数学代写|现代代数代写Modern Algebra代考|Math 4120

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|现代代数代写Modern Algebra代考|Operations on Binary Relations

Operations on pairs of binary relations arise in many occasions in the study of modern algebra. Such an operation is defined now.

Let $R: X \rightarrow Y$ and $S: Y \rightarrow Z$ be two binary relations. Then the composite $S \circ R$ of $R$ and $S$ is defined by $S \circ R={(x, z) \in X \times Z:$ if there exists $y \in Y$ such that $(x, y) \in R$ and $(y, z) \in$ $S} \subseteq X \times Z$.
So, $S \circ R$ is a binary relation from $X \rightarrow Z$.
If $R$ is a binary relation from $X$ to $Y$, the inverse $R^{-1}$ is defined by
$$R^{-1}={(y, x):(x, y) \in R} \subseteq Y \times X .$$
So, $R^{-1}$ is a binary relation from $Y$ to $X$.
Proposition 1.2.1 Let $R: X \rightarrow Y$ and $S: Y \rightarrow Z$ and $T: Z \rightarrow W$ be binary relations. Then
(i) $(T \circ S) \circ R=T \circ(S \circ R)$ (associative property);
(ii) $(S \circ R)^{-1}=R^{-1} \circ S^{-1}$.
Proof (i) Clearly, both $(T \circ S) \circ R$ and $T \circ(S \circ R)$ are binary relations from $X$ to $W$. Now $(x, w) \in(T \circ S) \circ R \Leftrightarrow$ there exists $y \in Y$ such that $(x, y) \in R$ and $(y, w) \in T \circ S$, where $x \in X, w \in W \Leftrightarrow$ there exists $y \in Y$ such that $(x, y) \in R$ and there exists $z \in Z$ such that $(y, z) \in S$ and $(z, w) \in T \Leftrightarrow$ there exist $y \in Y$ and $z \in Z$ such that $(x, y) \in R,(y, z) \in S$ and $(z, w) \in T$. Thus $(x, w) \in(T \circ S) \circ R \Leftrightarrow \exists z \in Z$ such that $(x, z) \in S \circ R$ and $(z, w) \in T \Leftrightarrow(x, w) \in T \circ(S \circ R)$.
Consequently, $(T \circ S) \circ R=T \circ(S \circ R)$.
(ii) Again $S \circ R: X \rightarrow Z \Leftrightarrow(S \circ R)^{-1}: Z \rightarrow X$.
Clearly, both $(S \circ R)^{-1}$ and $R^{-1} \circ S^{-1}$ are relations from $\angle$ to $X$.
Then for $z \in Z, x \in X,(z, x) \in(S \circ R)^{-1} \Leftrightarrow(x, z) \in S \circ R \Leftrightarrow \exists y \in Y$ such that $(x, y) \in R$ and $(y, z) \in S \Leftrightarrow \exists y \in Y$ such that $(y, x) \in R^{-1}$ and $(z, y) \in S^{-1} \Leftrightarrow$ $\exists y \in Y$ such that $(z, y) \in S^{-1}$ and $(y, x) \in R^{-1} \Leftrightarrow(z, x) \in R^{-1} \circ S^{-1}$.
Consequently, $(S \circ R)^{-1}=R^{-1} \circ S^{-1}$.

数学代写|现代代数代写Modern Algebra代考|Functions or Mappings

The concept of functions (mappings) is perhaps the single most important and universal notion used in all branches of mathematics. Sets and functions are closely related. They have the capacity for vast and intricate development. We are now in a position to define a function in terms of a binary relation.

Definition 1.2.15 Let $X$ and $Y$ be two non-empty sets. A function $f$ from $X$ to $Y$ is defined to be a binary relation $f$ such that
$$(x, y) \in f \quad \text { and }(x, z) \in f \quad \text { imply } \quad y=z$$
(i.e., $f$ is single valued).
The domain of $f$ denoted by dom $f$, range of $f$ denoted by range $f$ are defined by
\begin{aligned} \operatorname{dom} f &={x \in X:(x, y) \in f \text { for some } y \in Y} \subseteq X ; \ \text { range } f &={y \in Y:(x, y) \in f \text { for some } x \in X} \subseteq Y . \end{aligned}
Remark 1 Definition $1.2 .15$ means for each $x \in \operatorname{dom} f$, there exists a unique $y \in$ range $f \subseteq Y$ such that $(x, y) \in f$. Thus a function $f$ from a set $X$ to a set $Y$ is a correspondence which assigns to each $x \in \operatorname{dom} f$ exactly one element $y \in$ range $f \subseteq Y$. If $(x, y) \in f$, we write $y=f(x) ; y$ is called the image of $x$ under $f$ and $x$ is called a preimage of $y$ under $f$.

Remark 2 In elementary calculus, all the functions have the same range, namely, the real numbers, (depicted geometrically as $y$-axis), in algebra there are many different ranges, so that when we introduce a function it is important to specify both the domain and the range of the function as part of the definition of a function.

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