axiomatic sets 27 (real numbers 3)

In preceding posts, we have seen the properties of real numbers.

(1)arithmetics of addition
(2)arithmetics of multiplication
(3)distributive law
(4)ordered relation
(5)the completeness

Let us look at one of 'Dedekind cut' $\left\{ x\in\mathbb{Q} : x\lt a \right\}$  as one number $a$  ,and the set of all 'Dedekind cuts' as a set of numbers.
(you may remember the natural number $a+1=\left\{ a\cup\left\{ a \right\} \right\}$ . )

The set of all 'Dedekind cuts' $\mathcal{C}$ will satisfy from (1) to (4). (do you agree?)

Axiom for completeness (5) means that if a set of numbers is bounded above,
then it has a supremum. (a supremum must be in all numbers.)

We can not express $\sqrt{2}$ as a rational number and there is not $\sqrt{2}$ in $\mathcal{C}$ .

However, in $\mathcal{C}$  we are able to get a rational number close to $\sqrt{2}$
as much as you want. Because in $\mathcal{C}$  there is all rational numbers.

$1.4=\frac{14}{10}\in \mathcal{C}$
$1.41=\frac{141}{100}\in \mathcal{C}$
$1.414=\frac{1414}{1000}\in \mathcal{C}$
$1.4142=\frac{14142}{10000}\in \mathcal{C}$
$\cdots\cdots\cdots$
$1.41421356237=\frac{141421356237}{10000000000}\in \mathcal{C}$
$\cdots\cdots\cdots$

Here, we want you to remember that $0.99999\cdots=1$ in the preceding posts.
That is, the rational number series which becomes $1.41421356237\cdots$ will go to $\sqrt{2}$ .
Althogh $\sqrt{2}$  is not a rational number, there exists a 'Dedekind cut' $\left\{x\in\mathbb{Q} : x\lt\sqrt{2} \right\}$ .

Thus, $\mathcal{C}$ is almost enough for real numbers.

The problem is that there are many numbers which are not able to be expressed by rational numbers.

We call the numbers which are not rational numbers irrational numbers.

Thus, real numbers are constructed rational numbers and irrational numbers.


(Correctly, you must understand, all above is needed proofs. )