Let $\varphi : = (1 + \sqrt{5}) / 2$ be the D2137: Basic real golden ratio such that

(i)	\begin{equation} \varphi_* : = 1 - \varphi \end{equation}

We have \begin{equation} \begin{split} \varphi^2 - \varphi - 1 & = \left( \frac{1 + \sqrt{5}}{2} \right)^2 - \left( \frac{1 + \sqrt{5}}{2} \right) - 1 \\ & = \frac{(1 + \sqrt{5})^2}{4} - \frac{1 + \sqrt{5}}{2} - 1 \\ & = \frac{1 + 2 \sqrt{5} + 5}{4} - \frac{1 + \sqrt{5}}{2} - 1 \\ & = \frac{1}{4} + \frac{\sqrt{5}}{2} + \frac{5}{4} - \frac{1}{2} - \frac{\sqrt{5}}{2} - 1 \\ & = \frac{1}{4} + \frac{5}{4} - \frac{1}{2} - 1 \\ & = \frac{1}{4} + \frac{5}{4} - \frac{1}{2} - \frac{4}{4} \\ & = \frac{1}{4} + \frac{1}{4} - \frac{1}{2} \\ & = \frac{2}{4} - \frac{1}{2} \\ & = \frac{1}{2} - \frac{1}{2} \\ & = 0 \\ \end{split} \end{equation} The case for $1 - \varphi$ can similarly be directly computed. $\square$