Exercise 1.1.7 (Real numbers mod 1)

Proof. By definition of the floor function, for all $x,y\in G$

thus

so $x\star y\in G$: $\star$ is a well defined binary operation on $G$.

Moreover

• The operation $\star$ is commutative: for all $x,y\in G$,

  $$y\star x=y+x-\lfloor y+x\rfloor =x+y-\lfloor x+y\rfloor =x\star y.$$

• The operation $\star$ is associative.

  Note that if $n\in \mathbb{Z}$ and $a\in ℝ$, then

  $$\begin{array}{llll}\hfill & \lfloor a\rfloor \le a<\lfloor a\rfloor +1,& \hfill & \\ \hfill & \lfloor a\rfloor +n\le a+n<\lfloor a\rfloor +n+1,\text{where }\lfloor a\rfloor +n\in \mathbb{Z},& \hfill & \end{array}$$

  so

  $$\begin{array}{lll}\hfill \lfloor a+n\rfloor =\lfloor a\rfloor +n.& & \hfill \text{(1)}\end{array}$$

  Therefore

  $$\begin{array}{llllll}\hfill x\star (y\star z)& =x\star (y+z-\lfloor y+z\rfloor )& \hfill & & \hfill & \\ \hfill & =x+y+z-\lfloor y+z\rfloor -\lfloor x+y+z-\lfloor y+z\rfloor \rfloor & \hfill & & \hfill & \\ \hfill & =x+y+z-\lfloor x+y+z\rfloor & \hfill (\text{by }(1)).& & \hfill & & \hfill \end{array}$$

  Then, using this result and the commutativity of $\star$,

  $$\begin{array}{llll}\hfill (x\star y)\star z& =z\star (x\star y)& \hfill & \\ \hfill & =z+x+y-\lfloor z+x+y\rfloor & \hfill & \\ \hfill & =x+y+z-\lfloor x+y+z\rfloor & \hfill & \\ \hfill & =x\star (y\star z),& \hfill & \end{array}$$

  so the operation $\star$ is associative.

• For all $x\in G$, since $0\le x<1$,

  $$x\star 0=x+0-\lfloor x+0\rfloor =x,$$

  so $0$ an identity of $G$.

• Let $x\in G$. If $x=0$, then $0\star 0=0$, so $0$ is the inverse of $0$ for the operation $\star$.

  If $x\ne 0$, then $0<x<1$, thus $0<1-x<1$ so $1-x\in G$, and

  $$x\star (1-x)=x-1-x-\lfloor x+(1-x)\rfloor =0,$$

  so $1-x$ is an inverse of $x$ in $G$. Every element of $G$ has an inverse in $G$.

$G$ is a group under the operation $\star$. □