En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Relation d'équivalenceRelation (mathématiques)/Exercices/Relation d'équivalence », n'a pu être restituée correctement ci-dessus.
Soit
f
:
A
→
A
{\displaystyle f:A\to A}
injection . On définit sur
A
{\displaystyle A}
R
{\displaystyle {\mathcal {R}}}
x
,
y
∈
A
{\displaystyle x,y\in A}
x
R
y
⟺
∃
n
∈
N
y
=
f
n
(
x
)
ou
x
=
f
n
(
y
)
{\displaystyle x{\mathcal {R}}y\Longleftrightarrow \exists n\in \mathbb {N} \quad y=f^{n}(x){\text{ ou }}x=f^{n}(y)}
(Pour la notation
f
n
{\displaystyle f^{n}}
Puissances itérées d'une fonction .)
1° Montrer que (pour tous
x
,
y
∈
A
{\displaystyle x,y\in A}
x
R
y
⟺
∃
p
,
q
∈
N
f
p
(
x
)
=
f
q
(
y
)
{\displaystyle x{\mathcal {R}}y\Longleftrightarrow \exists p,q\in \mathbb {N} \quad f^{p}(x)=f^{q}(y)}
2° Montrer que
R
{\displaystyle {\mathcal {R}}}
A
{\displaystyle A}
3° Soit
C
{\displaystyle C}
R
{\displaystyle {\mathcal {R}}}
a) Montrer que
f
(
C
)
=
C
∩
Im
(
f
)
{\displaystyle f(C)=C\cap \operatorname {Im} (f)}
b) Montrer que si
a
∈
C
∖
f
(
C
)
{\displaystyle a\in C\setminus f(C)}
C
=
{
f
n
(
a
)
∣
n
∈
N
}
{\displaystyle C=\{f^{n}(a)\mid n\in \mathbb {N} \}}
C
∖
f
(
C
)
=
{
a
}
{\displaystyle C\setminus f(C)=\{a\}}
4° Montrer que toute partie
X
⊂
A
{\displaystyle X\subset A}
f
(
X
)
=
X
{\displaystyle f(X)=X}
5° Soit
f
:
Z
→
Z
{\displaystyle f:\mathbb {Z} \to \mathbb {Z} }
f
(
n
)
=
{
n
+
2
si
n
<
0
n
+
4
si
n
≥
0.
{\displaystyle f(n)={\begin{cases}n+2&{\text{si }}&n<0\\n+4&{\text{si }}&n\geq 0.\end{cases}}}
a) Montrer que
f
{\displaystyle f}
b) Déterminer
Im
(
f
)
{\displaystyle \operatorname {Im} (f)}
c) Décrire les classes d'équivalence de la relation
R
{\displaystyle {\mathcal {R}}}
f
{\displaystyle f}
Solution
1° Si
y
=
f
n
(
x
)
ou
x
=
f
n
(
y
)
{\displaystyle y=f^{n}(x){\text{ ou }}x=f^{n}(y)}
f
p
(
x
)
=
f
q
(
y
)
{\displaystyle f^{p}(x)=f^{q}(y)}
(
p
,
q
)
=
(
n
,
0
)
ou
(
0
,
n
)
{\displaystyle (p,q)=(n,0){\text{ ou }}(0,n)}
Réciproquement, si
f
p
(
x
)
=
f
q
(
y
)
{\displaystyle f^{p}(x)=f^{q}(y)}
p
,
q
∈
N
{\displaystyle p,q\in \mathbb {N} }
si
q
≤
p
{\displaystyle q\leq p}
f
p
−
q
(
x
)
=
y
{\displaystyle f^{p-q}(x)=y}
f
q
{\displaystyle f^{q}}
de même, si
q
≤
p
{\displaystyle q\leq p}
f
q
−
p
(
y
)
=
x
{\displaystyle f^{q-p}(y)=x}
Dans les deux cas,
x
R
y
{\displaystyle x{\mathcal {R}}y}
2° La réflexivité et la symétrie de
R
{\displaystyle {\mathcal {R}}}
f
0
{\displaystyle f^{0}}
R
{\displaystyle {\mathcal {R}}}
x
R
y
{\displaystyle x{\mathcal {R}}y}
y
R
z
{\displaystyle y{\mathcal {R}}z}
p
,
q
,
r
,
s
∈
N
{\displaystyle p,q,r,s\in \mathbb {N} }
f
p
(
x
)
=
f
q
(
y
)
{\displaystyle f^{p}(x)=f^{q}(y)}
f
r
(
y
)
=
f
s
(
z
)
{\displaystyle f^{r}(y)=f^{s}(z)}
f
p
+
r
(
x
)
=
f
q
+
r
(
y
)
=
f
q
+
s
(
z
)
{\displaystyle f^{p+r}(x)=f^{q+r}(y)=f^{q+s}(z)}
x
R
z
{\displaystyle x{\mathcal {R}}z}
3° a) Pour tout
