Why did they prove this amazing theorem in 200 different ways? Quadratic Reciprocity MASTERCLASS
ฝัง
- เผยแพร่เมื่อ 5 มิ.ย. 2024
- The longest Mathologer video ever, just shy of an hour (eventually it's going to happen :) One video I've been meaning to make for a long, long time. A Mathologerization of the Law of Quadratic Reciprocity. This is another one of my MASTERCLASS videos. The slide show consists of 550 slides and the whole thing took forever to make. Just to give you an idea of the work involved in producing a video like this, preparing the subtitles for this video took me almost 4 hours. Why do anything as crazy as this? Well, just like many other mathematicians I consider the law of quadratic reciprocity as one of the most beautiful and surprising facts about prime numbers. While other mathematicians were inspired to come up with ingenious proofs of this theorem, over 200 different proofs so far and counting, I thought I contribute to it's illustrious history by actually trying me very best of getting one of those crazily complicated proofs within reach of non-mathematicians, to make the unaccessible accessible :) Now let's see how many people are actually prepared to watch a (close to) one hour long math(s) video :)
0:00 Intro
4:00 Chapter 0: Mini rings. Motivating quadratic reciprocity
9:53 Chapter 1: Squares. When is a remainder a square?
16:35 Chapter 2: Quadratic reciprocity formula
24:18 Chapter 3: Intro to the card trick proof
29:22 Chapter 4: Picking up along rows and putting down by columns
29:21 Chapter 5: Picking up along columns and putting down along diagonals
45:16 Chapter 6: Zolotarev's lemma, the grand finale
55:47 Credits
This video was inspired by Matt Baker's ingenious recasting of of a 1830 proof of the LAW by the Russian mathematician Zolotarev in terms of dealing a deck of cards. Here is Matt's blog post that got me started (written for mathematicians):
mattbaker.blog/2013/07/03/qua...
If you want to read up on the properties of the sign of a permutation that I am using in this video, Matt also has a nice write-up of this.
mattbakerblog.files.wordpress...
The relevant Wiki articles are these:
en.wikipedia.org/wiki/Zolotar...
en.wikipedia.org/wiki/Quadrat...
Zolotarev's original paper lives here:
archive.numdam.org/ARCHIVE/NAM...
Here is a list of proofs of the law prepared by Franz Lemmermeyer
www.rzuser.uni-heidelberg.de/...
Franz Lemmermeyer is also the author of the following excellent book on everything to do with quadratic reciprocity (written for mathematicians):
Lemmermeyer, Franz (2000), Reciprocity Laws: from Euler to Eisenstein, Springer Monographs in Mathematics, Berlin
The first teaching semester at the university where I teach is about to start and all my teaching and lots of other stuff will happen this semester. This means I won't have much time for any more crazily time-consuming projects like this. Galois theory will definitely has to wait until the second half of this year :( Still, quite a bit of beautiful doable stuff coming up. So stay tuned.
Thank you to Marty for all the relentless nitpicking of the script, his wordsmithing and throwing cards at me in the video. Thank you to Eddie, Tristan and Matt for all your help with proofreading and feedback on the script and exposition.
Enjoy!
Burkard
If it isn't Euler it's Gauss. Remember folks, if a murderer comes into your house and asks who proved a theorem you never heard of the Best survival strategy is to say "Euler or Gauss"
You are damn right
Kkkkkkkkj
Then why Euler is not a part of trinity which comprises Archimedes, Newton and Gauss?
Such a shame
I don't know if this story ever reached Finland, but in the US, there was a mathematical genius who had Euler-like talent, but became a Ludditic terrorist. If anybody would have done something that deranged, it would have been this guy:
en.wikipedia.org/wiki/Ted_Kaczynski
Pragadeesh J - You can declare any trinity you like. What difference does it make?
WOW IT TOOK ME MY ENTIRE DEGREE TO MAKE SENSE OF THIS THEOREM AND HERE MATHOLOGER COMES SHUFFLING CARDS TO PROVE IT CLEARLY
It makes more sense the more proofs you read
I feel like there was an important lesson to be learned from this video:
NEVER LET THE MATHEMATICIAN DEAL THE CARDS!
OTOH it shows up on about page 20 of the first number theory book I ever saw, and the previous 20 pages largely deals with basic number theory (like unique prime factorization, greatest common divisor, etc.) and items interesting unto themselves whether or not quadratic reciprocity had ever been discovered, and each item typically takes under a paragraph to prove. Other books have briefer versions of the same proof than the 4 pages in that one, but that book covers the case for 2 and -1 all at the same time, and actually has two proofs of the more arduous section, the second one of which is a diagram that sums up what just took a couple of pages of algebra (provided you don't cafe about 2 or -1.)
I didn't want people to think this was going to be one of the hard things if they wanted to pursue number theory.
