Here’s the infamous “proof” that shows that  π is equal to 4.

The zig-zagged polygon, whose perimeter always remains 4, does appear to approach the circle as we repeat these steps to infinity. Ergo, π is equal to 4!

The problem, as Vi Hart explains in this amusing math doodle video, is that while the area of the polygon does approach the area of the circle, the actual perimeter of the polygon is much larger than the circumference of the circle. Confused? Think of it this way: if you put a jumbled up 10 feet long rope into a 1 foot long container, you wouldn’t say that the rope is now 1 foot long, would you? You would take out the rope, and extend it fully to measure its actual size.

See some more discussion here. Also, here’s another video that “proves” — using a similar approach — that π is equal to 2, and the square root of 2 (another one of my favorite irrational numbers) is also equal to 2!

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By the way, if π were actually equal to 4, all circles would be squares. What a terrible world that would be! Similarly, if π were equal to 3, all circles would be hexagons. (For a hexagon, the ratio of its circumference to its diameter is equal to 3.)

Oh, and happy pi day! Don’t forget to have some delicious pie, and as you eat it, marvel at the glory of this magnificent, transcendental, and most importantly, irrational number.

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Previously on this blog: Proof Without Words IVHappy Pi Day 20133.14A Sanskrit Mnemonic for πHappy Pi Day!A Mathematical Conundrum.

Posted in Recreational Math

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2 responses to “Pi Day Paradox”

1. Miguel Gómez Donoso

This is wrong. The sequence DOES converge to a circle.

If you look up the definition of convergence, it says (which is very intuitive) that a sequence approaches a given limit if every point can be made as close to the limit as desired.

And of course our sequence approaches the circle as much as we want, therefore in infinity it converges.

The issue here is that the limit of a sequence is not necessarily a member of the sequence, nor share any property with the members of the sequence.

The very typical example is that a sequence of continuous functions can converge to a discontinuous function. It makes clear that the argument “since all members of the sequence have x property… therefore the limit also has it” is wrong.