This is by far the most complete "simple explanation" video of the prove I've ever seen... It helped me alot to understand some connections that I knew existed but not quite get what they were about. Thank you so much for this.
Thanks Vinicious. Many of the CZcams videos on FLT are too simplistic with no/little real information about the magic of the proof. Some are way too hard for most people. So I wanted to make a video somewhere in the 'middle'.
Loved the video. When divulgations generally avoid saying anything significant about Modular Forms ("they are impossible to visualize"; " they lie in a hyperbolic space"), here we get a rather clear explanation (a special tipe of Complex Functions).
Thanks Manuel, part of the reason I do these intro videos is that many videos shy away from explaining things or use fairly useless analogies. So I really appreciate your reflecting that in your feedback.
Hello, I'm taking intro to elliptic curves this semester and my professor said we will see a mini-proof of FLT. So when I saw this video, I jumped (just a little!). Thank you for the great video. Can't wait for the other ones!
Hi Randell, Maths degree student here with an interest in Number Theory. There's not many informative youtube videos about FLT, but this is great. I may or may not have shook my head a little upon the realisation that I hadn't simplified powers and looked for primes! Thanks for the video!
Glad you liked it. I'll be uploading intro proof prime number theorem in a couple of weeks and there are lots of other number theory videos of mine at czcams.com/users/randellheyman
"The more simple the question is, the harder it gets to prove"!!!! i have a proof for this statement but since my class starts in 10 mins, I have to go now!!!have a good day!
This is a fantastic introductory video. The only part that remains to be shown is how the modular form is generated by the elliptic curve, or vice versa. This is probably a tricky step and would undermine the elegance of the video, so it makes sense why it wasn't detailed. edit: After some further thought, I have another question that was not elaborated in the video: why does the assumption of a^p + b^p = c^p (for a, b, c integers, p prime) imply that there is no associated modular form for the elliptic curve y^2 = x^3 + x^2(B^p - A^p) - x(AB)^p. Clearly if Fermat's theorem is false, then we can substitute B^p - A^p with another integer, C^p. Why is this impossible? Is it due to the fact that the coefficients of the elliptic curves must necessarily be irrational due to some relationship to pi?
Thanks for the feedback. In regard to your first paragraph it is, in my opinion, too tricky for the video (which was already one of my longest videos). A good starting point for your question might be at math.stackexchange.com/questions/1917984/the-corresponding-modular-form-of-an-elliptic-curve
In regard to your second paragraph I don't quite get what you are trying to say. Ribet in 1986 proved that if FLT is false then the elliptic curve y^2 = x^3 + x^2(B^p - A^p) - x(AB)^p has no associated modular form. The fact that you could replace B^p-A^p with C^p doesn't change the fact that this elliptic curve does not have an associated modular form. Hope that helps.
great work done... i have seen many videos about FLT the historical story of it.. but the mathematics of elliptical curves and modular forms are beautifully said here.. even a school level student can understand.. i would request to make a video on the ramanujan sum..
Ritam Chakraborty as a high school student who is considered as one of the best math people in the grade, I did not understand a single thing from the moment he talked about elliptic curves.
Thanks ,I understand , better the math involved. As for the proof. Power of prime 2 =assoc- modular elliptic curve. And powers to some prime >2 =non-assoc-modular elliptic curve,which don`t exist.
Super nice video! It sounds like you briefly labeled the square root of 2 as a rational though, but as you are surely aware, that isn't the case. But it's algebraic and real
Thanks for pointing out the mistake. CZcams doesn't seem to allow annotations post publication like it did before, so I don't think I can do much to point out the error.
Thanks for the suggestion. Cards are normally used to link to another video. Are you suggesting that the teaser for the card says there is a mistake or something else?
I believe that there is an x missing from the front of the middle expression and consequently also in front of the AB in the last (in slide at 7 minutes)
I have proved on 09/14/2016 the ONLY POSSIBLE proof of the Great Fermat's Theorem (Fermata!). I can pronounce the formula for the proof of Fermat's great theorem: 1 - Fermath's great theorem NEVER! and nobody! NOT! HAS BEEN PROVEN !!! and NEVER!!!! 2 - proven! THE ONLY POSSIBLE proof of Fermat's theorem 2A - Me opened : - EXIST THE ONLY POSSIBLE proof of Fermat's Great Theorem 3 - Fermat's great theorem is proved universally-proven for all numbers 4 - Fermat's great theorem is proven in the requirements of himself! Fermata 1637 y. 5 - Fermat's great theorem proved in 2 pages of a notebook 6 - Fermat's great theorem is proved in the apparatus of Diophantus arithmetic 7 - the proof of the great Fermat theorem, as well as the formulation, is easy for a student of the 5th grade of the school to understand !!! 8 - Me! opened the GREAT! Mystery! Fermat's theorem! (not "simple" - "mechanical" proof) !!!!- NO ONE! and NEVER!
Thank you so much for explain that I have been trying to understand it for a while. Also does the ad-bc come from putting the coefficients in a matrix and taking the determinant. Why does that have to be one, in linear algebra it just has to be non zero (I know this isn't really linear algebra but could you possibly explain)
You are correct that you can think of the restriction ad-bc=1 as a matrix determinant calculation. I try to avoid using terms like matrices for these videos so that they appeal to the widest audience. You can envisage functions that obey similar conditions but with the determinant is not equal to 1. But these are not modular forms and modular forms (with the determinant equal to 1) have proven to be the most useful for mathematicians.
@Fematika Thank you so much for that video suggestion. I'm halfway through it (I decided to start from the start, since I wasn't able to easily follow from where you've linked) and it very clearly (to a graduate level maths student) explains a lot of the concepts linking the elliptic curves to modular forms that were rightly glossed over in this video. The fact that the video is unlisted is a shame, and I would never have been able to find it without your comment.
If you mean by 'the equation' A^P+B^P=C^P, where P is a prime or A^N+B^N=C^N, where N is a counting number, then it has been proven that no numbers can be found that satisfy these equations.
