Is Pi = 4?

In Continuous Taxicab Geometry the blue shape and the grey shape have the same perimeter.

I was just showing this to some students as a 'proof' that $\pi = 4$ . One student shrieked and ran across the hall to hide behind another math teacher. I call that a success.

It does not make sense!!! When you cut the squares you're approaching the circle! Thus, the perimeter is the same as the circle... Somehow the question is "lost in translation"???

I understand that explanation but I'd don't get how the removal of the squares doesnt approach the perimeter of the circle? How is that different than a pixelated cirlce ? Does the pixelated circles perimeter have the same length as a square with legnths = diameter ? (Which doesn't make sense).

While it seems intuitively clear that the grey area will approach the area of the circle, I saw no way to prove this conclusively. Is there a simple way to do this, or as you said, to show that every point outside the circle will eventually be removed by a red square?

Let $B_{n}\subseteq\mathbb{R}^{2}$ denote the geometric object at stage $n\in\mathbb{N}$ and $\partial B_{n}$ its boundary. Similarly let $S$ denote the disk and $\partial S$ its boundary. Let $\mu(\cdot)$ denote the operation computing the 2-dimensional area $\ell(\cdot)$ the operation computing the lengths of objects. Clearly we have the following mere FACTS:

$\begin{array}{c}[c]{rcl} \mu(B_{n}) &\longrightarrow &\pi=\mu(S)\\ \ell(\partial B_{n})=4 &\not\longrightarrow &2\pi=\ell(\partial S)\\ \end{array}$

We need to answer two questions: 1. on what grounds could one even justify that $\ell(\partial B_{n})\longrightarrow\ell(\partial S)$ ; and 2. why does this fail? Ie, how could one almost be tricked, and why is that this reasoning is invalid.

Towards 1, we have in some not-yet-specified sense that the objects are approaching each other:

$\begin{array}{c}[c]{rcl} B_{n} &\longrightarrow &S\\ \partial B_{n} &\longrightarrow &\partial S\\ \end{array}$

Eg one could construct an homeomorphism that starts with the square (or its boundary) and approaches the disk (or its boundary) passing through each stage $n$ . The typical reasoning employed falls under the scheme: when a sequence of geometric objects converges to a particular object, then so too do (particular) attribute. We thus need to determine 1a do the geometric objects really converge and in what sense?; and 1b is there a theorem which guarantees that the desire attribute is continuous?

To answer this, we also need consider, just how exactly length (and for that matter, area) are being computed based on the geometric objects.

ANSATZ 1. Length computed via approximation.

Given a $C^{1}$ curve $\mathcal{C}$ , parameterised by say a $C^{1}$ function $f:[0,1]\longrightarrow\mathcal{C}$ and a metric $d$ on $\mathbb{R}^{2}$ , the Length of a the curve is given by

$\ell_{d}(\mathcal{C}) := \lim_{|\Xi|\longrightarrow 0}\underbrace{ \sum_{i<N(\Xi)}d(f(t^{\Xi}_{i+1}),f(t^{\Xi}_{i})) }_{=:\ell^{\Xi}_{d}(C)}$

where the limit is computed over all finite partitions

$\begin{array}{c}[c]{rcccccccc} \Xi &: &0=t^{\Xi}_{0} &\leq &t^{\Xi}_{1} &\leq &\ldots &\leq &t^{\Xi}_{N(\Xi)}=1\\ \end{array}$

of $[0,1]$ and directly ordered by refinement. (One also defines $|\Xi|:=\max\{|t^{\Xi}_{i+1}-t^{\Xi}_{i}|:0\leq i<N(\Xi)\}$ and requires this to converge to 0.)

Using this, and the metric $d(\mathbf{p},\mathbf{q}):=|q_{1}-p_{1}|+|q_{2}-p_{2}|$ , the so-called Manhattan-metric , we have that the measurements $\ell_{d}(\partial B_{n})$ correspond (∆) to computations $\ell^{\Xi_{n}}_{d}(\partial S)$ that arise in the process of computing the limit $\ell_{d}(\partial S)$ : one takes the points where the zig zag touches the circular boundary and views this as a partition of the curve. One also has $|\Xi_{n}|\longrightarrow 0$ . Hence, by definition , one has

$\ell_{d}(\partial B_{n})=\ell^{\Xi_{n}}_{d}(\partial S) \longrightarrow \ell_{d}(\partial S)$

So, the lengths of the zig-zag curves do approach the length of the disk… just under the wrong metric! We thus know, that the length of the boundary of the disk, under the Manhattan-metric is 4, and not $2\pi$ . But what about the Euklidean-metric —that is what we really want. Under this metric the correspondence in (∆) no longer holds. Thus there is no positive justification for the approach ‘the objects approach each other, hence the lengths approach each other’.

