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Show that every number is the limit of a sequence of irrational numbers


Proof:

Let $a\in\mathbb{R}$. Since $\mathbb{R}/\mathbb{Q}$ is dense of $\mathbb{R}$, then for the sequence of irrational numbers $\{a_n\}$, each members of $\{a_n\}\in(a-1/n,a+1/n)$ where $n\in\mathbb{N}$. Now let $\epsilon>0$, then by the Archimedean Property, there exists an index $N$ such that $1/N<\epsilon$. And for all $n\geq N$, we can get $1/n\leq1/N$. So $$|a_n-a|\leq 1/n\leq 1/N<\epsilon$$

Thus $\lim\limits_{n\rightarrow\infty}a_n=a$.


I found a same question Show that every number is the limit of a sequence of irrational numbers and seems I am doing the same thing from the hit provided by Kim Jong Un. Can anyone check my proof right or not ? Thanks

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The way you're trying to approach things is WAY overkill. Just give the sequences.

If $a$ is rational, let $a_n=a+\frac{\sqrt{2}}{n}$ (or pick your favorite irrational number).

If $a$ is irrational, let $a_n=a$.

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  • $\begingroup$ If I separated bu these two cases, I can easily apply Archimedean Property, right? $\endgroup$ – Simple Oct 11 '15 at 0:13
  • $\begingroup$ @Simple All you need to show is that the first converges to $a$, to do this you use Archimedian Property. You are 100% correct. :) $\endgroup$ – user223391 Oct 11 '15 at 0:14
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For any real $a$, let $a_n =\frac1{n}\lfloor n\,a \rfloor +\frac{\sqrt{2}}{n} $. Then

$\begin{array}\\ |a_n-a| &=|\frac1{n}\lfloor n\,a \rfloor+\frac{\sqrt{2}}{n}-a|\\ &=\frac1{n}|\lfloor n\,a \rfloor-n\,a+\sqrt{2}|\\ &\le \frac1{n}(1+\sqrt{2}) \end{array} $

since $|\lfloor n\,a \rfloor-n\,a| \le 1 $.

Note that, to get rational approximations, just remove the $\sqrt{2}$ part.

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    $\begingroup$ irrational numbers ;) Maybe just add a $\pi$ or something $\endgroup$ – user223391 Oct 10 '15 at 23:55
  • $\begingroup$ I noticed that and corrected it. $\endgroup$ – marty cohen Oct 10 '15 at 23:57

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