Pi is IRRATIONAL: animation of a gorgeous proof
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This video is my best shot at animating and explaining my favourite proof that pi is irrational. It is due to the Swiss mathematician Johann Lambert who published it over 250 years ago.
The original write-up by Lambert is 58 pages long and definitely not for the faint of heart (www.kuttaka.org/~JHL/L1768b.pdf). On the other hand, among all the proofs of the irrationality of pi, Lambert's proof is probably the most "natural" one, the one that's easiest to motivate and explain, and one that's ideally suited for the sort of animations that I do.
Anyway it's been an absolute killer to put this video together and overall this is probably the most ambitious topic I've tackled so far. I really hope that a lot of you will get something out of it. If you do please let me know :) Also, as usual, please consider contributing subtitles in your native language (English and Russian are under control, but everything else goes).
One of the best short versions of Lambert's proof is contained in the book Autour du nombre pi by Jean-Pierre Lafon and Pierre Eymard. In particular, in it the authors calculate an explicit formula for the n-th partial fraction of Lambert's tan x formula; here is a scan with some highlighting by me: www.qedcat.com/misc/chopped.png
Have a close look and you'll see that as n goes to infinity all the highlighted terms approach 1. What's left are the Maclaurin series for sin x on top and that for cos x at the bottom and this then goes a long way towards showing that those partial fractions really tend to tan x.
There is a good summary of other proofs for the irrationality of pi on this wiki page: en.wikipedia.org/wiki/Proof_t...
Today's main t-shirt I got from from Zazzle:
www.zazzle.com.au/25_dec_31_o...
(there are lots of places that sell "HO cubed" t-shirts)
lf you liked this video maybe also consider checking out some of my other videos on irrational and transcendental numbers and on continued fractions and other infinite expressions. The video on continued fractions that I refer to in this video is my video on the most irrational number: • Infinite fractions and...
Special thanks to my friend Marty Ross for lots of feedback on the slideshow and some good-humoured heckling while we were recording the video. Thank you also to Danil Dimitriev for his ongoing Russian support of this channel.
Merry Christmas!
Fantastic! One of the most accessible proofs of this fact I’ve ever seen.
2 of the best math channels on youtube
Glad you like it :)
Mathologer ft. Pi creature
Where's Brady??
Loved the crossover video you uploaded today about topology. And then there was a Mathologer video. Great Christmas presents :)
"Welcome to the last Mathologer video" - WHAT "... of the year" - phew
Lmfao
@@realbignoob1886I came to the video 6 years after that original comment, and had the same thought!
That shirt! 25 base 10 = 31 base 8. In other words, 25 Dec = 31 Oct. Merry Halloween!
Also, according to a certain Blink 182 song and a movie based on a similar premise, you can have Halloween on Christmas.
Nicely noted😉 Though, shouldn't it therefore be more like "Merry Halloween"? LAWllll🙂
😯😯😯
December = 12 & October = 10
Soundwave Yes, but deca=10 and oct=8
To answer the puzzle: none of those logs is rational. Not even the one claiming it is. I mean, come on, a piece of wood shouting out statements on its own rationality? That's completely bonkers!
That's quite the irrational statement
In fact, the log's statement itself sounds pretty irrational 😂
Logs usually display comments... I believe this is merely an old fashioned :P
Re. talking logs it sounds like the overlap between mathologer fans and twin peaks fans is the empty set - oh guess I fit that description, never mind...
Karl Young, that was a Twin Peaks reference? But yeah, you're right, I wouldn't know...
Regarding the first puzzle : log10(2) cannot be rational. The same method can be used, and it is trivial to show that no power of two can be divisible by a power of 10 (Except 10^0, of course). log7(8/9) cannot be rational either. log7(8/9) = log7(8) - log7(9); and log7(8) cannot be rational since 7 is odd and 8 is even. This leaves the woodlog. We need to go down two paths for this, assuming the drawing represents a real situation : 1. Either woodlogs are incapable of reason or speech. In which case, this one could be part of the few rational/speaking ones, but it is unlikely such a behavior would evolve so fast without intermediary steps. 2. Or woodlogs are capable of reason and speech, usually. Yet, they never speak. They get chopped, sawed, burnt, and they still don’t speak. If they are both rational and willing to go through this shutting up, they must have a damn good reason. Yet this log just broke millenias of omerta over a pun. I can’t imagine a situation in which that’s rational.
