16 22:11:01 UTC
Permalink
Maybe we should look at Bendixson:
Concerning solution of a polynomial equation by radicals Bendixson returned to Niels Henrik Abel's original contribution and showed that Abel's methods could be extended to describe precisely which equations could be solved by radicals.

But he was also sucked into Poincaré stuff:
The analysis problem which intrigued Bendixson more than all others was the investigation of integral curves to first order differential equations, in particular he was intrigued by the complicated behaviour of the integral curves in the neighbourhood of singular points. The Poincaré-Bendixson theorem, which says an integral curve which does not end in a singular point has a limit cycle, was first proved by Henri Poincaré but a more rigorous proof with weaker hypotheses was given by Bendixson in 1901.

https://en.wikipedia.org/wiki/Ivar_Otto_Bendixson
Post by b***@gmail.com
Can you prove this? You know Able as able (pun) to prove that
some equations don't have solutions in radicals.
Its not exactly the same question, but you seem to be expert
in constructible numbers, so can you prove or disprove the
transitivity of befriended numbers? (P.S. its already trivially
transitive for algebraic numbers, what about transcendental numbers)
Post by bassam king karzeddin
And you would never have any method available because there isn't any method for sure
bassam king karzeddin
2017-07-17 16:13:00 UTC
Permalink
Post by b***@gmail.com
Concerning solution of a polynomial equation by radicals Bendixson returned to Niels Henrik Abel's original contribution and showed that Abel's methods could be extended to describe precisely which equations could be solved by radicals.
The analysis problem which intrigued Bendixson more than all others was the investigation of integral curves to first order differential equations, in particular he was intrigued by the complicated behaviour of the integral curves in the neighbourhood of singular points. The Poincaré-Bendixson theorem, which says an integral curve which does not end in a singular point has a limit cycle, was first proved by Henri Poincaré but a more rigorous proof with weaker hypotheses was given by Bendixson in 1901.
https://en.wikipedia.org/wiki/Ivar_Otto_Bendixson
Observe how many new topics had been recently started since nearly a year in this regard and especially that we had started digging under their stony graves in many occasions, and once the doubt had been identified exactly, you would certainly see soon many alleged skilled heroes with much longer tongues than you and me, and much deeper throats as well, and with authority help bringing out all the buried secrets that were hidden purposely from the history

BKK
Post by b***@gmail.com
Post by b***@gmail.com
Can you prove this? You know Able as able (pun) to prove that
some equations don't have solutions in radicals.
Its not exactly the same question, but you seem to be expert
in constructible numbers, so can you prove or disprove the
transitivity of befriended numbers? (P.S. its already trivially
transitive for algebraic numbers, what about transcendental numbers)
Post by bassam king karzeddin
And you would never have any method available because there isn't any method for sure
Me
2017-07-16 13:30:32 UTC
Permalink
Post by bassam king karzeddin
And you would never have any method available because there isn't any method for sure
If you "draw" a(n) (ideal) circle with a compass, won't you think that the circumference of this circle has a certain length? And if so, won't the length be "measured" by a transcendental number (if the length of the diameter of the circle is = 1)?

See: https://en.wikipedia.org/wiki/Circumference
bassam king karzeddin
2017-07-16 15:13:14 UTC
Permalink
Post by Me
Post by bassam king karzeddin
And you would never have any method available because there isn't any method for sure
If you "draw" a(n) (ideal) circle with a compass, won't you think that the circumference of this circle has a certain length? And if so, won't the length be "measured" by a transcendental number (if the length of the diameter of the circle is = 1)?
See: https://en.wikipedia.org/wiki/Circumference
I saw that source, and it doesn't imply the fact they try to demonstrate in accordance with what was well established first in mathematics about (pi) being real as a real existing number

Of course, our ability of visibility is quite limited, and what you see as a circle is actually a regular constructible polygon with many sides, where then the sum of the sides (which is a constructible number only that is close to what you might think as (pi))

And this must be reflected as what a layperson thinks that is actual (pi), where as the absolute magnitude of (pi) remains as no more secret only in the perfect circle (that can never exist in any reality)

I had explained this point earlier in more details to KON, where he didn't object, especially that the vast majority of angles we know today are proven of impossible existence the same way that the majority of real numbers are fiction numbers too, were also continuity is impossible existence because infinity itself of impossible existence, sure

