anchor.diff:                  &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\dots}}}}\\
                   \phi &=[1;1,1,1,\dots]
                 \end{align}
  -            </p>
  -            <p class="M B !w:1o2">
  +            p.M.B(class="!w:1o2").
                 \begin{align}
                   \Phi = \frac{1}{\phi} &= \frac{1}{1+\Phi}\\
                   &= \frac{1}{1+\frac{1}{1+\Phi}}\\
  
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:                 &= 0+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\dots}}}}}\\
                  \Phi &= [0;1,1,1,\dots]
                \end{align}
        section
          h2 Theorems involving #[span.M \phi] and #[span.M \Phi]
          dl.Theorem
            dt: p
              | For Natural powers of #[span.M \phi],
              span.M.M--B \phi^{n+1} = \phi^{n+1} + \phi^n
            dd
              p: strong Proof (informal).
              p.
                Starting with #[span.M \phi]
                in quadratic form, #[span.M \phi^2 = \phi + 1], we multiply
                both sides by #[span.M \phi^n] and we get our result.
              p For a formal proof, use Mathematical Induction on #[span.M n].
          dl.examp
            dt: p
              | For Natural powers of #[span.M \Phi],
              span.M.M--B \Phi^n = \Phi^{n+1} + \Phi^{n+2}
            dd
              p: strong Proof (informal).
              p.
                Start with
                #[span.M \Phi^2 = 1 - \Phi] then multiply both sides by
                #[span.M \Phi^n] to get
                #[span.M \Phi^{n+2} = \Phi^n - \Phi^{n+1}], which yields
                #[span.M \Phi^{n+2} + \Phi^{n+1} = \Phi^n].
                #[i.qed(lang="la"): abbr(title="quod erat demonstrandum") q.e.d.]
              p For a formal proof, use Mathematical Induction on #[span.M n].
          p
            | Thus we have the following quite symmetrical equations:
            span.M.B \phi^n = \phi^{n+2} - \phi^{n+1}
            span.M.B \Phi^n = \Phi^{n+2} + \Phi^{n+1}
          dl.examp
            dt: p #[span.M \phi] is Irrational.
            dt: p.M.M--B \phi \in \mathbb{I}
            dd
              p: strong Proof.
              p.
                Assume to the contrary that #[span.M \phi] is rational. Let
                #[span.M \exists q \in \mathbb{Q} : q = \phi = \frac{1+\sqrt{5}}{2}].
                Then #[span.M.M--B 2q-1 = \sqrt{5}]
              p.
                We know that the field #[span.M \mathbb{Q}]
                is closed under multiplication and  subtraction, so the left side of the equation
                is rational. However, the right side is irrational. We have a contradiction,
                so #[span.M \phi] is not rational. What remains to show is that
                #[span.M \phi\in\mathbb{R}]
                #[del(datetime="2014-08-24"), which is obvious. #[i.qed(lang="la"): abbr(title="quod erat demonstrandum") q.e.d.]]
          dl.examp
            dt: p #[span.M \phi] is Algebraic.
            dt: p.M.M--B \phi \in \mathbb{A}
            dd
              p: strong Proof.
              p.
                By definition, an Algebraic number is a solution to a polynomial
                equation #[span.M 0=\sum_{k=0}^{n}a_kx^k] with integer
                coefficients #[span.M a_k]. Let this polynomial be
                #[span.M x^2-x-1].#[i.qed(lang="la"): abbr(title="quod erat demonstrandum") q.e.d.]
          dl.examp
            dt: p #[span.M \phi] is Constructible.
            dd
              p: strong Proof.
        section
          section
            h3 Connections to the Fibonacci Sequence
            dl.examp
              dt: p.
                A Natural power of #[span.M \phi] can be expressed as a linear
                function of #[span.M \phi] with integer coefficients.
              dt: p.M.M--B \forall n \in\mathbb{N} : \exists a,b \in\mathbb{Z} : \phi^n = a\phi + b
              dd
                p: strong Proof.
                p.
                  We use proof by induction. For
                  #[span.M n=0], we have #[span.M a=0]
                  and #[span.M b=1], so the base case is shown.
                  (Though #[span.M n=1] need not be a base case, it is
                  easy to see that #[span.M a=1] and
                  #[span.M b=0].)
                p
                  | Now let #[span.M n=k] and assume
                  | #[span.M \phi^k = a\phi + b] for some integers
                  | #[span.M a] and #[span.M b]. Then
                  span.M.M--B.
                    \begin{align}
                      \phi^{k+1} &= \phi^k\phi\\
                      &= (a\phi + b)\phi\\
  