Not rude at all! I'm glad you all caught it. I don't want anyone thinking the prime integers aren't prime any more! (Although, as we saw in the video, 2 isn't *always* prime....;) )
Since we are all being frank now, I feel I must suggest that the mistake is serious enough to require an edit to the video rather than just a note (whether onscreen or in the comments as you now have). I might be wrong -- but that is the way it seems to me. In a live class, someone would probably have noticed that mistake almost immediately and would have told you about it. Then you could agree, correct yourself and continue. Unfortunately a youtube video typically has no live audience, so the error becomes sticky, and I think this one is troublesome enough to require an edit.
Technically, we do not know that yet. Just because the best proof we could come up with until now is long, does not mean that there is no shorter one. It's just really, really unlikely.
I've got that case covered. "A proof" means at least one proof is too large to fit in the margin. Even if a shorter proof exists, Wiles's proof is too large to fit in the margin.
Fermat did not say "a proof" could not fit, but rather that his proof could not fit. We know that Fermat only thought he had a proof. I have a problem saying that a non-existent thing could not fit in that margin.
Don't usually comment on videos, but I just want to say: While the transition between hosts was (very) rough, I've really enjoyed the last few videos. I'm glad you've been able to find that sweet spot balancing rigor, entertainment, and video length. Thanks for all you do. =)
Jason Martin analytic number theory may not explain things as "neatly" as algebraic number theory, but sometimes we must resort to analysis (which I had always preferred since age 15) to prove theorems.
That was one of the best videos going off on a tangent out of nowhere and then actually answering the question in the titel while really showing the connection with the seemingly unrelated tangent
Lem Lordje Ko: Do you mean to learn the statement of the theorem, or a proof of it? The statement is simple; it is in the video, at the beginning. A proof was discovered in 1993, and it is very long and complicated.
Credit were credit is due! This was huge improvement with relation to the last video presented by the same host! It is clear, it presents an interesting topic without occulting it with metaphors and comparisons! I think you listened very effectively to the public criticism and corrected your course! Congratulations and keep up the great work!
English has this tendency to butcher anything Latin or Greek. While it is usual (a convention that I follow, too) to use the English pronunciation to refer to these letters and words within an English sentence, there's nothing wrong with the original Greek pronunciations. The ones seemingly every other language has adopted, along with most people studying Latin or Greek.
6:30: as soon as those factors popped up, I expected a look up at a second camera, a fist raised in frustrated rage, and “Phiiiiiiii!!!” (Okay, less “expected,” more “chuckled at the mental image of,” but, y’know.)
Wow, very well done video. It was complicated to be sure, but even though there where parts I didn't quite get (I need to watch this again), you taking us through those examples with the detail you gave let me at least follow along and understand your conclusions. I never realized that the properties of primes where so unique to the integers, and I ended up learning something fairly profound about Fermatt's last theorem that I didn't know before. Thank you for this one! :)
I can prove it one line if a prime number is a factor of a number then it must always be a part of factor the number since it can't be written by two numbers as 3.4 =12 3.2.2=12=3x2.2=12 which =3.2x2 you are writing the same thing since 3 can't be divided it must always be a factor. any non prime factor can be expressed as product of primes and hence can be changed but this can't happen. primes are building blocks of any number new number can be made from primes but all factors are just different way of writing product of primes. Its by the same logic that any non prime number must contain prime number in its factors the prime number which are a factor of a number can't be dived as they are prime. There you go it was pretty easy.
Perhaps Fermat thought he could generalize his technique of the infinite descent, which can be used to show the non-solvability of the equation x^4+y^4=z^4.
Hey Gabe, don't hold yourself back in the answers. Most of us indeed don't fully understand when people talk advanced maths, doesn't mean we're not interested in hearing what the discussion is about :-) - IMHO it's rather motivating to see how much is left for us to discover!
I have noticed how not a single episode yet has touched upon Coding Theory, to me a very interesting topic with interesting applications. Perhaps even a suggestion to cover for a future episode?
1 is the unique product of no primes, it’s like the multiplicative zero. Oh, and I f you pronounce the golden ratio “fee” then you have to pronounce the circumference diameter ratio “pee”. And I think I like that.
What a great presentation! Here is my theory. We know Fermat's note to himself is all the evidence there is, and he never returned to this problem again. This is behavior (not revisiting a "Marvelous" discovery) is associated with realising we are wrong, and is the single most ignored piece of evidence in FLT. I seem to be the only person who has dared to conjecture that the next time Fermat took a look at his "Marvelous" proof he realised it was a trivial proof to an even more trivial problem and never bothered with it again. Following discovery and publication of his private notes, mathematicians did not see the ambiguity of FLT as written, and misunderstood it as a far more complex problem. However, this assumption ignores the case of a "Marvelous" solution to a trivial problem, so trivial in fact, that it really wasn't that "Marvelous" after all. Fermat realised he was wrong (and he was also sometimes wrong without realising it!). All very human. The "trivial" FLT for which the solution is likewise trivial, is to assume the Pythagorean (h^2 = x^2 + y^2) as a given. The "trivial" proof is then one of substitution, requiring about 6 steps. How embarrassing!!! But it is still too long to squeeze into the available space of his margins. However after his death, the mathematicians assumed a non-trivial FLT, and set out to prove the Pythagorean a different way (for integers) rather than assume it. In doing so, it fails to take advantage of the particular case in order to prove the general case, but also sets out to prove the general case as if the Pythagorean (reals) was outside the scope of the Diophantine (integers). I fail to see the need to do this. Its quite easy to show graphically by setting h=1 for all N>2 that a right triangle cannot be formed and the Pythagorean no longer applies and this is true for all reals, not just integers. Only for N=2 is it possible to form a right triangle, and for greater or lesser values of N the right angle increases or decreases. The ration (h=1:x:y) then holds for any integer or real (h,x,y). The proposition that Fermat would simply walk away from such a claim as he made without realising he was wrong, has far less credibility than the proposition that he wittingly abandoned a "Marvelous" discovery 'just like that'. I, for one, am not convinced one iota that mathematicians ever understood the severe triviality of FLT, but embarked a different problem altogether. That is my position, and it is over to Andrew Wiles et. al. to show why my understanding and solution for FLT - as a trivial solution to a trivial problem - is wrong (my solution it is certainly not wrong). Emperor Fermat has no clothes!!!!
I can assure you: 1) The equation a^n+b^n=c^n does have a solution in positive reals and for n being an integer greater than 2 (it also has a solution where a, b, and c are positive integers and n is a real number greater than 2); 2) If a solution using trivial arithmetic existed, chance is that it would have been found ages ago.
As someone who really struggled with abstract algebra. I really wish this video came out 2 years ago. Would have helped a whole lot. Just giving that little nudge in the right direction, that one of the main motivations for rings is that, a ring is the least amount of information needed to talk about primality, would have been incredibly helpful.
