One of the first transcendental numbers -- Liouville's Constant 

Can you link that video? I cannot find one about catalans constant

@@nagoshi01 ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-XPBrFqU2Fxc.html here!! i had to search to dig it up but this was my introduction to it!

@@nagoshi01 that's because no one knows wether it's rational or not.

If it is the minimum, just put it in 18:54, and you'll get that the difference is bigger than b^-n, which contradicts.

I was going to comment the same. Maybe someone can correct us? Suppose 1/(2^(n-1)*b) < 1/b^n. Then 2^(n-1)*b = (2^(n)*b)/2 > b^n. Then 2^(n)*b > 2b^n. Then 2^(n-1) > b^(n-1) which is a contradiction, since 2^(n-1) ≤ b^(n-1). Or alternatively, a direct proof I just realised: b ≥ 2. Then b^(n-1) ≥ 2^(n-1). Then 2b^n ≥ 2^(n)*b. Then b^n ≥ 2^(n-1)*b. So 1/(2^(n-1)*b) ≥ 1/b^n. My proofs may be wrong. I'm quite an amateur

At least he used a different colour

@@angelmendez-rivera351 Hence why they said that they never thought proving *a* number to be transcendental is "easy"

@@angelmendez-rivera351 I see. I know that even we can now prove e and pi are transcendental, seems many (or perhaps all?) other known transcendental numbers are basically constructed to be in the first place. This shows how hard to be to prove the transcendentless of a number.

My guess is that they have been devised with the goal of creating transcendental numbers that are demonstrably so.

It's a number which can be arbitrarily close to some rational number without itself being rational. Imagine you pick a very large value of n: b > 1, so the difference between alpha and a satisfactory rational number a/b is extremely small, but it cannot be 0 because of the left inequality 0 < |alpha - a/b|. Now, alpha cannot itself be rational because of the proof shown in the video. Practically, this definition gives two necessary conditions for alpha; if either of them is violated in a proof, one of the assumptions was wrong, so you can easily show a contradiction. Naturally, these conditions can be a bit restrictive, so Liouville's constant is important because it's a concrete example of a Liouville number.

Louisville 😂

​@@neonlines1156It's probably because that's basically how Michael says it- he swaps the sounds for "ou" and "i."

Lucy Liou?

@@wernergamper6200 Miou-Miou ?

The function defined that way is not continuous: By the intermediate value theorem, it would have to produce a rational number because the rationals are dense in R.

@@reijerboodt8715 b can be any real number > 1. Continuity follows.

You can rewrite *L(b)* as a power series if you define *a_k* to be 1 if *k = n!* and zero otherwise: *L(b) = \sum_{k = 0}^\infty a_k * (1/b)^k =: f(1/b)* The function *f* is a power-series with coefficients *a_k* and convergence radius *R = 1* . It is continuous for all *|x| = |1/b| < 1* . By composition, the function *L(b)* is continuous as well for all *|b| > 1*

You can always construct a polynomial that has arbitrary other distinct roots. The constraint is on the coefficients, not on the roots.

Other roots are the troublemakers in this case. So, it is fine if there are no other roots.

I noticed that too. Reminded me of Darry Jenner from Jeepers Creepers ;)

You can, in general, calculate the first (n+1)!-1 digits with just n terms. (You know the next digits would all be zero, but might not know exactly where to put the n+1st 1.)

en.wikipedia.org/wiki/Joseph_Liouville

Very good question. So you would solve this by drawing a line that passes through sqrt x and ln x and checking where its intersection have the same derivates.

Hmm, thank you. Will try drawing it on a piece of paper to visualise it better

If it exists, it is zero or should occur along the common normals as the common normals give you local minimums for the distance between the two curves.( Given functions are well defined at those points, if not try to find the distance between the curves at those points by taking limits at those points.) Compare all those distances and take the least value.

The functions do not intersect. So the min. distance is not zero. That woučd br a trivial answer.

@@angelmendez-rivera351 This sounds cool, but I am not yet so familliar with vector calculus.

f(x) is a polynomial, therefore f'(x) is a polynomial (power rule) and polynomials are always finite on any finite interval (closed or open). A polynomial being finite means that there is always a finite maximum value. A strict, less handwavy, proof of this lies in the Extreme Value Theorem: polynomials are continuous, and the EVT states that functions continuous on a closed interval have a minimum and a maximum on that interval. *Here polynomial means a polynomial with finite coefficients and exponents obviously.

For sure, I was visiting faculty there for one wonderful year!

...more like "Loo-vall"

More like Lioo-veel, where Li is the same Li in Liberty.

@@GutReconIkaros Vous avez raison! Mais j'aime garder les choses simples pour les Américains...

@@reintsh Ça se tient haha !

While it is true that any transcendental number can be approximated with arbitrary precision by rational numbers this is not the best way to elucidate that fact. Not all transcendental numbers are Liouville numbers so the definition would not necessary imply that about the class as a whole. It will probably be a bit disappointing, but the fact that this holds for transcendental numbers comes from the fact that it holds for the wider class of real numbers, pretty much by definition. The real numbers can be defined as the set of values to which sequences of rationals converge. Given converging means that after some point all terms in the sequence lie within a given precision of the value you can always find rational numbers within arbitrary precision of any real number.

No, at most we can say that Liouville's numbers can be approximated by arbitrary precision by rationals. This because the video shows only one implication: If a number is a Liouville's number, then it's trascendental. However the inverse is not true, there are trascendental numbers that are not Liouville. In measurement theory it was showed that the set of all Liouville's numbers have Lebesgue's measure equal to 0, while trascendentals in [a;b] have Lebesgue measure b-a. For example "e" and "pi" are trascendental but not Liouville

@@Rendertk1 ah yeah that makes sense that it's a property of all real numbers and not just transcendental ones. That's because of the cauchy sequence definition of rational numbers, yeah?

@@alessiodaniotti264 Yeah I should have realized the reverse isn't necessarily true. Thanks for pointing that out!

No, it is correct as is.

As a French speaker, in French, it would be pronounced loo-ee-vil. Edit: I misread Liouville as Louiville. It should actually be pronounced lee-oo-vil.

@@youngmathematician9154 Also as a French native speaker, I'd pronounce it lee-ooh-vil or smt relatively closer

@@leofabregues5824 sorry I thought it was Louiville… You are right!

No

@@youngmathematician9154 alright mate that's fine, happens ig

Liouville wasn't that versed in binary it would appear. Any "base" (or, more specifically, any base for the exponent) statisfies

He underlined (for emphasis) the strict inequality with yellow chalk. Not the best way to emphasize something haha

The polynomial is allowed to be zero, just not evaluated at the Liouville number. It might possibly have a maximum of 0

You can pause the video you know?

It has been proven that almost all real numbers are normal to every base (i.e. the set of real numbers that are NOT normal has measure 0). But determining whether specific numbers (like pi) are normal is difficult (just like determining whether specific numbers like pi + e and pi × e are transcendental although we know almost all real numbers are transcendental). I think there are a few rather contrived numbers that have been proven to be normal in every base but I can't recall for certain. I am aware of two numbers that have been proven to be normal in base 10; the Champernowne constant and the Copeland-Erdos constant.

α being a Liouville number is the hypothesis. it is then shown that α being rational leads to a contradiction, so α is irrational. (you could phrase the theorem in a different way though; the proof shows "Liouville + rational => contradiction", so the theorem could also be stated as "All rational numbers are not Liouville numbers")