![如何对齐数组中的第一行](https://linux22.com/image/353213/%E5%A6%82%E4%BD%95%E5%AF%B9%E9%BD%90%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E7%AC%AC%E4%B8%80%E8%A1%8C.png)
我经常需要使用这种格式来输入练习,而且效果不错,但我想知道如何将括号中的字母与相邻练习的第一行对齐。因此,第一个问题是如何使用gathered
inside来实现它array
,而后续问题是是否有更合适的方法来实现它。
\documentclass{article}
\usepackage{amssymb,amsmath}
\begin{document}
\noindent \textbf{3.4}
\[
\begin{array}{cccc}
\noindent \textbf{(a)}&
\begin{gathered}
\mathbb{Z}^*_{7}=\left\{1,3,5\right\}\\
4^3=64=1
\end{gathered}&
\noindent \textbf{(b)}&
\begin{gathered}
\mathbb{Z}^*_{8}=\left\{1,3,5,7\right\}\\
3^4=81=1\\
\end{gathered}\\\\
\noindent \textbf{(c)}&
\begin{gathered}
\mathbb{Z}^*_{11}=\left\{1,,2,3,4,5,6,7,8,9,10\right\}\\
\begin{array}{c|cccccccccc}
\cdot & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\
\hline
{\rule{0pt}{2.6ex}}
5 & 5 & 10 & 4 & 9 & 3 & 8 & 2 & 7 & 1 & 6\\
\end{array}\\
5^{-2}=5^{-1}\cdot 5^{-1}=9\cdot 9=4\\
\end{gathered}&
\noindent \textbf{(d)}&
\begin{gathered}
\mathbb{Z}^*_{12}=\left\{1,5,7,11\right\}\\
\begin{array}{c|cccc}
\cdot & 1 & 5 & 7 & 11\\
\hline
{\rule{0pt}{2.6ex}}
5 & 5 & 1 & 11 & 7\\
\end{array}\\
5^{-4}\cdot 7^2=5^{-1}\cdot 5^{-1}\cdot 5^{-1}\cdot 5^{-1}\cdot 1=5^4=1\\
\end{gathered}
\end{array}
\]
\end{document}
答案1
像这样?
我使用包exams
和选项[t]
来描述aligned
和gathered
环境:
\documentclass{article}
\usepackage[margin=25mm]{geometry}
\usepackage{amsmath, amsfonts}
\usepackage{array}
\setlength\extrarowheight{2pt}
\newlength\labelwd
\settowidth\labelwd{\bfseries viii.)}
\usepackage{tasks}
\settasks{counter-format =tsk[a].),
label-format=\bfseries,
label-offset=1em,
label-align=right,
label-width=\labelwd,
item-indent=\dimexpr\labelwd+1em\relax,
before-skip =-0.5ex,
after-item-skip=\medskipamount}
\usepackage{enumitem}
\setlist[enumerate,1]{% (
leftmargin=*, itemsep=\baselineskip,
label={\textbf{\thesection.\arabic*}}
}
\begin{document}
\section{section title}
\begin{enumerate}
\item
\begin{tasks}(2)
\task $\begin{aligned}[t]
\mathbb{Z}^*_{7} & = \left\{1,3,5\right\}\\
4^3 & = 64 = 1
\end{aligned}$
\task $\begin{aligned}[t]
\mathbb{Z}^*_{8} & = \left\{1,3,5,7\right\}\\
3^4 & = 81 = 1
\end{aligned}$
\task $\begin{gathered}[t]
\mathbb{Z}^*_{11}=\left\{1,2,3,4,5,6,7,8,9,10\right\}\\
\begin{array}{c|cccccccccc}
\cdot & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\
\hline
5 & 5 & 10 & 4 & 9 & 3 & 8 & 2 & 7 & 1 & 6\\
\end{array}\\
5^{-2}=5^{-1}\cdot 5^{-1}=9\cdot 9=4\\
\end{gathered}$
\task $\begin{gathered}[t]
\mathbb{Z}^*_{12}=\left\{1,5,7,11\right\}\\
\begin{array}{c|cccc}
\cdot & 1 & 5 & 7 & 11\\
\hline
5 & 5 & 1 & 11 & 7\\
\end{array}\\
5^{-4}\cdot 7^2=5^{-1}\cdot 5^{-1}\cdot 5^{-1}\cdot 5^{-1}\cdot 1=5^4=1\\
