Rudin Theorem 1.20 Can someone please explain help which proposition or axiom the following step comes from.  I will highlight it in the proof.
$1.20\ (a)$ If $x \in R,\ y\in R$, and $x > 0$, then there is a positive integer $n$ such that $nx > y$.
Pf
Let $A = \{nx\ |\ n \in Z^{+}\}$.  Suppose on the contrary that there exists no such element that is greater than $y$.  Then $y$ is an upper bound and so there must exists an $\alpha = sup\ A$.  Then since $x > 0$, $\alpha - x < \alpha$, and $\alpha - x$ is not an upper bound of $A$.  Hence $$\alpha - x < mx$$ for some positive integer $m$.  But then $$\alpha < (m+1)x \in A.$$
It seems obvious that we can say that, but I don't see yet exactly which axiom or propositions allows us to make the last statements.  Can someone please help clear this up?
Thanks.
 A: Pf
Let $A = \{nx\ |\ n \in Z^{+}\}$.  Suppose on the contrary that there exists no such element that is greater than $y$.  Then $y$ is an upper bound and 

so there must exists an $\alpha = sup\ A$.
This follows from the least-upper-bound property and Theorem(1.19 in the book) which proves the existence of and ordered field $\mathbb{R}$ which has the least-upper-bound-property.

Then since $x > 0$, $\alpha - x < \alpha$, and $\alpha - x$ is not an upper bound of $A$. 

Hence $$\alpha - x < mx$$  for some positive integer $m$.

Definition Suppose $S$ is an ordered set, $E \subset S$ and E is bounded above. Suppose there exists an $\alpha\in  S$ with the following properties;


*

*$\alpha$ is an upper bound of $E$.

*If $\gamma < \alpha $ then $\gamma$ is not and upper bound of $E$


Then $\alpha$ is called as least upper bound of $E $ or simply supremum.

So that "hence" part comes from the 2nd condition for the definition of
  supremum.

(continuation of proof)

But then $$\alpha < (m+1)x \in A.$$

See that this statement contradicts the 1st condition of the definition of supremum.

How $\alpha-x<mx \implies \alpha <(m+1)x$?
Definition An ordered field is a field $F$ which is also an ordered set, such that


*

*$x+y<x+z$ if $x,y,z\in F$ and $y<z$

*$xy>0$ if $x\in F$, $y\in F$, $x>0$, and $y>0$


It follows from the 1st part of the above definition(1.17 in Baby Rudin)
