Let $G = \{x \in \mathbb{R}~|~0\leq x < 1\}$ and for $x,y \in G$ let $x*y$ be the fractional part of $x+y$ i.e $x*y = x + y - [x + y]$ where $[a]$ is the greatest integer less than or equal to $a$. I need help proving $*$ is a well defined binary operation on $G$ and that $G$ is an abelian group under $*$.
To prove that $*$ is well defined, I rely on the assumption that +,- is well defined in the set of real numbers. evaluating the brackets [] is also well-defined in the set of real numbers. (is this an okay assumption?)
To prove $*$ is associative, (I imagine because +,- is associative, so will *?) I show that $(x*y)*z = x*(y*z)$ $$(x*y)*z =(x+y-[x+y])*z = x+y-[x+y]+z - [x+y-[x+y]+z]$$ $$x*(y*z) = x*(y+z - [y+z]) = x + y + z - [y+z] - [x+y+z - [y+z]]$$ rearranging $$(x*y)*z = x+y+z-[x+y]-[x+y+z-[x+y]]$$ $$x*(y*z) = x+y+z-[y+z]-[x+y+z-[y+z]]$$ I am not so sure that these two are equal. I don't see how distributing the $-$ to remove the terms $[x+y]$ and $[y+z]$ is fair (ex: $x = y = \frac{1}{2}$ so $x*y = .5+.5-[.5]-[.5]$ yields a different answer than $x*y = .5+.5-[.5+.5]$
The identity element would be $0$, the inverse would have to be $-x$ which isn't in $G$. What am I doing wrong?
edit: commutativity would be proven by showing $x*y = y*x$ which is easy to show $x+y - [x+y] = y+x - [y+x]$ right?
You can think of your group $(G, *)$ as $(\mathbb{R}, +)/\mathbb{Z}$: That is, the reals under addition modded out by the integers.
This clearly works because we are essentially saying that is $x, y \in \mathbb{R}$, $x - y \in \mathbb{Z}$, then $x = \mathbb{Z} + y$ in $\mathbb{R} /\mathbb{Z}$. This has the same effect as “truncation”, but is now a nice subgroup Hence,
$$ (G, *) \simeq (\mathbb{R}/\mathbb{Z}, +) $$
To imagine this, realize that $\mathbb{R}$ just contains shifted copies of $[0, 1)$ all laid down one next to the other. So, we simply “collapse” all these copies together for this construction to work out.
Now, for this to be a group, we need the kernel of the map $$\phi: (\mathbb{R}, +) \to (\mathbb{R}, +) /\mathbb{Z}$$
to be a normal subgroup. Since the kernel of a quotient map is the quotienting subgroup itself, we simply need to show that $\mathbb{Z}$ is a normal subgroup of $\mathbb{R}$. This is obvious in abelian groups.