Liouville's number, the easiest transcendental and its clones (corrected reupload)

This is a corrected re-upload of a video from a couple of weeks ago. The original version contained one too many shortcut that I really should not have taken. Although only two viewers stumbled across this mess-up it really bothered me, and so here is the corrected version of the video, hopefully free of any more reupload-worthy mistakes. For those of you who already watched the previous version of this video, see whether you can figure out what required fixing :)
This video is all about convincing you that Liouville's number is really a transcendental number. I am presenting a proof for this fact that you won't find in any textbook and I am keeping my fingers crossed that people will agree that this is the most accessible proof of the transcendence of any specific number. Also part of this video is a nice way to create a clone of the real numbers using the Liouville's number as a template. This clone is a seriously paradoxical subset of the reals: it consists entirely of transcendental numbers (with one exception), just like the reals it is uncountably infinite AND of it is of measure 0, that is, it is hidden so well within the reals that in a sense it not even there.
The measure 0 extra video is at • Liouville's number: EX... on Mathologer 2.
Liouville's original paper is here:
Liouville, J. "Sur des classes très-étendues de quantités dont la valeur n'est ni algébrique, ni même réductible à des irrationelles algébriques." J. Math. pures appl. 16, 133-142, 1851.
sites.mathdoc.fr/JMPA/PDF/JMPA...
And if you are interested in having a look at this proof as it also appears in all the textbooks here is one possible reference:
people.math.sc.edu/filaseta/gr...
The proof that I am showing you in this video was inspired by Conway and Guy's take on the subject in their "Book of numbers". In particular, if you are familiar with this book you'll also recognise the 6th degree polynomial that I am using as one of the examples.
This week's t-shirt is from here: shirt.woot.com/offers/liars-p...
You can download the comments of the original video as a pdf file here: www.qedcat.com/misc/comments.pdf.
Thank you very much for my friends Marty Ross for his feedback on a draft of this video and Danil Dmitriev for his Russian subtitles.
Enjoy!

[@mmicoski]

When I was younger, I was amazed at the fact that some operations define new sets of numbers leaving the initial one, such as the subtraction that leads from Natural to Integer numbers, division leading to Rational numbers, square root that leads to Irrational and Imaginary numbers . For me it's like being able to "escape" our 3D space into a 4D space doing a special movement. Now with the Transcendental numbers, I realize probably there is an infinite amount of sets of non mapped numbers, depending on new operations yet to be defined. This is truly amazing!

[@katakana1]

Check out Taylor Series and his videos about tetration then!

[@petrie911]

This is true in a precise sense. To define a number, we need a finite description of it in some language. The number of finite strings in any language is countably infinite, so no matter how fancy our language gets, we can only describe countably many numbers. But the reals are uncountably infinite, so there will always be reals that are undefinable.

[@gocrazy432]

@@petrie911 Except that there there will always be undefined rationals and algebraic numbers too.

[@petrie911]

@@gocrazy432 Actually there aren't. Every algebraic number has a minimal polynomial that serves as its definition. Granted for some it will be inordinately long, but it is finite.

[@gocrazy432]

@@petrie911 But with that logic the summations for transendentals is also a finite defintion. I meant that with the set being "countably infinite", there will always be an infinite amount of numbers never written or used and thus undefined.

[@cetjberg]

You are an excellent expositor. When I was a grad student in Mechanical Engineering at MIT I took George Thomas' course in analysis, Thomas was a world renowned expositor of mathematics, so I do not use the term loosely. In today's lexicon I would say that you are one Hell of an expositor of mathematics. Thank you for these superior quality videos. Charles A Berg D.Sc MIT, Mechanical Engineering 1962

[@Mathologer]

Glad you think so :)

[@MisakaMikotoDesu]

Your videos force me to pay attention and think. Much better than other math channels. Thank you.

[@Mathologer]

This is a corrected re-upload of a video from a couple of weeks ago. The original version contained one too many shortcut that I really should not have taken. Although only two viewers stumbled across this mess-up it really bothered me, and so here is the corrected version of the video, hopefully free of any more reupload-worthy mistakes. For those of you who already watched the previous version of this video, have fun figuring out what required fixing :) The new bits start at 6:36 :) Oh and I also just uploaded the extra measure zero material that I mention at the end of the video to Mathologer 2: kzbin.info/www/bejne/apjEZmt_hZacgLc

[@IcySlime1]

Mathologer Thanks man, we appreciate it.

[@pieffe8]

Yeah negative coefficients were the main problem. I actually figured the issue as I was proving the theorem by myself. I had to make several changes to the theorem proof for that reason.

[@MichaelRothwell1]

Dear Mathologer, this new version is spot on, the punch line is now easier to follow. Thanks a lot! The extra footage on measure zero also appreciated. I'm definitely going to read Conway and Guy's The book of numbers. I was lucky enough to attend lectures by J. H. Conway in Cambridge in the 1970s.

[@Mathologer]

+Michael Rothwell Glad this works for you. I am a great fan of John Conway and if you are interested I've got two other (early) videos that were inspired by his writings and lectures. In particular his proof for Morley's miracle: kzbin.info/www/bejne/nZvLnptprryciqs and a very nice proof that root 2 is irrational. That one is not by him but he's been using it in popular lectures and write-ups for decades: kzbin.info/www/bejne/nGLcdXiug6Z4g8k

[@MichaelRothwell1]

Dear Mathologer, thanks, will take a look. Now, how about a video on surreal numbers? Actually, I think a series of videos would be needed to do the topic justice. There are several videos out there, but those that I have seen are rather superficial, as I found out when I started working through the material myself guided by Wikipedia.

[@giladreti]

If n is transcendental and k is a rational number their sum is transcendental: Suppose that the number isn't transcendental, by definition it implies that there's a polynomial with rational coefficients that n+k is one of his roots. Let's call that polynomial p(x). p(x+k)=q(x) is also a polynomial with rational coefficients, because any coefficient of p just got multiplied or increased by a rational number. Because n+k is a root of p, p(n+k)=0 which implies q(n)=0, which contrasts our assumption.

