%EXERCISE FIVE
          
\documentclass{amsart}
\thispagestyle{empty}
\newtheorem*{clm}{Claim}
\begin{document}
             
\begin{center}
	\textbf{Exercise Five: Alignment, Part Two - Labels and Annotation}
\end{center}  

\begin{clm}
The real number 2 is equal to the real number 1.
% Some people object to starting a sentence with a symbol or number, so don't just write 2=1.
\end{clm}

\begin{proof}
\begin{align}
a &= b \label{defineab} && \text{Suppose $a,b\in\mathbb{R}$ are non-zero and that $a=b$.} \\
a^2 &= ab && \text{What's the harm in multiplying each side by $a$?} \\
a^2-b^2 &= ab-b^2 && \text{And it's fair to subtract $b^2$ from each side, no?} \\
(a-b)(a+b) &= b(a-b) && \text{Just factoring. Nothing fishy here.} \\
a+b &= b && \text{Cancel the factor $(a-b)$ from each side.} \\
b+b &= b && \text{Recall from (\ref{defineab}) that $a=b$, so we replace $a$ by $b$.} \\
2b &= b && \text{Who could possibly object to this?} \\
2 &= 1 && \text{In (\ref{defineab}) we defined $b\neq 0$, so we can divide by $b$.} 
\end{align}
\end{proof}

% So... Is there a flaw in this proof?
% If you think there is, where is it?
% If you think there isn't, well ... think again.

\end{document}