Can you make a video about this on a slightly lower level? I'm a maths student and I find this quite interesting, I'd like to know more about it than just the surface view haha
@@RandellHeyman haha a little bit more detailed and harder ... something that would be more interesting for someone who's more experienced with math, but still not like a professor (like a maths student) lol
Ernst Kummer proved during the 1840s FLT for all regular primes. Unfortunately, an infinite number of irregular primes exist (37, 59, 67, 101, 103, 131, 149, 157 and so on),
Thanks for comment. The history of the last theorem is very interesting, and important in the development of mathematics. In my video I wanted to concentrate on the proof as the history is more readily accessible in other videos.
@@RandellHeyman ah i see. so i have 1 more question let E is an elliptic curve with an associated level 7 modular form. does the same series at 13:26 apply the same for E but with powers of multiples of 7? and does the solutions to E mod 7 cannot be found with the associated modular form?
Fermat's Last Theorem: 홀수 솟수 p에 대하여 x^p+y^p=z^p을 만족하는 자연수 x, y, z는 존재하지 않는다 (My Proof) x, y, z가 존재한다면 Fermat's Little Theorem에 의해, vw(v+w+2pk)F(v, w, p, k) = p^(p-1) k^p 으로 변형되며 그 해는 (x, y, z) = (v+pk, w+pk, v+w+pk) 꼴이다. 그런데 자연수 k= n인 경우 해가 존재한다면 n=1*n 이므로 k=1일 때의 해의 n배의 해를 가져야 하는데... k=1일 때 해가 존재한다면 '홀수=짝수'로 모순. 따라서 해는 존재하지 않음.
Excellent introduction. One part however I could not follow is around 13:26. I don't understand the q substitution and how the resulting long multiplication is related to the modular functions earlier where you had a,b upstairs and c,d downstairs of a z? is this supposed to be some short cut to that type of function of z? And I don't see the relationship to the elliptical curve above it. How is the the particular modular function you have related specifically to the elliptical curve equation about it? What ties them together? There may be a math step/s that are implied here to get from one to the other because it may seem obvious for more mathematically practiced individuals; however I see no connection between the two - I just see two different unrelated equations. Any clarification is appreciated in advance.
the f(z) at 13:26 is a modular form of level 11 and weight 1 because f(z) has the property that f((az+b)/(cz+d))=(cz+d)^1f(z) for all integers a,b,c,d with c a multiple of 11 and ac-bd=1. Here, you need to distinguish between the function that is the modular form and the relationship that is used to describe it. A simpler example which has nothing to do with modular forms.... f(x)=sin x is a function. It has the property that f(x)=f(x+2k*pi) for any integer k. The modular form and the elliptic curve are related as the modular form gives information about the number of modulo p solutions to the elliptic curve for each prime p. Also, nothing I say in the video explains how to get the modular form for a particular curve. All that I say is that every elliptic curve has a modular form that has the number of solutions to the elliptic curve embedded in the modular form.
Thanks for the quick reply. And thanks for the clarification of what you did and didn't say. That helps as I thought I might be missing how one derived the modular form from the curve. So as far as the f(z) function and it's relationship to the curve above it am I to understand without explanation that it is the right modular form for that curve? Following up on your comment about needing to "distinguish between the function that is the modular form and the relationship that is used to describe it" is the f(z) then the modular form - analogous to the f(x)=sin x and the line below it the relationship/property - analogous to "f(x)=f(x+2k*pi) for any integer k"? Finally as far as the equations below the f(z) line - titled Modulo Form Weight 1 Level 11 - how are those derived? For instance why 12z+1 versus 3z+2, 11z+1 versus 22z+15? Sorry if I am asking questions that I should know the answers from earlier parts of the video but until I came to this part I was following quite well I thought. Thanks again for any further elaboration.
am I to understand without explanation that it is the right modular form for that curve? .....it is the right modular form in the sense that f(z) has embedded in it, by its coefficients, information about the number of solutions to the elliptic curve.
is the f(z) then the modular form - analogous to the f(x)=sin x and the line below it the relationship/property - analogous to "f(x)=f(x+2k*pi) for any integer k"? .....yes
Modulo Form Weight 1 Level 11 - how are those derived?....it is level one because the power of (cz+d) in f((az+b)/(cz+d))=(cz+d)^1f(z) is 1. It is level 11 because c is divisible by 11. It is a modular form because f((az+b)/(cz+d))=(cz+d)^1f(z) for all integers a,b,c,d with ad-bc=1.
I have proved on 09/14/2016 the ONLY POSSIBLE proof of the Great Fermat's Theorem (Fermata!). I can pronounce the formula for the proof of Fermat's great theorem: 1 - Fermath's great theorem NEVER! and nobody! NOT! HAS BEEN PROVEN !!! and NEVER!!!! 2 - proven! THE ONLY POSSIBLE proof of Fermat's theorem 2A - Me opened : - EXIST THE ONLY POSSIBLE proof of Fermat's Great Theorem 3 - Fermat's great theorem is proved universally-proven for all numbers 4 - Fermat's great theorem is proven in the requirements of himself! Fermata 1637 y. 5 - Fermat's great theorem proved in 2 pages of a notebook 6 - Fermat's great theorem is proved in the apparatus of Diophantus arithmetic 7 - the proof of the great Fermat theorem, as well as the formulation, is easy for a student of the 5th grade of the school to understand !!! 8 - Me! opened the GREAT! Mystery! Fermat's theorem! (not "simple" - "mechanical" proof) !!!!- NO ONE! and NEVER!
