Seja
Z
{\displaystyle \mathbb {Z} }
+
{\displaystyle +}
⋅
{\displaystyle \cdot }
Γ
=
Z
×
(
Z
∖
{
0
}
)
{\displaystyle \Gamma =\mathbb {Z} \times (\mathbb {Z} \setminus \{0\})}
∼
{\displaystyle \sim }
Γ
{\displaystyle \Gamma }
(
m
,
n
)
∼
(
p
,
q
)
⇔
m
⋅
q
=
n
⋅
p
{\displaystyle (m,n)\sim (p,q)\Leftrightarrow m\cdot q=n\cdot p}
Mostre que
∼
{\displaystyle \sim }
Γ
{\displaystyle \Gamma }
reflexiva
∀
(
a
,
b
)
∈
Γ
,
(
a
,
b
)
∼
(
a
,
b
)
⇔
a
⋅
b
=
b
⋅
a
{\displaystyle \forall \;(a,b)\in \Gamma ,(a,b)\sim (a,b)\Leftrightarrow a\cdot b=b\cdot a}
a
,
b
∈
Z
{\displaystyle a,b\in \mathbb {Z} }
simétrica
Dado
(
a
,
b
)
,
(
c
,
d
)
∈
Γ
,
(
a
,
b
)
∼
(
c
,
d
)
⇔
a
⋅
d
=
b
⋅
c
{\displaystyle (a,b),(c,d)\in \Gamma ,(a,b)\sim (c,d)\Leftrightarrow a\cdot d=b\cdot c}
a
,
b
,
c
,
d
∈
Z
{\displaystyle a,b,c,d\in \mathbb {Z} }
d
⋅
a
=
c
⋅
b
⇔
(
c
,
d
)
∼
(
a
,
b
)
{\displaystyle d\cdot a=c\cdot b\Leftrightarrow (c,d)\sim (a,b)}
transitiva
Dado
(
a
,
b
)
,
(
c
,
d
)
,
(
e
,
f
)
∈
Γ
,
(
a
,
b
)
∼
(
c
,
d
)
e
(
c
,
d
)
∼
(
e
,
f
)
⇔
a
⋅
d
=
b
⋅
c
e
c
⋅
f
=
d
⋅
e
{\displaystyle (a,b),(c,d),(e,f)\in \Gamma ,(a,b)\sim (c,d)\;e\;(c,d)\sim (e,f)\Leftrightarrow a\cdot d=b\cdot c\;e\;c\cdot f=d\cdot e}
Como
a
⋅
d
⋅
f
=
b
⋅
c
⋅
f
e
b
⋅
c
⋅
f
=
b
⋅
d
⋅
e
⇒
a
⋅
d
⋅
f
=
b
⋅
d
⋅
e
⇒
a
⋅
f
=
b
⋅
e
⇒
(
a
,
b
)
∼
(
e
,
f
)
{\displaystyle a\cdot d\cdot f=b\cdot c\cdot f\;e\;b\cdot c\cdot f=b\cdot d\cdot e\Rightarrow a\cdot d\cdot f=b\cdot d\cdot e\Rightarrow a\cdot f=b\cdot e\Rightarrow (a,b)\sim (e,f)}
Descreva-se os elementos das classes de equivalência
[
(
m
,
1
)
]
{\displaystyle [(m,1)]}
m
∈
Z
{\displaystyle m\in \mathbb {Z} }
Dado
(
a
,
b
)
∈
[
(
m
,
1
)
]
⇒
(
a
,
b
)
∼
(
m
,
1
)
⇒
a
=
b
⋅
m
⇒
(
a
,
b
)
=
(
b
m
,
b
)
{\displaystyle (a,b)\in [(m,1)]\Rightarrow (a,b)\sim (m,1)\Rightarrow a=b\cdot m\Rightarrow (a,b)=(bm,b)}
Assim
[
(
m
,
1
)
]
=
{
x
∈
Γ
:
x
=
(
b
⋅
m
,
b
)
:
b
∈
(
Z
∖
{
0
}
)
,
m
∈
Z
}
=
{\displaystyle [(m,1)]=\{x\in \Gamma :x=(b\cdot m,b):b\in (\mathbb {Z} \setminus \{0\}),m\in \mathbb {Z} \}=}
=
{
x
∈
Γ
:
x
=
.
.
.
,
(
−
2
m
,
−
2
)
,
(
−
m
,
−
1
)
,
(
m
,
1
)
,
(
2
m
,
2
)
,
.
.
.
:
∀
m
∈
Z
}
{\displaystyle =\{x\in \Gamma :x=...,(-2m,-2),(-m,-1),(m,1),(2m,2),...:\forall \;m\in \mathbb {Z} \}}
Exemplo
[
(
5
,
1
)
]
=
{
.
.
.
,
(
−
15
,
−
3
)
,
(
−
10
,
−
2
)
,
(
−
5
,
−
1
)
,
(
5
,
1
)
,
(
10
,
2
)
,
(
15
,
3
)
,
.
.
.
}
{\displaystyle [(5,1)]=\{...,(-15,-3),(-10,-2),(-5,-1),(5,1),(10,2),(15,3),...\}}
Denotamos por
Q
{\displaystyle \mathbb {Q} }
Γ
/
∼
{\displaystyle \Gamma /\sim }
⊕
,
⊗
:
Q
×
Q
↦
Q
{\displaystyle \oplus ,\otimes :\mathbb {Q} \times \mathbb {Q} \mapsto \mathbb {Q} }
[
(
m
,
n
)
]
⊕
[
(
p
,
q
)
]
=
[
(
m
⋅
q
+
n
⋅
p
,
n
⋅
q
)
]
{\displaystyle [(m,n)]\oplus [(p,q)]=[(m\cdot q+n\cdot p,n\cdot q)]}
[
(
m
,
n
)
]
⊗
[
(
p
,
q
)
]
=
[
(
m
⋅
p
,
n
⋅
q
)
]
{\displaystyle [(m,n)]\otimes [(p,q)]=[(m\cdot p,n\cdot q)]}
Mostre que estas operações binárias não dependem do representante de classe, isto é, se
[
(
m
,
n
)
]
=
[
(
n
′
,
m
′
)
]
{\displaystyle [(m,n)]=[(n',m')]}
[
(
p
,
q
)
]
=
[
(
p
′
,
q
′
)
]
{\displaystyle [(p,q)]=[(p',q')]}
[
(
m
,
n
)
]
⊕
[
(
p
,
q
)
]
=
[
(
m
′
,
n
′
)
]
⊕
[
(
p
′
,
q
′
)
]
{\displaystyle [(m,n)]\oplus [(p,q)]=[(m',n')]\oplus [(p',q')]}
[
(
m
,
n
)
]
⊙
[
(
p
,
q
)
]
=
[
(
m
′
,
n
′
)
]
⊙
[
(
p
′
,
q
′
)
]
{\displaystyle [(m,n)]\odot [(p,q)]=[(m',n')]\odot [(p',q')]}
Prove-se as propriedades associativas e comutativas para as operações
⊕
{\displaystyle \oplus }
⊗
{\displaystyle \otimes }
A
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
x
,
y
,
z
∈
Q
{\displaystyle x,y,z\in \mathbb {Q} }
x
⊕
(
y
⊕
z
)
=
(
x
⊕
y
)
⊕
z
{\displaystyle x\oplus (y\oplus z)=(x\oplus y)\oplus z}
Sejam
x
=
[
(
a
,
b
)
]
,
y
=
[
(
c
,
d
)
]
,
z
=
[
(
e
,
f
)
]
,
a
,
b
,
c
,
d
,
e
,
f
∈
Z
,
c
o
m
b
,
d
,
f
≠
0
{\displaystyle x=[(a,b)],y=[(c,d)],z=[(e,f)],a,b,c,d,e,f\in \mathbb {Z} ,com\;b,d,f\neq 0}
Assim
x
⊕
(
y
⊕
z
)
=
(
x
⊕
y
)
⊕
z
⇔
1
[
(
a
,
b
)
]
⊕
(
[
(
c
,
d
)
]
⊕
[
(
e
,
f
)
]
)
=
(
[
a
,
b
]
⊕
[
(
c
,
d
)
]
)
⊕
[
(
e
,
f
)
]
⇔
2
{\displaystyle x\oplus (y\oplus z)=(x\oplus y)\oplus z\Leftrightarrow _{1}[(a,b)]\oplus ([(c,d)]\oplus [(e,f)])=([a,b]\oplus [(c,d)])\oplus [(e,f)]\Leftrightarrow _{2}}
⇔
2
[
(
a
,
b
)
]
⊕
[
(
c
f
+
d
e
,
d
f
)
]
=
[
