302.S2b: Simple Extensions

Поделиться
HTML-код
  • Опубликовано: 17 дек 2024

Комментарии • 20

  • @PunmasterSTP
    @PunmasterSTP 4 месяца назад +1

    I knew the rationals were countable, but didn't even consider the countability of the algebraic numbers before. Very interesting stuff!

    • @selahw2609
      @selahw2609 2 месяца назад +1

      Baby Rudin chapter 2 has a problem about this! Super cool proof. (Unexpected from an analysis book to have such an algebraic result :p )

    • @PunmasterSTP
      @PunmasterSTP 2 месяца назад

      @@selahw2609 Oh that's cool!

  • @osianrhys1257
    @osianrhys1257 2 года назад +4

    This guys is probably the best lecturer I’ve seen on RUclips. 👏🏻

    • @PunmasterSTP
      @PunmasterSTP 4 месяца назад

      I think he's right up there with Professor Dave and Sal Khan, and for me, Matt's puns and wackiness are the icing on the cake!

  • @sabarnahazra5018
    @sabarnahazra5018 6 лет назад +7

    Greetings from India Matthew. Your 20 mins of one video is really worth attending 1 month of algebraic structures class. God bless you

    • @PunmasterSTP
      @PunmasterSTP 4 месяца назад

      How are things going in India, and how did your algebraic structures class go overall?

  • @marievanraemdonck8291
    @marievanraemdonck8291 10 лет назад +9

    Matthew Salomone, your videos are really great!
    little note:
    in this video at 5:23 you say: alfa = Pi^2 satisfy t^2 - Pi = 0
    it must be alfa =sqrt(Pi) satisfy t^2 - Pi = 0
    I think then you can also say Pi is algebraic over R: it satisfies t^2 - Pi^2=0.
    But Pi should be transcendental. So we can only speak about algebraic over Q and not over R.

    • @Jkfgjfgjfkjg
      @Jkfgjfgjfkjg 4 года назад +1

      Your first point is correct. As to your second point, pi is algebraic over R, because it satisfies the polynomial x - pi = 0.

    • @PunmasterSTP
      @PunmasterSTP 4 месяца назад

      I agree that the number alpha was probably supposed to be sqrt(pi), but as far as labelling things as algebraic, why would that only be restricted to Q and not R?

  • @danteng5651
    @danteng5651 10 лет назад +1

    In 12:28, the "this always works" theorem, are we suppose to have that k

    • @Mrpallekuling
      @Mrpallekuling Год назад

      Yes, because otherwise there would be no reason to introduce "k" ("n" would be sufficient). So k less or equal n-1. I hope the professor can confirm.

  • @ich123binsimmernoch
    @ich123binsimmernoch 3 года назад +1

    Thank you so much!

  • @BillShillito
    @BillShillito 10 лет назад +4

    Tiny note ... I think you meant the square root of pi instead of pi squared.
    Great videos though!

    • @stapleman007
      @stapleman007 2 года назад +2

      Who eats square pies anyways.

  • @ouafieddinenaciri3783
    @ouafieddinenaciri3783 10 лет назад

    Your videos are great too Bill Shillito ....

  • @witness6347
    @witness6347 5 месяцев назад

    12:00 you can prove k=n-1 if p is irreducible.

  • @stapleman007
    @stapleman007 2 года назад

    I've never encountered anyone / text that's bothered properly explaining algebraic vs transcendental before. But it did take 1.5 Abstract Algebra to get here, so the 'price of admission' is very steep.

  • @carlacvekla1444
    @carlacvekla1444 4 года назад +1

    xd pi² doesn't satisfy t²-pi=0, +-sqrt(pi) does