Francesco Malhabile Just to point out if log7(9) is irrational then your proof for log7(8/9) no longer works. (x=Pi,y=Pi + 1, x-y=-1)
log7(9) is indeed irrational, you simply can’t find a rational number a/b that satisfies 7^a=9^b. And so, the proof is as follows: Assume log7(8/9) = a/b where a,b are whole numbers 7^(a/b)=8/9 | ^b 7^a=(8/9)^b *7^a=(8^b)/(9^b)* The left side is always a whole number when a is positive, whereas this doesn’t hold for any signed b on the right side. Turn this around: 1/(7^a)= *7^(-a)=(8^(-b))/(9^(-b))* =(9^b)/(8^b) The left side is always a whole number when a is _negative,_ whereas this doesn’t hold for any signed b on the right side. Since a can’t be positive or negative, it’s 0 by default, and when a=0, this means that b=0 too. 0/0 is indeterminate, which is a contradiction to the assumption that a/b is rational. => log7(8/9) can’t be rational! Proofs like these are why general statements about irrationals shouldn’t be used carelessly. With this proof format in mind, all that is needed to be accepted is basic algebraic manipulation, not assumptions that look like they’re begging for a counterexample. edit: spelling
jesna.j@akbartravels.in1 their first games. The first one. The 5@@korayacar1444
Koray Acar At 7^a = (8^b)/(9^b), you can re-arrange it to 7^a * 9^b = 8^b, which means that an odd number is equal to an even number; contradiction.
I did log7 8/9 differently. 7^u/v = 8/9 7^u = 8^v/9^v 7^u *9^v =8^v Odd / even Simpler 7^u*3^2v = 2^3v
This video is my best shot at animating and explaining my favourite proof that pi is irrational. It is due to the Swiss mathematician Johann Lambert who published it over 250 years ago. The original write-up by Lambert is 58 pages long and definitely not for the faint of heart (www.kuttaka.org/~JHL/L1768b.pdf). On the other hand, among all the proofs of the irrationality of pi, Lambert's proof is probably the most "natural" one, the one that's easiest to motivate and explain, and one that's ideally suited for the sort of animations that I do. Anyway it's been an absolute killer to put this video together and overall this is probably the most ambitious topic I've tackled so far. I really hope that a lot of you will get something out of it. If you do please let me know :) Also, as usual, please consider contributing subtitles in your native language (English and Russian are under control, but everything else goes). Today's main t-shirt I got from from Zazzle: www.zazzle.com.au/25_dec_31_oct_t_shirt-235809979886007646 (there are lots of places that sell "HO cubed" t-shirts) Merry Christmas, burkard
Mathologer Thanks for the video, and a merry christmas from me!
Mathologer hey, we too have a KZhead channel named ZORTHU-S.And there we made a video on why sum of all positive natural numbers is -1/12 .please check it out hope u like it
Mathologer You wrote "cox x" in the description in the part where you talk about the the French book
The link to the original write-up doesn't seem to work.
Loved it
7:26 Small mistake, the second term of the expansion of cos(x) should be of degree 2.
Yes, luckily one of those magical self-correcting typos :)
me too.
Maybe it's just written later
I wondered how many people would notice that!
Ye i came here to tell that
I can only imagine how awesome Lambert must have felt that night when he finished that proof!
Is this the same Lambert of the LambertW function?
@@carultch Yes
Can I just say that Mathologer is one of the most cranckiest, craziest, wackiest, nerdiest and most likable personalities in KZhead? Hello? Love these amazing videos!
SPLENDID! More please! You and 3B1B are both doing such great work with math-related animations.
No kidding shit
This proof is far more natural than any proof of this I've ever seen. Please make the video you mentioned at 15:56 :)
Wow that last part of the proof was really nice :)
Seeing you so cheerful about math on all your videos makes me happy. The joy you exude is infectious! Happy holidays!
