I think it might be worth making the last step more explicit. That is, since a_n*M_n + ... a_0*M is an integer not equal to 0, for all sufficiently large p, and the "epsilon sum" tends to 0 as p tends to infinity, you can pick some p such that the "epsilon sum" has magnitude less than 1, which contradicts the "M sum" and the "epsilon sum" adding together to give 0.
@@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*
As a 'big picture' to the proof- you start of by assuming the existence of a polynomial such that if you plug e in you get 0. You then rewrite e using some funky integrals, and then plug that rewrite version of e in. Quite spookily, you find that this thing isn't 0, which is a contradiction.
This is Hermite’s proof. It is interesting to note that he thought that the transcendence of pi was way more difficult to prouve. But Lindemann proved it using essentially the same tricks. If Hermite had had more confidence in his method, he would have got pi himself.
@@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.
@@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?
That was a beautiful proof. The moment you factorised the expression and said that part of it didn’t equal zero but the whole must, I realised where it was going, and watching the final few minutes lock into the contradiction was wonderful.
There are a couple of errors @17:00. The (u+k) term in the integral should be to the (p-1) power not the p power. On the next line, we need (p-1+j)! in the numerator of the sum, not (p+1-j)!.
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.
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.
@@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)
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 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. :-)
This is not the conceptually simplest proof, it is Hilbert's unpacking of Hermite's proof. The integrals involved are completely unmotivated here, they are the integrals to extract Pade approximants to exp(x).
it is interesting that at the heart of this proof lies the fact that the exponential function dominates the linear functions, in the sense that lim (x^n/e^x)=0 for all n. This comes up in the computation of M, M_n via the integration by parts formula.
the last step needs to be more refined, the second sum tends to zero is a better statement to make because we also know that the first sum is not only non zero but also an integer so it absolute value is greater than 1 which means the second sum can never cancel it out.
26:45. I'm not sure I completely understand how this argument work, if it's to be made rigorous, so let me rephrase it to make sure I get it: We've got a_n M_n + ... + a1 M1 + a0 M, *dependent on p*, that is NOT = 0. Then, we've got a_n ε_n + ... + a1 ε1, *also* dependent on p, that *tends to 0* as p -> inf. So, the argument should go like this: since their sum *always* is *exactly* zero, and the 2nd can become *as close to zero* as we like, the 1st one *should be exactly zero*. Well, how do we know the 1st one (the M-sum) doesn't *also* tend to 0 as p -> inf? Wouldn't that invalidate our reasoning?
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)
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).
I wouldn’t even understand how someone can arrived at this steps. In the beginning I was completely lost and it began to struck me near the end. So perhaps it would have been easier if we walk it thru backwards 😂
At 21:35 "each of these three things are integers by our construction" is false: M and Mk are integer but Ek is not (if Ek were integer, then the sum Mk+Ek would also be integer, which would imply that M*e^k is integer, thus e^k is rational, which is not). You can also check this with a counterexample: p=5, n=4, k=3 give Ek = 368.92 approx. In the end the contradiction arises because you have built a sequence Ap + Bp = 0 with Bp going to 0 as p goes to infinity (this is not a contradiction) which implies that Ap also goes to 0 - and this is a contradiction because Ap is always integer and different from 0.
One way of easing your way into this proof might be to show that e cannot be the root of a quadratic equation with integer coefficients. That way, you can introduce strategies of a less familiar nature.
The general idea is to assume e is algebraic of degree n and find a prime p larger than both n and the constant term of the minimal polynomial while also being large enough to force an integer not divisible by p to have absolute value less than 1.
In the expressions for M and M_k, at the bottom of the summing sign, you had n=0 and n=1. Those should have said j=0 and j=1 for those summations, respectively.
It's undeniable that mathematicians really do like to work BWOC. Let's do it BWOC. BWOC is the way to go. When you hear BWOC BWOC BWOC, a flock of mathematicians is the image that reflexively pops into your mind.
@@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". 😀
If instead of letting the integrals go to infinity we let them go to a very large integer, would the value then be a rational number? It would seem that if you assume that there is some last integer B for the infinite series Σ1/n! For N=0 to B, there will be B+1 terms that would need to be added together. But if there are B+1 terms to be added together then B cannot be the last integer, because B+1 would also be an integer. However if B were the last integer there would only be finitely many terms to be added together. And you would only get a rational answer if there are finitely many terms to be added together. However then you also do not have 1 specific value for e rather an infinite number of different finite values of e. So if e is defined to be an infinite series, but is only rational when a finite number of terms are added together. It would seem to be a contradiction.
