# Watermelon Paradox: A Line/Surface/Object/HyperObjext/... has infinite length/area/volume/hypervolume? [closed]

I discovered (I do not know if I'm the first one tho) that when cutting a watermelon in half, I added two surfaces while the volume of the two different pieces is the same.

(You gotta be silly with the name of the paradox or no one will remember it, but it need to contain some truth or it makes no sense)

So the surface area of the watermelon increased while the volume stays the same.

To simplify, I will just use a 2D square, the area acts as volume and the circumference of the square acts as the surface area.

Imagine a square with the length of 2:

|--2--|
-----
|     |
|     |
-----


(not a very accurate square)

$$A=2^2=4$$

$$C=2*4=8$$

Then, we split the square in two:

 --   --
|  | |  |
|  | |  |
--   --


Now, the long side, $$a$$ is 2, and the short side, $$b$$ is 1

$$a=2$$

$$b=2/2=1$$

$$A_1=a*b=2*1=2$$

$$A_2=a*b=2*1=2$$

$$A_{total}=A_1+A_2=4$$ (Same as the unsplitted square).

$$C_1=(a+b)*2=6$$

$$C_2=(a+b)*2=6$$

$$C_{total}=C_1+C_2=12$$ ($$C_{total}-C=4$$ because there is two more segments with length 2 added)

Then we split the two rectangles into 4 rectangles so that the circumference is

$$C_4=(\frac{2}{4}+2)*4=20$$

So, the formula for getting the total circumference of a square of 2x2 splitted into a Real Number of rectangles without changing the total area of the square is:

$$C_x=(\frac{2}{x}+2)*x=2x+2=2(x+1)$$

Funny thing is, it doesn't converge, meaning

$$\lim_{x \to \infty} 2(x+1) = \infty$$

If we want to see the total circumference of x (x is a Real Number) number of splits of a mxm square without changing the total area is:

$$C_x = (\frac{m}{x}+m)*x=mx+m=m(x+1)$$

Now, here's the catch:

If a rectangle's short segment is zero, it becomes a line.

in this formula: $$\lim_{x \to \infty} m(x+1) = \infty$$

It describes when a square of mxm size is cut into infinite pieces along the vertical axis. Therefore, the short length of the rectangle becomes infinitly close to zero.

Therefore, a line is infinitly long.

Although I counted a line as two segments, you can always divide the formula by 2, and a line is still infinitly long.

$$\lim_{x \to \infty} \frac{m(x+1)}{2} = \infty$$

It also works with 3 Dimensions (and N Dimensions?)

Question:

I know this is not quite right, since a line have a length.

Bonus:

I found that the larger the square, the smaller the ratio of $$m(x+1)$$ to the area.

$$\frac{m(x+1)}{m^2}=\frac{x+1}{m},m\ne0$$

Meaning, $$\frac{d}{dx} \frac{x+1}{m}$$ decreases when m gets bigger.

So

$$\lim_{m \to \infty} \lim_{x \to \infty} \frac{x+1}{m} = 0$$

More wierdly, how can $$\frac{\infty}{\infty}=0$$?

Can we conclude from the above that $$m(x+1) = m^2$$?

My Works in 2D splitting: https://www.desmos.com/calculator/rlrcirxmnx

• Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
– Community Bot
Commented Aug 2 at 15:45
• Not exactly what you are asking for, but have you looked up Gabriel Horn? Commented Aug 2 at 15:48
• @Srini Yes I did. But I'm sure its not related. I'm asking if a line is infinitly long (since a line indeed have a length) in the context of the text before the questions. The bonus acts as bonus questions that popped inside my mind when writing this. Commented Aug 2 at 15:51
• I think this is related to that old “proof” that $\pi=4$. Where you inscribe a circle in a square, and keep folding in the corners of the square. Commented Aug 2 at 15:54
• Hi! I'm not sure I see a paradox here either. Sticking with two dimensions and a unit square (to keep things simple), you've divided it into $n$ $1$-by-$1/n$ pieces, so they have a total area of $1$ (as before) and a total perimeter of $n(2+2/n) = 2n+2$. This does indeed increase without bound. However, the individual perimeter of each piece is still just $2+2/n$, which approaches $2$ as $n$ increases without bound. The only "paradox"—if you want to call it that—is the final ratio of $2$ to $1$; however, that's only because we've counted both "sides" of the rectangle approaching the segment. Commented Aug 2 at 16:11

This is a nice example, and getting clear about what's going on here requires getting clear about how limits in two variables work. The resolution is that you are trying to commute two limits that do not commute.