Seeing this video through to the end might give the 'truly marvelous demonstration' Fermat had in mind. But I stopped because my brain is 'too narrow to contain' it.
can't agree more, with a little bit of understanding of abstractions, things can go smooth and also abstraction is not a new cutting edge or Martian concept either
I think that this video was a missed opportunity. The main problem in solving equations like the x^n+y^n=z^n is that they contain both multiplication (the n-th power) and addition. Adding a root of unity to the mix helps a lot because then we can transform the equation to be multiplicative in nature. If we know that primes=irreducibles (which is the main subject of the video), then we can use results about unique factorization to attack this problem. This idea gives a good motivation for why we need this UFD definition, while the video mainly gives the definitions and a claim that it is somehow related to Fermat's last theorem. In particular, for x^2+y^2=z^2, every student should learn that sum of squares should almost always lead to the decomposition (x+iy)(x-iy)=z^2. Then, it is a simple exercise to use the unique factorization property in the Gaussian integers Z[i] in order to find all Pythagorean triples.
The video is nice but it contains an error: A ring being a UFD is not equivalent to atoms (= irreducibles) being prime. This is only true if the ring is atomic, i.e. if every non-zero non unit is the product of atoms. To be more specific: R is a UFD (R is atomic and every atom is prime), for every domain R.
Sure, asking for "unique" factorization only makes sense if you have "factorization" in the first place. But this video is an introduction to ring theory (actually, it's a *pre*-introduction---I don't even give the definition of a ring!). So I think, for pedagogical reasons, it's fine (and helpful, in fact) to not mention all the nuances that come with rings (and there are *many*) when sharing the subject with those who might be hearing it for the very first time. But I'm glad you're keeping a sharp eye out. Excellent!
If you want to learn more about rings I recommend "Elements of Number Theory" by John Stillwell It's a textbook I used when I took a number theory course in college.
Nice catch! I was watching carefully to see if she made that mistake ('cause it's really easy to make), but I didn't even notice she made it right at the beginning.
Here's a puzzle: If you scramble a rubik's cube with any arbitrary algorithm (a random set of moves). Knowing only the result of the scramble. How often do you have to do the same scramble to return to the orignal state?
Maybe I need more coffee. I got lost at 2:52. It seemed like she was making a statement about integers, said it was NOT true, but then said it WAS true for INTEGERS.
Thanks for the great video! Fermat never claimed to anyone that he had a proof. He wrote in the margin of his own book. This was not discovered until after he died. So now he's "the greatest troll ever." Just remember folks, some future archaeologist, digging some landfill, may read some stuff you wrote and threw away.
for any non americans confused by the term “foil” its an acranym for first outside inside last used to remember the distributive rule. americans use the anidote to refer to the action, where the phrase foiling a product comes from or foiled when done in past tense, this is the same as saying I added the numbers as per the law of addivtivity, or in more common terms it is analagious to saying i used long division to divide two numbers rather then simply saying i divided. This is just the weird nomenclature americans use to remember it in high school and the years after learning the anidote the students are quickly taught the more generalized rule for larger problems
Anthony Marantino correct. FOIL, was introduced after my time, so I only encountered it when I was tutoring math. Was convenient, perhaps. It’s funny to observe the things kids have now that didn’t exist 40 or 50 years ago. Car seats for kids. My favorite is the rope they use for really little kids when they’re walking in the street as a group; the kids are told to hold onto the rope so they stick together better.
This fairly trivially extends into more than two factors (with integers). No matter how you factorize a number n, a multiple of a prime p, p must exist in at least one of those factors because it is a priori a prime factor of n. Any factor of a number that is itself a prime is a prime factor of that number, again by definition.
A simpler way to express this is this: no matter how one groups the prime factors of a composite n, any of those prime factors p must be in one of those groups. these groups are factors of n.
Wait, so at 5:40 does this mean 3 is not a unit in Z but it is a unit in Q? For the number u=3 in Q, there is a v=1/3 also in Q, so that uv=1 Are all rationals considered units under their own set? Or what am I missing?
Yes! You're spot on! In fact, because every number in Q is a unit, we give Q a special name: it's called a *field*. More generally, if every non-zero number in an arbitrary ring has a multiplicative inverse (i.e. it's a unit), then we call that ring a field. So Q is a field, but Z is not!
Awesome. Other than having heard of it, I didn't know anything about rings (or fields) but I actually understood most of this video and all of this comment. You guys are the best.
I can give you a basic introduction: Rings are sets with a + and * operation on them that satisfy intuitive operations for + and *, including the existence of a 0 element and "negatives" in a sense. If it also has a 1 element, it is called a unital ring (or a ring with unity). If every nonzero element has a "reciprocal" (multiplicative inverse), then the ring is called a field (Actually it's technically a division ring; a field also assumes * is commutative, but non-field division rings are so rare, the distinction is often ignored).
The definitions of irreducible and prime elements should include the proviso that the element isn't a unit. Otherwise, 1 is irreducible and prime in Z. Elements in a UFD are not necessarily "uniquely" the product of irreducible elements. For example: in Z, 6 = (2)(3) = (-2)(-3). The definition of a UFD is: each non-unit of the UFD can be factored into irreducible elements and for any two factorizations of each non-unit of the UFD into irreducible elements, there is a 1-1 correspondence between the factors so that each factor of one the factorizations corresponds to an associate of itself (that is, that factor times a unit).
Hello, How about taking a break from abstract algebra and checking out something truly obvious but long overlooked. I designed 2 incredible algorithms which establish 1-to-1 correspondence between positive floats and positive integers. For details please access the UA-cam video “Pairing Floats and Integers” at ua-cam.com/video/tgVOrCo_5wE/v-deo.html You may see the step-by-step unfolding of one of the most exciting discovery, which up until now was believed to be mathematically impossible. It is amazing that a single action of digit-reversal made it possible to construct all possible floats using a pair of integers (Whole and Fractional). With these algorithms the understanding, teaching and the relevant literature of this very important subject will considerably improve. Please leave any feedback in the COMMENT section of the video. or email to Tamas Varhegyi at secondcause@gmail.com
Proof of def A if and only if def B: Let P = XY. Then P divides XY, so that P divides either X or Y. If P = XY divides X, say XYZ = X, then we are led to YZ = 1 and thus Y = 1 or Y = -1. Now let P divide XY. Suppose P does not divide Y. Suppose we know that GCD (P, Y) = 1. Then GCD (XP, XY) = X GCD (P, Y) = X. Since P divides both XP and XY, P must divide X. So all we need is GCD (P, Y) = 1. Clearly GCD (P, Y) must divide P, say Z GCD (P, Y) = P. Since GCD (P, Y) is positive we are led to either Z = 1 or GCD (P, Y) = 1. Assuming Z = 1, we are led to GCD (P, Y) = P which cannot happen since P does not divide Y.
