Infinite fractions and the most irrational number

[@ffggddss]

My answer to the riddle: Is it = 1 or = 2? In that continued fraction, as you write it, you alternate between ending with a numerator and ending with a denominator. If you stop to evaluate it, ending always with a numerator, that sequence converges to 2 - in fact, it IS just 2, every time! But if you always evaluate it ending with a denominator, the sequence converges to 1. Of course, if you stop at every numerator and every denominator, the sequence fails to converge. So the meta-answer is, you have to define how you mean to interpret the CF; the usual way being the 2nd - always stop at a denominator. That definition makes the answer, 1.

[@Mathologer]

Great answer :)

[@ffggddss]

Thanks! On further contemplation, it seems that you will run into this conundrum when and only when you allow numerators to exceed denominators, in a non-vanishing fraction of the terms. Or some condition close to that. Which may be one reason that the 'all numerators must = 1' case is of special interest.

[@KnakuanaRka]

Basically, no matter how far you carry the calculation, whether you stop at a 2 or a 3 radically affects the end value. This is basically saying that the fraction doesn’t converge, which is what Mathologer said is part of recognizing that the fraction is not meaningful, since the truly infinite version does not stop at any point.

[@aaronh8095]

So basically -1/12?

[@jaymangal1200]

Hey @@ffggddss , great answer above. Although I think the theory of numerators exceeding denominator might not be right. Consider (1/2)/((3/2)-((1/2)/(3/2)-((1/2)/(3/2)... ( This was the best font I could come up with in KZbin comments). This above sequence converges to either 1 or 1/2 depending on your theory

[@M0rph1sm55]

If you start with 1, the sequence of partial fractions is constantly 1, thus converges to 1. If you start with 2, same thing just with 2. If you just look at the infinite fraction, you get a quadratic equation by calculating its value. This equation has two solutions: 1 and 2. So the infinite fractions are basically the same, but their values depend on the value you start with.

[@PC_Simo]

Indeed 🎯.

[@m4gamer_syr414]

In the 1 infinite fraction, in the end, the denominator will be 3-1. But in the 2 infinite fraction, in the end, the denominator will be 3-2. Here is the difference between the infinite fraction of the 2 and the infinite fraction of the 1. So they are not equal because of this difference.

[@damienw4958]

Let the infinite fraction from the first problem equal x. Now replace the "infinite part" with x to get x=2/(3-x) Rearrange ==>. 3x-x^2=2 x^2-3x+2=0 (x-2)(x-1)=0 Therefore there are two separate solutions x=1 and x=2 but since they are separate, 1 does not equal 2! Maths is not broken! Rejoice!

[@edeneden97]

x=1 and x=2... still doesnt answer the question

[@WarrockSuX]

+Eden Lumbroso no, because a polinomial of nth degree has n solutions(zeros). Damien ended up with a polinomial of the 2nd degree with two solutions x1 = 1, x2 = 2.

[@edeneden97]

+My Name is Chef still doesn't answer the question how come 1 and 2 both equal the same thing (the fraction. not x). a guy wrote 2 is not a valid answer because the fraction does not get closer and closer to it like it does with 1.

[@damienw4958]

+Eden Lumbroso it does, substitute x=1 and x=2 into the original equation and they both work but not at the same time, it is either one or the other. This is due to the nature of quadratics and polynomials in that there will be n roots where n is the highest order of the polynomial (n=2 for quadratics) but those roots are not equal to each other and with the rearrangement I did, I proved the the infinite fraction is just a quadratic in disguise and will follow the same rules as polynomials.

[@gagaoolala9167]

It does; it shows that for any single-repeating-digit continued fraction there are two valid solutions. In general for q = 1/(a+(1/a+(1/a + ...))) we see that: 1/q - a = q so 1 - qa = q^2 q^2 + aq -1 = 0 Without factorising, we see that it has two solutions because it's order two.

[@Snakeyes244]

MAYBE 1 REALLY DOES EQUAL 2

[@birricforcella5459]

Being familiar with post-modernism, that sounds totally reasonable.

[@minimon796]

+Birric Forcella I wouldn't say something that is post-modernistic is reasonable. More like it's not reasonable but who cares it doesn't matter, if I feel it's right then it is right in some form that no one would even bother

[@birricforcella5459]

Gee, haven't you ever heard of something called humor or sarcasm? (Facepalm)

[@minimon796]

Birric Forcella it wasn't even sarcasm nor it was funny

[@CYXXYC]

Do you care? Math is broken anyway

[@brianconstantinescu484]

you never told us what the continued fraction in your thumbnail evaluates to...that one's the coolest one...

[@김지원-m8q]

I am so curious

[@landsgevaer]

That can be regarded as the most rational irrational number, since it can be approximated better and better by rationals the deeper you go. It equals I0(2)/ I1(2) = 1.4331274..., where the In are modified Bessel functions of the first kind.

[@idklol2566]

@@landsgevaer Hi, I’m not good at math and I know I’m a month late, but would you mind explaining it in a way where I wouldn’t need to know Bessel function is? Or is there a way of expressing this number as a radical?

[@aitorcazalis2307]

@@landsgevaer What about 2 + (1/(4 + 1/(8 + 1... with powers of 2? or powers of 3?

[@landsgevaer]

@@idklol2566 Although I knew the Bessel functions, I only found that result somewhere, so I have no proof or derivation.

[@Mathologer]

A couple of interesting answers to the puzzle already, but I'll refrain from making any comments myself until tomorrow (here in Australia). Have fun until then :) Also, thank you to Zacháry Dorris for contributing English subtitles to the last video (Riemann's paradox), Rodrigo Naranjo for contributing Spanish subtitles and Étienne Leb for his French subtitles! Added the next day (in Australia). Quite a few answers that are spot on. Before I give my own answer to the puzzle just two hints: 1) The continued fractions in the puzzle are a bit different from the simple continued fractions that I discuss in the rest of the video. 2) What I say in the interlude is crucial for getting to the bottom of what is going on here.

[@DaniPhii]

I think that the simplest answer is that a rational number can't be expressed as an infinite fraction, only irrational numbers can be expressed that way, so I think you did say the answer in the video, you tried to got us! 😆

[@tomirendo]

I don't think so. If you use the method explained in the video in order to get a continued fraction of either 1 or 2 the continued fraction will be finite. I don't think it means you cannot get an infinite fraction of a rational number.

[@DaniPhii]

+yotam vaknin But he himself says that for a rational number, there's no infinite fraction, it can be expressed as a fraction, but not an infinite one.

[@TheMusicJan]

+Ⲇⲁⲛⲓ Φi No he said that using the methode to create a fraction will end, not that any given fraction representing that number has to end (aus far as I understood)

[@Professor_Kaas]

The partial fractions do not converge.

[@BigDBrian]

I believe the answer to the puzzle is in the interlude. You can only place the equals sign if the partial sums converge to the answer. But in the expansion for 2 they don't, so you cannot say 2 = (the written infinite fraction) = 1, because it only converges to 1.

[@mesplin3]

that fact depends on how you write the equation. One sequence follows: {2/3, 6/7, 14/15, ... k/(k+1)...}. Another: {2, 2, 2, ...} So both converge.

[@BigDBrian]

Michael Esplin The partial sums were defined to include the entire fractions. I'm just using Mathologer's definitions.

[@mesplin3]

+mrBorkD Fair point.

[@JordanMetroidManiac]

Then why doesn't the substitution for two work? Obviously 2 does equal 2/(3-2), so anywhere you see a two, you can replace it with 2/(3-2).

[@RipleySawzen]

Both identities and infinite fractions are, in fact, correct. The nuance here is at 1:23 when he states the right sides are both identical; they are not, because in the case of 1 we have been replacing a 1 the entire time, in the case of 2, a 2. The "last digit" of this series is, in fact, quite important, because it is the number we have been substituting for this entire time. The "dot dot dot"s are not the same at all, as one has a value 1, the other, 2.

[@AnastasisGrammenos]

on 9:30 I noticed that after 333/106 comes 355/113 witch is (333 + 22)/(106 + 7)! Wow, weird!! [22/7 is the first one]

[@Mathologer]

That's an interesting observation. I should add that if you've got three consecutive best approximations they are always related like this a/b, c/d, (a+kc)/(b+kd) for some k. Try it :)

[@KnakuanaRka]

That’s because the last number in the finite fraction is a 1. For example, phi has a continued fraction of all 1s, and its finite approximations are ratios of consecutive Fibonacci numbers, which always have that property. In fact, this works for any pair of approximations, except you have to multiply top and bottom of one of them by the last number of the fraction before adding things up; it’s just more visible when that last number is 1.

