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 $a$ .∎