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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

$n\sqrt2=\sqrt2=1.4142...$

and its first digit to the dot is 1.

For $n=2$ we have

$n\sqrt2=2\sqrt2=2.8284...$

and its first digit left to the dot is 2.

For $n=3$ we have

$n\sqrt2=3\sqrt2=4.2426...$

and its first digit to the dot is 4.

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

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3  
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
1  
$a = 0.124578912456891245689123568912356890235679023567902346790234679013467801346780‌​134578013457801245780...$ –  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

1 Answer 1

up vote 12 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.

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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

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