[@jackweslycamacho8982]

It's kinda easy to see empirically, approximation fractions usually consist of numbers that don't share factors

[@ArjunSharma-vo2jq]

@@Mathologer you should definitely make a vid. explaining this

[@averagejohnson3985]

very circular

[@SeeTv.]

10:10 turn on captions

[@akshataggarwal4002]

Bruh! Plot Twist hahaha

yeah right? these captionmakers are so annoying

[@andrewbuchanan5342]

Very fluent and intuitive coverage of a topic which can be very algebraic and opaque. A lot of ground is covered in a very short talk.

[@rayvaldez8582]

this channel is incredible

[@Mohsens]

Some of the most educational videos are on this channel. What a wonderful way of making sometimes complex topics so accessible. Simply amazing.

[@-abigail]

Thanks for the interesting video! I've also been watching videos about "distributed mode loudspeakers" which are essentially a flat panel (of wood, foam, etc.) with a small speaker driver attached to the back. One of the recurring problems with DML construction is resonance - depending on the dimensions of the panel and the position of the driver certain wavelengths will cause the panel to resonate. For example, with a square 1m x 1m panel, and the driver placed in the middle, there will be resonance at frequencies that correspond to a wavelength of 50cm, as the wave will reflect off the edges of the panel and constructively interfere with itself. But it's worse than that, as as well as all these simple reflections along a single axis you also get more complex 2D resonances - all the different patterns you that Chladni plates show are visualizations of all the different resonances that we want to minimize. A rule of thumb I've heard from DML designers is to use a panel whose side lengths are not a simple ratio of each other, and to place the driver roughly ⅖ from two adjacent edges and ⅗ from the other two edges. I wonder if the golden ratio has applications here? Specifically, I hypothesize that the amplitude of resonance peaks may be minimised by a panel whose side lengths have the golden ratio to each other, and the driver is positioned according to the golden ratio - I think (1/φ) from two adjacent edges and (1 - 1/φ) from the others? It should be possible to simulate this but I can't find any existing Chladni plate simulators that allow for non-square plates, and I don't think they model the position of a 'driver' either.

[@dimicdragan5922]

I am rediscovering math by watching your videos. They are by far one of the top best math videos i follow and watch on youtube. Keep up the good work.

[@KipIngram]

I love this continued fraction way of "assessing irrationality." Sounds like a crazy question at first, but then when you look at it the right way it makes total sense that phi is "the most irrational."

[@saykindavid]

*Mathologer*, thank you so much for that introduction into *infinite fractions*. My answer to your puzzle is that *infinite fractions* need a _mathematically strict_ definition (just as *infinite sums* do). In the sense of definition you gave at 6:50 the fraction being considered converges to 1 and to 1 only. Again as one did with *infinite sums* he may define it as a limit of _sequence of numbers_ -- *partial fractions* (compare to *partial sums*). When you try to mess around with such objects only relying on , you may face (like with conditionally convergent series). However one may come up with another definition. When you calculate _N_ th partial fraction you may replace _diagonal dots_ with some particular number _d_ instead of zero. If you choose _d_ = 2 it converges to 2. For all other values of _d_ but 3, 7/3, ... when division by zero takes place (see comments below) it converges to 1. One may also notice that if such fraction has a value it _should be_ the root of x(3-x)=2. However such trick only works for recursive fraction expansions (e.g. not for pi). P.S. I'm looking forward to watching your video on Phyllotaxis phenomenon since I'm only familiar with Physicists' explanation of it. I would like to share it with you. Here are the lecture slides www.mit.edu/~levitov/FibonacciPhyllotaxis.pdf and here is the experimental proof arxiv.org/pdf/1002.0622.pdf.

[@Mathologer]

Great spot-on comment and thank you very much for the links. There was also a great feature article on the American math society site on all all this. Just remembered that I want one of those magnetic cacti !

[@kamlot0]

Small clarification: If you choose d = 2 it converges to 2. If you choose d = (2^(n+1)-1)/(2^n-1), that is { 3; 7/3; 15/7; 31/15 ... } it is division by 0. (If you define 2/0 = infinity and 2/infinity = 0, it still converges to 1) For all other d it converges to 1.

[@earthbind83]

+David Saykin Are you sure it diverges for d > 2? What about 1000? The resulting term of the first iteration is very close to 0, which leads to a term very close to 2/3, and for d < 2 it converges to 1, right? Also try d = 3 or d = 7/3 or d = 15 / 7 etc. :-) My favorite response to this video is the one from some guy who compared this proof to the following one: 1 = 0 + 1 substitue 1 by 0 + 1: 1 = 0 + 0 + 1 substitute endlessly: 1 = 0 + 0 + 0 +..... = 0

[@saykindavid]

+Kamil Musiał thank you. I wasn't careful enough. +earthbind83 I made a mistake. You're right. Your example shows how trick was done pretty well, I myself also find it similar to infinite sums.

[@saykindavid]

+TheExaltedPheonix that's how Mathologer calls them.

[@1paper1pen63]

In 1=2 riddle, it depends wether we end the continued fraction with 2 or 1. If continued fraction ends with 2 in its tail, it collapses to 2 while if continued fraction ends with 1 in its tail, it collapses. Since, the 3 dots doesn't tells us anything about term on tail, the continued fraction doesn't have a particular value. This is similar to infinite series 1-1+1-1+1... whose sum is either 1 or 0 depending on the last term and since this doesn't conclude that 0=1, our ICF paradox doesn't conclude 1=2.

[@nuklearboysymbiote]

Binomials expansion gives 1-1+1-1+… equals 0.5 lol!

[@1paper1pen63]

@@nuklearboysymbiote will you please elaborate? Expansion of which bi. expression?

[@nuklearboysymbiote]

@@1paper1pen63 (1 + x)^-1⠀⠀when x = 1 (which of course wouldn't be a "valid" expansion but it does work out to 0.5 if you just evaluate (1 + 1)^-1, but the expansion becomes 1-1+1-1+…)

[@gauravagarwal7508]

The partial fraction does converge to both 1 and 2 but it depends on the way one decides to evaluate the partial fraction. If we evaluate it by just considering the fraction before the negative sign then it will converge to 1, but if we were to evaluate it by removing everything below the denominator then the fraction will always be 2!

[@herrreinsch]

Oh man, your videos are so good, they must take forever to make, but I love them. I wached them all. keep up the good work.👍

[@SamraiCast]

herrreinsch why the fuck am I seeing you here lmao

[@SoopaPop]

The identity at the start of the video only holds for finite expansions of 1 and 2. If we do not end it with a 1 or 2, then we are essentially breaking the equal sign. On top of that, if you compute the sequence of partial fractions for these two sequences, you will find it converging to 1. For example: try 2/(3 - 2/(3 - 2/(3 -2/(3 - 2/3)))).

[@Maximilian_1]

Step 1: Turn into x = 2/(3-x) Step 2: Solve for x to get x = 1 or x = 2 Step 3: Cry because this simple method that you taught me doesn't work and I suck at maths

[@TanNguyen-sf5xi]

the funny thing is it actually is theright answer, i think :]

[@damienw4958]

Think about how you solved it (don't say calculator) the answer is in the method

[@Bordpie]

Both 1 and 2 work for that equation (and that infinite fraction above), but that doesn't mean they are equivalent. x=1 OR x=2. Not 1=x=2 => 1=2. I believe that is the problem answer.

[@christophem6373]

What you've done in step 1 and 2 is find the invariant point(s) of the sequence defined by u_{n+1} = 2/(3-u_{n})

[@kingbeauregard]

I think you nailed it: just because there are two solutions to the equation doesn't mean the two solutions are identical.

[@admiralbananas]

I think I've got it. Mathologer, I'd love to know if I am right or not. The interlude is really important because you explain what the "equals" means here. The infinite series has to converge to the number in question for us to label it as equals. If we start taking the partial sums by cutting off at the minus signs, we first get 2/3 (0.6667). The next term is 2/(3-2/3) which is 6/7 (0.8571) Third term is 2/(3-2/(3-2/3)), which is 14/15 (0.9333) The fourth term is really nasty to write out, but it gives us 30/31 (0.9677). The pattern emerges to show that each term can be determined by its preceding term. The numerator of any term n will be double the denominator of term n-1, and the denominator of term n will be 1 greater than that. While we see this series does converge to 1, we can prove it using calculus. This series is just a faster converging series of the form n/(n+1). The nth term test gives us infinity/infinity, but L'hopital's rule says we can take the derivative of the top and bottom, perform the nth term test again, and find out what the series converges to. The derivative of n is 1, and the derivative of n+1 is 1. The limit of 1/1 as n approaches infinity is 1, so the series n/(n+1) also converges to 1. We can use this information to also conclude that the infinite fraction also converges to 1. Thanks for leaving us this problem to work out. Great video!

