Section 5.1
edit
Remember the definitions:Definition 5.3.1 Let
A
{\displaystyle A}
B
{\displaystyle B}
R
{\displaystyle R}
A
{\displaystyle A}
B
{\displaystyle B}
x
∈
A
{\displaystyle x\in A}
relation class of
x
{\displaystyle x}
R
{\displaystyle R}
R
[
x
]
{\displaystyle R[x]}
R
[
x
]
=
{
y
∈
B
|
x
R
y
}
{\displaystyle R[x]=\{y\in B|xRy\}}
Definition 2.2.1 Let
a
{\displaystyle a}
b
{\displaystyle b}
a
{\displaystyle a}
divides the number
b
{\displaystyle b}
q
{\displaystyle q}
a
q
=
b
{\displaystyle aq=b}
a
{\displaystyle a}
b
{\displaystyle b}
a
|
b
{\displaystyle a|b}
a
{\displaystyle a}
b
{\displaystyle b}
b
{\displaystyle b}
a
{\displaystyle a}
Let
a
S
b
⇔
a
=
|
b
|
{\displaystyle aSb\Leftrightarrow a=|b|}
a
,
b
∈
Z
{\displaystyle a,b\in \mathbb {Z} }
S
[
3
]
=
{
b
∈
Z
:
3
S
b
}
{\displaystyle S[3]=\{b\in \mathbb {Z} :3Sb\}}
S
{\displaystyle S}
S
[
3
]
=
{
b
∈
Z
:
3
=
|
b
|
}
{\displaystyle S[3]=\{b\in \mathbb {Z} :3=|b|\}}
3
{\displaystyle 3}
−
3
{\displaystyle -3}
3
=
|
3
|
=
|
−
3
|
{\displaystyle 3=|3|=|-3|}
S
[
3
]
=
{
−
3
,
3
}
{\displaystyle S[3]=\{-3,3\}}
S
[
−
3
]
=
{
b
∈
Z
:
−
3
S
b
}
=
{
b
∈
Z
:
−
3
=
|
b
|
}
=
∅
{\displaystyle S[-3]=\{b\in \mathbb {Z} :-3Sb\}=\{b\in \mathbb {Z} :-3=|b|\}=\emptyset }
S
[
6
]
=
{
b
∈
Z
:
6
S
b
}
=
{
b
∈
Z
:
6
=
|
b
|
}
=
{
−
6
,
6
}
{\displaystyle S[6]=\{b\in \mathbb {Z} :6Sb\}=\{b\in \mathbb {Z} :6=|b|\}=\{-6,6\}}
Let
a
D
b
⇔
a
|
b
{\displaystyle aDb\Leftrightarrow a|b}
a
,
b
∈
Z
{\displaystyle a,b\in \mathbb {Z} }
D
[
3
]
=
{
b
∈
Z
:
3
D
b
}
=
{
b
∈
Z
:
∃
k
∈
Z
and
b
=
3
k
}
=
{
.
.
.
,
−
6
,
−
3
,
−
1
,
0
,
1
,
3
,
6
,
.
.
.
}
{\displaystyle D[3]=\{b\in \mathbb {Z} :3Db\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}b=3k\}=\{...,-6,-3,-1,0,1,3,6,...\}}
D
[
−
3
]
=
{
b
∈
Z
:
−
3
D
b
}
=
{
b
∈
Z
:
∃
k
∈
Z
and
b
=
−
3
k
}
=
{
.
.
.
,
−
6
,
−
3
,
−
1
,
0
,
1
,
3
,
6
,
.
.
.
}
{\displaystyle D[-3]=\{b\in \mathbb {Z} :-3Db\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}b=-3k\}=\{...,-6,-3,-1,0,1,3,6,...\}}
D
[
6
]
=
{
b
∈
Z
:
6
D
b
}
=
{
b
∈
Z
:
∃
k
∈
Z
and
b
=
6
k
}
=
{
.
.
.
,
−
12
,
−
6
,
0
,
6
,
12
,
.
.
.