x
∈
A
{\displaystyle x\in A}
f
(
x
)
R
x
{\displaystyle f(x){\mathcal {R}}x}
x
∈
C
⇔
f
(
x
)
∈
C
⇔
f
(
x
)
∈
C
∩
Im
(
f
)
{\displaystyle x\in C\Leftrightarrow f(x)\in C\Leftrightarrow f(x)\in C\cap \operatorname {Im} (f)}
b) Si
a
∈
C
∖
f
(
C
)
{\displaystyle a\in C\setminus f(C)}
R
{\displaystyle {\mathcal {R}}}
a
∉
Im
(
f
)
{\displaystyle a\notin \operatorname {Im} (f)}
∀
x
∈
A
x
∈
C
⇔
x
R
a
⇔
∃
n
∈
N
x
=
f
n
(
a
)
{\displaystyle \forall x\in A\quad x\in C\Leftrightarrow x{\mathcal {R}}a\Leftrightarrow \exists n\in \mathbb {N} \quad x=f^{n}(a)}
a
{\displaystyle a}
C
∖
Im
(
f
)
=
C
∖
f
(
C
)
{\displaystyle C\setminus \operatorname {Im} (f)=C\setminus f(C)}
4° Soit
X
{\displaystyle X}
X
=
∪
x
∈
X
x
¯
{\displaystyle X=\cup _{x\in X}{\overline {x}}}
x
¯
{\displaystyle {\overline {x}}}
x
{\displaystyle x}
x
∈
X
{\displaystyle x\in X}
x
¯
⊂
X
{\displaystyle {\overline {x}}\subset X}
y
R
x
⇒
y
∈
X
{\displaystyle y{\mathcal {R}}x\Rightarrow y\in X}
y
=
f
n
(
x
)
{\displaystyle y=f^{n}(x)}
n
∈
N
{\displaystyle n\in \mathbb {N} }
y
∈
f
n
(
X
)
=
X
{\displaystyle y\in f^{n}(X)=X}
x
=
f
n
(
y
)
{\displaystyle x=f^{n}(y)}
y
∈
(
f
n
)
−
1
(
X
)
=
X
{\displaystyle y\in \left(f^{n}\right)^{-1}(X)=X}
f
−
1
(
X
)
{\displaystyle f^{-1}(X)}
f
−
1
(
f
(
X
)
)
{\displaystyle f^{-1}(f(X))}
f
{\displaystyle f}
X
{\displaystyle X}
5° a)
f
{\displaystyle f}
Z
∖
N
{\displaystyle \mathbb {Z} \setminus \mathbb {N} }
A
=
{
…
,
−
2
,
−
1
,
0
,
1
}
{\displaystyle A=\{\dots ,-2,-1,0,1\}}
N
{\displaystyle \mathbb {N} }
B
=
{
4
,
5
,
6
,
…
}
{\displaystyle B=\{4,5,6,\dots \}}
A
{\displaystyle A}
B
{\displaystyle B}
f
{\displaystyle f}
b)
Im
(
f
)
=
A
∪
B
=
Z
∖
{
2
,
3
}
{\displaystyle \operatorname {Im} (f)=A\cup B=\mathbb {Z} \setminus \{2,3\}}
c) Il y a quatre classes :
2
¯
=
2
+
4
N
{\displaystyle {\overline {2}}=2+4\mathbb {N} }
3
¯
=
3
+
4
N
{\displaystyle {\overline {3}}=3+4\mathbb {N} }
0
¯
=
4
N
∪
−
2
N
{\displaystyle {\overline {0}}=4\mathbb {N} \cup -2\mathbb {N} }
1
¯
=
1
+
0
¯
{\displaystyle {\overline {1}}=1+{\overline {0}}}
Les relations
(
X
,
Γ
)
{\displaystyle (X,\Gamma )}
X
=
Z
{\displaystyle X=\mathbb {Z} }
Γ
=
{
(
x
,
y
)
∈
Z
2
∣
x
+
y
∈
2
Z
}
{\displaystyle \Gamma =\{(x,y)\in \mathbb {Z} ^{2}\mid x+y\in 2\mathbb {Z} \}}
X
=
R
{\displaystyle X=\mathbb {R} }
Γ
=
{
(
x
,
y
)
∈
R
2
∣
cos
(
2
x
)
=
cos
(
2
y
)
}
{\displaystyle \Gamma =\{(x,y)\in \mathbb {R} ^{2}\mid \cos(2x)=\cos(2y)\}}
X
=
R
{\displaystyle X=\mathbb {R} }
Γ
=
{
(
x
,
y
)
∈
R
2
∣
x
2
−
y
2
=
x
−
y
}
{\displaystyle \Gamma =\{(x,y)\in \mathbb {R} ^{2}\mid x^{2}-y^{2}=x-y\}}
X
=
Z
{\displaystyle X=\mathbb {Z} }
Γ
=
{
(
x
,
y
)
∈
Z
2
∣
x
2
−
y
2
∈
3
Z
}
{\displaystyle \Gamma =\{(x,y)\in \mathbb {Z} ^{2}\mid x^{2}-y^{2}\in 3\mathbb {Z} \}}
X
=
Z
2
{\displaystyle X=\mathbb {Z} ^{2}}
Γ
=
{
(
(
x
,
y
)
,
(
x
′
,
y
′
)
)
∈
X
2
∣
(
x
+
x
′
,
y
2
−
