Does Pi Equal 2?? (Spoiler: no)

In this short, we show a thought experiment that arises by doubling the number of semicircles drawn in a semicircle but halving their radii. The process produces an infinite collection of smaller and smaller chains of more and more semicircles that always have combined circumference length equal to pi. If we think about the limiting process, it seems like maybe this technique shows that pi = 2. But Pi can definitely be shown to be greater than 3 - so where does this argument fail?
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoffee.com/VisualPr.... Thanks!
Here is a related false proof that shows root 2 is equal to 2:
• Is square root of 2 eq...
This false argument was suggested to me by Jeff Stuart as a nice alternative to the more classic Pi = 4 argument.
#math #manim #visualproof #proofwithoutwords #circle #circumference #radius #pi #piday #shorts #infinite #falseproof
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[@KinuTheDragon]

Length of limit of curves =/= Limit of length of curves

[@boas_]

With GBoard you can hold the = sign to make ≠

[@boas_]

I think my comment got removed?

[@priangsunath3951]

​@@boas_all your comments are there don't sweat it

[@andreigebauer9893]

​@@boas_ it's a conspiracy! Google doesn't want us to know how to make the "not equal" sign 😱

[@aathisankar6432]

≠

[@FrozenLabRat]

There is only half a Pie. Conclusion; someone ate the other half.

[@KevinLivian]

I’ve been spotted

[@loapher]

​@@KevinLivian gettim boys

[@aww.funvids7589]

FBI OPEN UP

[@KevinLivian]

FBI: “sir you’re under arrest for stealing half a pie” me: “AND I’LL DO IT AGAIN”

[@HarisRehmanGG]

​@@KevinLivian: "but half a pi looks like tau, so really I multiplied it by 4 to make 2π= tau"

[@pablo_mates7884]

It never flattens. The derivative has to converge too. I have never seen this counterexample of that situation! Awesome

[@tedhorton6315]

Agree. In the video he says that the semicircles "squash down and flatten". As you state, no they don't. Never. That is where his argument fails.

[@mcqueen424]

@@tedhorton6315let’s be a little more clear here; of course they don’t ever “flatten,” but when he said that, he meant the semi circles converge pointwise to the diameter which is true. The argument fails because you are interchanging limits without justification

[@nanamacapagal8342]

Something similar happens with the schwarz lanterns: the normals must converge too!

[@JoseJimeniz]

Looks pretty flat to me. Keep going until the original 1 m diameter circule is made up of circles each smaller than the plank length.

[@nanamacapagal8342]

@@JoseJimeniz this is math, not physics. we are not bound by the restraints of our reality. we openly embrace infinity and all of its counterintuitive weirdness.

[@randomsandwichian]

Regardless of how small the semicircle is, it will never be a straight line. So π ≠ 2

[@AMGMineCrick]

True, because semicircles have length as well and breadth while a line only has length and no breadth.

[@yannld9524]

Of course, but that doesn't explain why this method doesn't work. The purpose of a limit is to approximate an object as precisely as we need, not necessarily to be equal to that object. Using the same reasoning as you, I can say that 1/n is never 0 so lim 1/n ≠ 0... which is very wrong.

[@randomsandwichian]

@@yannld9524 A diagonal length of two opposite points of a square is always going to be longer than the side, what more the length of a curve through said square with a radial length of that square's side. So using said rational, the diameter of a circle will always be shorter than 1/2 the circumference, even if the circumference is recreated with smaller and smaller dimensions of curves within that semicircle as shown in this example.

[@yannld9524]

@@randomsandwichian Good for you if you think this explanation is more convincing than the video. I was just pointing out that your first argument wasn't a good one

[@terranceparsons5185]

Yet the same people argue 0.99999 recurring is equal to 1

[@rodrigoqteixeira]

The famous "aproximates the area but not the permeter"

[@pxzzvc6357]

Can you elaborate?

[@pamplemoo]

​@@pxzzvc6357every iteration, the area of the semicircles is divided by 2. But the perimeter always stays equal to π

[@piercexlr878]

​@@pxzzvc6357Basically when things look close to each other generally area is also close. Perimeter is super deceptive though. Things that look close can have almost any perimeter including infinity.