I have seen this video over 10 times and I still get the chills when it gets proved that PI is irrational. Wonderful work. I know you worked very hard to make this animation and I must tell you, your hard work has been fruitful to many math lovers out there. I hope you never stop making such videos. Love from India.
log_10(2) = a/b [for some integers a, b, with b not zero] 2 = 10^(a/b) 2^b = 10^a 2^b = (5^a)(2^a) 5 divides 2 Contradiction. Therefore we conclude that log_10(2) is irrational. log_7(8/9) = a/b 8/9 = 7^(a/b) (8^b)/(9^b) = 7^a 8^b = (7^a)(9^b) = 2^3b 7 divides 2 Contradiction. Therefore we conclude that log_7(8/9) is irrational.
Not only is this video itself a great example of making a proof accessible, you are a great example of an educator that genuinely enjoys teaching others. You certainly make me feel more confident that I want to teach maths myself to others.
While watching this part of the video (4:56), I have come to the following remarkable theorem: 2L = 1E where L is the area measure of Lambert's nose and E is the area measure of Euler's nose.
What about their neck circumferences 😂
Beginning at 7:08 the series for "cos x" has "x/2" instead of "x²/2". When you expand the multiplications of x it is all solved, so it doesn't change the final result. Just pointing the typo if you want to correct it with a commentary on youtube. Great work ^^
Yes, luckily all under control in this respect :)
111aa
Great video, Mathologer! You really went all in on this one..! Very clear, light, intuitive and beautiful!! For years I'm avoiding looking into these irrationality proofs with fear from all the technical details... I value your great work very much! Thank you!
21:54 The Well Ordering Principle. Every non-empty subset of the natural numbers has a least element, so there can never be an infinitely decreasing sequence of natural numbers.
Here is an easier way: it is easy to show directly that every subset of the natural numbers which is bounded above is finite. Assuming an infinite, strictly decreasing sequence existed, the first term bounds the sequence above. Thus, there are only finitely many numbers in the sequences. But there are infinitely many numbers in the sequence because it is strictly decreasing. Contradiction.
Again you've made one of the best explanations with incredible animations just to teach people with the same interests on the internet. I'd like to take a moment and just thank you for your work this year all round. I hope you'll have great holidays and a happy new year. "Met een vriendlijke groet", (Dutch) Luc de Graaff.
Absolutely beautiful! Thank you for taking the time to do these videos! These are among the finest quality content I've watched!! Always excited to see them! Happy holidays from Venezuela!
This animation made following what was happening soooo much easier than just going page by page. Thanks so much.
I'm not a mathematician, and I love this channel. Thanks for all the sophisticated work.
What a great topic to analyze. Thank you so much for taking the time to create these magnificent animations.
These animations are hypnotic. Great stuff !
What software do you use to generate these wonderful animations?
Thank you for all the work you did in 2017! Looking forward for more exciting videos. Frohe Weihnachten, joyeux Noël ! 🎅🎄
i like how you present proofs sir. showing their sketch first and then filling in the gaps makes it both easy to understand them and remember, and leaves no room to get lost while we fill in the gaps later on. i recall countless times being totally lost after already like a half hour long proof done in a from a to z fashion what are we even proving in the first place..
Wow! Great presentation. Bad news is I don't think I could remember how to do this proof on my own in a million years. Good news is that when you were reviewing it, I could very easily follow the logic of each step. The animation definitely made tackling those nasty fractions much more palatable! I can't even begin to imagine what that proof looks like on paper. Ugh!
I love how much fun these guys always have filming their videos. It always makes me happy.
7:25 shouldn't the second term for cos x have x^2 instead of x?
hyperstone9 saw that too
Every good maths video needs at least one typo :) Luckily this one corrects itself a couple of seconds later.
hyperstone9 yes
He’s just checking to see if we are PAYING ATTENTION!! ;-)
Do we get gold stars for spotting errors?
Happy Christmas Mathologer x nice to see you again x
awesome video, thanks for your work! I imagine it must be super hard to create those videos and explain such complex topics in simple terms.
0:29 - I'd love to see Vihart's reaction to that sentence.
This is the first one of your videos that I've come across. Absolutely delightful. Thank you for making it.
