# Show that every irrational number in <mrow class="MJX-TeXAtom-ORD">

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.
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Tristan Ward
Let $\alpha$ be irrational. For each positive integer $n$ there is a least integer $k$ such that $\frac{k}{n}>\alpha$, 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 $\frac{k\sqrt{2}}{n}$ instead of $\frac{k}{n}$.
[You'll also need to prove that $\frac{k\sqrt{2}}{n}$ 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 $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$.