Seeking advice for my clumsy math homework template [closed]

Question

I typed my math homework with LaTeX. This was my first try on making some document with LaTeX. I can Google for math symbols, so it's okay, but the overall structure is really clumsy; especially spacing.

Any advice would be really appreciated.

\documentclass[11pt,letterpaper]{article}
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{9in}
\setlength{\oddsidemargin}{.125in}
\pagestyle{plain}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}

\makeatletter
\newcommand{\rmnum}[1]{\romannumeral #1}
\newcommand{\Rmnum}[1]{\expandafter\@slowromancap\romannumeral #1@}
\makeatother

\begin{document}

\title{Assignment#2\\MTH000}
\author{MY NAME\\student number}

\maketitle

\begin{theorem}{\textbf{Problem 4.1}} 
\end{theorem}

\begin{proof}
\vspace{1mm}
\qquad (1) $f$ is not a function because $f$ does not satisfy the condition that if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$. That is, $(1,4) \in f$ and $(1,2) \in f$, but $4 \neq 2$.

\vspace{1mm}
\qquad (2) $f$ is not a function because $f$ does not satisfy the condition that if $x \in A$ then there exists $y \in B$ such that $(x,y) \in f$.
\newline while $4 \in A$, there is no $y \in B$ such that $(4,y) \in f$.

\end{proof}
\vspace{5mm}
\begin{theorem}{\textbf{Problem 4.3}} 
Given: $A=\{1,2\}, B=\{1,2,3\}$
\end{theorem}

\begin{proof}
\vspace{1mm}
\qquad The set of all $f:A \to B$ is $f = \{(1,1),(1,2),(1.3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$
\end{proof}


\begin{theorem}{\textbf{Problem 4.5}} 
\end{theorem}

\begin{proof}
\vspace{1mm}
\qquad 
\vspace{1mm}
\qquad 

\end{proof}
\vspace{5mm}


\begin{theorem}{\textbf{Problem 4.6}} 
\end{theorem}

\begin{proof}
\vspace{1mm}  
\qquad (\rmnum{1}) The image of $f$ is $\{2,4,5\}$\\
\vspace{1.5mm}  
\qquad \quad (\rmnum{2}) $f^{-1}(\{3,4\})$ does not exist, as there is no $x$ such that $f(x)=3$\\
\vspace{.1mm}
\qquad \quad (\rmnum{3}) $f(\{1,2,4\})=\{2,4\}$\\
\vspace{2mm}
\qquad \quad (\rmnum{4}) $f^{-1}(\{3\})$ does not exist, as there is no $x$ such that $f(x)=3$.
\vspace{1mm}  
\qquad \quad \ (\rmnum{5}) $f(f^{-1}(\{2,3\}))$ does not exist, because $f^{-1}(\{2,3\})$ does not exist, as there is no $x$ such that $f(x)=3$.
\end{proof}
\vspace{5mm}


\begin{theorem}{\textbf{Problem 4.7}} 
\end{theorem}

\begin{proof}
\vspace{1mm}
\qquad $f(1)=2$, $f(2)=4$, $f(3)=3$, $f(4)=6$, $f(5)=1$, \\
\vspace{0.5mm}
\qquad and $g(1)=4,g(2)=4,g(3)=1,g(4)=3,g(5)=2,g(6)=2$.
\vspace{2mm}
\end{proof}

\begin{proof}
\vspace{1mm}
\qquad $g \circ f:A \to C$ is : \\
\vspace{1mm}
\qquad $g \circ f(1) = g(f(1)) = g(2) = 4$ \\
\vspace{1mm}
\qquad $g \circ f(2) = g(f(2)) = g(4) = 3$ \\
\vspace{1mm}
\qquad $g \circ f(3) = g(f(3)) = g(3) = 1$ \\
\vspace{1mm}
\qquad $g \circ f(4) = g(f(4)) = g(6) = 2$ \\
\vspace{1mm}
\qquad $g \circ f(5) = g(f(5)) = g(1) = 4$ \\
\vspace{5mm}
\end{proof}

\begin{theorem}{\textbf{Problem 4.}} 
\end{theorem}

\begin{proof}
\vspace{1mm}
\qquad 
\vspace{1mm}
\qquad 

\end{proof}
\vspace{5mm}



\end{document}

Answer 1

I think this is a great question, and applaud you for dabbling in LaTeX.

Here's what I have done to improve your code:

• loaded the geometry package to help with page dimenstions
• changed all the occurrences of \begin{theorem}{\textbf{Problem...}} to \subsection*{Problem...}; there's no point using a theorem environment for this because it seemed that you just wanted the heading
• used the enumerate environment for your lists, and used the enumitem package to help with the roman enumeration
• used the align environment (from the amsmath)for your last problem

• most importantly- removed all of the manual horizontal and vertical spacing. Page settings should be tweaked globally in your preamble.

• You should not use \\ to end a line (unless you are in a tabular or align, or similar environment); leave line breaking in paragraphs to TeX, or else use a blank line instead which will start a new paragraph.

Code

\documentclass[11pt,letterpaper]{article}
\usepackage[top=.5in,textheight=9in]{geometry}
\usepackage{amsmath}
\usepackage{enumitem}

\title{Assignment\#2\\MTH000}
\author{MY NAME\\student number}

\begin{document}

\maketitle

\subsection*{Problem 4.1}
\begin{enumerate}
  \item $f$ is not a function because $f$ does not satisfy the condition that if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$. That is, $(1,4) \in f$ and $(1,2) \in f$, but $4 \neq 2$.
  \item $f$ is not a function because $f$ does not satisfy the condition that if $x \in A$ then there exists $y \in B$ such that $(x,y) \in f$.
while $4 \in A$, there is no $y \in B$ such that $(4,y) \in f$.
\end{enumerate}

\subsection*{Problem 4.3}
Given: $A=\{1,2\}, B=\{1,2,3\}$

The set of all $f:A \to B$ is $f = \{(1,1),(1,2),(1.3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}$

\subsection*{Problem 4.5}

\subsection*{Problem 4.6}

\begin{enumerate}[label=(\roman*)]
  \item image of $f$ is $\{2,4,5\}$
    \item $f^{-1}(\{3,4\})$ does not exist, as there is no $x$ such that $f(x)=3$
    \item $f(\{1,2,4\})=\{2,4\}$
    \item $f^{-1}(\{3\})$ does not exist, as there is no $x$ such that $f(x)=3$.
    \item $f(f^{-1}(\{2,3\}))$ does not exist, because $f^{-1}(\{2,3\})$ does not exist, as there is no $x$ such that $f(x)=3$.
\end{enumerate}

\subsection*{Problem 4.7}
$f(1)=2$, $f(2)=4$, $f(3)=3$, $f(4)=6$, $f(5)=1$, 
and $g(1)=4,g(2)=4,g(3)=1,g(4)=3,g(5)=2,g(6)=2$.

$g \circ f:A \to C$ is 
\begin{align*}
 g \circ f(1)& = g(f(1)) = g(2) = 4 \\
 g \circ f(2)& = g(f(2)) = g(4) = 3 \\
 g \circ f(3)& = g(f(3)) = g(3) = 1 \\
 g \circ f(4)& = g(f(4)) = g(6) = 2 \\
 g \circ f(5)& = g(f(5)) = g(1) = 4 \\
\end{align*}

\subsection*{Problem 4.}

\end{document}