W.Rudin (1.22 decimals) I am reading the book of Rudin "Principles of mathematical analysis". 
In 1.22 decimals he wrote: Let $x>0$be real. Let $n_0$ be the largest integer such that $n_0\leqslant x$ (Note that the existance of $n_0$ depends on the archimedian property of $R$).
I have some questions:
1) Why he write about natural number but he did not defines them?
2) How he uses archimedian property in existance of $n_0$?
 A: 1) Isn't really answerable by anyone other than Rudin unless you can clarify what you want to know exactly (or what kind of answer you'd expect).  
For 2) can you write down the Archimedian property and then first try to prove that for all $x>0$ there is some $n\in\mathbb N$ such that $n> x$?
This proves that the set $\{m \in\mathbb Z \mid m \le x\}$ is bounded above and thus has a maximal element. This element is the $n_0$ from your text. (It's also called the floor of $x$, denoted by $\lfloor x \rfloor$)
A: For me, it is acceptable to take as an axiom the following statement:
Well-ordering principle: Every non-empty subset of $\mathbb{N}$ has a minimum element.
You are correct that Rudin never stated this property explicitly. To be fair to Mr. Rudin though, page 1 states:
"We shall not, however, enter into any discussion of the axioms that govern the arithmetic of the integers, but assume familiarity with the rational numbers...."
Using the axiom above, we can return to the decimal representation issue. For $x\in{}\mathbb{R}^+$ we have the set $S:=\{z\in{}\mathbb{Z}\,\,\,\wedge{}\,\,\,x<z\}$. This set is non-empty by the archimedean property* therefore it has a minimum element $m_0$ by the well-ordering principle. Then $m_0-1=n_0$.
*The archimedean property furnishes a natural number $n$ such that $1\cdot{}n>x$. Thus $n\in{}S$ and $S\neq{}\varnothing{}$.
