Pi is IRRATIONAL: animation of a gorgeous proof
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- Опубликовано: 14 авг 2024
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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.... 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/...
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.o...
Today's main t-shirt I got from from Zazzle:
www.zazzle.com...
(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!
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![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/1.gif)
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
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 RUclips 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
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
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
SPLENDID! More please! You and 3B1B are both doing such great work with math-related animations.
No kidding shit
Wow that last part of the proof was really nice :)
Can I just say that Mathologer is one of the most cranckiest, craziest, wackiest, nerdiest and most likable personalities in RUclips? Hello?
Love these amazing videos!
This proof is far more natural than any proof of this I've ever seen. Please make the video you mentioned at 15:56 :)
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.
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.
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.
I'm not a mathematician, and I love this channel. Thanks for all the sophisticated work.
Seeing you so cheerful about math on all your videos makes me happy. The joy you exude is infectious! Happy holidays!
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?
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..
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.*
✅
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!
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 😂
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.
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.
I love how much fun these guys always have filming their videos. It always makes me happy.
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.
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!.
3:45 Funnily enough; tan(60°) = tan(π/3) = √3; and tan(30°) = tan(π/6) = 1/√3; for both of which, it can be proven, why that’s the case; by cutting an equilateral triangle, in 2 congruent, right triangles. Then, letting the side length of the equilateral triangle = 2; which implies that the base, of one of the right triangles, is 2/2 = 1; while the height of either the equilateral triangle or one of the right triangles, is √(2²-1²) = √(4-1) = √3 (by Pythagoras’ theorem). Then, the side, with length: 1, is opposite to the 30° = π/6 rad angle, and adjacent to the 60° = π/3 rad angle; and the side, with length: √3, is opposite to the 60° = π/3 rad angle, and adjacent to the 30° = π/6 rad angle. Thus: tan(π/6) = 1/√3, and: tan(π/3) = (√3)/1 = √3.
1/√3 is, also, suspiciously close to γ (Euler-Mascheroni -constant). 🤔
These animations are hypnotic. Great stuff !
What software do you use to generate these wonderful animations?
0:29 - I'd love to see Vihart's reaction to that sentence.
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.
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: M = (-3+1)/(-2) = 1 &: M = (-3-1)/(-2) = 2. We, then, solve finite chunks of the infinite fraction; to see, which root it tends towards; and thus, which root we should pick (M = (-3+1)/(-2) = 1, when truncating, at the ”-”-signs, or: M = (-3-1)/(-2) = 2, when truncating, at the vincula).
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
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.
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!
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 RUclips community.
Also, this is a great video on the proof of irrationality of Pi. The 3 step approach is very cool.
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!
What a great topic to analyze. Thank you so much for taking the time to create these magnificent animations.
This animation made following what was happening soooo much easier than just going page by page. Thanks so much.
the first puzzle:
log10(2) is irrational:
10^u/v = 2
10^u = 2^v
5^u.2^u = 2^v
the right side and the left side have different primes because 5 is not on the right side, which breaks the uniqueness of prime factorization
log7(8/9) is irrational:
7^a/b = 8/9
7^a = (8^b)/(9^b)
7^a = (2^3b)/(3^2b)
3^2b * 7^a = 2^3b
again, this breaks the uniqueness of prime factorization
for the second log, I think that claiming you're rational is like saying you're smart, rational guys don't say they're rational, so I think it's irrational
Just finished my proof that shows
Mathologer=Awesome
that smile on mathologer's face at 10:43 is priceless lmao he just achieved the legendary proof by "et caetera"
before i discovered your channel, i was thinking that advanced maths are beyond a "common" individuals understanding.
Thanks for prooving me wrong Math.
