TITLE: Irrational and rational sequence proof
QUESTION [11 upvotes]: 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?

REPLY [7 votes]: 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 $\alpha$.