Ordered field proof

Question

Prove that in any ordered field F, $a^2+1>0$ for all $a\in \!\,F$.

There are 2 cases,

Suppose $x\ne 0$. Then $x^2>0$.

Suppose $x=0$. Then $x^2=0$.

In both cases, $x^2+1>x^2\ge\ 0$

Does my proof look ok? Is it missing anything?

Conclude from this that if the equation $x^2+1=0$ has a solution in a field, then that field cannot be ordered. (Thus it is not possible to define an order relation on the set of all complex numbers that will make it an ordered field).

Ok to be honest, I have no idea what the second part is saying. Could someone try to explain what it is I am supposed to conclude?

score 3 · Answer 1 · answered Nov 03 '14 at 23:23

3

The first part is mostly okay. It sounds like this is a first principles course, so I guess I would add step explaining why $x\neq 0$ implies $x^2 > 0$. Other than that, it looks good.

For the second part, the contrapositive of what you've shown is that if there is an element $x$ in your field such that $x^2+1 \leq 0$, then the field is not ordered. In particular, the complex number $i$ is such that $i^2 + 1 = 0$, so the complex numbers can not be an ordered field.

answered Nov 03 '14 at 23:23

Callus - Reinstate Monica

5,507

If $x\ne0$, $x^2>0$ follow from considering that if $x>0$, then $x^2>0$; if $x<0$, then $-x>0$ and so $(-x)^2>0$. The axioms of ordered fields include that $a>0$ and $b>0$ implies $ab>0$. – egreg Nov 03 '14 at 23:25
@egreg looks good to me. – Callus - Reinstate Monica Nov 04 '14 at 01:25

Ordered field proof

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