I dunno if it's clearly to the un-degreed mind but yes i believe he proved it
i love your username
(28 March) Really bizarre, this video was basically invisible for almost two weeks with hardly any recommendations going out to fans. Only now TH-cam has decided to actually show it to people. Who knows, maybe it was a mistake to mention cat videos in previous videos and the TH-cam AI is now under the impression that the target audience for these videos has changed :)
The longest Mathologer video ever, just shy of an hour (eventually it's going to happen :) One video I've been meaning to make for a long, long time. A Mathologerization of the Law of Quadratic Reciprocity. This is another one of my MASTERCLASS videos. The slide show consists of 550 slides and the whole thing took forever to make. Just to give you an idea of the work involved in producing a video like this, preparing the subtitles for this video took me almost 4 hours. Why do anything as crazy as this? Well, just like many other mathematicians I consider the law of quadratic reciprocity as one of the most beautiful and surprising facts about prime numbers. While other mathematicians were inspired to come up with ingenious proofs of this theorem, over 200 different proofs so far and counting, I thought I contribute to it's illustrious history by actually trying me very best of getting one of those crazily complicated proofs within reach of non-mathematicians, to make the unaccessible accessible. Now let's see how many people are actually prepared to watch a (close to) one hour long math(s) video :). Have a look at the description for relevant links and more background info.
The first teaching semester at the university where I teach just started last week and all my teaching and lots of other stuff will happen this semester. This means I won't have much time for any more crazily time-consuming projects like this. Galois theory will definitely has to wait until the second half of this year :( Still, quite a bit of beautiful doable stuff coming up. So stay tuned.
I can confirm that I've just watched the video for the first time today, even though I exclusively use subscriptions feed page and watch all of it thoroughly. It's very likely that it wasn't in the subscriptions feed at all at the moment of posting.
Yes. Very strange. I confirm that this video was hidden from me too. Anyway I'm glad I watched it. A lot of hard work to explain rings.
Ok...now what do I do for the next 23 hours today?
Yeah it just got recommended even though I've been waiting for this to be done on a big math channel for a while now
I only just saw it today.
People will stay at home
Time to make 1hour long video
I think this is it for me with one hour long videos for a while :) Just think about it. If a video like this takes one hour to watch how long does it take to make?
Nooooo!! It's a nice way to spend your time indoors... and humanity needs you right now... keep it up!
@@Mathologer Rest assured that your efforts are greatly appreciated. This is one of the (if not the) best mathematics channels on youtube and the reason is that people recognize quality and hard work when they see it. Happy π-day!
@@mannyc6649 *the* best by far AFAIC
Or start a series on class field theory :-)
I will never tire of 99999999 in a strong German accent
Isn't Mathologer's accent Austrian, or am I mistaken?
@@tracyh5751 ehh, German, Austrian, pretty close to the same account. Sie beide sprechen Deutsch.
Ja ja ja ja ja ja.....
*NEIN, NEIN, NEIN, NEIN, NEIN, NEIN, NEIN, NEIN, NEIN, NEIN, NEIN, NEIN, NEIN, NEIN !!!!!*
* repeatedly slams desk angrily *
note to self, it's at 2:58
I actually guessed Gauss! Take that Mathologer!
Me: 1
Mathologer: 387ish
Never underestimate the power of trial, error and gausswork.
@@Tubluer :P
Was gauss not in the thumbnail at the time of this comment? Maybe he's only recognized that easily by the countrymen who paid with bills with his face on it for years?
Please add a COMMA: "Take that, Mathologer!".
Here are some examples with/without a comma:
Take that Mathologer and keep him locked away.
Take that job and... shove it.
Take that, Job, said God. "Ouch", said Job.
Let's eat grandma [said the cannibal to his brothers].
Let's eat, grandma!
Etc.
Antinatalist It’s a matter of style. I’ve seen several writers who omit that comma.
-He says something about rings
-I think it will be another Mathologer analogy or something
-He is actually talking about damn rings from abstract Algebra
-I get excited
Wooow, Exactly what I have felt :-)
They aren't just any rings, they're abelian rings!
What's your favorite field of math
@@camerontankersley3184 if you mean field in the common way, I love topology, measure theory, functional analysis and differential geometry. If you mean THE field, then I'll go with the reals xd
@@jksmusicstudio1439 Omg su. But the reals do be poppin doe. Anyway, I'm in college and my parents freak out whenever I wanna take any math courses with big numbers in them. But they kinda be paying the bills doe, so I have to listen to them. So Anyway, since I can't take the juicy stuff in school I pretty much have to self-study. So my goal is to go through a textbook on Linear Alg this break. Since ur a math expert, do you think I'd be ready for Abby Alg after that? I took a foundations course this semester so I have a touch of set theory and I can prove some super basic stuff. I'm trying to sneak in a combinatorics course next semester with calculus three, so should I wait for that to finish? Like what's the verdict and what should I take after Linear and Abstract Alg?