Pika250: I am having a difficult time following your proof. It appears that in the first paragraph, you are assuming that P is a prime in the sense of Def. B and trying to prove that it satisfies Def. A. It seems more or less okay, but a statement of what you are trying to prove would help. But it's the second paragraph that has me scratching what I laughingly call my head. It starts out okay, for a proof that Def. A implies Def. B. You say basically, suppose P divides XY. So you need to prove, given that P satisfies Def. A, that either P divides X or P divides Y. Then you suppose P does not divide Y. So you need to prove that P divides X. So far, so good. Then you say, "Suppose we know that GCD (P, Y) = 1." Okay. Then a couple of sentences later you have "GCD (P, Y) = X". That would be consistent if X=1. A couple of sentences later you say, "So all we need is GCD (P, Y) = 1." How can you need that if you started out assuming that? And it goes downhill from there. I'm sorry. I can't keep track of what you're assuming or what you're trying to prove anymore.
I put X GCD (P, Y) = X, not GCD (P, Y) = X. Let's restructure the proof of A implies B as follows: Let P divide XY and suppose P does not divide Y. Clearly GCD (P, Y) must divide P, say Z GCD (P, Y) = P. Since GCD (P, Y) is positive we are led to either Z = 1 or GCD (P, Y) = 1 by def A. Assuming Z = 1, we are led to GCD (P, Y) = P which cannot happen since P does not divide Y. We therefore must have GCD (P, Y) = 1. Since P divides both XP and XY, P must divide GCD (XP, XY) = *X GCD (P, Y)* = X. Ergo, def B.
Pika250: OK, I think I see what happened. I mistook the space between X and GCD, and later between Z and GCD, as a separation between two sentences or phrases, with a missing period or semicolon as a typo, rather than an indication of multiplication. I would have used a symbol such as x or parentheses to make it clear I was multiplying, and maybe put the equation on a line by itself, like this: GCD (XP, XY) = X[GCD (P, Y)] = X It also helps that you didn't put "Suppose we know that GCD (P, Y) = 1" at the beginning of the argument. Now it is much clearer. Thanks for the clarification.
9:00 "According to abstract ring theory, irreducible and prime are equivalent concepts if and only if your ring is a UFD." Well, except for some pathological cases. See ramanujan.math.trinity.edu/rdaileda/teach/m4363s07/non_ufd.pdf for an example of a domain satisfying prime = irreducible, but not unique factorisation.
Yep, rings come in lots (seriously, I mean *lots*) of flavors, and the key is that your ring must have factorization in the first place. (And the nicer rings that mathematicians work with have this property!)
At the core is prime. Surrounding the core is irreducibility. And surrounding this is unique factorization. Then standard factorization comes next. This is the justifcation I needed to try and pry math away from the intangible and the abstract. Thank you.
There is a possibility of a very large reduction of the general proof of the Fermat theorem. This proof can only be obtained by analyzing the Pythagorean theorem.
There's a small mistake in the video. The video states that a (commutative, unital) ring is a UFD iff primality is equivalent to irreducibility. But this only gives you uniqueness of factorization if it exists, but you may not have existence of a factorization into irreducibles. So you need some other condition like that the ring needs to be Noetherian or whatever. This is obviously unimportant to viewers who don't know any ring theory, but just in case it confuses any students studying this who come across this video.
It's often tricky to balance accessibility with nuanced precision, but thanks for keeping a sharp eye out! Rings do indeed come in lots of flavors. But to avoid overloading viewers with heavy terminology on an introductory video on ring theory (in which I don't even provide the definition of a ring), I chose to omit many of the nuances of ring theory (and there are tons. I mean *tons*...), including refraining from using the phrase "commutative, unital ring." Fortunately, all of the nicer rings that mathematicians work with have factorization into irreducibles (so that it then makes sense to talk about *unique* factorization). Rings with this property are quite nicely called 'factorization domains' or sometimes 'atomic domains.' Noetherian rings show up all over the place, too, but they exhibit a different property.
One question about UTF: Let's say that a,b,c,d are unique primes. Why is ab≠cd at all times? This was obvious to me throughout university studies, but not anymore.
Apologies if this is already addressed, haven't finished video. But regarding the p|ab implies p|a or p|b thing... what if p is 2, a is 2, and b is 2? 2 divides ab (4), but it also divides both a and b. By this rule, that would mean 2 is not prime. What am I missing?
What about complex numbers? Do they form a ring? Can there be prime complex numbers and if so, can composite complex numbers be factorised into primes?
Fermat's last theorem is a fact. Fermat's did not claim that there are no whole solutions to equation. Fermat's claim that are no solutions to the equation in whole numbers.
A unique factorization domain (UFD) is not uniquely defined by primes being the same as irreducibles. Consider the ring Z + Q[x]. Not a UFD, but all irreducibles are prime.
I have a small quarry with that definition of prime. Assume we are in a integral domain, so that xy=0 implies x=0 or y=0. Clearly, 0|0, since we can take 0=0*1. But even more, assume that 0|x. Then for some y, we have x=0*y=0 and with that we prove that 0|x iff x=0. Now let's check the given definition for prime when p=0: If 0|ab, then ab=0. But we are in a domain, so a=0 or b=0, but that means that 0|a or 0|b, so 0 is prime. In particular, if we are talking about integers, that definition says that 0 is a prime integer, which is untrue (in fact, it is simple to show that 0 is not irreducible, with the given definition). The proper definition would be that p is a prime iff p is not 0 or a unit, and p|ab implies p|a or p|b.
I believe Pierre Fermet had a proof for the theorem. Since Fermat was a prominent person, he would had been putting his name at stake. I don't believe he would have risk it. Andrew Wiles may have just taken the long way around using the sheer power of modern methods. I can demonstrate very simple solutions for quadratic, cubic, and quartic equations. It's, well, if I may be so bold, marvelous.
How would he have risked his reputation by writing a note in a book that was in his private property, and never showing the note or mentioning the result to anybody? (And note that Fermat was a lawyer, not a professional mathematician.)
Z[sqrt(5)] includes a/2+b×sqrt(5)/2 where a and b are both odd, as well as when they're both even. So it includes Φ and 1/Φ, which are units. So 2×Φ×2/Φ is not essentially different from 2×2. This is true for all sqrt(p) where p≡1 mod 4. If p=-3, you get the Eisenstein integers. Z[sqrt(5)] is a UFD, but Z[sqrt(-5)] is not.
For those noticing that Definition B should be 'at least' instead of 'exactly': it's the same in this case, you could use 'exactly' without loss of generality because of associativity, you can always rearrange the product to mean exactly one and the definition holds true for any product, no matter how you associate. It's not that the observation is wrong, is just that you can use the definition to mean exactly the same, and keep it compact and elegant at the same time.
