Square of Golden Mean equals One plus Golden Mean

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Theorem

$\phi^2 = \phi + 1$

where $\phi$ denotes the golden mean.


Decimal Expansion

The decimal expansion of $\phi^2$ is given as:

$\phi^2 \approx 2 \cdotp 61803 \, 39887 \, 49894$


Thus the square of the golden mean is the unique number $n$ such that:

$\sqrt n = n - 1$


Proof

\(\ds \phi\) \(=\) \(\ds \frac 1 {\phi - 1}\) Definition 3 of Golden Mean
\(\ds \leadstoandfrom \ \ \) \(\ds \phi \paren {\phi - 1}\) \(=\) \(\ds 1\)
\(\ds \leadstoandfrom \ \ \) \(\ds \phi^2\) \(=\) \(\ds \phi + 1\)

$\blacksquare$


Sources

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