3. Gaia: Multzoak. Erlazioak. Funtzioak [ aldatu ] Lehendabizi, multzo-teoria ren oinarrizko kontzeptuak aztertzen hasiko gara; ondoren, multzo baten elementuen arteko erlazioak ikusiko ditugu; eta bukatzeko, bi multzoren arteko funtzioak landuko ditugu.
Logikaren kontzeptuek lotura handia dute multzo-teoriarekin, eta gai honetan ikasiko ditugun froga asko kontzeptu horietan daude oinarriturik.
∧
{\displaystyle \wedge }
∨
{\displaystyle \vee }
¬
{\displaystyle \neg }
⟹
{\displaystyle \implies }
p
⟹
q
{\displaystyle p\implies q}
q
{\displaystyle q}
p
{\displaystyle p}
p
{\displaystyle p}
q
{\displaystyle q}
p
{\displaystyle p}
q
{\displaystyle q}
p
{\displaystyle p}
q
{\displaystyle q}
p
{\displaystyle p}
q
{\displaystyle q}
q
{\displaystyle q}
p
{\displaystyle p}
⟺
{\displaystyle \iff }
p
⟺
q
{\displaystyle p\iff q}
(
p
⟹
q
)
eta
(
q
⟹
p
)
{\displaystyle (p\implies q){\mbox{ eta }}(q\implies p)}
p
{\displaystyle p}
q
{\displaystyle q}
p
{\displaystyle p}
q
{\displaystyle q}
∀
{\displaystyle \forall }
∀
x
{\displaystyle \forall x}
x
{\displaystyle x}
x
{\displaystyle x}
x
{\displaystyle x}
∃
{\displaystyle \exists }
∃
x
{\displaystyle \exists x}
x
{\displaystyle x}
x
{\displaystyle x}
∃
!
{\displaystyle \exists !}
∄
{\displaystyle \not \exists }
3.2.1 Multzoak eta azpimultzoak [ aldatu ] Senaz, multzo elementu
Ongi definitutako
Multzo bat bi eratan defini dezakegu:
Hedapenaren arabera: bere elementu guztiak banan-banan aipatuz,
A
=
{
0
,
1
,
2
,
…
,
20
}
.
{\displaystyle A=\{0,1,2,\ldots ,20\}.}
Ulermenaren arabera: Bere elementuen ezaugarriak erakusten dituzten propietateak aipatuz,
A
=
{
x
/
x
zenbaki osoa da eta
0
≤
x
≤
20
}
.
{\displaystyle A=\{x\;/\;x{\mbox{ zenbaki osoa da eta}}\;0\leq x\leq 20\}.}
Idazkera.
A
{\displaystyle A}
B
{\displaystyle B}
C
{\displaystyle C}
X
{\displaystyle X}
…
{\displaystyle \ldots }
(
a
,
b
,
c
,
x
,
…
)
{\displaystyle (a,b,c,x,\ldots )}
x
{\displaystyle x}
A
{\displaystyle A}
x
∈
A
{\displaystyle x\in A}
x
{\displaystyle x}
A
{\displaystyle A}
x
{\displaystyle x}
A
{\displaystyle A}
x
{\displaystyle x}
A
{\displaystyle A}
x
{\displaystyle x}
A
{\displaystyle A}
x
∉
A
{\displaystyle x\not \in A}
3.1 Adibidea.
A
=
{
0
,
1
,
2
,
…
,
20
}
{\displaystyle A=\{0,1,2,\ldots ,20\}}
11
∈
A
{\displaystyle 11\in A}
21
∉
A
{\displaystyle 21\not \in A}
Maiz erabiltzen diren multzo batzuk sinbolo bereziez adierazten dira:
Zenbaki arrunten multzoa:
N
=
{
0
,
1
,
2
,
…
}
{\displaystyle \mathbb {N} =\{0,1,2,\ldots \;\}}
Zenbaki osoen multzoa:
Z
=
{
…
,
−
2
,
−
1
,
0
,
1
,
2
,
…
}
{\displaystyle \mathbb {Z} =\{\;\ldots ,-2,-1,0,1,2,\ldots \;\}}
Zenbaki arrazionalen multzoa:
Q
=
{
x
/
x
=
a
b
,
non
a
,
b
∈
Z
eta
b
≠
0
}
{\displaystyle \mathbb {Q} =\left\{x\;/\;x={\frac {a}{b}},{\mbox{ non }}a,b\in \mathbb {Z} {\mbox{ eta }}b\neq 0\right\}}
Zenbaki errealen multzoa:
R
{\displaystyle \mathbb {R} }
Zenbaki konplexuen multzoa:
C
{\displaystyle \mathbb {C} }
0
{\displaystyle 0}
N
∗
=
Z
+
=
{
x
/
x
∈
N
eta
x
≠
0
}
=
{
x
/
x
∈
Z
eta
x
>
0
}
{\displaystyle \mathbb {N} ^{*}=\mathbb {Z} ^{+}=\{x\;/\;x\in \mathbb {N} {\mbox{ eta }}x\neq 0\}=\{x\;/\;x\in \mathbb {Z} {\mbox{ eta }}x>0\}}
Antzeko eran,
Z
∗
{\displaystyle \mathbb {Z} ^{*}}
Z
−
{\displaystyle \mathbb {Z} ^{-}}
Q
∗
{\displaystyle \mathbb {Q} ^{*}}
Q
+
{\displaystyle \mathbb {Q} ^{+}}
Q
−
{\displaystyle \mathbb {Q} ^{-}}
R
∗
{\displaystyle \mathbb {R} ^{*}}
R
+
{\displaystyle \mathbb {R} ^{+}}
R
−
{\displaystyle \mathbb {R} ^{-}}
C
∗
{\displaystyle \mathbb {C} ^{*}}
3.2 Definizioa.
A
{\displaystyle A}
A
{\displaystyle A}
A
{\displaystyle A}
kardinal deituko diogu eta honela adieraziko dugu:
|
A
|
{\displaystyle \vert A\vert }
card
(
A
)
{\displaystyle {\mbox{card}}(A)}
#
A
{\displaystyle \#A}
3.3 Adibidea.
A
=
{
0
,
1
,
2
,
…
,
20
}
{\displaystyle A=\{0,1,2,\ldots ,20\}}
card
(
A
)
=
21
{\displaystyle {\mbox{card}}(A)=21}
3.4 Definizioa.
A
{\displaystyle A}
B
{\displaystyle B}
azpimultzoa dela esango dugu, eta
A
⊆
B
{\displaystyle A\subseteq B}
A
{\displaystyle A}
B
{\displaystyle B}
x
∈
A
⟹
x
∈
B
{\displaystyle x\in A\;\implies \;x\in B}
betetzen bada.
Adierazpen hauek ere erabiliko ditugu:
A
{\displaystyle A}
B
{\displaystyle B}
partea da;
A
{\displaystyle A}
B
{\displaystyle B}
multzoan dago ;
B
{\displaystyle B}
A
{\displaystyle A}
barruan dauka .
A
{\displaystyle A}
B
{\displaystyle B}
A
⊈
B
{\displaystyle A\not \subseteq B}
A
{\displaystyle A}
B
{\displaystyle B}
∃
x
∈
A
,
non
x
∉
B
den
.
{\displaystyle \exists x\in A,{\mbox{ non }}x\not \in B{\mbox{ den}}.}
Ohar gaitezen,
A
{\displaystyle A}
A
⊆
A
.
{\displaystyle A\subseteq A.}
3.5 Adibidea.
A
=
{
0
,
1
,
2
}
{\displaystyle A=\{0,1,2\}}
B
=
{
−
2
,
−
1
,
0
,
1
,
2
}
{\displaystyle B=\{-2,-1,0,1,2\}}
A
⊆
B
{\displaystyle A\subseteq B}
B
⊈
A
{\displaystyle B\not \subseteq A}
3.6 Definizioa.
A
{\displaystyle A}
B
{\displaystyle B}
berdinak direla esango dugu, eta
A
=
B
{\displaystyle A=B}
x
∈
A
⟺
x
∈
B
{\displaystyle x\in A\;\iff \;x\in B}
betetzen bada; edo gauza bera dena,
A
⊆
B
{\displaystyle A\subseteq B}
B
⊆
A
{\displaystyle B\subseteq A}
3.7 Definizioa.
A
{\displaystyle A}
B
{\displaystyle B}
azpimultzo propioa da, eta
A
⊂
B
{\displaystyle A\subset B}
A
⊆
B
{\displaystyle A\subseteq B}
A
≠
B
{\displaystyle A\neq B}
x
∈
A
⟹
x
∈
B
,
{\displaystyle x\in A\;\implies \;x\in B,}
eta
∃
x
∈
B
,
non
x
∉
A
den
{\displaystyle \exists x\in B,{\mbox{ non }}x\not \in A{\mbox{ den}}}
betetzen badira.
Honela ere esan dezakegu:
A
{\displaystyle A}
B
{\displaystyle B}
propioa da, edo
A
{\displaystyle A}
zeharo dago
B
{\displaystyle B}
B
{\displaystyle B}
A
{\displaystyle A}
zeharo barruan dauka.
3.8 Adibidea.
A
=
{
0
,
1
,
2
}
,
B
=
{
−
2
,
−
1
,
0
,
1
,
2
}
eta
C
=
{
x
/
x
∈
Z
eta
0
≤
x
≤
2
}
{\displaystyle A=\{0,1,2\},\;B=\{-2,-1,0,1,2\}\;{\mbox{ eta }}\;C=\{x\;/\;x\in \mathbb {Z} {\mbox{ eta }}0\leq x\leq 2\}}
A
=
C
⊂
B
{\displaystyle A=C\subset B}
Multzoaren kontzeptua zabalduz, definizio hauek eman ditzakegu:
3.9 Definizioa. multzo huts deituko diogu.
∅
{\displaystyle \emptyset }
Ohar gaitezen
∅
≠
{
∅
}
{\displaystyle \emptyset \neq \{\emptyset \}}
∅
{\displaystyle \emptyset }
{
∅
}
{\displaystyle \{\emptyset \}}
∅
{\displaystyle \emptyset }
∅
≠
0
{\displaystyle \emptyset \neq 0}
∅
{\displaystyle \emptyset }
0
{\displaystyle 0}
∅
≠
{
0
}
{\displaystyle \emptyset \neq \{0\}}
∅
{\displaystyle \emptyset }
{
0
}
{\displaystyle \{0\}}
0
{\displaystyle 0}
Horrez gain,
A
{\displaystyle A}
∅
⊆
A
.
{\displaystyle \emptyset \subseteq A.}
Izan ere,
x
∈
∅
⟹
x
∈
A
{\displaystyle x\in \emptyset \implies x\in A}
betetzen da,
x
∈
∅
{\displaystyle x\in \emptyset }
∅
{\displaystyle \emptyset }
A
{\displaystyle A}
3.10 Definizioa. multzo unibertsal edo erreferentzi multzo deituko diogu.
U
{\displaystyle U}
Esate baterako, zenbaki errealen propietateak aztertzen ari bagara,
R
{\displaystyle \mathbb {R} }
A
{\displaystyle A}
A
⊆
U
{\displaystyle A\subseteq U}
betetzen da. Izan ere,
x
∈
A
⟹
x
∈
U
{\displaystyle x\in A\implies x\in U}
beti betetzen da,
x
∈
U
{\displaystyle x\in U}
3.11 Definizioa.
A
{\displaystyle A}
A
{\displaystyle A}
parteen multzo edo
A
{\displaystyle A}
potentzia multzo deituko diogu elementutzat
A
{\displaystyle A}
P
(
A
)
{\displaystyle {\mathcal {P}}(A)}
3.12 Adibidea.
A
=
{
0
,
1
,
2
}
{\displaystyle A=\{0,1,2\}}
P
(
A
)
=
{
∅
,
{
0
}
,
{
1
}
,
{
2
}
,
{
0
,
1
}
,
{
0
,
2
}
,
{
1
,
2
}
,
A
}
{\displaystyle {\mathcal {P}}(A)=\{\emptyset ,\{0\},\{1\},\{2\},\{0,1\},\{0,2\},\{1,2\},A\}}
izango da.
A
{\displaystyle A}
card
(
A
)
=
n
{\displaystyle {\mbox{card}}(A)=n}
card
(
P
(
A
)
)
=
∑
k
=
0
n
(
n
k
)
=
2
n
{\displaystyle {\mbox{card}}({\mathcal {P}}(A))=\sum _{k=0}^{n}{\binom {n}{k}}=2^{n}}
beteko da.
