Golden Ratio example not satisfying ratio

Question

Golden ratio says $\frac{a}{b}=\frac{a+b}{a}$ and it shows up in rectangle also. If we take rectangle of $a=4$ and $b=2$. then $\frac{a}{b}=\frac{4}{2}=2$ and $\frac{a+b}{a}=\frac{4+2}{4}=\frac{6}{4}=1.5$. But $1.5 \ne 2$, how does this hold true?

Answer 1

Golden rectangle is a rectangle with side ratio $1 : \frac{\sqrt{5} +1}{2} $ , not any arbitrary rectangle.

Answer 2

I think you're confused because you are thinking both $a$ AND $b$ are independent variables you can arbitrarily set. A golden rectangle is any rectangle whose side lengths $a$ and $b$ satisfy $$ \frac{a}{b} = \frac{a+b}{a} $$ If you decide to set $a = 2$, then $b$ must be restricted if you want the above equation to hold true. In fact, you'd get \begin{align} \frac{2}{b} &= \frac{b+2}{2} \\ b(b+2) &= 4 \\ b^2 + 2b - 4 &= 0 \end{align} Using the quadratic formula: \begin{align} b &= \frac{-2 + \sqrt{2^2-4(1)(-4)}}{2} \\ &= \frac{-2 + \sqrt{20}}{2} \\ &= \frac{-2 + 2\sqrt{5}}{2} \\ &= \sqrt{5} - 1 \end{align} Hence we get \begin{align} \frac{a}{b} &= \frac{2}{\sqrt{5}-1} \\ &= \frac{2(\sqrt{5}+1)}{5-1} \\ &= \frac{2(\sqrt{5}+1)}{4} \\ &= \frac{\sqrt{5}+1}{2} \end{align} which is the Golden Ratio.