Posted on Categories:Discrete Mathematics, 数学代写, 离散数学

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|离散数学代写Discrete Mathematics代考|FUNCTION

Let $\mathrm{A}$ and $\mathrm{B}$ be two non-empty sets. A relation $f$ from the set $\mathrm{A}$ to the set $\mathrm{B}$ is said to be a function if it satisfies the following two conditions.
(i) $\mathrm{D}(f)=\mathrm{A}$ and
(ii) if $\left(x_1, y_1\right) \in f$ and $\left(x_2, y_2\right) \in f$ then $y_1=y_2$.

In other words a relation $f$ from the set $A$ to the set $B$ is said to be a function if for each element $x$ in A there exists unique element $y$ in B. A function from A to B is some times denoted as $f: \mathrm{A} \rightarrow \mathrm{B}$.

Consider the following relations from the set $\mathrm{A}={1,2,3,4}$ to the set $\mathrm{B}={1,4,6,9,16,18}$.
\begin{aligned} & f_1={(1,1),(2,6),(4,9),(4,18)} \ & f_2={(1,1),(2,6),(3,9),(4,9),(4,16)} \ & f_3={(1,1),(2,4),(3,9),(4,16)} \ & f_4={(1,1),(2,4),(3,9),(4,9)} \end{aligned}
and
Now, $\mathbf{D}\left(f_1\right)={1,2,4} \neq \mathrm{A}$. Therefore $f_1$ is not a function from the set $\mathrm{A}$ to the set $\mathrm{B}$. Further $\mathrm{D}\left(f_2\right)={1,2,3,4}=\mathrm{A}$; but $(4,9) \in f_2$ and $(4,16) \in f_2$ with $9 \neq 16$. This implies $f_2$ can not be a function from the set $A$ to the set $B$.

Again $\mathrm{D}\left(f_3\right)={1,2,3,4}=\mathrm{A}$ and for every element $x \in$ A there exists unique $y \in \mathrm{B}$. Therefore $f_3$ is a function from the set A to the set B. Similarly $f_4$ is also a function. The arrow diagrams are given below.

Note: From the above discussions it is clear that One-Many and Many-Many relations are not functions.

## 数学代写|离散数学代写Discrete Mathematics代考|Domain and co-domain of a Function

Suppose that $f$ be a function from the set A to the set B. The set A is called the domain of the function $f$ where as the set $\mathrm{B}$ is called the co-domain of the function $f$.
Consider the function $f$ from the set $\mathrm{A}={a, b, c, d}$ to the set $\mathrm{B}={1,2,3,4}$ as
$$f={(a, 1),(b, 2),(c, 2),(d, 4)}$$
Therefore, domain of $f={a, b, c, d}$ and co-domain of $f={1,2,3,4}$. i.e. $\mathrm{D}(f)={a, b, c, d}$ and Co-domain $f={1,2,3,4}$.
4.1.2 Range of a Function
Let $f$ be a function from the set $\mathrm{A}$ to the set $\mathrm{B}$. The element $y \in \mathrm{B}$ which the function $f$ associates to an element $x \in \mathrm{A}$ is called the image of $x$ or the value of the function $f$ for $x$. From the definition of function it is clear that each element of A has an unique image on $B$. Therefore the range of a function $f: \mathrm{A} \rightarrow \mathrm{B}$ is defined as the image of its domain $\mathrm{A}$. Mathematically,
$$\mathrm{R}(f) \text { or } \operatorname{rng}(f)={y=f(x): x \in \mathrm{A}}$$
It is clear that $R(f) \subseteq B$.
Consider the function $f$ from $\mathrm{A}={a, b, c}$ to $\mathrm{B}={1,3,5,7,9}$ as $f={(a, 3),(b, 5),(c, 5)}$. Therefore $\mathrm{R}(f)={3,5}$.