I see your point for a number that is (for a lack of a better word) exactly a=p^n with n>1, and in that case i stand corrected. But for p|ab it holds true in the prior case for b different than one, and for any composite number where p|ab you can always associate conveniently. I think i gloss over your particular case because is the trivial case of the definition, but i do see your point. The channel made a correction as "typo". I don't think it necessarily is.
If p|ab and a or b are or contains a power of p you still can associate whatever you want to keep the assertion of 'exactly one', without losing generality. p|p^n(1) is the trivial case indeed. I stand my observation. I misread your comment but im gonna leave my mistaken prior comment. Specifically: you have ab = p^n(c)p^m(d) = p^n+m(c)d = c'd (where p doesnt devide c or d) which is exactly what im saying. The definition of 'exactly' is true for the last part of the equality so it holds true for the first one. You don't lost generality.
Wrong statement at 9:13. Irreducibles are primes if (but not only if) the ring is a UFD. But an IP ring gotta be atomic (aka any number has at least one factorization to irreducibles).
Fermat's Last Theorem Simplified Proof a^n +b^n =c^n odd+even=odd a^3+b^3=c^3 let c=2x+1 8x^3+12x^2+6x+1=c^3 let b=2x b^3+12x^2+6x+1=c^3 a^3=12x^2+6x+1 Only valid solution to a is when x=0 for all c=2x+1,2x+3,2x+5,... n=3,4,5,... No whole number solution for a,b,c when n>2
I was thinking about something. Could rings with UFDs be used like ℕ in the RSA algorithm? Is there something special about ℤ that make it work only for ℤ, or is there a property that makes it work for more? I was thinking about it because ℤ[i] is a UFD but super tricky to implement RSA in. Is it possible but would require a bit more computing power, or are the integers the *only* ring that RSA works for?
Fermat's Last Theorem: 홀수 솟수 p에 대하여 x^p+y^p=z^p을 만족하는 자연수 x, y, z는 존재하지 않는다 My Marvelous 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 변형. 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.
All Fermat did was construct an exponent table and calculate the differences between different power tables ^2 nets you an ‘even’ difference of odd values, making a nice a^2+b^2=c^2 equivalence with 3, 4, and 5, among other sets X^3 differences get more complicated with a total sum of 6•[0,x)+1, but it is still reasonable to work with, it even has an equivalence set directly based on the Pythagorean Triples, it’s just not a triple set From there on up, though exponential equivalences do persist to the infinite, perfect triplets all but stop existing and get replaced by more complex sums Put simply, the differences between subsequent powers of bases get exponentially more complicated as the power variable increases, preventing triples from being constructed past ^2 And that’s the _condensed_ version, it’s no wonder Fermat didn’t publish his proof, he probably had a textbook worth of numerical dynamics regarding this and other exponential concepts dancing in his dome when he passed
Yes! Sorry for the confusion. This is *perfectly* fine. 3 must divide *at least one* (not *exactly* one) of the factors -- I pinned this correction in a comment up top. So 3 is still a good ol' fashioned prime number ;)
Alexander F: If you classify 0 as composite, then you need to exclude it in the statement of the Unique Factorization Theorem, since 0 has infinitely many factorizations, being "divisible by every integer after all".
"Typo" at 1:10! The prime may divide *at least one* (not exactly one) of the two integers. (Thanks to some of you for spotting this!)
Yeah, we all noticed. But we thought it would be rude to mention it.
Sorry!
Not rude at all! I'm glad you all caught it. I don't want anyone thinking the prime integers aren't prime any more! (Although, as we saw in the video, 2 isn't *always* prime....;) )
Ya that was messing with my head for a while
Since we are all being frank now, I feel I must suggest that the mistake is serious enough to require an edit to the video rather than just a note (whether onscreen or in the comments as you now have).
I might be wrong -- but that is the way it seems to me.
In a live class, someone would probably have noticed that mistake almost immediately and would have told you about it. Then you could agree, correct yourself and continue. Unfortunately a youtube video typically has no live audience, so the error becomes sticky, and I think this one is troublesome enough to require an edit.
:D
Technically Fermat was correct, a proof was too large to fit in the margin of the book.
Technically, we do not know that yet. Just because the best proof we could come up with until now is long, does not mean that there is no shorter one. It's just really, really unlikely.
I've got that case covered. "A proof" means at least one proof is too large to fit in the margin. Even if a shorter proof exists, Wiles's proof is too large to fit in the margin.
Fermat did not say "a proof" could not fit, but rather that his proof could not fit. We know that Fermat only thought he had a proof. I have a problem saying that a non-existent thing could not fit in that margin.
Dr. Gerbils, technically it would still be vacuously true
No he wasn't. He said he _had_ a proof that was too large, which he did not.
Don't usually comment on videos, but I just want to say: While the transition between hosts was (very) rough, I've really enjoyed the last few videos. I'm glad you've been able to find that sweet spot balancing rigor, entertainment, and video length. Thanks for all you do. =)
Thanks for your kind words!
Algebraic number theory is like pure satisfaction... I'd love to see more abstract algebra on the channel!
Yes please, a thousand times! Algebraic number theory is so much more elegant than analytic number theory.
Couldn't agree more !!!
Jason Martin analytic number theory may not explain things as "neatly" as algebraic number theory, but sometimes we must resort to analysis (which I had always preferred since age 15) to prove theorems.
If you understand French you can have a look at my newborn channel : ua-cam.com/channels/QFaJoQu0TP7je5HchCNjHA.html
Ah cool je vais regarder, ca me semble bon! :)
That was one of the best videos going off on a tangent out of nowhere and then actually answering the question in the titel while really showing the connection with the seemingly unrelated tangent
Mission accomplished ;)
I'm sad that this channel has stopped posting new videos, but I always wish you guys good luck and happiness.
I know exactly what Fermat was thinking, but the explanation is too long for a UA-cam comment.
Overkillius lol
Lol
incredible.
"...if and only if your ring has a very special property."
The one part of this video I actually understood ;-)
Have you heard the Last Theorem of Lord Fermat, the Wise?
Of course not, is not a theorem that a mathematician would prove...
Patricio Borquez Math Lords are our speciality.
Is it possible to learn this theorem?
Lem Lordje Ko: Do you mean to learn the statement of the theorem, or a proof of it? The statement is simple; it is in the video, at the beginning. A proof was discovered in 1993, and it is very long and complicated.
Not from your school's math book.
Meanwhile, we see different alien(mysterious) complex functions dancing in front of us..
I just want to say congratulations. Your channel is awesome.
Credit were credit is due! This was huge improvement with relation to the last video presented by the same host! It is clear, it presents an interesting topic without occulting it with metaphors and comparisons!