One ring for each integer above 1.
And one ring to rule them all.
That's called Z
There is actually a ring for each integer, but the negatives just copy their positive counterparts, 0 gives the integers itself, and 1 gives a very boring very mini ring that only has one element. :)
Three Rings for the Math-kings under the sky,
Seven for the Physics-lords in their halls of stone,
Nine for Computer Programmers doomed to die,
One for The Professor on his dark throne
In the Land of Ideas where the Colossus lies.
One Ring to rule them all, One Ring to find them,
One Ring to bring them all, and in the darkness bind them,
In the World of Imagination where the Shadows lie
@Ron Maimon Great summary! I only know a little bit about abstract algebra, but you just summarized multiple hours of video lectures into one YT comment! 😄👍
Just one thing. You said, "There's only one integer divisible by 0, and that's zero." Wouldn't it be more accurate to say that, "There's only one integer that's a *multiple* of 0, and that's 0."? Even 0 is not divisible by 0, right?
@@robharwood3538 I think graphically you can see in the curve for 1/x there is often a vertical line at x=0 so you could ask, "if x=0, what is y?" (to which the answer is "it could be anything" or as others put, "undefined").
I watched this video 5 times and I have to watch 10 times more in order to really get it, but I just wanted to tell you how much I love your videos and the way you teach. Your are exceptional!
I can't believe it's actually finally here. I've been waiting for this since the joke at the end of pi is transcendental proof. Thank you, Mathologer. You've somehow outdone yourself yet again!
Why isn't this in my subscription feed... I had to search up the channel name to find it
Pankaj Chowdhury Partha same here. TH-cam didn’t recommend it to me somehow.
Same for me... I only go to my subscription feed, but this video hasn't been shown there.
TH-cam's recommendation is, those days more than ever, under authority in order to broadcast only OMS approved informations about Coronavirus (or to censor any criticizing information on the situation or the behaviour of governmemts and their links to pharmaceutical lobbies or finance : it depends of your consideration of free speech)
Pankaj Chowdhury Partha same
This just appeared in my Home tab. No way I could have missed this.
Mathologer: " Avoid negativity "
Computer science: " A void negativity "
LMFAO
-|x|
kerning 💥
"I'm not about to propose"
You just broke my (math) heart!
Same, I had a cardioid arrest
@@TheOnlyGeggleslegendre-y pun.
I cant believe I hadnt heard of Quadratic Reciprocity considering my honours dissertation was on finite fields. Granted, it was efficient computation of matrix opperations under GF2 on GPUs, but still, I cant believe I'd never come across any of this.
Same. Except everything after "I can't."
That’s interesting that the addition and multiplication tables for 2 are the logic tables for XOR and AND gates 7:00
That's why it's taught on Master's classes in Electronics (or Electrical) Engineering Programs(or sub fields).
I noticed the exact same thing 👌🏻😌👍🏻.
@@HiteshAH That makes sense 💡.
What makes Quadratic Reciprocity so special, and in particular, why was it so important to Gauss? Why is squaring integers so important in Number Theory? To understand the answers to these questions, you need to appreciate the work Gauss did in the theory of quadratic forms and their application to differential geometry. Gauss's "fundamental theorem" of differential geometry is about how the 2nd order partial derivates of a function that defines a "surface", e.g. `f(x,y)=height`, depends on a specified quadratic form called the "metric tensor" of the surface (and its first derivates), which gives a rule for how to measure distances "on the surface". This shows a profound and deep insight into the nature of the relationship between a large number of otherwise seemingly unrelated abstract concepts.
Squaring numbers plays a fundamental role in both Number Theory and Geometry. After watching this video twice, going on a few wild goose chases, and not being able to stop thinking about it: I finally think I understand why Gauss placed such a high degree of importance on this particular theorem, and feel like I have a lot of reading to do now!
I think I read (wiki?) that Fermat gave specific examples of the "law". So lots of mathematicians were motivated in this direction to prove/generalize another example of something Fermat stated without proof.
Very interesting video, far from anything I've ever been exposed to. (And a welcome distraction in these confusing times.)
No that's a bit of a stretch. It's safe to say the metric tensor and quadratic reciprocity have nothing to do with each other directly.
One ring to rule them all
that seems to ring true, but there's also wedding ring ... and suffering
My thoughts, exactly 🎯😅!
I understand every word you say, but your sentences are mostly beyond my understanding! I am amazed at how you can explain these issues (I last studied Math over 50 years ago, but remain fascinated).
20:33 “Where did that come from?”