}
{\displaystyle D[6]=\{b\in \mathbb {Z} :6Db\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}b=6k\}=\{...,-12,-6,0,6,12,...\}}
Let
a
T
b
⇔
b
|
a
{\displaystyle aTb\Leftrightarrow b|a}
a
,
b
∈
Z
{\displaystyle a,b\in \mathbb {Z} }
T
[
3
]
=
{
b
∈
Z
:
3
T
b
}
=
{
b
∈
Z
:
b
|
3
}
=
{
b
∈
Z
:
∃
k
∈
Z
and
3
=
b
k
}
=
{
−
3
,
−
1
,
1
,
3
}
{\displaystyle T[3]=\{b\in \mathbb {Z} :3Tb\}=\{b\in \mathbb {Z} :b|3\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}3=bk\}=\{-3,-1,1,3\}}
T
[
−
3
]
=
{
b
∈
Z
:
−
3
T
b
}
=
{
b
∈
Z
:
b
|
−
3
}
=
{
b
∈
Z
:
∃
k
∈
Z
and
−
3
=
b
k
}
=
{
−
3
,
−
1
,
1
,
3
}
{\displaystyle T[-3]=\{b\in \mathbb {Z} :-3Tb\}=\{b\in \mathbb {Z} :b|-3\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}-3=bk\}=\{-3,-1,1,3\}}
T
[
6
]
=
{
b
∈
Z
:
6
T
b
}
=
{
b
∈
Z
:
b
|
6
}
=
{
b
∈
Z
:
∃
k
∈
Z
and
6
=
b
k
}
=
{
−
6
,
−
3
,
2
,
−
1
,
1
,
2
,
3
,
6
}
{\displaystyle T[6]=\{b\in \mathbb {Z} :6Tb\}=\{b\in \mathbb {Z} :b|6\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}6=bk\}=\{-6,-3,2,-1,1,2,3,6\}}
Let
a
Q
b
⇔
a
+
b
=
7
{\displaystyle aQb\Leftrightarrow a+b=7}
a
,
b
∈
Z
{\displaystyle a,b\in \mathbb {Z} }
Q
[
3
]
=
{
b
∈
Z
:
3
Q
b
}
=
{
b
∈
Z
:
3
+
b
=
7
}
=
{
4
}
{\displaystyle Q[3]=\{b\in \mathbb {Z} :3Qb\}=\{b\in \mathbb {Z} :3+b=7\}=\{4\}}
Q
[
−
3
]
=
{
b
∈
Z
:
−
3
Q
b
}
=
{
b
∈
Z
:
−
3
+
b
=
7
}
=
{
10
}
{\displaystyle Q[-3]=\{b\in \mathbb {Z} :-3Qb\}=\{b\in \mathbb {Z} :-3+b=7\}=\{10\}}
Q
[
6
]
=
{
b
∈
Z
:
6
Q
b
}
=
{
b
∈
Z
:
6
+
b
=
7
}
=
{
1
}
{\displaystyle Q[6]=\{b\in \mathbb {Z} :6Qb\}=\{b\in \mathbb {Z} :6+b=7\}=\{1\}}
Let
S
{\displaystyle S}
(
x
,
y
)
S
(
z
,
w
)
⇔
y
=
3
w
for all
(
x
,
y
)
,
(
z
,
w
)
∈
R
2
{\displaystyle (x,y)S(z,w)\Leftrightarrow y=3w{\text{ for all }}(x,y),(z,w)\in \mathbb {R} ^{2}}
S
[
(
0
,
0
)
]
=
{
(
z
,
w
)
∈
R
2
:
(
0
,
0
)
S
(
z
,
w
)
}
=
{
(
z
,
0
)
}
{\displaystyle S[(0,0)]=\{(z,w)\in \mathbb {R} ^{2}:(0,0)S(z,w)\}=\{(z,0)\}}
y
=
3
w
⇒
0
=
3
w
{\displaystyle y=3w\Rightarrow 0=3w}
w
=
0
{\displaystyle w=0}
x
{\displaystyle x}
S
[
(
3
,
4
)
]
=
{
(
z
,
w
)
∈
R
2
:
(
3
,
4
)
S
(
z
,
w
)
}
=
{
(
z
,
4
3
)
}
{\displaystyle S[(3,4)]=\{(z,w)\in \mathbb {R} ^{2}:(3,4)S(z,w)\}=\{(z,{\dfrac {4}{3}})\}}
y
=
3
w
⇒
4
=
3
w
{\displaystyle y=3w\Rightarrow 4=3w}
w
=
4
3
{\displaystyle w={\dfrac {4}{3}}}
y
=
4
3
{\displaystyle y={\dfrac {4}{3}}}
Let
T
{\displaystyle T}
(
x
,
y
)
T
(
z
,
w
)
⇔
x
2
+
3
y
2
=
7
z
2
+
w
2
, for all
(
x
,
y
)
,
(
z
,
w
)
∈
R
2
{\displaystyle (x,y)T(z,w)\Leftrightarrow x^{2}+3y^{2}=7z^{2}+w^{2}{\text{, for all }}(x,y),(z,w)\in \mathbb {R} ^{2}}
T
[
(
0
,
0
)
]
=
{
(
z
,
w
)
∈
R
2
:
(
0
,
0
)
T
(
z
,
w
)
}
=
(
0
,
0
)
.
