**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.7**

Let $G = \{ x \in \mathbb{R} \ |\ 0 \leq x < 1 \}$ and for $x,y \in G$ let $x \star y$ be the fractional part of $x+y$, i.e., $$x \star y = x + y – \lfloor x+y \rfloor.$$ Prove that $\star$ is a well-defined binary operation on $G$ and that $G$ is an abelian group under $\star$.

Solution: Before we begin, note that $\lfloor x+n \rfloor = \lfloor x \rfloor + n$ for all $x \in \mathbb{R}$ and $n \in \mathbb{Z}$ as follows. For all integers $k$, we have $k \leq \lfloor x+n \rfloor$ iff $k \leq x + n$ iff $k-n \leq x$ iff $k \leq \lfloor x \rfloor + n$.

We first show that $\star$ is well defined. Suppose $x,y \in G$. There are two cases for $\lfloor x+y \rfloor$. If $\lfloor x+y \rfloor = 0$, then we have $0 \leq x+y < 1$ and so $x \star y \in G$. If $\lfloor x+y \rfloor = 1$, we have $x+y \geq 1$. But since $x,y < 1$, $x+y < 2$. Thus $x \star y \in G$. So $\star$ is indeed a binary operator on $G$.

We now show that $\star$ is associative. To that end, let $x,y,z \in G$. Then we have the following. \begin{align*} (x \star y) \star z = &\ (x \star y) + z – \lfloor (x \star y) + z \rfloor\\ = &\ x + y – \lfloor x + y \rfloor + z – \lfloor x + y – \lfloor x + y \rfloor + z \rfloor\\ = &\ x + y – \lfloor x + y \rfloor + z – \lfloor x + y + z \rfloor + \lfloor x + y \rfloor\\ = &\ x + y + z – \lfloor x + y + z \rfloor\\ = &\ x + y – \lfloor y + z \rfloor + z – \lfloor x + y + z \rfloor + \lfloor y + z \rfloor\\ = &\ x + y + z – \lfloor y + z \rfloor – \lfloor x + y + z – \lfloor y + z \rfloor \rfloor\\ = &\ x + (y \star z) – \lfloor x + (y \star z) \rfloor\\ = &\ x \star (y \star z). \end{align*} We now show that 0 is the identity element of $G$ under $\star$. If $x \in G$, then $\lfloor x \rfloor = 0$. Thus we have $$x \star 0 = x + 0 – \lfloor x + 0 \rfloor = x$$ and $$0 \star x = 0 + x – \lfloor 0 + x \rfloor = x.$$ We now show that every element has an inverse. If $x \neq 0$, then $1-x \in G$. Then $$x \star (1-x) = x + (1 – x) – \lfloor x + (1-x) \rfloor = 1 – 1 = 0$$ and similarly $$(1-x) \star x = (1 – x) + x – \lfloor (1 – x) + x \rfloor = 1 – 1 = 0.$$ Certainly, if $x = 0$ then $0 \star 0 = 0$. Thus $G$ is a group under $\star$. It is also clear that $x \star y=y \star x$. Hence $G$ is an abelian group under $\star$.