Proof That e Is Irrational - Proof

Proof

Towards a contradiction, suppose that e is a rational number. Then there exist positive integers a and b such that e = a/b where clearly b > 1.

Define the number

x = b!\,\biggl(e - \sum_{n = 0}^{b} \frac{1}{n!}\biggr)\!

To see that if e is rational, then x is an integer, substitute e = a/b into this definition to obtain

x = b!\,\biggl(\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\biggr) = a(b - 1)! - \sum_{n = 0}^{b} \frac{b!}{n!}\,.

The first term is an integer, and every fraction in the sum is actually an integer because n ≤ b for each term. Therefore x is an integer.

We now prove that 0 < x < 1. First, to prove that x is strictly positive, we insert the above series representation of e into the definition of x and obtain

because all the terms with n ≤ b cancel and the remaining ones are strictly positive.

We now prove that x < 1. For all terms with n ≥ b + 1 we have the upper estimate

\frac{b!}{n!} =\frac1{(b+1)(b+2)\cdots(b+(n-b))} \le\frac1{(b+1)^{n-b}}\,.\!

This inequality is strict for every n ≥ b + 2. Changing the index of summation to k = n – b and using the formula for the infinite geometric series, we obtain

x =\sum_{n = b+1}^\infty \frac{b!}{n!} < \sum_{n=b+1}^\infty \frac1{(b+1)^{n-b}} =\sum_{k=1}^\infty \frac1{(b+1)^k} =\frac{1}{b+1} \biggl(\frac1{1-\frac1{b+1}}\biggr) = \frac{1}{b} < 1.

Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so e must be irrational. Q.E.D.

The above proof can be found in Proofs from THE BOOK, where the stronger result that eq is irrational for any non-zero rational q is also proved.

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