Proving the rationals are dense in R I know this is a common proof. I'm following Rudin's proof and I'm following everything except for one step.
Suppose $x, y \in \Bbb R$ and $x < y$. Then there exists an $n \in \Bbb N$ such that $n(y-x) > 1$.
Again by the Archimedean property, there exist $m_{1}, m_{2} \in \Bbb N$ such that $m_{1} > nx$ and $m_{2} > -nx$, i.e.
$$
-m_{2} < nx < m_{1}
$$
From here, Rudin says there must be an $m \in \Bbb Z$ with $-m_{2} \le m \le m_{1}$ and that
$$
m-1 \le nx < m
$$
I'm confused about these two steps. If $-m_{2} < nx < m_{1}$, then isn't $-m_{2} < m_{1}$?
edit: to be clear, I follow everything up until the introduction of $m$.
 A: The whole proof seems to hinge on (1) Archimedian property, and (2) well ordering principle. While Rudin states and proves Archimedian property, just prior to the proof that we are discussing, well ordering principle is not mentioned by Rudin anywhere in his book, neither before nor after the current proof. Assuming something that is not stated anywhere in the book, is not a good proof strategy, especially in subjects like real analysis.
A: The proof seems rather oddly written. I find it more intuitive to note that $nx-ny>1$ implies there must be an integer between $ny$ and $nx$. Look at $m=\lfloor ny\rfloor+1$.  
A: Yes, $-m_2<m_1$, but we knew this anyway: $m_1$ and $m_2$ are positive integers, so $-m_2$ is negative and $m_1$ is positive.
Rudin introduces both $m_1$ and $m_2$ in order to avoid having to split the argument into cases depending on whether $x>0$, $x=0$, or $x<0$. If you simply take $m_1$ to be the minimal such that $m_1>nx$, you’re in trouble if $x$ is negative: $m_1=1$, which doesn’t do what you want.
Once you have $-m_2<nx<m_1$, you can use the well-ordering principle to set
$$k_0=\min\{k\in\Bbb N:-m_2+k>nx\}\;;$$
$\{k\in\Bbb N:-m_2+k>nx\}$ is non-empty, because it contains $m_1-(-m_2)$ so the well-ordering principle ensures that $k_0$ exists. Now let $m=-m_2+k_0$, and you can easily check that $m-1\le nx<m$.
A: I agree with Pedro above. 
Choose $n >1/(b-a)$ so that $an > bn +1$.
Then we can prove there is a k in Z with an < k < bn because:
Let $k = \min\{m \in Z : an \le m\}$ This is nonempty by Archimedian thingy
(and we can easily prove that nonempty bounded below sets of integers have a min element
using the fact that nonempty sets of $N$ have a min element)
Then $an \le k < bn$   (first $an\le k$ by def and $k < bn$ because  if $k \ge bn$ then $k\ge bn > an +1$ implies $k-1 >an$ which would contradict minimality)
Then $a\le k/n < b$  ($n \ge 1$) 
