Algebra Solutions: The reals mod the integers are a group

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 $\star$ .

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.

$(x \star y) \star z$ = $(x \star y) + z - \lfloor (x \star y) + z \rfloor$

= $x + y - \lfloor x + y \rfloor + z - \lfloor x + y - \lfloor x + y \rfloor + z \rfloor$

= $x + y - \lfloor x + y \rfloor + z - \lfloor x + y + z \rfloor + \lfloor x + y \rfloor$

= $x + y + z - \lfloor x + y + z \rfloor$

= $x + y - \lfloor y + z \rfloor + z - \lfloor x + y + z \rfloor + \lfloor y + z \rfloor$

= $x + y + z - \lfloor y + z \rfloor - \lfloor x + y + z - \lfloor y + z \rfloor \rfloor$

= $x + (y \star z) - \lfloor x + (y \star z) \rfloor$

= $x \star (y \star z)$ .
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 $\star$ .

Algebra Solutions

The reals mod the integers are a group

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$(x \star y) \star z$	=	$(x \star y) + z - \lfloor (x \star y) + z \rfloor$
	=	$x + y - \lfloor x + y \rfloor + z - \lfloor x + y - \lfloor x + y \rfloor + z \rfloor$
	=	$x + y - \lfloor x + y \rfloor + z - \lfloor x + y + z \rfloor + \lfloor x + y \rfloor$
	=	$x + y + z - \lfloor x + y + z \rfloor$
	=	$x + y - \lfloor y + z \rfloor + z - \lfloor x + y + z \rfloor + \lfloor y + z \rfloor$
	=	$x + y + z - \lfloor y + z \rfloor - \lfloor x + y + z - \lfloor y + z \rfloor \rfloor$
	=	$x + (y \star z) - \lfloor x + (y \star z) \rfloor$
	=	$x \star (y \star z)$ .