revisiting Irrationals, are there two types of irrationals
Archimedes Plutonium
Newsgroups: sci.math
Date: Sat, 19 Aug 2017 15:16:46 -0700 (PDT)
Subject: (tentative proof) that Irrationals come in only one type, one
flavor-- Algebraic Irrationals// Transcendental numbers are just illusions by
bigots of math
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Sat, 19 Aug 2017 22:16:46 +0000
(tentative proof) that Irrationals come in only one type, one flavor-- Algebraic Irrationals// Transcendental numbers are just illusions by bigots of math
Now the history of Transcendental numbers starts with the theory of polynomials where it was noticed that sqrt2 and other roots were Algebraic Irrationals because they showed up and were present in the polynomial solutions. And it was noticed that no matter how hard a goof ball mathematician tried to get pi or 2.71.... as a solution in polynomials, they never could and so the goofball ends up saying-- since we cannot get pi and 2.71... from out of these polynomials, we call them Transcendental Irrationals.
Very, very, very bad logic, indeed. Just because I can never get a human baby to be found in a litter of kittens from a cat, do we classify human babies as Transcendental babies.
So math, with idiots composing definitions using all sorts of Non logical thought, came up with the Transcendental number. If you want human babies, go to where they come from, don't go to cats, horses, sheep, etc. If you want pi or 2.71... go to circles and log spirals, and do not be running over to Polynomials expecting them as solutions to polynomials. Do not expect Champernowne's number .1234567.... to come falling out of a polynomial.
So, is pi and 2.71.... Algebraic Irrationals? If by Algebraic we mean-- are they constructable by the four operators of math, add, subtract, multiply, divide, especially divide, then yes, pi and 2.71... are Algebraic. So, here in math history, what we have is a case of goonclod minds absent of logic, wanting to think that a small corner of math-- Polynomials-- is going to be the overarching definer of what is a number. This is like having the President of the USA, defining who is good and who is not good, and expect everyone to respect his decisions.
So, when math went along with this crazy scheme of bifurcation of irrational into two types-- algebraic and transcendental, what if instead, someone came along and said-- hey, hey, Polynomial is not the deciding body of knowledge to decide, but rather, Circle and log spirals is the deciding body to define Algebraic and Transcendental Irrationals. So, then everyone crowds around the circle and log spiral auditorium and sees that every number in the unit circle is a multiple of 3.14159..... and only does 1, -1, 1/2, -1/2, 0 pop up that is not a multiple of 3.14.... So, numbers from the Circle and Log Spiral conferences then declare that 1, 1/2 are Transcendental while pi and all its multiples are Algebraic.
You see how fruitcake that is? It was never the case that mathematics had a legitimate classification of Algebraic versus Transcendental. Those two were as phony and fake as having the President decide who is good and who is bad. Artificial classification. Biology has a perfect example of artificial classification in the concept of "weeds". Weeds is not a part of biology, but a mere artificial scheme. To me, a milkweed is a weed, but to a monarch butterfly and those who like butterflies, a milkweed is a lovely plant. To me, a thorn tree like locust is a weed, for its thorns are dangerous, but to a person that wants trees and only locust can grow there, it is a godsend.
So, in mathematics, we have this artificial structure, built by mathematicians with a pinhead mind of logic, who said, I can't find pi or 2.71.... in polynomials, so I build a entire structure of Transcendental. Just like calling human babies Transcendental, because cats never produce them. You see, when a goofball lacking logic, expects something where he should not be expecting, then goes ahead and makes the horrible sin of a fabricated structure, a structure that really does not exist, only in the mind of the goofball. Delusional math, by people who have no logical abilities.
Enough said.
So, let us do a proof that a Transcendental Irrational is total fiction.
But before we do the proof, we need to make clear what a Rational Number is versus an Irrational Number. And this is truly a Logical Math item. It is logically true that a Rational number exists as well as a Irrational. But here, as in the case of the goofballs with Transcendental, here, also, the goofballs failed to adequately define the difference, the difference between a Rational and a Irrational.
What the goofballs did was say-- Rational means a fraction a Ratio of one counting number in the numerator and one counting number in the denominator, for a ratio of two counting numbers, and that is how we get the term "Rational". And all as fine and dandy. The irrational is then a number that can not be placed in a ratio of one counting number over another counting number. The right triangle whose one leg is 1 and other leg is 1 has a hypotenuse of sqrt2. And it was seen that sqrt2 is never a ratio of one counting number divided by another counting number. All was fine and dandy so far, for this was Ancient Greek times with their math by Pythagorus and Euclid. But trouble came circa 1202 when Fibonacci introduced the Decimal Representation of numbers to Western Civilization. Trouble starts after 1202, when decimals can focus in on what the number sqrt2 actually is and start writing it out as 1.41421356..... Here is where the goofballs of math enter again, in their failure of Logical Reasoning, they never made it clear that sqrt2 is irrational, but why is it irrational. So, what the goofballs said that screws things up is they said a Rational is a decimal that repeats is a digit or repeats in a block, whereas a irrational has no repeating of anything. Which, on the surface, looks okay, and is okay, provided you have a definition of infinity that is fakery-- where infinity is endlessness. So that if you go far enough, out there, something begins to repeat. But, if you define infinity precisely with a borderline, then, some big numbers as ratios, are not going to repeat in blocks of digits, before you reach the borderline. So, here again the goofballs strike out, fail.
What was missing, in the history of Rationals versus Irrationals, was the mechanism of why a number is not Rational. Why is sqrt2 irrational? That was missing in understanding the difference between Rational and Irrational.
Sqrt2 is irrational, not because it is "not Rational" for that is a goofball talking.
Sqrt2 is irrational because it is not a single number, a single solo number, but rather is two numbers acting as one, two different numbers working as one number.
In true math, sqrt2 is not 1.4142135.... for that is a single solo number.
In true math, sqrt2 is 1.41 with 1.42, then higher up, it is 1.414 with 1.415, and always a dance of two different numbers in multiplcation.
So, here, we finally get the real distinction, the real meaning of Rational as a single solo number, whereas, a Irrational number is not a single solo number, but two different Rational numbers working as one number.
If you had a rod of length of distance and it was a Rational number length such as 1.414, meaning 1.41400000..... then that distance length is fixed. But, if you had a rod whose length was irrational sqrt2, then at the end of the rod, it would be vibrating back and forth between two different numbers 1.414 and vibrating with 1.415
That is what Irrational means compared to Rational. Rational is one number, fixed, and irrational means you have two different numbers acting and behaving as though they were one number.
Now we get back to pi and 2.71... as irrationals. Many irrationals are like sqrt2, where you multiply two different numbers together, the sqrt10 is 3.16 x 3.17 = 10.0 in 10 Grid, in 3.162 x 3.163 = 10.00 in 100 Grid and on done the line. But the irrationals of pi and 2.71... involve division
Pi is 22/7 = 3.14 in 100 Grid
333/106 = 3.1415 in 10^4 Grid
355/113 = 3.141592 in 10^6 Grid
So instead of multiplication for sqrt2 irrational as that of 1.41 x 1.42 then the next progression 1.414 x 1.415, then the next progression
Instead of multiplication we have irrational by division, where we have two different numbers 22 with 7, as 22/7 then the next progression of two different numbers 333 with 106 as 333/106, then the next progression.
So with sqrt2 we have two different numbers in a PROGRESSION of multiplication, with pi we have two different numbers in a PROGRESSION of division.
But here it looks as though pi is a different type of Irrational, a Transcendental Irrational than is sqrt2, because of division.
