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!
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 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.
@@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.
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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4ga58IP1iJU.html
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.
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.
+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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-gjhmh3yWiTI.html 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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-f1yDExNAEMg.html
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.
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
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.
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!
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!
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 😅.
Many, many thanks for this. I loved Fourier's proof of the irrationality of e. Lambert's proof of the irrationality of Pi made Fourier's efforts look pretty ordinary, but a proof of anything being transcendental transcends both - and pretty much anything else at the same time. Never thought I'd see something like this. Many, many thanks.
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...
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 :)
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.
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.
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.
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.
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.
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.
+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 :)
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.
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...
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 :)
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
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)
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.
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?
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.
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.
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.
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) + ... ?
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 :)
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!
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 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.
I seems to me like at 7:15 you assume that L is the solution to the polynomial and then you look for a contradiction. I get that. What i dont get is why at the next step you say "If L is a solution then L5 must also be a solution and all L's bigger than 5 are also solutions". How?
At this point, he's basically giving you the outline of the proof. The detail you felt was missing requires a _lot_ more justification. He spends 10:24 - 17:02 justifying it.
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?
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.)
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...
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.
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!
It was my understanding that it is NOT true that if the limit of an equation has a property, there must exist an equation not quite at the limit with that property. Or another way of saying this is that sometimes you care about what happens AT the limit, not as you approach the limit as those are different. There is nothing wrong with the left side approaching a limit at a slightly different rate than the right side, as for any finite n, neither side will ever get there. Each side has a different highest power, so the number of digits will be different - they should all be correct as far as they go, but they should not go the same number of digits. So the assumption that leftside[n]=rightside[n] at the limit as n approaches positive infinity does not imply that they're equal for any finite n. You stated that it should as an assumption. I think if the equation leftside[n]=rightside[n] would be true for a finite n, then you can express this as truncate(leftside(infinity),a) = truncate(rightside(infinity),b) where a= the number of digits the left side is correct for and b is the number of digits the right side is correct for. But a can't equal b for a finite n. But how does this prove that the non-truncated numbers aren't equal?
It's true that if something happens in the limit, it doesn't have to happen before the limit. But it wasn't an _assumption_ that leftside[n] = rightside[n]. He _proved_ this (or at least, gave a sketch of the proof). This is the key step of the proof, so good job on recognizing this is important! The sketch is from 10:24 - 17:02.
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?
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.
+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 :)
I'm not a mathematician so maybe I'm just missing something all mathematicians know and accept as obvious. If I understood you right, you have asserted that if L is a solution to some polynomial equation, then some truncation of L must also be a solution to the same polynomial. But I don't believe you would never accept a truncation of root 2, say 1.4142, as a solution to the equation x^2 - 2 = 0. That number is irrational; a truncated version is not. "From a certain point onward" sounds to me like you've used approximation to prove an exact point. By the way, I had 1 = 0.9999... proved in 4th grade, so I'm comfortable with that one. But, note, this can only be approximately true for a truncation.
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
Why is the Gelfond-Schneider constant (2^sqrt(2)) transcendental when it CAN be written using roots and natural numbers? Is your definition not precise enough? It may be that the Gelfond-Schneider constant is never a solution to an algebraic equation, but you also said it cannot be written using natural numbers and roots (like pi and e). Isn't it what you call a "rooty" number?
It cannot be written as a formula of finite length using the operations of addition, subtraction, multiplication, division, and n-th roots (with integer constants). Observe that exponentiation is not among the list of operations; so if a and b can be expressed in this way, that doesn't mean that a^b can. Indeed, 2^√2 has been proven to be transcendental. A number is algebraic, if it can be expressed as a solution of a polynomial equation with integer coefficients. All "rooty" numbers are algebraic, but the converse is not true; for example, the real solution of the equation x^5 - x - 1 = 0 cannot be expressed as a "rooty" formula.
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)
"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.
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.
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?
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?
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.
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).
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.
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
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.
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.
If you could say that 1 = 0.9999. and 0 = 1 - 1 = 0.999999 - 1 then turning 0.999... -> 0.9900090000000009.... Would that complicate how you would treat 0?
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!
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.
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. >:)
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.
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 ?*
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
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.
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.
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. 🤔
13:00 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.
@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
