Let $\mathbb{P}$ be the set of prime numbers, and consider $m=\displaystyle\sum_{p\in\mathbb{P}} \frac{1}{2^{p}}$. Is $m$ irrational?

In the following paper, the author recalls several sufficient criteria for irrationality. When applying some of this criteria to $m$ I arrive that the condition of some of the criteria are not satisfied by means of Bertrand´s Postulate.

I found a similar result by Sándor:

Let $\lbrace a_m\rbrace$, $m\geq 1$ be a sequence of positive integers such that $$\text{lim sup}\frac{a_{m+1}}{a_1a_2\cdots a_m}=\infty \;\;\;\;\text{and}\;\;\;\;\;\;\text{lim inf}\frac{a_{m+1}}{a_m}>1$$ Then the series $\displaystyle\sum_{m} \frac{1}{a_m}$ is an irrational number.

I have yet to prove that if $f$ is a continuous function then $\text{lim sup} \;f(a_m)=f(\text{lim sup}\, a_m)$. Assuming this;

$$\text{lim inf}\; P_{m+1}-P_{m}>1$$ where $P_m$ is the $m$-th prime.

But I have trouble with $$\text{lim sup}\; P_{m+1}-\sum_{j=1}^{m}P_{j}$$ my guess is that $$\text{lim sup}\; P_{m+1}-\sum_{j=1}^{m}P_{j}\neq\infty$$ so $$\text{lim sup}\frac{2^{P_{m+1}}}{2^{P_1}\cdots 2^{P_m}}\neq\infty$$

and this theorem will result useless to tell if $m$ is irrational.

  1. How can I prove (or disprove) that $m$ is irrational (are there any other simpler criteria)?
  2. How can I use Dirichlet criterion or Hurwitz criterion?
  3. What is its irrationality measure?

Any help is highly appreciated.

Edit: That $m$ is irrational is clear by the criteria provided in the comments, namely that if $x$ is rational then the binary representation of $x$ is periodic. The number $m$ in binary expansion has $1$ in the $P$-th position and zero elsewhere. As there are arbitrarily large gaps between primes; then the binary representation fails to be periodic.

So the question that remains unsolved is:

  • What is its irrationality measure?
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    To answer your question, "is $m$ irrational", Yes it is. To see this just prove the following proposition: Let $x$ be a rational number. Then the binary representation of $x$ is eventually periodic. Note that this is true for rationals in any base. – Baby Dragon Feb 17 '14 at 02:32
  • Fot the irrationality measure, I make a vague conjecture. It is that the irrationality measure is high (the word high is what makes this vague). The reason for this is that the primes are sparse, so truncations of $m$ do a good job of approximating $m$ and so we have a high irrationality measure. I think that this certainly makes it highly intuitively plausible that $m$ is transcendental (although I do not have a formal proof to offer right now). – Baby Dragon Feb 17 '14 at 02:44
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    It's not just irrational, but also transcendental. – Lucian Feb 17 '14 at 02:48
  • @Lucian How would the sketch of such proof be? – Chazz Feb 17 '14 at 03:04
  • See [Liouville number](http://en.wikipedia.org/wiki/Liouville_number). – Lucian Feb 17 '14 at 03:06
  • @BabyDragon. So truncation of $m$ would work when applying Dirichlet criteria for irrationality? – Chazz Feb 17 '14 at 03:06
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    @Lucian The gaps between primes are too small for the usual arguments relevant to Liouville numbers to apply. What variant of the argument do you have in mind? – Andrés E. Caicedo Feb 17 '14 at 03:57
  • @AndresCaicedo: You're right. I was thinking of something more in terms of the [Copeland-Erdos](http://en.wikipedia.org/wiki/Copeland-Erdos_constant) or [Champernowne](http://en.wikipedia.org/wiki/Champernowne_constant) constants. – Lucian Feb 17 '14 at 04:40
  • https://oeis.org/A051006 – Charles Feb 17 '14 at 19:07
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    @BabyDragon: the primes are not nearly sparse enough to force the irrationality measure to be high. For example, the $k$th powers are far sparser than the primes, yet the truncations to the analogous series $\sum_{n=1}^\infty 2^{-n^k}$ are not even good enough to establish an irrationality measure greater than $1$. – Greg Martin Jan 04 '15 at 00:41

1 Answers1


This number is called the prime constant. Since it is irrational because it doesn't have a periodic binary representation it's irrationality measure is 2 or more according to this article.

Again according to the article If the irrationality number of the prime constant were more than 2 it would be transcendental but as far as I can tell it is not known to be transcendental although it is listed as a suspected transcendental number, see:


So the prime number constant is irrational, its irrationality measure is 2 or more, and it is an open problem whether or not it is transcendental and whether or not its irrationality number is more than 2.

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