The author of the article was referring to Niven's theorem, according to which cos(theta) and theta (in radians) are never both rational except for theta=0. The author of the article has completely misrepresented this theorem because in its construction theta can be irrational.
Thanks for that. It seemed totally bizarre that somebody would do all that fairly smart construction and then end with "But the cosine of an angle is never rational; contradiction." It's obvious by continuity that every rational in the range [-1,1] is the cosine of some angle.
Here I was keeping a lookout the whole time to try to pick where the proof falls apart, and the twist turns out to be that the problem with it is glaringly obvious! (Not, as Prof. Penn says, that there couldn't be a reasonable proof based on the premise that some derived quantity is rational when it couldn't be, but "any value whatsoever of cosine" isn't that.)
Yes, that was rather anticlimactic. I was expecting something that looked correct at first glance, not just a completely and obviously incorrect statement. I'm not really sure what the point of this video was...
@@thomasdalton1508 Yeah, also the "proof" was all about roots, inequalities, and cosines -- felt like there would be an error of a forgetting-a-second-solution kind.
Right? I paused a few times during the video to really make sure that every step was justified, only for the "contradiction" to be something totally ridiculous lol
@@christianaustin782 I also took a lot of time convincing myself that "shortening r to z" was actually a thing we were allowed to do and would result in a real triangle. I've seen errors come up in that sort of step often.
I wonder if Fermat's famous margin note was referring to a proof similar to this but he probably figured out his error quickly and didn't correct his margin note for whatever reason.
I'm more proud of the "m" sentence than my "w" sentence for sure. Glad you liked it haha. I write wild descriptions because it's helps get people to you know, read the stuff we want them to see lol there. now you know my secret. -Stephanie MP Editor
There is another issue with the shortening of triangle. For x,y,z to be sides of a triangle, they need to satisfy the triangle inequality. Now we know for sure that x,y,r are sides of a right triangle and we know that z < r, but if we assume wlog that x>=y, then it might very well be the case y+zx and z>y, and this tells us that z+x>y and z+y>x. Now we can use the fact that x+y>r>z to conclude that we can infact shorten it to be an appropriate triangle.
Tbh, this could be made a whole video series. All too much and too often we forget to look at past failures. But failure often teaches way more than success. On this example: I've definitely learned something from this whereas even if I tried to digest the correct proof of the theorem, I'd likely fail on page 1 and would need to do a two years preparation course to even start with it.
Imo the best false proof of FLT is the one by Lamé that falsely assumes unique factorization, and ultimately leads to proving FLT in the particular case when n is equal to a Sophie Germain prime
During my undergraduate I was solving a problem like "If x, y and z are sides of a right triangle such that z is the hypotenuse, show that z³ > x³ + y³". From that I generalized for all n≥3 natural, we have zⁿ > xⁿ + yⁿ, where x, y and z are a pythagorean triple. I mentioned this to my professor claiming that I proved Fermat's last theorem for a tiny subset of natural numbers =D
For this subset of the natural triples, namely the Pythagorean triples, you not only proved Fermat's last theorem, but also proved more: that the inequality > holds for this subset, since Fermat's theorem only states that it is different.
@@TonyHammitt i doubt he had a valid proof , if he did he would be the greatest mathematician in the history proving a statement which baffled people 100s of years after him.
Correct me if I'm wrong, but that doesn't show FLT is true for x,y,z? Because even though z^n > x^n + y^n, that's only true for a specific z. You haven't proved there exists no other number a such that a^n = x^n + y^n?
Hopefully, this comment will shed some light on the situation. TL;DR: author actually believed the proof to be correct, but he wasn't a number theorist. I decided to find the original article myself, which proved to be a challenge, since the link provided by the blogspot didn't lead anywhere. Fortunately, from the link I was able to identify the name of the newspaper, went to their archives, and searched via key words. The name of the article loosely translates to "May the humanity rest now?". It goes into history of Fermat's Last Theorem, how Euler and Gauss were unable to solve it. It also briefly mentions one russian mathematician who presumably went insane trying to solve it (I heard nothing about such a story). The article is rather poorly written and unpleasant to read, especially where there are any mathematical formulas involved (they are written 'as is', for example, FLT is "xn + yn = zn"). The article then tells the story of Alexander Ilin, author of the proof. Funnily enought, he studied at the university in my hometown. It is stated that Ilin worked on a space program, and then for no reason at all the article mentions how Alexander was taken to court for money laundering and how he defended himself. After that, he dedicated himself to proving FLT. Finally, the article states that Ilin came up with a proof and showed it to his collegues, who all agreed that they can't locate errors in it (while stating that they are not number theory specialists and are unqualified to say for certain). The article gives the proof, which is hard to understand because of how math is written, but it is at least almost identical to the one shown in the video. So, long story short, Alexander actually believed he had found the proof, but he was never a theoretical mathematician, let alone a number theory specialist, so the error is somewhat understable.
But the proof has nothing to do with number theory and, in particular, the claim that no angle has a rational cosine isn't a claim about number theory (it's real analysis). And that claim is so easily seen to be false that I'm just amazed that the author and his friends didn't spot it. Even in high school, we learn that cos(pi/3)=1/2. The fact that _every_ rational between +1 and -1 inclusive is the cosine of some angle is obvious from the intuitive meaning of continuity, even if you haven't done the first year of a university mathematics degree, so you don't know the words "intermediate value theorem".
2:36 "and that's because we definitely cannot fit the proof of Fermat's Last Theorem on this chalkboard". So you're saying that... you have a truly marvelous proof of this, which this chalkboard is too narrow to contain? 😅
This reminds me of when I proved Fermat's Last Theorem. I was using inequalities, trying to exploit Roth's Theorem by using the Taylor polynomials and remainders for y = (1-x^n)^(1/n). It taught me the danger of "working backwards" when dealing with inequalities. I found the error after hunting through it for at least an hour (assuming you're an idiot is always a safer bet than assuming you're brilliant). So fortunately I never showed anyone my "proof". (The essence of my error worked like this: If you only have that x > 10, and your desire is to prove that x > 20, then to see what logical steps it would take to get from x > 10 to x > 20 (using a few other things you know about x), start at the finish line by assuming x > 20 and then work backwards to see if you can't find the path to x > 10. Sure enough, if you start with x > 20, you can indeed connect all the dots back to x > 10, and so you have your proof!)
Hi - If you haven’t already made a video about it, could you do one explaining the ideas of Lamé’s incorrect proof from 1847? It was quickly determined to be incorrect due to a flawed assumption, and I don’t understand the details, but I’m curious about it because it seems like the technique he used wouldn’t necessarily have been beyond the reach of Fermat, implying that (perhaps) Fermat did in fact have an (incorrect) proof in mind when he wrote the famous note in the margin. I’d appreciate any insight you have into Lamé’s idea or whether Fermat might possibly have been thinking along the same lines, and only discovered the error in the reasoning later. Thanks in advance, and thanks for all your entertaining videos!
