Why is $-1>0$ not enough?

Question

Theorem: Prove that no order can be defined in the complex field that turns it into an ordered field.

Proof: Suppose complex field is an ordered field. So, either $i$ or $-i$ must be positive. Suppose $i>0$. Hence $i^2=-1>0$, but then $(-1)^2=1>0$. But this is contradiction as both $x$ and $-x$ cannot be true in an ordered field, where $x$ belongs to that ordered field.

My question: Why was $-1>0$ not itself a contradiction an end of the proof?

Answer 1

our argument would work if you already know there is only one ordered field over the real numbers! Many times textbooks forgo "understanding" proofs for more straightforward proofs. This is because they wish to help you in proof writing (something definitely teachable) over understanding (which is not as teachable or measurable).

*there being one field over the reals definitionally but not as obviously one ordering.

Answer 2

The fact that $-1\not>0$ may seem like something trivial that you learned when you were nine years old, but here we're dealing not with the number $-1$ but with the object that is the additive inverse of the multiplicative identity element of the field we're working with. That field need not be the field of real numbers, but could be any ordered field. An ordered field has an underlying set, and two binary operations called addition and multiplication, which need not be the usual addition and multiplication of real numbers, and it satisfies the axioms of ordered fields. How can one prove that $-1\not>0$ when $-1$ is not necessarily the number $-1$ but is the additive inverse of the multiplicative identity element of the field we're working with, and $0$ is not necessarily the number $0$, but is the additive identity element in the field we're working with? The point is that we must show that $-1>0$ conflicts with something in the axioms of ordered fields.

Answer 3

Let $0$ be the additive identity of a field, let $1$ be the multiplicative identity of the field, and let $-1$ be the additive inverse of $1.$

Proof that $\neg (-1>0)$ in an ordered field:

In any field we have $$(I).... \quad (-1)^2=1$$ because $$1=1+0=1+(-1)0=1+(-1)[1+(-1)]=1+[(-1)1+(-1)^2]= $$ $$= [1+(-1)]+(-1)^2=0+(-1)^2=(-1)^2.$$

In an ordered field we have $$(II).... \quad (a>b \land c>0) \implies ac>bc.$$ Applying (II) with $b=0$ and $c=a$ we have $$(III)....\quad a>0\implies a^2>0.$$ Applying (I) and (III) with $a=-1$ we have $$(IV)....\quad -1>0\implies 1=(-1)^2>0.$$ In an ordered field we also have $$(V)....\quad a>b\implies a+c>b+c .$$Applying (V) with $a=-1,b=0,c=1$ we have $$(VI)....\quad -1>0\implies 0= (-1)+1>0+1=1.$$ So by $(IV)$ and $(VI)$, we have a paradox if $-1>0$.