y
′
2
)
∈
(
2
Z
)
×
(
3
Z
)
}
{\displaystyle \Gamma =\{((x,y),(x',y'))\in X^{2}\mid (x+x',y^{2}-y'^{2})\in (2\mathbb {Z} )\times (3\mathbb {Z} )\}}
X
=
R
2
∖
{
(
0
,
0
)
}
{\displaystyle X=\mathbb {R} ^{2}\setminus \{(0,0)\}}
Γ
=
{
(
u
,
v
)
∈
X
2
∣
∃
λ
∈
R
v
=
λ
u
}
{\displaystyle \Gamma =\{(u,v)\in X^{2}\mid \exists \lambda \in \mathbb {R} \quad v=\lambda u\}}
Solution
Oui : c'est la relation « avoir même parité que ». Elle partitionne
Z
{\displaystyle \mathbb {Z} }
X
/
Γ
=
{
2
Z
,
2
Z
+
1
}
{\displaystyle X/\Gamma =\{2\mathbb {Z} ,2\mathbb {Z} +1\}}
Oui, comme toute relation dont le graphe est de la forme
{
(
x
,
y
)
∈
X
2
∣
f
(
x
)
=
f
(
y
)
}
{\displaystyle \{(x,y)\in X^{2}\mid f(x)=f(y)\}}
x
{\displaystyle x}
±
x
+
π
Z
{\displaystyle \pm x+\pi \mathbb {Z} }
[
−
1
,
1
]
{\displaystyle [-1,1]}
X
/
Γ
=
{
C
t
∣
t
∈
[
−
1
,
1
]
}
{\displaystyle X/\Gamma =\{C_{t}\mid t\in [-1,1]\}}
C
t
=
{
x
∈
R
∣
cos
(
2
x
)
=
t
}
{\displaystyle C_{t}=\{x\in \mathbb {R} \mid \cos(2x)=t\}}
Oui comme précédemment, car
x
2
−
y
2
=
x
−
y
{\displaystyle x^{2}-y^{2}=x-y}
x
2
−
x
=
y
2
−
y
{\displaystyle x^{2}-x=y^{2}-y}
x
↦
x
2
−
x
{\displaystyle x\mapsto x^{2}-x}
{
1
/
2
}
{\displaystyle \{1/2\}}
{
1
/
2
−
h
,
1
/
2
+
h
}
{\displaystyle \{1/2-h,1/2+h\}}
h
>
0
{\displaystyle h>0}
R
+
{\displaystyle \mathbb {R} _{+}}
Oui comme précédemment, en posant
f
(
x
)
=
{\displaystyle f(x)=}
x
2
mod
3
{\displaystyle x^{2}{\bmod {3}}}
Oui comme précédemment, en posant
f
(
x
,
y
)
=
(
x
mod
2
,
y
2
mod
3
)
∈
(
Z
/
2
Z
)
×
(
Z
/
3
Z
)
{\displaystyle f(x,y)=(x{\bmod {2}},y^{2}{\bmod {3}})\in (\mathbb {Z} /2\mathbb {Z} )\times (\mathbb {Z} /3\mathbb {Z} )}
2
Z
×
3
Z
{\displaystyle 2\mathbb {Z} \times 3\mathbb {Z} }
(
2
Z
+
1
)
×
3
Z
{\displaystyle (2\mathbb {Z} +1)\times 3Z}
2
Z
×
Y
{\displaystyle 2\mathbb {Z} \times Y}
(
2
Z
+
1
)
×
Y
{\displaystyle (2\mathbb {Z} +1)\times Y}
Y
{\displaystyle Y}
Oui, cf. Droite projective .
Montrer que sur
X
=
N
×
N
{\displaystyle X=\mathbb {N} \times \mathbb {N} }
R
{\displaystyle {\mathcal {R}}}
(
m
,
n
)
R
(
m
′
,
n
′
)
⟺
m
+
n
′
=
n
+
m
′
{\displaystyle (m,n){\mathcal {R}}(m',n')\Longleftrightarrow m+n'=n+m'}
Montrer que sur
X
/
R
{\displaystyle X/{\mathcal {R}}}
(
m
,
n
)
¯
+
(
s
,
t
)
¯
=
(
m
+
s
,
n
+
t
)
¯
{\displaystyle {\overline {(m,n)}}+{\overline {(s,t)}}={\overline {(m+s,n+t)}}}
(
m
,
n
)
¯
⋅
(
s
,
t
)
¯
=
(
m
s
+
n
t
,
n
s
+
m
t
)
¯
{\displaystyle {\overline {(m,n)}}\cdot {\overline {(s,t)}}={\overline {(ms+nt,ns+mt)}}}
Quel est l'intérêt de cette construction ?
On note
F
3
=
Z
/
3
Z
=
{
0
¯
,
1
¯
,
−
1
¯
}
{\displaystyle \mathbb {F} _{3}=\mathbb {Z} /3\mathbb {Z} =\{{\overline {0}},{\overline {1}},-{\overline {1}}\}}
3
{\displaystyle 3}
+
{\displaystyle +}
×
{\displaystyle \times }
Z
{\displaystyle \mathbb {Z} }
F
3
{\displaystyle \mathbb {F} _{3}}
anneau commutatif unitaire ).
