Proof that e is irrational
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In mathematics, the series expansion of the number e
- <math>e = \sum_{n = 0}^{\infty} \frac{1}{n!}<math>
can be used to prove that e is irrational.
Suppose e = a/b, for some positive integers a and b. Consider the number
- <math>x = b\,!\left(e - \sum_{n = 0}^{b} \frac{1}{n!}\right)<math>
We will show that x is a positive integer less than 1, and this contradiction will establish the irrationality of e.
- To see that x is an integer, note that
- <math>x = b\,!\left(e - \sum_{n = 0}^{b} \frac{1}{n!}\right) = b\,!\left(\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\right)<math>
- <math>= a(b - 1)! - \sum_{n = 0}^{b}b(b-1)\cdots(n+1)<math>
- Here, the last term in the final sum is to be interpreted as an empty product.
- To see that x is a positive number less than 1, note that
- <math>x = b\,!\sum_{n = b+1}^{\infty} \frac{1}{n!} \mbox{ and so}<math>
- <math>0 < x = \frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + \frac{1}{(b+1)(b+2)(b+3)} + \cdots <math>
- <math>< \frac{1}{b+1} + \frac{1}{(b+1)^2} + \frac{1}{(b+1)^3} + \cdots = \frac{1}{b} \le 1<math>
- Here, the last sum is a geometric series.
Since there does not exist a positive integer less than 1, we have reached a contradiction, and so e must be irrational. This completes the proof.