Root 2 is Irrational from Isosceles Triangle (visual proof)
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- čas přidán 31. 05. 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.
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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...
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But Terrence Howard told me that it is Rational... 🤣🤣🤣
Infinite descension of integers is impossible because 1 is the smallest positive integer.
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.
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.
I’m surprised that Mathologer hasn’t covered this particular proof, in his video of shrink proofs, or ”Visual Irrationals”. 😮
There is a lot here to digest. Thank you!
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 ... .
I didn't understand anything but this is fascinating
Proof by infinite descent
I prefer considering the smallest such triangle
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!
VISUALS Visuals visuals is Key! Thanks as usual fa da VISUALS
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
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.
Very nice. It's very interesting that you didn't need any divisibilty arguments.
Simple is the best, thank you for proving that.
Neatly done!
Thank you for the video.
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!
How do I decide initial values (a0, b0) for initial values to approx sqrt(2)?