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.
Interesting proof, but what I didn't like was that it was based on M, Mk, and Epsilonk which, for me, came without any motivation, but just given. Would be interesting to see where those came from.
In 18:30, the numerator is (p+1-j)!, but I obtained (j+p-1)!. The conclusion that M_k is divisible by p also follows from my result, so everything's good though. Has anyone else reached the same result?
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!!! 😲
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! ;)
I think you really rushed through the section around 18:30 because you knew how the argument went, and what came out on the board had at least one error and I think a couple more. If you just had said "this works the same as for M except we've made sure the constant term is 0", I would have understood right away, but as it was I had to flip back and forth a bunch of times to figure out what you meant there since it didn't resemble what we had done with M precisely enough for me to make the connection. I also find that this proof in general doesn't have enough motivation in terms of how you're using the properties of the number e. You end up adding it to an integral and then saying the integral is known to be the gamma function which is equivalent to the factorial, but that's the only place in the entire proof that the nature of the number e is used, and it's really hidden.
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)!
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.
@@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.
There's just one thing I don't get in these proofs. Okay, you proved e is a transcendental number BWOC, but I don't understand how e is what makes the difference, and what would change of the whole statement was made for another number like 7. The ONLY place I can see the difference is when he evaluated the integral e^-x dx = 1 but how does that make a difference?
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.
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!.
@@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.
Bought that Calc book when (I think it was your vid) used it as a reference for the definition of cosine and the proof of derivative of cos without using sinx/x limit.
1:25 As a Norwegian, I am offended. The insolubility of the quintic is due to Abel (as in abelian group, the name is pronounced something like "aah-bell"), and some times also attributed to Ruffini. Not Galois. Now that I got that out of my system, this is a cool proof. Always fun to see proofs of those facts you've heard so many times you take them for granted.
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 really like your videos, but you have a number of mistakes when writing on the board (and sometimes repeated when talking), sometimes you correct them but some, like the mistakes at 17:00 ((u+k)^p should be (u+k)^(p-1)) and at 18:20 ((p+1-j)! should be (p+j-1)!) you never correct. Unfortunately these mistakes make your argument quite hard to follow, you can't expect your viewers to catch all of the mistakes themselves. I've noticed similar mistakes in other videos as well. You really should take some time to double check your videos for mistakes like these and correct them before uploading.
Always nice to see hardworking people like professor Penn get sponsorships. Massively deserved sir :)
It is a shame Michael needs sponsorship imhp
I think it might be worth making the last step more explicit. That is, since a_n*M_n + ... a_0*M is an integer not equal to 0, for all sufficiently large p, and the "epsilon sum" tends to 0 as p tends to infinity, you can pick some p such that the "epsilon sum" has magnitude less than 1, which contradicts the "M sum" and the "epsilon sum" adding together to give 0.
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.
As a 'big picture' to the proof- you start of by assuming the existence of a polynomial such that if you plug e in you get 0. You then rewrite e using some funky integrals, and then plug that rewrite version of e in. Quite spookily, you find that this thing isn't 0, which is a contradiction.
Thanks :-)
This is Hermite’s proof. It is interesting to note that he thought that the transcendence of pi was way more difficult to prouve. But Lindemann proved it using essentially the same tricks. If Hermite had had more confidence in his method, he would have got pi himself.
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?
You're wrong. e=3
And pi=3 so e=pi
@@WorthlessWinner and e^pi = 10 so 10 = 27
That was a beautiful proof. The moment you factorised the expression and said that part of it didn’t equal zero but the whole must, I realised where it was going, and watching the final few minutes lock into the contradiction was wonderful.
There are a couple of errors @17:00. The (u+k) term in the integral should be to the (p-1) power not the p power. On the next line, we need (p-1+j)! in the numerator of the sum, not (p+1-j)!.
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.
27:24
Thumbnail: "The Simplest Proof"
Video Length: 27:25
I was just kidding, and thanks for the great video as always.
Great proof!
Spivak too is my favorite Calculus book.
Thank you, professor and congratulations on the sponsorship.
1:36 I checked, the Galois group of that 9th-degree polynomial is indeed S₉ (not solvable).
as in not a single solvable?
Sooo beautiful. I really wanted this video thank you!!!!!!!!
e is transcendental number proof - panik
the simplest proof - kalm
27 minutes long - panik
Hard stuff, but interestingly presented. Thank you Prof. Penn!