ANSATZ 2. Length as a 1-dimensional measure (mass).

If one thinks about area, one may think of the box-counting method. This is the primitive basis for measure theory. The d-dimensional measure computes the d-dimensional ‘mass’ of geometric objects. Speaking to 1b above, there is a convergence result for measures. Let $\mu$ me a measure and $X,X_{0},X_{1},\ldots$ be measurable sets. Then

$\mu(X\Delta X_{n})\longrightarrow 0 \Longrightarrow \mu(X_{n})\longrightarrow \mu(X)$

Considering the 2-dimensional measure $\mu=\mathcal{L}^{2}$ on $\mathbb{R}^{2}$ and the (measurable) sets $S,B_{0},B_{1},\ldots\subseteq\mathbb{R}^{2}$ , one may argue that $B_{n}\Delta S=B_{n}\setminus S$ is coverable by boxes whose total area converges to $0$ with $n$ . Hence the condition to the above result is satisfied, implying the conclusion: convergence of areas.

Consider now the 1-dimensional measure $\mu=\mathcal{L}^{1}$ on $\mathbb{R}^{2}$ and the (measurable) sets $\partial S,\partial B_{0},\partial B_{1},\ldots\subseteq\mathbb{R}^{2}$ . One has that $\partial B_{n}\Delta\partial S=(\partial B_{n}\cup\partial S)\setminus P_{n}$ , where $P_{n}:=(\partial B_{n}\cap\partial S)$ . But $P_{n}$ consists of merely countably (actually finitely) many points! Hence $\mu(P_{n})=0$ . Thus $\mu(\partial B_{n}\Delta\partial S)=\mu(\partial B_{n}\cup\partial S)-\mu(P_{n})\geq\mu(\partial S)-0 > 0$ . It follows that $\mu(\partial B_{n}\Delta\partial S)\not\longrightarrow 0$ . Hence the conclusion, the convergence of lengths, is not guaranteed.

Hence, once again, there is no grounds for justifying (†) ‘the objects approach each other, hence their measures approach each other’, since under this conception of length the first part of (†) does not hold.

REMARK.

This may seem unsatisfactory. All I’m doing is undercutting potential pseudo reasons for arguing for the ‘intuitive’ thing, and revealing the reasons themselves as humbug, without determining whether or not the conclusion be true. To do the latter is straightforward: just compute the values of the lengths and show that the limit does not correspond to the length of the the limit of the objects. But that to me is unsatisfying, as it does not actually deal with why the intuitive reason is itself, independently of the facts, subject to fallacies/errors.

But at the limit when the area of the circle is the same as the area of the grey area, the 'square' lines have surely become points of no length on the circumference of the circle, and hence the 'perimiter' is the same as the circumference. Maybe it's the definition of perimiter that is the problem.

The moderator comment fails to explain what we mean by a good approximation.

Let's observe what is going on when we slice a single square corner. We have removed a segment of the area outside of the circle. Notice that we will always cut further corners while still being outside of the circle. In doing so, we are removing more and more of the area outside the circle, and hence tending towards the circle's area.

Now let's look at what is going on with the perimeter. After the first "cut" is made, in the top left corner (for reference), you'll notice that the left and top side of the red square equal the new "added" length of the perimeter (basically the right and bottom sides of the red square). But these corresponding sides are equal in length! Hence the perimeter of the top left corner hasn't changed. This argument extends to all corners, and equally to all cuts made, as the process simply repeats itself, irrespective of position and size of cut made.

Keep it simple, stupid.

I didn't understand in the moderator note about why is the area formula as an integral over y whereas as it is a square shouldn't it be y^2 and if not why????Pls help!!!

I think both the options are false. Since keeping on folding squares like this would make us to fall in a condition where a square wil have an arc above its diagonal we can not fold the square at that time.....further if the area of square is approaching area of circle then at the moment when it is very close its perimeter will be same as that of circle but if that happend pi obtains a value 4 which is not possible hence both options should be wrong.

I don't understand the formulas, but I do understand graph paper. To me it's crazy, the perimeter part, but true. Duplicate the starting image on graphing paper and then start crossing out one square at a time until you get as close as the whole squares will allow you to get to the circle. The perimeter of what use to be a perfect square is the same as the starting square.

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