[@DitDede]

as other people have noticed, x=2/(3-x) has 2 solutions: x=1,2. however, for the iterative process x_{n+1}=2/(3-x_n) to converge to either 2 or 1, we need to look at the derivative of the iteration. the derivative of 2/(3-x) is 2/(3-x)^2, which is 0.51 for x=2. That is why we approach 1 with the continued fractions (derivative less than 1). if we play with the equation a bit and rewrite it as x=3-2/x, then the iterations (not similar to continued fractions so much anymore) would actually converge to 2: 3, 7/3, 15/7, 31/15,... this is because the derivative of 3-2/x is 2/x^2, which is 2(>1) for x=1 and 0.5(

[@Anonymous-pm7qc]

This was hugely informative. Thank you so much for your videos Mathologer. I love you!

[@jeremygrodhaus3584]

You have to remember that in one series, 3-2=1 and in the other 3-1=2. They both are technically correct in and of themselves as they go on towards infinity, but the part that is being replaced is different even though they appear the same at a higher level. Also, keeping the above in mind, you can end either series at any point and so they are not an irrational infinite series, more of a faked one.

[@jeffleung2594]

The author just uses dot dot dot to replace the series, but the dot dot dot in each series are not the same. You explained it in words much better of course.

[@TonyBai]

For the 1=2 thing, you can try to solve for what 2/(3-2/(3-... equals by setting it equal to x. So you get x=2/(3-x), or x(3-x)=2, and you get the quadratic x^2-3x+2=0, and when you solve for x you get x=1 or x=2, so both answers are correct. They differ depending on what you interpret the dot dot dot as

[@IuliusPsicofactum]

So that's not an eyelash on my screen, it is you?

[@KaviAmanTenguriyaShaurya]

Here , we can write x=2÷(3-x) Now , quadratic equation form... x^2 - 3x + 2 = 0 (×-2)(×-1)=0 ×=1,2 So, x=2÷(3-×) satisfies for both .

[@Hwd405]

Okay, I'm gonna try and answer this now. Here's the problem - the way we represent continued fractions and the numbers used obscure the problem a bit. The thing is, infinite fractions don't automatically have any sort of meaning, we need to define what they mean. Otherwise we can just throw whatever numbers we want in and have no idea whether it actually represents a real number or not. That's why we use convergence as our definition - in the fraction given the sequence we get is 2/3, 2/(3-(2/3)), 2/(3-(2/(3-(2/3)))) and so on, the nth term of the sequence evaluating to (2^(n+1)-2)/(2^(n+1)-1), which tends to 1 as n tends to ∞. Notice, however, that we took only the integer part of each denominator in our partial terms - why? It's purely definition. We take the integer parts as they're what defines a continued fraction, at least in the case that the numerator is only allowed to be 1, so we continue with this definition when extending to more general fractions. Furthermore, the (cont)

[@Hwd405]

fractional part should be overwhelmed by the integer part - if instead of just including the integer part, we include the numerator of the fractional part too, the terms we get aren't really representative of the actual continued fraction. In an infinite continued fraction with numerators set to 1, this shouldn't really matter, but when we're working with fractions which don't have this regular behaviour and you start treating what should be a fraction as an integer when looking at convergence? That doesn't really make sense. However, if we take this approach when looking at the given fraction we end up with the sequence 2/(3-1), 2/(3-(2/(3-1))), 2/(3-(2/(3-(2/(3-1))))) and so on, which evaluates to an sequence of 2's. This is pretty obvious as in this case the numerator of the fraction in question actually is equal to the fraction as a whole! (That is, 2=2/(3-1)). I think that for this reason the definition of what counts as a continued fraction is somewhat obscured, but it (cont)

[@Hwd405]

+Hwd405 could explain why two different values emerge. It's all about definition and which way we approach a continued fraction, but usually we take the former approach and the result is convergence to 1. I'm not sure either result is "more correct" than the other though, but it's important to choose a mode of convergence and stick with it so that we don't end up with concepts that aren't well defined.

[@olivermaclean8564]

I see what you're saying, of course you can define things however you like, although you don't have to add any definitions. Sometimes it's alright for things to equal two different values (example bellow) without it breaking maths. Both values of the continued fraction can be found without looking at convergence of partial sums (but I go into that more in my main reply to the video) y = f(x) = √x for x = 4 y = 2 or -2 f(2) = 2 f(2) = -2 f(2) = f(2) -2 = 2 Hopefully thats an illustration that the problem here isn't that the continued fraction doesn't equal 1 and it isn't that the continued fraction doesn't equal 2. The problem is in the fallacy that because it can equal 1 and 2, means that 1 equals 2.

[@Hwd405]

+oliver maclean that's not a valid example, a function can only have one output given any input. f(2) must take the value of 2 OR -2, not both. Indeed, we usually define the square root symbol to mean the principal root (the positive one). Furthermore, the repeated fraction notation isn't a function - it's notation. If it's consistent well defined notation it shouldn't be equal to two different numbers, otherwise that breaks the transitivity of the equals relation.

[@Hwd405]

+Hwd405 sorry, I meant to say f(4), sorry for any confusion

[@ffggddss]

@Mathologer: Very neat! Continued fractions are also an interest of mine, and here's another twist on "most irrational." What you've shown us is the traditional way to do CF's with unit numerators. But notice that when a "1" occurs in the string of denominators, this is a signal that the CF up to (just before) that point, is pretty crummy. (This can be quantified by defining a figure-of-merit (FOM), that compares the fractional error |x - (n/d)| / x with the "size" of both num. and denom. - actually, with the reciprocal of the product of the two, 1/[nd] ). The rationale here is that the product, nd, is the "outlay" you make to get the fractional precision you obtain. FOM = (1/[nd]) / (|x - (n/d)| / x) = x / (n |dx - n|) Those "crummy" approximations can be eliminated by allowing negative denominators, d, but restricting them to |d| ≥ 2, with the additional proviso that when d = ±2, the next d must have the same sign. On a calculator, e.g., these denominators are found iteratively: • subtract the *nearest* integer to the current result [that integer is the current denominator] • then invert that difference to get the next result This technique gives, for π, {3; 7, 16, -294, 3, -4, 5, ...} rather than {3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, ...} - these are given to the same rational value: 5419351/1725033; integer + 6 fraction coefficients vs integer + 11 fraction coefficients, for the same precision. For e (allowing a little exception to the rules for the first couple terms for the sake of mathematical beauty), it gives: {1; 1, -2, -3, 2, 5, -2, -7, 2, 9, -2, -11, ...} rather than {2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, ...} - these, too, are shown out to the same fraction: 566827/208524 - The sequence without the exception is {3, -4, 2, 5, -2, -7, 2, 9, -2, -11, ...} integer + 9 fraction coefficients vs integer + 15 fraction coefficients, for the same precision. For φ, it gives: {2, -3, 3, ...}, with [-3, 3] being the infinitely repeating part. This makes φ seem not so very irrational. But for √2, it gives (as does the traditional method): {1; 2, 2, 2, 2, ...}, and now √2 takes over the title of "most irrational," because under the new rules, this is the "crummiest" possible convergent sequence. This can be verified using the FOM, which actually just gives a value very close to the (abs value of the) next result in the iterative procedure described above ("On a calculator, ..."). Any thoughts? Addendum: Under the "new rules," the FOM for φ is φ+1 = φ² = 2.618... And for √2, the FOM is 1 + √2 = 2.414... So the contest for "most irrational" is actually pretty close! Also worth noting, is that the FOM for a general irrational number, bounces all over the place as you proceed down the CF. Heck, just look at π, e.g.!

[@destiny2135]

I am unaware if you are still looking at the comments, but with the 1=2 madness in the beginning, we can use the partial sums like you say in the interlude. If we do this, we get 2/3, 6/7, 14/15, 30/31, 62/63, so on so forth. This seems to converge to 1, so I believe the final answer is 1

[@powerdriller4124]

It is a recurrent infinite sequence of the form 2/(3 - X), where X

[@Zartymil]

I think that the problem occurs when you transform the fractions into infinite fractions. You took a really important part away. The fraction with 1 always had a 1 to be substituted again. And the fraction of 2 always had a 2 to be substituted. So the fraction of 1 has a 1 in the end (that is hidden by the elipses) and the fraction of 2 has a 2 in the end (which is also hidden by the ellipses). And with that in mind we can say that they aren't really equal.

