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As a complex number set/field isn't an ordered set/field.

Now $1 \in \mathbb{C} $ & $2 \in \mathbb{C} $ .

How is $2>1$ ?

  • $\mathbb{C}$ can't be given an order making it a totally ordered field. But it surely can be made a totally ordered set. Think of the lexicographic order, for instance. – Julien May 22 '13 at 12:29

1 Answers1

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Any two numbers that are elements of an ordered set can be compared to one another. $2$ and $1$ are elements of the set of integers, so they are comparable. The fact that they are also members of the set of complex numbers is immmaterial.

rurouniwallace
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    Any analogy to this would be, the number $\frac{1}{2}$ can be represented as a ratio of two numbers, even though it's a real number. – rurouniwallace May 22 '13 at 12:21
  • Then why not state in axioms in complex number systems , that certain numbers are comparable . –  May 22 '13 at 12:29
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    @dkbose Doing that wouldn't be particularly useful. It is already stated in axioms of real numbers that all real numbers are comparable, and it is already stated in the definition of complex numbers that the real numbers are a subset of the complex number system. – rurouniwallace May 22 '13 at 12:33