6
$\begingroup$

I am curious to know if the following number is irrational or transcendental:

$$\displaystyle A = \sum_p 2^{-p},$$ where the sum is over all positive primes. A similar question can be asked for any number $k$ other than 2.

Edit: In retrospect and as pointed out by some comments below that it is trivial that $A$ is irrational. The transcendence question seems still relevant.

If we define a random number

$$\displaystyle b = \sum_n 2^{\rho(n)},$$

where $\rho(n)$ is a Bernoulli random variable defined to be $\rho(n) = -n$ with probability $1/\log n$ and $-\infty$ with probability $1 - 1/\log n$ (so that $2^{\rho(n)} = 0$), for say $n \geq 2$ and define $\rho(1) = -1$.

Can it be proved that say $n$ is transcendental with probability 1?

$\endgroup$
3
  • 10
    $\begingroup$ For irrational: since the primes cannot be eventually periodic, A is irrational. Am I missing something? $\endgroup$
    – user9072
    Nov 29, 2012 at 17:40
  • 2
    $\begingroup$ Definition of the random nunmber $b$ apart, recall that algebraic numbers are countable, so they are a null set for any atomless probability measure. $\endgroup$ Nov 29, 2012 at 18:02
  • 3
    $\begingroup$ The first number (apparently called the "prime number constant") appeared in the following thread on math.SE and Jonas Meyer collected a few links in his answer: math.stackexchange.com/q/42697 It appears to be unknown whether it is transcendental or not. $\endgroup$ Nov 29, 2012 at 18:17

1 Answer 1

11
$\begingroup$

The number $A$ is irrational, since the characteristic function of the set of prime numbers is not eventually periodic.

It definitely should be transcendental as it is not a base $2$ normal number, ie the asymptotic frequency of digits is not the same (and algebraic numbers are conjectured to be normal). But this should be open (as also in the link mentioned by Theo Buhler, while I was composing this).

This conjecture would also answer the question on $b$, but Pietro Majer gave a direct argument already; it might however be good to keep this in mind for variations of the question.

Now, since the above is a bit terse some information around this. There is an interesting rather recent paper on more or less precisely this type of problem by Bailey, Borwein, Crandall, Pomerance "On the binary expansions of algebraic numbers"

Among others they show that the binary distribution of a real algebraic number cannot be too extremely biased more precisely both digit have to occur (up to $N$) at least order of $N^{1/d}$ times where $d$ is the degree of the number. They also mention that thus the degree of $$\sum_p \frac{1}{2^{p^k}}$$ is at least $k+1$ (with the convention that the degree of a transcendental is infinite).

There is more recent work on this by Adamczewski and Faverjon on this, too.

Finally, a good overview on this type of problem is given in this paper of Adamczewski and Bugeaud (see in particular Chapter 6 on lacunary numbers, that is precisley those of the form considered here, ie the digits are the characteristic function of some set).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.