[@Mathologer]

Full marks :)

[@fernandobarrera4569]

A shorter proof goes like: the sum of two algebraic numbers is algebraic.

[@TheBeardedMathMan]

these videos encourage me to introduce these types of beautiful math into my HS math classes. Thank you for sharing, educating and enteraining me, all at once!

[@Mathologer]

That's great :)

[@davidmeijer1645]

Same here. Can't wait to unleash some of these concepts on my Grade 9 IB students next week, when we review number systems.

[@pfeilspitze]

The nuance I'm wondering about: what's important about the factorial specifically? I figure it wouldn't work with a polynomial, but it's not intuitive that it wouldn't work with an exponential, which of course also has infinitely growing gaps and grows faster than the polynomials that make something algebraic...

[@giuliosf]

It is because for any algebraic number x of degree n, there exists M s.t. |z-p/q|>1/(Mq^n) for any p and q>0, the factorial is taken to let fail this so the number is trascendental :)

[@jasondecker1460]

I really like the way the visual encoding of these ideas communicate the concepts involved.

[@Dominik-jk4yq]

I really do love your videos. Fills me with le grand joy when i watch these. Keep it up!

[@tonkuche693]

6:35 your ventriloquism skills rule, man!

[@anon6514]

Showing L can't be the root of a polynomial because it would imply that polynomial has an infinite number of roots. Awesome proof by contradiction. And as always you make it so easy to understand.

[@BuggaUgga]

Thank you. Watched it. Nice. I think you are the best math channel on KZbin. Thanks for the excellent work.

[@Mathologer]

:)

[@skyvoux2686]

Man this channel is the best ! The quality and content are just amazing !!! Keep up with the good work !

[@przemekmajewski1]

Wow, I did not know that was so damn trivial. (And I am one of those pps that has read professional transcendence proofs). Great job!

[@trevorgalivan2012]

I actually gasped when you said explained the contradiction of the polynomial having infinite solutions. This is beautiful.

[@PC_Simo]

Also; interpreting a number at different bases does *_NOT_* change their ”fundamental properties”, namely the ones linked to the size/quantity of said number represents. L still represents the same quantity, in base 2, 3,…, as it does, in base 10. The ”base” is basically just the language we use to represent these numbers. Think of it this way: We can take a word, let’s say: ”Tree”, and its German translation: ”Baum”; very different words, but they both represent the same thing: A tall, woody plant that’s the most noticable building block of forests. Changing from one language to another, changes neither the concept the word represents, nor the properties of said concept, like the properties of an ideal tree. *EDIT:* Unless you meant numbers, like: (2^-(1!)) + (2^-(2!)) + (2^-(3!)) + … and (3^-(1!)) + (3^-(2!)) + (3^-(3!)) + … and so on.

[@rudilapa6569]

Spivak's introductory text "Calculus" chapter 21 "e is transcendental" probs 3, 4 introduce irrational algebraics and Liousville''s number, and show that L is 'too close' to some rationals (but not itself rational) to be algebraic. Your videos are both entertaining and very illuminating - thank you!

[@PC_Simo]

That really reminds me of the ”warm-up” proof that e is irrational, in Mathologer’s video on the transcendence of e and π; in that the proof relied on the fact that e can be approximated by fractions better, than fractions can be approximated by fractions 😅.

[@mmicoski]

I've just noticed we can write the Liouville's number as a short formula, _L = sum(10^(-(N!)))_ , for N in {1,...}, which has similarities to other expansions like the Taylor series for _e_ for example.

[@jackismname]

Lovely comment, that really put it into perspective to me!!

[@PC_Simo]

I’ve noticed the same thing.

[@KirilIliev_Utube]

Real pleasure watching your videos. So much math magic so little time

[@PraetorDrew]

Adding an integer to a transcendental number will always give you a transcendental number. Because if it didn't, then there would be some formula Transcendental + Integer that gives you an Algebraic number. (T + I = A) If you subtract the integer from both sides, then your transcendental number is equal to the algebraic number minus the integer (T = A - I) Since an algebraic number minus an algebraic number always gives you an algebraic number, it means that the transcendental number would also be algebraic. But a transcendental number is defined as non-algebraic.

[@giladzxc17]

I already knew all about these numbers, including the proof, but I still enjoyed watching you speak man

[@paradoxica424]

As previously mentioned... Use the process of Louville's Construction on the Louville Number to create the Louville-Louville Number. Then repeat to infinity. Now let the fun begin. >:)

[@stevethecatcouch6532]

+Jack Lam I must be missing something. Doesn't that sequence of (Louville)^n numbers converge to 0.11? Maybe we just have different ideas of fun. : )

[@Ricocossa1]

Transcendental-clone-of-the-real-numbers-ception!

[@leo17921]

louville

[@JezzaWest]

0.100009000000000000000009...

[@richardschreier3866]

The extra detail on polynomial expressions with positive coefficients is a positive improvement. You fit this in beautifully.

[@Mathologer]

Great:) Yes, I thought since I am at it I might as well clarify a couple of other points that people stumbled over in the first video :)

[@Lens98052]

A long time ago I ran into a proof that pi is transcendental that was a short paragraph of words with no complicated math in it. It was in Spanish, so I had to learn enough Spanish to translate the math and verify it.

[@Mathologer]

And once you learned Spanish and read it you figured out that it wasn't a proof after all. How sad. On the bright side you now know how Spanish :)

[@KnakuanaRka]

You have a copy?

[@kylesheng2365]

@@KnakuanaRka bro, stop assuming

[@KnakuanaRka]

Kyle Sheng Yeah, I guess it would be obvious you wouldn’t still have it.

