0

Good day, I am trying to end the subsection Proof on line ~65 and have the following section reset its indentation to be inline with the previous section, not the subsection Proof on line ~65. I will be doing this throughout my document so if you could not only troubleshoot this issue for me, but include instructions for subsequent areas I would be greatly appreciative.

Thank you for all your assistance,

jhayes

enter image description here

\documentclass{book}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\usepackage{gensymb}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{siunitx}
\usepackage{amsfonts}
\usepackage{textcomp}
\usepackage{indentfirst}

\usepackage{bm}
\usepackage{float}

\usepackage{tikz}
\usepackage{setspace}
\onehalfspacing

\newcommand{\Z}{\mathbb{Z}}
\newcommand{\inv}{^{-1}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\gen}[1]{\ensuremath{\langle #1\rangle}}
\newcommand{\vs}{\vspace{2mm}}
\newcommand{\nl}{\\\>}
\def\mod{\,mod\,}

\newenvironment{tab}
  {\begin{tabbing}
           \hspace{30pt}\=\hspace{30pt}\=\kill
  }
  {\end{tabbing}}

\title{MATH 570 Lecture Notes}
\author{Professor: Dr. Jon Corson\\
Prepared by: Justin Hayes}

\date{Spring 2019}

\begin{document}

\tableofcontents

\maketitle

\chapter{Review of Functions}
\section{Definition}
\indent A function $\Phi$ from a set A (domain) to a set B (co-domain) [also known as mapping] [if from set A to set A then known as transformation] is a rule/relation that assigns each a$\in$A a unique element $\Phi$(a) in B.

\vs
Two functions($\Phi$ \& $\Psi$) are equal (denoted $\Phi$=$\Psi$) if they nave the same domain $\&$ co-domain $\&$ exactly the same rule and $\Phi$(a)=$\Psi$(a) for each a in the common domain.

\section{Identity Function:}
For any set A, the identity function on A is the function I$_A$:A$\rightarrow$A given by I$_A$(a)=a for all a$\in$A.

\section{Composition of Functions:}
If $\alpha$:A$\rightarrow$B and $\beta$:B$\rightarrow$C are functions then the composite of functions:\\
$\beta\circ\alpha$=$\beta\alpha$:A$\rightarrow$C is defined by ($\beta\alpha$)(a)=$\beta$($\alpha$(a)).

\section{Associative Law:}
If $\alpha$:A$\rightarrow$B, $\beta$:B$\rightarrow$C; $\gamma$:C$\rightarrow$D,\\
then $\gamma$($\beta\alpha$)=($\gamma\beta$)$\alpha$

\subsection{Proof:}

\begin{center}
    Both composites have domain $A$ \& co-domain $D$.\\
    For each element in the domain ($\alpha\in$A),\\
    $\lbrack\gamma$($\beta\alpha\rbrack$(a)$\Rightarrow\gamma$(($\beta\alpha$(a))=$\gamma$(($\beta$($\alpha$(a)))\\
    $\lbrack$($\gamma\beta$)$\alpha\rbrack$(a)$\Rightarrow$($\gamma\beta$)($\alpha$(a))=$\gamma$($\beta$($\alpha$(a)))\\
    Therefore $\gamma$($\beta\alpha)$=($\gamma\beta$)$\alpha$.
\end{center}

\section{Identity Law:}
If $\alpha$:A$\rightarrow$B, then\\
I$_B\circ\alpha$=$\alpha$ and $\alpha$ $\circ$ I$_A$=$\alpha$

\section{One-to-One, Onto \& Bijections}
\subsection{One-to-One}
We say that $\alpha:A\rightarrow B$ is \textbf{1-to-1}(injective) $\iff\alpha$(a$_1$)=$\alpha$(a$_2$)$\Rightarrow$(a$_1$)=(a$_2$)

\subsection{Onto}
We say that $\alpha$:A$\rightarrow$B is \textbf{onto} (surjective) if for each element in the codomain B $\exists a \in A$ such that $\alpha$(a)=b.

\subsection{Bijection}
We say that $\alpha:A\rightarrow B$ is \textbf{bijective} if $\alpha$ is both 1-to-1 \textbf{\&} onto.

\subsection{Theorem of Bijection:} 
A map $\alpha$:A$\rightarrow$B is a bijection if and only if there exists a map $\beta$:B$\rightarrow$A such that $\beta\alpha$=I$_A$ and $\alpha\beta$=I$_B$.\\
\textbf{Remark:} In this event, the function $\beta$ is unique.

\subsection{Proof of Bijective Theorem:}
\begin{center}
    Suppose B':B$\rightarrow$A is another such function.\\
    Then: B'=I$_A\circ$B'=$\beta$($\alpha$B')=$\beta\circ$I$_B$=$\beta$\\
    \textbf{Notation:} we denote the unique function $\beta$ in the theorem by a$^{-1}$, called the inverse of the bijection of $\alpha$.\\
    $\alpha^{-1}\alpha$=I$_A$ \& $\alpha\alpha^{-1}$=I$_B\square$\\
\end{center}
  • To me the "indentation" of section 1.5 is the same as section 1.4. Perhaps I don't understand what you mean by "indentation" in this case. Any chance of a graphic showing what you get and what you want? – Peter Wilson Feb 9 '19 at 18:48
  • First of all, before you try to fix any other part, please, use math mode only once to type a formula. For example, replace $\gamma$($\beta\alpha)$=($\gamma\beta$)$\alpha$. by $\gamma(\beta\alpha)=(\gamma\beta)\alpha$. – Sigur Feb 9 '19 at 18:53
  • @PeterWilson as you can see on my pdf display of the file 1.5 is not inline with 1.4, it is inline with the subsection 1.4.1. Thank you for your assistance, JLH – Jhayes Feb 10 '19 at 20:13
  • @Jhayes: Maybe you are referring to the left margin of the page, which is (naturally) different on even and odd pages if you use the book class (in twoside mode). Anyway, you may add the option oneside to the book class (which makes the left (and the right margin) the same on even and odd pages), but I'd strongly discourage that. – zetaeffe Feb 10 '19 at 21:27
  • This document will never be printed as a book, I solely used the book option to get the TOC. Since it will never be printed in book form, is it okay to use the oneside option? or are there other issues that might arise? – Jhayes Feb 11 '19 at 0:07
1

From your graphic everything is aligned on the first page of the ToC and is also aligned on the second page; the difference is that the left margin on the two pages is different which you get by default with the book class. If you want the left margin to be constant throughout your document use the oneside option.

\documentclass[oneside]{book}

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.