That is not true, for if you know algebra, 22/7 is also multiplication 22 x 1/7
So, what I am going to attempt to do in a proof, is show that sqrt10 is Algebraic Irrational, everyone in Old Math, even the ones with 0 brains of logic ability recognize sqrt10 as Algebraic Irrational. What I am going to do is show that sqrt10 = 3.162277.... is both a Multiplication Irrational and then a Division Irrational.
Then I am going to show that not only is pi a Division Irrational but a Multiplication Irrational
Then show that sqrt2 is not only a Multiplication Irrational but a Division Irrational.
In other words, if a number is two different numbers acting as one number, is the defining feature of Irrational, then there cannot be two different types of Irrational.
Because Irrationals are formed from two different Rationals, whether it be multiplication or be division, means that you cannot have a category of Algebraic along with a category of Transcendental. The definition of irrational as two different Rationals working as one, allows only a Algebraic category.
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Sat, 19 Aug 2017 23:20:14 -0700 (PDT)
Subject: nearer to a proof all irrationals are algebraic // Transcendental
numbers are just illusions by bigots of math
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Sun, 20 Aug 2017 06:20:15 +0000
nearer to a proof all irrationals are algebraic // Transcendental numbers are just illusions by bigots of math
On Saturday, August 19, 2017 at 5:16:57 PM UTC-5, Archimedes Plutonium wrote:
(snip)
> Pi is 22/7 = 3.14 in 100 Grid
>
> 333/106 = 3.1415 in 10^4 Grid
>
> 355/113 = 3.141592 in 10^6 Grid
>
Alright the first 6 digits right of the decimal point in pi are 3.141592
Now to get the root irrationals like sqrt10 , 3.16227766...... we do multiplication
First we do 10.0 in 10 Grid is 3.16 x 3.17 gives 10.0
then we do 10.00 in 100 Grid is 3.162 x 3.163 gives 10.00
continuing
Now do pi, of 3.141592
First we do 10 Grid is 22 x 1/7, which is 22 x .142 is 3.1
then we do 3.14 is 22 x .1428 is 3.14
For 1/7 = .14285714
then we do 3.1415 all staying with just plan 22/7, not even bothering with 333/106 or 355/113. Can we go all the way out to pi digits with just sticking with 22/7 and tinkering with the 1/7 digits. If so, I have made a clear case example, that there is difference between a Algebraic irrational and a Transcendental irrational.
So we have 22 x .1428 is too large and 22x.14279 is too small, means I have to now jump to 333/106
1/106 is .00943396
So 333 x .00943 is 3.14, then 333 x .009433 is 3.141, then 333x .009434 is 3.1415. Moving on to 3.14159 we have 333x .00943421 is 3.14159
In this manner we slowly walk through the digits of pi as a Algebraic Irrational number, where two different Rationals, one the circumference number, the other the diameter number slowly grind out the digits of pi.
In the case of sqrt2 or sqrt10, the two different numbers are close nearby together of one another, varying only a tiny bit. In the case of pi the two different numbers are separated by a factor of about 3 times the size of one another.
So, in this viewpoint, why could not Polynomials eke out pi or 2.71...? They could not eke out pi because there is no machinery in the polynomials to concern themselves with a circumference or diameter of a circle. Circles are off topic when dealing with polynomials. But to the obtuse and nattering nutter math professor who does not see any pi or 2.71... emerge from the fog of polynomials, well, this nutter then goes and calls pi and 2.71... a whole brand new category of irrational. A total illogical reaction to not finding pi in polynomials. You cannot find pi in polynomials, because polynomials and circles are different topics.
But now that demonstration does not constitute a proof that all irrationals are of one category only, never two types. And here, a proof, I sense, will invoke a concept I never saw used before in all of math. The concept I am thinking of is that given a number A and a second number B, you can always find a unique third number C that multiply B*C = A. And the problem that this highlights about Old Math, is that they thought given a number A, that there exists a number B such that B^2 = A.
So here is the essential flaw of Old Math and their idea that irrationals can be of two types, that irrationals can be of Algebraic or Transcendental. Does mathematics warrant the idea that given a arbitrary number A, that a different number B, such that B^2 = A exists? Or, does math only allow for given arbitrary A, then given arbitrary B, there exists a C such that B*C =A.
So, I think that is the crux of a proof that Transcendentals are nonexistent. The structure of mathematics is such that given A and give B, there exists a C such that B*C=A, but math is not structured to all for a given A, there exists a B such that B^2 = A, or in some further cases a B^3 = A or a B^4 = A etc
In other words a proof that Transcendental is a pseudo category, is a pseudo idea, hinges on the fact that any A, is arrived at by a B*C, and so any irrational is a multiplication of B*C, no matter if it is pi, or sqrt10 or 2.71.... or phi, or .1234567891011.....
So, a proof that all irrationals are Algebraic Irrationals is a simple easy proof-- show the axioms of mathematics do not support a given A there exists a B^2, but the axioms do support given A, there exists a B*C= A.
Now, maybe that is difficult, and maybe that was a hidden assumption in Old Math that a B^2 always existed. And just maybe my newfound missing axioms of Complementarity in the Matrix Columns is called into action once again, not only into action for proving Goldbach, but for proving that mathematics does not warrant a B^2 = A given A as arbitrary.
And this makes alot of sense, alot of commonsense, because if you have a borderline of infinity where the only numbers you can use for valid mathematics, then as you get to larger numbers near the borderline, you are going to automatically run out of numbers to be a B^2 = A. In Old Math where infinity was just hazy, foggy and crappy, you could get by with a silly notion that given any A, a B^2 equals to A exists.
AP
Newsgroups: sci.math
Date: Sun, 20 Aug 2017 05:54:34 -0700 (PDT)
Subject: Re: nearer to a proof all irrationals are algebraic // Transcendental
numbers are just illusions by bigots of math
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Sun, 20 Aug 2017 12:54:34 +0000
Re: nearer to a proof all irrationals are algebraic // Transcendental numbers are just illusions by bigots of math
On Sunday, August 20, 2017 at 1:20:26 AM UTC-5, Archimedes Plutonium wrote:
(snipped)
> So, a proof that all irrationals are Algebraic Irrationals is a simple easy proof-- show the axioms of mathematics do not support a given A there exists a B^2, but the axioms do support given A, there exists a B*C= A.
>
Now some people worry about my health as being an insomniac. I can assure you I get more than 8 hours of sleep a day, for I take several naps during the day. I like working at night because there is absolutely nothing going on except my science to interrupt me.
> Now, maybe that is difficult, and maybe that was a hidden assumption in Old Math that a B^2 always existed. And just maybe my newfound missing axioms of Complementarity in the Matrix Columns is called into action once again, not only into action for proving Goldbach, but for proving that mathematics does not warrant a B^2 = A given A as arbitrary.
>
I think I have the proof worked out. Basically it says that given any arbitrary Rational number A, that there exists a B and C, such that B*C = A. But, there does not necessarily exist a B^2 or some other power of B such that A = B^2.
Why is that crucial for a proof that Transcendentals do not exist, and only Algebraic Irrationals exist?
It is crucial in the fact that all numbers, whether rational or irrational are the product of two different numbers. An Algebraic Irrational A is the product of two different rationals B and C. A number T, suspected of being Transcendental Irrational is also, from such a theorem, the product of B and C.
No number exists, which is not the product of two different Rationals.
This means that the Irrational number is of one type only, not two types.
