# 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 $x,y \in \mathbb{R} \text{ with } x < y$.

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

$x < r +$$\frac{\sqrt 2}{n}$ $< y$

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 $y$.

• So this is the proof? – kjkx Nov 5 '17 at 22:35
• I don't get it, how the consecutive multiples of 1/n prove that? – kjkx Nov 5 '17 at 23:11
• I added some additional explanation. – marty cohen Nov 5 '17 at 23:44

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?

• No, sadly not.. – kjkx Nov 5 '17 at 22:12