The following text appears in the Mill's constant definition at the Wikipedia:

There is no closed-form formula known for Mills' constant, and it is not even known whether this number is rational (Finch 2003).

And then refers to Finch: Library of Congress Cataloging in Publication Data , Finch, Steven R., 1959- Mathematical constants / Steven R. Finch. p. cm.- (Encyclopedia of mathematics and its applications; v. 94)

But the only reference in Finch's text to Mill's constant being rational or irrational is this excerpt (in the text Mill's constant is named $C$):

It is not known whether C must necessarily be irrational.

Just that and only that. No references, no explanation if I did not miss nothing when reading it.

Making some simple calculations it seemed that it should be only irrational, but according to Finch that is wrong. This is my logic (the questions are at the end):

  1. Any Mill's constant $A$ provides a prime-generating function $\lfloor A^{k^n} \rfloor$ for $k \gt 2$ where the constant is gradually calculated for each $n$ as $A_n=p^{\frac{1}{k^n}}$ for some selected on purpose prime $p \in [n^k,(n+1)^k]$ being the final unknown value of $A=lim_{n \to \infty}A_n$.

  2. If $A_n$ is rational then $A_n = p^{\frac{1}{k^n}} = \frac{a}{b}$ for $a,b \in \Bbb N$. Then:

$$b \cdot p^{\frac{1}{k^n}} = a$$

  1. If both sides are powered to ${k^n}$ then:

$$b^{(k^n)} \cdot p = a^{(k^n)}$$

  1. But that is only possible if $p=c^{(k^n)}$ for some natural number $c$. And that is a contradiction because $p$ is always a prime selected on purpose for each iteration when constructing the constant, so it can not have divisors apart from $1$ and $p$ by the very definition of prime numbers. So according to this if the logic is correct, $A_n$ could not be rational, they just could be irrational, $\forall n$ and $\forall k \gt 2$.

I would like to ask the following questions:

  1. Are the calculations correct? is the proof by contradiction wrong or it is not possible to apply it to $n \to \infty$?

  2. Why does Finch state that it is not known if $A$ must necessarily be irrational? Are there other references / papers regarding this point? Thank you!

  • 2
    $\begingroup$ $a_n$ is rational or irrational has nothing to do wtih $\lim_{n\to\infty} a_n$ is rational or irrational. e.g rational/rational $1-1/n \to 1$. rational/irrational $\sum_{k=1}^n \frac{1}{k^2} \to \frac{\pi^2}{6}$. irrational/rational $1-\pi/n \to 1$. irrational/irrational $\pi-\pi/n \to \pi$. $\endgroup$ Jun 25, 2016 at 14:30

1 Answer 1


(Answering 1. only:)

You have showed that the numbers $A_n$ cannot be rational. But Mills' constant is the limit of that sequence of numbers, not one of the numbers themselves.

There is nothing to prevent a sequence of irrational numbers from having a rational limit. For a simpler example, consider the sequence $c_n=2^{1/n}$, with $\lim_{n \to \infty} c_n=1$.

  • $\begingroup$ so that was the point! it is kind of "philosophic"' but in this case the limit whatever we dig into $\infty$ should define an hypothetic prime in an hypothetic far away interval being always irrational, in the case of your example there is no restriction, but in the case of Mill's constant would not the restriction affect to the very value of the limit? Its value if exists must define a prime via the prime-generating function, but there are infinite primes, so it would define the last existing prime, which is impossible, again a contradiction. $\endgroup$
    – iadvd
    Jun 25, 2016 at 14:48

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