## 数学代写|离散数学代写Discrete Mathematics代考|FUNCTION

(i) $\mathrm{D}(f)=\mathrm{A}$和
(ii)如果$\left(x_1, y_1\right) \in f$和$\left(x_2, y_2\right) \in f$，则$y_1=y_2$。

\begin{aligned} & f_1={(1,1),(2,6),(4,9),(4,18)} \ & f_2={(1,1),(2,6),(3,9),(4,9),(4,16)} \ & f_3={(1,1),(2,4),(3,9),(4,16)} \ & f_4={(1,1),(2,4),(3,9),(4,9)} \end{aligned}

## 数学代写|离散数学代写Discrete Mathematics代考|Domain and co-domain of a Function

$$f={(a, 1),(b, 2),(c, 2),(d, 4)}$$

4.1.2函数的范围

$$\mathrm{R}(f) \text { or } \operatorname{rng}(f)={y=f(x): x \in \mathrm{A}}$$

Posted on Categories:Discrete Mathematics, 数学代写, 离散数学

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|离散数学代写Discrete Mathematics代考|TYPES OF RELATIONS AND RELATION MATRIX

Let $\mathrm{A}=\left{a_1, a_2, \ldots, a_i, \ldots, a_j, \ldots \ldots, a_n\right}$ be a non-empty set and $\mathrm{R}$ be a relation defined on the set A. Hence the matrix of the relation $\mathrm{R}$ relative to the ordering $a_1, a_2, \ldots, a_i, \ldots, a_j, \ldots \ldots$, $a_n$ is defined as
\begin{aligned} \mathrm{M}(\mathrm{R}) & =\left[m_{i j}\right]{n \times n} \ m{i j} & = \begin{cases}1 & \text { If } a_i \mathrm{R} a_j \ 0 & \text { If } a_i \mathrm{R} a_j\end{cases} \end{aligned}
3.12.1 Reflexive Relations
The relation $\mathrm{R}$ is said to be reflexive if $m_{i i}=1 \forall 1 \leq i \leq n$ i.e. all elements of the main diagonal in relation matrix $\mathrm{M}(\mathrm{R})$ are 1 .
3.12.2 Symmetric Relations
The relation $\mathrm{R}$ is said to be symmetric if $m_{i j}=m_{j i} \forall 1 \leq i \leq n$ and $1 \leq j \leq n$.
In other words the relation $R$ is said to be symmetric if $M(R)=[M(R)]^T$. where $[M(R)]^{\mathrm{T}}$ represents the transpose of the relation matrix $M(R)$.
3.12.3 Transitive Relation
The relation $\mathrm{R}$ is said to be transitive if $m_{i j}=1$ and $m_{j k}=1$, then $m_{i k}=1$ for $1 \leq i \leq n ; 1 \leq j \leq n$ and $1 \leq k \leq n$.

In other words the relation $\mathrm{R}$ is said to be transitive if and only if $\mathrm{R}^2 \subseteq \mathrm{R}$. i.e. Whenever entry $i, j$ in $[\mathrm{M}(\mathrm{R})]^2$ is non-zero, entry $i, j$ in $\mathrm{M}(\mathrm{R})$ is also non-zero.
Let $R$ be a relation on the set $A$ and $R$ is transitive.
Let
$$(x, z) \in \mathrm{R}^2=\mathrm{R} . \mathrm{R} \text {. }$$
So, there exists $y \in \mathrm{A}$ such that $(x, y) \in \mathrm{R}$ and $(y, z) \in \mathrm{R}$
Thus $(x, z) \in \mathrm{R}[\because \quad \mathrm{R}$ is transitive $]$
i.e.
$$(x, z) \in \mathrm{R}^2 \Rightarrow(x, z) \in \mathrm{R}$$
Therefore
$$\mathrm{R}^2 \subseteq \mathrm{R}$$
Conversely,Suppose that $\mathrm{R}^2 \subseteq R$.
Let $\quad(x, y) \in \mathrm{R}$ and $(y, z) \in \mathrm{R}$
This implies
i.e.
$(x, z) \in \mathrm{R} . \mathrm{R}=\mathrm{R}^2$
i.e.
$(x, z) \in \mathrm{R}^2 \subseteq \mathrm{R}$
$(x, z) \in \mathrm{R}$
Therefore $\mathrm{R}$ is transitive.