[@oscargr_]

​@@pamplemooIt completely followed what you are saying. And then I thought.. but wait, if a circle is the shape that has the ultimate (fixed ) ratio of area/permiter, then how can one stay equal while the other divides by two every step. And then I realised that originally, the shape isn't a full circle, after one step the (total) shape is one full circle and every next iteration you keep adding full circles. (So you are not comparing similar aggregate shapes.)

[@DrSpecialful]

@@oscargr_ the argument the creator makes would also work if you drew every semicircle as a circle. the argument fails still then, this isnt the flaw.

[@TechnoBoy-yl9hq]

Okay so the reason why pi doesn't equal 2 here is because while the semicircles look like they are only 2 units long. It's instead similar to an accordian. Or an old plastic straw you'd get with a juice box. It looks smaller because it's compressed, but if you were to grab both ends, and pull. Then it would lengthen out to ~3.1416 units long.

[@lastchance8142]

Good analogy. Impossible to imagine the semicircles at the limit, but true.

[@TechnoBoy-yl9hq]

@@lastchance8142 thanks. It's not my analogy originally though. I got it from Vi Hart and her video about why the proof of pi=4 doesn't work. The video is titled "Rhapsody on the Proof of Pi = 4" if you wanna check it out

[@hasupoetry]

Best analogy/explanation

[@wyattstevens8574]

This reminds me of the classic pi=4 "proof!"

[@MathVisualProofs]

Yep! This is the other side where pi is way too small. Maybe we should just average the two proofs to get pi = 3? 😉

[@wibbliams]

Tbh thats pretty close so i'd take that​@MathVisualProofs

[@daybench]

​@@MathVisualProofs engineers:

[@emulgatorx]

@@MathVisualProofs We're getting closer. If we averaged the results of infinitely many false pi proofs, would we get actual pi?

[@MathVisualProofs]

@@emulgatorx 😆

[@juliavixen176]

This is a variation of the "Staircase Paradox".

[@oscarmartinpico5369]

But It is not a paradox.

[@Emily-fm7pt]

⁠@@oscarmartinpico5369Mathematicians have historically called anything that’s counterintuitive as “paradoxical,” e.g. the “birthday paradox”

[@johnny40coats4]

​@Emily-fm7pt I don't think this is counter intuitive. It just misunderstands what a limit is. To take a limit we need a function which approaches something. The 'function' for this doesn't approach anything it stays exactly where it is, pi at the first approximation and pi at the nth

[@Emily-fm7pt]

@@johnny40coats4 Well you at least have to accept that the area of the limiting shape approaches that of a circle, because it does get closer and closer to being a circular shape, but in doing so you’re not changing the perimeter, like you said, so you can’t use the perimeter to calculate pi. The counterintuitive part here is that a shape that looks and feels a lot like a circle can still be not even close in its perimeter.

[@ShadowPhoenix4798]

@@Emily-fm7ptnot just mathematicians, the definition of a paradox as something that counteracts itself is only a strict definition, albeit commonly used. There exists a more lax definition as anything that doesnt make immediate sense/doesnt feel like it would work even after thought. This definition is less common however.

[@RandomStuff0714]

Technically, it never flattens, but our eyes can't see that, so they get tricked into thinking it does

[@Allthenamesweretaken17]

exactly

[@-._.ChaLoe._.-]

Funnily enough, if you continue to create smaller semi-circles, the limit of each point on the surface of the semi-circles, does actually approach the exact line, and if the limit is taken to infinity, it genuinely does flatten, it doesn't make a bumpy approximation, the points exactly line up. The problem is the perimeter is not preserved when the process is done infinitely many times

[@nexuscross3233]

​​@@-._.ChaLoe._.-The limit cannot be taken to infinity. 🤦‍♂️

[@duckyoutube6318]

​@@nexuscross3233 Are you sure about that?

[@Sean-of9rs]

@@nexuscross3233It actually can be and the curves do converge to the diameter. But taking lengths is also calculated by a limit. Commuting two limits (the length of the limit of the curves vs the limit of the length of the curves) is a big no unless you have a theorem that allows it. And that's why a lot of early analysis is learning when you're allowed to do it :)

[@AryanKumar-vo1ic]

I've not yet studied calculus but i think something's wrong wtih the limiting process

[@daniel_77.]

yeah. limits are way different that absolute values

[@RMF49]

What’s wrong is that while the height of the semi circles above line does approach zero, the number of them goes to infinity. So the total length of them does not approach 2. It remains exactly pi with each iteration.

