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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There are several different versions of this particular paradox – like Holli’s Paradox, and The Unexpected Hanging Paradox. Before I present the surprise examination version, it behooves me to mention that no correct (or final) solution to this paradox has been established yet.

A teacher announces in class that an examination will be held on some day during the following week, and moreover that the examination will be a surprise.

The students argue that a surprise exam cannot occur. For suppose the exam were on the last day of the week. Then on the previous night, the students would be able to predict that the exam would occur on the following day, and the exam would not be a surprise. So it is impossible for a surprise exam to occur on the last day. But then a surprise exam cannot occur on the penultimate day, either, for in that case the students, knowing that the last day is an impossible day for a surprise exam, would be able to predict on the night before the exam that the exam would occur on the following day. Similarly, the students argue that a surprise exam cannot occur on any other day of the week either.

Confident in this conclusion, they are of course totally surprised when the exam occurs (on Wednesday, say). The announcement is vindicated after all. Where did the students’ reasoning go wrong?

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Similar posts: The Monty Hall Paradox, A Mathematical Conundrum, The Voting Paradox, Broken Clocks.

Posted in Puzzles

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