3.2.2 Multzoen arteko eragiketak [ aldatu ] Izan bitez
A
{\displaystyle A}
B
{\displaystyle B}
3.13 Definizioa.
3.14 Adibidea.
A
=
{
0
,
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
}
{\displaystyle A=\{0,1,2,3,4,5,6,7,8,9,10\}}
B
=
{
0
,
2
,
4
,
6
,
8
,
10
}
{\displaystyle B=\{0,2,4,6,8,10\}}
C
=
{
1
,
3
,
5
,
7
,
9
}
{\displaystyle C=\{1,3,5,7,9\}}
D
=
{
0
,
1
,
2
,
3
,
4
,
5
}
{\displaystyle D=\{0,1,2,3,4,5\}}
U
=
N
{\displaystyle U=\mathbb {N} }
C
A
(
B
)
=
A
∖
B
=
C
{\displaystyle C_{A}(B)=A\setminus B=C}
A
c
=
{
x
/
x
∈
N
eta
x
≥
11
}
{\displaystyle A^{c}=\{x\;/\;x\in \mathbb {N} {\mbox{ eta }}x\geq 11\}}
B
∩
D
=
{
0
,
2
,
4
}
{\displaystyle B\cap D=\{0,2,4\}}
B
{\displaystyle B}
C
{\displaystyle C}
B
∪
D
=
{
0
,
1
,
2
,
3
,
4
,
5
,
6
,
8
,
10
}
{\displaystyle B\cup D=\{0,1,2,3,4,5,6,8,10\}}
B
∪
C
=
B
∪
˙
C
=
A
{\displaystyle B\cup C=B{\dot {\cup }}C=A}
C
∖
D
=
{
7
,
9
}
{\displaystyle C\setminus D=\{7,9\}}
3.15 Propietateak.
Trukatze-legea:
A
∪
B
=
B
∪
A
{\displaystyle A\cup B=B\cup A\quad }
A
∩
B
=
B
∩
A
{\displaystyle A\cap B=B\cap A}
Elkartze-legea:
A
∪
(
B
∪
C
)
=
(
A
∪
B
)
∪
C
{\displaystyle A\cup (B\cup C)=(A\cup B)\cup C\quad }
A
∩
(
B
∩
C
)
=
(
A
∩
B
)
∩
C
{\displaystyle A\cap (B\cap C)=(A\cap B)\cap C}
Banatze-legea:
A
∪
(
B
∩
C
)
=
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle A\cup (B\cap C)=(A\cup B)\cap (A\cup C)\quad }
A
∩
(
B
∪
C
)
=
(
A
∩
B
)
∪
(
A
∩
C
)
{\displaystyle A\cap (B\cup C)=(A\cap B)\cup (A\cap C)}
De Morgan-en legeak:
(
A
∪
B
)
C
=
A
C
∩
B
C
{\displaystyle (A\cup B)^{C}=A^{C}\cap B^{C}\quad }
(
A
∩
B
)
C
=
A
C
∪
B
C
{\displaystyle (A\cap B)^{C}=A^{C}\cup B^{C}}
Idenpotentzia:
A
∪
A
=
A
{\displaystyle A\cup A=A\quad }
A
∩
A
=
A
{\displaystyle A\cap A=A}
A
∪
U
=
U
{\displaystyle A\cup U=U\quad }
A
∩
U
=
A
{\displaystyle A\cap U=A}
A
∪
A
C
=
U
{\displaystyle A\cup A^{C}=U\quad }
A
∩
A
C
=
∅
{\displaystyle A\cap A^{C}=\emptyset }
A
∪
∅
=
A
{\displaystyle A\cup \emptyset =A\quad }
A
∩
∅
=
∅
{\displaystyle A\cap \emptyset =\emptyset \quad }
A
⊆
A
∪
B
{\displaystyle A\subseteq A\cup B\quad }
A
∩
B
⊆
A
{\displaystyle A\cap B\subseteq A}
U
C
=
∅
{\displaystyle U^{C}=\emptyset \quad }
∅
C
=
U
{\displaystyle \emptyset ^{C}=U\quad }
(
A
C
)
C
=
A
{\displaystyle (A^{C})^{C}=A}
Froga.
Propietate horiek lokailu logikoen propietateak erabiliz frogatzen dira.
Adibide gisa, horietako bi frogatuko ditugu.
A
∪
(
B
∩
C
)
=
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle A\cup (B\cap C)=(A\cup B)\cap (A\cup C)}
Multzoen arteko berdintza frogatu behar dugu; hortaz, ondoko hau frogatu beharko dugu :
A
∪
(
B
∩
C
)
⊆
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle A\cup (B\cap C)\subseteq (A\cup B)\cap (A\cup C)}
(
A
∪
B
)
∩
(
A
∪
C
)
⊆
A
∪
(
B
∩
C
)
{\displaystyle (A\cup B)\cap (A\cup C)\subseteq A\cup (B\cap C)}
A
∪
(
B
∩
C
)
⊆
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle A\cup (B\cap C)\subseteq (A\cup B)\cap (A\cup C)}
x
∈
A
∪
(
B
∩
C
)
⟹
x
∈
(
A
∪
B
)
∩
(
A
∪
C
)
.
{\displaystyle x\in A\cup (B\cap C)\implies x\in (A\cup B)\cap (A\cup C).}
x
∈
A
∪
(
B
∩
C
)
⟹
(
x
∈
A
)
∨
(
x
∈
B
∩
C
)
(
∪
bilketaren definizioa)
⟹
{\displaystyle x\in A\cup (B\cap C)\implies (x\in A)\vee (x\in B\cap C){\mbox{ (}}\cup {\mbox{ bilketaren definizioa) }}\implies }
⟹
(
x
∈
A
)
∨
(
x
∈
B
∧
x
∈
C
)
(
∩
ebaketaren definizioa)
⟹
{\displaystyle \implies (x\in A)\vee (x\in B\wedge x\in C){\mbox{ (}}\cap {\mbox{ ebaketaren definizioa) }}\implies }
⟹
(
x
∈
A
∨
x
∈
B
)
∧
(
x
∈
A
∨
x
∈
C
)
(
∨
disjuntzioaren banakortasuna
{\displaystyle \implies (x\in A\vee x\in B)\wedge (x\in A\vee x\in C){\mbox{ (}}\vee {\mbox{ disjuntzioaren banakortasuna }}}
∧
konjuntzioarekiko)
⟹
(
x
∈
A
∪
B
)
∧
(
x
∈
A
∪
C
)
(
∪
bilketaren definizioa)
⟹
{\displaystyle \quad \wedge {\mbox{ konjuntzioarekiko) }}\implies (x\in A\cup B)\wedge (x\in A\cup C){\mbox{ (}}\cup {\mbox{ bilketaren definizioa) }}\implies }
⟹
x
∈
(
A
∪
B
)
∩
(
A
∪
C
)
(
∩
ebaketaren definizioa).
{\displaystyle \implies x\in (A\cup B)\cap (A\cup C){\mbox{ (}}\cap {\mbox{ ebaketaren definizioa). }}}
(
A
∪
B
)
∩
(
A
∪
C
)
⊆
A
∪
(
B
∩
C
)
{\displaystyle (A\cup B)\cap (A\cup C)\subseteq A\cup (B\cap C)}
x
∈
(
A
∪
B
)
∩
(
A
∪
C
)
⟹
(
x
∈
A
∪
B
)
∧
(
x
∈
A
∪
C
)
(
∩
ebaketaren definizioa)
⟹
{\displaystyle x\in (A\cup B)\cap (A\cup C)\implies (x\in A\cup B)\wedge (x\in A\cup C){\mbox{ (}}\cap {\mbox{ ebaketaren definizioa) }}\implies }
⟹
(
x
∈
A
∨
x
∈
B
)
∧
(
x
∈
A
∨
x
∈
C
)
(
∪
bilketaren definizioa)
⟹
{\displaystyle \implies (x\in A\vee x\in B)\wedge (x\in A\vee x\in C){\mbox{ (}}\cup {\mbox{ bilketaren definizioa) }}\implies }
⟹
(
x
∈
A
)
∨
(
x
∈
B
∧
x
∈
C
)
(
∨
disjuntzioaren banakortasuna
{\displaystyle \implies (x\in A)\vee (x\in B\wedge x\in C){\mbox{ (}}\vee {\mbox{ disjuntzioaren banakortasuna }}}
∧
konjuntzioarekiko)
⟹
(
x
∈
A
)
∨
(
x
∈
B
∩
C
)
(
∩
ebaketaren definizioa)
⟹
{\displaystyle {\mbox{ }}\wedge {\mbox{ konjuntzioarekiko) }}\implies (x\in A)\vee (x\in B\cap C){\mbox{ (}}\cap {\mbox{ ebaketaren definizioa) }}\implies }
⟹
x
∈
A
∪
(
B
∩
C
)
(
∪
bilketaren definizioa).
{\displaystyle \implies x\in A\cup (B\cap C){\mbox{ (}}\cup {\mbox{ bilketaren definizioa). }}}
A
⊆
A
∪
B
{\displaystyle A\subseteq A\cup B}
x
∈
A
⟹
(
x
∈
A
)
∨
(
x
∈
B
)
(erantsiz)
⟹
x
∈
A
∪
B
(
∪
bilketaren definizioa).
{\displaystyle x\in A\implies (x\in A)\vee (x\in B){\mbox{ (erantsiz) }}\implies x\in A\cup B{\mbox{ (}}\cup {\mbox{ bilketaren definizioa).}}}
◻
{\displaystyle \Box }
Bilketaren eta ebaketaren elkarkortasunak aukera ematen digu
A
∪
B
∪
C
{\displaystyle A\cup B\cup C}
A
∩
B
∩
C
{\displaystyle A\cap B\cap C}
I
{\displaystyle I}
i
∈
I
{\displaystyle i\in I}
A
i
{\displaystyle A_{i}}
⋃
i
∈
I
A
i
:=
{
x
/
∃
i
∈
I
,
non
x
∈
A
i
}
,
{\displaystyle \bigcup _{i\in I}A_{i}:=\{x\;/\;\exists i\in I,{\mbox{ non }}x\in A_{i}\},}
⋂
i
∈
I
A
i
:=
{
x
/
∀
i
∈
I
x
∈
A
i
}
.
{\displaystyle \bigcap _{i\in I}A_{i}:=\{x\;/\;\forall i\in I\;\;\;x\in A_{i}\}.}
3.16 Adibideak.
Izan bedi
A
i
=
{
1
,
…
,
i
}
{\displaystyle A_{i}=\{1,\ldots ,i\}}
i
∈
N
∗
{\displaystyle i\in \mathbb {N} ^{*}}
⋃
i
∈
N
∗
A
i
=
N
∗
eta
⋂
i
∈
N
∗
A
i
=
{
1
}
.
{\displaystyle \bigcup _{i\in \mathbb {N} ^{*}}A_{i}=\mathbb {N} ^{*}\quad {\mbox{ eta }}\quad \bigcap _{i\in \mathbb {N} ^{*}}A_{i}=\{1\}.}
i
∈
Z
{\displaystyle i\in \mathbb {Z} }
A
i
=
{
i
,
i
+
1
}
{\displaystyle A_{i}=\{i,i+1\}}
⋃
i
∈
Z
A
i
=
Z
eta
⋂
i
∈
Z
A
i
=
∅
.
{\displaystyle \bigcup _{i\in \mathbb {Z} }A_{i}=\mathbb {Z} \quad {\mbox{ eta }}\quad \bigcap _{i\in \mathbb {Z} }A_{i}=\emptyset .}
Froga genitzake De Morganen lege orokortuak
(
⋃
i
∈
I
A
i
)
c
=
⋂
i
∈
I
A
i
c
,
{\displaystyle \left(\bigcup _{i\in I}A_{i}\right)^{c}=\bigcap _{i\in I}A_{i}^{c},}
(
⋂
i
∈
I
A
i
)
c
=
⋃
i
∈
I
A
i
c
.
{\displaystyle \left(\bigcap _{i\in I}A_{i}\right)^{c}=\bigcup _{i\in I}A_{i}^{c}.}
3.2.3 Multzo baten partiketa [ aldatu ] 3.17 Definizioa.
A
{\displaystyle A}
A
{\displaystyle A}
partiketa bat
A
{\displaystyle A}
A
{\displaystyle A}
Multzo baten partiketa
P
=
{
A
i
/
i
∈
I
}
{\displaystyle {\mathcal {P}}=\{A_{i}\;/\;i\in I\}}
A
{\displaystyle A}
(
∀
i
∈
I
)
{\displaystyle (\forall i\in I)}
∅
≠
A
i
⊆
A
{\displaystyle \;\;\emptyset \neq A_{i}\subseteq A}
(
∀
i
,
j
∈
I
)
{\displaystyle (\forall i,j\in I)}
A
i
≠
A
j
⟹
A
i
∩
A
j
=
∅
{\displaystyle \;\;A_{i}\neq A_{j}\implies A_{i}\cap A_{j}=\emptyset }
⋃
˙
i
∈
I
A
i
=
A
{\displaystyle \displaystyle {{\dot {\bigcup }}_{i\in I}A_{i}=A}}
A
i
{\displaystyle A_{i}}
klase deituko diegu.
3.18 Adibideak.
A
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
}
{\displaystyle A=\{1,2,3,4,5,6,7,8,9,10\}}
A
1
=
{
1
,
3
,
5
,
7
,
9
}
{\displaystyle A_{1}=\{1,3,5,7,9\}}
A
2
=
{
2
,
4
,
6
,
8
,
10
}
{\displaystyle A_{2}=\{2,4,6,8,10\}}
A
{\displaystyle A}
P
=
{
A
1
,
A
2
}
{\displaystyle {\mathcal {P}}=\{A_{1},A_{2}\}}
B
1
=
{
1
,
2
,
3
}
{\displaystyle B_{1}=\{1,2,3\}}
B
2
=
{
4
,
6
,
8
,
10
}
{\displaystyle B_{2}=\{4,6,8,10\}}
B
3
=
{
5
,
7
,
9
}
{\displaystyle B_{3}=\{5,7,9\}}
A
{\displaystyle A}
P
′
=
{
B
1
,
B
2
,
B
3
}
{\displaystyle {\mathcal {P}}'=\{B_{1},B_{2},B_{3}\}}
P
=
{
Z
−
,
{
0
}
,
Z
+
}
{\displaystyle {\mathcal {P}}=\{\mathbb {Z} ^{-},\{0\},\mathbb {Z} ^{+}\}}
Z
{\displaystyle \mathbb {Z} }
Definizioaren ondorioa
A
{\displaystyle A}
P
=
{
A
i
/
i
∈
I
}
{\displaystyle {\mathcal {P}}=\{A_{i}\;/\;i\in I\}}
A
{\displaystyle A}
P
{\displaystyle {\mathcal {P}}}
∀
x
∈
A
∃
!
A
i
∈
P
,
non
x
∈
A
i
.
{\displaystyle \forall x\in A\quad \exists !A_{i}\in {\mathcal {P}},{\mbox{ non }}x\in A_{i}.}
3.2.4 Bi multzoren arteko biderketa kartesiarra [ aldatu ] 3.19 Definizioa.
a
{\displaystyle a}
b
{\displaystyle b}
a
{\displaystyle a}
b
{\displaystyle b}
(
a
,
b
)
{\displaystyle (a,b)}
bikote ordenatu deituko diogu.
