e is irrational (e∉ℚ)

[@timurpryadilin8830]

I've never seen this proof before, very elegant and rigorous. I love that!

[@eithan]

Beautiful proof indeed

[@malawigw]

it is so mindfuck. Like you can write e as a sum of rational numbers, and by closure of rational numbers one might think that e should be rational as well.

[@tomkamikaze]

@20100 leiN it's a field so it's closed under addition, just not closed for infinite summation

[@Hiltok]

@@malawigw The result shows that infinity is bigger than you think!

[@radadadadee]

@@malawigw it's as if you can escape rational prison by building a ladder using rational rungs!

[@sirlight-ljij]

I think that once we established that e=sum 1/n! , there is a much simpler proof of irrationality. Suppose e=p/q, then we have p/q=sum 1/n!; multiply both sides by q!, then we have p*(q-1)! = sum q!/n! . The LHS is integer, but the RHS consists of summands where nq. First part is integer, but the second is 2 it is impossible to have an equality.

[@sirlight-ljij]

@curioushegemon Not when you can prove that the sum in question is less than 1/q-1. The point is once we have established that fact, there is a much simpler way of proving the irrationality of e that does not involve a complicated sine test

[@Bayerwaldler]

That proof is correct. It was found by MacDivitt, A. R. G.; Yanagisawa, Yukio (1987), "An elementary proof that e is irrational", It can also be found in en.wikipedia". Proof that e is irrational". This proof is so simple, it should be taught in highschool!

[@sirlight-ljij]

@@srijanraghunath4642 The issue is that the proof in question is an unnecessary more complicated way of showing the same fact. It is not a different proof, it is the same proof but obfuscated with a bunch of relations. It's like you can write 1+1=2, or you can write ln(e)+sin^2(x)+cos^2(x)=sqrt_3(8)

[@jacoboribilik3253]

@@sirlight-ljij this is for fun, no one cares about facts. Besides your proof was first put forward by Fourier and is extremely well known. It is always fun and enriching to see new ways (more creative and original) to prove already proven results, it gives you insight into how to tackle the same problem fron different angles.

[@toddtrimble2555]

@@jacoboribilik3253 To me, Sir Light's (or Fourier's) proof is the first thing one should think of (I have a hard time believing Euler didn't know it), and I can't see anything in the video proof that makes it "more creative" than Fourier's. Indeed, no disrespect to Michael Penn, but the video proof seems long-winded and cumbersome by comparison, and doesn't furnish more insight.

[@thetajay392]

Using periodicity along with factorial to detect rationality is really a neat trick!

[@JonathanMandrake]

Yes, neat and simple, but still requires a good foundation in real analysis

[@mtaur4113]

Lemma 3 also needs that either a_n = a only finitely often, or that f(a)=L. Otherwise, f can be undefined or discontinuous and jump every time a_n = a.

[@glacagnina]

You need continuity.

[@malawigw]

Looking forward to real analysis course

[@ThePharphis]

D2 barb?

[@vikaskalsariya9425]

I think we should celebrate the 50k subscribers achivement with a livestream. It would be great to have something like a QnA session or something?

[@CrossMax122]

At the very last part, where we took the limit, isn't the limit an indeterminate form? What I mean is, I understand how after some point every n would give zero, is it enough to determine the limit? Lovely videos, as always!

[@il_caos_deterministico]

It's not indeterminate since sin(....) does not only tend to zero, but is exactly zero for all n>q, which does not give you an indeterminate form

[@Maniclout]

@@il_caos_deterministico What is q?

[@metakaolin]

@@Maniclout the assumed denominator for e if it were rational, ie e = p/q

[@goodplacetostop2973]

17:51

[@malawigw]

we need a compilation video

[@ilyafoskin]

Now he knows...

[@goodplacetostop2973]

Glenn Wouda And call it “The countable set of good places to stop” 😂

[@soranuareane]

Next, please do the proof that e is transcendental! I've had difficulty understanding some of these proofs and would love to see your take on it.