It's not true, it does not look true, there is no common misconception of it being true, it's not counterintuitive, it just comes how of nowhere from someone who didn't even bother to make his proof look based on anything maths-related. That really amazes me 🤔
Other commenters have speculated that the author may have misunderstood Niven's theorem, says that, for all rational a with 0 < a < 1/2, except for a=1/3, cos (a.pi) is irrational. But, even then, there's no contradiction: we know that cos(theta) is rational, so that just means that theta is either pi/3 or some irrational multiple of pi, either of which is fine.
This makes me think that probably Fermat didn't have the proof he claimed, because of how difficult the present proof is and how many efforts like this one have been made and failed
Et ,on rabaisse FERMAT encore une fois. Il a eu raison de ne pas révéler sa découverte, car il aurait été victime d'ingratitude comme il l'est avec de pareils propos !
For a moment there I expected we were going to try and (wrongly) prove that theta is greater than pi/2, thus the contradiction. Kind of a let down that the author went with what he did.
If anyone is interested, there are infinitely many solutions to to z^2=x^2+y^2-xy and z^2=x^2+y^2+xy. They are called Eisenstein triples. I actually did a talk about this a couple of days ago. I researched this topic years ago and my goal is to create a systematic tree of all solutions.
How would such a tree look? Obviously if (x,y,z) is an Eisenstein Triple, then (ax,ay,az) is one as well. Also, for all given triples (x,y,z) i z is given by the equation, so if we drop that, we only have to look at pairs (x,y). Then we only have to really consider pairs (x,y) which have a gcd of 1, for above reasons. So the question is, for what values of x, are there values for y with gcd(x,y) = 1, such that x^2+y^2-xy is a square.
Yes there is. z^n = x^n + y^n, so z > x and z > y and, furthermore, z+x > y and z+y > x (source: another comment). And (x+y)^n > x^n + y^n > z^n, so x+y > z. So every "side" is greater in length than the sum of the other two, so they are actual sides of an actual triangle.
It is acknowledged at 4:09 that every power is an integer squared multiplied by the integer raise to the power minus 2. Basic laws of indices. Therefore a Fermat triple is impossible for every power above 2 due to the Distributive Law of Multiplication. Q.E.D.. So yes, a complete and absolute proof does fit on the chalk board with space to spare.
Proof of Fermat's Last Theorem for Village Idiots (works for the case of n=2 as well) To show: c^n a^n + b^n for all natural numbers, a,b,c,n, n >1 c = a + b c^n = (a + b)^n = [a^n + b^n] + f(a,b,n) Binomial Expansion c^n = [a^n + b^n] iff f(a,b,n) = 0 f(a,b,n) 0 c^n [a^n + b^n] QED n=2 "rectangular coordinates" c^2 = a^2 + b^2 + 2ab Note that 2ab = 4[(1/2)ab] represents the areas of four right triangles) "radial coordinates" Lete p:= pi, n= 2 multiply by pi pc^2 = pa^2 + pb^2 + p2ab Note that pc^2, pa^2, and pb^2 represent areas of circles, wile p2ab = a(2pb) is the product of a radius (a) and a circumference (2pb). This proof also works for multi-nomial functions. Note: every number is prime relative to its own base: a = a(a/a) = a(1_a) a + a = 2a (Godbach's Conjecture (now Theorem.... :) (Wiles' proof) used modular functions defined on the upper half of the complex plane. Trying to equate the two models is trying to square the circle. c = a + ib c* - a - ib cc* = a^2 + b^2 #^2 But #^2 = [cc*] +[2ab] = [a^2 + b^2] + [2ab] so complex numbers are irrelevant. Note: there are no positive numbers: - c = a-b, b>a iff b-c = a, a + 0 = a, a-a=0, a+a =2a Every number is prime relative to its own base: n = n(n/n), n + n = 2n (Goldbach) 1^2 1 (Russell's Paradox) In particular the group operation of multiplication requires the existence of both elements as a precondition, meaning there is no such multiplication as a group operation) (Clifford Algebras are much ado about nothing) Remember, you read it here first) There is much more to this story, but I don't have the spacetime to write it here.
Just to add, it might be interesting (obvious) to note that for theta in (0,pi/2) we have infinite values of theta such that cos(theta) is rational, but the measure of this set of theta has measure 0.
This proof, on the other hand, is staggeringly obviously false. I mean, it's entirely plausible, and then it just makes the flat-out false claim that cosines are never rational.
Please specify the direct link to the article in the blogspot, then I will be able to evaluate what its author or the publisher meant. The proof seems so strange... The questions is as follows, "Is there some simpler proof?" I have read a book about the Wiles' proof, there was written that he spent more that a decade and made a correction after the first presentation. As to Fermat, here is a question whether he had found a correct proof, anyway according to his remark the proof had to be substantially shorter than more than 100 pages.
Yes. z^n = x^n + y^n, so z > x and z > y and, furthermore, z+x > y and z+y > x (source: another comment). And (x+y)^n > x^n + y^n > z^n, so x+y > z. So every "side" is greater in length than the sum of the other two, so they are actual sides of an actual triangle.
There are an infinite number of rational numbers (fractions) that can be cos(theta), like cos(theta)=1/2, 1/3, 1/4, 2/3 etc. as long as the fraction is between zero and one.
@@tcoren1 No he is making a geometric argument, where cos(theta) is a relation between two sides of a triangle, so it is between 0 and 1. Cos(theta) can be expanded by the unit circle to also include negative numbers between -1 and 0, but that doesnt apply here.
@@dariosilva85 either way, my point is, it's not an intimate number of rational numbers, it's all rational numbers. Every single rational number between -1 and 1 is the cosine of some angle
@@tcoren1 Yeah, that is my whole point in the comment. Except that I would not say all rational numbers between -one and one, because he is making a geometric argument, not a trigonometric argument. cos(theta) has been artifically expanded by mathematicians by using cos(theta) and sin(theta) as x-y-coordinates of a unit circle. Originally cosine and sine came from triangles, and therefore could not be negative numbers. In the false proof, cosine must be positive. It is an important detail.
A good thing is that this exercice uncovers triples that mimic pythagorean triples. In fact if x^n +y^n> z^n , let x^2 + y ^2 =r^2 as shown. Then also r>z. Let cos(theta)= a/b = (x^2 +y^2 _ z^2)/2xy , then the triplet ((by _ ax) , ax , bY ) is a pseudo pythagorean triplet i.e we have: (ax)^2 +(bY)^2 = (by _ ax)^2. Y= sqrt( z^2 _x^2) and ( z^2 _x^2) is never a perfect square. Example: z=13 , x=11 , y =9 , then a=1 , b=6 and Y=sqrt(13^2 _ 11^2)=4sqrt(3). So the triplet (1×11 , 24sqrt(3) , (6×9_ 1×11) is a pseudo pythagorean triplet. We have 11^2 + (24sqrt(3))^2=43^2 We can find as many of these triplets as we wish. Another example : z=17 , x= 13 , y=11. 13^2+11^2 > 17^2. We find a=1, b=286. Y=sqrt(17^2 _ 13^2)=sqrt(120). The triplet (13, 286sqrt(120) , 3133) is a pseudo pythagorean triplet.