Proof of the Theorem that every arbitrary Rational A is the product of two different Rationals B, C. The Rationals are dense and closed to multiplication and division. Given any A and B, by being dense, closed means a C exists. QED
Theorem Statement:: The irrationals of math exist in only one type-- Algebraic, and that the Transcendental Irrational is a fiction.
Proof Statement:: A irrational is the multiplication of two different Rationals. Can there be two different types of irrationals if all irrationals are the multiplication of two different Rationals. No, for if the definition of Irrational A = B*C, and by the theorem that all A have a B*C, then no irrational falls outside the scope of A= B*C. QED
Now that proof is probably going to need patchwork. Even though my critics want me to publish as is.
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Sun, 20 Aug 2017 14:53:49 -0700 (PDT)
Subject: the AXIOM of ALGEBRA Re: nearer to a proof all irrationals are
algebraic // Transcendental numbers are just illusions by bigots of math
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Sun, 20 Aug 2017 21:53:49 +0000
the AXIOM of ALGEBRA Re: nearer to a proof all irrationals are algebraic // Transcendental numbers are just illusions by bigots of math
On Sunday, August 20, 2017 at 7:54:46 AM UTC-5, Archimedes Plutonium wrote:
(snipped)
>
> > Now, maybe that is difficult, and maybe that was a hidden assumption in Old Math that a B^2 always existed. And just maybe my newfound missing axioms of Complementarity in the Matrix Columns is called into action once again, not only into action for proving Goldbach, but for proving that mathematics does not warrant a B^2 = A given A as arbitrary.
> >
>
Alright, just like in Goldbach, we have to reach to the first beginnings of mathematics, its core axioms that start off mathematics to prove Goldbach, and now, to prove that Irrationals are of only one type-- Algebraic and that Transcendental was a phony baloney category.
In Goldbach, we come to a point in the proof, that says, there exists a Bertrand prime on leftside of Matrix Column and a Bertrand prime that exists on rightside of Column
In the Columns below, starting with 4, it is Goldbach 2+2, but also it is Two Prime Composite 2*2, and then there is 8, which is 3+5 and also 3*5.
Old Math knew the Numbers were closed and complete under addition and multiplication, but, the people in Old Math never packaged this fact into an axiom that allows them to actually use the fact that Counting Numbers are Closed and Complete to addition and multiplication.
In order to prove Goldbach, a stage in the proof is where we say the Bertrand prime on leftside lines up with the Bertrand prime on rightside. In Old Math, this is true due to this axiom, that Old Math forgot to include, forgot to make prominent, simply forgot altogether.
So in the proof of Goldbach, taking 8 as example. We have a Bertrand prime on left and one on right and those two primes are the only ones we can validly know and speak about, so to prove Goldbach we need them to line up so that we have 3+5 = 8 and have 3*5=15. This is true because Algebra Math has Closure, has Completeness, has Complementarity of add to multiply. I call this fact the Algebra Axiom, or simply Matrix Column Axiom. When I invoke the Matrix Axiom I am invoking complementarity, closure, completeness
So the proof of Goldbach:: a simple short paragraph proof. There is a Bertrand prime on left, one on right. They line up due to Matrix Axiom. hence Goldbach, QED
Now I bring up this axiom because I need it today to prove that Irrational Numbers have only one classification possible-- Algebraic Irrational. The Transcendental Irrational was all a illusion in a delusion. And it begins with the first principles of Counting Numbers and Rationals. That given a Rational number A and another B, there always exists a C such that A = B*C, but the people in Old Math, bizarrely as if they were living in a nightmare all their lives doing math, believed that in math, given a A Rational Number, that there exists a B^2 always, and, given a A Real number, that there exists, bizarrely a Real B such that A= B^2 or any other power.
You see, in Old Math, they had almost a zero logical mind, for they knew that in Counting Numbers, only a few special Counting Numbers obeyed A = B^2 or B^3 etc. They all knew that.
And, they all knew that in Rationals only a few Rationals obeyed A = B^2, such as 1/4 = (1/2)^2
But, then, all of a sudden, as if all the logic in every brain of a mathematician had been sucked out of the mind, mathematicians believed that given a Real A, there exists a B^2. It was not true for Counting Numbers, not true for Rationals, yet all of a sudden, the dolts of math thought it true for Reals. And the reason they thought it true for Reals, is because they had no precision definition of infinity, no borderline on infinity, and they never realized that being Irrational, only meant two Rationals acting as one number. In such a toxic insane environment of Old Math, it is easy to see, they would jump to their next pathetic mistake-- that irrationals can be transcendental.
Axiom One of Advanced Algebra
Matrix Column of Numbers in Mathematics. Math's first axiom of Algebra for it entails Closure and Completeness of addition complementary to multiplication
0 1
0 2
1 1
0 3
1 2
0 4
1 3
2 2
0 5
1 4
2 3
0 6
1 5
2 4
3 3
0 7
1 6
2 5
3 4
0 8
1 7
2 6
3 5
4 4
0 9
1 8
2 7
3 6
4 5
0 10
1 9
2 8
3 7
4 6
5 5
0 11
1 10
2 9
3 8
4 7
5 6
0 12
1 11
2 10
3 9
4 8
5 7
6 6
.
.
.
.
> I think I have the proof worked out. Basically it says that given any arbitrary Rational number A, that there exists a B and C, such that B*C = A. But, there does not necessarily exist a B^2 or some other power of B such that A = B^2.
>
> Why is that crucial for a proof that Transcendentals do not exist, and only Algebraic Irrationals exist?
>
> It is crucial in the fact that all numbers, whether rational or irrational are the product of two different numbers. An Algebraic Irrational A is the product of two different rationals B and C. A number T, suspected of being Transcendental Irrational is also, from such a theorem, the product of B and C.
>
> No number exists, which is not the product of two different Rationals.
>
> This means that the Irrational number is of one type only, not two types.
>
> Proof of the Theorem that every arbitrary Rational A is the product of two different Rationals B, C. The Rationals are dense and closed to multiplication and division. Given any A and B, by being dense, closed means a C exists. QED
>
> Theorem Statement:: The irrationals of math exist in only one type-- Algebraic, and that the Transcendental Irrational is a fiction.
>
> Proof Statement:: A irrational is the multiplication of two different Rationals. Can there be two different types of irrationals if all irrationals are the multiplication of two different Rationals. No, for if the definition of Irrational A = B*C, and by the theorem that all A have a B*C, then no irrational falls outside the scope of A= B*C. QED
>
> Now that proof is probably going to need patchwork. Even though my critics want me to publish as is.
>
>
Alright, I do have a proof here.
Some are going to say, well, Mr. AP, what is your definition of Reals in New Math. And I would say, I do not need Reals, for all math needs only Rationals. For Reals are just the inclusion of Irrationals with Rationals, but then irrationals are just the combination of two Rationals acting as one number. So it all boils down that math is just Rationals.
In the proof that Transcendentals do not exist, we simply invoke the Axiom of Algebra above, that addition is complementary to multiplication. That Rationals always obey -- Given A and B, there exists a C such that A= B*C. So, every Irrational is covered by A=B*C, and all are one type-- Algebraic Irrational.
The fruitcakes of Old Math, realized Algebra had closure, had completeness with addition complement multiplication, but the fruitcakes never assembled that idea into a Axiom, which never would allow them to prove Goldbach or prove that Transcendentals were phony. If you are a mathematician, and lack logic, then your career in math is going to be disappointment. You should call yourself a Calculator, not a Mathematician. For that is what the majority of people in math are-- Calculators, not Mathematicians.