[@Lucashallal]

Well he did now lol

[@makmesanovic8568]

Hey Michael I really love your videos, especially those olympiad problems. I have a couple of interesting problems from my own country's math olympiad ( Bosnia and Herzegovina ), so I wanted to know if there is a way to send them to you because they might be good ideas for some videos. Wishing you all the best from Bosnia :)

[@xSvenCat]

Could you send them to me as well? Would love to see some Bosnian olympiad problems! (I have family living there)

[@makmesanovic8568]

@@xSvenCat yes of course I can send them to you as well. Do you want me to send it to you via mail or how exactly. And you will just have to give me some time to translate the problems to english so you can understand them more easily

[@dinnokoluh]

Hi, I am also from Bosnia watching Michael's videos

[@mohdaftab2819]

If you can then please write them down here itself(comment section)

[@vikaskalsariya9425]

You can upload to some website like mediafire or google drive and send us it's link. :3

[@mirahor1]

Hi, in 8:24 there is mistake it should be 0

[@schweinmachtbree1013]

yup, the third tool itself is false (a counterexample is given by the constant sequence (a, a, a, ...) and any function f which is not continuous at a). as you say, the hypothesis "lim_{x->a} f(x) = L" should be replaced with "f is continuous at a"

[@schweinmachtbree1013]

@Paul Hametner no my counterexample does work. As I said, take the constant sequence (a, a, ...) (or any sequence which is eventually constant with the value a) and take any function f which is not continuous at a, e.g. f: R->R defined by f(x) = 0 for all x =/= a and f(a) = 1. then this sequence and this function satisfy the hypotheses of the tool (with L=0), but lim_{n->oo} f(a_n) = f(a) = 1 =/= L = 0.

[@schweinmachtbree1013]

@Paul Hametner No. continuity at a point deals with a neighborhood of that point (that is, the behaviour at and around the point), whereas the limit at a point deals with a *punctured neighborhood* of that point (that is, the behaviour around the point but *not* at it). For example the limit of a function f at a point b is totally independent of the value f(b) (indeed, b does not even need to be in the domain of f to talk about the limit of f at b, so long as b is "infinitely close" to the domain of f, in a sense which can be made precise). Therefore your assertion that "lim_{x->a} f(x) = L lim_{n->oo} f(a_n) = L for every sequence (a_n) with a_n -> a" is false; the corrected version is "lim_{x->a} f(x) = L lim_{n->oo} f(a_n) = L for every sequence (a_n) such that a_n -> a and a_n =/= a for all n" (that is, the sequence (a_n) is in a punctured neighborhood of a, as opposed to an ordinary neighborhood). And for the record lim_{x->a} f(x) does exist (with my example for f); the limit is 0 - like I said, the value of f at a is entirely irrelevant when talking about the limit of f at a.

[@HarmonicEpsilonDelta]

@@schweinmachtbree1013 actually, a function is defined to be continuous at a point a if and only if lim_{x->a} f(x)=L. In more advanced calculus, this is established the next way. For any epsilon>0 there exists some delta verifying that for any points in the inverval (a-delta, a+delta), lets call that point, for example, x, then |f(x)-f(a)|

[@schweinmachtbree1013]

@@HarmonicEpsilonDelta Yes I am aware (and I assume you mean lim_{x->a} f(x) = f(a)). As I said, continuity at a point deals with a neighborhood whereas the limit at a point deals with a punctured neighborhood - the two notions are different. So for the eps-delta definition of the limit of f at a (to contrast with the definition you gave for continuity), we define lim_{x->a} f(x) = L if for all eps > 0 there exists some delta > 0 such that for all x in (a-delta, a) U (a, a+delta) we have |f(x) - L| < eps. Note that the point a is not included for limits, but it is included for continuity.

[@xaxuser5033]

now let s proof that is transcendental with another marvelous unknown proof

[@sirlight-ljij]

Can't help but to recommend Mathologer's video for that: kzbin.info/www/bejne/jarSeZKsnM6kjq8

[@spaghettiking653]

What about the case where n < q? Didn't this only prove e is irrational if the denominator is less than some n?

