Among any two integers or real numbers one is larger, another smaller. But you can't compare two complex numbers.
We must be cautious with this kind of general statements. It's actually possible to define the relationship
"<" between complex numbers. Let's agree that
(a + ib) < (c + id),|
provided either a < c or a = c and b < d.
This is what is called the lexicographic order.
This is how words are ordered in dictionaries. This relation, exactly like in the case of integers or real numbers, exhibits the following properties:
|No reflexivity||It's not true that x < x|
|No symmetry||x < y does not imply y < x|
|Transitivity||If x < y and y < z then x < z|
Also, since a real number x can be identified with the complex number x + i0, we immediately see that thus defined order conforms to the usual order of real numbers. Additionally, for real number, the relation "<" is related to the arithmetic operations "+" and "×". For example, if r < s and t < u then (r + t) < (s + u), where r, s, t, u are real numbers. The same is true for complex numbers as well. For example, (2 + 3i) < (3 + 4i) and (-1 + i) < (2 - 6i) imply
(1 + 4i) < (5 - 2i) which is of course true.
Still it's not very useful. To see what the problem is let's turn to the multiplication. The following is true for real numbers:
If a < b and 0 < c then ac < bc
Note that i may be written as 0 + i1 while 0 = 0 + i0. Therefore, by definition, 0 < i. Assume (1) holds for complex numbers and the lexicographic order. Obviously -1 < 1. Multiplying by i we get -i < i. Multiplying by i one more time produces -i2 < i2. By definition, this implies 1 < (-1) which is absurd.
The derivation is quite formal and does not use any particular properties of the lexicographic order. The crucial thing was that i satisfied the inequality 0 < i. Had we had i < 0 we would have gotten a contradiction in precisely the same way now applying 0 < (-i). The real problem stems from the sheer fact that i is somehow comparable to 0.
The law by which for any two numbers a and b, either a < b or b < a or a = b, is known as Trichotomy. Trichotomy holds for the usual order of real numbers. But, as we just saw, the order between complex numbers for which (1) holds, could not satisfy the law of trichotomy and vice versa. If the order between complex numbers is complete (i.e. satisfies the Trichotomy law) then (1) could not possibly hold.
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Copyright © 1996-2018 Alexander Bogomolny