How about expressing X^n and Z^n etc. as sums of integers and then showing that their difference can never match such a pattern? So that Y^n can't be equal Z^n - X^n
It is an interesting argument yeah. I like it because it's understandable why somebody could think it would work. It's complicated enough that it wouldn't be shocking that nobody had noticed it before if it had worked, but also uses methods which are understandable. I wonder if something like this was the proof Fermat himself claimed to have had.
@@CallMeIshmael999 thinking cosine of an angle equaling a rational number is a contradiction, is so obviously wrong you can see it either by a counter-example or the Intermediate value theorem or just apply the law of cosines to a triangle with all sides being rational numbers and cosine of that angle will also be a rational number
@@DendrocnideMoroides Fair. I was more referring to the process leading up to it which seemed like a reasonable approach. I agree that thinking cosine has no rational values is a silly mistake.
-- I've took two prime numbers and other two prime numbers multiplied them through and got the same answer -- And in which ring? -- what is a ring anyway?
3:45 "You might say, `Well, what about bigger than or equal to, since x and y could both be equal to one?' " No they can't. Suppose x=1. Then y^n and z^n are two n-th powers of integers that differ by one, which isn't possible for any integer n>1.
@@galoomba5559 Yup -- you just end up concluding that, by Niven's theorem, theta is irrational, which is fine, as there are plenty of irrational angles.
It totally feels like there was something lost in translation, that's way too big of a hole in this proof. What I find interesting is, there exists elementary proof that Fermat's last theorem is true for all even exponents. It fits on single page. And it works with trigonometric functions. I really wonder whether there isn't some similarity. Maybe, just maybe, this is part of a variant of that proof?
I saw proof by some Indian professor. It was based on modulo 2^n. As far as I can recall it was 8. So he wrote table of remainders and claimed they don't addup modulo 8. And thus they won't addup in non-modulo case.
For powers of 3 or higher, working mod 8, the evens are clearly 0 mod 8. The odds will be congruent to 1 if n is even and congruent to themselves if n is odd. So if n is odd, you could have 1+7= 0 (mod 8) or 3+5= 0 (mod 8). Or there could be a solution with all even xyz. So you have proven if there is a solution, z is even and x and y are odd and add to a multiple of 8. Proving further that there is no solution needs additional steps as far as I can tell.
@@wafelsen one can rely on fact that FGT is disproven for n=3,4 and that further proof for all other n only requires proof for primes >3. by this we simplify this argument to odd case. And since we assume that all (x,y,z) to be coprime to eachother. I guess there can only be 1+7= 0 (mod 8) or 3+5= 0 (mod 8). You said that. And after this I have no idea what's next. :)
What a great video! Fermat said: "I have a marvelous proof, but it will not fit in the margin." he did NOT say: "I have a marvelous proof, but I need to invent a new mathematics, prove some other thing and then write a 100+ page paper." Either a) He was mistaken or b) We have yet to discover his proof.
I recall a more interesting false proof I believe by Cauchy that uses complex numbers. Cauchy's mistake was that complex numbers with natural number components do not have a unique factorization.
4:29 is n is ODD, then having something like (x^2+y^2)^5.5 I can't simply use Binomial Theorem, as it works for natural numbers, but not for rational n. What do I do there? Use the floor function? Some other formula? How do I "expand" it? Why did you skip explanation on this part?
Yes, that argument seems to be bogus, but the claim that (a+b)^c > a^c + b^c is true. One proof is that, for all c>1, the function f(z) = z^c is convex for z>0, and f(0) = 0. From this, it follows that f(a+b) > f(a) + f(b) for all a,b > 0. See the Wikipedia page "Convex function" for the details.
Is it true that there are only a handful of people in the world that can understand the proof by Andrew Wiles ? I heard that many math professors themselves still do not understand it.
Yes it is. I guess nowadays elliptic curves theory is more avanced and used so you could argue that a mathematician specialized in this field could understand the proof... but most definitely it is accessible only to elliptic curves theorists. And even for specialists, to be able to understand properly a proof that takes hundreds of pages requires much time and dedication. I'd say it's not even a matter of difficulty but rather of specialization. I think it would take many years for the most brilliant minds to catch up with the amount of underlying knowledge required to understand Wiles proof
So, after sitting through the beginning the fallacy is that the "proof" depends on cos(theta) can't be rational! But it is trivial to prove that every rational number of absolute value
Other commenters have speculated that the author may have misunderstood Niven's theorem, says that, for all rational a with 0 < a < 1/2, except for a=1/3, cos (a.pi) is irrational. But, even then, there's no contradiction: we know that cos(theta) is rational, so that just means that theta is either pi/3 or some irrational multiple of pi, either of which is fine.
For people that has mathematics in their blood: Maybe mathematicians got tired and accepted a very complicated path and 100 pages so that only a minimum number of mathematicians understand what the proof is. Here's a different approach to the Last Theorem on a single page: czcams.com/video/-jpA-tr68ww/video.html
I keep telling you all I have solved this. Wiles is wrong. Here is the proof. 1.Add the 3 natural logs, 2,3,5 such that 2+3=5 . 2. Raise each exponent to the power of 4 such that 2^4+3^4=5^4. 3.Given the power rule of natural logs where q^n=nq then (4x2)+(4x3)=(4x5)=8+12=20. 4. QED. Tell the Abel people to send me my prize money.
Without using De Morgan’s Laws, P∧Q=¬(¬P∧¬Q) , by just using the distributive property. Let P=K≤2 and Q=∃A,B,C(A^K+B^K=C^K ). Also let A,B,C,K∈N . So we have P∧Q which means: When K is less than or equal to 2, there exists A,B, and C such that (A^K+B^K=C^K ). So we have P∧Q=¬(K>2∧∀A,B,C(A^K+B^K≠C^K ))
I'm at the 6:03 mark and I'm guessing the 'incorectness' of the proof is going to stem from assuming z satisfies the triangle inequality? EDIT: I guess it wasn't. By the way, what's with the video description?
No, z satisfies the triangle equality just fine. z^n = x^n + y^n, so z>x and z>y, so z+x > y and z+y > x. Also, (x+y)^n > x^n + y^n = z^n, so that side works, too.
I wonder whether n=3 works with three terms in the sum. This would be suitable for 3D instead of a complex plane. If so, I wonder whether giving n the number of terms in the sum as well as the exponent might yield infinite sets. This probably could be explored using Clifford Algebra.
Furthermore if we go to 5th powers it was (famously) found in 1966 that it can be done with only 4 terms, disproving a conjecture by Euler that you need at least n terms to make an nth power: 27^5 + 84^5 + 110^5 + 133^5 = 144^5
@ Fisher99 -- If you're going to ask about an intended value of cosine of pi/6 radians, then you are to put pi/6 inside of grouping symbols. You just ran a bunch of symbols together. *cos(pi/6)*
I believe Goldbach's conjecture is of a very different kind. We knew for a long time that Fermat's big theorem was linked to the different ways of decomposing x^n+y^n as a product in certain spaces enlarging the natural numbers. This decomposition problem has been studied for so many years and is one of the most prominent domain of modern mathematics, even though it is extremely hard. To my knowledge, Goldbach's conjecture comes out of nowhere and it is very hard to distinguish a link between it and any research field in mathematics nowadays. Not to diminish the absolutely incredible work if Wiles, but he found the proof thanks to the amazing development of his mathematical domain and using many tools developed by other mathematicians as well. I would be very interested to know where a possible proof of Goldbach's conjecture would come from, but for now I don't believe it's a possibility at all. I'd like to be surprised though!