Other comments appreciate the quality of the presentation, and say they understand.., but it's a typical example of "how to play the game" type application of empirical rules, that I have difficulty in remembering without a video of the mechanism at work. (Thanks for the example of the process of thinking) QM-TIMESPACE, the application of the basic principle of connection, Mathematical, Phys-Chem and in terms of self-defining relative meaning, (measures, proportions and associations), exists in an infinite-eternal duration of probability and sum-of-all-history superimposed on a point connection, as "The Wave" of all wave-package identities. That's the full measure/rule, of the potential possibilities in Polar-Cartesian Coordination, one ultimate continuous in-form-ation self-modulation, so Modular Forms of probability in possibility potential frequencies and amplitudes in parallel coexistence temporally, is The Calculus, wave-package integration at .dt, or 1-0 relative infinities (Cantor style) Coordinated by natural occurrence in the quantization e-Pi-i numerical multiple spectrum. A Modulation Mechanism of probabilities substantiation defines the dominant probability of the Origin and an infinite "mathematical" clock mechanism timing and spacing of the wave function at a distributed temporal superposition Singularity.., a temporal Singularity is the complex i-cause-effect of observable spacetime, a temporal reflection-projection screen of interference.., a similar property to a Radar, driven from the pulse generator of the Universal wave-package - clock. The mathematical rigour of substantiation in principle, ..the proof of a state-ment by naturally occurring form-ulation, must conform to the manifestation of the QM-Time Principle, so the terminology used for explanation of processes in elemental statements of math construction have a methodology established in the practical development of the Quantum Fields Modulation Mechanism. (?)
Er, maybe I’m wrong but I thought..... Tanayama-Shimura conjectured that all elliptic curves have corresponding modular forms. Frey/Ribet proved that the Fermat equation was elliptic, and didn’t have a corresponding modular form. Therefore, if T-S was true, Fermat must be true. Wiles gave up trying to prove Fermat and proved T-S instead, thus implying Fermat.
I don't think you are wrong. The only thing that I would disagree with is that Wiles gave up trying to prove Fermat. He wanted to prove Fermat and from the work of Ribet knew that if he proved T-S he would get Fermat for free.
Bit of trivia...I met one of the mathematicians whose office was next to Wiles. So Andrew comes in one day and says "I've solved Fermat's Last Theorem".
You probably need to be a good late high school or 1st year university student. It helps a lot to have seen complex numbers before. If you let me know exactly where you start to get lost I might be able to help.
It's at about 7:40 where you start to talk about 'complex value functions' and then the examples that follow. I was looking through the list of videos on your channel. Are there any you've done already that might be a help to me? (thanks for the quick reply btw and for posting this video!)
I would suggest you watch a video on complex mapping by a colleague of mine at the university of new south wales. Then ask me again if you are still unsure of something. czcams.com/video/uguhyTIHQRk/video.html
There have been a number of different proofs over the years. Not sure what your mathematical level is. This proof (which I haven't checked) might be accessible to you: fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html
@@RandellHeyman Dang, I haven't seen that web site over ten years. Good historical information regarding to Fermat's Last Theorem. Direct proof the theorem was very convoluted. I must keep your head clear. By the way there are many ways lead to the top of the mountain. One is easier than another. A 120+ pages proof couldn't hardly qualify as easy.
EVERY X^n = Z^2 - Y^2 so Z^2 = Y^2 + (X^n/2)^2 i.e. a Pythagorean triple with one root irrational unless the root is equal to an integer raised to an even power. So for example 25 = 13^2 -12^2 so that 13^2 = 12^2 + 5^2. Likewise 3^3 =27 = 14^2 - 13^2 and therefore 14^2 = 13^2 + (3^1.5)^2. There is no mathematical distinction between the two other than the irrationality of (3^1.5)^2. The distributive law of multiplication for the powers does not enable multiplication of all three terms by any common factor that can elevate the whole expression to a higher power so the Fermat triple is a contradiction Q.E.D. The Pythagoras theorem reigns SUPREME.
All elliptic curves have an associated modular form. The elliptic curve y^2=x (x-a^p... has no associated modular forms...a contradiction. Therefore, Fermat's Last Theorem is still most likely true.
Don't despair. Everyone finds (some) mathematics hard to learn. Just get what you can from the video and move on. If you look at it again in the future you will learn a little bit more.
I don't believe anyone is inherently incapable of understanding. You just have to work with math until you better understand. At that point, you could come back to this video and understand everything. That being said, I did not understand a damn thing; not at the fault of Randell, though.
-It doesnt explain the elliptic curve or modular relation with FLT (bc doing so would take hours) -It doesnt explain the a, b, c, d, k, z stuff, just stated it -This is extremely advanced mathematics which took centuries to prove--of course you probably won't understand it in a 15 minute video, especially if you've never studied elliptic curves or modular forme or number theory in general, as well as algebraic geometry I think, etc Don't feel bad.
i'm no mathematician whatsoever but i watched the video and got a sense of the logic behind it........the explanation of the proof was done in 15 minutes. ......since the proof can be explicated relatively well in 15 minutes it can't be all that extremely onerous to understand......my question is why did it take 250 years to come up with a 15 minute explanation?
I have only given an overview of the proof. I skip over how to prove, for example, that every elliptic curve has an associated modular form. The proof of this is incredibly intricate and high level.
Yes, the overview of the proof is not too complicated but that doesn't mean it's easy to complete the prove by having the overview. The statement of the theorem itself is even more simple, yet by this logic it should be easy to prove.
Many years ago I saw what I recall to be a documentary on Wiles proving Fermat's last theorum. I recall that his proof was controversial because he used string theory. Is that correct?
Randell Heyman I heard he used noneuclidean geometry maybe that's where confusion. Of mr Dwight comes in. Some people don't like noneuclidean geometry.
i see, 1 million dollar things were started from year 2000, so any proofs before that couldnt be given to. so fermat theorm proof could have gotten just as the same way as other bounty problems.
FERMAT n'a jamais utilisé tous ces calculs inconnus de son temps ! La solution est bien plus facile. Trouvez d'abord l'équation générale UNIVERSELLE cachée de FERMAT, valable quelle que soit la valeur de Zpuissance(N) et ENSUITE, vous en déduisez que Zpuissance(N) =Xpuissance(N) + Ypuissance(N) n'a pas de solution ! (conjecture de FERMAT).J'ajoute que c'est le plus beau CANULAR mathématique de tous les temps et c'est un mystère de voir autant de mathématiciens tomber dans le piège en formulant des théories inconnues de FERMAT .Il y en a même qui pensent que FERMAT n'a jamais eu la démonstration ! La démonstration est si courte et si facile que vous ne trouverez pas, cherchant une solution compliquée. Ce n'est pourtant pas difficile à comprendre ! Pour ceux que ça intéresse , afin de ne pas me répéter encore une fois, aller sur d'autres sites de FERMAT pour avoir plus d'explications de ma part..
