Let $G=\{x \in \mathbb{R}\mid 0 \leq x < 1 \}$ and for $x,y \in G$ let $x\star y$ be the fractional part of $x+y$ (i.e $x\star y=x+y-\lfloor x+y \rfloor$ where $\lfloor a \rfloor $ is the greatest integer less than or equal to a). Then, how do I prove that $\star$ is a well defined binary operation on $G$ and that $G$ is an abelian group under $\star$?
Thank you.I have just started group theory.My progress on this is minimal and next to 0.
Edited .
Best Answer
There is no issue of "well-definedness": the elements of $G$ are real numbers, and for every real number $r$, $r-\lfloor r\rfloor$ is a real number that lies in $[0,1)$; thus, $\star$ is a function from $\mathbb{R}\times\mathbb{R}$ to $[0,1)$, and hence by restriction $\star$ is a function from $G\times G$ to $[0,1)\subseteq G$.
(Note: The next paragraph was written when the definition of $G$ was that $G$ included $1$; this has since been changed.)
However, as given, $G$ is not a group under the operation: note that if $(G,\cdot)$ is a group, then in particular for every $g\in G$ there exist $x,y\in G$ with $x\cdot y = g$. However, there are no elements of your $G$ which satisfy $x\star y = 1$, even though $1\in G$.
If, however, you change $G$ to be $G=\{x\in\mathbb{R}\mid 0\leq x\lt 1\}$, then the set is indeed a group (It is, in fact, isomorphic to $\mathbb{R}/\mathbb{Z}$). Note that $0$ is a two-sided identity, since if $x\in G$ then $\lfloor x\rfloor = 0$; and that if $x\neq 0$ is in $G$, then $1-x\in G$ and $x\star(1-x) = 0$. And trivially, since addition of reals is commutative, $x\star y = y\star x$.
So the only thing that needs to be proven is associativity.
The key is to note that for all real numbers $r$ and all integers $n$, $$\lfloor r-n\rfloor = \lfloor r\rfloor - n.$$
So, if $x,y,z\in [0,1)$, then: $$\begin{align*} (x\star y)\star z &= \Bigl( x+y - \lfloor x+y\rfloor\Bigr)\star z\\ &= \Bigl( x+y-\lfloor x+y\rfloor + z\Bigr) - \lfloor x+y - \lfloor x+y\rfloor + z\rfloor\\ &= x+y+z - \lfloor x+y\rfloor -\Bigl( \lfloor x+y+z\rfloor - \lfloor x+y\rfloor\Bigr)\\ &= x+y+z - \lfloor x+y+z\rfloor;\\ x\star(y\star z) &= x\star\Bigl( y+z - \lfloor y+z\rfloor\Bigr)\\ &= x + y + z - \lfloor y+z\rfloor - \lfloor x+y+z-\lfloor y+z\rfloor\rfloor\\ &= x+y+z - \lfloor y+z\rfloor - \bigl( \lfloor x+y+z\rfloor - \lfloor y+z\rfloor\bigr)\\ &= x+y+z - \lfloor x+y+z\rfloor\\ &= (x\star y)\star z. \end{align*}$$
You can also go the longer route and consider the possibilities of $x+y\lt 1$, $y+z\lt 1$, $x+y+z\lt 1$; or $x+y\lt 1$, $1\leq y+z\lt 2$ and $1\leq x+y+z\lt 2$; $x+y\geq 1$, $y+z\geq 1$, and $1\leq x+y+z\lt 2$; and $x+y\geq 1$, $y+z\geq 1$, and $2\leq x+y+z\lt 3$. But the observation above is much easier.