How to prove that $\lim_{n\to \infty } \left(\sqrt{2} n-\left\lfloor \sqrt{2} n\right\rfloor \right) $ doesn't exist? I've tried to show that every $L\in\Re$ is not a limit. It's pretty easy for $L>1$ and $L<-1$. The main problem is when $-1\leq L\leq 1$ .

It is an homework question, so i should use the basic limit definition ($\epsilon$).


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    $\begingroup$ What have you tried? Note that the quantity you are looking at is the fractional part of $n\sqrt{2}$. How does this quantity change as you go from $n$ to $(n+1)$? $\endgroup$ – Aaron Nov 18 '13 at 7:51
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    $\begingroup$ Please edit so the question is in the body, not just the title. Also, please explain what you mean by the symbol, $L$, as you don't relate it to the question anywhere. Also, why does it matter whether the proof uses the basic limit definition --- is this a homework problem? $\endgroup$ – Gerry Myerson Nov 18 '13 at 8:01
  • $\begingroup$ I'll edit my question. $\endgroup$ – Lior Nov 18 '13 at 8:12
  • $\begingroup$ Note that $0 \leq (\sqrt{2}n - \lfloor \sqrt{2}n \rfloor) < 1$, hence limit L if it exists must satisfy $0\leq L \leq 1$. Now you need to show that for any L in this range the definition of limit can't be satisfied. $\endgroup$ – Paramanand Singh Nov 18 '13 at 8:34
  • $\begingroup$ there will some trouble with checking the cases $L = 0$ and $L = 1$, rest should not be that difficult. $\endgroup$ – Paramanand Singh Nov 18 '13 at 8:55

Let $\{x\} = x - \lfloor x\rfloor = x $ (mod 1) be the fractional part of $x$.

Suppose such a limit L exists. Then given any $\epsilon > 0$, there would have to exist an $N$ s.t. for all $n \ge N$, we would have $|\{n\sqrt{2}\} - L| < \epsilon$. Now, as Aaron said, calculate

$|\{(n+1)\sqrt{2}\} - \{n\sqrt{2}\}| = |\sqrt{2} - (\lfloor (n+1)\sqrt{2}\rfloor - \lfloor n\sqrt{2} \rfloor)| > 1/3$

This leads us to consider $\epsilon = 1/6$, say. Supposing our inequality holds for $n = N$, the above shows that even the very next term $\{(N + 1)\sqrt{2}\}$ must land farther away from $L$ than $\epsilon$. So no such $N$, and therefore no such limit, can exist.

Explicitly calculating the first few $\{n\sqrt{2}\}$ leads to the intuition that the sequence "moves around a lot"; this can be made precise and is a special case of the equidistribution theorem, which is itself an instance of the ergodic theorem.

  • $\begingroup$ Thank you. Although I understood Aaron hint I coulden't write in a formal way. Now it is much clearer. $\endgroup$ – Lior Nov 18 '13 at 21:21

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