Hi Arek, I don't think that is correct. Suppose I have provnd the theorem for all primes. You claim that I need to prove that there is no solution to a^35+b^35=c^35. But if that was true then (a^5)^7+(b^5)^7=(c^5)^7. Letting d=a^5, e=b^5 and f=c^5 we see that d,e and f are integers and d^7+e^7=f^7. But there are no solutions to this equation since 7 is a prime. So there can be no solution to a^35+b^35=c^35. Hope that helps.
For a number that has a third root, we give the name a green number. The sum of two green numbers, WILL NOT BE a green number It is not possible to prove a claim of the type "WILL NOT BE"" Aetzbar
I want someone to ask Andrew Wiles help me. Why use a hammer to kill a germ? With 400 pages if that was used to teach me become a Ph.D is good. Because i am too stupid. But to solve a three-letter equation is super foolish. x,y,z,n belong set N.n>2. Proving impossible. x^n+y^n=z^n. Suppose: x^n+y^n=z^n. Define n=na.1/a Found there exists at least an integer a ,such that z¹/a is an irrational number. Because x^n+y^n=z^n. so : x¹/a+y¹/a=z¹/a+d.Or: (x¹/a+y¹/a)^na=(z¹/a+d)^na development,simplify and make some changes that is absolute logic. x^n+nax^(na - 1)/a.y¹/a+….+nax¹/a.y^(na - 1)/a+y^n=z^n+naz^(na - 1)/a.d+….+naz¹/a.d^(na - 1)+d^na. So z^n=z¹/a.{nax^(na - 1)/a.y¹/a+….+nax¹/a.y^(na - 1)/a - [….+naz¹/a.d^(na - 1)+d^na]} /na. (x¹/a+y¹/a - z¹/a). Howeverr, found there exists at least an integer a ,such that z¹/a is an irrational number. So z¹/a.{nax^(na - 1)/a.y¹/a+….+nax¹/a.y^(na - 1)/a - [….+naz¹/a.d^(na - 1)+d^na]} /na. (x¹/a+y¹/a - z¹/a).is an irrational number. So z must be an irrational number.If want x^n+y^n=z^n.
This is by far the most complete "simple explanation" video of the prove I've ever seen... It helped me alot to understand some connections that I knew existed but not quite get what they were about. Thank you so much for this.
Thanks Vinicious. Many of the CZcams videos on FLT are too simplistic with no/little real information about the magic of the proof. Some are way too hard for most people. So I wanted to make a video somewhere in the 'middle'.
@@RandellHeyman brachistochrone hhhhh
Thank you for your time.
Brasuca?
I realize it is quite off topic but do anyone know a good place to stream new tv shows online?
@Mordechai Brooks i use Flixzone. You can find it on google =)
Loved the video. When divulgations generally avoid saying anything significant about Modular Forms ("they are impossible to visualize"; " they lie in a hyperbolic space"), here we get a rather clear explanation (a special tipe of Complex Functions).
Thanks Manuel, part of the reason I do these intro videos is that many videos shy away from explaining things or use fairly useless analogies. So I really appreciate your reflecting that in your feedback.
This is the best explanation I have heard so far. I can almost understand the idea of Fermat's proof now.
Thanks for taking the time to comment.
I have a proof of FLT by using elementary algebra. I'd show you, but it's a little too large for my current paper area...
I'm so tired of this kind of comments
Goddamn it
Vivek Jha I have a new got detonation of prime numbers
Vivek jhaat
Lol
Hello, I'm taking intro to elliptic curves this semester and my professor said we will see a mini-proof of FLT. So when I saw this video, I jumped (just a little!). Thank you for the great video. Can't wait for the other ones!
Thanks, enjoy your course this semester.
how was it?
First video on the whole web that clarifies the topic. Excellent, thanks
It always nice to know that I have helped someone understand this great theorem. Thanks for commenting.
Hi Randell, Maths degree student here with an interest in Number Theory. There's not many informative youtube videos about FLT, but this is great. I may or may not have shook my head a little upon the realisation that I hadn't simplified powers and looked for primes! Thanks for the video!
Glad you liked it. I'll be uploading intro proof prime number theorem in a couple of weeks and there are lots of other number theory videos of mine at czcams.com/users/randellheyman
Excellent presentation.
"The more simple the question is, the harder it gets to prove"!!!! i have a proof for this statement but since my class starts in 10 mins, I have to go now!!!have a good day!
This is excellent.
Very well explained.
Excellent.
Thanks for taking the time to comment.
Thank you so much ♥️
You’re welcome 😊
Right as he said it’s impossible to use the 100th power, I had just thought of it
Triangle form is modular form. Proof of x^n+y^n=z^n < triangle ratio.
This is a fantastic introductory video. The only part that remains to be shown is how the modular form is generated by the elliptic curve, or vice versa. This is probably a tricky step and would undermine the elegance of the video, so it makes sense why it wasn't detailed.
edit: After some further thought, I have another question that was not elaborated in the video: why does the assumption of a^p + b^p = c^p (for a, b, c integers, p prime) imply that there is no associated modular form for the elliptic curve y^2 = x^3 + x^2(B^p - A^p) - x(AB)^p. Clearly if Fermat's theorem is false, then we can substitute B^p - A^p with another integer, C^p. Why is this impossible? Is it due to the fact that the coefficients of the elliptic curves must necessarily be irrational due to some relationship to pi?