Suppose log_10 (2) is rational, then there exist integers p, q, such that log_10 (2) = p/q (with q nonzero), which is equivalent to 10^(p/q) = 2. Raising both sides to the power of q and simplifying, we obtain 10^p = 2^q, which can be written as 2^p * 5^p = 2^q. Then, we can subtract 2^p from both sides and obtain 5^p = 2^(q-p). Since any integer power of a number has the same parity as that number, the right hand side is odd while the left hand side is even, rendering a contradiction. Hence, log_10 (2) is irrational.
Suppose log_7 (8/9) is rational, then there exist integers p, q, such that log_7 (8/9) = p/q (with q nonzero), which is equivalent to 7^(p/q) = 8/9. Then, multiplying both sides by 9, we obtain 9 * 7^(p/q) = 8. Then, we can raise both sides to the power of q and get 9^q * 7^p = 8^q. 9^q and 7^p are both odd, and 8^q is even. However, since the product of two odd numbers is always odd, this is a contradiction and hence log_7 (8/9) is irrational.
Note that the algebra for these proofs doesn't hold if p or q are negative, however we can eliminate the case(s) where either p or q is negative since that would yield a negative output of the log function, which is impossible. For the case where both p and q are negative, we can raise both sides (for both proofs) to the power of -1, which effectively takes the absolute value of p and q and then everything holds.
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
√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
5:48 Proof that “this fraction” = sqrt(3)
Let’s ignore the leading “1 +” for now, and just compute the fraction part, let’s call it x
Notice that our x is periodic, with a repeating pattern every 2 division bars. In particular, this expression is self-similar, so that everything after (and included with) the 3rd division bar can be replaced with x itself:
X = 1 / [1 + 1/(2+x)]
If we invert both sides and rearrange a bit, we have the following:
1/x = 1 + 1/(2+x)
1/x - 1 = 1/(2+x)
(1-x)/x = 1/(2+x)
Now cross-multiply:
(1-x)(2+x) = x
2 + x - 2x - x^2 = x
2 - x - x^2 = x
Move everything to the right:
0 = x^2 + 2x - 2
Apply the quadratic formula:
X = [-2 +/- sqrt(4 - (-8))] / 2
= [-2 +/- sqrt(12)] / 2
= [-2 +/- 2*sqrt(3)] / 2
= -1 +/- sqrt(3)
Thus we have shown that x = -1 +/- sqrt(3)… But can our original expression for x really be both of these?
Well, long story short, both x-values we found are fixed points of the function
f(x) = 1 / [1 + 1/(2+x)].
However, that doesn’t mean we can actually arrive at both of these values when iterating our continued fraction (where we basically start from x = 1, then keep iterating f(x) from there)
In particular, notice that our expression for x must be a positive number (since it only involves plus signs, and the positive integers 1 and 2). Thus it cannot equal -1 - sqrt(3), which is negative. Therefore, our x = -1 + sqrt(3)
Lastly, recall the original expression was given by A = 1 + x, which is to say that A = sqrt(3), as desired
17:28 Proof that continued fraction = 1
Similar to above, we can set
x = 2 / (3 - x)
Which yields
x(3 - x) = 2
3x - x^2 = 2
0 = x^2 - 3x + 2
0 = (x - 1)(x - 2)
So that x = 1 or 2. However, just like before, these are simply 2 different fixed points of the function that our continued fraction is implicitly using, but which one is the correct value?
