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If I am not wrong then while we are at mathematics the statements like "$A$ is greater than $B$" or "$A$ is lesser than $B$" are meaningless unless and until we have defined what exactly we mean by "greater" or "smaller". Hence we need to define order relations . An order relation $R$ is defined as a relation on a set $A$ which has the following properties:

  1. If $x,y$ belong to $A$ then either $xRy$ or $yRx$;
  2. For no $x$ belonging to $A$, $xRx$.
  3. $R$ is transitive.

Now let us consider the two positive real numbers $x$ and $y$. Now let $x$ be greater than $y$ , with the order relation being defined as $a>b$ if $a$ lies to the right of $b$ on the number line. So $\frac{x}{y}>1$. Now this means $\frac{-x}{-y}>1$. But as per the definition of order relation $-y$ lies to the left of $-x$. So will I be right in concluding that these two are completely different order relations.

At the same time, I have heard that the whole set of complex numbers is not an ordered field.(I say , I have heard cause I don't have anything to prove or disprove). Can anyone provide me a proof that we can never define a relation on $\Bbb C$ in such a way that it obeys the conditions satisfied and required by an order relation?

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    You can find help with writing mathematics [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference).2012-12-21

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Your axioms define a strict linear order. It is possible to define strict linear orders on $\Bbb C$. For example, we can set $a+bi if and only if $a, or $a=c$ and $b. What we cannot do is define a strict linear order on $\Bbb C$ that makes it an ordered field. That would require finding a linear order on $\Bbb C$ that also satisfied the following two conditions:

  1. for any $x,y,z\in\Bbb C$, if $x\le y$, then $x+z\le y+z$, and
  2. for any $x,y\in\Bbb C$, if $0\le x$ and $0\le y$, then $0\le xy$.

To see why this is impossible, look at what happens when we try to decide whether $0 or $i<0$. Suppose that $0. Then by (2) we must have $0, and by (2) again we must have $0<(-1)^2=1$. But since $0<-1$, (1) implies that $1=0+1<-1+1=0$, and we have a contradiction.

Now suppose that $i<0$. Then $0=i+(-i)<0+(-i)=-i$ by (1), so by (2) $0<(-i)^2=-1$, and by (2) again $0<(-1)^2=1$. But this is exactly the situation that gave us the contradiction before: add $1$ to both sides of $0<-1$, and you find that $1<0$.

Thus, we can’t have $0 or $i<0$ and therefore cannot have a linear order satisfying (1) and (2).

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    @danny: My pleasure!2012-12-21