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 00:50:50 -0700 (PDT)
Subject: the Arithmetic Axiom for closure, completeness, uniqueness //
Transcendental numbers are just illusions by bigots of math
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Mon, 21 Aug 2017 07:50:50 +0000
the Arithmetic Axiom for closure, completeness, uniqueness // Transcendental numbers are just illusions by bigots of math
Call it the Axiom of Arithmetic Algebra
Or call it the Axiom of Algebra
And it is simply the Matrix Columns
It contains the axioms of Complementarity, that addition is complement of multiplication
It contains the axioms of Uniqueness of add operation and uniqueness of multiplication operation
It contains the axioms of Completeness and Closure which means given any two numbers in the Matrix, their addition or multiplication will be another number found in the Matrix
Now, sad to say, that the Peano Axioms for the Counting Numbers never had this Axiom of Algebra, and it was one of the biggest flaws of Peano Axioms, for the concepts of this Axiom were assumed in Peano.
Now to invoke the axiom, we simply state Matrix Column
If we do not count Pythagorean theorem as an axiom, then the Matrix Column is the first axiom in math history that is a picture, rather than a long spiel of words.
- hide quoted text -
AP
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 01:38:19 -0700 (PDT)
Subject: chugging along Re: nearer to a proof all irrationals are algebraic //
Transcendental numbers are just illusions by bigots of math
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Mon, 21 Aug 2017 08:38:19 +0000
chugging along Re: nearer to a proof all irrationals are algebraic // Transcendental numbers are just illusions by bigots of math
On Sunday, August 20, 2017 at 7:54:46 AM UTC-5, Archimedes Plutonium wrote:
(snipped)
>
> Why is that crucial for a proof that Transcendentals do not exist, and only Algebraic Irrationals exist?
>
> It is crucial in the fact that all numbers, whether rational or irrational are the product of two different numbers. An Algebraic Irrational A is the product of two different rationals B and C. A number T, suspected of being Transcendental Irrational is also, from such a theorem, the product of B and C.
>
> No number exists, which is not the product of two different Rationals.
>
> This means that the Irrational number is of one type only, not two types.
>
> Proof of the Theorem that every arbitrary Rational A is the product of two different Rationals B, C. The Rationals are dense and closed to multiplication and division. Given any A and B, by being dense, closed means a C exists. QED
>
> Theorem Statement:: The irrationals of math exist in only one type-- Algebraic, and that the Transcendental Irrational is a fiction.
>
> Proof Statement:: A irrational is the multiplication of two different Rationals. Can there be two different types of irrationals if all irrationals are the multiplication of two different Rationals. No, for if the definition of Irrational A = B*C, and by the theorem that all A have a B*C, then no irrational falls outside the scope of A= B*C. QED
>
> Now that proof is probably going to need patchwork. Even though my critics want me to publish as is.
>
>
Alright, it is a strange proof, and I am still reeling around it. One fun I have found in math proofs, is that they seem to all have different flavor of proof, different character each. Not as if you do one math proof and some cookie cutter is going to do the next proof.
So to stop my uneasyness, I repeat the proof over and over in my head, until satisfied.
So let me repeat the proof again:
We define Irrational number as the multiplication of two *different Rationals* So, square root of 2 is 1.414 x 1.415, and the pi is 22 x 1/7 in 100 Grid. So the key is Two Different Rationals acting as if they are one single number.
Now we ask, can there be a Transcendental Irrational? Something different from a Algebraic Irrational?
Both would still be defined as the Multiplication of two different rationals.
If all Irrationals are the multiplication of two different rationals, can that definition provide for two classification schemes of irrationals. Call the Irrational as P and P is R_1 X R_2. Can there be a classification scheme that marks apart a Algebraic Irrational from a Transcendental Irrational?
I say no, in that the axiom that says given Rational numbers R_1 and R_2 there is a unique R_3 such that R_1 = R_2 X R_3
I see no wiggle room in there to allow for a Transcendental classification. And so, it all hinges on the Uniqueness property of three Rationals.
Now what is perplexing to the reader, is that they envision that R_1 is rational, and I am saying it is Irrational, because it is composed of R_2 X R_3. Let me write that as sqrt2 to ease the reader.
So, sqrt2 is Irrational, but why is it irrational? Not because there is a single solo number 1.414.... out there that is sqrt2. No, there is no single solo number that is sqrt2. What there is, is two Rational numbers out there 1.414 with1.415 that goes to compose sqrt2.
And the proof above is trying to explain why no Transcendental Irrationals exist, why none of them exists. And the reason being in the proof above, is that all Rationals taken pairwise two different Rationals produce all possible Irrationals uniquely, and no distinction can be made of them as to Algebraic or Transcendental. The uniqueness of multiplication forbids a second type of irrational.
Nay, still not happy,,,,
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 14:32:24 -0700 (PDT)
Subject: call it the Stevin-Fibonacci Decimal Representation Re: the number
1/3 is really .3333..33(1/3), not .3333....
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Mon, 21 Aug 2017 21:32:25 +0000
call it the Stevin-Fibonacci Decimal Representation Re: the number 1/3 is really .3333..33(1/3), not .3333....
On Monday, August 21, 2017 at 2:38:59 AM UTC-5, Archimedes Plutonium wrote:
> Now the failure of mathematicians on .99999.... and missing the second-decimal point .9999..99(3*(1/3)) the suffix (3*(1/3)) or for 1/3 = .333..33(1/3) where the suffix is just (1/3) is not a rare thing.
>
> No, this suffix miss, or failure is throughout Old Math in everyday math.
>
> It is not just a failure on particular numbers but a failure throughout the structure of math.
>
> The only time it is not a failure is when a number ends in zeros quickly.
>
Stevin-Fibonacci, 1202 Decimal Representation of Numbers has been missing the suffix for over 800 years now, 815 years to be exact since the book Liber Abaci (Book of Calculation).
Now if someone had been bright enough-- had seen this
________
3 | 10,000 = 3,333+1/3 and knowing full well that it was not equal to 3,333. but had to include the remainder
Had been sharp, had been bright, instead of the run of the mill math professor that is never bright, never sharp when it concerns mathematics
Would have known that
_________
3 | 1.0000 = .3333(1/3)
But no, the dunces for 815 years kept it as 1/3 = .33333....
Math sucks, if you want to be a math professor but have no abilities of Logical Reasoning.
AP
END
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 15:08:25 -0700 (PDT)
Subject: changing course-- there maybe a classification of Transcendental
Irrational, after all, since no proof is forthcoming,,,,
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Mon, 21 Aug 2017 22:08:25 +0000
I left off last night with this nagging nonproof,,, which makes me think I need to investigate the other side-- a transcendental category exists.
Let me inject Polynomial theory into the argument::
So we come to the lines in Polynomial theory of
x^2 = 2
1 = 2/x*x
Now we inject the Axiom of Arithmetic, that A = B*C but seldom does a A = D^2 and we can see the birth of the Algebraic Irrational
Now, using Polynomial theory, can we see a birth of the Transcendental Irrational
C = 3.14D
C/D = 3.14
1 = 3.14/(C/D)
1 = 3.14 D / C
In the case of 2/x*x the x's have to be two different numbers
In the case of 3.14 D / C the D and C have to be two different numbers
But, what separates the two cases, is the second case is always a division case whilst the first is a multiplication. This "difference" allows for a whole entire category of irrationals as transcendental.
