taken from D&F, question 1.1.7:

Let $\displaystyle G = \{x \in \mathbb{R} | 0 \leq x < 1\}~\text{and for}~x,y \in G, \text{let }~x * y~\text{be the fractional part of } x + y$ (i.e. $\displaystyle x * y = x+y - [x+y]$). Show that $\displaystyle G$ is an abelian group.

The question asks for verification of the group axioms, although I'm a little confused with showing the existence of inverses. One can say that for an arbitrary $\displaystyle a \in G, a * (-a) = a + (-a) + [a-a] = 0$, which is the identity. But notice that I said $\displaystyle -a$, and that's confusing me. Is $\displaystyle (-a) \in G$? Is it literally the negative of $\displaystyle a$? I would presume as much since I actually took that into consideration when I added $\displaystyle a + (-a)$. But $\displaystyle -a$ is not part of $\displaystyle G$ by construction. So how do I show the existence of inverses?