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.

We use Dirichlet's Theorem on primes in arithmetic progressions. Let $A_k$ be the number whose decimal expansion consists of $k$ consecutive $1$'s. The numbers $A_k$ and $10^{k+1}$ are relatively prime. It follows by Dirichlet's Theorem that there are infinitely many primes of the shape $A_k +n\cdot 10^{k+1}$. Such a prime has $k$ consecutive $1$'s at the left end of its decimal expansion.

Thus your number has strings of consecutive $1$'s of arbitrary length. By another simpler application of Dirichlet's Theorem, the number has non-$1$'s arbitrarily far in its decimal expansion. Thus the decimal expansion cannot be ultimately periodic.

$0.23571113...$ 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 $N < p < 10N$.