About two 'negative' continued fractions whose sum equals $1$ Letting $a_1,a_2,\cdots,a_r$ be integers which are larger than or equal to $2$, let us define 
$$[a_1,a_2,\cdots,a_r]=\cfrac{1}{a_1-\cfrac{1}{a_2-\cfrac{1}{\ddots-\cfrac{1}{a_r}}}}$$
(Note that the negative signs are used)
Also, let $X, Y, Z$ be positive integers which satisfy
$$Z\lt X+Y,\  Z\gt X,\  Z\gt Y$$
and let 
$$\frac XZ=[a_1,a_2,\cdots,a_r],\ \frac YZ=[b_1,b_2,\cdots,b_s].$$
Then, here is my question.
Question : Is the following true?
"There exist $r^{\prime}\le r, s^{\prime}\le s$ such that 
$$[a_1,a_2,\cdots,a_{r^{\prime}}]+[b_1,b_2,\cdots,b_{s^{\prime}}]=1$$
for any $(X,Y,Z)$."
Remark : Observing the initial numbers is not sufficient because the nearer to $1$ the value $\frac XZ+\frac YZ$ is, the harder it is to find the answer (see example 2). 
This question has been asked previously on math.SE without receiving any answers. 
Examples :


*

*$\frac XZ=\frac 37=[3,2,2]$ and $\frac YZ=\frac 57=[2,2,3]$ leads $[3]+[2,2]=\frac 13+\frac 23=1$ where $\frac 37+\frac 57=\frac 87\approx 1.143$

*$\frac XZ=\frac{901}{2067}=[3,2,2,4,2]$ and $\frac YZ=\frac{1170}{2067}=[2,5,2,2,3]$ leads $[3,2,2,4]+[2,5,2,2]=\frac{10}{23}+\frac{13}{23}=1$ where $\frac XZ+\frac YZ=\frac{2071}{2067}\approx 1.002.$
Motivation : I've got an algorithm to find $b_1,b_2,\cdots,b_s$ such that 
$$1-x=[b_1,b_2,\cdots,b_s]$$
for any given $x=[a_1,a_2,\cdots,a_r]$.
Algorithm : Supposing that $2^r$ represents $r$-consecutive $2$s, I'm going to write 
$$[a_1,a_2,\cdots,a_r]=[2^{q_1},p_1,2^{q_2},p_2,\cdots,2^{q_s},p_s,2^{q_{s+1}}]$$
where $p_i\ge 3\in \mathbb N, q_i\ge 0\in \mathbb Z$. For example, $[2,2,5,3,2,4]=[2^2,5,2^0,3,2^1,4,2^0]$.
Then, the algorithm is 
$$1-[2^{q_1},p_1,2^{q_2},p_2,\cdots,2^{q_s},p_s,2^{q_{s+1}}]$$
$$=[(q_1+2),2^{(p_1-3)},(q_2+3),2^{(p_2-3)},(q_3+3),2^{(p_3-3)},\cdots,(q_s+3),2^{(p_s-3)},(q_{s+1}+2)].$$
After getting this algorithm, I reached the above expectation. I can neither find any counterexample even by using computer nor prove that the expectation is true. Can anyone help?
 A: The answer is YES even for numbers $\alpha$, $\beta$ of the form $\alpha=\frac X{Z_1}$, $\beta=\frac Y{Z_2}$. Suppose we look for convergents $\bar\alpha$, $\bar\beta$ to $\alpha$, $\beta$ such that $\bar\alpha+\bar\beta=1$.
If both numbers  $\alpha$, $\beta$ are not less when $1/2$ an answer is trivial:  $\bar\alpha=\bar\beta=1/2$. So we can assume that $\alpha<1/2$ and $\alpha+\beta>1$.
It means that $$\tag{1}\alpha=[a-\alpha']$$ for some $a\ge 3$ and $\alpha'\in[0,1)$. Then $\alpha<\frac1{a-1}$ and $\beta>1-\alpha>1-\frac1{a-1}$. From last inequality follows that continued fraction expansion (CF) for number $\beta$ has the form $$\beta=[\underbrace{2,\ldots,2-\delta}_{a-2}]=\frac{a-2-(a-3)\delta}{a-1-(a-2)\delta}\tag{2}$$ with $\delta\in[0,1)$. If we have one more digit $2$ in CF for $\beta$ then we have a solution, because
$$\bar\beta=[\underbrace{2,\ldots,2}_{a-1}]=\frac{a-1}{a}$$ and we can take $\bar\alpha=\frac 1a$.
If next digit in CF for $\beta$ is not equal to $2$, then $\delta=[b_1,b_2,\ldots]$, where $b_1\ge 3$. From (1) and (2) follows that inequality $\alpha+\beta>1$ equivalent to 
$$\alpha'+\beta'>1,$$ where $\alpha'$ is defined in (1) and $$\beta'=\frac{\delta}{1-\delta}=[b_1-1,b_2,\ldots].$$ The formula $\bar\alpha+\bar\beta=1$ is equivalent to
$\bar\alpha'+\bar\beta'=1$, so we reduced our problem from $\alpha$, $\beta$ to simpler numbers $\alpha'$, $\beta'$.
A: $\let\ds\displaystyle$Here is a different proof of the fact proven by Alexey Ustinov; it was found by Alexey Volostnov. 
Assume that $[a_1,\dots,a_r]=\frac ab$, $[b_1,\dots,b_s]=\frac cd$, $\frac ab+\frac cd>1$, and $d\leq b$. We will prove that if we delete $a_r$ the resulting fractions will sum up to at least one; the conclusion follows. Notice that the case $r=1$ is impossible: otherwise we would have $\frac ab=\frac1{a_r}$ and $\frac cd\leq 1-\frac 1d\leq 1-\frac 1b=1-\frac ab$. 
Let $\frac pq=[a_1,\dots,a_{r-1}]$, and assume that $\frac pq+\frac cd<1$. Then we have $\frac pq+\frac cd\leq 1-\frac 1{qd}$, $\frac ab+\frac cd\geq 1+\frac1{bd}$, and $aq-bp=1$. This implies 
$$
  \frac 1{bq}=\frac ab-\frac pq\geq \frac1{bd}+\frac1{qd},
$$
in particular, $bq<qd$ and $b<d$. A contradiction.
