Chapitre 1 - Pierre

N={0 ; 1 ; 2 ; 3 ; 4 ; . . .}.
Z={. . . −4 ; −3 ; −2 ; −1 ; 0 ; 1 ; 2 ; 3 ; 4 ; . . .}.
D
1,2∈D−1,578 ∈D3,333333 ∈D
1
3= 0,333333 . . . 6∈ D
a
baZbN\ {0}Q
R
N⊂Z⊂D⊂Q⊂R.
∗
N∗N\ {0}N0
N∗=N\ {0}={1 ; 2 ; 3 ; 4 . . .}.
∗

0
R∗=R\ {0},
Z∗=Z\ {0}, .
R+R−
R+R+= [0; +∞[R−=] − ∞; 0]
Z+
Z−
∀a∈R∗,0
a= 0.
∀a∈R,a
1=a.
∀a∈R,a
−1=−a.
∀a∈R,∀b∈R∗,a
−b=−a b=−a b.
∀a∈R,∀b∈R∗,a
b=a1 b=1 ba.
∀a, b ∈R∗,1
a
b
=b
a.
∀a∈R,∀b, k ∈R∗,a
b=a×k b×k.
∀a, c ∈R,∀b, d ∈R∗,
a
b
c
d
=a
b×d c.
A=
2
5
8
7
B= 3 ×7
15

3(x+ 1) = 3x+ 3
3x+ 3 = 3(x+ 1)
(a+b)2=a2+ 2ab +b2
(a−b)2=a2−2ab +b2
a2−b2= (a−b)(a+b)
(a+b)3=a3+ 3a2b+ 3ab2+b3
(a−b)3=a3−3a2b+ 3ab2−b3
a3−b3= (a−b)(a2+ab +b2)
a3+b3= (a+b)(a2−ab +b2)
(1 −x)3
(5 + 4)2

25 −x2
a∈R∗n∈Na n
an=a×a×. . . ×a
| {z }
n
n≥1
a0= 1
ana−n
a−n=1
an.
(456)1=. . .
(−999999)0=. . .
33=. . .
a b n m
am×an=am+nam
an=am−n(am)n=am×n
am×bm= (a×b)mam
bm=a bm
A=63×3−4×27
122.