[@zachbills8112]

The point is to show that the expression is equal to zero for all n>q if e=p/q. This immediately shows that the limit of the sequence equals zero as required.

[@satyapalsingh4429]

Very good method of proving that e is an irrational number . Marvellous , Outstanding , Extraordinary . Keep it up!!!

[@nielsmeesschaert477]

It seems like you could have skipped working with the sin function here. It's only purpose was to map multiples of 2*pi to zero. So in fact the proof can also go as follows. Assume e is rational. Write e as p/q. Now choose n>q large enough such that e*n! is not an integer (see argument made at the 12:00 minute mark). Then e*n! = p*n!/q and hence e*n! is an integer. This contradicts our assumption that it was not an integer. Am I wrong?

[@garrycotton7094]

Not following how e*n! is an integer.

[@nielsmeesschaert477]

We assumed that e was rational. Let’s say e equals p/q. Then e times n! is an integer for n larger than or equal to q.

[@garrycotton7094]

​@@nielsmeesschaert477 Gotcha. For some reason I kept thinking of e*n not e*n! in spite typing e*n! in my reply! Thanks for the response :) Given that, eyeballing the idea I can't see a reason it doesn't work although I'm not math prof :P. Provided the full argument is given I can it see working ie. setting n>q and e*n! = p*n!/q = sigma(m=0->inf, n!/m!) and then splitting it apart as in the video gives a residual quotient in spite assuming it was an integer. That said, you would need to prove the second series is in fact a quotient which Mike doesn't do in the video.

[@nielsmeesschaert477]

@@garrycotton7094 Well. He says that it is going to zero and hence it needs to be strictly smaller than 1 for values of n large enough. Combine that with the fact that it's still a sum of strictly positive integers, we have reached the conclusion that the 'tail' of e * n! for n large enough is strictly between 0 and 1, while the first part is an integer. Hence not an integer as a whole.

[@mFix09]

17:30 isn’t infinity*0 an indeterminate form?

[@eshnar]

It would be if the sin part tended to 0. Instead, we showed that the sin part IS 0. That's not a limit anymore, it is really equal to 0. As such, any n multiplied by it will give 0 as well.

[@malawigw]

as @eshnar wrote, we do not have sin(2 pi z_n) -> 0 we have sin(2 pi z_n) = 0 instead. Hence, the limit at 17:30 is well defined.

[@ДенисЛогвинов-з6е]

@@malawigw I reckon we should check firstly is there a zero under the limit and after try to plug in an infinity

[@IoT_]

Неа. Синус этого числа равен нулю не в пределе, а просто равен нулю , потому никакой неопределенности тут нет как таковой

[@ranjansingh9972]

Great vid - crisply explained. How does one motivate the construction of the original limit? What might inspire it?

[@stabbysmurf]

That's a tidy proof! And the components make excellent exercises to spring this as a surprise result in an intro analysis class. First have the kids prove the bounds on the sum as an unrelated warm-up, and have them solve a related limit like lim n*sin(2pi 1/(n+5)) as a standard application of lim anbn = (lim an)(lim bn). Then surprise! You have all the parts to prove e is irrational.

[@sudhanshumishra6482]

Very good proof. Really like your math videos. I'm very interested in seeing some videos involving complex analysis. Plz make some on complex analysis sometime. Again, really good explanation of problems.

[@brendanfan3245]

Great explaination!

[@zanti4132]

Superfast proof that e is irrational: In base factorial, the decimal positions for e are .111..., ad infinitum. (Moving away from the decimal point, the decimal positions have the values 1/2, 1/6, 1/24, ..., 1/n!, ...) Any rational number written in base factorial must terminate. As e does not terminate, it must be irrational.

[@Maxmuetze]

what digits are allowed in “base factorial” for each place value? It’s going to be complicated to define a system that makes representations unique (e.g. 1/2 = 1/3+1/6, so .100...=.01001). Maybe .1111... also converges to another, finite representation.