I think you listened very effectively to the public criticism and corrected your course! Congratulations and keep up the great work!
A drunk friend once told me:
"If Φ was really called 'phee', then π would be called 'pee' "
That's actually the correct pronunciation
That's how it's pronounced in Greek
It would be pronounced exactly the same as our corresponding letter P, which it is, in Greek.
This video kicked my butt. I didn't know about Definition B and UFDs. Thank you!
Eight years as a mathematician and I thought phi was one of the Greek letters we all pronounced the same way. But now here you are with "fee"
English has this tendency to butcher anything Latin or Greek. While it is usual (a convention that I follow, too) to use the English pronunciation to refer to these letters and words within an English sentence, there's nothing wrong with the original Greek pronunciations. The ones seemingly every other language has adopted, along with most people studying Latin or Greek.
I split the difference and call it "foh".
ikr!! I always pronounced it fi... as in hi!
English is literally the only spoken language in the entire world which doesn't pronounce 'i' as 'ee'
@Alexander F, I prefer "fumb"
Really enjoyed this episode! Great explanations. Keep up the good work :)
A lot of books stated the theorem: if p divides ab, then p divides either a or b.
So coooool
I'm learning Abstract Algebra and so excited to master all this subject
D'où vient ce ''copié_collé"" d'il y a 5 ans !??? !! c'est une usurpation d'identité.
Comment faites-vous pour copier mon texte et l'arranger à votre manière ! ! !
@@gilberttheisen9270 Calm down and go there eat your baguette
OK, I didn't understand a lot of things you said, but at least I learned in which space Fermat's theorem was being analyzed.
6:30: as soon as those factors popped up, I expected a look up at a second camera, a fist raised in frustrated rage, and “Phiiiiiiii!!!” (Okay, less “expected,” more “chuckled at the mental image of,” but, y’know.)
Wow, very well done video. It was complicated to be sure, but even though there where parts I didn't quite get (I need to watch this again), you taking us through those examples with the detail you gave let me at least follow along and understand your conclusions. I never realized that the properties of primes where so unique to the integers, and I ended up learning something fairly profound about Fermatt's last theorem that I didn't know before. Thank you for this one! :)
Fantastic! Glad you enjoyed it!
I’m glad you give the viewer pieces to prove. This material is not trivial, even for people with a pretty advanced understanding of mathematics.
Awesome episode on Number Theory !!! More ! MORE !!!
I support PBS Space Time, but all the work PBS does here is excellent.
Thanks for the reply and the great video as usual. You nailed that last name Gabe!
This episode is SO GOOD!
Maybe there was no marvellous proof & Fermat was just being mischievous.
TimeAndChance: Or maybe he was mistaken about the validity of his proof.
Erik Lönnrot He categorically, 100% did not have a correct proof. He either had no proof or some simplistic and incorrect algebraic one.
this seems very likely
In other words, Fermat was trolling
I can prove it one line
if a prime number is a factor of a number then it must always be a part of factor the number since it can't be written by two numbers as
3.4 =12
3.2.2=12=3x2.2=12 which =3.2x2 you are writing the same thing
since 3 can't be divided it must always be a factor. any non prime factor can be expressed as product of primes and hence can be changed but this can't happen. primes are building blocks of any number new number can be made from primes but all factors are just different way of writing product of primes.
Its by the same logic that any non prime number must contain prime number in its factors the prime number which are a factor of a number can't be dived as they are prime.
There you go it was pretty easy.
Perhaps Fermat thought he could generalize his technique of the infinite descent, which can be used to show the non-solvability of the equation x^4+y^4=z^4.
Nice video! Great job, Tai!!
Hey Gabe, don't hold yourself back in the answers. Most of us indeed don't fully understand when people talk advanced maths, doesn't mean we're not interested in hearing what the discussion is about :-) - IMHO it's rather motivating to see how much is left for us to discover!
hahaha this video ending is amazing XD replayed that more than 5 times haha
How we miss this channel.
I have noticed how not a single episode yet has touched upon Coding Theory, to me a very interesting topic with interesting applications. Perhaps even a suggestion to cover for a future episode?
Wow, I never thought field extensions would come into play with primality. Thank you.
Great video!
I find Gabe much more organic and enjoyable than the other presenter. Any one else agree?
1 is the unique product of no primes, it’s like the multiplicative zero. Oh, and I f you pronounce the golden ratio “fee” then you have to pronounce the circumference diameter ratio “pee”. And I think I like that.
Qermaq In some languages, like Norwegian, pi is pronounced exactly like "pee".
In some languages like Norwegian... or Greek. Except that the "i" in Greek is shorter than "ee" in English.
Took Ancient Greek classes. Most classicists do pronounce them phee and pee.
Wonderful presentation
What a great presentation! Here is my theory. We know Fermat's note to himself is all the evidence there is, and he never returned to this problem again. This is behavior (not revisiting a "Marvelous" discovery) is associated with realising we are wrong, and is the single most ignored piece of evidence in FLT. I seem to be the only person who has dared to conjecture that the next time Fermat took a look at his "Marvelous" proof he realised it was a trivial proof to an even more trivial problem and never bothered with it again. Following discovery and publication of his private notes, mathematicians did not see the ambiguity of FLT as written, and misunderstood it as a far more complex problem. However, this assumption ignores the case of a "Marvelous" solution to a trivial problem, so trivial in fact, that it really wasn't that "Marvelous" after all. Fermat realised he was wrong (and he was also sometimes wrong without realising it!). All very human.
The "trivial" FLT for which the solution is likewise trivial, is to assume the Pythagorean (h^2 = x^2 + y^2) as a given. The "trivial" proof is then one of substitution, requiring about 6 steps. How embarrassing!!! But it is still too long to squeeze into the available space of his margins.
However after his death, the mathematicians assumed a non-trivial FLT, and set out to prove the Pythagorean a different way (for integers) rather than assume it. In doing so, it fails to take advantage of the particular case in order to prove the general case, but also sets out to prove the general case as if the Pythagorean (reals) was outside the scope of the Diophantine (integers). I fail to see the need to do this. Its quite easy to show graphically by setting h=1 for all N>2 that a right triangle cannot be formed and the Pythagorean no longer applies and this is true for all reals, not just integers. Only for N=2 is it possible to form a right triangle, and for greater or lesser values of N the right angle increases or decreases. The ration (h=1:x:y) then holds for any integer or real (h,x,y).
The proposition that Fermat would simply walk away from such a claim as he made without realising he was wrong, has far less credibility than the proposition that he wittingly abandoned a "Marvelous" discovery 'just like that'. I, for one, am not convinced one iota that mathematicians ever understood the severe triviality of FLT, but embarked a different problem altogether. That is my position, and it is over to Andrew Wiles et. al. to show why my understanding and solution for FLT - as a trivial solution to a trivial problem - is wrong (my solution it is certainly not wrong).