Antzeko eran definitzen dira
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
(
a
,
b
,
c
,
d
)
{\displaystyle (a,b,c,d)}
(
a
1
,
…
,
a
n
)
{\displaystyle (a_{1},\ldots ,a_{n})}
n
{\displaystyle n}
Beraz,
(
a
,
b
)
≠
(
b
,
a
)
{\displaystyle (a,b)\neq (b,a)}
{
a
,
b
}
=
{
b
,
a
}
{\displaystyle \{a,b\}=\{b,a\}}
Bestalde,
(
a
,
b
)
{\displaystyle (a,b)}
(
c
,
d
)
{\displaystyle (c,d)}
berdinak dira
a
=
c
{\displaystyle a=c}
b
=
d
{\displaystyle b=d}
3.20 Definizioa.
A
{\displaystyle A}
B
{\displaystyle B}
A
{\displaystyle A}
B
{\displaystyle B}
biderkadura kartesiar deituko diogu, eta
A
×
B
{\displaystyle A\times B}
A
{\displaystyle A}
B
{\displaystyle B}
A
×
B
:=
{
(
a
,
b
)
/
a
∈
A
,
b
∈
B
}
.
{\displaystyle A\times B:=\{(a,b)\;/\;a\in A,\;b\in B\}.}
Antzeko eran definitzen da
A
1
,
…
,
A
n
{\displaystyle A_{1},\ldots ,A_{n}}
n
{\displaystyle n}
A
1
×
⋯
×
A
n
:=
{
(
x
1
,
…
,
x
n
)
/
x
i
∈
A
i
,
i
=
1
,
…
,
n
}
.
{\displaystyle A_{1}\times \cdots \times A_{n}:=\{(x_{1},\ldots ,x_{n})\;/\;x_{i}\in A_{i},\;i=1,\ldots ,n\}.}
Idazkera.
A
n
=
A
×
⋯
n
×
A
{\displaystyle A^{n}=A\times {\stackrel {n}{\cdots }}\times A}
3.21 Adibideak.
A
=
{
a
,
b
}
,
B
=
{
1
,
2
,
3
}
,
C
=
{
a
}
{\displaystyle A=\{a,b\},B=\{1,2,3\},C=\{a\}}
A
×
B
=
{
(
a
,
1
)
,
(
a
,
2
)
,
(
a
,
3
)
,
(
b
,
1
)
,
(
b
,
2
)
,
(
b
,
3
)
}
{\displaystyle A\times B=\{(a,1),(a,2),(a,3),(b,1),(b,2),(b,3)\}}
B
×
A
=
{
(
1
,
a
)
,
(
1
,
b
)
,
(
2
,
a
)
,
(
2
,
b
)
,
(
3
,
a
)
,
(
3
,
b
)
}
{\displaystyle B\times A=\{(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)\}}
A
×
B
×
C
=
{
(
a
,
1
,
a
)
,
(
a
,
2
,
a
)
,
(
a
,
3
,
a
)
,
(
b
,
1
,
a
)
,
(
b
,
2
,
a
)
,
(
b
,
3
,
a
)
}
{\displaystyle A\times B\times C=\{(a,1,a),(a,2,a),(a,3,a),(b,1,a),(b,2,a),(b,3,a)\}}
B
2
=
{
(
1
,
1
)
,
(
1
,
2
)
,
(
1
,
3
)
,
(
2
,
1
)
,
(
2
,
2
)
,
(
2
,
3
)
,
(
3
,
1
)
,
(
3
,
2
)
,
(
3
,
3
)
}
{\displaystyle B^{2}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}}
A
3
=
{
(
a
,
a
,
a
)
,
(
a
,
a
,
b
)
,
(
a
,
b
,
a
)
,
(
a
,
b
,
b
)
,
(
b
,
a
,
a
)
,
(
b
,
a
,
b
)
,
(
b
,
b
,
a
)
,
(
b
,
b
,
b
)
}
{\displaystyle A^{3}=\{(a,a,a),(a,a,b),(a,b,a),(a,b,b),(b,a,a),(b,a,b),(b,b,a),(b,b,b)\}}
C
7
=
{
(
a
,
a
,
a
,
a
,
a
,
a
,
a
)
}
{\displaystyle C^{7}=\{(a,a,a,a,a,a,a)\}}
3.22 Definizioa.
A
{\displaystyle A}
A
{\displaystyle A}
erlazio bitar bat
A
×
A
{\displaystyle A\times A}
biderkadura kartesiarraren
R
{\displaystyle {\mathcal {R}}}
R
⊆
A
×
A
.
{\displaystyle {\mathcal {R}}\subseteq A\times A.}
(
a
,
b
)
∈
R
{\displaystyle (a,b)\in {\mathcal {R}}}
a
{\displaystyle a}
b
{\displaystyle b}
R
{\displaystyle {\mathcal {R}}}
erlazionaturik
a
R
b
{\displaystyle a{\mathcal {R}}b}
(
a
,
b
)
∉
R
{\displaystyle (a,b)\notin {\mathcal {R}}}
a
{\displaystyle a}
ez dagoela
b
{\displaystyle b}
R
{\displaystyle {\mathcal {R}}}
erlazionaturik
a
R̸
b
{\displaystyle a{\mathcal {\not R}}b}
3.23 Adibideak.
A
=
{
1
,
2
,
3
,
4
}
{\displaystyle A=\{1,2,3,4\}}
R
1
=
{
(
1
,
1
)
,
(
1
,
2
)
,
(
1
,
3
)
,
(
2
,
1
)
,
(
2
,
2
)
,
(
2
,
3
)
,
(
3
,
1
)
,
(
3
,
2
)
,
(
3
,
3
)
,
(
4
,
4
)
}
.
{\displaystyle {\mathcal {R}}_{1}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,4)\}.}
A
=
Z
{\displaystyle A=\mathbb {Z} }
R
2
=
{
(
x
,
y
)
/
x
,
y
∈
Z
eta
x
≤
y
}
.
{\displaystyle {\mathcal {R}}_{2}=\{(x,y)\;/\;x,y\in \mathbb {Z} {\mbox{ eta }}x\leq y\}.}
x
,
y
∈
Z
{\displaystyle x,y\in \mathbb {Z} }
x
R
2
y
⟺
x
≤
y
{\displaystyle x{\mathcal {R}}_{2}y\iff x\leq y}
A
=
N
∗
{\displaystyle A=\mathbb {N} ^{*}}
R
3
=
{
(
x
,
y
)
/
x
,
y
∈
N
∗
eta
x
y
-ren zatitzailea da
}
.
{\displaystyle {\mathcal {R}}_{3}=\{(x,y)\;/\;x,y\in \mathbb {N} ^{*}{\mbox{ eta }}x\;y{\mbox{-ren zatitzailea da}}\}.}
x
,
y
∈
N
∗
{\displaystyle x,y\in \mathbb {N} ^{*}}
x
R
3
y
⟺
x
y
-ren zatitzailea da.
{\displaystyle x{\mathcal {R}}_{3}y\iff x\;y{\mbox{-ren zatitzailea da.}}}
x
∣
y
{\displaystyle x\mid y}
A
=
N
∗
{\displaystyle A=\mathbb {N} ^{*}}
R
4
=
{
(
1
,
1
)
,
(
2
,
2
)
}
.
{\displaystyle {\mathcal {R}}_{4}=\{(1,1),(2,2)\}.}
A
=
Z
{\displaystyle A=\mathbb {Z} }
x
,
y
∈
Z
{\displaystyle x,y\in \mathbb {Z} }
x
R
5
y
⟺
|
x
|
=
|
y
|
.
{\displaystyle x{\mathcal {R}}_{5}y\iff \vert x\vert =\vert y\vert .}
R
5
=
{
(
x
,
y
)
/
x
,
y
∈
Z
eta
|
x
|
=
|
y
|
}
.
{\displaystyle {\mathcal {R}}_{5}=\{(x,y)\;/\;x,y\in \mathbb {Z} {\mbox{ eta }}\vert x\vert =\vert y\vert \}.}
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
|
x
|
{\displaystyle \vert x\vert }
x
{\displaystyle x}
3.24 Definizioa.
A
{\displaystyle A}
R
{\displaystyle {\mathcal {R}}}
R
{\displaystyle {\mathcal {R}}}
bihurkorra dela edo bihurtze-propietatea betetzen duela esango dugu
(
∀
x
∈
A
)
(
x
,
x
)
∈
R
bada,
{\displaystyle (\forall x\in A)\quad (x,x)\in {\mathcal {R}}{\mbox{ bada,}}}
(
∀
x
∈
A
)
x
R
x
bada
.
{\displaystyle (\forall x\in A)\quad x{\mathcal {R}}x{\mbox{ bada}}.}
R
{\displaystyle {\mathcal {R}}}
simetrikoa dela edo simetria-propietatea betetzen duela esango dugu
(
∀
x
,
y
∈
A
)
(
x
,
y
)
∈
R
⟹
(
y
,
x
)
∈
R
bada,
{\displaystyle (\forall x,y\in A)\quad (x,y)\in {\mathcal {R}}\,\implies \,(y,x)\in {\mathcal {R}}{\mbox{ bada,}}}
(
∀
x
,
y
∈
A
)
x
R
y
⟹
y
R
x
bada
.
{\displaystyle (\forall x,y\in A)\quad x{\mathcal {R}}y\,\implies \,y{\mathcal {R}}x{\mbox{ bada}}.}
R
{\displaystyle {\mathcal {R}}}
antisimetrikoa dela edo antisimetria-propietatea betetzen duela esango dugu
(
∀
x
,
y
∈
A
)
(
x
,
y
)
∈
R
eta
(
y
,
x
)
∈
R
⟹
x
=
y
bada,
{\displaystyle (\forall x,y\in A)\quad (x,y)\in {\mathcal {R}}\,{\mbox{ eta }}\,(y,x)\in {\mathcal {R}}\,\implies \,x=y{\mbox{ bada,}}}
(
∀
x
,
y
∈
A
)
x
R
y
eta
y
R
x
⟹
x
=
y
bada
.
{\displaystyle (\forall x,y\in A)\quad x{\mathcal {R}}y\,{\mbox{ eta }}\,y{\mathcal {R}}x\,\implies \,x=y{\mbox{ bada}}.}
R
{\displaystyle {\mathcal {R}}}
iragankorra dela edo iragate-propietatea betetzen duela esango dugu
(
∀
x
,
y
,
z
∈
A
)
(
x
,
y
)
∈
R
eta
(
y
,
z
)
∈
R
⟹
(
x
,
z
)
∈
R
bada,
{\displaystyle (\forall x,y,z\in A)\quad (x,y)\in {\mathcal {R}}\,{\mbox{ eta }}\,(y,z)\in {\mathcal {R}}\,\implies \,(x,z)\in {\mathcal {R}}{\mbox{ bada,}}}
(
∀
x
,
y
,
z
∈
A
)
x
R
y
eta
y
R
z
⟹
x
R
z
bada
.
{\displaystyle (\forall x,y,z\in A)\quad x{\mathcal {R}}y\,{\mbox{ eta }}\,y{\mathcal {R}}z\,\implies \,x{\mathcal {R}}z{\mbox{ bada}}.}
3.25 Adibideak.
A
=
{
1
,
2
,
3
,
4
}
{\displaystyle A=\{1,2,3,4\}}
R
1
=
{
(
1
,
1
)
,
(
1
,
2
)
,
(
1
,
3
)
,
(
2
,
1
)
,
(
2
,
2
)
,
(
2
,
3
)
,
(
3
,
1
)
,
(
3
,
2
)
,
(
3
,
3
)
,
(
4
,
4
)
}
{\displaystyle {\mathcal {R}}_{1}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,4)\}}
R
2
=
{
(
x
,
y
)
/
x
,
y
∈
Z
eta
x
≤
y
}
{\displaystyle {\mathcal {R}}_{2}=\{(x,y)\;/\;x,y\in \mathbb {Z} {\mbox{ eta }}x\leq y\}}
Z
{\displaystyle \mathbb {Z} }
x
R
2
y
{\displaystyle x{\mathcal {R}}_{2}y}
x
≤
y
{\displaystyle x\leq y}
Bihurkorra da:
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
x
≤
x
{\displaystyle x\leq x}
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
x
R
2
x
{\displaystyle x{\mathcal {R}}_{2}x}
Ez da simetrikoa:
0
R
2
7
{\displaystyle 0{\mathcal {R}}_{2}7}
7
R̸
2
0
{\displaystyle 7{\mathcal {\not R}}_{2}0}
Antisimetrikoa da:
x
,
y
∈
Z
{\displaystyle x,y\in \mathbb {Z} }
x
R
2
y
eta
y
R
2
x
⟹
x
≤
y
eta
y
≤
x
⟹
x
=
y
.
{\displaystyle x{\mathcal {R}}_{2}y{\mbox{ eta }}y{\mathcal {R}}_{2}x\implies x\leq y{\mbox{ eta }}y\leq x\implies x=y.}
Iragankorra da:
x
,
y
,
z
∈
Z
{\displaystyle x,y,z\in \mathbb {Z} }
x
R
2
y
eta
y
R
2
z
⟹
x
≤
y
eta
y
≤
z
⟹
x
≤
z
⟹
x
R
2
z
.
{\displaystyle x{\mathcal {R}}_{2}y{\mbox{ eta }}y{\mathcal {R}}_{2}z\implies x\leq y{\mbox{ eta }}y\leq z\implies x\leq z\implies x{\mathcal {R}}_{2}z.}
R
3
=
{
(
x
,
y
)
/
x
,
y
∈
N
∗
eta
x
∣
y
}
,
{\displaystyle {\mathcal {R}}_{3}=\{(x,y)\;/\;x,y\in \mathbb {N} ^{*}{\mbox{ eta }}x\mid y\},}
zatigarritasunaren definizioaren arabera,
N
∗
{\displaystyle \mathbb {N} ^{*}}
x
R
3
y
{\displaystyle x{\mathcal {R}}_{3}y}
∃
k
∈
Z
{\displaystyle \exists k\in \mathbb {Z} }
y
=
k
x
{\displaystyle y=kx}
Bihurkorra da:
x
∈
N
∗
{\displaystyle x\in \mathbb {N} ^{*}}
x
R
3
x
{\displaystyle x{\mathcal {R}}_{3}x}
x
∣
x
{\displaystyle x\mid x}
k
=
1
∈
Z
{\displaystyle k=1\in \mathbb {Z} }
x
=
1
⋅
x
{\displaystyle x=1\cdot x}
Ez da simetrikoa:
2
R
3
8
{\displaystyle 2{\mathcal {R}}_{3}8}
8
R̸
3
2
{\displaystyle 8{\mathcal {\not R}}_{3}2}
2
{\displaystyle 2}
8
{\displaystyle 8}
8
{\displaystyle 8}
2
{\displaystyle 2}
∃
k
=
4
∈
Z
,
non
8
=
4
⋅
2
den, baina
∄
k
∈
Z
,
non
2
=
8
k
den
.