(
a
d
+
b
c
,
b
d
)
]
⊕
[
(
e
,
f
)
]
⇔
3
[
(
a
(
d
f
)
+
b
(
c
f
+
d
e
)
,
b
(
d
f
)
)
]
⇔
4
{\displaystyle \Leftrightarrow _{2}[(a,b)]\oplus [(cf+de,df)]=[(ad+bc,bd)]\oplus [(e,f)]\Leftrightarrow _{3}[(a(df)+b(cf+de),b(df))]\Leftrightarrow _{4}}
⇔
4
[
a
(
d
f
)
+
b
(
c
f
+
d
e
)
]
⋅
(
b
d
)
f
=
b
(
d
f
)
⋅
[
(
a
d
+
b
c
)
f
+
(
b
d
)
e
]
⇔
4
[
(
(
a
d
+
b
c
)
f
+
(
b
d
)
e
,
(
b
d
)
f
)
]
⇔
5
{\displaystyle \Leftrightarrow _{4}[a(df)+b(cf+de)]\cdot (bd)f=b(df)\cdot [(ad+bc)f+(bd)e]\Leftrightarrow _{4}[((ad+bc)f+(bd)e,(bd)f)]\Leftrightarrow _{5}}
⇔
5
a
(
d
f
)
+
b
(
c
f
)
+
b
(
d
e
)
=
a
(
d
f
)
+
b
(
c
f
)
+
b
(
d
e
)
{\displaystyle \Leftrightarrow _{5}a(df)+b(cf)+b(de)=a(df)+b(cf)+b(de)}
onde as duplas implicações 1,2,3,4 ocorrem pela definição em
Q
{\displaystyle \mathbb {Q} }
Z
{\displaystyle \mathbb {Z} }
Logo a soma
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
x
,
y
,
z
∈
Q
{\displaystyle x,y,z\in \mathbb {Q} }
x
⊗
(
y
⊗
z
)
=
(
x
⊗
y
)
⊗
z
{\displaystyle x\otimes (y\otimes z)=(x\otimes y)\otimes z}
Sejam
x
=
[
(
a
,
b
)
]
,
y
=
[
(
c
,
d
)
]
,
z
=
[
(
e
,
f
)
]
,
a
,
b
,
c
,
d
,
e
,
f
∈
Z
,
c
o
m
b
,
d
,
f
≠
0
{\displaystyle x=[(a,b)],y=[(c,d)],z=[(e,f)],a,b,c,d,e,f\in \mathbb {Z} ,com\;b,d,f\neq 0}
Assim
x
⊗
(
y
⊗
z
)
=
(
x
⊗
y
)
⊗
z
⇔
1
[
(
a
,
b
)
]
⊗
(
[
(
c
,
d
)
]
⊗
[
(
e
,
f
)
]
)
=
(
[
a
,
b
]
⊗
[
(
c
,
d
)
]
)
⊗
[
(
e
,
f
)
]
⇔
2
{\displaystyle x\otimes (y\otimes z)=(x\otimes y)\otimes z\Leftrightarrow _{1}[(a,b)]\otimes ([(c,d)]\otimes [(e,f)])=([a,b]\otimes [(c,d)])\otimes [(e,f)]\Leftrightarrow _{2}}
⇔
2
[
(
a
,
b
)
]
⊗
[
(
c
e
,
d
f
)
]
=
[
(
a
c
,
b
d
)
]
⊗
[
(
e
,
f
)
]
⇔
3
[
(
a
(
c
e
)
,
b
(
d
f
)
)
]
=
[
(
(
a
c
)
e
,
(
b
d
)
f
)
]
⇔
4
{\displaystyle \Leftrightarrow _{2}[(a,b)]\otimes [(ce,df)]=[(ac,bd)]\otimes [(e,f)]\Leftrightarrow _{3}[(a(ce),b(df))]=[((ac)e,(bd)f)]\Leftrightarrow _{4}}
⇔
4
a
(
c
e
)
⋅
(
b
d
)
f
=
b
(
d
f
)
⋅
(
a
c
)
e
{\displaystyle \Leftrightarrow _{4}a(ce)\cdot (bd)f=b(df)\cdot (ac)e}
as duplas implicações 1,2,3 ocorrem pela definição em
Q
{\displaystyle \mathbb {Q} }
Z
{\displaystyle \mathbb {Z} }
Z
{\displaystyle \mathbb {Z} }
Logo o produto
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
A soma
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
∀
x
,
y
∈
Q
:
x
⊕
y
=
y
⊕
x
{\displaystyle \forall x,y\in \mathbb {Q} :x\oplus y=y\oplus x}
Sejam
x
=
[
(
a
,
b
)
]
,
y
=
[
(
c
,
d
)
]
,
a
,
b
,
c
,
d
,
∈
Z
,
c
o
m
b
,
d
≠
0
{\displaystyle x=[(a,b)],y=[(c,d)],a,b,c,d,\in \mathbb {Z} ,com\;b,d\neq 0}
Assim
x
⊕
y
=
1
[
(
a
,
b
)
]
⊕
[
c
,
d
)
]
=
2
[
(
a
d
+
b
c
,
b
d
)
]
=
3
(
[
d
a
+
c
b
,
d
b
]
=
4
[
(
c
,
d
)
]
)
⊕
[
(
a
,
b
)
]
=
5
y
⊕
x
{\displaystyle x\oplus y=_{1}[(a,b)]\oplus [c,d)]=_{2}[(ad+bc,bd)]=_{3}([da+cb,db]=_{4}[(c,d)])\oplus [(a,b)]=_{5}y\oplus x}
onde as igualdades 1,5 ocorrem pela definição em
Q
{\displaystyle \mathbb {Q} }
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
Z
{\displaystyle \mathbb {Z} }
Logo a soma
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
A multiplicação
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
∀
x
,
y
∈
Q
:
x
⊗
y
=
y
⊗
x
{\displaystyle \forall x,y\in \mathbb {Q} :x\otimes y=y\otimes x}
Sejam
x
=
[
(
a
,
b
)
]
,
y
=
[
(
c
,
d
)
]
,
a
,
b
,
c
,
d
,
∈
Z
,
c
o
m
b
,
d
≠
0
{\displaystyle x=[(a,b)],y=[(c,d)],a,b,c,d,\in \mathbb {Z} ,com\;b,d\neq 0}
Assim
x
⊗
y
=
1
[
(
a
,
b
)
]
⊗
[
(
c
,
d
)
]
=
2
[
(
a
c
,
b
d
)
]
=
3
(
[
c
a
,
d
b
]
=
4
[
(
c
,
d
)
]
)
⊗
[
(
a
,
b
)
]
=
5
y
⊗
x
{\displaystyle x\otimes y=_{1}[(a,b)]\otimes [(c,d)]=_{2}[(ac,bd)]=_{3}([ca,db]=_{4}[(c,d)])\otimes [(a,b)]=_{5}y\otimes x}
onde as igualdades 1,5 ocorrem por definição em
Q
{\displaystyle \mathbb {Q} }
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
⊗
{\displaystyle \otimes }
Z
{\displaystyle \mathbb {Z} }
Logo a multiplicação
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
Prove a existência do elemento neutro
θ
{\displaystyle \theta }
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
x
⊕
θ
=
θ
⊕
x
=
x
{\displaystyle x\oplus \theta =\theta \oplus x=x}
x
∈
Q
{\displaystyle x\in \mathbb {Q} }
Sejam
x
,
θ
∈
Q
{\displaystyle x,\theta \in \mathbb {Q} }
x
=
[
(
a
,
b
)
]
,
θ
=
[
(
p
,
q
)
]
{\displaystyle x=[(a,b)],\theta =[(p,q)]}
Assim
x
⊕
θ
=
