In: Advanced Math
Logic & Sets (Proofs question)
Show that complex numbers cannot be ordered in a way that satisfies our axioms.
Axioms for order:
1. if x is less than/equal to y and w is greater than zero, then wx is less than/equal to wy
2. for w, x, y, z w is less than/equal to x, y is less than/equal to z then w + y = x + z if and only iff w = x and y = z
We will prove by contradiction that the set of all complex numbers cannot be ordered in a way that satisfies given two axioms .
Suppose that the set of all complex numbers can be ordered in a way that satisfies given two axioms .
As , and so either or .
Case 1 : If
, as and by property 1 .
, a contradiction to .
Case 1 : If
as and by property 1 .
, a contradiction to .
So we obtained a contradiction in each case .
Hence the set of all complex numbers cannot be ordered in a way that satisfies given two axioms .
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