# Golden Ratio

We say two quantities are in the **golden ratio** if their ratio is equivalent to the ratio of the sum of the quantities to the larger of the two quantities. It is obtained by dividing a line into two parts such that the longer part divided by the smaller part is also equal to the whole length divided by the longer part:

The golden ratio, also called the golden number, divine proportion, etc., has a very close association with the Fibonacci sequence. It is often represented by the Greek letter phi, $\varphi$ or $\phi$. For the sake of uniformity, we shall denote the golden ratio by $\phi$.

We say two quantities $a$ and $b$, where $a>b$, are in the

golden ratio, if$\frac{a}{b}=\frac{a+b}{a}.$

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## The Golden Number, $\phi$

In this section, we will try to evaluate the numerical value of the golden number or the golden ratio. We know that two quantities $a$ and $b$, where $a>b$, are in the golden ratio, if

$\phi=\frac{a}{b}=\frac{a+b}{a}.$

Then, we have

$\begin{aligned} \phi=\frac{a}{b} & =\frac{a+b}{a} \\ & =1+\frac{b}{a} \\ & =1+\frac{1}{\phi} \qquad \left(\text{since } \frac{a}{b}=\phi\right) \\ \Rightarrow \phi &=\frac{\phi+1}{\phi} \\ \Rightarrow \phi^2-\phi -1 &= 0. \end{aligned}$

The last equation, being a quadratic equation, has its roots given by the quadratic formula as

$\begin{aligned} \phi & =\frac{1\pm\sqrt{(-1)^2-4\cdot(-1)\cdot1}}{2} \\ & = \frac{1\pm\sqrt{5}}{2}. \end{aligned}$

But since the ratio is always positive,

$\phi=\frac{\sqrt5+1}{2}\approx 1.61803.$

## Golden Ratio in terms of an Infinite Continued Fraction

The golden ratio can be expressed as a beautiful infinite continued fraction:

$\displaystyle \phi = 1+ \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}$

## Golden Ratio in terms of Nested Functions

The golden ratio, or simply the golden number, is a very special number and can be found in some very beautiful nested functions. Some of them are discussed below.

$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$

Let

$\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}} =x.$

Then

$\begin{aligned} \sqrt{1+x} &= x \\ 1+x &=x^2 \\ x^2-x-1 &= 0. \end{aligned}$

Note that this is the equation which has roots $\frac{1\pm\sqrt{5}}{2}$. But since the radical is positive,

$x=\frac{\sqrt{5}+1}{2}=\phi. \ _\square$

$\phi = \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+ \cdots }}}}}$

Let $x=\sqrt{a-\sqrt{a+\sqrt{a-\sqrt{a+\cdots}}}}$ and $y=\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a-\sqrt{a+\cdots}}}}},$ then

$\begin{aligned} x^2&=a-y &\qquad (1)\\ y^2&=a+x. &\qquad (2) \end{aligned}$

Eliminating $a$ from (1) and (2), we get $x=y-1$. Now, putting this in (1) gives

$y^2-y+(1-a)=0 \implies y=\frac{1+\sqrt{4a-3}}{2}.$

Now, putting $a$ = 2, we get the given nested radical. $_\square$