Why There's 'No' Quintic Formula (proof without Galois theory)

[@alecbg919]

Great video! Beautiful idea explained very well. I was going to ask if there were functions that worked for the quintic, but Google has already told me that elliptic functions do the trick

[@notallwrong]

Thanks! Another, more minimal way to go about this is to use the fact there's a solution to the quartic to convert the general quintic to the form z^5 + z + t = 0. The (very multivalued!) solution to this is a function z = B(t) called the Bring radical, and introducing this single function to your set of allowed ingredients lets you solve a general quintic. A more fun way to look at this is to 'rearrange' the above equation as z = - (t+z)^(1/5) and then iterate to write z = - (t-(t-(t-... ^(1/5))^(1/5))^(1/5)) as an **infinitely** nested root! Given an appropriate starting guess for z, this even defines a convergent approximation for a root of B(t) for many values of t.

[@alexk6942]

@@notallwrong but why doesn't the same logic you applied to radicals in the video not apply to the Bring radical?

[@notallwrong]

The short answer is: because it has 4 branch points, not 2. Let me explain a little. The argument in the video relies entirely on one property of the usual radical: if the number of times the thing inside the radical winds around the origin is zero, the value of the radical returns to where it started. Technically, the radical has multiple *branches* but only at zero (and technically at infinity too) is there is a *branch point*, a place where encircling that point lets the function change onto a different branch. This means that commutator paths leave radicals unchanged because they automatically have no winding. But the Bring radical has four such points! Taking t around any of these 4 points lets you switch branches and change to one of the other possible values the function might have when you return. Now it isn't as simple as counting how many times you encircle the single finite branch point. You have to worry not only about how many times you encircle multiple branch points, but also about *in what order* you do so. Going around branch point A then branch point B then going back around A and back around B does *not* necessarily return you to your starting value for such a function. This means that even a commutator path might tie you in knots around multiple branch points! And that is exactly what happens with the Bring radical.

[@henrikljungstrand2036]

@@notallwrong Where do these 4 branch points lie for the standard Bring radical? Zero, Infinity, One, Minus One?

[@notallwrong]

The branch points are where the solutions of the quintic coincide. It depends exactly how you scale the parameters in the definition, but up to an overall constant they should be 1, - 1, i, - i. You can find the exact expression by looking up the discriminant of a quintic equation and looking for where that vanishes as a function of t.

[@adsoyad2607]

16:34 "Algebraic expressions are limited in how multi-valued they can be." This sentence was so eye-opening for me

[@joaofrancisco8864]

Exactly!!!

[@zeferreira8805]

@@joaofrancisco8864 Mesmo foda-se

[@Zxv975]

You might be interested in looking up "algebraic" and "transcendental" branch points if you want to develop this idea further. This the technical way in which the multivaluedness of the square root function (which has an associated a winding number of 2 around the origin, meaning you need 2 copies of the complex plane to be glued together to form the relevant Riemann surface) differs from, say, the complex logarithm (which requires infinite copies of the complex plane).

[@adsoyad2607]

@@Zxv975 will do!

[@skippyXG]

Same here 😊

[@jacoblojewski8729]

It wasn't until watching this that I realized why some books/texts etc. prefer to state the Fundamental Theorem of Algebra as "Every polynomial (with complex coefficients) with degree at least 1 has at least one root (in the complex numbers)". Makes it easier to not have to discuss the counting of repeated roots.

[@ingiford175]

And this shows that the complex numbers are enough to solve every (finite) polynomial. That means there is no polynomial where you had to go outside the comblex (to say a quaternion world) to find a solution . One thing I had trouble with when trying to prove FT of Algebra is how do you show that the complex was 'enough', that you did not need to add other axis to solve larger roots say a polynomial with n = 64. The 'continuous' near circles for given large magnitude of complex numbers gives a near circle that has the origin inside, and shrinking this magnitude to 0 (which lets gets you to the point a_0) at 'some time' you have to cross the origin is proof i find most often. But that requires continious transforamtions , etc that must be well defined before hand.

[@henrikljungstrand2036]

@@ingiford175 This is only true when working with the real numbers as base field though. Starting with the rational numbers instead, the algebraic conpletion of that field is countably infinite dimensional, and there are infinite fields of any dimension inbetween, extending the rational numbers. If we desire some notion of continuity, which rational numbers lack, then for each prime p, the p-adic numbers works similarly when extending this field with roots to algebraic equations, and possibly taking the metric closure, i.e. we get lots of finite dimensional extensions and an infinite dimensional closure. Finite fields (with p^n elements), work similarly to the rational numbers and their finite dimensional extensions, and their algebraic closure is also infinite dimensional. The field of functions in one formal variable over a finite field is somewhat similar to the p-adic numbers, in that it admits a notion of continuity, though i am not sure about the possible dimensions of its extensions.

[@aditya-u7s9v]

Zero not a root

[@duckymomo7935]

Problem with using that definition requires the reader to understand the assumptions of knowing the complex number system which is iffy at a high school level and knowing that the largest algebraically closed field possible is the complex numbers themselves (proof) Q -> R infinitely many extensions R-> C only one C -> C trivial Most people are uncomfortable with i x^2 + sqrt(3)x + 2 having a solution in C

[@voidify3]

At least 1 and at most n where n is the degree

[@MayawireCH]

As someone who’s been watching 3B1B, Numberphile, Michael Penn, and the likes for years, I can confidently say that this is one of the best math videos I have ever watched. The argument is elegant and presented in a way that makes it very accessible, the animations are fantastic, and you can clearly see that you have a passion for this subject. Please keep making these videos if you can, it’s a gift to all the people out there without access to a course in abstract algebra that we can still try to learn and understand this.

