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Topic: Harvard professors endorse proof of Fundamental Theorem of Algebra to arxiv
Replies: 9   Last Post: Sep 17, 2014 2:48 PM

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Registered: 3/31/08
Harvard professors endorse proof of Fundamental Theorem of Algebra to arxiv
Posted: Aug 18, 2014 12:03 AM
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Alright, I chose Harvard math professors to endorse the worlds first true proof of Fundamental Theorem of Algebra, because Harvard had Charles Sanders Pierce, a logician, but that Harvard seems vacant and vacuous of a logical person since Pierce. The Fundamental Theorem of Algebra requires a person of high logical order as seen by the proof below, and that the USA after Pierce had no-one of high logical order, but rather had illogical minds that kept muddying up the waters of mathematics. So Harvard can address the fact that its mathematics department needs shoring up or bulwarking of Logic.

Alright, let me continually improve and revise these Proofs.
Naturals Completeness Proof

We define Completeness by its most primitive Algebraic structure and that comes from the Axioms of the Naturals circa 1889 with Dedekind-Peano, specifically the Successor axiom with the Mathematical Induction axiom.

The Naturals are the set 0, 1, 2, 3, 4, 5, . . .

So, Completeness and being able to have Mathematical Induction on a set of numbers is identical in these proofs. In Old Math, they resorted to the opinion that completeness meant continuity. Strange, very strange twist of illogic, to think that Completeness meant continuity, when those two concepts are utterly foreign to one another. And, if anyone had asked those old rascals what does continuity have to do with the Naturals and Rationals and yet those two sets are Complete, would have realized how illogical it is for a mathematician to think that a pursuit of continuity means completeness.

Now in the proof of Fundamental Theorem of Algebra, I am going to equate Completeness with the concept of Algebraically-Closed.

So the proof for Naturals is very short and sweet in that it is just to restate those axioms of Successor with Mathematical Induction. But we have to include a detail of something that is more general than the "polynomial equation" used in the Old Math of the Fundamental Theorem of Algebra with complex numbers of D'Alembert in 1746 with Gauss in 1799 and also, indirectly with Dedekind cut for Reals in 1872.

Also, important is that the Naturals are axioms that require only the operator addition and needs no multiplication operator. When we include subtraction as an operator upon Naturals we get the Integers, and when we include division upon integers we get the Rationals. Finally, when we include multiplication and roots upon Rationals we get the Infinite-Numbers with their roots such as sqrt2 or sqrt-1 or sqrt-2 as examples.

So, what we do for Completeness proofs, going far beyond D'Alembert and Gauss who were stuck on polynomials and stuck on continuity is make the more general case of given the progression of Number sets:

1) Naturals from addition

2) Integers from subtraction on Naturals

3) Rationals from division on Integers

4) Infinite Numbers from multiplication and roots on Rationals

Now I used to write this proof discussing Reals and Complex numbers with imaginary "i". I no longer can do that because nearly all of that is phony math.

I have to discuss the Fundamental Theorem of Algebra from first beginnings-- the Naturals.

I have to also inject the infinity borderline upon those four listed sets above. So not only were the definitions illogical of completeness, algebraically-closed, continuity, but there was a problem of applying the infinity borderline to numbers.

The infinity borderline is 1*10^603 and it is derived from the fact that when pi has its first three zeroes in a row at floor-pi*10^603, the area of the tractrix equals the area of the associated circle for the first time, and since Huygens proved the area of tractrix equals associated circle at infinity, we have a borderline.

So that 1*10^603 is the last and largest finite number, and all numbers beyond are infinite-numbers and all numbers smaller than 1*10^-603 (except 0) are infinite-numbers since 1*10^-603 is the smallest nonzero positive finite number.

So, the proof that Naturals are a Complete set is that they have both Successor and Mathematical Induction. The Naturals have a Math-Inductor Element and it is the number 1, so given 0 and 1 adding 1 to 1 gives 2, adding 1 to 2 gives 3.

The next larger set is the Integers and we split them into the negative integers and 0 and the positive integers and we prove that Integers have a Math-inductor element of 1 and -1.

The next larger set on our list is the Rationals
Rationals Completeness Proof

To prove the Rationals form a Complete set in terms of Algebra, or algebraically closed set, we must show that a Successor function and a Mathematical Induction exists upon that set.

Here, the proof requires an Infinity borderline, which the Naturals did not require that.

We start with the Infinity borderline created by the Tractrix area equalling the corresponding circle area for the first time at 1*10^603 which gives us microinfinity to be 1*10^-603. That causes the Rationals to be only those points of this set in the first quadrant of the plane {0, 1*10^-603, 2*10^-603, . . , 1*10^603}. That set is 1*10^1206 members along the x-axis in 1st quadrant. The total number of coordinate points covered by the Rationals in 1st quadrant is a Grid of dots evenly spaced of 1*10^2412 points. And since there are 4 quadrants of these in full, there are 4*10^2412 members in the Euclidean Plane.

To prove Rationals are Complete or Algebraically closed, as these dots of finite points evenly spaced is to prove that this set has Mathematical Induction.

So, all I need to prove is that Rationals have a Successor with Mathematical Induction. Since the set of positive Rationals is
 {0, 1*10^-603, 2*10^-603, 3*10^-603,  . . , (1*10^603 -1), 1*10^603}

We see it has a Successor and Math Induction using the inductor element of
1*10^-603. All the Rationals are an induction of 1*10^-603. Same argument for the negative Rationals.

Fundamental Theorem of Algebra Completeness Proof or Algebraically Closed

First, let me note that the imaginary and complex number and the Complex Plane are just Infinite Numbers, just as the irrationals are also infinite-numbers such as sqrt-2 is just the conjugates {-1.4142, +1.4143} {1.4142, -1.4143} in pretend infinity border is 10^3 since 10^603 is too cumbersome to write.