Thanks for the feedback. In regard to your first paragraph it is, in my opinion, too tricky for the video (which was already one of my longest videos). A good starting point for your question might be at math.stackexchange.com/questions/1917984/the-corresponding-modular-form-of-an-elliptic-curve
In regard to your second paragraph I don't quite get what you are trying to say. Ribet in 1986 proved that if FLT is false then the elliptic curve y^2 = x^3 + x^2(B^p - A^p) - x(AB)^p has no associated modular form. The fact that you could replace B^p-A^p with C^p doesn't change the fact that this elliptic curve does not have an associated modular form. Hope that helps.
شكرا جزيلا
great work done... i have seen many videos about FLT the historical story of it.. but the mathematics of elliptical curves and modular forms are beautifully said here..
even a school level student can understand..
i would request to make a video on the ramanujan sum..
Thanks for the positive feedback. It is always pleasing when someone says I have made something more understandable.
Ritam Chakraborty as a high school student who is considered as one of the best math people in the grade, I did not understand a single thing from the moment he talked about elliptic curves.
Hai Ritam I'm also very interested in this
Sir kindly requested to go through the link czcams.com/video/o__QbEsF66c/video.html
Thanks ,I understand , better the math involved.
As for the proof.
Power of prime 2 =assoc- modular elliptic curve.
And powers to some prime >2 =non-assoc-modular elliptic curve,which don`t exist.
NICE video ; hop someon will do somme freach results for Riemann hypothesis
I have a video Intro Riemann Hypothesis if you are interested.
*Hello:Can Fermat's Last Theorem be poved by Mathematical Induction?*
Not that I am aware of. If there was a simple induction proof I think it would probably be known by now.
Other than just cracking into the proof itself, is there a good next step up to it from here? Thank you for making this.
Not that I know of.
Super nice video! It sounds like you briefly labeled the square root of 2 as a rational though, but as you are surely aware, that isn't the case. But it's algebraic and real
Thanks Mathias. I even have a video proving that the square root of 2 is irrational!
There is a mistake at 6:53. I believe it should say y^2 = x(x+A^P)(x-B^P) = x^3+x^2(A^P-B^P)-x(AB)^P
Thanks for pointing out the mistake. CZcams doesn't seem to allow annotations post publication like it did before, so I don't think I can do much to point out the error.
Put a card at the end!
Thanks for the suggestion. Cards are normally used to link to another video. Are you suggesting that the teaser for the card says there is a mistake or something else?
Randell Heyman
You can pin a comment
My simple and ingenious proof of FLT doesn't fit on my mini-Pad.
:)
I believe that there is an x missing from the front of the middle expression and consequently also in front of the AB in the last (in slide at 7 minutes)
I had another look around 7 minutes. There is an error. Thanks for pointing it out.
I have proved on 09/14/2016 the ONLY POSSIBLE proof of the Great Fermat's Theorem (Fermata!).
I can pronounce the formula for the proof of Fermat's great theorem:
1 - Fermath's great theorem NEVER! and nobody! NOT! HAS BEEN PROVEN !!! and NEVER!!!!
2 - proven! THE ONLY POSSIBLE proof of Fermat's theorem
2A - Me opened : - EXIST THE ONLY POSSIBLE proof of Fermat's Great Theorem
3 - Fermat's great theorem is proved universally-proven for all numbers
4 - Fermat's great theorem is proven in the requirements of himself! Fermata 1637 y.
5 - Fermat's great theorem proved in 2 pages of a notebook
6 - Fermat's great theorem is proved in the apparatus of Diophantus arithmetic
7 - the proof of the great Fermat theorem, as well as the formulation, is easy for a student of the 5th grade of the school to understand !!!
8 - Me! opened the GREAT! Mystery! Fermat's theorem! (not "simple" - "mechanical" proof)
!!!!- NO ONE! and NEVER!
Thank you so much for explain that I have been trying to understand it for a while. Also does the ad-bc come from putting the coefficients in a matrix and taking the determinant. Why does that have to be one, in linear algebra it just has to be non zero (I know this isn't really linear algebra but could you possibly explain)
You are correct that you can think of the restriction ad-bc=1 as a matrix determinant calculation. I try to avoid using terms like matrices for these videos so that they appeal to the widest audience. You can envisage functions that obey similar conditions but with the determinant is not equal to 1. But these are not modular forms and modular forms (with the determinant equal to 1) have proven to be the most useful for mathematicians.
This is way too late, but here's a nice video on it: czcams.com/video/A8fsU97g3tg/video.htmlm42s
@Fematika
Thank you so much for that video suggestion. I'm halfway through it (I decided to start from the start, since I wasn't able to easily follow from where you've linked) and it very clearly (to a graduate level maths student) explains a lot of the concepts linking the elliptic curves to modular forms that were rightly glossed over in this video.
The fact that the video is unlisted is a shame, and I would never have been able to find it without your comment.
This video is historic
Thanks
So am I correct in saying that FLT was solved but the numbers that fit the equation have not been found.
If you mean by 'the equation' A^P+B^P=C^P, where P is a prime or A^N+B^N=C^N, where N is a counting number, then it has been proven that no numbers can be found that satisfy these equations.
My understanding is that there are no solutions.
Can you make a video about this on a slightly lower level? I'm a maths student and I find this quite interesting, I'd like to know more about it than just the surface view haha
Do you want a video that is more detailed and harder? Or do you want a video that is less detailed and easier?
@@RandellHeyman haha a little bit more detailed and harder ... something that would be more interesting for someone who's more experienced with math, but still not like a professor (like a maths student) lol
How about this equation: *a^3 + b^3 + c^3 = d^3*
3^3+4^3+5^3=6^3. :)
Randell Heyman, *Congratulation!*
You got one solution for a^3 + b^3 + c^3 = d^3
It's interesting that: 3^2+4^2=5^2 and 3^3+4^3+5^3=6^3
@@RandellHeyman 8^2+6^2=10^2 how about this?
@@juraoncrack What about it?
Ernst Kummer proved during the 1840s FLT for all regular primes. Unfortunately, an infinite number of irregular primes exist (37, 59, 67, 101, 103, 131, 149, 157 and so on),
Thanks for comment. The history of the last theorem is very interesting, and important in the development of mathematics. In my video I wanted to concentrate on the proof as the history is more readily accessible in other videos.
