Densesness of irrational numbers Given $Q$: the aggregate of the rational numbers and $R$: the aggregate of the real numbers and $Qc$: the aggregate of the irrational numbers; Give $x,y \in \mathbb{R} \text{ with } x &lt; y$. 

Show that one $n \in \mathbb{N}$ and one $r \in \mathbb{Q}$ hast to exist, so: 

$x &lt; r  +  $$\frac{\sqrt 2}{n}$ $&lt; y$
The needed property is
provided by the
Archimedean axiom.
See, for example,
here:
https://en.wikipedia.org/wiki/Archimedean_property

One form states that,
for any real $z &gt; 0$,
there is an integer $n$
such that
$\dfrac1{n} &lt; z$.

Now apply this to
$\dfrac1{n} &lt;(y-x)/2$
and look at the
consecutive multiples
of $\dfrac1{n}$
or $\dfrac{\sqrt{2}}{n}$.



More detail.

Let $k$ be the first multiple of
$\dfrac1{n}$ such that
$\dfrac{k}{n} &gt; x$.
This means that
$\dfrac{k-1}{n} \le x$.
These mean
$k &gt; nx$
and
$k-1 \le nx$
so
$k = 1+\lfloor nx \rfloor
$.

Consider
$w
=\dfrac{k+1}{n}
=\dfrac{k}{n}+\dfrac1{n}
$.

From
$\dfrac1{n} &lt;(y-x)/2$
we get

$\begin{array}\\
w
&amp;=\dfrac{k}{n}+\dfrac1{n}\\
&amp;\lt\dfrac{k}{n}+\dfrac{y-x}{2}\\
&amp;\lt x+\dfrac1{n}+\dfrac{y-x}{2}\\
&amp;\lt x+\dfrac{y-x}{2}+\dfrac{y-x}{2}\\
&amp;=y\\
\end{array}
$

Therefore
$\dfrac{k}{n}$
and
$\dfrac{k+1}{n}$
are two rationals
that are between
$x$ and $y$.

If we specify $n$ by
$\dfrac1{n}
\lt \dfrac{y-x}{m}$,
where $m$ is a positive integer,
we can find
$\dfrac{k+j}{n}$
which are between
$x$ and $y$
for $j = 0$ to $m-1$.

If we choose $m=3$,
then
$\dfrac{k}{n}$,
$\dfrac{k+1}{n}$,
and
$\dfrac{k+2}{n}$,
are all between $x$ and $y$,
so that
$\dfrac{k+\sqrt{2}}{n}$
is between
$x$ and $y$.
Do you know how to prove that, if x&lt; y are two irrational numbers, then there exist an irrational number, r, between them: x&lt; r&lt; y?