Prove that if $ab > 0$, then $a$ and $b$ are either both positive or both negative 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
 A: Assume $b \lt 0 \lt a.$  Then $0 \lt -b$ by Axiom $8$, so $a(-b) \gt 0$ by Axiom $7$.
$ab+a(-b)=a(b+(-b))=a(0)=0$, so $a(-b) \gt 0 \Rightarrow ab \lt 0$ by Axiom $7$.
As Anonymous correctly points out, showing that $\forall a \in \Bbb R~(0a=0)$ is the converse of Theorem I.$11$, so you can't use the Theorem to prove it.  You need to prove it from the field axioms.  To prove $0a=a$, note that $0a=(0+0)a=0a+0a \Rightarrow 0=-(0a)+0a=(-(0a)+0a)+0a=0+0a=0a$.
A: 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.
