**RIGOROUS** FUNDAMENTAL THEOREM OF ALGEBRA

AT last he has done it. ONE guy saves all pure mathematics

Other urls found in this thread:

ncatlab.org/nlab/show/intermediate value theorem
youtube.com/watch?v=QSZsTeO-C1o
youtube.com/watch?v=qoSGtoy7DlI
twitter.com/AnonBabble

...

Watching that video yesterday made me very hopeful for the future of mathematics. In the past I have doubted if we can really do mathematics without reals but this just does it. This is how you replace reals with rationals.

And think about it for a second. Real numbers are defined as cauchy sequences of rational numbers (or as cuts which are very similar) but then through logic we can prove that basically lines are perfectly continuous and that means we have to be less rigorous in the future. We can simply call a number "root 2" and we know it is there.

Wildberger is making this more rigorous. Instead of making a set of cauchy sequences (reals), what he is asking is that in any problem you literally prove that such a sequence exist.

Think about it. You have a never ending family of [math] w_1 , w_2 , ... w_i , ... [/math] rational numbers. That never ending family is a cauchy sequence! And then [math] p ( w_i ) [/math] is a cauchy sequence which convergence to 0.

Wildberger did it! In the past cauchy sequences were a brainlet tool to make problems easier and that is where the problem with reals come from. When you say "lol there are just reals everywhere, don't worry about it ;^)" then you ignore a lot of the complications with real numbers but here Wildberger is very explicit. To solve a polynomial with a real root you must literally find the cauchy sequence and be very explicit about it and its quadrance.

It is simply amazing. I now definitely think this man can rewrite analysis in terms of rational numbers. It is amazing.

he just said that he is working on a calculus course.

No shit, that's word for word equivalent to having a complex root. If you admit the usual FTA, then you can find a cauchy sequence w_i of rational complex numbers that satisfies this and conversely, this form implies FTA due to the Bolzano-Weierstrass theorem.
I gave this exact formulation to wild in the comments of his last video, explained how it was equivalent and how real numbers are literally just a shorthand for statements about rationals and he completely dodged the argument.

Proofs of the FTA tend to use real analysis which is not properly formulated with rational numbers.

You do realize that the reason real numbers are defined the way they are is to prove certain properties that then make it really easy to make big statements about numbers? Proofs of the intermediate value theorem (need to prove the FTA) does not talk about explicit sequences, just uses the properties of the reals.

What analysts prove today is that: If we have a perfectly continuous set then it has these properties.

What wildberger wants: If we have the rational numbers and the freedom to manipulate them into various objects then we can have these properties.

>real analysis which is not properly formulated with rational numbers.
As I said, any statement about real numbers can be translated into an equivalent statement about rationals. Here's an example with the intermediate value theorem.
An equivalent form of the intermediate value theorem is: if f is a continuous function on a segment [a,b] with rational extremities, in the sense that for each (x_n), (y_n), Cauchy sequence of rationals in [a,b] such that (y_n - x_n) converges to 0, (f(x_n)) and (f(y_n)) are Cauchy and (f(x_n) - f(y_n)) converges to 0, then for each rational c between f(a), f(b) and N > 0, we can find a rational r in [a,b] s.t. |f(r) - c| < 1/N.
Now, to prove this, we can do the exact proof by dichotomy from your calculus textbook:
Assume f(a) < c < f(b):
Let [math]M > 0[/math] a positive integer.
Set [math]a_0 = a, b_0 = b[/math] and define [math]a_{n+1} = \frac{a_n+b_n}{2}, b_{n+1} = b_n[/math] if [math]f((a_n+b_n)/2) \le c[/math] and [math]a_{n+1} = a_n, b_{n+1} = \frac{a_n+b_n}{2}[/math] otherwise.
Now, (a_n) and (b_n) are Cauchy sequences of rationals since they are monotone and bounded (fun exercise since you can't talk about the sup or the inf) and (b_n - a_n) converges to 0 and hence (f(a_n)) and (f(b_n)) are also Cauchy and their difference converges to 0.
But [math]0 \le f(c) - f(a_n) \le f(b_n) - f(a_n)[/math], so there is an n such that [math]f(b_n) - f(a_n) < \frac{1}{M}[/math], qed.

Now you're going to say "b-but you assumed the function was continuous !" but IVT does require some assumptions but this sure as hell is satisfied by polynomials, so please stop with the silliness.
Let him have his fun but don't let him pollute your mind with his nonsense.

IVT requires some choice
ncatlab.org/nlab/show/intermediate value theorem

I'm not sure what kind of analysis you think you're going to do without dependent choice or at least countable choice

Veeky Forums educate me what is this about ?
where was the problem and why did he solve it ?