[@神のユージン]

Mathologer where

[@nbme-answers]

Can you mention more KZbin math channels you enjoy? I will add Leios and some other favorites I've collected in this playlist: kzbin.info/aero/PLUFeA6y-5sFlrBflZw8UWz5wOI_D5ngbu Happy mathing everyone!

[@charansai2476]

Khan academy

[@duckymomo7935]

Have you seen mathemaniac or aleph 0

[@nbme-answers]

@@duckymomo7935 No, but will check them out now.. Thanks!

[@PaulMurrayCanberra]

My first thought was, how can there be a quadratic equation? If you just use the positive root, then this gives you a smooth, single-valued function for one of the solutions. But that's exactly what this is all about: it doesn't work for complex numbers. There isn't a way of choosing "the positive root" that's smooth, because complex numbers form a plane. I heard a while ago that points on a plane "couldn't be ordered" - didn't really know what that meant. This "flipping the roots" animation … I have only just now got it. An ordering (in that sense) means that the greater than/less than relation of two points stays the same when you smoothly move them. Only works on a line. Nice when things gell.

[@DonkoXI]

There is sort of a way to get around this. Imagine taking the complex plane and splitting it open along the positive real line. If you do this with two copies of the plane, you can glue them together along that tear. The idea is that, once you wrap back around to the positive real line, instead of going back to where you started, you transition on to this second copy. This way you're technically at a new point. Once you go back around again, it's glued back to the original so you land where you started, but it takes two rotations to get there. If you want a visual of what I'm describing, search for pictures of "Riemann Surface". The wikipedia page shows exactly what I'm talking about. You can do this to handle any nth root by spiraling around n times before completing the gluing. If you take this spiraled version of the complex plane, you can define a legit smooth function which takes points to a consistent choice of root. Going to different layers of the spiral gives you your other roots. From here, you can capture exactly what this tame multivalued behavior is. The layers of the spiral correspond to each of the roots, and performing the commutator not only brings you back to the same complex value, but also on the same layer that you started. The commutator changes the root, but the layer is the same, so there's a problem.

[@CarrotCakeMake]

Points on a plane can be ordered, look up the Hilbert curve. And the issue isn't smoothness or single-valuedness or anything like that, it is about being path connected.

[@helman-tov]

I watched this video earlier this week, but I came back to say it *blew my mind*. I have never even heard about the existence of this proof before, and now I can't believe it isn't more famous. This video is great! It's so clear and interesting. Thank you so much for making this video. I really hope you continue to create this kind of videos once in a while.

[@druid7456]

Outstanding video. An unusual approach to the subject, appealing to the imagination. Solid presentation. I had the pleasure to evaluate your work in the peer-review stage of SoME1. And it turned out to be the best of more than 20 shown to me by the system. We are, in a sense, rivals in this competition. But competing in such a company is an honour for me.

[@notallwrong]

Very kind words, thank you! I must say there is potential for a great community evolving from SoME1, looking at the Discord, and the many nice comments on KZbin. Screw rivalry, it's better off as a constructive community-building exercise that's encouraging people to get stuck in and try and make something - more hackathon than competition! (I was moving house so haven't had the chance to engage as much as I'd like with peer review, but hope to go through more contributions, yours included, when they are all shared.)

[@GDigora]

@@notallwrong I could not say better than Druid, so just say my thanks to you.It’s a wonderful video. Looking forward for more

[@Adam_42_01]

What a hidden gem this video is - Great pacing, explanations, and graphics. Thank you for making it

[@DanielKRui]

When the commutators started showing up and the commutators of commutators my brain was forming thoughts that my mouth wasn’t ready to put into words, and then when the terminology “derived subgroup” was revealed EVERYTHING clicked. Absolutely brilliant, making this very abstract math intuitive. It’s like the notion of calculable constants in physics I just read about today on Stack Exchange; the more fundamental or deep a theory, the more "standard" constants are explained from more fundamental and deeper constants. Derived series were always one of those “unexplainable constants”, but now with this deeper and more fundamental understanding, it is “merely” a corollary to a clever trick doing and undoing paths. And my god I just got to the proof of the non-solvability of all S_n or A_n for n>=5. Absolutely stellar. That trick of expressing sigma and tau themselves as a commutator of 3-cycles felt like a magician dropping the bottom out of a box forming a well of infinite regress...I can't believe such a "simple" observation shows that all derived subgroups of S_n or A_n for n>=5 must contain a 3-cycles. I wonder if there is a nice proof of the simplicity of the A_n for n>=5 given these facts (since all proofs I know of that are still have somewhat of a "coincidental" quality to them). In 45 minutes, you have drawn back the curtain on a semester (or perhaps even semesters!) of graduate level abstract algebra, and indeed I have waited for a video explaining this theorem/argument for a very long time (I did not understand any existing videos/papers trying to explain it); and for that, this will rank in my mind as one of the greatest math videos/expositions of all time!

[@qo7052]

:nerd_face:

[@SRangarajan]

Hi Daniel, there is a part in this video where he proves that a single nth root isn’t enough since the commutator ensures that no phase change is picked up. How does nesting solve this issue? Since nothing picks up a phase change right? Can you please help me with this? I have drank the Internet trying to find this answer but it’s not mentioned anywhere. Thanks a lot!

