\documentclass{article}
\usepackage{amsmath}
\usepackage{amsthm}

\newtheorem{conj}{Conjecture}

\title{Proofs in Cases}
\author{Doug Baldwin}
\date{Spring 2021}


\begin{document}

\maketitle

For an example of what a proof by cases might look like when fully written out, suppose $f(x)$ is defined by the following:
\begin{equation}
f(x) = \begin{cases}
x^2 & \mathrm{if\ } x < 0 \\
x & \mathrm{if\ } 0 \le x \le 1 \\
2x - 1 & \mathrm{if\ } 1 < x
\end{cases}
\end{equation}
Then we have
\begin{conj}
For all real numbers $x$, $f(x)$ as defined by (1) is greater than or equal to $0$.
\end{conj}
\begin{proof}
We will prove that $f(x) \ge 0$ for all real numbers $x$ by analyzing each case in the definition in turn.

For the first case, consider $x < 0$. Then $f(x) = x^2$, which is greater than or equal to $0$ for all values of $x$.

For the second case, consider $0 \le x \le 1$. In this case $f(x) = x$ which is greater than or equal to $0$ from the constraint $0 \le x \le 1$.

Finally, consider $1 < x$. Here, $f(x) = 2x - 1$ which is an increasing function, i.e., $f(x) > f(1)$ for all $x > 1$. Since $f(1) = 1 \ge 0$, we see that $f(x) \ge 0$ for all $x > 1$.

We have now shown that $f(x) \ge 0$ for all cases in the definition of $f$, and thus that $f(x) \ge 0$ for all real numbers $x$.
\end{proof}
\end{document}