[@yuvalsela4482]

there isnt "at the end"

[@anthonypergrossi8454]

Well, right. The issue is that the "1" version needs a 1 at the end to maintain the identity. Filling it in with the ellipsis actually breaks the identity by changing it to a sequence of partial fractions which converges to 2, not 1.

[@matekon2]

The problem is that the notation of the first example with three dots isn't even rigorously defined at all. We should do: Lets define f(n)=2/(3-n). {a(n)} and {b(n)} are 2 recurrent sequences defined by a(n+1)=h(a(n)) and b(n+1)=h(b(n)), but a(0)=1 and b(0)=2, which means the 2 sequences aren't the same one. That implies that their limits aren't necessarily the same. An expression of the form in the video doesn't give us an clue of the original term/seed of the sequence in question and isn't well defined because of that. The notation works for the rest of the video because the sum (and so the limit of the sequence) is actually done in the other way: from left to right. So, it actually has a very well defined first term.

[@jacquespoulemer3577]

Burkard Polster. I studied Philosophy and Math at University in the 70's and look at at least one math video every day. I just recently discovered your channel and have subscribed and started to put thumbs up on your shows (although I suspect this is the reverse of what the ancient Romans meant by it) 😁 I'll be looking in on you soon JIM

[@ImaginaryHuman072889]

not sure if i'd agree that phi is "more irrational" than pi, although I guess it depends on the context of what you're asking. phi after all is algebraic and pi is transcendental so i would have thought that pi would have been "more irrational" in that sense.

[@nightshadyify]

imagine if you will two line graphs with a gradient of the natural number you want to test y = phi*x which can be compared to another, your example includes pi so y = pi*x the closer either line graph is to the cardinal points, the "more" rational it is. This happens when their convergent fractions come very close to approximating their value. ie in each convergent fractions second iteration ( pi ~ 22/7 as opposed to phi ~ 3/2 )

[@dbodow]

Absolutely correct. More formally, the Liouville-Roth irrationality measure of the golden ratio is 2, while that of any Liouville number would be infinite.

[@dbodow]

en.m.wikipedia.org/wiki/Liouville_number#Irrationality_measure

[@biggerthaninfinity7604]

@@dbodow Thank you!!

[@biteofbytes]

If we abstract the infinite fraction: x=2/(3-x); -x²+3x-2=0; Using the cuadratic formula it has two results: x=2 and x=1 (sorry for my bad english)

[@TrimutiusToo]

Well when you do this infinite fraction you wound up with square equation ( for example x = 2 / (3-x) ) which of course will have 2 solutions most of the time. But if you do substraction both of them might be positive and this fraction doesn't really converge... Even x = 1 + 1 / x will actually suffice for both positive ф and and negative golden ratio (-0.6180339887498948482), and just because we are searching for positive number we can exclude negative solution

[@MattMcIrvin]

My take on the puzzle: If you consider the function f(x) = 2/(3-x), it has two fixed points where f(x) = x: x=1 and x=2. However, x=1 is an attracting fixed point of the function, and x=2 is a repelling fixed point. That is, if you take some value close to x=1 and then iterate the function, calculating f(x), f(f(x)), f(f(f(x))) and so on, the values will get closer to 1; but if you start with some value close to 2, the values will run away from 2. This is because the function's slope is shallower than 1 at x=1, but greater than 1 at x=2. I am thinking this is a thing to be wary of when working with continued fractions.

[@MattMcIrvin]

...and the Mathologer's "interlude" about partial fractions about halfway through really gives the key. An infinite fraction expansion only really makes sense if the sequence of partial fractions converges.

[@roberttheiss6377]

no matter what value you pick the iterations converge to 1.

[@KnakuanaRka]

Similar to what 3Blue1Brown said in his “Things They Won’t Teach You in Calculus” video (kzbin.info/www/bejne/eZe6aWdrgbOCmK8 ) about why the proper solution for 1+1/1+1/1+1/1+1/.... is phi, rather than its negative counterpart, even if that also satisfies x=1+1/x.

[@xj-vn4eo]

Interesting perspective by modeling it with a dynamical system.

[@aldobernaltvbernal8745]

@@roberttheiss6377 unless you pick 2

[@PC_Simo]

11:40 I used to think you could 1-up φ, by choosing 0’s in all the denominators (because there’s still the infinite ”+(1/[stuff])” -tail, also, in each denominator; it’s not division by 0). You *_CAN_* do that. However; it turns out to be anything but irrational: 0+(1/(0+(1/(0+(1/(0+(1/(0+…)))))))) = 1/(1/(1/(1/(1/(1/(1/…)))))); which, actually, collapses to just 1 (because it’s just: ”1 over 1 over 1 …”, over and over again, it stays, at 1, through all the infinitely many truncations, we get out of it. So, trying to get too greedy, results in the complete opposite of an irrational number: An Integer; and a basic one, at that: *1.* 😅

[@bonbonpony]

02:54 It's quite interesting to extract out the repeating part to see what number it is alone. In this case, it is `√2 + 1 = [2; 2, 2, 2, ...]`. Its inverse is `√2 - 1 = [0; 2, 2, 2, ...]`, so when you add 1, you get the `√2` back: `√2 = [1; 2, 2, 2, 2, ...]` :) Similarly for `√3 = [1; 1, 2, 1, 2, 1, 2, ...]` we can take the repeating part out and it turns out to be `(√3 + 1)/2 = [1; 2, 1, 2, ...]`, which is the inverse of `√3 - 1 = [0; 1, 2, 1, 2, ...]`. In general, numbers and their inverses have the same continued fractions, just shifted by one position. Another interesting observation I made was when I tried to represent tangents of different angles as continued fractions one day (I used them to find closest fractional approximations so that I could easily draw those angles without a protractor ;> ). I noticed then that tangents of many angles have the same representations in continued fractions, just shifted, as if they were inverses of each other. And it turned out that the angles for which this is true were complementary (they add up to 90°)! The explanation for that mystery was simple: when one subtracts an angle from 90°, it's like rotating it by 90°, and the rectangular coordinates switch places, so the tangent ratio flips upside down, turning into its inverse, that is, a cotangent :) But there's also an interesting "problem": if I remove the integer part but I don't flip the fraction into its inverse yet, and calculate its arctangent, the resulting angle is no longer rational and it doesn't seem to have anything in common with the family of angles I studied :P 04:14 So let's extract the "pattern" from this number too and let's see what number it represents: `(e-2)/(3-e) = [2; 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, ...] = 2.549646778...` hmm.... Is this number in any way special or useful for something? To me, it looks like a ratio of the distances of `e` from its nearest integers, 2 and 3. If that is the case, it could be interesting to do this with `π` too :>

[@ritamchakraborty40]

i am following this channel for quite sometime... it is really fun to dig into this mystries of number...

[@Mathologer]

Glad you like what I am doing here and thank you for saying so :)

[@ritamchakraborty40]

+Mathologer sir, i would also like to know the prime number theorem of ramanujan of what proof he did... also the partition theorem .. please post a video on these topic sir... eagerly waiting..

[@Mathologer]

Well, there should be some more of Ramanujan's math in a video in the next couple of month, stay tuned :)

[@ritamchakraborty40]

thank you sir... there was no direct proof of his work in the internet.. so, number theory lovers are finding it hard to know about this great man.. please do it sir.. thank you

[@fishandchips8813]

You do such great work here!Have you ever considered doing something about Turing's halting problem and the associated Busy Beaver numbers?Just a thought- I eagerly await you next video, whatever the topic may be.Thanks again!

[@alexsantee]

Well, the problem is that you can't really keep the equal sign when you do that with two because the sum doesn't converge to 2, but to 1 [2/3;6/7;14/15;30/31;...]. If you call the infinite fraction x, you can get this equation because of the pattern: x=2/(3-x) and the equation roots are 1 and 2. But, even though 2 solves the equation, it breaks the condition of the partial fractions converging to the number, so it have to be discarded. The necessity of discarding one of the possible solutions becomes more obvious if you try to solve the square root of two equation, because you'll get sqrt(2) and -sqrt(2) and the negative answer is clearly false because you are summing only positive numbers.