[@Fire_Axus]

your feelings are irrational

[@DanielKRui]

@@SRangarajan The video argues this at 29:39-31:10. The point is that we previously said that if we revisit the same point over and over again is a special way (in the complex plane) that allows us to do and then undo some phase changes, resulting in a total phase change of 0 (i.e. the root revisits its original value in the Riemann surface/multivalued graph). But when we have nested roots, we do this "revisiting procedure" on the coefficients, but that only makes the *inner* root revisit its original value once. So to get the outer root to revisit its original value, we have to do more of this "revisiting procedure" to make the inner root *also* revisit its original value multiple times. I think the video explained it better than I did, but I can try again if that didn't make any sense either.

[@johnchessant3012]

This is absolutely brilliant!! So glad this got featured in 3b1b's recap video. This really gave me a sense for why commutators are important. If I'm understanding correctly, for any rational function r in the quintic's coefficients, we have a homomorphism S_5 -> Z which describes the phase difference of r as we swap the roots; the commutators are exactly the elements that map to 0 since Z is abelian!

[@pedro3005]

I'd say it maps to U(1)

[@johnchessant3012]

@@pedro3005 Technically yes, though the value of r doesn't change as we swap the roots so the phase difference is an integer multiple of 2pi.

[@SRangarajan]

Hi, there is a part in this video where he proves that a single nth root isn’t enough since the commutator ensures that no phase change is picked up. How does nesting solve this issue? Since nothing picks up a phase change right? Can you please help me with this? I have drank the Internet trying to find this answer but it’s not mentioned anywhere. Thanks a lot!

[@alphansahin]

That's a great one! For me, the critical aspect was around 27:20 in the video (i.e., Exercise 1 in Leo Goldmakher's paper, "the image of f(p)^a for any rational a as p traverses the commutator loop [γ1, γ2] is a loop in C.") In a way, a commutator allows us to discuss the existence of a single-valued function that includes a root operation. Some MATLAB code on commutator for anyone interested: clear all close all clc if(1) L = 4; % quartic L = 5; % quintic actions = perms(1:L); for nestLevel = 1:4 Nperm = size(actions,1); com = zeros(Nperm^2,L); for j = 1:Nperm for k = 1:Nperm L1 = actions(j,:); L2 = actions(k,:); com(Nperm*(j-1)+(k-1)+1,:) = fcnComm(L1,L2); end end actions = unique(com,'rows'); size(actions,1) end end function Lcomm = fcnComm(L1,L2) [~,L1inv] = sort(L1,'ascend'); [~,L2inv] = sort(L2,'ascend'); Lcomm = L2inv(L1inv(L2(L1))); end

[@janszwyngel4820]

I think the best part of this proof, and the main advantage over the Galois theory approach, is the fact that it doesn't use field theory. The only assumptions about a hypothetical quintic formula are, as far as I can tell, that it's made from continuous functions of the coeffitients which can be single-valued or multi-valued, but walking along a path that doesn't loop around the origin can't change the value (also by shifting the domain this can be extended to looping around any point, but not many points at once), which is much more general than the four operations together with natural degree radicals that you get from the Galois theory. It's beautiful that such a great result can be motivated by arguments from continuity and the limitation on multi-valuedness of the functions used only. The presence of the seemingly arbitrary boundary of fifth degree only makes it better. Great argument, clear presentation, and one of the weirdest, unexpected results.

[@MatthijsvanDuin]

You could easily flip that around though: the main advantage of Galoid theory over this approach is that it doesn't require a notion of continuity (which is absent in plenty of applications of Galois theory). Neither approach is strictly more general than the other, they're just different. Having said that, the approach in this video is excellent for creating a deeper and more intuitive understanding of what's going on and deserves to be taught prior to Galois theory.

[@gadam4254]

galois theory itself was not always based on theory of fields. you can check some older french books on the subject like wantzel's or galois' own writings.

[@SRangarajan]

Hi, there is a part in this video where he proves that a single nth root isn’t enough since the commutator ensures that no phase change is picked up. How does nesting solve this issue? Since nothing picks up a phase change right? Can you please help me with this? I have drank the Internet trying to find this answer but it’s not mentioned anywhere. Thanks a lot!

[@SRangarajan]

@@MatthijsvanDuin Hi, there is a part in this video where he proves that a single nth root isn’t enough since the commutator ensures that no phase change is picked up. How does nesting solve this issue? Since nothing picks up a phase change right? Can you please help me with this? I have drank the Internet trying to find this answer but it’s not mentioned anywhere. Thanks a lot!

[@christopherjackson2575]

In my studies, I had come to appreciate that the nonexistence of a quintic formula was somehow a corollary of the fact that A5 was simple. Now that you have connected the derived series to nested radicals for me, I am in bliss and will never be the same again. Thank you, so much! 😍🤩

[@SRangarajan]

Hi, there is a part in this video where he proves that a single nth root isn’t enough since the commutator ensures that no phase change is picked up. How does nesting solve this issue? Since nothing picks up a phase change right? Can you please help me with this? I have drank the Internet trying to find this answer but it’s not mentioned anywhere. Thanks a lot!

[@peterhall6656]

I'm a 70 year old functional analyst and I'm not knee deep in group theory but this is a really insightful exposition. I was not aware that Arnold cooked this approach up but in a sense it does not surprise me. His work on the geometry of dynamical systems (KAM theory etc) is full of really fundamental insights. Well done.

[@iwersonsch5131]

It was at 35:25 that I realized why that simple group in Galois theory is so important. It sounded completely unmotivated to me back then, thank you!!!