But it has a implication, which can be tested-- the implication is that every Algebraic Irrational is a multiplication root irrational. So if we look at phi, it involves sqrt5. However, we look at Gelfond Schneider a^b where b is root irrational purported to be transcendental yet we see no division.
So I think what will happen is the Gelfond-Schneider has a flaw in it, that is, if Transcendentals do exist afterall
--- yesterday's proof that does not work ---
chugging along Re: nearer to a proof all irrationals are algebraic // Transcendental numbers are just illusions by bigots of math
Archimedes Plutonium
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 18:13:42 -0700 (PDT)
Subject: C/D rather than x*x Re: changing course-- there maybe a
classification of Transcendental Irrational, after all, since no proof is forthcoming,,,,
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Tue, 22 Aug 2017 01:13:43 +0000
C/D rather than x*x Re: changing course-- there maybe a classification of Transcendental Irrational, after all, since no proof is forthcoming,,,,
The reason I shifted 180 degrees, is because the Axiom of Arithmetic-- closure and completeness of multiplication and division of Rationals does not give me a proof of **no transcendental category exists**. When faced with the reality that no proof can be fetched, leaves the wise person no alternative than to explore-- transcendentals do exist. By exploring whether transcendentals exist, I may as well do the logical thing-- start with Polynomials, where transcendental definition originates.
On Monday, August 21, 2017 at 5:08:37 PM UTC-5, Archimedes Plutonium wrote:
> I left off last night with this nagging nonproof,,, which makes me think I need to investigate the other side-- a transcendental category exists.
>
> Let me inject Polynomial theory into the argument::
>
> So we come to the lines in Polynomial theory of
>
> x^2 = 2
>
> 1 = 2/x*x
>
> Now we inject the Axiom of Arithmetic, that A = B*C but seldom does a A = D^2 and we can see the birth of the Algebraic Irrational
>
> Now, using Polynomial theory, can we see a birth of the Transcendental Irrational
>
> C = 3.14D
>
> C/D = 3.14
>
> 1 = 3.14/(C/D)
>
Let me pick that form as the primal form
> 1 = 3.14 D / C
>
So the primal form for Algebraic Irrational is
1 = 2/x*x
and the primal form for Transcendental Irrational is
1 = 3.14/(C/D)
Notice, in primal form one side of polynomial is 1 and the other side is the irrational number of interest
Notice that Algebraic Irrational is a multiplication of two different rationals x*x and the transcendental irrational is a division C/D
Now, when I was trying to prove Transcendentals do not exist, I was expecting that the form C/D
is the same as the form x*x for we all know C/D can be rearranged as C*(1/D)
However, it is apparent that C*(1/D) cannot replace or substitute in a form of x*x for the simple reason of a infinite progression-- approach to the infinity borderline.
For example: in x*x for sqrt2 we have 1.41 x 1.42 and then the progression is 1.414 x 1.415
Now for pi we have C/D we have 22/7 then in the progression we have 333/106 then we have 355/113
More later,,,,,
AP
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 18:53:17 -0700 (PDT)
Subject: Re: C/D rather than x*x Re: changing course-- there maybe a
classification of Transcendental Irrational, after all, since no proof is forthcoming,,,,
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Tue, 22 Aug 2017 01:53:18 +0000
Re: C/D rather than x*x Re: changing course-- there maybe a classification of Transcendental Irrational, after all, since no proof is forthcoming,,,,
Now of course, I would misspell "course" in the title as coarse, but there is nothing coarse in this topic.
Now, in this turnaround, let us graph the Irrational point sqrt2 in the Plane. Now taking a right triangle
with points (0,0) (1,0) (1,1)
Now the hypotenuse from (0,0) to (1,1) has length sqrt2
Now, if we were to find the point on x-axis of sqrt2 we run into trouble, because it is not a single solo point, or number.
The coordinate point (1,0) are two numbers involved but the coordinate point (sqrt2,0) is actually three numbers involved in 100 Grid as ((1.414,1.415), 0)
So in New Math, true math, all Irrationals when plotted in geometry on a graph, are really more points than realized.
So plotting the point of (sqrt2, sqrt5) would be ((1.414,1.415), (2.236, 2.237)), while in Old Math it would be (sqrt2, sqrt5).
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 21:31:15 -0700 (PDT)
Subject: we have 1 = 2/x*x versus 1 = 3.14/(C/D) Re: changing COURSE-- there
maybe a classification of Transcendental Irrational, after all
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Tue, 22 Aug 2017 04:31:16 +0000
we have 1 = 2/x*x versus 1 = 3.14/(C/D) Re: changing COURSE-- there maybe a classification of Transcendental Irrational, after all
On Monday, August 21, 2017 at 5:08:37 PM UTC-5, Archimedes Plutonium wrote:
(snip)
> >
> > Proof Statement:: A irrational is the multiplication of two different Rationals. Can there be two different types of irrationals if all irrationals are the multiplication of two different Rationals. No, for if the definition of Irrational A = B*C, and by the theorem that all A have a B*C, then no irrational falls outside the scope of A= B*C. QED
> >
I think I can muster up a proof tonight, that what distinguishes the Algebraic Irrational from the Transcendental Irrational is that the first is multiplication x*x, while the second is division C/D. And that both are sequences to the Infinity Borderline, for which both warrant their own classification.
A few days ago, I thought the C/D is just another form of x*x so that the Transcendental category is another Algebraic Irrational.
But, somehow the division separates the two types, so that they are distinct.
What this means, is that some of the Old Math proofs of transcendental are fake, have a flaw in their argument. The Gelfond-Schneider proof of a^b where b is algebraic-irrational and claiming a^b is transcendental irrational is a flawed proof because there is only multiplication, no division.
Of course pi is Transcendental since it involves Circumference divided by diameter. E as 2.71... is transcendental for it is the Circumference-arc/ radius of a Log spiral. Then Champernowne's number
.123456789101112.... is Transcendental Irrational because you can only build a Sequence a Progression of Terms from division
From .12 as 3/25, to 123/1000, to 617/5000, etc etc.
Now notice all the Algebraic Irrationals, they all bobb up and down over a constant number, which is reached by a Sequence of multiplication not by division.
So we pick sqrt10
Our sequence has to be 10.0 then 10.00 then 10.000, etc etc and it goes like this::
3.16 * 3.17 then 3.162 x 3.163 etc etc.
So, what should we notice about Algebraic Irrationals versus Transcendental Irrationals?
We notice that the Algebraic Irrational is a number bobbing up and down to reach 10.0000.... of bobbing from 9.999... to 10.000... and above 10.0000....
Whereas Transcendental Irrationals, notice that we have no root of a number to tinker around with, tinker about with. We have a root of sqrt2 or of sqrt10 which we keep tinkering around, raising the value of a digit by one more in order to lift the number from 999s to 000s. So we have no tinkering in pi or 2.71, but rather we have to assail it with division.
And notice, the numbers in sqrt2 such as 1.414 with 1.415 are close together, where as in 3.14.... the two complements are far apart as 22 with 7 or as 333 with 106 or as 355 with 113.
So now, Wikipedia claims 3.300330000000000330033..... is Transcendental
And that is alright with the idea that we need a division set up to extract two Rationals A and B to have A/B
What is not alright is the Gelfond-Schneider idea that A^B where B is algebraic irrational that the whole is Transcendental Irrational. It must have been a flawed proof because it has no division involved.
