# Do permutations on the decimal expansions of irrational numbers retain the property of irrationality?

Suppose we have an irrational number with the following decimal expansion:

$$A = a_0 \ a_1 \ a_2 \ a_3 \ a_4 \ a_5 \ a_6 \dots$$

Now, construct a new real number through a permutation on the decimals of the original number in the following manner:

$$A' = (a_1 \ a_0) \ (a_3 \ a_2) \ (a_5 \ a_4) \dots$$

Question 1: is $$A'$$ irrational, too?

We can generalize this question by considering other permutations as well. Let's consider the irrational number $$A$$ as above again. Let $$\sigma_{k}(\cdot)$$ be some permutation on a tuple of decimals of length $$k$$. Define

$$f_{k} (A) = \sigma_{k} (a_{0} \ a_{1} \dots \ a_{k-1} ) \ \sigma_{k} (a_{k} \ a_{k+1} \ \dots \ a_{2k-1} ) \ \sigma_{k} (a_{2k} \ a_{2k+1} \ \dots a_{3k-1} ) \dots$$

Question 2: is $$f_{k} (A)$$ necessarily irrational for all $$k \geq 2$$ and every possible permutation?

• Interesting (+1) , but probably out of reach. Irrationality proofs are in general extremely difficult as the open cases $\gamma$ , $e+\pi$ and $e\cdot \pi$ show. Aug 19, 2023 at 10:58
• Irrational is easy in this case, since rational $A'$ leads to a rational $A$ immediately. @Peter Aug 19, 2023 at 10:59

All decimal expansions involved below do not have any ambiguity of the shape $$0.999\dots=1.000\dots$$ - since we are dealing with an irrational $$A$$ as a start. (And the permutations used are also not disturbing the pattern, seen as a stationary pattern.)
Assume that $$A'$$ is rational. Then its decimal representation is periodic, let $$P=(d_1d_2\dots d_n)$$ be a period. We may and do assume that $$n$$ is a multiple of $$k$$, else replace $$P$$ by its repeated pattern $$\underbrace{PP\dots P}_{k\text{ times}}$$. Now go back from $$A'$$ to $$A$$ by using the inverse of $$\sigma_k$$. We obtain a periodic decimal representation for $$A$$, so $$A\in\Bbb Q$$. A contradiction.