A complete ordered field has a unique ordering.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $\langle A,+,\cdot,0,1 \rangle$ be a field and $<_1,<_2$ be a linear orders on $A$ such that $\langle A,<_1,+,\cdot,0,1 \rangle$, $\langle A,<_2,+,\cdot,0,1 \rangle$ are complete ordered fields.
Assume that $a \le_1 b$ and $b \le_ 2 a$. Then $0 \le_1 b-a$ and thus there exists a unique $0 \le_1 x$ such that $x^2=b-a$. Let $y=a-b$, then $0 \le_2 y$. We have $x^2+y=0$ and $0 \le_2 y$ $\implies x^2 \le_2 0$ $\implies x^2=0 \implies a=b$. Hence $\le_1 \space =\space \le_2$.