Sorry if that's not the right place for asking this, but didn't have anywhere else to go. I was cheking out some math problems in the Mathematical Olympiad site, and I found this one:
Let $\mathbb Q^+$ be the set of all positive rational numbers. Let $f:\mathbb Q^+ \mapsto R$ be a function satisfying the following three conditions:
- for all $x,y \in \mathbb Q^+$, we have $f(x)\cdot f(y)\geq f(x \cdot y)$
- for all $x,y \in \mathbb Q^+$, we have $f(x+y)\geq f(x)+f(y)$
- $\exists a \in \mathbb Q^+$ such that $a>1$ and $f(a)=a$.
Prove that $f$ is the identity function, i.e., that $\forall x \in \mathbb Q^+: f(x) = x$.
The solution I propose is the following: Consider $L$ some formal language with addition and multiplication. We split the axioms above in the following: $\Phi _{i} \cup \Phi_{ii} \cup \Phi_{iii} = \Phi$. So, what we have to do is just prove that $(f)(\vdash (f \models \Phi) .\supset. f=id)=\Psi$ is true, i.e, that $\vdash \Psi .\equiv. \sim$ Provable$(\sim \Psi)$. But, how exactly we do that? I tried to solve in the following way: suppose $f$ satisfies the axioms, but with $\sim (\exists a)(a \epsilon D(f) .\supset. f(a)=a) $, so, $f$ is not the identity function; then $f$ fails to satisfy $\Phi_{iii}=(a)(a\epsilon D(f) . a=1 :\supset. f(a)=a)$, because $f$ can't have the same image for the same argument. Does that is enough to demonstrate? I put "model theory" in the title because I wanted to ask, also, if there's a pratical way to determine wheter a set of axioms has a unique model (that would be helpfull in solving the problem above). Thanks for your time.