Question about a proof that $\mathbb{Q}$ is dense in $\mathbb{R}$ This is from Ross's elementary analysis book. The statement is if $a,b \in \mathbb{R}$ such that $a<b$ then there exists a rational $r \in \mathbb{Q}$ such that $a<r<b$.
I don't understand an important part of the proof which I will point out.
Here is the proof:

Since $b-a>0$ then by the archimidean property there exists a natural number call it $n$ such that $n(b-a)>1$. Now we must prove there exists an integer $m$ such that $an<m<bn$. By the archimidean property again there exists an integer $k$ such that $k>\max(|an|, |bn|)$ so that $-k<an<bn<k$. Then the set $$\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$$ is finite and nonempty and we can set $m = \min\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$. Then $an < m$ but $m - 1 \leq an$. Also, we have $m = (m - 1) + 1 \leq an + 1 < an + (bn - an) = bn$.

Comment: I get up to the point how the author sets$ k$ to be the integer that is larger than both $an$ and $bn$. And since $a<b$ we get the bounded inequality $-k<an<bn<k$. From here on this is where I get confused.  He creates a set  $\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$ which he calls finite which I see and nonempty (I'm guessing since this set has a least upper bound $k$ thus it is nonempty). But then he lets  $m = \min\{j \in \mathbb{Z}: {-k<j \leq k}\text{ and }an<j\}$ which follows $an<m$ but $m - 1 \leq an$. I don't get how the author gets to that point?   
 A: The set is finite because there are at most $2k$ elements in it. It is non-empty because by assumption $k$ is in it. By the least upper bound property for $\mathbb Z$, since this set is bounded below, it admits a minimum. The minimum $m$ is in that set, but by minimality, $m-1$ is not in that set, because otherwise $m-1$ would be an element of the set smaller than its minimum. This is why $an < m$ (i.e. $m$ is in the set) but $m-1 \le an$ (i.e. $m-1$ is not in that set).
Hope that helps,
A: By definition of $m$ you have $an<m$. Note that $-k<m-1$ because $-k<an<m$ and $m-1<m<k$. Assume $an<m-1$, then $m$ is not the minimal number of the set $\{j:-k<j\leq k,\; an<j\}$. Contradiction, so $m-1\leq an$.
A: $S_k := \{j\in Z:−k<j≤k \text{ and }an<j\}$ is finite clearly and it is non-empty because $an \lt k$.  Then you can find the smallest integer $m \in S_k$ such that $an \lt m$.  Once you have this $m$, you want to show that $an < m < bn$.  To do this, he says $m = (m-1)+1 \le an + 1$ because by definition $m-1 \le an$.  And $an + 1 \lt an + (bn -an) = bn$.  Hence $an < m < bn$. 