x
⇔
1
[
(
a
,
b
)
]
⊕
[
(
p
,
q
)
]
=
[
(
a
,
b
)
]
⇔
2
[
(
a
q
+
b
p
,
b
q
)
]
=
[
(
a
,
b
)
]
⇔
3
(
a
q
+
b
p
)
⋅
b
=
b
q
⋅
a
⇔
4
{\displaystyle x\oplus \theta =x\Leftrightarrow _{1}[(a,b)]\oplus [(p,q)]=[(a,b)]\Leftrightarrow _{2}[(aq+bp,bq)]=[(a,b)]\Leftrightarrow _{3}(aq+bp)\cdot b=bq\cdot a\Leftrightarrow _{4}}
onde a dupla implicação 1 ocorre por definição em
Q
{\displaystyle \mathbb {Q} }
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
∼
{\displaystyle \sim }
Γ
{\displaystyle \Gamma }
Z
{\displaystyle \mathbb {Z} }
Assim
a
q
+
b
p
=
q
a
⇔
6
b
p
=
0
⇔
5
=
0
⇒
b
=
0
{\displaystyle aq+bp=qa\Leftrightarrow _{6}bp=0\Leftrightarrow _{5}=0\Rightarrow b=0}
p
=
0
{\displaystyle p=0}
b
∈
(
Z
∖
0
)
,
{\displaystyle b\in (\mathbb {Z} \setminus {0}),}
p
=
0
{\displaystyle p=0}
Portanto existe elemento neutro
θ
{\displaystyle \theta }
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
θ
=
[
(
0
,
q
)
]
,
∀
q
∈
(
Z
∖
{
0
}
)
{\displaystyle \theta =[(0,q)],\forall \;q\in (\mathbb {Z} \setminus \{0\})}
Vamos mostrar que
θ
=
[
(
0
,
1
)
]
{\displaystyle \theta =[(0,1)]}
θ
=
[
(
0
,
q
)
]
=
[
(
0
,
1
)
]
⇒
0
⋅
1
=
q
⋅
0
{\displaystyle \theta =[(0,q)]=[(0,1)]\Rightarrow 0\cdot 1=q\cdot 0}
Portanto
θ
=
[
(
0
,
1
)
]
{\displaystyle \theta =[(0,1)]}
Prove a existência do elemento neutro
I
{\displaystyle \mathrm {I} }
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
x
⊗
I
=
I
⊗
x
=
x
{\displaystyle x\otimes \mathrm {I} =\mathrm {I} \otimes x=x}
x
∈
Q
{\displaystyle x\in \mathbb {Q} }
Sejam
x
,
I
∈
Q
{\displaystyle x,\mathrm {I} \in \mathbb {Q} }
x
=
[
(
a
,
b
)
]
,
I
=
[
(
p
,
q
)
]
{\displaystyle x=[(a,b)],\mathrm {I} =[(p,q)]}
Assim
x
⊗
θ
=
x
⇔
1
[
(
a
,
b
)
]
⊗
[
(
p
,
q
)
]
=
[
(
a
,
b
)
]
⇔
2
[
(
a
p
,
b
q
)
]
=
[
(
a
,
b
)
]
⇔
3
a
p
b
=
b
q
a
⇔
4
p
=
q
{\displaystyle x\otimes \theta =x\Leftrightarrow _{1}[(a,b)]\otimes [(p,q)]=[(a,b)]\Leftrightarrow _{2}[(ap,bq)]=[(a,b)]\Leftrightarrow _{3}apb=bqa\Leftrightarrow _{4}p=q}
onde a dupla implicação 1 ocorre por definição em
Q
{\displaystyle \mathbb {Q} }
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
∼
{\displaystyle \sim }
Γ
{\displaystyle \Gamma }
Z
{\displaystyle \mathbb {Z} }
Portanto existe elemento neutro para a multiplicação
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
I
=
[
(
p
,
q
)
]
=
[
(
q
,
q
)
]
{\displaystyle \mathrm {I} =[(p,q)]=[(q,q)]}
q
∈
(
Z
∖
0
)
,
{\displaystyle q\in (\mathbb {Z} \setminus {0}),}
Vamos mostrar que
I
=
[
(
1
,
1
)
]
{\displaystyle \mathrm {I} =[(1,1)]}
I
=
[
(
q
,
q
)
]
=
[
(
1
,
1
)
]
⇒
q
⋅
1
=
q
⋅
1
{\displaystyle \mathrm {I} =[(q,q)]=[(1,1)]\Rightarrow q\cdot 1=q\cdot 1}
Portanto
I
=
[
(
1
,
1
)
]
{\displaystyle \mathrm {I} =[(1,1)]}
Mostre que existe o elemento inverso para a soma
⊕
{\displaystyle \oplus }
x
∈
Q
{\displaystyle x\in \mathbb {Q} }
y
∈
Q
{\displaystyle y\in \mathbb {Q} }
x
⊕
y
=
y
⊕
x
=
θ
{\displaystyle x\oplus y=y\oplus x=\theta }
Sejam
x
,
y
∈
Q
{\displaystyle x,y\in \mathbb {Q} }
x
=
[
(
a
,
b
)
]
,
x
=
[
(
c
,
d
)
]
,
θ
=
[
(
0
,
1
)
]
{\displaystyle x=[(a,b)],x=[(c,d)],\theta =[(0,1)]}
Assim
x
⊕
y
=
θ
⇔
1
[
(
a
,
b
)
]
⊕
[
(
c
,
d
)
]
=
[
(
0
,
1
)
]
⇔
2
[
(
a
d
+
b
c
,
b
d
)
]
=
[
(
0
,
1
)
]
⇔
3
(
a
d
+
b
c
)
⋅
1
=
b
d
⋅
0
⇔
4
{\displaystyle x\oplus y=\theta \Leftrightarrow _{1}[(a,b)]\oplus [(c,d)]=[(0,1)]\Leftrightarrow _{2}[(ad+bc,bd)]=[(0,1)]\Leftrightarrow _{3}(ad+bc)\cdot 1=bd\cdot 0\Leftrightarrow _{4}}
⇔
4
a
d
+
b
c
=
0
=
a
d
−
a
d
⇔
5
{\displaystyle \Leftrightarrow _{4}ad+bc=0=ad-ad\Leftrightarrow _{5}}
onde a dupla implicação 1 ocorre por definição em
Q
{\displaystyle \mathbb {Q} }
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
∼
{\displaystyle \sim }
Γ
{\displaystyle \Gamma }
Z
{\displaystyle \mathbb {Z} }
Z
{\displaystyle \mathbb {Z} }
Assim
c
=
−
a
d
⇒
[
(
c
,
d
)
]
=
[
(
−
a
,
b
)
]
=
y
{\displaystyle c=-ad\Rightarrow [(c,d)]=[(-a,b)]=y}
Portanto existe elemento inverso para a soma
⊕
{\displaystyle \oplus }
Q
{\displaystyle \mathbb {Q} }
x
∈
Q
{\displaystyle x\in \mathbb {Q} }
=
[
(
−
a
,
b
)
]
,
∀
a
,
b
∈
Z
,
c
o
m
b
≠
0
{\displaystyle =[(-a,b)],\forall \;a,b\in \mathbb {Z} ,com\;b\neq 0}
Mostre que existe elemento inverso para a multiplicação
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
x
∈
Q
