By induction $ f(x)=\frac{x}{x+1}.$

Yes, your answer is correct. But the main problem is how to use induction, and it is not easy though.

We can think of a graph of $ \frac{n}{m} $ and $ m+n $ is costant. This is very crucial. ( Or, You can use $ Ord2 $ )

Rust wrote: By induction $ f(x)=\frac{x}{x+1}.$ Omid Hatami wrote: Yes, your answer is correct. But the main problem is how to use induction, and it is not easy though. No, his answer is wrong (not to mention useless "solution").

The answer is $$ f(\frac{n}{m})=\frac{n}{|m|+n}\text{ } (n>0)$$and $f(0)\in\{0,1\}$ This is a very nice problem

MNJ2357 wrote: The answer is $$ f(\frac{n}{m})=\frac{n}{|m|+n}\text{ } (n>0)$$and $f(0)\in\{0,1\}$ Well, after some fix of the problem statement, you're right. But if the proposer forgot to say that the domain of the first equation is $\mathbb Q \setminus \{0\}$, then pco or any other member has every right to say that the problem has no solution, like here. MNJ2357 wrote: This is a very nice problem I agree. So why no one posted a solution yet? There are at least 5 copies of this problem on the site, this particular topic is 11 years old. I guess some people think that knowing a buzzword like induction would won them maximal number of points at a real Olympiad automatically.

MNJ2357 wrote: The answer is $$ f(\frac{n}{m})=\frac{n}{|m|+n}\text{ } (n>0)$$and $f(0)\in\{0,1\}$ I thought the same thing for a while. But now it appeared that even this is wrong This answer is right for the case $f \colon \mathbb Q^+ \to \mathbb Q^+$ or at least $f \colon \mathbb Q_{\ge 0} \to \mathbb Q$. But for this particular variant of the problem, when $f \colon \mathbb Q \to \mathbb Q$, this answer is wrong! See the details at the revival attempt. The original statement presented at the 1991 Irish Olympiad contained $\mathbb Q^+$ as both domain and codomain.

For a solution