@@RandellHeyman Roger that. Thanks for your nice presentation. Good job! Keep it up! Regards from Denmark.
Checkmate simple structure 2 II = #
why exactly are we restricting ourselves to weight 11? are there any applications of weight different from 11 to elliptic curves?
Hi, I chose weight 11 as an example. I needed a concrete value of c to explain this part of the theorem.
@@RandellHeyman ah i see. so how can a level 11 mod forms expand to 13:26?
@@ferivertid what I say at 13:26 is that the function f(z) is a modular form of weight 11. Proving that is true is beyond the scope of this video.
@@RandellHeyman ah i see. so i have 1 more question let E is an elliptic curve with an associated level 7 modular form. does the same series at 13:26 apply the same for E but with powers of multiples of 7? and does the solutions to E mod 7 cannot be found with the associated modular form?
@@ferivertid There are different modular forms for different elliptic curves. So the modular form at 13:26 can't be used for every elliptic curve.
can it not be done by induction
?
I very much doubt it.
Sir kindly requested to go through the link czcams.com/video/o__QbEsF66c/video.html
Fermat's Last Theorem: 홀수 솟수 p에 대하여 x^p+y^p=z^p을 만족하는 자연수 x, y, z는 존재하지 않는다
(My Proof) x, y, z가 존재한다면 Fermat's Little Theorem에 의해, vw(v+w+2pk)F(v, w, p, k) = p^(p-1) k^p 으로 변형되며
그 해는 (x, y, z) = (v+pk, w+pk, v+w+pk) 꼴이다. 그런데 자연수 k= n인 경우 해가 존재한다면 n=1*n 이므로
k=1일 때의 해의 n배의 해를 가져야 하는데... k=1일 때 해가 존재한다면 '홀수=짝수'로 모순. 따라서 해는 존재하지 않음.
5:20 seems like a logical leap to me
You're either being sarcastic or a very good mathematician. Either case, thanks for the comments.
Excellent introduction. One part however I could not follow is around 13:26. I don't understand the q substitution and how the resulting long multiplication is related to the modular functions earlier where you had a,b upstairs and c,d downstairs of a z? is this supposed to be some short cut to that type of function of z? And I don't see the relationship to the elliptical curve above it. How is the the particular modular function you have related specifically to the elliptical curve equation about it? What ties them together? There may be a math step/s that are implied here to get from one to the other because it may seem obvious for more mathematically practiced individuals; however I see no connection between the two - I just see two different unrelated equations. Any clarification is appreciated in advance.
the f(z) at 13:26 is a modular form of level 11 and weight 1 because f(z) has the property that f((az+b)/(cz+d))=(cz+d)^1f(z) for all integers a,b,c,d with c a multiple of 11 and ac-bd=1. Here, you need to distinguish between the function that is the modular form and the relationship that is used to describe it. A simpler example which has nothing to do with modular forms.... f(x)=sin x is a function. It has the property that f(x)=f(x+2k*pi) for any integer k.
The modular form and the elliptic curve are related as the modular form gives information about the number of modulo p solutions to the elliptic curve for each prime p.
Also, nothing I say in the video explains how to get the modular form for a particular curve. All that I say is that every elliptic curve has a modular form that has the number of solutions to the elliptic curve embedded in the modular form.
Thanks for the quick reply. And thanks for the clarification of what you did and didn't say. That helps as I thought I might be missing how one derived the modular form from the curve. So as far as the f(z) function and it's relationship to the curve above it am I to understand without explanation that it is the right modular form for that curve? Following up on your comment about needing to "distinguish between the function that is the modular form and the relationship that is used to describe it" is the f(z) then the modular form - analogous to the f(x)=sin x and the line below it the relationship/property - analogous to "f(x)=f(x+2k*pi) for any integer k"? Finally as far as the equations below the f(z) line - titled Modulo Form Weight 1 Level 11 - how are those derived? For instance why 12z+1 versus 3z+2, 11z+1 versus 22z+15? Sorry if I am asking questions that I should know the answers from earlier parts of the video but until I came to this part I was following quite well I thought. Thanks again for any further elaboration.
am I to understand without explanation that it is the right modular form for that curve? .....it is the right modular form in the sense that f(z) has embedded in it, by its coefficients, information about the number of solutions to the elliptic curve.
is the f(z) then the modular form - analogous to the f(x)=sin x and the line below it the relationship/property - analogous to "f(x)=f(x+2k*pi) for any integer k"? .....yes
Modulo Form Weight 1 Level 11 - how are those derived?....it is level one because the power of (cz+d) in f((az+b)/(cz+d))=(cz+d)^1f(z) is 1. It is level 11 because c is divisible by 11. It is a modular form because
f((az+b)/(cz+d))=(cz+d)^1f(z) for all integers a,b,c,d with ad-bc=1.
Wait then what about the 4th power? Aren't they just squares?
Ah but A,B,C and C must be integers got it.
There are lots of A,B and C such that A^2+B^2=C^2 but in no case are A,B and C all square numbers. Does that make it clear?
very much! :D
if we take a=1, b=0 and c=1 then????
Hi. a,b and c have to all be positive numbers. The number zero is not positive.
Number forms is elliptic curves.dismodular.
I have proved on 09/14/2016 the ONLY POSSIBLE proof of the Great Fermat's Theorem (Fermata!).
I can pronounce the formula for the proof of Fermat's great theorem:
1 - Fermath's great theorem NEVER! and nobody! NOT! HAS BEEN PROVEN !!! and NEVER!!!!
2 - proven! THE ONLY POSSIBLE proof of Fermat's theorem
2A - Me opened : - EXIST THE ONLY POSSIBLE proof of Fermat's Great Theorem
3 - Fermat's great theorem is proved universally-proven for all numbers
4 - Fermat's great theorem is proven in the requirements of himself! Fermata 1637 y.
5 - Fermat's great theorem proved in 2 pages of a notebook
6 - Fermat's great theorem is proved in the apparatus of Diophantus arithmetic
7 - the proof of the great Fermat theorem, as well as the formulation, is easy for a student of the 5th grade of the school to understand !!!