So the Infinite Numbers include all the irrationals, the complex-numbers and the transcendental numbers and imaginary numbers. So in math, either a number is a Rational or a Infinite-Number.

Let me define Infinite Number-- an infinite number does not come as a single solo number as all the Rationals do, but rather, an Infinite Number comes in at least a pair of two numbers different from one another. For example the sqrt-1 is (-1,1) and the sqrt2 is the pair (1.4142, 1.4143) in pretend infinity border is 10^3 for 10^603 is too long to write.

Now if you look at the Cartesian Coordinate Plane it is just dots of finite points evenly spaced with mostly empty space between those dots, and empty space is the region occupied by infinite-numbers.

So in Pretend Infinity to be 10^3 because it is cumbersome for 10^603, that the Rationals would have points of 1.414 while lying just next to 1.414 are the Infinite-numbers of 1.4142 and 1.4143, what used to be called sqrt2 in Old Math as irrational-number and complex number of sqrt-2, are in New Math all just simply Infinite-numbers. The sqrt-1 is the infinite-number (-1,1).

Now all the numbers beyond the infinity borderline of 1*10^603 are infinite-numbers but what is their pair partner. For instance 2*10^603, and what would be its different number that forms a pair? The inverse of 2*10^603 is 1/2*10^603 and it is the pair (2*10^603, .5*10^603).

So, now a proof that FTA is Complete that given any polynomial has solutions, is too narrow, because we needed no polynomial test to show Naturals and Rationals are complete and algebraically closed, so why put that test to Infinite-numbers? The only real test is whether the Infinite Numbers have mathematical induction and thus an inductor-element.

An example will show what I mean.

Sqrt-10 in pretend infinity of 10^3 is -3.1623 with 3.1624 (or, alternatively is 3.1623 with -3.1624) which gives us -10.000, but to get -10.0000 we need to jump to a higher infinity of 10^4 Grid where we have -3.16228 x 3.16229.

So as we cannot get -10 with four zeroes in 10^3 Grid we jump to a higher Grid by Mathematical Induction.

So, is there a Successor and a Mathematical Induction upon all these numbers of Infinite numbers? Yes, because the 10^604 Grid, then the 10^605 Grid etc etc have their own inductors of 1*10^-604 and 1*10^-605 etc etc.

Now, the question comes, is there any higher set than the Infinite Numbers which in Old Math would be the irrationals and Complex-numbers together, or does Completeness stop with the Infinite Numbers? That is like asking is the Euclidean Plane full when it has Rationals and Infinite-Numbers? Is there anything more to the plane once we have dots of Rationals with their empty space in between the dots as Infinite-Numbers? The answer is the plane is full and there is no higher set than the Infinite Numbers.

Alright, let me summarize.

When we have an infinity borderline between finite and infinite. The finite numbers compose Grid Systems of Rational numbers with a macroinfinity and a microinfinity. So that all the numbers that exist for mathematics proper are these:

0, 1*10^-603, 2*10^-603, 3*10^-603, . . . 1*10^603, along with negative Rationals

That is a set of 1*10^1206 numbers for the positive x-axis and the Grid for the 1st quadrant is 1*10^2412  coordinate points. Those are all discrete numbers and coordinate points. They have gaps or holes in between successor points. Those holes are filled with infinite-numbers. Infinite Numbers come as pairs of two different numbers, unlike rationals which are single solo numbers. For example, in pretend infinity is 10^3 rather than actual 10^603 since it is too long to write a number out. The sqrt2 is the conjugate pair of 1.4142 with 1.4143 which when multiplied in 10^3 Grid is 2.000.

Now, is there a proof that Old Math Reals are a fake number system and that the Combo Number System of New Math of Rationals unioned with Infinite Numbers is the true number system? Well, yes, there is a easy proof, in that the Iterative Roots Function where you keep taking the square root of a positive number other than 0, and it always converges to 1. If the number is larger than 1 it converges down to 1, and if the number is between 0 and 1, it converges up to becoming 1.

So, how does that function prove Old Math Reals are fake and that New Math Combo is true? Because as you apply successive roots to numbers to converge, once that root is an irrational root, it remains irrational forever in converging to 1, for it never switches back to a rational number root. That means irrationals must have a Solid-Block feature. New Math can prove that the Iterative Roots once it is irrational remains irrational. However, Old Math has an impossible conjecture for them to prove using their Reals. In Old Math, you cannot prove the conjecture that once the root is irrational, it is forever irrational thereafter. New Math easily proves it-- in a sentence-- because irrationals are two different numbers and a further root would have 4, then 8 then 16 etc etc of irrational conjugates. Whereas in Old Math Reals, those people actually believed a irrational number was a single solo number, and for that belief are never able to prove the conjecture.
Another proof comes from Continued Fractions representation of numbers, in the idea that all irrationals repeat in digits, for example sqrt2 = [1; 2, 2, 2, 2, ...] and sqrt3 = [1; 1, 2, 1, 2, . . .]

So, Continued Fractions are two numbers together that compose one single number, for sqrt 2 the 1 and the number [2, 2, 2, 2, . . .] for sqrt3 the number 1 then the number [1,2,1, 2, 1,2. . .]

When the Continued Fraction gets to phi it is a perfect irrational number as [1; 1, 1, 1, 1, . . .] and that would be obvious in the fact it is (1+sqrt5)/2.


Recently I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of front page hogs, mockers and hatemongers.!forum/plutonium-atom-universe        
Archimedes Plutonium

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