e is irrational -- the best proof!! 

Well said

How can you say it's a very nice proof then?

@@Sir_Isaac_Newton_ maybe because there are a lot of degrees that require taking higher math courses?

@@Sir_Isaac_Newton_ The very same way I can say "Spem in Alium" is a very nice motet, or that "Regatta at Argenteuil" is a very nice painting. What confuses you about that? Are you suggesting that I'm incapable of appreciating elegance and beauty wherever it's found?

I would say that compared to many other proofs that e is irrational, its actually a very nice proof. Most proofs that e is irrational, which I have seen, rely on much more esoteric and unintuitive theorems then the nested interval theorem.

yeah you're too good at math, we get it

What other proof is better?

​@@El0meletteIf someone claims the supremum of a set (e.g., "this is the best"), the burden is not on others to prove or disprove the claim. I did not say that there is a better proof, I pointed out that the video title (as with any of his click-bait titles, and they have virtually all devolved to click-bait titles) makes an unsubstantiated claim which he does not prove.

@@xizar0rg Ok, but do you know another one? I thought from your comment that you know a better one and I was really interested, needless to say how subjective the "best" is.

​ @El0melette Taking your reply at face value, I'll have to think a bit. I don't have a favorite proof of e's irrationality (though I do appreciate how much easier it is to prove for e than for pi). If all you want is other proofs without my opinions, there's a page on wikipedia, as well as mathworld (or whatever wofram's co-opted it as). Numberphile, of course, has a video for it. Polster (mathologer) also has one. Off the top of my head, I'm certain they approach it from different vantages, as I recall thinking Penn's demonstration here was unique to my experience (which isn't saying much as I fell off the wagon after my MS and going into public school teaching).

Rather halves, six-ths, ..., n!-ths. I like the idea.

I think it was at the end when we determined that the intersection of the nested intervals and the rationals is the null set

@@soupisfornoobs4081 Where exactly?

@@Macieks300 pretty much the final step of the proof. I don't have a timestamp but it's right before the end

I don’t think it was used

If we didn't take for granted that e exists and is the sum of the series of factorial reciprocals, then the nested interval theorem would have been used to say that there is something in all of them which we would call e. But since we assume we already have e and showed that e is in the intersection, we never needed to use the theorem. I'm guessing Michael just wanted to introduce the audience to a very important theorem that is not as well known. It's also possible that the arxiv paper didn't assume the existence of e and used the theorem to do so.

Bro the video is 21 minutes long. It's only been 11 minutes since it came out

This may not be a good idea because we usually don’t have any limit or exponential function when we develop nested interval theorem from completeness axiom

This is more likely to appear on a number theory course.

@@expl0s10n Perhaps you have Kuert Goedel in mind, citing his Incompleteness. Theorem which states no solution, several solutions, or no closed form, such as for Quadratic, Biquadratic, or Cubic equations. Revisit Galios Proof for X^n+Y^n=Z^n for n>3 impossible to find real numbers. Search recent Paper by Prof Andrew Wiles on this subject.

ok

This explanation is way simpler than the one in the video, even though it is basically equivalent. It is intuitive and avoids that nested intervals stuff. Good job!

You can get a simpler similar proof by showing in the same way that 1/e is irrationnal, which is equivalent. The difference is that, as 1/e =e^-1, the series for 1/e is the obtained from the previous by replacing the + sign by a - sign preceding every term with an odd rank. So you have an alternate series. The simplification is in the majoration of the remainder. You avoid the geometric series because for such series, the remainder's magnitude is less than the magnitude of the first ommitted term, which here is 1/(q+1)!

@@robert.ehrlich8942 That's nice!

Well, e would not belong to any of the other sub-intervals. This is part of the proof, where it is shown that ∑1/m! < e < ∑1/m! + 1/n! (where the sums are computed for m from 0 to n). These inequalities prove that e is always in the second interval.

In fact it's supposed to be the contradiction symbol, which is an arrow pointing right touching an arrow pointing left, so it may look similar to an asterisk.

