# additive inverse of one element times another element is the additive inverse of their product

Let $R$ be a ring. For all $x,y\in R$

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

All we need to prove is that $(-x)\cdot y+x\cdot y=x\cdot(-y)+x\cdot y=0$

Now: $(-x)\cdot y+x\cdot y=((-x)+x)\cdot y$ by distributivity.

Since $(-x)+x=0$ by definition and for all $y$, $0\cdot y=0$ we get:

$(-x)\cdot y+x\cdot y=0\cdot y=0$ and thus $(-x)\cdot y=-(x\cdot y)$

For $x\cdot(-y)$, use the previous properties of rings to show that

$x\cdot(-y)+x\cdot y=x\cdot((-y)+y)=x\cdot 0=0$

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

Title additive inverse of one element times another element is the additive inverse of their product AdditiveInverseOfOneElementTimesAnotherElementIsTheAdditiveInverseOfTheirProduct 2013-03-22 15:43:40 2013-03-22 15:43:40 cvalente (11260) cvalente (11260) 8 cvalente (11260) Theorem msc 16-00 msc 20-00 msc 13-00