Well, after the first iteration, we have x = 2/3, which is less than 1. In fact, our iteration function f(x) = 2 / (3 - x) will always be less than 1 , for any x < 1:
x < 1
-x > -1
3 - x > 2
1 / (3-x) < 1/2
f(x) = 2 / (3-x) < 1
Thus we can see that our continued fraction must be approaching 1 from below (and could never reach 2), so x = 1 is the correct value
BONUS:
For any integer n > 0, it turns out that the continued fraction
n / [(n+1) - n / [(n+1) - … ]]
Also equals 1. The notation is a bit cumbersome here, but basically we are just replacing “2” and “3” from the fraction in the video with “n” and “n+1” instead
By a similar argument, we get
x = n / [(n+1) - x]
…
0 = x^2 - (n+1)x + n
0 = (x - 1)(x - n)
So that x = 1 or n. As before, we can easily show that f(x) = n / [(n+1) - x] < 1 for all x < 1. Thus, since we start our continued fraction iteration with x = n / (n+1) < 1, it must be approaching 1 from below (and could never reach n, unless n = 1, in which case our 2 solutions were the same anyway). Therefore, x = 1 is the correct value
I can see that these points are fixed points of the associated functions, but I'm missing why the iteration does converge. If there are two fixed points, not necessarily will one of them be a point of convergence. The "fixed point" quality is a necessary condition for convergence (only in case that the function is continuous at the fixed point!), but it's not sufficient. For example, the function 2/x has two fixed points, but none of them is a point of convergence.
Thank you for all the work you did in 2017! Looking forward for more exciting videos. Frohe Weihnachten, joyeux Noël ! 🎅🎄
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!
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.
14:42 BIG credit for emphasizing that the initial calculations only tell us about the equality at each finite step, but are insufficient to prove that tan(x) is equal to the continued fraction after 'infinite number of steps' - the final one. Many, many, many people don't understand that and would just stop at this point, claiming their proof of the formula tan(x) = [infinite continued fraction] is complete.
5:50 evaluating 1 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + ...
Denote the quantity by x.
Then we have 1/(x - 1) = 1 + 1/(2 + 1/(1 + 1/(2 + ...
Doing the same manipulation again gives
1 / (1/(x - 1) - 1) = 2 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + ...
Now notice the right side is equal to x + 1. This gives us an equation that leads to a closed form for x.
1 / (1/(x - 1) - 1) = x + 1
1 / (x - 1) - 1 = 1 / (x + 1)
(x + 1 - x^2 + x) / (x - 1) = 2
-x^2 + 2x + 1 = 2x - 2
-x^2 = -3
x = sqrt(3)
Love his laugh
Эльдар Гафаров даж
Fzabanaci
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.
2:40 Here’s my take on it: log(10) of 2 is irrational. Here’s, why: Assume log(10) of 2 is rational. That means it can be written as a ratio of two integers: ”a/b”. That means 10^(a/b) = 2. Since 10 > 2, log(10) of 2 cannot be an integer, itself; so, maybe (for the sake of simplicity), consider the opposite: log(2) of 10, which is just a reciprocal of log(10) of 2: ”b/a”; so, 2^(b/a) = 10. 10 > 2 also implies that a < b. Now, can ”b/a” be an integer value? Well, 2³ = 8 < 10, and 2^4 = 16 > 10. This implies that 3 < (b/a) < 4. Since there is no integer between 3 and 4, ”b/a” cannot be an integer. Since neither ”a/b” nor ”b/a” are integers, this means that neither a or b are equal to 1; so, we really need to do things the long way. Since a < b, and 3 < (b/a) < 4; these facts, together, imply that 3a < b < 4a. This implies that 2^(3a) < 2^b < 2^(4a). Let’s plug in a some values for a, and see, where that takes us. Let a = 2. This means that 6 < b < 8. Let’s, now, consider the upper and lower bounds of 2b: 6 and 8. First: 2^6 = 64 < 10² = 100. 2^8 = 256 > 10² = 100. Now, the denominator (in this case, a) just means that you take the ath root of ”2^b”, which should, then, give you 10. Now, in this case, that would be a square root. Since, 6 < b < 8, where b is integer, there’s only 1 candidate for b, if a = 2: b = 7. So, sqrt(2^7) should equal 10, right? Well, sqrt(2^7) = sqrt(128) ≈ 11,314… ≠ 10; so, ”b/a” ≠ 7/2. What about a = 3? That would mean that 9 < b < 12; so, b could be 10 or 11; and we would, now take cube roots of 2^b to try and get to 10. 2^10 = 1024 > 1000 = 10³; so, we don’t even need to check for 2^11, since: 2^11 > 2^10 > 10³. So, cbrt(2^11) > cbrt(2^10) > 10. For a = 4, 12 < b < 16 -> b equals either: 13, 14, or 15; and we would take 4th roots to try and get 10. 2^13 = 8192 < 10^4 < 2^14 < 2^15. I think y’all should get the point, by now, that 2 basically can’t be taken to any power b, such that the ath root of ”2^b” = 10. So, log(2)10 can’t be represented as a ratio of 2 integers (b/a), which means that log(10)2 can’t be represented as a ratio of 2 integers (a/b); so, log(10)2 is irrational. Not quite a mathematical proof, per se; but just my train of thought written out ;-).
