Irrational and rational sequence proof

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?

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 $\mathbb{R}$.

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 $\alpha$.