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A rectangle with side lengths proportional to $\varphi$ is a "golden rectangle". If we perform a procedure of subtracting the shorter side length from the longer side length, a golden rectangle keeps the same proportions.

We can represent this procedure mathematically. If the old rectangle's sides are of length $A$ and $B$, with $A$ shorter, then the new rectangle will have side lengths $A$ and $B-A$. Hence, we have a golden rectangle when $\frac A B = \frac{B-A}B$, or $A^2-B^2+AB = 0$, and we define the golden ratio $\varphi$ as $\frac B A$ in this case. $A$ and $B$ clearly cannot themselves be rational numbers. They must, I'm guessing, likewise be transcendental numbers.

The second key point is that if you have a rectangle with side lengths that are rational numbers, repeating the procedure of subtracting the shorter side length always leads to generating a square after a finite number of steps. Repeating this procedure on the square gives the ratio $\frac 0 0$, which is undefined.

So we have two interesting properties of the golden rectangle. More generally, it does not converge on a square when we iterate the procedure of subtracting the shortest side length from the longer side length. More specifically, it keeps the same proportions as we do this.

It seems to me that if a rectangle can have proportions of $\varphi$, then it can have proportions of other transcendental numbers as well. And it also seems intuitively like iterating this procedure would also not converge on a square.

What interests me is what it would converge on. Would it approximate a square? Jump around indefinitely with any range of proportions?

I don't quite know how to think about this, so I thought I'd ask Stack Exchange and see if anybody can point me to some work in this area. I'm sure there must be something out there...

Thanks for any suggestions!

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    $\begingroup$ $\varphi$ is not transcendental, it's algebraic. I guess you meant irrational. $\endgroup$
    – jjagmath
    Sep 5 '21 at 1:41
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    $\begingroup$ Also, from $A^2-B^2+AB=0$ you can't conclude that both are irrationals. It can happen that one of them use rational and the other irrational, or both irrational, or even both transcendental. $\endgroup$
    – jjagmath
    Sep 5 '21 at 1:45
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    $\begingroup$ I suggest you read about irrational numbers and transcendental numbers. $\endgroup$
    – jjagmath
    Sep 5 '21 at 1:52
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    $\begingroup$ And the geometric process you describe is directly related to continued fractions $\endgroup$
    – jjagmath
    Sep 5 '21 at 1:53
  • $\begingroup$ Thank you for all your comments! $\endgroup$ Sep 5 '21 at 2:17
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First of all, φ is not actually a transcendental number, but it is an algebraic irrational one. This is because φ can be expressed as the solution to an algebraic equation (specifically the one derived from the cutting rectangles), $x^2 - x - 1$. Assuming you were referring to irrational numbers, there is no other ratio that doesn't result in a square after truncation. This is because the golden ratio is the ONLY ratio of the side lengths that results in self-similarity, as it is the only positive solution to the equation above. It is easy to prove that the ratio cannot be rational; the Euclidean algorithm would guarantee that the sides would be the same length at some point. Regarding irrationals, the simplest argument would be that a lack of self similarity from one step to the next would imply that an infinite number of rectangles would be created, which would mean that at some point, the rectangle would be square. jjagmath's comments on the original post would be helpful for you to revise.

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  • $\begingroup$ Thanks I caught that after posting - you must have replied in a hurry :) $\endgroup$ Sep 5 '21 at 2:10
  • $\begingroup$ "Assuming you were referring to irrational numbers, there is no other ratio that doesn't result in a square after truncation." I'm actually not sure this is correct. I tried repeating the procedure with rectangles that have side lengths equal to square roots of primes, and instead of converging to a square, the proportions cycle. I'll give an example of a cycle for a rectangle with side lengths sqrt(2) and sqrt(5) in the next comment. $\endgroup$ Sep 5 '21 at 2:15
  • $\begingroup$ sqrt(5)/sqrt(2) -> sqrt(2)/(sqrt(5)-sqrt(2)) -> (sqrt(5) - sqrt(2))/(2sqrt(2)-sqrt(5)) -> (2sqrt(2)-sqrt(5))/(2sqrt(5)-3sqrt(3) -> sqrt(5)/sqrt(2) $\endgroup$ Sep 5 '21 at 2:16
  • $\begingroup$ "there is no other ratio that doesn't result in a square after truncation" : I disagree with your assertion. Assume that $\frac{A}{A+B} ~: ~A > B$ is irrational. Then, $\frac{A+B}{A}$ is irrational, and therefore, so is $\frac{B}{A} = \frac{A+B}{A} - 1.$ $\endgroup$ Sep 5 '21 at 2:26
  • $\begingroup$ Hmm... let me double check my work $\endgroup$ Sep 5 '21 at 2:42

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