BKK
konyberg
2017-07-16 18:31:13 UTC
Permalink
Do you belive that I agree with you about non existent angles? I don't! In mathematics any angle is possible. Just draw two crossing lines and there is an angle. Not all angles can be constructed the old way, but they nevertheless exist.
KON
bassam king karzeddin
2017-07-17 17:38:31 UTC
Permalink
Post by konyberg
Do you belive that I agree with you about non existent angles? I don't! In mathematics any angle is possible. Just draw two crossing lines and there is an angle.
KON wrote:
Not all angles can be constructed the old way, but they nevertheless exist.
Post by konyberg
KON
Yes for sure, they do exist in those so tiny skulls as fairy angles, especially that they are thought as the majorities, and with loud confession that we can't show EXACTLY only one of them

And people would ignore deliberately my irrefutable proofs, irrefutable conjectures regarding those very hot issues,(especially in number theory despite we are living funnily in a SUPERCOMPUTER era), really so funny and very understandable mathematicians behaviours for sure

But maybe the arbitrary choice would make such alleged existing angles, where with all our knowledge couldn't and will not make it, for sure, wonder!

The existing angles are only those angles found in any triangle with constructible sides, sure

BKK

Me
2017-07-17 12:37:08 UTC
Permalink
Post by bassam king karzeddin
the majority of real numbers are fiction numbers too
Indeed! Actually, all of them. :-)
bassam king karzeddin
2017-07-17 15:31:38 UTC
Permalink
Post by Me
Post by bassam king karzeddin
the majority of real numbers are fiction numbers too
Indeed! Actually, all of them. :-)
You are coming close, but Never all of them

Only constructible numbers are the truly real numbers, but the rest of the alleged real numbers are manufactured fictions for entertainments of very few abnormal and mad people,(100%) sure

BKK
bassam king karzeddin
2017-07-16 11:39:37 UTC
Permalink
Post by b***@gmail.com
I don't know whether its is somewhere in the works
of Kurt Mahler, that a ~ equivalence relation between
The equivalence classe of [1] = { a ~ 1 | a in } equals
the algebraic numbers. Maybe I have picked it up from
somewhere else.
Dont remember. So two real numbers a,b are friends,
if there exists a non-constant integer polynomial p
a ~ b iff exists p in Z[x] (deg(p)>0 & p(a)-p(b) = 0)
Is this relation transitive, so that we can form
equivalence classes? It is surely symmetric,
and reflexive.
For symmetric, if a ~ b then there was a p in Z[x],
with p(a)-p(b)=0. But then we also have p(b)-p(a)=0.
So we also have b ~ a.
For reflexiv, we need to show a ~ a. Which happens
for any polynomial, trivially p(a)-p(a)=0. What
a ~ b & b ~ c => a ~ c
Anybody up to a proof or disproof?
Post by Ross A. Finlayson
There's Mahler on transcendentals with a
descriptive approach, there are also ways
Are there real numbers which are the enemy of each other in mathematics, wonder!
BKK
bassam king karzeddin
2017-07-15 08:47:49 UTC
Permalink
Post by Ross A. Finlayson
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
The title is of a retro-finitist crankety-troll,
"transcendental numbers" are all the non-algebraic
real numbers, which exist.
And retro-finitist crankety trolls can get bent.
"Retro-finitism" is because mathematics already
has infinity and trying to take it back (instead
of, say, leaving it back) has the retro-finitism
is an ignorance (vis-a-vis foundations besides
models of numbers).
Transcendental numbers are where the algebraic
numbers don't yet have properties of being
continuous, as a body.
So, they exist. The real numbers are continuous.
They are: continuous, the real numbers.
There's Mahler on transcendentals with a
descriptive approach, there are also ways
to describe the transcendentals in terms
of say, completions after exhausting the
algebraics (or, for example, rationals,
rational numbers).
This points to the "signal continuity".
And when you ask those a retro-infinitism cranky-troll, why most of your numbers don't confine to reality, since you claim that they are reals, especially that some of them do obey with reality and exactly

Then, with their very well known replies in advance, they would tell the so innocents that numbers are only abstractions in mind, and every thing is absolutely correct since there aren't any contradictions to the definition, and so and so..., as if the human mind is the way to test the true science, Wonder!

