e is transcendental -- the best proof! 

It is a shame Michael needs sponsorship imhp

How would π pe proven similarly to this? I can't think of any integral which has π related to it like this?

@@anshumanagrawal346 Had it been obvious, Hermite would have found it! The connection with e is Euler's identity e^{i\pi} = - 1. Lindemann proved that, if a is an algebraic nonzero number, then e^a is transcendental, which implies that i\pi and thus \pi is transcendental. But yes, Lindemann's theorem is trickier to prove than Hermite's. But the integrals are the same as is the arithmetic-analytic conclusion by contradiction. Lindemann used no other tool than those certainly known to Hermite.

@@ahoj7720 Did hermite know about this e^a thing?

@@anshumanagrawal346 Obviously not, this is Lindemann's theorem and the transcendence of \pi is an immediate consequence of it. Hermite was well aware of Euler's formula for sure!

@@ahoj7720 It's just beautiful. If I got you right, here, Lindemann has extended the definition of algebraic numbers to the complex numbers, showing that e^z is transcendental if z is algebraic. Am I right?

Thanks :-)

Right-I think you have to use the fact that the M sum is an integer, regardless of p. Otherwise the M sum also might tend to 0

Alternatively (with less words and more limits), from (a_n M_n + ... + a_0 M) + (a_n eps_n + ... + a_1 eps_1) = 0 and using lim_{p→∞} eps_k = 0 for all 1 ≤ k ≤ p, taking limits on both sides we have lim_{p→∞}[(a_n M_n + ... + a_0 M) + (a_n eps_n + ... + a_1 eps_1)] = lim_{p→∞} 0, ⇒ lim_{p→∞}[a_n M_n + ... + a_0 M] + lim_{p→∞}[a_n eps_n + ... + a_1 eps_1] = 0, ⇒ lim_{p→∞}[a_n M_n + ... + a_0 M] = 0, which contradicts the fact that a_n M_n + ... + a_0 M ≠ 0 (for all p).

@@schweinmachtbree1013 I think it’s a bit more complicated than that. We should use the fact that it’s an integer bigger than 1 so it can never tend toward 0. The fact that a limit of a sequence is 0 and the sequence only has non-zero terms is not a contradiction in itself. For example, 1/n tends toward 0 when n tends to +∞ however 1/n is never equal to 0 for all n in ℕ*

Nice!

Thanks ! this is the claim that I miss.

Thank you, I was very confused by this step.

I noticed them too. Clearly he wanted us to remain attentive, that's why he threw a few errors in there so we could go back and double check. Also I found that he went pretty quickly on the fact that the d_j sum started at 1 instead of 0, because by construction d_0 = 0 since the polynomial contains a factor u. This is basically what makes M_k divisible by p while M isn't, pretty much the heart of the proof. I know he said all of it, but it went pretty fast, I had to pause the video to really understand that the sum started at 1 and that it wasn't just another error.

And pi=3 so e=pi

@@WorthlessWinner and e^pi = 10 so 10 = 27

as in not a single solvable?

Simplest doesn't mean shortest

You could prove it very fast probably by invoking some random pure math master level theorem. But this would be worse, wouldn't it be?

@@Noam_.Menashe I mean if you can prove it in 5 minutes with theorems from different fields of mathematics, I'd call that better

@@thatkindcoder7510 if you allow "theorems from different fields of mathematics" then just state lindemann weierstrass theorem and it's done in less than 10 seconds lol

@@okokok4321 How about proving that there're infinitely many primes with the PNT (if the Prime number theorem doesn't depend on having infinitely many primes)

I don't think it's hard, just plain geeky! it has no taste to it, it is a slog. The result is achieved but it looks rather that it was stepped in rather than derived.

Mathematics is often compared to science; however it is not just hard science, some of its "invention" is philosophical. That's what lets a mathematician question fundamental propositions, such as whether math is invented or discovered.

@@howwitty Mathematics actually is Art

@@howwitty I mean how do they even construct the proof. I try to prove a trivial statement for homework, and it takes me hours, and the proof is often simple. I'm just really envious of people that can go "First, let's use these variables and identities I pulled out from thin air, then prove through contradiction that the statement is false"

@@thatkindcoder7510 You have to account for the fact that what you're seeing is the polished final product. Whoever came up with the first version of this proof (apparently Hermite according to someone in this comments section) will have spent a long time coming up with this argument and figuring out why it should work and working through the details to make it work. However when finished proofs are presented a lot of the motivation and intuition for the proof isn't mentioned (mostly because these aren't needed for the correctness of the proof, and including them would often make the proof much longer).

@@thatkindcoder7510 They can't do that. What you have seen is a tidied up in hind sight approach. If Mathematics is an art then this proof was derived from joining up the dots in an impressionist nightmare such as one might have in a cave. :-)

True for quadratic, assume true for P_n(e)=0 then inductively for P_n+1(e)=0 then bingo, but would not know how to do that!

please do tell! who's the professor?

squarespace stealing our maths

unless you hang out with constructivists...

@@kumoyuki Very good point. Constructivist math is actually a branch I'd like to study a little more, in my copious free time.

@@kumoyuki Ha ha, yes. But even your everyday classical mathematician will usually prefer a direct proof if one is available. When was the last time you heard a claim that a certain diagram commutes, and the first step of the proof was "suppose it didn't". 😀

You are correct. That follows both from the observation on the lower left and the upper bound on the absolute value of ε_k (assuming they aren't 0). I also stumbled over this. Luckily, it is never used that ε_k is an integer, only that the M and M_k are. For those, it was indeed proven in the video.

