real number system 3 (Archimedean property)

For understanding the completeness axiom [41], we should find out the density of real numbers and Archimedean property.

[density of real numbers]
If $a$ and $b$ are real numbers with $a<b$, then there is a real number $x$ such that $a<x<b$.

The proof of this proposition is very simple. If we put $x=(a+b)/2$, the proposition is always satisfied. Moreover, you will immediately find that there are so many real numbers in open interval $(a,b)$. For example, let $x$ be $a+(b-a)/n, (n>1, n\in \mathbb{N})$. This property is called the density of real numbers.

 The equation $0.999\cdots=1$ which we have addressed at first post is consistent with the dense property. Because if $0.999\cdots\ne 1$, as $0.999\cdots<1$ is a natural relationship on the ordered field, then there is an $x$ such that $0.999\cdots<x<1$. However, we can never accept it.

[Archimedean property of real numbers]
If $a$ and $b$ are positive real numbers, then there is an $n\in \mathbb{N}$ such that $na>b$.

The proof is by contradiction. For any $n\in \mathbb{N}$, if $na\leq b$, the set $A=\left\{ na | n\in \mathbb{N} \right\}$ has an upper bound $b$. By axiom for completeness [41], there exists $\sup A=s $.

For a $n'\in \mathbb{N}$, if $s=n'a\in A$, $s<(n'+1)a$. It is impossible, since $s$ is an upper bound of $A$. Therefore for any $n\in \mathbb{N}$, $na<s$. That is, $s\notin A$.

However by the density of real numbers, for any $\epsilon>0$(no matter how small), as we can put $s-\epsilon<s$, $s-\epsilon$ is not an upper bound of $A$. Therefore, we can choose an $n$ such that $s -\epsilon<na$.

At this time, if $\epsilon=a$, $s - a<na$, and $s<(n+1)a$. As $s$ is the least upper bound, it is a contradiction. Thus the proposition is correct.

Real number system became a dense ordered field satisfied completeness.