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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 . A 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 on the number line. so X/Y>1. now this means (-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 C in such a way that it obeys the conditions satisfied and required by an order relation?

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(1) Please use LaTeX to write mathematics here. You can go the FAQ section for directions. (2) After your number (3) you seem to have made a complete mess between x,y,X,Y. Please try to correct this. – DonAntonio Dec 21 '12 at 10:23
You can find help with writing mathematics here. – Brian M. Scott Dec 21 '12 at 10:31
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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 $a+bi<c+di$ if and only if $a<c$, or $a=c$ and $b<d$. 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 $0<i$ or $i<0$. Suppose that $0<i$. Then by (2) we must have $0<i^2=-1$, 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 $0<i$ or $i<0$ and therefore cannot have a linear order satisfying (1) and (2).

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M scott ; Thank you sir! – danny gotze Dec 21 '12 at 10:49
@danny: My pleasure! – Brian M. Scott Dec 21 '12 at 10:50

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