Four back. The nature of the real number system

The first two constructs the real number system, and proves that the real numbers all constitute an ordered domain, and the rational numbers domain is the subdomain of the real domain in the isomorphic sense. So can real numbers describe the laws of nature that some rational numbers cannot describe? The answer is yes.

(property) Existence of real numbers $r >0$ makes
$ $r ^2=2.$$
(The meaning of the above is actually the existence of the rational number of the basic column $ (q_n) $ to make $ (q_n^2) = (2) $.)

Prove. The structural proof is used here. The following is a summary definition of the sequence: $q _1=2$,

$ $q _{n+1}=\frac{1}{2}q_n+\frac{1}{q_n},\quad N\geq 1.$$
The second time has explained that $ (q_n) $ is a basic column, which is described below:
$$ (q_n) ^2=2.$$
In fact
$ $q _{n+1}^2-2=\frac{(q_n^2-2) ^2}{4q_n^2}.$$
In this way, we know:
$ $q _n^2> 2,\quad \forall n\geq1.$$
So
$ $q _{n+1}^2-2\leq\frac{1}{8} (q_n^2-2) ^2,\quad \forall n\geq1.$$
$$\log (q_{n+1}^2-2) \leq 2\log (q_n^2-2)-\log8.$$
\begin{align*}
\log (q_{n+1}^2-2)-\log8&\leq2\left (\log (q_n^2-2)-\log8\right) \ \
&\leq\cdots\\
&\leq2^n\left (\log (q_1^2-2)-\log8\right) \ \
&=-2^n\log (4).
\end{align*}
So
$$0<q_{n+1}^2-2\leq 8\cdot4^{-2^n}.$$
That
$$ (Q_n) ^2= (2). $$


In our daily life, we are accustomed to using decimal decimals to represent a real number, this representation is essentially given the first $n $ decimal of this basic column, with the precision of the former $n $ decimal place at the $10^{-n}$ level.

Enter the following code in Python:

s=2

For I in range (100):

s=s/2+1/s

The results of the operation are as follows:

S

1.414213562373095

Usually take 1.414213562373 as an approximation of the side length of a square, and use $\sqrt{2}$ to denote its length.

The basic theorem of the real number system is given below, which is a motive for constructing the real number.

(completeness theorem) set $ (r_n) _{n\geq1}$ is the basic column in $\mathbb{r}$, there is a unique $r \in\mathbb{r}$ makes
$$\lim_{n\rightarrow\infty}r_n=r.$$

(Monotone has a defined rationale) Set the number of $ (a_n) $ monotonically increment, and there is an upper bound, then $\lim\limits_{n\rightarrow\infty}a_n$ exists.

(The existence theorem of certainty) sets the number set ~ $A $~ has an upper bound, there must be a definite boundary.

(Bolzano-weierstrass theorem) Set the number of columns $ (a_n) $ bounded, then there must be convergence of the child column $ (A_{n_k}) $.

(Heine-borel theorem) is provided with a family open interval $\{(A_\LAMBDA,B_\LAMBDA) \}_{\lambda\in \lambda}$ covers the closed interval $[a,b]$, i.e.
$$[a,b]\subset\bigcup_{\lambda\in \lambda} (A_\LAMBDA,B_\LAMBDA). $$
There must be a finite number of $\{(a_i,b_i) \}_{i=1}^n$ covering $[a,b]$ in this family open interval.

(closed interval set theorem) set $\{[a_n,b_n]\}_{n\geq1}$ is a closed interval set, i.e.

$$[a_{n+1},b_{n+1}]\subset [A_n,b_n],\quad n\geq1,$$
and has
$$\lim_{n\rightarrow\infty} (B_n-a_n) =0.$$
There is a unique $\xi$ that makes
$ $a _n\leq\xi\leq b_n,\quad\forall~n\geq1.$$

Four back. The nature of the real number system