Emperor Fermat has no clothes!!!!
I can assure you: 1) The equation a^n+b^n=c^n does have a solution in positive reals and for n being an integer greater than 2 (it also has a solution where a, b, and c are positive integers and n is a real number greater than 2); 2) If a solution using trivial arithmetic existed, chance is that it would have been found ages ago.
As someone who really struggled with abstract algebra. I really wish this video came out 2 years ago. Would have helped a whole lot. Just giving that little nudge in the right direction, that one of the main motivations for rings is that, a ring is the least amount of information needed to talk about primality, would have been incredibly helpful.
I'm delighted to hear the video was helpful! (Even if it was 2 years late...)
:o i haven't been watching for a while, and now there's new people!
Including the old spacetime host!
:O
Seeing this video through to the end might give the 'truly marvelous demonstration' Fermat had in mind. But I stopped because my brain is 'too narrow to contain' it.
Ring thoery was one of my favourite topics at university. It's such a shame that interesting maths like this isn't really presented at schools.
can't agree more, with a little bit of understanding of abstractions, things can go smooth and also abstraction is not a new cutting edge or Martian concept either
You made me lol at the end of the video, Gabe, haha.
My favorite episode from this new host.
Any episodes on Heegner numbers and UFDs coming up? Because please and thank you!
I think that this video was a missed opportunity. The main problem in solving equations like the x^n+y^n=z^n is that they contain both multiplication (the n-th power) and addition. Adding a root of unity to the mix helps a lot because then we can transform the equation to be multiplicative in nature. If we know that primes=irreducibles (which is the main subject of the video), then we can use results about unique factorization to attack this problem. This idea gives a good motivation for why we need this UFD definition, while the video mainly gives the definitions and a claim that it is somehow related to Fermat's last theorem. In particular, for x^2+y^2=z^2, every student should learn that sum of squares should almost always lead to the decomposition (x+iy)(x-iy)=z^2. Then, it is a simple exercise to use the unique factorization property in the Gaussian integers Z[i] in order to find all Pythagorean triples.
Great vid!!! Will comment more later, thanks.
Can you do a video about p-adic numbers?
Please do a video on an introduction and overview to ring theory! My apologies if you already have.
The video is nice but it contains an error: A ring being a UFD is not equivalent to atoms (= irreducibles) being prime. This is only true if the ring is atomic, i.e. if every non-zero non unit is the product of atoms. To be more specific: R is a UFD (R is atomic and every atom is prime), for every domain R.
Sure, asking for "unique" factorization only makes sense if you have "factorization" in the first place. But this video is an introduction to ring theory (actually, it's a *pre*-introduction---I don't even give the definition of a ring!). So I think, for pedagogical reasons, it's fine (and helpful, in fact) to not mention all the nuances that come with rings (and there are *many*) when sharing the subject with those who might be hearing it for the very first time. But I'm glad you're keeping a sharp eye out. Excellent!
@7:15 check out the references below for why 2 is irreducible- which reference?
If you want to learn more about rings I recommend "Elements of Number Theory" by John Stillwell
It's a textbook I used when I took a number theory course in college.
Cliff86 Stillwell is a great author!
Fermat videos will never not be uninteresting
1:10 nope, the prime p may divide both factors. The world 'exactly' shouldnt be there
Good catch! "At least one" is the correct quantifier.
Yes! Squares or multiples of squares come to mind.
Lol yea, of course p will divide both p^k and p^(n-k) for all 0
Nice catch! I was watching carefully to see if she made that mistake ('cause it's really easy to make), but I didn't even notice she made it right at the beginning.
Czeckie shouldn't*
Very well made video. Keep it up
Here's a puzzle:
If you scramble a rubik's cube with any arbitrary algorithm (a random set of moves).
Knowing only the result of the scramble.
How often do you have to do the same scramble to return to the orignal state?
I miss this channel...
Maybe I need more coffee. I got lost at 2:52. It seemed like she was making a statement about integers, said it was NOT true, but then said it WAS true for INTEGERS.
Thanks for the great video! Fermat never claimed to anyone that he had a proof. He wrote in the margin of his own book. This was not discovered until after he died. So now he's "the greatest troll ever." Just remember folks, some future archaeologist, digging some landfill, may read some stuff you wrote and threw away.
Super nice video.
"if you are looking for a buzzword", nice humor there
7:29 definition A is incorrect over the integers, as it would include 1 as a prime.
for any non americans confused by the term “foil” its an acranym for first outside inside last used to remember the distributive rule. americans use the anidote to refer to the action, where the phrase foiling a product comes from or foiled when done in past tense, this is the same as saying I added the numbers as per the law of addivtivity, or in more common terms it is analagious to saying i used long division to divide two numbers rather then simply saying i divided. This is just the weird nomenclature americans use to remember it in high school and the years after learning the anidote the students are quickly taught the more generalized rule for larger problems
Anthony Marantino correct. FOIL, was introduced after my time, so I only encountered it when I was tutoring math. Was convenient, perhaps. It’s funny to observe the things kids have now that didn’t exist 40 or 50 years ago. Car seats for kids. My favorite is the rope they use for really little kids when they’re walking in the street as a group; the kids are told to hold onto the rope so they stick together better.
Anthony Marantino
It’s not weird nomenclature.
Don’t be ignorant
This fairly trivially extends into more than two factors (with integers). No matter how you factorize a number n, a multiple of a prime p, p must exist in at least one of those factors because it is a priori a prime factor of n. Any factor of a number that is itself a prime is a prime factor of that number, again by definition.
A simpler way to express this is this: no matter how one groups the prime factors of a composite n, any of those prime factors p must be in one of those groups. these groups are factors of n.
Wait, so at 5:40 does this mean 3 is not a unit in Z but it is a unit in Q?
For the number u=3 in Q, there is a v=1/3 also in Q, so that uv=1
Are all rationals considered units under their own set? Or what am I missing?
Yes! You're spot on! In fact, because every number in Q is a unit, we give Q a special name: it's called a *field*. More generally, if every non-zero number in an arbitrary ring has a multiplicative inverse (i.e. it's a unit), then we call that ring a field. So Q is a field, but Z is not!
The rationals are a field. In any field, all numbers except 0 are units.
Awesome. Other than having heard of it, I didn't know anything about rings (or fields) but I actually understood most of this video and all of this comment. You guys are the best.
I can give you a basic introduction:
Rings are sets with a + and * operation on them that satisfy intuitive operations for + and *, including the existence of a 0 element and "negatives" in a sense. If it also has a 1 element, it is called a unital ring (or a ring with unity). If every nonzero element has a "reciprocal" (multiplicative inverse), then the ring is called a field (Actually it's technically a division ring; a field also assumes * is commutative, but non-field division rings are so rare, the distinction is often ignored).