Dresser les tables de ces deux opérations sur
F
3
{\displaystyle \mathbb {F} _{3}}
Soit
A
=
F
3
[
X
]
{\displaystyle A=\mathbb {F} _{3}[X]}
F
3
{\displaystyle \mathbb {F} _{3}}
B
⊂
A
{\displaystyle B\subset A}
2
{\displaystyle 2}
0
¯
{\displaystyle {\overline {0}}}
−
∞
{\displaystyle -\infty }
B
{\displaystyle B}
En notant
q
(
P
)
{\displaystyle q(P)}
r
(
P
)
{\displaystyle r(P)}
P
∈
A
{\displaystyle P\in A}
P
{\displaystyle P}
X
2
+
1
¯
{\displaystyle X^{2}+{\overline {1}}}
A
{\displaystyle A}
q
(
P
)
∈
A
,
r
(
P
)
∈
B
et
P
=
(
X
2
+
1
¯
)
q
(
P
)
+
r
(
P
)
{\displaystyle q(P)\in A,\quad r(P)\in B\quad {\text{et}}\quad P=\left(X^{2}+{\overline {1}}\right)q(P)+r(P)}
A
{\displaystyle A}
R
{\displaystyle {\mathcal {R}}}
P
R
Q
⇔
r
(
P
)
=
r
(
Q
)
{\displaystyle P{\mathcal {R}}Q\Leftrightarrow r(P)=r(Q)}
cl
(
P
)
{\displaystyle \operatorname {cl} (P)}
R
{\displaystyle {\mathcal {R}}}
P
∈
A
{\displaystyle P\in A}
r
¯
:
A
/
R
→
B
,
cl
(
P
)
↦
r
(
P
)
{\displaystyle {\overline {r}}:A/{\mathcal {R}}\to B,\quad \operatorname {cl} (P)\mapsto r(P)}
Montrer que
P
R
Q
{\displaystyle P{\mathcal {R}}Q}
X
2
+
1
¯
∣
P
−
Q
{\displaystyle X^{2}+{\overline {1}}\mid P-Q}
P
−
Q
{\displaystyle P-Q}
X
2
+
1
¯
{\displaystyle X^{2}+{\overline {1}}}
A
{\displaystyle A}
Montrer qu'il existe une unique application
⊕
:
A
/
R
×
A
/
R
→
A
/
R
{\displaystyle \oplus :A/{\mathcal {R}}\times A/{\mathcal {R}}\to A/{\mathcal {R}}}
⊗
:
A
/
R
×
A
/
R
→
A
/
R
{\displaystyle \otimes :A/{\mathcal {R}}\times A/{\mathcal {R}}\to A/{\mathcal {R}}}
∀
P
,
Q
∈
A
cl
(
P
)
⊕
cl
(
Q
)
=
cl
(
P
+
Q
)
et
cl
(
P
)
⊗
cl
(
Q
)
=
cl
(
P
×
Q
)
{\displaystyle \forall P,Q\in A\quad \operatorname {cl} (P)\oplus \operatorname {cl} (Q)=\operatorname {cl} (P+Q){\text{ et }}\operatorname {cl} (P)\otimes \operatorname {cl} (Q)=\operatorname {cl} (P\times Q)}
Calculer
r
(
−
X
2
)
{\displaystyle r(-X^{2})}
r
(
(
X
+
1
¯
)
(
X
−
1
¯
)
)
{\displaystyle r((X+{\overline {1}})(X-{\overline {1}}))}
∀
P
∈
B
∖
{
0
¯
}
∃
Q
∈
B
r
(
P
×
Q
)
=
1
¯
{\displaystyle \forall P\in B\setminus \{{\overline {0}}\}\quad \exists Q\in B\quad r(P\times Q)={\overline {1}}}
A
/
R
{\displaystyle A/{\mathcal {R}}}
Solution
+
0
¯
1
¯
−
1
¯
0
¯
0
¯
1
¯
−
1
¯
1
¯
1
¯
−
1
¯
0
¯
−
1
¯
−
1
¯
0
¯
1
¯
,
×
0
¯
1
¯
−
1
¯
0
¯
0
¯
0
¯
0
¯
1
¯
0
¯
1
¯
−
1
¯
−
1
¯
0
¯
−
1
¯
1
¯
{\displaystyle {\begin{array}{c|ccc}+&{\overline {0}}&{\overline {1}}&-{\overline {1}}\\\hline {\overline {0}}&{\overline {0}}&{\overline {1}}&-{\overline {1}}\\{\overline {1}}&{\overline {1}}&-{\overline {1}}&{\overline {0}}\\-{\overline {1}}&-{\overline {1}}&{\overline {0}}&{\overline {1}}\\\end{array}},\quad {\begin{array}{c|ccc}\times &{\overline {0}}&{\overline {1}}&-{\overline {1}}\\\hline {\overline {0}}&{\overline {0}}&{\overline {0}}&{\overline {0}}\\{\overline {1}}&{\overline {0}}&{\overline {1}}&-{\overline {1}}\\-{\overline {1}}&{\overline {0}}&-{\overline {1}}&{\overline {1}}\\\end{array}}}
B
{\displaystyle B}
9
{\displaystyle 9}
0
¯
,
1
¯
,
−
1
¯
,
X
,
X
+
1
¯
,
X
−
1
¯
,
−
X
,
−
X
+
1
¯
,
−
X
−
1
¯
{\displaystyle {\overline {0}},{\overline {1}},-{\overline {1}},X,X+{\overline {1}},X-{\overline {1}},-X,-X+{\overline {1}},-X-{\overline {1}}}
L'application
r
¯
{\displaystyle {\overline {r}}}
