4.6 作业
- 令
A
=
{
1
,
2
,
5
,
8
,
9
}
\mathbf{A} = \{1, 2, 5, 8, 9\}
A={1,2,5,8,9}, 写出
A
\mathbf{A}
A 上的 “模 2 同余” 关系及相应的划分.
答： R = { ( a , b ) ∈ A × A ∣ a m o d 2 = b m o d 2 } . \mathbf{R} = \{(a, b) \in \mathbf{A} \times \mathbf{A} \vert a \mod 2 = b \mod 2\}. R={(a,b)∈A×A∣amod2=bmod2}.
\space\space\space\space\space\space\space R = { ( 1 , 5 ) , ( 1 , 9 ) , ( 2 , 8 ) , ( 5 , 9 ) } . \mathbf{R} = \{(1,5),(1,9),(2,8),(5,9)\}. R={(1,5),(1,9),(2,8),(5,9)}.
\space\space\space\space\space\space\space P = { { 1 , 5 , 9 } , { 2 , 8 } } . \mathcal{P} = \{\{1,5,9\},\{2,8\}\}. P={{1,5,9},{2,8}}. -
A
=
{
1
,
2
,
5
,
8
,
9
}
\mathbf{A} = \{1, 2, 5, 8, 9\}
A={1,2,5,8,9},自己给定两个关系
R
1
\mathbf{R}_1
R1和
R
2
\mathbf{R}_2
R2, 并计算
R
1
R
2
\mathbf{R}_1 \mathbf{R}_2
R1R2,
R
1
+
\mathbf{R}_1^+
R1+,
R
1
∗
\mathbf{R}_1^*
R1∗.
答： R 1 = { ( 1 , 2 ) , ( 5 , 8 ) } \mathbf{R}_1=\{(1, 2), (5, 8)\} R1={(1,2),(5,8)}, R 2 = { ( 2 , 9 ) , ( 5 , 9 ) , ( 1 , 8 ) } \mathbf{R}_2=\{(2, 9), (5, 9),(1,8)\} R2={(2,9),(5,9),(1,8)}
\space\space\space\space\space\space\space R 1 R 2 = { ( 1 , 9 ) } \mathbf{R}_1 \mathbf{R}_2=\{(1,9)\} R1R2={(1,9)}
\space\space\space\space\space\space\space R 1 + = ⋃ i = 1 ∣ A ∣ R 1 i = ⋃ i = 1 5 R 1 i \mathbf{R}_1^+=\bigcup_{i = 1}^{|\mathbf{A}|} \mathbf{R}_1^i=\bigcup_{i = 1}^5\mathbf{R}_1^i R1+=⋃i=1∣A∣R1i=⋃i=15R1i
\space\space\space\space\space\space\space \space\space\space\space\space\space\space = R 1 ∪ R 1 R 1 ∪ R 1 R 1 R 1 ∪ R 1 R 1 R 1 R 1 ∪ R 1 R 1 R 1 R 1 R 1 =\mathbf{R}_1\cup\mathbf{R}_1\mathbf{R}_1\cup\mathbf{R}_1\mathbf{R}_1\mathbf{R}_1\cup\mathbf{R}_1\mathbf{R}_1\mathbf{R}_1\mathbf{R}_1\cup\mathbf{R}_1\mathbf{R}_1\mathbf{R}_1\mathbf{R}_1\mathbf{R}_1 =R1∪R1R1∪R1R1R1∪R1R1R1R1∪R1R1R1R1R1
\space\space\space\space\space\space\space\space\space\space\space\space\space\space = { ( 1 , 2 ) , ( 5 , 8 ) } ∪ ∅ ∪ ∅ ∪ ∅ ∪ ∅ =\{(1, 2), (5, 8)\}\cup\emptyset\cup\emptyset\cup\emptyset\cup\emptyset ={(1,2),(5,8)}∪∅∪∅∪∅∪∅
\space\space\space\space\space\space\space\space\space\space\space\space\space\space = { ( 1 , 2 ) , ( 5 , 8 ) } =\{(1, 2), (5, 8)\} ={(1,2),(5,8)}
\space\space\space\space\space\space\space R 1 ∗ = R + ∪ A 0 \mathbf{R}_1^*= \mathbf{R}^+ \cup \mathbf{A}^0 R1∗=R+∪A0
\space\space\space\space\space\space\space A 0 = { ( x , x ) ∣ x ∈ A } = { ( 1 , 1 ) , ( 2 , 2 ) , ( 5 , 5 ) , ( 8 , 8 ) , ( 9 , 9 ) } \mathbf{A}^0 = \{(x, x) \vert x \in A\}=\{(1,1),(2,2),(5,5),(8,8),(9,9)\} A0={(x,x)∣x∈A}={(1,1),(2,2),(5,5),(8,8),(9,9)}
\space\space\space\space\space\space\space R 1 ∗ = { ( 1 , 2 ) , ( 5 , 8 ) , ( 1 , 1 ) , ( 2 , 2 ) , ( 5 , 5 ) , ( 8 , 8 ) , ( 9 , 9 ) } \mathbf{R}_1^*=\{(1, 2), (5, 8),(1,1),(2,2),(5,5),(8,8),(9,9)\} R1∗={(1,2),(5,8),(1,1),(2,2),(5,5),(8,8),(9,9)}