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law of signs under multiplication in a ring
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(Derivation)
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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 $(-1)\cdot x = x \cdot (-1) = -x$ . 
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"law of signs under multiplication in a ring" is owned by alozano.
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Cross-references: associativity, inverse, additive, ring with unity
There are 2 references to this entry.
This is version 7 of law of signs under multiplication in a ring, born on 2004-03-09, modified 2006-03-10.
Object id is 5675, canonical name is XcdotYXcdotY.
Accessed 4850 times total.
Classification:
| AMS MSC: | 13-00 (Commutative rings and algebras :: General reference works ) | | | 16-00 (Associative rings and algebras :: General reference works ) | | | 20-00 (Group theory and generalizations :: General reference works ) |
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Pending Errata and Addenda
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