2000 years unsolved: Why is doubling cubes and squaring circles impossible?
Today's video is about the resolution of four problems that remained open for over 2000 years from when they were first puzzled over in ancient Greece: Is it possible, just using an ideal mathematical ruler and an ideal mathematical compass, to double cubes, trisect angles, construct regular heptagons, or to square circles?
00:00 Intro
05:19 Level 1: Euclid
08:57 Level 2: Descartes
16:44 Level 3: Wantzel
24:00 Level 4: More Wantzel
31:30 Level 5: Gauss
35:18 Level 6: Lindemann
40:22 Level 7: Galois
Towards the end of a pure maths degree students often have to survive a "boss" course on Galois theory and somewhere in this course they are presented with proofs that it is actually not possible to accomplish any of those four troublesome tasks. These proofs are easy consequences of the very general tools that are developed in Galois theory. However, taken in isolation, it is actually possible to present proofs that don't require much apart from a certain familiarity with simple proofs by contradiction of the type used to show that numbers like root 2 are irrational.
I've been meaning to publish a nice exposition of these "simple" proofs ever since my own Galois theory days (a long, long time ago.
Finally, today is the day :)
For some more background reading I recommend:
1. chapter 3 of the book "What is mathematics?" by Courant and Robbins (in general this is a great book and a must read for anybody interested in beautiful maths).
2. The textbook "Field theory and its classical problems" by Hadlock (everything I talk about and much more, but you need a fairly strong background in maths for this one).
Here is a great two-page summary by the mathematician Drew Armstrong of what is going on in this video
www.math.miami.edu/~armstrong/...
Here is a derivation of the cubic polynomial for the regular heptagon construction by (I think) the mathematician Reinhard Schultz math.ucr.edu/~res/math153/s10/...
(there is a little typo towards the bottom of the page. It should be 8 cos^3 theta + 4 cos^2 theta - (!) 4 cos theta -1 = 0. Replace cos theta by x and you get the cubic equation I mention in the video. )
Here is an interesting paper that explores why Wantzel's results did not get recognised during his lifetime www.sciencedirect.com/science...
Thank you to Marty and Karl for your help with creating this video. And thank you to Cleon Teunissen for pointing out that the picture of Pierre Wantzel that I use in this video is actually not showing Pierre Wantzel but rather Gustave Gaspard de Coriolis who was also a mathematician and lived around the same time as Pierre Wantzel. It appears that whenever there does not exist an actual picture of some person Google and other internet gods simply declare some more or less random picture to be the real thing. See also this page by the SciFi writer Greg Egan who made sure that no actual picture of himself is to be found on the internet: www.gregegan.net/images/GregE...
Enjoy :)
Burkard
Mathologer Patreon: / mathologer
Mathologer PayPal: paypal.me/mathologer
When I want to double a cube, I just pull out my complex ruler and e^i(pi) compass and get to work. Easy.
:)
No wait. You're supposed to use e^iπ to square the circle. Save it for that.
I just set pi to 113/355 and boom squered the circle.... even euclid says its fine xD
I just take the next dice out of my pocket and then I got 2 dice.
@@lukahutinski9075 Almost. That is 1/pi. pi itself is 355/113. :)
Hey. Compared to the amount of minutes in 2000 years, this video is remarkably short.
Indeed
no shi sherlock ;) sorry just wanted to say that
I'm sure that with extra circles and more extrapolation, these problems can be solved.
I was a physics student at the University of Arizona in the 1970s and stumbled on a retired gentleman (my age now) at a Swensons coffee shop who was prodigiously working on geometric constructions with a ruler and compass. He explained to me what he was doing, squaring a circle, trisecting angles. I looked at his work and was amazed at the number of books he had compiled and his efforts. We became friends over coffee and discussed his progress and life in general. He had many words of wisdom for a young man in very turbulent times and he was also certain he could solve these problems with a ruler and compass. Then suddenly, he disappeared one day never to return to the coffee shop. Later I was told by a friend that he returned to Long Island and passed away. Another traveler in time trying to tackle difficult problems. His name was Bruce Stegmann from Port Washington, New York. A good friend.
Damn, mad respect for you dude. Sometimes that’s the way it goes.
Gotta love those kind of run-ins.
Oh I had an internet version of this I was replying to some fellow about my kinks and happened to look at his about me and said that he could teach you anything about physics and I was just getting into physics and mathmatics(also this is quite recent just three or 4 months ago I think) so I asked him about books for mathematics and if I should read the Feynman physics books and he was true to his words so he then recommend me the book and gave advice to focus on algebra to not have problems.Also yeah edit:can't find his reply so rip I can't tell him about my progress
Time travel is far more comfortable than its abrupt halt. He should have left markers incase of a repetitious universe 🤷♂️
There are only two students attending a math lecture. Suddenly four of them got up and left. The professor thought with regret: "Well, now two more will come in and there will be no one left at all."
That happened all the time when I last attended math class almost 20 years ago. Students would make deals with one another to write several names on the attendance list, because too low attendance = no grades.
You are a good human my next clienr will be here. WHETHER us discuss math or earth, be Well. Be nice. Till next time.
Well, 2 2-4 =-2 If +2 then -2+2=0 But the part that puzzles me is how four left when there were only two? Do we have some Quantom debt or something?
