16-00-AdditiveInverseOfOneElementTimesAnotherElementIsTheAdditiveInverseOfTheirProduct.tex

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\pmtitle{additive inverse of one element times another element is the additive inverse of their product}
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\begin{document}
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 $x \cdot (-y) = - (x\cdot y)$
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