{\displaystyle \exists k=4\in \mathbb {Z} ,\;{\mbox{ non }}\;8=4\cdot 2{\mbox{ den, baina }}\not \exists k\in \mathbb {Z} ,\;{\mbox{ non }}\;2=8k{\mbox{ den}}.}
Antisimetrikoa da:
x
,
y
∈
N
∗
{\displaystyle x,y\in \mathbb {N} ^{*}}
x
R
3
y
eta
y
R
3
x
{\displaystyle x{\mathcal {R}}_{3}y{\mbox{ eta }}y{\mathcal {R}}_{3}x}
x
=
y
{\displaystyle x=y}
x
R
3
y
y
R
3
x
}
⟹
x
∣
y
y
∣
x
}
⟹
y
=
k
x
,
k
∈
Z
x
=
l
y
,
l
∈
Z
}
{\displaystyle \left.{\begin{array}{l}x{\mathcal {R}}_{3}y\\y{\mathcal {R}}_{3}x\end{array}}\right\}\implies \left.{\begin{array}{l}x\mid y\\y\mid x\end{array}}\right\}\implies \left.{\begin{array}{l}y=kx,k\in \mathbb {Z} \\x=ly,l\in \mathbb {Z} \end{array}}\right\}}
x
=
l
k
x
,
l
k
∈
Z
⟹
l
k
=
1.
{\displaystyle x=lkx,\;lk\in \mathbb {Z} \implies lk=1.}
l
,
k
∈
Z
{\displaystyle l,k\in \mathbb {Z} }
l
=
k
=
1
{\displaystyle l=k=1}
l
=
k
=
−
1
{\displaystyle l=k=-1}
x
=
y
{\displaystyle x=y}
x
=
−
y
.
{\displaystyle x=-y.}
x
,
y
∈
N
∗
⟹
x
,
y
>
0
⟹
x
=
y
.
{\displaystyle x,y\in \mathbb {N} ^{*}\implies x,y>0\implies x=y.}
Iragankorra da:
x
,
y
,
z
∈
N
∗
{\displaystyle x,y,z\in \mathbb {N} ^{*}}
x
R
3
y
y
R
3
z
}
⟹
x
∣
y
y
∣
z
}
⟹
y
=
k
x
,
k
∈
Z
z
=
l
y
,
l
∈
Z
}
{\displaystyle \left.{\begin{array}{l}x{\mathcal {R}}_{3}y\\y{\mathcal {R}}_{3}z\end{array}}\right\}\implies \left.{\begin{array}{l}x\mid y\\y\mid z\end{array}}\right\}\implies \left.{\begin{array}{l}y=kx,k\in \mathbb {Z} \\z=ly,l\in \mathbb {Z} \end{array}}\right\}}
z
=
l
k
x
,
l
k
∈
Z
⟹
x
∣
z
⟹
x
R
3
z
.
{\displaystyle z=lkx,\;lk\in \mathbb {Z} \implies x\mid z\implies x{\mathcal {R}}_{3}z.}
A
=
N
∗
{\displaystyle A=\mathbb {N} ^{*}}
R
4
=
{
(
1
,
1
)
,
(
2
,
2
)
}
{\displaystyle {\mathcal {R}}_{4}=\{(1,1),(2,2)\}}
A
=
Z
{\displaystyle A=\mathbb {Z} }
x
R
5
y
⟺
|
x
|
=
|
y
|
{\displaystyle x{\mathcal {R}}_{5}y\iff \vert x\vert =\vert y\vert }
3.3.2 Ordena-erlazioak [ aldatu ] 3.26 Definizioa.
A
{\displaystyle A}
R
{\displaystyle {\mathcal {R}}}
ordena-erlazioa da bihurkorra, antisimetrikoa eta iragankorra bada.
A
{\displaystyle A}
R
{\displaystyle {\mathcal {R}}}
R
{\displaystyle {\mathcal {R}}}
A
{\displaystyle A}
A
{\displaystyle A}
multzo ordenatua dela esango dugu.
3.27 Definizioa.
A
{\displaystyle A}
R
{\displaystyle {\mathcal {R}}}
ordena osoko erlazioa dela (ordena totala ) esango dugu
(
∀
x
,
y
∈
A
)
(
x
,
y
)
∈
R
{\displaystyle (\forall x,y\in A)\quad (x,y)\in {\mathcal {R}}\,}
(
y
,
x
)
∈
R
{\displaystyle \,(y,x)\in {\mathcal {R}}}
edo baliokidea dena,
(
∀
x
,
y
∈
A
)
x
R
y
{\displaystyle (\forall x,y\in A)\quad x{\mathcal {R}}y\,}
y
R
x
{\displaystyle \,y{\mathcal {R}}x}
Kasu horretan,
R
{\displaystyle {\mathcal {R}}}
A
{\displaystyle A}
osoki ordenatzen duela edo
A
{\displaystyle A}
multzo osoki ordenatua dela esango dugu.
Bestela,
R
{\displaystyle {\mathcal {R}}}
ordena partzialeko erlazioa izango da, eta
R
{\displaystyle {\mathcal {R}}}
A
{\displaystyle A}
partzialki ordenatzen duela edo
A
{\displaystyle A}
multzo partzialki ordenatua dela esango dugu.
3.28 Adibideak.
R
2
=
{
(
x
,
y
)
/
x
,
y
∈
Z
eta
x
≤
y
}
{\displaystyle {\mathcal {R}}_{2}=\{(x,y)\;/\;x,y\in \mathbb {Z} {\mbox{ eta }}x\leq y\}}
Z
{\displaystyle \mathbb {Z} }
R
2
{\displaystyle {\mathcal {R}}_{2}}
Z
{\displaystyle \mathbb {Z} }
x
,
y
∈
Z
{\displaystyle x,y\in \mathbb {Z} }
x
≤
y
{\displaystyle x\leq y}
y
≤
x
{\displaystyle y\leq x}
R
3
=
{
(
x
,
y
)
/
x
,
y
∈
N
∗
eta
x
∣
y
}
{\displaystyle {\mathcal {R}}_{3}=\{(x,y)\;/\;x,y\in \mathbb {N} ^{*}{\mbox{ eta }}x\mid y\}}
N
∗
{\displaystyle \mathbb {N} ^{*}}
R
3
{\displaystyle {\mathcal {R}}_{3}}
N
∗
{\displaystyle \mathbb {N} ^{*}}
3
R̸
3
7
{\displaystyle 3{\mathcal {\not R}}_{3}7}
7
R̸
3
3
{\displaystyle 7{\mathcal {\not R}}_{3}3}
3
{\displaystyle 3}
7
{\displaystyle 7}
3.3.3 Baliokidetasun-erlazioak [ aldatu ] 3.29 Definizioa.
A
{\displaystyle A}
R
{\displaystyle {\mathcal {R}}}
baliokidetasun-erlazioa da bihurkorra, simetrikoa eta iragankorra bada.
3.30 Adibideak.
R
1
{\displaystyle {\mathcal {R}}_{1}}
A
=
{
1
,
2
,
3
,
4
}
{\displaystyle A=\{1,2,3,4\}}
R
1
=
{
(
1
,
1
)
,
(
1
,
2
)
,
(
1
,
3
)
,
(
2
,
1
)
,
(
2
,
2
)
,
(
2
,
3
)
,
(
3
,
1
)
,
(
3
,
2
)
,
(
3
,
3
)
,
(
4
,
4
)
}
{\displaystyle {\mathcal {R}}_{1}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,4)\}}
Grafo moduan adieraz daiteke, non
A
{\displaystyle A}
R
1
{\displaystyle {\mathcal {R}}_{1}}
Z
{\displaystyle \mathbb {Z} }
x
R
5
y
⟺
|
x
|
=
|
y
|
{\displaystyle x{\mathcal {R}}_{5}y\iff \vert x\vert =\vert y\vert }
3.31 Definizioa.
R
{\displaystyle {\mathcal {R}}}
A
{\displaystyle A}
a
∈
A
{\displaystyle a\in A}
a
{\displaystyle a}
A
{\displaystyle A}
a
{\displaystyle a}
baliokidetasun-klase deituko diogu, eta
[
a
]
R
{\displaystyle [a]_{\mathcal {R}}}
[
a
]
R
:=
{
x
∈
A
/
(
x
,
a
)
∈
R
}
=
{
x
∈
A
/
x
R
a
}
.
{\displaystyle {\begin{array}{ll}[a]_{\mathcal {R}}&:=\{x\in A\;/\;(x,a)\in {\mathcal {R}}\}\\&\\&=\{x\in A\;/\;x{\mathcal {R}}a\}.\end{array}}}
Ez badu nahasmenik sortzen,
[
a
]
{\displaystyle [a]}
[
a
]
R
{\displaystyle [a]_{\mathcal {R}}}
3.32 Adibideak.
A
=
{
1
,
2
,
3
,
4
}
{\displaystyle A=\{1,2,3,4\}}
R
1
=
{
(
1
,
1
)
,
(
1
,
2
)
,
(
1
,
3
)
,
(
2
,
1
)
,
(
2
,
2
)
,
(
2
,
3
)
,
(
3
,
1
)
,
(
3
,
2
)
,
(
3
,
3
)
,
(
4
,
4
)
}
{\displaystyle {\mathcal {R}}_{1}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,4)\}}
[
1
]
=
[
2
]
=
[
3
]
=
{
1
,
2
,
3
}
eta
[
4
]
=
{
4
}
.
{\displaystyle [1]=[2]=[3]=\{1,2,3\}\quad {\mbox{eta}}\quad [4]=\{4\}.}
R
5
{\displaystyle {\mathcal {R}}_{5}}
Z
{\displaystyle \mathbb {Z} }
x
R
5
y
⟺
|
x
|
=
|
y
|
,
{\displaystyle x{\mathcal {R}}_{5}y\iff \vert x\vert =\vert y\vert ,}
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
[
x
]
=
{
−
x
,
x
}
dugu
.
{\displaystyle [x]=\{-x,x\}\,{\mbox{ dugu}}.}
3.33 Teorema.
A
{\displaystyle A}
R
{\displaystyle {\mathcal {R}}}
(
∀
x
∈
A
)
x
∈
[
x
]
.
{\displaystyle (\forall x\in A)\quad x\in [x].}
(
∀
x
,
y
∈
A
)
x
R
y
⟺
[
x
]
=
[
y
]
.
{\displaystyle (\forall x,y\in A)\quad x{\mathcal {R}}y\,\iff \,[x]=[y].}
(
∀
x
,
y
∈
A
)
[
x
]
≠
[
y
]
⟹
[
x
]
∩
[
y
]
=
∅
.
{\displaystyle (\forall x,y\in A)\quad [x]\neq [y]\,\implies \,[x]\cap [y]=\emptyset .}
⋃
x
∈
A
[
x
]
=
A
.
{\displaystyle \bigcup _{x\in A}[x]=A.}
Froga.
R
{\displaystyle {\mathcal {R}}}
x
∈
A
{\displaystyle x\in A}
x
R
x
{\displaystyle x{\mathcal {R}}x}
[
x
]
{\displaystyle [x]}
x
∈
[
x
]
{\displaystyle x\in [x]}
x
,
y
∈
A
{\displaystyle x,y\in A}
x
R
y
⟺
[
x
]
=
[
y
]
{\displaystyle x{\mathcal {R}}y\,\iff \,[x]=[y]}
⟹
{\displaystyle \implies }
Demagun
x
R
y
{\displaystyle x{\mathcal {R}}y}
[
x
]
=
[
y
]
{\displaystyle [x]=[y]}
[
x
]
⊆
[
y
]
{\displaystyle [x]\subseteq [y]}
[
x
]
⊇
[
y
]
{\displaystyle [x]\supseteq [y]}
⊆
{\displaystyle \subseteq }
z
∈
[
x
]
⟹
z
R
x
(
[
x
]
-aren definizioz
)
⟹
{
z
R
x
x
R
y
(hipotesiz)
}
⟹
⟹
z
R
y
(
R
iragankorra delako
)
⟹
z
∈
[
y
]
(
[
y
]
-aren definizioz
)
.
{\displaystyle {\begin{array}{l}z\in [x]\implies z{\mathcal {R}}x\,([x]{\mbox{-aren definizioz}})\implies \left\{{\begin{array}{l}z{\mathcal {R}}x\\x{\mathcal {R}}y{\mbox{ (hipotesiz) }}\end{array}}\right\}\implies \\\\\implies z{\mathcal {R}}y\,({\mathcal {R}}{\mbox{ iragankorra delako}})\implies z\in [y]\,([y]{\mbox{-aren definizioz}}).\end{array}}}
⊇
{\displaystyle \supseteq }
z
∈
[
y
]
⟹
z
R
y
(
[
y
]
-aren definizioz
)
⟹
{
z
R
y
x
R
y
(hipotesiz)
}
⟹
⟹
{
z
R
y
y
R
x
(
R
simetrikoa delako
)
}
⟹
z
R
x
(
R
iragankorra delako
)
⟹
⟹
z
∈
[
x
]
(
[
x
]
-aren definizioz
)
.