[@zanti4132]

@@Maxmuetze In base factorial (a.k.a. base !), the representation of any number is unique. For the units position, only the digits 0 and 1 are allowed. For the position to the left of the units, equal to 2!, the digits 0, 1 and 2 are allowed. To the left of that position, equal to 3!, the digits 0, 1, 2, and 3 are allowed. And so on - a new digit comes into play for each new position. The decimal positions work the same way. Aside from the redundancy you can have with infinite decimals that happens with any integer base (e.g. 1 = .99999... in base 10, 1 = .77777... in base 8, 1 = .12345... in base !) every number has a unique representation.

[@maxime_weill]

​@@zanti4132 but it is wrong to say that any number written in base factorial must terminate, since 1 can be written 0.12345... what i think is true is that a rational has to have a finite representation?

[@zanti4132]

@@maxime_weill I think your way of saying it is a little better, although you're going to run into the same situation with any integer base. For example, you can say the decimal example of 1/4 in base ten doesn't have to terminate because it can be written as 249999...

[@michakapustka9462]

Just imagine "overkill" of this

[@iagonewats5113]

Who is the author of this proof?

[@brainsify]

It would be interesting trying to approximate that limit numerically. I think your best chance of doing so would be to find a good approximation of e where the denominator is extremely large.

[@lorenzodavidsartormaurino413]

I haveone nagging doubt. In the end you said suppose that n is bigger than q, and then you derived a contradiction, what if n is smaller than q? It is clear that you get the same result if n is equal to q, but what about if it is smaller?

[@asklar]

sin(t) is bounded for real t, so why is n.sin(2pi.e.n!) not diverge since it is something that goes to infinity times a bounded quantity?

[@JoaoVictor-gy3bk]

That bounded quantity might be tending towards 0

[@redindian5768]

How you opened m! = m(m-1).....…........(n+2)(n+1)n! ?? Please explain 🙂

[@math-ch9zk]

In 10:38 I still did not understand how took n! Inside the sigma as n is some constant tending to infinity

[@garrycotton7094]

As it's some constant, you can pop it in or outside the series as you wish.

[@hymnsg2320]

Pretty nice proof, thanks for sharing with us :).

[@MegaGuiaGamer]

Amazing proof... Just one thing, why is not necessary to consider n less or equal than q? I know that if it's equal then you can do exactly the same method, but I'm not sure that you can do it if n is strictly less than q.

[@VerSalieri]

I think you should have addressed the indeterminante form you get st the end (n.sin(2kpi)) does not have a limit of 0 since it has the form (infinity)(0). Obviously, you can verify it by using Hopital's rule but for the first time watching your videos, I disagree with that being a "good place to stop". That being said, I love your videos and your content is amazing. Keep up the good work my friend.

[@arimermelstein9167]

“More things” = o(n^(m-n))

[@cdavid2200]

I read that More things equals om nom

[@arimermelstein9167]

C David lol.

[@arimermelstein9167]

C David i added more parentheses. I also realize now that it should be o and not O

[@MiroshnikovAlex]

I'm wondering if there was a function exact like sin(x) but its period would be 2*e, could we proove that pi is irrational by the same steps?...

[@porygonC]

writing pi as an infinite series is way different so no.

[@MiroshnikovAlex]

@@porygonC Penn used inf. series to find a limit. Maybe there would be another way to find it

[@MiroshnikovAlex]

@@angelmendez-rivera351 i mean lets suppose f(x) is periodic with t=2e and lim n*f(2pi*e*n!)=A!=0 where n ->inf, then we could do the same and suppose pi =p/q where p,q are in N, so 2pi*e*n! is also in N so we have limit is inf or 0 annd we have a contradiction

[@abhijitharakali]

Nice proof of a cool theorem! I'd love to see a proof for the fact that e is transcendental.

[@glacagnina]

Don't you need to say that f, for the last property, must be continuous?

