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

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

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

$e=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}.$

Initially e is assumed to be a rational number of the form Template:Sfrac. The idea is to then analyze the scaled-up difference (here denoted x) between the series representation of e and its strictly smaller b-th partial sum, which approximates the limiting value e. By choosing the scale factor to be the factorial of b, the fraction Template:Sfrac 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 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 e = Template:Sfrac. Define the number

$x=b!(e-\sum _{n=0}^b\frac{1}{n!}).$

Use the assumption that e = Template:Sfrac to obtain

$x=b!(\frac{a}{b}-\sum _{n=0}^b\frac{1}{n!})=a(b-1)!-\sum _{n=0}^b\frac{b!}{n!}.$

The first term is an integer, and every fraction in the sum is actually an integer because n ≤ b for each term. Therefore, under the assumption that e is rational, 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

$x=b!(\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}-\sum _{n=0}^b\frac{1}{n!})=\sum _{n=b+1}^{\mathrm{\infty }}\frac{b!}{n!}>0,$

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

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

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

$x=\sum _{n=b+1}^{\mathrm{\infty }}\frac{b!}{n!}<\sum _{n=b+1}^{\mathrm{\infty }}\frac{1}{(b+1{)}^{n-b}}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{(b+1{)}^k}=\frac{1}{b+1}\left(\frac{1}{1-\frac{1}{b+1}}\right)=\frac{1}{b}\le 1.$

And therefore $x<1.$

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

Alternative proofs

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

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

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

Still another proof can be obtained from the fact that

$\frac{1}{e}=e^{-1}=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{n!}.$

Define $s_n$ as follows:

$s_n=\sum _{k=0}^n\frac{(-1{)}^k}{k!}.$

Then

$e^{-1}-s_{2n-1}=\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{k!}-\sum _{k=0}^{2n-1}\frac{(-1{)}^k}{k!}<\frac{1}{(2n)!},$

which implies

$0<(2n-1)!(e^{-1}-s_{2n-1})<\frac{1}{2n}\le \frac{1}{2}$

for any positive integer $n$.

Note that $(2n-1)!s_{2n-1}$ is always an integer. Assume that $e^{-1}$ is rational, so $e^{-1}=p/q,$ where $p,q$ are co-prime, and $q\ne 0.$ It is possible to appropriately choose $n$ so that $(2n-1)!e^{-1}$ is an integer, i.e. $n\ge (q+1)/2.$ Hence, for this choice, the difference between $(2n-1)!e^{-1}$ and $(2n-1)!s_{2n-1}$ would be an integer. But from the above inequality, that is not possible. So, $e^{-1}$ is irrational. This means that $e$ is irrational.

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, which implies that e³ is irrational. More generally, e^q is irrational for any non-zero rational q.

Charles Hermite further proved that e is a transcendental number, in 1873, which means that is not a root of any polynomial with rational coefficients, as is e^α for any non-zero algebraic α.

See also

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