The defining property of the Golden Ratio is a self-replicating quality of rectangles:

*We say that a rectangle has a Golden Ratio, or that its dimensions are in a Golden Ratio if after taking away the biggest square possible, the remaining rectangle is proportional to the original one.*
This means that a rectangle like above will satisfy this condition if

$$\frac{a+b}{a}=\frac{a}{b}\,.$$

Having this property means in particular that it is possible to carry out this trick

*forever,*that is, since the leftover rectangle is similar to the original one, it is possible to take*another*square out and have a smaller leftover rectangle which is similar to the original one. Then, the process can be applied once more, and once again, indefinitely.
This

*recursion*does not hold for other ratios. For example, if the ratio between the dimensions is rational, this process actually ends.
For example, for a rectangle with dimensions $11\times 19$ we have that the process ends after 7 iterations

That is, after 7 times we had no leftover rectangle! The light blue $1\times 1$ leftover was a square already, so it is not possible to repeat the process. This happens because $\frac{19}{11}$ is a rational number. In the case of the golden ratio, it is an irrational number,

$$\phi=\frac{1+\sqrt{5}}{2}\,,$$

so we can never finish the process, as there will always be a leftover rectangle. A nice pattern arises when performing this procedure which is surprisingly related to

*continued fractions*. For the ratio $11\times19$ we have that its simple continued fraction expansion is
$$\frac{19}{11}=1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2}}}}\,,$$

or

$$\frac{19}{11}=[1;1,2,1,2]\,.$$

Notice that there is 1 $11\times 11$ square, 1 $8\times 8$ square, 2 $3\times 3$ squares, 1 $2\times 2$ square, and 2 $1\times 1$ square, the same as the continued fraction coefficients!

Looking at this more geometrically, the continuous fraction expansion tells a way to write down the area of the rectangle as a sum of squares, that is,

$$11\times 19=209=11^2+8^2+3^2+3^2+2^2+1^2+1^2\,.$$

This is also true if the dimensions of the rectangle are commensurable, were the coefficients of the continued fraction indicates the number of times a certain square is repeated. If we have a rectangle with dimensions $p\times q$ with $p/q$ is rational. Then these squares will have dimensions that are of the type $d=np-mq$ with $n,m$ natural numbers.

In the case of a rectangle with dimensions $\phi\times 1$ this process gives

$$\phi=1^2+(\phi-1)^2+(2-\phi)^2+(2\phi-3)^2+\dots\,,$$

which, not surprisingly enough, can be written using Fibonacci numbers,

$$\phi=\sum_{n=0}^\infty \left(F_n \phi-F_{n+1}\right)^2\,.$$

In general, for any real $x>1$ we can do this trick. Let the continued fraction of $x$ be given by

$$x=[a_0;a_1,a_2,a_3,\dots]\,,$$

then we have that considering a rectangle with dimensions $1\times x$ gives a square decomposition

$$x=\sum_{n=0}^\infty a_0(b_n x-c_n)^2\,,$$

which is finite if and only if $x$ is rational. An interesting problem would be to study the properties of the sequences $\{b_n\}$ and $\{c_n\}$, but I'll leave that for another time.

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