{\displaystyle {\begin{array}{l}z\in [y]\implies z{\mathcal {R}}y\,([y]{\mbox{-aren definizioz}})\implies \left\{{\begin{array}{l}z{\mathcal {R}}y\\x{\mathcal {R}}y{\mbox{ (hipotesiz)}}\end{array}}\right\}\implies \\\\\implies \left\{{\begin{array}{l}z{\mathcal {R}}y\\y{\mathcal {R}}x\,({\mathcal {R}}{\mbox{ simetrikoa delako}})\end{array}}\right\}\implies z{\mathcal {R}}x\,({\mathcal {R}}{\mbox{ iragankorra delako}})\implies \\\\\implies z\in [x]\,([x]{\mbox{-aren definizioz}}).\end{array}}}
⟸
{\displaystyle \Longleftarrow }
Demagun
[
x
]
=
[
y
]
{\displaystyle [x]=[y]}
x
R
y
{\displaystyle x{\mathcal {R}}y}
1.aren arabera
x
∈
[
x
]
{\displaystyle x\in [x]}
[
x
]
=
[
y
]
{\displaystyle [x]=[y]}
x
∈
[
y
]
{\displaystyle x\in [y]}
[
y
]
{\displaystyle [y]}
x
R
y
{\displaystyle x{\mathcal {R}}y}
Edozein
x
,
y
∈
A
{\displaystyle x,y\in A}
[
x
]
≠
[
y
]
{\displaystyle [x]\neq [y]}
[
x
]
∩
[
y
]
=
∅
{\displaystyle [x]\cap [y]=\emptyset }
Absurdora eramanez egingo dugu,
[
x
]
∩
[
y
]
≠
∅
{\displaystyle [x]\cap [y]\neq \emptyset }
[
x
]
∩
[
y
]
≠
∅
⟹
∃
z
∈
[
x
]
∩
[
y
]
⟹
{
z
∈
[
x
]
⟹
z
R
x
⟹
x
R
z
z
∈
[
y
]
⟹
z
R
y
}
⟹
x
R
y
⟹
[
x
]
=
[
y
]
(2.aren arabera)
.
{\displaystyle [x]\cap [y]\neq \emptyset \implies \exists z\in [x]\cap [y]\implies \left\{{\begin{array}{l}z\in [x]\implies z{\mathcal {R}}x\implies x{\mathcal {R}}z\\z\in [y]\implies z{\mathcal {R}}y\end{array}}\right\}\implies x{\mathcal {R}}y\implies [x]=[y]{\mbox{ (2.aren arabera)}}.}
[
x
]
≠
[
y
]
{\displaystyle [x]\neq [y]}
[
x
]
∩
[
y
]
≠
∅
{\displaystyle [x]\cap [y]\neq \emptyset }
[
x
]
∩
[
y
]
=
∅
{\displaystyle [x]\cap [y]=\emptyset }
⊆
{\displaystyle \subseteq }
Edozein
x
∈
A
{\displaystyle x\in A}
[
x
]
⊆
A
{\displaystyle [x]\subseteq A}
⋃
x
∈
A
[
x
]
⊆
A
{\displaystyle \bigcup _{x\in A}[x]\subseteq A}
⊇
{\displaystyle \supseteq }
z
∈
A
⟹
z
∈
[
z
]
(1.aren arabera)
⟹
z
∈
⋃
x
∈
A
[
x
]
(
∪
-ren definizioz
)
.
{\displaystyle z\in A\implies z\in [z]{\mbox{ (1.aren arabera) }}\implies z\in \bigcup _{x\in A}[x]\,(\cup {\mbox{-ren definizioz}}).}
◻
{\displaystyle \Box }
3.34 Korolarioa.
R
{\displaystyle {\mathcal {R}}}
A
{\displaystyle A}
P
=
{
[
x
]
/
x
∈
A
}
,
{\displaystyle {\mathcal {P}}=\{[x]\,/\,x\in A\},}
A
{\displaystyle A}
Froga.
3.32 Teoremaren 1, 3 eta 4 propietateak
P
{\displaystyle {\mathcal {P}}}
3.17 Definizioa. Partiketa ).
◻
{\displaystyle \Box }
Partiketaren definizioaren ondorioz,
A
{\displaystyle A}
∀
x
∈
A
∃
!
[
a
]
∈
P
/
x
∈
[
a
]
,
{\displaystyle \forall x\in A\quad \exists ![a]\in {\mathcal {P}}\quad /\quad x\in [a],}
eta
[
a
]
{\displaystyle [a]}
ordezkari deituko diogu.
3.35 Definizioa.
R
{\displaystyle {\mathcal {R}}}
A
{\displaystyle A}
A
{\displaystyle A}
gain
R
{\displaystyle {\mathcal {R}}}
zatidura multzo deituko diogu, eta
A
/
R
{\displaystyle A/{\mathcal {R}}}
A
/
R
:=
{
[
x
]
/
x
∈
A
}
.
{\displaystyle A/{\mathcal {R}}:=\{[x]\;/\;x\in A\}.}
3.36 Adibideak.
R
1
{\displaystyle {\mathcal {R}}_{1}}
A
=
{
1
,
2
,
3
,
4
}
{\displaystyle A=\{1,2,3,4\}}
R
1
=
{
(
1
,
1
)
,
(
1
,
2
)
,
(
1
,
3
)
,
(
2
,
1
)
,
(
2
,
2
)
,
(
2
,
3
)
,
(
3
,
1
)
,
(
3
,
2
)
,
(
3
,
3
)
,
(
4
,
4
)
}
,
{\displaystyle {\mathcal {R}}_{1}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,4)\},}
[
1
]
{\displaystyle [1]}
1
{\displaystyle 1}
2
{\displaystyle 2}
3
{\displaystyle 3}
[
4
]
{\displaystyle [4]}
4
{\displaystyle 4}
A
/
R
1
=
{
[
1
]
,
[
4
]
}
=
{
[
2
]
,
[
4
]
}
=
{
[
3
]
,
[
4
]
}
.
{\displaystyle A/{\mathcal {R}}_{1}=\{[1],[4]\}=\{[2],[4]\}=\{[3],[4]\}.}
R
5
{\displaystyle {\mathcal {R}}_{5}}
Z
{\displaystyle \mathbb {Z} }
x
R
5
y
⟺
|
x
|
=
|
y
|
,
{\displaystyle x{\mathcal {R}}_{5}y\iff \vert x\vert =\vert y\vert ,}
x
∈
Z
∗
{\displaystyle x\in \mathbb {Z} ^{*}}
[
x
]
{\displaystyle [x]}
x
{\displaystyle x}
−
x
{\displaystyle -x}
[
0
]
{\displaystyle [0]}
0
{\displaystyle 0}
Z
/
R
5
=
{
[
x
]
/
x
∈
Z
}
=
{
[
0
]
}
∪
{
[
x
]
/
x
∈
Z
+
}
=
{
[
0
]
}
∪
{
[
x
]
/
x
∈
Z
−
}
.
{\displaystyle \mathbb {Z} /{\mathcal {R}}_{5}=\{[x]\;/\;x\in \mathbb {Z} \}=\{[0]\}\cup \{[x]\;/\;x\in \mathbb {Z} ^{+}\}=\{[0]\}\cup \{[x]\;/\;x\in \mathbb {Z} ^{-}\}.}
3.3.4 n [ aldatu ] 3.37 Definizioa.
n
∈
Z
{\displaystyle n\in \mathbb {Z} }
n
>
1
{\displaystyle n>1}
a
,
b
∈
Z
{\displaystyle a,b\in \mathbb {Z} }
a
{\displaystyle a}
b
{\displaystyle b}
kongruentea da modulu
n
{\displaystyle n}
a
≡
b
(
mod
n
)
{\displaystyle a\equiv b{\pmod {n}}}
n
∣
a
−
b
{\displaystyle n\mid a-b}
∃
k
∈
Z
{\displaystyle \exists k\in \mathbb {Z} }
a
=
b
+
k
n
{\displaystyle a=b+kn}
4.1 Definizioa. Zatigarritasuna ).
3.38 Adibideak.
17
≡
2
(
mod
5
)
{\displaystyle 17\equiv 2{\pmod {5}}}
5
∣
15
{\displaystyle 5\mid 15}
∃
k
∈
Z
,
non
17
−
2
=
5
k
{\displaystyle \exists k\in \mathbb {Z} ,\;{\mbox{ non }}\;17-2=5k}
−
23
≡
−
5
(
mod
6
)
{\displaystyle -23\equiv -5{\pmod {6}}}
6
∣
−
18
{\displaystyle 6\mid -18}
∃
k
∈
Z
,
non
−
23
+
5
=
6
k
{\displaystyle \exists k\in \mathbb {Z} ,\;{\mbox{ non }}\;-23+5=6k}
−
7
≡
2
(
mod
3
)
{\displaystyle -7\equiv 2{\pmod {3}}}
3
∣
−
9
{\displaystyle 3\mid -9}
∃
k
∈
Z
,
non
−
7
−
2
=
3
k
{\displaystyle \exists k\in \mathbb {Z} ,\;{\mbox{ non }}\;-7-2=3k}
−
7
≢
2
(
mod
4
)
{\displaystyle -7\not \equiv 2{\pmod {4}}}
4
∤
−
9
{\displaystyle 4\not \mid -9}
∄
k
∈
Z
,
non
−
7
−
2
=
4
k
{\displaystyle \not \exists k\in \mathbb {Z} ,\;{\mbox{ non }}\;-7-2=4k}
3.39 Teorema.
n
∈
Z
{\displaystyle n\in \mathbb {Z} }
n
>
1
{\displaystyle n>1}
n
{\displaystyle n}
Z
{\displaystyle \mathbb {Z} }
Froga.
n
{\displaystyle n}
Z
{\displaystyle \mathbb {Z} }
Bihurkorra. Hau frogatu behar da:
(
∀
x
∈
Z
)
x
≡
x
(
mod
n
)
.
{\displaystyle (\forall x\in \mathbb {Z} )\quad x\equiv x{\pmod {n}}.}
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
x
−
x
=
0
=
0
⋅
n
{\displaystyle x-x=0=0\cdot n}
n
∣
x
−
x
{\displaystyle n\mid x-x}
x
≡
x
(
mod
n
)
{\displaystyle x\equiv x{\pmod {n}}}
Simetrikoa. Hau frogatu behar da:
(
∀
x
,
y
∈
Z
)
x
≡
y
(
mod
n
)
⟹
y
≡
x
(
mod
n
)
.
{\displaystyle (\forall x,y\in \mathbb {Z} )\quad x\equiv y{\pmod {n}}\;\implies \;y\equiv x{\pmod {n}}.}
x
,
y
∈
Z
{\displaystyle x,y\in \mathbb {Z} }
x
≡
y
(
mod
n
)
⟹
n
∣
x
−
y
⟹
∃
k
∈
Z
,
non
x
−
y
=
k
n
⟹
⟹
∃
(
−
k
)
∈
Z
,
non
y
−
x
=
(
−
k
)
n
⟹
n
∣
y
−
x
⟹
y
≡
x
(
mod
n
)
.
{\displaystyle {\begin{array}{l}x\equiv y{\pmod {n}}\implies \;n\mid x-y\implies \exists k\in \mathbb {Z} ,\;{\mbox{ non }}\;x-y=kn\implies \\\;\implies \;\exists (-k)\in \mathbb {Z} ,\;{\mbox{ non }}\;y-x=(-k)n\implies \;n\mid y-x\implies y\equiv x{\pmod {n}}.\end{array}}}
Iragankorra. Hau frogatu behar da:
(
∀
x
,
y
,
z
∈
Z
)
x
≡
y
(
mod
n
)
eta
y
≡
z
(
mod
n
)
⟹
x
≡
z
(
mod
n
)
.
{\displaystyle (\forall x,y,z\in \mathbb {Z} )\quad x\equiv y{\pmod {n}}\,{\mbox{ eta }}\,y\equiv z{\pmod {n}}\;\implies \;x\equiv z{\pmod {n}}.}
x
,
y
,
z
∈
Z
{\displaystyle x,y,z\in \mathbb {Z} }
x
≡
y
(
mod
n
)
y
≡
z
(
mod
n
)
}
⟹
n
∣
x
−
y
n
∣
y
−
z
}
⟹
∃
k
1
∈
Z
,
non
x
−
y
=
k
1
n
∃
k
2
∈
Z
,
non
y
−
z
=
k
2
n
}
eta
x
−
y
balioari
y
−
z
balioa gehituz, esan dezakegu
∃
(
k
1
+
k
2
)
∈
Z
,
non
x
−
z
=
(
k
1
+
k
2
)
n
⟹
n
∣
x
−
z
⟹
x
≡
z
(
mod
n
)
.
◻
{\displaystyle \quad \quad \quad {\begin{array}{l}\left.{\begin{array}{l}x\equiv y{\pmod {n}}\\y\equiv z{\pmod {n}}\end{array}}\right\}\;\implies \;\left.{\begin{array}{l}n\mid x-y\\n\mid y-z\end{array}}\right\}\;\implies \;\left.{\begin{array}{l}\exists k_{1}\in \mathbb {Z} ,\;{\mbox{ non }}\;x-y=k_{1}n\\\exists k_{2}\in \mathbb {Z} ,\;{\mbox{ non }}\;y-z=k_{2}n\end{array}}\right\}\\{\mbox{ eta }}x-y{\mbox{ balioari }}y-z{\mbox{ balioa gehituz, esan dezakegu }}\\\exists (k_{1}+k_{2})\in \mathbb {Z} ,\;{\mbox{ non }}\;x-z=(k_{1}+k_{2})n\;\implies \;n\mid x-z\implies \;x\equiv z{\pmod {n}}.\\\Box \end{array}}}
Baliokidetasun-erlazio honetarako zatidura-multzoa bi eratan adierazi ohi da,
Z
/
Z
n
{\displaystyle \mathbb {Z} /\mathbb {Z} n}
Z
n
{\displaystyle \mathbb {Z} _{n}}
Z
/
Z
n
=
Z
n
=
{
[
x
]
/
x
∈
Z
}
.