[@renedelatorre2138]

In calculus, arc length has a different formula for area. Arc length uses circular arcs while area uses rectangles. When I asked my prof why this is so, his answer was "It just is".😃

[@samueljuarez5132]

When you have two curves and you change one to make it "converge" to the other one, this will only happen when the shape and the derivative of curve 1 are being changed. In the case in the video, the derivative was not changing, and you could see "pointy" parts in between

[@Fire_Axus]

not always

[@britannic124]

This is like the coastline paradox.

[@vardhanshah2810]

Engineers might be loving these proofs

[@yannld9524]

In fact, it's a very good question and I don't think most people have ever heard the answer. If a curve C is parametrized by a (sufficiently regular) function f, then to compute the length of C we don't need f but f '. And when we approximate C (uniformly) by certain curves C_n as in the video, this corresponds to choosing a sequence of functions (f_n) that converges uniformly to f. Hence in the very convenient case where the sequence (f_n ' ) converges uniformely to f ', we will obtain that lim length(C_n) = length(C), but there's no reason for the sequence (f_n ' ) to converge uniformly to anything, there are a lot of counter examples.

[@matheusalmeidadamata]

Perfect.

[@pendragon7600]

Finally somebody who actually knows what theyre taking about. This is about uniform convergence

[@giuseppegaleotti9149]

We need uniform convergence of the derivatives not of the function, the function itself is uniformly convergent to the line in fact.

[@yannld9524]

@@giuseppegaleotti9149 Yep, that's what i said

[@giuseppegaleotti9149]

@@yannld9524 I was answering to pendragon7600 but it didn't tag him.

[@colinjava8447]

Its like diagonal of a square, and a step case approximating that diagonal. The step case will never approach the length of the diagonal, but 2L instead.

[@zenefu-waitt8228]

It was so trippy when he zoomed in.

[@Xyfria]

Fun fact: this is basically the exact design of a fractal vise, a vise which conforms to any object's shape

[@locheyoutube5252]

Reminds me of when I saw a “Proof” that the diagonal on a 3-4-5 triangle was actually 7 🤯 ! It took me a while to recover from that one too. Hopefully I will bounce back faster on this one 😅

[@Riftoo]

Each iteration, the circumference of each semicircle decreases by half, while the amount of semicircles doubles.This means, since the total perimeter of these semicircles is (circumfrence of smaller semicircle)x(amount of semicircles), that each iteration, the decrement of the circumfrence of each semicircle perfectly cancels out the increment of the quantity of semicircles, causing the total perimeter to remain the same. Here's what the math would look like circumfrence x amount of semicircles (π/2)x(2) = π (iteration 1) (π/16)x(16) = π (iteration 4) (π/1024)x(1024) = π (iteration 10) The number π is being divided by increases by taking 2 to the power of the iteration (e.g, 2^10 = 1024) and so does the number it is being multiplied by.This allows us to come up with a formula (n representing the number of iterations): Total perimeter = (circumfrence of smaller semicircle)x(number of smaller semicircle) Total perimeter = (π/(2^n))x(2^n) Total perimeter = π, since 2^n cancels out Since π is a constant, it is unchanging, so no matter the number of iterations, the total perimeter is always π.

[@archidebroy5442]

Finally a proof of common sense

[@hydropage2855]

An expression for the total circumference is (2^n)(π / 2^n), starting at n = 0. Each iteration’s semicircles have circumference π/2^n, and there are 2^n of them. The limit as n goes to infinity of this is π still

[@XBGamerX20]

the limit is 0 actually. but true π doesn't change, its just a constant and taking the limit to 0 completely obliterates any constant down to 0.

