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 $\mathbb{R}$.

  • $\begingroup$ This looks like fairly hard work. Presumably you require that the ordering satisfies certain conditions such as (1) $0<1$ and (2) $x<y$ implies $x+z<y+z$, and (3) $x<y$ and $0<a\le b$ implies $ax<by$. Then you show this must be the same as the usual ordering on $\mathbb{Z}$, then on $\mathbb{Q}$, and finally on $\mathbb{R}$. $\endgroup$
    – almagest
    Jan 11, 2020 at 14:17
  • $\begingroup$ For the requirement indicates the ordering, it is $\le$ and I do not realize how to indicate? $\endgroup$
    – Minh
    Jan 11, 2020 at 14:19
  • 1
    $\begingroup$ $\le$ and $<$ are closely related: if $\le$ is defined, then $a<b$ is just shorthand for $a\le b$ and $a\ne b$. $\endgroup$
    – almagest
    Jan 11, 2020 at 14:21
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    $\begingroup$ The real numbers has this amazing property: If $a \in \mathbb R$, the either there is $b$ with $a=b^2$ or there is $b$ with $a=-b^2$. $\endgroup$
    – GEdgar
    Jan 11, 2020 at 14:36

1 Answer 1


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.


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