{\displaystyle x\in \mathbb {Q} }
y
∈
Q
{\displaystyle y\in \mathbb {Q} }
x
⊗
y
=
y
⊗
x
=
I
{\displaystyle x\otimes y=y\otimes x=\mathrm {I} }
Sejam
x
,
y
∈
Q
{\displaystyle x,y\in \mathbb {Q} }
x
=
[
(
a
,
b
)
]
,
y
=
[
(
c
,
d
)
]
,
I
=
[
(
1
,
1
)
]
{\displaystyle x=[(a,b)],y=[(c,d)],\mathrm {I} =[(1,1)]}
Assim
x
⊗
y
=
I
⇔
1
[
(
a
,
b
)
]
⊗
[
(
c
,
d
)
]
=
[
(
1
,
1
)
]
⇔
2
[
(
a
c
,
b
d
)
]
=
[
(
1
,
1
)
]
⇔
3
a
c
⋅
1
=
b
d
⋅
1
⇔
4
a
c
=
b
d
{\displaystyle x\otimes y=\mathrm {I} \Leftrightarrow _{1}[(a,b)]\otimes [(c,d)]=[(1,1)]\Leftrightarrow _{2}[(ac,bd)]=[(1,1)]\Leftrightarrow _{3}ac\cdot 1=bd\cdot 1\Leftrightarrow _{4}ac=bd}
onde a dupla implicação 1 ocorre por definição em
Q
{\displaystyle \mathbb {Q} }
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
∼
{\displaystyle \sim }
Γ
{\displaystyle \Gamma }
Z
{\displaystyle \mathbb {Z} }
Assim
a
c
=
b
d
⇔
[
(
c
,
d
)
]
=
[
(
b
,
a
)
]
=
y
{\displaystyle ac=bd\Leftrightarrow [(c,d)]=[(b,a)]=y}
Portanto existe elemento inverso para a multiplicação
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
x
=
[
(
a
,
b
)
]
∈
Q
{\displaystyle x=[(a,b)]\in \mathbb {Q} }
y
=
x
−
1
{\displaystyle y=x^{-1}}
=
[
(
b
,
a
)
]
,
∀
a
,
b
∈
Z
,
c
o
m
b
≠
0
{\displaystyle =[(b,a)],\forall \;a,b\in \mathbb {Z} ,com\;b\neq 0}
Mostre que
x
⊗
θ
=
θ
⊗
x
=
θ
{\displaystyle x\otimes \theta =\theta \otimes x=\theta }
x
∈
Q
{\displaystyle x\in \mathbb {Q} }
Sejam
x
,
θ
∈
Q
{\displaystyle x,\theta \in \mathbb {Q} }
x
=
[
(
a
,
b
)
]
,
θ
=
[
(
0
,
1
)
]
{\displaystyle x=[(a,b)],\theta =[(0,1)]}
Assim
x
⊗
θ
=
θ
⇔
1
[
(
a
,
b
)
]
⊗
[
(
0
,
1
)
]
=
[
(
0
,
1
)
]
⇔
2
[
(
a
⋅
0
,
b
⋅
1
)
]
=
[
(
0
,
1
)
]
⇔
3
[
(
0
,
b
)
]
=
[
(
0
,
1
)
]
⇔
4
0
⋅
1
=
b
⋅
0
{\displaystyle x\otimes \theta =\theta \Leftrightarrow _{1}[(a,b)]\otimes [(0,1)]=[(0,1)]\Leftrightarrow _{2}[(a\cdot 0,b\cdot 1)]=[(0,1)]\Leftrightarrow _{3}[(0,b)]=[(0,1)]\Leftrightarrow _{4}0\cdot 1=b\cdot 0}
⇔
5
0
=
0
{\displaystyle \Leftrightarrow _{5}0=0}
onde a dupla implicação 1 ocorre por definição em
Q
{\displaystyle \mathbb {Q} }
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
∼
{\displaystyle \sim }
Γ
{\displaystyle \Gamma }
Como x foi tomado de modo arbitrário em
Q
{\displaystyle \mathbb {Q} }
x
⊗
θ
=
θ
⊗
x
=
θ
{\displaystyle x\otimes \theta =\theta \otimes x=\theta }
x
∈
Q
{\displaystyle x\in \mathbb {Q} }
Mostre em
Q
{\displaystyle \mathbb {Q} }
x
⊗
y
=
θ
{\displaystyle x\otimes y=\theta }
x
=
θ
o
u
y
=
θ
{\displaystyle x=\theta \;ou\;y=\theta }
Q
{\displaystyle \mathbb {Q} }
Sejam
x
,
y
∈
Q
{\displaystyle x,y\in \mathbb {Q} }
x
=
[
(
a
,
b
)
]
,
y
=
[
(
c
,
d
)
]
,
θ
=
[
(
0
,
1
)
]
{\displaystyle x=[(a,b)],y=[(c,d)],\theta =[(0,1)]}
Assim
x
⊗
y
=
θ
⇔
1
[
(
a
,
b
)
]
⊗
[
(
c
,
d
)
]
=
[
(
0
,
1
)
]
⇔
2
[
(
a
⋅
c
,
b
⋅
d
)
]
=
[
(
0
,
1
)
]
⇔
3
a
c
⋅
1
=
b
d
⋅
0
⇔
4
a
c
=
0
{\displaystyle x\otimes y=\theta \Leftrightarrow _{1}[(a,b)]\otimes [(c,d)]=[(0,1)]\Leftrightarrow _{2}[(a\cdot c,b\cdot d)]=[(0,1)]\Leftrightarrow _{3}ac\cdot 1=bd\cdot 0\Leftrightarrow _{4}ac=0}
onde a dupla implicação 1 ocorre por definição em
Q
{\displaystyle \mathbb {Q} }
⊗
{\displaystyle \otimes }
Q
{\displaystyle \mathbb {Q} }
∼
{\displaystyle \sim }
Γ
{\displaystyle \Gamma }
Vamos usar na equação
a
c
=
0
{\displaystyle ac=0}
a
=
c
o
u
a
<
c
o
u
a
>
c
{\displaystyle a=c\;ou\;a<c\;ou\;a>c}
Caso a<c, existe um inteiro k tal que
a
+
k
=
c
{\displaystyle a+k=c}
a
c
=
0
⇔
5
a
⋅
(
a
+
k
)
=
0
⇔
6
a
a
+
a
k
=
a
k
−
a
k
⇔
7
a
a
=
−
a
k
⇔
8
a
=
−
k
{\displaystyle ac=0\Leftrightarrow _{5}a\cdot (a+k)=0\Leftrightarrow _{6}aa+ak=ak-ak\Leftrightarrow _{7}aa=-ak\Leftrightarrow _{8}a=-k}
Assim
a
+
k
=
c
⇔
9
c
=
−
k
+
k
⇔
10
c
=
0
{\displaystyle a+k=c\Leftrightarrow _{9}c=-k+k\Leftrightarrow _{10}c=0}
y
=
[
(
c
,
d
)
]
=
[
(
0
,
d
)
]
=
[
(
0
,
1
)
]
=
θ
{\displaystyle y=[(c,d)]=[(0,d)]=[(0,1)]=\theta }
onde a dupla implicação 5 ocorre pela desigualdade dos inteiros, onde a dupla implicação 6 ocorre pela distributiva, pelo inverso aditivo e elemento neutro da soma dos inteiros, onde as duplas implicações 7,8 ocorrem pela lei do corte nos inteiros, a dupla implicação 9 ocorre pela substituição do valor de "a" e a dupla implicação 10 ocorre pelo inverso aditivo e elemento neutro da soma dos inteiros.
