Theorem 18.8

Return to Theorems, Glossary, Homework Problems.

Statement:

If $R$ is a ring with additive identity 0, then for any $a,b\in R$ we have
1.$0a=a0=0$,
2.$a(-b)=(-a)b=-(ab)$,
3.$(-a)(-b)=ab$.

Proof:

For Property 1, note that by axioms $\mathscr{R}_1$ and $\mathscr{R}_2$,

(1)

\begin{equation} a0+a0=a(0+0)=a0=0+a0. \end{equation}

Then by the cancellation law for the additive group $\langle R,+\rangle$, we have $a0=0$. Likewise,

(2)

\begin{equation} 0a+0a=(0+0)a=0a=0+0a \end{equation}

implies that $0a=0$.
By the left distributive law,

(3)

\begin{equation} a(-b)+ab=a(-b+b)=a0=0, \end{equation}

since $a0=0$ by Property 1. Likewise,

(4)

\begin{equation} (-a)b+ab=(-a+a)b=0b=0. \end{equation}

For Property 3,note that

(5)

\begin{equation} (-a)(-b)=-(a(-b)) \end{equation}

by Property 2. Again by Property 2,

(6)

\begin{equation} -(a(-b))=-(-(ab)), \end{equation}

and $-(-(ab))$ is the element that when added to $-(ab)$ gives 0. This is $ab$ by definition of $-(ab)$ and by the uniqueness of an inverse in a group. Thus, $(-a)(-b)=ab$.