Les nombres transcendants

α∈C Q P∈Q[X]P(α)=0, P 6= 0
C Q
Q C
α Pα
P(α)=0 P6= 0
α
α
α∃k∈Nkα
α den(α)k kα
p
qden(p q) = q
α d α1=α, α2, ..., αd
α α
|α|= max
i|αi|
Q[X]
R
QT[0; 1]
aij b= 0, b1b2...bn... bi6=aii b
k b k
bk6=akk

α d >2C > 0
p
q
α−p
q
>C
qd
P∈Z[X]α
P(X) = adXd+... +a1X+a0
∀p∈Z,∀q∈N∗
Pp
q=adp qd
+... +a1
p
q+a0
qdPp
q∈Z∗
Pp
q
>1
qd
M= supx∈[α−1;α+1] |P0(x)|M > 0 deg P>1
αp
q∈[α−1; α+ 1]
Pp
q
6M
p
q−α
P(α)=0
∀p
q∈[α−1; α+ 1] ,
α−p
q
>1
Mqd
p
q6∈ [α−1; α+ 1] 
α−p
q
>1>1
qd
C= min 1,1
M
α−p
q
>C
qd,∀p q∈Q

α∀n∈N∗∃pn
qn
qn>0|α−pn
qn|<1 qn
n
α=∞
X
k=0
ak
10k!
ak∈J0,9K
∀n∈N∗pn= 10n!Pk=0 nak
10k!qn= 10n!
06α−pn
qn
=∞
X
k=n+1
ak
10k!69
10(n+1)! 1 + 1
10 +1 100 +...610
(10n!)n10n!61
(qn)n
α
eΠ
α1, ..., αt, b1, ..., btαibi
b1eα1+... +bteαt6= 0
Q

α eα
b=eαeα−b6=
0
e e2,√e, e√2, ...
Π
ΠiΠeiΠ=−1
Π
x > 0x6= 1 xln(x)
xln(x)eln(x)=x
x
∀n∈N, n >2,ln(n) ln 2,ln 3
a6= 0 sin(a),cos(a) tan(a)
sin(a)ia, −ia, 1
2i
eia −e−ia
2i−sin(a)6= 0
cos(a) tan(a)
1 + tan 2(a) = 1
cos 2(a)cos(a)
sin (1) ,cos (1) ,cos √2,tan 1+√5
2, ...
αβeΠ
√2√2