Small detail: the numbers x, y, z belong to N* not N, if it is just N then 0 is allowed and you get infinite solutions of the form(s) (0 k, k) , (k, 0, k), with one of x or y being 0 and the other taking any value, which z also takes
Professor Penn is of the subset of mathematicians who do not consider 0 to be in N. (Personally, I'm with you, but we can't force others to be right.😂)
No, the set of natural numbers is not consistently defined as the set of nonnegative integers around the world. What Michael Penn should do each time instead for a problem that uses it, is to write/state that his variable(s) belong to "the set of positive integers" to clarify that immediately and not "N/the set of natural numbers."
So this does partition the possible solutions to FLT from an uncountable solution set to a countable (infinite) number of possible solution. So reducing the order of magnitude of a solution set is a good start I would say. But then that is already known since the set of x, y, n are countable, so the combination of them are countable.
Yes, but I was not thinking about that until the cos() was rational, then I started typing, and ended up half way realizing it was always countable, which was why the second half of the post.
Doesn't the argument fall apart earlier because there's no way we can shorten one side of a triangle while keeping the other two sides the same length (one would have to become longer)?
No, it might look that way, though. If you just rotate the side of length y a little ways towards the side of length x and connect the end point you get a new (acute) triangle with two of the same side lengths as you started with. There are restrictions on how small the third side length can get, though, i.e. the Triangle Inequality.
if you are so good to combat simple proofs of fermat last theorem then like you to show that the fermat last theorem proof shown in the pinned comment to the video title fermat last theorem 1637proof is correct to embarrass mr.anddrew wiles.george ice
cos = b/c in a right triangle so it definitely can be rational. If cos can't be rational, you can prove a whole bunch of nonsense. This is a total time waster.
It seems that you pronounce the name "Fermat" not quite correctly. You pronounce it as if it were "Furmat", the first syllable sounding like "fur". The name "Fermat" is correctly pronounced more like "Fairmat", the first syllable sounding like "fair" as in a short "air", with no emphasis on the "i", .
Hmm... , my approach would be: (a+b)^n = a^n + ... (n|n-j) a^n-j b^j ... + b^n The middle stuff to the other side: a^n + b^n = (a+b)^n - ... (n|n-j) a^n-j b^j ... And then showing somehow that right side can't be of the form: z^n for n>2 Know, sounds easier than it is. But was my first idea. And with more time, maybe ... 🙃
Meh. Was this supposed to be an April fool? The error in the proof (the claim that cosines can't be rational) is so egregious that I can't see any other reason for talking about the rest of the proof.
"We most definitely cannot fit the proof of Fermat's Last Theorem on this chalkboard." I see what you did there.
this meme is wildly overused in the math community, and I love it
#MyCommentButBetter
I didn’t catch that
That’s great😂
Only because he wrote to the edge of the board and didn't leave a big enough margin
Wait I don’t get it
I was expecting the false contradiction to be more subtle but they pretty much just gave up and shat their pants.
Yeah, exactly... I mean if it's obviously wrong, what is the point of talking about it?
The author of the article was referring to Niven's theorem, according to which cos(theta) and theta (in radians) are never both rational except for theta=0. The author of the article has completely misrepresented this theorem because in its construction theta can be irrational.
Thanks for the explanation. I knew it couldn't have been as simple as it was presented here.
Thanks for that. It seemed totally bizarre that somebody would do all that fairly smart construction and then end with "But the cosine of an angle is never rational; contradiction." It's obvious by continuity that every rational in the range [-1,1] is the cosine of some angle.
The description was great.
Also, showing errors is very instructive.
Thank you, professor.
Here I was keeping a lookout the whole time to try to pick where the proof falls apart, and the twist turns out to be that the problem with it is glaringly obvious!
(Not, as Prof. Penn says, that there couldn't be a reasonable proof based on the premise that some derived quantity is rational when it couldn't be, but "any value whatsoever of cosine" isn't that.)
Yes, that was rather anticlimactic. I was expecting something that looked correct at first glance, not just a completely and obviously incorrect statement. I'm not really sure what the point of this video was...
@@thomasdalton1508 Yeah, also the "proof" was all about roots, inequalities, and cosines -- felt like there would be an error of a forgetting-a-second-solution kind.
Right? I paused a few times during the video to really make sure that every step was justified, only for the "contradiction" to be something totally ridiculous lol
@@christianaustin782 I also took a lot of time convincing myself that "shortening r to z" was actually a thing we were allowed to do and would result in a real triangle. I've seen errors come up in that sort of step often.
I'm going to need a source for cos(pi/6)=1/2.
Well cos(pi/3) = 1/2 so cos(pi/6)=1/2 as pi/6 tends to pi/3. Seems convincing to me.
It's easy to prove in a universe where cos=sin.
I wonder if Fermat's famous margin note was referring to a proof similar to this but he probably figured out his error quickly and didn't correct his margin note for whatever reason.
If he did find this proof but found out it was wrong, maybe he left it in intentionally so that someone could make a better proof.
@@astralnekomimi C'est aujourd'hui fait !
to see fermat proof check the pinned comment to the video titltlefermat last theorem 1637prooof.do you see any error?
Description is amazing
Yes. Especially the last sentence, with the 'm' words.
Seems to be made with ChatGPT4 to me !
@speedsterh I'm sorry to tell you that I write all the descriptions, not ChatGPT4.
Stephanie
-MP Editor
I'm more proud of the "m" sentence than my "w" sentence for sure. Glad you liked it haha. I write wild descriptions because it's helps get people to you know, read the stuff we want them to see lol there. now you know my secret.
-Stephanie
MP Editor
@@MichaelPennMath Looked very much like something like ChapGPT4 could do ! Sorry if I sounded like I belittled your human handmade work !
There is another issue with the shortening of triangle. For x,y,z to be sides of a triangle, they need to satisfy the triangle inequality. Now we know for sure that x,y,r are sides of a right triangle and we know that z < r, but if we assume wlog that x>=y, then it might very well be the case y+zx and z>y, and this tells us that z+x>y and z+y>x. Now we can use the fact that x+y>r>z to conclude that we can infact shorten it to be an appropriate triangle.
nice observation
It is obvious we can satisfy the triangle inequality, no need to prove the trivial
If z^n = x^n + y^n, then trivially z > x and z > y, yielding both z+x > y and z+y > x.
That, people, is why we have:
- "proof is left to the reader"
- peer review
@@tetraedri_1834 x,y,z aren't random, hence it's trivial
One of my Professors once told me, that he received at least 10 (wrong) proofs of FLT (mostly from non-mathematicians ;-) ) a week.
Aujourd'hui, c'est la bonne !
