# Is a irrational number still irrational when we apply some mapping to its decimal representation?

This is a question I came up with in high school, but has not yet found a decent proof.

Let $f(n)=0$ if $n$ is even, and $f(n)=1$ if $n$ is odd.

Let the decimal representation of $\sqrt2$ be

$$\sqrt2=\sum_{i=0}^{+\infty}\frac{d_i}{10^i}$$

and, $$a=\sum_{i=0}^{+\infty}\frac{f(d_i)}{10^i}$$

Prove that $a$ is irrational.

I have tried to use the idea that all rational numbers have a finite or recurring decimal representation, but that doesn't seem to help much.

I also believe this is just a special case of a more general problem as the form of $f(n)$ can be more complicated and the base need not be $10$.

Thank you in advance for any help.

## 1 Answer

Not enough is known about the decimal representation of $\sqrt{2}$ to be able to prove this. For all we know, it could be that for sufficiently large $i$, $d_i$ is always even. That is almost certainly false, but in the current state of the art there is no way to disprove it.

On the other hand, there do exist irrational numbers (uncountably many of them) whose base $10$ digits are all even.

• So I have accidentally touched the limit of state of art when I was in high school? Interesting. And yes, the second paragraph is the reason why I say rational number always have a finite or recurring decimal representation does not help in the proof. Thank you very much. – Weijun Zhou Jan 23 '18 at 2:50