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
Canonical name	UniquenessOfAdditiveInverseInARing
Date of creation	2013-03-22 14:13:54
Last modified on	2013-03-22 14:13:54
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	7
Author	alozano (2414)
Entry type	Theorem
Classification	msc 20-00
Classification	msc 16-00
Classification	msc 13-00
Related topic	UniquenessOfInverseForGroups