This post contains a little more technical stuff, with less explanation, than many of the posts on this blog. If there is something you don’t understand, feel free to ask about it below. Or skim it – you won’t miss out on the main idea.

In the last post, we looked at this mystery number:

We left off with a question: what is the value of this sum, written as a “regular” number?

To keep things from getting too messy, I’m going to make up a new name for the mystery number. I’m going to call it “M”: that’s a nice short name, and it reminds me of the word “mystery.”

The letter M here is an example of notation. I am choosing to represent the mystery number as M, instead of as a big fraction. This type of neatness is generally useful. The more abstract an idea, the more important it is to have a good way to put it on paper.

Unfamiliar notation can make written math look intimidating, at least until you learn what each symbol stands for. But good notation makes things clearer. An eloquent discussion of notation can be found in a classic book on problem solving, George Pólya’s How To Solve It.

Now let’s talk about M.

I have two ways to approach it, though there are certainly more out there that I didn’t think of. My first method involves some algebra.

Method Number One

Take a look at this number again.

Notice that if we peel the outermost “1+1/” off this number, we end up with another copy of M, albeit in a slightly smaller font. Because of the fractal nature of M, the bottom half of the fraction is the same as the entire number. That gives us two different ways to write the mystery number: as M, or as 1 plus 1 over M. We can write that as an equation:

[latex] 1+\frac{1}{M} = M[/latex].

By writing M in two different ways, we learned something about it. This equation means that if you take 1 plus 1 over the mystery number, you get the mystery number again.

Now, if we can find a “regular” number that satisfies this equation, we know it must be the mystery number.

Well, if $latex 1+\frac{1}{M}$ and $latex M$ are equal, that means they both represent the same number. They are two different ways of writing the same thing. So, if I subtract 1 from each of those terms, I’ll get a new number, but both sides will still be equal. That’s the idea behind a lot of algebraic manipulation: when two terms are equal, they represent the same number, so if you do the same thing to both of them, they will still be equal.

[latex] 1+\frac{1}{M}-1= M-1[/latex].

Now, $latex 1+\frac{1}{M}-1$ is the same as $latex \frac{1}{M}$. So,

[latex] \frac{1}{M} = M-1[/latex].

Since these two terms are equal, we can multiply each by M and again get two equal terms. So,

[latex] M \times \frac{1}{M} = M \times (M-1)[/latex].

Multiply this out, and the M’s on the left cancel. On the right side, we multiply everything in the parenthesis by M. So,

[latex]1=M^2-M[/latex].

We still don’t know which number makes this true, but the equation is starting to look workable. Now, subtract 1 from both sides again.

[latex] 0 = M^2-M-1 [/latex].

This is a special type of equation called a quadratic equation, and it has a 0 on one side. Using something called the quadratic formula, which I will not explain in depth right now (though it does have good reasons behind it if you want to learn about it!), we find

[latex] M=\frac{1+\sqrt{5}}{2} [/latex].

Hmm, that’s interesting.

Before we talk about why that’s interesting, let’s take a different perspective on this number.

Method Number Two

Let’s look at a number that’s similar to the one we’re interested in: $latex 1+\frac{1}{1+\frac{1}{1}}$. It almost looks like a mini-version of M, but it’s not quite the same, because it stops repeating itself pretty quickly. There is no “…” in this number.

First, let’s look at the fraction in the second term of the denominator: $latex \frac{1}{1}$. This is equal to 1.

Next, the denominator as a whole: $latex 1+\frac{1}{1}$ is equal to $latex \frac{2}{1}$, or 2.

Finally, let’s look at the bigger fraction: $latex 1+\frac{1}{1+\frac{1}{1}}$ is equal to $latex \frac{3}{2}$.

What if we added another layer, another “1+1/”, to the outside of this fraction? We’d have $latex 1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}$. We know what most of this fraction looks like already. The whole thing is equal to $latex 1+\frac{1}{1+\frac{3}{2}}$, which is the same as $latex \frac{5}{3}$.

