You are one of a kind. Don't think I've ever come across an instructor with your level of knowledge, clarity and enthusiasm! Higher mathematics education could use more people like you.
Bravo..the finale! Your logic, enthusiasm, and great articulation are great assets to my learning. To do it over again, I would watch this video first...then, knowing what I need to learn, and then start methodically with some perspective that what is being taught and how it will be used. I have search RUclips high and low looking for a good understanding of Galois' s proof. This is clearly the best without smoke and mirrors. Congrats to you for your teaching ability and congrats to me for finding your lessons and patiently learning from them.
That. Felt. Good. Satisfying QED indeed. And the music helped. XD Great series of lectures - you have a style that's actually energizing to listen to. The quintic was the one topic from my Abstract Algebra book we never went over, and it always felt like a crime we never did so. Now I understand it, and it was so worth it. Thanks so much!
AWESOME ! thanks for sharing all this. I did this in group theory and it was tough stuff. What do you think of this as a lay summary ? ROOTS can be like permutations of objects, permutations are a really core fundamental type of group that contain all other groups (non-abelian), some groups play nice, they all have formulas, some groups don't play nice, if a polynomial type was to be associated with a group that didn't play nice it wouldn't have a formula.
If you allow also the matrix transformation sometimes you can solve more then the eye can tell. For example. If I have got the X^5 + 5*X^3 + 5*X +2 = 0 and I transform it with the quadratic X^2 + X + 2 + Z = 0 then the determinant of the matrix will give you the quintic Z^5 + 15*Z + 44 = 0 . And the X in this case is equal too ( -1 + (2)^(1/2) )^(1/5) + ( -1 + -(2)^(1/2) )^(1/5) is about equal to -0.354378367 now you can just fill it in the the quadratic to calculate the Z . Z is equal about -1.77120573 and this is the same zero of the Quintic in Z. Now in this case I wonder. What is the solable group of this quintic in Z?
This is a great introduction to Galois' Theory, but without good definitions and real proofs, you can't truly say you proved that fifth degree equations are unsolvable by radicals.
Hello. Thanks for this. I subscribed to you. Could you kindly put up a playlist of the videos from the beginning of group theory up to this video of quintic insolvability. Thanks.
Matthew Salomone Hello. I'm really poor in abstract algebra right now so I hope you can answer my question in a non-sophisticated manner (and hopefully a video). What is the direct connection of the solvability of a group & also the abelian part of the group tower in the solvability of the quintic ? Does my question make sense ?
Bravo ......wonderful videos........now the question is how do you generally find a solution to a 5th degree equation??? without using extensions Numerically?
+Damien Mitchell Actually, it could have the same root more than once, making it appear as though it has less than 5 roots. If you are interested in math, try math.stackexchange.com
It can be solvable if use i=(√-1). You can also use the sum or difference of 2 radicals in a root or radical. That would make it seem unsolvable but it can work if you try. And you can use substitution, ex. p=a+b, or you can combine theories to make the algebraic roots of the equation and have answers of the polynomial. Ex. x=(-b±(√(b^2-4ac))/2a
+Damien Mitchell That's not how it works. It won't work for "all" quintic polynomials, trust me. If the solution to all quintic polynomials existed in closed form, I would know it. You are confusing this with whether or not a "case-by-case" scenario is solvable.
+Peter Nguyen Indeed, the point of this video (and of the whole video series in general) is to prove that no formula exists for the quintic in general. The substitutions and radical tricks you describe would work for some quintics (in particular, if their Galois group is solvable, e.g. S4) but there is no hope for a one-size-fits-all formula for all quintic polynomials.
Do be pedantic, you don't actually *have* to find one whose Galois group is all of S5, you just have to prove that one exists. I don't know how you would do that without constructing one, but I bet you could.
Nice prove. This quintic is indeed not solvable in radicals. But what you can do is two make the last coeeficienten equal like this. X^5 + -6*X + 3 = 0 X = k*Y -6/k^4 = 3/k^5 (-1/96)*Y^5 + Y + 1 = 0 u= -1/96 k = -1/2 Y = -1.011002460811 Y = 3.341870528962 Y = -2.803283758532 Y = -1 + u + -5*u^2 + 35*u^3 + -285*u^4 + 2555*u^5 Y = -1 + -1/96 + -5/9216 + -35/884736 + -285/84934656 + -2555/8153726976 = = -1.01100243008 X = -1/2 * -1.01100243008 = 0.5050122 the quintic has three real zeros. The others are. X = -1.670935264 X = 1.4016418
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
You are one of a kind. Don't think I've ever come across an instructor with your level of knowledge, clarity and enthusiasm! Higher mathematics education could use more people like you.
Thank you so much for making & sharing these videos. You make a complex & abstract subject so accessible and exciting!
You are a genius teacher. Your videos are so funny and understandable at the same time.
Bravo..the finale! Your logic, enthusiasm, and great articulation are great assets to my learning. To do it over again, I would watch this video first...then, knowing what I need to learn, and then start methodically with some perspective that what is being taught and how it will be used. I have search RUclips high and low looking for a good understanding of Galois' s proof. This is clearly the best without smoke and mirrors. Congrats to you for your teaching ability and congrats to me for finding your lessons and patiently learning from them.
That. Felt. Good.
Satisfying QED indeed. And the music helped. XD Great series of lectures - you have a style that's actually energizing to listen to. The quintic was the one topic from my Abstract Algebra book we never went over, and it always felt like a crime we never did so. Now I understand it, and it was so worth it.
