law of signs under multiplication in a ring

Lemma 1.

Let $R$ be a ring with unity, which we denote by $\mathrm{1}$. For all $x\mathrm{,}y\mathrm{\in }R$:

$$(-x)\cdot (-y)=x\cdot y$$

where $\mathrm{-}x$ denotes the additive inverse of $x$ in $R$.

Proof.

Here we use the fact $(-1)\cdot a=-a$ for all $a\in R$. First, we see that:

$$(-1)\cdot (-1)\cdot a=(-1)\cdot \left((-1)\cdot a\right)=(-1)\cdot (-a)=a$$

since, clearly, the additive inverse of $-a$ is $a$ itself.

Hence:

$$(-x)\cdot (-y)=(-1)\cdot x\cdot (-1)\cdot y=(-1)\cdot (-1)\cdot x\cdot y=x\cdot y$$

where we have used several times the associativity of $\cdot$ and the fact that $(-1)\cdot x=x\cdot (-1)=-x$.∎