This is a pretty fundamental question, so it requires us to use even more fundamental ideas.
To attempt to prove this we need to go back to the basic axioms concerning the algebraic properties of real numbers.
I will attempt a proof here. (It is a proof, but being complicated, doesn't actually add much to understanding, I think)!
I will assume we are trying to prove that [MATH](-a)(-b)=ab[/MATH]
An abbreviated version:
Among other things, (1) uses that a number plus its negative, equals 0
(2) allows us to 'pull out' one negative in a multiplication,
which we then use twice to prove (3).
[MATH](1) \hspace1ex \boxed{-(-x)=x}\\
x+0=x\\
x+(-x + (-(-x))=x\\
(x+(-x))+(-(-x)))=x\\
0+(-(-x))=x\\
-(-x)=x\\
\text{ }\\
(2) \hspace1ex \boxed{x(-y)=-(xy)}\\
x(-y)+xy=x(-y+y)\\
=x\cdot 0\\
=0\\
\therefore x(-y)+xy=0=-(xy)+xy\\
\therefore x(-y)=-(xy)\\
\text{ }\\
(3) \hspace1ex \boxed{(-a)(-b)=ab}\\
(-a)(-b)=-((-a)\cdot b) \text{ by (2)}\\
=-(-(ab))\text{ by (2)}\\
=ab \text{ by (1)}[/MATH]
The full version - a proper, rigorous proof, directly from the axioms.
The basic axioms of the real numbers wrt adding and multiplication:
A1 [MATH] \hspace1ex a+b=b+a[/MATH]A2 [MATH] \hspace1ex a+(b+c)=(a+b)+c[/MATH]A3 [MATH] \hspace1ex a+0=a[/MATH]A4 [MATH] \hspace1ex a+(-a)=0[/MATH]M1 [MATH] \hspace1ex a\cdot b=b \cdot a[/MATH]M2 [MATH] \hspace1ex a\cdot(b\cdot c)=(a\cdot b)\cdot c[/MATH]M3 [MATH] \hspace1ex a \cdot 1 = a[/MATH]M4 [MATH] \hspace1ex a \cdot \frac{1}{a}=1, a≠0[/MATH]D [MATH] \hspace1ex a \cdot (b+c)=a\cdot b + a \cdot c[/MATH]
[MATH](1) \hspace1ex \boxed{-(-x)=x}\\
x+0=x \hspace2ex A3\\
x+(-x+(-(-x)))=x \hspace2ex A4\\
(x+(-x))+(-(-x))=x \hspace2ex A2\\
0+(-(-x))=x \hspace2ex A4\\
-(-x)+0=x \hspace2ex A1\\
-(-x)=x \hspace2ex A3\\
\text{ }\\
(2) \hspace1ex \boxed{x(-y)=-(xy)}\\
x(-y)+xy=x(-y+y) \hspace2ex D\\
=x(y+(-y)) \hspace2ex A1\\
=x\cdot 0 \hspace2ex A4\\
=x\cdot 0 +0 \hspace2ex A3\\
=x\cdot 0 + (x+(-x)) \hspace2ex A4\\
=(x\cdot 0 +x)+(-x) \hspace2ex A2\\
=(x\cdot 0+ x\cdot 1) + (-x) \hspace2ex M3\\
=x(0+1)+(-x) \hspace2ex D\\
=x(1+0)+(-x) \hspace2ex A1\\
=x\cdot 1 + (-x) \hspace2ex A3\\
=x+(-x) \hspace2ex M3\\
=0 \hspace2ex A4\\
\text{ }\\
\text{i.e. } x(-y)+xy=0 *\\
\text{ }\\
\text{Now}\\
x(-y)+0=x(-y) \hspace2ex A3\\
x(-y)+(xy+(-xy))=x(-y) \hspace2ex A4\\
(x(-y)+xy)+(-xy)=x(-y) \hspace2ex A2\\
0+(-xy)=x(-y) \hspace2ex \text{ by *}\\
-xy+0=x(-y) \hspace2ex A1\\
-(xy)=x(-y) \hspace2ex A3\\
\text{Q.E.D.}\\
\text{ }\\
(1) \hspace1ex \boxed{-(-x)=x}\\
(2) \hspace1ex \boxed{x(-y)=-(xy)}\\
\text{ }\\
\text{ }\\
(3) \text{ Finally } \hspace1ex \boxed{(-a)(-b)=ab}\\
(-a)(-b)=-((-a)\cdot b) \hspace2ex \text{ by (2)}\\
=-(b(-a)) \hspace2ex M1\\
=-(-(ba)) \hspace2ex \text{ by (2)}\\
=ba \hspace2ex \text{ by (1)}\\
=ab \hspace2ex M1\\[/MATH]
