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

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

avatest™

## 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

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