Here is the first post introducing the golden ratio.

What else is there to the Golden Ratio? We will see that there are infinite possibilities with it. Here are a few.

From the first notable equation mentioned in the previous post, we can take a square root of both sides. We just get another relation involving \(\phi\). The left hand side is \(\phi\) itself, but the right hand side is some other expression containing \(\phi\). What happens if I substitute the \(\phi\) on the right side with something else. I can then repeat. Here is the math:

\[\phi\] \[=\sqrt{1+\phi}\] \[=\sqrt{1+\sqrt{1+\phi}}\] \[=\sqrt{1+\sqrt{1+\sqrt{1+\phi}}}\] \[=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}}\]

We can extend this operation so that it goes on forever. This is what's called an infinite radical. But it still always equals the golden ratio. This number is starting to seem more special than just some ratio.

Now, using the second notable equation, let's apply the same idea:

\[\phi\] \[=1+\frac{1}{\phi}\] \[=1+\frac{1}{1+\frac{1}{\phi}}\] \[=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}}\]

This goes on forever too! And this one's called a continued fraction. Continued fractions are a well developed subject of mathematical study, where all the one's change to other numbers. Every number has it's own continued fraction. And, low and behold, the golden ratio has the simplest continued fraction of all.

Click here for the next post.

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## Comments

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TopNewestThis is AWESOME !!!!!!!!

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Nice..challenging!

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Prove that 0! equals 1

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How many ways are there to choose 0 objects from 0 objects?

One: DO NOTHING!

(I got this from AoPS. It's not really a rigorous proof of 0!=1 but it explains it well.)

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nCn is equal to 1. so (n!)/(n-n)!.(n!) = 1 therefore (n-n)!=0! must be equal to 1.

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(n-1)! * n = n! For n = 1: (1-1)! * 1 = 1! 0! *1 = 1 0! = 1

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ncn=1 ,implies n!/(n-n)! .n!=1.. so,0!=1 hence proved..

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Good..................

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Can you prove these equation chains?

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Because I derived them using facts about the golden ratio, there is no need to prove them. They are already proven.

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prove a^0=1

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See, Devansh.........\(a^{b}*a^{-b}=a^{b-b}=a^{0}=a*\frac{1}{a}=1\)

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