Let \(\phi=\dfrac{1+\sqrt{5}}{2}\approx1.61803\). For integers \(d_k, d_{k-1},\dots,d_1,d_0,d_{-1},\dots,d_{-r}\), each equal to \(0\) or \(1\), the expression \[(d_kd_{k-1}\cdots d_2d_1d_0.d_{-1}d_{-2}\cdots d_{-r})_{\phi}\] is called a base \(\phi\) expansion and represents the real number \[d_k\phi^k+d_{k-1}\phi^{k-1}+\cdots+d_1\phi+d_0+d_{-1}\phi^{-1}+d_{-2}\phi^{-2}+\cdots+d_{-r}\phi^{-r}\] The integers \(d_k\) through \(d_{-r}\) are called the digits of the expansion. For example, the base \(\phi\) expansion \(1101.011_{\phi}\) represents the real number \[(1\times \phi^3)+(1\times\phi^2)+(0\times\phi)+1+(0\times\phi^{-1})+(1\times\phi^{-2})+(1\times\phi^{-3})\] which can be simplified to get \[\begin{align*} \phi^3+\phi^2+1+\frac{1}{\phi^2}+\frac{1}{\phi^3} &= \left(\frac{1+\sqrt{5}}{2}\right)^3+\left(\frac{1+\sqrt{5}}{2}\right)^2+1+\left(\frac{2}{1+\sqrt{5}}\right)^2+\left(\frac{2}{1+\sqrt{5}}\right)^3 \\ &=\frac{16+8\sqrt{5}}{8}+\frac{6+2\sqrt{5}}{4}+1+\frac{4}{6+2\sqrt{5}}+\frac{8}{16+8\sqrt{5}} \\ &= (2+\sqrt{5})+\left(\frac{3}{2}+\frac{1}{2}\sqrt{5}\right)+1+\left(\frac{3}{2}-\frac{1}{2}\sqrt{5}\right)-(2-\sqrt{5}) \\ &= 4+2\sqrt{5}\end{align*}\] and so \(1101.011_{\phi}=4+2\sqrt{5}\).
What are the real numbers represented by \(1011_{\phi}\) and \(10000_{\phi}\)?
Find a base \(\phi\) expansion of the real number \(4+3\sqrt{5}\).
Show that \(\phi^2=\phi+1\) and use this to deduce that \(\phi^{n+1}=\phi^n+\phi^{n-1}\) for all integers \(n\).
Show that every positive integer has a base \(\phi\) expansion and find a base \(\phi\) expansion for each positive integer from \(1\) through \(10\). One approach is to prove and use the following two facts.
If a real number \(n\) has a base \(\phi\) expansion, then it has a base \(\phi\) expansion that does not have two consecutive digits equal to \(1\).
If a real number \(n\) has a base \(\phi\) expansion, then it has a base \(\phi\) expansion that has its units digit, \(d_0\), equal to \(0\).