1. ## 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.

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

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

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

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

3. 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.

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

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

Or:

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

Or:

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

THE ROOTS ARE:

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

6. 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?

7. Originally Posted by dwsmith
Why isn't the negative value considered in the question?