# 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 :

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

2. $\frac XZ=\frac{901}{2067}=[3,2,2,4,2]$ 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?

• What was the nature of your computer work? Did you do a systematic search of all fractions with denominator up to some $N$? – Gerry Myerson Oct 16 '13 at 22:44
• @GerryMyerson: Yes, but not all $N$ of course. – mathlove Oct 17 '13 at 8:23
• OK, so, how far up have you checked it? And did you see any patterns that might suggest that it's true in general? Also, are you assuming $\gcd(X,Z)=\gcd(Y,Z)=1$? – Gerry Myerson Oct 17 '13 at 12:04
• Just up to $N=100$. I noticed the following: If $[a_1,a_2,\cdots,a_r]+[b_1, b_2,\cdots,b_s]=1$, then $[a_1,a_2,\cdots,a_r,2]+[b_1,b_2,\cdots,b_s+1]=1$ though this may not be a reply to your question. Also, I'm not assuming that. – mathlove Oct 17 '13 at 14:50
• Thanks. The kind of pattern I had in mind is this: we may assume $X\lt Y$. If $X\ge Z/2$, then $1/2$ shows up as a convergent for both fractions, so we may assume $X\lt Z/2\lt Y$. Now, for certain ranges of $X$ and $Y$, $1/3$ will show up as a convergent for $X/Z$, and $2/3$ for $Y/Z$, so we can eliminate those ranges as possible sources of counterexamples. Then we can eliminate the ranges that lead to $1/4$ and $3/4$. And so on. Then we can look at what's left, and see which ranges of $X/Z$ and $Y/Z$ lead to which convergents summing to 1. And so on. – Gerry Myerson Oct 17 '13 at 22:04

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 $\alpha'$, $\beta'$.

• Are where any reasons beyond this question? – Alexey Ustinov Oct 19 '13 at 9:00
• Sorry, but I can't get what you mean. – mathlove Oct 21 '13 at 14:39
• Has this question arised as an auxiliary fact for something more serious? – Alexey Ustinov Oct 22 '13 at 1:41
• No. I found the property of this question just because I've been interested in continued fractions. – mathlove Oct 22 '13 at 14:28

$\let\ds\displaystyle$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.

• Very nice solution – Alexey Ustinov Nov 11 '13 at 2:49