B
e
c
a
u
s
e
0
2
+
3
⋅
0
2
=
7
z
2
+
w
2
⇒
Z
=
0
a
n
d
w
=
o
{\displaystyle T[(0,0)]=\{(z,w)\in \mathbb {R} ^{2}:(0,0)T(z,w)\}=(0,0).Because0^{2}+3\cdot 0^{2}=7z^{2}+w^{2}\Rightarrow Z=0andw=o}
T
[
(
3
,
4
)
]
=
{
(
z
,
w
)
∈
R
2
:
(
3
,
4
)
T
(
z
,
w
)
}
=
{
(
z
,
57
−
7
z
2
)
}
{\displaystyle T[(3,4)]=\{(z,w)\in \mathbb {R} ^{2}:(3,4)T(z,w)\}=\{(z,{\sqrt {57-7z^{2}}})\}}
3
2
+
3
⋅
4
2
=
7
z
2
+
w
2
⇒
w
=
57
−
7
z
2
{\displaystyle 3^{2}+3\cdot 4^{2}=7z^{2}+w^{2}\Rightarrow w={\sqrt {57-7z^{2}}}}
y
=
57
−
7
z
2
{\displaystyle y={\sqrt {57-7z^{2}}}}
Let
Z
{\displaystyle Z}
(
x
,
y
)
T
(
z
,
w
)
⇔
x
=
z
∨
y
=
w
, for all
(
x
,
y
)
,
(
z
,
w
)
∈
R
2
{\displaystyle (x,y)T(z,w)\Leftrightarrow x=z\vee y=w{\text{, for all }}(x,y),(z,w)\in \mathbb {R} ^{2}}
Z
[
(
0
,
0
)
]
=
{
(
z
,
w
)
∈
R
2
:
(
0
,
0
)
Z
(
z
,
w
)
}
=
{
(
0
,
w
)
∨
(
z
,
0
)
}
{\displaystyle Z[(0,0)]=\{(z,w)\in \mathbb {R} ^{2}:(0,0)Z(z,w)\}=\{(0,w)\vee (z,0)\}}
Z
[
(
3
,
4
)
]
=
{
(
z
,
w
)
∈
R
2
:
(
3
,
4
)
Z
(
z
,
w
)
}
=
{
(
3
,
w
)
∨
(
z
,
4
)
}
{\displaystyle Z[(3,4)]=\{(z,w)\in \mathbb {R} ^{2}:(3,4)Z(z,w)\}=\{(3,w)\vee (z,4)\}}
Let
A
=
{
1
,
2
,
3
}
{\displaystyle A=\{1,2,3\}}
A
×
A
{\displaystyle A\times A}
A
{\displaystyle A}
M
=
{
(
3
,
3
)
,
(
2
,
2
)
,
(
1
,
2
)
,
(
2
,
1
)
}
{\displaystyle M=\{(3,3),(2,2),(1,2),(2,1)\}}
N
=
{
(
1
,
1
)
,
(
2
,
2
)
,
(
3
,
31
)
,
(
1
,
2
)
}
{\displaystyle N=\{(1,1),(2,2),(3,31),(1,2)\}}
![{\displaystyle R[x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ce54622cb380383ab3a42441b056626ea0d2440)
![{\displaystyle R[x]=\{y\in B|xRy\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a86e8b1fae40a11b5b16b81c2adddfaba12ce905)
![{\displaystyle S[3]=\{b\in \mathbb {Z} :3Sb\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd6d26c8ae251b42348bfe1148d9cc173f66c1d1)
![{\displaystyle S[3]=\{b\in \mathbb {Z} :3=|b|\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/571656e3c13d7446f8845cbf5623042c0e02a2ca)
![{\displaystyle S[3]=\{-3,3\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbce293708374cfba0dfb537458ceebc947bc6ca)
![{\displaystyle S[-3]=\{b\in \mathbb {Z} :-3Sb\}=\{b\in \mathbb {Z} :-3=|b|\}=\emptyset }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b36096a42df57353a29cfe789d85a0fb922b79b6)
![{\displaystyle S[6]=\{b\in \mathbb {Z} :6Sb\}=\{b\in \mathbb {Z} :6=|b|\}=\{-6,6\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d16c62f1c0043e977101848217c3329d8cb4fc32)
![{\displaystyle D[3]=\{b\in \mathbb {Z} :3Db\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}b=3k\}=\{...,-6,-3,-1,0,1,3,6,...\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d06b892758005ff97eab2f4a72bac1a34f88c697)