{\displaystyle \mathbb {Z} /\mathbb {Z} n=\mathbb {Z} _{n}=\{[x]\;/\;x\in \mathbb {Z} \}.}
n
{\displaystyle \mathbf {\mathit {n}} }
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
n
{\displaystyle \mathbf {\mathit {n}} }
[
x
]
{\displaystyle [x]}
Horretarako, Euklidesen teorema erabiliz,
x
{\displaystyle x}
n
{\displaystyle n}
r
{\displaystyle r}
x
=
q
n
+
r
{\displaystyle x=qn+r}
q
,
r
∈
Z
{\displaystyle q,r\in \mathbb {Z} }
0
≤
r
<
n
{\displaystyle 0\leq r<n}
x
−
r
=
q
n
{\displaystyle x-r=qn}
n
∣
x
−
r
{\displaystyle n\mid x-r}
x
≡
r
(
mod
n
)
{\displaystyle x\equiv r{\pmod {n}}}
n
{\displaystyle \mathbf {\mathit {n}} }
x
R
r
{\displaystyle x{\mathcal {R}}r}
3.32 Teoremaren 2. propietatearen arabera,
[
x
]
=
[
r
]
{\displaystyle [x]=[r]}
r
{\displaystyle r}
[
x
]
{\displaystyle [x]}
Ondoren,
[
r
]
{\displaystyle [r]}
[
r
]
=
{
y
∈
Z
/
y
R
r
}
=
{
y
∈
Z
/
y
≡
r
(
mod
n
)
}
=
=
{
y
∈
Z
/
n
∣
y
−
r
}
=
{
r
+
k
n
/
k
∈
Z
}
=
=
{
⋯
,
r
−
2
n
,
r
−
n
,
r
,
r
+
n
,
r
+
2
n
,
⋯
}
=
=
{
y
∈
Z
/
r
da
y
zati
n
zatidura euklidearraren hondarra
}
.
{\displaystyle {\begin{array}{ll}[r]&=\{y\in \mathbb {Z} \;/\;y{\mathcal {R}}r\}=\{y\in \mathbb {Z} \;/\;y\equiv r{\pmod {n}}\}=\\&=\{y\in \mathbb {Z} \;/\;n\mid y-r\}=\{r+kn\;/\;k\in \mathbb {Z} \}=\\&=\{\cdots ,r-2n,r-n,r,r+n,r+2n,\cdots \}=\\&=\{y\in \mathbb {Z} \;/\;r\;{\mbox{ da }}\;y\;{\mbox{ zati }}\;n{\mbox{ zatidura euklidearraren hondarra}}\}.\end{array}}}
Horrela,
n
{\displaystyle n}
zatidura multzoa ) honela adieraziko dugu:
Z
/
Z
n
=
Z
n
=
{
[
0
]
,
[
1
]
,
⋯
,
[
n
−
1
]
}
.
{\displaystyle \mathbb {Z} /\mathbb {Z} n=\mathbb {Z} _{n}=\{[0],[1],\cdots ,[n-1]\}.}
3.40 Adibidea.
Z
/
Z
5
=
Z
5
=
{
[
0
]
,
[
1
]
,
[
2
]
,
[
3
]
,
[
4
]
}
.
{\displaystyle \mathbb {Z} /\mathbb {Z} 5=\mathbb {Z} _{5}=\{[0],[1],[2],[3],[4]\}.}
Baliokidetasun-klase horietan dauden elementu batzuk emango ditugu:
22
∈
[
2
]
{\displaystyle 22\in [2]}
22
=
2
+
5
⋅
4
⟹
22
−
2
=
5
⋅
4
⟹
∃
k
∈
Z
/
22
−
2
=
5
k
⟹
5
∣
22
−
2
⟹
22
≡
2
(
mod
5
)
⟹
22
R
2
⟹
22
∈
[
2
]
.
{\displaystyle 22=2+5\cdot 4\implies 22-2=5\cdot 4\implies \exists k\in \mathbb {Z} \;/\;22-2=5k\implies 5\mid 22-2\implies 22\equiv 2{\pmod {5}}\implies 22{\mathcal {R}}2\implies 22\in [2].}
Hortaz,
22
{\displaystyle 22}
22
{\displaystyle 22}
5
{\displaystyle 5}
2
{\displaystyle 2}
Modu berean,
−
22
∈
[
3
]
{\displaystyle -22\in [3]}
−
22
=
(
−
5
)
⋅
5
+
3
{\displaystyle -22=(-5)\cdot 5+3}
−
800
∈
[
0
]
{\displaystyle -800\in [0]}
−
800
=
(
−
160
)
⋅
5
+
0
{\displaystyle -800=(-160)\cdot 5+0}
3.41 Definizioa.
A
{\displaystyle A}
B
{\displaystyle B}
A
{\displaystyle A}
B
{\displaystyle B}
funtzio bat
A
{\displaystyle A}
B
{\displaystyle B}
f
{\displaystyle f}
f
{\displaystyle f}
A
{\displaystyle A}
B
{\displaystyle B}
f
:
A
⟶
B
edo
A
⟶
f
B
.
{\displaystyle f\,:\,A\longrightarrow B\quad {\mbox{ edo }}\quad A{\stackrel {f}{\longrightarrow }}B.}
A
{\displaystyle A}
f
{\displaystyle f}
abiaburu multzoa da eta
B
{\displaystyle B}
f
{\displaystyle f}
helburu multzoa da.
f
{\displaystyle f}
a
∈
A
{\displaystyle a\in A}
b
∈
B
{\displaystyle b\in B}
b
{\displaystyle b}
a
{\displaystyle a}
f
{\displaystyle f}
irudia , eta
a
{\displaystyle a}
b
{\displaystyle b}
f
{\displaystyle f}
aurreirudi bat. Eta honela adieraziko dugu:
f
(
a
)
=
b
edo
a
⟼
b
.
{\displaystyle f(a)=b\quad {\mbox{ edo }}\quad a\longmapsto b.}
3.42 Adibideak.
h
{\displaystyle h}
h
:
N
⟶
N
x
↦
x
+
1.
{\displaystyle {\begin{array}{cccc}h:&\mathbb {N} &\longrightarrow &\mathbb {N} \\&x&\mapsto &x+1.\end{array}}}
7
∈
N
{\displaystyle 7\in \mathbb {N} }
8
{\displaystyle 8}
h
(
7
)
=
8
{\displaystyle h(7)=8}
1
∈
N
{\displaystyle 1\in \mathbb {N} }
0
{\displaystyle 0}
h
(
0
)
=
1
{\displaystyle h(0)=1}
0
∈
N
{\displaystyle 0\in \mathbb {N} }
∄
x
∈
N
/
h
(
x
)
=
0
{\displaystyle \not \exists x\in \mathbb {N} \;/\;h(x)=0}
l
{\displaystyle l}
l
:
Z
⟶
Z
x
↦
{
0
,
x
bikoitia bada,
1
,
x
bakoitia bada.
{\displaystyle {\begin{array}{cccl}l:&\mathbb {Z} &\longrightarrow &\mathbb {Z} \\&x&\mapsto &\left\{{\begin{array}{l}0,\;x{\mbox{ bikoitia bada,}}\\1,\;x{\mbox{ bakoitia bada.}}\end{array}}\right.\end{array}}}
p
{\displaystyle p}
p
:
R
2
⟶
R
(
x
1
,
x
2
)
↦
x
1
.
{\displaystyle {\begin{array}{cccc}p:&\mathbb {R} ^{2}&\longrightarrow &\mathbb {R} \\&(x_{1},x_{2})&\mapsto &x_{1}.\end{array}}}
(
1
,
2
)
∈
R
2
{\displaystyle (1,2)\in \mathbb {R} ^{2}}
p
(
1
,
2
)
=
1
{\displaystyle p(1,2)=1}
Ondokoak ez
f
:
{
1
,
2
,
3
}
⟶
{
a
,
b
,
c
,
d
}
1
↦
a
1
↦
b
2
↦
c
3
↦
d
.
{\displaystyle {\begin{array}{cccc}f:&\{1,2,3\}&\longrightarrow &\{a,b,c,d\}\\&1&\mapsto &a\\&1&\mapsto &b\\&2&\mapsto &c\\&3&\mapsto &d.\\\end{array}}}
Ez da funtzioa, abiaburu multzoaren
1
{\displaystyle 1}
bi
1
{\displaystyle 1}
g
:
{
1
,
2
,
3
}
⟶
{
a
,
b
}
1
↦
a
2
↦
b
.
{\displaystyle {\begin{array}{cccc}g:&\{1,2,3\}&\longrightarrow &\{a,b\}\\&1&\mapsto &a\\&2&\mapsto &b.\\\end{array}}}
Ez da funtzioa, abiaburu multzoaren
3
{\displaystyle 3}
ez
3
{\displaystyle 3}
h
:
N
⟶
N
x
↦
x
−
1.
{\displaystyle {\begin{array}{cccc}h:&\mathbb {N} &\longrightarrow &\mathbb {N} \\&x&\mapsto &x-1.\end{array}}}
0
{\displaystyle 0}
j
:
N
⟶
R
x
↦
x
.
{\displaystyle {\begin{array}{cccc}j:&\mathbb {N} &\longrightarrow &\mathbb {R} \\&x&\mapsto &{\sqrt {x}}.\end{array}}}
Pentsa genezake
4
{\displaystyle 4}
−
2
{\displaystyle -2}
2
{\displaystyle 2}
+
{\displaystyle +}
−
{\displaystyle -}
f
(
x
)
=
+
x
{\displaystyle f(x)=+{\sqrt {x}}}
f
(
x
)
=
−
x
{\displaystyle f(x)=-{\sqrt {x}}}
+
{\displaystyle +}
j
(
4
)
=
+
4
=
2
{\displaystyle j(4)=+{\sqrt {4}}=2}
3.43 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
f
{\displaystyle f}
funtzioaren grafo deituko diogu multzo honi:
G
f
:=
{
(
x
,
f
(
x
)
)
/
x
∈
A
}
.
{\displaystyle G_{f}:=\{(x,f(x))\;/\;x\in A\}.}
Ohar gaitezen
G
f
⊆
A
×
B
{\displaystyle G_{f}\subseteq A\times B}
A
{\displaystyle A}
G
f
{\displaystyle G_{f}}
3.44 Adibideak.
h
{\displaystyle h}
G
h
{\displaystyle G_{h}}
h
:
N
⟶
N
x
↦
x
+
1.
G
h
=
{
(
x
,
x
+
1
)
/
x
∈
N
}
.
{\displaystyle {\begin{array}{cccc}h:&\mathbb {N} &\longrightarrow &\mathbb {N} \\&x&\mapsto &x+1.\end{array}}\quad \quad G_{h}=\{(x,x+1)\;/\;x\in \mathbb {N} \}.}
l
{\displaystyle l}
G
l
{\displaystyle G_{l}}
l
:
Z
⟶
Z
x
↦
{
0
,
x
bikoitia bada,
1
,
x
bakoitia bada.
{\displaystyle {\begin{array}{cccl}l:&\mathbb {Z} &\longrightarrow &\mathbb {Z} \\&x&\mapsto &\left\{{\begin{array}{l}0,\;x{\mbox{ bikoitia bada,}}\\1,\;x{\mbox{ bakoitia bada.}}\end{array}}\right.\end{array}}}
G
l
=
{
(
x
,
0
)
/
x
∈
Z
eta
x
bikoitia da
}
∪
{
(
x
,
1
)
/
x
∈
Z
eta
x
bakoitia da
}
.
{\displaystyle G_{l}=\{(x,0)\;/\;x\in \mathbb {Z} {\mbox{ eta }}x{\mbox{ bikoitia da}}\}\cup \{(x,1)\;/\;x\in \mathbb {Z} {\mbox{ eta }}x{\mbox{ bakoitia da}}\}.}
Beraz, funtzio bat izanik, grafo bat definitzen dugu. Alderantziz ere beteko da. Hau da,
G
⊆
A
×
B
{\displaystyle G\subseteq A\times B}
A
{\displaystyle A}
G
{\displaystyle G}
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
G
{\displaystyle G}
3.45 Definizioa.
f
{\displaystyle f}
g
{\displaystyle g}
berdinak direla esango dugu, eta
f
=
g
{\displaystyle f=g}
A
{\displaystyle A}
B
{\displaystyle B}
(
∀
x
∈
A
)
f
(
x
)
=
g
(
x
)
{\displaystyle (\forall x\in A)\quad f(x)=g(x)}
3.46 Adibidea.
f
:
Z
⟶
Z
x
↦
x
2
,
g
:
{
−
3
,
3
}
⟶
Z
x
↦
x
2
,
{\displaystyle {\begin{array}{cccc}f:&\mathbb {Z} &\longrightarrow &\mathbb {Z} \\&x&\mapsto &x^{2},\end{array}}\quad \quad {\begin{array}{cccc}g:&\{-3,3\}&\longrightarrow &\mathbb {Z} \\&x&\mapsto &x^{2},\end{array}}}
h
:
{
−
3
,
3
}
⟶
N
x
↦
x
2
,
k
:
{
−
3
,
3
}
⟶
Z
x
↦
9.
{\displaystyle {\begin{array}{cccc}h:&\{-3,3\}&\longrightarrow &\mathbb {N} \\&x&\mapsto &x^{2},\end{array}}\quad \quad {\begin{array}{cccc}k:&\{-3,3\}&\longrightarrow &\mathbb {Z} \\&x&\mapsto &9.\end{array}}}
f
≠
g
{\displaystyle f\neq g}
f
{\displaystyle f}
g
{\displaystyle g}
g
≠
h
{\displaystyle g\neq h}
g
{\displaystyle g}
h
{\displaystyle h}
g
=
k
{\displaystyle g=k}
g
{\displaystyle g}
k
{\displaystyle k}
{
−
3
,
3
}
{\displaystyle \{-3,3\}}
g
{\displaystyle g}
k
{\displaystyle k}
Z
{\displaystyle \mathbb {Z} }
(
∀
x
∈
{
−
3
,
3
}
)
g
(
x
)
=
9
=
k
(
x
)
{\displaystyle (\forall x\in \{-3,3\})\quad g(x)=9=k(x)}
3.47 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
A
1
⊆
A
{\displaystyle A_{1}\subseteq A}
f
{\displaystyle f}
A
1
{\displaystyle A_{1}}
murrizketa deituko diogu, eta
f
∣
A
1
{\displaystyle f\mid _{A_{1}}}
f
∣
A
1
:
A
1
⟶
B
x
↦
f
(
x
)
.
