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

## Fourier's proof

The most well-known proof is Joseph Fourier's proof by contradiction, which is based upon the equality
    $e=\sum _{n=0}^{\infty }{\frac {1}{n!}}.$
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.
Now for the details. If e is a rational number, there exist positive integers a and b such that e = ⁠a/b⁠. Define the number
    $x=b!\left(e-\sum _{n=0}^{b}{\frac {1}{n!}}\right).$
Use the assumption that e = ⁠a/b⁠ to obtain
    $x=b!\left({\frac {a}{b}}-\sum _{n=0}^{b}{\frac {1}{n!}}\right)=a(b-1)!-\sum _{n=0}^{b}{\frac {b!}{n!}}.$
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
    $x=b!\left(\sum _{n=0}^{\infty }{\frac {1}{n!}}-\sum _{n=0}^{b}{\frac {1}{n!}}\right)=\sum _{n=b+1}^{\infty }{\frac {b!}{n!}}>0,$
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
    ${\frac {b!}{n!}}={\frac {1}{(b+1)(b+2)\cdots {\bigb+(n-b){\big )}}}\leq {\frac {1}{(b+1)^{n-b}}}.$
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
    $x=\sum _{n=b+1}^{\infty }{\frac {b!}{n!}}<\sum _{n=b+1}^{\infty }{\frac {1}{(b+1)^{n-b}}}=\sum _{k=1}^{\infty }{\frac {1}{(b+1)^{k}}}={\frac {1}{b+1}}\left({\frac {1}{1-{\frac {1}{b+1}}}}\right)={\frac {1}{b}}\leq 1.$
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.

## Alternate 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
    ${\frac {1}{e}}=e^{-1}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}.$
Define $s_{n}$ as follows:
    $s_{n}=\sum _{k=0}^{n}{\frac {(-1)^{k}}{k!}}.$
Then
    $e^{-1}-s_{2n-1}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}-\sum _{k=0}^{2n-1}{\frac {(-1)^{k}}{k!}}<{\frac {1}{(2n)!}},$
which implies
    $0<(2n-1)!\left(e^{-1}-s_{2n-1}\right)<{\frac {1}{2n}}\leq {\frac {1}{2}}$
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 $q\neq 0.$ It is possible to appropriately choose $n$ so that $(2n-1)!e^{-1}$ is an integer, i.e. $n\geq (q+1)/2.$ 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 e$^2$ is irrational followed by a proof that e$^2$ is not a root of a second-degree polynomial with rational coefficients. This last fact implies that e$^4$ 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 e$^3$ 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.