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A relação ~ em
N
2
{\displaystyle \mathbb {N} ^{2}}
(
a
,
b
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∼
(
c
,
d
)
⇔
a
+
d
=
b
+
c
{\displaystyle (a,b)\sim (c,d)\Leftrightarrow a+d=b+c}
é uma relação de equivalência se, e somente se for:
(reflexiva)
∀
(
a
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b
)
∈
N
2
{\displaystyle \forall \;(a,b)\in \mathbb {N} ^{2}}
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a
,
b
)
∼
(
a
,
b
)
⇔
a
+
b
=
b
+
a
{\displaystyle (a,b)\sim (a,b)\Leftrightarrow a+b=b+a}
(simétrica)
∀
(
a
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b
)
,
(
c
,
d
)
∈
N
2
{\displaystyle \forall \;(a,b),(c,d)\in \mathbb {N} ^{2}}
(
a
,
b
)
∼
(
c
,
d
)
⇔
a
+
d
=
b
+
c
⇔
d
+
a
=
c
+
b
⇔
(
c
,
d
)
∼
(
a
,
b
)
{\displaystyle (a,b)\sim (c,d)\Leftrightarrow a+d=b+c\Leftrightarrow d+a=c+b\Leftrightarrow (c,d)\sim (a,b)}
(transitiva)
∀
(
a
,
b
)
,
(
c
,
d
)
,
(
p
,
q
)
∈
N
2
{\displaystyle \forall \;(a,b),(c,d),(p,q)\in \mathbb {N} ^{2}}
(
a
,
b
)
∼
(
c
,
d
)
e
(
c
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d
)
∼
(
p
,
q
)
⇔
a
+
d
=
b
+
c
e
c
+
q
=
d
+
p
⇔
a
+
d
+
c
+
q
=
b
+
c
+
d
+
p
⇔
a
+
q
=
b
+
p
⇔
(
a
,
b
)
∼
(
p
,
q
)
{\displaystyle (a,b)\sim (c,d)\;e\;(c,d)\sim (p,q)\Leftrightarrow a+d=b+c\;e\;c+q=d+p\Leftrightarrow a+d+c+q=b+c+d+p\Leftrightarrow a+q=b+p\Leftrightarrow (a,b)\sim (p,q)}
Podemos dividir o conjunto de
N
2
{\displaystyle \mathbb {N} ^{2}}
N
2
{\displaystyle \mathbb {N} ^{2}}
caso n=1, eles coincidem com
[
(
1
,
1
)
]
=
{
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a
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b
)
∈
N
2
}
:
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a
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b
)
∼
(
1
,
1
)
}
{\displaystyle [(1,1)]=\{(a,b)\in \mathbb {N} ^{2}\}:(a,b)\sim (1,1)\}}
caso em que
n
≠
1
{\displaystyle n\neq 1}
[
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1
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n
)
]
=
{
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a
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b
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∈
N
2
}
:
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a
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b
)
∼
(
1
,
n
)
}
{\displaystyle [(1,n)]=\{(a,b)\in \mathbb {N} ^{2}\}:(a,b)\sim (1,n)\}}
[
(
n
,
1
)
]
=
{
(
a
,
b
)
∈
N
2
}
:
(
a
,
b
)
∼
(
n
,
1
)
}
{\displaystyle [(n,1)]=\{(a,b)\in \mathbb {N} ^{2}\}:(a,b)\sim (n,1)\}}
Assim podemos dizer que
N
2
/
∼=
{
[
1
,
1
]
}
∪
{
x
∈
N
2
:
x
=
[
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1
,
n
)
]
:
n
∈
N
}
∪
{
x
∈
N
2
:
x
=
[
(
n