{\displaystyle {\begin{array}{cccc}f\mid _{A_{1}}:&A_{1}&\longrightarrow &B\\&x&\mapsto &f(x).\end{array}}}
3.48 Adibidea.
f
:
Z
⟶
Z
x
↦
x
2
;
f
∣
{
−
3
,
3
}
:
{
−
3
,
3
}
⟶
Z
x
↦
x
2
.
{\displaystyle {\begin{array}{cccc}f:&\mathbb {Z} &\longrightarrow &\mathbb {Z} \\&x&\mapsto &x^{2};\end{array}}\quad \quad {\begin{array}{cccc}f\mid _{\{-3,3\}}:&\{-3,3\}&\longrightarrow &\mathbb {Z} \\&x&\mapsto &x^{2}.\end{array}}}
3.49 Definizioa.
A
{\displaystyle A}
A
{\displaystyle A}
identitate funtzioa
i
d
A
{\displaystyle id_{A}}
1
A
{\displaystyle 1_{A}}
i
d
A
:
A
⟶
A
x
↦
x
.
{\displaystyle {\begin{array}{cccc}id_{A}:&A&\longrightarrow &A\\&x&\mapsto &x.\end{array}}}
3.50 Definizioa.
A
{\displaystyle A}
funtzio karakteristiko deituko diogu eta
f
A
{\displaystyle f_{A}}
f
A
:
U
⟶
{
0
,
1
}
x
↦
{
1
,
x
∈
A
bada,
0
,
x
∉
A
bada,
{\displaystyle {\begin{array}{cccl}f_{A}:&U&\longrightarrow &\{0,1\}\\&x&\mapsto &\left\{{\begin{array}{l}1,\;x\in A{\mbox{ bada,}}\\0,\;x\not \in A{\mbox{ bada,}}\end{array}}\right.\end{array}}}
non
U
{\displaystyle U}
3.51 Adibidea.
Z
{\displaystyle \mathbb {Z} }
f
N
:
Z
⟶
{
0
,
1
}
x
↦
{
1
,
x
≥
0
bada,
0
,
x
<
0
bada.
{\displaystyle {\begin{array}{cccl}f_{\mathbb {N} }:&\mathbb {Z} &\longrightarrow &\{0,1\}\\&x&\mapsto &\left\{{\begin{array}{l}1,\;x\geq 0{\mbox{ bada,}}\\0,\;x<0{\mbox{ bada.}}\end{array}}\right.\end{array}}}
3.52 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
A
1
⊆
A
{\displaystyle A_{1}\subseteq A}
A
1
{\displaystyle A_{1}}
A
1
{\displaystyle A_{1}}
f
{\displaystyle f}
irudi deituko diogu eta
f
(
A
1
)
{\displaystyle f(A_{1})}
f
(
A
1
)
:=
{
f
(
x
)
/
x
∈
A
1
}
.
{\displaystyle f(A_{1}):=\{f(x)\;/\;x\in A_{1}\}.}
Ohar gaitezen
f
(
A
1
)
⊆
B
{\displaystyle f(A_{1})\subseteq B}
Hitzarmena.
f
(
∅
)
:=
∅
{\displaystyle f(\emptyset ):=\emptyset }
3.53 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
f
(
A
)
{\displaystyle f(A)}
f
{\displaystyle f}
irudi multzo deitzen zaio eta Im
f
{\displaystyle f}
Im
f
:=
{
f
(
x
)
/
x
∈
A
}
.
{\displaystyle f:=\{f(x)\;/\;x\in A\}.}
3.54 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
funtzio injektiboa dela esango dugu
A
{\displaystyle A}
B
{\displaystyle B}
(
∀
x
1
,
x
2
∈
A
)
x
1
≠
x
2
⟹
f
(
x
1
)
≠
f
(
x
2
)
betetzen bada
,
{\displaystyle (\forall x_{1},x_{2}\in A)\quad x_{1}\neq x_{2}\quad \Longrightarrow f(x_{1})\neq f(x_{2}){\mbox{ betetzen bada}},}
edo, baliokidea dena,
(
∀
x
1
,
x
2
∈
A
)
f
(
x
1
)
=
f
(
x
2
)
⟹
x
1
=
x
2
betetzen bada
.
{\displaystyle (\forall x_{1},x_{2}\in A)\quad f(x_{1})=f(x_{2})\quad \Longrightarrow x_{1}=x_{2}{\mbox{ betetzen bada}}.}
3.55 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
funtzio surjektiboa dela esango dugu
B
{\displaystyle B}
A
{\displaystyle A}
I
m
f
=
B
{\displaystyle Imf=B}
∀
y
∈
B
∃
x
∈
A
,
non
f
(
x
)
=
y
den
.
{\displaystyle \forall y\in B\quad \exists x\in A,{\mbox{ non }}f(x)=y{\mbox{ den}}.}
3.56 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
funtzio bijektiboa dela esango dugu injektiboa eta surjektiboa bada.
3.57 Adibideak.
f
:
R
⟶
R
{\displaystyle f:\mathbb {R} \longrightarrow \mathbb {R} }
f
(
x
)
=
3
x
+
1
{\displaystyle f(x)=3x+1}
g
:
N
⟶
N
{\displaystyle g:\mathbb {N} \longrightarrow \mathbb {N} }
g
(
x
)
=
x
+
1
{\displaystyle g(x)=x+1}
|
⋅
|
:
R
⟶
R
{\displaystyle \vert \cdot \vert :\mathbb {R} \longrightarrow \mathbb {R} }
|
x
|
=
{
−
x
,
x
<
0
,
x
,
0
≤
x
.
{\displaystyle |x|=\left\{{\begin{array}{cc}-x,&\;x<0,\\\ \ x,&\;0\leq x.\end{array}}\right.}
p
:
R
2
⟶
R
{\displaystyle p:\mathbb {R} ^{2}\longrightarrow \mathbb {R} }
p
(
x
1
,
x
2
)
=
x
1
{\displaystyle p(x_{1},x_{2})=x_{1}}
p
{\displaystyle p}
ez
p
(
1
,
π
)
=
p
(
1
,
0
)
{\displaystyle p(1,\pi )=p(1,0)}
(
1
,
π
)
≠
(
1
,
0
)
{\displaystyle (1,\pi )\neq (1,0)}
p
{\displaystyle p}
∀
y
∈
R
∃
(
x
1
,
x
2
)
=
(
y
,
0
)
∈
R
2
,
non
p
(
x
1
,
x
2
)
=
y
den
.
{\displaystyle \forall y\in \mathbb {R} \quad \exists (x_{1},x_{2})=(y,0)\in \mathbb {R} ^{2},{\mbox{ non }}p(x_{1},x_{2})=y{\mbox{ den}}.}
Ohar gaitezen
y
∈
R
{\displaystyle y\in \mathbb {R} }
p
(
y
,
π
)
=
y
{\displaystyle p(y,\pi )=y}
3.58 Adibidea.
A
{\displaystyle A}
A
{\displaystyle A}
i
d
A
:
A
⟶
A
x
↦
x
,
{\displaystyle {\begin{array}{cccc}id_{A}:&A&\longrightarrow &A\\&x&\mapsto &x,\end{array}}}
injektiboa eta surjektiboa da. Hortaz, bijektiboa da.
3.4.3 Alderantzizko funtzioa [ aldatu ] 3.59 Teorema.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
f
bijektiboa da
⟺
∀
y
∈
B
∃
!
x
∈
A
/
f
(
x
)
=
y
betetzen bada
.
{\displaystyle f\;{\mbox{ bijektiboa da}}\quad \Longleftrightarrow \quad \forall y\in B\;\exists !x\in A\,/\,f(x)=y{\mbox{ betetzen bada}}.}
Froga.
⟹
{\displaystyle \implies }
f
{\displaystyle f}
∀
y
∈
B
∃
x
∈
A
/
f
(
x
)
=
y
.
{\displaystyle \forall y\in B\;\;\exists x\in A\,/\,f(x)=y.}
Ikus dezagun
x
∈
A
{\displaystyle x\in A}
∃
x
1
,
x
2
∈
A
{\displaystyle \exists x_{1},x_{2}\in A}
f
(
x
1
)
=
y
{\displaystyle f(x_{1})=y}
f
(
x
2
)
=
y
{\displaystyle f(x_{2})=y}
f
(
x
1
)
=
f
(
x
2
)
{\displaystyle f(x_{1})=f(x_{2})}
f
{\displaystyle f}
x
1
=
x
2
{\displaystyle x_{1}=x_{2}}
⟸
{\displaystyle \Longleftarrow }
Demagun
∀
y
∈
B
∃
!
x
∈
A
/
f
(
x
)
=
y
{\displaystyle \forall y\in B\;\exists !x\in A\,/\,f(x)=y}
∀
y
∈
B
∃
!
x
∈
A
/
f
(
x
)
=
y
{\displaystyle \forall y\in B\;\exists !x\in A\,/\,f(x)=y}
f
{\displaystyle f}
x
∈
A
{\displaystyle x\in A}
Ikus dezagun
f
{\displaystyle f}
Har ditzagun
x
1
,
x
2
∈
A
{\displaystyle x_{1},x_{2}\in A}
f
(
x
1
)
=
f
(
x
2
)
{\displaystyle f(x_{1})=f(x_{2})}
f
(
x
1
)
,
f
(
x
2
)
∈
B
{\displaystyle f(x_{1}),f(x_{2})\in B}
∃
!
x
∈
A
/
f
(
x
)
=
f
(
x
1
)
=
f
(
x
2
)
{\displaystyle \exists !x\in A\,/\,f(x)=f(x_{1})=f(x_{2})}
x
=
x
1
=
x
2
{\displaystyle x=x_{1}=x_{2}}
x
∈
A
{\displaystyle x\in A}
◻
{\displaystyle \Box }
Orduan,
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
∀
y
∈
B
∃
!
x
∈
A
/
f
(
x
)
=
y
{\displaystyle \forall y\in B\;\exists !x\in A\,/\,f(x)=y}
y
∈
B
{\displaystyle y\in B}
x
∈
A
{\displaystyle x\in A}
y
∈
B
{\displaystyle y\in B}
x
∈
A
{\displaystyle x\in A}
3.60 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
f
−
1
{\displaystyle f^{-1}}
f
{\displaystyle f}
alderantzizko funtzio deituko diogu honako funtzio honi:
f
−
1
:
B
⟶
A
y
↦
x
,
non
f
(
x
)
=
y
.
{\displaystyle {\begin{array}{cccl}f^{-1}:&B&\longrightarrow &A\\&y&\mapsto &x,{\mbox{ non }}f(x)=y.\end{array}}}
3.61 Adibidea.
f
:
R
⟶
R
{\displaystyle f:\mathbb {R} \longrightarrow \mathbb {R} }
f
(
x
)
=
3
x
+
1
{\displaystyle f(x)=3x+1}
y
∈
R
{\displaystyle y\in \mathbb {R} }
∃
!
x
∈
R
/
f
−
1
(
y
)
=
x
{\displaystyle \exists !x\in \mathbb {R} \,/\,f^{-1}(y)=x}
f
(
x
)
=
y
{\displaystyle f(x)=y}
3
x
+
1
=
y
{\displaystyle 3x+1=y}
x
=
y
−
1
3
{\displaystyle \displaystyle {x={\frac {y-1}{3}}}}
f
−
1
:
R
⟶
R
{\displaystyle f^{-1}:\mathbb {R} \longrightarrow \mathbb {R} }
f
−
1
(
y
)
=
y
−
1
3
{\displaystyle \displaystyle {f^{-1}(y)={\frac {y-1}{3}}}}
f
{\displaystyle \;f}
3.62 Ondorioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
f
{\displaystyle f}
f
−
1
:
B
⟶
A
{\displaystyle f^{-1}:B\longrightarrow A}
(
∀
x
∈
A
)
(
∀
y
∈
B
)
f
−
1
(
y
)
=
x
⟺
f
(
x
)
=
y
.
{\displaystyle (\forall x\in A)\;(\forall y\in B)\quad f^{-1}(y)=x\;\iff \;f(x)=y.}
(
∀
x
∈
A
)
f
−
1
(
f
(
x
)
)
=
x
.
{\displaystyle (\forall x\in A)\quad f^{-1}(f(x))=x.}
(
∀
y
∈
B
)
f
(
f
−
1
(
y
)
)
=
y
.
{\displaystyle (\forall y\in B)\quad f(f^{-1}(y))=y.}
3.63 Teorema.
3.64 Adibidea.
f
:
R
⟶
R
{\displaystyle f:\mathbb {R} \longrightarrow \mathbb {R} }
f
(
x
)
=
3
x
+
1
{\displaystyle f(x)=3x+1}
f
−
1
:
R
⟶
R
{\displaystyle f^{-1}:\mathbb {R} \longrightarrow \mathbb {R} }
f
−
1
(
y
)
=
y
−
1
3
{\displaystyle \displaystyle {f^{-1}(y)={\frac {y-1}{3}}}}
3.4.4 Funtzioen konposizioa [ aldatu ] 3.65 Definizioa.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
g
:
B
⟶
C
{\displaystyle g:B\longrightarrow C}
f
{\displaystyle f}
g
{\displaystyle g}
funtzio konposatu deituko diogu, eta
g
∘
f
{\displaystyle g\circ f}
g
∘
f
:
A
⟶
C
x
↦
g
(
f
(
x
)
)
.
{\displaystyle {\begin{array}{cccc}g\circ f\,:&A&\longrightarrow &C\\&x&\mapsto &g(f(x)).\end{array}}}
Ohar gaitezen
g
∘
f
{\displaystyle g\circ f}
f
{\displaystyle f}
g
{\displaystyle g}
g
∘
f
:
A
⟶
f
B
⟶
g
C
x
⟼
f
(
x
)
⟼
g
(
f
(
x
)
)
.
