# Inequalities of real and complex numbers

1. Apr 3, 2010

### Char. Limit

I was considering complex numbers (independently) and I came across an interesting question. Are any of these statements true?

$$1<i$$

$$1>i$$

$$i<-i$$

$$-1<i$$

$$-i<i<-1$$

2. Apr 3, 2010

### HallsofIvy

Staff Emeritus
The complex numbers are not a "ordered field". Neither "1< i" nor "i< 1" is true. In fact, you simply cannot write "a< b" for general complex numbers a and b.

3. Apr 3, 2010

### Gerenuk

Indeed you cannot order complex numbers.
I guess, the order relations need to obey some sensible rules. With complex numbers you would always get contradictions. You could try to define the order by the modulus of the complex number. But then the universal law
$$a<b\qquad\Rightarrow\qquad a+c<b+c$$
would not hold for all complex numbers. Therefore you wouldn't be able to use the most basic properties for rearranging equations with variables.

4. Apr 3, 2010

### HallsofIvy

Staff Emeritus
Specifically, in an ordered field, the order must obey
1) if a< b and c is any third member of the field then a+ c< b+ c.

2) if a< b and 0< c, then ac< bc

3) For any two a, b in the field one and only one must be true:
i) a= b
ii) a< b
iii) b< a.

Suppose the Complex number were an ordered field with order "<". Clearly i is not equal to 0 so by (iii) either i< 0 or 0< i.

If i< 0 then, adding -i to both sides, by (i) 0< -i. So by (ii) we must have i(-i)< o(-i) or $-i^2= 1< 0$. Since this is not necessarily the "usual" order on real numbers that is not yet a contradiction. However, that implies, as before, that 0< -1 and then, multiplying i< 0 by -1, we have -i< 0, contradicting 0< -i above.

Suppose 0< i. Then 0(i)< i(i)= -1. But then 0(-1)< i(-1) so that 0< -i and, adding i to both sides, i< 0, again a contradiction.