The big “whoa” moment for me! Thanks for making it so much easier to understand why this equation is significant. I could never understand why when I was taking a course in number theory.
In C↑D↓, the reason it only mixes up the rows and not the columns is because both C↑ and D↓ use the same vertical sequence (i.e., top, middle, button, top middle, bottom...). Only the horizontal sequence changes.
I prefer Flammable Math's avoid positivity shirt -|x|
Excuse me what the frick
*Mathologer walks into a party*
"Hey wanna see a card trick?"
You sure wouldn't want to play poker with him!
You always explain things so clearly that I know exactly where I stop understanding the maths. I often don't quite understand the last section or two of your long videos, but they're still fascinating.
"This video was so long, the Mathologer had hair when it started!"
. . . . .
"And so did I!!"
Fred
The story of how Euler got into formulating quadratic reciprocity is fascinating. It's contained in "Primes of the form x^2+n*y^2" by David A. Cox. HIGHLY RECOMMENDED because it's an historical recount.
This approach is awesome! We covered the law of quadratic reciprocity in number theory class, but the proof was omitted, and it came down to uninspired manipulation and flipping of Legendre symbols. But now I'm 95% convinced that I understand why it all works :)
I suppose something that could have been expanded upon for the benefit of other viewers is what the Legendre symbol is used for (which was briefly mentioned in the video), such as an example of solving a quadratic congruence over Z/pZ. Maybe that would make the whole LQR/Legendre business feel more motivated, but it's great as it is.
And maybe another visualisation of the quadratic residues/non-residues mod a larger prime (like 17 maybe?) so we can get a feel for the numbers that appear along the diagonal, in particular what to expect (somehow it is easy to forget obvious things like (49/p) = 1 for any prime p, no need to find remainder first, as the symbol can lose attachment to its definition when purely evaluating by rule).
i love your channel and all of your videos, but this must be the most mind blowing one in a positive way. your passion for math and didactics is just so fun to watch. in german there is the expression of a spark jumping “over” when someone successfully communicated something. at least in my case you made many sparks jump over/through this medium and ignited interest and sparked fires but in a forest clearing sense, burning ignorance/not-knowing. the best thing watching and rewatching the videos is the feeling of being part of an experience you and your team obviously planned to be captivating and entertaining but not compromising (maybe impossible to compromise) complexity. i love that! whilst making it seem light and easy to lay out al kinds of layered/interconnected topics and parallel to that showing the struggle with the task/empathizing with an audience and our common attention spans/rivaling media fun/game et cetera. self-referential in a postmodern sense but by god not so heady and dead serious! Vielen Dank und Grüße aus Deutschland.
Cycling answer: if you have a permutation of cards, looking at the last one, we see that all the other numbers are behind it. In particular, the lower ones are behind. Putting the last card in the front means creating inversions and flipping the sign that amount of times. But also all the numbers bigger than it are already behind it, and so there are already inversions, but putting the card at the front undoes these inversions and flips the sign by the amount of cards. So we see that combining both cases means that we flip the sign by the amount of cards behind tge last card. So if there are n cards, we flip n-1. If n is odd, n-1 is even, so the sign doesn't change. If n is even, n-1 is odd, so the sign changes.
Firstly, thanks. I'm looking forward to this. Before I get too far in and while I remember: Do you take requests? Can I suggest a Beginner's Guide to p-adic Numbers.
I will second that request.
Me too.
Well done with the video. I had to watch it a few times to understand it all! I was OK to the halfway point then started to struggle, but stayed until the end. Thank goodness for the chapters. Keep making them.
It was crystal clear for me. But I must admit I'm a french PhD in math, however not a specialist in Number Theory (in fact in Complex Geometry, Conformal Field Theory with a little extra knowledge in Non-Commutative Geometry, Intuitionist Logic and Measure Theory...what a conceited & pompous mess !). You've remembered me of my undergraduate years at the University and the wonderful mathematicians I had the chance to meet there. A pure joy !!!
*You've REMINDED me of my ...
se rapeller = remember
rapeller = remind
33:22 I just spotted another easy way to count the number of inversion-pairs, for this type of permutation: Notice that the top-left and bottom-right cards: 1 & 15, don’t feature in any inversion-pairs; while their ”opposites”, in a way, the top-right and bottom-left cards: 13 & 3, feature in the maximum number of inversion-pairs: 8; and all the other cards display a pretty nice pattern: The 1st row sees the number of inversion-pairs always incrementing by 2: 0, 2, 4, 6, 8; while the 2nd row features the constant number of inversion-pairs: 4. Finally; the 3rd row mirrors the pattern of the 1st row (not surprisingly). This means that the average number of inversion-pairs a single card features in, is 4. Then, multiplying 15 by 4 gives: 15*4 = 60; but that counts each inversion-pair twice; so, we need to divide by 2, to get: 15*2 = 30, which is also the number of inversion-pairs you get from multiplying the binomial coefficients: ”p choose 2” * ”q choose 2”. In general; the formula would be:
(pq*((((p+1)/2)-1)*(q-1)))/2. 🙂
I've never heard about quadratic reciprocity before … and I want to know more about this!
Thank you for providing good-quality post-bachelor math popularization!
I am a 9th grade from India and I would say the real pleasure and happiness by preparing for the mathematical Olympiads is this! If anyone is preparing for the Olympiad he/she can understand this very well!
AH! here is an olympian! you should check out 3b1b channel also . It is also quite dope
Love and respect to India, from Finland. 🇫🇮❤🇮🇳
I hope you did well, in those Mathematical Olympiads 😌.
Wow, this is the first mathologer video that I wasn't able to finish in one sitting. The longest and possibly also one of the hardest.
Still also one of the most useful I'd think. Prime number theorems are some of the hardest but most rewarding!
Thank you so much for continuing your amazing effort!!! You are a true gem in all this dimension.
37:25
Alternate proof from the one of eliya sne (for those who know more about permutations) :
Let Sigma be a random permutation, and Tau the cycled one.
We can get Tau by "multiplying" the permutation Sigma with a fixed cycle of length n, let us call it Epsilon = (1 n n-1 n-2 n-3 ... 3 2)
So the sign of the cycled permutation Tau is equal to the sign of Sigma times the sign of Epsilon, so we want the sign of Epsilon. It is easy to see that Epsilon has n-1 inversions, so if n is odd, then sgn(Epsilon) = 1 and the sign of Sigma is unchanged, and if n is even, then sgn(Epsilon) = - 1 and the sign of Sigma is changed.
End of proof.
I know this is a little bit more technical, but I hope it's a good alternative :)
So, basically invariance under conjugation ;)
This is the happiest classroom ever. Even though I don't practice mathematics (I'm Law and Philosophy student), I really appreciate your videos. You guys rock!
I liked your proof exercise near the beginning because it made me realize *why* 0 works the same in modular multiplication-it feels obvious to me now but it’s because multiplying by 0 here is like multiplying by the modulus in our familiar ring of integers, and when you multiply a number x by the modulus m, x * m is always going to be congruent to 0 (mod m).
So glad I stumbled on some of your videos today. I remember when we both worked at The University of Adelaide. Your enthusiasm and humour have inspired many, many people for several decades. I am thrilled to see you are still actively exciting people about Mathematics. I will have to view all your videos when I can!
Best Wishes, DB
What years did the Mathologer work at Adelaide Uni?
@@ccarson I guess it was back in the ‘90s. Gosh, I feel really, really old now.
Of course, in those days, it was still correctly referred to as “The University of Adelaide”, as it was formally named and incorporated in the 6th November, 1874 Act of South Australian Parliament. Yes, the name included “The” - with a capitol “T”. The age of the Internet saw variations informally introduced and eventually embraced, which can be evidenced on its own Web site. I am old enough to remember when the correct name was important … lol. Embracing the term Adelaide University was eventually accepted, if only because it elevated the university on alphabetically ordered lists.
@@Xubono For me, the usage of ”The” with a capital ”T”, in proper names seems quite natural. For example, the longer version of the name of my and my Best Friend’s micronation: ”The Forest”, includes ”The” - with a capital ”T”. 😅
Damn, this was, I think, your toughest masterclass yet. Almost all of it worked for me, but I was really stuck at the point where you used the powers of two and got the fact that the new permutation will be obtained by adding 3 to the new natural permutation. I would have benifited from a little step-by-step at that point.
This was a lot of fun, though. I'm looking forward to learning more about these fields and the inversions and other properties of permutations. Thanks for this!
39:16 tile an (ab)x(ab) grid with (a)x(b) rectangles and draw the top-left to bottom-right diagonal. If it intersects the bottom-right corner of a rectangle then a square is formed by (n)x(m) lots of (a)x(b) so na=bm, n!=b, m!=a so a,b cannot be prime.
This was a difficult video, but you made it much easier to understand. Thank you so much for your hard work.
I'm reading a proof-writing book and it's covering modular arithmetic right now so I'm going to give a proof that the diagonal placement covers the whole board (or try to).
Claim: If we're placing the cards down diagonally on a grid with prime dimensions, we will cover the entire board
Proof: Call the top left square (0,0) and the bottom right square (p-1,q-1) where p and q are distinct primes. Then the nth card we place down (starting indexing at 0) will have dimensions (n mod p, n mod q). Assume we are not covering the entire board, so there is a card we place down at some point, say, the kth card, that ends up in the same position as a previous card, the nth card, where k < pq and n
It essentially follows from the Chinese remainder theorem:
If
x = n mod p
and
y = n mod q,
then there is only one solution for n mod pq for a given pair (x, y), provided that p and q are coprime.
