I have a question about the irrationality of $e$:

In proving the irrationality of $e$, one can prove the irrationality of $e^{-1}$ by using the series $$e^x = 1+x+\frac{x^2}{2!} + \cdots + \frac{x^n}{n!}+ \cdots$$ So the series for $e^{-1}$ is $$s_n = \sum_{k=0}^{n} \frac{(-1)^{k}}{k!}$$ so that $$0 < e^{-1}-s_{2k-1} < \frac{1}{(2k)!}$$ or $$0 < (2k-1)!(e^{-1}-s_{2k-1}) < \frac{1}{2k} \leq \frac{1}{2}$$ If $e^{-1}$ were rational then we would have a difference of two integers which is not an integer (i.e. between $0$ and $\frac{1}{2}$).

Question: Is this "irrationality of $e^{-1}$ by alternating series" proof what motivated the definition of irrationality measure? It seems that this was born out of using alternating series.

Does the same method work for $\pi$ ?

  • 29
    $\begingroup$ I always thought the definition of irrationality measure was motivated from Liouville's theorem: en.wikipedia.org/wiki/… $\endgroup$ Aug 3, 2011 at 5:14
  • 8
    $\begingroup$ I think there is no connection. Incidentally, I prefer going through $e^{-1}$, as you did. Proofs I have seen of the irrationality of $e$ use the series for $e$ directly, which involves some additional work to produce the required estimate. $\endgroup$ Aug 3, 2011 at 5:30
  • 9
    $\begingroup$ As far as I know, the "series proof" for the irrationality of $e$ first appeared in the following 1815 book by Janot de Stainville (Article 232, pp. 339-341): books.google.com/books?id=5J0AAAAAMAAJ Stainville says he learned the proof from Poinsot and that the proof itself is due to Fourier. $\endgroup$ Aug 3, 2011 at 14:16
  • 9
    $\begingroup$ J.H.Lambert (1728-1777) gave a proof (c.1766) that $\pi$ is irrational. In the early to mid19th century preceding the Hermite-Lindemann transcendence theorem there were limited results,e.g. : If $A,B,C$ are rational and $Ae^2+Be+C=0$ then $A=B=C=0$ (easy). If $r\in \mathbb{Q}$ and $r\ne 0$ then $e^r\notin \mathbb{Q}$ (not easy). $\endgroup$ Nov 29, 2015 at 5:33
  • 14
    $\begingroup$ @openspace Some steps are skipped. If $e^{-1} = p/q$ for some coprime $p,q \in \mathbb N$ then $k$ can be chosen so that $2k-1 > q$, hence $(2k-1)! e^{-1}$ is an integer. $\endgroup$ Feb 12, 2017 at 16:25

1 Answer 1


The motivation for the definition of irrationality measure came from results in diophantine approximation. E.g., Dirichlet's Theorem states that for every real irrational $x$ there are infinitely many integer pairs $p,q$ with $|x-(p/q)|<q^{-2}$. If $x$ is a real quadratic irrational, then there's a positive constant $C$ such that for every integer pair $p,q$ we have $|x-(p/q)|>Cq^{-2}$. Liouville's Theorem was mentioned in the first comment. These can all be stated in terms of irrationality measure, and form an obvious motivation for that concept.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .