perkee's blog

LaTeX math with anotations that doesn’t look horrible. Of note is the use of align with multiple points of alignment, each denoted by a single ampersand. I’m pretty proud of this. Source is below, with some stuff taken out. Sorry I can’t post full source anymore.

\usepackage{commath}
\newcommand\cartesian\times
\newcommand\union\cup

…

Let us choose $a\in A$ and let $A'=A-\set a$.
Then $\absolute{A'}=k$ and so---by our inductive
hypothesis---$\absolute{A'\cartesian A'}=k^2$
\begin{align*}
  &\text{But}&\absolute{A\cartesian A}
  &=\absolute{\left(A'\union\set a\right)\cartesian
    \left(A'\union\set a\right)}&\\
  &&&=\absolute{\left(A'\cartesian A'\right)\union
    \left(\set a\cartesian A'\right)\union
    \left(A'\cartesian\set a\right)\union
    \left(\set a\cartesian\set a\right)}&
      \parbox{10em}{
        \raggedleft
          \scriptsize{%\noindent
          The cartesian product
          distributes over union.
          }
      }\\
  &&&=\underbrace{\absolute{A'\cartesian A'}}_
    {\parbox{4em}
      {
        \centering
          \scriptsize{
            $k^2$
            \begin{flushleft}
              by the inductive hypothesis
            \end{flushleft}
          }
        %centering ends here
      }
    }+
    \underbrace{\absolute{\set a\cartesian A'}}_
    {\parbox{4.5em}
      {
        \centering
          \scriptsize{
            $k$
            \begin{flushleft}
              because each one of the $k$
              elements in $A'$ matches up
              once with the single element
              in $\set a$
            \end{flushleft}
          }
        %centering ends here
      }
    }+
    \underbrace{\absolute{A'\cartesian\set a}}_
    {\parbox{4em}
      {
        \centering
          \scriptsize{
            $k$
            \begin{flushleft}
              likewise
            \end{flushleft}
          }
        %centering ends here
      }
    }+
    \underbrace{\absolute{\ordered{a,a}}}_1&
    \parbox{10em}{
      \raggedleft
      \scriptsize{
      The size of a union of sets is equal to the 
      sum of the sizes of the constituent sets.}
    }\\
  &&&=k^2+2k+1&\\
  &&&={(k+1)}^2&\text{as desired.}
\end{align*}