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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.

Victor Feitosa
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    Small typo I think you meant $0(-b) - a(-b)$ in the second paragraph. I'm not seeing any immediate problem. The most likely error is probably to mix up unary $-$ and binary $-$ but you seemed to have avoided that, good job. – DanielV Jul 25 '18 at 01:43
  • Thanks for the correction – Victor Feitosa Jul 25 '18 at 02:01
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    "so we do have 0−ba " Have you proven that $b0 = 0$? That's probably not* a given axiom. – fleablood Jul 25 '18 at 02:44
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    "by Th 1.2(already been proved)," If you quote a Theorem to us who do not know what book you are using, you need to tell use what Th. 1.2 says. We can't tell you if your proof is good if we don't know what axioms, theorems (or even what book!) you are using. – fleablood Jul 25 '18 at 02:47
  • @fleablood I will add those axioms, sorry for the mistake. As by "Have you proven that b∗0=0? That's probably not a given axiom", yes, I've already proven it – Victor Feitosa Jul 25 '18 at 14:17

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