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 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 second degree polynomial with rational coefficients; in particular, e2 is irrational.
Initially e is assumed to be a rational number of the form a⁄b. Note that b could not be equal to 1 as e is not an integer. It can be shown using the above equality that e is strictly between 2 and 3:
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.
Suppose that e is a rational number. Then there exist positive integers a and b such that e = a⁄b. 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 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 bx < 1. This is impossible, of course, since b and x are natural numbers.
Define as follows:
for any integer
Note that is always an integer. Assume is rational, so, where are co-prime and It's 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's impossible. So, is irrational. This means that is irrational.
In 1840, Liouville published a proof of the fact that e2 is irrational followed by a proof that e2 is not a root of a second degree polynomial with rational coefficients. This last fact implies that e4 is irrational. His proofs are similar to Fourier's proof of the irrationality of e. In 1891, Hurwitz explained how it is possible to prove along the same line of ideas that e is not a root of a third degree polynomial with rational coefficients. In particular, e3 is irrational.
More generally, eq is irrational for any non-zero rational q.
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