modifier
Exemple : f surjection de A sur B
f
:
A
↦
B
{\displaystyle f:A\mapsto B}
∀
b
∈
B
∃
a
∈
A
,
f
(
a
)
=
b
{\displaystyle \forall b\in B\exists a\in A,\,f(a)=b}
∀
(
x
,
y
)
∈
A
2
,
f
(
x
)
=
f
(
y
)
⇒
x
=
y
{\displaystyle \forall (x,y)\in A^{2},\,f(x)=f(y)\Rightarrow x=y}
f
(
A
)
=
{
f
(
a
)
/
a
∈
A
}
⊂
B
{\displaystyle f(A)=\{f(a)/a\in A\}\subset B}
B
′
⊂
B
{\displaystyle B'\subset B}
f
−
1
(
B
)
=
{
a
∈
A
/
f
(
a
)
∈
B
′
}
{\displaystyle f^{-1}(B)=\{a\in A/f(a)\in B'\}}
Revoir
N
{\displaystyle \mathbb {N} }
Z
{\displaystyle \mathbb {Z} }
Exemple : E ensemble fini à n éléments
P
p
(
E
)
{\displaystyle {\mathcal {P}}_{p}(E)}
p
≤
n
{\displaystyle p\leq n}
C
a
r
d
(
P
p
(
E
)
)
=
(
n
p
)
=
C
n
p
=
n
!
p
!
(
n
−
p
)
!
{\displaystyle Card({\mathcal {P}}_{p}(E))={\binom {n}{p}}=C_{n}^{p}={\frac {n!}{p!(n-p)!}}}
Exemple : Formule du triangle de Pascal
(
n
+
1
p
+
1
)
=
(
n
p
+
1
)
+
(
n
+
)
p
{\displaystyle {\binom {n+1}{p+1}}={\binom {n}{p+1}}+{\binom {n}{+}}{p}}
Méthodes de démonstration :
formule du binôme de Newton
u
,
v
∈
A
,
u
×
v
=
v
×
u
,
n
∈
N
{\displaystyle u,v\in A\,,u\times v=v\times u\,,n\in \mathbb {N} }
(
u
+
v
)
n
=
∑
k
=
0
n
(
n
k
)
u
k
v
n
k
{\displaystyle (u+v)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}u^{k}v^{n_{k}}}
(
1
+
X
)
n
+
1
=
(
1
+
X
)
(
1
+
X
)
n
=
(
1
+
X
)
(
∑
(
n
k
)
x
k
)
.
.
.
{\displaystyle (1+X)^{n+1}=(1+X)(1+X)^{n}=(1+X)\left(\sum {\binom {n}{k}}x^{k}\right)\,...}
Revoir les groupes, anneaux, corpsStructures de référence :
Groupes :
(
R
,
+
)
,
(
Z
,
+
)
,
(
C
,
+
)
,
(
Q
,
+
)
,
(
R
∗
,
×
)
,
(
Z
/
n
Z
,
+
)
,
(
M
n
(
R
)
,
+
)
,
(
R
R
,
+
)
,
(
U
n
=
{
z
∈
C
/
z
n
=
1
}
,
×
)
,
(
S
n
=
B
i
j
(
[
[
1
,
n
]
]
,
[
[
1
,
n
]
]
)
,
∘
)
{\displaystyle (\mathbb {R} ,+),(\mathbb {Z} ,+),(\mathbb {C} ,+),(\mathbb {Q} ,+),(\mathbb {R} ^{*},\times ),(\mathbb {Z} /n\mathbb {Z} ,+),({\mathfrak {M}}_{n}(\mathbb {R} ),+),(\mathbb {R} ^{\mathbb {R} },+),(\mathbb {U} _{n}=\{z\in \mathbb {C} /z^{n}=1\},\times ),({\mathcal {S}}_{n}=Bij([[1,n]],[[1,n]]),\circ )}
Corps : Quaternions
H
{\displaystyle \mathbb {H} }
Z
/
p
Z
{\displaystyle \mathbb {Z} /p\mathbb {Z} }
K
(
X
)
=
{
P
Q
/
P
,
Q
∈
K
[
X
]
2
,
Q
≠
0
}
{\displaystyle \mathbb {K} (X)=\{{\frac {P}{Q}}/P,Q\in \mathbb {K} [X]^{2},Q\neq 0\}}
modifier
(
K
[
X
]
,
+
,
×
,
⋅
)
{\displaystyle (\mathbb {K} [X],+,\times ,\cdot )}
bases...
division euclidienne
∀
(
A
,
B
)
∈
K
[
X
]
2
,
B
≠
0
,
∃
!
(
Q
,
R
)
∈
K
[
X
]
2
/
A
=
B
Q
+
R
,
d
∘
R
<
d
∘
B
{\displaystyle \forall (A,B)\in \mathbb {K} [X]^{2},B\neq 0,\exists !(Q,R)\in \mathbb {K} [X]^{2}/A=BQ+R,d^{\circ }R<d^{\circ }B}
décomposition en éléments simples
modifier
H
=
{
(
a
−
b
¯
b
a
¯
)
,
(
a
,
b
)
∈
C
2
}
{\displaystyle \mathbb {H} =\left\{{\begin{pmatrix}a&-{\bar {b}}\\b&{\bar {a}}\end{pmatrix}},(a,b)\in \mathbb {C} ^{2}\right\}}
(
H
,
+
,
×
)
{\displaystyle (\mathbb {H} ,+,\times )}
n
∈
N
∗
,
n
≥
2
{\displaystyle n\in \mathbb {N} ^{*},n\geq 2}
P
n
=
∏
k
=
1
n
−
1
sin
(
k
π
n
)
{\displaystyle P_{n}=\prod _{k=1}^{n-1}\sin({\frac {k\pi }{n}})}
...
