Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The number $a=0.12457...$ is defined as follows: The digit on the $n$-th place after the dot is the first digit left to the dot of the number $n\sqrt2$.

For example, for $n=1$ we have


and its first digit to the dot is 1.

For $n=2$ we have


and its first digit left to the dot is 2.

For $n=3$ we have


and its first digit to the dot is 4.

I'd like to show that $a$ is irrational.

share|cite|improve this question
Do you know for a fact that $a$ is irrational? – Michael Albanese Dec 19 '13 at 4:41
Good question. No, I only assumed that it is. I did some calculations up to $n=14$ with $a=0.12457891245689$ and it seemed to me that it seems to go back to similar numbers but not identically the same. – PandaMan Dec 19 '13 at 4:46
$a = 0.124578912456891245689123568912356890235679023567902346790234679013467801346780‌​134578013457801245780...$ – Dan Dec 19 '13 at 4:48
@PeterPanda: How did you come up with the algorithm? – user99680 Dec 19 '13 at 4:56
This is very interesting, indeed if $\alpha$ is irrational, then the number $a$ you construct is also irrational. Of course, if $\alpha$ is rational, then $a$ also is rational. – i707107 Dec 19 '13 at 6:17
up vote 14 down vote accepted

Consider the cases $n=10^k$. Then we get that the $n$th digit of $a$ is the $k$th digit of $\sqrt{2}$. Now, if $a$ is rational, then it repeats with some frequency, $f$. But then we can find $d$ so that $f\mid 10^{k+d}-10^k$ for $k$ large enough. Therefore, for large enough $k$, the $10^{k+d}$th digit of $a$ and the $10^k$th digit of $a$ must be the same.

But that means that $k+d$th digit of $\sqrt{2}$ is the same as the $k$th digit of $\sqrt{2}$ for large enough $k$, and therefore $\sqrt{2}$ repeats, and therefore $\sqrt{2}$ is rational, which is a contradiction.

share|cite|improve this answer
Is this sufficient? What if one of the digits is periodic but the rest are not? Say $0.1x1x1x\dots$. There are only 10 combinations for $1x$, but it seems to me that these sequences need not be periodic. – Yong Hao Ng Dec 19 '13 at 5:32
@Yong Hao Ng: This starts with assuming $a$ is rational, then derives contradiction. Thus, $a$ is irrational. – i707107 Dec 19 '13 at 6:09
@i707107 $0.1xxx1xxx1xxx\dots$ refers to an embedding into $\sqrt 2$ from $a$, assuming that $a$ is periodic. Anyway, the rest of the sequence should also be periodic since $f| 10^{k+d}-10^k$ implies $f| 10^{(k+1)+d}-10^{k+1}$ so that $\sqrt 2$ has period $\leq d$. Somehow I interpreted the answer as only for a specific $k$ when it meant all $k\geq N$ for some $N$. – Yong Hao Ng Dec 19 '13 at 6:33
Yes, "for sufficient large $k$" means for $k\geq N$ for some $N$. – Thomas Andrews Dec 19 '13 at 6:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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