Consider rational number $(Q,+,\times,>)$, its multiplication, addition, and order are usual. Can we find a order field $(Q,+,\times,>')$(addition and multiplication are usual) such that it is not order isomorphic to $(Q,+,\times,>)$, if we replace $Q$ by $R$, what conclusion will we have?
Remark: When I real Rudin's Principle of mathematical analysis, page 7 order field, I found he only give definitions and some theorem but lack examples, if we can find some examples above, I think it will be funny.
The order on the rationals is determined by the addition/multiplication, since for any rationals $a$ and $b$, we have $a\ge b$ if and only if $a$ is equal to $b$ plus a sum of four squares (of rationals). This follows from Lagrange's Theorem that every non-negative integer is the sum of four squares. (There are other ways to show that the order on the rationals is determined by the addition and multiplication.)
The same is true for the reals, since $a\ge b$ if and only if there exists an $x$ such that $a=b+x^2$.