Hermite's Proof that e is Transcendental (A Thorough Explanation!)

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  • Published on Aug 4, 2018
  • * Prerequisites include Calculus I and the knowledge of basic definition of algebraic numbers. Of course, more experience in or stronger intuition over mathematics always helps.
    Although I tried my best to elucidate each step as clearly as possible, I cannot hope to adequately cover every tiny detail in such a magnificent proof. If you have any questions or notice any errors, I would appreciate it if you could comment it below.
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Comments • 58

  • LetsSolveMathProblems

    For a much easier proof that e is irrational: usclip.net/video/K2Y05rHJ7tM/video.html
    *I point out that we did prove the irrationality of e in this video by proving that it is transcendental.

  • Edward Szla
    Edward Szla 5 months ago

    Please do more number theory videos, this is so interesting!!!

  • wat am i doing?
    wat am i doing? 7 months ago

    This is amazing!
    How does one even start a proof like this?
    Do they consider special properties of e? Maybe think how would a trascendental number would behave?

  • saitaro
    saitaro Year ago +4

    Hermite on this photo looks very unhappy about the e's transcendence.

  • Martin Epstein
    Martin Epstein Year ago +2

    Aren't the M and A unnecessary? The maximum of value of e^-x when 0 < x < n is just 1, and the maximum value of x on that interval is just n. No need to introduce new letters.

    • Martin Epstein
      Martin Epstein Year ago +1

      LetsSolveMathProblems Thinking about it more an upside of not using specific values for the max and min is that doing so can give the impression that they are essential to the proof, when really all that matters is that some upper bound exists.
      Btw I had a fun time re-expressing your reasoning about the right side of the inequality with Sigma notation. It was doable after I abbreviated the unimportant parts of the polynomials with q(x) and remembered the binomial theorem for derivatives.

    • LetsSolveMathProblems
      LetsSolveMathProblems  Year ago +1

      You are right. I glossed over the simplification you mentioned just to save a minute of explanation, but it probably looks like an unnecessary overcomplication for more astute viewers.

  • Alejandro Duque
    Alejandro Duque Year ago +1

    Why P has to be prime? I think i missed the reason

    • Alejandro Duque
      Alejandro Duque Year ago +2

      LetsSolveMathProblems oh okok i got It.
      Pd: you explain really well, im Happy i found this Channel while studying. Keep the good work :)

    • LetsSolveMathProblems
      LetsSolveMathProblems  Year ago +3

      We have to make sure p is not in the prime factorization of (-1)^p*(-2)^p*...*(-n)^p. If p is prime, p > n implies what we want. However, if p is not prime, p > n is insufficient. Consider p = 4 > n = 2. Then, (-1)^4*(-2)^4 is divisible by p = 4 because p divides 2^4. In summary, we need p to be prime because prime numbers cannot be broken down into smaller integer factors that may appear in (-1)^p*(-2)^p*...*(-n)^p.

  • ahoj7720
    ahoj7720 Year ago

    To add to my previous comment, I refer to an article on planetmath.org, which gives concise yet complete and (almost) elementary proofs of Hermite and Lindemann theorems as well as Lindemann-Weierstrass theorem, which generalizes both. Prerequisites are elementary calculus, some notions on algebraic numbers (they are closed under addition and multiplication, their minimal polynomial and their conjugates) and the "fundamental theorem on symmetric polynomials".
    archive.is/ruQFk
    Lots of corollaries can be derived from Lindemann-Weierstrass. Unfortunately, nothing is known about e+Pi or e*Pi, except that at least one of them is transcendental...

  • Walt S
    Walt S Year ago +1

    Great pair of videos on e!

  • Gaurav Goel
    Gaurav Goel Year ago +1

    Wow! I loved this. Thanks, man.

  • Albert Einstein
    Albert Einstein Year ago +3

    Hey ,could you recommend me good books for Amc10/12 ,Aime ,USAMO and stuff like that? :D Maybe AoPS Vol 1/2?

    • LetsSolveMathProblems
      LetsSolveMathProblems  Year ago +5

      AoPS Volume 1 and 2 are both outstanding. I would first recommend those two books to any high schoolers new to problem solving. Paul Zeitz's Art and Craft of Problem Solving also contains superbly explained problem solving strategies and tactics. For USAMO, I have heard Engel's Problem Solving Strategies is excellent.

  • James Wilson
    James Wilson Year ago +1

    You are awesome for doing this!

  • Tom Yao
    Tom Yao Year ago +1

    very nice.

  • Michal Dvořák
    Michal Dvořák Year ago +2

    Great! I was actually only aware of the proof using mvt. Gotta dislike the part where you start to put in concrete numbers which makes your proof less rigorous.

    • Michal Dvořák
      Michal Dvořák Year ago

      @Andy Arteagamath.stackexchange.com/q/2868170/504245

    • Andy Arteaga
      Andy Arteaga Year ago

      Michal Dvořák Could you provide the proof please?

    • Michal Dvořák
      Michal Dvořák Year ago

      divisbyzero.com/2010/09/28/the-transcendence-of-e/
      There it is, check it out. I can provide the proof of the lemma which is left to the reader in the OP

    • Michal Dvořák
      Michal Dvořák Year ago

      LetsSolveMathProblems, For me it was also quite work to go through the proof itself and make sure that I understand everything. There are some interesting lemmas, which can be elementarily proven and which can be used to prove the second part of the proof, that is, a_kF_k is a non-zero polynomial divisible by p. I can share it if you are interested!

