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?

  • 2
    $\begingroup$ 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

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 $\alpha$.

  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @Q.matin Practice, practice, practice. $\endgroup$ – Pedro Tamaroff Mar 22 '13 at 23:35

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 $\mathbb{R}$.

  • $\begingroup$ (+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

For the first, take more and more digits of the decimal expansion.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.