E is irrational, Mathematics

If e were rational, then e = n/m for some positive integers m, n. So then 1/e = m/n. But the series expansion for 1/e is

1/e = 1 - 1/1! + 1/2! - 1/3! + ...

Call the first n values of this alternating series S(n). How good is this approximation to e? Well, the error is bounded by the next part of the alternating series:

0 < | 1/e - S(n) | = | m/n - S(n)| < 1/(n+1)!

But multiplying through by n!, you can see that

0 < | integer - integer | < 1/(n+1) < 1.

But there is no integer between 0 and 1, so this is a contradiction; e must be irrational.

Posted Date: 3/30/2013 2:49:36 AM | Location : United States







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