我想在表格中列出一些练习的答案。我成功地做到了这一点,只需使用两个数组(并分解方程式)和以下代码:
\documentclass{article}
\usepackage{amssymb,amsmath}
\begin{document}
\noindent \textbf{1.3 (4)}
$$\begin{array}{cccccc}
\textbf{(a)} & 13\equiv b \pmod{8} & \textbf{(b)} & 52\equiv b\pmod{9} & \textbf{(c)} & -8\equiv b\pmod{10}\\
& 13=8\cdot 1 +5 & & 52=9\cdot 5 +7 & & -8=10\cdot (-1) +2\\
& \therefore 13\bmod 8= 5 & & \therefore 52\bmod 9= 7 & & \therefore -8\bmod 10= 2 \\
& \therefore \text{ s{\"a}tt } b=5 & & \therefore \text{ s{\"a}tt } b=7 & & \therefore \text{ s{\"a}tt } b=7\\
\end{array}$$
$$\begin{array}{ccccccc}
\textbf{(d)} & 14\equiv b\pmod{14} & \textbf{(e)} & 8\equiv b\pmod{12} & \textbf{(f)} & -23\equiv b\pmod{11}\\
& 14=14\cdot 1 +0 & & 8=12\cdot 0 +8 & & -23=11\cdot (-3) +10\\
& \therefore 14\bmod 14= 0 & & \therefore 8\bmod 12= 8 & & \therefore -23\bmod 11= 10\\
&\therefore \text{ s{\"a}tt } b=0 & & \therefore \text{ s{\"a}tt } b=8 & & \therefore \text{ s{\"a}tt } b=10\\
\end{array}$$
\end{document}
但我想知道是否可以通过将收集的方程式嵌入到一个数组中来实现这一点(因为这对我来说更有意义,这样我就可以将方程式放在一起而不是将它们分开)。我尝试了以下代码,但无法使其工作:
\documentclass{article}
\usepackage{amssymb,amsmath}
\begin{document}
\noindent \textbf{1.3 (4)(a)}
$$\begin{array}{ccc}
\begin{gather*}
13\equiv b \pmod{8}\\
13=8\cdot 1 +5\\
\therefore 13\bmod 8= 5\\
\therefore \text{ s{\"a}tt } b=5\\
\end{gather*} &
\begin{gather*}
52\equiv b\pmod{9}\\
52=9\cdot 5 +7\\
\therefore 52\bmod 9= 7\\
\therefore \text{ s{\"a}tt } b=7\\
\end{gather*} &
\begin{gather*}
-8\equiv b\pmod{10}\\
-8=10\cdot (-1) +2\\
\therefore -8\bmod 10= 2\\
\therefore \text{ s{\"a}tt } b=7\\
\end{gather*}\\
\begin{gather*}
14\equiv b\pmod{14}\\
14=14\cdot 1 +0\\
\therefore 14\bmod 14= 0\\
\therefore \text{ s{\"a}tt } b=0\\
\end{gather*} &
\begin{gather*}
8\equiv b\pmod{12}\\
8=12\cdot 0 +8\\
\therefore 8\bmod 12= 8\\
\therefore \text{ s{\"a}tt } b=8\\
\end{gather*}&
\begin{gather*}
-23\equiv b\pmod{11}\\
-23=11\cdot (-3) +10\\
\therefore -23\bmod 11= 10\\
\therefore \text{ s{\"a}tt } b=10\\
\end{gather*}
\end{array}$$
\end{document}
答案1
您的数组已处于数学模式,因此您无法再次嵌套数学模式。尝试:
\documentclass{article}
\usepackage{amssymb,amsmath}
\begin{document}
\noindent \textbf{1.3 (4)(a)}
\[\setlength\arraycolsep{12pt} % more space between columns
\begin{array}{ccc}
\begin{gathered}% it is designed for use inside math environment
13\equiv b \pmod{8}\\
13=8\cdot 1 +5\\
\therefore 13\bmod 8= 5\\
\therefore \text{ s{\"a}tt } b=5\\
\end{gathered} % <---
& \begin{gathered} % <---
52\equiv b\pmod{9}\\
52=9\cdot 5 +7\\
\therefore 52\bmod 9= 7\\
\therefore \text{ s{\"a}tt } b=7\\
\end{gathered} % <---
& \begin{gathered} % <---
-8\equiv b\pmod{10}\\
-8=10\cdot (-1) +2\\
\therefore -8\bmod 10= 2\\
\therefore \text{ s{\"a}tt } b=7\\
\end{gathered} % <--- \\
% second row
\begin{gathered} % <---
14\equiv b\pmod{14}\\
14=14\cdot 1 +0\\
\therefore 14\bmod 14= 0\\
\therefore \text{ s{\"a}tt } b=0\\
\end{gathered} % <---
& \begin{gathered} % <---
8\equiv b\pmod{12}\\
8=12\cdot 0 +8\\
\therefore 8\bmod 12= 8\\
\therefore \text{ s{\"a}tt } b=8\\
\end{gathered} % <---
& \begin{gathered} % <---
-23\equiv b\pmod{11}\\
-23=11\cdot (-3) +10\\
\therefore -23\bmod 11= 10\\
\therefore \text{ s{\"a}tt } b=10\\
\end{gathered} % <---
\end{array}
\]
\end{document}