Joseph Liouville (1809 – 1882) was one of the leading French
mat...

Note that Liouville doesn't use the factorial symbol "!". Although ...

If we multiply both sides by $n!$ and we use $e^{x}= \sum_{l=0}^{\i...

Liouville proves that $e$ is not an algebraic number of degree 2 (i...

This is the original paper published in the French journal "Journal...

About the irrationality of e=2.718...

J. Liouville

1840

We prove that e, basis of the neperien logarithm, is not a rational number.

The same method can also be used to prove that e cannot be a root of a second

degree equation with rational coeﬃcients, i.e. we cannot ﬁnd ae +

b

e

= c,such

that a is a positive integer and b, c are positive or negati ve integers. As a matter

of fact, if we replace e and 1/e or e

1

with their power series deduced from e

x

,

and given that we multiply both sides of the equation by 1 ⇥ 2 ⇥ 3 ⇥ ... ⇥ n,we

ﬁnd that

a

n +1

✓

1+

1

n +2

+ ...

◆

±

b

n +1

✓

1

1

n +2

+ ...

◆

= µ

where µ is an integer. We can make it so that

±

b

n +1

is positive; we just need to suppose that n is even when b<0 and t h at n is

odd when b>0; for big values of n the equation above leads to a contradiction;

the ﬁrst term of the equation is positive and very small, with values between 0

and 1, and as such can never be equal to an integer µ. So, etc.

1

This is the original paper published in the French journal "Journal de Mathématiques" in 1840.
Joseph Liouville (1809 – 1882) was one of the leading French
mathematicians in the generation between Évariste Galois and Charles Hermite. In all, Liouville
wrote over 400 mathematical papers, 200 in number theory alone.
![](https://www.ecured.cu/images/thumb/c/c8/JosephLiouville.png/260px-JosephLiouville.png)
Liouville proves that $e$ is not an algebraic number of degree 2 (in other words it's not a solution of 2nd degree polynomial with rational coefficients) but it's not difficult to see that if $c=0$ we get for free the proof that $e^2$ is irrational.
$$
e^2=\frac{b}{a} \equiv ae=be^{-1}
$$
If we multiply both sides by $n!$ and we use $e^{x}= \sum_{l=0}^{\infty} \frac{x^k}{k!} $ we get
$$
n!a\left(\sum_{k=0}^{n} \frac{1}{k!}+\sum_{k=n+1}^{\infty} \frac{1}{k!} \right) + n!b\left(\sum_{k=0}^{n} \frac{(-1)^{k}}{k!}+\sum_{k=n+1}^{\infty} \frac{(-1)^{k}}{k!} \right) = n!c \\
\sum_{k=n+1}^{\infty} \frac{n!a}{k!} + \sum_{k=n+1}^{\infty} \frac{(-1)^{k}n!b}{k!} = \underbrace{n!c + \sum_{k=0}^{n} \frac{n!a}{k!} + \sum_{k=0}^{n} \frac{(-1)^{k}n!b}{k!}}_{\mu}
$$
Note that since $k<n$ all the terms on the right hand side are integers and so $\mu$ is also an integer.
$$
\sum_{k=n+1}^{\infty} \frac{n!a}{k!}=\frac{a}{n+1}\left(1+\frac{1}{n+2}+...\right) < a\left(\frac{1}{n+1}+\frac{1}{(n+1)^2}+...\right)=\frac{a}{n}
$$
At the same time
$$
\sum_{k=n+1}^{\infty} \frac{(-1)^{k}n!b}{k!}=\frac{b(-1)^{n+1}}{n+1}\left(1- \frac{1}{n+2}+...\right)
$$
Since
$$
\frac{1}{n+1}\left(1- \frac{1}{n+2}+...\right) < \frac{1}{n+1}\left(1+ \frac{1}{n+2}+...\right) < \frac{1}{n}
$$
if we choose a combination of $b$ and $n$ so that $b(-1)^{n+1} > 0$ we get
$$
0<\sum_{k=n+1}^{\infty} \frac{n!a}{k!} + \sum_{k=n+1}^{\infty} \frac{(-1)^{k}n!b}{k!}< \frac{a+b}{n}
$$
In particular if we pick a large enough $n$ we can get
$$
\frac{a+b}{n}<1
$$
which generates a contradiction since on the left hand side of the equation we have something that is less than 1 and greater than 0 and on the right hand side we have an integer.
Note that Liouville doesn't use the factorial symbol "!". Although the notation n! was first introduced by Christian Kramp (1760-1826) in 1808 in his book "Éléments d'arithmétique universelle" it hadn't still been widely adopted by the mathematical community.