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$is not an algebraic number of degree 2 (i... This is the original paper published in the French journal "Journal... About the irrationality of e=2.718... J. Liouville 1840 We prove that e, basis of the neperien logarithm, is not a rational number. The same method can also be used to prove that e cannot be a root of a second degree equation with rational coecients, i.e. we cannot ﬁnd ae + b e = c,such that a is a positive integer and b, c are positive or negati ve integers. As a matter of fact, if we replace e and 1/e or e 1 with their power series deduced from e x , and given that we multiply both sides of the equation by 1 2 3 ... n,we ﬁnd that a n +1 1+ 1 n +2 + ... ± b n +1 1 1 n +2 + ... = µ where µ is an integer. We can make it so that ± b n +1 is positive; we just need to suppose that n is even when b<0 and t h at n is odd when b>0; for big values of n the equation above leads to a contradiction; the ﬁrst term of the equation is positive and very small, with values between 0 and 1, and as such can never be equal to an integer µ. So, etc. 1 This is the original paper published in the French journal "Journal de Mathématiques" in 1840. Joseph Liouville (1809 – 1882) was one of the leading French mathematicians in the generation between Évariste Galois and Charles Hermite. In all, Liouville wrote over 400 mathematical papers, 200 in number theory alone. ![](https://www.ecured.cu/images/thumb/c/c8/JosephLiouville.png/260px-JosephLiouville.png) Liouville proves that$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. $$e^2=\frac{b}{a} \equiv ae=be^{-1}$$ 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. 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.