[@alexsantee]

Here is how I solved the sqrt(2) equation: x = 1+1/2+1/2+1/2+... x-1 = 1/2+1/2+1/2+... x-1 = 1/2+(x-1) x-1 = 1/x+1 x^2 - 1 = 1 x^2 = 2 x = +-sqrt(2)

[@Colony28]

I love this video! So my approach was as follows: Besides the fact that 1 and 2 are both natural, hence not rational, which means that writing them like that seems a bit cumbersome, let's take a look at convergence of the partial fractions of the expansion. We can write the sequence like so: a_1=2/3 a_n=2/(3-a_{n-1}) From a_n follows, that if a_{n-1} < 1, then also a_n < 1. Also, for 0 < a_{n-1} < 1 is a_n > a_{n-1}. I couldn't figure out how to write it as a limit, even though from the first few members 2/3, 6/7, 14/15, 30/31, 62/63, 126/127 it is obvious that it converges to 1. Of course, the expansion cannot converge to two different numbers, so the result is that 1 = expansion which is not = 2. Cheers!

[@Mathologer]

Good answer :)

[@bonbonpony]

+Mathologer I have some questions, though: 1. As you said, irrational numbers which are algebraic, still have that repeating pattern in their continuous fraction expansions, while transcendental numbers don't have any repeating pattern. My first question is then: How to detect if there is a pattern without the need of collecting all the integer parts and checking if they start to repeat? (Because that pattern might sometimes be long and starts to repeat after maaaaany terms :P) 2. Every number can be converted to decimal expansion, and every decimal expansion can be then converted to continuous fraction form. But what if I _don't_ have the decimal form? Or if calculating it would take too much time? Let's say I have an *infinite series* representation of some number. I could calculate its decimal form, but it turns out it converges very slowly :P so it takes too much time until I get any meaningful digits. But even then, this is only an approximation, because I end it sooner than infinity :P so my calculation of the continuous fraction form might be inaccurate as well :P So my next question is: Is there any way to skip the decimal expansion step altogether and go straight to the continuous fraction from an infinite series representation or some other algebraic representation? (There's nothing special in the decimal representation anyway :P so we shouldn't depend on it when calculating continuous fractions for numbers, right?) 3. When I cut the continuous fraction at some point and reduce it to a simple fraction, I need to "denest" it, starting from the point I cut it at, which is the most nested one, and go outwards. But this means that if I want to calculate the next closest fractional approximation, I need to start the entire process again, because the starting number changes :P I cannot easily reuse the fraction I calculated earlier. Or can I? And this is the third question: Is there any way to reuse the fraction I calculated so far, to get the next one, without the need of recalculating the entire series from scratch? :P

[@ffggddss]

NB: The term is "continued" not "continuous" fractions. NB2: Mathologer may need to step in here, to amend/extend/correct my remarks; but here's what I think I know. 1. Actually, it's only algebraic numbers of 2nd degree (solutions of integer quadratic equations that aren't solutions of any lower-degree integer polynomial) that have infinitely repeating CF's. All higher-degree algebraic numbers have non-infinitely-repeating CF's. And if you know the coefficients of the quadratic for your solution, you can find the CF terms from those. Not sure of the method, but it can be done. E.g., for the simple special case of the form x² - n = 0, where n is a non-square > 0, the CFE is found by a process for solving Pell's Equation, and the number of terms in the repeating part varies wildly as n steps up: 2, 3, 5, 6, 7, 8, 10, 11, ... 2. Not an expert in this, but I know enough to know that there's no general technique that will convert any convergent series into a CF. Pretty much every case is special. I'm guessing that it's a lot like solving ODE's; there are some techniques that cover certain subclasses of the problem, but no general technique for all of them. 3. From a truncated CF, you can work it from the far end by de-nesting; or you can work it from the front end by linear recursion. For example: [2; 2, 3, 2, 5, ...] can be worked out from the front end by: i: -2 -1 0 1 2 3 4 r: 2; 2, 3, 2, 5 p: 0 1 2 5 17 39 212 q: 1 0 1 2 7 16 87 where the i = -2 and i = -1 terms are always as shown here, and the recursion rule is: (pᵢ, qᵢ) = rᵢ(pᵢ₋₁, qᵢ₋₁) + (pᵢ₋₂,qᵢ₋₂) And at each stage, your fraction is p/q.

[@davidpaganin3361]

Many thanks! I learned a great deal from this, and enjoyed it immensely!

[@Mathologer]

That's great :)

[@mash8742]

X=2/(3-X) 3X-X^2=2 (x^2)-3X+2=0 Factorise for: (X-2)(X-1)=0 Therefore X=2,1 That was a fun one! Hidden polynomials are great puzzlers ;)

[@nicreven]

ohhh, so it's not that 1=2, it's that both of them are the correct answer?

[@PC_Simo]

My answer was exactly the same (just, via the quadratic formula; rather, than factorization). 🎯👍🏻

[@PC_Simo]

@@nicreven Exactly 🎯!

[@brighamlucero5543]

if 2=(2/(3-2)), then why don't you substitute the right side of the equation to both numerator and denominator?

[@edgarmatias]

This is SUCH a great video. We really should teach infinite fractions better. This video does that incredibly well. Great work!

[@miles6875]

Great video. It was after watching this video that I decided to buy a couple of your books. Love what you put out

[@kresnadevaihza5577]

glad you upload it dude. appreciate it

[@Mathologer]

:)

[@niboe1312]

The infinite fraction 2/(3-2/(3-2/... is divergent. The sequence of partial sums alternates between numbers approaching 2 and numbers approaching 1.

[@NotBroihon]

Actually that's not the case. 2/3 = 2/3 2/(3-2/3) = 6/7 2/(3-2/(3-2/3)) = 2/(3-6/7) = 14/15 2/(3-2/(3-2/(3-2/3))) = 2/(3-14/15) = 30/31 62/63...124/125...etc

[@TheMusicJan]

+Not Broihon I think that if you use the partial fractions for the &quot;=1&quot; part you have to end with 1 and for the other one you have to end with 2 I will give try when I arrive at home

[@BigDBrian]

2/3 = 2/3 2/(3-2) = 2 2/(3-2/3) = 6/7 2/(3-2/(3-2))= 2 etc Depending on how you take the partial sums, it&#39;ll either converge to 1 or always be 2. EDIT, I believe the way to take the partial sums that lead to 2 are false.

[@niboe1312]

+Not Brohion I did the math again and you're correct. Don't know how I messed that up.

[@mrfrech2191]

There are no partial sums but fractions ;). He also mentions that you chop the seq. before the further addition for partial fractions, which only makes sense as you can´t add the -2 without aknowledging that it functions as a numerator thus cant be added without adding the denominator, as mult. has precedence over add.

[@MiahooJunk]

Because 1 and 2 are rationals, so the diversion is finite..? - which means we can choose if to 2 or 1 at the end of it. Is that the answer?

[@barutjeh]

This argument only holds if the continued fraction only has 1's as numerators.

[@MiahooJunk]

I miss the logic in it. divide or multiply to change it around.

[@simon24h]

Okay, (x-2)(x-0.5) = 0 -> x = 1/(2.5 - x) is true for x element of {2,1/2}

[@ffggddss]

Another interesting aspect of CF's is their use in finding solutions to Pell's Equation (it's a Diophantine problem: given the integer n, find integers p,q): p² = nq² + 1 for any non-square n > 0. I sometimes call this equation the "near-squares" problem, because the fact that such an n can't have a rational square root, is equivalent to showing that no (non-0) perfect square is n times another; while a solution to Pell's Equation is equivalent to finding a perfect square as close as possible to n times another. So this equation is asking, "How close can you get? Can n times a square be just one unit away from some other square?" And a Pell's Equation solution answers that on the "one unit less" side, which turns out to be soluble for every non-square n. (There are solutions on the "one unit more" side for infinitely many n's; but no such solution for infinitely many others).

[@PC_Simo]

4:15 Well, the beginning part of the decimal expansion of e is actually quite beautiful, featuring a repeated string of 4 digits, and a few multiples of 9 that actually form the angles of a ”unit right-angled triangle” / ”unit half-square”, where the short sides are of length: 1, and the hypotenuse is of length: √2: 45°, 90°, 45°: 2,718281828459045…

[@yeremiafrans9425]

"22 over 7" BOOM dead giveaway

[@paulchapman8023]

Can’t the square root of 5 be approximated by finding two perfect squares whose ratio is approximately 5 (e.g. 81 and 16)? Based on that, isn’t 9/4 a fairly good approximation of sqrt(5), and thus 13/8 a good approximation of phi?

