Prove the irrationality of $0.235711131719...$ How can I prove that the number formed by concatenating the primes in order i.e. $0.235711131719...$ is irrational.
I know I have to demonstrate that it has no period, but I'll be so thankful if someone can explain very clear, including all cases.

$0.23571113...$ is known as the Copeland–Erdős constant and it is given by: $$
    \displaystyle \sum_{n=1}^\infty p_n 10^{-\left( n + \sum\limits_{k=1}^n \left\lfloor \log_{10}{p_k}\right\rfloor \right)} .$$ Here are two proofs given in Hardy's book An Introduction to the Theory of Numbers (p. 113) that shows that it is irrational:
$\phantom{}1)$ Let us assume that any arithmetical progression of the form $$k10^{\,s+1}+1\quad(k=1,2,3,\ldots)$$ contains
primes. Then there are primes whose expressions in the decimal
system contain an arbitrary number $s$ of $\rm O$ 's, followed by a $1$. Since
the decimal
contains such sequences,
it does not terminate or recur.
$\phantom{}2)$ Let us assume that there is a prime between $N$ and $10N$ for every
$N \geqslant 1$. Then, given $s$, there are primes with just $s$ digits. If the decimal recurs, it is of the form  $$... a_
1 a_ 2 ...a_ k |a_ 1 a_ 2 ...a_ k |...$$ the bars indicating the period, and the first being placed where the
first period begins. We can choose $l\gt1$ so that
all primes with $s = kl$
digits stand later in the decimal than the first bar. If $p$ is the first such
prime, then it must be of one of the forms  $$p = a_ 1 a_ 2 ...a_ k |a_ l a_ 2 ...a _k| ...|a _1 a_ 2 ...a_ k$$
or $$p
=
a_ {m+1} ..a_ k| a _1 a_ 2 ...a_ k| ...|a_ l a _2 ...a _k| a _l a _2 ...a_ m$$
and is divisible by $a$, $a_2 ...a_k$ or by $a_{m+1} ..a_k a$, $a_2 ...a_m$
: a contradiction.
The first proof follows from a special case of Dirichlet's theorem and the second from a theorem stating that for every $N\geqslant1$ there is at least one prime satisfying $N <
p\leqslant2N$. It
follows, a fortiori, that $N < p < 10N$.