[@ultrio325]

you are the 3rd to last person I expected to see here

[@Zxv975]

Same! The derived group stuff was the point at which I was totally lost in my Galois theory course, whereas here it makes perfect sense why it emerges as something to study.

[@sofiagoy6125]

What a great video! Such a clever argument, and the animations are on point. It's really nice to see the topology of complex numbers come into play in the proof. Also, I've had a course on group theory before, and it really helped me to understand the usefulness of commutators. Now I'm looking forward to learning some Galois theory. Great job! :D

[@nsfeliz7825]

i have found a formula for for quintics but the margins of the comments section wont fit.

[@PC_Simo]

Nice one 😅👍🏻.

[@Jonathan-ij2ky]

Well played

[@AlienoW-t8n]

Ok Fermat.

[@yeetuwu-fx6rv]

unfunny fermat joke

[@deleted-something]

Welp you can only apply this to a right theorem 🤓

[@jamesthomas6228]

Excellent video. Great pace and clarity. More than 35 years since I studied Galois theory and all I remember is it being hard to understand. This proof and your explanation shows the fundamental result without all the baggage. What a beautiful proof.

[@tejing2001]

This not only makes it intuitive why there's no quintic formula, it also makes galois theory itself much more approachable to me. Well done!

[@mCoding]

Fantastic explanation demonstrating a deep understanding of the material. I would argue that you did use Galois theory in this presentation, but done so intuitively as to present the concepts while avoiding the usual symbolic definitions.

[@divyansharora6788]

I am a high school student and your coding videos have been wonderful for me thanks a lot!

[@duckymomo7935]

Yea it’s literally Galoisvthrory idk why people keep denying it

[@columbus8myhw]

@@duckymomo7935 Because Galois theory is all about field extensions and their automorphisms. This argument doesn't use a single field extension

[@jonathanhuang2299]

​@@divyansharora6788 l

[@prakharnigam9757]

Fantastic! Never thought I could understand Insolvability of quintic(Which was like a forbidden fruit to me) due to my background in Electrical and Computer engineering that involves only some introductory group theory, but you made it possible to make me understand it without galois theory and higher mathematics. Thank you so much! Congrats for being featured in SoME1 too!

[@hitarthk]

I can't count how many times I've watched this video. But at this time, I feel like I completely understood. Earlier, what I didn't understand was the section that proves "Only rational functions and nth roots are not enough for writing an algebraic equation for the roots". Specifically, how does the rational function not picking up any phase while permuting roots, comes into play? I then thought of actually writing the nth roots of exp(i*t) as exp((i*t + 2*k*pi)/n) where k=0,1,2,..(n-1). Each value of k can be made to correspond to one root as we'll see further. Let's say that we have an expression that uses an nth root of some rational function "r" of coefficients. Let's focus on one root z1 = (arg(r)^1/n) * exp(i*phase(r) / n). Looking at the expression of z1, it is extremely easy to see that if "r" does not pick up any phase (when the roots are continuously moved using commutators), all the inputs of z1 remain exactly the same so z1 cannot change. But we just moved the roots around via the commutator. Hence the contradiction! So I think the key insight here was to see where phase(r) shows up while writing the nth root. Hope this helps someone!

[@fawzibriedj4441]

Thanks sir! What I didn't understand, is why we assume that the permutation of roots "changes" z1. z1 is an "nth" root, so it can have "n" different values, for all we know, n could be = 5, and the 5 values of the 5th root are z1, z2, z3, z4 and z5. So permuting 3 roots will not really change the value of z1 as the 3 roots are already included in the 5 values of z1. It seemed to me that for the quadratic, what saved us is the fact that square root can has 2 values, so permuting these 2 roots do not affect it. Why does this same argument not apply to the nth root? If you have any idea, it would be of great help!

[@comet1072]

Beautiful. I have an exam about group theory in a couple days and was watching some yt to be distracted from studying and here I am hearing about permutations again haha. I guess I can not escape.

[@eswee6780]

Hey, die naam herken ik! :~P hopelijk was je examen goed gegaan.

[@comet1072]

@@eswee6780 Das toevallig :p. Examen was zeker goed gelukkig!

[@Celastrous]

Please make more like these. This was amazing, I've wanted an accessible proof for the lack of a quintic formula since college.

[@alphakrab5022]

9:25 I was skeptical when you said you were going to prove the fundamental theorem of algebra in a visual way, I had never seen the intuition behind this proof. This is oustanding, thank you!

[@adsoyad2607]

wow that was beautiful, such a unique and intuitive way of approaching a complicated subject like this

[@NinjaOfLU]

I can't express quite how pleasantly surprised I was by this. This is explained beautifully, paced well, and, whilst I tried to pick holes in it, I couldn't manage to: when I thought I'd found one, I paused and found that, thinking about what you'd said a second time, it had already closed itself. Could this be done in a shorter video? Totally! And pretty easily: I bet. But I _don't_ think it could be done in a much shorter video without losing something of what makes this video so great, which is that all of the steps are motivated and intuitive, rather than simply stated. Could it be cut down? Yes. Should it be? _Definitely not._ Understanding the unsolvability of the quintic has been on my bucket list for _years_, and I thank you for being the one to finally help to get me there!

[@PC_Simo]

If you were going for Italics, in ”_years_,” (as I suspect), you might want to flip the order of the comma and the second underscore, into: ”_years,_”; so, it becomes: _years,_ just saying 🙂. That being said, though; I absolutely agree with you that this video *_SHOULD NOT_* be cut down 👍🏻.

