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 $\pi$ ?

YuiTo Cheng
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    I always thought the definition of irrationality measure was motivated from Liouville's theorem: http://en.wikipedia.org/wiki/Diophantine_approximation#Approximation_to_algebraic_numbers – Qiaochu Yuan Aug 03 '11 at 05:14
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    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. – André Nicolas Aug 03 '11 at 05:30
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    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): http://books.google.com/books?id=5J0AAAAAMAAJ Stainville says he learned the proof from Poinsot and that the proof itself is due to Fourier. – Dave L. Renfro Aug 03 '11 at 14:16
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    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). – DanielWainfleet Nov 29 '15 at 05:33
  • But why $(2k-1)! e^{-1}$ should be integer? – openspace Sep 26 '16 at 17:22
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    @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. – A Blumenthal Feb 12 '17 at 16:25
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    It must be an integer in order to preclude a circular-reasoning proof. If you include it as a term of infinite floating point value you end up with a similar situation as with trying to prove there are no repeating sequences in pi. – SDsolar Jun 02 '17 at 18:23
  • Would it be correct to say that since a rational number is expressed as a fraction of two integers, then if $e^-1$ is irrational then its reciprocal $e$ must be as well? – Gil Keidar Mar 05 '18 at 16:38
  • @Gil Reciprocal of irrational number $x$ is always irrational. If it were rational, the $x$ would be rational as well. – Oldboy Apr 09 '18 at 14:23
  • There are many proofs about π irrationality in https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational – Paulo Buchsbaum May 02 '18 at 18:10
  • You may be interested in the book "Auxiliary Polynomials" recently written by David Masser. This example is the subject of chapter one, and in chapter 8 he talks about irrationality measures. It's very well written and easy to follow. (You do need to fell in the gaps sometimes with your own calculations though.) – JonHales Oct 04 '18 at 05:32

1 Answers1


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.

Gerry Myerson
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