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I was wondering which is the effect of changing additions by substractions on a simple continued fraction, and its mathematical interpretation (if any). For instance,

$$\sqrt{2}=[1;2,2,2,2,\dots]=1+ \cfrac{1}{ 2+\cfrac{1}{ 2+\cfrac{1}{ 2+\cfrac{1}{ 2+\ddots } } } }$$

And if we change additions for substractions,

$$x=1- \cfrac{1}{ 2-\cfrac{1}{ 2-\cfrac{1}{ 2-\cfrac{1}{ 2-\ddots } } } }$$

If we take the partial continued fractions up to $a_k$, we get we get exactly $x=\frac{1}{k}$, and thus the value of this continued fraction would be $0$.

Every simple infinite continued fraction with positive signs is an irrational number, but it seems that the simple infinite continued fractions with negative signs are always rational. Is this true? I would like to know a proof of it in this case, or some counterexample.

Any light on this topic (mathematical interpretation of the change described, and rationality of infinite continued fractions with negative signs) would be welcomed. Thanks in advance!

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    $\begingroup$ Note that your negative continued fraction is also $[1; -2, 2, -2, 2, -2, ...]$ $\endgroup$ – Paul Sinclair Jan 26 at 19:07
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Regarding the question of rationality of simple negative continued fractions, I have found a counterexample: I was able to represent $\phi$ as a simple negative continued fraction as follows:

$$\phi=2- \cfrac{1}{ 3-\cfrac{1}{ 3-\cfrac{1}{ 3-\cfrac{1}{ 3-\ddots } } } }$$

As $\phi$ is irrational, the question is solved.

Additionally, I found this interesting research that I share with you in case you are interested in the relationship between the negative and positive continued fractions:

https://scholarship.claremont.edu/cgi/viewcontent.cgi?filename=2&article=1183&context=hmc_theses&type=additional

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