[@trogdorstrngbd]

Amazing video! Just a tiny nitpick: at 16:33, you haven't shown that both sides (i.e., the top and bottom of the right side) match each other in the rate of growth of new digits. In fact, they don't because Ln^4 will always have more digits than Ln^3. The proof is easily fixed by adding Ln^5 to both sides, however.

[@Desmodious]

Got it. The only way i could see this getting easier is showing the intercepts for the polynomial. Great video.

[@huizilin65]

Very well done, a "lichtvolle Darstellung"! Sadly enough some people are not able to show the "big picture" and thus fail to motivate the reader or listener. These people hide their incompetence behind terminologism. thanks!

[@cbranalli]

general gist: IF WE ASSUME THAT the infinitely long number L (Liouville's Number - aka the Louisville Slugger) satisfies a particular polynomial equation of particular degree - say D - then there is an infinitely large set of TRUNCATIONS of L of various lengths - which will ALSO satisfy this polynomial equation (due to the detailed explanation given by Mr Mathologer). given the preceding hypothetical - we find that we have reached an impossible situation - where we have an "infinite number" of roots of this equation - which - by standard theory - can have ONLY D ROOTS. this proves that OUR ASSUMPTION IS FALSE - and L CANNOT BE a root of this particular polynomial equation - or (due to the generality of Mr M's disputation) ANY OTHER polynomial equation - which is the DEFINING CHARACTERISTIC of a transcendental number. hence - (q.e.d.) L MUST BE a transcendental number. (only had to watch the video 3 or 4 times !) *am i trailing in your wake - preferably head above water - Mr Mathologer ?*

[@thej3799]

Yes because of the shifting it's always going to be kind of like an oscillating between a binary alpha and Omega concepts as in everything that there is has to be bound by something the spectrum of all that there is has to be bound by something and at some point we cannot infer or use a method from from where we are at now to go and infer any information beyond that boundary because that boundary doesn't contain the third thing

[@adandap]

Oh, that's just a lovely argument. Love your work Mathologer! I've just discovered these and am working my way through the back issues. Just one question about this one - is the factorial spacing necessary to be able to prove the result? (It's obviously sufficient.) Or could we use a spacing like 1,4,9,16,25... which also grows arbitrarily large? I shall investigate...

[@jursamaj]

The factorial spacing itself isn't necessary, but the spacing has to grow arbitrarily large as you move down the digits. Factorial spacing gets you there quicker than square spacing, as factorials grow so fast.

[@jeremyredd4232]

I love your shirt! I have a neat conjecture based on that paradox (I have Pinocchio say "My nose will grow" but same diff). I argue that because there is a measurable outcome in the length of Pinocchio's nose, and the paradox itself does not limit my ability to measure the length of the nose before and after the statement, this suggests that any universe with measurable outcomes cannot contain paradoxes. I know it's a bit shaky but still fun!

[@jursamaj]

What about the simple statement "I am lying"?

[@prdoyle]

Wow, very cool. Some feedback from a first-time viewer... 16:29 was where I got snagged on my first pass through the video. I didn't notice at first the visual language established starting at 14:34; perhaps it might have helped to call out those black lines verbally? Or maybe have a way to animate the transition from the 14:34 section to 16:29 to show how the concepts relate.

[@HearTruth]

You are such a gift.. thank you for freely giving what you know to us.. Math is the Universal Language. ( I am still not even close but I so enjoy watching these tutorials)

[@Mathologer]

Just keep watching these videos and you'll get there :)

[@Juarqua]

I must say I'd have to watch the older version to figure out the fixed part. So, this re-upload convincingly conceales the taken out shortcut.

[@Mathologer]

The new bits start at 6:36 :)

[@Juarqua]

Mathologer Thanks a lot! I would really not have found out without this hint.

[@AttilaAsztalos]

Pretty much exactly where I got derailed on the previous version, so the extra material is much appreciated, it does make a difference... :)

[@BigDBrian]

Heya mathologer, long time fan. I've recently come up with a (pseudo?) proof for e^ipi = -1 using high school level physics and a bit of calculus (circular motion, derivates and their connection to position/speed/acceleration). I am confident this would make for a good video that could help provide some intuition for complex numbers, so I want to create such a video. However I don't really know how to get started. How do you write a script? how to plan it all? I've tried just starting with a powerpoint but that didn't really get me very far. What I have is a fairly good idea of how the 'proof' works, and I've written it out in a comments section once. Any advice, based on how you started out making videos, and put them together? I admire your work. Cheers.

[@deanyona6246]

I'd watch that. Long wanted to easily show people why that works.

[@strengthman600]

mrBorkD PM him the proof, if you can easily display the intuition in sure he'd be glad to turn it into a video

[@BigDBrian]

Sammy He already has a video on the topic w/ a different approach, and I want to create my own video Dean Yona cool :D

[@strengthman600]

mrBorkD Oh I see. I'd like to see the proof at least though

[@Mathologer]

+mrBorkD Well, I've been explaining maths for most of my life and I tend to think about mathematics in a very visual way. This means that the transition to doing the explaining on video wasn't a big deal. What made a real difference for me was that I met somebody who knew everything about shooting and editing videos and he helped me get up to speed in this respect for the first couple of months of doing these videos. Now I do everything myself. One thing to keep in mind is that even for me whose been doing these videos for a while now putting together a video like the one you've just been watching takes forever (have not kept track of the hours but I'd imagine at least 50 hours spread over many days). I'd say if this is your first video, really take your time and start by explaining your proof to a couple of people until you've found just the right words. Maybe record yourself while you are explaining your proof and then base your video on the best way of approaching the topic that you've been able to come up with when testing it on real people :)

[@pseudolullus]

Got it after 2-3 watchings, amazing video!

[@Wojciech940]

It's been a while since I felt so stupid. Thanks Mathologer!