[@ludvigm]

Beautiful approach! I think this really sheds light on what the central idea of Galois theory is, without going into any of the heavier machinery :)

[@amaarquadri]

This is insane! I only had to watch the video twice and I think I fully understand it! For any other Rubik's cube nerds, the commutator of commutators that is shown at 39:30 in the videos is exactly what happens when you do when do RL'R'L on a pyraminx. Each turn is a 3 cycle of the edges, and the 4 move algorithm is a commutator. The net result of the algorithm is itself a 3 cycle, just like in the video!!! This let's you do commutators of 3 cycles resulting in more 3 cycles.

[@DeadJDona]

so you can do RLR'L' or FBF'B' and get back, but you can't do FLTF'L'T' because dimensions will interest

[@samsonblack]

What a lovely video communicating such a lovely result. I have proved all the lemmas and propositions in order to demonstrate the theorem in an algebra class, but this video still sheds so much light on the argument. Fantastic!

[@noahzuniga]

it's taken be years to find a satisfying explanation of this proof, thank you for finally getting this off my chest

[@charlied.4683]

Yes! Finally! I always saw people saying that there was no quintic formula, and I always thought, why can’t you just construct an equation somehow out of eulers approximation method. But now I realise it just means algebraicly. Great video!

[@prettyigirl1]

Incredible video - despite already having learnt the Galois theoretic proof, I feel like I'm only now understanding how it works.

[@DonkoXI]

I agree. It really felt to me like this, in essence, is the Galois theory proof, but that's exactly what makes it good. It's just the essence of what makes the proof work, stripped from it's framework. I used to joke with people by saying the quintic was unsolvable just because 5 was too big of a number. This video somehow made that very precise. It's the smallest number that's at least 2 more than 3!

[@Your_choise]

@@DonkoXI why can’t a quartic or cubic not be a commutator of commutator ... of a commutator , why only 5th

[@digxx]

@@Your_choise That's what he said at the end. Because For cubic or quartic commuator of commutator of.... eventually always ends up at the trivial permutation.

[@duckymomo7935]

I’m confused why people say that It’s literally still galois

[@voidify3]

@@DonkoXImight want to remove that exclamation point or put a word between the 3 and it, your comment is a time bomb for guys who mistakenly think they’re super smart and funny to go “hehehe but 5 is actually 1 less than 3 factorial gotchaaaaa”

[@CommanderdMtllca]

KZbin just recommended your video to me. I absolutely love your voice and your explanation!

[@Lucashallal]

As someone who was trying to turn the argument presented on this video into a rigorous proof, I feel like there were a couple of things that were not made as clear as they should have been. This is still an excellent video of course, but I had some difficulties in understanding some parts. Now this may seem like a dumb question at first, but what *exactly* do you mean when you say "multivalued" throughout the video? Are you banning using +/- symbols as in the traditional quadratic formula? The reason I ask that is that otherwise the +/- symbols themselves act as a sort of "disambiguation agent" for the idea introduced in the 19:54 proof, for example. But then at 21:02, you say to imagine a formula for z1 that involves a n-th root, so what would that even mean if the definition for the n-th root you used made it so that it emcompasses all possible solutions to "z^n=w" instead of a specific one that gets distinguished for all the roots by "disambiguating agents" as mentioned before? It makes the sentence "without changing the n-th root" that you use later seemingly non-sensical, since there isn't 1 n-th root. This is *really* *really* hard to phrase and it probably will make no sense to anyone reading, but it feels like there is some ambiguity in the definition of "multivalued" used that makes the argument confusing to follow 100% rigorously.

[@irrelevant_noob]

44:38 damn, if you "made any mistakes"... Well, me with a partial uni-level background could barely keep up, so if anyone manages to correct you they must have published papers of their own, so kudos to them for still watching the video. ^^

[@nickellis999]

Thank you so much. I've always been resigned to not really understanding the quintic result because of the hurdle of learning Galois theory. But you've made me understand it using elementary techniques. Brilliant.

[@YitzharVered]

An amazingly beautiful video and proof. Should've subsrcibed earlier, but now I have! Please don't quit!

[@mrcpu9999]

WHat the heck? I was just about to post about how much I enjoyed this, and at least for me, it made more sense than some of the others I have watched on math-stuff, only to find out this is it? 2 vids? Is there another channel I don't know about? Because this is some good stuff...

[@Notthatkindofdr]

Brilliant! This is the most intuitive explanation of the unsolvability of the quintic that I have ever seen.

[@ItsJustAstronomical]

Brilliant! I read a few books on this trying to figure it out, but it never clicked for me. I thought explaining it in a 45-minute video was an impossible task. Well done!

[@unakade]

What an amazing video! You have a real talent for explaining things and I hope you keep making more of these!

[@tmhood]

26:18 - the precise point at which I decided to subscribe

[@alexsmart2612]

From the age of 17, since when I had first heard of the impossibility of the quintic formula and Galois' proof of it, I have had the unfufilled intellectual desire to build some sort of understanding for it. The statement is just so elegant, so beautiful. However, since I didn't take math in university, I never got the opportunity to learn algebra formally and all my attempts to self-learn failed because of the large built-up all the books take in order to arrive at the result. Somehow I would always fail to persevere and fall short of gaining the much desired insight. Thanks for finally helping me understand this. :)

[@jostpuur]

I had always found all material on Galois theory extremely confusing, and I had already accepted that I would never understand why there's no quintic formula. This video changed everything. Very nice.

