Prime constant

The number ${\displaystyle \rho }$ is irrational.^[2]

Suppose ${\displaystyle \rho }$ were rational.

Denote the ${\displaystyle k}$th digit of the binary expansion of ${\displaystyle \rho }$ by ${\displaystyle r_{k}}$. Then since ${\displaystyle \rho }$ is assumed rational, its binary expansion is eventually periodic, and so there exist positive integers ${\displaystyle N}$ and ${\displaystyle k}$ such that ${\displaystyle r_{n}=r_{n+ik}}$ for all ${\displaystyle n>N}$ and all ${\displaystyle i\in \mathbb {N} }$.

Since there are an infinite number of primes, we may choose a prime ${\displaystyle p>N}$. By definition we see that ${\displaystyle r_{p}=1}$. As noted, we have ${\displaystyle r_{p}=r_{p+ik}}$ for all ${\displaystyle i\in \mathbb {N} }$. Now consider the case ${\displaystyle i=p}$. We have ${\displaystyle r_{p+i\cdot k}=r_{p+p\cdot k}=r_{p(k+1)}=0}$, since ${\displaystyle p(k+1)}$ is composite because ${\displaystyle k+1\geq 2}$. Since ${\displaystyle r_{p}\neq r_{p(k+1)}}$ we see that ${\displaystyle \rho }$ is irrational.