[@Mathologer]

:)

[@Dimiranger]

So, the key here is that n! != O(n^x), where x is any constant. So whatever polynomial we choose, the gaps between the 1s in Liouville's number grow faster at some point than the polynomial. Following that logic, could we just put in 1s at every 2^n spot? So 1s at indices: 1, 2, 4, 8, 16, 32 etc... Since 2^n != O(n^x), which again means that the gaps between the 1s (because of the exponential function 2^x) will out-grow the polynomials, this should work, too, correct?

[@gralha_]

Now I wonder if we could build the whole set of trancendental by simply doing linear combinations of different constants created by this method (considering every function that is not O(n^x))

[@thorodinson7553]

n! grows 'Faster' than any polynomial funktion for large n. which seems to be the key here, that was my first thought aswell. but he didnt mention exponential functions at all. for example in the definition that transcendental numbers are all those left over, which cannot solve any polynomial equation... but he never said anything about exponential ones. like, can a transcendental number then solve exp functions? or is that also a criteria? and why? and if the definition would also require this, then why is he only talking about polynoms? in anycase it seems either you are right with your idea of creating a transcendental number based on x^n or he should mention them in the vid, why this is also not possible/defined.

[@irrelevant_noob]

Dimiranger i think the issue with 2^n-based expansions is that squaring them fills up too much of the gaps. Maybe 3^n (or even higher bases) might work better? Also, Thor Odinson : the issue with "exp functions" is that they're not the basis of an algebraic structure like polynomials are (rings, ideals, modules, fields, etc). They don't neatly define any category of numbers, so it's not much use to even go there...

[@arekkrolak6320]

as always - great content!

[@laugernberg4817]

absolutely fantastic.

[@PC_Simo]

18:15 Well; Liouville’s number is basically just the clone of 0,111…, which is algebraic (1/9), as it solves this non-trivial linear equation with integer coefficients: 9x - 1 = 0.

[@Clear_thinker678]

Would please put a video about the up arrow notation and graham's Number. I am having a hard time to grasp its meaning. I know your explanation is going to make it a lot easier.

[@user-kh5tv9rb6y]

Multiplication is iterated addition. Exponentiation is iterated multiplication. In exactly the same way, two arrows represent iterated exponentiation, and each new arrow represents an iterated form of the one before it. That's all there is to it.

[@suryaraju9496]

He said that he is probably not never going to get there. *Mathologer: Primary School Edition coming to a theatre near you!*

[@CalamityInAction]

Surya Raju It’s been too long

[@abu3qab]

There is a flaw in the explaination of the equality of the two approxiamations. We have yet to demonstrate that the gap of zeros lie in the same order of magnitude in the two expressions. Or may be I missed that.

[@PC_Simo]

4:45 I guess we should strive for looking for a long gap of 0’s, in the decimal expansion of π, which (assuming π is a ”normal number”) probably is there, at some point. Though, I understand that our computation power might be nowhere near the level needed to find such a stretch of 0’s. 🤔

[@ugignadl]

You're super awesome Burkard!

[@Mathologer]

:)

[@mmicoski]

Fantastic video and explanation!!

[@Mathologer]

Glad you like it, took a long time to get it right :)

[@mmicoski]

Mathologer Yes, I can see in your videos the care with each detail. It certainly costs a lot of effort and time, bit the result is very good. I always liked Mathematics as a tool, but you make it fun either. Congratulations!

[@Lolwutdesu9000]

Easily the best maths channel on KZbin.

[@Mathologer]

Glad you think so :)

[@socraticmathtutor1869]

Nice video. I assume that if a_n is a sequence of natural numbers satisfying the "factorial inequality" a_{n+1} \geq (n+1) a_n, then (sum i = 1 to infinity : 10^{a_i}) will be transcendental by essentially the same argument? More generally, I think that if b_n is an unbounded, non-decreasing sequence, then sequences a_n satisfying a_{n+1} >= b_n a_n should have this property.

[@roros2512]

Thanks for the video, Is there any applications of this numbers to other fields, like physics or maybe programming? Thanks again, the video is a really good work

[@kleinim]

I'm just a bit confused on the very last part of the proof. Why do all truncations from a certain point on work? Great video by the way, you are amazing!

[@electric7487]

He talks about this starting at 10:24

[@ivanzivkovic7572]

19:15 There isn't an obvious 1-to-1 correspondence between the set of the real numbers and the clone it definitely exists since the set is uncountably infinite (very easy to prove properly.... if you only take a subset of it where you only put non-zero digits in place of the 1s in Liouville's number, there's 9 possibilities for each of the non-zero digits, and there are as many of those as there are positive integers (so the cardinality is 9^aleph-null = cardinality of the continuum)) but it's not obvious just from the construction itself (or the fact it's a "clone" of the reals, as it depends on the rule by which the reals are "cloned")

[@jursamaj]

How is it non-obvious? Take any real number (with repeating 9s is needed) and space it's digits out in the Liouville pattern. Conversely, take any Liouville type number and just throw out the Liouville pattern 0s. This directly maps the reals and the clones both directions.

[@coolkusti]

There seems to be a small gap in the argument presented for the correctness of the digits of the mth power of the nth truncation. Applying the argument in base 2 tells us that for 2^(-1)+2^(-2)+2^(-5), taking the cube should not be a problem, since 3*(-2) > (-1)+(-1)+(-5), however not all the digits of the second truncation are correct. It seems that without quantifying the 'danger' a bit more rigorously, the argument is incomplete. It looks like the danger can be decreased by decreasing the power, increasing the base or increasing the gaps in the powers.

[@thej3799]

And you just dropped into the rhineman hypothesis my friend

[@joelkositsky4263]

So, the factorial "spacings" of the ones is actually not critical: all we need is a number with increasing spacings after the decimal point so that after a certain point the number of zeroes is sufficient to keep the Mth order polynomial from spilling over into the position of the previous one. E.g. the Nth one could be at position N^2, or even 2N, or (for more rapid increases of zeroes) factorial(factorial(N)). Correct?