## 数学代写|离散数学代写Discrete Mathematics代考|Anti-Reflexive Relations

The relation $\mathrm{R}$ is said to be anti-reflexive if $m_{i i}=0 \forall 1 \leq i \leq n$ i.e. All elements of the main diagonal in relation matrix $\mathrm{M}(\mathrm{R})$ are 0 (zero).
3.12.5 Asymmetric Relations
The relation R is said to be asymmetric if $m_{i j}=1$, then $m_{j i}=0$ and $m_{i i}=0$.
3.12.6 Anti-Symmetric Relations
The relation R is said to be anti-symmetric if $a_i \neq a_j$ then either $m_{i j}=0$ or $m_{j i}=0$ and $m_{i j}=1$ $=m_{j i}$ implies $a_i=a_j$.
Consider the following relations on the set $\mathrm{A}={1,3,5,7}$
\begin{aligned} & \mathrm{R}1={(1,1),(1,3),(1,7),(3,3),(3,7),(5,5),(5,7),(7,7)} \ & \mathrm{R}_2={(1,1),(1,5),(1,7),(3,5),(3,7),(5,1),(5,3),(7,1),(7,3)} \ & \mathrm{R}_3={(1,1),(1,3),(1,5),(1,7),(3,1),(3,3),(3,5),(3,7),(5,7)} \ & \mathrm{R}_4={(1,3),(1,7),(3,7),(5,7),(7,1)} \ & \mathrm{R}_5={(1,3),(3,5),(5,7),(7,1),(7,3)} \ & \mathrm{R}_6={(1,1),(1,7),(7,5),(7,3),(5,3)} \end{aligned} Relative to the ordering $1,3,5,7$ we get $$\begin{array}{ll} M\left(R_1\right)=\left[\begin{array}{llll} 1 & 1 & 0 & 1 \ 0 & 1 & 0 & 1 \ 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 \end{array}\right] ; & \mathbf{M}\left(\mathbf{R}_2\right)=\left[\begin{array}{llll} 1 & 0 & 1 & 1 \ 0 & 0 & 1 & 1 \ 1 & 1 & 0 & 0 \ 1 & 1 & 0 & 0 \end{array}\right] ; \ M\left(R_3\right)=\left[\begin{array}{llll} 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 \end{array}\right] ; & \mathbf{M}\left(\mathrm{R}_4\right)=\left[\begin{array}{llll} 0 & 1 & 0 & 1 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 \end{array}\right] ; \ \mathbf{M}\left(\mathrm{R}_5\right)=\left[\begin{array}{llll} 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ 1 & 1 & 0 & 0 \end{array}\right] ; & \mathbf{M}\left(\mathrm{R}_6\right)=\left[\begin{array}{llll} 1 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 1 & 1 & 0 \end{array}\right] ; \end{array}$$ From the above matrices it is clear that $m{i i}=1$ in $\mathbf{M}\left(\mathrm{R}1\right)$ and $m{i i}=0$ in $\mathbf{M}\left(\mathrm{R}_4\right)$ and $\mathbf{M}\left(\mathrm{R}_5\right)$. Thus the relation $R_1$ is reflexive where as the relations $R_4$ and $R_5$ are anti-reflexive. Again
$$\left[M\left(R_2\right)\right]^T=\left[\begin{array}{llll} 1 & 0 & 1 & 1 \ 0 & 0 & 1 & 1 \ 1 & 1 & 0 & 0 \ 1 & 1 & 0 & 0 \end{array}\right]=M\left(R_2\right)$$
So, the relation $R_2$ is symmetric. Also $\left[M\left(R_1\right)\right]^T \neq M\left(R_1\right)$, and hence the relation $R_1$ is not symmetric. Similarly it can be shown that the relations $R_3, R_4, R_5$ and $R_6$ are not symmetric.
Now in $M\left(R_1\right), M\left(R_2\right), M\left(R_3\right)$ and $M\left(R_6\right)$, we see that $m_{i i} \neq 0$, so the relations $R_1, R_2, R_3$ and $R_6$ are not asymmetric. In $\mathrm{M}\left(\mathrm{R}4\right)$ we see that $m{i i}=0$, but $m_{14}=1=m_{41}$. This violate the conditions of asymmetric relation hence not asymmetric. It is also observed that in $\mathbf{M}\left(\mathrm{R}5\right), m{i i}=0 ; m_{12}$ $=1, m_{21}=0 ; m_{23}=1, m_{32}=0 ; m_{34}=1, m_{43}=0 ; m_{41}=1, m_{14}=0$ and $m_{42}=1, m_{24}=0$. Thus the relation $R_5$ is asymmetric. Again
$$\left[M\left(R_3\right)\right]^2=\left[\begin{array}{llll} 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 \end{array}\right]\left[\begin{array}{llll} 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 \end{array}\right]=\left[\begin{array}{llll} 2 & 2 & 2 & 3 \ 2 & 2 & 2 & 3 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{array}\right]$$
We see that whenever $i, j$ in $\left[\mathrm{M}\left(\mathrm{R}_3\right)\right]^2$ is non-zero, entry $i, j$ in $\mathrm{M}\left(\mathrm{R}_3\right)$ is also non-zero. So the relation $\mathrm{R}_3$ is transitive. It is also cleared that $\left[\mathrm{M}\left(\mathrm{R}_i\right)\right]^2 \nsubseteq \subset \mathrm{M}\left(\mathrm{R}_i\right)$ for $i=1,2,4,5,6$. Thus the relations $R_1, R_2, R_4, R_5$ and $R_6$ are not transitive. Also it can be shown that the relation $R_6$ is anti-symmetric.

