changing coarse-- there maybe a classification of Transcendental Irrational, after all, since no proof is forthcoming,,,,

[Archimedes Plutonium]

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

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

[Me]

On Tuesday, August 22, 2017 at 12:08:37 AM UTC+2, Archimedes Plutonium wrote:

> I think [...] the Gelfond-Schneider has a flaw in it

Sure, Archie!

[j4n bur53]

Let you inject nonsense in your nonsense.
What can happen? Square nonsense?

Algebraic numbers are closed over +, -, *, /.
https://en.wikipedia.org/wiki/Resultant#Number_theory

You don't get transcendental numbers by division.

Archimedes Plutonium schrieb:

[Me]

On Tuesday, August 22, 2017 at 12:08:37 AM UTC+2, Archimedes Plutonium wrote:

> So, sqrt2 is Irrational, but why is it irrational?

Because there are no two integers p,q such that sqrt(2) = p/q. (Knowing the definition of /rational/ and /irrational/ numbers is helpful here.)

See https://www.math.utah.edu/~pa/math/q1.gif
for a proof.

Note though that it is not that obvious that there actually ARE transcendental numbers!

"Joseph Liouville first proved the existence of transcendental numbers in 1844." (Wikipedia)

Later Cantor could prove that the set of transcendental numbers is uncountably infinite.

In fact, "[t]he algebraic numbers are spotted over the plane like stars against a black sky; the dense blackness is the firmament of the transcendentals." (E. T. Bell, Men of Mathematics)

[Me]

On Tuesday, August 22, 2017 at 12:08:37 AM UTC+2, Archimedes Plutonium wrote:

> One fun I have found in math proofs, is that they seem to all have different
> flavor of proof, different character each.

Right. So where's your *proof* that certain conic sections are ovals but not (never) ellipses, Archie?

Actually, it's not that obvious (see)!
https://etc.usf.edu/clipart/76100/76194/76194_conic_ellips_lg.gif
http://2.bp.blogspot.com/-t3AuajLcGBo/UUT_kq_ArzI/AAAAAAAAAE4/gZgYZMSL0Is/s1600/Cone_w_Ell_Section.PNG
http://www.mathedpage.org/conics/three-d/cone-1_html.png

[Archimedes Plutonium]

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,,,,,

[Archimedes Plutonium]

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]

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

[Archimedes Plutonium]

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 Adsurdum 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]

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]

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

[Archimedes Plutonium]

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

[j4n bur53]

If a is a root of the polynomial

P(x) = a0 + ... + an*x^n

Then 1/a is a root of the polynomial

Q(x) = a0*x^n + .. + an

The later polynomial is just taken the former polynomial
and reversing the coefficient. We also have:

x^n*P(1/x) = Q(x)

Example:

sqrt(2) is root of 0 = x^2 - 2 = a2*x^2 + a1*x + a0
where (a2,a1,a0)=(1,0,-2)

1/sqrt(2) is root of 0 = 1 - 2*x^2 = b0 + b1*x + b2*x^2
where (b2,b1,b0)=(-2,0,1)

reverse((a2,a1,a0))=reverse((1,0,-2))=(-2,0,1)=(b2,b1,b0)

See also:
"Combined with the fact that 1/y is a root of ..."
https://en.wikipedia.org/wiki/Resultant#Number_theory

Archimedes Plutonium schrieb:

[Archimedes Plutonium]

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 conlusions 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

[burs...@gmail.com]

Profound confusion. Yes, polynoms are not closed
over division. For example there is no polynomial

in x, for the value:

1/(1+x)

Although there is a series:

1/(1+x) = 1 - x + x^2 - x^3 + - ...

Polynomials have to have a finite degree. But there
is a vexing property, although polynomials only

form a ring, and not a field, nevertheless the
algebraic numbers (roots of rational polynomials)

form a field. See here:
https://en.wikipedia.org/wiki/Resultant#Number_theory

[burs...@gmail.com]

Polynomial Ring:
https://en.wikipedia.org/wiki/Ring_%28mathematics%29

Algebraic Number Field:
https://en.wikipedia.org/wiki/Field_%28mathematics%29