![{\displaystyle D[-3]=\{b\in \mathbb {Z} :-3Db\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}b=-3k\}=\{...,-6,-3,-1,0,1,3,6,...\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5f629f70ccf889b5a68be2c44cfe066c18dcdd9)
![{\displaystyle D[6]=\{b\in \mathbb {Z} :6Db\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}b=6k\}=\{...,-12,-6,0,6,12,...\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfe55274a2257dcf23821f5c0bb2639ca59acabb)
![{\displaystyle T[3]=\{b\in \mathbb {Z} :3Tb\}=\{b\in \mathbb {Z} :b|3\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}3=bk\}=\{-3,-1,1,3\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9c7267b137eee2975e85d4706cacf2b4912b58a)
![{\displaystyle T[-3]=\{b\in \mathbb {Z} :-3Tb\}=\{b\in \mathbb {Z} :b|-3\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}-3=bk\}=\{-3,-1,1,3\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e8d918cac72298eeec69432e85804b0030fe53c)
![{\displaystyle T[6]=\{b\in \mathbb {Z} :6Tb\}=\{b\in \mathbb {Z} :b|6\}=\{b\in \mathbb {Z} :\exists k\in \mathbb {Z} {\text{ and }}6=bk\}=\{-6,-3,2,-1,1,2,3,6\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68035017d732f196c195b7a82469d9fc5e4ff989)
![{\displaystyle Q[3]=\{b\in \mathbb {Z} :3Qb\}=\{b\in \mathbb {Z} :3+b=7\}=\{4\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bef41568b2c9953acde5982d3f30c874f0c52abe)
![{\displaystyle Q[-3]=\{b\in \mathbb {Z} :-3Qb\}=\{b\in \mathbb {Z} :-3+b=7\}=\{10\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/337a88e09a8264267398777dcb9fda92f26951fc)
![{\displaystyle Q[6]=\{b\in \mathbb {Z} :6Qb\}=\{b\in \mathbb {Z} :6+b=7\}=\{1\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/754d4728e2fc4bb1652748ec28954ac849f80078)
![{\displaystyle S[(0,0)]=\{(z,w)\in \mathbb {R} ^{2}:(0,0)S(z,w)\}=\{(z,0)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2276764c14a51b061f1db29da0145fd0b9cd5e9d)
![{\displaystyle S[(3,4)]=\{(z,w)\in \mathbb {R} ^{2}:(3,4)S(z,w)\}=\{(z,{\dfrac {4}{3}})\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af2cc07e2e75ca62fd02c0f92909b6c6b919a476)
![{\displaystyle T[(0,0)]=\{(z,w)\in \mathbb {R} ^{2}:(0,0)T(z,w)\}=(0,0).Because0^{2}+3\cdot 0^{2}=7z^{2}+w^{2}\Rightarrow Z=0andw=o}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19e086c230a8c0f17b2521ae930b32091353c19b)
![{\displaystyle T[(3,4)]=\{(z,w)\in \mathbb {R} ^{2}:(3,4)T(z,w)\}=\{(z,{\sqrt {57-7z^{2}}})\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ad63827bd32eea7ae9c7395970e556c16244898)
![{\displaystyle Z[(0,0)]=\{(z,w)\in \mathbb {R} ^{2}:(0,0)Z(z,w)\}=\{(0,w)\vee (z,0)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6f909fecdad3f343c28fb9dde4cf9a6a88b3e8e)
![{\displaystyle Z[(3,4)]=\{(z,w)\in \mathbb {R} ^{2}:(3,4)Z(z,w)\}=\{(3,w)\vee (z,4)\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcb745c0581d9b9693078dda18ff211f81184809)
