# 数学代写|离散数学作业代写discrete mathematics代考|Reflexive Closure

#### Doug I. Jones

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## 数学代写|离散数学作业代写discrete mathematics代考|Reflexive Closure

Let $\mathrm{R}$ be a relation defined on the set $\mathrm{A}$. Then the reflexive closure $r(\mathrm{R})$ is defined by
(i) if $(x, y) \in \mathrm{R}$ then $(x, y) \in r(\mathrm{R})$
(ii) If $x \in \mathrm{A}$, then $(x, x) \in r(\mathrm{R})$
(iii) Nothing is in $r(\mathrm{R})$ unless it is so follows from (i) and (ii).
Consider the following relation on the set $\mathrm{A}={2,4,6,8}$
Therefore
\begin{aligned} \mathrm{R} & ={(2,2),(2,4),(6,8),(6,6),(6,4)} \ r(\mathrm{R}) & ={(2,2),(2,4),(6,8),(6,6),(6,4),(4,4),(8,8)} \end{aligned}
3.16. 2 Symmetric Closure
Let $\mathrm{R}$ be a relation defined on the set $\mathrm{A}$. Then the symmetric closure $s(\mathrm{R})$ is defined by
(i) if $(x, y) \in \mathrm{R}$ then $(x, y) \in s(\mathrm{R})$
(ii) If $(x, y) \in \mathrm{R}$, then $(y, x) \in s(\mathrm{R})$
(iii) Nothing is in $\mathrm{s}(\mathrm{R})$ unless it is so follows from (i) and (ii).
Consider the following relation on the set $\mathrm{A}={2,4,6,8}$
$$\mathrm{R}={(2,2),(2,4),(2,6),(4,2),(4,6),(6,4),(6,8),(8,2)}$$
Therefore $s(\mathrm{R})={(2,2),(2,4),(2,6),(4,2),(4,6),(6,4),(6,8),(8,2),(6,2),(8,6),(2,8)}$
3.16.3 Transitive Closure
Let $\mathrm{R}$ be a relation defined on the set $\mathrm{A}$. Then the transitive closure $t(\mathrm{R})$ is defined by
(i) if $(x, y) \in \mathrm{R}$ then $(x, y) \in t(\mathrm{R})$
(ii) If $(x, y) \in \mathrm{R},(y, z) \in \mathrm{R}$ then $(x, z) \in t(\mathrm{R})$
(iii) Nothing is in $t$ (R) unless it is so follows from (i) and (ii).
Consider the following relation on the set $\mathrm{A}={2,4,6,8}$
$$R={(2,2),(2,4),(4,6),(4,8),(2,8)}$$
Therefore $t(\mathrm{R})={(2,2),(2,4),(4,6),(4,8),(2,8),(2,6)}$

## 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

$$M\left(R_1 R_2\right)=M\left(R_1\right) M\left(R_2\right)$$

$$M\left(R_1\right)=\left[\begin{array}{lllll} 1 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 \end{array}\right] \text { and } M\left(R_2\right)=\left[\begin{array}{llll} 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \end{array}\right]$$

\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}

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