Is square root of 2 equal to 2? (Spoiler : no)

This is a short animation of an oft-claimed visual proof that the square root of 2 is equal to 2. Can you see why the two lengths aren't equal? Can you give a good, brief explanation? #math #manim #mathshorts #mathvideo #visualproof #proofwithoutwords #false #mathfail

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Music in this episode:

[xavierburval]

In the words of 3B1B, “The length of the limit is not the same as the limit of the lengths”

[boas_]

Not really mathematical, but I like to think that on an infinitasimally small scale, the line is not straight, and the side of the triangle is straight

[tamirerez]

THE ULTIMATE EXPLANATION:
No matter how small the triangles are, still, in each triangle, its hypotenuse is equal to its side times √2

[TheEulerID]

This is similar to the "proof" than pi is equal to four by drawing a square round a circle of unit diameter. The square will have a perimeter of 4 of course. However, introduce steps into the corners to make it appear to fit better, and it still has a periphery of 4. No matter how many "steps" you introduce into this to make an (apparently) better fit, the periphery is still 4.

The answer is simply that these apparent better fits are no such things. The geometrical approximation technique introduced simply doesn't reduce the cumulative error. This is a lesson for anybody using an approximation technique to generate an integral with a limit. Your approximation technique does have to reduce the cumulative errors as the delta approaches zero.

[paulquinones]

This example proves that the uniform convergence of a sequence of functions isn't related at all (without further assumptions) to the uniform convergence of the sequence of the derivatives. Here, the sequence of jagged curves converges uniformly to the diagonal. But the length of the jagged curves is related to the derivatives of these functions, and not these functions themselves (the int of the sqrt of 1 + the derivative² for those who are interested), and because these curves have corners, they are not differentiable everywhere ... the sequence of the derivatives is not converging uniformly (it does not even converges pointwise), and therefore we cannot use the double limit theorem, which permits us to swap the limits. Be careful, I have just proven that we don't have enough assumptions to swap the limits ... This was the explanation, I haven't proven that the proof presented is false. In fact, that's not so difficult. We can compute the length of all the jagged lines (that is 2), and the length of the diagonal (sqrt(2)), and we know that 2 is not equal to sqrt(2) ... That's as simple as that. This example teaches us to be careful when applying a theorem, we must check all hypotheses.

[hashasquiowo]

This is because the line converges to the diagonal, but it's derivate doesn't because it doesn't transform smoothly

[vnpgyef]

If there are a lot of corners, it‘s not a line

[offtheradar]

Distance (path taken) =/= Displacement (direct distance between 2 points)

[SanePerson]

If n is the number of times you folded over the steps to generate the( n+1)th set of stairs, the path length L(n+1) is still equal the the path length L(n). Taking the limit as n → ∞, L(n) is still 2 and the path never converges to the straight hypotenuse line.

[kookeekwisp]

It's kinda like a cheat code, it's like giving a 2 dimensional object a thickness that doesn't lose thickness no matter how infinitely thin it gets

[shankylezapanta]

A good analogy of this is about a high frequency and low frequency wave or just a plain straight line. From endpoint to endpoint they have equal distance but if you consider the frequency of waves like actually measuring from through to crest repeatedly not just measuring the shortest path, you will still get a length of 2 and the straight line is still equal to sqrt of 2 proving the statememt false

[staticostrich]

The perimeter never gets shorter. Visually, it only looks shorter because it is a fractal. The actual reason it isn't shorter is that it is given in the question that the two side lengths are still present in the angle, the question is just how many segments were made to make the staircase. Funnily enough, you can rebuild the little segments in any order, any combination you would like and as long as they maintain their previous orientation and connection angles to like pieces it would still reach the same spot.

Without going into the nitty gritty, I like to solve these puzzles by imagining them as something I just built. What if I could grab each end and pull on the segments in order to straighten them, like a spring being reduced to it's original wire shape? Eventually the line pulled would equal 2. If I tried the same thing with the hypotenuse it would not stretch because there are no angles to straighten. The only difference between each recursion you showed us before infinity is that technically if I built it it would just add friction and make it harder to pull out. But, in my mind, if everything was infinitely strong, it would pull out to be 2 every time.

[slipperynodd]

This is the same as the pi equals to four thing. If you zoom in you’ll see tiny little folds that if broken up again and put together with their individual axises, it adds back to 2 and clearly not the same as a square.

[BananaBLACK]

This is the coastline paradox, right?

[mhoover]

I discovered this in the '90s while trying to find a shortcut through a parking lot. I call it Hoover's paradox.😊

[antoopl]

like the "paradox" of the seaside of a country, the more you zoom in the more the length gets bigger

[raffimolero]

"the limit is not equal to the result of any finite number of steps"

[mananagrawal]

No, cause lenght of hypotenuse will be integral(sqrt(1+(dy/dx)^2)dx) from 0 to 1 which is root 2. This is because even at that scale you are applying the Pythagoras theorem. This is exactly have we find the lenghts of most curves. Well the curves for which that lenght is integrable.

[xevento]

*Actual Proof for anyone interested:*

Let's define the curve, that you want to fold inward at the corners, as a parametric function called c_0(t), where t ranges from 0 to 1 and describes every point on that curve. Now each iteration of the folding gets it'S own function, meaning the next iteration gets c_1(t) and the following c_2(t).

This video suggests considering the limit, when c_n(t) goes to infinity. Now consider any point on that curve, let's say c_n(0.2), as n approaches infinity. This point IS on the side of the triangle. If we define c_∞(t) = lim{n->∞}(c_n(t)), then each point given an input t will lie on the triangles side. This means that c_∞(t) precisely describes the triangles side, it is not an approximation.

Now comes the tricky bit. If we consider the limit of the lengths of our curves, meaning lim{n->∞}(len(c_n(t)), it will equal 2. Does this now prove that sqrt(2) is actually 2?! No, and here is why: In this example we can't assume that the limit of the length is equal to the length of the limit. If we calculate len(lim{n->∞}(c_n(t))), we will - in fact - get sqrt(2).

This example is a good example for a key takeaway of calculus, that something that is true for a sequence, might not be true for the limit. So be mindful of that when dealing with a visual proof, suggesting a limit.

(This explanation is inspired by a video of 3Blue1Brown on lying with visual proofs. I'd suggest searching it up, it uses a similar example to this to explain what I just exlpained in more detail. Also sorry for the bad readability of this chunk of text, youtube does not have LaTeX formatting.)

[thedudethatneveruploads]

I'm not certain if this is a mathematically rigorous (or even correct) explanation, but my guess is that it's because even as you approach the hypotenuse, you're merely approximating a straight line by using a bunch of steps, but the lengths of the steps still add to the same value: 2.

In other words, even if it _looks_ like the steps fit the hypotenuse, the distances away from the hypotenuse and back over and over again will add up to 2 anyways.