e is irrational

[@bouteilledargile]

"There ain't no integer between 0 and 1." -Dr. Peyam, 2018

[@JorgetePanete]

35cut that means there is

[@nullplan01]

Regarding this, I actually have a question: Is it possible to prove that 1/n! > sum(i=n+1..inf, 1/i!) for n > 0? Because if so I have a nice proof for e's irrationality. Let l_n = sum(i=0..n, 1/i!). All l_n < e (Proof: e - l_n = sum(i=n + 1..inf, 1/i!) > 0). Let u_n = l_n + 1/n!. All u_n > e, except u_0 (Proof: u_n - e > 0 is equivalent to sum(i=n+1..inf, 1/i!) > 1/n!, and that's the part I still need). All l_n are rational, since only rational numbers are used in the formula, and only operations the rational numbers are closed against, and only finitely many times. In fact, there exists an integer a, such that l_n = a/n! and u_n = (a+1)/n!. Proof: When adding two rational numbers m/n + c/d, it is always possible to write its denominator as the scm(n,d). If a < b, then gcd(a!, b!) = a!, since both factorials share all factors up to a. Therefore scm(a!, b!) = (a! b!)/a! = b!. Therefore, adding two rational numbers with factorial denominators gives a new number with a factorial denominator. The calculation of l_n always starts with a factorial denominator (0!), and only adds numbers with a factorial denominator, up to n!. Therefore l_n is a rational number with n! as denominator. u_n only adds exactly 1 to the numerator. This means that for all n>0, there is no rational number between l_n and u_n with a denominator at or below n!. So, to review, e is between all l_n and u_n (for n > 0), but no rational number with a denominator

[@antimatter2376]

Get a keyboard that let's you do ᵗₕⁱₛ. Superscripts and subscripts. It will make things so much easier to understand

[@Xrelent]

Or use a free online LaTeX editor and post a link :)) tex.stackexchange.com/questions/3/compiling-documents-online for options

[@Xrelent]

Proof of your assumption: www.overleaf.com/read/hvrdnqwdhnkk Edit: Totally misread part of your proof. Hit me while I was at work today. Had to come back to say that your proof totally works.

[@Myuri3146]

can you do a video of how to prove pi/e are transcendental numbers?

[@drpeyam]

I’ll think about it :)

[@Xrelent]

[@100najaja]

@@Xrelent There is a proof it's just not discovered yet :)

[@numbermathematics4137]

1 day before I made a video on understanding Euler's number in 2 parts and today you are making video on irrationality of Euler's number. When I saw this I got excited. ~Rajarshi maiti

[@snakespeak]

A very rational argument to prove irrationality, almost Godellian.

[@AvinashtheIyerHaHaLOL]

This is a really nice proof. I liked the method, and it was easy to follow. Never stop uploading, Dr. Peyam.

[@hellgate25000]

seriously one of my favorite channels. you inspire me to give these presentations and fully understand proofs.

[@martinepstein9826]

Nice! I'm guessing the transcendental proof is quite hard, but I would love to see a derivation of the continued fraction expansion.

[@gabrielmello3293]

Just watch a video on what a taylor series is (you need to know first principles or the CHEN-LU) and use it on e^x.

[@martinepstein9826]

Gabriel Mello Yes, power series is one kind of expansion that usually involves fractions. But it's not a continued fraction expansion.

[@pco246]

There is a general method to find the continued fraction expansion for any real number. Mathologer has a great video on the topic if you're intereseted :)

[@martinepstein9826]

PCreeper394 What I mean is that the continued fraction expansion of e follows a certain pattern; [2;1,2,1,1,4,1,1,6,1,1,8,1,1,...] and I don't know why that is or how to prove that the pattern holds indefinitely.

[@SefJen]

For the end of the proof (21:52) , it's not necessary to precise that X < sigma because further (23:37) , you will be able to say that 1/b

[@wojtek9395]

The first video I understood

[@JuanDeLaCruz-wx2pf]

Yeah!, Starting very well the year, thank you Dr.Peyam. Happy New year and now a proof of π please

[@pawenowak2938]

Juan De La Cruz Dr if u dont mind u could also explain some formulas for calculating pi in aproximation. Love videos like this

[@karstenmeinders4844]

Great proof!Could you present videos on the Fourier Transformation? Thx and BR

[@hassanalihusseini1717]

Nice way to proove that! Thank as lot , Dr. Peyam.

[@masheroz]

Have you done a proof of the taylor expansion?

[@drpeyam]

It’s basically just the definition of the Taylor series, as well as the fact that the n th derivative of e^x is e^x, hence for x = 0 it’s 1.

[@ImPresSiveXD]

Very nice! Could you also show that e and pi are transcendental numbers?

[@drpeyam]

I’ll think about it :)

[@ImPresSiveXD]

Ok, thank you :D

[@manla8397]

Absolutely love it. Very well explained. Thank you and happy new year.

[@Zeboss321]

If you are interested, Dr Peyam, there is a belgian statistician making great videos in french on probabilities, RNG, distributions, election processes, etc. It is called "La statistique expliquée à mon chat" and the aim is for the author to explain statistics to Albert, his cat. I expect you will not learn much new content, but you know, for pedagogical purposes.

