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I can think of of examples where a sequence of irrationals converges to $0$. But if we pick any rational will there always exist a sequence of irrationals which converges to it?

I cannot find a straight answer to this question.

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    $\begingroup$ Let $r$ be our rational. Look at $r+\frac{\sqrt{2}}{n}$. This may be the example you had in mind, "shifted" by $r$. $\endgroup$ Mar 11, 2014 at 18:47

5 Answers 5

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Assume your number is $\frac{p}{q}$. Then the sequence $$a_n=\frac{\pi}{n}+\frac{p}{q}$$ converges to the given number and is irrational (any irrational number in the place of $\pi$ would do).

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  • $\begingroup$ this is assuming $\lim_{n\to\inf}$, right? $\endgroup$
    – Cole Tobin
    Mar 11, 2014 at 23:47
  • $\begingroup$ @ColeJohnson Yes, right! $\endgroup$
    – Jimmy R.
    Mar 11, 2014 at 23:49
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Yes, take a sequence consisting of your sequence of irrationals converging to $0$ plus your desired rational limit.

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Yes: If $r\in\Bbb Q$, then $\forall n\in\Bbb N$: ${rn\over n+\sqrt2}\in{\Bbb Q}^c$ and $$\lim_{n\to\infty}{rn\over n+\sqrt2}=r.$$

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For any rational number $x=\frac{p}{q}$ with $\gcd(p,q)=1$, just consider: $$ x_n = \frac{p}{q}\cdot\frac{n}{\sqrt{n^2+1}}.$$ Clearly any $x_n$ belong to $\mathbb{R}\setminus\mathbb{Q}$ and we have $\lim_{n\to +\infty} x_n = x$.

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Yes. Consider

$\frac{p}{q} - \frac{\sqrt{2}}{n}$

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