Pure Mathematics proof for $(-a)b$ =$-(ab)$ How can I prove this without using the Multiplicative Property of zero? $(a*0=0)$
$$\begin{array}{rll}
(-a)b 
&amp;= (-a)b + 0 &amp;0 \text{ is identity for addition}\\
&amp;= (-a)b + (ab + (-ab))&amp;-ab \text{ is additive inverse of } ab\\
&amp;= ((-a)b+ab) + (-ab)&amp;\text{addition is associative}\\
&amp;= ((-a+a)b) + (-ab)&amp;\text{multiplication is distributive}\\
&amp;= \color{blue}{(0b)} + (-ab)&amp;-a \text{ is additive inverse of } a\\
&amp;= \color{blue}{(0b + 0)} + (-ab)&amp;0 \text{ is identity for addition}\\
&amp;= \color{blue}{(0b + (0b + (-0b)))} + (-ab)&amp; -0b \text{ is addition inverse of } 0b\\
&amp;= \color{blue}{((0b + 0b) + (-0b))} + (-ab)&amp; \text{addition is associative}\\
&amp;= \color{blue}{((0+0)b + (-0b))} + (-ab)&amp;\text{multiplication is distributive}\\
&amp;= \color{blue}{(0b + (-0b))} + (-ab)&amp;0 \text{ is identity for addition}\\
&amp;= \color{blue}{0} + (-ab)&amp;-0b \text{ is additive inverse of } 0b\\
&amp;= -ab &amp;0 \text{ is identity for addition}
\end{array}$$

Please note that the part in $\color{blue}{\text{blue}}$ is a sub-proof of the statement $0b = 0$.
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}