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

ϕ2=ϕ+1\phi^2=\phi +1

And is equal to 1+52\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:



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 1+52\frac{1+\sqrt{5}}{2} is equal to the golden ratio, but can you express 152\frac{1-\sqrt{5}}{2}, 1+52\frac{-1+\sqrt{5}}{2}, and 152\frac{-1-\sqrt{5}}{2} in terms of the golden ratio? Post your answers and work below.

Note by Bob Krueger
7 years, 4 months ago

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

Bob Krueger - 7 years, 4 months ago

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That wasn't hard.

152=1ϕ=1ϕ=11+11+1()\frac{1 - \sqrt{5}}{2} = 1 - \phi = -\frac{1}{\phi} = - \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{(\cdots)}}} 1+52=ϕ1=1ϕ=11+11+1()\frac{-1 + \sqrt{5}}{2} = \phi - 1 = \frac{1}{\phi} = \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{(\cdots)}}} 152=ϕ=111+11+1()\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 - 7 years, 4 months ago

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