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 $x,y \in \mathbb{R} \text{ with } x < y$.
Show that one $n \in \mathbb{N}$ and one $r \in \mathbb{Q}$ hast to exist, so:
$x < r + $$\frac{\sqrt 2}{n}$ $< y$
On
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 $z > 0$, there is an integer $n$ such that $\dfrac1{n} < z$.
Now apply this to $\dfrac1{n} <(y-x)/2$ 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 $\dfrac{k}{n} > x$. This means that $\dfrac{k-1}{n} \le x$. These mean $k > nx$ and $k-1 \le nx$ so $k = 1+\lfloor nx \rfloor $.
Consider $w =\dfrac{k+1}{n} =\dfrac{k}{n}+\dfrac1{n} $.
From $\dfrac1{n} <(y-x)/2$ we get
$\begin{array}\\ w &=\dfrac{k}{n}+\dfrac1{n}\\ &\lt\dfrac{k}{n}+\dfrac{y-x}{2}\\ &\lt x+\dfrac1{n}+\dfrac{y-x}{2}\\ &\lt x+\dfrac{y-x}{2}+\dfrac{y-x}{2}\\ &=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< y are two irrational numbers, then there exist an irrational number, r, between them: x< r< y?