[@hydropage2855]

@@XBGamerX20 What in the world are you talking about… (2^n)(π / 2^n) simplifies to π because of 2^n / 2^n, so it doesn’t matter what n is anyway, and the limit to infinity is still not affected. Do you understand algebra by chance? Lmao

[@XBGamerX20]

@@hydropage2855 i wasn't talking about your formula. I was talking about π/2^n which gives the circumference of a single semicircle and if you take the limit to infinity the fraction becomes zero which means there's no circumference due to infinite amount of semicircles. and no circumference means no semicircle so theres nothing else besides the diameter. what you said is the most obvious thing and there's no point to take the limit of a constant. don't act like you know everything.

[@hydropage2855]

@@XBGamerX20 I’m not acting like I know everything lol I literally said it’s adding up all of those circumferences, and you’re talking about something totally different. The value of any given circumference goes to 0, but their sum remains as π regardless because there’s 2^n of them, making the value of n itself not actually matter as you take the amount of slices to infinity. Adding up infinite things that all go to 0, ever heard of an integral? Pretty sure those work pretty well. Also there’s absolutely a point to take the limit of a constant, the fact that the constant doesn’t depend on n anymore is the entire point, it’s why the limit is still π as n goes to infinity, because n cancels out in the first place

[@XBGamerX20]

@@hydropage2855 you don't have to take the limit because you simplify the expression anyway and simplifying means its no longer dependable to n on any circumstance. I really don't have to continue this any further, because I'm not taking more of my time to prove that I indeed know algebra or calculus to someone on youtube.

[@microsoft7000]

Should it not be the infinite series of the summation of the diameters of the smaller circles equal 2?

[@daniman7256]

Ye i think so

[@YG-kk4ey]

That actually will be the case. Since they original diameter was 2

[@yiutungwong315]

π = 1 + 1 = 1 + 1/2 + 1/4 + 1/8 + ... = 2 In the Riemann Paradox and Sphere Geometry System Incorporated

[@inevitable1222]

If the semicircles are infinitely small to match the diameter there are an infinite number of semicircles(taking the limit) so it always adds up to pi

[@yiutungwong315]

1 + 1/2 + 1/4 + 1/8 + ... = 1 + 1 = π = 2 in the Riemann Paradox and Sphere Geometry System Incorporated

[@Nico-vm9xp]

More interesting is that when you turn your phone to the side, you have a sleeping snorlax

[@alexkelberman2307]

Love that

[@cordongrouch9323]

You shall never die... + unless you change your mind.

[@DebashishGhoshOfficial]

The conundrum is resolved when you realize that not all infinities are equal.

[@RetroGDGamer]

While the curves get infinitely smaller, there is also infinitely more curves. The ratio of how much smaller they are to how many there are stays the same.

[@Pinch0fEverything]

No because however small the circles may be there is a diameter/radius of said circle and adding that together makes pi

[@MisterSnail1234]

Exacly and the limit is pi, not 2, since the final line length never approaches 2, but stays at pi

[@yiutungwong315]

π = 2 in Riemann Paradox and Sphere Geometry System Incorporated

[@yiutungwong315]

Tau ÷ π = 2 Alright 👌👍

[@meccamiles7816]

The total arc lengths of the semicircles does not converge. It remains the same.

[@theadamabrams]

A constant sequence _does_ converge (to that same constant). To me that's obvious, but if you want to use the official definition of lim_(n→∞) a_n on the constant sequence a_n = C, that works too: for any ε there exists N (in fact, N = 1) so that if n > N then |a_n - C| < ε (in fact, |a_n - C| = 0).

[@meccamiles7816]

@@theadamabrams ah, yes, true, true. Hopefully you understood the spirit of my words rather than the literal meaning. That said, your correction is valid. Thanks.

[@LawrenceMalhame]

It’s like Australia’s coastline. No matter how small the bumps are the ratio still makes a difference

[@user-tl9wo2xo2v]

I'm an undergrad, and for quite some time (4years to be exact) I've been going the same route to my uni, which incorporates some diagonals and thus considered popular among students. But there was a similar argument in my head for a while - if a diagonal is just a hypothenuse of a 1 sided triangle as the radius in this example, then by applying the same logic in limit the right-angles path would be the same length as the diagonal, which is counterintuitive. This for me was stemming from misunderstanding of the limits on my first year. The argument shatters if we consider an object moving along both lines at the same rate. As this limiting line doesn't change its length, it would still require longer time to traverse than the diagonal. This sorta reminds of integrating over the length of the line which is neat. I hope this all made sense 😅

[@Drittux]

There is infinite distortion. This reminds me of the coastline paradox.

