Proof that e is irrational
The number e 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 e 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 e is irrational in 1737 (but the text was only published seven years later). He computed the representation of e as a simple continued fraction, which is Since this continued fraction is infinite and every rational number has a terminating continued fraction, e is irrational. A short proof of the previous equality is known. Since the simple continued fraction of e is not periodic, this also proves that e is not a root of a quadratic polynomial with rational coefficients; in particular, e2 is irrational. Fourier's proof The most well-known proof is Joseph Fourier's proof by contradiction, which is based upon the equality Initially e is assumed to be a rational number of the form . The idea is to then analyze the scaled-up difference (here denoted x) between the series representation of e and its strictly smaller partial sum, which approximates the limiting value e. By choosing the scale factor to be the factorial of b, the fraction and the partial sum are turned into integers, hence x must be a positive integer. However, the fast convergence of the series representation implies that x is still strictly smaller than 1. From this contradiction we deduce that e is irrational. Now for the details. If e is a rational number, there exist positive integers a and b such that . Define the number Use the assumption that e = 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 e is rational, x is an integer. We now prove that . 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 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 This last inequality is strict as . Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so e is irrational, Q.E.D. Alternate proofs Another proof can be obtained from the previous one by noting that and this inequality is equivalent to the assertion that bx < 1. This is impossible, of course, since b and x 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