))))@@xXJ4FARGAMERXx Math is not afraid of negative values
@@Ma_X64 _mathematics_ is not afraid of negative values, but what about physics? Chemistry? Biology?
I have been watching this while eating a breakfast bowl of cereal. I was thinking about how much work has gone into preparing the video. Even with a very knowledgeable presenter, it must have taken hours. I was pondering this when the answer popped out. About 200 hours. I am not surprised. The video is an incredible achievement. Thank you Burkard and Marty. This is a video I will be revisiting. A lot.
:)
No way I'm also watching this whilst having my breakfast 😅
The infinite series 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 ..., conditionally converges to 1/3. This means that with an infinite number of angle bisections, you can trisect an angle! Normally it takes a while, but my friend Zeno has a magic tortoise that can bisect an angle in half the time as the previous bisection. So the tortoise completes the trisection in a finite amount of time. Pretty spiffy!
If you are allowed infinitely many steps you can construct any number whatsoever :)
Mathologer Including several numbers that aren’t in the real numbers, including a number that’s larger than any real number and a number that’s larger than 0 and smaller than any positive real number!
@@cephalosjr.1835 With infinite step, constructing some numbers that aren't real number like infinity and infinitesimal isn't a problem, but I don't think there's a way to construct complex number
@@Noname-67 imagination! Add imagination and complex number (like pink unicorns) isn't a problem anymore :)
@@Mathologer And also, within a finite number of steps you can construct any number, to within a satisfactory precision. I'd love to see good, well-convergent constructions of some of these.
I had a Rubik's cube, and then I went on Amazon and ordered another one and in less than 24 hours later, BAM! I just doubled my cube. Solved in less than a day. Next!
...but did you solve both rubk's cubes?
Amazon disagrees - you did not purchase a ruler and compass with that.
i cut a circle into square. BAM! i just squared the circle. solved in less than 2 minutes. next!
Yah, nowadays when you need to solve a puzzle you can pay instead of think! Awesome.
If that joke were any cheesier, you'd have to contend with a lynch mob of lactose intolerants.
They said it couldn't be done, But he went right to it And took that thing that couldn't be done And couldn't do it.
:)
And...?
I just figured it out. No kidding, even though I’m a kid. I’m also a geek. A total geek.
But with a trick, no ruler & compass here
@@thatssomethingthathappened9823 woah then you must be from Greek
I will have to come back to this video when my head is fresher :)
@MathPencil - Mathematical sense Put one of your MathPencils inside each of DanJan09's ears and make his head fresher!
For this you need to be bald
I was like #250!
Maybe in a next life, for me.
Why on earth would you do that?
Finally, a crazily busy semester here in Australia is almost over. Just some end-of-semester exams left to finalise next week, a short trip to Japan and then I should have a bit more time for making Mathologer videos for the rest of the year. Can’t wait. Anyway, today's video is about the resolution of four problems that remained open for over 2000 years from when they were first puzzled over in ancient Greece: Is it possible, just using an ideal mathematical ruler and an ideal mathematical compass, to double cubes, trisect angles, construct regular heptagons, or to square circles? Towards the end of a pure maths degree students often have to survive a "boss" course on Galois theory and somewhere in this course they are presented with proofs that it is actually not possible to accomplish any of those four troublesome tasks. These proofs are easy consequences of the very general tools that are developed in Galois theory. However, taken in isolation, it is actually possible to present very accessible proofs that don't require much apart from a certain familiarity with simple proofs by contradiction of the type used to show that numbers like root 2 are irrational. I've been meaning to publish a nice exposition of these "simple" proofs ever since my own Galois theory days a long, long time ago. Finally, today is the day :) If you make it to the end of this 40 minute video please leave a comment below and let me know how well this explanation worked for you.
There is no problem at all!! Take any time you need. I prefer videos of quality, maybe sparse in years, than not as good videos every month. Thank you very much for your time and effort to share with us this amazing "visually enhanced" maths.
question: can you draw a parabola or a graph of x^k (k > 1) with ruler and compass? I dont think so...
Yeah it has been really long time since that turtle video but your content is worth the wait as always. It surely was a bit tough due to lot of algebraic proof in level 3 that you need to follow but it was easy in following levels.
another question: Suppose we break the compass & ruler rule and we will devise a brand new tool, how would that tool work? Can there be a new tool that would give us more power?
How did you find time to put five forty-hour work weeks worth of effort into this video, in addition to your other obligations?
Not enough comments here are talking about how amazingly accessible you made these advanced topics. Very well done.
Glad to see you too are a Mathologer viewer, MrCheeze. Also, I agree immensely - Mathologer is amazingly accessible
:)
Note to self: If Mathologer says something was explained in a previous video, actually watch the previous video before continuing to watch the video. I made this a lot harder for myself than it needed to be :')
Yup - Pro Tip FTW :-)
These are really the only long videos I can stand; and I actually enjoy them a lot; you explain very clearly and I understand every thing you do. Thank you for such good content
Every time I see a video on this channel I secretly hope it is a long one. I feel happy now :)
Well, this one is the second longest Mathologer video so far :)
@@Mathologer Time that record fell :-)
Wow. Such an amazing and carefully constructed video. Also, the important warning about ‘changing the rules’ shows (to me in my opinion) how seriously and solemnly the ethic of education is handled on this channel. You have a very satisfied and excited new subscriber!
