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So, I realise that the area of a circle is actually $\pi r^2,$ so basically half of $2\pi r^2,$ however, after trying to prove the area of a circle for myself, I came up with a really convincing and intuitive theorem for why it is 2πr^2 (even though it is not).

By the way, I've set the radius to 100/2π so that the circumference would be 100, as I wanted to mark the percentage moved across the circumference for any given angle. There is no reason for this with respects to (my attempt of) this so-called 'proof', it is just for convenience.

If you take the circumference of a circle and you flatten it out into a straight line, it would look something like this:

enter image description here

From here, I reason that a line segment can be drawn across any, indeed every point along the line equal to the radius of the circle, since said line is equal to the circumference of the circle to begin with, and thus this step just falls out of the definition for how a circle is even defined. Here are a few examples:

enter image description here

If you did this for every point across the circumfence and its corresdonding line, you would have filled in the area of the circle on the left, and created a rectangle on the right, whose length (x-axis) is the circumference of the circle and whose height (y-axis) is the radius:

enter image description here

It's not obvious what the circumference multiplied by the radius would be, but since we know that the diameter goes into the circumference π times, we can just rewrite the circumference as π•diameter, and in turn rewrite the diameter as 2•radius, so 2πr. Having rewrote the circumference as 2πr, we just multiply this by r to derive 2πr^2.

As you can see, this attempt of a proof involves only three steps. It is short, simple, intuitive, and, dare I say, eloquent. Above all else, however, it is also wrong. You can even see this visually, just by eyeballing the image.

Now, I need no convincing on what the area of a circle actually is. I've looked up actual proofs online showing why it is πr^2, and I also simply trust the likes of Archimedes, as well as Pythagoras, Newton, the team at NASA, etc. What I need convincing of, however, is that my proof is incorrect. By "convincing", I don't mean it in the usual sense, but in quite a literal one. As in, I can consciously accept that my attempted proof is incorrect (again you can even see this visually by comparing the areas of the circle/rectangle), but my heart and soul cannot, because I've managed to construct such a simple, easy to follow, and intuitive proof that ended up being false. It's clear that the mistake made along the way was not a technical one. I mean sure, the mistake is that my derived equation is off by a factor of exactly 2, but there's something very fundamental about the nature of maths itself that I clearly have not grasped, and I have absolutely no starting point to work from in trying to figure out what that is. This is, to me, like trying to understand why two plus two is two instead of four, for I cannot wrap my head around it.

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    $\begingroup$ Well, when $r=1,$ the circle centered at $(0,0)$ of radius $1$ is contained in the square with corners $(\pm1,\pm1),$ which has area $4.$ $2\pi>4,$ and the circle can't have area more than that square. $\endgroup$ Feb 21, 2023 at 17:18
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    $\begingroup$ Hint: area is not an invariant for all transformations... Second hint: what happens to your center with your transformation? $\endgroup$
    – Martigan
    Feb 21, 2023 at 17:20
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    $\begingroup$ This is a word for word duplicate of the other question no? $\endgroup$ Feb 21, 2023 at 17:55
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    $\begingroup$ I'd say the flaw in your "proof" @user2901512 , as already expressed in one of the answers below, is that you are assuming bijections preserve area. Yes, there is indeed a surjective bijection from every point in your circle to every point in your rectangle. But that still does NOT mean they have equal area. $\endgroup$
    – Mike
    Feb 21, 2023 at 18:18
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    $\begingroup$ Amusingly enough, you could alter your fake proof to get any value for the area between $0$ and $2 \pi r^2$. Say you want to get an area of $.01 \pi r^2$. Instead of drawing your dots along the circumference of the circle itself, draw them around a circle of radius $.005r$. Then, instead of laying the circumferential circle out along the $x$ axis from $x=0$ to $x=2\pi r$, laying that little circle of radius $.005r$ out along the $x$ axis from $0$ to $.01 \pi r$. Now line up your radii. You'll get a rectangle with side lengths $.01 \pi r$ and $r$. Voila! Area $.01 \pi r^2$. $\endgroup$
    – Lee Mosher
    Feb 21, 2023 at 18:36

3 Answers 3

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This argument. fails (as you know) because you can't just think of the circle as a collection of radii, then move the radii into a rectangle to get the area. You are trying to add up the areas of infinitely many segments each of which has $0$ area - it's no surprise that you get conflicting answers.

But I congratulate you on your clever attempt. Here's how to make it work. Instead of cutting the circle into infinitely many radii, cut into pie slices and think about what happens as they get thinner and thinner (but always real slices).

enter image description here

The picture is from https://www.colorado.edu/csl/2017/03/23/slices-pi .