@Mathologer I am happy that you are back, I love you..
Professor, I am looking forward to you aiming at presenting the Galois theory the Mathologer way in the future. I have always considered the inability to solve quintics by radicals mysterious, hopefully your video will shed some light on the concept to non-mathematicians like me. Thank you very much for all the stuff you create, it is unimaginable to comprehend the amount of thought, time and effort to get such things done, with the above video being an excellent example.
Meanwhile the following may interest you: Galois theory is not necessary for proving the existence of a polynomial of degree five whose roots cannot be expressed by radicals; the prior Abel-Rufini theorem already establishes that; But Galois theory is helpful in finding an example of such a polynomial.
@@loicetienne7570 Thank you.
Thank you for making this AND taking the time to caption it! It made it so much easier for me to follow, I really liked this one and can't wait for permutations!
Took me almost four hours just to make the captions ! :)
@@Mathologer perfect timing too
this is the only if not one of channels that i watch all its content. amazing as always!!
1:25 scares me. But it's indeed the only surviving "portrait" of Legendre, the mathematician (previous ones actually showed _Louis_ Legendre, a politician)
He looks like Cruella De Vil 😅.
I'm a simple math lover. Mathloger uploads. I click.
Wow! That was awesome. Thoroughly enjoyed this new way to think about an equation I was already familiar with.
The pink identity can also be demonstrated as follows. Looking at the position of the cards after the permutation, each card makes an inversion with all other cards that are either higher and to the right or lower and to the left of that card. Such cards form smaller rectangles of cards that go either up to the top right corner or down to the bottom left corner.
Counting all cards in all these smaller rectangles, we would count every inversion twice, so we only consider say the top right rectangles. To get the total number of inversions we sum up the number of cards in all possible smaller top right rectangles, which is
Sum{i=1..p-1; j=1..q-1} i.j = Sum{i=1..p-1} i.(q.(q-1)/2) = (p.(p-1)/2)(q.(q-1)/2).
Legendre: >:[
Wherever he is now, I'm sure he is much happier knowing that there is such a great video discussing his work!
or he is just dead
@@toniokettner4821 Wow! Blunt atheist is blunt. 😅
Finally, one of my favorite theorems! I want to leave you an exercise about this theorem that made me appreciate it even more:
Let p be a prime such that p is either equal to 2 or 3 modulo 5. Prove that the sequence n!+n^p-n+5 has at most a finite number of squares
Thank you so very much. This is already becoming one of my favourite mathematics videos on TH-cam!
This video is quite the Magnum Opus! It makes a result from deep in the heart (bowels?) of mathematics accessible to mere mortals, and introduces a bunch of mathematical constructs (squares in Zn, rings, sign of a permutation) along the way. For me, watching this video felt like being a tourist on an eloquent expertly-guided tour of a hidden room inside a massive museum. I am left in awe of the inventiveness of the mathematical minds of yore and supremely appreciative of Mathologer's efforts to spread his enthusiasm for mathematics.
In response to Mathologer's query of what worked for me, with the first viewing I felt I was on top of the material until maybe the last 10 minutes. I expect that a second viewing will fix that, so... Congratulations Mathologer! I think you achieved your goal.
"Oh, I'll just watch one quick Mathologer video before sleep." I don't know why I don't look at duration before making these decisions....
You should Mind Your Decisions.
37:25
Proof:
Let n be the number of cards,
Let k be the number of cards bigger than the last one.
By transferring the last card to the other side i eliminated k instances in which sometimes bigger than it is before it but also added (n-1)-k instances in which its bigger the other numbers (because the rest of the numbers are smaller then it).
So eventually i just added n-1-2k .
Since its a power of two i can ignore the -2k (its even).
Now, i am left with just n-1 added to the power.
If n is even then n-1 is odd and therefore the sine changes.
If n is odd then n-1 is even and therefore the sine doesn't change.
[|||]
Great! You beat me to it.
Because this operation can be decomposed into (n-1) transpositions. And each transposition changes the sign.
(-1)^(n-1) = 1 when n is odd and -1 when n is even.
Once again you @Mathologer nailed it...gotta love mathematics more than before
I am teaching math my thirty-odd y-o daughter, who had forgotten much of what math she had learnt at school. And we just ended a long chapter on variations, permutations, and combinations (with or w/o repetition), their formulae, and their equivalences to certain functions. Now this video is coming to be the top cherry to that chapter! Many thanks, Mathologer.
I already love math, but the passion that you have is contagious
Same. I used to solve Maths problems, as a pastime, in kindergarten 🙂.
This didn't show up for me. Thought you might want to know that youtube seema to do some weird stuff again.