I am truly amazed by what you have achieved in this video. This is the hardest maths I have ever been able to understand, and I only needed to watch the video once. Masterpiece!
I proved that infinite fraction at 6:00! Remember the fraction is 1+(1/(1+(1/(2+(1/(1+(1/(2+(1/...). we set the whole fraction equal to x so now x = (the fraction) we now subtract 1 on both sides, and then take the reciprocal of both sides . Leaving us with 1/(x-1)= 1+(1/(2+... Now we set the whole thing equal to z so now z =1/(x-1)= 1+(1/(2+... If you look at the fraction it's just a repeating pattern of 1 +...and 2+..., if we cover the first 1 +... and 2+... It's still the same pattern of 1+... and 2+.... Which is precisely our z! We can actually plug in z so now it's z=1+(1/(2+(1/z))) plug in z for the 1/(x-1) we get 1/(1-x)=1+(1/(2+(1/(1/(x-1))))) unfold bottom to top we get 1/(x-1)= (x+2)/(x+1) Cross multiply and rewrite as a quadratic in standard form gets you 0=x^2-3 add 3 on both sides and take the square root gives you x= ±sqrt(3). Throw out the negative sqrt(3) because it isn't a solution for the original equation. Finally by substitution, 1+(1/(1+(1/(2+(1/(1+(1/(2+(1/...).= sqrt(3) This proof would be better with actual visuals and math speak but it works.
Lovely proof, but wrong timestamp :^) The fraction is correct, though At 6:18, the fraction shown is equal the e. The one equal to root 3 is at 6:00
@@user-rv9vk8by5i Thanks! I changed the time stamp.
One fatal flaw in your proof: You let x be a number equal to that object. The problem is, you haven't yet proved that object really is a number! Here is why that matters: I will prove -1 = 0 1) Let x be the number 1 + 1 + 1 + ... 2) Note that we have -1 + x = -1+(1 + 1 + 1 + ...) = 1 + 1 + 1 + ... = x 3) That is x-1=x 4) Subtracting x from both sides yields -1 = 0 as needed.
@@travellcriner6849 Nice spot! Fortunately it's east to show convergence in the first case whereas 1+1+1+... doesn't converge
Absolutely amazing!And excellent explained.Uao Lambert was a genius not less than Euler if he succeeded to make all these steps.
Yes, did not know much about him before making this video, but the more I find out about him the more interesting it gets :)
Yes, very nice proof but “not less than Euler” might be a little strong...
You're quick to forget that it was Euler's work that inspired him to utilize infinite fractions, then he used like 2 clever tricks and basic calculation... It definitely is very beautiful, but not genius
Best Christmassy Video. Thank you. Love your T Shirt as always. Merry Christmas to you and your team. A big Thank You for all the lessons and hard work which you and your team have put into making Mathematics interesting for Humans. ♥️
Thanks a lot for uploading this video to share information. Your efforts are not wasted but your efforts taught others that there is a lot to explore in mathematics. And this video will also make people who hated mathematics to love it. This all would not have possible without your animations. Please keep making videos and share your knowledge with others.
YES! I found a Mathologer Easter Egg in the intro. 1010011010 = 666. 666 = (36*37)/2 which means it half of a pronic number, which means 1 + 2 + 3 + 4... + 35 + 36 = 666
You are only the third person to comment on this since the channel got going :)
I love it how you almost pointlessly factored out the "Only" in that statement. Its like equivalent to factoring out 100% in this statement. There is a 100% chance of a 1% chance of "ME" being mathologers favorite fan. By the way I know more digits of pi and √2 then you(2091 digits of pi( Age World Record) and 1024(2^10) digits of 2^1/2) :P.
Thank you, sorry I forgot some of my mathematical vocabulary, I will try to improve it when I have some time.
@@SpiffyCheese2 it's all in the name lol
He probably knows when to use "than," though.
Just finished my proof that shows Mathologer=Awesome
I get so excited whenever you post a video!
Another glorious gem made both accessible and entertaining for modern armchair math enthusiasts around the world. I continue to be amazed how just a few twists and turns of reasoning can illuminate a claim as surprising and seemingly unfathomable as the statement that tan(rational>0) is guaranteed to be irrational. Truly amazing! Thank you for making these delightful videos. If you were to set up Patreon or some other means for us fans to show our support, I'd gladly participate!.