A simpler and *_(SLIGHTLY)_* more rigorous ”proof” might be that since both 2 and 10 are integers, and one is not a power of the other, since: 10 = 2*5, and 2^b = 2*2*2*…*2; there exists no power of 10 that would also be a power of 2; since: 2 and 5 are coprime, which implies that no power of 2 is divisible by 5; whereas *_ALL_* powers of 10 *_ARE_* divisible by 5, since: 10^a = (2^a)*(5^a). In other words; since 2 and 10 are both integers, and log(2) of 10 is not integer, it cannot be rational; implying that log(10) of 2 cannot be rational; so, log(10) of 2 is irrational.
Wow; that was a mouthful 😮💨!
*TL;DR:* log(10) of 2 is irrational.
”However; here’s a bit of bad news.” At these words, I saw The Matholantern’s eyes (pardon the pun) really *_LIGHT_* up 🎃😅.
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?
So a myth-loger shows us a math-speaking log
A circle is the equation πr² a sphere is πr³. A circle is the ring or symbol of "marriage" for it is to last or continue 'forever. A form of infinity you and your spouse's rings. This is because the circle can be split equal three times across before the calculation into each section becomes "askew" of the shape. In this the formula to go around the circle must be forever, or it would have an end. No circle has an end. Even our longitudinal lines are off in this sequence. 0° , 180° 359.999999999999....° approaching the zero, north or south pole. 3.14 is just the approx yet it has every single numerical value or sequence. Yet data, or datum, is not meta until it has a store value assigned to it. Every address you lived at by number will occur somewhere, yet only you would know it.
By the way, the "pi/4 thing". In many programming languages, pi is not defined, so you see "pi=3.1415926" statement in the program. There is a better way! "pi = 4*atn(1)". This gives you pi to the precision of the number system used in the program!
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. 🙂👍🏻
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.
My proof is chord based as the diameter of all circumferences is the last and largest chord found inside a circle. 3.14159ect. Is a ratio that is numerically based only, whereas the circumference has numbers which describe its diameter the value of a circumference ratio is 1.5702. There are 100 ratios to a circle’s circumference. They are found by both simple division of a limited numbers of values between 1-10 and by use of chords. Also, the corruption of time by the addition of fractions of a second is the second proof. Minus all decimals in the clock Hours, Minutes, and Seconds are all chords which are ratio of the circumference. This explains the end of General Relativity as the relationships is fix and is Intentional Relativity.
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'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*
That T-shirt is awesome!!!! Btw, I'm still watching the video.
Lambert's portrait looks like he's about to pull off a really tricky prank, and waiting for his high school bully to fall for it.
log10(2) being irrational:
Assume log10(2) can be written as p/q, where p and q are integers. This implies that 2 = 10^(p/q) = 2^(p/q)*5^(p/q), so 2^(1-p/q) = 5^(p/q). Taking both sides of the previous equation to the q'th power, we have that 2^(q-p) = 5^p. Since 2 is an even number and 5 is an odd number, and both q-p and p are integers, there can't be any set for p and q that satisfy the equation log10(2) = p/q, therefore log10(2) is IRRATOINAL.
