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Yes, but that is not as easy as Pi. ...nt | 756 comments | Create New Account
Comments belong to whoever posts them. Please notify us of inappropriate comments.
Yes, but that is not as easy as Pi. ...nt
Authored by: tknarr on Saturday, July 21 2012 @ 06:32 PM EDT

Indeed, it has. And yes, that supposition is the basis for the Limit Theorem that underlies a lot of calculus. But it can also be (ab)used to prove that the square root of 2 is 2.

Take a 1-unit square. The length of the diagonal is the square root of 2, but the length of the path following only the grid lines is 2, right? Now, divide the sides of the square in half, and draw a 1/2-unit grid. The length of the path from corner to the other is still 2, but the area between the diagonal and the path (the error in the approximation of the diagonal by the path) is less. Repeat the subdivision as often as needed to reduce the error below the acceptable limit k. If we accept that any path with an error less than k is equivalent to the diagonal, then the length of the path must equal the length of the diagonal. We know the length of the path is 2, no matter now fine the grid is. We know the length of the diagonal is the square root of 2. Therefore, the square root of 2 must be 2. Which is patently false, but there aren't any mathematical errors in how we got to our result. The only error is a subtle one: there isn't any k greater than zero that would result in our path being equivalent to the true diagonal. The supposition is valid, just not valid in the context of this problem.

[ Reply to This | Parent | # ]

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