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The Golden Ratio: Beginning the Algebra of

Here is the previous post concerning the Golden Ratio. For a collection of all the posts concerning the Golden Ratio, click #GoldenRatio below.

Today we start delving into the algebra around the golden ratio. Remember that the golden ratio is derived from the equation

$\phi^2=\phi +1$

And is equal to $$\frac{1+\sqrt{5}}{2}$$. Also, remember that we touched base a little on some interesting identities in one of the first posts. They were:

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

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

To move on, we need to first introduce Fibonacci numbers. But before we do this, I have a challenge for you. We already know that $$\frac{1+\sqrt{5}}{2}$$ is equal to the golden ratio, but can you express $$\frac{1-\sqrt{5}}{2}$$, $$\frac{-1+\sqrt{5}}{2}$$, and $$\frac{-1-\sqrt{5}}{2}$$ in terms of the golden ratio? Post your answers and work below.

Note by Bob Krueger
2 years, 9 months ago

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That wasn't hard.

$\frac{1 - \sqrt{5}}{2} = 1 - \phi = -\frac{1}{\phi} = - \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{(\cdots)}}}$ $\frac{-1 + \sqrt{5}}{2} = \phi - 1 = \frac{1}{\phi} = \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{(\cdots)}}}$ $\frac{-1 - \sqrt{5}}{2} = - \phi = -1 - \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{(\cdots)}}}$

PS: This is what I was doing before going to sleep... I literally did the same exercises :D · 2 years, 9 months ago