[@pegasus567]

3b1b did a video on this. The partial, finite perimeters don't actually approach a value because the perimeters shown aren't approaching a straight line; they'll always be curved, no matter how many times we recur the pattern

[@farrankhawaja9856]

It does approach a straight line in the limit though. This explanation is incomplete. The limit of the points is exactly on the line.

[@mirkotorresani9615]

Actually the semicircles are converging to the diameter.

[@CMT_Crabbles]

The derivative must also converge, those curves will never flatten out.

[@Konzon]

The problem is when you said the circles "squash down and flatten". Small circles aren't flatter than large circles. Just cause you don't have the resolution doesn't mean it's not still curvy.

[@KaliFissure]

The fractal density is Pi - 2 Since the path is always Pi long whether the frequency is 1 or 128. But the travel distance is 2 So actual travel Pi minus minimum path 2 = 1.14159.... is the fractal dimension of a flat< circle. ??

[@DavidAW27]

Cantor dust is not length zero. Same idea.

[@int2006]

If you solve using limits i.e Limit nXπ/n N times circumference on small circle which is π/n So answer is π

[@robyngwendolynshiloh5277]

Instructions unclear, I got an empanada

[@czar6203]

I remember another argument that pi = 4 lol

[@ralphinoful]

The intuition here comes from the theorem, if Sn is a sequence that converges to L, and f is continuous, then f(Sn) converges to f(L). In this case, our sequence Sn, is a sequence of semi-circles. And L is the line. But now we need to look at what f is. f would be some sort of function, computing the length of the object. Fundamentally, the function f, that computes arc length, with respect to path, is discontinuous. To be more precise, if I have a function f, mapping C-->R, where C is the set of all paths, and R the reals. If I take two paths c1 & c2, both infinitely close together, it is possible that the arc length f(c1) is wildly different than the arc length f(c2). Small changes in path, does not necessarily mean that there will be small changes in path length. For intuition on why two paths can be infinitely close, but different lengths. I would tell you that "direction" is one of the most important things to consider, when computing arc lengths. The direction of the two paths are extremely different. The derivative of one, does not converge to the other. The infinite case of the semi-circles is alternating between pointing left and right, while also pointing mostly down. The line, is only pointing down. For intuition on why discontinuity matters, you can look at simpler examples when considering infinite processes. lim{3, 3.1, 3.14, 3.141, 3.1415, ...} converges to pi. But consider the function f(x) = 0 if x is rational, 1 if x is irrational, a clearly discontinuous function. lim f({3, 3.1, 3.14, 3.141, 3.1415, ...}) = lim {0,0,0,0,0, ... } = 0 != 1 = f(pi) = f(lim{3, 3.1, 3.14, 3.141, 3.1415, ...}). The limit of the function is only equal to the function of the limit, when the function is continuous.

[@oscarmartinpico5369]

The sum is always the same: pi. It those not matter what tiny are the ripples, there are that many as the divisions are. and there is not reason to not take them into account. And, by the way, the limit is not a line, It is a ripple, lim|n->inf| n (pi 1/n) = lim|n->inf|| pi = pi. In limits of this types, what takes out of consideretions are the fixed amount of something, whitch those not happens in this example.

[@DragonHeart53]

This process makes me think of how to create huge amounts of surface area on otherwise tiny structures. Like the pads on gecko feet, where despite having such a small area the whole pad takes up, but it's entire surface area is many times greater than the diameter of each toe on the gecko because of so many tiny little hills increasing the total surface area.

[@blacklight683]

At the end... he is using the curved part to replace the stright part

[@Yokuyin]

This is why the length of a curve is defined as the integral of the derivative of the parameterization. This way, any limiting process has to ensure the derivative converges, which is not the case in this video.

[@jorgitus9939]

"Un chingo de webaditas es lo mismo que una webadota" -Aristótoles

[@FireyDeath4]

Don't even get me started on infinitesimal angles

[@Serbravez]

It’s the same concept as the staircase paradox (I’m not sure if that’s its name)

[@jens6076]

That’s some lovely ideal triangles in the half plane model of hyperbolic geometry you’ve got there

[@thewoodpeckers655]

It’s like a spring compressing itself to appear smaller, but when fully expanded out, is twice as long.

