I am obsessed with decimal representations (and the number $2$). Here is another question related to them.

For $n\in\mathbb N^+$, define $b_n$ as the leading digit in the decimal representation of $2^n$.

Define $a$ as $$a=\sum_{n=1}^{\infty}\frac{b_n}{10^n},$$

So the first few digits of $a$ are $$a=0.2481361\ldots$$

Is $a$ rational?

I think this will be a real challenge as the definition of $b_n$ implicitly includes the floor function and the irrational number $\log_210$. However I am going to post this question anyway to see if anyone has any smart ideas on it.

Last question: Is a irrational number still irrational when we apply some mapping to its decimal representation?

Edit: I thought over it again, this can actually be quite general. If $k>1$ is a number (doesn't even have to be algebraic) such that $\lg k$ is irrational, and $c_n$ is the leading digit in the decimal representation of $k^n$. Define $a'$ as $$a'=\sum_{n=1}^{\infty}\frac{c_n}{10^n},$$ Then $a'$ is irrational. The proof in the accepted answer apply.


$a$ is irrational. If it were rational the leading digits of $2^n$ would have to repeat, so (for any $n$ greater than some $N$) the first digit of $2^{n+k}$ is the same as the first digit of $2^n$, where $k$ is the length of the repeat. This would require that $2^k$ be a power of $10$, which it is not. If $2^k$ is not a power of $10$ there is some $m$ such that the leading digit of $2 ^{km}$ is $2$ and $2^{n+km}$ will not have the same leading digit as $2^n$

  • $\begingroup$ Thank you. Quite elegant proof. Just the type I am seeking. $\endgroup$ Jan 29 '18 at 3:37

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