Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to
$$
\frac{f(m)}{f(n)}\in F
$$
where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then
$$
n=\underbrace{1+1+\dots+1}_{\text{$n$ times}}
$$
and therefore $0\prec n$. Conversely, if $n<0$, then
$$
n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,)
$$
and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields,
$$
0\prec n\cdot\frac{m}{n}=m
$$
and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
You should have a couple of examples in your mind throughout this post (and generically). A field is roughly something that behaves like the rationals $\mathbb{Q}$, the reals $\mathbb{R}$, or the complex numbers $\mathbb{C}$. There are other fields, but these are the ones which you are probably most familiar with.
More generally, a field is a set of things in which it makes sense to add two things together and multiply two things together. Both of these operations should be commutative, meaning that $a+b = b+a$ and $ab = ba$. Further, there is a distributive law connecting addition and multiplication, so that you have $a(b+c) = ab + ac$.
A field should also have two distinguished elements, that we might suggestively call $0$ and $1$, so that $a+0 = a$ for all $a$ and $ b*1 = 1$ for all $b \neq 0$.
Finally, for every $b \neq 0$, there should be another element $b^{-1}$ such that $bb^{-1} = 1$.
In other words, a field is a set where addition and multiplication work essentially as they work in $\mathbb{R}, \mathbb{Q}$, or $\mathbb{C}$.
Now one should note that not all fields really do look like $\mathbb{R}, \mathbb{Q}$, or $\mathbb{C}$. Some fields have could have weird definitions of "multiplication" or "addition" --- what's important isn't that they're exactly the same, but that they behave in ways similar to the fields we're very familiar with.
Returning to the question at hand, we now know what a field is. An ordered field is a field where you can say, for any two elements $a$ and $b$, whether $a < b$, $a = b$, or $a > b$. And this should be in such a way that the typical ideas we have about inequalities hold true. These ideas include
- $1 > 0$
- if $a > b$ and $b > c$, then $a > c$
- if $a > 0$ and $b > c$, then $ab > ac$
and things of that nature.
Best Answer
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.