Problem: Show that the real field $\mathbb{R}$ has a unique ordering and indicates that ordering.
My question: We knew that $\le$ is an ordering on $\mathbb{R}$. Do we need to prove that $\le$ is an ordering on $\mathbb{R}$? How to show that the uniqueness? Thank all!
EDIT 1: Suppose $\le$ be an ordering on $\mathbb{R}$, so it satisfies $\le$ is a total order relation on $\mathbb{R}$, $\forall z \in \mathbb{R}, x \le y \Rightarrow x + z \le y + z $, $0 \le x, 0 \le y \Rightarrow 0 \le xy$. Suppose $<$ be another ordering on $\mathbb{R}$ then $x<y$ defined by $x \le y$ and $x \ne y$. Hence there is a unique ordering on $\mathbb{R}$.
Best Answer
Denote by $\leq$ the usual ordering on $\mathbb{R}$ and suppose $\preceq$ is some ordering on $\mathbb{R}$. So $(\mathbb{R},\preceq)$ is a linearly ordered field. Suppose that $x \neq 0$ is a real number. Then we can prove that $0 \preceq x^2$ and since $x\neq 0$, you deduce that $0\prec x^2$ and also $-x^2\prec 0$. As a consequence we derive that $$[0,+\infty) = \{x\in \mathbb{R}\,|\,0\preceq x\}$$ Thus orderings $\leq$ and $\preceq$ have the same subset of nonnegative elements. This implies that they are the same.