Here are a couple of very basic questions. Note that "basic" means "foundational", not necessarily easy.
If $F$ is a field and $x\in F$, how do we prove that $-x=(-1)x$?
I feel this should follow directly from the field axioms.
If $F$ is an ordered field, how do we prove that a negative times a negative is a positive?
Now we have answered question one and have shown that the product of any $x \in F $ and $(-1)$ is equal to the additive inverse of $x$. So now suppose we have the product $(-x)(-y)$ where $x,y \in F$ whose additive inverses are $(-x)$ and $(-y)$ respectively, also suppose $x \geq y>0$. From the result of question one and the field axiom which gives us commutativity of multiplication this product can be re-written.
$$(-x)(-y)=(-1)x(-1)y=(-1)(-1)xy$$
Now note, by the result of question one, that that the product $(-1)(-1)$ on the right side must be equal to the additive inverse of $(-1)$. Since this additive inverse is $1$, which is the multiplicative identity, we can conclude that,
$$(-x)(-y)=xy.$$
Now from the multiplication law in the order axioms we know that $xy <0 $ if and only if $x$ or $y$ is less than $0$, this is not the case since we have assumed $x,y >0$. It follows that $xy>0.$ Thus the product of two negatives is positive.
I think this fixes the error from earlier.
Got number one during dinner.
Note that $(-1)$ is the additive inverse of $1$. Thus $1+(-1)=0$. Now since Distributivity of $x \in F$ over sums of elements of $F$ is guaranteed by the field axioms; and since x(1)=x by definition of the multiplicative identity, we have,
$$x(1+(-1))=x+(-1)x=0$$
By commutativity of addition in the field we can now write,
$$x+(-1)x=(-1)x+x=0.$$
Thus $(-1)x$ behaves exactly as the additive inverse of $x$ is defined to behave. Since inverses are unique, we can conclude that
$$(-1)x=-x$$
I think number one can be proven in this way:
Suppose $F$ is a field that has an additive identity $0$ and a multiplicative identity $1$, and that $x \in F$. By $-x$, we mean the inverse of $x$ under addition. That is,
$$x+ (-x) = 0.$$
It is a theorem of abstract algebra that the additive inverse of an element is unique. Thus, it suffices to show that $(-1)x + x = 0$ to prove the result.
Notice that we can write $(-1)x+ x = (-1)x + (1) x$, and use the distributive property (taken as a field axiom) in reverse to write
$$(-1)x + (1) x = (-1+1)x.$$
But now, by definition this is just $(0)x$ since $1$ plus its inverse is $0$. It is another axiom that $0x = 0$ for any $x$ in the field, thus we have shown that
$$(-1)x + x = 0,$$
and so $(-1)x = -x$.
EDIT: Apparently during my typing, Spiff figured it out.
Hey Spiff, I like your #1 and even hit the little heart button for it! I don't like your #2 so much. Just because $xy$ has no explicit minus sign doesn't mean it's positive. We could have $x=-1$ and $y=1$, for example. If you want to show that $xy$ is positive, you'll need to show that $xy>0$; that's what it means for something to be positive. So I don't think this problem can be done without the order axioms.