'carefully constructed' _groans_
He doesn't want the deluge of mail claiming solutions to the "unsolvable" problems!
There were quite a few places where I just went, “that makes sense,” without proving it to myself, but I felt like I followed the whole video.
Nice Easter egg at 13:25 with captions on. Well done Karl.
I'll have to have a chat with Karl tomorrow :)
@@Mathologer If you cannot find him: last time I saw him he was at 13:25... :-)
@@Mathologer Haha your kid is cool.
This wouldn't be Karl Kilroy, would it?
@@NuisanceMan No, this is Karl . kzhead.info/sun/rMeeYN6sr4B_d4E/bejne.html . except he is now taller than me :)
The time you spent putting this together was VERY WELL SPENT. Since the result was: BRILLIANT! I followed each and every level enjoying everyting!. I eagerly anticipate the follow up on Galois Theory
I dont even know math but your personality and enthusiasm has drawn me to watch
Read the rules (they're called axioms) and hope that you don't encounter impossible problem
This is the most healthy, uncorrupted commend section on the whole of KZhead edit: and positive
Thus proving the effectiveness of irrational roots as an Eedjit Deterrent (TM) :-)
This comment section isn't just positive: it's far more complex, so be real. Otherwise we won't be able to have a rational conversation.
Here is a true story regarding your comment: In the year 1864, professional arm wrestler Stanley Von Armbro was challenged by an unknown wanderer. The wanderer sat at the table in front of him and removed his coat unveiling massive bulging biceps that seemed gelatinous yet firm like weathered ball sacks. He uttered a single word to him: “poopoo.” He then grabbed Stanley’s arm and slammed it through the table and 6 feet into the ground. His pulsating veins exploded, filling the hole with poop and burying Stanley alive. The wanderer then slowly walked out of the bar and flapped away with his massive deflated arm sacks, never to be seen again. This is where we get the phrase “Work your biceps full of poop or you’ll die probably.” Have a wonderful day
At 28 minutes you say we feel a little rooted as they say in Australia. In Australia rooted means you had sex
F*ck you
I want to share something uncommon (maybe, in fact I don't know) I am a french student in master's degree (Paris XI) who recently went through a course in Galois Theory (soberly named "Algèbre"). It was tough and waaaaay too short to really understand the ideas. The exam went pretty poorly for me because of that. These kind of videos are therefore precious resources for me: I can recognize all the tools we had to use (extension fields etc.) in a way that clearly go into why we need them and how they work. That part was rather cryptic before. Since I also want to be a teacher, that video (with some others, and on many subjects !) is a great gift in that respect. I really wanted to thank you for that. You are a great professor.
@@benoitavril4806 Hm Tu extrapoles beaucoup ce que je dis mais bon pourquoi pas XI c'est 11 hein... donc Paris 11 (ou Paris Sud)
quelle partie avez-vous considéré comme rare?
Nice work, I remember when I studied Galois Theory, good times, well it was actually last semester but seems like eons ago.
What's happened since to make it seem so distant? Love? Sickness? A new, nasty branch of mathematics you have to learn?
It’s gotta be employment.
Absolutely love this. Although I’ve come across these before and spent much time in the past wrapping my head around this, I couldn’t stop listening to your amazing explanation. You’re a very gifted teacher sir!
looking forward to the continuation this was pretty cool
At first I was terrified by a video that is 40 minutes long, but it seems like the right amount of time to explain this stuff. Thanks for a wonderful video, super excited and looking forward to Level 7 :)
Thanks for this video. It was a wonderful presentation on a usually very stuffy and difficult subject. I honestly cannot say that I (PhD applied maths) was able to follow everything while I was watching the video, but it showed me one very important thing: that I skipped numebr theory, Galois theory, extension fields and so on for all the wrong reasons when I was studying maths at uni. If they had explained to me then that it would give a lot of insight into the boundaries of what is possible with Euclid's geometry I would have been all over it.
Lovely video, great speaker, great graphs, and most importantly great mathematician. Good Job!
These videos always go way over my head by about half way through! sometimes something clicks later, but i thoroughly enjoy following the logical progression
Thanks for this amazing effort! Just 200 hours to make this (..) seems time well spent, given how many viewers enjoy the results. Keep 'm coming!
I'm not gonna lie, I spend the whole video waiting for a Galois level, I'm glad I wasn't disappointed!
Thanks Mr. Mathologer, I'm taking a course on fields and modules and this video gave a brilliant motivation as to why we're studying fields like Q (-/a) and algebraic field extensions. We're doing Galois groups soon so I look forward to that :)
One of the hardest subject to compile in the shortest possible video. very very carefully structured and beautifully rendered as always.
Beautifully done. I can't wait for level 7.
This video is full of some lovely gems. The construction of a regular pentagon using 4 intermediate circles and one intermediate line was my personal favourite and I have shared this with friends. (To prove the method works, I solved for the coordinates of the key points and used the fact that cos(pi/5) is (1+root(5))/4 and that cos(2*pi/5) is (-1+root(5))/4. Not elegant, but it got me there.) The description of the rooty-expression subfield and the consequence that certain geometric constructions are not possible provides closure for more decades-old gaps in my math knowledge. (I am ashamed to admit that I forgot the rational root theorem and had to re-watch that video.) Thanks for putting this video together. Your enthusiasm for the subject combined with your depth of knowledge is a wonder to behold. No one can get the the "root" of the problem like you can!