Q.E.D.
log7(8/9) being irrational:
Again, assume log7(8/9) can be written as p/q, where p and q are integers and q is different than 0.
log7(8/9) = p/q
=> 7 = (8/9)^(p/q) = 8^(p/q)/9^(p/q)
=> 7*3^(2*p/q) = 2^(3*p/q)
=> 7^q * 3^(2*p) = 2^(3*p)
The only solution, for p and q is both of them being 0 (hence 1=1), but since we assumed q != 0, and that that would imply that log7(8/9) = 0/0, which is not even a number, there are no integer solutions for this - considering the fact that the prime factors are different - meaning that log7(8/9) is IRRATIONAL.
Q.E.D.
Big animations for you
You're a big animation
For you
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?
An easy solution for the sqrt(3) riddle is not to transform the term to get rid of the double "+1" in the beginning, but to rip apart the 2 into 1+1:
x = 1+(1/(1+1/(2+1/...))) = 1+(1/(1+1/(1+1+1/...))) = 1+(1/(1+1/(1+x))) = 1+(1/((2+x)/(1+x))) = 1 + (1+x)/(2+x).
That directly leads to x^2 = 3 x = ±sqrt(3), where only the positive solution makes sense.
The result is correct but think about what would happen if you applied the same logic to the sum in 17:29 (1=2/(3-2/(3-2/...)))
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.
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
JAWOHL!
Mathologer: «cringe»
I laughed, probably too hard.
I've always been bad at logarithms (I kinda have to work the notation out in my head), but I decided to give your puzzle a try.
log(10) 2 is not rational. Write 10^(x/y) = 2, and raise each to the power of y to get 10^(x) = 2^(y). However, there can be no power of two which is divisible by 10, because if 2^y is divisible by 10, then it must also be divisible by 5 as well, and the prime factorization for powers of two consist only of twos, so there can be no power of two whose prime factorization contains 5. So, there is no power of 2 which is a multiple of 5, thus there is no power of two which is a multiple of 10, thus there is no power of two which is also a power of 10.
Not sure about the second one, but it seems to me that if a wood log walks up to me and claims to be rational, then I think it's time to see a psychiatrist because perhaps it is I who is not rational...
log(7) 8/9 must also be irrational. Again, if we have 7^(x/y) = 8/9, again we can raise both sides to y to get 7^(x) = (8/9)^(y). However, all powers of 8/9 are less than 1, and all powers of 7 are greater than or equal to 1, thus there can be no power of 8/9 which is equal to a power of 7.
I hope that all makes sense. I'm not sure if these are right, because like I said, logarithms have always been a weak point for me, but I'm always looking to learn, even in failure :)
17:45 - Solving the equation a=2/(3-a) indicates that the continued fraction, if it exists, is either 1 or 2. Clearly it can't be both, so which one is it? It turns out that the continued fraction can't be equal to 2. To see this, calculate successive approximations using the recursive formula f(n+1)=2/(3-f(n)) and f(0) = 0. Now it is clear that if f(n) is less than 1, then f(n+1) must also be less than 1, since then the denominator for f(n+1) is greater than 2. We also know that f(0)=0, which is less than 1. Thus f(n)a. So the series f(n) is monotonically increasing and bounded by 1. Therefore, by the least upper bound property, it must converge in the real numbers. Then the initial quadratic equation indicates that answer must be 1.
I saw below some people posting proofs which basically go like:
1=2/(3-1)
1=2/(3-2/(3-1))
1=2/(3-2/(3-2/(3-1))
...
However this proof isn't correct by itself. To see this note that it also works with 2:
2=2/(3-2)
2=2/(3-2/(3-2))
2=2/(3-2/(3-2/(3-2))
...
The moral is that one needs to be careful when dealing with infinite series.
Awesome....
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
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.
Absolutely beautiful proof. Thank you and happy new year
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?