[@GamerX-2000]

I’d imagine it’s similar to the idea of how a fractal, like the Mandelbrot set, as an infinite perimeter, but a finite area. While we know that these semi circles are confined to a small space and their form is infinitely small, we cannot discount that they have this form. We have to account for every bit of their infinitely small circumference for their perimeter to be considered accurate, just like we can’t assume the circumference of the Mandelbrot set reaches a finite value despite it’s infinitely small crevices and shapes.

[@YG-kk4ey]

Just like there's an infinite number of fractions between 1 and 2

[@nikitakipriyanov7260]

I love this. I can also suggest another instance of such conundrum The explanation is, the sequence of curves does converge to the vertical line, but it doesn't converge to it *uniformly* ; this notion of uniform convergence was actually coined to help such cases. That is, all points should move and approach the line with relatively same speed, but they aren't: the middle point takes a big jump and then it stays stationary, various points at p/2^n do series of a smaller jumps before jumping right to the end on the n'th stage and stay still, and all other points perform decreasing jumps indefinitely. This is not uniform convergence, so the taking limit operation can not be swapped with the sum (of lengths of little half-circles) operation.

[@enderoftime2530]

If I’m saying it right, outside of the ends of the semicircles, the curvature (second derivative) grows in magnitude for each smaller division. If it were actually approaching the value of two, the curvature and the slope/gradient (2nd and 1st derivatives) should go to zero indicating a straighter and straighter line.

[@autumn948]

this just goes to show that limits can often be misleading, and should not be taken as true values of the functions they work on

[@randallbrittain]

Because the line now looks like a coiled telephone cable. Stretch it out and it’ll be back to its old irrational self.

[@dawnbeckett9179]

This is similar to the issue with measuring coastlines accurately. When calculating smaller measurements, you will end up getting a much larger length than when estimating from a larger scale, so while it looks like two, all the tiny curves still add up to π

[@NightSkyJeff]

Since the perimeter of a semicircle is known to be pi * r, and the perimeter of a semicircle with r=1 must be longer than 2r, pi must be greater than 2.

[@GMPranav]

This argument never works of length, but does work for areas.

[@Flinxy]

π = 2 All of universe collapsed

[@yiutungwong315]

In English and Desutch... Eleven and Twelve that means π = 2 In the Universe 🌠🌠 and University and Universal ♾️💍 That Means π = 2

[@diamondmemer9754]

If you approximate enough shiny pokemon stop existing so don't do that

[@CalimeroDu64]

That "almost" doing a lot of work lol

[@stevenmcmaster8219]

The small semi circles look flat, but they aren't.

[@johnknowledge4064]

Phew, you scared me for a second. I thought you were going full blown Terrence Howard on me. Anyway, the definition of a line states that it has "no width or curvature", so any equation that includes line width is erroneous.

[@truenova1594]

Formula of the circumference of a semicircle is πr + 2r. So, here the circumference will be π+2.

[@MathVisualProofs]

Yea. My bad. I meant to say arclength

[@xcheesyxbaconx]

My first thought was the coastline paradox. I'm not really sure how these sorts of things are resolved, but we all know that coastlines are not infinitely long, and that pi is not 2.

[@user-gh9qm5bd6o]

Each semicircle with length pi/n (where n is 2^k) is longer than two sides of a triangle with sides sqrt2*R/n, sqrt2* R/n and 2R/n, so pi > 2sqrt(2) > 2. So the length of the combination of semicircles is bounded below with some number. It would converge to the straight line if there had not been such number at least after some N

[@calebturner1015]

This actually reminds me a lot of the coastline paradox!