[@Mathologer]

Yes, there is a lot of room for variation here :)

[@electric7487]

@@Mathologer 1+1/(2^1)+1/(3^2^1)+1/(4^3^2^1)+1/(5^4^3^2^1)+1/(6^5^4^3^2^1)+... = 1.611114925808376736 [183213 ones] 272243682859... Is also a Liouville number.

[@Aufenthalt]

Best channel on KZbin

[@himanshumallick2269]

at 17:17, what happens if we replace the ones with the digits of liouville's number itself, does it start approaching a terminating decimal on iterations?

[@Mathologer]

Yes, should approach 0.11 :)

[@KnakuanaRka]

17:24 Trivial. A number is algebraic if it solves some polynomial equation, like ax^b+cx^d+fx^g+hx^j...=0. Now assume the opposite of the conclusion and imagine there is a transcendental number T that turns into the algebraic number S when you add an integer Z to it. That means that S solves a polynomial of the form ax^b+cx^d+fx^g+hx^j...=0. Now, since S=(T+Z), substitute that in; a(T+Z)^b+c(T+Z)^d+f(T+Z)^g+h(T+Z)^j...=0. Expanding out via the binomial theorem creates lots of terms that are the product of integer coefficients, powers of T, and powers of Z (an integer); combining like terms gets you another polynomial with integer coefficients which has T has a solution, meaning T is not transcendental. This is a contradiction, and means that T meeting our conditions does not exist. Therefore, adding an integer and a transcendental number always gives you another transcendental number.

[@joojitaold]

Can I take a transcendental number that is L but with ones in the place of the zeroes and vice versa, and add it to L to make an algebraic(10/9)?

[@danielfarbowitz671]

I thought this was neat: Showing 0.0011101111... was transcendental pretty easy. (I called that number the Liouville complement.) I proved you can add any integer to a transcendental and still get a transcendental. This extends to rationals, and so a rational minus a transcendental is transcendental. Thus 1/9 (or .1111111...) minus Liouville’s number gives 0.001110111111..., another transcendental.

[@ryanhollstein4164]

@Mathologer because there are so many non zero elements in the speech my question is whether an irrational number like pi can only only show it's digits in non zero spaces can pi's 30th digit which i believe is a zero be described in the 30th factorial place or do you have to use the 31st digit

[@PC_Simo]

18:22 Well, Liouville’s constant, itself, is really just a ”Liouville-clone” of the algebraic (and rational) number 0,111…, which is just 1/9.

[@philbeaudoin]

Is there a reason why the gaps between the 1s in the decimal expension of Liouville number must increase so quickly? Couldn't you just as well use your proof with a number where each gap was "1 larger than the previous gap" that is 10^-1 + 10^-3 + 10^-6 + 10^-10 + ... + 10^(-i(i+1)/2) + ... ?

[@giuliosf]

It is because for any algebraic number x of degree n, there exists M s.t. |z-p/q|>1/(Mq^n) for any p and q>0, the factorial is taken to let fail this so the number is trascendental :)

[@RebelleLaFleur]

Thank you very much for this video! There are not many explanations regarding how to create Liouville numbers, so I'm glad you explained it to us so well. It was really helpful for my homework. The question was: Be (a_n) a strictly monotonically increasing progression of natural numbers. We create a number alpha = infinite sum of 10^-a_n. The task was to find a criterium for the progression so that alpha is transcendental. If I got you right, the progression (|a_n - a_n+1|) also has to be strictly monotonically increasing, true?

[@raifegeozay687]

Can we use different gaps? I mean instead of using 1 in every n! th digit and 0 otherwise,Can we use 1 in every (n^n) th digit is 1 and 0 otherwise and prove this number is transcentendal the same way? (because n! can make gaps of 0s as big as we want, so do n^n (or anything that grows faster than n!) And also, we can add an algebraic number to transcentendal clones to create more transcentendal numbers because algebraic+transcentendal =transcentendal

[@vernanonix]

This isn't related to the video but rather a request. I know you like doing videos on math-related subjects involving cartoons and such. So what I'm asking about is, in Season 2, Episode 13 of Gravity Falls (so towards the end of the series in spoiler territory if you care to take time to sit down to watch it), there's an item introduced called the Infinity-Sided die which is supposed to have truly infinite outcomes even beyond numbers. They way it's described, if it was rolled, "our faces could melt into jelly, the world could turn into an egg, or you could just roll an 8." My question about this is, what are the odds of rolling an event (the world turning into an egg, for instance) versus rolling a number. I've thought of events as being discreet and capable of being numbered via the natural numbers, implying they are an infinitely smaller set, but I'm unsure. By my thoughts, there's a ~0% change of rolling events but I want you take on it.

[@worldnotworld]

What arithmetical and algebraic properties are preserved by the clone of the reals? Order is preserved, as is addition, multiplication - but do operations on the clones always stay in the universe of clones? Also, one can imagine iterating the cloning, mapping the reals R to their transcendental clones C(R), and then these transcendental clones to C(C(R)), then C(C(C(R))), and so on...

[@giuliosf]

Not the sum nor the roduct are internal to the clones. Interesting idea to iterate the construction. I think that the intersection of all that classes is all numbers like 0,** when for * i mean a digit :)

[@alecpiazza3857]

great proof, really nice

[@mauricioachigar]

nice proof! congrats!!

[@boba9170]

Number theory seems to be more 'puzzles' and 'parlor tricks'.

[@Mathologer]

Why? :)

[@crancowan523]

Very interesting. Can this be extended to say: sum(from n=1 to inf) 1/(A^(n!)) is transcendental for all |A| >1?

[@trogdorstrngbd]

Definitely for all rational A = p/q; just do the proof in base q. Not sure about the rest of the reals.

[@kdSU30]

How about a number where ones are placed tree(1), tree(2), tree(3)… digits away from the decimal place?

