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I am trying to prove the following property for a ring $R$ $$a\cdot(-b)=(-a)\cdot b=-(a\cdot b)$$ where $a,b\in R$ and let $-x$ denote the additive inverse of $x\in R$. By definition, we know that such an inverse exists for all ring's elements. Also let $1$ be the multiplicative identity which does not necessarily exists in $R$. Thus, I cannot figure out a rigorous proof for the above result.

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  • $\begingroup$ What this means is that if you add $a\cdot (-b)$ to $ab$ you should get zero; so add the two together, and rearrange using the axioms of arithmetic in a ring. $\endgroup$ – User0112358 Aug 31 '17 at 6:45
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We have $$ a\cdot b + a\cdot (-b) = a\cdot (b + (-b)) = a\cdot 0 = 0 $$ so we see that $a\cdot (-b)$ fulfills the defining property of $-(a\cdot b)$. Since the addition in a ring gives a group, we know that additive inverses are unique, so we must have $a\cdot (-b) = -(a\cdot b)$. A very similar proof shows that $(-a)\cdot b = -(a\cdot b)$.

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