# law of signs under multiplication in a ring

###### Lemma 1.

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

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

where $-x$ 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 $(-1)\cdot x=x\cdot(-1)=-x$. ∎

Title law of signs under multiplication in a ring LawOfSignsUnderMultiplicationInARing 2013-03-22 14:14:03 2013-03-22 14:14:03 alozano (2414) alozano (2414) 10 alozano (2414) Derivation msc 20-00 msc 16-00 msc 13-00 $(-x)\cdot(-y)=x\cdot y$ Ring