Tbh, this could be made a whole video series. All too much and too often we forget to look at past failures. But failure often teaches way more than success. On this example: I've definitely learned something from this whereas even if I tried to digest the correct proof of the theorem, I'd likely fail on page 1 and would need to do a two years preparation course to even start with it.
Imo the best false proof of FLT is the one by Lamé that falsely assumes unique factorization, and ultimately leads to proving FLT in the particular case when n is equal to a Sophie Germain prime
Very interesting video about how mathematical proofs are elaborated. Small correction: I believe that cos(pi/3) = 0.5, not cos (pi/6) :)
Maybe Michael was using pi = tau
:^)
During my undergraduate I was solving a problem like "If x, y and z are sides of a right triangle such that z is the hypotenuse, show that z³ > x³ + y³".
From that I generalized for all n≥3 natural, we have zⁿ > xⁿ + yⁿ, where x, y and z are a pythagorean triple.
I mentioned this to my professor claiming that I proved Fermat's last theorem for a tiny subset of natural numbers =D
For this subset of the natural triples, namely the Pythagorean triples, you not only proved Fermat's last theorem, but also proved more: that the inequality > holds for this subset, since Fermat's theorem only states that it is different.
And that would, in fact, fit in the margin of a page... Wonder what Fermat thought he had??
@@TonyHammitt i doubt he had a valid proof , if he did he would be the greatest mathematician in the history proving a statement which baffled people 100s of years after him.
@@TonyHammitt that's only for pythagorean triples bruh
Correct me if I'm wrong, but that doesn't show FLT is true for x,y,z? Because even though z^n > x^n + y^n, that's only true for a specific z. You haven't proved there exists no other number a such that a^n = x^n + y^n?
Hopefully, this comment will shed some light on the situation.
TL;DR: author actually believed the proof to be correct, but he wasn't a number theorist.
I decided to find the original article myself, which proved to be a challenge, since the link provided by the blogspot didn't lead anywhere. Fortunately, from the link I was able to identify the name of the newspaper, went to their archives, and searched via key words.
The name of the article loosely translates to "May the humanity rest now?". It goes into history of Fermat's Last Theorem, how Euler and Gauss were unable to solve it. It also briefly mentions one russian mathematician who presumably went insane trying to solve it (I heard nothing about such a story). The article is rather poorly written and unpleasant to read, especially where there are any mathematical formulas involved (they are written 'as is', for example, FLT is "xn + yn = zn").
The article then tells the story of Alexander Ilin, author of the proof. Funnily enought, he studied at the university in my hometown. It is stated that Ilin worked on a space program, and then for no reason at all the article mentions how Alexander was taken to court for money laundering and how he defended himself. After that, he dedicated himself to proving FLT.
Finally, the article states that Ilin came up with a proof and showed it to his collegues, who all agreed that they can't locate errors in it (while stating that they are not number theory specialists and are unqualified to say for certain). The article gives the proof, which is hard to understand because of how math is written, but it is at least almost identical to the one shown in the video.
So, long story short, Alexander actually believed he had found the proof, but he was never a theoretical mathematician, let alone a number theory specialist, so the error is somewhat understable.
But the proof has nothing to do with number theory and, in particular, the claim that no angle has a rational cosine isn't a claim about number theory (it's real analysis). And that claim is so easily seen to be false that I'm just amazed that the author and his friends didn't spot it. Even in high school, we learn that cos(pi/3)=1/2. The fact that _every_ rational between +1 and -1 inclusive is the cosine of some angle is obvious from the intuitive meaning of continuity, even if you haven't done the first year of a university mathematics degree, so you don't know the words "intermediate value theorem".
@@beeble2003 Michael kind of straw-manned the argument from the original paper here see comment by @VideoFusco
In Soviet Russia Fermat's Last Theorem proves you!
,
Ok
2:36 "and that's because we definitely cannot fit the proof of Fermat's Last Theorem on this chalkboard".
So you're saying that... you have a truly marvelous proof of this, which this chalkboard is too narrow to contain? 😅
LOL
Great lecture in Trondheim, it was a pleasure to see you live
This reminds me of when I proved Fermat's Last Theorem. I was using inequalities, trying to exploit Roth's Theorem by using the Taylor polynomials and remainders for y = (1-x^n)^(1/n). It taught me the danger of "working backwards" when dealing with inequalities. I found the error after hunting through it for at least an hour (assuming you're an idiot is always a safer bet than assuming you're brilliant). So fortunately I never showed anyone my "proof".
(The essence of my error worked like this: If you only have that x > 10, and your desire is to prove that x > 20, then to see what logical steps it would take to get from x > 10 to x > 20 (using a few other things you know about x), start at the finish line by assuming x > 20 and then work backwards to see if you can't find the path to x > 10. Sure enough, if you start with x > 20, you can indeed connect all the dots back to x > 10, and so you have your proof!)
Hi - If you haven’t already made a video about it, could you do one explaining the ideas of Lamé’s incorrect proof from 1847? It was quickly determined to be incorrect due to a flawed assumption, and I don’t understand the details, but I’m curious about it because it seems like the technique he used wouldn’t necessarily have been beyond the reach of Fermat, implying that (perhaps) Fermat did in fact have an (incorrect) proof in mind when he wrote the famous note in the margin. I’d appreciate any insight you have into Lamé’s idea or whether Fermat might possibly have been thinking along the same lines, and only discovered the error in the reasoning later. Thanks in advance, and thanks for all your entertaining videos!
I don't know if Michael will see it so if he doesn't I might be able to take a look at this if I get some bored time do you have a link to the proof
I don't quite get it: Why would anyone believe that cosine of some angle being rational leads to a contradiction?
It's not true, it does not look true, there is no common misconception of it being true, it's not counterintuitive, it just comes how of nowhere from someone who didn't even bother to make his proof look based on anything maths-related.
That really amazes me 🤔
Other commenters have speculated that the author may have misunderstood Niven's theorem, says that, for all rational a with 0 < a < 1/2, except for a=1/3, cos (a.pi) is irrational. But, even then, there's no contradiction: we know that cos(theta) is rational, so that just means that theta is either pi/3 or some irrational multiple of pi, either of which is fine.
This makes me think that probably Fermat didn't have the proof he claimed, because of how difficult the present proof is and how many efforts like this one have been made and failed
(First!)
Et ,on rabaisse FERMAT encore une fois. Il a eu raison de ne pas révéler sa découverte, car il aurait été victime d'ingratitude comme il l'est avec de pareils propos !
since like prof.wiles you do not know the high school algebra you think flt was had for fermat
Fermat decided not providing a proof was a good place to stop.
It would be nice if you covered some wrong proofs for the Collatz Conjecture
Look in any YT comment section for videos that cover Collatz. You'll find plenty! 🤣
Lmao there’s too many
2:39 And certainly not in its margins.
For a moment there I expected we were going to try and (wrongly) prove that theta is greater than pi/2, thus the contradiction. Kind of a let down that the author went with what he did.
If anyone is interested, there are infinitely many solutions to to z^2=x^2+y^2-xy and z^2=x^2+y^2+xy. They are called Eisenstein triples. I actually did a talk about this a couple of days ago. I researched this topic years ago and my goal is to create a systematic tree of all solutions.