Assuming Log2=a/b and a,b are positive integers also b>a because log2 is not bigger than 1 but bigger than 0, then 2=10^(a/b) 2^b=10^a 2^b=(2^a)(5^a) 2^(b-a)=5^a 2 is an even number while 5 is odd. So this equation cant be correct while (b-a) and a are postive integers. Therefore log2 cant be rational, it is irrational
Sımişka Zırıhta you could have ended the proof at 2^b=10^a since a power 2 never ends in 0
But this proof proofs both statements
... or use modular arithmetic. 2^b = 10^a implies 2^b = 0 (mod 5)... and use the principle of uniqueness of representation of a number through the product of its prime numbers: If it is possible to have 2^n = 0 (mod 5), some integer would have TWO possible representations, one without a 5 as its primes ( 2^n) and one with a 5 among its prime ( to be 0 mod 5). So 2^b = 10^a cannot hold with integers.
that smile on mathologer's face at 10:43 is priceless lmao he just achieved the legendary proof by "et caetera"
Absolutely beautiful proof. Thank you and happy new year
FFS. I couldn't help but notice that 25 Dec = 31 Oct T-shirt and kept thinking about what it meant. I finally figured it out (and banged my head on the dining table few times enough to worry my parents). You have the best collection of T-shirts of all KZhead community. Also, this is a great video on the proof of irrationality of Pi. The 3 step approach is very cool.
Love his laugh
Эльдар Гафаров даж
Fzabanaci
√2 and π were fighting outside. 8 tried to calm them down. But they are irrational so they kept fighting. 8 came between them, but everything got worse because, unfortunatly √2 ate π.
Lol
Amazing video! We can see you worked a lot to make it! Good job, really good job! It blows my mind how deeply you're devoted to explaining all these maths stuff. I study Maths and work for the University of Geneva in popularization (mainly) and I must say not everyone has your talent for explaining stuff so cleverly Bravo! Can't wait for the next ones!
Glad you enjoy my videos so much and thank you very much for saying so :)
While eveeyone does try to show the toughest concepts to be simpler, you on other hand, mathologer, make us fall in love with it..Millons thanks for all your great efforts.
That T-shirt is awesome!!!! Btw, I'm still watching the video.
that T-SHIRT
"No it doesn't... wait... wait... oh shit it's true"
I put a link to where I got it from in the description :)
Thanks
I had been racking up my brain then it hit me. Brilliant!
Another awesome video, thank you very much! Happy holidays!
Great animations. They definitely make the maths more accessible. Keep them coming please, they're certainly worth the effort 👍
At 17:27, if we ignore the 1 and set it to x, doesn't that equation have two answers, being both 1 and 2, as the equation simplifies to x = 2/(3-x), and solving the equation results in valid answers for x being equal to both 1 and 2?
My exact thoughts. I don't how we could show it is not 2 though. Any ideas?
This was surprisingly easy to follow; even for me, who took Intermediate Maths in high school, and stopped there. Thank you for making this beautiful proof accessible even for amateur mathematicians, like me. 🙂👍🏻
17:34 Well; since the bit getting subtracted from the first denominator is equal to the whole infinite fraction, due to its fractal-like nature, we can write the whole thing as: x = 2/(3-x). Now, we can solve for x: x = 2/(3-x) | *(3-x) 3x - x² = 2 x² - 3x = -2 | +2 x² - 3x + 2 = 0 *[Enter quadratic formula]* x = (3 +/- sqrt((-3)² - 4*1*2))/(2*1) = (3 +/- sqrt(9-8))/2 = (3 +/- sqrt(1))/2 = (3 +/- 1)/2 x = (3-1)/2 = 2/2 = 1 *OR* x = (3+1)/2 = 4/2 = 2 Of these, only x = 1 will expand out to give the infinite fraction (as you can verify with your calculator, as the truncations will converge to 1). Therefore, we discard the ”x=2”-solution; and therefore: *x = 1.* ✅
Lovely !!! I enjoyed every single minute of this video ! And all those animations must have took so much time and effort to make! Incredible ! Thank you !
