# uniqueness of additive inverse in a ring

###### Lemma.

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 (http://planetmath.org/Ring) for $a$.

###### Proof.

Let $a$ be an element of $R$. By definition of ring, there exists at least one additive inverse (http://planetmath.org/Ring) 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

 $a+b_{2}=0$

where $0$ is the zero element (http://planetmath.org/Ring) 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

 $\displaystyle b_{1}$ $\displaystyle=$ $\displaystyle b_{1}+0\quad\text{(definition of zero)}$ $\displaystyle=$ $\displaystyle b_{1}+(a+b_{2})\quad(b_{2}\text{ is an additive inverse of }a)$ $\displaystyle=$ $\displaystyle(b_{1}+a)+b_{2}\quad(\text{associativity in }R)$ $\displaystyle=$ $\displaystyle 0+b_{2}\quad(b_{1}\text{ is an additive inverse of }a)$ $\displaystyle=$ $\displaystyle b_{2}\quad\text{(definition of zero)}.$

Therefore, there is a unique additive inverse for $a$. ∎

Title uniqueness of additive inverse in a ring UniquenessOfAdditiveInverseInARing 2013-03-22 14:13:54 2013-03-22 14:13:54 alozano (2414) alozano (2414) 7 alozano (2414) Theorem msc 20-00 msc 16-00 msc 13-00 UniquenessOfInverseForGroups