[@ffggddss]

You can do that trick with any √n, where n > 0 is a non-square. It's a Diophantine problem (solving algebraic equations in integers) called Pell's Equation: b² = na² + 1 Every +ve non-square n has a "primitive" solution (a₁, b₁), from which an infinite number of larger solutions can be generated by the recursion a₊ = b₁a + a₁b b₊ = na₁a + b₁b For n = 5, (a₁, b₁) = (4, 9); the one you quote. The next couple of solutions are (72, 161), (1292, 2889). Of some interest, you can get more "best fractions" for n=5 because it's one of the n's that also has a solution to the "Associated Pell's Equation": b² = na² - 1 namely, (1, 2), which can be used to generate more "best fractions," every other one being in the above list, so that they alternate between satisfying P.E. and the A.P.E. : (1, 2), (4, 9), (17, 38), (72, 161), (305, 682), (1292, 2889), ... For general n, the primitive solution, (a₁, b₁) jumps around erratically. Some are easy to find by trial-and-error; some are next to impossible that way. n = 13 is somewhat challenging; n = 61 is very hard. Luckily, there's a procedure involving - wait for it - Continued Fractions! - that will always yield the first solution. Fred

[@_DD_15]

It would be interesting to see continued partial fractions of complex numbers. Awesome video btw, too bad our brain is used to simple summation, and not fractional one. I have read about a guy who could do them as easily as we normal people do ordinary summation. Incredible!

[@bonbonpony]

04:48 There's a nice connection of this with the Euclidean algorithm for calculating the greatest common divisor ;> Let's start with 1473. How many times we can subtract 50 from it? After subtracting it *29* times we end up with 23 and cannot subtract any more. And guess what: *29* is the integer part in the continuous fraction representation ;> Then we "flip it around" and try to subtract that remainder of 23 from 50. How many times it could be done? Only *2*, and we end up with 4. And surprise, surprise: *2* is the next integer part in the continuous fraction :> Then we flip it around again and test how many times 4 fits into 23. As it turns out, it fits *5* times, leaving a remainder of 3. And guess what: *5* is the next integer part in the continuous fraction ;> Flip around again: how many times we can subtract 3 from 4? Only *once*. And *1* is in fact the next integer part in the continuous fraction. The remainder is 1. Flip it around. How many times we can subtract it from 3? *3* times, of course. And this is the last integral part in the continuous fraction :> The remainder is 0, so we stop there. (And since the last remainder was 1, we also know that 1473 ⊥ 50, that is, they're coprime.) Neat, eh? :) This is also the reason behind what you've said in 10:24: the Euclidean algorithm works the slowest for Fibonacci numbers, because there is too much back-and-forth with subtracting one from another (we need to flip it after every single subtraction :P ). So if a number is approximated by partial fractions which are based on Fibonacci numbers, the continuous fraction will converge the slowest as well. And that happens for the golden number.

[@ffggddss]

Yes! That connection is completely true.

[@Khwartz]

Thanks for Both your Clarity and Rigour! The Best Pearl imho in this video is at 7:12 when You talk about THE NEED TO DEFINE WHAT WE MEAN BY "=" :)

[@rkcst6503]

I think the problem is that 1 is the loneliest number, and since 2 can OBVIOUSLY be as bad as 1 (because its the loneliest number since number 1), there is no way to say they are completely equal. Not to mention the that fact that 1 + 2 = buckle my shoe, and shoes don't have buckles anymore, so it's clear that something weird is going on. Either way, it's been 1 week since you looked at me, and even though I TOLD YOU I had 2 tickets to paradise, you INSIST that I start back at 1, just to make your dreams come true.

[@kurchak]

hahahaha omg best response ever.

[@mikenewman7375]

lol I thought it was because they were both rational numbers that couldn't be applied to an infinite fraction, but after reading your post I realize that you were right all along. It just makes so much sense!

[@thetofinator8605]

Can you support this logic with a few more son.. I mean, uh... historical mathematical... pieces?

[@rkcst6503]

You should have listened to your heart.

[@mikenewman7375]

Well, you are right, the heart will go on. I would say it's a shot to the heart, and that you're just a bit too late, but I would just be left with an akey-breaky heart, which is never a good thing. But if you could just look into your heart shaped box and see that a total eclipse of the heart is bound to happen when you check in at the heartbreak hotel, then you would see that you're terrin up my heart when I'm with you, and that sgt pepper and his lonely band just want to have fun.

[@TheMaxtimax]

Hey Mathologer ! Nice video, as usual. I was wondering if there's a "true definition" (conventional definition, rather) of "more irrational" ? If not, don't you think the definition we invent should be such that phi is less irrational than pi, since the latter is a transcendental number, whereas phi is one of the "most algebraic" numbers there are, so in this regard, phi is "more rational", than pi ?

[@urthoperator3126]

Pi is both trascendental and irrational; if it's trascendental it needs to be irrational, but it doesn't mean the trascendentality makes it more irrational (trascendentality only speaks of our inability to express that number algebraically, this of course means we can't express it as a finite fraction, but it says nothing about how badly we can approximate it). If we extrapolate the definition of irrationality to be different depending on which irrational number you chose, you'd have the definition seen in the video: an irrational number m is more irrational than n if it takes more digits in the denominator to approximate the same number of digits in m as in n. This makes Φ the most irrational number (over any other irrational *and* trascendental number).

[@urthoperator3126]

Also, here's an excerpt from the description: "If you are reasonably clued up mathwise have a look at the following VERY nice textbook chapter on infinite fractions by Professor Paul Loya from Binghampton University: [LINK] In particular, check out section 7.5.1. The mystery of π and good and best approximations. *I use the definition of "best rational approximation" given there.*"

[@ryan2-518]

7:33 "well that's only justified if we pin down what we mean for it to be equal at that point." Can someone explain to me what this means

[@soko-ban]

it's the argument wether a converging infinite Series/Sequence etc is really equal to the Number it converges to, basically if you can really treat that infinite series as a number itself.

[@ryan2-518]

recryed thank you

[@soko-ban]

@@ryan2-518 sure mate

[@jeffleung2594]

Most appreciate the author's puzzle 1=2 at the beginning to let us put our minds to work

[@fawzibriedj4441]

I can answer the puzzle using sequences : Un+1 = 2/(3 - Un) We have two stable points : 1 and 2 (where Un+1 = Un) When we try to calculate the infinite fraction, we always suppose that U0 = 0 for example : U1 = 2/3 = 0.666... U2 = 2/(3-2/3) = 0.85... U3 = 2/(3-2/(3-2/3)) = 0.9333... In all these cases, we assume that U0 = 0 and we calculate U1, U2 and U3 So, we can deduce two things : The infinite fraction does not converge (in the sense that the limit of (Un) depend on U0, so it is not unique) The infinite fraction seem to converge to 1 because when we calculate the "partial fractions" we assume that U0 = 0, so it converges to the closest attractive stable point.

[@firefly618]

The answer to the puzzle lies in the jump from finite to infinite. You could replace that fraction 10^10^10^10^10 times (just to mention a staggering number) and it would still equal 1 or 2, depending on the innermost digit. But as soon as you jump to an infinite sequence, it becomes an irrational number, which cannot be either 1 or 2. This reminds me of geometrical paradoxes. Take a square of side x and draw a stepped diagonal through it, like a staircase from one corner to its opposite. No matter how many steps-and how small-you draw, its length will always be 2x. Now jump to infinite steps, each infinitely small, and you get the usual diagonal, whose length is, unsurprisingly, an irrational number: x√2

[@firefly618]

Because that's what irrational (computable) numbers are: they are "placeholders" for what you would get it you could run a given algorithm forever. π, e, √2 are all such objects. In this sense irrational numbers don't really exist in the real world, because you can't run computations forever. They also have intrinsic fallacies. For example you can't always compare two algorithms for equality or ordering (it's undecidable in the general case) so it follows that equality and ordering aren't always defined for irrational numbers as well. But wait, there's more. According to current orthodox math, 100% of "real" numbers are not only irrational, they aren't even computable. So you can't even give them a name or identify them, except in a few rare cases (such as Ω, defined over the halting problem, which is uncomputable.) Conclusion: don't believe what math books tell you about numbers and only use fractions!

