e is irrational -- the best proof!!

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2023-11-24に共有
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コメント (21)
  • That's a lovely proof. A concrete example of set theory and abstract definitions. Thanks, Michael!
  • It's not only a very nice proof, but it was also very well presented. I'm neither a mathemtician nor a math student, but I love your videos because you make even complex topics accessible to ordinary people like me. Please keep up the good work!
  • @noahtaul
    I’m not as impressed with this proof as everyone else, since it morphs directly into the standard proof. But what you can glean from this is that every rational number eventually ends up as the left endpoint of one of these intervals, and every irrational number can’t possibly. So you can use this to create a ton of irrational numbers: every number can be written as an integer plus the sum from 2 to infinity of a_i/i! where a_i is between 0 and i-1, and the irrational numbers are those that have a_i not all eventually 0 and not all eventually i-1. This isn’t super enlightening; it’s basically like saying that the irrationals are those that don’t have a periodic decimal expansion. It’s just that this representation captures all possible denominators, not just factors of powers of 10, so we can say a rational number’s expansion must terminate instead of just maybe repeating; and it’s also suited for applying to e because of the series formula for e.
  • @xizar0rg
    I don't know if it's the "best" proof, but it's certainly very digestible, making it a fine aperitif for thanksgiving leftovers.
  • @StevenHodder
    That is literally the most straightforward proof that e is irrational I have ever seen. Well done with the presentation.
  • @landsgevaer
    I like the construction of e from "second subintervals"; that was new to me! 💛 The rest of the proof of irrationality is essentially the same as other proofs I have seen; that I find less special (and much less geometric...).
  • @Nikolas_Davis
    Brilliant! Another way to look at this proof is that we're writing e in this strange mixed base system, where each successive digit represents halves, thirds, fourths, ..., n-ths, etc. And in this system, e has an expansion made up of all 1s (the second interval in each subdivision).
  • @Chalisque
    One detail I had to think for a moment in, is that while s_n needn't be a natural, since 1/k! is divides m! for k <= m, it follows that every term in the sum in the definition of s_n becomes an integer in the sum for m!s_n. So we do have a in the open interval between consecutive integers.
  • @Macieks300
    Very nice proof. The only thing I don't understand is how or even did we use the nested interval theorem? Since we directly proved that e is in the intersection of the nested intervals we didn't need that theorem really.
  • @curtiswfranks
    Really bloody nice. This is my new proof of the irrationality of e. High school students could follow it.
  • @curtiswfranks
    I was actually going to look up such a proof before going to bed just last night. Thanks!
  • @tcmxiyw
    This is so elegant. Summarizing (heavily): there is a set I that has e as an element and that has no rational elements. It is innovative in that it doesn’t follow the expected pattern beginning with “suppose e is rational”. Nice presentation Professor Penn!
  • @Jonasz314
    Cool stuff. I think for clarity, you could start by giving the definition of e which suits this approach. It gets explained much later in the video, and as a result, it's really not clear why we're taking this approach in the beginning. For someone not very well versed in math, it will look like a wild goose chase, where you miraculously end up on the result you wanted. Math can be frustrating for a lot of folks, if they do not see what approach to take in order to tackle a problem.
  • @s4623
    Perhaps include this in the next iteration of your real analysis course because it is a very nice application of the nested interval from there.
  • @WolfgangFeist
    Yes. Very lovely proof. It does not only show e is irrational - it make one understand WHY e has to be irrational. By the very definition of e.
  • @Happy_Abe
    Now for a proof it’s transcendental
  • Shouldn't one explain at 19:45 why m! times s_m is a natural number? It only needs a short explanation, but since this is a crucial point in the proof, I think one shouldn't omit it.
  • @woody442
    The construction of e in this way is a damn pretty visual representation of evaluating the powerseries of the exponential function at x=1. Personally found this even more interesting than the actual proof of irrationality. I've never realised that the sum over reciprocals of factorials implied this. Thanks!