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.
Simplest proof be like 27mins and 10mins of prepared board
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)
How on earth does one even begin to invent this proof
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. :-)
wonderful proof
this proof is pure technique, hard work and calculations, I always wondered how did they come up with the expressions M, M_k and E_k to begin with
This is not the conceptually simplest proof, it is Hilbert's unpacking of Hermite's proof. The integrals involved are completely unmotivated here, they are the integrals to extract Pade approximants to exp(x).
it is interesting that at the heart of this proof lies the fact that the exponential function dominates the linear functions, in the sense that lim (x^n/e^x)=0 for all n. This comes up in the computation of M, M_n via the integration by parts formula.
Fantastic video Michael. Perfect to watch alongside Ch. 21 in Spivak.
the last step needs to be more refined, the second sum tends to zero is a better statement to make because we also know that the first sum is not only non zero but also an integer so it absolute value is greater than 1 which means the second sum can never cancel it out.
26:45. I'm not sure I completely understand how this argument work, if it's to be made rigorous, so let me rephrase it to make sure I get it:
We've got
a_n M_n + ... + a1 M1 + a0 M,
*dependent on p*, that is NOT = 0.
Then, we've got
a_n ε_n + ... + a1 ε1, *also* dependent on p, that *tends to 0* as p -> inf.
So, the argument should go like this: since their sum *always* is *exactly* zero, and the 2nd can become *as close to zero* as we like, the 1st one *should be exactly zero*.
Well, how do we know the 1st one (the M-sum) doesn't *also* tend to 0 as p -> inf? Wouldn't that invalidate our reasoning?
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).
The cinematographic purist in me would prefer black bars on both top and bottom centering the matted wide screen. (As opposed to just at the top)
I wouldn’t even understand how someone can arrived at this steps.
In the beginning I was completely lost and it began to struck me near the end. So perhaps it would have been easier if we walk it thru backwards 😂
At 21:35 "each of these three things are integers by our construction" is false: M and Mk are integer but Ek is not (if Ek were integer, then the sum Mk+Ek would also be integer, which would imply that M*e^k is integer, thus e^k is rational, which is not). You can also check this with a counterexample: p=5, n=4, k=3 give Ek = 368.92 approx.
In the end the contradiction arises because you have built a sequence Ap + Bp = 0 with Bp going to 0 as p goes to infinity (this is not a contradiction) which implies that Ap also goes to 0 - and this is a contradiction because Ap is always integer and different from 0.
One way of easing your way into this proof might be to show that e cannot
be the root of a quadratic equation with integer coefficients. That way, you
can introduce strategies of a less familiar nature.
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!
Enlightning! Thank you!😍
The general idea is to assume e is algebraic of degree n and find a prime p larger than both n and the constant term of the minimal polynomial while also being large enough to force an integer not divisible by p to have absolute value less than 1.
16:33. Why is 1
Never mind, he keeps this restriction at 18:50
Just got Michael's Squarespace sponsorship segment interrupted by a YT ad 😅 - and now back to math!
In the expressions for M and M_k, at the bottom of the summing sign, you had n=0 and n=1. Those should have said j=0 and j=1 for those summations, respectively.
It's an overstatement to say its the best proof
Spivak's Calculus was the text used for my honors calculus I class at Purdue in 1987, but we used the Second Edition.
It's undeniable that mathematicians really do like to work BWOC. Let's do it BWOC. BWOC is the way to go. When you hear BWOC BWOC BWOC, a flock of mathematicians is the image that reflexively pops into your mind.
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". 😀
That is an amazing piece of art!
I have the Spivak book. Another favorite is Apostol
Ok, great!
I should reconcile with inequalities because they allow you to find a lot of stuff.
Excellent exposition
If instead of letting the integrals go to infinity we let them go to a very large integer, would the value then be a rational number?
It would seem that if you assume that there is some last integer B for the infinite series Σ1/n! For N=0 to B, there will be B+1 terms that would need to be added together. But if there are B+1 terms to be added together then B cannot be the last integer, because B+1 would also be an integer.
However if B were the last integer there would only be finitely many terms to be added together. And you would only get a rational answer if there are finitely many terms to be added together. However then you also do not have 1 specific value for e rather an infinite number of different finite values of e.
So if e is defined to be an infinite series, but is only rational when a finite number of terms are added together. It would seem to be a contradiction.
