Exercice - Jean

n m
n m
n m
n m
A
•
a
B
•
b
C
•
c
A B A
•
a
B
•
a
[AB]
A
A
A
•
a
B
•
b
←→
•
c
A
•
a0
•
c
B
•
b
a0+ 2c=a
a= 5
b= 3
fR R ∀n∈N∗,∃x∈R,|f(x)| ≤ x
n.
fR
n∈N∗x∈R|f(x)| ≤ x
n
nlimn→∞
x
n= 0 |f(x)| ≤ 0f(x)=0 f
x
p∈ P b∈Nc∈Np b2−4c x p

x2+bx +c
p∈ P b∈Nc∈Np b2−4c x ZP(x) = x2+bx +c
4P(x) = 4x2+ 4bx + 4c= (2x+b)2+ 4c−b2x∈Zp2x+b
p b x =p−b
2x=2p−b 2
x p (2x+b)2p4c−b2p4P(x)
p x2+bx +c
n∈N∗A1, A2, . . . , Ann
n
n
n= 1 A1
n= 2 A1A2
(AB)
n n + 1
A1, A2, . . . , An+1 n+ 1 A1, A2, . . . , An
n
D A2, A3, . . . , An+1 n
A2, A3, . . . , AnD An+1
n+ 1 D
n

ABC AB =AC
O BAC
[AB]P O (AB)Q O (AC)
O BAC OP OQ
AP O AQO P Q
OP =OQ OA =OA
AP =AQ
O[AB]OB OC
OBP OCQ BP =CQ
AB =AP +BP =AQ +CQ =AC AB AC

a b
a+b ab
1023
262 = 3 + 1
1 + 1
9+ 1
2+ 1
12
1023
262
n n
101,10101,1010101,...,101 · · · 101 k+1
k
n11n

A, B C E
P(A∩B) = P(A)∩ P(B)
P(A∪B) = P(A)∪ P(B)
A∩(B∪C)⊂A∪B
(A∪B)∩(¯ A∪B) = B
A⊂B⇔A∪B=B
A∪B=A∩C⇔B⊂A⊂C
A∩B=A∩C⇔A∩¯ B=A∩¯ C
f E F
A B E f(A∪B) = f(A)∪f(B)
f A B E f(A∩B) = f(A)∩f(B)
f A B E f(¯ A) = f(A)
R R
∀x∈R, f(x) = x3+ 2, g(x) = ln(x2+ 1), h(x) = 5 −2e−x, l(x) = x+ex.
fNP(N)
A={n∈N|n /∈f(n)}A f
P(N)
n∈Nn