## 数学代写|离散数学代写Discrete Mathematics代考|TYPES OF RELATIONS AND RELATION MATRIX

\begin{aligned} \mathrm{M}(\mathrm{R}) & =\left[m_{i j}\right]{n \times n} \ m{i j} & = \begin{cases}1 & \text { If } a_i \mathrm{R} a_j \ 0 & \text { If } a_i \mathrm{R} a_j\end{cases} \end{aligned}
3.12.1自反关系

3.12.2对称关系

3.12.3传递关系

$$(x, z) \in \mathrm{R}^2=\mathrm{R} . \mathrm{R} \text {. }$$

$$(x, z) \in \mathrm{R}^2 \Rightarrow(x, z) \in \mathrm{R}$$

$$\mathrm{R}^2 \subseteq \mathrm{R}$$

$(x, z) \in \mathrm{R} . \mathrm{R}=\mathrm{R}^2$

$(x, z) \in \mathrm{R}^2 \subseteq \mathrm{R}$
$(x, z) \in \mathrm{R}$

## 数学代写|离散数学代写Discrete Mathematics代考|Anti-Reflexive Relations

3.12.5非对称关系

3.12.6反对称关系

$$\left[M\left(R_3\right)\right]^2=\left[\begin{array}{llll} 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 \end{array}\right]\left[\begin{array}{llll} 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 \end{array}\right]=\left[\begin{array}{llll} 2 & 2 & 2 & 3 \ 2 & 2 & 2 & 3 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{array}\right]$$

Posted on Categories:Discrete Mathematics, 数学代写, 离散数学

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|离散数学代写Discrete Mathematics代考|RELATION MATRIX (MATRIX OF THE RELATION)

A matrix is a convenient way to represent a relation $\mathrm{R}$. Such a representation can be used by a computer to analyze the relation.
Let
\begin{aligned} & \mathrm{A}=\left{a_1, a_2, a_3, \ldots ., a_i, \ldots ., a_k\right} \ & \mathrm{B}=\left{b_1, b_2, b_3, \ldots, b_j, \ldots, b_l\right} \end{aligned}
and
be two finite sets and $R$ be a relation from the set $A$ to the set $B$. Then the matrix of the relation $R$, i.e. $M(R)$ is defined as
\begin{aligned} \mathrm{M}(\mathrm{R}) & =\left[m_{\mathrm{Ij}}\right] \text { of order }(k \times l) \ m_{\mathrm{Ij}} & = \begin{cases}1 ; & \text { if } a_i \mathrm{R} b_j \ 0 ; & \text { if } a_i \mathrm{R} b_j\end{cases} \end{aligned}
In other words label the rows of rectangular array by the elements of $A$ and the columns by the elements of B. Each position of the array is to be filled with $a 1$ (one) or 0 (zero) according as $a \in \mathrm{A}$ is related or not related to $b \in \mathrm{B}$. Consider the example

Let $\mathrm{A}={1,2,3} ; \mathrm{B}={a, b, c, d, e}$ and $\mathrm{R} \subseteq(\mathrm{A} \times \mathrm{B})$ such that $\mathrm{R}={(1, a),(1, d),(2, b),(3, c),(3, d)}$.
So the matrix of the above relation $\mathrm{R}$ is given as
$$M(R)=\left[\begin{array}{ccccc} a & b & c & d & e \ 1 \ 2 & 0 & 0 & 1 & 0 \ 3 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \end{array}\right]$$

## 数学代写|离散数学代写Discrete Mathematics代考|COMPOSITION OF RELATIONS

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 is defined by
$$\mathrm{R}_1 \mathrm{R}_2=\left{(x, z) \in(\mathrm{A} \times \mathrm{C}) \mid \text { for some } y \in \mathrm{B},(x, y) \in \mathrm{R}_1 \text { and }(y, z) \in \mathrm{R}_2\right}$$
Consider the example: Let $\mathrm{A}={1,2,4,5,7}$;
\begin{aligned} & \mathrm{B}={a, b, c, d, e} \ & \mathrm{C}={1,4,16,25} . \end{aligned}
and
Consider the relations $R_1: A \rightarrow B$ and $R_2: B \rightarrow C$ as