AP
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 22:08:42 -0700 (PDT)
Subject: easy to find the holes in Gelfond- Schneider Re: changing COURSE--
there maybe a classification of Transcendental Irrational, after all
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Tue, 22 Aug 2017 05:08:43 +0000
easy to find the holes in Gelfond- Schneider Re: changing COURSE-- there maybe a classification of Transcendental Irrational, after all
Yes, I think I spotted the Flaw of the Gelfond-Schneider Attempt of A^B transcendental if B is Algebraic Irrational.
They made a silly, very silly assumption. They assumed that if in A^B that A^B is algebraic irrational and if B is a element of the Rationals, that such would overturn their assumption A^B is algebraic.
That is as ludicrous as trying to prove the Brandenberg Gate is in Berlin, by proving the Sieg Victory Column is in Berlin. So that if Sieg is in Berlin, then Brandenberg is in Berlin.
As ridiculous as trying to prove Europa is a moon of Jupiter, if we an show that Ganymede is a moon of Jupiter.
As ridiculous as trying to prove birds fly, if we only we can prove that fish fly.
The crass logical error of Gelfond Schneider was to think that if something A is a "neighbor" of B, that when we show A, we have shown B.
But, they make another error of Logic, they make a double Reductio Absurdum Argument. They assume two falsehoods. They assume -- Suppose A^B is algebraic Irrational simultaneously they assume B is Rational. This is like saying Suppose Big Ben is in Paris and suppose Eiffel tower is in New York city. Once you show Eiffel tower is not in New York City, does not allow you to assert Big Ben is not in Paris.
So, the Gelfond Schneider episode is a pitiful example of barb wired logic, hoop de do logic.
Memory is slowly coming back to me that a long time ago I berated this fakery of Gelfond-- was it the 1990s?
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Mon, 21 Aug 2017 22:38:07 -0700 (PDT)
Subject: pity that math professors never had to take Logic in school Re: easy
to find the holes in Gelfond- Schneider Re: changing COURSE-- there maybe a
classification of Transcendental Irrational, after all
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Tue, 22 Aug 2017 05:38:08 +0000
pity that math professors never had to take Logic in school Re: easy to find the holes in Gelfond- Schneider Re: changing COURSE-- there maybe a classification of Transcendental Irrational, after all
Yes, the failure of the Gelfond-Schneider attempt is a use of a Double Reductio ad Absurdum. In New Math, true math, we cannot even use the Single Reductio ad Absurdum for it is only a probability truth, not a math required-- guarantee of truth-- proof.
So the Gelfond-- Schneider makes two assumptions at the outset. They assume A^B is algebraic, plus, in addition, they assume B is rational. The only thing obvious here, is that neither Gelfond nor Schneider ever had a handle on Logic.
Now, I know that Wiles used Reductio Ad Absurdum in his fakery of Fermat's Last Theorem. In New Math, his offering would never merit even a glance look, but just thrown into the trash. But, since Gelfond-Schneider did a Double Reductio Ad Absurdum, the question is, did Wiles do a Double Reductio Ad Absurdum.
And we should look to see if any mathematician got away with doing a Triple Reductio Ad Absurdum. What that is, is where you make three assumptions in the opening or middle of the proof argument and where you find one of the assumptions is false, you then go ahead and discharge the other two assumptions all in one fell swoop, landing you smack in the middle of a sheer fakery of math.
What logic teaches us, is that a math proof is never a Reductio Ad Absurdum. And for those that want to try Reductio ad Absurdum for the curiosity sake, they know, to be correct, that you cannot run two assumptions together. That you have to discharge a assumption, before you make another assumption.
But hey, try telling that to the mathematics community, that insists its education system where professors graduate as "professors of math" yet they know not one stitch of Logic Reasoning, logical argument. They are bred to do a calculation, not bred to think straight to think clear.
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Tue, 22 Aug 2017 00:16:17 -0700 (PDT)
Subject: is x_1*x_2 different from C*(1/D) Re: changing COURSE-- there maybe a
classification of Transcendental Irrational, after all
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Tue, 22 Aug 2017 07:16:17 +0000
is x_1*x_2 different from C*(1/D) Re: changing COURSE-- there maybe a classification of Transcendental Irrational, after all
Alright, I want to go to bed tonight, knowing the truth behind Transcendental and Algebraic Irrationals.
Brief recap:: I suspected the Transcendentals did not exist, for the reason that a division C/D can be placed as C*(1/D), imitating the x*x. A axiom of arithmetic says given any Rational A and B, there is a C such that A = B*C, but that axiom also says, given A,D. there is not always a A = D^2. This axiom would then constitute a proof that the Algebraic Irrational exists and would be where the definition of Irrational is-- two different numbers behaving as if they are one and the same number. In other words, a Rational number is a single solo number, whereas the Irrational number is defined as two different numbers acting as one.
So, I thought such a axiom and definition would make all Irrationals be Algebraic Irrationals, thinking that x*x is C/D when we turn that into C*(1/D).
So, trying to prove it yesterday, I just could not overcome the hurdles. Thus, I reverse gears and prove that x*x is altogether different from C*(1/D).
But, one last obstacle still stood in the way-- Gelfond-Schneider fake proof of A^B transcendental when B is algebraic irrational. That would be a transcendental in the form of x*x and no division.
So, had to disproof Gelfond-Schneider. Luckily it was a cinch, and needed not read far to realize what a klutz set-up they had. Their travesty was to do a Reductio Ad Absurdum, but compounding that into a Double Reductio ad Absurdum. Their set-up was to assume A^B was algebraic and then also, simultaneously assume B is a Rational. Huge huge flaw, and shows that Logic was really not in vogue when this proof attempt was assembled.
So, everything is coast clear, for a proof that Algebraic Irrational are all those numbers of form x*x, root irrationals not just confined to sqrt but extending to cube roots and beyond. While Transcendental Irrationals are all those of form C/D, division sequence rather than multiplication sequence.
Now I have to link in Polynomial theory, because this concept of Transcendental Irrational comes from Polynomial theory.
So here the set-up is this::
x^2 = 2
1 = 2/x*x by Polynomial theory
C = 3.14D by Polynomial theory
C/D = 3.14 by Polynomial theory
1 = 3.14/ (C/D)
So, strictly in keeping with Polynomial theory we have two means, two methods of getting square root of 2 or 3.14. We can multiply x*x where the x is two different numbers by Axiom of Arithmetic or we can obtain 3.14 by division.
So, we have two options of getting a solution, we multiply two different numbers, or we divide by two different numbers, and the first is Algebraic Irrational, the second is Transcendental Irrational.
Now, I still have not solved in my mind why C*(1/D) is not the same as x*x (keeping in mind they are x_1 different from x_2).
So, what is the POLYNOMIAL theory say about that?? Is the hitch in Polynomial theory itself that the
C*(1/D) cannot be a x*x (x_1*x_2)
Is it Polynomial theory causing the hitch, the glitch, the snag?
So let me go back to the polynomials
x^2 - 2 = 0
x^2 = 2
1 = 2/ (x^2)
1 = 2 / x*x
Nothing adverse there unless you want to invoke Rationals with the axiom of Arithmetic that says, you are always guaranteed a A = BC, but rarely given a A = D^2, paving the way to the idea that roots are irrationals, without the need for Pythagorean theorem, and that more clearly, we see a root is two different numbers acting as one number.
So now, the division in Polynomial theory
C - 3.14 D = 0 (where here I have c circumference and d diameter)
could have written it for the over fastidious algebra nerds 3.14x - k = 0 who couldn't recognize a polynomial otherwise
C = 3.14 D
C/D = 3.14
1 = 3.14 / (C/D)
So, in Polynomial theory, can we say x_1 *x_2 is C*(1/D) or are they fundamentally different.
