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

###### Lemma 1.

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

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

where $-1$ is the additive inverse of $1$ and $-a$ 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 $a$.

 $\displaystyle a+(-1)\cdot a$ $\displaystyle=$ $\displaystyle 1\cdot a+(-1)\cdot a,\quad\text{ by the definition of }1$ $\displaystyle=$ $\displaystyle(1+(-1))\cdot a,\quad\text{ by the distributive law}$ $\displaystyle=$ $\displaystyle 0\cdot a,\quad\text{ by the definition of }-1$ $\displaystyle=$ $\displaystyle 0,\quad\text{ as a result of the properties of zero}$

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 MinusOneTimesAnElementIsTheAdditiveInverseInARing 2013-03-22 14:14:00 2013-03-22 14:14:00 alozano (2414) alozano (2414) 9 alozano (2414) Theorem msc 16-00 msc 13-00 msc 20-00 $(-1)\cdot a=-a$ 0cdotA0