$\mathrm{R}_1={(1, a),(1, c),(2, d),(2, e),(5, d)}$ and $\mathrm{R}_2={(c, 1),(d, 4),(e, 25)}$. The arrow diagram is given as
So,
$$\mathrm{R}_1 \mathrm{R}_2={(1,1),(2,4),(2,25),(5,4)}$$

## 数学代写|离散数学代写Discrete Mathematics代考|RELATION MATRIX (MATRIX OF THE RELATION)

\begin{aligned} & \mathrm{A}=\left{a_1, a_2, a_3, \ldots ., a_i, \ldots ., a_k\right} \ & \mathrm{B}=\left{b_1, b_2, b_3, \ldots, b_j, \ldots, b_l\right} \end{aligned}

\begin{aligned} \mathrm{M}(\mathrm{R}) & =\left[m_{\mathrm{Ij}}\right] \text { of order }(k \times l) \ m_{\mathrm{Ij}} & = \begin{cases}1 ; & \text { if } a_i \mathrm{R} b_j \ 0 ; & \text { if } a_i \mathrm{R} b_j\end{cases} \end{aligned}

$$M(R)=\left[\begin{array}{ccccc} a & b & c & d & e \ 1 \ 2 & 0 & 0 & 1 & 0 \ 3 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \end{array}\right]$$

## 数学代写|离散数学代写Discrete Mathematics代考|COMPOSITION OF RELATIONS

$$\mathrm{R}_1 \mathrm{R}_2=\left{(x, z) \in(\mathrm{A} \times \mathrm{C}) \mid \text { for some } y \in \mathrm{B},(x, y) \in \mathrm{R}_1 \text { and }(y, z) \in \mathrm{R}_2\right}$$

\begin{aligned} & \mathrm{B}={a, b, c, d, e} \ & \mathrm{C}={1,4,16,25} . \end{aligned}

$\mathrm{R}_1={(1, a),(1, c),(2, d),(2, e),(5, d)}$ 还有$\mathrm{R}_2={(c, 1),(d, 4),(e, 25)}$。箭头图为

$$\mathrm{R}_1 \mathrm{R}_2={(1,1),(2,4),(2,25),(5,4)}$$

Posted on Categories:Discrete Mathematics, 数学代写, 离散数学

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|离散数学代写Discrete Mathematics代考|APPLICATION OF SET THEORY

Let $\mathrm{A}$ and B be finite sets. Let $n(\mathrm{~A})$ be the number of distinct elements of the set $\mathrm{A}$. Then
$$n(\mathrm{~A} \cup \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B}) .$$
Further if $A$ and $B$ are disjoint, then
$$n(\mathrm{~A} \cup \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})$$
Proof: A and B be finite sets and $n(\mathrm{~A})$ represent the number of distinct elements of the set $\mathrm{A}$.
From the above Venn diagram it is clear that
and
\begin{aligned} n(\mathrm{~A}) & =n(\mathrm{~A}-\mathrm{B})+n(\mathrm{~A} \cap \mathrm{B}) \ n(\mathrm{~B}) & =n(\mathrm{~B}-\mathrm{A})+n(\mathrm{~A} \cap \mathrm{B}) \ n(\mathrm{~A} \cup \mathrm{B}) & =n(\mathrm{~A}-\mathrm{B})+n(\mathrm{~A} \cap \mathrm{B})+n(\mathrm{~B}-\mathrm{A}) \ & =n(\mathrm{~A})-n(\mathrm{~A} \cap \mathrm{B})+n(\mathrm{~A} \cap \mathrm{B})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B}) \ & =n(\mathrm{~A})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B}) \end{aligned}
and
i.e.
$$n(\mathrm{~A} \cup \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B})$$
If $\mathrm{A}$ and $\mathrm{B}$ are disjoint, then $(\mathrm{A} \cap \mathrm{B})=\phi$ i.e. $n(\mathrm{~A} \cap \mathrm{B})=0$ Therefore, $n(\mathrm{~A} \cup \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})$.

