Prime constant

{{Short description|Real number whose nth binary digit is 1 if n is prime and 0 if n is composite or 1}}

{{distinguish|Twin prime#First Hardy–Littlewood conjecture{{!}}Hardy–Littlewood's twin prime constant|Brun's theorem{{!}}Brun's twin prime constant}}

The prime constant is the real number $\rho$ whose $n$ th binary digit is 1 if $n$ is prime and 0 if $n$ is composite or 1.{{Cite web |last=Weisstein |first=Eric W. |title=Prime Constant |url=https://mathworld.wolfram.com/PrimeConstant.html |access-date=2025-01-31 |website=mathworld.wolfram.com |language=en}}

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: $\rho = 0.414682509851111660248109622\ldots$ {{OEIS|A051006}}

The beginning of the binary expansion is: $\rho = 0.011010100010100010100010000\ldots_2$ {{OEIS|A010051}}

Irrationality

The number $\rho$ is irrational.{{Cite book|last=Hardy|first=G. H.|url=https://www.worldcat.org/oclc/214305907|title=An introduction to the theory of numbers|date=2008|publisher=Oxford University Press|others=E. M. Wright, D. R. Heath-Brown, Joseph H. Silverman|isbn=978-0-19-921985-8|edition=6th|location=Oxford|oclc=214305907}}

= Proof by contradiction =

Suppose $\rho$ 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 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

Category:Irrational numbers

Category:Prime numbers

Category:Articles containing proofs

Category:Mathematical constants