Root 2 is Irrational from Isosceles Triangle (visual proof)
ฝัง
- เผยแพร่เมื่อ 22 ส.ค. 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/10.108...
To learn more about the convergents argument and the relationship between this proof and convergents, see this wonderful article by Doron Zeilberger:
sites.math.rut...
and here you can learn more about convergents:
en.wikipedia.o...
#irrationalnumbers #realnumbers #manim #math #mtbos #animation #theorem #visualproof #proof #iteachmath #mathematics #irrational #triangle #righttriangle #isoscelestriangle #proofbycontradiction #root2 #algebra #infinitedescent #numbertheory #convergents
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Infinite descension of integers is impossible because 1 is the smallest positive integer.
That’s literally the point of the proof
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 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!
Very nice. It's very interesting that you didn't need any divisibilty arguments.
1:15 we are assuming that those tangents are equal (which they are, no argument there), but I missed the proof. th-cam.com/video/r5BIjv6jHSQ/w-d-xo.html explains why the tangents from the same external point are of the same length, and since the arc is less than semicircular, we know there will be such a point, which in this case it is the rightmost vertex of the smaller triangle with bottom angles 45deg.
2:08 "infinite list of decreasing positive integer ratios". Yeah, but I missed the explanation there too. The assumption that those lengths are integers comes from the assumption that a and b are integers, so then the expressions "a - b" and "2b - a" will also be integers (positive integers, by construction) because the operations applied here will leave us in the same number set.
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.
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
😎
VISUALS Visuals visuals is Key! Thanks as usual fa da VISUALS
:)
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.
Except for 1, right?
@@BenWard291+1=2 sqrt(2) is not an integer
What is interesting about that implication?
@@majorproblem8796 Yeah you're right- I'm an idiot. I misread your comment.
@@BenWard29 no problem with that mate, owning up to your mistakes already puts you ahead of 90% of the online community
I’m surprised that Mathologer hasn’t covered this particular proof, in his video of shrink proofs, or ”Visual Irrationals”. 😮
Simple is the best, thank you for proving that.
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!
But Terrence Howard told me that it is Rational... 🤣🤣🤣
what?
oh waut nvm
@@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. :)
@@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 ...
@@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.
The isocoslis is 3°~3•1=1/2 which equals 0
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 :)
7 × (2, 2, 3, 5, 9)
Thank you for the video.
Thanks for watching!
Neatly done!
Thanks!
Great videos. what software do you use?
I use manimgl. That’s the python library developed by 3blue1brown
@@MathVisualProofs great thank you
That is a very cool proof
👍😀
How do I decide initial values (a0, b0) for initial values to approx sqrt(2)?
1/1 is the best approximation with denominator 1. Also the simplest solution to 2b^2-a^2=1. I should have mentioned it. :)
A sequence that has a much better covergence is
an=(an+2÷an)÷2
The 3rd term is already equal to 577÷408
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.
Yet you get the same terms from the other sequence, just skipping intermediate steps (I think you double the index).
@@MathVisualProofs ik
I just wanted to add
I prefer considering the smallest such triangle
I didn't understand anything but this is fascinating
is that seq relates with pell equation
with approximates root n
Yes. For sure !
Beautiful
Thank you
I don't follow the jump of ever decreasing = proof that root 2 is irrational. You can ever decrease any ratio, but it doesn't mean the resulting numerator and denominator are integers. 1/2 ==> .5/1 ==> .25/.5 ==> etc... All of these are still a factored representation of 1/2
But remember here that both the numerator (2a-b) and the denominator (b-a) are in fact integers and those are what are shrinking. I can't keep shrinking those since they are integers.
Haha, amazing! Simple and nice ❤❤❤
Visualising is the best way to understand math
Do you agree 👍
This is Pell's equation in disguise, right?
Yes. This is related to Pell’s equation for sure :)
nice
Thanks!
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)
sqrt2 is clearly positive
Proof by infinite descent
amazing
Thanks!
Sir I said that this channel has a conic section playlist ?
I don’t have many about conic sections.
@@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
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.
I note explicitly that shrinking the triangles would produce an infinite list of decreasing positive integers. This contradicts the well-ordering principle.
why is everyone saying positive integers instead of natural numbers
if you say positive integers, you can avoid the arguments about whether or not 0 is a natural number :)
@@MathVisualProofs yea, that was one of my guesses, thanks :)
Sorry, I think I believe Toward Herrence who says that it is rational
Easy: 0/0 = √2
This "non-standard" proof might be the oldest one.
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.
@@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