Showing that $-2(-b)=2b$ Multiplication of:


*

*$3\times 4=12$, obvious

*$-3 \times 4=-12$, obvious

*$-3 \times -4=12$, not so obvious. We are just told to memorises it.
I would like to demonstrate that the multiplication of two negatives numbers is positive.
Here is my steps:
$5-3=2$
let $a-b=1$
$5-2(a-b)=3$
$5-2a-2(-b)=3$
$-2(-b)=3-5+2a$
$-2(-b)=2(a-1)$ where $a-1=b$
$-2(-b)=2b$
Are my steps correct? If not valid, I would like to see your method of showing the multiplication of two negative numbers is positive.
 A: $\overbrace{\bf\ Law\ of\ Signs}^{\rm\Large {(-x)(-y)}\ =\ xy} $ proof: $\rm\,\ (-x)(-y) = (-x)(-y) + \color{#c00}x(\overbrace{\color{#c00}{-y} + y}^{\Large =\,0}) = (\overbrace{-x+\color{#c00}x}^{\Large =\,0})(\color{#c00}{-y}) + xy = xy.$
A: 
I would like to see your method

Multiplication of $p$(suppose in $\mathbb{Z}$) by $i=\sqrt{-1}$ can be interpreted geometrically as rotating the line joining $(p,0)$ and $(0,0)$ by $90$ deg. without scaling the line. Suppose we are multiplying $3$ and $-1$. As, multiplying by $-1=i^2$ is same as multiplying by $i$ twice, hence we can interpret multiplication by $-1$ as rotating the line made by $(0,0)$ and $(3,0)$ by $180$ deg. without scaling it and looking at the point where the line cuts the imaginary axis. 
Now, if you multiply $3$ with $-2$, think it like $3\times 2\times -1$, i.e; multiply the multiplicand by the magnitude of the multiplier and then rotate the line $(6,0)$ by $180$ deg. 
Same for $-3\times -2$. We can rewrite like $3\times 2\times -1\times -1$. Multiplying $3$ and $2$ gives $6$, and multiplying by $-1$ twice means rotating the line joining $(0,0)$ and $(6,0)$ by $180$ deg twice, or a total $360$ rotation and you will land on $(6,0)$ again!
I think this way will convince you. 
