Proof that e is irrational
Summary of the proof
This is Joseph Fourier's proof by contradiction. Initially e is assumed to be a rational number of the form a/b. We then analyze a blown-up difference x of the series representing e and its strictly smaller b th partial sum, which approximates the limiting value e. By choosing the magnifying factor to be the factorial of b, the fraction a/b and the b th partial sum are turned into integers, hence x must be a positive integer. However, the fast convergence of the series representation implies that the magnified approximation error x is still strictly smaller than 1. From this contradiction we deduce that e is irrational.
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.
Another proof can be obtained from the previous one by noting that
and this inequality is equivalent to the assertion that :. This is impossible, of course, since b and x are natural numbers.
- Characterizations of the exponential function
- Transcendental number, including a proof that e is transcendental
- Lindemann–Weierstrass theorem