Processing math: 0%

ページ

2013/01/09

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.

0 件のコメント:

コメントを投稿