Thread: Golden Ratio

The Golden ratio is the ratio $\displaystyle b:a \ (b>a)$ so that $\displaystyle b:a \ =(a+b):b$ or $\displaystyle \frac{b}{a}=\frac{a+b}{b}$.

Show that if $\displaystyle b:a$ is the Golden ratio, then $\displaystyle \phi=b/a=\frac{1+\sqrt{5}}{2}$

I am confused on what do for this problem.

$\displaystyle \frac{b}{a} = \varphi $

$\displaystyle \frac{a+b}{b} = \frac{a}{b} + 1 = \frac{1}{\varphi} + 1 $

so $\displaystyle \varphi = \frac{1}{\varphi} + 1 $

Now just solve solve for $\displaystyle \varphi $ and ignore the negative root.

Originally Posted by Random Variable

$\displaystyle \frac{b}{a} = \varphi $

$\displaystyle \frac{a+b}{b} = \frac{a}{b} + 1 = \frac{1}{\varphi} + 1 $

so $\displaystyle \varphi = \frac{1}{\varphi} + 1 $

Now just solve solve for $\displaystyle \varphi $ and ignore the negative root.

I don't understand why you even did that. With this question, I have no idea what they were even asking.

Two numbers are in the golden ratio if $\displaystyle \frac{b}{a} = \varphi $ and $\displaystyle \frac{a+b}{b} = \varphi $

Were interested in the ratio $\displaystyle \varphi $. But notice that $\displaystyle \varphi = \frac{a+b}{b} = \frac{a}{b} + 1 = \frac{1}{\varphi} + 1 $

$\displaystyle \phi=\frac{1}{\phi}+1$

Or:


$\displaystyle \phi^2=\phi+1$

Or:


$\displaystyle \phi^2-\phi-1=0$

THE ROOTS ARE:

$\displaystyle \phi_{1,2}=\frac{1\pm \sqrt{1+4}}{2}$

Originally Posted by Random Variable

Two numbers are in the golden ratio if $\displaystyle \frac{b}{a} = \varphi $ and $\displaystyle \frac{a+b}{b} = \varphi $

Were interested in the ratio $\displaystyle \varphi $. But notice that $\displaystyle \varphi = \frac{a+b}{b} = \frac{a}{b} + 1 = \frac{1}{\varphi} + 1 $

Why isn't the negative value considered in the question?

Originally Posted by dwsmith

Why isn't the negative value considered in the question?

Read about this ratio here:

Golden ratio - Wikipedia, the free encyclopedia