Nice pun! And thank you for the mathematical insight! I still haven't fully worked it out, but I searched the comments for a comment like yours and am glad to have found it!
Watched the entire video and LOVED all 40 minutes of it ... although I am not a mathematician, I am a high maths fan and was surprised at how much of this at least made logical sense -- great presentation! I will have to go back to school to study math again to really understand ... but if I do, this may be part of the inspiration!
I love your T-shirt. I can't take my eyes off it. I need one just like it.
Construct one using only ruler and compass for your homework
@@mavwavesify have done! I created it digitally and am sending it off to get it printed.
Same! Just got to the comments to say that but you beat me to it lol. Quite adequate election of shirt too.
If I stare at the shirt pattern for more than 30 seconds, I can see the 12th dimension.
Check Dodecahedron - Kwintessens album cover
Your videos enrich my life. Thanks so much!! This was so great.
hello mathologer. your content brings me such incredible joy and I am so grateful for the effort you put into making this beautiful mathematics accessible. although I had to replay the level 4 section once or twice to understand the recursive field logic (I am also rather drunk), I feel I have taken a relaxing tour and come out with a strong barebones understanding of the proof. at the very least, I leave this video with a deeper appreciation for our human struggle which unites us over time and place to evolve our logical tools. what an incredible triumph for wantzel, whom I have learned died only at 33 but will live eternally in legacy. (although he had more time than galois.) let us learn not only from his proof but also from his example, and practice moderation in excitatory drugs and late night mathematics!
Great video Mathologer, as usually. I'll have to rewatch it again though because solving equations was a little bit to fast for me to get in the first viewing. But, don't blame it on yourself. You were top notch, and I have to take a lot of new info in so it takes time.
Your passion, presentation is impeccable, graphics sync with explaination,3D overlaying effect Voice is so soothing,... Totally impeccable content... Kudos to your effort in making enthusiastic way...
Great vid!! Respect from Greece
Me: I am terrible at math and want nothing to do with it. KZhead: We recommend this video about impossible to solve mathematical equations. Me: ok.
I think it's because your profile picture looks like Euclid
@@BroArmyCommander It's Filthy Frank.
@@table2392 Close enough
@@TheGamingCats I mean Frank is a master mathemetician, his ability to count to autism is unsurpassed.
Same here 😂
At 29:30 you said "extension" and suddenly I learnt something I was chasing to learn for a good time now. Many thanks for that eureka moment.
You sir, are my teacher away from school, each time a new video comes out, I’m eager to listen and eager to learn, thank you for all the lessons you do
You can't square a circle but you can turn math into gold 💖
kzhead.info/sun/bJerg7x5b5Obemw/bejne.html
Ok Midas
I like how with each level came increasing brevity. Level 7 - just a tease of it's existance!
There's so much work, passion and clarity. Can't thank enough.
Thank you for this pure gold content!
I suspected I was brain dead. It took 15 minutes for "Mathologer" to confirm this. Now I pull over and watch some sumo wrestling
I love how popular this channel has become. Great work.
I used to play with constructing shapes using a ruler and compass in grade school. I loved the heptagon and heptagram because they were hard to construct without a protractor. I leaned to use a heptagram in place of a pentagram when asked to place a star on homework. This video takes me back and explains well a curiosity I had as a child. Thank you!
OMG these proofs are so beautiful!
"Of course, faced with the task of constructing numbers such as the cube root of 2 and root pi, it's natural to just stare helplessly into space. Which is pretty much all that happened for 2000 years." I legitimately laughed not just out load but very loudly and for a few minutes. Wonderful.
5:30 This is simply stunning, great animation.
This is as clear an explanation of Galois theory as I have seen and I feel like it's the perfekte Einführung für Autodidakten! I can't tell you how often I've banged by head against the wall with this. Danke! Now, to prove that knifflige pentagon construction ...
awesome video ! im excited for the next(s) level(s) !!!
As one of my professors used to say, the unsolved problems aren't unsolved because they're hard, but because the established methods don't work for them. Once you solve them they're often very obvious - hence many problems are solved by ppl who are new to the field. He says he circumvents this by getting into a new topic every couple of years XD I sure did like old Professor Paul.
Exactly, this girl solved the trisecting an angle problem within the rules. XD kzhead.info/sun/i76JiZxunKFpi40/bejne.html
@@realitycheck3363 This looks promising, but in the step where she trying draw a line by gues'timating it shows it will have tiny inaccuracies, so the angle will be not trisected but only nearly trisected with tiny error. I mean it is the same thing you could do by guessing what angle is right, you can do it to astonishing precision with only ruler but it won't be precise. I might be wrong there is maybe legitimate way to trisect and angle with only ruler and compass in a way she did it, it could be that she did not present her method clearly.
@@MrWorldOfQuests Lets be honest, how accurate you can draw depends on your tools, and your ability to use them. It does not take anything away from the fact that that distance is exactly the distance you need to trisect a triangle. I have tried it myself, and it works on every random angle I have tried it on. And not sort of, but exactly the right way. I'm sorry if mathematicians are jealous that they did not think of it before some random girl, but the fact remains, she found a way to do it, using the rules as stipulated. That deserves some kudos.
