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数学代写|离散数学作业代写discrete mathematics代考|Composition of Relations and Relation Matrix

Let $R_1$ be a relation from the set $A$ to the set $B$ and $R_2$ be a relation from the set $B$ to the set C. That is $R_1$ is a subset of $(A \times B)$ and $R_2$ is a subset of $(B \times C)$. Then the composition of $R_1$ and $R_2$ is given by $R_1 R_2$ and the matrix of the composition $R_1 R_2$ is defined as
$$
M\left(R_1 R_2\right)=M\left(R_1\right) M\left(R_2\right)
$$
And replace all nonzero entries by 1 in $M\left(R_1 R_2\right)$ where $M\left(R_1\right)$ is the matrix of the relation $R_1$ and $M\left(R_2\right)$ is the matrix of the relation $R_2$.
Consider the same example stated above; we have
$$
M\left(R_1\right)=\left[\begin{array}{lllll}
1 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0
\end{array}\right] \text { and } M\left(R_2\right)=\left[\begin{array}{llll}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 1
\end{array}\right]
$$
So,
$$
\begin{aligned}
M\left(R_1 R_2\right) & =M\left(R_1\right) \mathbf{M}\left(R_2\right) \
& =\left[\begin{array}{lllll}
1 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{llll}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 1
\end{array}\right]=\left[\begin{array}{llll}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 1 \
0 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 0
\end{array}\right]
\end{aligned}
$$
Therefore $R_1 R_2={(1,1),(2,4),(2,25),(5,4)}$.

数学代写|离散数学作业代写discrete mathematics代考|Theorem

Let $R_1$ and $R_2$ are relations from the set $A$ to the set $B$. Let $R_3$ and $R_4$ are relations from the set B to the set C. If $R_1 \subseteq R_2$ and $R_3 \subseteq R_4$ then $R_1 R_3 \subseteq R_2 R_4$.
Proof: Given $R_1$ and $R_2$ are relations from the set $A$ to the set $B$. $R_3$ and $R_4$ are relations from the set $\mathrm{B}$ to the set $\mathrm{C}$.
Suppose that $\mathrm{R}_1 \subseteq \mathrm{R}_2$ and $\mathrm{R}_3 \subseteq \mathrm{R}_4$.
Let $(x, z) \in \mathrm{R}_1 \mathrm{R}_3$
Then for some $y \in \mathrm{B}$, we have $(x, y) \in \mathrm{R}_1$ and $(y, z) \in \mathrm{R}_3$. Therefore we have $(x, y) \in \mathrm{R}_1 \subseteq$ $\mathrm{R}_2$ and $(y, z) \in \mathrm{R}_3 \subseteq \mathrm{R}_4$.
$$
\begin{array}{ll}
\text { i.e. } & (x, y) \in \mathrm{R}_2 \text { and }(y, z) \in \mathrm{R}_4 . \
\text { This implies } & (x, z) \in \mathrm{R}_2 \mathrm{R}_4 . \
\text { Hence } & (x, z) \in \mathrm{R}_1 \mathrm{R}_3 \Rightarrow(x, z) \in \mathrm{R}_2 \mathrm{R}_4 \
\text { i.e. } & \mathrm{R}_1 \mathrm{R}_3 \subseteq \mathrm{R}_2 \mathrm{R}_4 .
\end{array}
$$
3.10.3 Theorem
Let $R_1$ be relation from the set $A$ to the set $B$ and $R_2$ be a relation from the set $B$ to the set $C$. Then.
$$
\left(R_1 R_2\right)^{-1}=R_2^{-1} R_1^{-1} \text {. }
$$
Proof: Let $R_1$ be a relation from the set $A$ to the set $B$ and $R_2$ be a relation from the set $B$ to the set C.
Our claim: $\quad\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}=\mathrm{R}_2^{-1} \mathrm{R}_1^{-1}$.
i.e.
$$
\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \subseteq \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \text { and } \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \subseteq\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}
$$
Let
$$
(x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}
$$
This implies $(z, x) \in \mathrm{R}_1 \mathrm{R}_2$. Then for some $y \in \mathrm{B}$ we have
$$
\begin{aligned}
& (z, y) \in \mathrm{R}_1 \text { and }(y, x) \in \mathrm{R}_2 \
& \Rightarrow \
& (y, z) \in \mathrm{R}_1^{-1} \text { and }(x, y) \in \mathrm{R}_2^{-1} \
& \text { i.e. } \
& (x, y) \in \mathrm{R}_2^{-1} \text { and }(y, z) \in \mathrm{R}_1^{-1} \
& (x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \
& (x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \Rightarrow(x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \
& \left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \subseteq \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \
&
\end{aligned}
$$
This implies
Therefore
i.e.
$$
(x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \Rightarrow(x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1}
$$
$$
\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \subseteq \mathrm{R}_2^{-1} \mathrm{R}_1^{-1}
$$
Again let $(x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1}$. Then for some $y \in \mathrm{B}$ we have
$\Rightarrow$
i.e.
This implies
i.e.
Therefore
i.e.
$$
\begin{aligned}
& (x, y) \in \mathrm{R}_2^{-1} \text { and }(y, z) \in \mathrm{R}_1^{-1} \
& (y, x) \in \mathrm{R}_2 \text { and }(z, y) \in \mathrm{R}_1
\end{aligned}
$$
$(z, y) \in \mathrm{R}_1$ and $(y, x) \in \mathrm{R}_2$
$(z, x) \in \mathrm{R}_1 \mathrm{R}_2$
$(x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}$
$(x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \Rightarrow(x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}$
$\mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \subseteq\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}$
Thus from equations ( $i$ ) and (ii) we get $\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}=\mathrm{R}_2^{-1} \mathrm{R}_1^{-1}$.

