anchor.diff:                      &= a\phi + a + b\phi\\
                       &= (a+b)\phi + a
                     \end{align}
  -                </span></p>
  -              <p>Since <span class="M">(a+b) \in \mathbb{Z}</span>, the
  -                property holds for <span class="M">k+1</span>.</p>
  -              <p>Since the property holds for the base case, and the fact that it
  +              p.
  +                Since #[span.M (a+b) \in \mathbb{Z}], the
  +                property holds for #[span.M k+1].
  +              p.
  +                Since the property holds for the base case, and the fact that it
                   holds for some natural number implies that it holds for the next natural number,
  -                the property must hold for all natural numbers. <i class="qed" lang="la"><abbr title="quod erat demonstrandum">q.e.d.</abbr></i></p>
  -            </dd>
  -          </dl>
  -          <p>The consequence of this theorem is that we need not deal with any powers
  -            of <span class="M">\phi</span>, because we can always express it in the first degree.
  -            <mark class="site">This
  -            property will be useful when I talk about the layout structure of this Website.</mark></p>
  -          <p>Let <span class="M">\phi^k = a_k\phi + b_k</span>, and consider
  -            a few examples.</p>
  -          <p class="M M--B">
  +                the property must hold for all natural numbers.
  +                #[i.qed(lang="la"): abbr(title="quod erat demonstrandum") q.e.d.]
  +          p.
  +            The consequence of this theorem is that we need not deal with any powers
  +            of #[span.M \phi], because we can always express it in the first degree.
  +            #[mark.site This property will be useful when I talk about the layout structure of this Website.]
  +          p Let #[span.M \phi^k = a_k\phi + b_k], and consider a few examples.
  +          p.M.M--B.
               \begin{align}
                 \phi^0 &= 0\phi + 1\\
                 \phi^1 &= 1\phi + 0\\
  
positive.code:                 &= 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">
                \begin{align}
                  \Phi = \frac{1}{\phi} &= \frac{1}{1+\Phi}\\
                  &= \frac{1}{1+\frac{1}{1+\Phi}}\\
  
negative.code:               &= f_{n}(1) + f_{n-1}(0)\\
                &= f_{n}\\
              \end{align}
            p.
              Here's an example: Take #[span.M f_{10}], which
              is 55. 55 is made of one 34 and one 21. Now since a 34 is made of one 21 and one 13,
              55 is made of a 21, a 13, and another 21, which, more simply put, is the same as two 21s
              and one 13.
            p.
              Since a 21 is made of one 13 and one 8, 55 is made of two 13s, two 8s, and another 13,
              which again is three 13s and two 8s.
              Since a 13 is made of one 8 and one 5, 55 is made of three 8s, three 5s, and two more 8s,
              which is five 8s and three 5s.
            p.
              We can keep breaking down Fibonacci numbers into more Fibonacci numbers, and we obtain
              eight 5s and five 3s, or thirteen 3s and eight 2s, or twenty-one 2s and thirteen 1s,
              or thirty-four 1s and twenty-one 1s, or fifty-five 1s. In conclusion, one 55 is
              the same as fifty-five 1s.
            pre(class="!ff:ms").
              55=
                = 1*( 34  ) + 1*21
                = 1*(21+13) + 1*21 = 1*21 + 1*13 + 1*21 = 2*( 21 ) + 1*13
                                                        = 2*(13+8) + 1*13 = 2*13 + 2*8 + 1*13 = 3*(13 ) + 2*8
                                                                                              = 3*(8+5) + 2*8 = 3*8 + 3*5 + 2*8 = 5*( 8 ) + 3*5
                                                                                                                                = 5*(5+3) + 3*5 = 5*5 + 5*3 + 3*5 = 8*( 5 ) + 5*3
                                                                                                                                                                  = 8*(3+2) + 5*3 = 8*3 + 8*2 + 5*3 = 13*( 3 ) + 8*2
                                                                                                                                                                                                    = 13*(2+1) + 8*2 = 13*2 + 13*1 + 8*2 = 21*( 2 ) + 13*1
                                                                                                                                                                                                                                         = 21*(1+1) + 13*1 = 21*1 + 21*1 + 13*1 = 34*1 + 21*1
                                                                                                                                                                                                                                                                                = 55*1
            p.
              Notice the similarities between the sequence
              #[span.M f_n = f_{n-1} + f_{n-2}] and the property
              #[span.M \phi^n=\phi^{n-1}+\phi^{n-2}].
            dl.examp
              dt: p.
                Corollary: A natural power of #[span.M \phi] can be expressed as
                a linear function of #[span.M \phi] with coefficients corresponding to the
                Fibonacci Sequence.
              dt: p.M.M--B \phi^n = f_n\phi + f_{n-1}
              dd
                p: strong Proof.
                p.
                  This proof is a specification of the
                  previous. It says that not only are the coefficients of #[span.M \phi] integers,
                  but they are terms of the Fibonacci Sequence in sequential order. The proof
                  follows directly from the previous, with but a few changes.
                p
                  | Again, we use induction. For #[span.M n=0],
                  | #[span.M f_n = 0] and #[span.M f_{n-1}]
                  | does not exist. Set it to 1, and
                  | #[span.M \phi^1 = f_1\phi + f_0 = 1\phi+0 = \phi]. For the
                  | inductive hypothesis, suppose
                  | #[span.M \phi^k = f_k\phi + f_{k-1}]. Then,
                  span.M.M--B.
                    \begin{align}
                      \phi^{k+1} &= \phi^k\phi\\
                      &= \left(f_k\phi + f_{k-1}\right)\phi\\
  