If fundamentally different we have a proof that Algebraic is a multiplication irrational while division is a transcendental irrational.
AP
Newsgroups: sci.math
Date: Tue, 22 Aug 2017 01:24:13 -0700 (PDT)
Subject: Re: is x_1*x_2 different from C*(1/D) Re: changing COURSE-- there
maybe a classification of Transcendental Irrational, after all
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Tue, 22 Aug 2017 08:24:13 +0000
Re: is x_1*x_2 different from C*(1/D) Re: changing COURSE-- there maybe a classification of Transcendental Irrational, after all
What I am trying to resolve, is whether a Polynomial does not have the structure of division, only the structure of multiplication.
So take a bigger polynomial
x^3 + x^2 +5x -3.14 = 0
can that be turned into a division like 1 = 3.14/(C/D)
x^3 + x^2 +5x = 3.14
1 = 3.14/ (x^3 + x^2 + 5x)
Do Polynomials prevent division
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Tue, 22 Aug 2017 03:13:50 -0700 (PDT)
Subject: polynomial structure cannot afford a:: 1 = k/(x/y) Re: changing
COURSE-- there maybe a classification of Transcendental Irrational, after all
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Tue, 22 Aug 2017 10:13:53 +0000
polynomial structure cannot afford a:: 1 = k/(x/y) Re: changing COURSE-- there maybe a classification of Transcendental Irrational, after all
On Tuesday, August 22, 2017 at 3:24:26 AM UTC-5, Archimedes Plutonium wrote:
> What I am trying to resolve, is whether a Polynomial does not have the structure of division, only the structure of multiplication.
>
>
> So take a bigger polynomial
>
> x^3 + x^2 +5x -3.14 = 0
>
> can that be turned into a division like 1 = 3.14/(C/D)
>
So, here, I think I solved this problem, solved it cleanly, and altogether.
There is no doubt that
x^2 = 2 is a polynomial and that it can be turned into
1 = 2/x*x
and further using the Axiom of Arithmetic the x*x is in general x_1*x_2
> x^3 + x^2 +5x = 3.14
>
> 1 = 3.14/ (x^3 + x^2 + 5x)
>
> Do Polynomials prevent division
However the circumference of circle as C = pi *d
1 = pi*d/C
1 = pi/(C/d)
is not a polynomial, never was a polynomial
C = pi *d
is
Y = pi*x
And no polynomial is of two variables.
So, the reason Irrationals of Algebraic Irrationals exist from Polynomial structures, is because you have a structure of
1 = k/x^n
And no polynomial can be of a division structure
1 = k/(x^n/y^m)
In other words, no polynomial can be a constant divided by a division of (x/y)
For this reason, pi or e, or Champernowne's number can ever show up in a polynomial, for a polynomial has no division structure.
And, at long last, Irrationals exist as either a multiplication irrational the familiar root irrationals or exist as a transcendental irrational such as pi, 2.71.....
Now, as for the bonehead conclusions that transcendentals exceed in cardinality those of other number sets is all just pure bogus garbage, all crafted with infinity having no borderline-- all pure crap.
AP
On Tuesday, August 22, 2017 at 6:00:00 PM UTC-5, Archimedes Plutonium wrote:
Alright, so, looking at the Gelfond Schneider theorem for it is a fakery. If we have sqrt2 being two different Rationals acting as one number 1.414 Rational times 1.415 Rational to equal to 2.000 in 1000 Grid. Then we look at 2^sqrt2 and ask, is that algebraic irrational or transcendental?
So, 2^sqrt2 is piecewise 2^1.414 and it is algebraic irrational and then 2^1.415 which is again algebraic irrational.
This example bolsters my claim that the Gelfond-Schneider proof is riddled with Logical holes-- namely, they used a Double Reductio Ad Absurdum argument. What is double-RAA? That is where make a so called proof by having two assumptions at the outset. Now, a crazy prize should be awarded to any mathematician who has foisted upon the math community a so called proof, that contains 3 or more assumptions at the outset, never discharging any of them until the end. I am guessing Wiles is such a bonehead argument of Fermat's Last Theorem.
The math community at present is in crisis management, for it is so dumb that it never saw the Conic was a oval, never an ellipse, so dumb that it never saw 1/3 is not .333333... but rather is .3333..33(1/3). So dumb that it thought sine was a sinusoid, when it is really a semicircle wave when you define unit circle and sine = opposite/hypotenuse. So dumb that their calculus had rectangles of 0 width.
This is what happens to math, when it passes professors of math in colleges without ever taking a formal course in Logic. They all come out Logic brain dead. They all come out not knowing it is a sin of Logic to make several assumptions without discharging any of them until the end of a proof.
Not only are 99% of math professors at present blind, but logically dumb.
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Tue, 22 Aug 2017 17:01:33 -0700 (PDT)
Subject: explaining 22*1/7 Re: There are two irrational classifications--
algebraic, transcendental, but, the Gelfond-Schneider input is a fakery
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Wed, 23 Aug 2017 00:01:33 +0000
explaining 22*1/7 Re: There are two irrational classifications-- algebraic, transcendental, but, the Gelfond-Schneider input is a fakery
On Tuesday, August 22, 2017 at 6:00:00 PM UTC-5, Archimedes Plutonium wrote:
(snipped)
>
> Alright, so, looking at the Gelfond Schneider theorem for it is a fakery. If we have sqrt2 being two different Rationals acting as one number 1.414 Rational times 1.415 Rational to equal to 2.000 in 1000 Grid. Then we look at 2^sqrt2 and ask, is that algebraic irrational or transcendental?
>
> So, 2^sqrt2 is piecewise 2^1.414 and it is algebraic irrational and then 2^1.415 which is again algebraic irrational.
>
> This example bolsters my claim that the Gelfond-Schneider proof is riddled with Logical holes-- namely, they used a Double Reductio Ad Absurdum argument. What is double-RAA? That is where make a so called proof by having two assumptions at the outset. Now, a crazy prize should be awarded to any mathematician who has foisted upon the math community a so called proof, that contains 3 or more assumptions at the outset, never discharging any of them until the end. I am guessing Wiles is such a bonehead argument of Fermat's Last Theorem.
>
> The math community at present is in crisis management, for it is so dumb that it never saw the Conic was a oval, never an ellipse, so dumb that it never saw 1/3 is not .333333... but rather is .3333..33(1/3). So dumb that it thought sine was a sinusoid, when it is really a semicircle wave when you define unit circle and sine = opposite/hypotenuse. So dumb that their calculus had rectangles of 0 width.
>
> This is what happens to math, when it passes professors of math in colleges without ever taking a formal course in Logic. They all come out Logic brain dead. They all come out not knowing it is a sin of Logic to make several assumptions without discharging any of them until the end of a proof.
>
> Not only are 99% of math professors at present blind, but logically dumb.
>
>
Now my critics are going to jump all over me on this example of why Gelfond-Schneider is wrong.
So I take 2^sqrt2 and piecewise break that into 2^1.414 and 2^1.415 and since each alone is algebraic irrational that 2^sqrt2 is still algebraic irrational.
The critics will then say, how do I handle 2^pi as that of 22/7 are we to say 2^22 with 2^7 for each is clearly a Rational, not even a irrational.
Here is where I need to argue the Polynomial origins again, of how polynomials are defining algebraic irrational versus transcendental irrational.