[@ethanbartiromo2888]

I was able to follow everything of the first two tools, but the last one I didn’t really understand, I’m taking reals this upcoming semester so maybe that might help

[@starwarsjk99]

The following geometric argument might give some intuition. Draw some curve f with a hole at x=a. The coordinates of that hole would be (a, L), where L is defined to be the limit L = lim_(x->a) f(x). Now look at the x-axis as a number line, and imagine some sequence of points a_n, (all positioned on the x-axis), that happen to converge around a. Perhaps it would be helpful to draw out some sequence. Now look at the corresponding y coordinates for those x coordinates in that sequence. You will get a sequence of points on the curve, given by (a_n, f(a_n)). Ask yourself what point is this sequence of points converging towards? It is (a, L). As the a_n's get closer and closer to a, your f(a_n) must get closer to L, because L = lim_(x->a) f(x). If you draw this out it might make sense. So (a_n, f(a_n)) -> (a, L). Thus lim_{n-> infinity} f(a_n) = L.

[@ethanbartiromo2888]

jk991234 that actually helps a whole lot!!! Thank you!

[@presidentworld5360]

Something seems uncertain about this proof: Would the use of the power expansion of e make this proof flawed (or at least follow circular reasoning)? I know that Professor Penn tried to follow a conclusion by contradiction. But, wouldn't the infinite power expansion of e directly imply it is irrational? I say this because he considers the base case of the sine function, which includes the sum expansion, to be correct and the consequence of the rational definition to be incorrect. Of course there is a proof for the infinite summation of e, but this would imply by itself that e is irrational. Thanks in advance. Note: Recall that circular reasoning is defined by having some conclusion C for which its premises adapt assuming the correctness of such C .

[@chuckaway6580]

I don't think it's circular at all. For example, we can write $2 = \sum_{k = 0}^\infty \frac{1}{2^k}$, but that doesn't mean that 2 is irrational. The definition of e is what it is. You might say that every definition of e makes it look like it has to end up being irrational, but it's still a fact that has to be proved.

[@l.3ok]

why isnt the limit of n sin(2πen!) indeterminate ?

[@Maniclout]

That's what I'm confused about as well

[@l.3ok]

@@Maniclout oh yeah, I forgot to remove this comment. It's because n is an integer 😅! (Try to test for n=100, 250, and 1000 in wolfram alpha). Well, at least it helped someone.

[@juanjodoblas6954]

And what would happen if n

[@brunojani7968]

n tends towards infinity. if q>n, q would tend towards infinity as well. This makes p/q tends towards 0.

[@willsawyerrrr]

In order to prove the first limit, could we have used the fact that n!/(m!) is equal to n!/(m•n!), which is equal to 1/m? As n tends to infinity, so too does m, forcing 1/m towards 0.

[@randomthingsandstuff2843]

2!/4! = 2/24=1/12 2!/(4*2!)=2/(8)=1/4 In fact m! = m*(m-1)*...*(n+1)*n!, which is what he wrote on the board

[@willsawyerrrr]

théophane cengiz I agree with this; however, m is defined as n+1, meaning that there would only be the one term prior to n! in your expansion, namely m. I have not studied Sigma notation, so I do not know its technicalities - as m increases to infinity, does n also increase? My method would rely on both increasing, meaning that the sum is equal to the sum of all 1/m. I think my mistake may have been that in the next term m may be n+2, then n+3 and so on, as n may not increase.

[@ribhuhooja3137]

@@willsawyerrrr Yes, n does not increase. Your method only works for the first iteration. Then m becomes n +2, n+3 and so on

[@willsawyerrrr]

Ribhu Hooja This really helps, thank you!

[@ribhuhooja3137]

@@angelmendez-rivera351 Hm? But Pi product is for multiplying a series, and here we're adding up a series? Or do you mean in the denominator?

[@backyard282]

How about pi is irrational?

[@11kravitzn]

Beautiful proof

[@gaeb-hd4lf]

Indeed.

[@Supernova799]

Do a series on kurepa function

[@MrRitwikSarkar]

What about e is transcendental?

[@innokentiyromanchenko1450]

is there an irrational x such that lim n*sin(2pi*x*n!) = 0 ?