Exactly. Because primes are infinite, the series defining e cannot be simplified to a finite fraction

@@cbraidotti It has little to do with primes. There are plenty of series of fractions where the partial sums get larger and larger denominators that can't be simplified, but the limit is entirely rational (the sum of 1/(n^2+n) comes to mind). It's more a matter of the series converging "too fast" to be rational.

@@MasterHigure you might want another example because the one you gave simplifies pretty neatly (in fact, that is a way to show its convergence to a rational)

@@cbraidotti The partial sum is (n-1)/n. Which gets a larger and larger denominator, and can't ever be simplified, just like the series of e. But my example series converges to something rational, as opposed to the series of e. You can't use the fact that partial sums get larger and larger denominators as a proof that the end result is rational or irrational. That was the point of my example.

​@@MasterHigure Denominators get larger for all converging series. But in the series for e, new primes are factored into the denominator repeatedly and infinitely, not simplified by the numerator

It's a valid proof and a similar technique would _not_ actually be able to show other numbers are irrational. The key to the proof is in pretty much the last step - showing that no rational number can be in the intersection of I, which fundamentally relies on the series expansion of e. However, the proof that the intersection contains no rationals is pretty much identical to the "standard" proof that e is irrational (Fourier's proof). And this is why I and a couple other commenters haven't been as keen on this proof as the rest of the comments seem to be. In the end, it's just the same as the "standard" proof that e is irrational, but with a ton of extra setup. (And to boot, I'm fairly sure that it doesn't actually use the Nested Interval Theorem at all.) Looking at the paper on arXiv that this proof was taken from, it appears that the point was to give a _geometric_ argument that e is irrational. So the point is not necessarily to be super clear or efficient. It's more to explain e's irrationality _geometrically,_ because there aren't many examples of geometric irrationality arguments (apart from sqrt(2)).

Nested intervals is the most basic property of real numbers. In fact, in many constructions it is one of the axioms.

its one of simplest proof of a fundamental result I have seen, much better than proofs of irrationality of pi.

You're exactly right! Since we have an intersection of nested intervals, if two real numbers a and b are in the intersection, then [a,b] is in the intersection as well, and therefore (a,b) is in the intersection as well. So if some other irrational besides e were in the intersection, then all real numbers between e and that other irrational would be in the intersection, which would necessarily imply a rational was in the intersection, by density of the rationals.

By definition, “Q is dense in R” means that the closure of Q in R is exactly R. The closure of Q in R consists of Q itself along with any real number which may be obtained as the limit of a sequence of rational numbers. But such a limit might not lie in Q. The element of the intersection of the nested intervals can be written as the limit of a sequence with its nth term in the ith interval. If you take the sequence to be a sequence of rational numbers, then as above its limit need not lie in Q.

In any particular interval you can find a rational, that is why Q is dense in R. However, their is no rational, that lies in ALL intervals at once. In fact you can prove, that if intervals get infinitely small, their is always exactly one number that lies in all of them.

Thanks for reminding about this awesome book! This book was recommend to me while ago by my algebra professor. I was simply amazed when I opened it. I highly recommend it as well!

@@cpitrada You're welcome! This book is truly phenomenal. It ranges over many mathematical branches and every proof I've read in this book involved at least one clever idea. And the passion of the authors shines through too!

I agree - the proof that the intersection contains no rationals is pretty much, step by step, identical to Fourier's proof that e is irrational. But from looking at the paper on arXiv, the goal of the proof is to give a _geometric_ argument of irrationality. The paper claims that, while many famous constants have been proven irrational, very few constants have a _geometric_ proof of irrationality. So being able to sort of "narrow in" on e with smaller and smaller intervals that force out every rational number gets at that geometric flavor.

What does “infinite factor in the denominator” even mean? Because if it just mean “e can be written as a sum of infinitely many fractions with distinct integer denominators”, then it doesn’t constitute a proof as there are counterexamples such as 2=1/1+1/2+1/4+1/8+…, where 1,2,4,8, etc. are obviously distinct integers, but 2 is clearly rational. And if you need other properties, how can you be sure that the terms in the series formula for e won’t happen to combine in a way that satisfies those properties?