{\displaystyle {\begin{array}{cccccc}g\circ f\,:&A&{\stackrel {f}{\longrightarrow }}&B&{\stackrel {g}{\longrightarrow }}&C\\&x&\longmapsto &f(x)&\longmapsto &g(f(x)).\end{array}}}
g
∘
f
{\displaystyle g\circ f}
Abiaburu multzoko elementu bakoitzak irudi bat du helburu multzoan:
x
∈
A
⟹
f
(
x
)
∈
B
⟹
g
(
f
(
x
)
)
∈
C
.
{\displaystyle x\in A\implies f(x)\in B\implies g(f(x))\in C.}
Abiaburu multzoko elementu bakoitzaren irudia bakarra da:
x
1
,
x
2
∈
A
{\displaystyle x_{1},x_{2}\in A}
x
1
=
x
2
⟹
f
(
x
1
)
=
f
(
x
2
)
⟹
g
(
f
(
x
1
)
)
=
g
(
f
(
x
2
)
)
.
{\displaystyle x_{1}=x_{2}\implies f(x_{1})=f(x_{2})\implies g(f(x_{1}))=g(f(x_{2})).}
3.66 Adibideak.
f
{\displaystyle f}
g
{\displaystyle g}
g
∘
f
{\displaystyle g\circ f}
f
:
Z
⟶
N
x
↦
x
2
,
g
:
N
⟶
R
x
↦
x
+
3
,
g
∘
f
:
Z
⟶
R
x
↦
x
2
+
3.
{\displaystyle {\begin{array}{cccc}f:&\mathbb {Z} &\longrightarrow &\mathbb {N} \\&x&\mapsto &x^{2},\end{array}}\quad {\begin{array}{cccc}g:&\mathbb {N} &\longrightarrow &\mathbb {R} \\&x&\mapsto &x+3,\end{array}}\quad {\begin{array}{cccc}g\circ f:&\mathbb {Z} &\longrightarrow &\mathbb {R} \\&x&\mapsto &x^{2}+3.\end{array}}}
(
g
∘
f
)
(
x
)
=
g
(
f
(
x
)
)
=
g
(
x
2
)
=
x
2
+
3.
{\displaystyle (g\circ f)(x)=g(f(x))=g(x^{2})=x^{2}+3.}
Ohar gaitezen ezin dela
f
∘
g
{\displaystyle f\circ g}
g
{\displaystyle g}
f
{\displaystyle f}
h
{\displaystyle h}
j
{\displaystyle j}
j
∘
h
{\displaystyle j\circ h}
h
∘
j
{\displaystyle h\circ j}
h
:
Z
⟶
Z
x
↦
x
2
,
j
:
Z
⟶
Z
x
↦
x
+
3.
{\displaystyle {\begin{array}{cccc}h:&\mathbb {Z} &\longrightarrow &\mathbb {Z} \\&x&\mapsto &x^{2},\end{array}}\quad \quad {\begin{array}{cccc}j:&\mathbb {Z} &\longrightarrow &\mathbb {Z} \\&x&\mapsto &x+3.\end{array}}}
j
∘
h
:
Z
⟶
Z
x
↦
x
2
+
3
,
h
∘
j
:
Z
⟶
Z
x
↦
(
x
+
3
)
2
.
{\displaystyle {\begin{array}{cccc}j\circ h:&\mathbb {Z} &\longrightarrow &\mathbb {Z} \\&x&\mapsto &x^{2}+3,\end{array}}\quad \quad {\begin{array}{cccc}h\circ j:&\mathbb {Z} &\longrightarrow &\mathbb {Z} \\&x&\mapsto &(x+3)^{2}.\end{array}}}
(
j
∘
h
)
(
x
)
=
j
(
h
(
x
)
)
=
j
(
x
2
)
=
x
2
+
3.
(
h
∘
j
)
(
x
)
=
h
(
j
(
x
)
)
=
h
(
x
+
3
)
=
(
x
+
3
)
2
.
{\displaystyle (j\circ h)(x)=j(h(x))=j(x^{2})=x^{2}+3.\quad (h\circ j)(x)=h(j(x))=h(x+3)=(x+3)^{2}.}
Beraz, nahiz eta
h
∘
j
{\displaystyle h\circ j}
j
∘
h
{\displaystyle j\circ h}
h
∘
j
≠
j
∘
h
{\displaystyle h\circ j\neq j\circ h}
Adibideetan ikusi dugu, oro har, funtzioen konposizioa ez dela trukakorra.
3.67 Propietateak.
Funtzioen konposizioa elkarkorra da.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
g
:
B
⟶
C
{\displaystyle g:B\longrightarrow C}
h
:
C
⟶
D
{\displaystyle h:C\longrightarrow D}
h
∘
(
g
∘
f
)
=
(
h
∘
g
)
∘
f
.
{\displaystyle h\circ (g\circ f)=(h\circ g)\circ f.}
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
f
∘
i
d
A
=
f
,
i
d
B
∘
f
=
f
.
{\displaystyle f\circ id_{A}=f,\quad id_{B}\circ f=f.}
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
f
−
1
∘
f
=
i
d
A
,
f
∘
f
−
1
=
i
d
B
.
{\displaystyle f^{-1}\circ f=id_{A},\quad f\circ f^{-1}=id_{B}.}
Funtzio surjektiboen funtzio konposatua surjektiboa da.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
g
:
B
⟶
C
{\displaystyle g:B\longrightarrow C}
f
,
g
surjektiboak
⟹
g
∘
f
surjektiboa.
{\displaystyle f,g\;{\mbox{ surjektiboak }}\quad \implies \quad g\circ f\;{\mbox{ surjektiboa.}}}
Funtzio injektiboen funtzio konposatua injektiboa da.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
g
:
B
⟶
C
{\displaystyle g:B\longrightarrow C}
f
,
g
injektiboak
⟹
g
∘
f
injektiboa.
{\displaystyle f,g\;{\mbox{ injektiboak }}\quad \implies \quad g\circ f\;{\mbox{ injektiboa.}}}
Funtzio bijektiboen funtzio konposatua bijektiboa da.
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
g
:
B
⟶
C
{\displaystyle g:B\longrightarrow C}
f
,
g
bijektiboak
⟹
g
∘
f
bijektiboa.
{\displaystyle f,g\;{\mbox{ bijektiboak }}\quad \implies \quad g\circ f\;{\mbox{ bijektiboa.}}}
f
:
A
⟶
B
{\displaystyle f:A\longrightarrow B}
g
:
B
⟶
C
{\displaystyle g:B\longrightarrow C}
g
∘
f
{\displaystyle g\circ f}
(
g
∘
f
)
−
1
=
f
−
1
∘
g
−
1
.
{\displaystyle (g\circ f)^{-1}=f^{-1}\circ g^{-1}.}
Oharra.
h
∘
g
∘
f
{\displaystyle h\circ g\circ f}
Idazkera.
f
n
=
f
∘
⋯
n
∘
f
{\displaystyle f^{n}=f\circ {\stackrel {n}{\cdots }}\circ f}
Froga.
Ikusteko zeintzuk diren
h
∘
(
g
∘
f
)
{\displaystyle h\circ (g\circ f)}
g
∘
f
{\displaystyle g\circ f}
g
∘
f
:
A
⟶
f
B
⟶
g
C
,
{\displaystyle g\circ f\,:\,A{\stackrel {f}{\longrightarrow }}B{\stackrel {g}{\longrightarrow }}C,}
h
∘
(
g
∘
f
)
{\displaystyle h\circ (g\circ f)}
h
∘
(
g
∘
f
)
:
A
⟶
g
∘
f
C
⟶
h
D
.
{\displaystyle h\circ (g\circ f)\,:\,A{\stackrel {g\circ f}{\longrightarrow }}C{\stackrel {h}{\longrightarrow }}D.}
(
h
∘
g
)
∘
f
{\displaystyle (h\circ g)\circ f}
h
∘
g
{\displaystyle h\circ g}
h
∘
g
:
B
⟶
g
C
⟶
h
D
,
{\displaystyle h\circ g\,:\,B{\stackrel {g}{\longrightarrow }}C{\stackrel {h}{\longrightarrow }}D,}
(
h
∘
g
)
∘
f
{\displaystyle (h\circ g)\circ f}
(
h
∘
g
)
∘
f
:
A
⟶
f
B
⟶
h
∘
g
D
.
{\displaystyle (h\circ g)\circ f\,:\,A{\stackrel {f}{\longrightarrow }}B{\stackrel {h\circ g}{\longrightarrow }}D.}
h
∘
(
g
∘
f
)
=
(
h
∘
g
)
∘
f
{\displaystyle h\circ (g\circ f)=(h\circ g)\circ f}
f
∘
i
d
A
:
A
⟶
i
d
A
A
⟶
f
B
,
f
:
A
⟶
B
{\displaystyle f\circ id_{A}\,:\,A{\stackrel {id_{A}}{\longrightarrow }}A{\stackrel {f}{\longrightarrow }}B,\quad f:\,A\longrightarrow B}
i
d
B
∘
f
:
A
⟶
f
B
⟶
i
d
B
B
,
f
:
A
⟶
B
{\displaystyle id_{B}\circ f\,:\,A{\stackrel {f}{\longrightarrow }}B{\stackrel {id_{B}}{\longrightarrow }}B,\quad f:\,A\longrightarrow B}
f
−
1
∘
f
:
A
⟶
f
B
⟶
f
−
1
A
,
i
d
A
:
A
⟶
A
{\displaystyle f^{-1}\circ f\,:\,A{\stackrel {f}{\longrightarrow }}B{\stackrel {f^{-1}}{\longrightarrow }}A,\quad id_{A}:\,A\longrightarrow A}
f
∘
f
−
1
:
B
⟶
f
−
1
A
⟶
f
B
,
i
d
B
:
B
⟶
B
{\displaystyle f\circ f^{-1}\,:\,B{\stackrel {f^{-1}}{\longrightarrow }}A{\stackrel {f}{\longrightarrow }}B,\quad id_{B}:\,B\longrightarrow B}
g
∘
f
:
A
⟶
f
B
⟶
g
C
{\displaystyle g\circ f\,:\,A{\stackrel {f}{\longrightarrow }}B{\stackrel {g}{\longrightarrow }}C}
f
{\displaystyle f}
g
{\displaystyle g}
Frogatu behar dugu
g
∘
f
{\displaystyle g\circ f}
∀
z
∈
C
∃
x
∈
A
/
(
g
∘
f
)
(
x
)
=
z
betetzen dela
.
{\displaystyle \forall z\in C\quad \exists x\in A\,/\,(g\circ f)(x)=z{\mbox{ betetzen dela}}.}
Izan bedi
z
∈
C
{\displaystyle z\in C}
g
{\displaystyle g}
∃
y
∈
B
/
g
(
y
)
=
z
{\displaystyle \exists y\in B\,/\,g(y)=z}
y
∈
B
{\displaystyle y\in B}
f
{\displaystyle f}
∃
x
∈
A
/
f
(
x
)
=
y
{\displaystyle \exists x\in A\,/\,f(x)=y}
Orduan,
z
=
g
(
y
)
=
g
(
f
(
x
)
)
=
(
g
∘
f
)
(
x
)
{\displaystyle z=g(y)=g(f(x))=(g\circ f)(x)}
∃
x
∈
A
/
(
g
∘
f
)
(
x
)
=
z
{\displaystyle \exists x\in A\,/\,(g\circ f)(x)=z}
g
∘
f
:
A
⟶
f
B
⟶
g
C
{\displaystyle g\circ f\,:\,A{\stackrel {f}{\longrightarrow }}B{\stackrel {g}{\longrightarrow }}C}
f
{\displaystyle f}
g
{\displaystyle g}
Frogatu behar dugu
g
∘
f
{\displaystyle g\circ f}
(
∀
x
1
,
x
2
∈
A
)
(
g
∘
f
)
(
x
1
)
=
(
g
∘
f
)
(
x
2
)
⟹
x
1
=
x
2
betetzen dela
.
{\displaystyle (\forall x_{1},x_{2}\in A)\quad (g\circ f)(x_{1})=(g\circ f)(x_{2})\implies x_{1}=x_{2}{\mbox{ betetzen dela}}.}
x
1
,
x
2
∈
A
{\displaystyle x_{1},x_{2}\in A}
(
g
∘
f
)
(
x
1
)
=
(
g
∘
f
)
(
x
2
)
⟹
g
(
f
(
x
1
)
)
=
g
(
f
(
x
2
)
)
⟹
⟹
f
(
x
1
)
=
f
(
x
2
)
,
(
g
injektiboa delako)
⟹
x
1
=
x
2
,
(
f
injektiboa delako).
{\displaystyle {\begin{array}{l}(g\circ f)(x_{1})=(g\circ f)(x_{2})\implies g(f(x_{1}))=g(f(x_{2}))\implies \\\implies f(x_{1})=f(x_{2}),(g{\mbox{ injektiboa delako) }}\implies x_{1}=x_{2},(f{\mbox{ injektiboa delako).}}\end{array}}}
Aurreko bi propietateen ondorio zuzena da.
g
∘
f
:
A
⟶
f
B
⟶
g
C
,
(
g
∘
f
)
−
1
:
C
⟶
A
,
f
−
1
∘
g
−
1
:
C
⟶
g
−
1
B
⟶
f
−
1
A
{\displaystyle g\circ f\,:\,A{\stackrel {f}{\longrightarrow }}B{\stackrel {g}{\longrightarrow }}C,\quad (g\circ f)^{-1}:\,C\longrightarrow A,\quad f^{-1}\circ g^{-1}\,:\,C{\stackrel {g^{-1}}{\longrightarrow }}B{\stackrel {f^{-1}}{\longrightarrow }}A}
◻
{\displaystyle \Box }
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![{\displaystyle [x]\cap [y]\neq \emptyset \implies \exists z\in [x]\cap [y]\implies \left\{{\begin{array}{l}z\in [x]\implies z{\mathcal {R}}x\implies x{\mathcal {R}}z\\z\in [y]\implies z{\mathcal {R}}y\end{array}}\right\}\implies x{\mathcal {R}}y\implies [x]=[y]{\mbox{ (2.aren arabera)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3e5067c1e42a749b375028688a66b9d41a4a7ab)
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![{\displaystyle -22\in [3]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa1ab918c23a71332251b0b118f26cc99e9bdd98)
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