m1.tex

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\title{MATH 4338 Main Problem 1}
\date{}
\author{Andy Lu}

\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amsthm}
\newtheorem{theorem}{Theorem}

\begin{document}
  \maketitle
  \begin{theorem}
    If $a, b \in \mathbb{R}$  and $a \neq 0$, then there is one and only one
     number $x$ s.t. $a \cdot x = b$. The number is given by $x = ba^{-1}$.
  \end{theorem}
  
  \begin{proof}
    Assume $a, b \in \mathbb{R}$ and $a \neq 0$. Let $x = ba^{-1}$. As $a
     \in \mathbb{R}$, $a^{-1}$ is a number by the existence of a reciprocal 
     axiom. By the closure property, $x \in \mathbb{R}$. Using the
    axioms,
    \begin{align*}
      ax & = \\
         & = a(ba^{1}) & [\text{by substitution}]\\
         & = a(a^{-1}b) & [\text{by commutative property}]\\
         & = (aa^{-1})b & [\text{by associativity}]\\
         & = (1)b & [\text{since } a^{-1} \text{ is the reciprocal of } a]\\
         & = b & [\text{since } 1 \text{ is the multiplicative identity}]
    \end{align*}
    Hence, $x=ba^{-1}$. Assume $y$ is a number such that $a \cdot y = b$.
    Using the axioms,
    \begin{align*}
      ay & = b & [\text{by assumption}]\\
      (a^{-1})ay & = (a^{-1})b & [\text{multiply by} a^{-1}]\\
      (a^{-1}a)y & = (a^{-1}b) & [\text{by associativity}]\\
      (aa^{-1})y & = (ba^{-1}) & [\text{by commutativity}]\\
      (1)y & = ba^{-1} & [\text{since } a^{-1} \text{ is the reciprocal of } a]\\
      y & = ba^{-1} & [\text{since } 1 \text{ is the multiplicative identity}]
    \end{align*}
    Thus, $y=ba^{-1}=x$ and $x$ is unique.
  \end{proof}
\end{document}