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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
3 years, 10 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

Guilherme Dela Corte - 3 years, 10 months ago

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Can an admin edit this note so that the tags #CosinesGroup, #GoldenRatio, and #Algebra appear?

Bob Krueger - 3 years, 10 months ago

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