Solution

Answer: The desired set of rational numbers is \(\left\{\frac{n+1}{n}: n \in \mathbb{Z}, n \neq 0\right\}\).

Let \(Z\) be the set of all rational numbers \(q\) such that for every function \(f \in \mathcal{F}\), there exists some \(z \in \mathbb{R}\) satisfying \(f(z)=q z\). Let further

\[S=\left\{\frac{n+1}{n}: n \in \mathbb{Z}, n \neq 0\right\}\]

We prove that \(Z=S\) by showing the two inclusions: \(S \subseteq Z\) and \(Z \subseteq S\). We first prove that \(S \subseteq Z\). Let \(f \in \mathcal{F}\) and let \(P(x, y)\) be the relation \(f(x+f(y))=f(x)+\) \(f(y)\). First note that \(P(0,0)\) gives \(f(f(0))=2 f(0)\). Then, \(P(0, f(0))\) gives \(f(2 f(0))=3 f(0)\). We claim that

\[f(k f(0))=(k+1) f(0)\]

for every integer \(k \geqslant 1\). The claim can be proved by induction. The cases \(k=1\) and \(k=2\) have already been established. Assume that \(f(k f(0))=(k+1) f(0)\) and consider \(P(0, k f(0))\) which gives

\[f((k+1) f(0))=f(0)+f(k f(0))=(k+2) f(0)\]

This proves the claim. We conclude that \(\frac{k+1}{k} \in Z\) for every integer \(k \geqslant 1\). Note that \(P(-f(0), 0)\) gives \(f(-f(0))=0\). We now claim that

\[f(-k f(0))=(-k+1) f(0)\]

for every integer \(k \geqslant 1\). The proof by induction is similar to the one above. We conclude that \(\frac{-k+1}{-k} \in Z\) for every integer \(k \geqslant 1\). This shows that \(S \subseteq Z\). We now prove that \(Z \subseteq S\). Let \(p\) be a rational number outside the set \(S\). We want to prove that \(p\) does not belong to \(Z\). To that end, we construct a function \(f \in \mathcal{F}\) such that \(f(z) \neq p z\) for every \(z \in \mathbb{R}\). The strategy is to first construct a function

\[g:[0,1) \rightarrow \mathbb{Z}\]

and then define \(f\) as \(f(x)=g({x})+\lfloor x\rfloor\). This function \(f\) belongs to \(\mathcal{F}\). Indeed,

\[\begin{aligned} f(x+f(y)) & =g(\{x+f(y)\})+\lfloor x+f(y)\rfloor \\ & =g(\{x+g(\{y\})+\lfloor y\rfloor\})+\lfloor x+g(\{y\})+\lfloor y\rfloor\rfloor \\ & =g(\{x\})+\lfloor x\rfloor+g(\{y\})+\lfloor y\rfloor \\ & =f(x)+f(y) \end{aligned}\]

where we used that \(g\) only takes integer values. **Lemma 1:** For every \(\alpha \in[0,1)\), there exists \(m \in \mathbb{Z}\) such that

\[m+n \neq p(\alpha+n)\]

for every \(n \in \mathbb{Z}\). *Proof:* Note that if \(p=1\) the claim is trivial. If \(p \neq 1\), then the claim is equivalent to the existence of an integer \(m\) such that

\[\frac{m-p \alpha}{p-1}\]

is never an integer. Assume the contrary. That would mean that both

\[\frac{m-p \alpha}{p-1} \text { and } \frac{(m+1)-p \alpha}{p-1}\]

are integers, and so is their difference. The latter is equal to

\[\frac{1}{p-1}\]

Since we assumed \(p \notin S, 1 /(p-1)\) is never an integer. This is a contradiction. Define \(g:[0,1) \rightarrow \mathbb{Z}\) by \(g(\alpha)=m\) for any integer \(m\) that satisfies the conclusion of Lemma 1. Note that \(f(z) \neq p z\) if and and only if

\[g(\{z\})+\lfloor z\rfloor \neq p(\{z\}+\lfloor z\rfloor)\]

The latter is guaranteed by the construction of the function \(g\). We conclude that \(p \notin Z\) as desired. This shows that \(Z \subset S\).