[@Mathologer]

What a strange conclusion. Maybe have another look at the interlude kzbin.info/www/bejne/eZLEpJWcma-XgMkm11s

[@firefly618]

Mathologer Yes, the conclusion was a bit of a hyperbole. But I'm not the only one to thing the current view of real numbers is nonsense. See for example Prof. N. J. Wildberger here: njwildberger.com/2016/01/01/uncomputable-decimals-and-measure-theory-is-it-nonsense/ «This is modern pure mathematics going beyond parody. Future generations are going to shake their heads in disbelief that we happily swallowed this kind of thing without even a trace of resistance, or at least disbelief.»

[@Mathologer]

Very smart guy and definitely a real mathematician. However, I think it is important to stress that he is really only speaking for himself when he criticises the foundations of mathematics and in particular the real numbers. He is a professor of mathematics and so am I and neither I nor any other professional mathematician I know agrees with him in this respect. Having said that, I find a lot of what he has to say very interesting and thought provoking. As well, his videos are a great resource for learning a lot of great mathematics. So, despite of me disagreeing with him in many way I consider him to be one of the "good guys" :)

[@firefly618]

Fair enough. In any case I was wrong: the "infinite" fraction 2/(3 - 2/(3 - ...)) is nothing but a fancy way to write the equation x = 2/(3 - x) which has exactly two solutions, 1 and 2. Nothing irrational here.

[@robinbfh5893]

Assume that the infinite fraction converges. Then there exists a limit a. Then a = 2/(3-a). Solve for a to get a1=1,a2=2. This contradicts that the limit is unique and therefore the infinite fraction diverges.

[@aldobernaltvbernal8745]

wrong

[@DanielHernandez-hg5ey]

That shirt is amazing.

[@en0x86]

I am no math guy and i cannot articulate exactly what the issue is but i have a couple of observations that i would love to hear some comments on to help me understand. 1. The each step in the denominator brings you closer to the approximation in the phi and pi case. but those numbers had additions in the denominator. 1=2/(3-1) has a subtraction and it is under 2. I get the feeling this denominator does little in getting us closer to the actual value in much the same way that phi's denominator did little in zeroing in on the actual value. 2. all rational numbers must have an end to this chain. clearly the end is very important in this case. perhaps this in combination with my first observation is the nature of the problem I would love to hear some feedback on this! Great video!

[@brocpage4204]

I'm not familiar with infinite sums but i still enjoy watching your videos! For nonmathologers, you make them (relatively) easy to follow. Some things you go over pretty quickly but I assume that's based on a presumption of knowledge by your viewers. So some stuff I just don't understand without the formal education. but you do a well enough job with most of your videos that I can follow the logic you imply! edit: grammar

[@Mathologer]

Cool, just keep watching this and other maths channels and in no time you'll be able to understand everything :)

[@taimao2]

one and two can written as rational numbers; therefore the fractional expansions are finite

[@SKyrim190]

But he is not doing continued fractions in the form of a_0 + 1/(a_1 + 1/(a_2 + ...)

[@Johann757]

The problem with the identity is that the second part of the identity is false. 2 is not equal to 2/(3-2/(3-... If we look at the partial fractures, 2/3, 2/(3+2/3), 2/(3+2/(3+2/3), ... , we see them converge to 1. The whole fracture only becomes 2 if you put a 2 at the end. In the infinite fracture, there is no 2 at the end, because there is no end, so the infinite fracture converges to 1. So the second identity claimed is false.

[@Mathologer]

Good answer :)

[@westmech3846]

Mathologer, once again great video. But can you explain how you get from "partial fractures" to partial sums?

[@DimitriosKalemis]

Hi, Mathologer. Why do you say: "Good answer"? Yes, it is a very good answer, but I think that you say this because this answer, as thoughtful as it is, does not reveal the real answer. And you want the secret, the real answer, kept hidden and the discussion to keep going. So you say "Good answer" because this answer hints at the answer and does not really reveal it. And you want that. And I do not like that you want that (if, of course, I am correct). I guess that others that talked about attractors, basins, attracting and repelling fixed points, recursive calculations, nonlinear recursion x(n+1) = 2/(3-x(n)), f(f(f(x))), etc. really have the correct answer. The correct answer as to why Mathematicians choose to equate this to 1, instead of: a) 2 b) either 1 or 2 c) not convergent, because it alternates between 1 and 2 d) something else, like the average of 1 and 2. So, am I correct or have I misjudged you?

[@prateekbajpai1978]

Thanks for the video! One nitpick - Everytime I heard "square of 2" instead of "square root of 2", a little piece of me died inside. :(

[@PC_Simo]

Same here 😢.

[@3Max]

In the description of "how irrational is it" around 9:20, you mentioned the number of digits in the denominator gives you a hint at "how close should you expect to be". But at 10:44, you compare fractions based on the index in the sequence of expansion of the infinite fraction, rather than based on the number of digits in the denominator. I wonder if these two approaches yield different "levels of irrationality" in the gray zone. It's also interesting that an algebraic number is "more irrational" than a transcendental number!

[@faustobarbuto]

I'm amazed that such an educational and well-elaborated video got only 15K thumbs-up in almost eight years. That makes less than 2K thumbs-up per year.

[@beirirangu]

Well, if we replace the "solutions" with x, we can find _a_ reason using algebra: x=2/(3-x) → 3x-x^2=2 → -x^2+3x-2=0 → x^2-3x+2=0 → (x-2)(x-1)=0

[@natecat33]

Very clever

[@無料駐車場]

amazing! x is 1 OR 2 so 1=2 is wrong

[@beirirangu]

Exactly! It even "works" the same if you change the numbers: (x-3)(x-4) → x^2-7x+12 → x=12/(7-x) ∴ 3=4 lol

[@Evan490BC]

Well, no... x = 1 or x = 2, in the context of the above equation. Period. No other implication.

[@beirirangu]

Evan490BC but if you noticed anything past "numbers", you'd see that you can make any version of this "problem" with any numbers you want, like I did where I changed it from 1 and 2 to 3 and 4, making it ?=12/(7-?)

[@massimilianotron7880]

Maybe the problem is that you created an infinit fraction for a rational number, and that only works with irrationals?

[@jesusnthedaisychain]

That's what I'm thinking.

[@bryanarnold4288]

Using the simple form: 1's in the numerator and addition of positives in the denominator --ur assertion is true. That is, a simple continued fraction terminates if and only if it is an expansion of a rational number. However the form here is not simple. And so has nothing to do with that fact.

[@TieJote]

That seems legit to me. Also, choosing 2 over (3-1) seems somewhat cheaty to me. Like, you could possibly do this with any number and get an identity. Like in the video, 3 = 3 over (4-3) having the 3 in the denominator again, thus also the infinite fraction.

[@massimilianotron7880]

aa Maybe if you cared to explain why it is dead wrong instead of simply complaining, then that wouldn't be a problem.

[@jesusnthedaisychain]

aa Well, that was sad. Just go around like Homer and proclaim, "I am so smart! I am so smart! S-M-R-T!"

[@paaaaaaaaq]

You can't ... it since there is - not a +. This means you don't get smaller more meaningless numbers out. Only number you want to know is the last number 1 or 2.

[@tsterTV]

Others seem to have said the same thing, but this is what I came up with. So long as you terminate the infinite fractions with either a 1 or 2, the expression is valid. The problem is that you are treating the continued fraction of a rational number like that of an irrational number, allowing it to iterate to infinity. If you stop after a finite number of iterations, the identity still holds -- meaning that the problem is trying to continue to infinity. I mean, I have no idea why exactly what you did is wrong (I don't see the flaw in the algebraic logic -- somebody please point it out!), but I just understand that the error lies in treating the rational number like it's irrational and continuing the fraction forever. By the way, this is the first I've heard of this channel. This was probably one of the most enjoyable maths videos I've ever watched. Do you have a donation link somewhere?

[@chre3611]

I heard soething about the most irrational number and was looking for the explanation in internet. So I found this video. You explained this for someone who wants to understand the meaning without getting in to much mathematik details. Isn't this great?! The continued fractions show us the deap structure of numbers inside, that we can't see in decimal written. When we think of continued fractions we get to the idea that there must be a number that hat only 1s inside. And for sure ist is a special number. But this number ist not pi or e. It is the golden ratio! This is amazing! And besides that, it is the number with the slowest approximation and so it is the most irrational one. Very interesting!

[@vladidobro]

same approach: 1=0+1=0+0+1=0+0+0+1=0+0+0+...=0

[@earthbind83]

This one is my favorite answer! :-) I hope Mathologer makes a video about the most interesting answers given here, instead of just commenting.