[@aidengrubbs4655]

Yeah it reminds me of the pi=4 thing where you take a square and fold the corners until it almost perfectly resembles the circumference of the circle

[@sumcarson]

Speaking more formally, this is an indertiminant limit (ie. pi/h as h approaches infinity is not determininant)

[@eliseocarizo6774]

Here's the formula ive made with that. It's like this Let n= number of semicircles 1st semicircle the lengt of the curve is (pi×1)(1) =pi For 2 semicircles inside, new r becomes half of original r, therefore (pi × ½)(2) =pi If u can see the pattern, the formula would be (pi× new radius) times the number of semicircles. Lc=(pi×r)(n) If you follow the pattern you will see that n will always be equal to the reciprocal of r. Take for example n=100, if you draw 100 identical semi circles inside, you will get r=0.01 as the radius of each semicircle. Following (pi×r)(n) youll still get pi. Even if you make an infinite amount of semicircles, you will also make a value of r = 1.0×10^(-infinity) which will result to a length of curve being pi So pi is not equal to 2.

[@Telruin]

This is sort of like an inverted "Coastline paradox"... Come to think of it. This sounds like something that would have a name of it's own. Edit: It did. "Staircase paradox"

[@fatimaalaa2659]

This is the same thing as the pi=4 argument where you put a circle inside a square & progressively fold the square onto the circle's circumference

[@attempt58]

By the same argument, I can also state that √2=1 can't I? Take a square, with the sides being 1 unit each. Divide two of the sides in half, and connect them in the middle, essentially creating an L shape. This new shape has the same circumference as the starting square. Repeat a few more times, and you should have a staircase, still the same circumference as the starting square. If we repeat this an infinite number of times, we'll eventually get a triangle. But the 45-45-90 triangle's sides are 1, 1 and √2. Yet, we just found our "staircase" shape's sides to be 1 unit each. Does this make sense? If it doesn't I might make a short video and link it here.

[@cesarjom]

Take one semicircle that has been shrunk down, such that its half circumference S= pi/n and n is the number of semicircles along the vertical line length 2. So here the semicircle radius is 1/n and thus the diameter length is d= 2/n. We can say for each small semi-circle that S > d. Now summing up n many semi-circles, S, and diameters, d... we have n*S > n*d By substitution, n * (pi/n) > n * (2/n) Hence pi > 2 still holds!

[@BikerFromSpace]

The term "infinitly squiggly" comes to mind. It will never be truely flate

[@alfgarcia965]

if the circumferences are measurable, then it is evident that it is not a flat line. this presents the same type of stipulation as measuring the perimeter coastline of an island, where the length depend on the size and precision of the measuring stick and the detail that we want to include, since the perimeter is not a distinct edge. another way of understanding the discrepancy is to realize that there are an infinite amount of real numbers that are in between the whole numbers 1 and 2, or the number can depend on how much detail or increment you deem as considerable

[@ricardiumhues]

Coastline paradox

[@bogdanalexandruluchian6628]

The basic explanation here is that you would have to interchange a limit with an integral, so some very well-known results such as Lebesgue's Dominated Convergence Theorem give you the desired outcome as long as the limit of the considered sequence is the function you integrate (you would also have to check that the sequence is bounded by a measurable function). The key fact here is that the length of a curve is actually defined in terms of the derivative of the curve, so even if a sequence of curves converges to a predefined one in the usual sense (in the sup-norm, i.e. C^0 norm), that would not be enough for the sequence of lengths to converge to the length of the original one- for this you would need the sequence to converge in the C^1 norm, and by some very immediate calculations one may show that it's enough to check the convergence of the derivatives in the sup-norm when the sequence of curves and the given curve all coincide in at least one point (the same point for all) - you can look at this argument as having a sequence of approximating solutions for an ordinary differential equation.

[@codatheseus5060]

It appears you're creating a sideways stars over Babylon inside that semicircle

[@jeffking]

This is why the coastline of California is infinite.

[@apope2087]

Many, many things here…. The smaller circles don’t “squash” or get distorted..they are in proportion just smaller. You could apply this argument with any shape (circles, squares, pentagons etc. nothing unique about circles) and you can “prove” π to be a lot of different values depending on the shape you used. What it boils down to is you’re “splitting” the line into “n” amount of semicircles with π/n arc length. So what is the limit as n-> ∞ of n*π/n? Well, π is a constant so you want the limit of n/n. If you want to cheat? The answer is 1 so the overall limit is 1*π or just π. If dividing by ∞ doesn’t sit right, then use L’hopital’s rule for evaluating limits of the form 0/0 and ∞/∞ (which we clearly have here). Differentiate with respect to the variable (n) top and bottom separately and find the limit of that. It is 1/1 which is 1, back to 1*π which is π. The infinite limit is π of this construction. Hands down. So why doesn’t it match 2? Simple: you’re using a 2D shape to represent a 1D shape.

