Requesting feedback on proof of theorem I'm trying to self study my way through Apostol's calculus and have just started.  Having completed an undergraduate degree in physics some time ago, the math courses I took mostly focused on calculation and much less on proofs, so I'm a bit unsure of myself when doing proofs.  Any feedback on correctness, form, style, or otherwise would be most appreciated.
The theorem is from Volume 1, page 20:
Theorem 1.24:  If $a b>0$, then both $a$ and $b$ are positive or both are negative.
Proof: Assume the contradiction: if $a b>0$ if one of $a$ or $b$ is positive and the other is negative.  Without loss of generality we will assume $b$ is positive and $a$ is negative.  This means by assumption that $(-1 \cdot a b) > 0$.  Because $a b$ is positive this would mean that $-1 > 0$, which is a contradiction.  Therefore $a$ and $b$ must both be positive or negative for $a b>0$. $\square$
Axioms for the real numbers, introduced to this point, have been (ref Apostol):


*

*Commutative laws

*Associative laws

*Distributive laws

*Existence of identity elements

*Existence of negatives

*Existence of reciprocals

*If $x$ and $y$ are in $R^+$, so are $x+y$ and $xy$.

*For every real $x \neq 0$, either $x \in R^+$ or $-x \in R^+$, but not both.

*$0 \notin R^+$

 A: I'm reading Apostol's calculus for self study too.Here's my proof.
My approach is first to prove that $a$ and $a^{-1}$ $(a \neq 0)$ are both negative or positive.  
Suppose $a>0$ and $a^{-1}<0$, and then by axiom 8 we have $ (-a)^{-1}>0$ (I'm using a fact $-(a/b)=(-a/b)=a/(-b) \textbf{ if } b \neq 0.$ from exercise 9 in the $*I \space 3.3 \space Exercises$.This fact influences many step below.)   
Then we have $a * (-a)^{-1} = -1 > 0$ by axiom 7. But in the theorem 1.21 we have $1>0$,this against the axiom 8 that For every real x≠0, either x∈R+ or −x∈R+, but not both. so $a^{-1} must > 0$ if $a>0$.
same thing if we suppose $a<0$ and $a^{-1}>0$. and use the expression $(-a)*a^{-1}$ to lead to the contradiction.   
having the lemma above in hands.we can prove the theorem now.
suppose $a>0$,so we have $a^{-1}>0$,then by axiom 7 we have $ab*a^{-1} = a*a^{-1}*b =b>0$
if $a<0$,we have $-a^{-1}>0$,then $ab*(-a)^{-1} = -b >0$ so $ b < 0$.
Hope it helps.
A: You can prove the contrapositive: if $a,b$ have opposite signs, then $ab<0$.