Caso c<a, analogamente teríamos
x
=
[
(
a
,
b
)
]
=
[
(
0
,
b
)
]
=
[
(
0
,
1
)
]
=
θ
{\displaystyle x=[(a,b)]=[(0,b)]=[(0,1)]=\theta }
Caso c=a, suponhamos por contradição que
a
=
c
≠
0
⇒
a
=
c
≥
1
o
u
a
=
c
≤
1.
{\displaystyle a=c\neq 0\Rightarrow a=c\geq 1\;ou\;a=c\leq 1.}
Caso
a
=
c
≥
1
⇒
a
≥
1
,
c
≥
1
⇒
a
c
≥
1.
{\displaystyle a=c\geq 1\Rightarrow a\geq 1,c\geq 1\Rightarrow ac\geq 1.}
Caso
a
=
c
≤
−
1
⇒
a
≤
−
1
,
c
≤
−
1
⇒
a
c
≥
1.
{\displaystyle a=c\leq -1\Rightarrow a\leq -1,c\leq -1\Rightarrow ac\geq 1.}
Logo
a
=
0
o
u
c
=
0
⇒
x
=
[
(
0
,
1
)
]
=
θ
o
u
y
=
[
(
0
,
1
)
]
=
θ
{\displaystyle a=0\;ou\;c=0\Rightarrow x=[(0,1)]=\theta \;ou\;y=[(0,1)]=\theta }
Portanto
x
=
θ
o
u
y
=
θ
{\displaystyle x=\theta \;ou\;y=\theta }
(a1) Mostre que
Ψ
{\displaystyle \Psi }
Ψ
{\displaystyle \Psi }
∀
Ψ
(
p
)
,
Ψ
(
q
)
∈
Ψ
(
Z
)
,
Ψ
(
p
)
=
Ψ
(
q
)
⇒
p
=
q
{\displaystyle \forall \;\Psi (p),\Psi (q)\in \Psi (\mathbb {Z} ),\Psi (p)=\Psi (q)\Rightarrow p=q}
Mas por definição de
Ψ
,
Ψ
(
p
)
=
[
(
p
,
1
)
]
,
Ψ
(
q
)
=
[
(
q
,
1
)
]
.
{\displaystyle \Psi ,\Psi (p)=[(p,1)],\Psi (q)=[(q,1)].}
Tomando
Ψ
(
p
)
=
Ψ
(
q
)
⇒
[
(
p
,
1
)
]
=
[
(
q
,
1
)
]
⇒
(
p
,
1
)
∼
(
q
,
1
)
⇒
p
⋅
1
=
1
⋅
q
⇒
p
=
q
{\displaystyle \Psi (p)=\Psi (q)\Rightarrow [(p,1)]=[(q,1)]\Rightarrow (p,1)\sim (q,1)\Rightarrow p\cdot 1=1\cdot q\Rightarrow p=q}
Portanto
Ψ
{\displaystyle \Psi }
(a2) Mostre que
Ψ
(
Z
)
=
Z
{\displaystyle \Psi (\mathbb {Z} )={\mathfrak {Z}}}
Ψ
(
Z
)
=
Z
{\displaystyle \Psi (\mathbb {Z} )={\mathfrak {Z}}}
Ψ
(
Z
)
⊂
Z
e
Z
⊂
Ψ
(
Z
)
{\displaystyle \Psi (\mathbb {Z} )\subset {\mathfrak {Z}}\;e\;{\mathfrak {Z}}\subset \Psi (\mathbb {Z} )}
Vamos mostrar que
Ψ
(
Z
)
⊂
Z
{\displaystyle \Psi (\mathbb {Z} )\subset {\mathfrak {Z}}}
y
∈
Ψ
(
Z
)
⇒
∃
x
∈
Z
,
t
a
l
q
u
e
y
=
Ψ
(
x
)
=
[
(
x
,
1
)
]
,
{\displaystyle y\in \Psi (\mathbb {Z} )\Rightarrow \exists \;x\in \mathbb {Z} ,tal\;que\;y=\Psi (x)=[(x,1)],}
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
y
=
[
(
x
,
1
)
]
∈
Z
{\displaystyle y=[(x,1)]\in {\mathfrak {Z}}}
Suponha que
Z
⊂
Ψ
(
Z
)
{\displaystyle {\mathfrak {Z}}\subset \Psi (\mathbb {Z} )}
y
∈
Z
⇒
y
=
[
(
x
,
1
)
]
{\displaystyle y\in {\mathfrak {Z}}\Rightarrow y=[(x,1)]}
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
x
∈
Z
⇒
Ψ
(
x
)
∈
Ψ
(
Z
)
,
o
n
d
e
Ψ
(
x
)
=
[
(
x
,
1
)
]
=
y
,
l
o
g
o
y
∈
Ψ
(
Z
)
{\displaystyle x\in \mathbb {Z} \Rightarrow \Psi (x)\in \Psi (\mathbb {Z} ),onde\;\Psi (x)=[(x,1)]=y,logo\;y\in \Psi (\mathbb {Z} )}
Portanto,
Ψ
(
Z
)
=
Z
{\displaystyle \Psi (\mathbb {Z} )={\mathfrak {Z}}}
(b1) Mostre que
Ψ
(
x
+
y
)
=
Ψ
(
x
)
⊕
Ψ
(
y
)
{\displaystyle \Psi (x+y)=\Psi (x)\oplus \Psi (y)}
∀
x
,
y
∈
Z
⇒
x
+
y
∈
Z
{\displaystyle \forall x,y\in \mathbb {Z} \Rightarrow x+y\in \mathbb {Z} }
Ψ
(
a
)
=
[
(
a
,
1
)
]
,
∀
a
∈
Z
{\displaystyle \Psi (a)=[(a,1)],\forall \;a\in \mathbb {Z} }
Logo
Ψ
(
x
+
y
)
=
[
(
x
+
y
,
1
)
]
=
[
(
x
⋅
1
+
1
⋅
y
,
1
⋅
1
)
]
=
[
(
x
,
1
)
]
⊕
[
(
y
,
1
)
]
∈
Z
{\displaystyle \Psi (x+y)=[(x+y,1)]=[(x\cdot 1+1\cdot y,1\cdot 1)]=[(x,1)]\oplus [(y,1)]\in {\mathfrak {Z}}}
Dados
x
,
y
∈
Z
⇒
Ψ
(
x
)
,
Ψ
(
y
)
∈
Ψ
(
Z
)
{\displaystyle x,y\in \mathbb {Z} \Rightarrow \Psi (x),\Psi (y)\in \Psi (\mathbb {Z} )}
Ψ
(
x
)
=
[
(
x
,
1
)
]
e
Ψ
(
y
)
=
[
(
y
,
1
)
]
{\displaystyle \Psi (x)=[(x,1)]\;e\;\Psi (y)=[(y,1)]}
Assim
[
(
x
,
1
)
]
⊕
[
(
y
,
1
)
]
=
Ψ
(
x
)
⊕
Ψ
(
y
)
{\displaystyle [(x,1)]\oplus [(y,1)]=\Psi (x)\oplus \Psi (y)}
(b2) Mostre que
Ψ
(
x
⋅
y
)
=
Ψ
(
x
)
⊗
Ψ
(
y
)
{\displaystyle \Psi (x\cdot y)=\Psi (x)\otimes \Psi (y)}
∀
x
,
y
∈
Z
⇒
x
⋅
y
∈
Z
{\displaystyle \forall x,y\in \mathbb {Z} \Rightarrow x\cdot y\in \mathbb {Z} }
Ψ
(
a
)
=
[
(
a
,
1
)
]
,
∀
a
∈
Z
{\displaystyle \Psi (a)=[(a,1)],\forall \;a\in \mathbb {Z} }
Logo
Ψ
(
x
⋅
y
)
=
[
(
x
⋅
y
,
1
)
]
=
[
(
x
⋅
y
,
1
⋅
1
)
]
=
[
(
x
,
1
)
]
⊗
[
(
y
,
1
)
]
∈
Z
{\displaystyle \Psi (x\cdot y)=[(x\cdot y,1)]=[(x\cdot y,1\cdot 1)]=[(x,1)]\otimes [(y,1)]\in {\mathfrak {Z}}}
Dados
x
,
y
∈
Z
⇒
Ψ
(
x
)
,
Ψ
(
y
)
∈
Ψ
(
Z
)
{\displaystyle x,y\in \mathbb {Z} \Rightarrow \Psi (x),\Psi (y)\in \Psi (\mathbb {Z} )}
Ψ
(
x
)
=
[
(
x
,
1
)
]
e
Ψ
(
y
)
=
[
(
y
,
1
)
]
{\displaystyle \Psi (x)=[(x,1)]\;e\;\Psi (y)=[(y,1)]}
Assim
[
(
x
,
1
)
]
⊗
[
(
y
,
1
)
]
=
Ψ
(
x
)
⊗
Ψ
(
y
)
{\displaystyle [(x,1)]\otimes [(y,1)]=\Psi (x)\otimes \Psi (y)}
Conclusão: Estas propriedades mostram que
Z
{\displaystyle {\mathfrak {Z}}}
Z
{\displaystyle \mathbb {Z} }
Em
Q
{\displaystyle \mathbb {Q} }
⊘
:
Q
×
(
Q
∖
{
θ
}
)
↦
Q
{\displaystyle \oslash :\mathbb {Q} \times (\mathbb {Q} \setminus \{\theta \})\mapsto \mathbb {Q} }
x
⊘
y
=
x
⊗
y
−
1
{\displaystyle x\oslash y=x\otimes y^{-1}}
y
−
1
{\displaystyle y^{-1}}
Q
=
{
[
(
m
,
1
)
]
⊘
[
(
n
,
1
)
]
:
m
,
n
∈
Z
,
n
≠
θ
}
{\displaystyle \mathbb {Q} =\{[(m,1)]\oslash [(n,1)]:m,n\in \mathbb {Z} ,n\neq \theta \}}
Mostre que a equação
2
⋅
x
=
1
{\displaystyle 2\cdot x=1}
Q
