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Show that every irrational number in $\mathbb{R}$ is the limit of a sequence of rational numbers. Every rational number in $\mathbb{R}$ is the limit of a sequence of irrational numbers.

How can I prove this?

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    $\begingroup$ Do you know that both $\Bbb Q$ and $\Bbb R - \Bbb Q$ are dense in $\Bbb R$? $\endgroup$ – Dylan Yott Mar 22 '13 at 22:04
  • $\begingroup$ Given an irrational number $x$, can you think of a set of interval $I_i$ containing $x$ and such that the end points of $I_i$ are always rational and the lengths of the intervals $I_i$ tend to 0 as $i$ tends to infinity? $\endgroup$ – Dan Rust Mar 22 '13 at 22:04
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All you need is to prove that $\Bbb Q$ and $\mathbb R\setminus\mathbb Q$ are dense in $\Bbb R$. This means (equivalently) that for every pair of reals $x,y$ there exist $r\in\Bbb Q$ and $\ell \in \Bbb R\setminus \Bbb Q$ such that $$x<r<y$$ and $$x<\ell<y$$

For the first, I give you some hints: Hover over the grey areas for extra, possibly spoiling, hints.

$(1)$ Assume that $y-x>1$. Prove there exists an integer $m$ between $x,y$

Look at $\lfloor x\rfloor +1$.

$(2)$ Now, let $x,y$ be such that $y-x>0$. The archimedean property of $\Bbb R$ means that there exists $n$ such that $n(y-x)=ny-nx>1$. Use $(1)$

There exists an integer $m$ between $ny,nx$, from where $nx<m<ny$ or $$x<\frac m n z y$$ and we have found our rational number.

$(3)$ Here, we might use that, say $\sqrt 2$ is irrational. Then since $\sqrt 2 <2$, $\frac{\sqrt 2}2<1$. Then we start with an irrational $\mu\in(0,1)$. Given two rationals, $r,s$ with $r-s>0$, we have that $$0<\mu(r-s)<r-s$$ so that $$s<s+\mu(r-s)<r$$

It suffices to prove that $s+\mu(r-s)$ is irrational. Can you do this?

All the above proves that $\Bbb Q$ and $\Bbb R\setminus \Bbb Q$ are dense in $\Bbb R$. Can you see why?

Now, if I give you an irrational (or irrationals) $\lambda$, look at the intervals of the form $$\left(\alpha-\frac 1 n,\alpha +\frac 1n \right)$$

and build a sequence of rationals (resp. irrationals) converging to $\alpha$.

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    $\begingroup$ @Q.matin You're welcome. Glad it helped. $\endgroup$ – Pedro Tamaroff Mar 22 '13 at 22:42
  • $\begingroup$ Peter, so to prove this I just have to essentially show that rational and irrational are dense, and that will complete the proof? $\endgroup$ – Q.matin Mar 22 '13 at 23:12
  • $\begingroup$ @Q.matin If by "this" you mean that any irrational is the limit of a rational sequence and vice versa, yes. $\endgroup$ – Pedro Tamaroff Mar 22 '13 at 23:17
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    $\begingroup$ Well, since $r_n\in \left(\alpha-\frac 1 n,\alpha+\frac 1n\right)$, we have that $\alpha-\frac 1 n<r_n<\alpha+\frac 1n$. This means that $-\frac 1 n<r_n-\alpha<\frac 1n$, i.e. $|r_n-\alpha|<1/n$. But for any $\epsilon >0$, we can find $N\in\Bbb N$ such that $1/N<\epsilon$ (and thus $1/n<\epsilon$ for $n\geq N$). By the definition of limit of a sequence, $$r_n\to\alpha$$ since we can make $|r_n-\alpha|<\epsilon$ for all $n\geq N$ for any given $\epsilon$. $\endgroup$ – Pedro Tamaroff Mar 22 '13 at 23:30
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    $\begingroup$ @Q.matin Practice, practice, practice. $\endgroup$ – Pedro Tamaroff Mar 22 '13 at 23:35
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Let $\alpha$ be irrational. For each positive integer $n$ there is a least integer $k$ such that $\dfrac{k}{n} > \alpha$, and then this number is rational. You need to prove that this sequence tends to $\alpha$.

Now let $\alpha$ be rational, and repeat the above argument with something like $\dfrac{k\sqrt{2}}{n}$ instead of $\dfrac{k}{n}$. [You'll also need to prove that $\dfrac{k\sqrt{2}}{n}$ is irrational whenever $k \ne 0$, and cook up a way of avoiding having $0$ in the sequence.] Edit: Or consider $\alpha+\frac{\sqrt{2}}{n}$ or something like that, as Cameron Buie suggests in the comments.

The moral of the story is that $\mathbb{Q}$ and $\mathbb{R} \smallsetminus \mathbb{Q}$ are dense in $\mathbb{R}$.

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  • $\begingroup$ (+1): It's even easier than that in the second case. If $\alpha$ us rational, then $\alpha+\frac\pi n$ is irrational for all positive integers $n$, and this sequence converges to $\alpha.$ $\endgroup$ – Cameron Buie Mar 22 '13 at 22:07
  • $\begingroup$ Oh yes, that's much more simple, thanks! $\endgroup$ – Clive Newstead Mar 22 '13 at 22:08
  • $\begingroup$ Alright, thanks guys! I am going to try and attempt now. $\endgroup$ – Q.matin Mar 22 '13 at 22:17
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For the first, take more and more digits of the decimal expansion.

For the second, if your number is $a$, take $a+e, a+e/10, a+e/100, \ldots$

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