It's about the fundamental theorem of algebra, which states that every polynomial with complex coefficients has a complex root.
He disagrees with this statement on the grounds of it not being constructive (which is true, although his argument is fallacious):
youtube.com/watch?v=QSZsTeO-C1o
Now, he has found an equivalent formulation of the theorem that is purely in terms of rationals:
youtube.com/watch?v=qoSGtoy7DlI

thanks man. But i guess that is too much maths for me, cause i dont even know what the fundamental theorem is for and what it is implying i cant get myself to care about what the guy is talking about.

Literally just the same as FTA without limits. It implies the same shit, but is less far reaching and useful. Congratulations.

Literally all he's done is take a limit but deny that the limit actually approaches a value

Basically the fundamental theorem just tells you that any equation of degree n, ie. of the form [math]x^n + a_{n-1}x^{n-1} + \dots + a_0 = 0[/math] has a solution in complex numbers (ie. you can find a complex x that satisfies the equation).
It's a very important result and is the main reason why we even care about complex numbers. It also has many important implications in many areas of math.
What he's claiming is that it's "wrong" because the theorem and its proof don't say anything about what the solutions should look like, or how to find them, but give theoretical arguments as to why they have to exist.

"You have a never ending family of w..."

if that's true then Wildberger has already contradicted himself.

No, you dummy. Me saying never ending was just for simplicity. Notice that in this statement he says that k is an element of Nat.

So for every k in nat, there needs to be a w.

So the family of w is actually just about 10^200 long. As after 10^200 there are no more k to find w's for.

>RatCom
>Nat
>'+' signs are indistinguishable from 't's

REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

>RatCom
Rational complex numbers

>Nat
Primitive natural numbers

Please take your cult jargon to

It is not cult jargon. It is enlightened jargon.

boiunp

Enlightened by your own intelligence?

Just image, if mathematicians had been complete morons they might have ended up defining "Nat"
without bounding it above. Then k could become arbitrarily distant from 0, 1/k could become
arbitrarily close to 0, and this completely intuitive theorem, perverted by what those poor fools
might have refered to as "infinity", would have state that all -- yes, all -- polynomials with rational complex
coefficients have roots among the rational complex numbers; an obviously false result.

Praise Wildberger.

so we are allowed to invent a solution to x^2 + 1 = 0

but x^2 - 2 = 0 is off limits?

Because [math] \mathbb{Q}^2 = \overline{\mathbb{C}} [/math]

Well, i is not an infinite decimal. It is just an algebraic constant.

Square root of 2 could also be just an algebraic constant... but you faggots didn't want to leave it as that. You started saying: No, look at this. It is actually 1.41...

NO! It isn't.

Root 2 exists in the 2 dimensional Rational extended (root 2) plane. It does not exist in the number line. That is unrigorous bullshit.

what if I told you that by defining a series of standard operators, and then taking their reverse functions, you could reach all real numbers, given no upper bound on the order of operator.
Take integers, these are found by inverse addition (1)
Rationals, these are found by inverse multiplication (2)
Irrationals, these are found by inverse exponentiation (3)
Some unnamed set, found by inverse tetration (4)

This is unrigorous because it assumes rational numbers exist.

Rational numbers can be constructed, taking assumption out of it

>refuse reals
>accept complex numbers
I won't never get this guy, reals, even if some of them can be constructed with just straight edge and compass don't exist, but i, which has no physical representation, does not exist in real world exists and is perfectly valid an object?

Read

He just accepts algebraic numbers, be them "real" or complex

If he rejects reals, what does he think the ratio between a circle circumference and its radius is? Or the ratio between a square diagonal and its side?

Integers are group-ified naturals
Rationals are field-ified integers
Reals are Cauchy complete rationals
So yeah, you're basically right.

I've never watched Wildberger, but in, say, intuitionism you can approximate it as precisely as you want. So I guess he has somewhat similar results, because he, from what I've seen in these threads, practices an extreme form of constructivism.

Oh shit, that's interesting. What's this construction of the reals called? Or rather, how do you prove this is an equivalent construction to the other versions?

Interesting. So in effect, then, he doesn't actually believe in circles or square diagonals.

Only algebraic numbers can be constructed like this. Transcendentals like e and pi can not.

So how would something like measure theory be built with Wildberger's system if every set is countable?

Existence of circles is counterintuitive, so your question is logically incoherent

\frac{a}{b}

[math]\frac{a}{b-1}[/math]

you didnt prove all polynomials are continuous!

>True Fundamental Theorem of Algebra
He is just taking the fucking Fundemental theorem of Algebra, and elaborating "complex number" with "Cauchy sequence of rational complex numbers." There is literally no difference.

