Circle group addition of fractional parts is well-defined

Question

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 .

Answer 1

I'm expanding Bill Dubuque's note:

Forget about fractional parts for the moment. Call two real numbers $x$, $y$ equivalent if $y-x\in{\mathbb Z}$. Denote the equivalence class of $x\in{\mathbb R}$ by $[x]$ and the set of all equivalence classes by $F$. The definition $$[x]+[y]\ :=\ [x+y]$$ defines addition uniquely on $F$ (check this!), $[0]$ is the neutral element, and the addition inherits from ${\mathbb R}$ the familiar properties: commutativity, associativity and existence of (additive) inverses. Therefore $F$ is an abelian group.

Now each equivalence class $[x]\in F$ contains exactly one element of your set $G:=[0,1[\ \subset{\mathbb R}$, namely the (well defined!) number $x-\lfloor x\rfloor$. It then follows that $G$ is a complete set of representatives for $F$. So instead of dealing with classes you can just add their representatives, and that's what has been set up in your problem.

Answer 2

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.

Answer 3

Hint $ $ The only difficult part is proving associativity. Denote the fractional part of a real by $\, f(r) = r - \lfloor r\rfloor = r\bmod 1.\,$ Note that $\,x\star y = f(x+y).\,$ CLAIM: $\ f(x+f(y)) = f(x+y)\,$
Proof: let $\,f(y) = r,\ y = r + n,\ n\in \mathbb Z.\,$ The claim is equivalent to $\,f(x+r) = f(x+r+n)\,$
which is true: adding an integer $\,n\,$ doesn't alter the fractional part. Hence applying claim twice

$$ x\star(y\star z)\ =\ f(x+f(y+z))\ =\ f(x+y+z)\ =\ f(f(x+y)+z)\ =\ (x\star y)\star z\qquad $$

Note $ $ The same proof yields associativity of integer addition $\!\bmod m,\,$ with $\,f(k) = k\bmod m.\,$ The essence of the matter will be much clearer after you learn about quotient groups. Then you will see that this group is simply the real circle group $\,\Bbb R/\Bbb Z = \Bbb R\bmod\Bbb Z = $ reals modulo $1\,.\,$ That the group laws are preserved in every modular image becomes a triviality from this general standpoint.

Generally ring $\rm\color{#c00}{identities}$ (e.g. associative, commutative, distributive laws) that are true in $\,\Bbb Z\,$ are preserved in $\,\Bbb Z_m = \Bbb Z\bmod m\,$ simply because the map $\,h\,:\, a\mapsto a_m\,$ is a surjective (onto) ring homomorphism. Similarly for other quotient structures such as the above circle group. Follow the link for details.