The prime constant is the real number ρ {\displaystyle \rho } whose n {\displaystyle n} th binary digit is 1 if n {\displaystyle n} is prime and 0 if n {\displaystyle n} is composite or 1. In other words, ρ {\displaystyle \rho } is the number whose binary expansion corresponds to the indicator function of the set of prime numbers. That is,
ρ = ∑ p 1 2 p = ∑ n = 1 ∞ χ P ( n ) 2 n {\displaystyle \rho =\sum _{p}{\frac {1}{2^{p}}}=\sum _{n=1}^{\infty }{\frac {\chi _{\mathbb {P} }(n)}{2^{n}}}}
where p {\displaystyle p} indicates a prime and χ P {\displaystyle \chi _{\mathbb {P} }} is the characteristic function of the set P {\displaystyle \mathbb {P} } of prime numbers. The beginning of the decimal expansion of ρ is: ρ = 0.414682509851111660248109622 … {\displaystyle \rho =0.414682509851111660248109622\ldots } (sequence A051006 in the OEIS) The beginning of the binary expansion is: ρ = 0.011010100010100010100010000 … 2 {\displaystyle \rho =0.011010100010100010100010000\ldots _{2}} (sequence A010051 in the OEIS)
Irrationality The number ρ {\displaystyle \rho } is irrational.
Proof by contradiction Suppose ρ {\displaystyle \rho } were rational. Denote the k {\displaystyle k} th digit of the binary expansion of ρ {\displaystyle \rho } by r k {\displaystyle r_{k}} . Then since ρ {\displaystyle \rho } is assumed rational, its binary expansion is eventually periodic, and so there exist positive integers N {\displaystyle N} and k {\displaystyle k} such that
r n = r n + i k {\displaystyle r_{n}=r_{n+ik}} for all n > N {\displaystyle n>N} and all i ∈ N {\displaystyle i\in \mathbb {N} } . Since there are an infinite number of primes, we may choose a prime p > N {\displaystyle p>N} . By definition we see that r p = 1 {\displaystyle r_{p}=1} . As noted, we have r p = r p + i k {\displaystyle r_{p}=r_{p+ik}} for all i ∈ N {\displaystyle i\in \mathbb {N} } . Now consider the case i = p {\displaystyle i=p} . We have r p + i ⋅ k = r p + p ⋅ k = r p ( k + 1 ) = 0 {\displaystyle r_{p+i\cdot k}=r_{p+p\cdot k}=r_{p(k+1)}=0} , since p ( k + 1 ) {\displaystyle p(k+1)} is composite because k + 1 ≥ 2 {\displaystyle k+1\geq 2} . Since r p ≠ r p ( k + 1 ) {\displaystyle r_{p}\neq r_{p(k+1)}} we see that ρ {\displaystyle \rho } is irrational.
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