Equivalence of Definitions of Golden Mean

Theorem

The following definitions of the concept of Golden Mean are equivalent:

Definition 1

Let a line segment $AB$ be divided at $C$ such that:

$AB : AC = AC : BC$

Then the golden mean $\phi$ is defined as:

$\phi := \dfrac {AB} {AC}$

Definition 2

The golden mean is the unique positive real number $\phi$ satisfying:

$\phi = \dfrac {1 + \sqrt 5} 2$

Definition 3

The golden mean is the unique positive real number $\phi$ satisfying:

$\phi = \dfrac 1 {\phi - 1}$

Proof

Definition 1 is equivalent to Definition 3

Let $AB : AC = AC : BC$.

Let $\dfrac {AB} {AC} = \dfrac {AC} {BC} = \phi$.

Then:

\(\ds \phi\)

\(=\)

\(\ds \frac {AC + BC} {AC}\)

as $AB = AC + BC$

\(\ds \)

\(=\)

\(\ds 1 + \frac {BC} {AC}\)

\(\ds \)

\(=\)

\(\ds 1 + \frac 1 \phi\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \phi - 1\)

\(=\)

\(\ds \frac 1 \phi\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \frac 1 {\phi - 1}\)

\(=\)

\(\ds \phi\)

$\Box$

Definition 2 equivalent to Definition 3

\(\ds \phi\)

\(=\)

\(\ds \frac 1 {\phi - 1}\)

Definition 3 of Golden Mean

\(\ds \leadstoandfrom \ \ \)

\(\ds \phi \paren {\phi - 1}\)

\(=\)

\(\ds 1\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \phi^2 - \phi - 1\)

\(=\)

\(\ds 0\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \phi\)

\(=\)

\(\ds \frac {1 \pm \sqrt {1^2 - 4 \times 1 \times \paren {-1} } } 2\)

Quadratic Formula

\(\ds \)

\(=\)

\(\ds \frac {1 \pm \sqrt 5} 2\)

Of these two roots, only $\dfrac {1 + \sqrt 5} 2$ is positive.

$\blacksquare$