@@realitycheck3363 Oh I found wiki about this method, it seems it was known before. en.wikipedia.org/wiki/Neusis_construction . And it seems the ruler has to be with marked unit, so it is against rules. And she did not think it by herself, she studied Neusis construction as stated in her video description. But still it is legitimate way to trisect and angle.
beautiful beautiful beautiful!!!!!!!!! :D love this channel, great work
I probably have to rewatch this 100 times, like I do all your videos, but at the end it's always worth it when you get this huge AHA moment :D
Really amazing. You explain with flair. Worth my time.
x^3 - 2 is an increasing function. That is if a < b then a^3 -2 < b^3 -2. Hence there is only ONE real root. But as you demonstrated if a + b root(7) were a root then so is a - b root(7). Just another way to get to your contraction.
Yes that part was a bit fast. There are 2 cases b is not 0 in that case there's 2 real solutions and as you explained it's impossible Or b=0, in that case there's no condradiction yet (there's only 1 real solution). The contradiction comes from the fact that if b=0 it means a=3root (2) and that is impossible. (The +- b something van be discarted)
c muller vb
OMG 40 minutes from Mathologer, am i in heaven on a dream-lecture?
No, you have OCD and unrealistic perceptions and expectations of reality.
@@russianbot4418 I'm just meta for this.
Fantastic! Thank you! ✊
What a beautiful video. Thank you very much ! ❤
I’m basically 10 years old and this seems impossible, but I’m trying to understand the details, I love this channel
Good for you! He does make it fascinating!
Hey, this is a challenging video. I've been through 7+ years of highschool and college math and don't understand this very well. My major is mostly calculus. Remind yourself that algebra is not just a school topic, but an 8+ year PHD. Also math is important but will not ever make you important to but the smallest amount of people. Make sure you treat everyone in your life so that they might think of you as important regardless of your life accomplishments. Good luck man, math is a shit show for us all!
Good job! You're gonna go far in life
Thx y’all!
I would love to see your presentation of the heptadecagon construction, including the algebraic part of the proof :)
the Mathologer video format is absolutely brilliant!!!!!!
I don't usually watch a video this long, but this video was so interesting I could not stop watching it
Wow! I managed to follow along right to the end, I sure did not totally understand everything and I'd guess I kept up 70% without spending days on just. I have to say as someone who flunked out of maths and physics in year 11 due to frustration with algebra that this was the lesson I needed back then. I'm still really into Math and Physics and always wanted to master Math and get a real grasp on relativity and I'm a few percent of the way there but this video just gave me whole new insight into geometry and the beauty of the universe that I almost missed out on had I not stumbles across it, thousands of years in the making and a true work of art this clip is and getting it to 40mins is just mind boggling and I wish I had another lifetime and could meet you in my younger years when the grey matter could absorb all this even better..... Thank You so much for taking time to make this, I can't get as much out of it as I wish I could but someone will and WOW! What that could inspire in a young mind is amazing to think about 👏 Great Work....... P.S I'm in Tasmania 👍 was legit sitting there part way through thinking "Man, just imagine if this bloke was in Australia and I could of learned from him during my younger years or could chew ya ear over a beer one day 😅 I instantly just assumed that you must be based in the UK or US given teaching style and accent but serves my argument true that Australia has some of the most talented folk about.
Oh my gods! I can't believe it! I actually made it through the video! I mean, I had to repeat some parts, a couple times, but I made it through! Although... I did have to just take your word for it on a couple of things because, unfortunately, I didn't learn _all_ the math things in high school, and I never came across some of them in college (I don't exactly dream of taking trig, lol). But for the most part, I was right there with you! And honestly that surprises me! (I've never been that good at higher-level maths)
My Mathematical seatbelt is ON!!! I like the way this gentleman explains things and does his videos. Lucky to have found this channel just recently (days ago! ~ 11/22/2019)
This is one of the best mathematics channels out there. Keep it up. Long live the Queen of the Sciences!
In Spain, when someone gives an absurd and rondabout explanation for something we say he's "squaring the circle"
Really? That's awesome! I know a little bit of Spanish, but I hadn't heard that idiom before.
@@yf-n7710 It's a bit on disuse, but in cultured circles it's still used.
In German this is still very much in use: "Quadratur des Kreises".
Curioso, acá en Latinoamérica no he escuchado a alguien decir que está "Cuadrando el círculo"
I'm so glad you're making this - I've been trying to study group theory in my free time (as a prerequisite to Galois theory) - the other major problem I know of in that field is the proof that there is no basic formula for the roots of a fifth degree polynomial. Is it possible to 'Mathologerise' that proof as well, or does it require some of the more heavy-duty parts of Galois theory?
Presenting a Mathologerised proof that there is no solution in radicals to general polynomial equations of degree five or higher is also on my to my to-do list. This is actually very doable in a very visual and accessible way using a "topological Galois theory" approach. But first I'll have to recover from this latest video making marathon for a bit :)
I hope to see that soon. I've been fascinated by the unsolvability of polynomials of 5th degree or higher after reading about it in Yanofsky's "The Outer Limits of Reason". However, I haven't been able to find a decent explanation for someone who has not had several semesters of group theory.
