Prime constant explained

\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} (1)/(2 ^{p}) = \sum ^{inf ty} _{n =1} (\chi _{P} (n))/(2 ^{n}) where p indicates a prime and \chi _{P} is the characteristic function of the set 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.^[1] 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 thatr _{n} = r _{n+ik} for all n > N and all i \in 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 N . Now consider the case i = p . We have r _{p+i} = r _{p+p} = r _{p(k+1)} =0 , since p(k +1 ) is composite because k +1 \geq 2 . Since r _{p} ≠ r _{p(k+1)} we see that \rho is irrational.

References

1. Book: Hardy, G. H.. An introduction to the theory of numbers. 2008. Oxford University Press. E. M. Wright, D. R. Heath-Brown, Joseph H. Silverman. 978-0-19-921985-8. 6th. Oxford. 214305907.