Prime constant

The prime constant is the real number $$\rho$$ whose $$n$$th binary digit is 1 if $$n$$ is prime and 0 if $$n$$ is composite or 1.

In other words, $$\rho$$ is the number whose binary expansion corresponds to the indicator function of the set of prime numbers. That is,
 * $$ \rho = \sum_{p} \frac{1}{2^p} = \sum_{n=1}^\infty \frac{\chi_{\mathbb{P}}(n)}{2^n}$$

where $$p$$ indicates a prime and $$\chi_{\mathbb{P}}$$ is the characteristic function of the set $$\mathbb{P}$$ of prime numbers.

The beginning of the decimal expansion of ρ is: $$ \rho = 0.414682509851111660248109622\ldots$$

The beginning of the binary expansion is: $$ \rho = 0.011010100010100010100010000\ldots_2 $$

Irrationality
The number $$\rho$$ can be shown to be irrational. To see why, suppose it were rational.

Denote the $$k$$th digit of the binary expansion of $$\rho$$ by $$r_k$$. Then since $$\rho$$ is assumed rational, its binary expansion is eventually periodic, and so there exist positive integers $$N$$ and $$k$$ such that $$r_n = r_{n+ik}$$ for all $$n > N$$ and all $$i \in \mathbb{N}$$.

Since there are an infinite number of primes, we may choose a prime $$p > N$$. By definition we see that $$r_p=1$$. As noted, we have $$r_p=r_{p+ik}$$ for all $$i \in \mathbb{N}$$. Now consider the case $$i=p$$. We have $$r_{p+i \cdot k}=r_{p+p \cdot k}=r_{p(k+1)}=0$$, since $$p(k+1)$$ is composite because $$k+1 \geq 2$$. Since $$r_p \neq r_{p(k+1)}$$ we see that $$\rho$$ is irrational.