No, and never was the human mind was a laboratory to test the absolute facts, it is the physical experiment reality that can decide only since positive real
numbers are not only a matter of mathematics but basically physics because every real number must represent a physical distance and EXACTLY

So, the final word in this regard must be for physics, especially for real numbers, other wise let the mad people of failures mathematicians also run the real physics and produce more interesting fiction stories about the universe and many dimensional universes too, where then a Hollywood fiction film makers can't even produce for sure

BKK
Markus Klyver
2017-07-15 15:03:05 UTC
Permalink
Post by bassam king karzeddin
Post by Ross A. Finlayson
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
The title is of a retro-finitist crankety-troll,
"transcendental numbers" are all the non-algebraic
real numbers, which exist.
And retro-finitist crankety trolls can get bent.
"Retro-finitism" is because mathematics already
has infinity and trying to take it back (instead
of, say, leaving it back) has the retro-finitism
is an ignorance (vis-a-vis foundations besides
models of numbers).
Transcendental numbers are where the algebraic
numbers don't yet have properties of being
continuous, as a body.
So, they exist. The real numbers are continuous.
They are: continuous, the real numbers.
There's Mahler on transcendentals with a
descriptive approach, there are also ways
to describe the transcendentals in terms
of say, completions after exhausting the
algebraics (or, for example, rationals,
rational numbers).
This points to the "signal continuity".
And when you ask those a retro-infinitism cranky-troll, why most of your numbers don't confine to reality, since you claim that they are reals, especially that some of them do obey with reality and exactly
Then, with their very well known replies in advance, they would tell the so innocents that numbers are only abstractions in mind, and every thing is absolutely correct since there aren't any contradictions to the definition, and so and so..., as if the human mind is the way to test the true science, Wonder!
No, and never was the human mind was a laboratory to test the absolute facts, it is the physical experiment reality that can decide only since positive real
numbers are not only a matter of mathematics but basically physics because every real number must represent a physical distance and EXACTLY
So, the final word in this regard must be for physics, especially for real numbers, other wise let the mad people of failures mathematicians also run the real physics and produce more interesting fiction stories about the universe and many dimensional universes too, where then a Hollywood fiction film makers can't even produce for sure
BKK
So give me a ruler of exactly one meter. Can you do it?
bassam king karzeddin
2017-07-16 09:07:39 UTC
Permalink
Post by Markus Klyver
Post by bassam king karzeddin
Post by Ross A. Finlayson
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
The title is of a retro-finitist crankety-troll,
"transcendental numbers" are all the non-algebraic
real numbers, which exist.
And retro-finitist crankety trolls can get bent.
"Retro-finitism" is because mathematics already
has infinity and trying to take it back (instead
of, say, leaving it back) has the retro-finitism
is an ignorance (vis-a-vis foundations besides
models of numbers).
Transcendental numbers are where the algebraic
numbers don't yet have properties of being
continuous, as a body.
So, they exist. The real numbers are continuous.
They are: continuous, the real numbers.
There's Mahler on transcendentals with a
descriptive approach, there are also ways
to describe the transcendentals in terms
of say, completions after exhausting the
algebraics (or, for example, rationals,
rational numbers).
This points to the "signal continuity".
And when you ask those a retro-infinitism cranky-troll, why most of your numbers don't confine to reality, since you claim that they are reals, especially that some of them do obey with reality and exactly
Then, with their very well known replies in advance, they would tell the so innocents that numbers are only abstractions in mind, and every thing is absolutely correct since there aren't any contradictions to the definition, and so and so..., as if the human mind is the way to test the true science, Wonder!
No, and never was the human mind was a laboratory to test the absolute facts, it is the physical experiment reality that can decide only since positive real
numbers are not only a matter of mathematics but basically physics because every real number must represent a physical distance and EXACTLY
So, the final word in this regard must be for physics, especially for real numbers, other wise let the mad people of failures mathematicians also run the real physics and produce more interesting fiction stories about the universe and many dimensional universes too, where then a Hollywood fiction film makers can't even produce for sure
BKK
So give me a ruler of exactly one meter. Can you do it?
This is not any fiction for film makers, many can do it for sure

And the way you keep asking the question is just to escape from answering the real tragedy I had encased you inside, but basically to liberate your minds from so many obvious illusions, or maybe to wake you up from many sweet dreams which are so annoying for sure

I simply tell you, here is a real IRRATIONAL number say sqrt(3), which is a diagonal of a cube with UNITY side in the physical reality (since we are describing + REAL numbers), that is unique, physically comprehensible in few seconds, and also not any magic, isn't it?,
of course, many more of alike examples an average school student can provide

So, give us a real physical representation for only a SINGLE example of your alleged real Transcedential numbers (since you claim they are also the majority of real numbers)

Another choice to make it much easier for you, provide a single form that shows how do you present only ONE real transcendental number for others in order to believe in the logic that official mathematics follow

And if you really can't, we are not asking you to applause of the hands, but at least keep silent and learn silently, or bring back a rigorous point, not of the type (every thing is in mind), but in physical REALITY for sure

BKK
b***@gmail.com
2017-07-16 10:25:57 UTC
Permalink
Geometrically friend numbers are easily to spot.
Just draw a polynomial p(x) in Z[x], and then
intersect it at height y=p(a)=p(b)=u, with a line
parallel to the x-axis, it should at least intersect

at x=a and at x=b. The question of transitivity is
now if we have 3 points x=a, x=b and x=c, the
x=a and x=b resulting from p1, and the x=b and x=c
resulting from p2, is there automatically a polynomial

p3 in Z[x] which can make x=a and x=c?

