2

I quite often need to use this kind of format for typing up exercises and it is satisfactory but I am wondering how I can align the letter in brackets with the first line of the exercise adjacent to it. So the first question would be how do I accomplish it using gathered inside array and the follow up question would be if there isn't a more appropriate way of accomplishing it.

\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}

enter image description here

1

like this?

enter image description here

I use package exams and options [t] for aligned and gathered environments:

\documentclass{article}
\usepackage[margin=25mm]{geometry}
\usepackage{amsmath, amsfonts}
\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*}}
                    }
\usepackage{array}
\setlength\extrarowheight{2pt}

\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}
  • Thanks! I'll definitely take a closer look at the exams package. – ifearthenight Apr 18 '17 at 7:28
2

Just use the [t] option for gathered. However, I suggest using the tasks environment, from the homonymous package, which is done for horizontal lists of exercises. I also suggest using aligned where relevant, and I defined a \set command to have a lighter code:

\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} 

enter image description here

  • Thank you! I'll study up on tasks a bit and see if I can use them more often. – ifearthenight Apr 18 '17 at 7:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.