P
n
=
n
2
n
−
1
{\displaystyle P_{n}={\frac {n}{2^{n-1}}}}
Solutions complètes à venir.
modifier
Définition Soient
(
G
,
×
)
{\displaystyle (G,\times )}
(
H
,
+
)
{\displaystyle (H,+)}
f
:
G
↦
H
{\displaystyle f:G\mapsto H}
f morphisme de groupes si
∀
(
g
,
g
′
)
∈
G
,
f
(
g
×
g
′
)
=
f
(
g
)
+
f
(
g
′
)
{\displaystyle \forall (g,g')\in G,f(g\times g')=f(g)+f(g')}
Propriété Dans ce cas :Si G' sous groupe de G, f(G') sous groupe de H
Si H' sous groupe de H,
f
−
1
(
H
′
)
{\displaystyle f^{-1}(H')}
Démonstration du point 2
f
−
1
(
H
′
)
⊂
G
{\displaystyle f^{-1}(H')\subset G}
f
(
1
G
)
=
0
H
∈
H
′
⇒
1
G
∈
f
−
1
(
H
′
)
{\displaystyle f(1_{G})=0_{H}\in H'\Rightarrow 1_{G}\in f^{-1}(H')}
(
a
,
b
)
∈
f
−
1
(
H
′
)
,
f
(
a
×
b
−
1
)
=
f
(
a
)
∈
H
′
−
f
(
b
)
∈
H
′
∈
H
′
⇒
a
×
b
−
1
∈
f
−
1
(
H
′
)
{\displaystyle (a,b)\in f^{-1}(H'),\,f(a\times b^{-1})={\underset {\in H'}{f(a)}}-{\underset {\in H'}{f(b)}}\in H'\Rightarrow a\times b^{-1}\in f{-1}(H')}
En particulier
I
m
f
=
f
(
G
)
{\displaystyle Im\,f=f(G)}
K
e
r
f
=
f
−
1
(
{
O
H
}
)
{\displaystyle Ker\,f=f^{-1}(\{O_{H}\})}
Remarque f injective si... ?
f
(
x
)
=
f
(
y
)
⇔
f
(
x
×
y
−
1
)
=
0
{\displaystyle f(x)=f(y)\Leftrightarrow f(x\times y^{-1})=0}
⇔
K
e
r
f
=
{
1
G
}
{\displaystyle \Leftrightarrow Ker\,f=\{1_{G}\}}
Exemples :ln isomorphisme de
(
R
+
∗
,
×
)
{\displaystyle (\mathbb {R} _{+}^{*},\times )}
(
R
,
×
)
{\displaystyle (\mathbb {R} ,\times )}
e
x
p
=
(
l
n
)
−
1
{\displaystyle exp=(ln)^{-1}}
σ
:
S
n
→
{
−
1
,
1
}
;
σ
(
f
)
=
(
−
1
)
N
f
{\displaystyle \sigma :{\mathcal {S}}_{n}\to \{-1,1\}\,;\,\sigma (f)=(-1)^{N_{f}}}
N
f
=
C
a
r
d
{
(
i
,
j
)
∈
[
[
1
,
n
]
]
2
/
i
<
j
,
f
(
i
)
>
f
(
j
)
}
{\displaystyle N_{f}=Card\{(i,j)\in [[1,n]]^{2}/i<j,f(i)>f(j)\}}
det
:
G
L
n
(
C
)
→
C
∗
{\displaystyle \det :GL_{n}(\mathbb {C} )\to \mathbb {C} ^{*}}
![{\displaystyle (\mathbb {R} ,+),(\mathbb {Z} ,+),(\mathbb {C} ,+),(\mathbb {Q} ,+),(\mathbb {R} ^{*},\times ),(\mathbb {Z} /n\mathbb {Z} ,+),({\mathfrak {M}}_{n}(\mathbb {R} ),+),(\mathbb {R} ^{\mathbb {R} },+),(\mathbb {U} _{n}=\{z\in \mathbb {C} /z^{n}=1\},\times ),({\mathcal {S}}_{n}=Bij([[1,n]],[[1,n]]),\circ )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20dd442e9ca7799082ed7c6a511029141fd82a06)
![{\displaystyle \mathbb {K} (X)=\{{\frac {P}{Q}}/P,Q\in \mathbb {K} [X]^{2},Q\neq 0\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e342ac7370872d50656ce8e8b45ffe1d97c5777)
![{\displaystyle (\mathbb {K} [X],+,\times ,\cdot )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54dae6fef599b107a1f9752d50dd1d7bdedf2bf1)
![{\displaystyle \forall (A,B)\in \mathbb {K} [X]^{2},B\neq 0,\exists !(Q,R)\in \mathbb {K} [X]^{2}/A=BQ+R,d^{\circ }R<d^{\circ }B}](https://wikimedia.org/api/rest_v1/media/math/render/svg/947a107acc8389691745afabdb6b5a2b94a1165d)
![{\displaystyle N_{f}=Card\{(i,j)\in [[1,n]]^{2}/i<j,f(i)>f(j)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ab9d25d6842c3e420b2050eef459ba451584778)