[@maxime_weill]

That's a good question. So you need to find a number a irrationnal so that (fractionnal part of (n!*a)) = o(1/n)

[@maxime_weill]

I ve been searching for a while and i'm pretty sure the number sum(1/(k^2)!) Works.

[@pankajkumarbehera8169]

Interesting.

[@dinnokoluh]

At 1:32 shouldn't the m! go down to a 1 and not stop at n!

[@ps200306]

It _does_ go all the way down to 1 ... the n! takes it the rest of the way from (n+1).

[@dinnokoluh]

Yeah, my bad

[@wafizariar8555]

Could you make videos on Galois' cohomology?

[@xshortguy]

Wouldn't the last limit (in the BWOC) be an indeterminate form and require further examination? Not sure how you immediately concluded it was zero.

[@dariobarisic3502]

Because sin(2pi e n!) is EXACTLY zero. This would be indeterminate only if sin(2pi e n!) --> 0 as n--> oo.

[@MrEddrool]

@@dariobarisic3502 Thanks. I missed that, too.

[@inakiarias7465]

@@angelmendez-rivera351 I would suggest you post that beautiful explanation somewhere it can get more traction, not in a random YT comment where it would be only read by like me and some small number of other people before it reaches irrelevancy c: (That's ofc, if you haven't done so already)

[@haytamelhayani2369]

Heeyy , that's a good vedio , but i have just a small remark. During the vedio you have just explained the proof , not the way how can we find it by our selves ,cause i don't think that there's any one who can think like this , so i would prefer to see you explaining the way of thinking that can lead us to the solution on next vedios , ;-)

[@blackbomber72]

Well, sin(2πx) is 0 whenever x is an integer. So, if you substitute x by e*n!, if indeed e is rational, then n! should cancel out the denominator. So you go on a quest to find the value of the limit, using definitions of e.

[@nafisalzahid2597]

Looking forward to undergraduate mathematics course

[@דףאחדעלמתמטיקה]

Why this video isnt a part of tge overkill series? Its should be

[@Pklrs]

@@angelmendez-rivera351 no, there are much simpler proofs but ths is a good one as well

[@wesleydeng71]

@@angelmendez-rivera351 I'd say Fourier's proof is simpler.

[@wesleydeng71]

@@angelmendez-rivera351 Yeah, don't think this is an overkill. A rather smart proof.

[@דףאחדעלמתמטיקה]

@@angelmendez-rivera351 you are wrong. Fourier's proof is much easier and dont need any lemmas at all

[@maxime_weill]

@@angelmendez-rivera351 see this proof i copied from the comments : "Suppose e=p/q, then we have p/q=sum 1/n!; multiply both sides by q!, then we have p*(q-1)! = sum q!/n! . The LHS is integer, but the RHS consists of summands where nq. First part is integer, but the second is 2 it is impossible to have an equality." i think you have to be insane to think this proof is more complicated than the one in the video.

[@edwardjcoad]

Cant the n!/m! = 1/n+1 directly...and because 1/n+1 tends to 0 for part 1??

[@anthonyserafini]

I understand the limit part, but, just looking at the limit, it should not exist, right? Because sin (n!) is going to diverge between 1 and -1, and n->infinity shouldn't make the limit oscillate between -infinity and infinity?

[@half_pixel]

The difference is that we're looking at sin(2πen!), not sin(n!). That way, we pick out very specific "nice" outputs of sin, that end up being very close to 2π/n. The limit of sin(2πen!) does in fact converge to 0, rather than diverging between -1 and 1 :)

[@ogasdiaz]

That's what we call an elegant proof. I'd love to se a proof of the irrationality of pi as well.

[@brunojani7968]

I believe the pi proof was a lot harder. Would still be interested to see one.

[@sabhrant5533]

How do you prove that e = sum_{m = 0}^{\infty} 1/m! ?

[@azhakabad4229]

Thumbnail brings most of your viewers.

[@usernameisamyth]

Nice proof.