P.S.: Here is an example in Wolfram Alpha:

The red dots are -pi-1 and pi:
https://www.wolframalpha.com/input/?i=x^2%2Bx-pi^2-pi%3D0
BKK, is a trademark for stupidity
b***@gmail.com
2017-07-16 13:07:12 UTC
Permalink
At the moment, I don't know a more coordinate free
formulation of the problem. If you accept polynomial
curves as some geometric objects,

its already kind of coordinate free. But I don't
know how this could be improved. Some polynomials
might appear in chains of compass/ruler construction.
Post by b***@gmail.com
Geometrically friend numbers are easily to spot.
Just draw a polynomial p(x) in Z[x], and then
intersect it at height y=p(a)=p(b)=u, with a line
parallel to the x-axis, it should at least intersect
at x=a and at x=b. The question of transitivity is
now if we have 3 points x=a, x=b and x=c, the
x=a and x=b resulting from p1, and the x=b and x=c
resulting from p2, is there automatically a polynomial
p3 in Z[x] which can make x=a and x=c?
https://www.wolframalpha.com/input/?i=x^2%2Bx-pi^2-pi%3D0
BKK, is a trademark for stupidity
bassam king karzeddin
2017-07-15 12:01:46 UTC
Permalink
Post by Ross A. Finlayson
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
The title is of a retro-finitist crankety-troll,
"transcendental numbers" are all the non-algebraic
real numbers, which exist.
And retro-finitist crankety trolls can get bent.
"Retro-finitism" is because mathematics already
has infinity and trying to take it back (instead
of, say, leaving it back) has the retro-finitism
is an ignorance (vis-a-vis foundations besides
models of numbers).
Transcendental numbers are where the algebraic
numbers don't yet have properties of being
continuous, as a body.
So, they exist. The real numbers are continuous.
They are: continuous, the real numbers.
And when do you ask those retro-infinitist crankety-trolls,

What is the least rational number that is greater than say sqrt(3)? they say shamelessly that DOESN'T exist

Or: What is the largest rational number that is less than sqrt(3)? they say also and so confidently it doesn't exist, but they say loudly continuity oddly exists

It doesn't much matter if you repeat the questions in reals as here

What is the least real number that is greater than say sqrt(3)? they say shamelessly that DOESN'T exist


Or: What is the largest real number that is less than sqrt(3)? they say also and so confidently it doesn't exist, but they say loudly continuity oddly exists

So, yes it exists only in there too tiny minds for sure

But continuity doesn't exist since infinity doesn't exist

BKK
Post by Ross A. Finlayson
There's Mahler on transcendentals with a
descriptive approach, there are also ways
to describe the transcendentals in terms
of say, completions after exhausting the
algebraics (or, for example, rationals,
rational numbers).
This points to the "signal continuity".
Markus Klyver
2017-07-15 15:08:25 UTC
Permalink
Post by bassam king karzeddin
Post by Ross A. Finlayson
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
The title is of a retro-finitist crankety-troll,
"transcendental numbers" are all the non-algebraic
real numbers, which exist.
And retro-finitist crankety trolls can get bent.
"Retro-finitism" is because mathematics already
has infinity and trying to take it back (instead
of, say, leaving it back) has the retro-finitism
is an ignorance (vis-a-vis foundations besides
models of numbers).
Transcendental numbers are where the algebraic
numbers don't yet have properties of being
continuous, as a body.
So, they exist. The real numbers are continuous.
They are: continuous, the real numbers.
And when do you ask those retro-infinitist crankety-trolls,
What is the least rational number that is greater than say sqrt(3)? they say shamelessly that DOESN'T exist
Or: What is the largest rational number that is less than sqrt(3)? they say also and so confidently it doesn't exist, but they say loudly continuity oddly exists
It doesn't much matter if you repeat the questions in reals as here
What is the least real number that is greater than say sqrt(3)? they say shamelessly that DOESN'T exist
Or: What is the largest real number that is less than sqrt(3)? they say also and so confidently it doesn't exist, but they say loudly continuity oddly exists
So, yes it exists only in there too tiny minds for sure
But continuity doesn't exist since infinity doesn't exist
BKK
Post by Ross A. Finlayson
There's Mahler on transcendentals with a
descriptive approach, there are also ways
to describe the transcendentals in terms
of say, completions after exhausting the
algebraics (or, for example, rationals,
rational numbers).
This points to the "signal continuity".
What? Since when did Archimedean fields become a problem for you? And how do you draw the conclusion that functions cannot have continuity over a field A if infinity is not in A?
bassam king karzeddin
2017-07-15 07:55:53 UTC
Permalink
Post by b***@gmail.com
The thread title says transcendental number, what do
we know about them?
Nothing we must know about nothingness or ghost numbers for sure!

