Cfrgtkky

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 

$Psigma(P)$.}



item[b)]{Symmetry line}



item[c)]{$taucircsigma(A) = tau(sigma(A)) =  tau(A)$ since $sigma(A) 

= 

A$. Since $A$ is on the line $AB$, reflection of $A$ in $AB$ again gives 

$A$, so 

$tau(A) = A$, hence $taucircsigma(A) = A$ and $taucircsigma$ fixes 

$A$. 

Analogously, we find for $B$ that $taucircsigma(B) = B$ and therefore 

that 

$taucircsigma$ also fixes $B$ (since $sigma(B) = B$ and $B$ is on $AB$). 

For 

$C$ we have $taucircsigma(C) = tau(sigma(C))$. $sigma(C)neq C$, 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 $taucircsigma$ also fixes C. Because 

$taucircsigma$ fixes $A$, $B$ and $C$, $taucircsigma$ is the identity. 

Furthermore, since $tau$ is the inverse of $sigma$, $sigma = tau$.}



end{description}



end{document}

Thanks in advance!

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

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 

$Psigma(P)$.}



item {Symmetry line}



item {$taucircsigma(A) = tau(sigma(A)) =  tau(A)$ since $sigma(A) 

= 

A$. Since $A$ is on the line $AB$, reflection of $A$ in $AB$ again gives 

$A$, so 

$tau(A) = A$, hence $taucircsigma(A) = A$ and $taucircsigma$ fixes 

$A$. 

Analogously, we find for $B$ that $taucircsigma(B) = B$ and therefore 

that 

$taucircsigma$ also fixes $B$ (since $sigma(B) = B$ and $B$ is on $AB$). 

For 

$C$ we have $taucircsigma(C) = tau(sigma(C))$. $sigma(C)neq C$, 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 $taucircsigma$ also fixes C. Because 

$taucircsigma$ fixes $A$, $B$ and $C$, $taucircsigma$ 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?

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

$Psigma(P)$.



item Symmetry line



item $taucircsigma(A) = tau(sigma(A)) = tau(A)$ since $sigma(A)

=

A$. Since $A$ is on the line $AB$, reflection of $A$ in $AB$ again gives

$A$, so

$tau(A) = A$, hence $taucircsigma(A) = A$ and $taucircsigma$ fixes

$A$.

Analogously, we find for $B$ that $taucircsigma(B) = B$ and therefore

that

$taucircsigma$ also fixes $B$ (since $sigma(B) = B$ and $B$ is on $AB$).

For

$C$ we have $taucircsigma(C) = tau(sigma(C))$. $sigma(C)neq C$, 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 $taucircsigma$ also fixes C. Because

$taucircsigma$ fixes $A$, $B$ and $C$, $taucircsigma$ is the identity.

Furthermore, since $tau$ is the inverse of $sigma$, $sigma = tau$.



end{enumerate}



end{document}