Proof of density of $\mathbb{Q}$ in $\mathbb{R}$ I am trying to understand the proof of density of $\mathbb{Q}$ in $\mathbb{R}$ presented in a video lecture by Winston Ou and inspired by Baby Rudin.
The claim is that given $x,y \in \mathbb{R}$ with $x < y$, there exists $q \in \mathbb{Q}$ such that $x < q < y$. I am going to replicate his proof, stopping where I get stuck. We are taking the Archimedean property for granted.

If $x < y$, $y - x > 0$, so by the Archimedean property, there exists $n \in \mathbb{N}$ such that $n(y-x) > 1$.

I'm ok so far, but I do wonder why I'm not allowed at this point to say $ny - nx > 1$, so there exists $z \in \mathbb{Z}$ such that $nx < z < ny$, so $x < \frac{z}{n} < y$, and then the proof is complete.  Is there something wrong with finishing the proof there?

Continuing with the proof. Applying the Archimedean property twice more, we find $m_1, m_2 \in \mathbb{N}$ so that $m_1 \cdot 1 > nx$ and $m_2 \cdot 1 > - nx$, so $-m_2 < nx$. So $-m_2 < nx < m_1$. Let $S = \{z \in \mathbb{Z} \mid nx < z \leq m_1\}$.

I am confused twice at this point. First, why define the set this way? That is, why do I require $z$ to be strictly greater than $nx$? Why would it not suffice to say that $nx \leq z \leq m_1$? I'm not seeing the full link. Second, we want to eventually find a least element. The second is finite since $nx$ and $m_1$ are finite -- that makes enough sense intuitively -- but how do I know $S$ is nonempty? What if $nx = 4.1$ and $m_1 = 5$? Then, there are no integers $z$ with this property and I certainly cannot find a least element. On the flip side, if I required $nx \leq z \leq m_1$, I can take $z = 5$.

Proceeding, assuming that I have found such a least element $m$ which is the smallest element of $S$ satisfying $nx < m \leq n_1$, I know that $m-1 < m$ surely does not satisfy this property. Certainly $m - 1 < m \leq n_1$, so we must have $m-1 \leq nx$. So $m \leq nx + 1$. So we have $nx < m \leq nx + 1 < ny$ (by choice of $n$, so $nx < m < ny$, so $x < \frac{m}{n} < y$, and the proof is complete.

I'd appreciate any help. I'm still trying to put the pieces together in my head so that this proof makes sense.
 A: 
I'm ok so far, but I do wonder why I'm not allowed at this point to say $ny - nx > 1$, so there exists $z \in \mathbb{Z}$ such that $nx < z < ny$, so $x < \frac{z}{n} < y$, and then the proof is complete.  Is there something wrong with finishing the proof there?

I'm just going to copy my comment:

When proving basic properties, you have to be careful about what you assume. I agree that if $r,s$ are real numbers such that $r−s>1,$ then there's some integer $z$ such that $s<z<r,$ but this is not definitionally true, so we'd need to prove it. What proof would you like to use?

I've not thought through the proof, but I think it would be similar to some of the steps in the proof you're currently asking about.


First, why define the set this way? That is, why do I require $z$ to be strictly greater than $nx$? Why would it not suffice to say that $nx \leq z \leq m_1$?

If we did so, then later in the proof we'd be able to prove only $x \leq \frac{m}{n} < y$, which doesn't suffice to prove density.  For that, we need strict inequality between each of $x,y$ and the rational number $\frac{m}{n}$.

but how do I know $S$ is nonempty? What if $nx = 4.1$ and $m_1 = 5$? Then, there are no integers $z$ with this property and I certainly cannot find a least element.

I think you're confused.  If $nx = 4.1$ and $m_1 = 5$, then $4.1 < 5 \leq 5$ means that $5 \in S$.  In fact, since the inequality on the right is non-strict, then we always have $m_1 \in S$, so it's not empty.
