Does this map send every algebraic number to a periodic 2-adic?

Question

I worked out recently that every algebraic number $x$ greater than or equal to one has a corresponding 2-adic number. To find it, simply create a $1:x$ rectangle. Then repeatedly remove a square from the shortest side ad infinitum.

Write a $1$ for the first square, and each time you remove a square from the same side, mark the same digit as last time i.e. a $1$ if the last digit was a $1$. Each time you remove a square from the other side, mark a zero. The binary number which you write is the corresponding 2-adic.

This has some potentially interesting consequences for polynomials. For example, I'm pretty sure any number whose 2-adic is periodic of order $n$ should be the solution to a polynomial of degree no more than $n+1$.

Additionally, the $n-$periodic 2-adics enumerate a set of radicals or roots which can be added to $\Bbb Q$ to form the basis for all solutions to polynomials, and they number no more than the number of base 2 Lyndon words of length less than $n+1$.

All very interesting, but I assume these are not new results.

Question

Does some algebraic real number greater than $1$ exist, whose corresponding binary string or $2$-adic number found by this method is never periodic?

Edit

I just thought I'd add this in case it helps somebody answer. Each 2-adic number, through the process of truncating rectangles, is essentially converted into a continued fraction representation of the number in question. I know continued fractions have a lot to do with the definition of being algebraic and rational.

Answer 1

ok I managed to work out for myself that the eventually periodic representations actually correspond with the eventually periodic simple continued fraction representations with integer coefficients.

Therefore by a theorem of Lagrange (and its converse by Euler) the n-periodic 2-adics, and the set of rectangle truncation orbits, enumerate all the real quadratic roots.

Tantalisingly, the fact we are essentially generating a set of rational affine transformations which represents the space of rectangle shapes hints at the suggestion that truncating $p-$hyperrectangles would generate the set of all roots of polynomials of degree $p$