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[parent] uniqueness of additive inverse in a ring (Theorem)
Lemma 1   Let $ R$ be a ring, and let $ a$ be any element of $ R$. There exists a unique element $ b$ of $ R$ such that $ a+b=0$, i.e. there is a unique additive inverse for $ a$.
Proof. Let $ a$ be an element of $ R$. By definition of ring, there exists at least one additive inverse of $ a$, call it $ b_1$, so that $ a+b_1=0$. Now, suppose $ b_2$ is another additive inverse of $ a$, i.e. another element of $ R$ such that
$\displaystyle a+b_2=0$
where 0 is the zero element of $ R$. Let us show that $ b_1=b_2$. Using properties for a ring and the above equations for $ b_1$ and $ b_2$ yields
$\displaystyle b_1$ $\displaystyle =$ $\displaystyle b_1+0$   (definition of zero)  
  $\displaystyle =$ $\displaystyle b_1+(a+b_2) \quad (b_2$    is an additive inverse of $\displaystyle a)$  
  $\displaystyle =$ $\displaystyle (b_1+a)+b_2 \quad ($associativity in $\displaystyle R)$  
  $\displaystyle =$ $\displaystyle 0+b_2 \quad (b_1$ is an additive inverse of $\displaystyle a)$  
  $\displaystyle =$ $\displaystyle b_2$   (definition of zero)$\displaystyle .$  

Therefore, there is a unique additive inverse for $ a$. $ \qedsymbol$



"uniqueness of additive inverse in a ring" is owned by alozano.
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See Also: uniqueness of inverse (for groups)


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Cross-references: equations, properties, inverse, additive, ring
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This is version 4 of uniqueness of additive inverse in a ring, born on 2004-03-09, modified 2006-10-26.
Object id is 5672, canonical name is UniquenessOfAdditiveIdentityInARing.
Accessed 2836 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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