Never mind, he keeps this restriction at 18:50

Yeah he wasn't super clear with this - at 21:30 he says "but it is an integer" as sort of an off-hand comment, when really it's one of the most important observations in the proof. Since the first thing is always an integer, it can't converge to 0 unless it eventually equals 0 exactly.

Also right after this he says "note that each of these things are integers by construction," which is false - the epsilon_k terms are definitely not integers. But the fact that M and M_k are integers can be seen from the expressions he eventually finds for them (note that there are no denominators)

Good question. It's because the M-sum is an integer (so being non-zero means it has absolute value at least 1).

Yeah, at 26:58 when he states that a_n M_n + ... + a1 M1 + a0 M = 0 that is outright incorrect. a_n ε_n + ... + a1 ε1 tends to zero as p tends to infinity (and therefore so does a_n M_n + ... + a1 M1 + a0 M), but neither even actually equals zero. I think the important point is that there is some sufficiently large p where a_n ε_n + ... + a1 ε1 becomes less than 1 and so it is impossible for a_n M_n + ... + a1 M1 + a0 M to be a positive integer I think I would find things a bit clearer if M, M_k, and ε_k were also subscripted by p. This might clarify what is going on near the end. Also a reminder at the top of the board as to what is or isn't an integer.

@@japanada11 Thanks! I hadn't realized. That makes it clear, then, since the M's and the coefficients of the polynomial are integers (but the ε_k aren't).

lol, just think about it! 2 is root of x-2=0, therefore, 2 is algebraic. But wait a second! If 2 is transcendental, x-2 is not a polynomial in integers, and we cannot use that! ;)

just the ordinary binomial theorem - he's using (a+b)^p = a^p + ... + b^p (where the terms in the middle would have binomial coefficients, but we don't care about them) where a = x^n and b = ±n!.

Yes, and you are right: observe the integral - the minimum power of u is p and the maximum power is np+p-1 (well technically it looks like np +n but this is because Michael has made an error when copying down the (u+k)^(p-1) term.) Now in the two sums, if the numerator is (p-1+j)! then we get a sum of terms between (p-1)! and (p-1+np)!, which is what you should get when evaluating the sum of all the gamma function integrals. If you use (p+1-j)! instead you get a sum of decreasing factorials between (p+1)! and (p+1-np)! which not only doesn't match the integral the last one might be the factorial of a negative number.

@@StanleyDevastating Thanks for your answer! You're absolutely right about the minimum and maximum powers of u. The only detail I think you missed in your explanation (although it is still valid) is that if the numerator is (p-1+j)! then the first factorial in the sum is actually p! and not (p-1)!, because the initial value for j is 1 and not 0. Likewise, when using the (p+1-j)! numerator instead, the first factorial in the sum is also p! and not (p+1)!, for the same reason. Other than that, it was well observed! The point you made on the last factorial when using the incorrect numerator is also quite interesting. To elaborate on that a little, I think that only when n=1 do we get the factorial of a positive number for (p+1-np)!, which in this case becomes (p+1-1*p)! = 1!. Otherwise, it becomes the factorial of a negative number. This happens because in this case n>=2, so p>=3 (because p>n and p is prime, as defined in 8:39) and therefore p+1-np

@@navilistener Ah yes thanks for that correction, I added that p-1 power back by accident! I think Michael has just made a sign error when writing his notes. I see from your work that it nearly every prime number produces negative factorials if we write (p+1-j)!. But the proof would not be valid with a negative factorial because integrals of the form: int_0^(infinity) x^k * e^(-x) dx are only gamma functions equal to (k+1)! when k> -1. You can't just feed it a negative power and have it spit out some "negative factorial" in your proof. So a negative factorial suddenly appearing without explanation would be problematic! Maybe e isn't transcendental after all!!! 😲

Use exactly the same steps as before when investigating the integral M.

But what is D_j?

@@Lucashallal the coefficients of the np-degree polynomial in u you get by expanding out all the brackets. We only care that integrating u^t*e^(-u) gives t! ; and so use d_j to throw away constants whose exact value doesn't matter.

Ohhhh thanks

Wait but there is still the (u+k)^p-1 (which he wrote wrong). If it was just u^p-1 i would understand, but now i am still confused

What do you mean?

What exactly are you proposing to construct? If you mean a proof not by contradiction then the only way to proceed would be to show that for every integer polynomial, e is not a root of it. This is not hard for linear and quadratic polynomials, but gets ugly as the degree increases, and from degree 5 onwards there don't even exist algebraic formulas for the roots of a general polynomial.

@@japanada11 without using the law of excluded middle

@@ayernee I guess the comment below mine makes it more clear, but I'll rephrase: how do you even define "transcendental" without the law of excluded middle, let alone prove that something is transcendental?

Every proof by contradiction (which itself doesn't use existence theorems inside) can be made into a direct proof by starting from the bottom and using: A -> B = not B -> not A Don't quote me I'm not a logician.

You also need the integral \int_0^\inf x^t y^(-x) dx to be an integer, and that only happens when -y = e^k, for any positive integer k.- Edit: missed a minus sign in my computation, in fact it only happens when y = e.

Apparently you have mixed up some things... The integral which is used is equal to Gamma(j+p) and hence equal to (j+p-1)! On the other hand, Gamma(j+p-1) = (j+p-2)!, not (j+p)!

3:14

Taking myself seriously, the proof would consist of showing that e is between (but not equal to) 2 and 3, yet there are no integers in that open interval

i love all the tensor algebra stuff too