cl
(
P
)
=
cl
(
P
)
⇔
P
R
Q
⇔
r
(
P
)
=
r
(
Q
)
{\displaystyle \operatorname {cl} (P)=\operatorname {cl} (P)\Leftrightarrow P{\mathcal {R}}Q\Leftrightarrow r(P)=r(Q)}
∀
P
∈
B
P
=
r
(
P
)
=
r
¯
(
cl
(
P
)
)
{\displaystyle \forall P\in B\quad P=r(P)={\overline {r}}(\operatorname {cl} (P))}
X
2
+
1
¯
∣
P
−
Q
⇔
r
(
P
−
Q
)
=
0
⇔
r
(
P
)
−
r
(
Q
)
=
0
{\displaystyle X^{2}+{\overline {1}}\mid P-Q\Leftrightarrow r(P-Q)=0\Leftrightarrow r(P)-r(Q)=0}
La formule
cl
(
P
)
⊕
cl
(
Q
)
=
cl
(
P
+
Q
)
{\displaystyle \operatorname {cl} (P)\oplus \operatorname {cl} (Q)=\operatorname {cl} (P+Q)}
⊕
:
A
/
R
×
A
/
R
→
A
/
R
{\displaystyle \oplus :A/{\mathcal {R}}\times A/{\mathcal {R}}\to A/{\mathcal {R}}}
⊗
{\displaystyle \otimes }
⊕
{\displaystyle \oplus }
⊗
{\displaystyle \otimes }
cl
(
P
1
)
=
cl
(
P
2
)
{\displaystyle \operatorname {cl} \left(P_{1}\right)=\operatorname {cl} \left(P_{2}\right)}
cl
(
Q
1
)
=
cl
(
Q
2
)
{\displaystyle \operatorname {cl} \left(Q_{1}\right)=\operatorname {cl} \left(Q_{2}\right)}
cl
(
P
1
+
Q
1
)
=
cl
(
P
2
+
Q
2
)
{\displaystyle \operatorname {cl} \left(P_{1}+Q_{1}\right)=\operatorname {cl} \left(P_{2}+Q_{2}\right)}
cl
(
P
1
×
Q
1
)
=
cl
(
P
2
×
Q
2
)
{\displaystyle \operatorname {cl} \left(P_{1}\times Q_{1}\right)=\operatorname {cl} \left(P_{2}\times Q_{2}\right)}
r
(
P
1
)
=
r
(
P
2
)
{\displaystyle r(P_{1})=r(P_{2})}
r
(
Q
1
)
=
r
(
Q
2
)
{\displaystyle r(Q_{1})=r(Q_{2})}
pour
⊕
{\displaystyle \oplus }
r
(
P
1
+
Q
1
)
−
r
(
P
2
+
Q
2
)
=
r
(
P
1
)
+
r
(
Q
1
)
−
r
(
P
2
)
−
r
(
Q
2
)
=
0
{\displaystyle r\left(P_{1}+Q_{1}\right)-r\left(P_{2}+Q_{2}\right)=r\left(P_{1}\right)+r\left(Q_{1}\right)-r\left(P_{2}\right)-r\left(Q_{2}\right)=0}
pour
⊗
{\displaystyle \otimes }
X
2
+
1
¯
{\displaystyle X^{2}+{\overline {1}}}
P
1
−
P
2
{\displaystyle P_{1}-P_{2}}
Q
1
−
Q
2
{\displaystyle Q_{1}-Q_{2}}
P
1
Q
1
−
P
2
Q
2
=
P
1
(
Q
1
−
Q
2
)
+
(
P
1
−
P
2
)
Q
2
{\displaystyle P_{1}Q_{1}-P_{2}Q_{2}=P_{1}\left(Q_{1}-Q_{2}\right)+\left(P_{1}-P_{2}\right)Q_{2}}
X
2
+
1
¯
{\displaystyle X^{2}+{\overline {1}}}
−
X
2
=
−
(
X
2
+
1
¯
)
+
1
¯
{\displaystyle -X^{2}=-(X^{2}+{\overline {1}})+{\overline {1}}}
X
2
−
1
¯
=
(
X
2
+
1
¯
)
+
1
¯
{\displaystyle X^{2}-{\overline {1}}=(X^{2}+{\overline {1}})+{\overline {1}}}
1
¯
+
1
¯
+
1
¯
=
3
¯
=
0
¯
{\displaystyle {\overline {1}}+{\overline {1}}+{\overline {1}}={\overline {3}}={\overline {0}}}
r
(
−
X
2
)
{\displaystyle r(-X^{2})}
r
(
(
X
+
1
¯
)
(
X
−
1
¯
)
)
{\displaystyle r((X+{\overline {1}})(X-{\overline {1}}))}
1
¯
{\displaystyle {\overline {1}}}
P
=
±
X
,
±
(
X
+
1
¯
)
,
±
(
X
−
1
¯
)
{\displaystyle P=\pm X,\pm (X+{\overline {1}}),\pm (X-{\overline {1}})}
B
∖
{
0
¯
}
{\displaystyle B\setminus \{{\overline {0}}\}}
Q
∈
B
{\displaystyle Q\in B}
r
(
P
×
Q
)
=
1
{\displaystyle r(P\times Q)=1}
P
=
±
1
¯
{\displaystyle P=\pm {\overline {1}}}
Q
=
P
{\displaystyle Q=P}
A
/
R
{\displaystyle A/{\mathcal {R}}}
∀
x
∈
A
/
R
∖
{
cl
(
0
¯
)
}
∃
y
∈
A
/
R
x
⊗
y
=
cl
(
1
¯
)
{\displaystyle \forall x\in A/{\mathcal {R}}\setminus \{\operatorname {cl} ({\overline {0}})\}\quad \exists y\in A/{\mathcal {R}}\quad x\otimes y=\operatorname {cl} ({\overline {1}})}
(
A
/
R
,
⊕
,
⊗
)
{\displaystyle (A/{\mathcal {R}},\oplus ,\otimes )}
corps à 9 éléments.
