Staircases and Diagonal Lines

There are a pair of linked paradoxes that tend to do the rounds on the internet: one that proves that pi is equal to two, and one that proves that the square root of two is equal to two. Obviously this isn’t true which is why they are called paradoxes. What follows is my rather small addition to the paradox.

The Staircase Paradox
Imagine a 2-dimensional square of side-length 1. A diagonal line drawn between opposite corners has an easily calculable length using Pythagoras’ Theorem:

Figure 1: A variety of corner-to-corner paths through a 1x1 square. The diagonal path is in red, followed by a variety of staircases with n=2 (green), n=4 (yellow), and n=8 (blue)

x^2+y^2=z^2
1+1=z^2
z=\sqrt{2}\approx 1.414

Of course there are other ways to get from one corner to the other… We could go halfway along one side, then take a ninety-degree turn, go halfway again (to the middle), then repeat to get to the far corner. Or we could take even more steps! Figure 1 shows a couple of these “staircase” options, in addition to the direct diagonal path.

The paradox then goes on that as the number of steps in the staircase (n) is increased it converges on the diagonal line. Therefore as n approaches infinity the staircase still has length 2, but it also converges infinitely close on the diagonal line.

Does it Converge?

Figure 2: The area between the staircase and the diagonal (light orange) is a proxy measure for the subjective "distance" between the two. Ai is the area of one triangle and is indicated by the dark orange shading. The black square is once again 1x1 units.

The paradox typically ends there and the solution claimed is that the lines aren’t actually converging, it just looks like they are. I was wondering if there was a way to show that geometrically they were actually converging and this is what I came up with. The total area between the staircase and the diagonal line can be used as a proxy for the total “difference” between the two lines. Figure 2 shows the n=4 staircase and the 4 areas being measured. Since each area is a simple right-angled triangle, calculation of the area of one triangle is easily accomplished:

A_i = \frac{1}{2}\cdot h \times w
A_i = \frac{1}{2}\cdot \frac{1}{n} \times \frac{1}{n}
A_i = \frac{1}{2n^2}

From this, we can calculate the total area between the staircase and the diagonal line as:

A = n\times A_i
A = \frac{1}{2n}

And now applying limits we see that as the number of steps increases, the area of the difference between the two paths does actually converge on zero, while the distance traversed by each path remains different!

\underset{n \to \infty}{\lim} \frac{1}{2n}=\emptyset

Same Thing With Circles!

Figure 3: A series of subdividing semicircles. The largest semicircle (red) has n=1 and a radius of 2. Green (n=2, r=0.5), yellow (n=4, r=0.25) and blue (n=8, r=0.125) then follow as progressive divisions of the horizontal line.

It turns out that the exact same thing happens with semi-circles! Figure 3 shows a series of semicircles that subdivide the semicircle larger than it. The length of the first semicircle (red) is given by:

\ell_1=\frac{2\pi r}{2}=\pi r
\ell_1=\pi

The length of the n=2 red semicircle is:

\ell_2=2\times\frac{2\pi r}{2}
\ell_2=2\times\frac{2\pi (0.5)}{2}=\pi

If these calculations are repeated for n=4 and n=8:

\ell_4=\pi
\ell_8=\pi

At this point the paradox argument goes (like in the staircase above) that the semicircles eventually converge on the straight line. Since the straight line has a fixed length of 2 and the circles always have a length of π, that therefore π = 2!

In fact, the same area-based logic can be applied to this problem as well. First we find a general formula for the total area under the semicircles in terms of the number of semicircles along the line, n.

r=\frac{1}{n}

A_i=\pi r^2
A_i=\frac{\pi}{n^2}

A = n\times A_i
A = \frac{\pi}{n}

This is actually the same form of equation as in the staircase-case, except that the triangle coefficient in the numerator (1/2) is replaced by the circle coefficient (π)! Once again the limit shows that this area converges to zero as we increase the number of semicircles:

\underset{n \to \infty}{\lim} \frac{\pi}{n}=\emptyset

So How Does This Resolve Anything?!
It doesn’t, I’ve just made it more confusing… But Gödel did point out that paradoxes have to exist… For that matter, we can always prove that 0.9999[…] is exactly equal to 1:

\frac{1}{9}=0.\overline{1}
\frac{1}{9}\times 9=0.\overline{1}\times 9
\frac{9}{9}=1=0.\overline{9}

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