Yes, very strange, pretty much did not get recommended for two week :(
Great video, Mathologer! Thank you for your efforts in communicating amazing theorems like this. But may I dare to ask for more applications of the reciprocity law?
Love this proof, I started thinking of it at the beginning of the video, and then you went into it! I remember working out how this card trick worked years ago (although I learned it as 3 x 7, but now I see the same rules apply and its a more general thing, not just a relationship between 3 and 7!). I love how that same thing generalizes .
16:26 the only squares mod 5 are 1 and 4. -1= 4(mod5) so the negative of the square 1 is also a square, and -4=1(mod5) so the negative of the square 4 is also a square, thus all the square on z/5z are squares.
Q.E.D
46:28 okay, now that is the sneakiest conjugation I think I've ever seen
Isn't it soo nice :) ? You are the first one to remark on this.
@@Mathologer I didn't get it. Pls explain
This is no doubt one of the very best Mathloger videos. Thanks for creating this, I thoroughly enjoyed watching it.
This didn’t show up in my feed and I have been subscribed for a long time, the good thing is that now I have something to do for 1 hour
My 10yr boy watched this with me
- trying to inspire him- teach the language of math. 99.99999 he liked that part about who knows knows
99.999999...
So, 100? 😆
I used the "9.9999... really is equal to 10" to inspire my 11 yr old. Mathologer is great for all ages!
8:28 "actually, my first BOOK!" ... He says BOOK so loudly, I jumped.
Thanks for warning me, I checked the time in the video when I saw this comment, it was 3 seconds before, which was enough time for me to prepare.
R.I.P., people with headphones 😔.
Thank you for making this video. It is fantastic! I couldn’t stop watching it just to see what was next.
33:00 That ”Positively Sloped = Inversion” -trick really makes sense, for our Row-Up -> Column-Down -permutation; because, in any positively sloped pair, the upper card is the earlier one, in the row-wise order, and the later one, in the column-wise order (and vice versa, for the lower card); because the row-wise order is: ”Left -> Right; Top -> Bottom”, whereas the column-wise order is: ”Top -> Bottom; Left -> Right”. So, any card: A that’s up and right, comes before any other card: B that’s down and left, in the row-wise order; and vice versa, in the column-wise order: The down-&-left card: B comes before the up-&-right card: A. Also; in this Row-Up
-> Column-Down -permutation, we’re essentially switching rows and columns, which means that the internal order of any such pair gets flipped. Then, because we started with the cards in natural (row-wise) order (meaning: No Inversions.), the flipping of the internal order of any such pair amounts to an inversion. Whereas; in any horizontally or vertically aligned, or negatively sloped pair, the card that comes earlier, in the row-wise order, also comes earlier, in the column-wise order; thus, no inversions would emerge. 😌
Finally. Amazing.
My proof for the little thing:
Let’s look at the following order of cards: 4/5/3/1/2 lets say that the number of inversions here is る
We do the cycling thing: 2/4/5/3/1
Consider this section: (4/5/3/1): notice how the number of inversions between these 4 numbers’ orders doesn’t change, let’s say that number is X
So the only thing that changes after the cycling is the number of inversions for the cycled number that takes the first place after the cycling (2) in this case, the number of inversions it had in the initial order being logically る-X. Let’s look at the second order now: it’s number of inversions now becomes [( (5) -1 ) - (る-X)] with (5) being the number of cards we are playing with and (る-X) being it’s number of initial inversions
Explanation: when it comes to ( (5) - 1 ): this is the max number of inversions a single member ( (2) in our case) can do with the other members because a single member cannot have an inversion with itself. When it comes to the subtraction i did: that’s because when we did the cycling we took out member from last place to first place cancelling out all the inversions it initially had with other members (because they are in growing order now) AND the it now has an inversion with members it did not have inversions with initially, basically the number of inversions it has now has become the compliment of the initial number of inversions ( their sum is equal to (5) - 1)
Knowing all this, We can now move on to a more general case with (n) number of cards placed horizontally which have る inversions and the section of cards that did not change order after the cycling has X inversions ( we can do this in a more general case because all the explanations i did apply there too): the number of inversions after the cycling is therefore X + [n-1 - (る-X)] = 2X + n -1 - る ( lets call it N )
Notice how 2X is ever therefore it does not determine the parity of N so the parity of N in comparison to る is only determined by the parity of n-1: N and る have the same parity if n-1 is even aka if n is odd and the opposite is true as well. So the sign of the permutation doesn’t change if n is odd but it changes if n is even.
Tell me if i have a mistake because im not experienced with this type of math
@Mathologer This is definitely 1 of the all-time gems. Thank You, Professor Burkard Polster. 😌
Thank you for all the hard work you put into that video!!! I will need to watch this one a few more times if I hope to understand it. Could you movtivate me with a really cool example of how quadratic reciprocity might be used to study pendulums or something?
