Need help checking and verifying that my proofs are correct. Using these field axioms: 1.
[2.][2.]


*Prove that for all $a,\, b\in\mathbb R$, $ab>0$ if and only if $a>0$ and $b>0$, or $a<0$ and $b<0$.
In other words, $ab$ is positive if and only if either $a$ and $b$ are both positive or both negative.
There are two directions, starting with the first direction: 
Assume $a>0$ and $b>0$.  Then it is clear $ab>0$ by (B4).  Similarly, if $a<0$ and $b<0$, then $ab=(-1)(-1)|a||b|=|a||b|>0$ also by (B4). 
In handling the harder direction:
Assume $ab>0$.  
Clearly, neither $a=0$ or $b=0$, otherwise we would have $ab=0$ by (B4) which is not the case.  So $a\neq 0$ and $b\neq 0$.  
Now assume that $a$ and $b$ have opposite signs (without loss of generality, we can let $a>0$ and $b<0$, both by (problem2)).  Then $ab=(-1)|a||b|=-|a||b|<0$ which is not the case. 
Thus, we either have $a$ and $b$ be both positive, or both negative. 
 A: Part 1 is good, except you did too much! You didn't need to prove the converse.
In part 2, I think you should justify why, if $c < 0$ and $b - a > 0$, then $c(b - a) < 0$. This is not part of the order axioms. I also think that $c(b - a) = cb - ca$ needs to be further justified (did you really mean to refer to (A3), the existence of a unique additive identity?).
I understand that there may be propositions that you've proven earlier that will help you (e.g. maybe you know that if $x < 0$ and $y > 0$ then $xy < 0$), but you should refer to them, not vaguely to the axioms that will help prove them.
Part 3 has some similar issues. Do you know that $-a = (-1)a$? You have to remember that $-a$ is the unique number $x$ such that $x + a = 0$. How do you know that the number that satisfies this equation is $(-1)a$? You also seem to assume that $(-1)(-1) = 1$ and $a0 = 0b = 0$ (I don't see how axiom (B4) obviously proves this). You finally refer back to problem 2 to justify that $x < 0$ and $y > 0$ imply $xy < 0$, even though this was an assumption used to prove part 2!
So, you've got a few holes to plug.
