Prime constant

The prime constant is the number $ho$ whose $n$th binary digit is 1 if $n$ is prime and 0 if it is composite.

In other words, $ho$ is simply the number whose binary expansion corresponds to the indicator function of the set of prime numbers. That is, :$ho\; =\; sum\_\{p\}\; frac\{1\}\{2^p\}\; =\; sum\_\{n=1\}^infty\; frac\{chi\_\{mathbb\{P(n)\}\{2^n\}$where $p$ indicates a prime and $chi\_\{mathbb\{P$ is the characteristic function of the primes.

The beginning of the decimal expansion of "ρ" is: $ho\; =\; 0.414682509851111660248109622...$

Irrationality

The number $ho$ is easily shown to be irrational. To see why, suppose it were rational.

Denote the $k$th digit of the binary expansion of $ho$ by $r\_k$. Then, since $ho$ is assumed rational, there must exist $N$, $k$ positive integers 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 $r\_p\; eq\; r\_\{p(k+1)\}$ we see that $ho$ is irrational.