{\displaystyle \mathbb {Q} }
Q
{\displaystyle \mathbb {Q} }
Considere que
Z
=
{
y
:
y
=
[
(
m
,
1
)
]
:
m
∈
Z
}
{\displaystyle {\mathfrak {Z}}=\{y:y=[(m,1)]:m\in \mathbb {Z} \}}
p
=
[
(
x
,
1
)
]
{\displaystyle p=[(x,1)]}
x
∈
Z
{\displaystyle x\in \mathbb {Z} }
Assim
2
⋅
x
=
1
⋅
1
⇔
(
x
,
1
)
∼
(
1
,
2
)
⇔
(
x
,
1
)
∈
[
(
1
,
2
)
]
{\displaystyle 2\cdot x=1\cdot 1\Leftrightarrow (x,1)\sim (1,2)\Leftrightarrow (x,1)\in [(1,2)]}
Pela questão 6 temos que
[
(
1
,
x
)
]
⊗
[
(
x
,
1
)
]
=
[
(
1
,
1
)
]
,
∀
x
∈
Z
{\displaystyle [(1,x)]\otimes [(x,1)]=[(1,1)],\forall \;x\in \mathbb {Z} }
[
(
1
,
2
)
]
⊗
[
(
2
,
1
)
]
=
[
(
1
,
1
)
]
{\displaystyle [(1,2)]\otimes [(2,1)]=[(1,1)]}
Mas
(
x
,
1
)
∈
[
(
1
,
2
)
]
⇒
[
(
x
,
1
)
]
=
[
(
1
,
2
)
]
⇒
[
(
x
,
1
)
]
⊗
[
(
2
,
1
)
]
=
[
(
1
,
2
)
]
⊗
[
(
2
,
1
)
]
⇒
[
(
x
,
1
)
]
⊗
[
(
2
,
1
)
]
=
[
(
2
,
2
)
]
=
[
(
1
,
1
)
]
⇒
x
=
2
,
o
u
s
e
j
a
{\displaystyle (x,1)\in [(1,2)]\Rightarrow [(x,1)]=[(1,2)]\Rightarrow [(x,1)]\otimes [(2,1)]=[(1,2)]\otimes [(2,1)]\Rightarrow [(x,1)]\otimes [(2,1)]=[(2,2)]=[(1,1)]\Rightarrow x=2,ouseja}
11. Seja a classe
[
(
m
,
n
)
]
{\displaystyle [(m,n)]}
m
⋅
n
>
0
{\displaystyle m\cdot n>0}
(
p
,
q
)
∈
[
(
m
,
n
)
]
{\displaystyle (p,q)\in [(m,n)]}
p
⋅
q
>
0
{\displaystyle p\cdot q>0}
D
⊂
Q
,
D
=
{
[
(
m
,
n
)
]
:
m
⋅
n
>
0
}
{\displaystyle D\subset \mathbb {Q} ,D=\{[(m,n)]:m\cdot n>0\}}
Q
{\displaystyle \mathbb {Q} }
(a) Se
x
,
y
∈
D
{\displaystyle x,y\in D}
x
⊕
y
{\displaystyle x\oplus y}
x
⊗
y
∈
D
{\displaystyle x\otimes y\in D}
(b) Os subconjuntos
D
,
D
−
=
{
−
x
:
x
∈
D
}
{\displaystyle D,D-=\{-x:x\in D\}}
{
θ
}
{\displaystyle \{\theta \}}
(c)
Q
=
D
∪
D
−
∪
{
θ
}
{\displaystyle Q=D\cup D^{-}\cup \{\theta \}}
Mostre-se que, se
x
≠
0
{\displaystyle x\neq 0}
Mostre-se que a equação
x
2
=
x
⊗
x
=
[
(
2
,
1
)
]
{\displaystyle x^{2}=x\otimes x=[(2,1)]}
Q
{\displaystyle \mathbb {Q} }
![{\displaystyle [(m,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e7279f2e128c8e1fa84ea72d5168134cc6b5843)
![{\displaystyle (a,b)\in [(m,1)]\Rightarrow (a,b)\sim (m,1)\Rightarrow a=b\cdot m\Rightarrow (a,b)=(bm,b)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/040e425419f6c2f1118662bb341147437070eea5)
![{\displaystyle [(m,1)]=\{x\in \Gamma :x=(b\cdot m,b):b\in (\mathbb {Z} \setminus \{0\}),m\in \mathbb {Z} \}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff0c342733bd7d98e8d46cfff486df2068cebd17)
![{\displaystyle [(5,1)]=\{...,(-15,-3),(-10,-2),(-5,-1),(5,1),(10,2),(15,3),...\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3043b416999458f799b7417e926a64fb3867aa44)
![{\displaystyle [(m,n)]\oplus [(p,q)]=[(m\cdot q+n\cdot p,n\cdot q)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f4dbf5ec1623452f082f3067be4baca23abd2ff)
![{\displaystyle [(m,n)]\otimes [(p,q)]=[(m\cdot p,n\cdot q)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67c6be9b77f47b1e0eda8d09b154b6e4dd83482f)
![{\displaystyle [(m,n)]=[(n',m')]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/107cfdda43032d9ee393b518af6198b09dbe868d)
![{\displaystyle [(p,q)]=[(p',q')]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f5c29c59416fdfefd43aafc82c505a24d68fc1e)
![{\displaystyle [(m,n)]\oplus [(p,q)]=[(m',n')]\oplus [(p',q')]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93affed525081f3dd5ccfcc9a8f9ca8bcec78a1d)
![{\displaystyle [(m,n)]\odot [(p,q)]=[(m',n')]\odot [(p',q')]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aebdd1936bbcd64212f2782a8608b91d9c7b5633)
![{\displaystyle x=[(a,b)],y=[(c,d)],z=[(e,f)],a,b,c,d,e,f\in \mathbb {Z} ,com\;b,d,f\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec1cd46fd6ed7bf279a3ce143fdf2ad50c40adac)
![{\displaystyle x\oplus (y\oplus z)=(x\oplus y)\oplus z\Leftrightarrow _{1}[(a,b)]\oplus ([(c,d)]\oplus [(e,f)])=([a,b]\oplus [(c,d)])\oplus [(e,f)]\Leftrightarrow _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/844fcdf8a29b39bc436c3faf4f75faf26c6c677e)
![{\displaystyle \Leftrightarrow _{2}[(a,b)]\oplus [(cf+de,df)]=[(ad+bc,bd)]\oplus [(e,f)]\Leftrightarrow _{3}[(a(df)+b(cf+de),b(df))]\Leftrightarrow _{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f83b19d5261f4ffa3dfe1bd2f0170061f2bb54e0)
![{\displaystyle \Leftrightarrow _{4}[a(df)+b(cf+de)]\cdot (bd)f=b(df)\cdot [(ad+bc)f+(bd)e]\Leftrightarrow _{4}[((ad+bc)f+(bd)e,(bd)f)]\Leftrightarrow _{5}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d13d7fff61ebad2499c019641bbc2dd8abc212c)
![{\displaystyle x\otimes (y\otimes z)=(x\otimes y)\otimes z\Leftrightarrow _{1}[(a,b)]\otimes ([(c,d)]\otimes [(e,f)])=([a,b]\otimes [(c,d)])\otimes [(e,f)]\Leftrightarrow _{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3121515692ae67ea8ae60e864df14a7263b7276c)
![{\displaystyle \Leftrightarrow _{2}[(a,b)]\otimes [(ce,df)]=[(ac,bd)]\otimes [(e,f)]\Leftrightarrow _{3}[(a(ce),b(df))]=[((ac)e,(bd)f)]\Leftrightarrow _{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5bc44cafaedb5c9b97675c39db2e960f3ff7c5e)