The definitions of irreducible and prime elements should include the proviso that the element isn't a unit. Otherwise, 1 is irreducible and prime in Z.
Elements in a UFD are not necessarily "uniquely" the product of irreducible elements. For example: in Z, 6 = (2)(3) = (-2)(-3). The definition of a UFD is: each non-unit of the UFD can be factored into irreducible elements and for any two factorizations of each non-unit of the UFD into irreducible elements, there is a 1-1 correspondence between the factors so that each factor of one the factorizations corresponds to an associate of itself (that is, that factor times a unit).
I do love abstract algebra!
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Proof of def A if and only if def B:
Let P = XY. Then P divides XY, so that P divides either X or Y. If P = XY divides X, say XYZ = X, then we are led to YZ = 1 and thus Y = 1 or Y = -1.
Now let P divide XY. Suppose P does not divide Y. Suppose we know that GCD (P, Y) = 1. Then GCD (XP, XY) = X GCD (P, Y) = X. Since P divides both XP and XY, P must divide X. So all we need is GCD (P, Y) = 1. Clearly GCD (P, Y) must divide P, say Z GCD (P, Y) = P. Since GCD (P, Y) is positive we are led to either Z = 1 or GCD (P, Y) = 1. Assuming Z = 1, we are led to GCD (P, Y) = P which cannot happen since P does not divide Y.
Pika250: I am having a difficult time following your proof. It appears that in the first paragraph, you are assuming that P is a prime in the sense of Def. B and trying to prove that it satisfies Def. A. It seems more or less okay, but a statement of what you are trying to prove would help. But it's the second paragraph that has me scratching what I laughingly call my head. It starts out okay, for a proof that Def. A implies Def. B. You say basically, suppose P divides XY. So you need to prove, given that P satisfies Def. A, that either P divides X or P divides Y. Then you suppose P does not divide Y. So you need to prove that P divides X. So far, so good. Then you say, "Suppose we know that GCD (P, Y) = 1." Okay. Then a couple of sentences later you have "GCD (P, Y) = X". That would be consistent if X=1. A couple of sentences later you say, "So all we need is GCD (P, Y) = 1." How can you need that if you started out assuming that? And it goes downhill from there. I'm sorry. I can't keep track of what you're assuming or what you're trying to prove anymore.
I put X GCD (P, Y) = X, not GCD (P, Y) = X. Let's restructure the proof of A implies B as follows: Let P divide XY and suppose P does not divide Y. Clearly GCD (P, Y) must divide P, say Z GCD (P, Y) = P. Since GCD (P, Y) is positive we are led to either Z = 1 or GCD (P, Y) = 1 by def A. Assuming Z = 1, we are led to GCD (P, Y) = P which cannot happen since P does not divide Y. We therefore must have GCD (P, Y) = 1. Since P divides both XP and XY, P must divide GCD (XP, XY) = *X GCD (P, Y)* = X. Ergo, def B.
Pika250: OK, I think I see what happened. I mistook the space between X and GCD, and later between Z and GCD, as a separation between two sentences or phrases, with a missing period or semicolon as a typo, rather than an indication of multiplication. I would have used a symbol such as x or parentheses to make it clear I was multiplying, and maybe put the equation on a line by itself, like this:
GCD (XP, XY) = X[GCD (P, Y)] = X
It also helps that you didn't put "Suppose we know that GCD (P, Y) = 1" at the beginning of the argument. Now it is much clearer. Thanks for the clarification.
9:00 "According to abstract ring theory, irreducible and prime are equivalent concepts if and only if your ring is a UFD." Well, except for some pathological cases. See ramanujan.math.trinity.edu/rdaileda/teach/m4363s07/non_ufd.pdf for an example of a domain satisfying prime = irreducible, but not unique factorisation.
Nicolas Daans thanks for bringing this up!
Yep, rings come in lots (seriously, I mean *lots*) of flavors, and the key is that your ring must have factorization in the first place. (And the nicer rings that mathematicians work with have this property!)
Then agreed.
At the core is prime. Surrounding the core is irreducibility. And surrounding this is unique factorization. Then standard factorization comes next. This is the justifcation I needed to try and pry math away from the intangible and the abstract. Thank you.
There is a possibility of a very large reduction of the general proof of the Fermat theorem. This proof can only be obtained by analyzing the Pythagorean theorem.
Or Fermat was the biggest troll in history of humanity
There's a small mistake in the video. The video states that a (commutative, unital) ring is a UFD iff primality is equivalent to irreducibility. But this only gives you uniqueness of factorization if it exists, but you may not have existence of a factorization into irreducibles. So you need some other condition like that the ring needs to be Noetherian or whatever.
This is obviously unimportant to viewers who don't know any ring theory, but just in case it confuses any students studying this who come across this video.
It's often tricky to balance accessibility with nuanced precision, but thanks for keeping a sharp eye out! Rings do indeed come in lots of flavors. But to avoid overloading viewers with heavy terminology on an introductory video on ring theory (in which I don't even provide the definition of a ring), I chose to omit many of the nuances of ring theory (and there are tons. I mean *tons*...), including refraining from using the phrase "commutative, unital ring."
Fortunately, all of the nicer rings that mathematicians work with have factorization into irreducibles (so that it then makes sense to talk about *unique* factorization). Rings with this property are quite nicely called 'factorization domains' or sometimes 'atomic domains.' Noetherian rings show up all over the place, too, but they exhibit a different property.
One question about UTF:
Let's say that a,b,c,d are unique primes. Why is ab≠cd at all times?
This was obvious to me throughout university studies, but not anymore.
Alex Johansson a divides the left side but not the right side
@@shvoregavim9435 Je n'ai jamais écrit ça ! ! !
Apologies if this is already addressed, haven't finished video. But regarding the p|ab implies p|a or p|b thing... what if p is 2, a is 2, and b is 2? 2 divides ab (4), but it also divides both a and b. By this rule, that would mean 2 is not prime. What am I missing?
The "or" is inclusive. So 2 is prime.
Ah. She'd explicitly said "exactly one" in the video and I'd missed the comment addressing that. Thanks!
You're welcome!
What about complex numbers? Do they form a ring? Can there be prime complex numbers and if so, can composite complex numbers be factorised into primes?
oh no this is bringing back memories from my algebra class
Fermat's last theorem is a fact.
Fermat's did not claim that there are no whole solutions to equation.
Fermat's claim that are no solutions to the equation in whole numbers.
@4 :12 b has to belong to Q* otherwise if b =0 you just have Q
A unique factorization domain (UFD) is not uniquely defined by primes being the same as irreducibles. Consider the ring Z + Q[x]. Not a UFD, but all irreducibles are prime.
Isn’t the def of unit number wrong in this video. u is unit number if uv = v right?
