Let $\mathfrak{Q}=\langle \Bbb Q,<,+,\cdot,0,1 \rangle,\mathfrak{A}=\langle A,<,+,\cdot,0',1' \rangle$ be ordered fields where $\Bbb Q$ is the set of rationals, and $f,g$ be isomorphisms from $\mathfrak{Q}$ to subfields of $\mathfrak{A}$. Then $f=g$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
It suffices to prove that $f(p)=g(p)$ for all $p\in\Bbb Q$ and $p>0$.
It is easy to prove that $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$.
For $p\in\Bbb Q$ and $p>0$, $p=\dfrac{m}{n}$ for some $m,n\in\Bbb N$.
$f(p)=f\left(\dfrac{m}{n}\right)=\dfrac{f(m)}{f(n)}=\dfrac{f(\underbrace{1+\cdots+1}_{m\text{ times}})}{f(\underbrace{1+\cdots+1}_{n\text{ times}})}=\dfrac{\underbrace{f(1)+\cdots+f(1)}_{m\text{ times}}}{\underbrace{f(1)+\cdots+f(1)}_{n\text{ times}}}=\dfrac{\underbrace{1'+\cdots+1'}_{m\text{ times}}}{\underbrace{1'+\cdots+1'}_{n\text{ times}}}$
$g(p)=g\left(\dfrac{m}{n}\right)=\dfrac{g(m)}{g(n)}=\dfrac{g(\underbrace{1+\cdots+1}_{m\text{ times}})}{g(\underbrace{1+\cdots+1}_{n\text{ times}})}=\dfrac{\underbrace{g(1)+\cdots+g(1)}_{m\text{ times}}}{\underbrace{g(1)+\cdots+g(1)}_{n\text{ times}}}=\dfrac{\underbrace{1'+\cdots+1'}_{m\text{ times}}}{\underbrace{1'+\cdots+1'}_{n\text{ times}}}$
It follows that $f(p)=g(p)$ and thus $f=g$.