Exercice

A⇒B A B
A B
A⇔B A ⇒B B ⇒A
∅
{1,2,3}
N
Z
Z∗
x∈E x E
{x∈E|. . .}x E
. . .
(E)E
∪
∩
A\B A B
∀x∈E x E
∃x∈E x E
A⊆B∀x∈A, x ∈B
A=B A ⊆B B ⊆A
A⊆B⇔A∩B=A⇔A∪B=B
A=B⇔A∪B=A∩B
A×B={(x, y)|x∈A y ∈B}
A×B=∅
•(A∩B)×(C∩D) =?(A×
C)∩(B×D)
•(A∪B)×(C∪D) =?(A×C)∪(B×D)
P(E) = {A|A⊆E}
P({1,2})P({1})P(∅)
P(P({1})) P(P(P(∅)))
x∈
P(E)x⊆P(E)x⊆E
E n
Ck
n
E k
Σn
k=0Ck
n
P n
n≥3P
P
A A
A
∀x∈R,∀ > 0,∃α > 0∀y∈R|x−y|< α
|f(x)−f(y)|< 
√2

• P(0)
• ∀n∈NP(n)⇒
P(n+ 1) n∈NP(n)
P(n+ 1)
• ∀n∈N,P(n)
f(x) = 1
3x3+ax2+bx
•a b ∀x∈Rf(x+ 1) −
f(x) = x2
• ∀n∈Nf(n)
•n∈NΣn
k=0k2=f(n+
1) = n(n+1)(2n+1)
6
A
A
A
n≥
2
(S, A)S
A⊆S×S
∀s, t ∈S(s, t)∈A(t, s)∈A
∀s∈S(s, s)6∈ A
G= (S, A)
s∈S s d(s)
R(s) = {t∈S|(s, t)∈A}
Px∈Sd(x)
R⊆E×E
E
•R∀x∈E, xRx
•R∀x, y, z ∈E xRy yRz
xRz
•R∀x, y ∈E xRy
yRx
R
Z×Z∗(a, b)R(x, y)ay =bx
f:A→B A B
f:A→B∀b∈B, ∃a∈A|
b=f(a)
f:A→B∀a1, a2∈
A, f(a1) = f(a2)⇒a1=a2
f:A→B∀b∈B, ∃!a∈A|
b=f(a)
f f
E
f:N→E
P(N)