Licence Math-Info Exercices d`Arithmétique

n≥1Pnn
n= 1 P1
Pnn≥1n+ 1
n
Pn
n
Pn
Pn+1 Pnn≥1
36
N≥2
(P) (∃n∈N)(∀p∈N), p ≤n
(P) (P)
IRf:I→RI
f
f
f

f
f
f
IRf:I→R
I
∀x∈I, f(x)6= 0
∀y∈R,∃x∈I, f(x) = y
∃M∈R,∀x∈I, |f(x)|6M
∀x, y ∈I, x 6y⇒f(x)6f(y)
∀x, y ∈I, f(x) = f(y)⇒x=y
∀x∈I, f(x)>0⇒x60
f1, f2, f3R R
(∀j∈ {1,2,3})(∃a∈R), fj(a)=1
(∃j∈ {1,2,3})(∀a∈R), fj(a)=1
(∃a∈R)(∀j∈ {1,2,3}), fj(a) = 1
(∀a∈R)(∃j∈ {1,2,3}), fj(a)=1
A, B E
A∩X=B
X∈ P(E)
A∪X=B
A E A
f E {0,1}f(x) = 0 x /∈A f(x) = 1 x∈A
A B E f g
1−f
fg
f+g−fg
E F f :E→F
f g :F→E g ◦f=IdE
f g :F→E f ◦g=IdF
f g :F→E g ◦f=IdEf◦g=IdF
f:E→F g :F→E f ◦g◦f
f g
P(P({a})) a

f E F A E B F
∀A∈ P(E)A⊂f−1(f(A))
A∈ P(E)f−1(f(A)) = A]⇐⇒ f
B∈ P(F)f(f−1(B)) ⊂B
B∈ P(F)f(f−1(B)) = B]⇐⇒ f
A, B E f :P(E)−→ P(A)× P(B)
∀X∈ P(E)f(X) = (X∩A, X ∩B)
f⇐⇒ A∪B=E
E E
f:E→E A ⊂E E f(A)⊂A
{fn(x)|n∈N∗}
fN∗N∗nN∗f(n+ 1) > f(f(n))
∀n∈N∗, f(n) = n
n x ≥n⇒f(x)≥n

A B E
sup({A, B}) inf({A, B})
n≥1x y n
n x −y x ≡y[n]
(E, ≤) (F, ≺)f
∀x, y ∈E x ≤y⇔f(x)≺f(y)X E
b E f(b)F f(X)
I[a, b]Ra≤b
Isup{[a, b],[c, d]}
[a, b] [c, d]I
E F X E
F(X, F )X F Φ Φ = {f|f∈
F(X, F ) et X∈ P(E)}.
ΦfRf0⇐⇒ X⊂X0f0∈ F(X0, F )f
f0X
RΦ
4R+?
x4y⇔ ∃n∈N, y =xn
4