Polynomials give
x^2 = 2
1 = 2/x*x
which applying Axiom of Arithmetic is 1 = 2/x_1*x_2 that x is algebraic irrational involving two different rational numbers x_1 with x_2
But, no Polynomial is of form
Circum = 3.14...diameter
C/d = 3.14...
1 = 3.14..../ (C/d)
No Polynomial is of that form because no Polynomial is of
Y = 3.14.... x
That is what C = 3.14...d is when turned into a polynomial-- two variables, Y and x, and no polynomial is of two variables.
It is the reason that you can have two classifications of Irrational, the Polynomial irrational of x_1*x_2 and then the transcendental irrational of C/d
So, sqrt2 is 1.414 with 1.415 in 1000 Grid
So, is pi in 100 Grid 22 with 1/7, and how do I explain the difference.
AP
Newsgroups: sci.math
Date: Tue, 22 Aug 2017 22:40:06 -0700 (PDT)
Subject: Re: explaining 22*1/7 Re: There are two irrational classifications--
algebraic, transcendental, but, the Gelfond-Schneider input is a fakery
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Wed, 23 Aug 2017 05:40:06 +0000
Re: explaining 22*1/7 Re: There are two irrational classifications-- algebraic, transcendental, but, the Gelfond-Schneider input is a fakery
- hide quoted text -
On Wednesday, August 23, 2017 at 12:13:37 AM UTC-5, Archimedes Plutonium wrote:
> This happens often in science, not just math or physics but almost all science, where, the moment you resolve some crisis problem, you resolve it satisfactorily, is the moment another vexing problem jumps up and out, at you. The only good thing I can say about this phenomenon, which most people never experience for they are not in doing science, the best I can say, is that if this arise of a next vexing problem occurs-- is a good sign you are on the right track of doing true science.
>
> If no vexing problem comes your way in solving an issue-- could signal that you escaped doing real science.
>
> So, I solved the problem of whether Irrationals come in two flavors or only one. It turns out that Irrationals come in two flavors-- Algebraic Irrational and Transcendental Irrational, and what distinguishes the two flavors is that the Algebraic has the format of x_1* x_2 whereas the Transcendental has the format of C/D, one is multiplication, the other is division.
>
> The best two examples are sqrt2 is algebraic for it is two different Rationals 1.414 with 1.415 when multiplied produces 2.000 in 1000 Grid. But pi as 3.141592.... is transcendental because it is a division C/D such as 22/7 yielding 3.14 in 100 Grid.
>
> Where the definition of Irrational whether Algebraic or Transcendental is defined as **TWO DIFFERENT RATIONALS ACTING AS ONE NUMBER**
Sorry, I hit the wrong key and off it flew as a post, prematurely.
Let me complete my thoughts there, above, ,,,,,,,,,
As I was saying, the Algebraic Irrational fits the bill of two different Rationals in Multiplication to arrive at a number such as 2.00000... In Old Math, their single solo number 1.414.... never reaches 2.0000... always below it at 1.99... never managing to surface at 2.0000.... as wanted. So that is the reason for why Irrationals exist at all, is because roots cannot have a single solo number 1.414.... that produces 2.000.... but requires two different numbers acting as one to produce 2.0000.... that is why irrationals are different from rationals, because it takes two rationals to compose 2.0000.....
But we come to the Transcendental irrational and does the same definition of Two Different Rationals when divided act as one number. Is the 22 like the 1.414 and the 1/7 like the 1.415 ?
So we have a problem. What are the two different Rationals upon dividing that composes 3.14159....
And here, I have to review the division.
At first I would want to say that the 22 is like the 1.414 and the 1/7 is like the 1.415, but then that ruins the explanation of why A^B where B is algebraic irrational is still itself algebraic irrational, because if 22 and 7 are the two rationals divided to yield a transcendental, is not going to work.
So, here, I need to renovate, to solve a new problem.
One solution is to say, that a Algebraic Irrational is the composition of two different Rationals, acting as one number. And then the Transcendental Irrational is the composition of two different Rational numbers in division and in a sequence of division.
So the Algebraic Irrational is two different Rationals in a sequence of multiplication
Transcendental Irrational is two different Rationals in a sequence of division.
As shown here::
For sqrt10, we have in 10 Grid is 3.16 x 3.17 gives 10.0
then we do 10.00 in 100 Grid is 3.162 x 3.163 gives 10.00
continuing
Now do pi, of 3.141592
First we do 10 Grid is 22 x 1/7, which is 22 x .142 is 3.1
then we do 3.14 is 22 x .1428 is 3.14
Pi is 22/7 = 3.14 in 100 Grid
333/106 = 3.1415 in 10^4 Grid
355/113 = 3.141592 in 10^6 Grid
So, what I think is happening is that I have the definition correct-- TWO DIFFERENT RATIONALS acting as one number, and I have the multiplication for algebraic and the division for transcendental, but what I am missing is the inclusion of the sequence or progression of division.
AP
Archimedes Plutonium
Newsgroups: sci.math
Date: Wed, 23 Aug 2017 00:49:25 -0700 (PDT)
Subject: Re: explaining 22*1/7 Re: There are two irrational classifications--
algebraic, transcendental, but, the Gelfond-Schneider input is a fakery
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Wed, 23 Aug 2017 07:49:25 +0000
Re: explaining 22*1/7 Re: There are two irrational classifications-- algebraic, transcendental, but, the Gelfond-Schneider input is a fakery
So, does this get me out of my troubles completely, that the irrational is a sequence of Rationals
Algebraic irrational as sequence of multiplication
For example sqrt2 = 1.41*1.42, 1.414*1.415, ....
Transcendental irrational as sequence of division
For example pi = 22/7, 333/106, 355/113, .....
So, what is different now, than before? Before i was harping and harping two different Rationals acting as one number. Here, to satisfy the Gelfond-Schneider, i am saying an entire Sequence of Rationals is forming a new number.
Definition of Rational -- two counting numbers in a ratio
Definition of Algebraic Irrational -- sequence of Rationals in multiplication
Definition of Transcendental Irrational -- sequence of Rationals in division
Now, does that solve everything on the issue of irrational?
It surely solves Gelfond-Schneider error of A^B, for that is still algebraic irrational
Now my definitions above reflect and remind one of the Cauchy sequence definition of number, and shows them superior in logic to the latter day Dedekind Cut of number.
Let me sleep on this,,,,
AP
Newsgroups: sci.math
Date: Wed, 23 Aug 2017 01:54:18 -0700 (PDT)
Subject: Re: explaining 22*1/7 Re: There are two irrational classifications--
algebraic, transcendental, but, the Gelfond-Schneider input is a fakery
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Wed, 23 Aug 2017 08:54:18 +0000
Re: explaining 22*1/7 Re: There are two irrational classifications-- algebraic, transcendental, but, the Gelfond-Schneider input is a fakery
Alright a new concept in Irrational number theory. I am going to call it the Waterline, until i find something better. I need it to explain the mistake of Gelfond- Schneider.
The waterline for sqrt2 is 1.41421356...
Concept of Bob-up, bob-down for the true sqrt2
Bobdown 1.41, bobup 1.42, then bobdown 1.414, bobup 1.415, etc
Now we review Gelfond-Schneider error thinking 2^sqrt2 was transcendental when in truth it was algebraic irrational
Bobdown 2^1.41 = 2.6573716, bobup 2^1.42 = 2.6758551
So, the Waterline is between those two bobs,,,
Will pick up tomorrow to show that the number that represents 2^sqrt2 is a Algebraic Irrational number.
AP