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[parent] law of signs under multiplication in a ring (Derivation)
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 $(-1)\cdot x = x \cdot (-1) = -x$ . $ \qedsymbol$




"law of signs under multiplication in a ring" is owned by alozano.
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See Also: ring

Other names:  $(-x)\cdot (-y)= x\cdot y$

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product of negative numbers (Derivation) by pahio
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Cross-references: associativity, inverse, additive, ring with unity
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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 MSC13-00 (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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prove needs prove by mj on 2005-01-27 15:20:38

It seems to me that in this prove the part (-1)*(-a)=a needs to be prove dand I'm not sure if it is...
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(-x).(-y)=x.y by perucho on 2004-08-05 03:04:52

Also the additive inverse of (-1) is 1.

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entry title by drini on 2004-03-09 13:50:53

shouldn't this be better titled "law of signs" or some other common phrasing?
 f
G -----> H G
p \ /_ ----- ~ f(G)
 \ / f ker f
 G/ker f 
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