At dinner the other night, our table wobbled. While the waitress fetched a wedge to stabilize it, I told my fellow diners about an applicable mathematical proof: If the legs of a table are even, but the table wobbles because of an uneven floor surface, then rotating the table will always find a position where all four legs are touching the ground.

They were skeptical, but it is true. The ground must be continuous, in the theory of functions sense: no cliff-like changes in height. As proof, here’s an intuitive thought experiment, and the same concept, but full of mind-bending rigor which I don’t understand.

## Comments

Wyatt9:58 PM on 9 Jan 2007I love "mind-bending rigor." Not necessarily the actual thing, but the phrase.

Paul Downs9:21 AM on 12 Jan 2007I have my doubts. The theorem is making assumptions about the condition of the floor between the four original points based upon the rotation of 90 degrees - which is actually a non-event, since the legs are all the same. It seems to imply that the floor condition is only changing from the original state under one of the legs, and that at some point in that rotation all four legs will be on a single plane. In reality it would be changing under all four legs, arbitrarily, and there is no guarantee that the 4 points will ever be in a single plane. In real life, unless the table is round, it doesn't help much, as the orientation of the table relative to the room is rarely arbitrary. I have a feeling that you are toying with us, presenting another flawed proof to see who will catch it.

Paul Downs

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