Root 2 is Irrational from Isosceles Triangle (visual proof)

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  • Опубликовано: 31 май 2024
  • In this short, we use a famous argument by Tom Apostol to prove that the square root of two is irrational by infinite descent using a right isosceles triangle. We also go a bit further and show how this proof hints at the number theoretic construction of the convergents of the square root of two, which are the best rational approximations of root 2.
    If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoffee.com/VisualPr.... Thanks!
    For an alternate visual proof of this fact, see this video: • Visual irrationality p...
    This animation is based on an argument due to Tom Apostol from issue 9 of the 2000 American Mathematical Monthly: doi.org/10.1080/00029890.2000...
    To learn more about the convergents argument and the relationship between this proof and convergents, see this wonderful article by Doron Zeilberger:
    sites.math.rutgers.edu/~zeilb...
    and here you can learn more about convergents:
    en.wikipedia.org/wiki/Continu...
    #irrationalnumbers #realnumbers​ #manim​ #math​ #mtbos​ ​ #animation​ #theorem​​ #visualproof​ #proof​ #iteachmath #mathematics #irrational #triangle #righttriangle #isoscelestriangle #proofbycontradiction #root2 #algebra #infinitedescent #numbertheory #convergents
    To learn more about animating with manim, check out:
    manim.community

Комментарии • 126

  • @skvttlez1263
    @skvttlez1263 Месяц назад +16

    Infinite descension of integers is impossible because 1 is the smallest positive integer.

    • @error_6o6
      @error_6o6 17 дней назад

      That’s literally the point of the proof

  • @ed.puckett
    @ed.puckett Месяц назад +28

    There is a lot here to digest. Thank you!

  • @terdragontra8900
    @terdragontra8900 Месяц назад +6

    I am fascinated that I’ve never seen this particular proof before, wow, pretty cool. In fact, a/b = sqrt(n) implies (nb - a)/(a - b) = sqrt(n), but these new integers are smaller than the originals only for n < 4… but at least it works for 3!

  • @bmx666bmx666
    @bmx666bmx666 Месяц назад +4

    Wow, this is the simplest way to find the approximate value of sqrt(2). So easy to remember - start with 1 / 1, double the bottom and sum with the top = save on top 3, sum top and bottom - save at the bottom 2, got 3 / 2, etc ... .

  • @guessundheit6494
    @guessundheit6494 Месяц назад +3

    Just like Fibonacci numbers, there's and easy calculation for subsequent numbers. The difference is, the initial value for the numerator must start at 1, and the denominator start at 0.
    Numerators: 1 { 1x2+1 = 3 > 3x2+1 = 7 > 7x2+3 = 17 > 17x2+7 = 41 > 41x2+17 = 99 > etc.
    Denominators: 0 { 1x2+0 = 2 > 2x2+1 = 5 > 5x2+2 = 12 > 12x2+5 = 29 > 29x2+12 = 70 > etc.

  • @alanthayer8797
    @alanthayer8797 Месяц назад +7

    VISUALS Visuals visuals is Key! Thanks as usual fa da VISUALS

  • @sankojuprithvi
    @sankojuprithvi Месяц назад +2

    Very genius and informative and spectacular video on mathematics and illustrations in this video very magnificient a I deeply congratulate the content maker of this video he is a spectacular mathematician

  • @SeanSkyhawk
    @SeanSkyhawk Месяц назад +6

    This also has an interesting implication: there is no number such that taking its square and adding that square to itself results in a square of a different integer.

    • @BenWard29
      @BenWard29 Месяц назад

      Except for 1, right?

    • @majorproblem8796
      @majorproblem8796 Месяц назад +2

      @@BenWard291+1=2 sqrt(2) is not an integer

    • @landsgevaer
      @landsgevaer Месяц назад

      What is interesting about that implication?

    • @BenWard29
      @BenWard29 Месяц назад +3

      @@majorproblem8796 Yeah you're right- I'm an idiot. I misread your comment.

