If e were rational, then e = n/m for some 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 terms of this alternating series S(n). How good is this approximation to e? Well, the error is bounded by the next term of the alternating series:
0 < | 1/e - S(n) | = | m/n - S(n)| < 1/(n+1)!
But multiplying through by n!, you will see that
0 < | integer - integer | < 1/(n+1) < 1.
But there is no integer strictly between 0 and 1, so this is a contradiction; e must be irrational.
Originally posted by idoru
I finished the first sentence and wanted to shoot you in the face. I'm not even joking, my stress level just shot through the roof. God, I hate your post.
Registered: Nov 2003
Location: The corner where 'l' resolves into '<'
Re: e is irrational
quote:
Originally posted by Pett
If e were rational, then e = n/m for some 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 terms of this alternating series S(n). How good is this approximation to e? Well, the error is bounded by the next term of the alternating series:
0 < | 1/e - S(n) | = | m/n - S(n)| < 1/(n+1)!
But multiplying through by n!, you will see that
0 < | integer - integer | < 1/(n+1) < 1.
But there is no integer strictly between 0 and 1, so this is a contradiction; e must be irrational.
Of course i'm going to use the colour red as an example. It's at the end of the spectrum so the rational of clipping it doesn't require a significant leap to link the two data points into a single cohesive logical unit.
Toronto! Ya give em the CN tower and they mistake it for the Peace Tower.
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