![{\displaystyle x=[(a,b)],y=[(c,d)],a,b,c,d,\in \mathbb {Z} ,com\;b,d\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d49962c3acefee4485b0e68db1b72334ae57280)
![{\displaystyle x\oplus y=_{1}[(a,b)]\oplus [c,d)]=_{2}[(ad+bc,bd)]=_{3}([da+cb,db]=_{4}[(c,d)])\oplus [(a,b)]=_{5}y\oplus x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc6f03ffa9aa3c686f3fc91a77dd8c9fa9992486)
![{\displaystyle x\otimes y=_{1}[(a,b)]\otimes [(c,d)]=_{2}[(ac,bd)]=_{3}([ca,db]=_{4}[(c,d)])\otimes [(a,b)]=_{5}y\otimes x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfe7d23c15f0ed7336525e9538d877b15e13c4d6)
![{\displaystyle x=[(a,b)],\theta =[(p,q)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0da651fa4b07e6fa6364c95783c1b4be03153fbe)
![{\displaystyle x\oplus \theta =x\Leftrightarrow _{1}[(a,b)]\oplus [(p,q)]=[(a,b)]\Leftrightarrow _{2}[(aq+bp,bq)]=[(a,b)]\Leftrightarrow _{3}(aq+bp)\cdot b=bq\cdot a\Leftrightarrow _{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c3c27cd2cfe1597444d71155136149ef020c61b)
![{\displaystyle \theta =[(0,q)],\forall \;q\in (\mathbb {Z} \setminus \{0\})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c85793add87a8f2f09692b68c5c07724e7562b65)
![{\displaystyle \theta =[(0,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39e232eda4277126c3a7b71eb3ef911ac6ab0b5e)
![{\displaystyle \theta =[(0,q)]=[(0,1)]\Rightarrow 0\cdot 1=q\cdot 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e64ebd2d8c04bbd747bd96d14b7b05e701882781)
![{\displaystyle x=[(a,b)],\mathrm {I} =[(p,q)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a38a451977a842302dea5eebfd1a65d189db45c)
![{\displaystyle x\otimes \theta =x\Leftrightarrow _{1}[(a,b)]\otimes [(p,q)]=[(a,b)]\Leftrightarrow _{2}[(ap,bq)]=[(a,b)]\Leftrightarrow _{3}apb=bqa\Leftrightarrow _{4}p=q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89b1184fb218c903a068673a0b825456dad728c8)
![{\displaystyle \mathrm {I} =[(p,q)]=[(q,q)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e02ff8caec0a7af41ec1c76ce52a9fb60faf7666)
![{\displaystyle \mathrm {I} =[(1,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12ef1487916881e861a30eb928bb7a78c106e955)
![{\displaystyle \mathrm {I} =[(q,q)]=[(1,1)]\Rightarrow q\cdot 1=q\cdot 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f221c1314895ddcd0073434013557e1c5a131c64)
![{\displaystyle x=[(a,b)],x=[(c,d)],\theta =[(0,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6492b4508ea8a3d0b77ee398ec2cfb76ae6046f3)
![{\displaystyle x\oplus y=\theta \Leftrightarrow _{1}[(a,b)]\oplus [(c,d)]=[(0,1)]\Leftrightarrow _{2}[(ad+bc,bd)]=[(0,1)]\Leftrightarrow _{3}(ad+bc)\cdot 1=bd\cdot 0\Leftrightarrow _{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a843e99df09102f23ce7a307e127f0844956f37)
![{\displaystyle c=-ad\Rightarrow [(c,d)]=[(-a,b)]=y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a4482c3f72be1652dd3eaee44de7270a3c3dbcb)
![{\displaystyle =[(-a,b)],\forall \;a,b\in \mathbb {Z} ,com\;b\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3be3083e39aeecd23c5a571d0c8fc37ae7c67073)
![{\displaystyle x=[(a,b)],y=[(c,d)],\mathrm {I} =[(1,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d75073b5fb8c1a0ddbcb309b0c2955e71ac1967e)
![{\displaystyle x\otimes y=\mathrm {I} \Leftrightarrow _{1}[(a,b)]\otimes [(c,d)]=[(1,1)]\Leftrightarrow _{2}[(ac,bd)]=[(1,1)]\Leftrightarrow _{3}ac\cdot 1=bd\cdot 1\Leftrightarrow _{4}ac=bd}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6c06149d110e42927904e22049f1ae42c45b4eb)
![{\displaystyle ac=bd\Leftrightarrow [(c,d)]=[(b,a)]=y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42dd42e573ac3488a771f7dc1c7eab175befed3e)
![{\displaystyle x=[(a,b)]\in \mathbb {Q} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/4698451f79685962d56ced586cdf3bae81738591)
![{\displaystyle =[(b,a)],\forall \;a,b\in \mathbb {Z} ,com\;b\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74a72bf9c226e7d6dac30ae6782a2ee218a84650)
![{\displaystyle x=[(a,b)],\theta =[(0,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/855356c9e02d0e831901fc3b855dd06ccc834e4a)
![{\displaystyle x\otimes \theta =\theta \Leftrightarrow _{1}[(a,b)]\otimes [(0,1)]=[(0,1)]\Leftrightarrow _{2}[(a\cdot 0,b\cdot 1)]=[(0,1)]\Leftrightarrow _{3}[(0,b)]=[(0,1)]\Leftrightarrow _{4}0\cdot 1=b\cdot 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f105ec44a434a93985d842afd63f232df8dbd029)
![{\displaystyle x=[(a,b)],y=[(c,d)],\theta =[(0,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7dd80d610c83f28923ef53b8331ccf0a63dd5b8)
![{\displaystyle x\otimes y=\theta \Leftrightarrow _{1}[(a,b)]\otimes [(c,d)]=[(0,1)]\Leftrightarrow _{2}[(a\cdot c,b\cdot d)]=[(0,1)]\Leftrightarrow _{3}ac\cdot 1=bd\cdot 0\Leftrightarrow _{4}ac=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e68e68e1e69e88f4b3941bf420769b3b5ad093f8)