BKK
Post by b***@gmail.com
Post by Markus Klyver
A real number is a member of the unique complete totally ordered field (ℝ ; + ; · ; <), up to an isomorphism.
bassam king karzeddin
2017-07-16 11:36:30 UTC
Permalink
Post by b***@gmail.com
The algebraic numbers are like living on a sphere,
you can always find a polynomial p(x) in Z[x], so
that p(r)=0 (you can reel in the leash).
For transcendental numbers this is not possible.
You always get stuck, take any polynomial p(x) in
Z[X], you will have p(r)<>0.
if you have two transcendental numbers r1 and r2,
we might indeed have p(r1)-p(r2)=0 for some polynomial,
call such transcendental numbers friends.
The friend relationship is transitive, if p1(r1)-p1(r2)=0
and p2(r2)-p2(r3)=0, then there is a polynomial p3 such
that p3(r1)-p3(r3)=0.
Proof: Left as an exercise.
Now look at the set of befriended equivalence classes,
i.e. { [r] | r in R }. How many equivalence classes form
the algebraic numbers, and how many equivalence classes
form the transcendental numbers?
Post by b***@gmail.com
Get a (two-dimensional) dog and a very long (one-dimensional) leash.
Send your dog out exploring, letting the leash play out. When the
dog returns, try to pull in the leash. (Meaning, you try to reel in
the loop with you and the dog staying put.) On a sphere, the leash
can always be pulled in; on a torus, sometimes it can't be.
Post by b***@gmail.com
Which wont make it probably into wikipedia.
https://math.stackexchange.com/a/854394/4414
Post by b***@gmail.com
Stackexchange and Wikipedia are two different things.
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
Don't you feel any shame you Wiki Piki (burr...)
This question is so irrelevant to my question since it talks about the existence of the uncountability of trans. numbers, whereas my question was about the existence os such a legendary numbers
Did you recognize the huge difference
Post by b***@gmail.com
https://math.stackexchange.com/q/2070246/4414
So is the other irrelevant nonsense topic
Maybe your Enclopidea-Wikipedia didn't help you so far in this regard!wonder!
Or maybe you don't comprehend any of my questions/or answers, wonder!
But most likely you don't understand at all for sure
BKK
Complete hallucinations, he (bursegan) is a true victim of nonsense, why to go to higher polynomials, he trusts the existence of continuity so blindly,(which had been refuted)

Ok, here is much simpler Question, prove the real existence of one real root of this simple polynomial and take into account my rigorous published proof especially for you, (x^11 = 81), but in REALITY and not as a notation in MIND, also provide how can you present them if you can't show their reality since there is a real root that you claim?

No need to read all other nonsense posts about befriend numbers unless you solve this simple puzzle by real demonstration and not any kind of APPROXIMATION

But, I swear you would never want to understand

Also, read or google (karzeddin mathematics crises),

BKK
bassam king karzeddin
2017-07-15 12:31:25 UTC
Permalink
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
https://math.stackexchange.com/q/2070246/4414
Very Poor Modern Mad mathematical mind, with no common sense at all, for sure.

He wants to get it by endless probability (of course in mind only and never in reality, since it is a mere FICTION in mind for sure)

But I showed you in the physical reality one real irrational number (without all that hogwash mad maths in mind) and in few seconds only, where also you can see many others for the rest of your life

So, do confess so openly that you can only approximate it and not even as much as you like, and don't forget to tell others the TRUE form of your approximations, which is ONLY CONSTRUCTIBLE FORM, and sure forever....

Now, where are those current dump dishonest professional historianS shamelessly hiding from such an event?

Or do they prefer to learn it from one of those meaningless alleged most REPUTABLE journals or UNIVERSITIES, Wonder!