Soit
R
{\displaystyle {\mathcal {R}}}
R
{\displaystyle \mathbb {R} }
x
R
y
⟺
x
2
−
y
2
=
x
−
y
{\displaystyle x{\mathcal {R}}y\Longleftrightarrow x^{2}-y^{2}=x-y}
Montrer que
R
{\displaystyle {\mathcal {R}}}
Décrire ses classes d'équivalence, et l'ensemble quotient
R
/
R
{\displaystyle \mathbb {R} /{\mathcal {R}}}
Donner un exemple de partie
X
{\displaystyle X}
R
{\displaystyle \mathbb {R} }
Soit
R
{\displaystyle {\mathcal {R}}}
R
{\displaystyle \mathbb {R} }
x
R
y
⟺
x
3
−
y
3
=
3
(
x
−
y
)
{\displaystyle x{\mathcal {R}}y\Longleftrightarrow x^{3}-y^{3}=3\left(x-y\right)}
Montrer que
R
{\displaystyle {\mathcal {R}}}
Montrer que chaque
R
{\displaystyle {\mathcal {R}}}
Pour préciser le résultat précédent, on pose
A
k
:=
{
x
∈
R
∣
la classe de
x
a exactement
k
éléments
}
{\displaystyle A_{k}:=\{x\in \mathbb {R} \mid {\text{la classe de }}x{\text{ a exactement }}k{\text{ éléments}}\}}
A
3
{\displaystyle A_{3}}
A
2
{\displaystyle A_{2}}
A
1
{\displaystyle A_{1}}
A
0
{\displaystyle A_{0}}
−
1
R
2
{\displaystyle -1\;{\mathcal {R}}\;2}
Donner un exemple de partie
X
{\displaystyle X}
R
{\displaystyle \mathbb {R} }
Montrer qu'une telle partie
X
{\displaystyle X}
puissance du continu .
Solution
x
R
y
⇔
f
(
x
)
=
f
(
y
)
{\displaystyle x{\mathcal {R}}y\Leftrightarrow f(x)=f(y)}
f
:
R
→
R
{\displaystyle f:\mathbb {R} \to \mathbb {R} }
f
(
t
)
=
t
3
−
3
t
{\displaystyle f(t)=t^{3}-3t}
La classe d'équivalence d'un réel
x
{\displaystyle x}
f
{\displaystyle f}
f
(
x
)
{\displaystyle f(x)}
R
{\displaystyle \mathbb {R} }
corps commutatif ), un polynôme de degré 3 ne peut pas prendre 4 fois la même valeur .
A
0
=
∅
{\displaystyle A_{0}=\varnothing }
A
1
,
A
2
,
A
3
{\displaystyle A_{1},A_{2},A_{3}}
Méthode analytique : le tableau des variations de
f
{\displaystyle f}
x
−
∞
−
2
−
1
1
2
+
∞
+
∞
↗
2
2
f
(
x
)
↗
↘
↗
−
2
−
2
↗
−
∞
{\displaystyle {\begin{array}{c|ccccccccccc|}x&-\infty &&-2&&-1&&1&&2&&+\infty \\\hline &&&&&&&&&&&+\infty \\&&&&&&&&&&\nearrow &\\&&&&&2&&&&2&&\\f(x)&&&&\nearrow &&\searrow &&\nearrow &&&\\&&&-2&&&&-2&&&&\\&&\nearrow &&&&&&&&&\\&-\infty &&&&&&&&&&\\\hline \end{array}}}
A
3
=
]
−
2
,
2
[
∖
{
−
1
,
1
}
,
A
2
=
{
−
2
,
−
1
,
1
,
2
}
et
A
1
=
]
−
∞
,
−
2
[
∪
]
2
,
+
∞
[
{\displaystyle A_{3}=\left]-2,2\right[\setminus \{-1,1\},\quad A_{2}=\{-2,-1,1,2\}\quad {\text{et}}\quad A_{1}=\left]-\infty ,-2\right[\cup \left]2,+\infty \right[}
Méthode algébrique : déterminons la classe d'un réel
a
{\displaystyle a}
x
{\displaystyle x}
x
3
−
a
3
=
3
(
x
−
a
)
{\displaystyle x^{3}-a^{3}=3(x-a)}
(
x
−
a
)
(
x
2
+
a
x
+
a
2
−
3
)
=
0
{\displaystyle (x-a)(x^{2}+ax+a^{2}-3)=0}
a
{\displaystyle a}
x
{\displaystyle x}
x
2
+
a
x
+
a
2
−
3
=
0
{\displaystyle x^{2}+ax+a^{2}-3=0}
Δ
:=
a
2
−
4
(
a
2
−
3
)
=
3
(
4
−
a
2
)
{\displaystyle \Delta :=a^{2}-4(a^{2}-3)=3(4-a^{2})}
>
0
{\displaystyle >0}
Δ
=
0
{\displaystyle \Delta =0}
Δ
<
0
{\displaystyle \Delta <0}
a
{\displaystyle a}
a
{\displaystyle a}
x
2
+
a
x
+
a
2
−
3
=
0
{\displaystyle x^{2}+ax+a^{2}-3=0}
a
=
±
1
{\displaystyle a=\pm 1}
Δ
>
0
{\displaystyle \Delta >0}
a
{\displaystyle a}
2
{\displaystyle 2}
2
+
1
=
3
{\displaystyle 2+1=3}
a
∈
A
3
{\displaystyle a\in A_{3}}
Δ
>
0
{\displaystyle \Delta >0}
a
≠
±
1
{\displaystyle a\neq \pm 1}
a
∈
A
2
{\displaystyle a\in A_{2}}
Δ
=
0
{\displaystyle \Delta =0}
a
=
±
1
{\displaystyle a=\pm 1}
a
∈
A
1
{\displaystyle a\in A_{1}}
Δ
<
0
{\displaystyle \Delta <0}
A
3
=
]
−
2
,
2
[
∖
{
−
1
,
1
}
{\displaystyle A_{3}=\left]-2,2\right[\setminus \{-1,1\}}
A
2
=
{
−
2
,
−
1
,
1
,
2
}
{\displaystyle A_{2}=\{-2,-1,1,2\}}
A
1
=
]
−
∞
,
−
2
[
∪
]
2
,
+
∞
[
{\displaystyle A_{1}=\left]-\infty ,-2\right[\cup \left]2,+\infty \right[}
D'après l'étude ci-dessus,
X
=
]
−
∞
,
−
2
[
∪
[
1
,
+
∞
[
{\displaystyle X=\left]-\infty ,-2\right[\cup \left[1,+\infty \right[}
L'application
X
→
f
(
R
)
=
R
,
x
↦
f
(
x
)
{\displaystyle X\to f(\mathbb {R} )=\mathbb {R} ,\;x\mapsto f(x)}
Soit
R
{\displaystyle {\mathcal {R}}}
R
{\displaystyle \mathbb {R} }
x
R
y
⟺
x
3
−
y
3
=
3
(
x
2
−
y
2
)
{\displaystyle x{\mathcal {R}}y\Longleftrightarrow x^{3}-y^{3}=3\left(x^{2}-y^{2}\right)}
Montrer que
R
{\displaystyle {\mathcal {R}}}
Montrer que les
R
{\displaystyle {\mathcal {R}}}
k
∈
N
{\displaystyle k\in \mathbb {N} }
A
k
:=
{
x
∈
R
∣
la classe de
x
a exactement
k
éléments
}
{\displaystyle A_{k}:=\{x\in \mathbb {R} \mid {\text{la classe de }}x{\text{ a exactement }}k{\text{ éléments}}\}}
Donner un exemple de partie
X
{\displaystyle X}
R
{\displaystyle \mathbb {R} }
Solution
x
R
y
⇔
f
(
x
)
=
f
(
y
)
{\displaystyle x{\mathcal {R}}y\Leftrightarrow f(x)=f(y)}
f
:
R
→
R
{\displaystyle f:\mathbb {R} \to \mathbb {R} }
f
(
t
)
=
t
3
−
3
t
2
{\displaystyle f(t)=t^{3}-3t^{2}}
La classe d'équivalence d'un réel
x
{\displaystyle x}
f
{\displaystyle f}
f
(
x
)
{\displaystyle f(x)}
R
{\displaystyle \mathbb {R} }
k
≥
4
{\displaystyle k\geq 4}
A
k
=
∅
{\displaystyle A_{k}=\varnothing }
A
0
=
∅
{\displaystyle A_{0}=\varnothing }
A
1
,
A
2
,
A
3
{\displaystyle A_{1},A_{2},A_{3}}
Méthode analytique : le tableau des variations de
f
{\displaystyle f}
x
−
∞
−
1
0
2
3
+
∞
+
∞
↗
0
0
f
(
x
)
↗
↘
↗
−
4
−
4
↗
−
∞
{\displaystyle {\begin{array}{c|ccccccccccc|}x&-\infty &&-1&&0&&2&&3&&+\infty \\\hline &&&&&&&&&&&+\infty \\&&&&&&&&&&\nearrow &\\&&&&&0&&&&0&&\\f(x)&&&&\nearrow &&\searrow &&\nearrow &&&\\&&&-4&&&&-4&&&&\\&&\nearrow &&&&&&&&&\\&-\infty &&&&&&&&&&\\\hline \end{array}}}
A
3
=
]
−
1
,
3
[
∖
{
0
,
2
}
,
A
2
=
{
−
1
,
0
,
2
,
3
}
et
A
1
=
]
−
∞
,
−
1
[
∪
]
3
,
+
∞
[
{\displaystyle A_{3}=\left]-1,3\right[\setminus \{0,2\},\quad A_{2}=\{-1,0,2,3\}\quad {\text{et}}\quad A_{1}=\left]-\infty ,-1\right[\cup \left]3,+\infty \right[}
Méthode algébrique : déterminons la classe d'un réel
a
{\displaystyle a}
x
{\displaystyle x}
x
3
−
a
3
=
3
(
x
2
−
a
2
)
{\displaystyle x^{3}-a^{3}=3(x^{2}-a^{2})}
(
x
−
a
)
(
x
2
+
a
x
+
a
2
−
3
x
−
3
a
)
=
0
{\displaystyle (x-a)(x^{2}+ax+a^{2}-3x-3a)=0}
a
{\displaystyle a}
x
{\displaystyle x}
x
2
+
(
a
−
3
)
x
+
(
a
2
−
3
a
)
=
0
{\displaystyle x^{2}+(a-3)x+(a^{2}-3a)=0}
Δ
:=
(
a
−
3
)
2
−
4
(
a
2
−
3
a
)
=
−
3
(
a
2
−
2
a
−
3
)
{\displaystyle \Delta :=(a-3)^{2}-4(a^{2}-3a)=-3(a^{2}-2a-3)}
>
0
{\displaystyle >0}
−
1
<
a
<
3
{\displaystyle -1<a<3}
Δ
=
0
{\displaystyle \Delta =0}
Δ
<
0
{\displaystyle \Delta <0}
a
{\displaystyle a}
a
{\displaystyle a}
a
{\displaystyle a}
x
2
+
(
a
−
3
)
x
+
(
a
2
−
3
a
)
=
0
{\displaystyle x^{2}+(a-3)x+(a^{2}-3a)=0}
a
=
0
{\displaystyle a=0}
2
{\displaystyle 2}
Δ
>
0
{\displaystyle \Delta >0}
a
{\displaystyle a}
2
{\displaystyle 2}
2
+
1
=
3
{\displaystyle 2+1=3}
A
3
=
]
−
1
,
3
[
∖
{
0
,
2
}
{\displaystyle A_{3}=\left]-1,3\right[\setminus \{0,2\}}
A
2
=
{
−
1
,
0
,
2
,
3
}
{\displaystyle A_{2}=\{-1,0,2,3\}}
A
1
=
]
−
∞
,
−
1
[
∪
]
3
,
+
∞
[
{\displaystyle A_{1}=\left]-\infty ,-1\right[\cup \left]3,+\infty \right[}
D'après l'étude ci-dessus,
X
=
]
−
∞
,
−
1
[
∪
[
2
,
