If $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x<y$,then there exists a rational $r$ such that $x<r<y$

Question

I am trying to prove that if $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x<y$,then there exists a rational $p$ such that $x<r<y$.

Here is what I tried to do:

I am trying to show that given $x$,$y$,there exist $m$ and $n$ such that $x<r=\frac{m}{n}<y$ where $m$ and $n$ are naturals or $$nx<m<ny$$

By the Archimedean property, as $x<y$,

$y-x>0$ so $n(y-x)>1$ for some $n\in \mathbb{N}$.So, $ny>1+nx$.

There must also be an integer $m$ between $nx$ and $1+nx$ so that $nx<m\leq 1+nx$

Which gives $nx<m<ny$, which completes the proof.

Is this proof correct?

I just read Arkeet's comment and I think the statement "There must also be an integer $m$ between $nx$ and $1+nx$" needs a formal proof.I don't think I have one at hand.

Answer 1

The part of the proof that you are missing (mentioned in Arkeet's comment) is essentially the existence of the integer part, that is if $x\in \mathbb{R}$ then $\exists k\in \mathbb{N}:k<x< k+1$. (apply this to $nx$ and you are done).

Proof of the existence of integer part:

Let $x\in \mathbb{R}$ and $S=\left\{ k\in \mathbb{Z} :k>x \right\}$. Then $S \subseteq \mathbb{Z}$ is bounded below by $x$ and so $S$ has a least element, $k_0\in \mathbb{Z}$. Then $k_0-1\notin S\Rightarrow k_0-1\le x$ while $k_0\in S\Rightarrow k_0>x$. Therefore by letting $k=k_0-1$ we have that $k<x<k+1$. That $k$ is called the integer part of $x$

Answer 2

This seems valid for me as long as you rely on knowledge about the embedding of natural numbers into the real numbers and $x$ and $y$ are positive. You still need to say, what happens if $x < y \leq 0$ or $x < 0 < y$.

As an alternative to this approach, you could have used that $\mathbb Q$ is dense in $\mathbb R$ or that real numbers are suprema of subsets in $\mathbb Q$. It depends a little how you defined the real numbers or what you already assume to know.