[@yanshengang4324]

Amazingly, there's a proof of the impossibility of angle trisection using exactly the same ideas! (Taken from Terry Tao's blog post, "A geometric proof of the impossibility of angle trisection by straightedge and compass") Theorem: There is no straightedge-and-compass construction that gives the trisector of any arbitrary angle. Suppose A,B,C are points in the plane, and we want to construct the trisector of angle ABC. Consider what happens if we move A one round around B and back to where it started: the trisector of angle ABC rotates at 1/3 the speed that A does, so it would end up 120 degrees away from its original position. Can we construct the trisector by only using straight lines? Note that anything you can do with straight lines (drawing the line joining two points, or the point of intersection of two lines) is "single-valued": there is no ambiguity in these constructions. So when A goes one round around B, the entire construction would end up back where it began; in particular, we couldn't have constructed the angle trisector. Can we construct the trisector using only one circle in our construction? Now some of our objects can be "multi-valued": for example, a line can intersect this circle in 2 points, and when we move A once around B, it could happen that these 2 intersection points swap places with each other. However, this means that when we move A *twice* around B, these intersection points swap back to their original places; similarly, the entire construction would also end back where it began. However, the trisector would have moved by 240 degrees, so we couldn't have constructed the trisector in this case either. Now the general case: we could have circles constructed using intersection points from other circles, and so on for several layers. But just as before, when we double the number of rounds that A moves around B, the next layer of the construction is guaranteed to end up where it started. Hence by moving A some power of 2 number of rounds around B, the entire construction ends back where it began. But *no power of 2 is a multiple of 3*, so the trisector would end up either 120 degrees or 240 degrees from its original position. So there's no way that we could have constructed the trisector. QED

[@notallwrong]

Thanks for reminding me about that post! Yes, it's a lovely argument too.

[@thiagokimgarcia8705]

I'm not even into the "meat" of the video, and it's already great. My favorite proof of the fundamental theorem of algebra has always been through Liouville's theorem (for reasons unrelated to the theorem itself). I knew that topological proofs existed, but when I was shown one in "intro to algebraic topology" class, I didn't truly understand it. For your proof at the start, the visuals plus your explanation made it very intuitively clear. I kept thinking about it in my head, trying to really understand it/kinda formalize it, and in the end it boils down to: if the polynomial didn't have any roots, that would contradict what we know about homotopies between functions defined on the circle. This was probably the core idea behind the proof I didn't understand as well, so I'm happy that I've finally "seen" it. Also feels like I understand homotopy a bit better now.

[@nghiaminh7704]

Could you help me with the proof? I don’t understand why the path of the function has to *wrap around* the origin. It could just be any closed path.

[@TheBouli]

@@nghiaminh7704 The point is to consider z large enough so that only the term az^n has a significant influence on the value of p(z). E.g. for p(z)=z^2+10z+100 when you take z on the order of magnitude 1000 or larger, z^2~=1 000 000, 10z~=10 000 and 100=100, meaning that 10z+100 makes up only about 1% of p(z). This is highly informal, but I think intuitively clear and formally all you need to prove is that |a_n*z^n|>|a_(n-1)*z^(n-1)+a_(n-1)z^(n-2)...+a_1z+a_0| for any a_0,...,a_n and |z| large enough. Once you have that, take z large enough so that p(z)=az^n * (1+e(z)), where |e| is very small, say |e|

[@ingiford175]

@@TheBouli For very large magnitude, a circle of input of that magnitude into p(z), the graph of output will lie near the circle created by a_n*z_n (and actually wrap around n times). with a Magnitude of 0, you get an output of p(0)=a_0. if you look at the continuity of the circles of input as it slowly 'shrinks', it will have to cross 0+0i at some magnitude to get from the large circle (extremely large magnitude) with 0 inside to the point when the magnitude is 0 which gives you the a_0 value. Quick example if you have p(z) = x^5+2x+i and prove it has a root in C, if you look at an input circle in C of say radius of 10^100, the output will almost look like a circle with a radius of 10^500 and clearly p(0)=i is within that circle. Shrink said circle continuously and it will deform as the number gets smaller, but its both the output is a continuous loop and the shrinking is continuous and eventually will cross the 0 point in the range (output) plane.

[@EricSeverson]

This Fundamental Thm of Algebra proof was new to be, and felt so much more visual and intuitive. It also seems like it directly gives you that there are n solutions up to multiplicity, since the loop that wraps around n times will hit the origin n times while it contracts to zero. I found that to be a lot more satisfying that just pulling out 1 root and then dividing the polynomial inductively.

[@diplomaticfish]

This video is crazy good. I saw the proof of this in a Galois theory class I took, but it completely went over my head. Every step in your argument is so well motivated it seems obvious, my mind is blown

[@suleimanbas]

Marvelous. I am fascinated with the clues and the presentation. Thank you a lot.

[@fan4every1lol89]

It's so frustrating seeing these golden channels release one great video and then promptly just disappears

[@notapplicable7292]

I didn't listen to a word you said and have no intrest or unstanding of commutators, group theory, or anything else you talked about however still watched the whole video cause your voice is great ambient sound.

[@brettaspivey]

One of the greatest math videos of all time

[@project.eutopia]

This reminds me a bit of the Futurama theorem. In that you have a permutation generated by some set of non-repeating swaps, and the goal is to, using non-repeating swaps, to permute the points back to their original position. The theorem shows how this is possible if you have two addition points to work with. Here 5 points is enough here to get around finite numbers of commutators because you can use 3 cycles and the additional two points as a kind of scratch pad.

