Prove $(-x)y=-(xy)$ using axioms of real numbers Working on proof writing, and I need to prove
$$(-x)y=-(xy)$$
using the axioms of the real numbers. I know that this is equivalent to saying that the additive inverse of $xy$ is $(-x)y$ but I am unsure how to prove it.
 A: You could prove that $(-x)y$ and $-(xy)$ are both the additive inverse of $xy$. Then use its uniqueness ($5^{th}$ axiom).
$$xy + (-x)y = (x+(-x))y = 0y = 0$$
Notice that $0y = 0$, because :
$$0y = (0+0)y = 0y + 0y$$
and if you add the additive inverse of $0y$ to both of the sides you come up with :
$$(-0y) + 0y = (-0y) + 0y + 0y$$
i.e. $$0 = 0y$$
A: Associative property: $$(xy)z=x(yz)$$
Then:
$$(-x)y=-(xy)\Longrightarrow(-1x)y=-1(xy)$$
When: $z=-1$
A: First start by proving : -x = (-1)x.
Proof: x+(-1)x = (1+(-1))x = 0x = 0. The proposition follows from uniqnes off negative off a number.
Now you can use this to prove your statement.
Proof: (−x)y = ((-1)x)y = (-1)(xy) = -xy 
A: $$ 0 = 0 $$
$$ 0y = 0 $$
$$ (x+(-x))y  = 0 $$
Distributive:
$$ xy + (-x)y = 0 $$
Add $ −(xy) $ to both sides:
$$ - (xy) + (xy + (-x)y) = -(xy) + 0 $$
Associative of +:
$$ (- (xy) + xy) + (-x)y = -(xy) $$
$$ (-x)y = -(xy) $$
The master key is the distribution.
The axiom of distribution implies precisely what we want to demonstrate. 
Without distribution, nothing links the operation + with the operation *. You would only have two groups on the same set.
