Irrationality of "primes coded in binary"

That $\ell$ is irrational is clear. There are arbitrarily large gaps between consecutive primes, so the binary expansion of $\ell$ cannot be periodic. Any rational has a periodic binary expansion.

The fact that there are arbitrarily large gaps between consecutive primes comes from observing that if $n>1$, then all of $n!+2, n!+3, \dots, n!+n$ are composite.

If it was rational, then the binary expansion would eventually repeat -- but the distribution of primes doesn't repeat.

To wit: if the repeat had period $n$ and $p$ was a prime large enough to be inside the repeating part, then $p+pn=(1+n)p$ would have to be prime, which obviously isn't the case. On the other hand, the repeating period cannot consist of all zeroes, because there are infinitely many primes.

Well to prove that this number is irrational you must know that a rational number has a periodic sequence of digits in any basis after a fixed digit. That is if $r=0.f_{1}f_{2}...f_{n}...$ then $f(n+T)=f(n)$ for some $T \in \mathbb{N}$ and $\forall n \geq k_{0} \in \mathbb{N}$. (check that!).

Then supposing by absurd, let $n_{0}$ be a natural such that $n_0\geq k_0$ and $n_{0}$ is prime then $1=f(n_{0})=f(n_{0}+T)=f(n_{0}+2T)=...=f(n_{0}+n_{0}T)=1$ absurd because $n_{0}+n_{0}T$ is not prime.

An interesting question arises is that number Algebraic, i.e., that is solution of a polinomial equation with rational coefficients?