If $\mathfrak{A}=\langle A,<,+,\cdot,0,1 \rangle,\mathfrak{B}=\langle B,<,+,\cdot,0',1' \rangle$ are ordered fields and $f:A \to B$ is an isomorphism between $\mathfrak{A}$ and $\mathfrak{B}$, then $f(0)=0'$ and $f(1)=1'$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$a+0=a \implies f(a+0)=f(a) \implies f(a)+f(0)=f(a)\implies f(a)+f(0)=f(a)+0'$. Cancel out $f(a)$ from both sides, we get $f(0)=0'$.
Take $a\in A$ such that $a>0$. Then $f(a)>0'$. We have $a\cdot 1=a \implies f(a\cdot 1)=f(a) \implies f(a)\cdot f(1)=f(a)\implies f(a)\cdot f(1)=f(a)\cdot 1'$. Since $f(a)>0'$, we cancel out $f(a)$ from both sides and get $f(1)=1'$.