[@mFix09]

What if n

[@luigimarigliano3576]

Since he's taking the limit as n --> ∞, you are only interested in what happens for large values of n. Since q is fixed, you can choose n > q.

[@malawigw]

It is irrelevant since we should take the limit n -> oo so we can always assume that at some point n > q will hold (remember, q is a FIXED natural number).

[@mrl9418]

Bravo

[@elipersky1591]

I choose to see this as proving that if e is rational then pi=0 and you can't stop me.

[@txikitofandango]

great

[@pablojesusmolinaconcha4504]

ok, great. This blew my mind hahahaha

[@faycalraghibi9662]

in the last line of the proof in the video you an indeterminate form of type [infinity*0]

[@connorhorman]

I'd just like to point out that the inequalities in the first two proofs should be

[@connorhorman]

Angel Mendez-Rivera So what your saying is that if a < lim[x->Inf] f(x) < a, then lim[x->Inf]f(x)=a. That violates trichotomy, which states that **exactly one** of x < y. x = y, x> y, is true, so its a contradiction. You can always introduce

[@otakurocklee]

Can you do a video on non-computable real numbers? This topic has always puzzled me.

[@FedeMumble]

Wow

[@jesusalej1]

Good morning Master, could you calculate... int from 0 to 1 of { -ln(1-zy)/zy *zdy }. Called dilogarithm?

[@MichaelPennMath]

I have some videos related to the dilogarithm already. Here is a link: kzbin.info/aero/PL22w63XsKjqwcLlI-HXw15sPytjIJbDwY

[@jesusalej1]

@@MichaelPennMath Thanks a lot!

[@s4623]

If you would just swap the first 2 lemmas/tools you would probably save up to a minute.

[@roopesh732]

2πen -> 2pien -> 2 Penn :D

[@mathadventuress]

What level of math is this 😯

[@anggalol]

Maybe you should start patreon

[@MichaelPennMath]

Thanks for the suggestion, but in my current situation (I have a nice job already), I feel this is not necessary. My goal with the channel is truly to popularize "hard" math. The best thing that you can do to help the channel is to share the videos and help me reach more people.

[@mohdaftab2819]

Are you a professor by profession?

[@maxjackson6616]

I love it cos the maths is awesome. I hate it cos I'd never come up with something this cool

[@anksssssssss]

If n>q , and e=p/q , then function is infinitely x zero . This can have finite ans

[@SlidellRobotics]

That's an epsilon? Looks like a tiny capital sigma.

[@elie.makdissi]

e to i : be real! i to e : be rational!

[@timurpryadilin8830]

i is also irrational, as it cannot be expressed as the ratio of 2 integers. It is rational in the Z[i] though

[@elie.makdissi]

@@timurpryadilin8830 okay nerd, don't take it seriously it's a joke, and show us that you are smart, cz the donkey know the fact that i is also irrational

[@rokus1145]

@@elie.makdissi don't get so defensive lmao, you recycled a cheap joke. Also calling someone a nerd for correcting you is really childish, especially since we're on a maths channel focused on educating and cool proofs

[@elie.makdissi]

@@rokus1145 Dude, how old r u 5? Tell me in your schools, the ones who love maths aren't called nerds? because that's the fact, it's not childish, it's the truth, and you can cry in the corner if you want. And this joke isn't recycled, i've just invented dickhead, because can yiu prove me that i saw the original joke, if there is, no you can't . So i invented this joke and you can't do a shit about it. I think Timur is indian, don't get me wrong, i have nothing for Indians but respect, but they seem to obsessed with maths, they think they are the smartest and they try to prove that by there nerdy comments,and i hate that. So again, i respect and love india but this nerdy smart Indian people must stop. And you kid frank, go in the corner and cry if you want.

[@malawigw]

@@elie.makdissi wtf is your problem?

[@wafizariar8555]

Could you make videos on Galois' cohomology?

[@toddtrimble2555]

That would be funny. Although there is a nice way of characterizing Pythagorean triples using Hilbert's Theorem 90.