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
To see that if e is rational, then x is an integer, substitute e = a/b into this definition to obtain
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
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
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.
Read more about this topic: Proof That e Is Irrational
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