This is an exercise from Tom Apostol's Calculus, and what I'm trying to do is to prove from the field and order axioms that if $a > 0$ then $b > 0$, and if $a < 0$ then $b < 0$, since I have already proved that $a$ and $b$ can't be $0$ by theorem 1.11 given in the book.
I have seen someone suggest proving it by contradiction. However, I would like to prove it directly. I have found a solution from a blog, but I don't understand how does $a(-b) < 0$ imply that $-b < 0$. I have also thought about using axiom 7 to prove that $ab(\frac{1}{a}) = b$ is positive if $a$ is positive, but I would need to prove that $\frac{1}{a}$ is positive if $a$ is positive, and I also get stuck there. I will leave an image of the solution I found and another of the axioms.
Theorem 1,11
Blog solution
Order axioms
Best Answer
This is Exercise 1 in I 3.5 asking you to prove Theorem I.24, and you are permitted to use earlier theorems and axioms.
Firstly, your argument to prove that $a$ and $b$ are nonzero is incorrect. This follows directly from one of the earlier theorems, but not from Theorem I.11 as you say. Can you identify the right one?
Next, assume for a contradiction that $a$ and $b$ are not both positive or both negative, that is that they have opposite signs. For example, since $a$ and $b$ play similar roles, we may assume that $a > 0$ and $b < 0$. (When I say that $a$ and $b$ play similar roles, I am relying on the fact that $ab = ba$.)
Now you can apply Theorem I.19 to the inequality $b < 0$, multiplying by $a$, which is positive. (We again need the same earlier theorem here.) This will yield a contradiction, proving what you want.
The reason a proof by contradiction is best, in my opinion, is that it avoids introducing consideration of inverses. In other words, you make certain assumptions about the signs of $a$ and $b$, and you determine the sign of $ab$. What really matters here are the cases where $ab$ is negative. Since your assumption is $ab > 0$, this means you must argue by contradiction.
More precisely, you are proving the contrapositive of the original statement, but this is a technicality.