![{\displaystyle y=[(c,d)]=[(0,d)]=[(0,1)]=\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/01bbc49f63c9c54bae6c5188372bbe6728599b41)
![{\displaystyle x=[(a,b)]=[(0,b)]=[(0,1)]=\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/29d3f9c678e3773c68357c374c245e0186068802)
![{\displaystyle a=0\;ou\;c=0\Rightarrow x=[(0,1)]=\theta \;ou\;y=[(0,1)]=\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d251fadb10584fe70d1a013e4dc33e3462562a8d)
![{\displaystyle {\mathfrak {Z}}\subset \mathbb {Q} ,onde\;{\mathfrak {Z}}=\{[(m,1)]:m\in \mathbb {Z} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c75cb3ef8e7e1ad5616e0f41002a3d8590ce55a)
![{\displaystyle \Psi (m)=[(m,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bdedd0429550de4909a39076878d5a260e122bd)
![{\displaystyle \Psi ,\Psi (p)=[(p,1)],\Psi (q)=[(q,1)].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95316b02d0daa0ced221009a472f9b4bf9b84f0b)
![{\displaystyle \Psi (p)=\Psi (q)\Rightarrow [(p,1)]=[(q,1)]\Rightarrow (p,1)\sim (q,1)\Rightarrow p\cdot 1=1\cdot q\Rightarrow p=q}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5afdce8cc533110cff5a5c4c0acba3ef843fb052)
![{\displaystyle y\in \Psi (\mathbb {Z} )\Rightarrow \exists \;x\in \mathbb {Z} ,tal\;que\;y=\Psi (x)=[(x,1)],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1be5762da4884fe2801eb2e813968529c89b5948)
![{\displaystyle y=[(x,1)]\in {\mathfrak {Z}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b01b8208dc949557a45996dee05c6d1eeb9d4d51)
![{\displaystyle y\in {\mathfrak {Z}}\Rightarrow y=[(x,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/690180c966c419c49301dd09a4f5bef6928c49e8)
![{\displaystyle x\in \mathbb {Z} \Rightarrow \Psi (x)\in \Psi (\mathbb {Z} ),onde\;\Psi (x)=[(x,1)]=y,logo\;y\in \Psi (\mathbb {Z} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca104465c5f54a1df88b48055a4c4678321f6c72)
![{\displaystyle \Psi (a)=[(a,1)],\forall \;a\in \mathbb {Z} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f45743cc6d284f5a24d18fb264e4db365e1a525)
![{\displaystyle \Psi (x+y)=[(x+y,1)]=[(x\cdot 1+1\cdot y,1\cdot 1)]=[(x,1)]\oplus [(y,1)]\in {\mathfrak {Z}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec25e603ebfcaa6aaba5d78c74783bc311aa0f5f)
![{\displaystyle \Psi (x)=[(x,1)]\;e\;\Psi (y)=[(y,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fd6e8b21b625b4c77139f23878063e026244b00)
![{\displaystyle [(x,1)]\oplus [(y,1)]=\Psi (x)\oplus \Psi (y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8983ec0e3a6027535744962a9e5329e65307671e)
![{\displaystyle \Psi (x\cdot y)=[(x\cdot y,1)]=[(x\cdot y,1\cdot 1)]=[(x,1)]\otimes [(y,1)]\in {\mathfrak {Z}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/095e314933faf5bec18410fc3e56963c8c235532)
![{\displaystyle [(x,1)]\otimes [(y,1)]=\Psi (x)\otimes \Psi (y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/236eaaa54d48ce578346ea9e2d04d293b5f80dec)
![{\displaystyle \mathbb {Q} =\{[(m,1)]\oslash [(n,1)]:m,n\in \mathbb {Z} ,n\neq \theta \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/545c10861dde7f0417cd8ea2d816da3fb22646ab)
![{\displaystyle {\mathfrak {Z}}=\{y:y=[(m,1)]:m\in \mathbb {Z} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/572bfab96bb9386d358625d2ba8724517665e1ef)
![{\displaystyle p=[(x,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86f69138d3fcde99daa9440b39b0430959882c99)
![{\displaystyle 2\cdot x=1\cdot 1\Leftrightarrow (x,1)\sim (1,2)\Leftrightarrow (x,1)\in [(1,2)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dbc2ecfd8b03eb05f45142fbcde3b3694129a1b)
![{\displaystyle [(1,x)]\otimes [(x,1)]=[(1,1)],\forall \;x\in \mathbb {Z} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fda6232288ec5f1450c165712640e1232546b9d)
![{\displaystyle [(1,2)]\otimes [(2,1)]=[(1,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98f58a3c97cc67f9f68c010e74b5c68cb3ea72cc)
![{\displaystyle (x,1)\in [(1,2)]\Rightarrow [(x,1)]=[(1,2)]\Rightarrow [(x,1)]\otimes [(2,1)]=[(1,2)]\otimes [(2,1)]\Rightarrow [(x,1)]\otimes [(2,1)]=[(2,2)]=[(1,1)]\Rightarrow x=2,ouseja}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e896dbedf8b6d4b9f15bc03c86308e8191c8cad)
![{\displaystyle [(m,n)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ec481a97ed305a03487636f4eb201f3386c5e78)
![{\displaystyle (p,q)\in [(m,n)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b138965313c84f5a031effa590aa890ae309ffbe)
![{\displaystyle D\subset \mathbb {Q} ,D=\{[(m,n)]:m\cdot n>0\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4599b7be0d9ccfc08075ef9d33f7c0746c0bc614)
![{\displaystyle x\preceq y\Leftrightarrow [y\oplus (-x)\in D\;ou\;x=y]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d46483b0e68f55eb7b3cda147d41326b30032085)
![{\displaystyle x^{2}=x\otimes x=[(2,1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7a0e5a9fd197f8ae0d4b5c89e45a3f3ad984a92)