Let's begin with a rectangle of length $$\ell$$ and height $$h$$, so initially its perimeter is $$p(1, \ell, h) = 2 \ell + 2h$$. If we split the rectangle into $$n$$ smaller rectangles lengthwise, we get $$n$$ rectangles of length $$\frac{\ell}{n}$$ and height $$h$$, whose total perimeter is

$$p(n, \ell, h) = n \left( 2 \frac{\ell}{n} + 2 h \right) = 2 \ell + 2nh.$$

Normally I would have just written $$p_n$$ but in this context we need to be clear that $$p$$ is a function of multiple variables. Now you've observed that:

1. If we take $$n \to \infty$$ then for any $$h \neq 0$$ (note this!), the total perimeter diverges: $$\lim_{n \to \infty} p(n, \ell, h) = \infty$$.

2. As $$h \to 0$$, the rectangles become lines and the total perimeter converges to the total length of the lines, which is constant: $$\lim_{h \to 0} p(n, \ell, h) = 2 \ell$$.

Both of these are true. But to conclude that "all lines are infinitely long" what you are trying to do (I think, I don't quite follow this part of your argument since you never make $$h$$ explicit) is to argue that

\begin{align*} \lim_{n \to \infty} \lim_{h \to 0} p(n, \ell, h) &= 2 \ell \\ &\color{red}{\stackrel{?}{=}} \lim_{h \to 0} \lim_{n \to \infty} p(n, \ell, h) \\ &= \infty \end{align*}

and that middle step where you try to commute the limit $$n \to \infty$$ and the limit $$h \to 0$$ simply doesn't work (and of course the fact that $$2 \ell \neq \infty$$ is the proof that it doesn't work).

If you never explicitly talk about limits but only talk about what happens for specific, finite values of $$n$$ and $$h$$ then there is no paradox and it's clear what happens: the total perimeter is just $$2 \ell + 2 nh$$, and the size of this number simply depends on the relative sizes of $$n$$ and $$h$$. If $$n$$ is very large, say $$n = 10^{100}$$, and $$h$$ is only kinda small but not that small, say $$h = 10^{-10}$$, then $$nh = 10^{90}$$ remains large. But if $$n$$ is only kinda large, say $$n = 10^{10}$$, and $$h$$ is very small, say $$h = 10^{-100}$$, then $$nh = 10^{-90}$$ is very small.

So there is no well-defined two-variable limit $$\lim_{n \to \infty, h \to 0} p(n, \ell, h)$$; the behavior of $$p(n, \ell, h)$$ for large $$n$$ and small $$h$$ depends on how quickly $$n$$ gets large compared to how quickly $$h$$ gets small, and that's all. Typically the first time you'd see an example of this is in multivariable calculus where it's relevant to questions like when partial derivatives commute, but the underlying phenomenon is pretty elementary if you think of it in terms of how big and small various things get.

Incidentally, this fact that the surface area of a watermelon (or anything else) gets larger when you chop it up into smaller bits is an important observation in several parts of science, for example it is related to why we chew our food (this increases the available surface area for enzymes to digest it!); see surface-area-to-volume ratio for more on this.

• This is about cutting squares, so $h=ℓ$. $p(n,ℓ,h)=n(2\frac{ℓ}{n}+2h)=2ℓ+2nh=2ℓ+2nℓ=2ℓ(n+1)$. Also, because we are cutting in the vertical axis, the height doesn't change. While the bottom and up size segments get cut into n segments as denoted by $\frac{ℓ}{n}$. Commented Aug 3 at 1:02
• @Ivan: sure, we can consider that case. Do you also want to take the limit as $\ell \to 0$ then? If you take that limit first you get $0$, and if you take the limit as $n \to \infty$ you get $\infty$, and again these limits don't commute. So the phenomenon is the same. Commented Aug 3 at 1:05
• @Ivan: but I don't understand how you're trying to conclude that "a line is infinitely long" in this setup. Which line are you referring to? Commented Aug 3 at 1:07
• When splitting a square into infinitly many slices, until $\frac{ℓ}{n} = 0$, the short side becomes 0, and you get a line. Now, by dividing the output by two (because the surface area actually counts the two "sides" of the line), we get $\frac{m(x+1)}{2}$. But it does diverges into infinity. Therefore, the length (original height of the square) of a line is infinite (according to the formula). Commented Aug 3 at 1:14
• @Ivan: I'm afraid I still don't follow. I agree that if you cut the square up into infinitely many slices you get $\lim_{n \to \infty} p(n, \ell, \ell) = \lim_{n \to \infty} 2 \ell (n + 1) = \infty$. This doesn't imply anything about $\ell$. Are you perhaps thinking about the average perimeter? This is $\frac{p(n, \ell, \ell)}{n} = \frac{2 \ell (n + 1)}{n}$, and as $n \to \infty$ it converges to $2 \ell$, twice the length, which is also as expected. There's still no infinite length here. Commented Aug 3 at 1:33