Since there exists an ordered field R which has the least-upper-bound property , what property or axiom of R makes this possible? I think it best for a little explanation but I really did not have the space to do it in the question sorry. Allow me first to state a theorem in real analysis and then ask the question. 
Theorem 1: There exists an ordered field R which has the least-upper-bound property , Moreover , R contains Q as a subfield.  This is straight out of Walter Rudin's book on real analysis.  
Now earlier on it was pointed out Q does not have a least- upper-bound property so the natural question to ask is since Q is a subfield of R and R does have the least-upper-bound property then what special property does R have that Q does not have that makes this possible?
There is an Appendix that I am struggling with to prove ( Theorem 1 ) , I have to repeatedly read it to get more details, but I need a little intuition going in to make it easier for me to get through the proof.  Thank you. 
 A: Suppose you want to enlarge $\mathbb Q$, in a reasonable way, so as to achieve the least upper bound (lub) property. Now $\mathbb Q$ has lots of counterexamples to the lub property, i.e., lots of subsets $X$ that are nonempty and bounded above but lack a lub in $\mathbb Q$. So what you want to do is to adjoin new elements to $\mathbb Q$ to serve as lubs for all these sets $X$.
But then you notice that often some distinct $X$ and $Y$ ought to have the same lub. In fact, if $X\subseteq\mathbb Q$ as above has no lub in $\mathbb Q$, then you can modify it, replacing each element $x\in X$ by all smaller rational numbers, to get $\hat X=\{y\in\mathbb Q:(\exists x\in X)\,y<x\}$ such that $\hat X$ is nonempty, bounded above, and without a lub in $\mathbb Q$, and furthermore $\hat X$ "ought to" have the same lub as $X$. 
That makes life a little easier: you only need to adjoin lubs for sets like $\hat X$, i.e., for subsets of $Q$ that are leftward closed (i.e., if the set contains some $y$ then it also contains every rational $z<y$), nonempty, not all of $\mathbb Q$, and having no largest element in the set, nor a smallest element in the complement. Such sets are sometimes called Dedekind cuts, and I'll use this terminology here. (For some people, a Dedekind cut consists of two sets, one like my $\hat X$ and another that is the complement of $\hat X$. These people would call my $\hat X$ the left half of a Dedekind cut, but for brevity, I'll just say "Dedekind cut".)
So we essentially build the desired extension of $\mathbb Q$ by adjoining a lub for each Dedekind cut. Now there are a lot of things to check: 
(1) The resulting set, linearly ordered in the natural way, has the lub property. This isn't obvious because we only adjoined lubs for subsets of $\mathbb Q$, not for subsets of the enlarged set.
(2) It is possible to define addition and multiplication on the enlarged set, so that it becomes an ordered field.  (That involes checking quite a few axioms.)  Moreover, the newly defined addition and multiplication, when applied to rational numbers, agree with the original addition and multiplication on $\mathbb Q$.
The checking becomes somewhat easier if one makes the construction more uniform. Instead of keeping the original rational numbers and adding a new element for each Dedekind cut, use only elements associated to Dedekind cuts, but allow the situation where the complement of a Dedekind cut has a smallest element. The elements associated to these newly allowed Dedekind cuts will correspond to the rational numbers. 
Finally, there's the matter of deciding, for each Dedekind cut $C$, what the associated element should be. Here, mathematicians have adopted a rather cheap but accurate solution: The entity that serves as the new element associated to $C$ is $C$ itself.  This takes some getting used to; the elements of the new, enlarged field, including those that correspond to the rational numbers, are sets of rational numbers (Dedekind cuts), so they don't "look like" numbers at all.  But they serve perfectly well as numbers, once one checks items (1) and (2) above.  And this is the usual construction of the real numbers as Dedekind cuts.
A: The "special property" you're looking for is that $\Bbb{R}$ is the completion of $\Bbb{Q}$. You can get this from dedekind cuts, as the previous answered gave, but you can also get it by viewing $\Bbb{R}$ as the set of equivalence classes of Cauchy sequences in $\Bbb{Q}$. If the details aren't also in Rudin I'd be surprised.