[@electric7487]

19:35 The entire set of Liouville numbers, of which this clone is a subset of, _also_ has measure 0.

[@DieselBoulder]

Are there any two transcendental numbers that when added, subtracted, multiplied or divided to or from each other that equals an algebraic number?

[@MikeRosoftJH]

Sure there are, and it's really easy to come up with an example.

[@DieselBoulder]

@@MikeRosoftJH don't hold out on me, let's see the example! :)

[@MikeRosoftJH]

@@DieselBoulder (1-pi)+pi=1. Here you go: sum of two transcendental numbers is algebraic.

[@MrRyanroberson1]

I think I was one of them, but I kind of accepted your skipping ahead as fact. thanks anyways XD at least now I know the intuition

[@Mathologer]

Cool :)

[@qwertz12345654321]

So there seems to be some problem with subtracting terms. But I can't quite figure out why

[@Mathologer]

Basically in addition to these exploding stretches of zeros you also get exploding stretches of 9s. These can be dealt with directly, but I think they way I do it in the video is more elegant :)

[@danielgrass9881]

I still don't understand the original proof with all positive coefficients. Couldn't there be a polynomial that gets closer and closer to a Louisville number, but the approximations still aren't equal to exactly 0?

[@mrmimeisfunny]

Daniel Grass We don't have to get to 0. We have to solve the polynomial. Which means the 2 sides are equal. And are equal from x truncation onwards.

[@danielgrass9881]

I'm refering to a polynomial with only positive coefficients. That would be equal to 0 because there are no negative coefficients.

[@danielgrass9881]

What if the coefficient on the constant term was negative?

[@danielgrass9881]

If that's the case, then why are there infinite solutions? The approximations could make the equation on the left close to equaling the constant on the right, but only the true Liouville number is a 0 for the equation. For example take an equation like 661*x^2 -8. Pretend that the true Liouville number is a 0, but .110001 only makse the equation close to 0. Each successive approximation gets arbitrarily close to 0, but only the Louisville number actually makes the equation equal to 0.

[@Mathologer]

+Daniel Grass Did you watch the whole video? You really have to watch it all the way to the end of the technical discussion around the 16:20 mark to understand why the truncations will always satisfy a given polynomial from some point on :)

[@SquirrelASMR]

Omg that would be so cute, a mathologer channel for preschoolers hahaha (I'd still watch)

[@mistou26]

Brilliant !!!

[@deusexmaximum8930]

SPOT ON

[@agnekost5015]

the 10x^6+21=75x^3+190x first one is aprox [1.22,53.92] and next one is aprox [2.104,888.58]

[@jasonrejman1956]

9:42. Ousted is Gauss' Fundamental Theorem of Algebra.

[@user-ug2ij5qx2v]

Отбросьте запятую и ставьте 1 на "факториальной" и 0 на "нефакториальной" позиции. Вот нечётные числа такого вида: 1 = 1 11 = 3 110001 = 49 110001000000000000000001 = 12845057 Формула перехода от одного нечётного числа к следующему: 1*2^(2!-1!)+1 = 3 3*2^(3!-2!)+1 = 49 49*2^(4!-3!)+1 = 12845057 Дальнейшие такие числа содержат 37, 217, 1518. 12138 цифр...

[@ajforbes77]

You know, it would be really helpful if you, or someone else, translated that into English. I mean, most of us who comment on his videos comment in English, and I, and most other Americans, can't read greek.

[@Kualinar]

What if the gaps are defined differently, like : N^2, N^3, N^4, ... ?

[@electric7487]

But then it wouldn't be Liouville.

[@MikeRosoftJH]

@@electric7487 But it would have been an interesting question whether or not such numbers can be proven to be transcendental.

[@electric7487]

@@MikeRosoftJH Hint: In that case, that number would have all 0's and 1's, with no occurrences of 2, 3, 4, 5, 6, 7, 8, or 9. It has been conjectured that all irrational numbers that are not normal (a normal number has an even distribution of digits) are transcendental.

[@MikeRosoftJH]

@@electric7487 Or, equivalently, that all algebraic irrational numbers are normal (in every base). But this has not been proven: no algebraic irrational number has been proven to be (or not to be) normal in any base.

[@MCGeorgeMallory]

Wait, if this clone is to match up to the real numbers 1:1, doesn’t that mean it must have something mapped to zero as well? I know this is easy enough to do with countably infinite sets, but how would this mapping be done in an uncountably infinite set?

[@trogdorstrngbd]

The "clone" matches up 1:1 with (0, +inf), not the whole of R. Any interval of the reals still has the same cardinality as R, i.e., uncountably infinite, thus so does the clone. EDIT: The clone construction also works for (-inf, 0), but my point still stands.

[@MikeRosoftJH]

You can take some countably infinite subset of reals and shift them by one position. For example, let the set in question be 1, 1/2, 1/3, 1/4, 1/5, ... Every non-zero real number has its associated transcendental number. Map the transcendental number of 1 (of course represented by the alternate decimal expansion 0.999...) to 0, the number of 1/2 to 1, of 1/3 to 1/2, and so on; the mapping of the remaining numbers will remain unchanged (for example √2 will be mapped with its own transcendental number).

[@robertgumpi7235]

Good video. What I don’t understand is the last part. We take any real and construct a transcendental within the reals. So far so good. But since we took any real we could have taken a constructed transcendental and construct a transcendental. Or the other way round we could take a constructed transcendental and apply the algorithm again. And then again. This a confusing imagination. New uncountable sets of numbers pop up again and again but still are completely hidden within the original reals. I think this only works because one can only apply the algorithm countable times. Isn’t it - in a way - the same idea as: double all integers and get the countable set of even numbers which is within the original integers (but not with measure 0)

[@trogdorstrngbd]

"Recursively" applying the construction an infinite number of times is allowed. The reason why the "clones" have (Lebesgue) measure 0 is because none of them are densely packed enough to form a continuous interval on the reals.

