The prime constant is the real number <math>\rho</math> whose <math>n</math>th binary digit is 1 if <math>n</math> is prime and 0 if <math>n</math> is composite or 1.
In other words, <math>\rho</math> is the number whose binary expansion corresponds to the indicator function of the set of prime numbers. That is,
:<math> \rho = \sum_{p} \frac{1}{2^p} = \sum_{n=1}^\infty \frac{\chi_{\mathbb{P(n)}{2^n}</math>
where <math>p</math> indicates a prime and <math>\chi_{\mathbb{P</math> is the characteristic function of the set <math>\mathbb{P}</math> of prime numbers.
The beginning of the decimal expansion of ρ is: <math> \rho = 0.414682509851111660248109622\ldots</math>
Proof by contradiction
Suppose <math>\rho</math> were rational.
Denote the <math>k</math>th digit of the binary expansion of <math>\rho</math> by <math>r_k</math>. Then since <math>\rho</math> is assumed rational, its binary expansion is eventually periodic, and so there exist positive integers <math>N</math> and <math>k</math> such that
<math>r_n = r_{n+ik}</math> for all <math>n > N</math> and all <math>i \in \mathbb{N}</math>.
Since there are an infinite number of primes, we may choose a prime <math>p > N</math>. By definition we see that <math>r_p=1</math>. As noted, we have <math>r_p=r_{p+ik}</math> for all <math>i \in \mathbb{N}</math>. Now consider the case <math>i=p</math>. We have <math>r_{p+i \cdot k}=r_{p+p \cdot k}=r_{p(k+1)}=0</math>, since <math>p(k+1)</math> is composite because <math>k+1 \geq 2</math>. Since <math>r_p \neq r_{p(k+1)}</math> we see that <math>\rho</math> is irrational.