Thanks so much!
Bill Shillito Are you going to make new videos??
Very energetic !
Your explanation is brilliant!
Eureka!!! Man love your energy and style... Hats off!!!
Thank you very much for these videos!
I love his enthusiasm
the best video I've ever watched !!!!
Bravo sir! Bravo.
AWESOME ! thanks for sharing all this. I did this in group theory and it was tough stuff. What do you think of this as a lay summary ?
ROOTS can be like permutations of objects, permutations are a really core fundamental type of group that contain all other groups (non-abelian), some groups play nice, they all have formulas, some groups don't play nice, if a polynomial type was to be associated with a group that didn't play nice it wouldn't have a formula.
14:33 Exactly how Galois felt about 5th degree polynomial roots.
14:36 Peak math education 👌
If you allow also the matrix transformation sometimes you can solve more then the eye can tell. For example. If I have got the X^5 + 5*X^3 + 5*X +2 = 0 and I transform it with the quadratic X^2 + X + 2 + Z = 0 then the determinant of the matrix will give you the quintic Z^5 + 15*Z + 44 = 0 . And the X in this case is equal too ( -1 + (2)^(1/2) )^(1/5) + ( -1 + -(2)^(1/2) )^(1/5) is about equal to -0.354378367 now you can just fill it in the the quadratic to calculate the Z . Z is equal about -1.77120573 and this is the same zero of the Quintic in Z. Now in this case I wonder. What is the solable group of this quintic in Z?
This is a great introduction to Galois' Theory, but without good definitions and real proofs, you can't truly say you proved that fifth degree equations are unsolvable by radicals.
Nice , can you show too that for example X^5 + 15X + 12 = 0 has a solvable group?
Hello. Thanks for this. I subscribed to you. Could you kindly put up a playlist of the videos from the beginning of group theory up to this video of quintic insolvability. Thanks.
Start here: Exploring Abstract Algebra II: Contents
Playlist: ruclips.net/p/PLL0ATV5XYF8DTGAPKRPtYa4E8rOLcw88y
Matthew Salomone
Hello. I'm really poor in abstract algebra right now so I hope you can answer my question in a non-sophisticated manner (and hopefully a video). What is the direct connection of the solvability of a group & also the abelian part of the group tower in the solvability of the quintic ? Does my question make sense ?
+Uta Shiori My recommendation for questions like these is to ask them in a math forum, particularly math.stackexchange.com
Bravo ......wonderful videos........now the question is how do you generally find a solution to a 5th degree equation??? without using extensions Numerically?
I believe numerical approximations are the only way to find "solutions" to the general 5th degree equation.
There exists serie expansions.
Elliptic curves
And some other methods
If approximations are acceptable, use a computer
The quintic equation can have 5 solutions but you need to try roots in roots or try to factor out the roots you know from the polynomials.
+Damien Mitchell Actually, it could have the same root more than once, making it appear as though it has less than 5 roots. If you are interested in math, try math.stackexchange.com
The quintic is this form ( Bring Gerard ) can have a maximum of three real roots. Three or one this is right.
It can be solvable if use i=(√-1). You can also use the sum or difference of 2 radicals in a root or radical. That would make it seem unsolvable but it can work if you try. And you can use substitution, ex. p=a+b, or you can combine theories to make the algebraic roots of the equation and have answers of the polynomial. Ex. x=(-b±(√(b^2-4ac))/2a
+Damien Mitchell That's not how it works. It won't work for "all" quintic polynomials, trust me. If the solution to all quintic polynomials existed in closed form, I would know it. You are confusing this with whether or not a "case-by-case" scenario is solvable.
+Peter Nguyen what about 3rd power equation and 4th power equation theory
+Peter Nguyen Indeed, the point of this video (and of the whole video series in general) is to prove that no formula exists for the quintic in general. The substitutions and radical tricks you describe would work for some quintics (in particular, if their Galois group is solvable, e.g. S4) but there is no hope for a one-size-fits-all formula for all quintic polynomials.
+Matthew Salomone It could be be possible to start with Ax^5+Bx+C=0 then divide A on both sides and then make a perfect quint is and go from there.
+Damien Mitchell i=(-1)^(1/2) would just be part of it
Sir chandrakant here from India sir ple teach us how to find onto ,into, homomorphism.
Except if you're using a 10-line bisection algorithm...
Do be pedantic, you don't actually *have* to find one whose Galois group is all of S5, you just have to prove that one exists. I don't know how you would do that without constructing one, but I bet you could.
Check it off of my "Bucket List".
Nice prove. This quintic is indeed not solvable in radicals. But what you can do is two make the last coeeficienten equal like this.
X^5 + -6*X + 3 = 0 X = k*Y -6/k^4 = 3/k^5
(-1/96)*Y^5 + Y + 1 = 0 u= -1/96 k = -1/2
Y = -1.011002460811 Y = 3.341870528962 Y = -2.803283758532
Y = -1 + u + -5*u^2 + 35*u^3 + -285*u^4 + 2555*u^5
Y = -1 + -1/96 + -5/9216 + -35/884736 + -285/84934656 + -2555/8153726976 =
= -1.01100243008
X = -1/2 * -1.01100243008 = 0.5050122
the quintic has three real zeros. The others are. X = -1.670935264 X = 1.4016418