Problem 4 Show That The Field Of Complex Numbers Cannot Be Ordered That Is Show That There Does Not Exist A Relation 1 (52.68 KiB) Viewed 27 times
Problem 4. Show that the field of complex numbers cannot be ordered. That is, show that there does not exist a relation ≤ on C which gives C the structure of an ordered field. (Hint: If i > 0, then what must be true of i²? Argue similarly if i < 0.)
Let us say that a field F is an ordered field if there is a relation ≤ on F which satisfies the following properties for each a, b, c ¤ F: (i) (Reflexive) a ≤ a (ii) (Transitive) a ≤ b and b ≤ c implies a ≤ c (iii) (Antisymmetric) a ≤ b and b ≤ a implies a = b (iv) (Strongly connected) a ≤ b or b ≤ a (v) If a ≤ b, then a+c<b+c. (vi) If 0 < a and 0 ≤ b, then 0 < ab. We say that a is strictly less than b, denoted a < b, if a ≤ b and a ‡ b.
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