[@mauricioachigar]

maybe that argument works with smaller gaps of zeroes. what about gaps growing linearly?

[@schizoframia4874]

When you drink too much and make a new number

[@zwishking6032]

What if you multiple like an irrational but algebraic constant like square root two

[@MikeRosoftJH]

A product of a transcendental number and an algebraic number (other than 0) can't be algebraic.

[@thomaskember4628]

Alright, are all transcendental numbers derived from pi, e or Louisville numbers or are there other transcendental numbers that have nothing to do with them?

[@giuliosf]

There are uncountably many :)

[@Novastar900]

Let's say f() is the function to switch a number into the clone of Liouville's number, so f(sqrt(2)) = f(1.414...) = 1.410004000000000000000002... Now let's calculate f(2) = 2.000... and also f(1.999...) = 1.990009000000000000000009... we have 2 = 1.999... (as shown in another video), but f(2) ≠ f(1.999)... how is that possible? We input the same numbers, but get different results. Is that function non-deterministic or how is that called?

[@MikeRosoftJH]

2 is not a Liouville number. (It's obviously algebraic.) So if you start with a number with a finite decimal expansion, you use its alternate representation which ends in an infinite string of nines. The only number for which the construction doesn't work is 0. So this proves that there are at least as many Liouville numbers, as positive real numbers. (Positive real numbers can be mapped one-to-one with real numbers as a whole, e.g. x -> ln(x).) Going backwards, Liouville numbers are a subset of reals, so there are at most as many of them as real numbers. (Theorem: If set X can be mapped one-to-one with a subset of Y, and Y with a subset of X, then there also exists a one-to-one function between the two sets - the sets have the same cardinality.)

[@cristybiakke9853]

A visual way to shoe a number is transcendental? Sounds like a contraddiction

[@user-kh5tv9rb6y]

Isn't this technically an infinite family of numbers, one for each integer base? Instead of the sum of 10^-n! for all n, the sum of b^-n! for all n, which should work for all b>1

[@giuliosf]

Yes :)

[@MrFermiMr]

This clone business is not surprising though. All the reals are contained in the the interval [0,1] and in the cantor set, which is also of measure zero.

[@randomnobody660]

I saw a comment asking this sort of confusingly and i don't think it was answered in replies. If somebody can explain to me why this is wrong that would be fantastic and much appreciated: For example say we have something like Ax^7 + Bx^3 = Cx^321 + D i understand if L is a solution then some approximation Ln will have the exact same decimal digits as well as eveything else, thus also be a solution to this equation, and similarly Ln+1, Ln+2 will also be solutions. My problem is if you think about it a different way, a zero is when f(x) = Ax^7 + Bx^3 - Cx^321 - D =0, and obviously more likely than not f(L)!=f(Ln)!=f(Ln+1) and so forth, doesn't that mean that if L is a zero Ln+1, Ln+2 are not?

[@danildmitriev5884]

I'm not sure that I fully understand your question, but I will try to answer it (or help you answer it yourself). Assume f(L) = 0. Let's suppose that, starting from some n, A*(L^7) is equal to A*(Ln^7) in the first M digits, B*(L^3) = B*(Ln^3) in the first M digits, and similar equalities hold for the other terms in the polynomial f(x), also for the first M digits. M is the number of non-zero digits in approximations to these terms, i.e. number of non-zero digits in A*(Ln^7), B*(Ln^3), and so on. We know that such M exists due to the widening gap of zeroes, which was mentioned in the video towards the end of the proof. Moreover, we know that the same equalities are going to be true if we take any larger n; the difference will be only in the corresponding value of M (it's going to be larger for higher-order approximations). I know that this paragraph is not very clear, so if you struggle with it, please tell me, and I will try to explain it clearer. All in all, I'm just repeating what Mathologer shows about these approximations in the video. Then, if we tried to plug in x=Ln, and calculate f(Ln), we would get that all terms in the polynomial feature the same digits up to M as in the case if we plugged in x=L instead. In other words, all terms in f(Ln) feature correct digits up to the M-th digit. Therefore, f(Ln), the resulting value, will itself feature only correct digits of f(L) up to the M-th digit. In addition, since all terms in f(Ln) don't feature non-zero digits after the "benchmark" M, we can guarantee that f(Ln) will feature only zero digits after the M-th digit. Now we recall that f(L) = 0, which means that the first M digits in the "correct" value are all zeroes. This implies that both the first M digits of f(Ln) and all its digits after that point are zeroes, i.e. f(Ln) = 0 = f(L). The same can be proved in exactly the same way for all truncations L_{n+1}, L_{n+2}, and so on. They will each have a different upper bound on non-zero digits, M, but the argument will still be valid. So we can prove that f(L) = 0 = f(Ln) = f(Ln+1) = f(Ln+2) = ..., and so on. Therefore, f(L) has infinitely many roots, but this is impossible due to the fact that f(x) is a polynomial of a finite degree (in our case, 321). So our assumption that f(L) = 0 is wrong. Note that the argument also works for any polynomial of a finite degree, so the specific form of f(x) wasn't important here. In essence, your way of thinking about the problem is pretty similar to what Mathologer is talking about in the video, and you can convince yourself that f(L) = f(Ln) = f(Ln+1) = ..., starting from some n, in the same way as proposed in the video. I hope that I've answered your question.

[@randomnobody660]

Danil Dmitriev so to simplify my question from the logic in this video f(Ln) should have all the same decimals as f(L) up to a certain point, and be all zeros after that. This I understand. but on the other hand f(L)=f(Ln) doesn't make any sense. From what i gather this IS the contradiction that disproves our assumption of f(L)=0? It's just somehow meeting our contradiction here (that both f(L)=f(Ln) AND f(L)!=f(Ln)) seems make me think it's less a successful proof by contradiction and more by we used something we should not have. It's like instead of assuming something and showing that our assumption implies itself to be false, we have instead shown by assuming something we can derive both something else and something else's compliment. The latter somehow feels like we did something wrong is all.