[@froyocrew]

Holy shit it finally clicked for me, thanks for pointing this out!

[@arnabpaul8079]

This is one of the most beautiful math-exposition videos that I found on youtube. I've been a big fan of 3Blue1Brown channel for a long time, and now I am your fan too. Please add more...

[@belgaer4943]

I watched this about a year ago, and took an abstract algebra course in the interim, so it’s quite fun to return to it with more background!

[@all_so_frivolous]

I honestly intended to study Galois theory just in order to understand the insolubility of the quintic, which always seemed to me kind of impossible to prove. Thanks for this beautiful argument and for saving me a lot of time (though I probably will eventually study Galois theory at some point, it seems pretty fun.)

[@bcthoburn]

Well, I intended to study group theory to study vector spaces to study tensor algebra to study tensors to study coordinate transforms to study multivariable integrals. It turns out, multivariable integrals are easy to do & understand if you think geometrically rather than algebraically.

[@nice3294]

This is a beautiful arguement and much simpler to follow than other explanations

[@yinq5384]

Great explanation! 36:32 All 1-order commutators are {(1), (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)}. All 2-order commutators are {(1), (12)(34), (13)(24), (14)(23)}. Then all 3-order ones are trivial. 43:36 All commutators are even permutations, which are the half we're looking for.

[@pladselsker8340]

To be fair, I lost you at ~35 minutes into it, but I'm proud I could understand up to that point nonetheless. I get the gist of why it's not possible to have a solution of some fifth degree polynomial equations in closed algebraic form now, so thanks for the time you've spent making this video :)

[@gaminikokawalage7124]

I think I lost him after the review of complex numbers. And then briefly had a light bulb moment at commutators. As someone with very little math knowledge, I'm still happy I vaguely understood this in the end

[@ingiford175]

I got lost late in the video also, but it gives me a place to start looking and digging into.

[@sangraampatwardhan1573]

Great video! One thing I couldn't really convince myself of, though, at 23:40, is why swapping two roots means a 2 * pi * m change in phase for r(b,c,d,e,f).

[@rr51019]

Echoing everyone else, great pacing and great explanation. Thanks for making this! Makes me want to go back to the book store and get that Intro to Galois Theory book, but I bought the differential geometry book, so I'm already on the hook for that one. 😬I'm envious of the kids just growing up who will have so many incredible resources. (finally, stop hiding behind that cute pi and show your face, 3blue1brown! :p )

[@Your_choise]

3blueonebrown has shown his face in the lockdown math

[@ingiford175]

Got 2 Galois books for christmas. Yay

[@joaofrancisco8864]

Now that was just great. Exquisite video. I hope there are more to come!

[@levk4321]

Finally, I get it! Really great presentation. It does require work, so it is not 45.03 minutes, but rather some hours to fully digest and absorb it :)

[@ZZZ1368]

This is super cool! Your argument is incredibly compact. There's a bit of subtlety surrounding how nested roots can remain multiple valued after a commutator -- because a commutator is in general a loop in itself. After I figured that out I can play back the whole argument in a flash in my mind. Now I finally *grasp* the whole story. Thank you!

[@tongjunhui8420]

"There's a bit of subtlety surrounding how nested roots can remain multiple valued", how do you figured this out? I have the same confuse now 😞

[@emadmustafa7294]

Thank you very much for this great video. I watched your video the first time couple of years ago, but I couldn't understand it. Then I recently came back to this video, and repeatedly watch it several times until I finally understood it. The amount of satisfaction once I understand the idea of this video is unlimited!

[@mikesteele5935]

Marvelous. An amazingly beautiful exposition.

[@Ganerrr]

3:52 anyone else freak out at that noise lol

[@spiffyavatar3611]

yeah

[@TranSylvainie]

The use of the rubixcube was really helpful to instantly grap what you are explaining. Your lecture allows us to imagine what young Galois forsaw during his "méditations" (sic - letters to Auguste Chevalier). I was brought here by 3b1b, congratulations for your well deserved selection.

[@a__f]

I loved this video! One of my favorite professors in college mentioned this in a lecture once, but I never had the time to study it myself. This is so satisfying as somebody who doesn't like Galois theory.

[@harrymattah418]

Just discover this marvel. Subscribed directly. Such a perfect mathematical construction as well as a perfect explanation. I feel your English to be particularly distinct, soft, and clearly understandabe to French ears.

[@mediaaccount8390]

Really fantastic walkthrough, good small tangents, clear, and good balance of hand-holding and try-at-home! Thank you. (note: The background music was a bit of a challenge to ignore as I was followint the video)

[@rutgermoody7204]

Thanks for this great video! Your explanation is 100% clear and to the point. In my opinion every algebra course should start with such a clear explanation of Abel Ruffini. To me Galois theory is such a vast amount of theorems that you're never really sure you haven't missed anything. This 45 min. video tells you exactly what it's about without obfuscating the proof in any way. Thanks again!

[@Mutual_Information]

Watched the whole thing. So good ! Well done

[@Asakoto1849]

This is the most intuitive proof of the Abel-Ruffini theorem I've ever seen and ngl I don't think anyone could understand the theorem while not understanding the content of this video. Btw, if Bring radical can be used to solve quintics, can it be used to solve sextics, septics and so on, or any such tools can be used for finitely many orders of equations?

