Proof that e is irrational

Fourier's proof

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

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

Initially e is assumed to be a rational number of the form ⁠a/b⁠. 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 ⁠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 x is still strictly smaller than 1. From this contradiction we deduce that e is irrational.

The detailed proof goes as follows:

If e is a rational number, there exist positive integers a and b such that e = ⁠a/b⁠. Define the number

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

Use the assumption that e = ⁠a/b⁠ to obtain

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

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

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

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

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

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

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

And therefore ${\displaystyle x<1.}$

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

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

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

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

Define ${\displaystyle s_n}$ as follows:

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

Then

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

which implies

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

for any positive integer ${\displaystyle n}$.

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