Exercise 3.2-6

Show that the golden ratio \(\phi\) and its conjugate \(\hat\phi\) both satisfy the equation \(x^2 = x + 1\).

\[\begin {aligned} \phi^2 & = \left(\frac {1 + \sqrt 5} 2\right)^2 \\ & = \frac {1 + 2\sqrt 5 + 5} 4 \\ & = \frac {3 + \sqrt 5} 2 \\ & = \frac {1 + \sqrt 5} 2 + 1 \\ & = \phi + 1 \end {aligned}\] \[\begin {aligned} \hat{\phi^2} & = \left(\frac {1 - \sqrt 5} 2\right)^2 \\ & = \frac {1 - 2\sqrt 5 + 5} 4 \\ & = \frac {3 - \sqrt 5} 2 \\ & = \frac {1 - \sqrt 5} 2 + 1 \\ & = \hat\phi + 1 \end {aligned}\]