21:40 "Each of these three things are integer by our construction". This clearly contradicts to written observation if k is not equal to zero
Finally! I am going to use this proof to proove that 2 is not a transcendental number
I knew I recognized the professor whose site was “stuck in the 90’s”. I’d be surprised if he didn’t find that bit hilarious
please do tell! who's the professor?
21:36 Sorry but I don't get why epsilon_k will be an integer, from the obs it doesn't seem like that it will be the case. Am I right?
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.
Interesting proof, but what I didn't like was that it was based on M, Mk, and Epsilonk which, for me, came without any motivation, but just given. Would be interesting to see where those came from.
In 18:30, the numerator is (p+1-j)!, but I obtained (j+p-1)!. The conclusion that M_k is divisible by p also follows from my result, so everything's good though. Has anyone else reached the same result?
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!!!
😲
Am I the only one who noticed the e^(-x) disappeared on the first board after the sponsorship?
squarespace stealing our maths
Beautiful.
Awesome video!
Shouldn't it be (j+p-2)! because Gamma(n+1)=n!
I didn't really followed every details because a little bit too hard for me but... crazy proof for e !
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! ;)
14:52 I do not understand how this transformation came to be. I understand that there's a typo and the sum is on J, not on N; but still i'm stuck.
What an interesting proof!
WHAT IS COMPONENT NUMBER
Since this is the simplest, can you do a very complex one? Not only because it would be a nice flex, but I'm genuinely curious.
neat
I think you really rushed through the section around 18:30 because you knew how the argument went, and what came out on the board had at least one error and I think a couple more. If you just had said "this works the same as for M except we've made sure the constant term is 0", I would have understood right away, but as it was I had to flip back and forth a bunch of times to figure out what you meant there since it didn't resemble what we had done with M precisely enough for me to make the connection.
I also find that this proof in general doesn't have enough motivation in terms of how you're using the properties of the number e. You end up adding it to an integral and then saying the integral is known to be the gamma function which is equivalent to the factorial, but that's the only place in the entire proof that the nature of the number e is used, and it's really hidden.
13:18 But Gamma(j+p-1) = (j+p-1+1)! = (j+p)! not (j+p-1)! . And then the whole argument doesn't work like that
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)!
is there a constructive proof of this result?
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.
Is mathematics explaining the universe, or are the equations simply getting larger and larger with more and more x, y, and zzzzzzzz's?
I think that this sponsor is cool and makes sense
Don't take it personal, but there is a hell lot of notational errors on the page at 18:50 :P
There's just one thing I don't get in these proofs. Okay, you proved e is a transcendental number BWOC, but I don't understand how e is what makes the difference, and what would change of the whole statement was made for another number like 7. The ONLY place I can see the difference is when he evaluated the integral e^-x dx = 1 but how does that make a difference?
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.
what rule he used to expand x^n....+n! to the p??? some newtons binomial extention or what?
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!.
17:50 can someone explain that in more detail?
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
Bought that Calc book when (I think it was your vid) used it as a reference for the definition of cosine and the proof of derivative of cos without using sinx/x limit.
Too hard to understand....
> the simplest proof
> 20+ minutes
1:25 As a Norwegian, I am offended. The insolubility of the quintic is due to Abel (as in abelian group, the name is pronounced something like "aah-bell"), and some times also attributed to Ruffini. Not Galois.
Now that I got that out of my system, this is a cool proof. Always fun to see proofs of those facts you've heard so many times you take them for granted.
👏👏👏
What
*E*
Simplest proof only 27 minutes
This is a bit much for me, can we start with a proof that e is not an integer? ;^)
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
> Simplest proof
> 27 minutes video
xD
woooo sponsership!
Very good. Thank you so much!
Please linear algebra course
i love all the tensor algebra stuff too
At first sight it seems the proof used by Spivak in his calculus book
3:14
would be bearable if it was spoken normally and not like a housewife of beverly hills
I really like your videos, but you have a number of mistakes when writing on the board (and sometimes repeated when talking), sometimes you correct them but some, like the mistakes at 17:00 ((u+k)^p should be (u+k)^(p-1)) and at 18:20 ((p+1-j)! should be (p+j-1)!) you never correct. Unfortunately these mistakes make your argument quite hard to follow, you can't expect your viewers to catch all of the mistakes themselves. I've noticed similar mistakes in other videos as well. You really should take some time to double check your videos for mistakes like these and correct them before uploading.
GHULAAM HAIN TUMLOG...