But the real true history can't be moderated as usual anymore, and for sure

BKK
Markus Klyver
2017-07-15 15:09:21 UTC
Permalink
Post by bassam king karzeddin
Post by b***@gmail.com
Well to begin with this is irrelevant nonsense. The
finite decimal representations, i.e. those representations
d/10^n
So they are a subset of the rational numbers. And the
irrational numbers, are all those real numbers, which
are not rational numbers.
I union Q = R
I intersect Q = {}
So they form a partition of the real numbers into
two kind of numbers. Which can be also written
R \ Q = I
Now the algeraic irrational numbers (AI) and the
transcendental numbers (T) form a further partition,
R \ Q \ AI = T
Or if we use algebraic numbers (A) as an umbrella
for algeraic irrational numbers (AI) and rational
R \ A = T
Now R is uncountable, A is countable, so T will
be uncountable as well. So the probability, that
The probability is 0
Hint A is countable because the polynomials
Z[X] are countable. Now P(B | C) = P(B n C) / P(C) =
P(B n C) / (P(B n C) + P(~B n C)). We can simulate
the event that we fix a prefix,
by using C=[0,1), this is the same as a prefix "0.",
Further use Rn=[-n,n) and B_n = Rn intersect T, we
P(B_n | C) = 1/(2n) / (1/(2n) + (2n-1)/n)
= 1/2n
P(T | C) = lim n->oo 1/2n = 0
The problem is that we allow an arbitrary transcendental
number. We have also readily from algebraic numbers the
P(A | C) = 0
So it doesn't say something about transcendental numbers.
If we restrict ourselfs from the beginning to lets say
transcendental numbers in the interval [0,1) different
probabilities.
https://math.stackexchange.com/q/37121/4414
https://math.stackexchange.com/q/2070246/4414
Very Poor Modern Mad mathematical mind, with no common sense at all, for sure.
He wants to get it by endless probability (of course in mind only and never in reality, since it is a mere FICTION in mind for sure)
But I showed you in the physical reality one real irrational number (without all that hogwash mad maths in mind) and in few seconds only, where also you can see many others for the rest of your life
So, do confess so openly that you can only approximate it and not even as much as you like, and don't forget to tell others the TRUE form of your approximations, which is ONLY CONSTRUCTIBLE FORM, and sure forever....
Now, where are those current dump dishonest professional historianS shamelessly hiding from such an event?
Or do they prefer to learn it from one of those meaningless alleged most REPUTABLE journals or UNIVERSITIES, Wonder!
But the real true history can't be moderated as usual anymore, and for sure
BKK
So construct a ruler of exactly one meter. Do it!
Markus Klyver
2017-07-11 15:09:14 UTC
Permalink
Post by bassam king karzeddin
What is the probability of picking up a random length such that it represents a transcendental number relative to any arbitrary chosen constructible unit length?
Big Hint: the probability exactly equals to zero, sure
Regards
Bassam King Karzeddin
7 July 2017
Let's formalize the problem, because the terminology you use is pretty vague and handwavy.

Let V = ℝ be a real vector space. Let 𝐵 = {r} be the basis for V, where r ≠ 0 is a real constructible number. Let ℙ be a probability measure and let X ~uniform[a, b] be a stochastic variable with a, b ∈ ℝ and a < b. Define T := { x ∈ ℝ : x constructible }. Let the tuple (c) be the coordinate vector for X in 𝐵. You ask: what is ℙ(c ∈ T)?

Consider a function f : ℝ -> ℝ such that f(x) = x*r. Recall that every constructible number is algebraic. If x*r were constructible (and thus algebraic), then r * (x*r) would also be algebraic (in fact, the only way for x*r to be constructible is for x to be constructible as well). So if we restrict the domain of f to T, the range of f will also be T, so T is closed under multiplication by a constructible number.

The coordinate vector for X in the standard basis will be the tuple (c*r). Butt we know that c*r is transcendental. We also know that [a, b] can be mapped bijectively to [a/r, b/r]. So ℙ(c ∈ T) = ℙ(X ∈ T) = 1, since |T| = |ℝ| and |ℚ| = |ℕ|.
mathman1
2017-07-12 00:47:15 UTC
Permalink
Algebraic numbers are countable, but real numbers are not. It appears that almost all numbers are
reals-algebraic(=transcendental)
bassam king karzeddin
2017-07-12 12:59:36 UTC
Permalink
Post by mathman1
Algebraic numbers are countable, but real numbers are not. It appears that almost all numbers are
reals-algebraic(=transcendental)
So, mathman 1, show them only and EXACTLY one real transcendental number in reality by any comprehensible Geometrical shape by avoiding many kinds of very well known APPROXIMATIONS methods such as (Limits, Infinity, Epsilon, Delta, Intermediate theorem, Newton’s Approximations, Famous cuts, Cauchy sequences, Floorings, Approaching fast, …etc)

