minus one times an element is the additive inverse in a ring

Lemma 1.

Let $R$ be a ring (with unity $\mathrm{1}$) and let $a$ be an element of $R$. Then

$$(-1)\cdot a=-a$$

where $\mathrm{-}\mathrm{1}$ is the additive inverse of $\mathrm{1}$ and $\mathrm{-}a$ is the additive inverse of $a$.

Proof.

Note that for any $a$ in $R$ there exists a unique “$-a$” by the uniqueness of additive inverse in a ring. We check that $(-1)\cdot a$ equals the additive inverse of $a$.

$a+(-1)\cdot a$ $=$ $1\cdot a+(-1)\cdot a,\text{ by the definition of }1$

$=$ $(1+(-1))\cdot a,\text{ by the distributive law}$

$=$ $0\cdot a,\text{ by the definition of }-1$

$=$ $0,\text{ as a result of the properties of zero}$

Hence $(-1)\cdot a$ is “an” additive inverse for $a$, and by uniqueness $(-1)\cdot a=-a$, the additive inverse of $a$. Analogously, we can prove that $a\cdot (-1)=-a$ as well. ∎

Title	minus one times an element is the additive inverse in a ring
Canonical name	MinusOneTimesAnElementIsTheAdditiveInverseInARing
Date of creation	2013-03-22 14:14:00
Last modified on	2013-03-22 14:14:00
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	9
Author	alozano (2414)
Entry type	Theorem
Classification	msc 16-00
Classification	msc 13-00
Classification	msc 20-00
Synonym	$(-1)\cdot a=-a$
Related topic	0cdotA0