@@benoitavril4806 I’m not sure if that’s how it goes, but that’s an interesting take on it. From what my professor has explained to me (in my group theory class), Galois theory deals with extension groups. Essentially, there’s ways to know if a group has a “nice” extension or not (which deals with factor groups), and the “niceness” of the extension has to do with the simple subgroups. Then, there’s an easy way to relate each polynomial of degree n to the alternating group of size n, and a well known result in group theory is that any alternating groups of size 5 or larger have no nontrivial subgroups, so there’s no “nice” extension. About the easiest part of all of that though is the proof of the alternating groups having no nontrivial subgroups, and is a relatively simple proof. As you can tell though, this approach takes a good grasp on group theory, and I’m not educated enough (yet) to know if there’s a better approach.
@@tomwill77 I don’t know if there is a decent explanation for that person. Maybe there is, and he’ll do a video on it, but that explanation will have to provide a crash course for it at least
In mechanical drawing I learned how to divide a line into any number of equal segments. Now you have me wondering which rule or rules I am breaking. Thanks for the video.
Oh, but dividing a segment into a given number of smaller segments, or into a given ratio, is not a problem with a ruler and compass. No rules need to be broken :) Check out the Intercept theorem: en.wikipedia.org/wiki/Intercept_theorem
@@AV-ws2rz Then neither is dividing an angle into three. Make a triangle. Then divide the side opposite the angle into three equal segments. Then connect the ends of the segments to the angle to be divided. You now have three equal triangles and equal angles.
@@johnchestnut5340 unfortunately, such approach doesn't work, unless your triangle is isosceles. Check it out for yourself; in a generic triangle, a bisector divides an angle into two equal angles, but not necessarily the side it crosses. Conversely, a median didvides a side into halves, but not the angle it originates at.
@@AV-ws2rz That's what I get for being out of school and not working with math. I forget things. Thanks for the respectful interaction. I still like math and science. I just don't get to work with it beyond a very basic level. John.
@@johnchestnut5340, it's my pleasure to have a civilised discussion like this one, bro, so thanks to you as well :) And it's normal to not remember things when you don't use them on a daily basis. I myself first started in mechanical engineering and after a while switched to maths. And believe me, most of professional mathematicians aren't much familiar with mathematics outside of their own field. Often they don't know what others might consider 'basic', so there's no shame in not knowing. Maths is such a tremendously vast subject, it's very non-linear (concepts are connected more in a cobweb-like fashion). You could literally pick an area of interest (be it geometry, algebra, knot theory or complex analysis - you name it!) and start discovering the chosen field with little to no knowledge of other stuff. Of course, basic understanding of how logic works is necessary, but you also get that understanding by practice. Cheers! Anton
Thanks! I particularly liked that the demonstration highlighted the consequences of the symmetry of polynomials. This was overlooked in the book I read on the subject that used the language of abstract algebra. A standard approach, I assume, and certainly more powerful and general one. But definitely a tough cookie for this poor engineer.
2:13 how I feel during a calculus test
I love how you and 3d1brown explain maths
3blue1brown? Yeah, he's amazing, too! Fred
This is much better. 3B1B gets too much pleasure out of making things more complicated (if sometimes also beautiful) - his objective is not to make difficult things understandable, but rather to provide new and imaginative ways to describe difficult, or even simple, things. I can enjoy him sometimes, sometimes I think he is too far up his own fundament for his and our good. But mathologer really does want us to understand, not just to admire his technique - which is wonderful, really.
hello mathologer very informative videos love it by the way have u made the video of squaring a circle with changed rules looking forward to it
This guy and his videos is absolutely awesome!
Great vid, as always. And another great shirt, as well 😁
you dont even know what a heptagon is , without looking for a dictionary or google
I can construct point cube root of 2 away from origin. If I have infinite time.
these videos are very graphically helpful and easy. God bless you.
This video and the explanations are incredible as your channel
this is great. the galois theory focusing on the constructions in my algebra book is like, another 350 pages forward...
Where was this video when I was writing my paper for college geometry
What cruel math professor makes there students write an essay in a math course?
@@TheInevitableHulk I just moved to the IB system and apparently we have to write an extended essay for every selected subject (might be less). So I guess that includes a maths essay.
TheInevitableHulk The course focused on the development of Euclidean and Hyperbolic geometry. The professor wanted us to explore topics outside of pure Euclidean and Hyperbolic geometry so I focused on why squaring the circle was impossible using algebra. Also it was a 400 course so time to write ;)
Mohan Sandal lit
Mohan Sandal hope you realize I’m not a scientist :)
Thank You. Watched with my seven year old. We spent the morning mapping out the five platonic solids with compass and ruler. The video homework should be lots of fun. I'm excited I'll let you know when if we get stuck. Correction noted.
Thank you for the video! All of you friends are super awesome!
What would happen if you had a 4 dimensional ruler and compass and you were drawing on the 3D "paper" ? Would that change anything to the constructable set of numbers? If yes, what numbers are constructable on n-D "paper" that aren't possible on 2D paper?
The set of equations you need to solve will contain three variables, but the degree stays at 2. Meaning you still wouldnt be able to extend Q with a cube root, sadly.
