Let $a=[a_1,a_2\dots]$ and $b=[b_1,b_2\dots]$ be two real numbers and their continued fraction representations. They may be infinite or finite.
Let us define a thing $+^*$ so that $a+^*b=[a_1,a_2\dots]+^*[b_1,b_2\dots]=[a_1+b_1,a_2+b_2\dots]$. If either $a_i$ or $b_i$ is undefined, ie. one has shorter representation than the other, we assume the undefined one to be $0$.
For example $$1/3 +^* 1/5 = [0,3+5]=1/8.$$ Other one: $$1/3 +^* 5/3 = [0,3]+^* [1,1,2] =[1,4,2]=11/9. $$
I am sure there are lots of problems with such a definition (like it is not well-defined, as we can split the last number of the expansion into sum of it minus one and one), but I am interested if this is true (regardless of the representation): $$\text{If } a+b \text{ is rational , then }a+^*b \text{ is also rational.}$$ Or in other words: $$\text{If } a+^*b \text{ is not rational , then }a+b \text{ is also not rational.}$$
This is not true with infinite summation, as $\sum_{n=1}^{\infty} 1/n^2 = \pi^2/6$ but the $+^*$ yields $\lim_{k \rightarrow\infty} [0,\frac{1}{6} k (1 + k) (1 + 2 k)] = [0,\infty]=0,$ so assume that we only sum two numbers.
Well, no.
Let $\phi = \frac{1 + \sqrt 5}{2}.$ The continued fraction for $\phi$ is $$[1;1,1,1,1,1,1,\ldots] $$
Then take $3 - \phi = \frac{5 - \sqrt 5}{2}.$ The continued fraction is$$[1;2,1,1,1,1,1,\ldots] $$
So, while the sum of the numbers is $3,$ the continued fraction your operation gives is $$[2;3,2,2,2,2,2,\ldots] $$
In case it matters, this one is the simple continued fraction for $$ 3 - \frac{1}{\sqrt 2}. $$