Your shirt is too distracting 😂😂😂
Proof that the fraction 1+(1/(1+1/(2+1/(1+...)))) = +-sqrt 3
First subtract 1 from both sides.
Now you can see, that the fraction is repeated in itself and because it is infinite, you can replace the first repetition with x-1. If you do that you get x-1=1/(1+1/(2+(x-1))).
Then you simplify: x-1=1/(1+1/(x+1))
Get a common denominator of x+1 and add the fractions: x-1=1/((x+2)/(x+1))
Then get rid of the 1/... part: x-1=(x+1)/(x+2))
Multiply with (x+2): (x-1)(x+2)=x+1
Get rid of the brackets: x^2+x-2=x+1
Subtract x and add 2: x^2=3
At this point you can see that both +sqrt(3) and -sqrt(3) solve the equation.
Q.E.D.
5:50
Take x = 1/(1+1/(2+1/(1+ 2/.... and so on
The recursive structure should be clear...
x = 1/(1+1/(2+x))
1/x = 1+1/(2+x)
2+x = x^2 +2x + x
x^2 + 2x -2 = 0
x = -1 (+-) sqrt3
So
x + 1 = (+-)sqrt3
The negative solution is to be rejected as the continuous fraction is made only by positive numbers, so it cannot result in a negative number.
Finally sqrt3 = 1 + 1/(1+1/(2+1/(1+ 2/.... and so on
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!
Wow.
Perhaps this is only video that I didn't see any advertisement before it. Thank God
This is one of the best video on RUclips, and it doesn't even have 1M views. We really don't value the right things.
(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
Math is irritational
Proof?
Lmao
Beautiful proof! One extra kick that fell out of it for me was the fact that you can't reverse the sequence of positive integers (1, 2, 3, ...) to, well, let's write it (..., 3, 2, 1). As a computer scientist, I know that reverting a list has O(n) time complexity, so reverting an infinite list takes forever but I still would've thought that it's not a _fundamental_ problem on the mathematical side. Well, turns out it is. Thanks for that side-insight!
Saying that 10^n
eq 3^m doesn't need to be an argument about even/odd. Claiming one side is even and one is odd is a special case of a more general argument about how its not possible to get the same prime factors on the left and right - the even odd argument is this argument, but with 2 as the chosen prime number. With this generalization in mind, we can easily see that log_10 2 and log_7 9/8 are both irrational. In the former, the side with 2^n will never have a five in its prime factors. In the latter, the simplest way to express it is 7^n8^m=9^m. The left side will never have a factor of three, and the right will never have factors of 7 or 2. As a sidenote, it's worth mentioning that in all these cases, m=n=0 actually solves the equation, but you have to assume m
eq 0 in an earlier step in the proof - so in a formal proof you would need to explain why it is necessarily true that m and n aren't zero
Very good. Actually in the original proof for log 3 you don't have to worry about v=0 because 3^0 is still odd (or because it's the denominator of a fraction and so cannot be 0) and of course u is not 0 because log 3 isn't. Can't hurt to say these things, but I don't think many mathematicians would bother, even in a formal proof :)
great video... very well explained. I hope your channel doesn't become more popular so you can keep responding to fan comments. As a physicist I have preconceptions about mathematicians being overconcerned with such small details. It's good to hear they don't always worry about such trifles.
GAAAAH
GAAAHHHH
1+1=5
POTATO is a COMPLEX NUMBER
SQURAE ROOT OF SANDWICH
lmao
zeroth
I get so excited whenever you post a video!
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 :)
That shirt is amazing! It took me a few minutes to work out that it was a base-conversion holiday pun. I'd wear that shirt all year long.
Another awesome video, thank you very much!
Happy holidays!
Thank you for working so hard on those slides for the infinite fraction, I finally understand how to create them now! I've been trying to learn for months
Also I absolutely searched for this video because it's pi day
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.
👍👍