Cannot follow proof that between two reals there is a rational. I am following a textbook and as is good practice, I am only skipping things I can do. this is self-learning
I always struggle with chapter 1, the one that builds the "axioms", hate it.
Anyway, I am struggling to follow some proofs, and it's supposed to not require the application of later things.
There are two parts, firstly:
a) if $x,y\in\mathbb{R}$ and $x>0$ then $\exists$ a positive integer $n$ such that $nx>y$
b) if $x,y\in\mathbb{R}$ and $x<y$ then there $\exists$ a $p\in\mathbb{Q}$ such that $x<p<y$
Both things I can do, but not the way it is laid out.
Part A
For part a) it goes as follows: (I follow this one)
Let $A$ be the set of all $nx$ (I would have used a sequence.... but it's not a huge problem) if a) were false (negating yields: $\forall n,\ nx\le y$ ) then y is an upper bound. Then $A$ has a least upper bound in $\mathbb{R}$. I am happy with this.
Let $\alpha = \sup(A)$, since $x>0,\ \alpha-x<\alpha$ and $\alpha-x$ is not an upper bound of $A$ - a little confused here with the since part,I don't see how since $x>0$ $\alpha-x$ is not an upper bound. I agree with the conclusion though. 
This is why I would have used a sequence, I'm not sure how to show $\alpha-x$ is not a lower bound with the equipment to hand. But I am happy with the process.
Here's where the author looses me:
Hence $\alpha-x<mx$ for some positive integer m
Then I am happy again:
but then $\alpha<mx+x=(m+1)x$ which is a contradiction that $\alpha$ is an upper bound.
Part B
Since $x<y$ we have $y-x>0$ and using a) we can furnish ourselves with a positive integer, $n$, such that $n(y-x)>1$ (I am happy with this, the 1 is the "y" from part a))
Apply a) again to obtain positive integers $m_1$ and $m_2$ such that $m_1>nx$ and $m_2>-nx$ okay,in this case "x" is 1 (from part a) and "y" is nx, so I am happy with this. 
I'm also happy with orders, so $-m_2<nx<m_1$
Here is where I get lost.
Hence there is an integer $m$ (with $-m_2\le m\le m_1$) such that $m-1\le nx<m$ 
I'm not sure why it says "hence" nor what has been applied here. 
If we combine these inequalities we obtain
$nx<m\le 1+nx<ny$ 
Since n > 0 it follows that 
$x<\frac{m}{n}<y$ this proves b) with $p=\frac{m}{n}$
I "sort of" see it, in that $-m_2<m_1$ and as they're integers.... but I wouldn't say I am confident in it. Nor do I see why.
Hence usually means "thus we can get to" and I'm uncomfortable because I cannot see how one can think to do this given what's at hand, which is why I am doing the chapter, I should be able to think for myself, rather than recall proofs
Addendum
Thinking about it, $-m_2<nx<m_1$ is useful, as the set $\{-m_2,-m_2+1,-m_2+2,...,m_1-1,m_1\}$ is finite.
not quite sure how to use this though.
 A: Let's deal with part A. First you need to understand very clearly the definition of least upper bound.
A number $\alpha$ is said to be the least upper bound of a non-empty set of real numbers A if every member $x \in A$ satisifes $x \leq \alpha$ and for any number $\beta < \alpha$ we have at least one member $y \in A$ with $y > \beta$.
So all members of A are less than or equal to the least upper bound $\alpha$. Buf if we go lower/less than $\alpha$, say $\beta$, then there are members of $A$ which are greater than $\beta$.
Now to the confusion in part A. We have $\alpha = \sup A$. And since $x > 0$ we have $\beta = \alpha - x < \alpha$ and hence we must have some member of $A$ which is greater than $\beta = \alpha - x$. And since all members of $A$ are of the form $nx$, it follows that there is some integer $m$ with $mx > \beta = \alpha -x $ or $\alpha < (m + 1)x$. 
Part B is not that difficult once you get part A. You have situation where there are two integers $m_{1}, m_{2}$ with $-m_{2} < nx < m_{1}$. Effective you see that the quantity $nx$ lies between two integers $-m_{2}$ and $m_{1}$. Hence there must be two consecutive integers say $m - 1$ and $m$ such that $nx$ lies between them. The idea is that you start from $-m_{2}$ and keep on increasing by $1$ so that you don't exceed $nx$. This way you get an integer which we write as $m - 1$. Clearly when you add $1$ you exceed the quantity $nx$ and hence $nx < m$. We finally have $m - 1 \leq nx < m$.
A: Clarifying what Paramanand Singh says.
Since you have $-m_2 < n x < m_1$, keep decreasing $m_1$ without violating $nx <m_1$. Similarly keep decreasing $m_2$ (increasing $-m_2$ until you can't anymore. Then $m_1=m$.