So far, by working outwards, we’ve got a little sequence: there’s $latex \frac{1}{1}$, then $latex \frac{2}{1}$, then $latex \frac{3}{2}$, then $latex \frac{5}{3}$. If you add another layer to the outside of the fraction, you get $latex \frac{8}{5}$. Add another, and you get $latex \frac{13}{8}$.

You might notice a pattern here. These fractions involve a sequence of positive integers. Each fraction is made up of some integer in the sequence, divided by the previous integer in the sequence. And the integers are: 1, 1, 2, 3, 5, 8, 13…

It’s the start of the Fibonacci sequence.

That’s weird.

Maybe this pattern will stop after a little while – there have certainly been stranger coincidences in mathematics. Or maybe the pattern continues forever. But there’s no way to directly check – no matter how many fractions we evaluate, there is always room to add another “1+1/” on the outside of the fraction.

But I think the Fibonacci pattern continues forever.

Actually, I don’t just think it. I know it. And I can prove it.

Here is my proof.

To make things easier to write, I’m going to call the first Fibonacci number $latex F_1$, the second Fibonacci number $latex F_2$, and so on. I’m also going to start with the fact that for some number of layers, the fraction is equal to a Fibonacci number divided by the Fibonacci number before it. I’m going to call that number of layers $latex n$. So when there are $latex n$ layers, the fraction is equal to $latex \frac{F_n}{F_{n-1}}$.

We just calculated this for a bunch of values of $latex n$, and we know it’s true for all of those. But for right now, we’re just assuming that $latex n$ is some specific number of layers for which the fraction is $latex \frac{F_n}{F_{n-1}}$. We’re not specifying which number $latex n$ actually is.

Now for some more algebra. By adding another layer to $latex \frac{F_n}{F_{n-1}}$, we get

[latex] 1+\frac{1}{\frac{F_n}{F_{n-1}}} [/latex].

This is equal to

[latex] 1+\frac{F_{n-1}}{F_{n}} [/latex]

Which is equal to

[latex] \frac{F_n+F_{n-1}}{F_{n}} [/latex].

So far so good. Now there’s one important thing we know about Fibonacci numbers: each is the sum of the previous two Fibonacci numbers. In other words, $latex F_{n-1}+F_n=F_{n+1}$. That tells us that the fraction above is equal to

[latex] \frac{F_{n+1}}{F_{n}} [/latex].

Let’s review what we just did here. We showed that if a fraction with $latex n$ layers is equal to $latex \frac{F_n}{F_{n-1}}$, then a fraction with $latex n+1$ layers is equal to $latex F_{n-1}+F_n=F_{n+1}$. That’s the same formula as $latex \frac{F_n}{F_{n-1}}$ for $latex n$ layers, but with $latex n+1$ in place of $latex n$.

We already know that when $latex n=2$, this works: when there are two layers, the fraction is equal to $latex \frac{F_2}{F_{1}}$, or $latex \frac{1}{1}$. Now, since the fraction works for $latex n$, it works for $latex n+1=3$ as well. And since it works when $latex n=3$, it must work for $latex n=4$. And since it works for $latex n=4$, it must work for $latex n=5$… and so on. This proves that for any positive integer $latex n \geq 2$, this fraction with $latex n$ layers is equal to $latex \frac{F_n}{F_{n-1}}$. The pattern continues forever.

As $latex n$ gets bigger and bigger, the value of $latex \frac{F_n}{F_{n-1}}$ gets closer to a particular number. That number is known as the Golden Ratio.

The number M is what you would get if you added infinitely many layers to this fraction. You’d get the number that $latex \frac{F_n}{F_{n-1}}$ approaches as $latex n$ gets bigger and bigger.

In other words, M is the Golden Ratio. As we showed with Method One, M is $latex \frac{1+\sqrt{5}}{2}$.

I don’t know about you, but I find this connection pretty amazing.

Well, there you have it. We took a strange-looking math idea, played with it, and ended up finding connections to some totally different ideas. Hopefully it was a different experience from that of the drills and rules you might associate with math.

There are a lot of follow-up questions I could ask on this topic.

But I’ll let you ask the questions yourself.