\end{gathered}$
\end{tasks}
\item
\begin{tasks}(2)
\task 1
\task 2
\task 3
\task 4
\end{tasks}
\end{enumerate}
\end{document}
附录:
从tasks
2020-03-21 开始,MWE 的版本 V1.2 不再按预期工作。其序言(重现问题中的图像)应更改为:
\documentclass{article}
\usepackage[margin=25mm]{geometry}
\usepackage{enumitem}
\setlist[enumerate,1]{leftmargin=*,
itemsep=\baselineskip,
label={\textbf{\thesection.\arabic*}}
}
\usepackage{amsmath, amssymb}
\usepackage{array}
\setlength\extrarowheight{2pt}
\usepackage{tasks}
\settasks{label=(\alph*),
label-format=\bfseries,
label-offset=1em,
before-skip =-0.5ex,
after-item-skip=\medskipamount
}
答案2
只需使用[t]
选项即可gathered
。但是,我建议使用tasks
来自同名包的环境,该环境用于练习的水平列表。我还建议aligned
在相关的地方使用,并且我定义了一个\set
命令以获得更轻量的代码:
\documentclass{article}
\usepackage{amssymb,mathtools}
\usepackage{tasks}
\usepackage[showframe]{geometry}
\DeclarePairedDelimiter\set\{\}
\begin{document}
\noindent \textbf{3.4}
\[
\begin{array}{cccc}
\noindent \textbf{(a)}&
\begin{gathered}[t]
\mathbb{Z}^*_{7}=\set{1,3,5}\\
4^3=64=1
\end{gathered}&
\noindent \textbf{(b)}&
\begin{gathered}[t]
\mathbb{Z}^*_{8}=\set{1,3,5,7}\\
3^4=81=1\\
\end{gathered}\\\\
\noindent \textbf{(c)}&
\begin{gathered}[t]
\mathbb{Z}^*_{11}=\set{1,2,3,4,5,6,7,8,9,10}\\
\begin{array}{c|*{10}{c}}
\cdot & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\
\hline
{\rule{0pt}{2.6ex}}
5 & 5 & 10 & 4 & 9 & 3 & 8 & 2 & 7 & 1 & 6\\
\end{array}\\
5^{-2}=5^{-1}\cdot 5^{-1}=9\cdot 9=4\\
\end{gathered}&
\noindent \textbf{(d)}&
\begin{gathered}[t]
\mathbb{Z}^*_{12}=\set{1,5,7,11}\\
\begin{array}{c|*{4}{c}}
\cdot & 1 & 5 & 7 & 11\\
\hline
{\rule{0pt}{2.6ex}}
5 & 5 & 1 & 11 & 7\\
\end{array}\\
5^{-4}\cdot 7^2=5^{-1}\cdot 5^{-1}\cdot 5^{-1}\cdot 5^{-1}\cdot 1=5^4=1\\
\end{gathered}
\end{array}
\]
\vspace{1cm}
\begin{tasks}[counter-format = (tsk[a]), label-format=\bfseries, label-width=1.5em, label-offset=0.5em, column-sep=1em](2)
\task \centering$ \begin{aligned}[t]
\mathbb{Z}^*_{7} & =\set{1,3,5}\\
4^3 & =64=1
\end{aligned} $
\task $ \begin{aligned}[t]
\mathbb{Z}^*_{8} & =\set{1,3,5,7}\\
3^4 & =81=1\\
\end{aligned} $
\task $ \begin{gathered}[t]
\mathbb{Z}^*_{11}=\set{1,2,3,4,5,6,7,8,9,10}\\
\begin{array}{c|*{10}{c}}
\cdot & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\
\hline
{\rule{0pt}{2.6ex}}
5 & 5 & 10 & 4 & 9 & 3 & 8 & 2 & 7 & 1 & 6\\
\end{array}\\
5^{-2}=5^{-1}\cdot 5^{-1}=9\cdot 9=4
\end{gathered} $
\task $ \begin{gathered}[t]
\mathbb{Z}^*_{12}=\set{1,5,7,11}\\
\begin{array}{c|*{4}{c}}
\cdot & 1 & 5 & 7 & 11\\
\hline
{\rule{0pt}{2.6ex}}
5 & 5 & 1 & 11 & 7\\
\end{array}\\
5^{-4}\cdot 7^2=5^{-1}\cdot 5^{-1}\cdot 5^{-1}\cdot 5^{-1}\cdot 1=5^4=1
\end{gathered}$
\end{tasks}
\end{document}