# Is this a proof that $\mathbb Q$ is dense in $\mathbb R$?

I read a proof that $\mathbb Q$ is dense in $\mathbb R$ which seemed unnecessarily complex. I was wondering if the following is also a valid proof.

Say for the sake of argument that the existence of $\mathbb Q$ and $\mathbb R$, and the greatest-lower-bound property of the latter, have already been established. Also, that for any $u$ and $v$ in $\mathbb R$ such that $u>0$, there exists some positive integer $n$ such that $un>v$.

Now let $x\in \mathbb R,y\in \mathbb R$ and $x<y$. I'll attempt to show that $(x,y)\cap \mathbb Q\neq\emptyset$.

Let $n$ be a positive integer such that $y-x>\frac{1}{n}$. Let $S=\{z\in \mathbb Z:xn<z\}$. $S$ cannot be empty, because analogously to the above, there must exist a positive integer $z$ such that $\frac{1}{n}z>x$.

Then $xn$ is a lower bound of $S$, and by the greatest-lower-bound property, $z=\inf S$ exists in $\mathbb R$. Since $S\subseteq\mathbb Z$, $z\in S$, and so $\frac{z}{n}>x$. It remains to show that $\frac{z}{n}<y$. Suppose $\frac{z}{n}\geq y$. Then $\frac{z}{n}\geq x+(y-x)$, $\frac{z}{n}>x+\frac{1}{n}$, and $\frac{z-1}{n}>x$. But then $z\neq \inf S$, which yields a contradiction. Thus $x<\frac{z}{n}<y$. Since $x$ and $y$ were arbitrary, $\mathbb Q$ is dense in $\mathbb R$.

Edit: Removed incorrect assertion that $x$ was a lower bound of $S$.

• it is not totally correct, observe that if $S:=\{z\in\Bbb Z:z>nx\}$ for some $n\in\Bbb N_{>0}$ then its perfectly possible that there is some $z\in S$ such that $nx<z<x$ when $x<0$ and $n>1$, then $x$ is not a lower bound of $S$. However you can use $nx$ instead. Dec 15, 2017 at 11:19
• Thanks @Masacroso, fixed. :) If you could have another look that would be great. Dec 15, 2017 at 12:28
• yes, it seems correct now. Dec 15, 2017 at 12:59
• Thanks for your time @Masacroso. Dec 15, 2017 at 13:01

I think this proof is perfectly fine, indeed the Archimedean property and the so called "axiom of completeness" are two huge facts giving the density of $\mathbb{Q}$ in $\mathbb{R}$. It is worth to note that one can prove the density of $\mathbb{Q}_F$ in $F$ Archimedean field without assuming completeness, but using the well-ordering principle for $\mathbb{N}_F = \mathbb{N}_{\mathbb{Q}_F}$.
• In fact the proof given in the question works almost exactly as stated for the general Archmidean case, because the proof does not make full use of the "greatest-lower-bound" property for $\mathbb{R}$. It uses only the greatest-lower-bound property for $\mathbb{Z}$ which has a simple proof using induction. Dec 15, 2017 at 16:29