These mini rings are what music theorists call modifications. Basically, you think of a clock with n values and you count around the clock face until you find the remainder
37:25
this operation is the same as performing the permutation given by n 1 2 ... n-1 to the previous permutation
Since this permutation clearly has n-1 inversions and signs of permutations multiply, when n is even it will change the sign and n is odd it will preserve the sign
Alternatively: The inversions in the cards that just moved over are obviously unaffected. The inversions involving the card that switched ends all get swapped. The parity of the number of swaps is therefore the parity of one less than the number of cards, so the sgn switches if the total is even and not if it is odd.
@@iabervon Exactly 🎯!
oh man i am so happy! to see another mathologer video!!
Mister
you gave me a strong passion for the maths
thanks a lot for your work
Gauss and euler and reimann A JACKPOT!!!!!!!!!
It's like the best spin on the mathematician slot machine 😄
"I'll never do this again, promise" Actually, this was my favorite video of yours, so, please do it again?
Thank you, this sparks so much interest into looking into maths more again!
Excelent video Mathologer!
TH-cam didn't show me this video until now and I'm annoyed so I'm commenting to try to convince the Almighty Algorithm to show it to everyone else.
Same
Mind blown at the reordering property. I felt my brain was reordering too...
The ordering property is not mysterious. Apply a reodering first, then your permutation, then the inverse of the reordering. The inverse of a reordering has the same parity as the reordering (inverse of a product of two-cycles is the same product but backwards), so the final result has the same parity as the original permutation.
That was simply amazing. Such a nice proof, such clean presentation.
This is a most amazing presentation of the most amazing theorem in Mathematics! A must see for all budding mathematicians.
As an aside, you gave tantalizing glimpse of finite geometry. Why don't you make a video on these topics? Particularly, the prime power conjecture for finite projective planes is a sadly neglected topic in TH-cam. With its beautiful links with orthogonal Latin squares, this ought to be an eminently suitable candidate for a mathology video.
What
An hour?
*hand hovers indecisively over play button*
and it's a masterclass?
*hand drifts away from the play button*
and I need to do my flute practice
*hand retreats to lap* (oh very funny, plz don't go there)
OTOH it's a video with a funny bald German in it
*hand approaches keyboard*
but there are probably no cats in it being cute
*hand falls to side*
*crickets*
*more crickets*
reciprocity prime what the? wat IS dat?
*Mathologer smiles impishly and wiggles the hook*
Command voice:" Lock all targets and fire at will! All ahead Warp 6! Take us in Mr Sulu!"
*Punches the big red button...*
I felt that
Legendre looks angry!
Read the fascinating story about this portrait here en.wikipedia.org/wiki/Adrien-Marie_Legendre#Mistaken_portrait :)
@@Mathologer Thank you very much for the link! I think I now understand why he is angry -- they were using the incorrect portrait!
I swear he looks like a classic Final Fantasy character portrait
@@Mathologer it's fascinating, and also kind of sad because that is a terrible drawing to be remembered by! 😅
This quarantine just got a bit better thanks to you. Amazing video as always
Just: WOW! 💪👍
Thanks a lot for your motivation but much more for your dedication! Was tough for me to follow and had to stop to rethink few times but then I grasped it in the end!
... still searching for the applause button (; 👏👏👏
Nice shirt, did you know that Flammable Maths has an "avoid positivity" shirt?
This is amazing.
For a quick explanation of the sign of a permutation: You can represent each switch of two cards with an identity matrix with as many rows and columns as cards (15 cards means 15×15 matrix) EXCEPT that the corresponding rows are switched. This is an elementary matrix. The determinant of this matrix equals the sign of the switch. We know from linear algebra that switching two rows flips the sign of the determinant. The identity has a determinant of 1, so every switch flips its sign, and therefore the sigb of your permutation is either 1 or -1.
What a insight, it reignite my spirit to study the quadratic reciprocity again which I did not get it in my number theory course in my university study in 15 years ago.
25:29 Your whimsy is boundless ^_^
200 hundred different proofs. Me, an intellectual: Sorry this margin is too small to contain a proof 🤷🏻♂️
Fermat didn't have a proof, or if he did have one, it had mistakes since it was never checked by others
*prove me wrong*
'200 hundred'
That is a weird way of saying 20,000.
Awesome video, first one with my name in as patreon 😁
Heavy stuff, but loved it. I will need your next video to fully understand stuff, so I'll be rewatching the 56 minutes again.
Keep up the good work and thanks for making these video's
Thank you so much for your effort in creating these enjoyable and accessible videos! As a technical writer who is entering the world of mathematics late, I find these really help me to internalize so much of what I need to go study. Cheers, Rick Lehtinen