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?

  • $\begingroup$ 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. $\endgroup$
    – Peter
    Aug 19, 2023 at 10:58
  • 2
    $\begingroup$ Irrational is easy in this case, since rational $A'$ leads to a rational $A$ immediately. @Peter $\endgroup$
    – dan_fulea
    Aug 19, 2023 at 10:59

1 Answer 1


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.


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