[@stevenglowacki8576]

I was spending most of this video translating what you were saying into Galois theory, and there's a very clean mapping. This gives a much better understanding for what exactly is happening inside the typical proof using Galois.

[@agranero6]

Wow! This is so much simpler to understand than Galois way. Very well explained. You can see the relation with Galois even. I will take the dust of my abstract algebra book that ends with Abel-Ruffini proof and now it will make sense in a new and richer way. I knew Arnold by his fantastic book on Classical Mechanics, but I didn't know of this argument.

[@moocowpong1]

Ohhh, this is a really interesting argument. I think I never quite got what the big deal was with solvability in the standard argument before now.

[@KipIngram]

This was a very valuable video to me. You did a great job putting these ideas at a level that made them "consumable." Once you showed us how the roots of a square "swap places" on one trip around the circle, I kind of "got it" that we were on the trail of something important. The details that followed were valuable, but you really "made the point" quite early on. Thanks very much for this!

[@thatapollo7773]

This is definitely one of the best SoME videos out there I will need to watch it a few more times to understand it fully but I have understood enough and finally understand why quintics don't have a general *algebraic* formula

[@gavintillman1884]

I never got my head around Galois theory. I can’t say I followed everything you said, but it’s helped motivate some of the Galois theory content and maybe I will get my head around it. Interesting viewing.

[@levpotomkin1769]

Amazing video! I loved how explanation is so clear seems so simple, even though the theorem itself is pretty complex. Please make more!

[@vs-cw1wc]

How i wish every mathematics textbook is like your video, full of intuition and ideas that build up naturally.

[@MichaelRothwell1]

Amazing and beautiful! I plan to study Galois theory one of these days, but it was wonderful to see this key result proved in such a visual and "elementary" fashion without needing to!

[@Unchained_Alice]

Never seen an approach for it like this. I tried to get my head round Galois theory but it requires a lot more than just the BSc I have! But this is much easier to understand. Great explanation of it.

[@AlexanderB41]

Beautiful video! This is the perfect visualisation of what i roughly remember from the algebra lecture back then :)

[@valentinofshteyn9246]

There is an error at the time 20:12. The proof should say that we changing the phase from 0 to 2Pi, not from 0 to Pi.

[@Math4e]

What a great piece of math and what a great presentation. You made it much more accessible. Who knews if I would haver ever fond the time to personally go and read the proof? I mean I'm specialised in a different subject of mathematics. So, to see the core ideas involved visually, was a blessing. Thanks for sharing this beautiful argument with us! ❤

[@Tony-cm8lg]

We definitely need more videos! Brilliant presentation

[@MarkPilloff]

Thank you so much for the video! I found it extremely helpful in understanding this concept. I do have a question about one point in your presentation. Around 30:55 you say that the n-th root of the rational expression (boxed in green) might have made a loop around the origin. But how can this be? The expression inside the inner root has undergone no net phase change because of the commutator construction; doesn’t this mean that the n-th root cannot have undergone any phase change either? Perhaps related to this, since the claim is that unnested roots alone can’t solve the cubic, shouldn’t our argument make explicit use of the other (rational) term under the outer root, without which this would be just an overall unnested root? It seems we could argue about the two terms under the outer root (i.e., a rational expression and a root of a rational expression), that neither of these picks up a net phase under the commutator action, but that we cannot guarantee the same about their sum. Is this a valid line of reasoning?

[@richardbloemenkamp8532]

I think you are right. Without any further operation under the outer root, you would just have the outer n-th root of the inner m-th root of the inner expression which would make an (n*m)-th root, which is not a nested root.

[@tongjunhui8420]

All understood, except for 30:55, why ABA'B' doesn't work for nested roots? In my understanding, no matter where these coefficients are, how many levels they lay in roots, 1. after A they will draw a closed path X and back to original value 2. after B they will draw a closed path Y and back to original value 3. after A' they will draw a closed path X reversly and back to original value 4. after B' they will draw a closed path Y reversly and back to original value So effectivey each coefficients draw 0 "circles" and the corresponding roots shoudn't be changed... what's wrong for my above statements?

[@wesleyyu1394]

I must remember to go on my monthly pilgrimage to this masterpiece for the rest of my life.

[@azursmile]

Good video - I think a bit of a build up explaining relationship between inner and outer roots at 31:00 would help a bit..

[@science_gang]

Your content is extraordinary. We admire your talent both in math and video creating 😉 Thank you for the great content.

[@siddharth1591]

Really well-explained! I wish people gave such a treatment to all aspects of abstract algebra.

[@wuzhenrong9539]

Hello, at 20:19 I'm a little confused by why z_1 must change by continuity. Would appreciate it if someone could provide some clarity thanks!

[@jimvj5897]

Wonderful! Is the last part (comms of comms of comms w 5 roots produces an existing comm) your contribution? The referenced pdf & other explanations of Vladimir Arnold's use a brute force technique of evaluating all 120x120 perms of perms to find those that are the result of comms of comms.

[@秦强-q7o]

Great video! To make arnold's argument clear I searched some essays and videos .Yours is the best.

[@MathVisualProofs]

Fantastic video. Very well done.

[@onedez]

3:51 there's a very ear piercing and startling click that happens around the left speaker here.. it made me jump. I thought there was something wrong with my headset, but it was actually part of the video

[@dudelookatree]

funny example of a commutator: My friend and I have different computers. both can open and close a single app normally. If I open word, open Excel, close word, close Excel, I get back to where I started. If he does the same steps, his screen rotates 90°.