+ - × ÷

So, you've seen (perhaps on some other pages) that the golden ratio is somewhere around 1.618, or so. Why that number, instead of just 1.6, or 3/2?

Well, instead of an answer, we'll look at a couple more questions first.

1+1/(1+1/(1+1/(1+1/(... = ?

sqrt(1+sqrt(1+sqrt(1+sqrt(1+... = ?

You can use computer programs to get approximations which hint that the answers are probably the same number we keep seeing over and over again. If these questions all have the same answer, they probably have something in common, right? Lets see if we can turn them all into algebraic problems.

The Fibonacci sequence

Remember that a term in the Fibonacci sequence is the sum of the two previous terms. This can be expressed as xn = xn-1 + xn-2. The ratio of two consecutive terms can be expressed as r = xn / xn-1. Combining the two equations yields r = (xn-1 + xn-2) / xn-1 which can be simplified to r = 1 + xn-2 / xn-1. Now this is where we play some fancy tricks: we're going to assume that the ratio is the same, regardless of how far down the sequence you are. it's not quite true, but as was shown in the demonstration, it gets pretty close pretty fast. So, if we make that assumption, then xn-2 / xn-1 = xn-1 / xn = 1 / (xn / xn-1) = 1 / r. That in turn (going back to the orginal equation for r) yields r = 1 + 1/r.

Huh. All of a sudden there's something that looks looks familiar. Now, just as we've found something in common, lets put it on hold for a bit, and look at some of the others.

The Golden Rectangle

The definition of a golden rectangle is that it has an aspect ratio such that if you stick a square on to the long side of the triangle, you get a new rectangle with the same aspect ratio. Lets start with a rectangle that has a short side that is 1 long. (one what? inch, foot? what? does it matter?) Then the length of the other side will be r. In order to stick a square onto the long side, the square needs to have sides that are r long as well. This will make the short side of the new rectangle r, and the long side will be the sum of the side of the square, and the short side of the old rectangle (which was 1), or r + 1. We can now write expressions for the aspect ratios of the two rectangles, and set them equal to each other: r / 1 = (r + 1) / r. Hmm. It's not the same as the equation above, exactly, but one of the things you learn in algebra is that things that look different may actually be the same, if you manipulate them properly. however, if you split the fraction in the right side, you get r = r/r + 1/r = 1 + 1/r.

The Golden Triangle

With the golden triangle, remember that the small isosceles triangle along the base was also a golden triangle. Lets call the length of that base 1, and the length of the legs r. Since the leg of that small triangle forms the base of the large one, the only thing we have to figure out is the length of the long leg. Notice, however, that the other triangle is isosceles also, so that the section extending beyond the base of the small triangle is the same as the base of the big triangle. so the length of the long leg is r + 1. Setting the ratios of the sides equal then gives us: r / 1 = (1 + r) / r, which is exactly where the golden rectangle problem ended up.

Infinite expression #1

At the top of the page, we wondered what 1+1/(1+1/(1+1/(1+1/(... was equal to.

Well, how about if we call it x for now? Then we get x = 1+1/(1+1/(1+1/(1+1/(... . Because it's a repeating expression, x shows up in its own definition. Here: x = 1+1/(1+1/(1+1/(1+1/(... ; notice that the bold part of that expression exactly matches what we orginally defined x as! So we can plug x in for the bold part of the equation (and you should be expecting this part now) we get x = 1+1/x. It's deja vu all over again!

Infinite expression #2

This is a different expression, but we can use exactly the same technique we just used. let x = sqrt(1+sqrt(1+sqrt(1+sqrt(1+.... Looking for the expression in itself gives x = sqrt(1+sqrt(1+sqrt(1+sqrt(1+sqrt(1+... = sqrt(1+x).

Well, that's certainly not what we expected, right? Or maybe not. Remember that equations can look different, but still mean the same thing. We'll simplify this one first, and then see what we can do with the rest of them.

x = sqrt(1+x)
get rid of the radical by squaring both sides.
x2 = 1 + x
x2 - x - 1 = 0
There we go - a simple quadratic equation with one variable in standard form.

Pulling it all together!

Now to get the other equation (since it was the same equation for all four problems) into standard form.

x = 1 + 1/x
hmm - get rid of the fraction by multilpying everything by x.
x2 = x + 1
x2 - x - 1 = 0
Wah Laaa! It's the same equation!

This is one of the big reasons we study math: there are often problems that look completely different, that should have nothing in common at all, except that once you express them in mathematical terms, you can find similarities between the problems. Even though there is no direct connection between the two real world situations, because they can be described in similar mathematical terms, you can combine the experiences from the two situations and know more afterwards about both of them that you ever would have had you not been able to draw the parallels between them. The final answer here isn't what's important, it's that all of the problems end up being identical.

Oh - the answer? You can simply plug the parameters into the quadratic formula, which will give you the answer of of x = (1+sqrt(5))/2 ~= 1.6180339887499. But we knew that already, didn't we?