arb246 wrote:Someone tell me if this is wrong:

Because the third panel involves an infinite recursion, every value represented there has to be indeterminate. As long as we assume the third panel is accurate, it will involve an infinite addition of infinitely small values to the third bar of the bar graph (in drawing an infinity of infinitely small diagrams) and the same would go for the pie, which would involve infinite additions of infinitely small values to the black wedge. If we assume the third panel is just an approximation, then we have to treat the other two as approximations as well.

Your first mistake is highlighted here. A sum of an infinite geometric series (ie, the next term is a constant % size of the last) can possibly be a finite number, especially if that % is between 0 and 1 and if it is extended all the way to infinity.

If the image was not discrete, and the third panel did continue to infinity, the black would have a finite, rational area.

Here's a proof by example (taken from wikipedia):

In this series, the recursive terms are (last term * 2/3).

[imath]s \;=\; 1 \,+\, \frac{2}{3} \,+\, \frac{4}{9} \,+\, \frac{8}{27} \,+\, \cdots[/imath] --

divide both sides by 2/3[imath]\frac{2}{3}s \;=\; \frac{2}{3} \,+\, \frac{4}{9} \,+\, \frac{8}{27} \,+\, \frac{16}{81} \,+\, \cdots[/imath] --

This new series is exactly the same as the first one, but one less. So it can be represented as [imath]s-1[/imath][imath]\frac{2}{3}s \; = \; s \,-\, 1[/imath] --

commence algebra magic...[imath]s \; = \; 3[/imath]

And there you have it; an infinite geometric series totaling to a rational number.