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 can not 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

$e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, \ldots, 2n, 1, 1, \ldots]. \,$

Since this continued fraction is infinite and rational numbers can't be written as infinite continued fractions, 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.

Fourier's proof

The most well-known proof is Joseph Fourier's proof by contradiction, which is based upon the equality

$e = \sum_{n = 0}^{\infty} \frac{1}{n!}\cdot$

Initially e is assumed to be a rational number of the form ^a⁄_b. Note that b couldn't be equal to one as e is not an integer. It can be shown using the above equality that e is strictly between 2 and 3.

$2 = 1 + \tfrac{1}{1!} \ \ < \ \ e = 1 + \tfrac{1}{1!} + \tfrac{1}{2!} + \tfrac{1}{3!} + \cdots \ \ < \ \ 2 + \tfrac{1}{2} (1 + \tfrac{1}{3} + \tfrac{1}{3^2} + \tfrac{1}{3^3} + \cdots) \ \ < \ \ 3$

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

$x = b!\,\biggl(e - \sum_{n = 0}^{b} \frac{1}{n!}\biggr)\!$

To see that if e is rational, then x is an integer, substitute e = ^a⁄_b into this definition to obtain

$x = b!\,\biggl(\frac{a}{b} - \sum_{n = 0}^{b} \frac{1}{n!}\biggr) = a(b - 1)! - \sum_{n = 0}^{b} \frac{b!}{n!}\,.$

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

$x = b!\,\biggl(\sum_{n = 0}^{\infty} \frac{1}{n!} - \sum_{n = 0}^{b} \frac{1}{n!}\biggr) = \sum_{n = b+1}^{\infty} \frac{b!}{n!}>0\,,\!$

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

$\frac{b!}{n!} =\frac1{(b+1)(b+2)\cdots(b+(n-b))} <\frac1{(b+1)^{n-b}}\,.\!$

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

$x =\sum_{n = b+1}^\infty \frac{b!}{n!} < \sum_{n=b+1}^\infty \frac1{(b+1)^{n-b}} =\sum_{k=1}^\infty \frac1{(b+1)^k} =\frac{1}{b+1} \biggl(\frac1{1-\frac1{b+1}}\biggr) = \frac{1}{b} < 1.$

Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so e must be irrational. Q.E.D.

Alternate proofs

Another proof can be obtained from the previous one by noting that

$(b+1)x=1+\frac1{b+2}+\frac1{(b+2)(b+3)}+\cdots<1+\frac1{b+1}+\frac1{(b+1)(b+2)}+\cdots=1+x,$

and this inequality is equivalent to the assertion that bx < 1. This is impossible, of course, since b and x are natural numbers.

Still another proof can be obtained from the fact that

$\frac1e=e^{-1}=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\cdot$

Generalizations

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, e^q is irrational for any non-zero rational q.

See also

• Characterizations of the exponential function
• Transcendental number, including a proof that e is transcendental
• Lindemann–Weierstrass theorem