Theorem proof of this equation

Question

How would you prove the theorem $(-a)\cdot (-x)=ax$? If you used multiplication and addition axioms.

Answer 1

$$ bx + b(-x) = b(x+(-x)) = b\cdot 0 = 0 \Rightarrow b(-x) = -(bx) $$ Now take $b=-a$

Answer 2

$$(-a)(-x) + (-ax) = (-a)(-x) + (-a)x = (-a) [ (-x) + x] = (-a)( x + (-x) ) = (-a) 0 = 0 $$

Answer 3

If $\alpha : A \to B$ is a map between (additively written) groups such that $\alpha(a+a')=\alpha(a)+\alpha(a')$, then $\alpha(0)=0$ (plug in $a=a'=0$ and cancel) and $\alpha(-a)=- \alpha(a)$ (plug in $a'=-a$ and use $\alpha(0)=0$). Thus, if $\beta : A \times A' \to B$ is additive in each variable (for example a multiplication $A \times A \to A$), then $\beta(-a,-x)=- \beta(a,-x)=- - \beta(a,x) = \beta(a,x)$.