If there exist any $c, d$ so that $\frac{a_{new}}{b_{new}}\rightarrow \frac{c}{d}$ but never reaches $\frac{c}{d}$, I think

Question

I have an algorithm to solve this problem * given two valid simplifying fractions $\dfrac{a}{b}$ and $\dfrac{c}{d},$ each transformation is adding both $a$ and $b$ by $+1,$ simplifying it, again and again until we meet $\dfrac{c}{d}.$ Count the number of steps used, $0$ if none" so that $\dfrac{a}{b}, \dfrac{c}{d}< 1$ and $0< a< b\leq 10^{5}, 0< c< d\leq 10^{5}.$ Easily, I see that the steps only continue if $ad< bc.$ Because we have that $\dfrac{a}{b}< \dfrac{a+ 1}{b+ 1}\overset{\div gcd(a, b)}{\rightarrow}\dfrac{a_{new}}{b_{new}}\rightarrow \dfrac{c}{d}.$ That sounds good.

But my test is not perfect, I wonder if there exist any $c, d$ so that $\dfrac{a_{new}}{b_{new}}\rightarrow \dfrac{c}{d}$ but never reaches $\dfrac{c}{d}.$ I need to your helps.

I can't imagine it. On the other hand, my algorithm is maybe infinite that way, I think I missed one or two conditions. — , Dec 20 '20 at 08:38
@ThomasSablik that's not the equation $\dfrac{a+ x}{b+ x}= \dfrac{c}{d}$ so that you needn't try by computer I still test by it.... but not any onservations. — , Dec 20 '20 at 08:39
Notice all elements of the sequence are of the form $(a+x)/(b+x)$ for $x\ge0$. — Kenta S, Dec 20 '20 at 08:40
@KentaS no, i.e., $\dfrac{1}{6}\rightarrow \dfrac{2}{7}\rightarrow \dfrac{3}{8}\rightarrow \dfrac{4}{9}\rightarrow \dfrac{5}{10}:=\dfrac{1}{2}\rightarrow \dfrac{2}{3}.$ — , Dec 20 '20 at 08:43
You just said what I was going to say; $2/3$ is of the desired form. This is enough to answer your question. — Kenta S, Dec 20 '20 at 08:45
What I meant is the above game, what you meant is the below equation, at all.. thank you. — , Dec 20 '20 at 08:47
$c/d = 1 - e$, for $c, d \in IN$. $\lim_{x -> \infty}\frac{a + x}{b+x} = 1$. That means that there exists one x so that $\frac{a}{b} >= \frac{c}{d}$ — Thomas Sablik, Dec 20 '20 at 08:56
There's a lot of info about Farey sequences in Allen Hatcher's Topology of Numbers, especially in the early chapters. See https://pi.math.cornell.edu/~hatcher/TN/TNpage.html — PM 2Ring, Dec 20 '20 at 09:13
Note that $b-a$ is non-increasing and decreases by division. So as soon as $d-c$ does not divide $b-a$ any more, you know that you cannot succeed — Hagen von Eitzen, Dec 20 '20 at 10:15

score 1 · Accepted Answer · answered Dec 20 '20 at 16:02

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Find $x$ such that $\frac{a+x}{b+x}=\frac cd$, i.e. $x=\frac{bc-ad}{d-c}$. If this is negative or not an integer, fail immediately. Otherwise, of all prime divisors $p$ of $b-a$, find the one minimising $y:=p-(b \bmod p)$. If $y\ge x$, the answer is $x$, otherwise it is $y$ plus the result for $\frac{a+y}{b+y}$ (simplified) abd$\frac cd$.

answered Dec 20 '20 at 16:02

Hagen von Eitzen

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Thks in advance – Dec 21 '20 at 03:49

If there exist any $c, d$ so that $\frac{a_{new}}{b_{new}}\rightarrow \frac{c}{d}$ but never reaches $\frac{c}{d}$, I think

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