How would such a tree look? Obviously if (x,y,z) is an Eisenstein Triple, then (ax,ay,az) is one as well. Also, for all given triples (x,y,z) i z is given by the equation, so if we drop that, we only have to look at pairs (x,y). Then we only have to really consider pairs (x,y) which have a gcd of 1, for above reasons. So the question is, for what values of x, are there values for y with gcd(x,y) = 1, such that x^2+y^2-xy is a square.
@@spankasheep For example, with the + version. (7,8,13) is connected to (55,57,97).
@@spankasheep Maybe something analogous to the trees of Pythagorean triples? en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples
@@iang0th That's a really cool property
I was able to follow this one! Thanks!
When we shorten r to z, there's no guarantee we still have a triangle
Yes there is. z^n = x^n + y^n, so z > x and z > y and, furthermore, z+x > y and z+y > x (source: another comment). And (x+y)^n > x^n + y^n > z^n, so x+y > z. So every "side" is greater in length than the sum of the other two, so they are actual sides of an actual triangle.
I think the problem is not in rationality of cosine, but in the fact you cannot build this triangle because of the triangle inequality.
cos(pi/3)=1/2
I wish some math channel would cover various open problems, progress on them to date, perhaps some incorrect proofs/attempts
It is acknowledged at 4:09 that every power is an integer squared multiplied by the integer raise to the power minus 2. Basic laws of indices. Therefore a Fermat triple is impossible for every power above 2 due to the Distributive Law of Multiplication. Q.E.D.. So yes, a complete and absolute proof does fit on the chalk board with space to spare.
2:37 "We most definitely cannot fit the proof of Fermat's last theorem on this chalkboard"
lol #IseeWhatYouDidThere
What did he do there
Proof of Fermat's Last Theorem for Village Idiots
(works for the case of n=2 as well)
To show: c^n a^n + b^n for all natural numbers, a,b,c,n, n >1
c = a + b
c^n = (a + b)^n = [a^n + b^n] + f(a,b,n) Binomial Expansion
c^n = [a^n + b^n] iff f(a,b,n) = 0
f(a,b,n) 0
c^n [a^n + b^n] QED
n=2
"rectangular coordinates"
c^2 = a^2 + b^2 + 2ab
Note that 2ab = 4[(1/2)ab] represents the areas of four right triangles)
"radial coordinates"
Lete p:= pi, n= 2
multiply by pi
pc^2 = pa^2 + pb^2 + p2ab
Note that pc^2, pa^2, and pb^2 represent areas of circles, wile p2ab = a(2pb) is the product of a radius (a) and a circumference (2pb).
This proof also works for multi-nomial functions.
Note: every number is prime relative to its own base: a = a(a/a) = a(1_a)
a + a = 2a (Godbach's Conjecture (now Theorem.... :)
(Wiles' proof) used modular functions defined on the upper half of the complex plane. Trying to equate the two models is trying to square the circle.
c = a + ib
c* - a - ib
cc* = a^2 + b^2 #^2
But #^2 = [cc*] +[2ab] = [a^2 + b^2] + [2ab] so complex numbers are irrelevant.
Note: there are no positive numbers: - c = a-b, b>a iff b-c = a, a + 0 = a, a-a=0, a+a =2a
Every number is prime relative to its own base: n = n(n/n), n + n = 2n (Goldbach)
1^2 1 (Russell's Paradox)
In particular the group operation of multiplication requires the existence of both elements as a precondition, meaning there is no such multiplication as a group operation)
(Clifford Algebras are much ado about nothing)
Remember, you read it here first)
There is much more to this story, but I don't have the spacetime to write it here.
Just to add, it might be interesting (obvious) to note that for theta in (0,pi/2) we have infinite values of theta such that cos(theta) is rational, but the measure of this set of theta has measure 0.
Alfred Kempe gave an incorrect proof of the four-color theorem in 1879, but it wasn't shown to be incorrect until 1890, over 10 years later
This proof, on the other hand, is staggeringly obviously false. I mean, it's entirely plausible, and then it just makes the flat-out false claim that cosines are never rational.
What on earth is that in your description?
I'm terribly sorry, but wouldn't any triangle with natural number sides have all of its angles have rational cosines?..
Yes. Another reason why it's stunningly obvious that the claim of a contradiction is in error.
Through that cosine formula one can see that the angles in a triangle with rational sides will have rational cosines, that's interesting
Please specify the direct link to the article in the blogspot, then I will be able to evaluate what its author or the publisher meant. The proof seems so strange... The questions is as follows, "Is there some simpler proof?" I have read a book about the Wiles' proof, there was written that he spent more that a decade and made a correction after the first presentation. As to Fermat, here is a question whether he had found a correct proof, anyway according to his remark the proof had to be substantially shorter than more than 100 pages.
check the pinned comment to the video fermat last theorem 1637proof which is less than 2 pages and is correct proof.ask the professor t confirm it
Can you be sure that the triangle with sides x, y, and z is a valid triangle?
Yes. z^n = x^n + y^n, so z > x and z > y and, furthermore, z+x > y and z+y > x (source: another comment). And (x+y)^n > x^n + y^n > z^n, so x+y > z. So every "side" is greater in length than the sum of the other two, so they are actual sides of an actual triangle.
There are an infinite number of rational numbers (fractions) that can be cos(theta), like cos(theta)=1/2, 1/3, 1/4, 2/3 etc. as long as the fraction is between zero and one.
And that infinite number is "all of them" (between -1 and 1)
@@tcoren1 No he is making a geometric argument, where cos(theta) is a relation between two sides of a triangle, so it is between 0 and 1. Cos(theta) can be expanded by the unit circle to also include negative numbers between -1 and 0, but that doesnt apply here.
@@dariosilva85 either way, my point is, it's not an intimate number of rational numbers, it's all rational numbers. Every single rational number between -1 and 1 is the cosine of some angle
@@tcoren1 Yeah, that is my whole point in the comment. Except that I would not say all rational numbers between -one and one, because he is making a geometric argument, not a trigonometric argument. cos(theta) has been artifically expanded by mathematicians by using cos(theta) and sin(theta) as x-y-coordinates of a unit circle. Originally cosine and sine came from triangles, and therefore could not be negative numbers. In the false proof, cosine must be positive. It is an important detail.
A good thing is that this exercice uncovers triples that mimic pythagorean triples.
In fact if x^n +y^n> z^n , let x^2 + y ^2 =r^2 as shown. Then also r>z.
Let cos(theta)= a/b = (x^2 +y^2 _ z^2)/2xy , then the triplet ((by _ ax) , ax , bY ) is a pseudo pythagorean triplet i.e we have:
(ax)^2 +(bY)^2 = (by _ ax)^2.
Y= sqrt( z^2 _x^2) and ( z^2 _x^2) is never a perfect square.
Example: z=13 , x=11 , y =9 , then a=1 , b=6 and Y=sqrt(13^2 _ 11^2)=4sqrt(3).
So the triplet (1×11 , 24sqrt(3) , (6×9_ 1×11) is a pseudo pythagorean triplet.
