If $x$ is an integer and $m$ is an element from a ring prove that $x(-m) = -(xm)$

Question

So my approach to this was to break it into 3 cases. Where x is =,>,< 0. The cases where x = 0 and x > 0 are easy. But I'm struggling with x < 0.

Here is what I have so far:

Let $x = -y$ where $y \in Z$. Then

$x(-m) = -(xm) \rightarrow x(-m)+(xm) = 0$

$= -y(-m)+(-ym)$

Now at this point I know the final result will be $ym + (-ym) = 0$ But i'm not entirely sure how to handle the $-y(-m)$

Thanks for the help, I realize this is quite a simple question!

Answer 1

The element $-m$ is defined as the additive inverse of $m$, i.e., $m+(-m)=0$ where $0$ is the identity element relative to $+$. Hence, if $x\in\mathbb{Z}$, then $x(m+(-m))=x0=0$. But by distributive property, $0=x(m+(-m))=xm+x(-m)$. This implies that $x(-m)$ is the additive inverse of $xm$. (Noting, of course, that the additive identity element is unique.) Therefore $x(-m)=-xm$.

Hoping it helps.