@@ruinenlust_ Exactly! The limitation comes from the shapes (and thus the equivalent polynomial functions that appear) and not the dimensions. The problems would be solvable If you were to use some drawing instrument able to produce cubic trajectories, for instance. It's not clear to me how one would use or operate with a 4D ruler, but I can envision some instrument that is able to find a plane given by 3 non-aligned points in 3D space.
I was wondering this in the sense of what space *would* be needed: perhaps a space with a Minkowski metric of order p allows construction up to p-root-y numbers? But that seems far to easy, I suspect many of the mentioned "tricks" stop working.
At the very least, I'm aware that you can use folding to trisect an angle, which I suppose means you could solve it with a ruler and compass that make 3D measurements rather than 2 and 1D
@@DeRien8 3d isn't why origami can trisect angles or double a cube: 3d distances are still square roots. Actually all the maths is in the 2d plane for those proofs anyway, but while the math doesn't really look too tricky, I can't find a good well-contained description of why it results in a cubic solution and I'm not confident enough to try it myself!
I did the first challenge of proving that the construction for the pentagon is actually true. Most of it was a bunch of trial and error figuring out what to do and in what order. That said, I did it! I'd post the full proof but frankly it was way too long. Even just the summary I did at the end to try and condense it down into something that could be read as a step-by-step process took a full page of paper. I also wasn't sure what exactly the ancient Greeks would have been able to use in their own proofs, so I didn't bother trying to emulate them too much. They obviously wouldn't use square roots or what have you, but they did know about certain special numbers that aren't rational that are essential for the construction. I restricted myself to having to do everything by hand and in exact form, so no decimal approximations. I'll try and convey the gist of the proof, though. My main concern is that I did things using a much more complicated method that did work, but required the identity for the difference of inverse cosines. I doubt the Greeks had this, but while I originally thought I found a much simpler way to get the same answer, I didn't realize until I was typing it out here that this simpler method requires an assumption that I can't make. Nonetheless, I am pretty satisfied with this proof. If anyone has any suggestions for a better way, let me know! So first, lets break down the construction. We start with a circle centered on the intersection of two perpendicular lines, with four points of intersection. Lets call one of these points A, and the opposite point B. We won't worry about the other two. In the end, we have two circles centered on the same point, lets say A, with different radii, and each of them intersect the original circle at two new points. Lets say the larger circle gets points named C and D, while the smaller gets E and F, such that the line segment CE doesn't intersect the line AB. We claim BCDEF forms a regular pentagon. Regardless of the method used to make these two new circles, it's obvious they're the key here, so it'd be nice to know what their radii are. Using the construction, we can use a little Pythagorean Theorem and some basic geometry to find them. Assuming the original circle is radius 1, then the larger of our two new circles has radius (sqrt(5)+1)/2, and the smaller has radius (sqrt(5)-1)/2. You might recognize these as the Golden Ratio, and the Golden Ratio - 1. Anybody who's studied regular pentagons knows that the Golden Ratio is integral to it, so this is promising. Using the method by which the points are constructed, it's easy to show that our pentagon has a mirror symmetry, which helps us significantly. Points C and D are formed by the intersections of two circles whose centers lie on the line AB, and thus CD is perpendicular to AB and CD is bisected by AB. The same is true for the line EF using the same argument. As points C and D are equidistant from the line AB, they are equidistant to all points on the line, so line segments BC and BD are equal, and AC and AD are equal. Similarly, AE = AF, and BE = BF. Notice that line CD is parallel to EF as they are both perpendicular to AB, and the midpoints of CD and EF both lie on AB. Let us call the midpoint of CD the point G, and the midpoint of EF the point H. It is easy to see that, being parallel lines, we can create more line segments that are equal between these points, such as CE and DF, or CF and DE. Notice that our pentagon is composed of the following line segments: BC, CF, FE, ED, and DB. We have seen that BC = DB, and ED = CF. We also have a number of diagonals within the pentagon, the line segments: BF, BE, CD, CE, DF. We have seen BF = BE, and CE = DF. It is easy to show that various angles are equivalent as well, so I will stop here and claim that it is fact that the pentagon has mirror symmetry over AB. Our goal is to prove that all the line segments on the edge are equal. With that knowledge we can easily show that all internal angles are also equal. First, we need some numbers. Take the triangle ABC. We know that AB = 2 and AC = Golden Ratio, hereafter called P as that is the most similar symbol on my keyboard to the proper symbol. Using Thales Theorem we know angle ACB is pi/2. We use the Pythagorean Theorem to find the length of BC as sqrt(1 - P^2). Next, take the triangle ABE. We know AB = 2 and AE = P - 1, and angle AEB = pi/2. We