We have 11^2 + (24sqrt(3))^2=43^2
We can find as many of these triplets as we wish.
Another example :
z=17 , x= 13 , y=11.
13^2+11^2 > 17^2.
We find a=1, b=286.
Y=sqrt(17^2 _ 13^2)=sqrt(120).
The triplet (13, 286sqrt(120) , 3133) is a pseudo pythagorean triplet.
How about expressing X^n and Z^n etc. as sums of integers and then showing that their difference can never match such a pattern?
So that Y^n can't be equal Z^n - X^n
It is an interesting argument yeah. I like it because it's understandable why somebody could think it would work. It's complicated enough that it wouldn't be shocking that nobody had noticed it before if it had worked, but also uses methods which are understandable.
I wonder if something like this was the proof Fermat himself claimed to have had.
this is extremely stupid Fermat would never have done this, but it could be something much less stupid
@@DendrocnideMoroides Why do you believe it's stupid? I mean besides the fact that it's wrong. Why do you think this type of thing is stupid?
@@CallMeIshmael999 thinking cosine of an angle equaling a rational number is a contradiction, is so obviously wrong you can see it either by a counter-example or the Intermediate value theorem or just apply the law of cosines to a triangle with all sides being rational numbers and cosine of that angle will also be a rational number
@@DendrocnideMoroides Fair. I was more referring to the process leading up to it which seemed like a reasonable approach. I agree that thinking cosine has no rational values is a silly mistake.
-- I've took two prime numbers and other two prime numbers multiplied them through and got the same answer
-- And in which ring?
-- what is a ring anyway?
3:45 "You might say, `Well, what about bigger than or equal to, since x and y could both be equal to one?' "
No they can't. Suppose x=1. Then y^n and z^n are two n-th powers of integers that differ by one, which isn't possible for any integer n>1.
I'm an undergrad math student but can nivens theorem be applied about rational inputs to rational outputs for sin?
There's no reason that theta would be rational
@@galoomba5559 Yup -- you just end up concluding that, by Niven's theorem, theta is irrational, which is fine, as there are plenty of irrational angles.
It totally feels like there was something lost in translation, that's way too big of a hole in this proof.
What I find interesting is, there exists elementary proof that Fermat's last theorem is true for all even exponents. It fits on single page. And it works with trigonometric functions. I really wonder whether there isn't some similarity. Maybe, just maybe, this is part of a variant of that proof?
Hi Dr. Penn!
9:04 I love how this "proof" literally doesn't hold water 😛
thank you for all your helpful work. creaky voice very painful for me.
This happens when you try to find the proof of Fermat’s last theorem after having a bottle of vodka 🥴🥴🥴
I saw proof by some Indian professor. It was based on modulo 2^n. As far as I can recall it was 8. So he wrote table of remainders and claimed they don't addup modulo 8. And thus they won't addup in non-modulo case.
For powers of 3 or higher, working mod 8, the evens are clearly 0 mod 8. The odds will be congruent to 1 if n is even and congruent to themselves if n is odd. So if n is odd, you could have 1+7= 0 (mod 8) or 3+5= 0 (mod 8). Or there could be a solution with all even xyz.
So you have proven if there is a solution, z is even and x and y are odd and add to a multiple of 8.
Proving further that there is no solution needs additional steps as far as I can tell.
@@wafelsen one can rely on fact that FGT is disproven for n=3,4 and that further proof for all other n only requires proof for primes >3. by this we simplify this argument to odd case.
And since we assume that all (x,y,z) to be coprime to eachother. I guess
there can only be 1+7= 0 (mod 8) or 3+5= 0 (mod 8). You said that.
And after this I have no idea what's next. :)
@@danielmilyutin9914 I thought about it more. even if we assume n is odd, we also have odd + even = same odd (mod 8)
@@wafelsen one more case.
i…don't even understand what the contradiction is supposed to be tbh.
What a great video! Fermat said: "I have a marvelous proof, but it will not fit in the margin." he did NOT say: "I have a marvelous proof, but I need to invent a new mathematics, prove some other thing and then write a 100+ page paper." Either a) He was mistaken or b) We have yet to discover his proof.
to see fermat proof check the pinned comment to the video fermat last theorem 1637 proof
Cos(Pi/6) = sqrt(3)/2
Can someone explain why showing r > z does not complete the proof ?
Because x^n + y^n = z^n > r^2 = x^2 + y^2
why would r > z solve the theorem?
Did Michael run out of things to prove?
Haha no. sometimes it’s illustrative to look at something like this.
-Stephanie
MP Editor
One moment please, if a mod,, then a remainder. And if a remainder, then not an integer..and thereby Wiles-Taylor outcome is a cheat.-Ernie Moore Jr.
Is it true that Euler proved FLT for n=3?
Yes
I like how if his proof was correct, all it would do is prove that x^2+y^2=z^2 has no solutions
That was neat.
7:58: “pi over three”.
Putin was putting out disinfo by putting out that proof ...😄
I recall a more interesting false proof I believe by Cauchy that uses complex numbers. Cauchy's mistake was that complex numbers with natural number components do not have a unique factorization.
Its also speculated that Fermat made the same mistake when he claimed to have a proof (too long to write in the margin of his notes)
2:35 LOL
4:29 is n is ODD, then having something like (x^2+y^2)^5.5 I can't simply use Binomial Theorem, as it works for natural numbers, but not for rational n.
What do I do there? Use the floor function? Some other formula? How do I "expand" it? Why did you skip explanation on this part?
Yes, that argument seems to be bogus, but the claim that (a+b)^c > a^c + b^c is true. One proof is that, for all c>1, the function f(z) = z^c is convex for z>0, and f(0) = 0. From this, it follows that f(a+b) > f(a) + f(b) for all a,b > 0. See the Wikipedia page "Convex function" for the details.
Is it true that there are only a handful of people in the world that can understand the proof by Andrew Wiles ? I heard that many math professors themselves still do not understand it.
Yes it is. I guess nowadays elliptic curves theory is more avanced and used so you could argue that a mathematician specialized in this field could understand the proof... but most definitely it is accessible only to elliptic curves theorists. And even for specialists, to be able to understand properly a proof that takes hundreds of pages requires much time and dedication.
I'd say it's not even a matter of difficulty but rather of specialization. I think it would take many years for the most brilliant minds to catch up with the amount of underlying knowledge required to understand Wiles proof
You should ask AI to make up some theorem and then try to prove them. Just a guess, might help u with the algo
Can't fit in the chalkboard, let alone the margin.
3:45 Why can't x = y = 1, please?
Because that reduces to 2=z^n, and z^n is trivially either 1 (for z=1) or >2 (for all other natural numbers z, given n>2).
So, after sitting through the beginning the fallacy is that the "proof" depends on cos(theta) can't be rational! But it is trivial to prove that every rational number of absolute value
Other commenters have speculated that the author may have misunderstood Niven's theorem, says that, for all rational a with 0 < a < 1/2, except for a=1/3, cos (a.pi) is irrational. But, even then, there's no contradiction: we know that cos(theta) is rational, so that just means that theta is either pi/3 or some irrational multiple of pi, either of which is fine.