[@danildmitriev5884]

No, the contradiction that the video (and my attempted proof above) is working towards is: assuming f(L)=0 and getting that there are infinitely many other, different numbers that also satisfy f(x)=0, which cannot be the case. Speaking in terms of an analogy, we are assuming that something like x^2-1=0 is solved by one number and getting that infinitely many other numbers also solve it -- which is clearly impossible. So we're working strictly within the assumptions we have made, no fishy business here. In fact, we are working hard to prove f(Ln)=0; we're not trying to show f(Ln)=f(L) in the proof. It may seem weird at first, since these two equations are equivalent, but this is, maybe, the most important trick in the argument. Let me try to explain the argument once again, emphasizing the important points. We know that f(Ln) coincides with f(L) in the first several digits until a certain point, and f(Ln) contains only zero digits after that point. Let's say for definiteness that f(Ln) coincides with f(L) in the first M digits, after which f(Ln) features ONLY zero digits. The latter fact is going to be important at the end of the next paragraph. Now we reconcile our initial assumption and the fact that f(Ln) coincides with f(L) in the first M digits (let's call this fact the "accuracy condition"). We have initially assumed that f(L)=0, i.e. all digits of f(L) are zeroes, f(L)=0.0000...0... . So, from this and the accuracy condition we get that the first M digits of f(Ln) also MUST be zeroes. Combining it with the fact that the digits after that point are also zeroes, we arrive at the conclusion that f(Ln)=0.0000000...0... = 0. In other words, Ln also solves equation f(x)=0. And similarly, we can show that any truncation of a higher order will also solve this equation; hence, we arrive at the desired contradiction.

[@randomnobody660]

Thanks for the your patience and detailed answers. I understand how the proof is derived, plz believe me i do. While i understand the proof however, i see the fact that ultimately we are still implying f(Ln) = f(n) as a slight problem. I am under the impression that given the same premise if two contradiction conclusions can be derived using two approaches, either the method used is flawed or the premise is self contradictory. That is my problem. Upon further inspection I suppose this isn't that big a problem when we ARE attempting to prove our premise self contradictory.

[@danildmitriev5884]

I think I'm starting to see what you're talking about. Indeed, we do accidentally prove along the way that f(Ln)=f(L), which is a contradiction in and of itself to our intuition, since Ln and L are very different numbers after the first several digits. However, it's not easy to prove mathematically why f(Ln)=f(L) is itself a contradiction; note that nowhere in the proof we showed that f(Ln) is, in fact, different from f(L). On the contrary, we have proved that if f(L)=0, then f(Ln)=0 and f(Ln)=f(L). But this is NOT yet a desired result, since it's not obvious (to me, at least), why f(L)=f(Ln) is an immediate contradiction. My gut feeling is with you on this, it must be a contradiction, but it's hard to show this. This is why we are going a slightly longer way of repeating the proof for L(n+1), L(n+2) and so on.

[@900ml5]

Can you do a video about, Base=1, Log1=What?

[@alexanderm5728]

Pretty sure that the digit 1 doesn't exist in base 1. Don't quote me though, I'm not certain.

[@frechjo]

I don't think that base 1 logarithms are defined. Any power of one is 1. Notice that log_b(a) = log_x(a) / log_x(b), and if b=1, log_x(b)=0. Therefore you get log_1(a)=log_x(a)/0 OTOH if you notice the logarithm is an approximation of the digits it takes to write a number in some base, then log_1(x) should be less than or equal to x, and log_1(x)>log_1(x-1) (if you're willing to consider 1 as a proper base for a positional number system). But that's just an interpretation, not a definition.

[@frechjo]

The digit you use for "base 1" is 1, because it should respect this equality: 111 = 1.(1^2) + 1.(1^1) + 1.(1^0) = 1+1+1 = 3 If you chose 0, which many people think is the reasonable digit: 000 = 0.(1^2) + 0.(1^1) + 0.(1^0) = 0+0+0 = 0

[@alexanderm5728]

I suppose that makes sense. Of course it doesn't leave any room for 0, but base 1 doesn't make much sense anyway. And now that I read your comment I realise that HailMary was of course talking about the base of the logarithm, not base as in binary, ternary etc. Ignore my comment.

[@frechjo]

But I don't want to ignore it, it would feel rude! :P And yeah, base 1 doesn't make sense as a positional number system. It's got no difference in positional value, kind of a central point I'd say... It's maybe the intersection of positional and additive number systems, a nice curiosity if you ask me.

[@stantheman719]

Very interesting.

[@mr.cheese5697]

Is there a function that converts number into amount of symbols in it? (1-9)→1 (10-99)→2 (100-999)→3 You get the idea.

[@MikeRosoftJH]

The log10 function? (Round down and add 1.)

[@okboing]

I like to believe that upon finishing that sentence, pinocchio's nose became a jackhammer.

[@amermelaoroma1432]

nice T-shirt!!

[@PC_Simo]

Well; because the powers of 10, in L, are -(n!)’s, and n! is the product of all positive integers up to and including n; as we increase n, the new factors will always get bigger and bigger, forever and ever after; so, the gaps of 0’s will also grow larger and larger, forever and ever after.

[@PC_Simo]

17:37 Well; if adding an integer (like the 5 Burkard gave as an example) made L (transcendental) into an algebraic number, we could, by definition, find a non-trivial polynomial equation with integer coefficients that L+5 solves. But this would, then, imply that we could tweak that polynomial equation, so that L solves the new polynomial equation; which would, by definition, imply that L is algebraic; which contradicts the well-known fact that L is transcendental. This works for *_ANY_* transcendental number (π, e,… you name it). *QED 𑀩*