5

I have worked out an exercise consisting of multiple parts in LaTeX, as a descriptive list consisting of part a, b and c. However, it aligns in a very weird way and I can't get it to look better. Is there anything I can try to align everything nicely?

\documentclass[a4paper, 11pt]{article}
\usepackage[english]{babel}
\usepackage{amsmath}
\usepackage{a4wide}

\title{Homework Symmetry}
\date{Quartile 2, Week 1}
\author{Benjamin Caris, Luuk Reijnders, Tom Jacobs}

\begin{document}

\maketitle
\pagebreak

\section*{Excercise 2.6}

\begin{description}

\item[a)]{$d(\sigma(A),\sigma(P)) = d(A, P)$, since $\sigma$ is a symmetry. 
Furthermore, since $\sigma(A) = A$, we have $d(\sigma(A),\sigma(P)) = 
d(A,\sigma(P))$. From these 2 observations we clearly see that $d(\sigma(A), 
\sigma(P)) = d(A,\sigma(P)) = d(A, P)$. So $d(A,\sigma(P)) = d(A, P)$, hence 
we 
can conclude that $A$ is on the perpendicular bisector of the line segment 
$P\sigma(P)$.}

\item[b)]{Symmetry line}

\item[c)]{$\tau\circ\sigma(A) = \tau(\sigma(A)) =  \tau(A)$ since $\sigma(A) 
= 
A$. Since $A$ is on the line $AB$, reflection of $A$ in $AB$ again gives 
$A$, so 
$\tau(A) = A$, hence $\tau\circ\sigma(A) = A$ and $\tau\circ\sigma$ fixes 
$A$. 
Analogously, we find for $B$ that $\tau\circ\sigma(B) = B$ and therefore 
that 
$\tau\circ\sigma$ also fixes $B$ (since $\sigma(B) = B$ and $B$ is on $AB$). 
For 
$C$ we have $\tau\circ\sigma(C) = \tau(\sigma(C))$. $\sigma(C)\neq C$, but 
since 
$\sigma$ is a reflection in $AB$ and $\tau$ is again a reflection in $AB$, 
$\tau(\sigma(C))$ is just the inverse of the reflection $\sigma(C)$. So 
$\tau(\sigma(C)) = C$, hence $\tau\circ\sigma$ also fixes C. Because 
$\tau\circ\sigma$ fixes $A$, $B$ and $C$, $\tau\circ\sigma$ is the identity. 
Furthermore, since $\tau$ is the inverse of $\sigma$, $\sigma = \tau$.}

\end{description}

\end{document}

it looks like this.

Thanks in advance!

  • Welcome to TeX SX! Please post a full compilable code. – Bernard Nov 14 '18 at 20:58
  • @Bernard you mean, the whole document? – Benjamin Caris Nov 14 '18 at 20:59
  • No, but the minimal code to make it compile without errors, not just a snippet. – Bernard Nov 14 '18 at 21:00
  • @Bernard alright, should be good now – Benjamin Caris Nov 14 '18 at 21:01
3

With the enumitem package ane an enumerate environment instead of the description you can get the following:

enter image description here

\documentclass[a4paper, 11pt]{article}
\usepackage[english]{babel}
\usepackage{amsmath}
\usepackage{a4wide}

\usepackage{enumitem}

\begin{document}



\section*{Excercise 2.6}

\begin{enumerate}[label=\textbf{\alph*)}]

\item {$d(\sigma(A),\sigma(P)) = d(A, P)$, since $\sigma$ is a symmetry. 
Furthermore, since $\sigma(A) = A$, we have $d(\sigma(A),\sigma(P)) = 
d(A,\sigma(P))$. From these 2 observations we clearly see that $d(\sigma(A), 
\sigma(P)) = d(A,\sigma(P)) = d(A, P)$. So $d(A,\sigma(P)) = d(A, P)$, hence 
we 
can conclude that $A$ is on the perpendicular bisector of the line segment 
$P\sigma(P)$.}

