Complex numbers ordering: Is there a consistent order for complex numbers?

[ashraf.yusoff]

i have 1 question..
the question is: Given any 2 distinct real numbers a and b, exactly one a<b or b<a must be true. The real numbers are said to be ordered. Show that there is no ordering of the complex numbers.

my problems is not understand that orders~~anybody help me?

 

[planck42]

Well, complex numbers are defined in a plane, while real numbers are defined along a line. See how there could be a problem in ordering complex numbers?

 

[HallsofIvy]

An "ordered field" is a field (so we have the usual rules for addition and multiplication) together with an order, x< y, such that:
1) if a< b, then a+ c< b+ c
2) if a< b and 0< c, then ac< bc
3) For any a and b one and only of
i) a= b
ii) a< b
iii) b< a
is true.

Clearly $i\ne 0$ so by (3) we must have either i> 0 or i< 0.

If i> 0 then, multiplying both sides by i, by (2), i(i)> i(0) or -1> 0. That is not, by itself a contradiction since this is not necessarily the "regular" order on the real numbers. But since -1> 0, multiplying both sides of i> 0 by -1, by (2) again, (-1)(i)> (-1)(0) so that -i> 0. Now add i to both sides- by (1), -i+ i> 0+ i or 0> i which contradicts i> 0.

If i< 0, then, adding -i to both sides, by (1), i- i< 0- i or 0< -i. Multiplying both sides of i< 0 by -i, by (2), (-i)(i)< (-i)(0) or 1< 0. Adding -1 to both sides, by (1), 0< -1. Again that is not itself a contradiction but multiplying both sides of i< 0 by -1, by (2), i(-1)< 0(-1) or -i< 0. Adding i to both sides, -i+ i< 0+ i so 0< i which does contradict i< 0.

 

[planck42]

Multiplying an inequality by a negative number inverts the direction of the inequality

 

[HallsofIvy]

Multiplying an inequality by a negative number inverts the direction of the inequality

In the usual order relation on the real numbers, yes, but that has nothing to do with the problem here.