,
1
)
]
:
n
∈
N
}
{\displaystyle \mathbb {N} ^{2}/\sim =\{[1,1]\}\cup \{x\in \mathbb {N} ^{2}:x=[(1,n)]:n\in \mathbb {N} \}\cup \{x\in \mathbb {N} ^{2}:x=[(n,1)]:n\in \mathbb {N} \}}
De forma geral dizemos que o conjunto quociente:
N
2
/
∼
=
{
[
(
a
,
b
)
]
∈
N
2
}
:
(
a
,
b
)
∈
N
2
}
{\displaystyle \mathbb {N} ^{2}/\sim \;=\{[(a,b)]\in \mathbb {N} ^{2}\}:(a,b)\in \mathbb {N} ^{2}\}}
Ao somarmos dois elementos de
N
2
{\displaystyle \mathbb {N} ^{2}}
⊕
:
N
2
×
N
2
↦
N
2
{\displaystyle \oplus :\mathbb {N} ^{2}\times \mathbb {N} ^{2}\mapsto \mathbb {N} ^{2}}
[
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b
)
]
⊕
[
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d
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]
=
[
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+
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b
+
d
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]
{\displaystyle [(a,b)]\oplus [(c,d)]=[(a+c,b+d)]}
Vamos mostrar que
⊕
{\displaystyle \oplus }
Tome
(
a
′
,
b
′
)
∈
[
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a
,
b
)
]
e
(
c
′
,
d
′
)
∈
[
(
c
,
d
)
]
⇒
(
a
′
,
b
′
)
∼
(
a
,
b
)
e
(
c
′
,
d
′
)
∼
(
c
,
d
)
⇒
{\displaystyle (a',b')\in [(a,b)]\;e\;(c',d')\in [(c,d)]\Rightarrow (a',b')\sim (a,b)\;e\;(c',d')\sim (c,d)\Rightarrow }
⇒
a
′
+
b
=
b
′
+
a
e
c
′
+
d
=
d
′
+
c
⇒
(
a
+
c
)
+
(
b
′
+
d
′
)
=
(
a
′
+
c
′
)
+
(
b
+
d
)
⇒
{\displaystyle \Rightarrow a'+b=b'+a\;e\;c'+d=d'+c\Rightarrow (a+c)+(b'+d')=(a'+c')+(b+d)\Rightarrow }
⇒
(
a
+
c
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b
+
d
)
∼
(
a
′
+
c
′
,
b
′
+
d
′
)
⇒
[
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a
+
c
,
b
+
d
)
]
=
[
(
a
′
+
c
′
,
b
′
+
d
′
)
]
{\displaystyle \Rightarrow (a+c,b+d)\sim (a'+c',b'+d')\Rightarrow [(a+c,b+d)]=[(a'+c',b'+d')]}
Ex.
[
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1
,
2
)
]
⊕
[
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3
,
4
)
]
=
[
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1
+
3
,
2
+
4
)
]
=
[
(
4
,
6
)
]
{\displaystyle [(1,2)]\oplus [(3,4)]=[(1+3,2+4)]=[(4,6)]}
(
4
,
6
)
∼
(
1
,
n
)
⇔
4
+
n
=
1
+
6
⇔
4
+
n
=
4
+
3
⇔
n
=
3
⇔
(
4
,
6
)
∼
(
1
,
3
)
{\displaystyle (4,6)\sim (1,n)\Leftrightarrow 4+n=1+6\Leftrightarrow 4+n=4+3\Leftrightarrow n=3\Leftrightarrow (4,6)\sim (1,3)}
N
2
{\displaystyle \mathbb {N} ^{2}}
(exercício) Prove que
(
a
,
b
)
∼
y
{\displaystyle (a,b)\sim y}
y
∈
{
(
1
,
x
)
,
(
1
,
1
)
,
(
x
,
1
)
}
{\displaystyle y\in \{(1,x),(1,1),(x,1)\}}
N
2
{\displaystyle \mathbb {N} ^{2}}
Ao multiplicarmos dois elementos de
N
2
{\displaystyle \mathbb {N} ^{2}}
⊙
:
N
2
×
N
2
↦
N
2
{\displaystyle \odot :\mathbb {N} ^{2}\times \mathbb {N} ^{2}\mapsto \mathbb {N} ^{2}}
[
(
