The prime constant is the real number \rho whose nth binary digit is 1 if n is prime and 0 if n is composite or 1.[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 [1]
The beginning of the binary expansion is: \rho = 0.011010100010100010100010000\ldots_2
Irrationality
The number \rho is irrational.[2]
Proof by contradiction
Suppose \rho were rational.
Denote the kth 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.
References
- ^ Weisstein, Eric W. "Prime Constant". mathworld.wolfram.com. Retrieved 2025-01-31.
- ^ Hardy, G. H. (2008). An Introduction to the Theory of Numbers. 6th ed. Oxford: Oxford University Press. ISBN 978-0-19-921985-8. OCLC 214305907