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  • $\begingroup$ Aye, I've seen this proof, as well as a second one showing how a circle can be thought of as circles within circles that when unflattened form a triangle. But here's the thing: Proofs like this seem as intuitive to me as the one I constructed. The ultimate goal for me, however, is to understand why my proof fails, not why others succeed. $\endgroup$ Feb 21, 2023 at 17:40
  • $\begingroup$ You mention that the mistake could be trying to add up areas of infinitely many segments, but what about calculus? Isn't that the main idea behind trying to find the area under a curve, adding up infinitely small rectangles? When you have a rectangle of zero width, you basically get a line, but couldn't that be said to be the radius of a circle? $\endgroup$ Feb 21, 2023 at 17:40
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    $\begingroup$ In calculus and in the above proof, the shapes having some dimension equaling $0$ are the end product of some limiting process. However, in your fake proof, this is not the case for your rectangle; where is the limiting process here? You never started with some shape having some non-zero dimension converging to 0. $\endgroup$
    – Zuy
    Feb 21, 2023 at 17:47
  • $\begingroup$ @Zuy, Ah, fair enough. Thank you $\endgroup$ Feb 21, 2023 at 17:55
  • $\begingroup$ @user2901512 The whole point of integral calculus is figuring out correct ways to add up infinitely many infinitely small pieces. Archimedes began it. Newton and Leibniz took the next big steps, but still with some handwaving. It was the 1800s when it was made rigorous enough for modern standards. $\endgroup$ Feb 21, 2023 at 18:50
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Consider using your strategy to compute the area of a 1x1 square: Place the square on the $x$-axis between 0 and 1, divide it into uncountably many vertical line segments, and move each segment horizontally from its initial position $x$ to position $5x$. The moved segments fill a 5x1 rectangle, so the area of the square must be 5.

Of course this is wrong, because when you move the segments, you're stretching the shape in a way that doesn't preserve its area.

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  • $\begingroup$ But to me, moving those horizontal slices of the 1x1 square would be like moving infinite radius's of a circle. By the way, I don't see why the slices would be duplicated along the way, you just move one slice of the square from left to right, over and over, until every slice has shifted 5 along to the right. To me, that is like splitting a circle into infinite slices, each slice being a radii, and rearranging it into a rectangle. A bit more complex than what we did with the square, but the idea seems the same. $\endgroup$ Feb 21, 2023 at 17:53
  • $\begingroup$ @user2901512 You object to taking line segments whose endpoints span a segment of length $1$ and placing them perpendicular to a segment of length $5$, and yet you do not hesitate to take line segments whose endpoints are all at one point (the center of a circle) and set those endpoints along a segment of length $100.$ You should have thought twice about that. $\endgroup$
    – David K
    Feb 21, 2023 at 17:57
  • $\begingroup$ @user2901512 Exactly. So do you find this to be a convincing proof that 1=5? $\endgroup$
    – Karl
    Feb 21, 2023 at 17:58
  • $\begingroup$ @Karl, No, but to me it seems like you just moved 1 along four to the right. OK so it's like, if one was an actual box, it's like you just moved the many slices of the box four meters right from where it was, not duplicated it four times for every metre. $\endgroup$ Feb 21, 2023 at 18:20
  • $\begingroup$ That's the point. You aren't duplicating things, you're just moving them, but because there are uncountably many pieces, the area is not preserved. Formally speaking, measures are only guaranteed to be countably additive. $\endgroup$
    – Karl
    Feb 21, 2023 at 18:30
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You have exploited the fact the each radius is infinitely thin. Taking something finite, cutting it up into an infinite number of infinitely small things, and then reassembling them does not preserve area. There are lots of paradoxes like yours based on this fact. Check out the Banach-Tarski paradox for one of the most mind blowing. (Although actually, the B.T. paradox only breaks things into two pieces, but I'd argue that they're infinitely complicated. But it is another case of area (okay, volume) not being preserved when cutting up sets.)

If you want to get a visceral feel for what went wrong, try reversing your process: cut out that strip of paper, and start packing it in to your circle. You'll notice that it starts to get all bunched up, with the bunching getting worse the closer you get to the center.

The proper way to add up an infinite number of infinitely small things is with integrals. What your two different ways of slicing up the area amounts to is using two different coordinate systems, and when changing coordinates there is a scaling factor, the Jacobian, that you multiply by that exactly compensates for the stretching or bunching you've noticed.

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    $\begingroup$ Banach-Tarski uses only finitely many pieces (!), so I'd say it's more about the weirdness of non-measurable sets than about dividing things into too many pieces. $\endgroup$
    – Karl
    Feb 21, 2023 at 17:55
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    $\begingroup$ @Karl - I just realized that upon giving my answer a copy edit pass! Do you think my additional comment fixes it, or should I just take it out? I figured the OP would enjoy seeing it, but maybe it's just too different. $\endgroup$
    – JonathanZ
    Feb 21, 2023 at 17:58
  • $\begingroup$ Ah, I've never heard of changing coordinates (at least not like this) or Jacobian, it sounds very interesting, I'll be sure to check it out! Hopefully it'll get me one step closer to understanding why my proof fails. $\endgroup$ Feb 21, 2023 at 18:00
  • $\begingroup$ @user2901512 - Your 'unrolled' version is very close to polar coordinates, where $x$ is the angle, and $y$ is $1-r$, so make sure to check out the Jacobian for polar coordinates in particular. $\endgroup$
    – JonathanZ
    Feb 21, 2023 at 18:04

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