8 - Me! opened the GREAT! Mystery! Fermat's theorem! (not "simple" - "mechanical" proof)
!!!!- NO ONE! and NEVER!
@@user-me5vv9wh3u Yes proved
@@kenichimori8533 - !!!! Me opened : - EXIST THE ONLY POSSIBLE proof of Fermat's Great "last" Theorem
@@kenichimori8533 - I have proved on 09/14/2016 the ONLY! - ONE! - POSSIBLE proof of the Great Fermat's Theorem (Fermata!).
Other comments appreciate the quality of the presentation, and say they understand.., but it's a typical example of "how to play the game" type application of empirical rules, that I have difficulty in remembering without a video of the mechanism at work. (Thanks for the example of the process of thinking)
QM-TIMESPACE, the application of the basic principle of connection, Mathematical, Phys-Chem and in terms of self-defining relative meaning, (measures, proportions and associations), exists in an infinite-eternal duration of probability and sum-of-all-history superimposed on a point connection, as "The Wave" of all wave-package identities. That's the full measure/rule, of the potential possibilities in Polar-Cartesian Coordination, one ultimate continuous in-form-ation self-modulation, so Modular Forms of probability in possibility potential frequencies and amplitudes in parallel coexistence temporally, is The Calculus, wave-package integration at .dt, or 1-0 relative infinities (Cantor style) Coordinated by natural occurrence in the quantization e-Pi-i numerical multiple spectrum.
A Modulation Mechanism of probabilities substantiation defines the dominant probability of the Origin and an infinite "mathematical" clock mechanism timing and spacing of the wave function at a distributed temporal superposition Singularity.., a temporal Singularity is the complex i-cause-effect of observable spacetime, a temporal reflection-projection screen of interference.., a similar property to a Radar, driven from the pulse generator of the Universal wave-package - clock.
The mathematical rigour of substantiation in principle, ..the proof of a state-ment by naturally occurring form-ulation, must conform to the manifestation of the QM-Time Principle, so the terminology used for explanation of processes in elemental statements of math construction have a methodology established in the practical development of the Quantum Fields Modulation Mechanism. (?)
Proof 3 2 1 action point.
Er, maybe I’m wrong but I thought.....
Tanayama-Shimura conjectured that all elliptic curves have corresponding modular forms.
Frey/Ribet proved that the Fermat equation was elliptic, and didn’t have a corresponding modular form.
Therefore, if T-S was true, Fermat must be true.
Wiles gave up trying to prove Fermat and proved T-S instead, thus implying Fermat.
I don't think you are wrong. The only thing that I would disagree with is that Wiles gave up trying to prove Fermat. He wanted to prove Fermat and from the work of Ribet knew that if he proved T-S he would get Fermat for free.
@@RandellHeyman You are right of course. Gave up was a clumsy choice of words. I meant switch his attack from Fermat to T-S.
Bit of trivia...I met one of the mathematicians whose office was next to Wiles. So Andrew comes in one day and says "I've solved Fermat's Last Theorem".
@@RandellHeyman Wow, that’s quite a thing to be able to tell people. All the best
Zero is point.零点。
The beginning was good but I got lost once you started to explain modular forms. What level of maths should I be at to understand the rest?
You probably need to be a good late high school or 1st year university student. It helps a lot to have seen complex numbers before. If you let me know exactly where you start to get lost I might be able to help.
It's at about 7:40 where you start to talk about 'complex value functions' and then the examples that follow. I was looking through the list of videos on your channel. Are there any you've done already that might be a help to me? (thanks for the quick reply btw and for posting this video!)
I would suggest you watch a video on complex mapping by a colleague of mine at the university of new south wales. Then ask me again if you are still unsure of something. czcams.com/video/uguhyTIHQRk/video.html
OK, will do. Thanks again!
9:43 for me
How do we know cubics don’t work?
There have been a number of different proofs over the years. Not sure what your mathematical level is. This proof (which I haven't checked) might be accessible to you:
fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html
Randell heyman
@@RandellHeyman Dang, I haven't seen that web site over ten years. Good historical information regarding to Fermat's Last Theorem. Direct proof the theorem was very convoluted. I must keep your head clear. By the way there are many ways lead to the top of the mountain. One is easier than another. A 120+ pages proof couldn't hardly qualify as easy.
I recently made a video on the cubics case. See my video Corona help - Fermat's last theorem, n=3.
EVERY X^n = Z^2 - Y^2 so Z^2 = Y^2 + (X^n/2)^2 i.e. a Pythagorean triple with one root irrational unless the root is equal to an integer raised to an even power. So for example 25 = 13^2 -12^2 so that 13^2 = 12^2 + 5^2. Likewise 3^3 =27 = 14^2 - 13^2 and therefore 14^2 = 13^2 + (3^1.5)^2. There is no mathematical distinction between the two other than the irrationality of (3^1.5)^2. The distributive law of multiplication for the powers does not enable multiplication of all three terms by any common factor that can elevate the whole expression to a higher power so the Fermat triple is a contradiction Q.E.D. The Pythagoras theorem reigns SUPREME.
Alastair Bateman you should look at my proof ‘one page proof of Fermat’s last thm’
All elliptic curves have an associated modular form.
The elliptic curve y^2=x (x-a^p... has no associated modular forms...a contradiction.
Therefore, Fermat's Last Theorem is still most likely true.
wait, is it the full proof
I hope so
It's exactly what the title says. It's an introduction to the proof.
X^n+Y^n=Z^(n+1)
x^n+y^n=z^n is natural number combination three take in set theory are the three theory did a graph undercover
Proofmon
Pythagorean = 3 III 2 II 1 I
算術０線也。
Zero is not number.
I agree
Gadgets.....
I have proof I don't exist
Good. So who posted the above comment? :-)
I'm pretty sure it's a nice video and a clear explanation but I'm too dumb
Don't despair. Everyone finds (some) mathematics hard to learn. Just get what you can from the video and move on. If you look at it again in the future you will learn a little bit more.
