I'm looking into expanding my knowledge in ways to show that some constant is irrational. I'm gonna give some examples of irrationality proofs, and I'm interested in learning what strategies you guys know and if such strategies can be expanded to other constants.

I'll ignore algebraic numbers such as $n^{1/m}$ where $n,m$ are integers or rationals because I believe such proofs abuse properties of integers and these proofs cannot be expanded to other constants. The strategies below concern constants that are transcendental or conjectured to be transcendentals.

  1. (Fast converging series) One way to show that a constant is irrational is to use a series that converges really fast to it. Example: We can use the truncated power series for $e$ to show that $$n!(e-\sum_{k=0}^n 1/k!) <1/n$$ for every positive integer $n$. If we assume $e=p/q$ we find an integer in $(0,1)$. This shows that the power series is so fast that as $n$ goes to infinity, $e-\sum_{k=0}^n 1/k!$ reaches $0$ faster than $n!$ goes to infinity. This can be expanded to $e^r$, where $r$ is rational. The other proof that uses a fast converging series is Apery proof of the irrationality of $\zeta(2),\zeta(3)$. Although, I don't remember seeing this approach to other constants.

  2. (Niven's polynomial) This approach uses the fact that the derivative of $e^x$ is itself to show that $e$ is irrational. Since the second derivative of $\sin x$ is minus itself this approach also works to show that $\pi$ is irrational, and $\sin, \cos, \exp$ are irrational in rational arguments, and so are the hyperbolics functions. I believe this approach only works for these constants because it uses the fact that the derivatives of said functions are themselves. For example see Niven's paper 'A simple proof that $\pi$ is irrational'.

  3. (Beukers integrals) I'll call a Beukers integral, the integral $$I_n=\int_0^1 x^nf(x)dx = \frac{a_n\xi+b_n}{d_n}$$ where $\xi$ is the number we want to show that is irrational and $a,b,d$ are sequence of integers. The challenge in using this approach is to find a suitable function $f(x)$. The strategy is to show that $I_n$ is not null, and as $n\to\infty$, $d_nI_n$ goes to $0$. Note that, we can change $x^n$ for a polynomial with degree $n$ and integer coefficients that the integral will still have the same form. Thus we can choose the Legendre polynomial. More details here Legendre polynomials in irrationality proofs. by Beukers We can use this approach to show that $\ln 2, e^r,\pi^2, \zeta(2),\zeta(3)$ are irrational. I don't think this approach have been used to other constants, but it seems it is the most promising strategy so far.

In the book Making Transcendence Transparent, it contains a proof that $e^\pi$ is irrational (also transcendental) which I did not know existed without using the Gelfond theorem. However I'm still learning about it, but wanted to mention it.

Please, share other methods and strategies that you're aware of, or share a link to a paper that demonstrates the strategy.

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    $\begingroup$ Here's one that's a bit of a cheat: Show that the binary digits of the number are an uncomputable function. For example, take some reasonable enumeration of Turing machines T_n and let x_n be 1 when T_n halts on the blank tape and 0 when it does not. Then 0.x_1x_2x_3... is irrational. The objection here is that this gives you classes of irrationals, but does not in general give you any way to show a constant is irrational. (I don't know of any example where a constant already of interest was shown to be irrational this way, and almost all constants we care about are by nature computable.) $\endgroup$
    – JoshuaZ
    Nov 25, 2022 at 13:30
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    $\begingroup$ I feel like the question is pretty much asking for a summary of all of transcendental number theory. Of course it's still restricting to irrationality, but it still seems to me overly broad. $\endgroup$
    – Wojowu
    Nov 25, 2022 at 13:57
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    $\begingroup$ The author's name is Niven. Ivan is his first (given) name. The accepted way to mention him is either "Niven", or "I. Niven", or "Ivan Niven", but certainly not "Ivan":-) $\endgroup$ Nov 25, 2022 at 14:01
  • 1
    $\begingroup$ And the polynomial can be called "Niven's polynomial", not "Ivan polynomial":-) $\endgroup$ Nov 25, 2022 at 14:05
  • 6
    $\begingroup$ @AlexandreEremenko "I, Niven", on the other hand, was not quite the hit with his mathematical reader base that Asimov had hoped for. $\endgroup$ Nov 25, 2022 at 19:46

4 Answers 4


Don't forget the obvious: If the digits of $\alpha$ in any base are not eventually periodic, then $\alpha$ is irrational.

Mel Nathanson and I (Integers 14 (2014), A40, pp. 1--11) used this to prove the following. With integers $k,s$, both at least 2, let $g_k^{(s)}(n)$ be the largest cardinality of a subset of $\{1,2,\ldots,n\}$ not containing any $k$-term geometric progression whose common ratio is a power of $s$. Then $$\lim_{n\to\infty} \frac{g_k^{(s)}(n)}{n}$$ exists and is irrational. A key part of the argument, that the digits of the limit are not periodic, rests on Szemerèdi's Theorem on arithmetic progressions.


Another technique, which is discussed in the Mathoverflow question Irrationality proof technique: no factorial in the denominator is to show that $n!$ cannot be the denominator for any $n$. In principal you could use this technique not for $n!$ but for any sequence of denominators $a_n$ where for any $k$, $\operatorname{lcm}(1,2,\dotsc,k)\mid a_n$ if $n$ is sufficiently large. Douglas Zare in that thread noted that this technique gives a nice proof that values of certain Bessel functions are irrational with a specific chosen set that is not $n!$ but arises from the series. It seems like this works mainly for fast-converging series, so this may be somehow that technique in disguise?

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    $\begingroup$ this is a very cool example!! great link $\endgroup$ Nov 25, 2022 at 14:53
  • $\begingroup$ I think you mean "for any $k$, $\operatorname{lcm}(1, 2, \dotsc, k) \mid a_n$ if $n$ is sufficiently large" (otherwise, with "$a_n$ sufficiently large" in place of "$n$ sufficiently large", the condition would be satisfied by a bounded sequence). What is the difference between that condition and "for every $k$, $k \mid a_n$ if $n$ is sufficiently large"? Maybe it's something subtle about the order of the quantifiers. $\endgroup$
    – LSpice
    Nov 26, 2022 at 4:23
  • $\begingroup$ @LSpice. You are correct. Fixed. $\endgroup$
    – JoshuaZ
    Nov 26, 2022 at 13:03

The original and some subsequent proofs of the irrationality of $\pi$ implicitly use the following lemma:

If there is a sequence $P_n(x) \in \mathbb{Z}[x]$ such that $P_n(\alpha)>0$ and $P_n(\alpha)=o(c^{\deg{P_n} })$ for all $0<c<1$, then $\alpha$ is irrational.

  • $\begingroup$ Can you provide a link where this lemma is used? $\endgroup$
    – Pinteco
    Nov 27, 2022 at 4:23
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    $\begingroup$ From all the implicit usages, it's most explicit in Cartwright's proof, appearing on Wikipedia: en.m.wikipedia.org/wiki/… $\endgroup$ Nov 27, 2022 at 7:24

Methinks that the following criterion has not been mentioned so far:

If $\{a_{i}\}_{i \in \mathbb{N}}$ is a strictly increasing sequence of natural numbers such that the series $\sum_{i=1}^{\infty} \frac{1}{a_{i}}$ diverges, then the unending decimal fraction $\alpha$ formed by juxtaposing the successive terms of the sequence $\{a_{i}\}_{i \in \mathbb{N}}$ represents an irrational number.

You can find more information regarding its provenance and proof here.


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