MODS
MODS
MODS

this meme must stop

They do not exist. Find me a perfect circle in nature. A polygon with [math]\displaystyle 10^{10^{10^{\vdots }}}[/math] sides does not count.

>perfect
what does perfect mean?

we all know that a circle is defined by a point and a distance : it is the set of points which are at this distance from this point.
this definition is perfect in some way.

what do you mean by perfect?

>what do you mean by perfect?
Precisely your definition. Not all points on a polygon with 1 000 000 sides are equidistant from the polygon's centre.

He just hates irrational numbers and uses rationals to approximate. Historically, irrational numbers were derived from rational numbers

i cant wait to see his publication of this discovery get peer reviewed

oh wait

I would peer review it

What privileges a "heap notation" (such as the theoretical counting of a bunch of objects, like planck volumes, side by side) over a "codified notation" that expresses amounts via predefined operational symbols (such as exponential notation)? Why should the former define a hard limit on mathematics? Both are abstract representations of an abstract concept (a single object is *not* [the number 1] just like all objects in existence are *not* [the biggest number]).

>DURR what are transcendental numbers???

>Why should the former define a hard limit on mathematics?

I don't know, ask him. He is usually active on his comment sections in the minutes after he publishes a video. That is why I am always looking out and try to post the first comment telling him that I love him. He never replies to my comments though as I just tell him I want him sexually. He answers to people who make questions or try to debunk his theory.

So ask him. Subscribe and then wait for him to upload, then post this question.

>Irrationals, these are found by inverse exponentiation (3)

It is funny but that comment validates Wildberger's view.

One of his biggest claims is that most people who work closely with real numbers don't know shit about them. And when these threads are made and some retard tries to argue for why real numbers are "so obvious lol" it shows. It shows a lot.

He said in a video that he believes pi is a meta-number which needs further mathematical development to explain fully

> Or the ratio between a square diagonal and its side?
2
Lengths don't exist. only quadrature
> what does he think the ratio between a circle circumference and its radius is?
circles don't belong in trig.
They're continuous curves, which requires at least analysis to understand properly

>Lenght don't exist. Only quadrature
Why? Because wildberger says so?

Because you have to call on the powers of analysis to find that length, and you really don't need to. You can just use some simple algebra and geometry if you use quadrature.

*sigh* Hello. My name is Simon.

BR Ratio of 0.6

Would you like further assistance? You can call me.

You can get transcendental numbers like this, you just can't get all of them. Basic cardinality issues dictate that you will obtain almost none of the real numbers by doing that process.

No, you fucking can't.

If irrational numbers are obtained via "inverse exponentiation" of rational numbers then guess what, all those numbers are algebraic.

Pretty sure any x such that x^x=2 is transcendental... do you even understand what is being discussed?

>x^x

Retard.

By that logic then when that guy said "inverse multiplication" then you could say that this would generate numbers such that x*x = 2, topkek.

You are definitely retarded.

>Some unnamed set, found by inverse tetration (4)

This is some pretty low level trolling friend.

I don't give a shit about that. It is said explicitly
>Irrationals, these are found by inverse exponentiation (3)

Not all irrationals can be found through fucking roots. You trying to damage control this shit is pathetic.

I didn't write that, although I have had the same idea myself before. That poster used the word 'irrational' incorrectly, but do you really care more about that than the interesting set of numbers they are pointing to?

>Things autists care about.

This is just your opinion. There are analyses without countable choice.

Do you have any example in mind ?
I get why people wouldn't want to use the full axiom of choice and, in most situations, you can work without it.
Doing analysis without being able to construct sequences is a different business.

inverse tetration is still algebraic numbers, since you are just inverse exponentiating a bunch of times

>without being able to construct sequences
You can construct sequences without countable choice. Are you dumb?

Well duh, that's not what I meant. What I mean is that in proofs, you're often lead to do things like "pick a_0 satisfying such and such, then assuming a_0,...a_n are constructed, pick a_n satisfying such and such" (for example think of the proof of Baire's theorem or Bolzano-Weierstrass, or Heine-Borel, or basically anything in your topology class about metric spaces) which, to actually define a sequence, require dependent choice.
Seeing how omnipresent things like this become makes me doubt that we can just shun countable choice.

my intelligence provides me with euphoria

do you not fucking understand that THE REALS ARE ALREADY CONSTRUCTED FROM THE RATIONALS
have you never heard the phrase "equivalence class of cauchy sequences"
fucking hell

this
is
already
how
analysis
is
done

read a goddamn textbook

What are you talking about? In Analysis quadrance doesn't even exist. Such a concept could only be conceived by the Wild man himself.

Have you ever heard of the phrase: Infinity is not fucking rigorous.

Not only are traditional cauchy sequences infinite, but equivalence classes of cauchy sequence based on convergence ARE FUCKING INFITE TOO.

Double infinity

FUCK YOU.

But the Wildman has it all solved. In his version of cauchy sequences that I talk about in , it is clear that his sequences are finite. With less than 10^200 elements.

He has made analysis FINITE. You got FUCKED. We don't even need infinity anymore, YOU GOT BTFO. GET OUT OF MATHEMATICS. YOU GOT REK'D. OUT. THE FUCK OUT. GET THE FUCK OUT. WE WON.

[math]
f : \mathbb N \to \mathbb N\\
f(n) = n^2
[/math]

oops i accidentally defined an infinite sequence in finite space, guess i broke the universe =(

if you wanna be an obnoxious constructivist who DOES know what they're talking about, go learn type theory. it's fun.

analysis has ALWAYS been finite since logicism ate mathematics, you idiot. it's traditionally formalized over ZFC, which is a system for manipulating finite strings of symbols, whose INTUITIVE INTERPRETATION deals with "infinite" objects

It's always funny how people call wildberger rigorous, have you watched his video about naturals, where he defines them as strokes on board, and his rigorous definition of addition is "just add them together lol"?

>oops i accidentally defined an infinite sequence in finite space, guess i broke the universe =(

Actually not. That is a finite sequence. The last term is the biggest perfect square smaller than 10^200

literally it doesn't matter whether you think there are "finitely many" "natural numbers" "in real life", the rules of set theory work exactly the same no matter how you decide to interpret what they "mean". if I define a sequence by [math]a_n = 2n[/math], it DOESN'T FUCKING MATTER whether I think the domain "is finite", the rules LET ME PROVE THE SAME THINGS.

I will if Wildberger starts a series on it. Else I see no point in wasting my time with unrigorous garbage.

then tell me what it is. if you're so obsessed with things that you can write down instead of "imaginary ideal things that SHOULD exist because it's intuitive", tell me what the last term of that sequence is. thanks in advance

In one of his videos he, when solving some problem, got the answer with quadrance involving sqrt(6), so his theory is as shitty as the regular one

OH, GLADLY! Thank you for asking.

Every number of the form 10^2n where n is a primitive natural is a perfect square.

Therefore, 10^200 is that perfect square. 10^200 IS THE LAST ELEMENT OF THAT SEQUENCE.

You badly misunderstand the AC. AC is required whenever there is no canonical choice of something. There are numerous ways to make canonical choices work whence no AC is required.

I'd very much like to see you prove Baire's theorem without dependent choice

That doesn't relate to our discussion. You said "you can't define muh sequences without ACC", I answered in details why you were wrong. You see you have no arguments so you asked an allegedly counter-question which doesn't make sense in context of the discussion. I didn't claim that every classical theorem can be translated into constructive framework without some choice axioms.

It does, it was one of the examples I used in my previous post.
And again, I didn't say you needed AC to define ANY sequence at all (I have stressed that already) but that, in a number of situations where you have to construct sequences *abstractly*, you have to use some sort of countable choice (again, Baire's theorem or Bolzano-Weierstrass which are fundamental theorems of real analysis).

That's not like constructive analysis is derived. There is no meaningful counterpart of Bolzano-Weierstrass' theorem in constructive analysis. In fact, it's a meaningless argument -- it's computational content is vacuous. If you interested in specific derivations -- look at the literature. If you didn't heat about it -- traying to project classical arguments directly onto constructive setting is silly.

what about 10^400 user?

>10^400

I am sorry but I am certain that such an expression does not evaluate to a primitive natural number, therefore we cannot do arithmetic with it.

>a primitive natural number
What does that even mean?

Primitive naturals are airthmetic numbers.

They are the algebraic structure Papa Wildberger invented to describe naturals numbers. Theoretically it can be described by 2 things: A container and something to count within that container. In paper the number 3 is usually denoted as [III].

If you say this is unrigorous then please watch the video where he lays out his axioms and operations. It is very, very clear.

From this definition you get two things: If you have a bigger container, you can make more marks.. And if you make smaller marks then you have more room to make marks. So what is the biggest number in the universe?

Simple. Take the biggest container in the universe... the universe itself. And take the smallest possible length to make your mark. If you do this then you can make a primitive natural that evaluates to 10^200.

This means that no bigger numbers exist in our universe and therefore it makes no sense to make arithmetic with those numbers, that Papa Wild calls "dark numbers".

That seems a little ridiculous considering we've already factored RSA-768 which is 232 digits long. That's bigger than 10^200