Proof verification for "$\Bbb{Q}$ is dense in $\Bbb{R}$" (without the well-ordering principle) I've come across a lot of websites that were giving a proof of the density of $\mathbb{Q}$ in $\mathbb{R}$, but it seems to me that none of them was using only what we had already covered in my lecture, so I tried to prove it by myself (with what we had already seen). The thing is I don't know if it's correct or not, since I didn't find the same proof somewhere else.
So here we go: we want to show that for all $x, y \in \mathbb{R}$ such that $x<y$, there exists $q \in \mathbb{Q}$ such that $x<q<y$.
We're going to use:


*

*the Archimedean property: for all $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ such that $n > x$.

*a consequence of the Archimedean property: if $x > 0$, then there exists $n \in \mathbb{N}$ such that $0 < \frac{1}{n} < x$.

*the "completeness axiom": any non-empty subset of $\mathbb{R}$ that is bounded below has a greatest lower bound (infimum).


And here is my proof:
Since $x<y$ with $x,y \in \mathbb{R}$, we have that $0<y-x \in \mathbb{R}$. Because of the consequence of the Archimedean property, we have that $0 < \frac{1}{n} < y-x$.
Let's now consider the subset $A \subset \mathbb{R}$ defined by:
$A:=\{p \in \mathbb{Z} : p > nx \}$ 
We know that $nx \in \mathbb{R}$ since $x \in \mathbb{R}$ and $n \in \mathbb{N}$. Therefore, because of the Archimedean property, there exists $N \in \mathbb{N}$ such that $N > nx$, which means that $A$ is non-empty. By definition of $A$, there exists $\inf(A) = nx$. Therefore, by the completeness axiom, we have:
$\forall \epsilon > 0$, $\exists m \in A$ such that $nx+\epsilon > m \ge nx$
And because $nx$ does not belong to $A$, we actually have
$\forall \epsilon > 0$, $\exists m \in A$ such that $nx+\epsilon > m > nx$.
Because this is true for all $\epsilon$, let's consider, for example, $\epsilon = 1$. We have that $x > \frac{m-1}{n}$.
And here comes the conclusion: as $\frac{1}{n} < y-x$ and $m>nx$, we have:
$x < \frac{m}{n} = \frac{m-1+1}{n} = \frac{m-1}{n} + \frac{1}{n} < x + (y-x) = y$
$\Rightarrow x < \frac{m}{n} < y$, with $m \in \mathbb{Z}$ and $n \in \mathbb{N}$. So there exists $q = \frac{m}{n} \in \mathbb{Q}$ such that $x<q<y$.
I don't know if everything is correct or if I missed something that I souldn't have. Any comment or correction would be welcome! (And sorry if this is the nth time someone asks something about this topic ; I just couldn't find exactly what I was looking for elsewhere.)
 A: There are, imo, several inaccurate things in the above after "And here is my proof": 
(1) In the second line, I think you wanted to say  "..and thus there exists $\;n_0\in\Bbb N\;$ s.t. $\;\cfrac1{n_0}<y-x\;$"
(2) Your definition of $\;A\;$ is, I think, depending on the specific $\;n_0\in\Bbb N\;$  whose existence was mentioned in (1),  and on the generic $\;x\in\Bbb R\;$ you began with , so it should be 
$$\;A_{0,x}:=\left\{\,p\in\Bbb Z\;/\;p>n_0x\,\right\}$$
(3) The infimum of $\;A_{0,x}\;$ is going to be its minimum as $\;A_{0,x}\subset\Bbb Z\;$ in fact, and we can point out directly this minimum:
$$\inf A=\lfloor n_0x\rfloor+1$$
Take into account thge above and try now to continue from here.
A: Without giving a definition of $\mathbb{R}$, it is not obvious if the part "We are going to use" uses circular logic or not.
A cleaner method is this: A metric space is dense in its completion, and $\mathbb{R}$ is by definition the completion of $\mathbb{Q}$ with respect to the usual metric (see "Construction from Cauchy sequences" in Wikipedia's "Construction of the real numbers").