And the question is also applicable for also one real algebraic number (that is not a constructible)

This question is not necessarily and certainly not only for you but for everyone thinking himself as a true mathematician, but since you are a math man one who certainly loves mathematical challenge, that also I am searching for, then it is hopeful that you can save the mathematics now before anyone else

And to facilitate your very easy task, consider a real existing irrational number as square root of say (29), which is simply a diagonal of a rectangle of sides (5, 2) units, (finished)

So, you see here not a single nonsense was used in representing that irrational number a [sqrt(29)], sure

And remember to do that easy task since you claim that most of your real numbers are transcendental! Wonder!

But, please avoid telling us that is only in mind, since this is really a biggest ever JOKE in the history of mathematics, for sure

Big Hint: never ask any professional mathematician, for more than sure

Good luck and regards

Bassam King Karzeddin
July 12, 2017
Post by mathman1
Algebraic numbers are countable, but real numbers are not. It appears that almost all numbers are
reals-algebraic(=transcendental)
bassam king karzeddin
2017-07-13 15:38:23 UTC
Permalink
Post by bassam king karzeddin
Post by mathman1
Algebraic numbers are countable, but real numbers are not. It appears that almost all numbers are
reals-algebraic(=transcendental)
So, mathman 1, show them only and EXACTLY one real transcendental number in reality by any comprehensible Geometrical shape by avoiding many kinds of very well known APPROXIMATIONS methods such as (Limits, Infinity, Epsilon, Delta, Intermediate theorem, Newton’s Approximations, Famous cuts, Cauchy sequences, Floorings, Approaching fast, …etc)
And the question is also applicable for also one real algebraic number (that is not a constructible)
This question is not necessarily and certainly not only for you but for everyone thinking himself as a true mathematician, but since you are a math man one who certainly loves mathematical challenge, that also I am searching for, then it is hopeful that you can save the mathematics now before anyone else
And to facilitate your very easy task, consider a real existing irrational number as square root of say (29), which is simply a diagonal of a rectangle of sides (5, 2) units, (finished)
So, you see here not a single nonsense was used in representing that irrational number a [sqrt(29)], sure
And remember to do that easy task since you claim that most of your real numbers are transcendental! Wonder!
But, please avoid telling us that is only in mind, since this is really a biggest ever JOKE in the history of mathematics, for sure
Big Hint: never ask any professional mathematician, for more than sure
Good luck and regards
Bassam King Karzeddin
July 12, 2017
Post by mathman1
Algebraic numbers are countable, but real numbers are not. It appears that
almost all numbers are
Post by bassam king karzeddin
Post by mathman1
reals-algebraic(=transcendental)
So, matman one, countable or uncountable, algebraic or trans., finite or infinite, positive or negative, complex or whatever, ....etc

All those mathematical nonsense classifications are so meaningless as long as you didn't make the task you were asked to

And I am so sorry to ask you to do such an IMPOSSIBLE task because I know in advance that is IMPOSSIBLE task, for more than sure

And this is not the only one impossible task that I asked the world mathematicians to do, but so many others in my POSTS

And so, unfortunately, this would certainly torture the many coming generations, exactly the same way that Fermat (the non-professional mathematician) had been torturing them since many centuries even after a proof, for sure

And those many around will be shamed and forgotten forever as their preceding mathematickers very soon, for sure

BKK
Jan
2017-07-12 20:05:59 UTC
Permalink
Post by bassam king karzeddin
What is the probability of picking up a random length such that it represents a transcendental number relative to any arbitrary chosen constructible unit length?
Your question is not even wrong.

--
Jan
Markus Klyver
2017-07-13 00:14:33 UTC
Permalink
Post by Jan
Post by bassam king karzeddin
What is the probability of picking up a random length such that it represents a transcendental number relative to any arbitrary chosen constructible unit length?
Your question is not even wrong.
--
Jan
As I interpret it: Let V = ℝ be a real vector space. Let 𝐵 = {r} be the basis for V, where r ≠ 0 is a real constructible number. Let ℙ be a probability measure and let X ~uniform[a, b] be a stochastic variable with a, b ∈ ℝ and a < b. Define T := { x ∈ ℝ : x constructible }. Let the tuple (c) be the coordinate vector for X in 𝐵. What is ℙ(c ∈ T)?