+
∞
[
{\displaystyle X=\left]-\infty ,-1\right[\cup \left[2,+\infty \right[}
On définit, sur l'espace
R
4
[
X
]
{\displaystyle \mathbb {R} _{4}[X]}
≤
4
{\displaystyle \leq 4}
R
{\displaystyle {\mathcal {R}}}
A
R
B
⇔
A
−
B
∈
R
2
[
X
]
{\displaystyle A{\mathcal {R}}B\Leftrightarrow A-B\in \mathbb {R} _{2}[X]}
R
{\displaystyle {\mathcal {R}}}
c
l
R
(
X
k
)
{\displaystyle \mathrm {cl} _{\mathcal {R}}(X^{k})}
k
=
0
,
1
,
2
,
3
,
4
{\displaystyle k=0,1,2,3,4}
Démontrer que la relation
∁
E
A
⊂
B
{\displaystyle \complement _{E}A\subset B}
E
{\displaystyle E}
De même pour la relation
B
⊂
∁
E
A
{\displaystyle B\subset \complement _{E}A}
Soient
n
1
,
…
,
n
k
∈
Z
{\displaystyle n_{1},\dots ,n_{k}\in \mathbb {Z} }
R
i
{\displaystyle {\mathcal {R}}_{i}}
x
≡
y
mod
n
i
{\displaystyle x\equiv y{\bmod {n_{i}}}}
Z
{\displaystyle \mathbb {Z} }
R
i
{\displaystyle {\mathcal {R}}_{i}}
Solution
On a
x
≡
y
mod
n
i
{\displaystyle x\equiv y{\bmod {n_{i}}}}
i
∈
{
1
,
…
,
k
}
{\displaystyle i\in \{1,\dots ,k\}}
x
≡
y
mod
ppcm
(
n
1
,
…
,
n
k
)
{\displaystyle x\equiv y{\bmod {\operatorname {ppcm} (n_{1},\dots ,n_{k})}}}
Arithmétique/PPCM ).
Sur l'ensemble
E
N
{\displaystyle E^{\mathbb {N} }}
∼
{\displaystyle \sim }
(
u
n
)
n
∈
N
∼
(
v
n
)
n
∈
N
⇔
∃
N
∈
N
∀
n
≥
N
u
n
=
v
n
{\displaystyle (u_{n})_{n\in \mathbb {N} }\sim (v_{n})_{n\in \mathbb {N} }\Leftrightarrow \exists N\in \mathbb {N} \quad \forall n\geq N\quad u_{n}=v_{n}}
Démontrer que
∼
{\displaystyle \sim }
Solution
La réflexivité et la symétrie sont immédiates. Vérifions la transitivité. Soient
u
,
v
,
w
{\displaystyle u,v,w}
u
∼
v
{\displaystyle u\sim v}
v
∼
w
{\displaystyle v\sim w}
N
,
N
′
∈
N
{\displaystyle N,N'\in \mathbb {N} }
∀
n
≥
N
u
n
=
v
n
{\displaystyle \forall n\geq N\quad u_{n}=v_{n}}
∀
n
≥
N
′
v
n
=
w
n
{\displaystyle \forall n\geq N'\quad v_{n}=w_{n}}
∀
n
≥
max
(
N
,
N
′
)
u
n
=
w
n
{\displaystyle \forall n\geq \max(N,N')\quad u_{n}=w_{n}}
u
∼
w
{\displaystyle u\sim w}
Quels intervalles sont des ensembles de représentants pour la relation de congruence modulo
2
π
{\displaystyle 2\pi }
R
{\displaystyle \mathbb {R} }
Solution
Les intervalles de la forme
[
a
,
a
+
2
π
[
{\displaystyle \left[a,a+2\pi \right[}
]
a
,
a
+
2
π
]
{\displaystyle \left]a,a+2\pi \right]}
![{\displaystyle [-1,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51e3b7f14a6f70e614728c583409a0b9a8b9de01)
![{\displaystyle X/\Gamma =\{C_{t}\mid t\in [-1,1]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cc4116754d49842fc198ba0ad1282527716b9c8)
![{\displaystyle A=\mathbb {F} _{3}[X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b808b8e1dfc44bc910696e4f54911fa2fe52a2b)
![{\displaystyle A_{3}=\left]-2,2\right[\setminus \{-1,1\},\quad A_{2}=\{-2,-1,1,2\}\quad {\text{et}}\quad A_{1}=\left]-\infty ,-2\right[\cup \left]2,+\infty \right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1df9fa472c15d216362b8b52305724369f10020)
![{\displaystyle A_{3}=\left]-2,2\right[\setminus \{-1,1\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9943d7f0b2fc587939595cd9d8dff977619c88eb)
![{\displaystyle A_{1}=\left]-\infty ,-2\right[\cup \left]2,+\infty \right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4900f270328f09748018322dfd690ec734b40395)
![{\displaystyle X=\left]-\infty ,-2\right[\cup \left[1,+\infty \right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bea852b134219a376aab0bed1339ebc496806f2)
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