![证明通常是如何在写作中构建的,一行还是分成多行?](https://linux22.com/image/402235/%E8%AF%81%E6%98%8E%E9%80%9A%E5%B8%B8%E6%98%AF%E5%A6%82%E4%BD%95%E5%9C%A8%E5%86%99%E4%BD%9C%E4%B8%AD%E6%9E%84%E5%BB%BA%E7%9A%84%EF%BC%8C%E4%B8%80%E8%A1%8C%E8%BF%98%E6%98%AF%E5%88%86%E6%88%90%E5%A4%9A%E8%A1%8C%EF%BC%9F.png)
这是以两种不同方式格式化的相同证明。我的问题是,像家庭作业这样的写作的标准是什么?我认为第一个看起来更好,但我不认为这是我通常看到的方式。如果有人能说出哪一个更好,我将不胜感激。
这是第一个输出的乳胶代码。
\subsection*{13.}
\subsubsection*{(a)}
In order to show $f(A_1 \cup A_2) = f(A_1)\cup f(A_2)$, we want to show $f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)$ and $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$.
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Let $y \in f(A_1 \cup A_2)$.
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Then there exists a $x \in A_1 \cup A_2$ such that $f(x) = y$.
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If $x \in A_1$, then $y \in f(A_1)$.
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As a result, $y \in f(A_1) \cup f(A_2)$.
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If $x \in A_2$, then $y \in f(A_2)$.
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So, $y \in f(A_1) \cup f(A_2)$.
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Therefore, $f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)$.
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Moreover, we want to show $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$.
\\
Let $y \in f(A_1) \cup f(A_2)$.
\\
If $y \in f(A_1)$, then there exists a $x \in A_1$ such that $f(x) = y$.
\\
Since $x \in A_1$, then $x \in A_1 \cup A_2$.
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If $y \in f(A_2)$, then there exists a $x \in A_2$ such that $f(x) = y$.
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Since $x \in A_2$, then $x \in A_1 \cup A_2$.
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As a result $y \in f(A_1 \cup A_2)$.
\\
Therefore, $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$.
\\
Since $f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)$ and $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$, we can conclude that $f(A_1 \cup A_2) = f(A_1)\cup f(A_2)$.
答案1
- 不要使用
\\
转到新行,如果需要新行,请使用\newline
,如果需要新段落,请留空行。\\
仅在表格中使用。 - 为什么要自己命名
\subsection
而\subsubsection
不是让 LaTeX 替你命名?是为了匹配作业编号还是只是因为你不能修改编号布局?
这就是我要做的。当然,我不了解这个主题,所以你可以改进它,把 displaymath 放在真正需要的地方。
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\subsection*{13.}% Why are you numbering the section and don't let LaTeX do it for you?
\subsubsection*{(a)}
In order to show \[f(A_1 \cup A_2) = f(A_1)\cup f(A_2)\] we want to show \[f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2) \quad \text{and} \quad f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\,\text{.}\]
Let $y \in f(A_1 \cup A_2)$.
Then there exists a $x \in A_1 \cup A_2$ such that $f(x) = y$.
If $x \in A_1$, then $y \in f(A_1)$.
As a result, \[y \in f(A_1) \cup f(A_2)\,\text{.}\]
If $x \in A_2$, then $y \in f(A_2)$.
So, $y \in f(A_1) \cup f(A_2)$.
Therefore, \[f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)\,\text{.}\]
Moreover, we want to show \[f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\,\text{.}\]
Let $y \in f(A_1) \cup f(A_2)$.
If $y \in f(A_1)$, then there exists a $x \in A_1$ such that $f(x) = y$.
Since $x \in A_1$, then $x \in A_1 \cup A_2$.
If $y \in f(A_2)$, then there exists a $x \in A_2$ such that $f(x) = y$.
Since $x \in A_2$, then $x \in A_1 \cup A_2$.
As a result \[y \in f(A_1 \cup A_2)\,\text{.}\]
Therefore, \[f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\,\text{.}\]
Since $f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)$ and $f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)$, we can conclude that \[f(A_1 \cup A_2) = f(A_1)\cup f(A_2)\,\text{.}\]
\end{document}