image of positive real number under field isomorphism

Question

Claim: If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a field isomorphism, then $f(x) >0$ for any $x>0$.

I know $f(x)$ can be determined by $f(1)$ the generator of $\mathbb{R}$ but not sure how to work in inequalities.

Answer 1

$\Bbb{R}$ admits no non-trivial field automorphisms, so your $f$ has to be the identity function. To see this: any automorphism of $\Bbb{R}$ must fix the rational numbers $\Bbb{Q}$. Also, the order on $\Bbb{R}$ is determined by its algebraic structure, because $x > y$ iff $\exists z(x - y = z^2)$. So any automorphism of $\Bbb{R}$ must fix $\Bbb{Q}$ and be order-preserving. As any element of $\Bbb{R}$ is the limit of a bounded ascending sequence of rationals and the limits of such sequences are unique, an automorphism has no choice but to fix every element of $\Bbb{R}$.