İs 𝑆=0.[2𝑘][3𝑘][5𝑘][7𝑘].... rational?

[littlemathquark]

Homework Statement

For any given positive integer 𝑘, let the decimal representation of a number m be denoted by [𝑚], and let 𝑘 be multiplied by successive prime numbers to form the number 𝑆=0.[2𝑘][3𝑘][5𝑘][7𝑘]....
For example, for 𝑘=2, we get 𝑆=0.461014...
For which values of k is the number S rational?

Relevant Equations

For any given positive integer 𝑘, let the decimal representation of a number m be denoted by [𝑚], and let 𝑘 be multiplied by successive prime numbers to form the number 𝑆=0.[2𝑘][3𝑘][5𝑘][7𝑘]....
For example, for 𝑘=2, we get 𝑆=0.461014...
For which values of k is the number S rational?

I know if k=1 then S=0,235711.. is Copeland-Erdös number and it's irational. Can you give me an idea?

 

[fresh_42]

I would bet that the proof for $k=1$ can be generalized to $k\ge 1.$

 

[WWGD]

Isn't f(nk) for n Natural, (I)rational if k is? Not sure , given the fact that we're not working with any mod system.
Edit: Which brings up an interesting problem on whether, given 2 decimal expansions , not obviously Rational , whether one is an integer multiple of the other.

 

[littlemathquark]

I would bet that the proof for $k=1$ can be generalized to $k\ge 1.$

Hardy's book I found that proof about irrationality of Copeland-Erdos number:
"Let us assume that any arithmetical progression of the form $k. 10^{s+1}+1$, (k = 1, 2, 3...) contains primes. Then there are primes whose expressions in the decimal system contain an arbitrary number $s$ of O’s, followed by a 1. Since the decimal contains such sequences, it does not terminate or recur."

 

[littlemathquark]

My solution like this:

According to Dirichlet's theorem, there are infinitely many prime numbers of the form $p = a + n \cdot 10^{r+1}$ (primes of the form $a00000...1$). These prime numbers can be found somewhere within the number $S$. In other words, within $S$, there exist primes containing arbitrarily many consecutive zeros. Therefore, since the decimal expansion of $S$ contains such sequences that neither terminate nor repeat at any point, $S$ cannot be a rational number.

 

[littlemathquark]

Another idea:

The number formed by multiplying consecutive prime numbers with a given positive integer can only be rational if its decimal representation is either finite or contains a periodically repeating block of digits. However, since there are infinitely many prime numbers, the number can be thought of as having an infinite sequence of digits, and as each consecutive prime number is larger than the previous one, their products also grow larger. This causes the number of digits in the decimal representation to continuously increase. In a periodic sequence, there must be blocks of fixed length with repeating digits, but since the number of digits in the decimal representation of S keeps increasing, this is not possible. Therefore, the digit sequence of S cannot be periodic, and hence, S must be an irrational number.​

 

[littlemathquark]

Another solution:
İf S is rational, then its decimal expansion ultimately repeats in blocks of length r. Fix a prime p large enough that [kp] lies entirely within these repeating blocks; suppose that p has n digits and choose t such that tr > n. Two applications of Bertrand's postulate show that there are two consecutive primes q, q' satisfying $10^{tr-1}/k \le q \lt q' \lt 10^{tr}/k.$ Then kq, kq' each have exactly tr digits and [kq], [kq'] abut within the repeating blocks of S. Because the pattern repeats after tr digits, kq=kq' contradicting q<q'.