Ilan Bernstein

Golden Fraction!

I haven’t much to say today, but hopefully this is a fun read.

In a previous post, I provided a proof that where $F_n$ is the nth term of the Fibonacci sequence, and where $\varphi$ is the golden ratio,

$\lim\limits_{n \to \infty } \frac{F_n}{F_{n-1}} = \varphi$

But I recently learned another fun argument, and I plan to show you it!

The Start

Another well known way to represent the golden ratio is as a continued fraction.

$\varphi = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}}$

So, to give us an idea of what’s going on lets estimate this continued fraction.

$\begin{aligned} & 1+\frac{1}{1}&=\frac{2}{1} \\ & 1+\frac{1}{1+\frac{1}{1}}&=\frac{3}{2} \\ & 1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}&=\frac{5}{3} \\ & 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}}&=\frac{8}{5} \\ \end{aligned}$

Now I’m sure some are starting to see a pattern emerge! But why might this be? We’ll before you read on, I suggest you think about it, because I’m sure you can figure it out (if you haven’t already).

The Why

There are two important parts to understanding the why. First, let’s flip a fraction:

$\frac{1}{\frac{n}{m}}=\frac{m}{n}$

Okay great. Now for the second part:

$\frac{n}{m}=\frac{n}{m}+\frac{m}{m}=\frac{n+m}{m}$

Alright—amazing. And for my final act, I’ll be simplifying from the bottom up:

$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}} = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+1}}} = 1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}} = 1+\frac{1}{1+\frac{1}{\frac{3}{2}}} = 1+\frac{1}{1+{\frac{2}{3}}} = 1+\frac{1}{\frac{5}{3}} = 1+\frac{3}{5} = \frac{8}{5}$

Whelp, sorry for that headache of a simplification. We use our numerator to “store” the last Fibonacci number, and then we add one to the fraction in order to add the current Fibonacci number to the last one, effectively making our new current in the numerator, and our old Fibonacci number in the denominator. The process continues.

I wish you luck in striking gold,
Ilan Bernstein