I don't believe anyone is inherently incapable of understanding. You just have to work with math until you better understand. At that point, you could come back to this video and understand everything. That being said, I did not understand a damn thing; not at the fault of Randell, though.
-It doesnt explain the elliptic curve or modular relation with FLT (bc doing so would take hours)
-It doesnt explain the a, b, c, d, k, z stuff, just stated it
-This is extremely advanced mathematics which took centuries to prove--of course you probably won't understand it in a 15 minute video, especially if you've never studied elliptic curves or modular forme or number theory in general, as well as algebraic geometry I think, etc
Don't feel bad.
i'm no mathematician whatsoever but i watched the video and got a sense of the logic behind it........the explanation of the proof was done in 15 minutes. ......since the proof can be explicated relatively well in 15 minutes it can't be all that extremely onerous to understand......my question is why did it take 250 years to come up with a 15 minute explanation?
I have only given an overview of the proof. I skip over how to prove, for example, that every elliptic curve has an associated modular form. The proof of this is incredibly intricate and high level.
Yes, the overview of the proof is not too complicated but that doesn't mean it's easy to complete the prove by having the overview. The statement of the theorem itself is even more simple, yet by this logic it should be easy to prove.
ネットワーク Metwork
You can't proof it just for primes numbers above two, you can't get powers of two prooven that way
sghaier mohamed As I say at about 2:00 the fourth power case was proven. This then proves the cases involving all higher powers of 2.
Many years ago I saw what I recall to be a documentary on Wiles proving Fermat's last theorum. I recall that his proof was controversial because he used string theory. Is that correct?
I am not aware of any string theory being used in Wiles's proof.
Randell Heyman I heard he used noneuclidean geometry maybe that's where confusion. Of mr Dwight comes in. Some people don't like noneuclidean geometry.
if fermat theorm was important or difficult, then how come it didnt have 1 million dollar bounty on it?
The Clay Mathematical Institute's $1 million problems were announced in 2000. By then Fermat's last theorem had been proven.
i see, 1 million dollar things were started from year 2000, so any proofs before that couldnt be given to. so fermat theorm proof could have gotten just as the same way as other bounty problems.
In 2016, Andrew Wiles was awarded the 650,000 Euro Abel prize.
Fermat's First theorem 0 + 0 = n^n
A tango 英単語
FERMAT n'a jamais utilisé tous ces calculs inconnus de son temps ! La solution est bien plus facile. Trouvez d'abord l'équation générale UNIVERSELLE cachée de FERMAT, valable quelle que soit la valeur de Zpuissance(N) et ENSUITE, vous en déduisez que Zpuissance(N) =Xpuissance(N) + Ypuissance(N) n'a pas de solution ! (conjecture de FERMAT).J'ajoute que c'est le plus beau CANULAR mathématique de tous les temps et c'est un mystère de voir autant de mathématiciens tomber dans le piège en formulant des théories inconnues de FERMAT .Il y en a même qui pensent que FERMAT n'a jamais eu la démonstration !
La démonstration est si courte et si facile que vous ne trouverez pas, cherchant une solution compliquée.
Ce n'est pourtant pas difficile à comprendre !
Pour ceux que ça intéresse , afin de ne pas me répéter encore une fois, aller sur d'autres sites de FERMAT pour avoir plus d'explications de ma part..
no, it is not sufficient to prove it for primes, you also need to prove it for any composite number, whose factors are not 2 and 3 :)
Hi Arek, I don't think that is correct. Suppose I have provnd the theorem for all primes. You claim that I need to prove that there is no solution to a^35+b^35=c^35. But if that was true then (a^5)^7+(b^5)^7=(c^5)^7. Letting d=a^5, e=b^5 and f=c^5 we see that d,e and f are integers and d^7+e^7=f^7. But there are no solutions to this equation since 7 is a prime. So there can be no solution to a^35+b^35=c^35. Hope that helps.
It isn't enough to prove it for prime numbers, it also needs to be proven for exponent 4 (which was proven by Fermat himself).
I swear to God with Me right solution
More Gravitiy.
Un MacBook.
X^n = N Y^n= N Z^n=N XYZnUN Polution.
For a number that has a third root, we give the name a green number.
The sum of two green numbers, WILL NOT BE a green number
It is not possible to prove a claim of the type "WILL NOT BE""
Aetzbar
Nonsense. Let's mark all odd numbers in red. I can easily prove that a sum of two red numbers will not be red.
I want someone to ask Andrew Wiles help me. Why use a hammer to kill a germ?
With 400 pages if that was used to teach me become a Ph.D is good. Because i am too stupid. But to solve a three-letter equation is super foolish.
x,y,z,n belong set N.n>2. Proving impossible. x^n+y^n=z^n.
Suppose: x^n+y^n=z^n. Define n=na.1/a
Found there exists at least an integer a ,such that z¹/a is an irrational number.
Because x^n+y^n=z^n. so : x¹/a+y¹/a=z¹/a+d.Or: (x¹/a+y¹/a)^na=(z¹/a+d)^na
development,simplify and make some changes that is absolute logic.
x^n+nax^(na - 1)/a.y¹/a+….+nax¹/a.y^(na - 1)/a+y^n=z^n+naz^(na - 1)/a.d+….+naz¹/a.d^(na - 1)+d^na.
So z^n=z¹/a.{nax^(na - 1)/a.y¹/a+….+nax¹/a.y^(na - 1)/a - [….+naz¹/a.d^(na - 1)+d^na]} /na. (x¹/a+y¹/a - z¹/a).
Howeverr, found there exists at least an integer a ,such that z¹/a is an irrational number.
So z¹/a.{nax^(na - 1)/a.y¹/a+….+nax¹/a.y^(na - 1)/a - [….+naz¹/a.d^(na - 1)+d^na]} /na. (x¹/a+y¹/a - z¹/a).is an irrational number.
So z must be an irrational number.If want x^n+y^n=z^n.
Web degree infinity loops take from alphabet injection choice prime number equal power os p. power os is point operating system.Allah.