Golden Ratio

**dwsmith** · June 11th 2011, 05:06 PM

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

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

I am confused on what do for this problem.

**Random Variable** · June 11th 2011, 05:19 PM

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

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

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

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

**dwsmith** · June 11th 2011, 05:20 PM

Originally Posted by Random Variable

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

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

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

Now just solve solve for $\varphi$ 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.

**Random Variable** · June 11th 2011, 05:30 PM

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

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

**Also sprach Zarathustra** · June 11th 2011, 05:40 PM

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

Or:

$\phi^2=\phi+1$

Or:

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

THE ROOTS ARE:

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

**dwsmith** · June 11th 2011, 05:53 PM

Originally Posted by Random Variable

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

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

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

**Also sprach Zarathustra** · June 11th 2011, 06:00 PM

Originally Posted by dwsmith

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

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