Was the music the FTL soundtrack?
I have a small quarry with that definition of prime. Assume we are in a integral domain, so that xy=0 implies x=0 or y=0. Clearly, 0|0, since we can take 0=0*1. But even more, assume that 0|x. Then for some y, we have x=0*y=0 and with that we prove that 0|x iff x=0. Now let's check the given definition for prime when p=0: If 0|ab, then ab=0. But we are in a domain, so a=0 or b=0, but that means that 0|a or 0|b, so 0 is prime. In particular, if we are talking about integers, that definition says that 0 is a prime integer, which is untrue (in fact, it is simple to show that 0 is not irreducible, with the given definition). The proper definition would be that p is a prime iff p is not 0 or a unit, and p|ab implies p|a or p|b.
Yes, exactly. To be extra careful, we'd want to toss out both units and 0 when making the definition. (Good catch.)
I believe Pierre Fermet had a proof for the theorem. Since Fermat was a prominent person, he would had been putting his name at stake. I don't believe he would have risk it. Andrew Wiles may have just taken the long way around using the sheer power of modern methods.
I can demonstrate very simple solutions for quadratic, cubic, and quartic equations. It's, well, if I may be so bold, marvelous.
How would he have risked his reputation by writing a note in a book that was in his private property, and never showing the note or mentioning the result to anybody? (And note that Fermat was a lawyer, not a professional mathematician.)
Z[sqrt(5)] includes a/2+b×sqrt(5)/2 where a and b are both odd, as well as when they're both even. So it includes Φ and 1/Φ, which are units. So 2×Φ×2/Φ is not essentially different from 2×2.
This is true for all sqrt(p) where p≡1 mod 4. If p=-3, you get the Eisenstein integers. Z[sqrt(5)] is a UFD, but Z[sqrt(-5)] is not.
For those noticing that Definition B should be 'at least' instead of 'exactly': it's the same in this case, you could use 'exactly' without loss of generality because of associativity, you can always rearrange the product to mean exactly one and the definition holds true for any product, no matter how you associate. It's not that the observation is wrong, is just that you can use the definition to mean exactly the same, and keep it compact and elegant at the same time.
I don't understand what you wrote there. "Exactly one" would not be the case for any composite number that contained p^n with n>1.
I see your point for a number that is (for a lack of a better word) exactly a=p^n with n>1, and in that case i stand corrected. But for p|ab it holds true in the prior case for b different than one, and for any composite number where p|ab you can always associate conveniently. I think i gloss over your particular case because is the trivial case of the definition, but i do see your point. The channel made a correction as "typo". I don't think it necessarily is.
If p|ab and a or b are or contains a power of p you still can associate whatever you want to keep the assertion of 'exactly one', without losing generality. p|p^n(1) is the trivial case indeed. I stand my observation. I misread your comment but im gonna leave my mistaken prior comment.
Specifically: you have ab = p^n(c)p^m(d) = p^n+m(c)d = c'd (where p doesnt devide c or d) which is exactly what im saying. The definition of 'exactly' is true for the last part of the equality so it holds true for the first one. You don't lost generality.
What about complex numbers with this regards to these definition
Please name the axioms "cheese" and "coco puffs" next time.
"Behave a lot like integers" is not exactly how I would informally describe a ring.
Its a generalization of addition and multiplication on the integers. Certainly the clearest metaphor and a good motivating example.
Define π Sum check solution.
Sum(n) = 1/2 n (n+1)
Sum(1) = 1/2 n × 1 × (1+1)
Sum(1) = 1/2 × 1 × 2
Sum(1) = 1
Wrong statement at 9:13. Irreducibles are primes if (but not only if) the ring is a UFD. But an IP ring gotta be atomic (aka any number has at least one factorization to irreducibles).
Between irreducible vs prime, does one imply the other?
I'm tempted to think that all primes are irreducible (probably a proof by contradiction?)
Fermat's Last Theorem Simplified Proof
a^n +b^n =c^n
odd+even=odd
a^3+b^3=c^3
let c=2x+1
8x^3+12x^2+6x+1=c^3
let b=2x
b^3+12x^2+6x+1=c^3
a^3=12x^2+6x+1
Only valid solution to a is when x=0 for all
c=2x+1,2x+3,2x+5,...
n=3,4,5,...
No whole number solution for a,b,c when n>2
I was thinking about something. Could rings with UFDs be used like ℕ in the RSA algorithm? Is there something special about ℤ that make it work only for ℤ, or is there a property that makes it work for more? I was thinking about it because ℤ[i] is a UFD but super tricky to implement RSA in. Is it possible but would require a bit more computing power, or are the integers the *only* ring that RSA works for?
Fermat's Last Theorem: 홀수 솟수 p에 대하여 x^p+y^p=z^p을 만족하는 자연수 x, y, z는 존재하지 않는다
My Marvelous 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 변형.
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.
Im quite sure that since the proof for every power, that is a multiple of 2, is quite easy Fermat thought he could generalize this for every number
« I already get a huge pile of cash, that's why my clothes look so nice » hahahahahaha!!
Gabe: Was just about to call Elon to convince him to patreon a mil, but upon observing how your threads are as flashy as mine....
All Fermat did was construct an exponent table and calculate the differences between different power tables
^2 nets you an ‘even’ difference of odd values, making a nice a^2+b^2=c^2 equivalence with 3, 4, and 5, among other sets
X^3 differences get more complicated with a total sum of 6•[0,x)+1, but it is still reasonable to work with, it even has an equivalence set directly based on the Pythagorean Triples, it’s just not a triple set
From there on up, though exponential equivalences do persist to the infinite, perfect triplets all but stop existing and get replaced by more complex sums
Put simply, the differences between subsequent powers of bases get exponentially more complicated as the power variable increases, preventing triples from being constructed past ^2
And that’s the _condensed_ version, it’s no wonder Fermat didn’t publish his proof, he probably had a textbook worth of numerical dynamics regarding this and other exponential concepts dancing in his dome when he passed
1:20 Wait, I don't get it.
We have 54=6*9. 3 divides both of the factors 6 and 9, and yet 3 is prime, so what did I miss? Thanks.
Yes! Sorry for the confusion. This is *perfectly* fine. 3 must divide *at least one* (not *exactly* one) of the factors -- I pinned this correction in a comment up top. So 3 is still a good ol' fashioned prime number ;)
3 must divide either 6 or 9, but it can also divide both.
Ah, I see what you mean now. No worries, these things are quite easy to misphrase in a youtube video. Nice video - keep it up.
What about zero? Is it prime, composite, nothing or both?
Zero is composite. It is divisible by every integer after all.
Alexander F: If you classify 0 as composite, then you need to exclude it in the statement of the Unique Factorization Theorem, since 0 has infinitely many factorizations, being "divisible by every integer after all".