Big animations for you
You're a big animation For you
So a myth-loger shows us a math-speaking log
It was a great, accessible video. Thanks to that, i was able to focus my attention to finding out flaws. And having seen an other of your videos, i'm now very sceptical about the entire adding and subtracting the "infinite series" denominator"; that too after re-arranging the terms.
Merry Christmas mr Math-person, and your funny heckler in the background
The continued fraction at 17:45 could equal either 1 or 2
No, it can't. You have to refute 2.
@@mathislove3722 How do you refute 2?
@@ekz9479 For example, you can show that the fraction is always smaller than a number smaller than 2. However, I proved the limit to be 1 directly. If you look at the terms of the sequence obtained from the continued fraction, you'll see that it follows the pattern (2^n-2)/(2^n-1). This pattern can be proved to be valid. So, the general term for the sequence simultaneously proves the limit to be equal to 1.
@@mathislove3722 Thanks a lot. It was bugging me.
Awesome....
The last part was really nice, very precise and strong, still simple. Thank you!
This is really great and accessible. I always show this channel to people who are interested in math, and have some grasp of it. Thanks alot mathologer. Your work is appreciated.
That's great :)
Dir auch fröhliche Weihnachten!
hahaha
(ha)^3 :)
Mathologer H*a*h*a*h*a:[Null]) (ha)^3/[Null]) (ha)^3/[Null] (ha)^3/0 (ha)^3=0 (Because if not it is=∞) h=0 or a=0 Q.E.D
I've always found it amusing that a ratio can be irrational. I guess I'm easily entertained.
That shows how we can't really understand infinity or at least the pythagoreans didn't
RATIOnal: can be expressed as a ratio You can then simply apply an "ir" to provide the converse, ir-ratio-nal *when a dead violin god applies English to math*
Thanks sir for these kinds of videos, now I'm getting addicted of watching your videos, it's really increasing my curiosity. Thanks for the wonderful and unique animation.
Merry Christmas mate. Thanks for the videos. You're my hero.
:)
JAWOHL!
Mathologer: «cringe» I laughed, probably too hard.
Wow.
What a beautiful Christmas present! Thank you
Oh and thank you! Merry christmas & happy holidays!
At 17:40 it is equal to 1 and 2 both.
vinod kumar Was just about to comment that :D
ReconFX, quadratic :)
so what does that mean? it is impossible to solve? It can't be both at the same time, right?
(HO)³
Marco Zz Great pun!
I find the 25 dec = 31 oct better tho
Yeah, cute. But to be rigorous (!), it should be 3HO. :-)
3ho = ho+ho+ho (ho)^3 = (ho).(ho).(ho) = hohoho [ 1 1 1 ] . ho = [ ho ho ho ] I think the vector makes more sense, but it's less "t-shirt friendly" ;)
@@frechjo idk but I can't stop laughing xD
Merry Christmas, Mathologer. Thanks for sharing that great proof, and your wonderful pun about an infinite descent into mathematical hell ;) Also, I loved your solution to 3Blue1Brown's mug puzzle. It's so very "you." It's great.
This channel is wonderful. You making proving things for the sake of proving things fun and exciting. That's hard to do!
The proof of the infinite fraction that yields one is simple: 1 = 2 divided by 2 2/2 = 2/(3-1) Now substitute repeatedly 1 with 2/2 and the 2 at the denominator with 3-1 QED
Davide Morgante 2=2/1, 2=2/(3-2), 2=2/(3-2/(3-2)), then again and again and you get the same formula as for 1, so 1=2 :)
if you solve it out algebraically you get S = 2/(3-S) , this yields a quadratic with two solutions: S = 1 , and S = 2. Makes sense given both of your proofs.
Harlequin314159 yep
That's quite nice!
Quantum I too arrived at that conclusion, so what does it really means? Are we doing this wrong?
Math is irritational
Proof?
Lmao
Helpful! Thank you for your way of explaining and for the animations
That is wonderful to see. This way of rendering mathematics is very very pleasant. Well, you make it look joyous and effortless to yourself. The animations allow the imagination and working - mathologising - brain to gen and enjoy. Thank you.
but first, please explain WHERE DO YOU GET THESE T-SHIRTS ??
