La propriété des nombres et les fractions continues

9.999 ···
1
3= 0,333 ··· 3
1=0,999 ··· ::::::::::::::
K≡0,999 ···
10K= 9,99 ··· = 9 + K K = 1 K≡0,999 ··· = 1
::::::::::::::
10K= 9,999 ···
−)K= 0,999 ···
9K= 9 K= 1
10K∗= 9,999 ···
+) K∗= 0,999 ···
11K∗= 10,999 ··· K∗= 0,999 ···
K K∗
1
3=1 3×10n 10n=
n
z }| {
999 ···9 +1
3×10n= 0,
n
z }| {
333 ···3 + 1
3×10n
= 0,
n
z }| {
999 ···9 + 10−n(n= 1,2,3,··· , n n
10−(n+1) <−0,
n
z }| {
999 ···9 = 10−n
n0,
n
z }| {
999 ···9 1

ana
an→a(n→ ∞)lim
n→∞
an=a
|an−a|< ε (N5n)
anε∗<|an−a|< ε, (N5n)
a n
a
ε ε∗
n
an
an≡A
n
P
k=1
σk=A(σ+σ2+··· +σn)
A σ σ 0< σ < 1
σ σ ×an=A(σ2+···+σn+σn+1)
(1 −σ)×an=Aσ −Aσn+1
an=Aσ
1−σ−Aσn+1 1−σ
Aσ
1−σn
an
ε∗<¯
¯
¯an−Aσ
1−σ¯
¯
¯=Aσn+1
1−σ<ε (N5n)
N
log A
(1 −σ)ε<(n+ 1) log 1 σN+ 1 = hlog A/ (1 −σ)ε log 1/σ i+ 1
A= 9 σ= 0,1an
an= 0,
n
z }| {
999 ···9an
Aσ
1−σ=9×0,1 1−0,1=0,9 0,9= 1

N=hlog 9/0.9ε
log 1/0.1i=hlog(10/ε) log 10 i=h1 + log 1/ε log 10 i= 1 + hlog 1/ε log 10 i
1
10 n+1 <¯
¯
¯0,
n
z }| {
999 ···9−1¯
¯
¯=1
10 n<1
10 (n−1) (N5n)
0,
n
z }| {
999 ···9 1
0,z }| { 999 ···
0,z }| { 999 ··· = 1
43
30 = 1 + 13 30 = 1 + 1
30
13
= 1 + 1
2 + 4
13
= 1 + 1
2 + 1
3 + 1
4
43
30 ≡1 + 1 2
1
3
1
4
++
43
30 ≡(1; 2,3,4)
r
r≡r(n) = (k(0) ;k(1) , k(2) ,··· , k(n−1) , k(n))k(0)
k(1), k (2) ,··· , k (n−1), k(n)
n ω
n
ω≡ω(n) = (k(0);k(1), k(2),··· , k (n−1), ξ (n))ξ(n)
ω
ω= (k(0);k(1), k(2),··· , k (n−1), k(n),···)
1)
√2 = 1 + 1
2
1
2= (1; 2 ,2,2···) = (1; ˙
2)
+ + ···

1+ 1
2 + 1
2 + 1
2 +
√2
···
x≡1 + 1
2 + 1
2 + 1
2 +
1)
···
1
x−1= 2 + 1
2 + 1
2 +
= 2 + 1
1
x−1
= 2 + (x−1) = x+ 1
···
1 = x2−1x=√2
1 = 0,999 ···
√2 = 1 + 1
√2+1 = (1; √2 + 1) = 1 + 1
2 + 1
√2+1
= (1; 2,√2 + 1)
1 + 1
√2+1 = 1 + 1
2 + 1
√2+1
z z ≡1+ 1
√2+1
z= 1 + 1
1 + zz2−1 = 1 . z
z=√2
√2
√2 = (1; √2 + 1) = (1; 2,√2 + 1) = (1; 2,··· ,2,√2 + 1)
··· 2√2 = √2
√2
√2+1
√2
n√2
ω(n)≡(1;
n−1
z }| {
2,2,··· ,2,√2 + 1) = √2n: 1,2,··· , n

√2 + 1 2
r(n)≡(1;
n−1
z }| {
2,2,··· ,2,2)
ω(n) = √2r(n)
1
16n<|ω(n)−r(n)|<1 4nn: 1,2,··· , n
r(n)√2
ω(∞)≡(1;
∞
z }| {
2,2,··· ,2,√2 + 1)
r(∞)≡(1;
∞
z }| {
2,2,··· ,2,2 )
√2 + 1 ω(∞) 2
(1; 2,2,···)√2
q123 ··· q
p
2
3
4
p
r(2,1)
r(3,1) r(3,2)
r(4,1) r(4,2) r(4,3)
r(p,1) r(p,q)
r(p,q)≡q
p
p= 2 ,3,··· , p
q: 1 5q5p−1
r(p,q)
0< r(p,q)<1
r(n) = (1;
n
z }| {
2,··· ,2)
r∗(n)r∗(n)≡r(n)−1 = (0;
n
z }| {
2,··· ,2)
r∗(n) 0 < r∗(n)<1
r∗(∞) = (0 ;
∞
z }| {
2,2,··· ,2,2 )