find BE = sqrt(1-(P-1)^2). Now take the triangle BEH. We know BE = sqrt(1-(P-1)^2), and angle BHE = pi/2. We know that in a right triangle, the sine of an angle is equal to the ratio of the length of the opposite leg divided by the length of the hypotenuse. Thus, sine of angle EBH = EH / BE. The angle EBH is equivalent to the angle EBA. In the triangle ABE, the sine of EBA is AE/AB. Thus, BH/BE = AE/AB. AE = P-1, AB = 2, so EH/BE = (P-1)/2. We know BE, so EH = (P-1)/2 sqrt(1-(P-1)^2). As EH is half of the segment EF, EF = (P-1) sqrt(1-(P-1)^2). It is not immediately obvious that EF = BC, as it requires (P-1) sqrt(1-(P-1)^2) = sqrt(1-P^2). However, with this value of P, it can be shown algebraically to be true. Thus, EF = BC = BD. We have shown three edges to be equivalent, and the remaining two equivalent to each other, as CE = DF. Similarly, we take the triangle BCG. We know BC = sqrt(1-P^2) and angle BGC = pi/2. The angle CBG is equal to the angle CBA. The sine of CBG = CG/BC, and the sine of CBA = AC/AB, so CG/BC = AC/AB. AC = P, AB = 2, so AC/AB = P/2. BC = sqrt(1-P^2), so CG = P/2 sqrt(1-P^2). As CG is half of CD, CD = P sqrt(1-P^2). It is not immediately obvious this is equal to BE = sqrt(1-(P-1)^2), but it can be shown algebraically to be true. Thus, CD = BE = BF. We have shown three diagonals to be equivalent, with the remaining two equivalent to each other, as CF = DE. We wish to show angle BCD = CBE. Recall G is the midpoint of CD, and BG is a perpendicular bisector of CD. Take the triangle BCG. We know the cosine of an angle in a right triangle is equal to the length of the adjacent leg divided by the hypotenuse. CG = P/2 sqrt(1-P^2). BC = sqrt(1-P^2). We know triangle BCG to be a right triangle, so the cosine of angle BCG = CG/BC = P/2. We know angle BCG equals angle BCD, so the cosine of BCG must be equal to the cosine of BCD. So, cosine of BCD = P/2. If the cosine of BCD equals the cosine of CBE, we shall assume this is enough to prove the angles equal. While it is possible for two different angles to have the same cosine, these angles would have significantly different appearances in the construction if this were to be the case. Sadly, finding the cosine of CBE is a difficult affair. We know angles CBE + EBA = CBA. We use the appropriate triangles for these angles. The cosine of EBA = sqrt(4-(P-1)^2)/2. Thus, EBA = arccos[sqrt(4-(P-1)^2)/2]. The cosine of CBA = sqrt(4-P^2)/2. Thus, CBA = arccos[sqrt(4-P^2)/2]. Thus, CBE = arccos[sqrt(4-P^2)/2] - arccos[sqrt(4-(P-1)^2)/2]. Note the following formula: arccos(x) - arccos(y) = arccos(xy + sqrt((1-x^2)(1-y^2))). Suffice to say that through a lengthy calculation we get the pleasant answer that angle CBE = arccos(P/2). Therefore, the cosine of CBE = P/2. Therefore, cosine of CBE = cosine of BCD, and we claim angle CBE is equal to angle BCD. It is worth noting that if we take a point I to be the midpoint of the segment BE in triangle BCE, and assume the line CI is perpendicular to BE, we can find that the cosine of angle CBE = P/2. However, this assumption is equivalent to assuming triangle BCE is isosceles, which is what we are trying to prove. Take the triangles BCD and CBE. We know BC = BC, CD = BE, and angle BCD = CBE. Thus, triangles BCD and CBE are congruent by Side-Angle-Side. Thus CE = BD = BC, and so, we have shown that all five edges are equivalent. We must now finally show that all internal angles are equivalent, but this is relatively straightforward. The points B, C, D, E and F are all distance 1 from the point O, the origin, as they lie on a circle of radius 1 centered on O. Take the isosceles triangle BOC and call the angle BOC = x. Take the isosceles triangle COE. We know BO = CO = EO, and BC = CE. Thus the two triangles are congruent by Side-Side-Side. Thus the angle COE is equal to x. More importantly, we know angle CBO = BCO and ECO = CEO because isosceles triangles, and angles CBO = BCO = ECO = CEO by congruent triangles. We also know the angles BCO + ECO = BCE. We can do this same construction for each of the five isosceles triangles within our pentagon and show they are all congruent, and thus all internal angles equal, and find their measure to be 108 degrees. We conclude that, as all edges are equal and all internal angles are equal, BCEFD is a regular pentagon.
Н. Wеуl, Symmetry, Princeton Univ. Press, 1952 - ?
very informative. thank you.
I can honestly say I was lost within moments of level 3 starting but I think I might have found my new favourite channel.
i think the basic principle was fairly understandeable. The crux is basically the idea that constructable numbers are square root expressions everything else follows naturally enough
Circle proof: a bit easier than the pentagon: The circle satisfies (x-(a+1)/2)^2+y^2=((a+1)/2)^2 and we plug in x=a y^2=((a+1)/2)^2-((1-a)/2)^2=a
@ Mathologer your videos are great! Any chance you could make one about proof that there are no more convex regular polyhedra beyond icosahedron?
Brilliant, thanks so much!
For those interested in constructions, like shown here: Euclidea is a great app introducing that by letting you play around with the allowed operations.
Me: "Oh cool, another Mathologer video" Burkard: "In this short 40 minute video" Me: "oh fuck yeah..."
Quite fascinating and very enjoyable, thanks :)
(My maximum formal math is circular trig [in high school].) I can do, at least, one more function in straight-edge/compass. I can plot a line from a point to the two tangents of a circle. From there, I can: 1. plot tangent lines on top & bottom of two dissimilar circles, and 2. plot tangent lines between any two circles.