I'm studying by Apostol and I need to proof both statements:
(1): $(-a)b = -(ab)$
(2): $(-a)(-b) = ab$
My try at (1):
We can rewrite $(-a)$ as $(0 -a)$ by Th 1.2(already been proved), so we can rewrite (1) as $(0 - a)b$. By Commutative Laws, we have $(0 - a)b = b(0 - a)$ and by Th 1.5(which state that $a(b - c) = ab - ac$, and have already been proven) we have $b(0 - a) = b0 - ba$, so we do have $0 - ba$ which is $-ba$ or $-(ba)$. By Commutative Laws, we can rewrite $ba as ab$, so we have our desired value of $-(ab)$.
My try at (2):
We can rewrite $(-a)$ as $(0 - a)$ by Th 1.2(already been proved). So we now have $(0 - a)(-b)$, which can be, by Th 1.5, rewritten as $0(-b) - a(-b)$, which is equal to $0 - a(-b)$ which is simple -a(-b). We can rewrite it as $-(-b)a$ by Commutative Laws. But by (1) we have that $(-b)a = -(ba)$, so we end with $-(-(ba))$. And again by Commutative Laws we can rewrite $ba$ as $ab$ so we now have $-(-(ab))$. But we have also proved that $-(-a) = a$(Th 1.4) so we have $-(-(ab)) = (ab)$. Hence we got our desired expression.