A lot I make myself but I am also constantly on the lookout for new maths t-shirts. I put a link to the place where I got the t-shirts in this video from in the description :)
Mathologer omg i didn’t expect your answer but thank you 🤗✨
I can't stop laughing. That shirt is hilarious!
zeroth
This was amazing, thank you for visualising such an interesting proof.
Great ivideo! Merry Christmas and happy new year!
what if b=a/2 c=b/2 d=c/2 e=d/2 and so on? every term would be less than the previous term and still be rational: a/2^n. PS: I am NOT saying pi is rational. I felt that this explanation was a bit incomplete to my expectation.(not saying that it is, I just felt so)
The numbers A, B, C, D, ... are positive integers :)
yes, but infinitely many, no?
i actually was having the same misunderstanding. the fact the A, B, C, ... are integers solves it for me. :) thx
Sushant Poudel they have to be positive integers
GAAAAH GAAAHHHH 1+1=5 POTATO is a COMPLEX NUMBER SQURAE ROOT OF SANDWICH
lmao
About the infinite fraction at 17:27: For any 2 positive real numbers with difference 1 the limit goes to 1. Let the numbers be x and x+1. Then it can be shown by induction that the partial fraction is (x^n+x^(n-1)+...+x) / (x^n+x^(n-1)+...+1) which goes to 1 as n goes to infinity.
At 5:52, let 2+1/(1+1/(2+1/(1+...)=x. Then 2+1/(1+(1/x))=x. Simplifying results in the quadratic x^2-2x-2 which has the positive root 1+sqrt(3). The desired value is x-1=sqrt(3). At 17:30, let 2/3-(2/3-(2/3-...)=x. Then 2/(3-x)=x which results in the quadratic x^2-3x+2 which has roots 1 and 2. In general, consider the infinite fraction k/(k+1)-(k/(k+1)-(k/(k+1)-(...) which becomes the quadratic x^2-(k+1)x+k which has roots 1 and k. Note this holds for all real k. Side note: I don't know how to prove why the fraction must equal 1 instead of k. If somebody can prove this, that would be great.
17:41 Since the infinite fraction really *_IS_* infinite, and has a simple, self-similar pattern; we can simplify it, by representing the infinite tail, as the fraction, itself; like so: M = 2 / (3 - M). Then; multiplying both sides by ”3 - M”, gives 3M - M² = 2, which can be rearranged to the form: -M² + 3M - 2 = 0, which can then be solved with the quadratic formula, to give 2 roots: 1 (M = (3 - 1) / 2) & 2 (M = (3 + 1) / 2). We then solve a finite chunk of the infinite fraction; to see, which root it tends towards; and thus, which root should we pick (M = (3 - 1) / 2 = 1).
Very nice proof! I was able to follow all the logic, though I'd have to muddle through the details in order to reproduce the proof. I've looked at other proofs of the irrationality of pi, but I was never able to follow them. Good job!
Wow! Thank you for uploading this video. I'm not very good with mathematics, and I'm currently struggling with it in college, but I am so interested by these "mysteries" (for me they are still mysteries lol), and your videos make me want to learn more about it. I wish they taught like you in my University. Love from Brazil.
You are awesome and the internet appreciate the hours you spend of this presentation...
Awesome, continue doing videos like this, don't know how much you helped me.
I appreciate the effort to make these over 100 slides thanks! I wonder if the final proof was around 40 pages long how many pieces of paper he went through coming up with it? the economics of the availability of paper might actually have been an issue so long ago! im going through all your videos now thx
Your videos have the same effect as reading a detective novel. The revelation at 22:46 is really mind blowing like a climax of a detective novel. 👍👍
One of the best explanations I've seen for pi's irrationality that doesn't include assumptions that seem to come from nowhere. The animation was great in showing this too. I'd love to see more videos simplifying proofs that seem too complicated to understand. Maybe one about pi's transcendence?
All the proofs of the transcendence of pi I know are really, really scary :) Anyway, will definitely give it a try eventually :)