\item {Symmetry line}

\item {$\tau\circ\sigma(A) = \tau(\sigma(A)) =  \tau(A)$ since $\sigma(A) 
= 
A$. Since $A$ is on the line $AB$, reflection of $A$ in $AB$ again gives 
$A$, so 
$\tau(A) = A$, hence $\tau\circ\sigma(A) = A$ and $\tau\circ\sigma$ fixes 
$A$. 
Analogously, we find for $B$ that $\tau\circ\sigma(B) = B$ and therefore 
that 
$\tau\circ\sigma$ also fixes $B$ (since $\sigma(B) = B$ and $B$ is on $AB$). 
For 
$C$ we have $\tau\circ\sigma(C) = \tau(\sigma(C))$. $\sigma(C)\neq C$, but 
since 
$\sigma$ is a reflection in $AB$ and $\tau$ is again a reflection in $AB$, 
$\tau(\sigma(C))$ is just the inverse of the reflection $\sigma(C)$. So 
$\tau(\sigma(C)) = C$, hence $\tau\circ\sigma$ also fixes C. Because 
$\tau\circ\sigma$ fixes $A$, $B$ and $C$, $\tau\circ\sigma$ is the identity. 
Furthermore, since $\tau$ is the inverse of $\sigma$, $\sigma = \tau$.}

\end{enumerate}

\end{document}

Further information on how the indentation of the label and the distance between label and text can be changed, are described here: can someone please explain the enumitem horizontal spacing parameters?

2

You'd better do that with an enumerate environment, which you can customise with package enumitem.

Unrelated: a4wide is obsolete and shouldn't be used anymore, according to 2tabu`.

\documentclass[11pt]{article}

\usepackage[showframe]{geometry}
\usepackage{enumitem}

\begin{document}

\section*{Exercise 2.6}

\begin{enumerate}[label =\alph*), font=\bfseries, wide=0pt, leftmargin=*]

\item $d(\sigma(A),\sigma(P)) = d(A, P)$, since $\sigma$ is a symmetry.
Furthermore, since $\sigma(A) = A$, we have $d(\sigma(A),\sigma(P)) =
d(A,\sigma(P))$. From these 2 observations we clearly see that $d(\sigma(A),
\sigma(P)) = d(A,\sigma(P)) = d(A, P)$. So $d(A,\sigma(P)) = d(A, P)$, hence
we
can conclude that $A$ is on the perpendicular bisector of the line segment
$P\sigma(P)$.

\item Symmetry line

\item $\tau\circ\sigma(A) = \tau(\sigma(A)) = \tau(A)$ since $\sigma(A)
=
A$. Since $A$ is on the line $AB$, reflection of $A$ in $AB$ again gives
$A$, so
$\tau(A) = A$, hence $\tau\circ\sigma(A) = A$ and $\tau\circ\sigma$ fixes
$A$.
Analogously, we find for $B$ that $\tau\circ\sigma(B) = B$ and therefore
that
$\tau\circ\sigma$ also fixes $B$ (since $\sigma(B) = B$ and $B$ is on $AB$).
For
$C$ we have $\tau\circ\sigma(C) = \tau(\sigma(C))$. $\sigma(C)\neq C$, but
since
$\sigma$ is a reflection in $AB$ and $\tau$ is again a reflection in $AB$,
$\tau(\sigma(C))$ is just the inverse of the reflection $\sigma(C)$. So
$\tau(\sigma(C)) = C$, hence $\tau\circ\sigma$ also fixes C. Because
$\tau\circ\sigma$ fixes $A$, $B$ and $C$, $\tau\circ\sigma$ is the identity.
Furthermore, since $\tau$ is the inverse of $\sigma$, $\sigma = \tau$.

\end{enumerate}

\end{document} 

enter image description here

  • Thanks, that looks a lot better! Why shouldn't I use a4wide? I have only recently started working with LaTeX and at the beginners course at my university they showed us some packages which one should always use. One of those was a4wide. Now I'm curious for the reason why it is obsolete now. – Benjamin Caris Nov 14 '18 at 21:19
  • It's been deprecated for many years. Read §1.1 in l2tabu, which sets it in the mortal sin category. The original is in german, and there are english, french and italian translations. To sum up its conclusions, in standard classes, it's replaced with the a4paper class option, and anyway, you can customise the parameters with geometry ery easily. – Bernard Nov 14 '18 at 21:29

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