    • @cheeseeater4434
      @cheeseeater4434 Месяц назад +5

      @@BenWard29 no problem with that mate, owning up to your mistakes already puts you ahead of 90% of the online community

  • @benjaminbailey3907
    @benjaminbailey3907 Месяц назад +3

    Very nice. It's very interesting that you didn't need any divisibilty arguments.

    • @PeachCrusher69
      @PeachCrusher69 Месяц назад +1

      but he did, didn't he? when he said that a infinite sequence of decreasing numerators & denominators denoting a rational no. is impossible
      it's impossible because all those num-denom pairs would make equivalent fractions, and the next term in the sequence is always made of smaller integers than the previous, but they can't get smaller than 1.

  • @user-dv5sn2xv2y
    @user-dv5sn2xv2y Месяц назад +3

    Simple is the best, thank you for proving that.

  • @TupperWallace
    @TupperWallace Месяц назад +3

    Neatly done!

  • @eonasjohn
    @eonasjohn Месяц назад +2

    Thank you for the video.

  • @PC_Simo
    @PC_Simo 24 дня назад +1

    I’m surprised that Mathologer hasn’t covered this particular proof, in his video of shrink proofs, or ”Visual Irrationals”. 😮

  • @CristianBaeza-rh7zq
    @CristianBaeza-rh7zq Месяц назад +1

    Haha, amazing! Simple and nice ❤❤❤

  • @tarekmesto
    @tarekmesto 21 час назад +1

    Great videos. what software do you use?

    • @MathVisualProofs
      @MathVisualProofs  18 часов назад

      I use manimgl. That’s the python library developed by 3blue1brown

    • @tarekmesto
      @tarekmesto 18 часов назад

      @@MathVisualProofs great thank you

  • @charliestein9350
    @charliestein9350 Месяц назад +3

    Very clever! I don't think I ever heard an argument like this before, but it is very sound. If I can repeat a process infinitely many times, and reduce the size of some variable in each iteration, the variable must eventually reach zero. Nice!

  • @michaelgonzalez9058
    @michaelgonzalez9058 Месяц назад +1

    The isocoslis is 3°~3•1=1/2 which equals 0

  • @DanielPoupko
    @DanielPoupko 21 день назад +1

    That is a very cool proof

  • @cupatelj
    @cupatelj Месяц назад +50

    But Terrence Howard told me that it is Rational... 🤣🤣🤣

    • @Larsbutb4d
      @Larsbutb4d Месяц назад +2

      what?

    • @Larsbutb4d
      @Larsbutb4d Месяц назад +1

      oh waut nvm

    • @cupatelj
      @cupatelj Месяц назад +14

      ​@@Larsbutb4d This is a joke. There is a Hollywood actor, Terrence Howard, who recently on podcasts and other platforms, has been making ridiculous scientific claims, including one that the square root of 2 is a rational number. :)

    • @samueldeandrade8535
      @samueldeandrade8535 Месяц назад

      ​@@cupatelj why tf are you invoking Terrence Howard ... Ffs, why humans need so much to demean other humans, using them as objects of jokes? You know what, yeah, he said 1×1=2 and √2 is rational. But ... I guess that considering
      a*b := 2ab
      those statements may be true? So ... not a big deal at all? He would just be changing the product ...

    • @cupatelj
      @cupatelj Месяц назад +5

      ​@@samueldeandrade8535As if he was speaking about a topic that isn't as strict as mathematics. Of course, he should be seen as a joke for making such childish claims.

  • @mauisstepsis5524
    @mauisstepsis5524 Месяц назад +2

    How do I decide initial values (a0, b0) for initial values to approx sqrt(2)?

    • @MathVisualProofs
      @MathVisualProofs  Месяц назад +1

      1/1 is the best approximation with denominator 1. Also the simplest solution to 2b^2-a^2=1. I should have mentioned it. :)

  • @Aman_iitbh
    @Aman_iitbh Месяц назад +2

    is that seq relates with pell equation
    with approximates root n

  • @Smartas599
    @Smartas599 Месяц назад +2

    Beautiful

  • @orisphera
    @orisphera Месяц назад +1

    I prefer considering the smallest such triangle

  • @monkey6114
    @monkey6114 Месяц назад +1

    A sequence that has a much better covergence is
    an=(an+2÷an)÷2
    The 3rd term is already equal to 577÷408

    • @MathVisualProofs
      @MathVisualProofs  Месяц назад +1

      I didn't mean (or say) that this sequence is the best convergent sequence to root 2. I said the convergents are the best rational approximations to root 2. So each number in this collection of rationals will always be the closes to root 2 compared to any other rational with same or smaller denominator.

    • @landsgevaer
      @landsgevaer Месяц назад +1

      Yet you get the same terms from the other sequence, just skipping intermediate steps (I think you double the index).

    • @monkey6114
      @monkey6114 Месяц назад +1

      @@MathVisualProofs ik
      I just wanted to add

  • @pelayomedina2174
    @pelayomedina2174 Месяц назад +2

    Congratulations, you just got a succesion of rational numbers that converges to an irrational number
    This is as if I say:
    The square root of 2 is 1.4, 1.41, 1.414, 1.4142,...
    Also is impressive because a and b can be any number so if you follow that rules you are going to approach square root of 2
    Oh it is explained later :)

  • @durpymcscurpy5169
    @durpymcscurpy5169 Месяц назад +1

    amazing

  • @IIScarletKingII
    @IIScarletKingII Месяц назад +1

    nice

  • @sagarsaha935
    @sagarsaha935 Месяц назад

    Sir I said that this channel has a conic section playlist ?

    • @MathVisualProofs
      @MathVisualProofs  Месяц назад

      I don’t have many about conic sections.

    • @sagarsaha935
      @sagarsaha935 Месяц назад

      @@MathVisualProofs Sir it is very good to make video on conic section
      I am currently having a lot of trouble understanding conic sections....... please sir give some thought to this

  • @tazepatates4805
    @tazepatates4805 Месяц назад +3

    I didn't understand anything but this is fascinating

  • @Eta_Bokas_slave
    @Eta_Bokas_slave Месяц назад

    Why do you say " where a and b are positive integers".
    I know it makes sense because you are talking about lengths but its not the definition of a rational number.(They can be negative)

  • @avijitpramanik5425
    @avijitpramanik5425 Месяц назад +1

    Visualising is the best way to understand math
    Do you agree 👍

  • @gonzus1966
    @gonzus1966 Месяц назад +2

    This is Pell's equation in disguise, right?

    • @MathVisualProofs
      @MathVisualProofs  Месяц назад +2

      Yes. This is related to Pell’s equation for sure :)

  • @pythontron8710
    @pythontron8710 Месяц назад +1

    Proof by infinite descent

  • @qwertek8413
    @qwertek8413 Месяц назад

    why is everyone saying positive integers instead of natural numbers

    • @MathVisualProofs
      @MathVisualProofs  Месяц назад

      if you say positive integers, you can avoid the arguments about whether or not 0 is a natural number :)

    • @qwertek8413
      @qwertek8413 Месяц назад +1

      @@MathVisualProofs yea, that was one of my guesses, thanks :)

  • @tired5925
    @tired5925 Месяц назад +1

    This video does not explain why it is a problem, if i have a triangle and take half of it then half of what is left then it will go forever, why is it a problem and a contradiction?. And even if it is, that just shows ur method creates a loop and doesnt arrive at an answer.

    • @MathVisualProofs
      @MathVisualProofs  Месяц назад +1

      I note explicitly that shrinking the triangles would produce an infinite list of decreasing positive integers. This contradicts the well-ordering principle.

  • @johnkardier6327
    @johnkardier6327 Месяц назад

    This "non-standard" proof might be the oldest one.

    • @MathVisualProofs
      @MathVisualProofs  Месяц назад +1

      The citation is from 2000. But I guess maybe this idea was known before though I haven’t seen it written down before the Apostol article.

    • @rivenoak
      @rivenoak Месяц назад

      @@MathVisualProofs afaik the "usual" numeral proof was made by Euclid ? we know he was busy and tinkered with pythagorean triangles back in ancient times, so perhaps the visual proof was known to him as well ? math lessons ~300 bc in alexandria must have been interesting times :D