Prove the irrationality of $0.235711131719...$

Question

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.

Answer 1

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.

Answer 2

Not really formal but should point you in the right direction. Well let's suppose there is a period. Then the period would be $n$ digits long. Now I can find a prime that is longer than $n$ and therefore I cannot have a finite period.

Answer 3

Suppose this were to start repeating after some point with a period $N$. Beyond this point, for any $k$ there are at most $N$ different strings of length $k$ appearing in the repeating part. So it is enough to show that for large values of $k$ we have more than $N$ primes with $k$ digits. The prime number theorem tells us to expect the number of primes with $k$ digits to be on the order of $\frac{10^k}{k}$, so in particular for $k >> 0$ we can make this bigger than any $N$.

Answer 4

Suppose this number is eventually periodic with a period length of $L$. Then eventually we have infinitely many primes in the sequence such that $P > 10^L$, and for some $P$, $10^{nL} > P \geq 10^{nL - 1}$ (in other words we can find a prime whose digit length is nL, a multiple of L)$^{[1]}$. Then since P has a digit length which is a multiple of L, we can write:

$$P = R \cdot 10^{L(n - 1)} + R \cdot 10^{L(n - 2)} + \cdots + R$$

Where $R$ is the repeating sequence. But then $R|P$, and $P$ is not prime, a contradiction.

[1] This is justified by Bertrand's postulate, see the comments in this answer.

Answer 5

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