Let $R$ be any ring. Prove that the following equations must hold in $R$. By letting $R$ be any ring I have to show these following equations hold in $R$: 
a) $x\cdot 0=0$
b) $0\cdot x=0$
c) $-(-x)=x$
d) $(-x)\cdot(-y)=x\cdot y$ 
I'm not really looking for the answer just for a hint to get me started. I guess I'm confused on how to show these properties hold within $R$. 
 A: *

*$x \cdot 0 = 0$: Observe that $x \cdot 0 = x \cdot (0 + 0) = x \cdot 0 + x \cdot 0$ Hence we have $x \cdot 0 = x \cdot 0 + x \cdot 0$. If we subtract $x \cdot 0$ from both sides, we get $0 = x \cdot 0$.

*$0 \cdot x = 0$: Similar reasoning.

*$-(-x) = x$: Observe that since $R$ is a ring, it is an additive group, so elements have additive inverses. Thus for all $x \in R$, $x + (-x) = 0$. Subtract $(-x)$ from both sides and get $x = -(-x)$.

*$(-x)(-y) = xy$: Before we show this, observe that $a(-b) = (-a)b = -ab$. Now observe that $0 = x \cdot 0 = x(y + (-y)) = xy + x \cdot(-y)$. Subtracting $x\cdot (-y)$ from both sides yields $xy = -x(-y) = (-x)(-y)$.


Why does $a(-b) = (-a)b = -ab$? Well observe that $0 = a \cdot 0 = a(b + (-b)) = ab + a(-b)$ which implies $a(-b) = -ab$. Similarly $0 = 0 \cdot b = (a + (-a))b = ab + (-a)b$ which implies $(-a)b = -ab$. Hence $a(-b) = (-a)b = -ab$.
A: Hint : $-1$ is the additive inverse of $1$ and $-1$ is the multiplicative inverse of $-1$.
$$ 1 -1 = 0$$ $$(-1)\times (-1) = 1$$
Then use distributivity.