[@blargblarg5657]

BSJ actually built a mail sorting machine using gears where pi was exactly three for neatness.

[@pauldacastello3266]

Fun fact: No matter how small those circles get, they always have an non zero area. A line always has a zero area.

[@Neptoid]

Ey guys in the 1st iteration from the semi circle you can see that the arc length sum of the two smaller semi circles is longer that the largest and original circle, if you imagine those lines being bendable cords because those preserve length and put it along the largest original semi circle. If you got intuition from doing technical drawing and art at least

[@iwsifify]

This actually kinda gives the idea of what it means that "some infinities are bigger than others"

[@klingonxxxxxx]

Even after 1milion subdivision steps, in each semicircle you can inscribe a right triangle with hypotenuse = Diameter. The length of each semicircle C will be > than the sum of the legs, C > Sqrt(2) * Diameter. By adding all the small C together we will get LTotCirc > Sqrt(2) * 2 Thus PI > 2.83

[@Xolareclipxe]

The semicircles never "flatten down" to a line. Zooming in will always reveal semicircles

[@sylph4252]

If we start with a radius 0.5, we'll end up with π=1, and the same goes for any radius: π=2*r. Moreover, if we start creating semicircles that are one third of the size of the original, we'll get π=3*r, and the same goes for any other fraction of size. In the end, we get π=x*r, which is as far from a constant as I can think of

[@axelinedgelord4459]

you don’t even have to explain it, i recognized it immediately

[@wilsonwilson3674]

The ratio of the circumference to the diameter, pi/2, remains constant, no matter how small a semicircle is. So adding up all the smaller circumferences and diameters (or equivalently, multiplying each by n) gives us the same exact ratio.

[@origamiowen5870]

Yes the curves are still there and affect the distance even if they can not be noticed

[@williamjaramillo5105]

I didn’t even see where the argument was

[@boium.]

People keep saying "the circles have width" or "it's not straight" and they are almost correct in their resoning. The reason why this argument fails is because the tangent vectors of the path that's made of the smaller circles, doesn't approach the tangent vector of the line. Recall that in high school you learned the formula of the length of a curve f(x). The length was given by L = int √ ( 1 + f'^2 ). You see the derivative of f in that functions. This gives you a hint that velocity vectors and tangent vectors matter.

[@Poyni]

By taking limits of semicircles we see that π=2 By taking limits of steps going along the outside of the circle we see that π=4 Therefore π=2=4 2=4 QED

[@alexbulza50]

Plot twist: the original semicircle is a part of another semicircle

[@nathanpatty6020]

That’s like saying a polygon will turn into a circle if given enough sides.

[@Drawoon]

if you do the math just right, you can make any number equal any other number

[@pizza8725]

Yeah bc those curves still add length,small but they add up enough times to equal pi

[@Connor_Crain]

This is like shorelines the more accurate you measure them the more infinitely long they get

[@Edmundajw]

It's because, although you might suppose that the semi circle that matches a tiny section of the line, to be only slightly longer than it, so it might as well be the same line- that ratio of curve length to the line is always 3.14 to 1. So no matter how far you zoom in, whether it's 1000 times, or 2 times, or just the one semi circle, the relationship between the length of the curves, and the total length of their diameters/their associated line, is always 3.14 to 1.

[@patrickschott265]

Jokes on you, I didn’t understand this video so you can’t trick me into thinking Pi = 2

[@missoulasam]

Imagine two paths: one that never deviates into another dimension and another that deviates along a single dimension. If they share an origin, the second path will have a length of pi, and the straight path will have a length of 2. If the wandering path moves in positive and negative directions on the added dimension, the length becomes 2 pi.

[@joaquinruggiero3997]

No matter how many times you divide the circles, you will always be able to zoom down and see tiny half circles, never actually forming the straight line that is equal to 2

[@haroldmurphy4791]

Simple, no local linearity.