(The answer is of course ℙ(c ∈ T) = 1)
mathman1
2017-07-13 00:37:58 UTC
Permalink
Circumference of circle of radius 1 is 2pi, which is transcendental.
mathman1
2017-07-14 01:00:12 UTC
Permalink
What is ? supposed to be? In any case algebraic numbers are define as roots of polynomials with integer coefficients. Your example doesn't appear to be that, although it is confusing looking.
j4n bur53
2017-07-15 10:01:49 UTC
Permalink
You find already an argument for this
little subproblem here:
https://math.stackexchange.com/a/2357906/4414
Post by b***@gmail.com
On a similar line of though, it is probably possible
to show if a is algebraic and b is transcendental,
a in A and b in T implies ~ (a ~ b)
Post by b***@gmail.com
But it seems also that algebraic and transcendental
numbers cannot be befriended. Whats the full
proof here?
https://groups.google.com/d/msg/sci.math/0jeLoUPAWeQ/RKQ04nUHBwAJ
Post by b***@gmail.com
For transcendental numbers, you cannot plug them
in, for example if you take a bivariate polynomial
p(x,y) in Z[x,y]
If you plug in a real value "a", first you get a
p(x,a) in R[x]
Only if a is algebraic, then you can find a polynomial
p_a(x) in Z[x]
p_a(x) = p(x,a)
https://math.stackexchange.com/a/1432275/4414
Now if a is transcendental, we dont have this method available.
j4n bur53
2017-07-15 14:00:57 UTC
Permalink
Hi,

Here is an example of two transcendental numbers which
are not friends, one number is a function from the
other number,

but where the corresponding function applied to an
algebraic number doesn't give a friend. So for an
algebraic number we have:

a ~ 1/a

This is since all algebraic numbers are befriended,
and if a is an algebraic number, root of A(x), so is
1/a an algebraic number, root of A(1/x).

But for a transcendental number we have:

~(t ~ 1/t)

Assume t and 1/t were friends, then there would
be a polynomial of degree n:

p(t) - p(1/t) = 0

Now since p is polynomial it will have the form:

p(t) = a0 + a1*t + .. + an*t^n

Now if we multiply both sides of the friends
relationship by t^n we get:

0 = p(t)*t^n - p(1/t)*t^n

= a0*t^n + a1*t^(n+1) + ... + an*t^(2n)

- a0*t^n - a1*t^(n-1) - .. - an

= q(t)

For some polynomial q of degree 2n. So this would
mean that t is algebraical, a contradiction. Does
this already lead to an insight,

which would allow us to construct a counter example
to the friends relationship, a counter example in
the transcendental numbers,

or is the friends relationship nevertheless transitive.

Have a Nice day!
Post by j4n bur53
You find already an argument for this
https://math.stackexchange.com/a/2357906/4414
Post by b***@gmail.com
On a similar line of though, it is probably possible
to show if a is algebraic and b is transcendental,
a in A and b in T implies ~ (a ~ b)
Post by b***@gmail.com
But it seems also that algebraic and transcendental
numbers cannot be befriended. Whats the full
proof here?
https://groups.google.com/d/msg/sci.math/0jeLoUPAWeQ/RKQ04nUHBwAJ
Post by b***@gmail.com
For transcendental numbers, you cannot plug them
in, for example if you take a bivariate polynomial
p(x,y) in Z[x,y]
If you plug in a real value "a", first you get a
p(x,a) in R[x]
Only if a is algebraic, then you can find a polynomial
p_a(x) in Z[x]
p_a(x) = p(x,a)
https://math.stackexchange.com/a/1432275/4414
Now if a is transcendental, we dont have this method available.
j4n bur53
2017-07-15 14:01:54 UTC
Permalink
Post by j4n bur53
but where the corresponding function applied to an
algebraic number doesn't give a friend.
Corr.:
but where the corresponding function applied to an
algebraic number gives a friend.
j4n bur53
2017-07-15 14:08:18 UTC
Permalink
The polynomial q is from Z[x], it has the
coefficients among {a0,..,an,0} which are

from Z since the coefficents {a0,..,an} are
from q in Z[x], a witness for the contradictarily

assumed friendship between t and 1/t.
Post by j4n bur53
For some polynomial q of degree 2n.
b***@gmail.com
2017-07-15 14:24:56 UTC
Permalink
Corr.:
1/a an algebraic number, root of A(1/x)*x^n.
Post by j4n bur53
1/a an algebraic number, root of A(1/x).
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