4
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

$a$ 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 $2^n$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.