[@drpeyam]

Thanks for the suggestion :D

[@Zeboss321]

My pleasure =)

[@fernandoraphael95]

Happy new year Dr. Peyam!!!

[@mattikemppinen6750]

your sigma looks like a cool designer vase :D with infinity flower in it!

[@MrRyanroberson1]

so 2

[@drpeyam]

Yep :)

[@MrRyanroberson1]

how to prove e^pi, pi^e, ln(pi), log pi(e)=1/ln(pi), etc (binary operations with e, pi as operands) are irrational?

[@drpeyam]

I’m guessing that’s pretty hard because even proving e + pi is irrational or not is still an open problem! :O

[@MrRyanroberson1]

really?? but if e+pi IS rational, we could trivially calculate pi as R-e, just memorizing R, with e being one of the easiest-to-calculate irrational numbers out there....

[@MrRyanroberson1]

idea: imagine two circles. one of diameter 1, and another of d=d such that the circumferences are pi, e. a circle with diameter = d+1 has a rational circumference. as such, its diameter of c/pi must be irrational. but c=e+pi; c/pi=(e/pi)+1, therefore e/pi is irrational since 1 is rational and their sum must be irrational (by the theory irr+rat=irr) if e/pi is rational, the converse can be proven: imagine a circle with diameter D such that c=e+pi. the diameter is therefore (e/pi)+1, but e/pi is rational therefore the diameter is rational. next, since the diameter is rational, any multiple of a rational number by an irrational is irrational (except by 0, but e/pi is not 0 as e>pi>0) and c=pi*D therefore c=irrational=e+pi proposing one is rational will prove the other cannot be. maybe you could do a video relating all of these dependencies, and find some *circular* logic proving that none can be rational? (pun not intended but gladly embraced)

[@mohammadrehan8564]

Again a lovely proof❤️

[@jwyliecullick8976]

Nature is beautiful, and has a whimsical sense of humor.

[@michelsantana2642]

e=a/b where a and b are integers such that gcd(a,b)=1; a>0, b>1

[@dolevgo8535]

8:10 why cant a,b be negative?

[@drpeyam]

First of all, e > 0, so a and b must have the same sign. Also if a and b are both negative, then -a and -b are both positive and e = a/b = -a/-b, so without loss of generality we can assume a and b to be positive!

[@dolevgo8535]

okay, thanks!

[@JorgetePanete]

dolev goaz can't*

[@dolevgo8535]

Jorgete Panete just don't

[@MrRyanroberson1]

If you prove e=[2;1,2,1,1,4,1,1,6,1...] then already p,q are indefinitely large and therefore do not fit the definition of some finite p,q where R=p/q

[@gabrielmello3293]

What do you mean by "e=[2;1,2,1,1,4,1,1,6,1...]"?

[@MrRyanroberson1]

continued fraction notation [a;b,c,d...]=a+1/(b+1/(c+1/(d+1/...))) there exists a proof that e is exactly that continuation, but either way you can get the expansion of a number by repeatedly taking modulo 1 and reciprocating it, writing down the amount the modulo removed. 35/101=0+1/(101/35); 101/35=2+1/(35/31); 35/31=1+1/(31/4); 31/4=7+1/(4/3); 4/3=1+1/3. thus the continued fraction is [0;2,1,7,1,3]

[@Xrelent]

@@angelmendez-rivera351 Actually, this was how Euler proved e's irrationality and it was even the first such proof of e's irrationality!

[@helixkirby]

Could you have ignored the fact that b!=1 and still (1/b)

[@drpeyam]

I think that should be ok, as long as one strict inequality remains. Of course we could have n = 1, which might be problematic!

[@SlingerDomb]

i just love Dr.Peyam jokes

[@OonHan]

pi?

[@drpeyam]

Coming soon-ish :)

[@김간디17]

The proof is juicy! Thank you!

[@quickmath8290]

Well done you have astonished me 😊

[@camiv61]

We love you back

[@michelsantana2642]

Beautiful proof!!!

[@michalbotor]

beautiful! it was very easy to follow and understand. thank you. ;)

[@broccoloodle]

It would be nicer if the proof was a direct proof. Great presentation, btw

[@ysapir84]

Great proof, as always!

[@eliascaeiro5439]

Pretty neat, very nice!

[@nickstenerson6310]

I

[@Rolljack]

Thank you

[@manu_j_]

Fantastic

[@znhait]

This again shows why infinity is only considered a concept. One can never treat it as any sort of number. The series of e^1 if stopped at whatever large value of n will be that of a rational number since we are just adding rational numbers together. However, we know that e is irrational and can't be written as a ratio of integers.

[@alihajmomeni2751]

Perfect🤌🤌🤌

[@AnotherZoruaAmongUs]

That feeling when Dr. Peyam accidentally puts down that 2^3=6...

[@jussibusy]

5:23 I understood it was 3!=6 because he kept the four first terms of Σ 1/n! intact.

[@drpeyam]

No, I wrote 1/3! = 1/6, I replaced all the terms starting with n = 4 with 2^n since the formula is valid for n greater than or equal to 4. JSonic1000 is correct about this

[@bonbonpony]

07:08 Unless it's bleem ;J