Pi Day Paradox

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!


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.

i ate some pie***

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


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.

    Please don’t post things about mathematics without contrasting them…

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Enter your email address to subscribe to this blog and receive notifications of new posts by email.

Join 66 other followers

On Twitter

Error: Twitter did not respond. Please wait a few minutes and refresh this page.


%d bloggers like this: