Proof that e is irrational
The number was introduced by Jacob Bernoulli in 1683. More than half a century later, Euler, who had been a student of Jacob's younger brother Johann, proved that is irrational; that is, that it cannot be expressed as the quotient of two integers.
Euler's proof
Euler wrote the first proof of the fact that is irrational in 1737. He computed the representation of as a simple continued fraction, which isSince this continued fraction is infinite and every rational number has a terminating continued fraction, is irrational. A short proof of the previous equality is known. Since the simple continued fraction of is not periodic, this also proves that is not a root of a quadratic polynomial with rational coefficients; in particular, is irrational.
Fourier's proof
The most well-known proof is Joseph Fourier's proof by contradiction, which is based upon the equalityInitially is assumed to be a rational number of the form. The idea is to then analyze the scaled-up difference between the series representation of and its strictly smaller partial sum, which approximates the limiting value. By choosing the scale factor to be the factorial of, the fraction and the partial sum are turned into integers, hence must be a positive integer. However, the fast convergence of the series representation implies that is still strictly smaller than 1. From this contradiction we deduce that is irrational.
Now for the details. If is a rational number, there exist positive integers and such that. Define the number
Use the assumption that to obtain
The first term is an integer, and every fraction in the sum is actually an integer because for each term. Therefore, under the assumption that is rational, is an integer.
We now prove that. First, to prove that is strictly positive, we insert the above series representation of into the definition of and obtain
because all the terms are strictly positive.
We now prove that. For all terms with we have the upper estimate
This inequality is strict for every. Changing the index of summation to and using the formula for the infinite geometric series, we obtain
And therefore
Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so is irrational, Q.E.D.
Alternative proofs
Another proof can be obtained from the previous one by noting thatand this inequality is equivalent to the assertion that. This is impossible, of course, since and are positive integers.
Still another proof can be obtained from the fact that
Define as follows:
Then
which implies
for any positive integer.
Note that is always an integer. Assume that is rational, so where are co-prime, and It is possible to appropriately choose so that is an integer, i.e. Hence, for this choice, the difference between and would be an integer. But from the above inequality, that is not possible. So, is irrational. This means that is irrational.
Generalizations
In 1840, Liouville published a proof of the fact that is irrational followed by a proof that is not a root of a second-degree polynomial with rational coefficients. This last fact implies that is irrational. His proofs are similar to Fourier's proof of the irrationality of. In 1891, Hurwitz explained how it is possible to prove along the same line of ideas that is not a root of a third-degree polynomial with rational coefficients, which implies that is irrational. More generally, is irrational for any non-zero rational.Charles Hermite further proved that is a transcendental number, in 1873, which means that is not a root of any polynomial with rational coefficients, as is for any non-zero algebraic.