a
,
b
)
]
⊙
[
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c
,
d
)
]
=
[
(
a
c
+
b
d
,
a
d
+
b
c
)
]
{\displaystyle [(a,b)]\odot [(c,d)]=[(ac+bd,ad+bc)]}
Vamos mostrar que
⊙
{\displaystyle \odot }
Tome
(
a
′
,
b
′
)
∈
[
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a
,
b
)
]
e
(
c
′
,
d
′
)
∈
[
(
c
,
d
)
]
⇒
(
a
′
,
b
′
)
∼
(
a
,
b
)
e
(
c
′
,
d
′
)
∼
(
c
,
d
)
⇒
{\displaystyle (a',b')\in [(a,b)]\;e\;(c',d')\in [(c,d)]\Rightarrow (a',b')\sim (a,b)\;e\;(c',d')\sim (c,d)\Rightarrow }
... (exercício)
⇒
(
a
c
,
b
d
)
(
a
′
c
′
,
b
′
d
′
)
⇒
[
(
a
c
,
b
d
)
]
=
[
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a
′
c
′
,
b
′
d
′
)
]
{\displaystyle \Rightarrow (ac,bd)~(a'c',b'd')\Rightarrow [(ac,bd)]=[(a'c',b'd')]}
![{\displaystyle [(1,1)]=\{(a,b)\in \mathbb {N} ^{2}\}:(a,b)\sim (1,1)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28d66582008fe47a2669df3db6f72a0a81b038c7)
![{\displaystyle [(1,n)]=\{(a,b)\in \mathbb {N} ^{2}\}:(a,b)\sim (1,n)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db5b1cfd08fffbe160ba779f171e5a8cb95b7c5d)
![{\displaystyle [(n,1)]=\{(a,b)\in \mathbb {N} ^{2}\}:(a,b)\sim (n,1)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30aaeacfc29ee3c89fc87ab0fc9b224a04fcf482)
![{\displaystyle \mathbb {N} ^{2}/\sim =\{[1,1]\}\cup \{x\in \mathbb {N} ^{2}:x=[(1,n)]:n\in \mathbb {N} \}\cup \{x\in \mathbb {N} ^{2}:x=[(n,1)]:n\in \mathbb {N} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fff506399136a57c7a7e4ead515ce9e385997021)
![{\displaystyle \mathbb {N} ^{2}/\sim \;=\{[(a,b)]\in \mathbb {N} ^{2}\}:(a,b)\in \mathbb {N} ^{2}\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9147706c014a5f61bb34f410f25d104b67b02516)
![{\displaystyle [(a,b)]\oplus [(c,d)]=[(a+c,b+d)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa46d5f7a18591cbe1916448db78ee9214c7ab62)
![{\displaystyle (a',b')\in [(a,b)]\;e\;(c',d')\in [(c,d)]\Rightarrow (a',b')\sim (a,b)\;e\;(c',d')\sim (c,d)\Rightarrow }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f4f64202b1938789b5ef80f468adee899092bbe)
![{\displaystyle \Rightarrow (a+c,b+d)\sim (a'+c',b'+d')\Rightarrow [(a+c,b+d)]=[(a'+c',b'+d')]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dfcf914928acf6540043d621e3ceea09766dd12)
![{\displaystyle [(1,2)]\oplus [(3,4)]=[(1+3,2+4)]=[(4,6)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60f9c1139d65f0b45e50803b6e7dd9b39bf69fcc)
![{\displaystyle [(a,b)]\odot [(c,d)]=[(ac+bd,ad+bc)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b912f29ee0938e01cbc54c1f4ea0868e49b5d7d1)
![{\displaystyle \Rightarrow (ac,bd)~(a'c',b'd')\Rightarrow [(ac,bd)]=[(a'c',b'd')]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72cd72a543613f9043c20c74038d3e7e9d2d750c)