For people that has mathematics in their blood: Maybe mathematicians got tired and accepted a very complicated path and 100 pages so that only a minimum number of mathematicians understand what the proof is. Here's a different approach to the Last Theorem on a single page: czcams.com/video/-jpA-tr68ww/video.html
People should start calling it Wiles' Theorem.
This is a good place to stop the still unending presentations/"proofs" that Fermat's Last Conjecture now Theorem is false!
I keep telling you all I have solved this. Wiles is wrong. Here is the proof.
1.Add the 3 natural logs, 2,3,5 such that 2+3=5 .
2. Raise each exponent to the power of 4 such that 2^4+3^4=5^4.
3.Given the power rule of natural logs where q^n=nq then (4x2)+(4x3)=(4x5)=8+12=20.
4. QED. Tell the Abel people to send me my prize money.
Without using De Morgan’s Laws, P∧Q=¬(¬P∧¬Q) , by just using the distributive property. Let P=K≤2 and Q=∃A,B,C(A^K+B^K=C^K ). Also let A,B,C,K∈N . So we have P∧Q which means: When K is less than or equal to 2, there exists A,B, and C such that (A^K+B^K=C^K ). So we have P∧Q=¬(K>2∧∀A,B,C(A^K+B^K≠C^K ))
the proposition Q as an implication of the proposition P[ k
I'm at the 6:03 mark and I'm guessing the 'incorectness' of the proof is going to stem from assuming z satisfies the triangle inequality?
EDIT: I guess it wasn't. By the way, what's with the video description?
No, z satisfies the triangle equality just fine. z^n = x^n + y^n, so z>x and z>y, so z+x > y and z+y > x. Also, (x+y)^n > x^n + y^n = z^n, so that side works, too.
I wonder whether n=3 works with three terms in the sum. This would be suitable for 3D instead of a complex plane. If so, I wonder whether giving n the number of terms in the sum as well as the exponent might yield infinite sets. This probably could be explored using Clifford Algebra.
3^3 + 4^3 + 5^3 = 6^3
Furthermore if we go to 5th powers it was (famously) found in 1966 that it can be done with only 4 terms, disproving a conjecture by Euler that you need at least n terms to make an nth power:
27^5 + 84^5 + 110^5 + 133^5 = 144^5
@@btd6vids I still have good instincts; good to know! Now, if I could have an original idea we would be in business...
And what about using a geometric proof for a mathematical theorem? I'm not sure it's quite legit, am I wrong?
You're wrong. Geometry is a part of mathematics.
cospi/6 is 1/2??
I think pi/3 was meant
@ Fisher99 -- If you're going to ask about an intended value of cosine of pi/6 radians, then you
are to put pi/6 inside of grouping symbols. You just ran a bunch of symbols together.
*cos(pi/6)*
@@robertveith6383 it's commonly understood, I was just lazy
We know cos theta can be rational because Pythagorean triples exist.
Speaking of classic problems... when is someone going to be the Andrew Wiles who actually comes up with a formal proof of Goldbach's conjecture?
I believe Goldbach's conjecture is of a very different kind. We knew for a long time that Fermat's big theorem was linked to the different ways of decomposing x^n+y^n as a product in certain spaces enlarging the natural numbers. This decomposition problem has been studied for so many years and is one of the most prominent domain of modern mathematics, even though it is extremely hard.
To my knowledge, Goldbach's conjecture comes out of nowhere and it is very hard to distinguish a link between it and any research field in mathematics nowadays. Not to diminish the absolutely incredible work if Wiles, but he found the proof thanks to the amazing development of his mathematical domain and using many tools developed by other mathematicians as well.
I would be very interested to know where a possible proof of Goldbach's conjecture would come from, but for now I don't believe it's a possibility at all. I'd like to be surprised though!
Are there any known alternative statements that imply Goldbach’s conjecture? Prove one of them not Goldbach directly, that’s how Wiles proved Fermat.
Small detail: the numbers x, y, z belong to N* not N, if it is just N then 0 is allowed and you get infinite solutions of the form(s) (0 k, k) , (k, 0, k), with one of x or y being 0 and the other taking any value, which z also takes
Professor Penn is of the subset of mathematicians who do not consider 0 to be in N. (Personally, I'm with you, but we can't force others to be right.😂)
No, the set of natural numbers is not consistently defined as the set of nonnegative integers around the world. What Michael Penn should do each time instead for a problem that uses it, is to write/state that his variable(s) belong to "the set of positive integers" to clarify that immediately and not "N/the set of natural numbers."
It's worth knowing that there are these two definitions of ℕ as part of knowing _any_ definition of ℕ, to avoid just such confusion.
Original "proof" was written in Russian, and Russian mathematicians assume that N doesn't consist 0 in it.
So this does partition the possible solutions to FLT from an uncountable solution set to a countable (infinite) number of possible solution. So reducing the order of magnitude of a solution set is a good start I would say. But then that is already known since the set of x, y, n are countable, so the combination of them are countable.
It was always countable. The variables x,y,z are integer-valued.
What are you talking about? This is a diophantine equation, the set of solutions is a most countably infinite by essence of the problem itself
Yes, but I was not thinking about that until the cos() was rational, then I started typing, and ended up half way realizing it was always countable, which was why the second half of the post.
With luck and more power to you.
Honestly, even though the proof is wrong, that's an interesting approach. Nice try.
Not Zee but Zed
Red pants.
Skip to 2:50 to save some time.
Doesn't the argument fall apart earlier because there's no way we can shorten one side of a triangle while keeping the other two sides the same length (one would have to become longer)?
No, it might look that way, though. If you just rotate the side of length y a little ways towards the side of length x and connect the end point you get a new (acute) triangle with two of the same side lengths as you started with. There are restrictions on how small the third side length can get, though, i.e. the Triangle Inequality.
Not in the general case, but can’t you shorten the unequal side of an isosceles triangle while keeping the other two sides the same?
you could say that the authors claim was ... irrational? LOL
if you are so good to combat simple proofs of fermat last theorem then like you to show that the fermat last theorem proof shown in the pinned comment to the video title fermat last theorem 1637proof is correct to embarrass mr.anddrew wiles.george ice
Meh
cos = b/c in a right triangle so it definitely can be rational. If cos can't be rational, you can prove a whole bunch of nonsense. This is a total time waster.
It seems that you pronounce the name "Fermat" not quite correctly. You pronounce it as if it were "Furmat", the first syllable sounding like "fur". The name "Fermat" is correctly pronounced more like "Fairmat", the first syllable sounding like "fair" as in a short "air", with no emphasis on the "i", .
Hmm... , my approach would be: (a+b)^n = a^n + ... (n|n-j) a^n-j b^j ... + b^n
The middle stuff to the other side: a^n + b^n = (a+b)^n - ... (n|n-j) a^n-j b^j ...
And then showing somehow that right side can't be of the form: z^n for n>2
Know, sounds easier than it is. But was my first idea. And with more time, maybe ... 🙃
Meh. Was this supposed to be an April fool? The error in the proof (the claim that cosines can't be rational) is so egregious that I can't see any other reason for talking about the rest of the proof.