L13: Diagonalization, Countability and Uncountability

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  • Опубликовано: 15 ноя 2024

Комментарии • 32

  • @neel_epoch7580
    @neel_epoch7580 5 лет назад +5

    The girl at 1:01:02 said my mind. LOVED IT!! Gonna see if I am now satisfied by the answer.

  • @viveksen3247
    @viveksen3247 8 лет назад

    This guy's INCREDIBLY helpful. Thank you.

  • @JohnDoe-kh3hy
    @JohnDoe-kh3hy Год назад

    what is the name of this class/subject? where can I find this playlist?

  • @aydinahmadli7005
    @aydinahmadli7005 4 года назад

    Great lecturer! thanks a lot, funny and educative at the same time :)

  • @laveenabachani
    @laveenabachani 7 лет назад

    professor answered the last student problem really well..

  • @owlattack1971
    @owlattack1971 11 лет назад +1

    This guy is cool, very down to earth.

  • @colinjava8447
    @colinjava8447 2 года назад

    Not sure about finite sets being countable, but its just semantics really.
    Although having just said that, there is no 1-1 correspondence from {1,2,3} to {1,2,3,...}
    So why is {1,2,3} classed as countable?

  • @hannahmorgan573
    @hannahmorgan573 9 лет назад

    Serious question: what was the reference 8:20?

  • @warnford
    @warnford 6 лет назад

    excellent lecture on diagonalisation

  • @kbstudios8402
    @kbstudios8402 5 лет назад

    Well, I surely won't pronounce Diagon Alley wrong after watching this lecture! :p

  • @olavisjo
    @olavisjo 10 лет назад

    At 51:00 , did he call pi a rational number?

    • @chriswatts5921
      @chriswatts5921 10 лет назад +1

      ℝ is for Real numbers, ℚ is Rational (quotient)

    • @colinjava8447
      @colinjava8447 2 года назад

      Slip of the tongue, that's all, he knows pi is not rational.

  • @raahilbadiani7289
    @raahilbadiani7289 3 года назад

    Answer to why we can't use diagonalisation on natural numbers to prove they aren't countable is still not satisfactory as natural numbers form an unbounded set and hence there is no number which we can say is biggest natural number so i dont understand why do we need finite digits in a natural number. let it have infinite digits and it seems to prove that natural numbers are uncountable.
    If anyone understands the answer pls do comment

    • @eggynack
      @eggynack 3 года назад

      Just cause there's no biggest natural number that doesn't mean there's a number infinity. And this number of infinite digits would decidedly just be infinity. The fundamental nature of the naturals is that you can get from any natural number to any later natural number via a series of +1's. A number like 111..., it's not reachable in this way. Also, it's not even clear how such a number works. How do you add it? Or multiply it? It definitely doesn't work the way naturals work. And, on top of all that, it's also just the definition. Natural numbers got finite digits.
      In any case, if the naturals had infinite digits going off to the left, then sure, there would be uncountably many of them. But they don't, so they don't.

  • @wenruima6431
    @wenruima6431 11 лет назад

    Great. Thanks.

  • @andywright8803
    @andywright8803 8 лет назад +1

    If you allow leading zeros to your real numbers, you can have as many of them as you like, so 3 could be written as 03, 003, or even ...0003, ie every natural number can be written as a unique infinite string of digits, just like the real numbers. What's stopping anybody then from proving, using Cantor's diagonalisation argument that the infinity of the naturals is as large as the infinity of the reals? BTW, I don't have a degree in Maths, so could any explanations be put in terms a layman could understand please?

    • @profonsager
      @profonsager 8 лет назад +2

      03, 003, and 0003 are all the same number. In the diagonalization argument, it is assumed you're attempting to enumerate unique real numbers, which is why some versions of Cantor's argument take care to state that you disallow any decimal representation ending in 999999.... (because 0.999999.... = 1). 03 and 003 are distinguishable *strings* but they are not distinguishable numbers. In fact, you can use the diagonalization argument to show that the set of infinite strings is uncountably infinite.

    • @MuffinsAPlenty
      @MuffinsAPlenty 6 лет назад

      Let's try it. Let's start with the most famous list of positive integers there is.
      ...001
      ...002
      ...003
      ...004
      etc.
      Now, applying Cantor's diagonal argument to this (sort of the diagonal in the opposite direction), the digit in the ones place should differ from 1. Let's say we make it 2.
      From then on, it's not difficult to notice that every other digit must differ from 0 (since the digit in the 10^n's place in the nth number on the list is definitely 0). Say we make all of them 1.
      Then the new number we have constructed is: ...1111112 (where the 1's extend infinitely to the left).
      So the positive integers are uncountable, right? We have found a new positive integer which is not on the list.
      Well, no. Not quite. ...1111112 is not an integer. Each individual integer has only finitely many nonzero digits (each integer is finite in value). So since ...1111112 has infinitely many nonzero digits (and hence would be infinite in value), it is not an integer. So we haven't actually found a positive integer which is not on our list. So we cannot conclude that the list of positive integers is incomplete.

  • @AdityaLandge1994
    @AdityaLandge1994 6 лет назад

    Any Potterheads hear who got the reference?

  • @tlaloc1525
    @tlaloc1525 11 лет назад +3

    Diagonalization is the same as the old argument from Cantor not a proof, the issue here is that is used as a definite proof but is not, is an "argument" presented by mathematician Georg Cantor 1891. This does not proves reals are uncountable, it only give a reference to defend your stand on reals as uncountable.

    • @tlaloc1525
      @tlaloc1525 9 лет назад +1

      This cantor's theorem is the one of the main B$$s used by poor mathematic teachers who lack self creation. This proof has no single application in the really world and is used by 1000's of trolling teachers.

    • @川新-k4i
      @川新-k4i 5 лет назад

      It’s totally a mathematically valid proof. And it’s of importance for a whole set of problems regarding paradox and infinity, plus questions regarding Completeness of axiomatic systems?

    • @colinjava8447
      @colinjava8447 2 года назад

      It does prove they are uncountable, it may not be hugely practical but you could say that about the study of black holes.
      Its still extremely interesting in its own right, as we now know there are lots of infinities (but to ask how many apparently makes no sense, but it would be at least countable infinity if you had to give an answer).

  • @mallikasen3796
    @mallikasen3796 9 лет назад

    7:32 This guy's funny

  • @chriswatts5921
    @chriswatts5921 11 лет назад

    0 is a natural number! Any sane person can show this!

    • @mohammadjuned8524
      @mohammadjuned8524 10 лет назад +3

      chutiya

    • @rodrigocientista
      @rodrigocientista 10 лет назад

      no, not necessarily, 0 and also all the naturals are too rationals (e.g. of the form n/n [but zero, in fact 0 is of nthe form 0/n forall n>0]), reals (e.g. the ones which forms the algebraic partition of the reals along with other non-natural reals, etc) and complex (n=a+bi putting a natural and b=0), on the other hand a number like n=5+(3/7)i is not a natural by definition;

    • @viveksen3247
      @viveksen3247 8 лет назад

      +Chris Watts (CJ) To prove that they are actually one-to-one he has to show that 0 is actually a natural number because 0 is also an even number. If he assumes that 0 is not a natural number then his proof that they're one-to-one is wrong.