anchor.diff:                    \left(\frac{f_n}{f_{n+1}}\right) &=\bigl([0],[0;1],[0;1,1],[0;1,1,1],[0;1,1,1,1],[0;1,1,1,1,1],\dots\bigr)\\
                     &\to[0;1,1,1,1,1,\dots]=\Phi\\
                   \end{align}
  -              </p>
  -              <p>&nbsp;<i class="qed" lang="la"><abbr title="quod erat demonstrandum">q.e.d.</abbr></i></p>
  -              <p>Of course for a formal proof, we would need to show that given any
  -                positive <span class="M">\varepsilon</span>, we can find at least one <span class="M">N</span> such that
  -                whenever <span class="M">n > N</span>, it is the case that
  -                <span class="M">\left| \frac{f_n}{f_{n+1}} - \Phi \right| \lt \varepsilon </span>.</p>
  -            </dd>
  -          </dl>
  -          <p>Interestingly,
  -            <span class="M">\left(\frac{f_n}{f_{n+1}}\right) \to \Phi</span> in an
  +              p &nbsp;#[i.qed(lang="la"): abbr(title="quod erat demonstrandum") q.e.d.]
  +              p.
  +                Of course for a formal proof, we would need to show that given any
  +                positive #[span.M \varepsilon], we can find at least one #[span.M N] such that
  +                whenever #[span.M n > N], it is the case that
  +                #[span.M \left| \frac{f_n}{f_{n+1}} - \Phi \right| \lt \varepsilon].
  +          p.
  +            Interestingly,
  +            #[span.M \left(\frac{f_n}{f_{n+1}}\right) \to \Phi] in an
               alternating manner: the sequence's odd terms underestimate the limit, and the even terms
  -            overestimate it. A decimal representation makes this easier to visualize.</p>
  -          <p class="M M--B">
  +            overestimate it. A decimal representation makes this easier to visualize.
  +          p.M.M--B.
                 \left(\frac{0}{1},\frac{1}{1},\frac{1}{2},\frac{2}{3},\frac{3}{5},\frac{5}{8},\frac{8}{13},\frac{13}{21},\frac{21}{34},\frac{34}{55},\dots,\frac{f_n}{f_{n+1}}\right)\\
                 =\left(0,1,0.5,0.\overline{6},0.6,0.625,0.\overline{615384},
                   0.\overline{619047},0.6\overline{1764705882352941},0.6\overline{18},
                   %0.\overline{61797752808988764044943820224719101123595505},0.6180\overline{5},
                   \dots\right)\\
                 \to 0.61803\ldots = \Phi
  -          </p>
  -        </section>
  -      footer.
  -        <h2>Further Reading</h2>
  -        <p>Read about how <span class="M">\phi</span> is the
  -          <a rel="external"
  -          href="http://www.ams.org/samplings/feature-column/fcarc-irrational1">most irrational number</a>!
  -          If that's true, what do you think about <span class="M">\Phi</span>?</p>
  +      footer
  +        h2 Further Reading
  +        p.
  +          Read about how #[span.M \phi] is the
  +          #[a(rel="external" href="http://www.ams.org/samplings/feature-column/fcarc-irrational1") most irrational number]!
  +          If that's true, what do you think about #[span.M \Phi]?
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positive.code:                     &= \left(f_k+f_{k-1}\right)\phi + f_k\\
                      &= \left(f_{k+1}\right)\phi + f_k
                    \end{align}
                  </span></p>
                <p><i class="qed" lang="la"><abbr title="quod erat demonstrandum">q.e.d.</abbr></i></p>
              </dd>
            </dl>
            <p>You might suspect that
              <span class="M"> \lim \frac{f_{n+1}}{f_n} = \phi </span>. Equivalently,</p>
            <dl class="examp">
              <dt><p class="M M--B">\lim \frac{f_n}{f_{n+1}} = \Phi</p></dt>
              <dd>
                <p><strong>Proof (informal).</strong></p>
                <p class="M M--B">
                  \begin{align}
                    \left(\frac{f_n}{f_{n+1}}\right) &= \left(\frac{0}{1},\frac{1}{1},\frac{1}{2},\frac{2}{3},\frac{3}{5},\frac{5}{8},\frac{8}{13},\frac{13}{21},\frac{21}{34},\frac{34}{55},\frac{55}{89},\frac{89}{144},\dots\right)\\
                    &=\left(0,\frac{1}{1},\frac{1}{2},\frac{1}{\frac{3}{2}},\frac{1}{\frac{5}{3}},\frac{1}{\frac{8}{5}},\dots\right)\\
  
negative.code:                 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\dots}}}}\\
                  \phi &=[1;1,1,1,\dots]
                \end{align}
              p.M.B(class="!w:1o2").
                \begin{align}
                  \Phi = \frac{1}{\phi} &= \frac{1}{1+\Phi}\\
                  &= \frac{1}{1+\frac{1}{1+\Phi}}\\
  