离散数学代写

数学代写|离散数学作业代写discrete mathematics代考|Composition of Relations and Relation Matrix

设$R_1$是集合$A$到集合$B$的关系，$R_2$是集合$B$到集合c的关系，即$R_1$是$(A \times B)$的一个子集，$R_2$是$(B \times C)$的一个子集。那么$R_1$和$R_2$的组合由$R_1 R_2$给出，组合$R_1 R_2$的矩阵定义为
$$
M\left(R_1 R_2\right)=M\left(R_1\right) M\left(R_2\right)
$$
并将$M\left(R_1 R_2\right)$中的所有非零项替换为1，其中$M\left(R_1\right)$是关系$R_1$的矩阵$M\left(R_2\right)$是关系$R_2$的矩阵。
考虑上述相同的例子;我们有
$$
M\left(R_1\right)=\left[\begin{array}{lllll}
1 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0
\end{array}\right] \text { and } M\left(R_2\right)=\left[\begin{array}{llll}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 1
\end{array}\right]
$$
所以，
$$
\begin{aligned}
M\left(R_1 R_2\right) & =M\left(R_1\right) \mathbf{M}\left(R_2\right) \
& =\left[\begin{array}{lllll}
1 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{llll}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 1
\end{array}\right]=\left[\begin{array}{llll}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 1 \
0 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 0
\end{array}\right]
\end{aligned}
$$
因此$R_1 R_2={(1,1),(2,4),(2,25),(5,4)}$。

数学代写|离散数学作业代写discrete mathematics代考|Theorem

设$R_1$和$R_2$是集合$A$到集合$B$的关系。设$R_3$和$R_4$是集合B到集合c的关系。如果$R_1 \subseteq R_2$和$R_3 \subseteq R_4$则$R_1 R_3 \subseteq R_2 R_4$。
证明:给定$R_1$和$R_2$是集合$A$到集合$B$的关系。$R_3$和$R_4$是集合$\mathrm{B}$到集合$\mathrm{C}$的关系。
假设是$\mathrm{R}_1 \subseteq \mathrm{R}_2$和$\mathrm{R}_3 \subseteq \mathrm{R}_4$。
让$(x, z) \in \mathrm{R}_1 \mathrm{R}_3$
对于一些$y \in \mathrm{B}$，我们有$(x, y) \in \mathrm{R}_1$和$(y, z) \in \mathrm{R}_3$。因此我们有$(x, y) \in \mathrm{R}_1 \subseteq$$\mathrm{R}_2$和$(y, z) \in \mathrm{R}_3 \subseteq \mathrm{R}_4$。
$$
\begin{array}{ll}
\text { i.e. } & (x, y) \in \mathrm{R}_2 \text { and }(y, z) \in \mathrm{R}_4 . \
\text { This implies } & (x, z) \in \mathrm{R}_2 \mathrm{R}_4 . \
\text { Hence } & (x, z) \in \mathrm{R}_1 \mathrm{R}_3 \Rightarrow(x, z) \in \mathrm{R}_2 \mathrm{R}_4 \
\text { i.e. } & \mathrm{R}_1 \mathrm{R}_3 \subseteq \mathrm{R}_2 \mathrm{R}_4 .
\end{array}
$$
3.10.3定理
设$R_1$是集合$A$到集合$B$的关系，$R_2$是集合$B$到集合$C$的关系。然后。
$$
\left(R_1 R_2\right)^{-1}=R_2^{-1} R_1^{-1} \text {. }
$$
证明:设$R_1$是集合$A$到集合$B$的关系，$R_2$是集合$B$到集合C的关系。
我们的索赔:$\quad\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}=\mathrm{R}_2^{-1} \mathrm{R}_1^{-1}$。
例如:
$$
\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \subseteq \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \text { and } \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \subseteq\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}
$$
让
$$
(x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}
$$
这意味着$(z, x) \in \mathrm{R}_1 \mathrm{R}_2$。对于一些$y \in \mathrm{B}$，我们有
$$
\begin{aligned}
& (z, y) \in \mathrm{R}_1 \text { and }(y, x) \in \mathrm{R}_2 \
& \Rightarrow \
& (y, z) \in \mathrm{R}_1^{-1} \text { and }(x, y) \in \mathrm{R}_2^{-1} \
& \text { i.e. } \
& (x, y) \in \mathrm{R}_2^{-1} \text { and }(y, z) \in \mathrm{R}_1^{-1} \
& (x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \
& (x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \Rightarrow(x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \
& \left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \subseteq \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \
&
\end{aligned}
$$
这意味着
因此
例如:
$$
(x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \Rightarrow(x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1}
$$
$$
\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1} \subseteq \mathrm{R}_2^{-1} \mathrm{R}_1^{-1}
$$
再次让$(x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1}$。对于一些$y \in \mathrm{B}$，我们有
$\Rightarrow$
例如:
这意味着
例如:
因此
例如:
$$
\begin{aligned}
& (x, y) \in \mathrm{R}_2^{-1} \text { and }(y, z) \in \mathrm{R}_1^{-1} \
& (y, x) \in \mathrm{R}_2 \text { and }(z, y) \in \mathrm{R}_1
\end{aligned}
$$
$(z, y) \in \mathrm{R}_1$和$(y, x) \in \mathrm{R}_2$
$(z, x) \in \mathrm{R}_1 \mathrm{R}_2$
$(x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}$
$(x, z) \in \mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \Rightarrow(x, z) \in\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}$
$\mathrm{R}_2^{-1} \mathrm{R}_1^{-1} \subseteq\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}$
因此，由式($i$)和(ii)我们得到$\left(\mathrm{R}_1 \mathrm{R}_2\right)^{-1}=\mathrm{R}_2^{-1} \mathrm{R}_1^{-1}$。

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