## 数学代写|离散数学代写Discrete Mathematics代考|PRODUCT OF SETS

The product of sets is defined with the help of an order pair. An order pair is usually denoted by $(x, y)$ such that $(x, y) \neq(y, x)$ whenever $x \neq y$. The product of two sets A and B is the set of all those order pairs whose first coordinate is an element of A and the second coordinate is an element of $\mathrm{B}$. The set is denoted by $(\mathrm{A} \times \mathrm{B})$. Mathematically,
$$(\mathrm{A} \times \mathrm{B})={(x, y) \mid x \in \mathrm{A} \text { and } x \in \mathrm{B}}$$
Consider the example
Let
\begin{aligned} & \mathrm{A}={1,2,3,5,7} \ & \mathrm{B}={4,9,25} \end{aligned}
So, $(\mathrm{A} \times \mathrm{B})={(1,4),(1,9),(1,25),(2,4),(2,9),(2,25),(3,4),(3,9),(3,25),(5,4),(5,9),(5,25)$, $(7,4),(7,9),(7,25)}$

Note : The product of sets can be extendable for $n$ sets $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3, \ldots \ldots ., \mathrm{A}_n$. Thus $\mathrm{A}_1 \times \mathrm{A}_2$ $\times \mathrm{A}_3 \times \ldots . . \times \mathrm{A}_n$ can be defined as
$$\mathrm{A}_1 \times \mathrm{A}_2 \times \mathrm{A}_3 \times \ldots . . \mathrm{A}_n=\left{\left(x_1, x_2, x_3, \ldots ., x_n\right) \mid x_1 \in \mathrm{A}_1 \text { and } x_2 \in \mathrm{A}_2 \text { and } x_3 \in \mathrm{A}_3 \text { and } \ldots \text { and } x_n \in\right.$$
$\left.\mathrm{A}_n\right}$ where $\left(x_1, x_2, x_3, \ldots ., x_n\right)$ is called as $n$-tuple of $x_1, x_2, x_3, \ldots, x_n$. To explain this consider the example in which $\mathrm{A}={a, b, c} ; \mathrm{B}={1,2}$ and $\mathrm{C}={\alpha, \beta}$. Therefore
$$\mathrm{A} \times \mathrm{B} \times \mathrm{C}={(a, 1, \alpha),(a, 1, \beta),(a, 2, \alpha),(a, 2, \beta),(b, 1, \alpha),(b, 1, \beta),(b, 2, \alpha),(b, 2, \beta),(c, 1, \alpha),$$
$(c, 1, \beta),(c, 2, \alpha),(c, 2, \beta)}$.

From the above example it is very clear that $|\mathrm{A} \times \mathrm{B} \times \mathrm{C}|=|\mathrm{A}| \times|\mathrm{B}| \times|\mathrm{C}|$. In general, $\left|\mathrm{A}_1 \times \mathrm{A}_2 \times \mathrm{A}_3 \times \ldots \ldots \times \mathrm{A}_n\right|=\left|\mathrm{A}_1\right| \times\left|\mathrm{A}_2\right| \times\left|\mathrm{A}_3\right| \times \ldots \times\left|\mathrm{A}_n\right|$.

## 数学代写|离散数学代写Discrete Mathematics代考|APPLICATION OF SET THEORY

$$n(\mathrm{~A} \cup \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B}) .$$

$$n(\mathrm{~A} \cup \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})$$

\begin{aligned} n(\mathrm{~A}) & =n(\mathrm{~A}-\mathrm{B})+n(\mathrm{~A} \cap \mathrm{B}) \ n(\mathrm{~B}) & =n(\mathrm{~B}-\mathrm{A})+n(\mathrm{~A} \cap \mathrm{B}) \ n(\mathrm{~A} \cup \mathrm{B}) & =n(\mathrm{~A}-\mathrm{B})+n(\mathrm{~A} \cap \mathrm{B})+n(\mathrm{~B}-\mathrm{A}) \ & =n(\mathrm{~A})-n(\mathrm{~A} \cap \mathrm{B})+n(\mathrm{~A} \cap \mathrm{B})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B}) \ & =n(\mathrm{~A})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B}) \end{aligned}

$$n(\mathrm{~A} \cup \mathrm{B})=n(\mathrm{~A})+n(\mathrm{~B})-n(\mathrm{~A} \cap \mathrm{B})$$

## 数学代写|离散数学代写Discrete Mathematics代考|PRODUCT OF SETS

$$(\mathrm{A} \times \mathrm{B})={(x, y) \mid x \in \mathrm{A} \text { and } x \in \mathrm{B}}$$

\begin{aligned} & \mathrm{A}={1,2,3,5,7} \ & \mathrm{B}={4,9,25} \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。