    • LetsSolveMathProblems
      LetsSolveMathProblems  Year ago +4

      I personally thought using a concrete example (p = 3, n = 2) would make the proof much clearer by providing an intuitive point of view, but I apologize if it had an adverse effect. I point out that I still explained everything rigorously. After working with a concrete example, I explained the general case by applying the same logic to f(x). Also, it might be argued that the second part of the proof (where we examine derivatives of f) is simple enough in its essence that a lengthy formal proof is unnecessary. In fact, I have seen a version of this proof that did not even formally explain the second part of the proof at all, simply stating that it is obvious.

  • Adrián Hernández

    What's the program you use for these videos?

  • Portsmouth FC
    Portsmouth FC Year ago +1

    Absolutely wonderful! More proofs please!

  • Fabio Calderón
    Fabio Calderón Year ago +1

    If Rudin was USclipr... Haha. Great explanation!

  • sideral
    sideral Year ago +1

    Beautiful!

  • anonimo aninimus
    anonimo aninimus Year ago +1

    really nice video!

  • just_another_Guy
    just_another_Guy Year ago +3

    Wonderful explanation, even for such a hard proof, i got the main idea, keep it up!

  • Martin Epstein
    Martin Epstein Year ago

    YES

  • Will Mcloughlin
    Will Mcloughlin Year ago +3

    Thanks for the video, please keep making more great content!

  • ImanBudi
    ImanBudi Year ago +1

    now Lindemann's Proof !!

  • leoitshere
    leoitshere Year ago +13

    Interesting that to prove that e is transcendental (algebraic property) you need to apply an analytic property of e.

    • TheGamer583
      TheGamer583 Year ago

      leoitshere in general, e is a very analytic number, as it lies at the heart of analysis and calculus, so to prove that it is in fact not algebraic is a huge stepping stone to connect those two fields in a meaningful way.

    • leoitshere
      leoitshere Year ago

      Well, it is not that simple. Look at the proof of the irrationality of e. For that, you had to use explicitly what e was. Here you use the fact that e^x is the solution to a certain differential equation. I think that is amusing, and you don't have to define e like that. e, as a number, is better known as either the infinite series or the limit. Typically the limit is consider the definition.

    • ein idiot 1
      ein idiot 1 Year ago

      leoitshere That property is the definition of e, so you have to use it to prove something about e.

  • Abidur Rahman
    Abidur Rahman Year ago +37

    Woah, this is the type of proof where I'd forget what I was trying to prove in the first place 😂

  • Abidur Rahman
    Abidur Rahman Year ago +21

    Finally, I get to see a proof for transcendence! 😀

  • ahoj7720
    ahoj7720 Year ago +29

    You explained Hermite's original proof very clearly. It's at times tricky, but on the whole it's elementary, requiring no advanced calculus.
    You mention that it does not extend to Pi. Hermite's proof is from 1873. He made the same remark and added that the proof of the transcendence of Pi would be extremely difficult. Actually, Lindemann proved the transcendence of Pi only 9 years later, following basically the same general idea. In fact he proved that if x is algebraic and nonzero, then e^x is transcendental. As a consequence, Pi is transcendental because e^{i*Pi}=-1. Similarly, if x is algebraic, then ln(x) is transcendental...

    • Aamir Zainulabadeen
      Aamir Zainulabadeen Year ago

      ahoj7720 lovely remarks! Thank you :)

    • Carl Fels
      Carl Fels Year ago +2

      Zusammenarbeit für Erfolg thanks i will read it :)

    • Zusammenarbeit für Erfolg
      Zusammenarbeit für Erfolg Year ago +4

      +Carl Fels
      The algebraic numbers are indeed closed and form a field. There seems to be no intuitive reason why that is, at least in my research, but there are several proofs of this fact, mostly using polynomial properties and linear algebra.
      One proof using linear algebra and matrices: empslocal.ex.ac.uk/people/staff/rjchapma/notes/algn.pdf (On page 3) A bit simplified and small, but the important points are covered.
      Sincerely,
      ZfE

    • Carl Fels
      Carl Fels Year ago +1

      Zusammenarbeit für Erfolg an thanks did not think enough :) but why can a product of algebraic numbers not be trans. Is (algebraic number, +, *) closed or why?

    • Zusammenarbeit für Erfolg
      Zusammenarbeit für Erfolg Year ago +3

      +Carl Fels
      e^(i*Pi) is _not_ transcendental, but algebraic (it's -1). That's the whole point. Because we know that e^x is transcendental if x is algebraic and nonzero, every algebraic value of y = e^x means that x is transcendental.
      Basically: e^something = algebraic => "something" must be transcendental, because if "something" was algebraic, the value of e^something wouldn't be algebraic, but transcendental.
      Now we know that e^(i*Pi) is algebraic, therefore i*Pi must be transcendental. The product of two algebraic numbers does never yield a transcendental number, so either i or Pi must be transcendental. i is an algebraic number because it is the solution to the equation x^2 + 1 = 0. Therefore, Pi must be transcendental.
      If you want me to go into more detail, I'd be happy to do so.
      Sincerely,
      ZfE

  • Quwertyn
    Quwertyn Year ago +4

    29:25

  • Oliver Hees
    Oliver Hees Year ago +7

    May you link to his actual proof in the description?

    • LetsSolveMathProblems
      LetsSolveMathProblems  Year ago +5

      Here is a well-explained proof I found online: www.math.utk.edu/~freire/m400su06/transcendence%20of%20e.pdf
      I include one more website that also includes the proof that pi is transcendental: sixthform.info/maths/files/pitrans.pdf
      *The proof in the video is not exactly the same as the ones from these websites because I modified (or even re-developed) many parts of it to fit my style of explanation or to increase the clarity for the viewers. However, the viewers with more experience may like the less intuitive but more systematic method of relying on taylor expansion in the first website. I used both websites above in the research stage of this video.

  • Two Spoon One Cup

    100th view

  • Problematic(puzzle channel)

    First comment!