[@louisjx8009]

I just checked. The serie doesn t converge to 2, it converge to one. It goes like this 2/3, 6/7, 14/15, 30/31...

[@TacoDude314]

+Argentarii Homini that wasn't the method he should to approximate them infinite fractions. You included the final -1 or -2 which drastically changes the answer. By leaving out the last number, the limit is more meaningful in a way.

[@TacoDude314]

+Argentarii Homini 2/(3), 2/(3-2/(3)), 2(3-2/(3-2/(3)))...

[@louisjx8009]

he said the limit of the series of partial fraction must always converge to the number you say it is equal to. So, in this case, it does not converge to 2. I don't think this is relavent to speak about the last number in an infinite serie.

[@TacoDude314]

Yes exactly

[@louisjx8009]

+Argentarii Homini this is an infinite serie, there is no last digit or last part, or else it would mean this is finite.

[@willemkossen]

Well. 1 isnt 2.

[@alperisler5353]

You solved the goddamn puzzle

[@matrixarsmusicworkshop561]

Good

[@picturelzt]

I consider this as a sequence: start from a_0, cook up a_1 as a_1= 2/(3-a_0), and in general, a_{n+1} = 2/(3-a_n). This sequence has the general term a_{n} = ( 2^{n+1} - 2 - (2^n - 2) a_0 ) / ( 2^{n+1} - 1 - (2^n - 1) a_0 ) = ( 2^n (2 - a_0) + 2 a_0 - 2 ) / ( 2^n (2 - a_0) + a_0 - 1 ). So, when a_0 != 2, the sequence converges to 1, and when a_0 = 2, the sequence converges to 2. The 'usual' way to chop off a continuous fraction is to take a_0 = 0, and consider the continuous fraction as the limit of the above-mentioned sequence. So the 'correct' value of the continuous fraction is lim_{n->\infty} a_n (with a_0 = 0), which is 1. Sorry if it appears twice, I cannot see my initial post.

[@reidflemingworldstoughestm1394]

Phi also pops up in the Reid Fleming numbers. Starting with the two values 1x10^-11 and 11, and following the same rule as the Fibonacci sequence -- adding the previous two numbers to get then next number -- as the sequence progresses the ratio between Reid Fleming(n+1)/Reid Fleming(n) approaches (1+sqrt5)/2.

[@closetofaraway]

x=2/(3-x) x(3-x)=2 3x-x^2=2 0=x^2-3x+2 0=(x-2)(x-1) x=2,1 x has 2 possible solutions

[@Mariek1975]

If 1 equals 2 that means I’ll never be alone cuz I can date myself :)

[@replicaacliper]

the fraction is meaningless because it converges to 1, but since the fraction is infinite there's always going to be a 2 above the 1, which will result in a quotient of 2

[@hahahasan]

i'm gonna stake my imaginary bets on this being right

[@awesomedude7576]

hahahasan lol

[@VideoFunForAll]

Can we deduce using infinite fractions not only if a number is irrational but also if it is transcendent? For instance, is there a repeating pattern in the "squares" of the fractions? And would Pi in this respect be an outer category as, unlike many other transcendent numbers that have a non-repeating pattern, Pi has none? Or is there nothing to say about that and it is merely coincidental?

[@johnredberg]

I haven't watched past 1:48 as I'm writing this (didn't want to be biased, hehe). The problem with the apparent identity of the two infinite fractions is the '...' part. This part is actually different in the two equations. +Mathologer was just a bit sneaky to have written them exactly the same way ;-) Let's call these parts x1, for the equation 1 = 2/(3 - x1), and x2, for the equation 2 = 2/(3 - x2). Solving the above equations is trivial and yields of course x1 = 1, and x2 = 2. If you play around with more successive iterations of the fraction where you stop at a certain "depth", say 1 = 2/(3 - 2/(3 - 2/(3 - x1))), you will easily see that you always recover x1 = 1 and x2 = 2. This is because, as you "unravel" the equation, you can repetitively "multiply over" the nasty denominator and carry back the constants. (This process is self-similar, due to the self-similar nature of the continued fraction.) The sort of mind-bending thing here, as compared to infinite sums, is that with infinite sums, you work your way *towards* infinity by making successive partial sums. Here, you kind of need to work your way backwards *from* infinity. Imagine you are given the question in reverse: 2/(3 - 2/(3 - 2/(3 - ... ))) = ? In this case, you can't simply take a calculator and guess the solution (as you might be able to do with simple converging infinite sums), because you don't know what "..." actually evaluates to. After some potentially painful trial and error, you would then realize that this expression does not have a unique value. If you assume the "..." to be 1, you get a self-consistent solution evaluating to 1. And similarly for the value 2. In this sense, looking at the infinite sum analogy, this expression would correspond to a bounded diverging sum, for example 1 - 1 + 1 - 1 + 1... I would assume there are some criteria as to whether the infinite fraction has a unique solution (as it is the case for the sums), but I don't know them. But hey, maybe there's something about that in the rest of the video, I'll continue watching now. :-)

[@puceno]

walltext XD

[@sarthakjamdar3023]

I want to write a report on this video can someone help me pls 🙂

[@ajaymane3572]

Tera hone ke baad mujhe send kar

[@sarthakjamdar3023]

@@ajaymane3572 😂😂

[@veronx2005]

What help you want , i can ...

[@abhishekvinchu838]

Mereko bhi send kr

[@sarthakjamdar3023]

@@veronx2005 Write a report on this video and send me ASAP XD XD

[@nazgulXVII]

It depends on what you end your fraction with. If 1, the whole thing equals 1, if 2, it equals 2. This answer does not feel satisfying because it does not seem to apply to the "infinitly long fraction", but it actually applies to the infinite case too! Think about it: it depends on what is your procedure to generate the fraction. One procedure is to replace the 1 in the denominator with 2/(3-1), while the other is to replace the 2 in the denominator with 2/(3-2). Since you cannot hand me the infinite fraction, you must hand me part of it and then exactly define what the three dots mean. The three dots mean different things in the two cases, and that is: which procedure are you using?

[@ffggddss]

This was essentially my answer, 4 years ago. The ellipsis hasn't been precisely defined. Therein hides the ambiguity. Fred

[@mahletbekele8727]

In 4:53 can't you replace 3 by its continued fraction

[@hubbertfarnsworth4338]

I had a course about continued fractions and some other things in number theory. Funily, in this course phi would have been presented as the most RATIONAL irrational number (This interpretation is somewhat a consequence of the dirichlet's approximation theorem). Nice video !

[@williamtachyon2630]

Very interesting channel you have here. I subscribed. Keep up the good work!

[@ZoeTheCat]

I think there are two valid points. 1) Both 1 & 2 are already factored as (1/1 & 2/1) and cannot (should not?) be reduced further. This is analogous to the (1473/50) example where the terminating CFE is 1/3. Therefore, it may not make any sense to apply this method (ie: we were done before we started) ;-) 2)Interlude cautions about the "=" sign. It is only applicable if the sequence converges to the number. This sequence converges as (2n*A_i+2)/(2n*B_i+1) for (n=1,2,3,...) So the converging sums are = {2/3, 6/7, 14/15, 30/31, 62/63, 126/127,...) which clearly converges to 1. But we still shouldn't have applied this technique in the first place (See point #1 ;-)

[@lowagner]

the sequence of truncated fractions converges to 1, not to 2: to get the truncated fraction, ignore the last 1: 2/(3-1) -> 2/3 2/(3-(2/(3-1))) -> 2/(3 - 2/3) = 6/(9-2) =6/7 2/(3-(2/(3-(2/(3-1)))) -> 2/(3 - 2/(3 - 2/3)) = 2/(3 - 6/7) = 14/(21-6) = 14/15 and so on... if you put in the last number as 2 (instead of 1), it always gets truncated anyways, so it can't make the difference it's supposed to. i suspect it has also something to do with standard form (1's everywhere except in the denominator of the next term) being a unique way to represent a number.

[@masterprogcoder4632]

Fantastic video sir... Keep on doing such interesting things... I am now you new SUBSCRIBER...

[@ozymandias3456]

I'm extremely bad at math, like almost failed pre calculus yet I find this guys videos really interesting

[@DerToasti]

is it better to include the 292 for pi or is it better to stop at the 1 (in the continued fraction i mean)? relatively speaking of course. at 8:16, including the 292 nets 3 additional digits, which seems like a lot. logically that 1/292 shouldn't add that much precision, because it's so small, so ignoring it would be better?