Complete the proof of $a(-b) = -(ab)$ Prove $a(-b)=-(ab)$, using axioms of equality, addition, and multiplication. 
I have tried: Let $a$ and $b$ be integers.
$(-a)+a= 0$ by the inverse axiom
$a=-(-a)$ by (If $a=b$ and $c=d$ then $a+c=b+d$ and $ac=bd$)
$a(-b)=-(-a)(-b)$ by (If $a=b$ and $c=d$ then $a+c=b+d$ and $ac=bd$)
$a(-b)=-(ab)$ by associative axiom.
End of proof.
Not quite certain if I did this correctly, I think I may have done some steps without an axiom to prove the step. I also, think I may have done some 'illegal' steps that don't actually fit with the amxioms I chose. 
 A: I'm not sure how $a=-(-a)$ follows from the fact that addition and multiplication are operations (that is, from the fact that $a+b=c+d$ and $ab=cd$ whenever $a=c$ and $b=d$). Rather, this follows from the definition of additive inverses, since $-(-a)$ is the unique number $c$ such that $-a+c=0,$ and clearly, $-a+a=0$ by (commutativity and by) definition of $-a.$
I also can't see how you've used the associative axiom, here, so your conclusion that $-(-a)(-b)=-(a\cdot b)$ has not been justified.
Remember that, axiomatically, $-(a\cdot b)$ is the unique number $c$ such that $a\cdot b+c=0.$ So, what you're being asked to show is that $a\cdot b+a\cdot(-b)=0.$ Readily, $$a\cdot b+a\cdot(-b)=a\cdot(b+-b)=a\cdot 0$$ by distributivity and additive inverses. If you've already proved (or taken as an axiom) that $a\cdot 0$ is necessarily $0,$ then you're done. Otherwise, all that remains is to show that $a\cdot 0=0$ for all $a.$ We can get started on this as follows: $$a\cdot 0=0+a\cdot 0=-a+a+a\cdot0=-a+a\cdot1+a\cdot0$$
Can you see the axioms we've used so far? Can you take it from there?
A: The intended method is to use the additive inverse property: $-x=(-1)x$.
\begin{align*}
a(-b)&=a((-1)b)&\quad\text{Additive Inverse Property}\\ 
&=a(b(-1))&\quad\text{Commutativity}\\
&=(ab)(-1)&\quad\text{Associativity}\\
&=(-1)(ab)&\quad\text{Commutativity}\\
&=-(ab)&\quad\text{Additive Inverse Property}\\
\end{align*}
