Pure Mathematics proof for $(-a)b$ =$-(ab)$ How can I prove this without using the Multiplicative Property of zero? $(a*0=0)$ 
 A: $-a = (-1)\cdot a$,
and you have LHS $= \big((-1)\cdot a \big)\cdot b = (-1)\cdot (a \cdot b)$ by associative property = RHS
A: $$\begin{array}{rll}
(-a)b 
&= (-a)b + 0 &0 \text{ is identity for addition}\\
&= (-a)b + (ab + (-ab))&-ab \text{ is additive inverse of } ab\\
&= ((-a)b+ab) + (-ab)&\text{addition is associative}\\
&= ((-a+a)b) + (-ab)&\text{multiplication is distributive}\\
&= \color{blue}{(0b)} + (-ab)&-a \text{ is additive inverse of } a\\
&= \color{blue}{(0b + 0)} + (-ab)&0 \text{ is identity for addition}\\
&= \color{blue}{(0b + (0b + (-0b)))} + (-ab)& -0b \text{ is addition inverse of } 0b\\
&= \color{blue}{((0b + 0b) + (-0b))} + (-ab)& \text{addition is associative}\\
&= \color{blue}{((0+0)b + (-0b))} + (-ab)&\text{multiplication is distributive}\\
&= \color{blue}{(0b + (-0b))} + (-ab)&0 \text{ is identity for addition}\\
&= \color{blue}{0} + (-ab)&-0b \text{ is additive inverse of } 0b\\
&= -ab &0 \text{ is identity for addition}
\end{array}$$
Please note that the part in $\color{blue}{\text{blue}}$ is a sub-proof of the statement $0b = 0$.
A: We have that $a+(-a)=0$
$\implies b(a+(-a))=0\cdot b$
$\implies ba+b(-a)=0$
$\implies b(-a)=-(ba)$
A: In process of rewriting. Will be back soon.
A: Suppose we only assume structure properties of distribution, closure, and existance/uniqueness of additive inverse.
$$(\forall a,b,c)\, (a + b)c = ac + bc\tag {R1}$$
Choose $b = -a$.
$$(\forall a,c)\, 0 \cdot c = ac + (-a)c \tag{R2}$$
Now let's consider the possibility that both are true:
$$(\exists x)\, 0\cdot x \ne 0 \tag{P1}$$ 
$$(\forall y,z)\, (-y)z = -(yz) \tag{P2}$$
Apply (P1) to (R2):
$$(\exists x \forall a)\, 0 \ne ax + (-a)x$$
Apply (P2):
$$(\exists x \forall a)\, 0 \ne ax + -(ax)$$
Which clearly contradicts closure and existance of an inverse.  So for this limited structure, $\lnot (P1) \lor \lnot (P2)$.
A: by Munna shigri
a(-b)=-(ab)
proof:
0=0
b+(-b)=0.   {b+(-b)=0}
a{b+(-b)}=a0.      {multiplying by (a )on both side}
ab+a(-b)=a0.       {by distributive property}
ab+a(-b)=0.         {a0=0}
{ab+a(-b)}+(-(ab)=0+(-ab). { by adding (-ab) on both side}
{a(-b)+ab}+(-ab)=(-ab)+0.    {by comutative property}
a(-b)+{ab+(-ab)}=(-ab)+0.    {by associative property}
a(-b)+0=(-ab)+0.                    {a+(-a)=0}
a(-b)=-ab.     proved                             {a+0=a}
